algebra
difference of squares
special factorizations

Determine a positive constant $ c$ such that the equation

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

has precisely three solutions $ (x,y)$ in positive integers.

EDIT> Ignore this post, I missed the exponent in the first term.

Factor:

$ (y \minus{} 1)(x \minus{} y) \equal{} c$

Set $ c \equal{} a^2$.

$ y \equal{} 2$
$ x \equal{} a^2 \plus{} 2$

$ y \equal{} a \plus{} 1$
$ x \equal{} 2a \plus{} 1$

$ y \equal{} a^2 \plus{} 1$
$ x \equal{} a^2 \plus{} 2$

So any c where c is the square of a prime number should be valid.

Doesn't $ (y\minus{}1)(x\minus{}y)\equal{}yx\minus{}y^2\minus{}x\plus{}y$? and the original equation is $ xy^2\minus{}y^2\minus{}x\plus{}y$

[hide="Hint"]
Multiply by 4, complete the square
[/hide]
[hide="Hint 2"]
Use difference of squares[/hide]
[hide="Hint 3"]
Now that it's factored, find c such that when you solve the systems for the possible factors, there's three solutions[/hide]


if$a^2+b^2+c^2=1$,then($a,b,c$are nonenegative.)
the value of the below expression is at most $4.5$:
$\frac{1}{1-ab}+\frac{1}{1-ac}+\frac{1}{1-bc}$.

[quote="dna"]if$a^2+b^2+c^2=1$,then($a,b,c$are nonenegative.)
the value of the below expression is at last $4.5$:
$\frac{1}{1-ab}+\frac{1}{1-ac}+\frac{1}{1-bc}$.[/quote]
I think it is a vasc's inequality.

[quote="arqady"]
Search does not work, therefore I shall plan the proof:
[/quote]
Here's it is:
http://www.mathlinks.ro/Forum/viewtopic.php?t=55869

Can anyone prove it by n-1 EV? Its shape suggests us that it is made by (n-1)EV.
Many thanks! :)

In "Algebraic Inequalities" I proved by EV-Method the inequality
$\frac 1{p-(x+y)^2}+\frac 1{p-(y+z)^2}+\frac 1{p-(z+x)^2} \leq \frac 9{3p-4}$
for $p \geq \frac{10}{3}$ and $x^2+y^2+z^2$=1.
For $p=\frac{10}{3}$, equality occurs again when $x=1/3$ and $y=z=2/3$.
This inequality implies
$\frac 1{q-xy}+\frac 1{q-yz}+\frac 1{q-zx} \leq \frac 9{3q-1}$
for $q \geq \frac 5{6}$. For $q=1$ we get the original inequality.


Inspired by http://slashdot.org/pollBooth.pl?qid=1518&aid=-1

Also, if you actually wanted to maximize the probability of getting the question right, which option would you pick?

To maximize, wouldn't it be to vote 76-100, because somehow everyone else is going to think that everyone else will vote that...?  :ninja:

I don't know if I make sense.

Only the first two options make sense...

This is a sort of paradox. I pick 0-25% at the start, because there have been no other posters. So, another person picks 100%, but is wrong, so picks 0%, but is wrong, so picks 100%......

Lol.
So far my choice has been highest. The second one. It's close to incorrect, though. By 'bout 8%. :P

umm.. usually 51%-100%.... well... not this time

I chose 26-50 which seems to be the most correct.  I wonder if theirs a logical way to think about this.

Then again, if too many people try to be logical and choose 26-50...

I chose 26-50. It works right now. Theoretically I would chose 76-100, but theory != reality.

EDIT: I remember a poll like this. It wasn't really a poll though. We had to enter one integer, from 1 to 100, as the percent. The correct choice turned out to be 3%. I voted 2% I think... *grumble*

@math_explorer:lol, I think I can guess where that "poll" or more like "conundrum" is from. Sorry if I'm not making any sense to you.

My thoughts on this:
I would probably pick between 0-50 since it is the best choice if people voted randomly. Then again, everyone would pick it if they knew so, but then no one would pick it if everyone knew everyone would pick it.

I hate circular logic.

I knew somebody would know!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

There being the age poll that was surprisingly surprising...

But I can't find the correct round...

 :(

And, I can't tell you my account because my account states my name way too clearly.

I combed all the rounds before 64 and couldn't find it. I also tried to use Google Advanced Search to get it but was unable to.

Anybody else know?

I put 26-50% and was right!!!!

*feeling good about self*

Ya, say it doesn't matter. then 1/4 of the people pick the same one as you, so it would be one of the first two. but because we're all pretty smart, we all do that so the most logical one would be 26-50.

That was easy... unless the second option changes to over 50%, which it might if people look at the Slashdot thing first.  (or if they just view the results :) )

There really isn't any logic to this because another person's logic is another person's logic. So...Say I use reverse phycology (which I did) and another person does then your going to be wrong. It's like predicting whats going to happen next in a game of chess when that person has 4 or 5 legitimate moves, its all luck. So really unless your telepathic there isn't any logical way to step up to that poll.


induction
inequalities unsolved
inequalities

The Karamata Ineq is posted in
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=14975[/url]
However, how to prove that?

You can use the idea of compression introduced [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=17288]here[/url].


Ένας οδηγός βρίσκεται με το αυτοκίνητο του σε μια κυκλική πίστα. Το αυτοκίνητό του αρχικά δεν έχει καθόλου βενζίνη. Στην πίστα υπάρχουν 100 σταθμοί ανεφοδιασμού και το σύνολο της βενζίνης που υπάρχει σε αυτούς τους σταθμούς αρκεί για ένα ακριβώς γύρο της πίστας.

Να δειχθεί ότι υπάρχει (τουλάχιστον) ένας σταθμός ανεφοδιασμού ώστε αν ο οδηγός ξεκινήσει από αυτόν θα μπορέσει να κάνει τον γύρο της πίστας.

Ωραιο προβλημα!! Δεν εχω πολυ χρονο και για συτο βαζω μονο μια προχειρη εκδοχη της λυσης μου(πιο πολυ σε μορφη υποδειξης). Sorry για το attachment αλλα δεν εχω μαθει Latex :blush:

Φιλοσοφημένε, μπορείς να εξηγήσεις πιο αναλυτικά την λύση σου;

Αν διαλέξεις τον α_1 όπως λές, τότε ίσως να μην μπορείς να πας από τον α_2 στον α_3.

Συγνωμη, για την καθυστερηση. Οριστε η λυση ολοκληρωμενη(το μονο που δεν εχω εξηγηση ειναι μια λεπτομερεια στο τελος που νομιζω ειναι trivial. Οπως και να' χει αν θες εξηγηση πες μου).

Νομίζω πως η απάντηση είναι λάθος (αλλά είσαι στον σωστό δρόμο). Για παράδειγμα, αν όλες οι αποστάσεις έχουν μήκος 1 και έχουμε βενζίνη 10 στον σταθμό 1, 20 στον σταθμό 2 και 70 στον σταθμό 95, τότε η απάντηση σου επιτρέπει να ξεκινήσουμε από τον σταθμό 1. Αλλά τότε δεν δουλεύει. (Πρέπει να ξεκινήσουμε από τον σταθμό 95)

Λοιπόν

Επειδή το "100 σταθμοί" ανεφοδιασμού μου φαίνεται ψιλοαυθαίρετο έχω μια υποψία πως λύνεται επαγωγικά...

 Έστω ότι η βενζίνη που απαιτείται για τον γύρο της πίστας μοιράζεται σε N σταθμούς. Θα δείξουμε πως για κάθε N, υπάρχει (τουλάχιστον) ένας σταθμός ανεφοδιασμού ώστε αν ο οδηγός ξεκινήσει από αυτόν θα μπορέσει να κάνει τον γύρο της πίστας.

  Αν είχαμε έναν σταθμό ανεφοδιασμού, δηλαδή N=1, τότε προφανώς το αυτοκίνητο θα μπορούσε να ξεκινήσει από αυτόν και να ολοκληρώσει τον γύρο της πίστας.

Έστω ότι η πρόταση αληθεύει για N σταθμούς ανεφοδιασμού.

Θα δείξουμε ότι τότε αληθεύει και για N+1.

  Ανάμεσα στους N+1 σταθμούς ανεφοδιασμού αναγκαστικά θα υπάρχει τουλάχιστον ένας, έστω Σ1, που έχει αρκετή βενζίνη για να φτάσει το αυτοκίνητο στον διπλανό του σταθμό Σ2. Αν δεν υπήρχε τέτοιος σταθμός τότε η συνολική βενζίνη δεν θα επαρκούσε για έναν πλήρη γύρο.  Συνεπώς αν φανταστούμε πως παίρνουμε όλη την βενζίνη του σταθμού Σ2 και την αδειάζουμε στον Σ1, πλέον έχουμε N σταθμούς που έχουν συνολικά αρκετή βενζίνη για την ολοκλήρωση ενός γύρου. Ανάμεσα σε αυτούς τους N σταθμούς όμως υπάρχει ένας από τον οποίο μπορεί να ξεκινήσει το αυτοκίνητο και να ολοκληρώσει τον γύρο της κυκλικής πίστας. Από τον ίδιο σταθμό μπορεί να ξεκινήσει και στην περίπτωση που έχουμε Ν+1 σταθμούς (αφού για να διανύσει την απόσταση ανάμεσα στους σταθμούς Σ1 και Σ2 επαρκεί η βενζίνη του σταθμού Σ1, ενώ για την υπόλοιπη διαδρομή έχει όση βενζίνη είχε και στην περίπτωση των Ν σταθμών  :wink: )

[quote="thanaseas"]Λοιπόν

Επειδή το "100 σταθμοί" ανεφοδιασμού μου φαίνεται ψιλοαυθαίρετο έχω μια υποψία πως λύνεται επαγωγικά...

 Έστω ότι η βενζίνη που απαιτείται για τον γύρο της πίστας μοιράζεται σε N σταθμούς. Θα δείξουμε πως για κάθε N, υπάρχει (τουλάχιστον) ένας σταθμός ανεφοδιασμού ώστε αν ο οδηγός ξεκινήσει από αυτόν θα μπορέσει να κάνει τον γύρο της πίστας.

  Αν είχαμε έναν σταθμό ανεφοδιασμού, δηλαδή N=1, τότε προφανώς το αυτοκίνητο θα μπορούσε να ξεκινήσει από αυτόν και να ολοκληρώσει τον γύρο της πίστας.

Έστω ότι η πρόταση αληθεύει για N σταθμούς ανεφοδιασμού.

Θα δείξουμε ότι τότε αληθεύει και για N+1.

  Ανάμεσα στους N+1 σταθμούς ανεφοδιασμού αναγκαστικά θα υπάρχει τουλάχιστον ένας, έστω Σ1, που έχει αρκετή βενζίνη για να φτάσει το αυτοκίνητο στον διπλανό του σταθμό Σ2. Αν δεν υπήρχε τέτοιος σταθμός τότε η συνολική βενζίνη δεν θα επαρκούσε για έναν πλήρη γύρο.  Συνεπώς αν φανταστούμε πως παίρνουμε όλη την βενζίνη του σταθμού Σ1 και την αδειάζουμε στον Σ2, πλέον έχουμε N σταθμούς που έχουν συνολικά αρκετή βενζίνη για την ολοκλήρωση ενός γύρου. Ανάμεσα σε αυτούς τους N σταθμούς όμως υπάρχει ένας από τον οποίο μπορεί να ξεκινήσει το αυτοκίνητο και να ολοκληρώσει τον γύρο της κυκλικής πίστας. Από τον ίδιο σταθμό μπορεί να ξεκινήσει και στην περίπτωση που έχουμε Ν+1 σταθμούς (αφού για να διανύσει την απόσταση ανάμεσα στους σταθμούς Σ1 και Σ2 επαρκεί η βενζίνη του σταθμού Σ1, ενώ για την υπόλοιπη διαδρομή έχει όση βενζίνη είχε και στην περίπτωση των Ν σταθμών  :wink: )[/quote]

Σωστά. (Νομίζω πως στην περίπτωση των Ν καινούργιων σταθμών ξεκινάμε από το Σ2, τότε στην περίπτωση των Ν+1 παλαιών σταθμών ξεκινάμε από το Σ1.)

Μια πιο "συνδιαστική" λύση: Έστω ότι δεν ισχύει. Τότε για κάθε σταθμό S_i υπάρχει σταθμός S_j ώστε να μην αρκεί η βενζίνη από την S_i στον S_j. Άρα μπορούμε να βρούμε $ i_1 \leq i_2 \leq \ldots$ ώστε η βενζίνη από τον $ S_{i_n}$ στον $ S_{i_{n \plus{} 1}}$ δεν αρκέι. Αλλά (pigeonhole!) υπάρχουν m<n ώστε $ i_m \equal{} i_n$. Αυτό μας λέει ότι υπάρχει κάποιο κ ώστε αν ο κάθε σταθμός μπορεί να χρησιμοποιηθεί κ φορές τότε η βενζίνη δεν αρκεί για κ γύρους, άτοπο.

[quote="Demetres"]

Σωστά. (Νομίζω πως στην περίπτωση των Ν καινούργιων σταθμών ξεκινάμε από το Σ2, τότε στην περίπτωση των Ν+1 παλαιών σταθμών ξεκινάμε από το Σ1.)

[/quote]

Ουπς! Τον Σ2 αδειάζουμε στον Σ1!Το διόρθωσα και πάνω!
Οπότε στην περίπτωση των Ν+1 σταθμών ξεκινά όπως είπα και παραπάνω από τον σταθμό που θα ξεκινούσε στην νέα περίπτωση των Ν σταθμών για την οποία ξέρουμε από την επαγωγική υπόθεση ότι υπάρχει λύση.
Σημειώνω ότι  αυτός ο σταθμός δεν είναι απαραίτητο να είναι ο Σ1...

[quote="thanaseas"]
Ουπς! Τον Σ2 αδειάζουμε στον Σ1!Το διόρθωσα και πάνω!
[/quote]

Ωραία. Νομίζω πως η απόδειξή σου τώρα είναι πλήρης.

Παιδια εγω δε μπορω να ολοκληρωσω λυση. Οποτε Δημητρη αν μπορεις εξηγησε λιγο καλυτερα τη λυση σου. Σε χανω καπου(τι ειναι τα i_1,i_2..... το i_m=i_n απο περιστερωφολια??).

Υπαρχει αραγε ευθεια αποδειξη??

[quote="filosofimenos"]Παιδια εγω δε μπορω να ολοκληρωσω λυση. Οποτε Δημητρη αν μπορεις εξηγησε λιγο καλυτερα τη λυση σου. Σε χανω καπου(τι ειναι τα i_1,i_2..... το i_m=i_n απο περιστερωφολια??).
[/quote]

Παίρνουμε $ i_1 \equal{} 1$. Ξέρουμε ότι υπάρχει σταθμός $ i_2$ ώστε η βενζίνη δεν αρκεί για να πάει από τον σταθμό $ i_1$ στον σταθμό $ i_2$. Ομοίως υπάρχει σταθμός $ i_3$ ώστε η βενζίνη δεν αρκεί για να πάει από τον σταθμό $ i_2$ στον σταθμό $ i_3$ κτλ. Από περιστεροφωλιά θα υπάρχει m < n ώστε $ i_m \equal{} i_n$. Αλλά

Δεν υπάρχει αρκετή βενζίνη από τον  $ i_m$ στον $ i_{m\plus{}1}$
Δεν υπάρχει αρκετή βενζίνη από τον  $ i_{m\plus{}1}$ στον $ i_{m\plus{}2}$
.
.
.
Δεν υπάρχει αρκετή βενζίνη από τον  $ i_{n\minus{}1}$ στον $ i_{n} \equal{} i_{m}$

Συνολικά κάθε σταθμός εμφανίζεται κ φορές στα παραπάνω διαστήματα για κάποιο κ. (Δεν μετράω την τελευταία εμφάνιση του  $ i_{m}$.) Άρα, αν επιτρέπεται να χρησιμοποιήσουμε την βενζίνη του κάθε σταθμού κ φορές, συνολικά δεν αρκεί για να κάνω κ γύρους της πίστας από τον $ i_{m}$ στον $ i_{m}$. Αυτό όμως είναι άτοπο.


A round-robin tournament has $n$ teams, and each team plays at most once in a round. Determine the number of rounds $f(n)$ needed to complete the tournament.

can you explain more clearly what round-robin tournament is???

Round-robin:  Every team plays every other team exactly once.

In World Cup Soccer, USA lost out last week in the 4-team Group E round-robin tournament.   :(  (Only the top 2 teams in each group advanced to the round of 16 which is single-elimination.)

A round-robin tournament is one where each team plays against every other team exactly once.

[hide="Solution"]There must be a total of $\left(\begin{array}{cc}n\\2\end{array}\right) = \frac{n(n-1)}{2}$ games.

If n is odd, there can be at most $\frac{n-1}{2}$ games per round, so our answer is $\frac{n(n-1)}{2} \times \frac{1}{\frac{n-1}{2}} = n$ rounds.

If n is even, there can be at most $\frac{n}{2}$ games per round, so our answer is $\frac{n(n-1)}{2} \times \frac{1}{\frac{n}{2}} = n-1$ rounds.[/hide]


[quote="alex__ib"]How do i find first decimal of:

$\sqrt{n^{2}+7n+11}$

n is a natural number.[/quote]
You can use the binomial theorem extended.

First decimal of  7,8 is 8, that is what i mean  :)

For bpms: That's too complicated  :maybe: , i think there might be an easier way.

How about using Newton's method:

To approximate $\sqrt{n}$, pick a number $x$ and compute $\frac{x+\frac{n}{x}}{2}$.  Repeat the process with this number as $x$ until you get close enough to the root.

[quote="alex__ib"]How do i find first decimal of:

$\sqrt{n^{2}+7n+11}$

n is a natural number.[/quote]

[code]{INT[10 * sqrt(n^2 + 7n + 11)]} MOD 10
[/code]


How to convert bromocyclohexane into 6-oxononanal?

Here's one method:
[hide="click to reveal the hiddden content"]
1.aq.KOH-->cyclohexanol
2.PCC-->cyclohexanal
3.CF3CO3H-->e-lactone
4.hydrolysis-->6-hydroxy hexanoic acid
5.protect -OH by TMS-Cl
6.SOCl2-->acyl chloride
7.(n-Pr)2Cd
8.H3O+ to remove the TMS-->Reqd.Pdt
[/hide]
I know its very long I will another short synthesis :maybe:

That seems to be correct, although a final oxidation step is required.


quadratics
number theory unsolved
number theory

Determine all naturals x,y such that x^2+y^2 and x^2+103y^2 are both prefect squares..

Bomb

You need $x^2 + 103y^2 = z^2$, where $x,y,z \in \mathbb{N}$. Obviously $x = y = z = 0$ is a solution. Hence let $x^2 + y^2 \neq 0$ in the following. Since $103$ is a prime and hence a squarefree number, you can admit $\gcd(103,xz) = 1$ and argue  $(xz^{-1})^2 \equiv -1 \bmod 103$, which is impossible since $-1$ is not a quadratic residue of any prime $\equiv 3 \bmod 4$. Write your own conclusions...


AMC USA(J)MO USAMO trigonometry probability CEMC logarithms

Hello Everybody, 3 days till Euclid...

I was preparing by doing some problems from past years, and found an error in #4(b) in Euclid 2000. I got the following answers:
w=68/3   d=-7, a=b=1

Whereas the offical solutions differ for values of w and d. I read the solutions posted on the http://www.cemc.uwaterloo.ca website and found that they have made a mistake when they get equation (4) becasue I think equation (4) should have been  6w+3d=115 instead of 6w+3d=105. Anyway, I think this is a typo, I would like to know what you ppl think.
BTW, what do they do when typos come up? What are the chances that there might be some typos on Euclid 2006?  :)

Inspired By Nature
i like your signature.
it's one of the most famous quotes by einstein

I did 4(a) quick and got w=16, d=3, a=b=1

Btw, what kind of calculators do you guys use?

I've used a Ti-83 for hte past few years for math and hate using normal sci calcs. I can deal not being able to graph stuff, and loosing my newton's method and quad programs, but I like the way you can deal and store numbers with it.

I was just checkin answers and I saw what you ment, so I think it must be a typo.

Math + acid ftl?  ;)

I think thye shoulda got 4w+7d=85 for that 4th one.

I've noticed looking over tests the last few days how stupid I can do some questions though haha. It seems none of my solutions look like what they have.

I'll do a question with half a page of work, and then look at solution.....they'll have like 1 line with a simple solution and i'll just be like, "Oh..."  :blush:  /lowerheadinshame

With less than 24 hours remaining in the Euclid Contest, I would like to wish good luck to all students writing the Euclid Contest. :)

Good luck to everybody who is going to be writing the Euclid contest in just over 12 hours now! :lol:

[quote="Inspired By Nature"]Thanks Nirvantear, but what do you think about #4 b of Euclid 2000? I think there is a typo in the solutions, which then leads to incorrect values for w and d. Please ppl tell me what you think?
And feel free to use this topic to discuss other Euclid questions esp #9s and 10s. Let's get something going here.  :lol:  :lol:[/quote]
oh that question
compare sussex and som first,get B value
then compare Derbys and Som

btw good luck on euclid!

[quote="Chocoboom"]omfg, my class is going on a field trip to disneyland tomorrow  :mad:  so mad, i miss euclid and hypatia >_< freaking peice of lard  :mad:  angry...[/quote]

Dude, I'd take the DisneyLand anyday...

Sigh...good times. :(

yea disneyland is fun, gonna get myself drunk... but still, MATH CONTESTS :cursing:

I found Euclid retarded this year. The problems were nice but they were waaaaaaaaaaaaaaaaaaaaaaaaaaaay too easy. I finished it it a bit over 2 hours and i think i did quite well. I AM SO PUMPED FOR USAMO DAY 2 :)

I heard a rumour that some *acclaimed* Mathies got 0 for 10a in this year's euclid. Are these people right or are they just stupid?

ps this is Mihai

dunno if we could start posting answers here now... :?

Did you ask your teachers for the results?

Yeah...results are sent to school...
Just cornered my teacher. I got 90, and our school champ is 93.  :(
Lol...My score is exactly the same as last year-what's the probability of that? :rotfl:

[quote="Singular"][quote="Singular"]rem .. you made IMO for sure.  I'd bet $\$$500 on this.[/quote]

OOPS.
[/quote]

So if rem doesn't go to IMO, he gets 500 bucks - that's cool. Either way he gets something. :D

the UBC luncheon was today.
it was so boring, only the CS presentation was exciting.

Just got my result today, 77, highest in my school.

Btw. Congratulation to mathgod for winning the Euclid.

Wow this year some ppl pwned the euclid..like two 100's thats crazy...the scores were much higher than last years

I only got 72  :( ,

[quote="Singular"]Omfg.  My scores were 39/29/23  for the first 4 questions, next 3 questions, and last 3 questions, totalling 91.  The book prize cutoff appears to be 92.  wtf at them taking off marks for the front 7.

 :roll:[/quote]

I though the book prize cut-off was at 95.

[quote="YawningLion"][quote="YY"][quote="Sunny"]Bleh.  My teacher forgot the school code AND the password. I guess I'll have to wait and see.[/quote]

Again?
This is getting old... :rotfl:  :rotfl:  :rotfl:

So how are you going to find out?[/quote]

Hi Yiyi! It looks like we tied for Euclid Group 3, so yay! I actually did worse than last year. Moo ...

And special congratulations to Mathgod! Winning Euclid in Grade 9 is rockage. You fit your AoPS username quite well. Good job![/quote]

Thanks :P, I just made the cutoff.  :P

[quote="nirvanatear"]man you don't be so modest..
everyone knows how well you did on COMC and other contests[/quote]Don't feed his ego.  :roll:


algorithm
combinatorics unsolved
combinatorics

I don't know if this is actually a combinatorics problem, but i don't know where else to ask. Here it goes: Given $k, R \in N$ how many solutions $\{x_1, x_2...x_k\}, x_i \in N$ does the equation $ x_1 + 2\cdot x_2 + 3 \cdot x_3 + \cdots + k \cdot x_k = R $ have, such that $ x_i \neq x_j$ for every $i \neq j$

Edited by pbornsztein : to make it readable...

It has all chances to be bloody problem  ;)

[quote="hamedkazerooni"]it is very hard
but this problem solve with dynamic algorithms[/quote]
What do you mean? Computation solution?


trigonometry
analytic geometry
graphing lines
slope

$x-r=\cos \theta$ and $\frac{y+2}{r}=\sin \theta$. :wink:(deduce)
you schould learn about LATEX.

[quote="bos1234"]Find the cartesian eqn of the curve $x = r+cos/theta$, $y =-2+rsin/theta$, and describe in geometrically if:

$/theta$ is constant and r is variable[/quote]

It's like this:

$x = r+\cos(\theta)$
$y =-2+r \sin(\theta)$

[hide="Solution"]We must get rid of $r$. So from the first equation,
\[r = x-\cos(\theta)\]

Substituting into the second equation,
\[y =-2+(x-\cos(\theta))\sin(\theta)\]

Simplifying,
\[y =-2+x\sin(\theta)-\sin(\theta)\cos(\theta)\]

\[y+2+\sin(\theta)\cos(\theta) = x\sin(\theta)\]

$\textrm{Thus, the equation is a line with slope}\quad \sin(\theta)\quad \textrm{with y-intercept}\quad (0,-(2+\sin(\theta)\cos(\theta)))$[/hide]


Is there a reason there isn't a My Classes link on the AoPSWiki sidebar?

Long story.  We might be in a position to fix it now, though.  We'll look at it in the next month or two.


calculus
derivative
function
LaTeX
calculus computations

What is the meaning of "the (n+1)-th derivative of the function $ f$ vanishes identically on $ \mathbb R$?

Moreover, the original problem is:

We denote the collection of all continuous real-valued functions on $ \mathbb R$ by $ C$. Take any non-negative integer $ n$. For any $ f,g\in C$, we write $ f$~$ _n g$ if the $ (n\plus{}1)$-th derivative of the function $ f\minus{}g$ given by $ (f\minus{}g)(x)\equal{}f(x)\minus{}g(x)$ for any $ x\in\mathbb R$ vanishes identically on $ \mathbb R$.

(a) i. Prove that $ f$~$ _nf$ for any $ f\in C$.
ii.  Prove that for any $ f,g\in C$, if $ f$~$ _ng$ then $ g$~$ _nf$.
iii. Prove that for any $ f,g,h\in C$, if $ f$~$ _ng$ and $ g$~$ _nh$ then $ f$~$ _nh$.

(b) Let $ f,g\in C$, and $ c_0,c_1, \cdots, c_n$ be $ n\plus{}1$ distinct real numbers. Suppose $ f(c_0)\equal{}g(c_0), f(c_1)\equal{}g(c_1), \cdots , f(c_n)\equal{}g(c_n)$. Prove that if $ f$~$ _ng$ then $ f(x)\equal{}g(x)$ for any $ x\in\mathbb R$.

Actually, how to type "~" with latex?

"Vanishes identically" means: $ f^{(n\plus{}1)}(x)\equal{}0$ for all $ x\in\mathbb R$.

Use \sim for ~: $ f\sim g$.


Out of a set of 11 natural-number scores on a 10-point quiz, the mean
is 8, the median is 8, and the mode is 7. Find the maximum number
of perfect scores possible on this test.

By the median, the set looks something like this:

_ _ _ _ _ 8 _ _ _ _ _

If we only had 2 7's, then we can have at most 1 10. And 3 7's doesn't seem to be enough (because that would only mean 2 10's), so we try:

_ 7 7 7 7 8 _ _ 10 10 10

This could easily be completed like this (by keeping track of how much over/under the mean)

6 7 7 7 7 8 8 8 10 10 10

So would 5 7's work?

7 7 7 7 7 8 _ _ _ _ _

For the mean to be 8, there can be at most 2 10's. Therefore, the maximum possible number of 10's is 3.


[quote="hotheadhacker_mathslover"]How many integer pairs (m,n) are there that satisfy the equation

m^3+6m^2+5m-27n^3-9n^2-9n-1=0

please include steps thank you[/quote]

Assume that such pair of $ (m,n)$ exists.
Then $ m^3\plus{}6m^2\plus{}5m\equal{}m(m\plus{}1)(m\plus{}5)\equal{}m(m\plus{}1)(m\plus{}2)\equal{}0(\mod 3)$
So in the equation, $ L.H.S.\equal{}\minus{}1(\mod 3)$ and $ R.H.S.\equal{}0(\mod 3)$, contradiction.

Therefore no such integer pairs exist.

[quote="hotheadhacker_mathslover"]err, wat abt n?sry if i seem naive bt im nt familiar with mod and its usages. and how did m(m+1)(m+5) become m(m+1)(m+2)?

pls teach me[/quote]
$ m \plus{} 5 \equiv m \plus{} 2 (mod 3)$
In other words $ m \plus{} 5$ and $ m \plus{} 2$ leaves the same remainder when dividing by $ 3$...
So stephencheng proved you that such an equality cannot occur :wink:
Do you understand it now???

[quote="hotheadhacker_mathslover"]err, wat abt n?sry if i seem naive bt im nt familiar with mod and its usages. and how did m(m+1)(m+5) become m(m+1)(m+2)?

pls teach me[/quote]

See page 2 , Theorem 1.1 for some basic properties of mod.
[b]http://db.math.ust.hk/notes_download/elementary/number/ne_N2.pdf[/b]

$ m\plus{}5 \equiv m\plus{}2(\mod 3) \Rightarrow m(m\plus{}1)(m\plus{}5) \equiv m(m\plus{}1)(m\plus{}2) (\mod 3)$ from Property (2).


How many integers from 100 to 199 inclusive have no consecutive identical digits? ( For example, 121 has no consecutive identical digits, but 322 does)

A consecutive identical digit is like a digit that repeats itself consecutively in a number. For instance:

322
422
442
533
335. Like that.

It's 1*9*9.

The 1 is the hundreds digit because there can only be 1 digit for the hundred's digit because of 100-199.

The 9 is for the tens digit because it can't be a 1 because of the consecutive digit rule.

The 9 is for the one's digit because of the consecutive digit rule.

$ 1*9*9\equal{}81$.

first of all, the first digit must be 1. however, if the number is 1ab, there could be a=1; then there would be 10 possibilities for b:
110,111,112,113,114,115,116,117,118,119.
then, if a=b; there would be 9 possibilities, excluding 111:
100,122,133,144,155,166,177,188,199.
so the answer would be 100-19=81.

EDIT: oops  :blush: 100-19=81.

[quote="k00lperson"]How many integers from 100 to 199 inclusive have [b]no[/b] consecutive identical digits? ( For example, 121 has no consecutive identical digits, but 322 does)[/quote]

:wink:


polynomial
quadratics
number theory unsolved
number theory

solve the equation in natural numbers
 x^2+xy+y^2=7^n

Let $ q\equal{}exp(\frac{2\pi i}{3})$, then $ q^3\equal{}1, \ q^2\plus{}q\plus{}1\equal{}0$.
Consider numbers $ a\equal{}xq\minus{}yq^2, \ x,y\in Z$, let $ s(a)\equal{}xq^2\minus{}yq$. Then $ as(a)\equal{}x^2\plus{}xy\plus{}y^2$, and for $ a_1\equal{}x_1q\minus{}y_1q^2,a_2\equal{}x_2q\minus{}y_2q^2$ we get $ a_1a_2\equal{}y_1y_2q\plus{}x_1x_2q^2\minus{}(x_1y_2\plus{}x_2y_1)\equal{}(x_1y_2\plus{}x_2y_1\plus{}y_1y_2)q\minus{}(\minus{}x_1x_2\minus{}x_1y_2\minus{}x_2y_1)q^2$ and $ s(a_1a_2)\equal{}s(a_1)s(a_2)$.
Because $ a\equal{}2q\minus{}q^2$ satisfyed $ as(a)\equal{}7$, we get $ a^n\equal{}x_nq\minus{}y_nq^2$ satisfyed $ a^ns(a^n)\equal{}x_n^2\plus{}x_ny_n\plus{}y_n^2\equal{}7^n$.

Hey Rust. You have solved it very beautifuly!:)
How did you come up with that kind of idea? Do you have any links to practising this "method"?

Rust is using the [url=http://mathworld.wolfram.com/EisensteinInteger.html]Eisenstein integers[/url].  The generalization in dealing with representation by quadratic forms is well-known; similar ideas can be used on the [url=http://mathworld.wolfram.com/GaussianInteger.html]Gaussian integers[/url] to characterize numbers representible as a sum of two squares and on (the rings of integers in) [url=http://mathworld.wolfram.com/RealQuadraticField.html]real quadratic fields[/url] to solve [url=http://en.wikipedia.org/wiki/Pell_equation]Pell equations[/url].


We know that $ \frac{1}{3} \leq xy\plus{}yz\plus{}zx \leq 3$ .
Find The Max and Min of :
1) $ xyz$
2) $ x\plus{}y\plus{}z$

 :)

hello, my calculations give
$ \frac{1}{27}\le xyz \le 1$
the minimum will be obtained for $ x\equal{}y\equal{}z\equal{}\frac{1}{3}$ and the maximum for $ x\equal{}y\equal{}z\equal{}1$
and  we have
$ 1\le x\plus{}y\plus{}z \le 3$, the minimum holds for $ x\equal{}y\equal{}z\equal{}\frac{1}{3}$ and the maximum for $ x\equal{}y\equal{}z\equal{}1$.
Sonnhard.

$ 1 \le 3(xy \plus{} xz \plus{} yz) \le (x \plus{} y \plus{} z)^2$
$ \sqrt [3]{xy \cdot xz \cdot yz} \le \frac {xy \plus{} xz \plus{} yz}{3} \le 1$

[b]navidj[/b], are $ x,y,z$ nonnegative/positive real numbers?


calculus
integration
trigonometry
calculus computations

[quote="kunny"]For real numbers $ a,\ b$ such that $ |a|\neq |b|$, let $ I_n \equal{} \int \frac {1}{(a \plus{} b\cos \theta)^n}\ (n\geq 2)$.

Prove that : $ \boxed{\boxed{I_n \equal{} \frac {a}{a^2 \minus{} b^2}\cdot \frac {2n \minus{} 3}{n \minus{} 1}I_{n \minus{} 1} \minus{} \frac {1}{a^2 \minus{} b^2}\cdot\frac {n \minus{} 2}{n \minus{} 1}I_{n \minus{} 2} \minus{} \frac {b}{a^2 \minus{} b^2}\cdot\frac {1}{n \minus{} 1}\cdot \frac {\sin \theta}{(a \plus{} b\cos \theta)^{n \minus{} 1}}}}$[/quote]

Solution:$ I_{n \minus{} 2} \equal{} \int \frac {1}{(a \plus{} b\cos \theta)^{n \minus{} 2}}d\theta$

$ \implies I_{n \minus{} 2} \equal{} \int \frac {(a \plus{} b\cos \theta)^2}{(a \plus{} b\cos \theta)^n}d\theta$

$ \implies I_{n \minus{} 2} \equal{} \int \frac {a^2 \plus{} 2ab\cos \theta \plus{} (b\cos \theta)^2}{(a \plus{} b\cos \theta)^n}d\theta$

$ \implies I_{n \minus{} 2} \equal{} \int \frac {b^2 \minus{} a^2 \plus{} 2a(a \plus{} b\cos \theta) \minus{} (b\sin \theta)^2}{(a \plus{} b\cos \theta)^n}d\theta$

$ \implies I_{n \minus{} 2} \equal{} (b^2 \minus{} a^2)\int \frac {1}{(a \plus{} b\cos \theta)^{n}}d\theta \plus{} 2a\int\frac {1}{(a \plus{} b\cos\theta)^{n \minus{} 1}}d\theta \minus{} \int\frac {(b\sin \theta)^2}{(a \plus{} b\cos\theta)^{n}}d\theta$

$ \implies I_{n \minus{} 2} \equal{} (b^2 \minus{} a^2)I_{n} \plus{} 2aI_{n \minus{} 1} \minus{} \int(b\sin \theta)\frac {(b\sin \theta)}{(a \plus{} b\cos \theta)^{n}}d\theta$

$ \implies (a^2 \minus{} b^2 )I_n \equal{} 2aI_{n \minus{} 1} \minus{} I_{n \minus{} 2} \minus{} \left(b\sin\theta\frac {1}{(n \minus{} 1)(a \plus{} b\cos\theta)^{n \minus{} 1}} \minus{} \int b\cos \theta\frac {1}{(n \minus{} 1)(a \plus{} b\cos\theta)^{n \minus{} 1}}d\theta\right)$

$ \implies(a^2 \minus{} b^2 )I_n \equal{} 2aI_{n \minus{} 1} \minus{} I_{n \minus{} 2} \minus{} \frac {b\sin\theta}{(n \minus{} 1)(a \plus{} b\cos\theta)^{n \minus{} 1}} \plus{} \frac {1}{n \minus{} 1}\int\frac { \minus{} a \plus{} a \plus{} b\cos\theta}{(a \plus{} b\cos\theta)^{n \minus{} 1}}d\theta$

$ \implies(a^2 \minus{} b^2 )I_n \equal{} 2aI_{n \minus{} 1} \minus{} I_{n \minus{} 2} \minus{} \frac {b\sin\theta}{(n \minus{} 1)(a \plus{} b\cos\theta)^{n \minus{} 1}} \minus{} \frac {a}{n \minus{} 1}\int\frac {1}{(a \plus{} b\cos\theta)^{n \minus{} 1}}d\theta \plus{} \frac {1}{n \minus{} 1}\int\frac {1}{(a \plus{} b\cos\theta)^{n \minus{} 2}}d\theta$

$ \implies(a^2 \minus{} b^2 )I_n \equal{} 2aI_{n \minus{} 1} \minus{} I_{n \minus{} 2} \minus{} \frac {b\sin\theta}{(n \minus{} 1)(a \plus{} b\cos\theta)^{n \minus{} 1}} \minus{} \frac {a}{n \minus{} 1}I_{n \minus{} 1} \plus{} \frac {1}{n \minus{} 1}I_{n \minus{} 2}$

$ \implies(a^2 \minus{} b^2)I_n \equal{} a\left(\frac {2n \minus{} 3}{n \minus{} 1}\right)I_{n \minus{} 1} \minus{} \frac {n \minus{} 2}{n \minus{} 1}I_{n \minus{} 2} \minus{} \frac {1}{n \minus{} 1}\left(\frac {b\sin\theta}{(a \plus{} b\cos\theta)^{n \minus{} 1}}\right)$

$ \implies \boxed{\boxed{I_n \equal{} \frac {a}{a^2 \minus{} b^2}\cdot \frac {2n \minus{} 3}{n \minus{} 1}I_{n \minus{} 1} \minus{} \frac {1}{a^2 \minus{} b^2}\cdot\frac {n \minus{} 2}{n \minus{} 1}I_{n \minus{} 2} \minus{} \frac {b}{a^2 \minus{} b^2}\cdot\frac {1}{n \minus{} 1}\cdot \frac {\sin \theta}{(a \plus{} b\cos \theta)^{n \minus{} 1}}}}$


[hide="Hint"]Consider what happens when $ x\equal{}0,\ 1$.[/hide]

is this true???
[hide][img]http://img411.imageshack.us/img411/5537/mathfffsrv6.gif[/img][/hide]

The first one is right...check the last two...
[hide]
In $ [0, 1)$, it should be $ - \frac {x^2 + 1}{x + 1}$, and for $ ( - \infty, 0)$, it should be $ x - 1$.[/hide]


number theory
greatest common divisor
number theory unsolved

It seems that the problem has a little problem.
Maybe it is
$A$=($a_{1}$,...$a_{m}$) is a residue $mod$ $m$

Maybe not hard.
there are at least $\frac{m}{d}+\frac{n}{d}-\frac{m}{d}\frac{n}{d}$ number x of the form $a_{i}b_{j}$ such that $x\equiv 0(\mod d)$
but if $C(a_{i}b_{j})$ is a residue $\mod mn$
there should be only$\frac{mn}{d}$ number x such that $x\equiv 0(\mod d)$
 contraduction.

[quote="Hawk Tiger"]Maybe not hard.
there are at least $\frac{m}{d}+\frac{n}{d}-\frac{m}{d}\frac{n}{d}$ number x of the form $a_{i}b_{j}$ such that $x\equiv 0(\mod d)$[/quote]

Why we have this?


Give all values of each of the following in the form $ a \plus{} ib$.

$ [( \minus{} 1)^4]^{\frac {1}{6}}$

$ [( \minus{} 1)^{\frac {1}{6}}]^4$

OK.  For the first one, $ ( \minus{} 1)^4$ is clearly one, so the values are simply $ cos(2k \pi/6) \plus{} i sin(2k \pi/6)$ with $ k \equal{} 0, 2, 3, 4, 5$ or $ cis(0), cis(\pi/3), cis(2\pi/3), cis(\pi), cis(4\pi/3), cis(5\pi/3)$.

So there are 6 distinct complex roots.

However, using de Moivre's formula, we get

$ ( \minus{} 1)^4$ = $ cis(4\pi)$ and $ [( \minus{} 1)^4]^{\frac {1}{6}} \equal{} cis((4\pi \plus{} 2k\pi)/6) \equal{} cis((2\pi \plus{} k\pi)/3), k \equal{} 0, 1, 2.$  So the 3 values are $ cis(2\pi/3), cis(\pi), cis(4\pi/3)$... now there are only 3!  What's wrong with this?  Why does de Moivre's formula betray us?  :huh:

I think we are regarding $ [( \minus{} 1)^4]^{\frac {1}{6}}$ as $ ( \minus{} 1)^{\frac {4}{6}}$ ...

Also, consider these two possible definitions of $ z^{\frac {m}{n}}$:
$ z^{\frac {m}{n}} \equal{} (z^{\frac {1}{n}})^m$
$ z^{\frac {m}{n}} \equal{} (z^m)^{\frac {1}{n}}$.

a) Assume $ m/n$ is an irreducible fraction.  Do the two definitions always yield  the same numerical results?  Support your claim.
b) Assume $ m/n$ is a reducible fraction.  Do the two definitions always yield  the same numerical results?  Support your claim.

[quote="mathwizarddude"]
However, using de Moivre's formula, we get

$ ( \minus{} 1)^4$ = $ cis(4\pi)$ and $ [( \minus{} 1)^4]^{\frac {1}{6}} \equal{} cis((4\pi \plus{} 2k\pi)/6) \equal{} cis((2\pi \plus{} k\pi)/3), k \equal{} 0, 1, 2.$  So the 3 values are $ cis(2\pi/3), cis(\pi), cis(4\pi/3)$... now there are only 3!  What's wrong with this?  Why does de Moivre's formula betray us?  :huh: [/quote]

Why did you decide that $ k$ can't go any higher than 3?  Inherently in your explanation, $ k$ can be any integer.

Write in modulus-argument notation $ z \equal{} (r,\theta)$.  Then obviously the modulus in both definitions is equal to $ r^\frac {m}{n}$.  Now consider the argument.  Now, the argument of $ z^\frac {1}{n}$ is clearly $ \frac {\theta \plus{} 2k\pi}{n}$, where $ k$ takes any integer value (although of course only $ n$ of them produce different results), so the argument in the first definition is $ \frac {m\theta}{n} \plus{} \frac {2km\pi}{n}$, while in the second definition the result is clearly $ \frac {m\theta}{n} \plus{} \frac {2k'\pi}{n}$, where $ k'$ takes all integer values.  We only need to prove that when $ k,k'$ take all possible values in $ \{0,1,2,...,n \minus{} 1\}$, the sets of values in both definitions are the same (modulus $ n$, of course).  If $ \gcd(m,n) \equal{} 1$, the result is trivially true since the remainder of $ mk$ modulus $ n$ takes all possible values; otherwise, $ k_1 > k_2$ exist such that $ n|m(k_1 \minus{} k_2)$, or $ n|k_1 \minus{} k_2$, absurd since $ \gcd(m,n) \equal{} 1$ and $ n > k_1 \minus{} k_2 > 0$.  In the case where $ \gcd(m,n)\neq1$, the result is not true, since any two values of $ k$ which leave the same remainder modulus $ \frac {n}{\gcd(m,n)}$ in the first definition will yield the same final result modulus $ n$.  So the second definition produces $ n$ roots, while the first definition produces only $ \frac {n}{\gcd(m,n)}$ roots.  This is just due to the fact that $ (z^p)^\frac {1}{p}\neq z$, in the sense that the LHS has $ p$ possible values, but the RHS has a single, well defined value...

IMHO, this is no paradox, just the consequence of the multi-valued nature of the definition of the $ n$-th root in $ \mathbb{C}$...

Lesson to be learned: be careful when you simplify exponents in $ \mathbb{C}$!!!  :lol:


The product $ (1.8)(40.3\plus{}.07)$ is closest to

\[ \textbf{(A)}\ 7 \qquad
\textbf{(B)}\ 42 \qquad
\textbf{(C)}\ 74 \qquad
\textbf{(D)}\ 84 \qquad
\textbf{(E)}\ 737
\]

Estimate, the product is clearly more than 7 and less than 84.  42 would seem more reasonable with something like 1.1, so our answer is C.

[hide="Click for solution"]
The desired product is about $ 2(40)\minus{}.2(40)\equal{}80\minus{}8\equal{}72$, so $ \boxed{\textbf{(C)}}$ is correct.
[/hide]

Expanding, we get $1.8*40.3+1.8*0.07$. Note that $1.8*0.07$ is very small, so we can ignore it. Just eyeballing, we get that $1.8*40.3$ is closest to $74$ and not $84$ because 1.8 would have to be at least like 2 or something.


A body moves with velocity 5m/s towards east and then in the 10th second,it moves 5m/s north.Find the average velocity of the body from t=0 to t=10seconds.

haha, old post, but since this problem is easy, i thought i would take a crack at it.

So from the information given in the problem, i will assume that the body moved 45 meters east and 5 m north. Thus, the total distance from the starting point is given by the pythagorean theorem, which gives (45^2 + 5^2)^1/2 = about 45.3 m. The direction is given by arctan (5/45) = 6.34 degrees. So the average magnitude is 4.53 m/s.

Thus the final answer is: 4.53 m/s @ 6.34 degrees north of east.

actually we had to find the acceleration,i wrote average velocity bymistake

just note that $\bar{a}=\frac{d\vector{v}}{dt}$


calculus
derivative
limit
real analysis
real analysis unsolved

I need to show that lim[cos(x^2)/x] as x goes to infinity is zero and the limit of its derivative does not exist.

Thank you for any help

$ \minus{} 1 \leq \cos(x^2) \leq 1$

$ \equal{} > \minus{} \frac1x \leq \frac {\cos(x^2)}{x} \leq \frac1x$

Since $ \lim_{x \to \infty} \minus{} \frac1x \equal{} \lim_{x \to \infty} \frac1x \equal{} 0$, then by the Sandwich Theorem we also have $ \lim_{x \to \infty} \frac {\cos(x^2)}{x} \equal{} 0$.

On the other hand, $ \lim_{x \to \infty} \left(\frac {\cos(x^2)}{x}\right)' \equal{} \lim_{x \to \infty} \frac{\minus{}2x^2 \sin(x^2) \minus{} \cos(x^2)}{x^2} \equal{} \lim_{x \to \infty} \left(\minus{}2 \sin(x^2) \minus{} \frac {\cos(x^2)}{x^2}\right)$ which do not exist since $ \lim_{x \to \infty} \sin(x^2)$ do not exist.


[quote="m4467"]find $ x,y,z\in\mathbb{N}$ (greater than 1) such that $ xyz \minus{} 1$ is divisible by $ (x \minus{} 1)(y \minus{} 1)(z \minus{} 1)$[/quote]
This is a very famous problem(and very hard, of course)! You can find its solution by google it. Dr. Kiran Kedlaya also has an article about this problem.It's also a PEN Problem(number A1)(find the solution in PEN forum)
@Moderator: This problem should be moved to Olympiad Number Theory.


$n>1$ $a_1,....,a_n,b_1,....,b_n$ are 2n real numbers. $\sum_{i<j}a_ia_j > 0$.
Prove that:$(\sum_{i\neq j} a_ib_j)^2\ge (\sum_{i\neq j} a_ia_j)(\sum_{i\neq j} b_ib_j)$

I have a solution:
 the ineq is equivalent to:
$((\sum_{i,j=1}^{n}{ a_ib_j})^2-(\sum_{1}^{n}{a_i})(\sum_{1}^{n}{b_j}))\geq (\sum_{1}^{n}{a_i^2}-(\sum_ {1}^{n}{a_i})^2)(\sum_1^n b_i^2-(\sum_1^nb_i)^2)$
seperat in two case:
case1:$\sum_1^n b_i^2-(\sum_1^nb_i)^2)\geq 0$
so the left side of the ineq $\geq 0$,but the left $\leq 0$
case 2:$\sum_1^n b_i^2-(\sum_1^nb_i)^2)\leq 0$
 the ineq is equivalent to
$(\sqrt{\sum a_i^2} \sum b_i-\sqrt{\sum b_i^2} \sum a_i)^2+(\sqrt{\sum a_i^2 \sum b_i^2}-\sum a_ib_i)(2(\sum a_i)^2(\sum b_i)^2-(\sqrt{\sum a_i^2 \sum b_i^2}+\sum a_ib_i))\geq 0$
we note that  $\sum a_i^2\leq (\sum a_i)^2,\sum b_i^2\leq (\sum b_i)^2$
thus
$\sqrt{\sum a_i^2 \sum b_i^2}+\sum a_ib_i\leq 2\sqrt{\sum a_i^2 \sum b_i^2}\leq 2(\sum a_i)^2(\sum b_i)^2$
so we prove it


modular arithmetic
number theory proposed
number theory

Let $X_{0}=1$, $X_{1}=3$ and $X_{n}=6X_{n-1}-X_{n-2}$ for all $n\ge 2$

Prove that:

A) if p=5 then for every natural n $p\nmid X_{n}$ (from the Engel)

B) if $p\equiv 5\pmod 8$ then for every natural n $p\nmid X_{n}$ (my own generalization)

Define $Y_{n}$ by $Y_{n}=0,\ Y_{n}=2,\ Y_{n}=6Y_{n-1}-Y_{n-2}$ for all $n\geq 2$. Then we have
\[X_{n}^{2}-2Y_{n}^{2}=1 \]
for all $n$. In particular, -2 must be a square modulo $p$ if $p$ divides $X_{n}$, so $p\equiv 1,3\pmod{8}$. Therefore $p\equiv 5\pmod{8}$ (and $p\equiv 7\pmod{8}$) are not possible.

[quote="jb05"][quote="scorpius119"]In particular, 2 must be a square modulo $p$ if $p$ divides $X_{n}$, so $p\equiv \pm 1\pmod{8}$.[/quote]

Don't you mean $-2$ must be a square modulo $p$ so $p\equiv 1$ or $3 \pmod{8}$, so it is impossible if $p\equiv 5$ or $7 \pmod{8}$?[/quote]
You're right. Edited. (After all, of course 3 divides some $X_{n}$...   :blush:  )


Explain the stability of $ \ce{CLi6}$. Can $ \ce{CH6}$ exist? Why or why not.

Li being highly electropositive is compenstating for the eg value of C. Also Li has a high oxdn potential and hydration energy. So all these compensate for the C. :maybe:

CH6 does not exist because C-H bond if formed will be covalent and so carbon with no d-orbital cannot extend its covalency beyond 4.

i really don't find  a point of having such a question...if a compound exists then it does exist...ok u may ponder over how we can make it but why it exist will take us a t some point into some kind of conjecturng whch is equivalent to conjetcuring the existance of the molecule

BanishedTraitor, can you describe the compound $ CLi_6$?

[quote="BanishedTraitor"]Sorry Pardesi clearly we have different areas of interest  :P . I find these questions fascinating. I even find things like why does $ \ce{BeH2}$ adopt a linear rather than a bent geometry intriguing. That's also because I am brushing up on Molecular Orbital Theory.[/quote]
ye surely... :D
except for the inequalities part :P
i would rather beleive on netonian mechanics that assumed the existance of mass and force and then put in the laws of motion to ...that is tell that some force must be acting on a body because it was moving than argue the other way round :)

[quote="Carcul"]Well, you stated above "Explain the stability of $ CLi_6$", so you must know from the start what is the structure of such compound. I am asking this because I have never heard of such species, and would like to know at least what it looks like.[/quote]

It adopts an octahedral geometry. It belongs to a class of compounds called hyperlithiated molecules.

Very interesting. Can you provide a link with some relevant information?

I am trying to find some myself. This was part of a problem concerning hypervalent species.

as far as i know, another species(CH5+), has been detected in solution by Olah. and it adopts a trigonal bipyramidal geometry and c has 8 electrons in its valence shell.

Some other compounds have been detected with multi center bonds having higher coordination number-
Al2(CH3)6                                                                         5
B10C2H10R2                                                                     6
Li4(CH3)4                                                                          7
[Co8C(CO)18]2-                                                                8
(Source: J.D. LEE)

And Banished, are you sure of the structure, I was beginning to think that probably there was (Li3)+ involved in the compound.


There are AMC Problems and Solutions on AoPSWiki. I would prefer to print it when I am doing the test.

But I have trouble to print those LeTaX symbols.

See this url for 2008 AMC 10A Problems: http://www.artofproblemsolving.com/Wiki/index.php/2008_AMC_10A_Problems

After I print preview or print, all the LeTaX symbols were missing.

Anybody knows how to print ?

[quote="worthawholebean"]I'd suggest printing from the [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182]resources section[/url] instead. Use the PDF download.[/quote]

But those pdf version does not show graph, I tried 2008 AIME. Instead, it converted graph as:
[asy]import math; unitsize(5mm); defaultpen(fontsize(9pt)+Helvetica()+linewidth(0.7));
pair O=(0,0); pair A=(0,sqrt(17)); pair B=(sqrt(17),0); pair C=shift(sqrt(17),0)*(sqrt(34)*dir(75));
etc, etc


geometry
trigonometry
perimeter
area of a triangle
Heron\u0027s formula
trig identities
Law of Cosines

Find the area of a triangle with sides 11,12,13.

There's definately something I haven't learned yet in this one.  I can draw out the triangle and make an altitude to the hypotenuse.  I've no clue how to get the measure of it though.  Am I on the right track?

One solution is to use heron's formula:

$ \text{[ABC]}\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$ where $ s\equal{}\frac{a\plus{}b\plus{}c}{2}$

another is to use the law of cosines to find one of the angles:
$ \frac{a^2\minus{}b^2\minus{}c^2}{\minus{}2bc}\equal{}\cos{\angle{A}}$
Then you can plug into the formula
$ \text{[ABC]}\equal{}\frac{bc\sin{\angle{A}}}{2}$

[hide="The answer is"]
$ \sqrt{18(7)(6)(5)}\equal{}6\sqrt{105}$[/hide]

Ah thanks.  I only had half the formula.  I had no clue what s was (semiperimeter now that I think of it).

The semiperimeter is half of the perimeter of the triangle.

for those who don't know heron's formula this could be done like this ....

a=11=AC
b=12=BC
c=13=AB

let v be the hight on c

let x be the distance from the point where height intersects with c and A

then we have
121=v^2+x^2
144=v^2+169-26x+x^2

we get
23=169-26x....

from here we can calculate the x and then v ...


function
LaTeX
continued fraction
real analysis
real analysis unsolved

Hello,
I have this sequence which I want to show that converges:
given as a function, beta, of positive integer variable d
the image of the formula is at the link:
[url]http://img152.imageshack.us/my.php?image=seminarbx4.gif[/url]

and beta(3) = 4/3
I have shown it's bounded between 1 and 2 and that if a limit exists then it must be (1+Sqrt(5))/2

hence, if I show it's monotinic, It converges and i am done.
I do not succeed showing that it's monotonic (though numeric calculation shows it probably is) and I haven't succeeded in any other way as well.

I would appreciate any help.

Thank you

You can write $ \beta(d)$ in terms of $ \beta(d\minus{}m)$ (reducing the numerator and the denominator by $ d\minus{}2$ every time) for a large fixed $ m$ and look at the resulting expression as $ d\to\infty$. Then it is basically the $ m$-th continued fraction for $ \frac{1\plus{}\sqrt 5}2$ in which the tail is replaced by something between $ 1$ and $ 2$. The standard proof of convergence for continued fractions works just fine after that (if you forgot it, all you need for this particular case is that $ x\mapsto \frac 1{1\plus{}x}$ is a contraction on $ (0,\plus{}\infty)$).

Hello, Thank you for the quick reply, I am sorry I could not respond earlier.
I have never studied continued fractions, so I dont know what is the standard way of proving thier convergence.
I have managed to find on the internet some lecture notes about the subject, I tried reading them, but I still don't see how to apply things to my case.

I tried something like this, as you suggested:
$ \beta(d)\equal{}1\plus{} \frac{d\minus{}2}{d\minus{}1 \plus{} \frac{d\minus{}2}{\beta(d\minus{}1)}} \equal{}
1\plus{} \frac{1}{\frac{d\minus{}1}{d\minus{}2} \plus{} \frac{1}{\beta(d\minus{}1)}} \equal{}
1\plus{} \frac{1}{\frac{d\minus{}1}{d\minus{}2} \plus{} \frac{1}{1\plus{} \frac{1}{\frac{d\minus{}2}{d\minus{}3} \plus{} \frac{1}{\beta(d\minus{}2)}}}}\equal{}...$

So If I understand correctly I get something along the lines of:
$ 1\plus{} \left[\frac{d\minus{}1}{d\minus{}2},1,\frac{d\minus{}2}{d\minus{}3},1,\frac{d\minus{}3}{d\minus{}4},1,....,1,\frac{d\minus{}m\plus{}1}{d\minus{}m},K\right]$ when $ 1\le K \le2$

How do I show convergence to $ \frac{1\plus{}\sqrt{5}}{2}$ ?
I am really lost with this one...

Thank you again.

By The Way,
 Thanks for suggesting the LaTeX, it's much more rapid to write math this way

OK, once you wrote that, you can safely conclude that every partial limit of the sequence is of the form $ F^m(K)$ where $ F(x)\equal{}1\plus{}\frac 1{1\plus{}\frac 1x}\equal{}2\minus{}\frac 1{1\plus{}x}$ and $ K\in [1,2]$ for arbitrarily large $ m$ (just pass to the limit when $ d\to\plus{}\infty$ and choose a convergent subsequence of $ K$). But $ F^m$ converge to the desired limit uniformly on $ [1,2]$ as $ m\to\infty$.

Also, this argument can be restated in the following way. Suppose that we have a sequence of positive numbers $ s_d\to 1$ and $ x_d\equal{}1\plus{}\frac {1}{s_d\plus{}\frac 1{x_{d\minus{}1}}}$. We need to prove that $ x_d\to \frac{1\plus{}\sqrt 5}2$. Generalize like this: assume that $ F_d: X\to X$ is a set of uniform contractions such that $ F_d\to F$ uniformly on $ X$. Then the sequence defined by $ x_d\equal{}F_d(x_{d\minus{}1})$ tends to the fixed point of $ F$. Now, to follow this line, you need to know the proof of the Banach fixed point theorem. Are you comfortable with that?

Hello, Thanks for helping,
As I am more familiar with banach fixed point theorem than with continued fractions I choose to use this way of proving.
well, I know how to prove banach theorem, but I guess to dont alter the scheme well enough since it does not work for me.

what I tried is to fix some $ \epsilon>0$, then
now, As I could check in this case, all functions $ F_d$ and $ F$ involved are lipsitz with M$ \le\frac{1}{4}$
From Uniform convergence, the series is uniformly cauchy so let $ N>0$ be the place where the elements get closer than $ \epsilon$ (uniformly on $ [1,2]$) for ease of writing let's truncate the series at this point and regard Nth place as 0th place.

for some $ k>0$
now,
 $ \left|x_k\minus{} x_{k\minus{}1}\right|\equal{}\left|f_{k\minus{}1}(x_{k\minus{}1}) \minus{} f_{k\minus{}2}(x_{k\minus{}2})\right|\le\left|f_{k\minus{}1}(x_{k\minus{}1}) \minus{} f_{k\minus{}1}(x_{k\minus{}2})\right|\plus{}\epsilon<\frac{1}{2}\left|x_{k\minus{}1} \minus{} x_{k\minus{}2}\right|\plus{}\epsilon \equal{} \frac{1}{2}\left|f_{k\minus{}2}(x_{k\minus{}2}) \minus{} f_{k\minus{}3}(x_{k\minus{}3})\right|\plus{}\epsilon\le\frac{1}{2}\left|f_{k\minus{}2}(x_{k\minus{}2}) \minus{} f_{k\minus{}2}(x_{k\minus{}3})\right|\plus{} \left(1\plus{}\frac{1}{2}\right)\epsilon\le\left(\frac{1}{2}\right)^2\left|x_{k\minus{}2} \minus{} x_{k\minus{}3}\right| \plus{} \left(1\plus{}\frac{1}{2}\right)\epsilon\le ... \le\left(\frac{1}{2}\right)^k\left|x_{1} \minus{} x_{0}\right| \plus{} \left(1\plus{}\frac{1}{2} \plus{} \frac{1}{2}^2 \plus{} \frac{1}{2}^3 \plus{} ... \plus{} \frac{1}{2}^{k\minus{}1}\right)\epsilon<\frac{1}{2}^k\left|x_{1} \minus{} x_{0}\right|\plus{}2\epsilon$

therefore inductivley, in the same process such as in banach thoerem proof i can conclude:
let k > m
$ \left|x_k\minus{}x_m\right| \le\left|x_k\minus{}x_{k\minus{}1}\right| \plus{} \left|x_{k\minus{}1}\minus{}x_{k\minus{}2}\right| \plus{} ... \plus{}\left|x_{m\plus{}1}\minus{}x_{m}\right| \le C\frac{1}{2}^{k} \plus{} 2(k\minus{}m)\epsilon$
(C is a constact containing all that come out during the expansion, does not depend on k)
the thing is that now epsilong gets multiplied by k-m, and by this stage epsilon is fixed, (in fact k-m depend of epsilon since we chose N according to epsilon)

What did I do wrong?
Can you please guide me further?

Have a nice day.


calculus
integration
number theory unsolved
number theory

Can we find infinitely many integer solutions to equation  $ 8c^3 \plus{} 1 \equal{} a^2$ ?

[url=http://en.wikipedia.org/wiki/Siegel%27s_theorem_on_integral_points]Siegel's theorem[/url] says no and [url=http://en.wikipedia.org/wiki/Mih%C4%83ilescu%27s_theorem]Mihailescu's theorem[/url] states that the only solution is $ c \equal{} 1, a \equal{} \pm 3$.


factorial
number theory unsolved
number theory

Hello,

1000 000! write   two based form, how many digit numbers?

ihsan YÜCEL

I think you missread the question.
6! has 10 digits.

6-4=2, but 110 (6?) is clearly not the number of digits in 6!.

Can you do it with sterlings formula? A friend of mine is something of an expert on estimating factorials, and there a number of ways of doing so.

I should have know better than to miss the factorial sign!

In that case, I have no clue how to convert the decimal factorial into binary, or more importantly how to find the number of digits in the binary form of 1000000!

I think I'm still new enough to get away with blaming this one on naivity; hope my simple errors don't persist.


Are we assuming x is an integer? (it's impossible otherwise)

$ 51 \equiv 3 \bmod{12}$, not $ 5 \bmod{12}$.

krishkoushik was correct: $ 3x \equiv 0,3,6$ or $ 9 \bmod{12}$.

We can also note that the difference between $ 3x$ and $ 5$ must be a multiple of $ 12.$ Thus, we have $ 3x\minus{}5\equal{}12y,$ for some integer $ y.$ Rearranging gives: $ 3x\minus{}12y\equal{}5 \implies 3(x\minus{}4y)\equal{}5.$ This means that $ 5$ can also be broken down into a product of $ 3$ and some other number. But that is not the case since $ 5$ is prime. Thus, the given congruence has $ \boxed{\text{no solutions}}.$


AMC
AIME
USA(J)MO
USAMO
USAMTS
trigonometry
AMC 10

What did you guys get? I didn't take it though.
Feel free to discuss.

can someone post the answers? we don't know

Same. I took the AIME I. I'm going to take a look at the problems now.

10 here. Made a bunch of mistakes. Unless i'm missing something relatively trivial, 12 was by far the hardest problem on the test - 13 was relatively straightforward with complex numbers and roots of unity, 14 was easy [b]if[/b] you saw the trick, and 15 required some massage to get it into a managable form but little else.

Got an 8. Hopefully I didn't misbubble any of them >_>

I made like 5 or 6 careless mistakes, I got a 6 or 7.

I got a 7. I mad a careless mistake on #4, got #12 wrong, and didn't realize that I had already seen #13.

It seemed harder than the I. Thoughts?

8. Not as good as I would of hoped, but good enough for USAMO (216.5 index). It definitely seemed harder than I (especially the first questions). 1-6,8,10 correct, 11,13 wrong.

[quote="#H34N1"]ERm I got a 5 since I made 8 stupid mistakes, maybe since I was like sick yesterdya and came just to take AIME....mieh

I thought this was easier but w/e

Also I spent a very long time on #12 and didn't get it
I brute forced #14 (that was teh sequence right?) and got it right xD[/quote]
thats too bad it seems like you could have got much higher

I thought this test was way harder than the AIME I. Numbers 6-9 on AIME I were so much easier than 6-9 on this one.

It seems AIME II is much harder than AIME I (though I haven't looked at the problems for the sake of future practice)...prediction for split index?

I did way less than what I should get. In practices I got like around 10s consistently. Taking the 2009 AIME I as practice, I answered like around 11. So some how I get 96 on USAMTS and 112.5 and 4 on AMC/AIME...(7 last year). This has been a terrible year for me.

I got a 9. Should have gotten [1,11], but only got [1,8] and 10.

Got a 10. Got 1-11 except 4. I thought this test was actually relatively easy, easier than the AIME I.

Didn't take the AIME I (yes I know, scandalous from a preparation standpoint) so I can't say comparatively but I got an 8 (#2-4, #6-10). Arithmetic errors on #1 and #11 (164+188 is not 354 and 102/3 is 34 respectively). I should have spent the last hour brute forcing 14 or at least read every problem during the last hour so I wouldn't have kept going sequentially.

6. I got 1, 3, 4, 6, 7, 8. I guessed #8 right. Yay me.

I spent more than an hour on #5 because I misread it the first time and thus proceeded to try a variety of things before using trig, which took forever and gave me an irrational answer. Then I read it correctly. I still got it wrong because I fail the Law of Cosines.

On #4, I thought k was just a specific contestant so that took me forever.

Between those, I didn't get to do very many problems. I also shouldn't have missed #2, but whatever.

[quote="blootzboy"]you took a nap during aime?[/quote]
Lol.

I made a bunch of mistakes, and I should have been more careful.

I got a $ 10$. I consider myself very lucky to have guessed a question right.

I'm really pissed off. I continued the tradition of making retarded mistakes on the AIME, so I got an 8, even though I did 12 problems. With a 200 index, it looks like no USAMO for me unless its 2007 all over again.

[quote="funtwo"]dude, i will kill myself. i made 2 careless mistakes and ended up with a 10  :mad:[/quote]Chill out. 10 is a very good score.

[quote="Ignite168"][quote="MCNational!"]i got a 5!!!!!!!!!! :o
PHAIL PHAIL PHAIL PHAIL

i got 1, 2, 3, 4, 9....

shoulda got 5.... i miscalculated and i had 7 but time ended.... Should've also got 8.... UBER PHAIL EPIC PHAIL UBER PHAIL EPIC PHAIL... I think i might've had some eraser marks ugh!

(shouldnt have taken that nap)  :P[/quote]

5 is not fail; many people I know got scores like a 2 and a 4. It honestly comes off as rude to complain about high scores..[/quote]

Substitute "10" for "5" in Ignite's post.

Nd yes i did take nappy nap nap.... :D

its 3 hours long i only used like 1 hour lol

but my nap was only like 1/2 hour

I got an 8, but only 2 better than last year (where I didn't even know how many questions were on AIME, or what it stood for, until the day of the test.)

I guess that's fine, but I missed number 5 (got 38/7 instead of 27/5), messed up number 10 (think I was careless), blanked out on number 7, and spent too long on number 8 before I got it, to get 5,10, and 7.

My index for AMC10+AIME II is 224 so I guess I'm okay, but I wish I'd been able to take AIME I, because I thought it was much easier and I got an 11 (my goal) in practice.

I got a 6...but an 8 last year...But it was much harder than AIME I

I did #1-4, 7, 8, and I liked how many of the answers contained 4 and 1 (4/1/09)...like #1, 3, 7, 8, and something else I think

This AIME was much harder than the AIME 1. I estimate I would have gotten ~10-11 on the I, since I got a 13 in practice. I got an 8 on this. I honestly had to skip #2 and come back later--I know it wasn't difficult, but it was more difficult than a standard #2. I made a careless oversight on #5 (Sigh the circles weren't tangent at the points of intersections of A and B, C and D...), was simply dim on #6, screwed up #11 (wrong rounding function, tried to solve equation which I believe has no solutions...), was just silly on #12, didn't have time to work on 13-15 much. Though I wish I'd forgone working on #12 for #13...I knew that identity...

Some dumbass in our school decided to pull the fire alarm so our school will probably be disqualified.　I think that I would have gotten a 4-6．Blah, ended up with 673 for number 13, gross...I suck at counting and number theory, which dominated this AIME.

@myrealname: that sucks. It's actually not that bad that you think you only got 4-6. If you had gotten higher, you would've been even more pissed.

I took this AIME II unofficially and got a 7.

Meh...I'm still sad that both this and my score from the AIME I (a horrible 6) wouldn't get me to USAMO :(

lol no one here should be talking bout 'epic phailing' :P 'cept me. lol fourr! >< woot....
*sighs* oh wells. thats what i get for studying for 3 hrs the night before. -.-
however, i did double my score. lmfao. =___='' embarrassing. i'll deal. X_X

Stop using the word "fail." Everybody. Seriously. It's really, really inconsiderate to those who did not do as well as you. This is not a forum rule, just something to think about when you post - try not to make others feel bad.

I got 11...no stupid mistakes!!!!  :lol:

spent a lot of time on 12-15, but didn't end up actually getting any of them...

I made 3 stupid mistakes, got lucky on one, and got a score of 5.

On the rectangle on I solved for AE, instead of AD. On the dice one I found that it was the sum of one sequence + 2 times another sequence. I forgot to multiply by two. On the one about the subset of 14 I missed one case out of 18.

So, my methodology was correct on 7, but I forgot a detail on three of them.

I got 26 off my USAMO index because I failed to multiply or divide by 2 during a problem.

Actual: 120/5 Without tiny errors: 132/8

I also think this one was much harder.  I got a 6 (same as last year), with #1-4,6,8.  Made a stupid mistake on #7 and didn't realize that $ 244\plus{}251\neq 295$, so I was 200 off in my computation of $ a$, didn't know how to find $ b$ but realized that I could only get an answer I could bubble if $ b\equal{}1$.  Wasted plenty of time on 9 and 10 - on 10, I definitely overcomplicated it.  But I just went through it twice and got $ 40\sqrt{13}/11$ both times, which does not agree with the official answer I've seen.  Anyway, going to look up full solutions.

I did the AIME I for practice and also got a 6 but only due to stupid mistakes.  Take out all my algebra and arithmetic errors and I would have had 11 easy, with possibly another 2 points - one question, with the guessing the values of the prizes, I forgot to factor in that the prizes were only up to 4 digits so didn't subtract out the 5 digit prices.  And then the choosing points on a line to get triangles with integer areas, first I missed the 1/2 in my shoelace and then I found how many points I could choose on my line.  I think I accidently did it as only positive integer coordinates or nonnegative or something, when it should have been the other, but then forgot I needed to find the ways to choose  2 points.

Anyway, I have a feeling that I would have done a bit better on the 2 if I hadn't been taking it officially, but not nearly as well as I could have done on the I.  My guess is that the scores on the II will be about 2 or 3 points lower than on the I.

anyway, my index is $ 115.5\plus{}60\equal{}175.5$.  Probably didn't make it, but will see.  depends on how much harder they decide it was than the AIME I.

facis what grade are you in??
so for people that are 10th grade or below, do they still use the index thingy (if so, it would be advantageous for those that took amc10...)???

I'm a senior (only for another month!)

btw, I figured out my error on #10.  for some reason I wrote 12/15 where I was supposed to have 12/5.  In any case, I wasn't even close to solving it on the contest.  I tried stewarts after finding a few things, and then massively failed with that.  Got an answer that wasn't the correct format.

For 10th and below, they use the AIME score compared to the floor, which is the advantage, but the index comes into play if too many people qualify from the floor.

OK, but what's "floor" exactly? I really hate it when they tell you the minimum score in order to qualify *after* the exam... (for example, it turns out that the minimum to qualify the aime through amc12 A is 97.5, which i thought to be 100 previous the exam... so annoying!!! also, i thought for the aime is 8 for freshman... btw, i just suck at exams where they gave you so little time limits... luckily not for aime with 3 damn good hours)
Ugh, why don't they just gave the same minimum score every year??? It makes no sense...

BTW, for next year, do you think it would be better for me to take amc 10 or amc 12? frankly i totally suck at tests with many problems and little time limits... that's why i took amc 12 this year (only answered 14 and just barely barely qualify) since i gotta get 20 more pts if i take the amc 10... the difficulty is not the problem... only the time

if you really think that you'll get the same score no matter which test you take, then the 12 would be better.  But that just doesn't seem reasonable.  Anyway, I would suggest that you do some practice on your own and try to figure it out for yourself.  You can find old AMCs in the resource section [url]http://www.artofproblemsolving.com/Forum/resources.php?c=182[/url].  Beware that AMC's given at the same time often overlap for a few problems.

As for the qualification scores - for the AIME, qualification scores vary very little.  It's just top 5% or 100 for AMC 12, whichever's lower.  It just so happens that the last 2 years top 5% has been 97.5.  but 2.5 points on the AMC is not much.

On the other hand, on the AIME 1 or 2 questions is an enormous variance in qualification for USAMO.  The main issue is that their goal is to get 500 USAMO competitors, so they set the cutoffs to produce this group.  Since AIME's and AMC's vary in difficulty (AIMEs more than AMC's it seems), I don't think its reasonable to have a specific limit.

As for time... I think that the 3 hours on the AIME is only really useful if you don't waste time.  I find it tempting to try to brute force some of the problems and then realize it will waste too much time and likely produce an incorrect answer anyway.  Since the problems are so long, arithmetic and algebra errors can easily screw me up.  Idk.

10 here: 1 through 11 without 8.
I thought this test was relatively hard conceptually but also easy to avoid mistakes (I am typically extremely careless and have been all the way up through this year.) I somehow miscalculusized #15 (got 8-2rt15), but that's okay.


trigonometry
invariant
geometry
arithmetic sequence

It's well known that the sum of the distances from any interior point of a regular polygon to it's sides is a constant.

If we loose the condition a little bit, the polygon has equal interior angles but not equal sides, is the statement still true?

Polygons with even numbers of sides:

[hide]If we are dealing with an even-sided polygon, the condition stated is equivalent to saying that opposite pairs of sides are parallel.  In this case, we see that no matter where the point is placed inside the polygon, the sum of the distances between it and that pair of sides is constant.  Adding around the polygon, we see that the sums of the distances from any point to the edges is fixed.[/hide]

The odd case is not so easy.

The statement is true for equiangular polygons.  I will outline one possible argument, whose central theme is the equation shown below:

[hide]Suppose the sides are S1, ... , Sn.  First, verify that from any point P in the interior the angles formed by dropping perpendiculars to the Si are all congruent.  Consider another point P' in the interior of the polygon and make a trig argument based on the fact that

[tex]\sum_{i=1}^{n} \cos(\alpha + \frac{2i\pi}{n}) = 0[/tex]

to show that the sums of the projection distances from P and P' are equal, which is sufficient to prove the claim.[/hide]

[quote="Mildorf"] Consider another point P' in the interior of the polygon and make a trig argument based on the fact that
[tex]\sum_{i=1}^{n} \cos(\alpha + \frac{2i\pi}{n}) = 0[/tex]
to show that the sums of the projection distances from P and P' are equal, [/quote]

Mildorf, the proof I have doesn't use trig functions. However I really like to see a different solution. Unfortunately, I have trouble to understand your argument since too much details were left out. Would you please provide more details about how to express the sums of distances from two different points in terms of the invariants of the ploygon?

Let [tex]$d_{i}$[/tex] denote the distance from [tex]$P$[/tex] to [tex]$S_{i}$[/tex] and define [tex]$d_{i}'$[/tex] analogously. We wish to show that

[tex]\displaystyle{\sum_{i=1}^{n} d_{i} = \sum_{i=i}^{n} d_{i}'}[/tex]

Let [tex]$\alpha_{i}$[/tex] denote the angle between line segment [tex]$PP'$[/tex] and the perpendicular to [tex]$S_{i}$[/tex] from [tex]$P$[/tex].  Write [tex]$d_{i} = d_{i}' + PP' \cdot \cos(\alpha_{i})$[/tex] and cancel like terms.  The [tex]$\alpha_{i}$[/tex] form an arithmetic sequence that leaves us with the previously mentioned identity times [tex]PP'[/tex].

Mildorf, thanks, I got it. Nice proof! I don't know where this problem originally come from. A friend of mine challenged me this problem a while ago. Here is my proof (sketch only):

[hide]Make a big regular polygon to contain the original polygon such that the corresponding sides are parallel. This is possible since the original polygon has equal interior angles.  Since the corresponding sides of the two polygons are parallel, the sum of distances between the corresponding sides is a constant. Then we can make the conclusion by using the well known fact that the sum of distances from any interior point of a regular polygon to it's sides is a constant (2 * area / side length).[/hide]


function
vector
analytic geometry
real analysis
real analysis unsolved

N>1 integer

E(N) = { be the C-vector space of functions from Z^2 ---------> C

such that for any (x,y) in Z^2, u(x+N,y+N)=u(x,y) }

Find a basis of E(N) ?

There are many bases, of course. It's an $ N^2$ dimensional complex vector space. The two most natural choices would be:

(1) The pointwise basis. For $ 0\le j,k\le N\minus{}1,$ let $ e_{jk}(x,y)\equal{}1$ for $ x\equal{}j,y\minus{}k,$ zero for other $ x,y$ between $ 0$ and $ N\minus{}1,$ continued to be periodic.

(2) The Fourier basis. For $ 0\le j,k\le N\minus{}1,$ let $ e_{jk}(x,y)\equal{}e^{2\pi i(jx\plus{}ky)/N}.$

If we define the inner product \[ \langle f,g\rangle\equal{}\frac1{N^2}\sum_{x\equal{}0}^{N\minus{}1}\sum_{y\equal{}0}^{N\minus{}1}f(x,y)\overline{g(x,y)}\] then both of these bases are orthonormal. To change coordinates between these two basis is to perform either a Fourier transform or an inverse Fourier transform.


[quote="Bicameral"]$ 24)$ The first four terms in an arithmetic sequence are x+y, x-y, xy, and x/y, in that order. What is the fifth term?

a) -15/8  b) -6/5  c) 0  d) 27/20  e) 123/40

$ 25)$ How many distinct four-digit numbers are divisible by $ 3$ and have $ 23$ as their last two digits?

a) 27  b) 30  c)33  D)81  e) 90

I don't know how to do these, so help would be appreciated![/quote]

[hide="hint for 25"]
hint for 25:
To be divisible by $ 3$, the sum of the number's digits has to be divisible by $ 3$. That way, the first $ 2$ digits are easily determined.[/hide]


factorial
function
binomial coefficients

Is there such thing as negative factorial? If so, could you evaluate -1 factorial?

Look up the Gamma function.  Not really appropriate for this forum though.

no ways ,the concept of factorial was made to simply calculations of permutations and combinations,so there exists no such thing as -n! , however we have 0!=1

$ \Gamma(z)$ is defined as $ (z\minus{}1)!$ for $ z\in\mathbb{C}.$ So $ (\minus{}1)!\equal{}\Gamma(0).$ You can find a way to evaluate that [url=http://en.wikipedia.org/wiki/Gamma_function]here.[/url]

No, you can't: the gamma function is not defined at nonpositive integers.  Which is good, because it obeys the same recurrence the factorials do, namely $ z\Gamma(z) \equal{} \Gamma(z \plus{} 1)$.  (Also a minor semantic point: $ \Gamma(z)$ is not defined as $ (z \minus{} 1)!$.  Those two expressions are equal when $ z$ is a positive integer, but that's not a definition.)

The gamma function is defined on all of $ \mathbb{C}$ [i]except[/i] the nonpositive integers, though.

One way to circumvent this difficulty is to take the [i]residue[/i] of the Gamma function at the non-positive integers, which gives you the [url=http://mathworld.wolfram.com/RomanFactorial.html]Roman factorial[/url].  This is the definition that agrees with the definition of negative binomial coefficients, for example.

Think about this: Is there a way to arrange -1 things?


ok people, i need your opinion.

For #4, i did use a "middle leg" approach, with the leg in the center having length P, which instantly showed that k_1 + k_3 = k_2+k_4 = 2P. So i let s_p be the number of ways that k_2 + k_4 = 2P and found that the number of stable tables is s_p^2. i had the number of stable tables be a summation of these s_p's, HOWEVER, i was totally stupid and only looked at when P was an integer when 2P couldve been. so that left me 10 minutes to correct 3 pages of false calculations towards the right answer and i made quite a mess fixing some of those 3 pages.

How many points will they try cutting off?

I would guess 2.  But I haven't seen your messy 3 pages.

Did you get the right answer?

If you did, that should automatically increase their leniency toward your proof.

If you left the proof as is (that is, considering P as an integer instead of 2P), you'll probably get 5 or even 6, as I don't think that's a massively critical mistake. However, I could be wrong in which case you may only get 3 or 4. This is all assuming your calculations were correct (even with the incorrect assumption)

What about if you got the right answer except you didn't really prove that $k_1+k_3=k_2+k_4$ but rather explained why it had to be so but all the rest of the steps were sound.

I think it will be 1 point but maybe 2 if your explanation was hard to understand.

On my #4 I proved k_1 + k_3 = k_2 + k_4 by coordinates(the 4 points where you cut the legs at are coplanar, get the equation, substitute/simplify to get k_1 + k_3 = k_2 + k_4), but I was so stupid I forgot when k_1 + k_3 exceeds n we lose solutions so I ended up with 1^2 + 2^2 +... (2n-1)^2, how much deduction will I get?

2-3 points?...maybe?...it's not really that big of a mistake.

Ugh, I feel so bad about #4. I had the problem solved but just didn't finish it. I had already gotten k1+k3=k2+k4, and then saying all of them are between 0 and n inclusive, but just forgot to figure out how many answers work. That's why I'm waiting nervously...

i didnt get the write answer. i did put in something like n(n-1)(2n-1)/3 + n^2 (i wasnt sure of problem statement so i assumed that cut of length n was not allowed) but i wasnt sure cause i was super panicky so i scribbled the n^2 out and messed up the signs on the other term

I wish I had explained that if that statement wasn't true then the four points of the legs would not be coplanar....booo :(  :(

Yeah, that coplanar explanation on the official solution makes a lot more sense than mine. Oh well, I hope mine was reasonable enough.

I'll kill myself if that causes me to miss MOP. Well, not really kill myself, but probably hit my head against the wall for an hour.

If I miss MOP because of #4, I will take a long knife and cut my wrist

Ow. If I don't make MOP because of #4, I will... do nothing.

something tells me i will miss MOP cause of #4, and maybe #2 and 3..

[quote="DPopov"]whoah...calm down feiqi. :?[/quote]

Oops.. I think this waiting is making me suicidal :(

If i miss MOP because of #4... then i will turn into a  :alien:

IS CUTTING THE WRIST MOST PAINLESS WAY TO DIE?

No. I would say the most painless is to turn the gas on before goingto sleep in a closed room.

Perhaps compass in the ear would be a good way to go. It is interesting the themes present in this thread...

Would I lose a lot of points if I interpreted "A cut leg of length 0 is permitted" to mean "Cutting off 0 from a leg is permitted, but cutting off n from a leg is not permitted"?
I noted that I interpreted it that way and said that substituting n+1 wherever there is an n would correct the problem.  (My answer is correct by my interpretation)

[quote="darktreb"]1. The graders are far smarter than you are. If your answer was wrong because of a misinterpretation, the graders will be able to figure out that was the case, and will adjust your score accordingly.[/quote]

Unlike AP graders :P.

[quote="darktreb"]I think you all are neglecting several facts.

1. The graders are far smarter than you are. If your answer was wrong because of a misinterpretation, the graders will be able to figure out that was the case, and will adjust your score accordingly.

2. Very rarely does someone get all the problems they were "capable" of on a contest. Only the select few elite are so well trained that they don't get extremely nervous, and don't make many dumb mistakes (and even then they'll still do those things). It's not easy to solve problems. It's even harder to solve problems under both a time constraint, and the pressure of knowing what's at stake.

I think everybody needs to relax a little. If you don't make a MOP, it's not the end of the world. Honestly.[/quote]

Great observations, Albert.

I don't need to chill. I'm not nearly as nervous or excited as I was during the ALCS this year.  :D

[quote="yif man12"]I think dying in your sleep is less painful, but I'm not sure how you'd manage that.[/quote]

I would like to quote a few more, but do not know how.

Wow, death!  I hope your school counselors are not reading this. Otherwise your parents will end up doing tons of paperwork to get you out off trouble.

Do you think any points would be lost for not knowing the sum of squares formula and just putting the final answer in sigma notation? (with a little n^2 attached or whatever it was of course)

[quote="bubala"]Wow guys, let's not get suicidal or I might have to contact an adult psychologist to get professional help. (lol)

Anyhow, I did the exact same thing as yif man 12.  How many points will I get (see I have a positive atitude and think about how may points I get rather than how many I lose!)[/quote]

That's because you're 8th grader.  ;)

You'll have a terrific chance of going to MOSP next year, so that's probably one reason you're so much calmer than all us crazy people.

Anyways, awesome! We made the same mistake!  :D  ;)

I don't think leaving it in sigma notation will be a problem.

[quote="lotrfan"]good, thanks for the reassurance!

if wondering how many you get is optimism and wondering how many you lose is pessimism, then what are you if you aren't wondering at all because you are pretty sure of what you got? an oppossumist?

lol, that was a horrible joke  :D[/quote]

You're just dying for me to post that limerick, aren't you?  :D  :P


I was on Alcumus on our living room computer, and when i last checked my Alcumus score was 6o-something percent. I saw some new videos, but I couldn't watch them because that computer didn't have flash installed(and i was to lazy to install it).
So anyways, I got off that computer and started using my mom's laptop computer that had flash installed, but when i got on i saw that some videos were missing and my score was about 47 percent(i didn't do any questions in between). I'm wondering what had happened and if i can get those videos back anytime soon.

I'm not very sure about the vids, but Alcumus seemed to have several problems yesterday, and as for the alcumus score, it could be the same problem as the vids, but also note the it is not updated in real-time.

maybe they do since that happened to me
i use a school computer an it is up
then when i use my own one it is lowered
 :mad:


[quote="Smartguy"]No, Stay in your own forum!
if your asking for this adoption then other states will start asking!
Why not create a new england forum then?[/quote]

Yes we should just make a separate New England forum...Although I wouldn't really mind merging...

If you want to merge with other somewhat inactive/non-existent forums, try Vermont, New Hampshire or Maine. It could be like the $ NE \otimes (MA \plus{} CT)$ forum or something.

[quote="mathemonster"]Mass is our closest neighbor[/quote]

Connecticut's just as close, try there.

Or just get a New England forum up.

But this is Massachusetts' forum. We shed our own Massachusetts-bred flesh and blood to make this work. :yankchain:

watever i still post in this forum so I guess it doesnt matter  :P


search
combinatorics proposed
combinatorics

Numbers 1,2,...,2004 are arranged in a row such that each number is either greater than all numbers to its left or less than all numbers to its right. How many such arrangements are there?

I'm surprised I haven't seen this characterization before -- these permutations are exactly those that "avoid the pattern 321."  That is, they contain no three entries among which the largest is first, the middle is second and the smallest is last.  (It's not too hard to show that this is equivalent to the given description.)  Here are some past threads on permutation pattern avoidance (for culture ;) ):
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=162478
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=424080401&t=199042
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=424080401&t=244959
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=424080401&t=223676


actually nice and elegant are words that go together so a but between them is not justified
and the equation you give to factorize is as good as $ y^{2}\minus{}5y\plus{}6$

i guess there is a typo

there should be 8 instead of 6

and if u think it is easy then u r wrong

post the complete factorization

as d<0 for this quadratic...... :wink:

[quote="Jaswinder"]factorize

x^6 - 5x^3 + 8

 :wink:[/quote]

$ (x^2\plus{}x\plus{}2)(x^4\minus{}x^3\minus{}x^2\minus{}2x\plus{}4)$
is it factorizing further?

great problem
sorry for misreading it :oops:

here is the soln to it

this can be written like

x^6 + x^3 + 8 - 6x^3

let x^2 = a
    x=b
     2=c

then we have to factorize

a^3 + b^3 +c^3 - 3abc    :P

and for that we know

(a+b+c)(a^2 +b^2 +c^2 - ab - bc - ca)

i.e

(x^2 + x +2)(x^4 - x^3 - x^2 - 2x + 4)

and we have done......... :wink:

oh no

rajiv

i am not at all a genius

u all here are geniuses

i am no one infront of u people

and plz rajiv post some nice problems if u have any

and i see u r still online

so plz reply

and once again forgive me for my rudeness

sorryyyyyyyy :oops_sign:


summer program
PROMYS
USAMTS
number theory

I just wanted to clarify because they weren't very specific in their directions (unless I am missing something  :blush: )

In order to apply to PROMYS, we have to send in the application, the recommendation, and the problems all together, right? They didn't really say for sure anywhere whether we had to send in the problems with our application or not.

Thanks in advance!

Where else would you send them?  (That is, yes.)

Okay, I was just making sure so I don't mess this up!
Thank you!

But I got the question while I was starting the problems, and I have to start now so I have time to finish by then!

yea im starting after usamts r4 is due  :D

[quote="buzzer11"]I think its ascending, but I can't be sure.[/quote]

It feels like it's getting harder.

[quote="buzzer11"]To tell you the truth, I haven't started working on them yet because I'm pretty busy at the moment... :) I hope I get them done. But from my brief overview of the problems, they seem to get harder.[/quote]idk i disagree but it depends on the perso i guess

how many have you guys gotten? i have 7 ish more like 5-6

The problems are the exact same as when I applied three years ago, and presumably will stay that way for a little while, so it would probably be better not to discuss them.

[quote="236factorial"][quote="mathemonster"]are the problems supposed to be in a ascending difficulty, or is it just random?[/quote]

I don't agree with those who say the difficulty is increasing. Proving the second part of number three, for example, is much harder than finding solutions to number 8.[/quote]
I agree with you mellowmelon + 236factorial
i pwned 3 tho  :D
EDIT: 236 are you currently a sophmore or junior?

[quote="mathemonster"]EDIT: 236 are you currently a sophmore or junior?[/quote]

Junior this year.

I'm not applying to the program, but I did the problems anyway because they are really very interesting.  And because many of them involve number theory  :D .

My brother is applying though.

[quote="mathemonster"]How old is your bro? is he good at math?[/quote]

He is a sophomore, and his math level is all right (I know, vague, but he doesn't like to participate in many math competitions). Don't worry; I won't leak any of the answers to him  :P

why not scared of failure is he? ya thanks haha


geometry incenter conics ellipse parabola

Given a line AB and a point C that varies on a line l parallel to AB, show the following:

(1) The locus of the incenter of ABC is half of an ellipse with major axis AB, but not containing A and B.
(2) The locus of the orthocenter of ABC is a parabola whose directrix is line l.

The first statement looks very fishy to me; I made a scetch myself and the locus looks strange but not like an elipse at all  :maybe:


abstract algebra
superior algebra
superior algebra unsolved

Let R be commutative ring with 1, M be a (left) R-module and f:M-->M be a R-module homomorphism such that f o f = f. Show that $ M\equal{}Ker(f)\oplus Im(f)$

I have shown using the f o f =f property that the intersection of Ker(f) and Im(f) is trivial, and then used the first theorem of isomorphism combined with the second to conclude that Ker(f)+Im(f)=M, hence proving the assertion. But is there a more direct approach to this problem?

Thanks.

The direct approach would be to write $ x \equal{} (x\minus{}f(x)) \plus{} f(x)$, but there's still some work to be done from there. (I think that you end up duplicating most or all of the work of your solution anyhow)

Why do you say there is still work to be done?

We have every x written as an element of Ker and an element of Im. And Ker n Im = {0}. So we have the result. No?


[quote="Ramanujan"]Αφού ο cmad δεν δείχνει να ενδιαφέρεται να γράψω τη λύση ???  :roll:[/quote]

Ο cmad την ελυσε!  :D
Απλα δεν εβαλε τη λυση.
Feel free to post yours....!
 :)


algebra
polynomial
quadratics
quadratic formula
system of equations

Find a process for synthetic substitution when dividing a polynomial f(x) by a quadratic instead of a linear factor.

you can still do the synthetic division just factor with the quadratic formula, and divide by both roots, but this can get messy, so the long division is easier

Say you are dividing $p(x)$ by $ax^2+bx+c=a(x-r_1)(x-r_2)$.  You'll probably want to use \[ p(x)=a(x-r_1)(x-r_2)q(x)+r(x)  \]

where $r(x)$ is the remainder.  The first thing to notice is that if you are dividing by a quadratic, the remainder must be linear.  So $r(x)=sx+t$.  Then, plug in $r_1$ and $r_2$ and you should get a system of equations for $s$ and $t$.  Solve that and you're done.

I'm pretty sure this still works even for $r_1,r_2\in\mathbb{C}$.  Take, for example, $p(x)=x^9+x^8+3x^4-12$ and divide it by $x^2+1=(x+i)(x-i)$.  Then we have \[ x^9+x^8+3x^4-12=(x+i)(x-i)q(x)+ax+b.  \]

Plugging in $x=i$ and $x=-i$, respectively gives

\begin{eqnarray*} (a-1)i+b+8=0\\ -(a-1)i+b+8=0\\ \end{eqnarray*}

So $a=1$ and $b=-8$ making the remainder $x-8$.  Hmm, I'm not sure about what exactly happens if either $p$ or $q$ has imaginary parts of it.  Does someone know if it always works or only for real or something?


Suppose four circles can be fit inside a quadrilateral (which has an incircle) such that each circle is externally tangent to two other circles and tangent to two sides of a quadrilateral. Prove that among the four circles at least two are congruent.

wow this is so cool!!!!!!!!!!!!!!!!!!!! i had heard about this here forum, but had never actually been here b4!!!!!!!! this is way KOOLER than other forums.

can someone tell me if the only solution is a square? and do the points of tangency have to lie along the apothems?

This is an IMOTC postal problem. DO NOT post solutions. Mod lock this topic.

Dear proposers of IMOTC postal problems,

Some creativity would be appreciated. This problem has been discussed long ago on this forum (I am not providing a link since this would help people like Mnrc_2). As for some other questions, I also remember having seen them before. Nowadays most of the students have internet access, so you might end up comparing their skills in searching rather than in problem-solving. Borrowing a few problems out of older competitions might be a way if there is no chance of getting enough self-made questions, but then make sure that these are either [i]really[/i] old or your competition is a written exam rather than a homework round.

  Darij

yes well said Darij.
most often it so happens and mostly quetsions froma certain shorlist(u know what i am referring to  :D ) get copied in our TST's which is understandable.but here such open cheating when everyone is told right at the top of their question papers to answer individually and not to indulge in any kind of groupism still people post such problems and that too at a place where they are assured of solutions so what's the point of have a postal "exam"
Well if everyone does so then no problem  :P
but there are some who don't get acess to internet or don'e even know about this iste so what about them???
so nothing but on moral grounds please stop posting the problems

[quote="pardesi"]but there are some who don't get acess to internet or don'e even know about this iste so what about them???[/quote]

This makes cheating more unfair, but makes stealing problems even more unfair because those with access to internet can rather easily locate them along with their solutions. This one, for instance, is not from the shortlist. It is from a pretty old competition. I am not saying more.

  darij

Probably the question-'makers' want people who have the best acess to internet with good memory power  :P


I don't have any solution these problem, could you help me
problem1
Fin min , max
$\sqrt{x^{2}+xy+y^{2}}+\sqrt{y^{2}+yz+z^{2}}+\sqrt{z^{2}+zx+x^{2}}$

Problem2:
x>=0, y>=0,3x+y<=6,x+y<=4
fin max and min f=2x-3y

Problem3:
x,y,z>0
x+y+z=1
x<=y
y<=z
find max and min f=2x+y-z

i think many proof for one problem to be method to solve, so could you solve these with many proof


Does anyone know how to make the 'bracket' that goes below something? It looks like a bracket { that's rotated 90 degrees... It's that thing you use when you have, for example, a bunch of 1's, and you want to indicate that there are n of them.

Thanks!

Take a look at http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1818096174&t=178121


what the heck just happened? I couldnt even make my skool team cuz i totally screwed up during practices. But get this. I somehow beat 3/4 of em at chapters just as an individual. wonder how my coach is feeling now... hehe

Good job lifesapain.  One of my individuals beat half my team at chapter too.  You never know how it's going to happen.  Did you qualify for states on your own?

Good luck

[quote="lifesapain"]what the heck just happened? I couldnt even make my skool team cuz i totally screwed up during practices. But get this. I somehow beat em all at chapters just as an individual. wonder how my coach is feeling now... hehe[/quote]
That happened to me too. I got 3rd in my chapter individually and I wasn't on the team. I'm going to state as an individual but some people from the team might not be able to go so I might be on the team.

by not being chosen onto the school team in tryouts

that mean this person was 5-8 in their school or for even larger schools 5-12

2 years ago we had like 11 alternates. yeah.

-jorian

I got seventh in my school competition (mainly for stupid mistakes) so I wasn't on the team, but I went to Chapter for individual competition and I got 3rd today. (Also, I got 2nd in Countdown :lol: )

[quote="jhredsox"]NICE JOB QUEVVY

do you live in CA by any chance?

jorian[/quote]
No, I live in Illinois, in the "DuKane" Chapter (Naperville)

This is what happened at the school I'm helping to coach (my alma mater, although that sounds kind of weird) this year. Sometimes, things just work out differently, heh.

It just happens that I got kicked out....... and that I've been to Naperville before. We drove there to visit one of our relatives.

[quote="bpms"][quote="jhredsox"]NICE JOB QUEVVY

do you live in CA by any chance?

jorian[/quote]
Dude, he knows fanatic.
He either lives in IN or IL, judging by his post IL.[/quote]
Or you could just look at my "Location"
<---

Yeah, I got kicked off the team because I missed the second tryout because of an afterschool soccer game.

So my coach thought it was unfair if I did the same tryout as everyone else (because that would give me a chance to cheat), and gave me the 2001 National Sprint Round. I still don't get how that was fair for me.

This really kinda sucked. We had three kids for our team and were trying to decide our fourth. There were two kids who could have filled the last place. I vouched for one, and our coach went with him. THe other guy went as an individual. The guy on the team made 27, the individual got 35. That probably dropped our team from the state.


[quote="elastiboysai"]Which of the following gives phthalic anhydride on reaction with conc. H2SO4 in the presence of Hg?
(a) Napthalene         (b) Phenol           (c) p-xylene          (d) m-xylene
  can any 1 give d ans and mechanism?[/quote]
the answer is napthalene da i have posted the answer with corresponding explanation in goiit da check it out there!!! :D