[hide="Solutions"]The midpoint for a segment with endpoints $ (x_1,y_1)$ and $ (x_2,y_2)$ is $ \left(\frac {x_1 \plus{} x_2}{2}, \frac {y_1 \plus{} y_2}{2}\right)$.  Plugging in $ 6 \equal{} x_1$, $ 12 \equal{} y_1$, $ 0 \equal{} x_2$, and $ \minus{} 6 \equal{} y_2$, we have $ \left(\frac {6}{2}, \frac { 6}{2}\right) \equal{} (3,3)$.  Thus, our answer is $ 3 \plus{} 3 \equal{} \boxed{6}$

OR
We are looking for $ \frac {x_1 \plus{} x_2}{2} \plus{} \frac {y_1 \plus{} y_2}{2} \equal{} \frac {x_1 \plus{} x_2 \plus{} y_1 \plus{} y_2}{2}$.  Since $ x_1 \plus{} x_2 \plus{} y_1 \plus{} y_2 \equal{} 6 \plus{} 0 \plus{} 12 \plus{} \minus{} 6 \equal{} 12$, our final answer is $ \frac {12}{2} \equal{} \boxed{6}$.  In my opinion, this solution is slightly faster than the first.[/hide]

[quote="GameBot"]What is the sum of the coordinates of the midpoint of the
segment with endpoints $ (6, 12)$ and $ (0, \minus{}6)$?[/quote]
first u take
the 12 then divide by 2
= 6

[quote="GameBot"]What is the sum of the coordinates of the midpoint of the
segment with endpoints $ (6, 12)$ and $ (0, \minus{}6)$?[/quote]

We can first find the midpoint with the Midpoint Formula.

M((6+0)/2, ((12+(-6))/2) =====> M( 3, 3)

So, the sum of the coordinates is 3+3 = 6

ANSWER = 6


percent probability counting distinguishability ratio FTW symmetry

If a ten question True/False test is given and $ 60\%$ is a passing grade, what is your percent change, to the nearest whole percent, of passing if you guess on each question?

The probability that you get $ 60 \%$ or higher is $ \frac{1}{2} \left (1\minus{}\binom{10}{5} \times \frac{1}{1024} \right )\equal{} \frac{1}{2} \left ( \frac{193}{256} \right )\equal{}\boxed{\frac{193}{512}}$.

Let's first assign some variables:

$ R$=getting question correct
$ N$=getting question incorrect

First, let's find the probability fraction, then convert it to a percent. I'll present two ways to solve this problem; the first is for those who know combinatorics, and the second is to clarify for those that don't know as much about probability.

Okay, since it is as likely for one to get a question right as wrong, we have probabilities of $ R \equal{} N \equal{} 0.5$, and there are only $ 2$ variables, we have a normal binomial distribution, and since a probability of $ 0.5$ makes the right side of the binomial distribution pointless (in the formula $ (R \plus{} N)C(R)*R^(0.5)*(1 \minus{} R)^(0.5)$, the part after the main combination is redundant because $ R \plus{} (1 \minus{} R)$ always equals $ 1$, and thus the right-hand side is the same in all calculations of choices of $ R$ and $ N$), we could just evaluate the left side out of total possible to obtain the answer, which is $ \frac {10C6 \plus{} 10C7 \plus{} 10C8 \plus{} 10C9 \plus{} 1}{2^{10}} \equal{} \frac {193}{512}$.

For those who might not know what a binomial distribution is, we could think:

Probability=Successes/Possible
How many ways are possible?
Well, we have $ 2$ possible outcomes for each question, $ R$ or $ N$, and we have $ 10$ questions, so therefore, there are $ 2^10$ possible different outcomes.
Now for successes. Since the extremes are usually the easiest, let's consider the extremes first:
In order to have $ 10$ of $ R$ (correct), we must have the right answer every time, as we can have no incorrect answers. Thus, there is only [u]$ 1$[/u] way to get all $ 10$ questions right. How about just $ 9$ right? Well gee, thinking in the opposite world, $ 9$ right means $ 1$ wrong, so we need exactly $ 1$ wrong to have $ 9$ right. There are ten questions as choices to get wrong. Thus, there are [u]$ 10$[/u] ways to get $ 9$ right. How about $ 8$? Well, thinking about the opposite again, we need exactly $ 2$ wrong to get $ 8$ right. Well, we have $ 10$ choices for the first problem wrong and $ 9$ choices for the second problem wrong. But then, notice that we have counted each choice twice, for example, on the test, there is no distinguishable difference between getting #4 and #7 wrong and getting #7 and #4 wrong. Thus, we divide our computed answer of $ 9\times10 \equal{} 90$ by $ 2$ to get [u]$ 45$[/u] ways to get $ 8$ correct. Repeating the process we did for getting $ 7$ right and getting $ 6$ right, we find there are [u]$ 120$[/u] and [u]$ 210$[/u] ways, respectively. (This is known as combinatorics, by the way). Since we have figured out all five cases of passing, we finally know our total number of successes: $ 1 \plus{} 10 \plus{} 45 \plus{} 120 \plus{} 210 \equal{} 386$. Thus, the probability, or ratio of successes over possible, is $ \frac {386}{1024}$, which reduces to $ \frac {193}{512}$.

We can finally convert our answer to a percent!
Making the fraction into a percent, $ \frac {193}{512}$ is approximately $ \fbox{38}\%$. If you actually successfully carried out the second method of solving this problem in an actual FTW game, I declare that you are certainly amazing. Thus, my second solution is given only as an explanation for solving the problem; in the real game, just know probability and use the first method.

EDIT 1: LOL I just realized that AIME15 posted a solution, an incorrect solution, while I was writing mines.

EDIT 2: Whoopsies! Thx for catching the error, math154! I fixed it now.

AIME15's solution is perfectly rigorous and much slicker. By symmetry, it is clearly equally likely for a person to get $ k$ questions right and $ 10 \minus{} k$ questions wrong as it is for a person to get $ k$ questions wrong and $ 10 \minus{} k$ questions right. However, we must be careful to account for the fact that we can't just directly divide by $ 2$, since $ 50\%$ could occur. After subtracting this probability from $ 1$, we can divide by $ 2$ to distinguish between the cases where $ 6\le k\le10$ and $ 6\le10 \minus{} k\le10$.

Or, as this will probably be a calculator problem, we can own with a TI-84.

The syntax here will be 1-binomcdf(10,1/2,5) which will effectively do 1-(P(0)+P(1)....+P(5))=38 percent.

[quote="r15s11z55y89w21"]Excuse me, math154:

1) please notice that not everybody who goes to this website would know the combinatorics of probability, and those who do know it could always look at my first solution

2) AIME15 does indeed have an incorrect solution, as the question asked for a percent, and AIME15 gave a fraction[/quote]

Sorry, I did not mean for my comment to be offensive. Also, I believe that your calling AIME15's solution incorrect without further distinction between [i]answer[/i] and [i]solution[/i] was ambiguous (plus, the solution is always more important than the answer :wink: ), which prompted my post with more explanation of his post (I thought you thought his content was incorrect). Similarly, I could say something like, "r15s11z55y89w21's solution was incorrect because it should be $ \frac {10C6 \plus{} 10C7 \plus{} 10C8 \plus{} 10C9 \plus{} 1}{2^10} \equal{} \frac {193}{512}$, not $ \frac {10C6 \plus{} 10C6 \plus{} 10C8 \plus{} 10C9 \plus{} 1}{2^10} \equal{} \frac {193}{512}$."

And what the heck (oops, sorry for cussing, mods) are the combinatorics of probability? :wink:  :P

Please do not take this facetious post too seriously.

[sarcasm]I'm so sorry that I forgot to use my calculator to change the fraction to percent.  And I don't really understand what the heck you did anyway.[/sarcasm]


[hide="partial solution"]y(i)=x(i)+6
z(k)=x(k)+12

x+y+z=x+x(i)+x(j)+18=32
x+x(i)+x(j)=14

How many ways can 3 6-sided dice be rolled to sum to 14.

Don't want to solve that part right now.[/hide]

[quote="rnwang2"]Let x be a member of the set {1,2,3,4,5,6}. Let y be a member of the set {7,8,9,10,11,12}. Let z be a member of the set {13,14,15,16,17,18}. How many distinct ordered triples (x,y,z) exist such that x+y+z=32?[/quote]

[hide]Max sum is 36, since you only have to have 4 less you can subtract these four from sets in different ways.

So the answer is the number of solutions to $a+b+c=4$ in non-negative integers, so it's $\binom{6}{2}=\boxed{15}$?[/hide]

[quote="ch1n353ch3s54a1l"][hide]Max sum is 36, since you only have to have 4 less you can subtract these four from sets in different ways.

So the answer is the number of solutions to $a+b+c=4$ in non-negative integers, so it's $\binom{6}{2}=\boxed{15}$?[/hide][/quote]
Interesting approach. That's the right answer.  :)

[quote="rnwang2"]Let x be a member of the set {1,2,3,4,5,6}. Let y be a member of the set {7,8,9,10,11,12}. Let z be a member of the set {13,14,15,16,17,18}. How many distinct ordered triples (x,y,z) exist such that x+y+z=32?[/quote]
[hide]x=a+1
y=b+7
z=c+13
$a+b+c=11$
$0 \leq a \leq 5$
$0 \leq b \leq 5$
$0 \leq c \leq 5$
Normally the answer would be $\binom{13}{2}=78$
Hockey stick, where a=6/7/8/9/10/11=$\binom{7}{3}-1=34$
Overlap=7/2/2, 9/1/1, 11/0/0
$78-34+3=47$[/hide]
I probably made a mistake :(

[hide="solution"]We try out a few cases: When z=18, there are 5 choices. When z=a has b choices, z=a-n has b-n choices. 5+4+3+2+1+0=15[/hide]

Subtract 6 from y and 12 from z, and 18 from the sum, to make it 14.

Now, x, (y-6), (z-12) are all integers from 1 through 6, and their sum is 14.

Questions, please:

[quote]How many ways can 3 6-sided dice be rolled to sum to 14.[/quote]

Is there any way to solve this without using case-by-case or brute force methods?

and also,

[quote]So the answer is the number of solutions to (a + b+ c = 4)  in non-negative integers, so it's 6C2 = 15? [/quote]

I followed you through a + b + c = 4, but how did you get 6 choose 2? It seems the 6 is from the 6 number of elements in the sets x, y, and z, but where did the 2 come from?

[quote="p4fn2w"][quote]So the answer is the number of solutions to (a + b+ c = 4)  in non-negative integers, so it's 6C2 = 15? [/quote]

I followed you through a + b + c = 4, but how did you get 6 choose 2? It seems the 6 is from the 6 number of elements in the sets x, y, and z, but where did the 2 come from?[/quote]

This is a well-known formula.  The number of solutions to $x_{1}+x_{2}+...+x_{k}= n$ in non-negative integers is given by

${n+k-1 \choose k-1 }$

I don't remember why at the moment.

[quote="p4fn2w"]I followed you through a + b + c = 4, but how did you get 6 choose 2? It seems the 6 is from the 6 number of elements in the sets x, y, and z, but where did the 2 come from?[/quote]

This is the "balls-and-urns" argument. If you want to place $n$ balls in $k$ urns, it is the same as trying to place $k-1$ divider elements among the $n$ balls. The result is $\binom{n+k-1}{k-1}$; search "balls-and-urns" on the forum to find ample discussion on it. It applies here if we consider $a, b, c$ as urns and $1, 1, 1, 1$ as balls.

EDIT: bah, beaten to it.

Ah... I see

I know next to nothing about combinatorics, so excuse me for asking that easy question...

[quote]How many ways can 3 6-sided dice be rolled to sum to 14. [/quote]

Using the balls and urns argument, would the answer be (13C2) - 3 (7C2) = 15?

Reasoning:
[hide]
I used the argument:
[quote]If there is the additional restriction that there must be at least one ball in each urn, the number of ways is (b-1)C(u-1).[/quote]
(b is # of balls and u is # of urns)
And basically, each dice is the urn, and 14 is the balls, so there are 14 balls and 3 urns, so there are 13C2 = 78 ways to fill it up.

However, there are cases where the urn (the dice) has more than 6 balls, so assume the case where urn#1 has 6 balls in it already, which means that any "balls-and-urn" argument in this case should be false. If urn#1 has 6 balls, then there are 14-6=8 balls left, so all possible permutations in this case is 7C2. Multiply this by 3 to substitue urn#1 with urn#2 and urn#3, and subtract all of these by 13C2 because these cases are false. This gives:

13C2 - 3(7C2) = 15

[/hide]

This is such a nifty tool!

this is just to clear up why p4fn2w uses 13 and 7

torajirou's formula is for non-negative integers.  for positive integers, it is the same as the number of solutions in non-negative integers of $\sum_{i=1}^{k}y_{i}= n-k$ which is just ${n-1 \choose k-1}$ which is where he gets 13 and 7

[quote="p4fn2w"]Questions, please:

[quote]How many ways can 3 6-sided dice be rolled to sum to 14.[/quote]

Is there any way to solve this without using case-by-case or brute force methods?[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=802346&highlight=dice#802346
[hide]$\binom{7}{4}$[/hide]


LaTeX
Online Math Open
inequalities
modular arithmetic
algebra
polynomial
Vieta

Πώς τα πήγατε παιδιά στον Αρχιμήδη?

πόσα λύσατε? ...και με πόσα θέματα πιστεύετε ότι περνάμε?

εγώ προσωπικά νομίζω ότι ελυσα 3 :P

Y.Γ. Θα μπορούσε κανείς να αναρτήσει τις λύσεις του γιατί εγώ δεν έχω Latex :blush:

Παιδιά ας βάλει κάποιος τα θέματα.
Καλά αποτελέσματα πλέον!!!!!!!!!

Λογικά μέχρι το απόγευμα θα έχουν μπει στον ιστιοτόπο της ΕΜΕ

Παντως συνοπτικά στο πρώτο αρκεί να ήξερες τα υπόλοιπα που αφήνει το κ^2 διαιρούμενο με το 4(0,1) και κατεληγες ότι δεν υπάρχουν τέτοια ν.

στο δεύτερο έθεσα χ=α+γ-β κ.τ.λ και έπειτα Andrescu και μετά πράξεις και έβγαινε.

Στο τέταρτο παίρνωντας την περίπτωση να είναι όλα 1 έβγαινε το άθροισμα 4014 που δν είναι πολ/σιο του 4. Έπειτα παρατήρησα ότι τοποθετώντας -1 το άθροισμα είτε μεταβαλλόταν κατά 4 είτε παρέμενε σταθερό άρα δεν γινόταν να γίνει 0 αφού πάντα θα έπαιρνε τη μορφή 4λ+2.

Σε ότι αφορά το τρίτο (δυστυχώς) δεν μπόρεσα να το βρω :oops:

Loipon oson afora tin anisotita twn megalwn:

Gia olous osous den tin exoun dei parathetw tin ekfwnisi:

Ean $a,b,c$ einai metra pleurwn trigwnou na apodeiksete oti:

$\frac{(a+c-b)^{4}}{a(a+b-c)}+\frac{(a+b-c)^{4}}{b(b+c-a)}+\frac{(b+c-a)^{4}}{c(a+c-b)}\geq ab+bc+ca$

[hide][color=red]LYSH:[/color]

Katarxin oti ta $a,b,c$ einai metra pleurwn trigwnou dinetai mono kai mono gia na eksasfalisei oti ta $a+c-b,a+b-c,b+c-a$ einai thetika (opote mporoume na efarmosoume tin anisotita Andreescu, tin opoia kai efarmozoume stin prwti anisotita pou emfanizetai sta parakatw diadoxika vimata.)
Loipon exoume diadoxika:

\begin{eqnarray}\frac{(a+c-b)^{4}}{a(a+b-c)}+\frac{(a+b-c)^{4}}{b(b+c-a)}+\frac{(b+c-a)^{4}}{c(a+c-b)}&=& \frac{[(a+c-b)^{2}]^{2}}{a(a+b-c)}+\frac{[(a+b-c)^{2}]^{2}}{b(b+c-a)}+\frac{[(b+c-a)^{2}]^{2}}{c(a+c-b)}\nonumber \\ &\geq& \frac{[(a+c-b)^{2}+(a+b-c)^{2}+(b+c-a)^{2}]^{2}}{a(a+b-c)+b(b+c-a)+c(a+c-b)}\nonumber \\ &=& \frac{[3(a^{2}+b^{2}+c^{2})-2(ab+bc+ac)]^{2}}{a^{2}+b^{2}+c^{2}}\nonumber \\ &\geq& \frac{[(a+b+c)^{2}-2(ab+bc+ac)]^{2}}{a^{2}+b^{2}+c^{2}}\nonumber \\ &=& \frac{(a^{2}+b^{2}+c^{2})^{2}}{a^{2}+b^{2}+c^{2}}\nonumber \\ &=& a^{2}+b^{2}+c^{2}\geq ab+bc+ac \nonumber \end{eqnarray}

H isotita profanws isxiei otan $a=b=c$[/hide]

Alexandros

Αρκετά εύκολη η ανισότητα..

Ήταν πρώτο θέμα μήπως;

Έχω την εντύπωση πως αυτή που έπεσε στον Ευκλείδη στη Β' Λυκείου ήταν πιο δύσκολη

Gia opoion itan ipopsiasmenos kai gnwrize na xeirizetai tin Andreescu nomizw oti itan eukoloteri apo ekeini tis B' Likeiou (i opoia na simeiwthei oti h diskolia tis itan to oti den emoiaze me tis klassikes). Opoios omws den gnwrize tin Andreescu, eprepe na kanei Cauchy ('h den kserw pio allo tropo) gia na parei ousiastika tin Andreescu, kai auto den einai poli poli aplo na to skeftei kapoios. Htan to 2o thema tou diagwnismou. To prwto itan poli poli eukolo. Mallon mpike (poli kalws) mono kai mono gia na min apogoiteuthoun oi diagwnizomenoi.

To parathetw:
[color=red]
Thema 1o Megalwn:[/color]
Na prosdiorisete tous fisikous arithmous $n$ gia tous opoious o arithmos $2007+4^{n}$ einai teleio tetragwno.

Alexandros

8a eprepe omos na pareis k tin periptosi v=0 gia na einai olokliromeno :wink:

[quote="bilot4"]mipos moreite na mou peite pos lynotan

to 3 8ema stoys megalous?[/quote]

[color=red]3o Thema Megalwn:[/color]
Se kikliko daktilio me aktines R kai R-2r me R=11r, topothetoume kiklous aktinas r efaptomenous twn kiklwn pou orizoun to daktilio kai ana dio mi epikaliptomenous. Na prosdiorisete to megisto arithmo twn kiklwn pou mporoume na topothetisoume mesa sto daktilio (Dinetai oti 9,94 < \sqrt{99} <9,95).

[hide]Tha to katharograpsw kai tha to steilw se kaliteri morfi otan mporesw.

H idea omws einai i eksis:

Ean $O$ einai to kentro twn dio kiklwn , $A$ einai to kentro kapoiou kiklou mesa ston daktilio kai $OB, OC$ einai oi efaptomenes tou mikrou kiklou (mesa ston daktilio) tote to profanws $AB+AC= r+r=2r$ kai ean $m$ einai to plithos olwn twn kiklwn pou xwrane mesa ston daktilio, tote to megisto mikos pou katalamvanoun einai $m\cdot 2r$ kai profanws auto to mikos einai mikrotero apo to mikos tou kiklou me kentro to $O$ kai aktina to $B$ (diladi apo to $2\pi OB=2\pi\sqrt{99}r$ (Pithagoreio sto $OAB$ kai vriskoume to $OB$)) kai megalitero apo to mikos tou kiklou me kentro $O$ kai aktina $A$ (diladi apo to $2\pi (R-2r+r) =20\pi r$).

Ara $2\pi\sqrt{99}r < 2mr<20\pi$ ap'opou $31,235<m<31,4$ opote mporoume na valoume to poli 31 kiklous.
[/hide]
Alexandros

Η περίπτωση $n=0$ δεν ισχύει .

pos ginetai OB=r?
mallon dn ktlva kala ti apoteloun ta OB kai OC...

Evala to sxima mou sto post parapanw!! Episis scannara kai ta themata mikrwn kai megalwn kai ta vazw edw!

apla na dieykriniso pos i apantisi dn einai 62

alla 31

e3allou to emvado tou kyklikou daktyliou einai 40pr^2kai to emvadon tou ka8e kyklou einai pr^2
ara m<40 sigoura :wink:

Δεν νομίζω κανένας να έχει παρεξηγηθεί με τη συζήτηση που έγινε... ίσα ίσα, προσωπικά γέλασα με το σχόλιο του Ράλλη   :P

Και πιστεύω ότι όταν δει αυτός τα περί... "Corina" θα γελάσει και ο ίδιος! :lol:

Αυτά  :D

σημερα εγιναν οι απονομες βραβειων εδω στην πατρα...ησιχα ητανε...απο οτι μου ειπε ενας φιλος ηταν ενας τυπος που εδωσε ενα λογο τεραστιο(το μισο ακροατηριο ειχε "αποκοιμηθει") τον οποιο ευτυχως προλαβα μονο στο τελος...ειδαμε πολλα παιδια πολλοι απο τη 2α λυκειου...ο nick οπως και περσυ πηρε κατι κιριλε μπιχλιμπιδια που ζηλεψε ολη η αιθουσα,επεισης καπου ειπωθηκε οτι θα παει αμερικη το κατολκαιρι...τι παιζει με αυτο(αν επιτρεπεται)?
αυτα απο την πατρα
σορρυ που ποσταρα εδω αλλα δε βρηκα πουθενα αλλου ςστε να ειμαι εστω κ λιγο ον τοπικ...
darksaga

@nick rapanos [hide]nice shirt you were wearing[/hide]

Molis twra paratirisa oti to 4o provlima twn megalwn iparxei autousio sto vivlio tou Engel " Problem Solving Strategies" selida 28 provlima 29.

Alexandros

Το είχα πει σε αυτούς που βρήκα στην Αθήνα αλλά ας μην θύγουμε τέτοια θέματα .Η πραγματική πηγή του προβλήματος προέρχεται από την αστίρευτη πηγήπροβλημάτων , την Ρωσσία.
Δείτε την άσκηση 17 από το 1961 στο βιβλίο των Πανενωσιακών και θα καταλάβετε  :wink: .

[quote="silouan"]Αφού ο Αλέξανδρος έθιξε δύο θέματα ας βάλλω και εγώ κάποιες διαφορετικές λύσεις που βρήκα .
Για το 3 των μικρών .
Θα πρέπει $n^{2}+24=k^{2}$ (1) και $n^{2}-9=m^{2}$ (2)
ή $(k-m)(k+m)=33=3\cdot 11$  και μετά απλό.

Για το 4 των μεγάλων
Έστω ότι $A_{1}+...+A_{2007}+B_{1}+...+B_{2007}=0$
Τότε  το πλήθος των - είναι 2007
Είναι όμως $A_{1}...A_{2007}B_{1}...B_{2007}=1$ άρα το πλήθος των - είναι άρτιο , άτοπο  :)[/quote]
συγνωμη για το σχολιο μετα απο μερες αλλα Σιλουανε για το τριτο θεμα στον μικρο αρχιμηδη θες να πεις οτι τα ριζικα ειναι ακεραιοι ? δεν νομιζω οτι ισχυει

Κοίτα αν και οι δύο είναι άρρητοι αφού συνδέονται με + το άθροισμα θα είναι άρρητος (αν δεν κάνω λάθος) .
Αν ένας μόνο είναι ακέραιος και ο άλλος άρρητος τότε το άθροισμα είναι ρητός+άρρητος =άρηητος άρα και οι δύο πρέπει να είναι ακέραιοι .
Κάνω λάθος . Αν ναι πες μου για να διορθωθώ .(Πάντα είχα ενδοιασμούς με αυτά. :?: )

Φίλε Σιλουανέ και Στέλιο.
Άρρητος + άρρητος δεν είναι απαραίτητα άρρητος.

Π.Χ. sqrt(2) + {1-sqrt(2)} = 1

Υ.Γ. Συγγνώμη που δεν έχω το πρόγραμμα για να γράφω μαθηματικά. sqrt σημαίνει τετραγωνική ρίζα αν και μάλλον θα το ξέρετε.

Αυτό που γράφεις όμως εσύ απαλείφεται , οπότε δεν έχει και κάποιο ουσιαστικό νόημα... επίσης στην άσκηση έχει $+$ μεταξύ των κλασμάτων, όχι $-$ !

Αυτά  :D

PS: Μου θυμίζεις λίγο το όνομά σου γιατί το... ξέχασα! :( Δε μου αρέσει να απευθύνομαι με το nick. Άλλωστε όλοι μία οικογένεια είμαστε εδώ  :)

Siluane..nomizo oti ayto 8eli ligo megaliteri eksigisi..(otan to ixa protodiavasi..,ayto skeftika +ego..(ayto pu ipe o protonios.+ine sosto..+o $\sqrt{2}$ ala kai o $1-\sqrt{2}$ ine ariti +omos exun a8risma rito..)ala stin sigekrimeni periptosi iparxi mia leptomeria...epidi to $n$ ine akereos..
an kai ta dio rizika ine akerei..telos...(vlepe lisi siluan..)
an ena mono ine akereo...telos...(profanes..)
an kanena riziko dn ine akereos...8a ine aritos..(dn mpori na ine ritos..gt to n ine akereos..)+ara o aritos aytos 8a grafete sti morfi $a\sqrt{b}$opu b ine square free akereos..(prosoxi..ayto isxii epidi to iporizo ine akereos...gt diaforetika p.x.$\sqrt{3+2\sqrt{2}}=1+\sqrt{2}$to opio dn mpori na grafi sti morfi $a\sqrt{b}$..)
epomenos..se ayti tin periptosi...to telioko a8risma 8a ine tis morfis
$a\sqrt{b}+c\sqrt{d}$ me $c,d$ square free akereus, to opio ine pleon profanes oti dn mpori na ine akereos....
simperasma..
sosti i lisi tu siluan...ala i8ele ligo mpla mpla..pio prin..
sosto oti to aritos+aritos mpori an mas kani rito...

 :D

Ράλλη δεν είδα σχόλια για το ''τεστ'' που έβαλα...  :rotfl:  :rotfl:  :P

Protonios είναι το όνομά μου. Τόσο ασυνήθιστο είναι? :|
Αστειεύομαι φυσικά :wink: Κώστας.

λοιπον αυτη συζητηση θα επρεπε να μπει στο βιβλιο,
γενικα αν  $a$,$b$ ακεραιοι οχι τετραγωνα ακεραιων τοτε η παρασταση $\sqrt{a}+\sqrt{b}$  ειναι αρρητος τοιδιο και για την διαφορα τους αποδεικνυεται απο το http://www.mathlinks.ro/Forum/viewtopic.php?t=130231 αρα ο σιλουαν εχει δικιο  :)

Ράλλη ευχαριστώ πολύ . Το έγραψες πολύ καλά... :)
Άρα η λύση μου είναι σωστή  :D
Πάντως αυτά που έγραψες είναι πολύ χρήσιμα αφού πολλοί τα μπερδεύουν .Πρέπει να μπει στο βιβλίο (νομίζω ότι δεν αναφέρεται στα κλασσικά βιβλία άλγεβρας )

[quote="mostel"]Νομίζω πως έχεις δίκαιο Σιλουανέ..

Έτσι για τεστ:

1) Να δειχθεί ότι μεταξύ δύο ρητών υπάρχει πάντα ακόμη ένας τουλάχιστον ρητός

2) Να δειχθεί ότι μεταξύ δύο οποιωνδήποτε αριθμών υπάρχει ρητός

3) Να δειχθεί ότι μεταξύ δύο οποιωνδήποτε αριθμών υπάρχει άρρητος[/quote]
 1)το ημιαθροισμα τους ειναι ενας ρητος
 2)αν x,y πραγματικοι αριθμοι απο αξιωμα αρχιμηδη εχουμε $n(x-y)>1$ ara $x>\frac{ny+1}{n}>\frac{[ny]+1}{n}$


Let $ x,y,z>2$ and $ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\equal{}1$.

Prove that $ (x\minus{}2)(y\minus{}2)(z\minus{}2)\leq 1$

there exists a triangle with sides $ a,b,c$ so that $ x\equal{}\frac{a\plus{}b\plus{}c}{a},\, y\equal{}\frac{a\plus{}b\plus{}c}{b},\, z\equal{}\frac{a\plus{}b\plus{}c}{c}$
so we have to prove $ (a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)\le abc$
Either use AM-GM or expand it and find $ a^3\plus{}b^3\plus{}c^3\plus{}3abc\geq a^2b\plus{}b^2a\plus{}a^2c\plus{}c^2a\plus{}b^2c\plus{}c^2b$ which is Schur's inequality.  :wink:

In fact there's a shorter alternate ending:
since $ a,b,c$ are sides of triangle we can find $ x,y,z$ with $ x\plus{}y\equal{}a$,etc.
Then $ \prod (a\plus{}b\minus{}c) \equal{} 8xyz \leq \prod (x\plus{}y) \equal{} abc$ :)

[quote="me@home"]In fact there's a shorter alternate ending:
since $ a,b,c$ are sides of triangle we can find $ x,y,z$ with $ x \plus{} y \equal{} a$,etc.
Then $ \prod (a \plus{} b \minus{} c) \equal{} 8xyz \leq \prod (x \plus{} y) \equal{} abc$ :)[/quote]
That's why I said use AM-GM OR expand  :P


[hide="hint"] Let $ p$ be the probability that he rolls more 1s than 6s. Then, $ p$ is also the probability he rolls more 6s than 1s. We can hence find the probability that he gets the same numbers of 1s and 6s, subtract from 1, and divide by 2.[/hide]

[hide="xpmath's hint"]find the probability that you have the same number of 1's and 6's, subtract from 1, and divide by 2[/hide]

you can go straight to it


I forgot what a perfect square is defined as.
If I knew, probably I could get this problem correct:

The number n is a perfect square. What is the next perfect square bigger than n?

Reply ASAP

A perfect square is a number that can be described as $x \times x$, where x is an integer.

For the problem, the next square would be $\left(1+\sqrt n\right)^{2}$.

And, of course, there are perfect cubes ($x^{3}$), perfect fourth powers ($x^{4}$), etc.

Hm, could a fourth power be a HYPERCUBE?  :)

A sixth power is a hyper cube. a 4th is a hyper square.

Isn't this getting off topic?

[quote="Temperal"]A sixth power is a hyper cube. a 4th is a hyper square.[/quote]

Something of a misuse of terms.  A "hypercube" refers to a generalized cube (the geometric object, not the function) in dimensions $ n \ge 4$.

[b]Nacho, Silver Surfer:[/b]  This forum is not your personal playground.  If you want to talk about math, then that's fine, but don't carry conversations in threads.  That's what personal messages are for.  And if you're friends and/or the same person, cut it out.

[quote="i_like_pie"]For the problem, the next square would be $ \left(1+\sqrt n\right)^{2}$.[/quote]
that would be true but it could also be $ \sum_{x=0}^{x=n+1}{2x+1}$ because the sum of all the first $ n$ odd numbers is $ n^{2}$

Well, no, it would actually be $ \sum_{k = 1}^{\sqrt{n}+1}2k-1$, which is hardly any simpler.


Toni picks a number and multiplies it by three.  She then adds four to the result and finally divides this new numbrer by two.  Her final result is fourteen.  What number did she start with?

pretty easy...i did it in my head
[hide="wut i got"]8[/hide]

i wonder where you got that answer from :roll:

I remember this problem from a second grade workbook. it's full of inverse operations.

[hide="my answer"]
8   (yea my first post)[/hide]


Look also http://www.mathlinks.ro/Forum/viewtopic.php?highlight=austria&t=89886

It was part of this years final round in Austria.


analytic geometry
graphing lines
slope
function

[quote="gauss202"]Draw the line y = x. Do you see how this helps?[/quote]

At (1,1), (2,2) etc, y=x

The possibilities are infinite.

If I am on the right track, what would the answer be?

An infinite number?

I drew lines at (1,1), (2,2), etc and at (0,0).

If I did it correctly, There are 2 points that touch the "S" on each line.

At (0,0), there are 3 points. I'm still confused as to what the actual total is and why.

[quote="cf249"]The answer is not infinite.   How many times does your graph touch the line y=x?

[hide]You may have graphed y=x incorrectly.  Note that y=x is the same thing as y=1x+0.[/hide][/quote]

Oy am I getting more confused.

I plotted (0,0) and (1,1). The line between these points is y = x. It extends out infinitely past both points.

y=1x+0

If I plug in numbers:

If x = 0 then y = 1

If x = 1 then y = 2

I don't understand. I'm plotting, but I don't understand what I am doing.

Is it 3?

Actually if y=1x+0

Plug in 0: 1(0)+0=0
Plug in 1: 1(1)+0=1

anyway the point is that the line y=x has slope 1 and y-intercept 0. So after plotting some points draw the line and see how many times it intercepts the other graph.

[quote="i_like_pie"][quote="Star00"]On the graph, for how many values of x does y=x?

The obvious answer is (0,0).
There have to be more. I don't know how to go about finding this answer.[/quote]
You can't get the exact answers unless you know what that graph is. Obviously it's not a function, so there won't be one simple equation.[/quote]

Well the units are given (1) and since it goes up by 1, there are 3 intersections. Notice the graph doesn't continue off screen and it doesn't have an arrow.

Yes, but you won't be able to find the [i]exact[/i] values.

[quote]
On the graph, for how many values of x does y=x? [/quote]

It doesn't ask for exact values

[quote="jli"]ok you guys are making this too complicated.  just plot x=y and it crosses 3 times...nobody cares about exact values or what happens after because they would tell us if they wanted to[/quote]
Yep. It's 3 times. I don't know the exact values of x and y, but I could give out some approximations if you want. y = (1)x + 0. How hard is it to graph that?  :P

Even if the S shaped graph would have arrows at the ends, it would still be 3 intersections. Imagine the S continuing, further and further away from the line y=x.

I posted the question as it is.

Thank you for all of the help.


calculus
integration
geometry
calculus computations

[quote="kunny"]Find the area of the part enclosed by the following curve.

\[x^2+2axy+y^2=1\ (-1<a<1)\][/quote]

By the given condition on 'a', it is clear that the curve represents an ellipse.

By rotating the axes by $45^0$, it is easy to find the lengths of semi major and semi minor axes as
$\frac{1}{{\sqrt {1 - a} }}$ and $\frac{1}{{\sqrt {1 + a} }}$ respectively.

Hence the area is $\frac{\pi }{{\sqrt {1 - a^2 } }}$


function
number theory
modular arithmetic

When do you use the definition symbol with three bars and when do you use a normal equal sign?  For example, if you want to express "we define the variable theta to be the length of segment AB", would you use the 3-bar symbol or the 2-bar?  Could someone explain the difference?  Thanks.

As far as I know, $\equiv$ is not used for definitions. It is used for [url=http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic/Introduction]modular arithmetic[/url]. You can use $: =$ to define functions, but I cannot think of a symbol used to define things in general. For example, you could say \[f(x): =x^{2}+1.\] In your example, there is no need to use any symbol at all; you could just say, "Let $\theta$ be the length of $\overline{AB}$." I guess if you want you could also say, "Let $\theta = AB$," but the former method is better.

Personally, in geometric context I'd use neither. For the definition I'd use column-equal sign, i.e. $\theta : = |AB|$, and the three-bar sign I restrict for coinciding objects, i.e. $A\equiv B$ denotes coinciding points, $l_{1}\equiv l_{2}$ means that the lines $l_{1}$ and $l_{2}$ are identical etc.

Generally, for defining in any context I use column-equal: $g(x) : = f(2x+1)-5$ means "we define function $g(x)$ to be..."


inequalities
real analysis
real analysis unsolved

Let $A_1$ and $A_2$ be a measurable subsets of $\mathbb{R}^n$ such that
\[ m(A_1\cap B(x,1/i))\leq m(A_2\cap B(x,1/i)), \]
as $x\in A_1$ and $i=1,2,\ldots$. Prove that $m(A_1)\leq m(A_2)$.

The given inequality tells us any Lebesgue point of $A_1$ is also a Lebesgue point of $A_2$. Since a.e. point in $A_1$ is a Lebesgue point, this means $A_1 \subset A_2$ up to a set of measure 0. The desired inequality follows.


Ice-cream-o-rama is eager to advertise how many flavors it has. But it really only has three basic flavors: chocolate, vanilla, and strawberry. However, they can make "new" flavors by taking four scoops of ice cream of those basic flavors and blending them together. Different proportions of the basic flavors give different new flavors. How many flavors can Ice-cream-o-rama now advertise?

we call this a stars and bars, but I like sticks and dots.

A visualization of the sticks and balls dividing the flavors might help.

example:

//OOOO  all straw
O/OOO/  1 van, 3 choc, no straw
OOOO//   all van

[hide]The problem asked how many "NEW" flavors Ice-Cream-O-Rama can advertise, not how many flavors can they advertise in total. Therefore the answer should be 15-3 = 12 because the "NEW" flavors would not include the original three flavors.  The problem as stated in the forum does not match the problem as asked in Alcumus.[/hide]

no the question asks for the total number of Ice cream flavors, not the "new" ones...

Hint:
[hide]Use the ‘Balls and Boxes’ formula, which is $\binom{n + k - 1}{k - 1}$, where $n$ is the balls and $k$ is the boxes. In this problem, there are $4$ balls and $3$ boxes.
[/hide]

[quote=wlpj11]The problem as stated in the forum does not match the problem as asked in Alcumus.[/quote]

Here, it says:

[quote=GameBot]How many flavors can Ice-cream-o-rama now advertise?[/quote]

But the question:
[quote=Alcumus]How many [b]total[/b] flavors can Ice-cream-o-rama now advertise?[/quote]

So you're right.

[hide=solution] There are 4 different types of ways of blending the flavors: (4, 0, 0), (3, 1, 0), (2, 2, 0), (2, 1, 1).
For (4, 0, 0), there are $\frac{3!}{2!}=3$ ways.
For (3, 1, 0), there are $3!=6$ ways.
For (2, 2, 0), there are $\frac{3!}{2!}=3$ ways.
For (2, 1, 1), there are $\frac{3!}{2!}=3$ ways.
Therefore, there is a total of $3*3+6=15$ ways.

is 4 vanilla balls is same as basic vanilla? Also, I haven't learned sticks and dots, so would be a good idea to bash with casework?

It's usually called Stars and Bars, or Balls and Urns, and you can learn it [url=https://artofproblemsolving.com/wiki/index.php/Ball-and-urn]here[/url]. However, yes, you can use casework for it. I did that when I hadn't learned about Stars and Bars either.

Edit: just looking at this giant time bumps in this thread:  :o

I'm so confused. Why isn't it $84$? There are $3$ choices for each flavor giving $3^4$ and we add back in the original flavors to get $84$?

Notice that "Chocolate Chocolate Vanilla Strawberry" is the same as "Chocolate Vanilla Strawberry Chocolate."

[hide=Sol]Let the number of chocolate scoops added be $c$, the number of vanilla scoops added be $v$, and the number of strawberry scoops added be $s$. We have $c+v+s=4$. If $c=0$ (0 chocolate scoops added to the new flavor), $v$ can be $0, 1, 2, 3$ or $4$, and $s$ will follow accordingly. So there are $5$ possible ways to have a unique sum without any chocolate scoops. Similarly, $1$ chocolate scoop will have $4$ possible ways to have a unique sum, and so on. So the number of possible flavors is $5+4+3+2+1=\boxed{15}$.[/hide]

Bruh I couldn’t think of stars and bars so I just counted:
[hide=solution ~1 min]
CCCC
WWWW
SSSS
SSSW
SSWW
SWWW
CWWW
CCWW
CCCW
SSSC
SSCC
SCCC
SCWW
SCWS
SCWC
[/hide]

I did 3^4 = 81 because you have 3 choices 4 times. Then I thought it was 4^3 because for the flavors each slot has 3 choices. :furious:  :(


calculus
integration
advanced fields
advanced fields solved

Find all circles C in the plane such that the integral of the differential form
w = x^2dy + y^2dx on C is 0
 \int _{C} w = 0

apply Green's theorem, it becomes

2\int \int _S (x-y) = 0

I guess all the circle must have origin as its center.

You can also  paramatrize a circle
x= x_0 + r.cost
y= y_0 + r.sint
plug in the integral   \int  _{C} w=0
you get after some calculus 2 \pi r^2(x_0 - y_0)=0

oh, it must be like that..
i didn't try that way.


algebra
polynomial
trigonometry
algebra unsolved

Let $f(k)$ be a coefficient of $x^k$ in the expansion of polynomial $(1+x+x^2)^n$.
Calculate $\sum_{k=0}^{2n} f(k)^2$

well after noticing that it is a symetric plynomial and calculating we find that $\sum_{k=0}^{2\,n} \left( f \left( k \right) \right) ^{2}=\sum_{k=0 }^{2\,n} \left( {2\,n\choose k} \right) ^{2}\cos \left( 4/3\, \left( n -k \right) \pi \right)$ im not sure we can find somthin more simple.

i think that your solution is correct, and i couldnt find anything better


vector
linear algebra
matrix
linear algebra unsolved

Hi would appreciate help with this question

P1) i) What is the dimension of the vector space of all upper triangular n x n matrices? ( The matrices with entries=0 below the diagonal) . Write down a basis for it

ii)In the vector space of all upper triangular n x n matrices ($a_{i}_{,}_{j})$
  let M be the set of all matrices with
$\sum_{i,j}^{}a_{i}_{,}_{j}=0$ . Prove that M is a vector space . Write down a basis for it and so find dim M.

Thanks in advance

1) In an upper triangular matrix you are free to choose $n(n+1)/2$ elements, and this is the dimension of the space. A natural basis is formed by matrices that have one entry equal to $1$ and the rest equal to $0$.

2) By adding a linear relation between entries, you decrease the dimension by $1$ compared to 1). A basis for this space is formed by upper-triangular matrices that have $-1$ in the upper left corner, $1$ at some other place, and $0$s everywhere else.


geometry
3D geometry
tetrahedron
calculus
integration
function
inequalities

[hide="Solution (#2)."]Given that $ \sqrt {a \plus{} x} \plus{} \sqrt {a \minus{} x} \equal{} a$, then $ \minus{} a \leq x \leq a$.  Getting rid of the radicals yields:

$ a \plus{} x \plus{} 2\sqrt { a^2 \minus{} x^2 } \plus{} a \minus{} x \equal{} a^2$
$ 2\sqrt { a^2 \minus{} x^2 } \equal{} a ^ 2 \minus{} 2a$
$ 4(a^2 \minus{} x^2) \equal{} a^2(a^2 \minus{} 4a \plus{} 4)$
$ x^2 \equal{} \minus{} \frac {a^2(a^2 \minus{} 4a)}{4}$

The right-hand side must be non-negative, so one of its terms must be negative or zero.

$ a(a \minus{} 4) \leq 0$.

The only possibilities to consider are {$ 0, 1, 2, 3, 4$}.  These correspond to the $ x$ values of {$ 0, \frac {\sqrt {3}}{2}, 2, \frac {3\sqrt {3}}{2}, 0$}, all of which work when plugged into the original equation, except for $ 1$ (since the $ x > a$, which is impossible).  Consequently, there are $ 4$ solutions.[/hide]

[quote="zhou_qiang"]1.In a tetrahedron PQRS, angle PSR=30 degree and angle QSR=40 degree. If the size of angle PSQ is an integral number of degrees, how many possible values can it have?

2.For how many positive integer values of a does the equation

                 sqrt(a+x)+sqrt(a-x)=a

have a real solution for x.[/quote]

[hide="Solution to #2"]Let $ f(x)\equal{}\sqrt{a\plus{}x}\plus{}\sqrt{a\minus{}x}$.  Since $ f(x)$ is an even function, we can assume WLOG that $ x$ is nonnegative.  Then $ f'(x)\equal{}\frac{1}{2\sqrt{a\plus{}x}}\minus{}\frac{1}{2\sqrt{a\minus{}x}}\le 0$ for $ 0\le x\le a$.  Therefore, $ f(a)\le f(x)\le f(0)$.  We are given that $ f(x)\equal{}a$, so we have $ \sqrt{2a}\le a\le 2\sqrt{a}$.

Squaring this equation, we get $ 2a\le a^{2}\le 4a$.  Dividing by $ a$, we get $ 2\le a\le 4$, so there are three solutions for $ a$.[/hide]

Another approach:
Applying Cauchy-Schwartz's inequality for LHS we have: LHS<=2square root(a)
Therefore, if a>4 then we easily have:a>2square root(a). So, the equation is inconsistent
Now, we just have to examine 4 cases which correspond to 4 values of a, that is: {1,2,3,4}
Easily to show that a=2,3,4 then the equation has at least one real solution(0,5/2,square root(3), respectively) and when a=1 the equation also becomes inconsistent.
So, the final result is:[b] 3[/b]

For #1, considering each of the angles at vertex $ S$, $ \angle PSQ$ must be strictly greater than $ (40 \minus{} 30) \equal{} 10$ degrees and strictly less than $ 40 \plus{} 30 \equal{} 70$ degrees. So it can take all values $ 11, 12, \ldots, 69$, giving us $ \boxed {59}$ integral values.


ARML
USAMTS
MATHCOUNTS
Gauss
AMC
AIME
AIME I

Is so dead here.

Why is the IL forum dead for so long? Where are all the middle school kids who used to post here?
They're gone :o

[quote="avec_une_h"]Yes, this forum is dead.

And all of us (or at least me) are in high school.[/quote]
Oops, sorry to mention the high schoolers :o
Other than you, everyone is dead! :o
Thus the forum is too! help! :o

Here's an idea: perhaps they're doing math?

Gah, I wish I could stop going on AoPS too much, but I can't stand it.

All the state forums are dead, except for states like FL or OH...

Well, I'll keep checking this forum even if no one else does.

hs season's started up. yay.
my school (nchs) is doing all right ish, but i don't know how stevenson is #1  :huh:

you guys are starting what contest? ARML? ICTM?

can you do arml or ictm in 8th grade? if so how

probably both.
ictm: ask your teacher to register.
arml: try out?

we started NSML.

We had our first math meet (NSML) a little while ago.  We did...okay.  Not great.  The freshman team won.  The school as a whole was nowhere close.

Is anyone here doing USAMTS or AMC? I am, along with MATHCOUNTS, and I'm in 8th grade :D  :D  :D

Triple cheers!

And I already finished USAMTS Round 1!

And I'm taking USAMTS and AMC for the very first time! :D

<_< bah out of smilies

I was going to do USAMTS.  Then, I got sidetracked.  But I'm definitely doing AMC.

[quote="Aryth"]btw, in your location, did you intend to misspell "guess" or "gauss" or whatever you were trying to spell?  :|[/quote]
yes :D :D  :D

How about start a math marathon?

At a party, 40% of the people like to dance. 60% of the people are girls. If 20 more boys, all of whom like to dance, joined the party, the party would be 58% girls. How many people at the party like to dance? (after the 20 boys joined)

Source: 2008 AIME I #1

I mean, really. This forum needs to stay alive. Maybe some math from the ILLINOIS people, with nice state spirit, can do it.

I'm sorry RussianRocket.  I know you, and you're not Michael Phelps.  By the way, how's MathCounts been so far this year?  My dad says I'm not allowed to go to practices until alumni day, when we're going to have some of the team from last year come.

*gasp*

HEY DUDE RUSSIANROCKET WHERE HAVE YOU BEEN LATELY???

SO GREAT UR BACK!!!! :D

[quote="avec_une_h"]I think he's had cross-country...[/quote]
lol but still won't he have at least some time to post on aops????

I've been in Florida where huricane 'Ike' like scared me really bad
Then I had swimming for 2 hours and Cross Country for 1 hour almost daily.
Plus MC for me only started like 2 weeks ago so I m still not in the MC grove.

huh? you've been in florida on vacation during school days?

[quote="gauss1181"]oh.

And why isn't ANYONE reading my thread known as "blah"?[/quote]

Because it sounds SSSSSSSSSSSOOOOOOOOOOOOOOOOOOOOOO interesting.


1/x+1/y+1/z=1 so  \sqrt xyz= \sqrt (xy/z)+ \sqrt (yz/x)+ \sqrt (zx/y)

 \sqrt( x+yz)>= \sqrt x+ \sqrt (yz/x)
  x+yz>=z+yz/x+2 \sqrt yz
  yz-yz/x>=2 \sqrt yz
  yz(1-1/x)>=2 \sqrt yz
  yz(1/y+1/z)>=2 \sqrt yz
  z+y>=2 \sqrt yz
  ( \sqrt z- \sqrt y)^2>=0  true
  In the same way we prove that:
  \sqrt (y+zx)>= \sqrt y+ \sqrt (zx/y)
    \sqrt (z+xy)>= \sqrt z+ \sqrt (xy/z)
  Finally we add the 3 ineq


conics
ellipse
quadratics
algebra
real analysis
real analysis unsolved

Find the points at which the following ellipses intersect?

x^2 - xy + y^2 = 3
x^2 + xy + y^2 = 6

All I know is that there are 4 points of intersection.

Subtract the equations: $2xy=3$. Now $y=\frac3{2x}$; substitute this in to one of the equations to get a quadratic equation for $x^2$.

[Edit- fixed a mistake]

I did that and got a whole bunch of points but how do I know which one are the intersecting points?
there are only 4.

Using that substitution you gave me I got the following:

for the equation x^2 - xy + y^2 = 3, I got:

when y=1.3093 x= 1.9640 and x= -.6547
when y=-1.3093 x=-1.9640 and x=.6547

for the equation x^2 + xy + y^2 = 6, I got:

when y=1.1239  x= -2.8097 and x= 1.6858
when y= -1.1239 x=2.8097 and x= -1.6858

See I have two possible x values to choose from in each case and dont know which ones to pick??

There was a mistake in what I wrote- look again. We have $y=\frac3{2x}$ so there is only one solution for each value of $x$.


http://en.wikipedia.org/wiki/Tarski%27s_circle-squaring_problem
I don't get it, how can we get rid of the curve lines?

The pieces don't have curved lines.
See th Banach-Tarski-Paradoxon, too.

what do u mean? it's a circle it has curved lines...

the axiom of choice creates some odd results...the thing is that the pieces are not simple geometrically...


trigonometry function LaTeX geometry counting distinguishability integration

The following problem was given:
[quote]Produce $\pi$ to four decimal places (i.e., 3.1416 = 3927/1250) using just four fours and standard mathematical symbols met at school level. The solution can be written on the back of a postage stamp.

Note/hint: The full interpretation of these symbols requires undergraduate level mathematics.[/quote]
The following answer was given:
[quote][hide]$\frac{{4.4\choose4}}{\sqrt{.\bar{4}}}=\frac{\Gamma(5.4)}{\Gamma(5)\Gamma(1.4)}\cdot\frac{3}{2}=3.1416$[/hide][/quote]
However, I came up with a way to produce $\pi$ [i]exactly[/i] using only three fours (although I don't think it's standard notation). Can you figure out how to do this?

[quote="JBL"]Why not just do it with 2 fours; [hide]$\cos^{-1} -\frac{4}{4}$[/hide][/quote]

[hide="Good one JBL but..."]You can't say $\cos^{-1}$! You must say $\arccos$.  :huh: :P  :rotfl:[/hide]

okay, i have no idea what $\text{cos}^{-1}$ means

It's the inverse cosine function.  Thus, $\cos\left(\cos^{-1} t\right) = t$ for $-1 \leq t \leq 1$.

Bictor, I always thought that the functions $\cos^{-1}$ and Arccos (with a capital A, which Latex doesn't recognize) were the principle value functions (that is, in the range $(-\pi, \pi]$), while the function $\arccos$ was the multiple-valued one.  In any event, I'm correct up to an epsilon of notation :-)

Also, mine is certainly nicer: to say that the gamma function is "met at school levels" is just silly, unless you're only talking about the letter gamma itself (in which case it's even sillier).

Inverse Cosine Function for Principal Values: $\text{Arccos}$ or $\text{Cos}^{-1}$
Inverse Cosine Function for All Values: $\arccos$ or $\cos^{-1}$

[quote="jmadsen"]I forgot to say that functions like cos are not allowed, only what are referred to as "standard mathematical symbols," e.g. $+,-,\times,\div,\sqrt{},!,.$ are allowed.[/quote]

Huh? I don't have much experience in this area, but how do you make it transcendental, then? Unless you're talking about a summation? Or a clever way of actually writing "$\pi$", like making the 4s really thin and indistinguishable from regular pencil marks and arranging them to form the pi symbol.[hide="Or is it this?"]
$\displaystyle\left(\int^{+\infty}_{-\infty}e^{-x^{\sqrt 4}}\;dx\right)^{\frac{4}{\sqrt 4}}$
Not exactly standard, but whatever. [/hide]

Regarding the inverse cosine, I think standard notation is the way NoSoupForYou described, although it's not very often that I see the capitalized ones used when they should be used.

[quote="math92"]i have no idea what this means:
$\displaystyle \displaystyle\left(\int^{+\infty}_{-\infty}e^{-x^{\sqrt 4}}\;dx\right)^{\frac{4}{\sqrt 4}}$[/quote]

That represents the square of the total area under the curve of $e^{-x^2}$ (and is indeed equal to $\pi$)

Although it can also be done with only two occurences of the number 4, as $\sqrt{4} = \frac{4}{\sqrt{4}}$

This was my solution:

[hide]$((-\frac{\sqrt{4}}{4})!)^{\sqrt{4}}=\Gamma(\frac{1}{2})^2=\pi$[/hide]


When I compile this document, I can't view it because it says (in a new window) "Yap. Not all new fonts can be loaded".
Any idea what to do? cmr10, I think.

\documentclass{article}
\title{2005 MMPC Part 2 Solutions}
\begin{document}

1. WLOG, let the intersection of the points of the circle be at $(0,0)$. So one chord has endpoints at $(2,0)$ and $(-6,0)$, the other at $(0,3)$ and $(0,-4)$. Let the center of the circle be at $(x,y)$

We note that the distance between the center and any of the points on the edge is equal.

So, by the distance formula:
$(x-2)^2+y^2=(x+6)^2+y^2=x^2+(y-3)^2=x^2+(y+4)^2$

To find $x$, we solve $(x-2)^2+y^2=(x+6)^2+y^2$. We disregard the extraneous solutions and find $x=-2$.

To find $y$, we solve $x^2+(y-3)^2=x^2+(y+4)^2$. We disregard the extraneous solutions and find $y=-\frac{1}{2}$

So the radius of the circle is the distance between the center of the circle $(-2,-\frac{1}{2})$ and any point on the edge of the circle. We pick $(0,3)$ as this point on the edge. So this distance is $\sqrt{(-2-0)^2+(-\frac{1}{2}-3)^2}$
$=\frac{\sqrt{65}}{2}$

So the distance is two times the radius, or $\boxed{\sqrt{65}}$

\end{document}

After the document class try putting in a \usepackage{amsmath}

[quote="zabelman"]There's nothing wrong with your LaTeX compiling; there's something going on with your YAP previewer. Unless there's a particular reason you want it in DVI, the easiest fix is to make PDFs instead (acrobat is much more reliable than YAP, anyway).

I compiled the document you posted, and it worked fine in both PDF and DVI. There's only one code error: the command \boxed is in the amsmath package, and it isn't included in standard LaTeX. All you have to do is insert [color=darkred]\usepackage{amsmath}[/color] in your preamble.[/quote]

How do Ichange it from DVI to PDF, then?

Are you using TeXnicCenter? If so, there should be a tab at the top of the screen that says [color=darkred]LaTeX=>DVI[/color]. Change that to [color=darkred]LaTeX=>PDF[/color].

Some more help, please...... :blush:

I can't get this to come out nice and with spaces. Can anyone correct this for me?

We note that in a triangle $A+B+C=180^\circ$

So we may rewrite the LHS of the equation as
$\cos^2{A}+\cos^2{B}+\cos^2{(180^\circ-A-B)}+2\cos{A}\cos{B}\cos{(180^\circ-A-B)}$

Now we use the trignometric identity $\cos{(180^\circ-x)}=-\cos{x}$ to show that $\cos{(180^\circ-A-B)}=-\cos{(A+B)}$.

So now our LHS is
$=\cos^2{A}+\cos^2{B}+\cos^2{(A+B)}-2\cos{A}\cos{B}\cos{(A+B)}$

$=\cos^2{A}+\cos^2{B}+(\cos{A}\cos{B}-\sin{A}\sin{B})^2-2\cos{A}\cos{B}(\cos{A}\cos{B}-\sin{A}\sin{B})$
by the identity $\cos{(x+y)}=\cos{x}\cos{y}-\sin{x}\sin{y}$

Expanding, we have:

$=\cos^2{A}+\cos^2{B}+\cos^2{A}\cos^2{B}-2\cos{A}\cos{B}\sin{A}\sin{B}+\sin^2{A}\sin^2{B}-2\cos^2{A}\cos^2{B}+2\cos{A}\cos{B}\sin{A}\sin{B}$

$=\cos^2{A}+\cos^2{B}-\cos^2{A}\cos^2{B}+\sin^2{A}\sin^2{B}$

$=\cos^2{A}+\cos^2{B}-\cos^2{A}\cos^2{B}+(1-\cos^2{A})(1-\cos^2{B})$

$=\cos^2{A}+\cos^2{B}-\cos^2{A}\cos^2{B}+1- \cos^2{A}-\cos^2{B}+\cos^2{A}\cos^2{B}= 1$, as desired. QED.

Thanks!

Okay, latex is driving me crazy now......

Now when I try to view, it says:

[quote][DocOpen("%bm.pdf")][FileOpen("%bm.pdf")]

Cannot execute command[/quote]

I'm confused.

First thing, use the align environment:

[code]
\begin{align*}
&=\cos^2{A}+\cos^2{B}+\cos^2{A}\cos^2{B}-2\cos{A}\cos{B}\sin{A}\sin{B}\\
&\qquad\qquad\qquad+\sin^2{A}\sin^2{B}-2\cos^2{A}\cos^2{B}+2\cos{A}\cos{B}\sin{A}\sin{B}\\
&=\cos^2{A}+\cos^2{B}-\cos^2{A}\cos^2{B}+\sin^2{A}\sin^2{B}\\
&=\cos^2{A}+\cos^2{B}-\cos^2{A}\cos^2{B}+(1-\cos^2{A})(1-\cos^2{B})\\
&=\cos^2{A}+\cos^2{B}-\cos^2{A}\cos^2{B}+1- \cos^2{A}-\cos^2{B}+\cos^2{A}\cos^2{B}= 1
\end{align*}[/code]

That's just one part of your document.  As you can see, to begin the align environment type \begin{align*} or \begin{align} (the latter gives your equations numbers) and to end it use \end{align*} or \end{align}.  NOTE: don't put dollar signs before the \begin{align*} nor after the \end{align*}.

Also, those \qquad command just add space.  You can put more or less (or none) suiting your preferences.  I did this since the first equation was too big for one line.  I basically broke it up over two lines and indented the second line to emphasize it was not a new equation.

You also may want to use double dollar signs more often.  Using $\$$\$ [Put LaTeX here] $\$$\$ puts the LaTeX in displaystyle and centers it on its own line.  Just use it when you have important or large equations that don't look right in text.

Do you have the pdf open when you are trying to view it?  You have to close the pdf file before recompiling your LaTeX.  This is the major drawback of using PDF's to preview your documents.  Usually, I use DVI for previewing everything (unless it contains images..which I then need to use PDF....) until I have my final copy ready to be compiled into a PDF.  Don't close all of Adobe, just the file you are viewing.  Closing and reopening all of Adobe takes a LOT more time.

Thanks for all the tips Joml88. I needed that!


What is the least positive integer $n$ such that $n^2-n$ is divisible by some but not all integer values of $k$ when $1\le k \le n$?

[quote="pianoforte"][hide]guess and check and you get 5
2^2-2=2
3^2-3=6
4^2-4=12
5^2-5=20, which is not divisible by 3, and is divisible by 2 and 4[/hide][/quote]

[hide]That is exactly the way I did

$5$[/hide]

[hide="wow"]
Er guess and check?
you can factor $n(n-1)$
If $n=3$
Then $n(n-1)=6$
6 is divisible by all $k$
If $n=4$
Then $n(n-1)=12$
12 is divisible by all
If $n=5$
Then $n(n-1)=20$
20 is divisible by 1,2,4,5 but not 3
Thus the answer is $5$[/hide]


algebra
polynomial
superior algebra
superior algebra unsolved

Prove that if $\mathfrak{a}_{1},\mathfrak{a}_{2},...,\mathfrak{a}_{n}$ are ideals of a ring $A$ such that $\mathfrak{a}_{1}+\mathfrak{a}_{2}+...+\mathfrak{a}_{n}=A$ then $\mathfrak{a}_{1}^{v_{1}}+\mathfrak{a}_{2}^{v_{2}}+...+\mathfrak{a}_{n}^{v_{n}}=A$ for all positive integers $v_{1},v_{2},...,v_{n}$.

that's cute :-).

pick $x \in \mathfrak{a}_{1}$ with $x \equiv 1$ mod $\mathfrak{a}_{2}+...+\mathfrak{a}_{n}$. then also $x^{v_{1}}\equiv 1$, thus $\mathfrak{a}_{1}^{v_{1}}+\mathfrak{a}_{2}+...+\mathfrak{a}_{n}=A$. continue the same way with the other ideals. finally we get $\mathfrak{a}_{1}^{v_{1}}+\mathfrak{a}_{2}^{v_{2}}+...+\mathfrak{a}_{n}^{v_{n}}=A$

I don't see what's wrong with that proof...

Except some possibility not considered in the first solution, too: there could ba a lack of $1$...

[quote="-oo-"]no (and YOU have to be sure of your proof)[/quote]
Yes, I am true and I used polynomials theorem :P

@NT: ok! then could you please explain the last $\in$ ?
@ZetaX: ?

Don't know what N.T.TUAN talks about in his second post (polynomials theorem¿)...

But expanding $(a_{1}+...+a_{n})^{v_{1}+...+v_{n}}$ is a sum of monomial terms of type $a_{1}^{w_{1}}\cdot ... \cdot a_{n}^{w_{n}}$ with $w_{1}+...+w_{n}=v_{1}+...+v_{n}$, and we cant have $w_{k}< v_{k}$ for all $k$, thus w.l.o.g. $w_{1}\geq v_{1}$ giving ${a_{1}^{w_{1}}\cdot ... \cdot a_{n}^{w_{n}}}\in \mathfrak{a}_{1}^{v_{1}}$.

[quote="ZetaX"]

But expanding $(a_{1}+...+a_{n})^{v_{1}+...+v_{n}}$ is a sum of monomial terms of type $a_{1}^{w_{1}}\cdot ... \cdot a_{n}^{w_{n}}$ with $w_{1}+...+w_{n}=v_{1}+...+v_{n}$, and we cant have $w_{k}< v_{k}$ for all $k$, thus w.l.o.g. $w_{1}\geq v_{1}$ giving ${a_{1}^{w_{1}}\cdot ... \cdot a_{n}^{w_{n}}}\in \mathfrak{a}_{1}^{v_{1}}$.[/quote]
This is what i said  :wink:

[quote="ZetaX"]
[quote="N.T.TUAN"]Prove that if $ \mathfrak{a}_{1},\mathfrak{a}_{2},...,\mathfrak{a}_{n}$ are ideals of a ring $ A$ such that $ \mathfrak{a}_{1}+\mathfrak{a}_{2}+...+\mathfrak{a}_{n}=A$ then $ \mathfrak{a}_{1}^{v_{1}}+\mathfrak{a}_{2}^{v_{2}}+...+\mathfrak{a}_{n}^{v_{n}}=A$ for all positive integers $ v_{1},v_{2},...,v_{n}$.[/quote]

But expanding $ (a_{1}+...+a_{n})^{v_{1}+...+v_{n}}$ is a sum of monomial terms of type $ a_{1}^{w_{1}}\cdot ... \cdot a_{n}^{w_{n}}$ with $ w_{1}+...+w_{n}=v_{1}+...+v_{n}$, and we cant have $ w_{k}< v_{k}$ for all $ k$, thus w.l.o.g. $ w_{1}\geq v_{1}$ giving ${ a_{1}^{w_{1}}\cdot ... \cdot a_{n}^{w_{n}}}\in \mathfrak{a}_{1}^{v_{1}}$.[/quote]

More radically  $ {\sqrt}({\mathfrak{a}_{1}^{v_{1}}+\mathfrak{a}_{2}^{v_{2}}+\cdots+\mathfrak{a}_{n}^{v_{n}}) \supset \mathfrak{a}_{1}+\mathfrak{a}_{2}+\cdots+\mathfrak{a}_{n}=1 \implies{\mathfrak{a}_{1}^{v_{1}}+\mathfrak{a}_{2}^{v_{2}}+\cdots+\mathfrak{a}_{n}^{v_{n}}= 1.$


floor function
logarithms
algebra
binomial theorem

Is there a relatively simple way to prove that the first four digits of $ 6^{1991}$ are $ 1991$?

You can show with ugly log stuff that the number has $ 1550$ digits. Then either find a way to compute $ \left\lfloor\frac{6^{1991}}{10^{1546}}\right\rfloor$, or perhaps expand $ (10\minus{}4)^{1991}$ or $ (1\plus{}5)^{1991}$ with the Binomial Theorem, and then attempt to find the terms that contribute to the first four digits.

I can't think of anything relatively simple, though.

$ x \equal{} 6^{1991}$. $ y \equal{} \log_{10} x$. $ z \equal{} y \minus{} \lfloor y \rfloor$. $ 10^z \approx 1.9913$.

$ x \equal{} 10^{\lfloor y \rfloor} \times 10^z \equal{} 19913...$


Solve the equation

2^{x^{2}+x}+\log_{2}x = 2^{x+1}

Latexified

$ 2^{x^{2}\plus{}x}\plus{}\log_{2}x \equal{} 2^{x\plus{}1}$

[hide="solution"]Note that $ x\equal{}1$ is a solution. For $ x\leq0$, $ \log_2 x$ is undefined. For $ 0<x<1$, $ x<1 \Rightarrow x^2<1 \Rightarrow x^2\plus{}x<x\plus{}1 \Rightarrow 2^{x^2\plus{}x}<2^{x\plus{}1}$, and $ \log_2 x<0$, so the equation cannot be true. Similarly for $ x>1$, $ x>1 \Rightarrow 2^{x^2\plus{}x}>2^{x\plus{}1}$, and $ \log_2 x>0$, so the equation cannot be true. Thus, the only solution is $ x\equal{}1$.[/hide]


Rules are [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=295442]here[/url]

[b]25[/b]

[b]8[/b]

lalalala ELMO'S WORLD!!!!!!!!! XD

[b]11[/b]

Apparently for this particular thread it's +2 or -2 for boys or girls.


What fraction of this square region is shaded? Stripes are equal in
width, and the figure is drawn to scale.
[asy]size(1inch);
pen p = red;

path ptop = (0,71)--(71,71)--(71,60)--(0,60)--cycle;
path pright = (60, 0)--(71,0)--(71,71)--(60,71)--cycle;
path pmiddletop = (0,47)--(47,47)--(47,36)--(0,36)--cycle;
path pmiddleright = (47,47)--(47,0)--(36,0)--(36,47)--cycle;
path pbottomtop = (0,23)--(23,23)--(23,12)--(0,12)--cycle;
path pbottomright = (23,23)--(23,0)--(12,0)--(12,23)--cycle;

fill(ptop, p);
fill(pright,p);
fill(pmiddletop, p);
fill(pmiddleright, p);
fill(pbottomtop, p);
fill(pbottomright, p);

draw((0,0)--(71,0)--(71,71)--(0,71)--cycle);[/asy]

Divide the square into 36 little squares, red takes 21 of them, so$ \frac{21}{36}\equal{}\frac{7}{12}$

Or note that the red stripes have two more "little squares" than the smaller white stripes, so there are $ 6$ more little squares, so we have $ 21/36\equal{}7/12$


There should have been better announcements.
I think I have to email everybody again.

??  is this for the relay practice?  i thought they handed out the sheet u typed up during the training session announcing the date and time of the practice  :?

yes, they did.
but im afraid many people didnt bother to read them.

i didn't get anything at the practice :( (well, it's because i didn't go, but who cares about such trivialities) so naturally i didn't get what was handed out there. can y9ou please post it here or via email?

[b]New York City Math Team[/b]
[b]Online Relay Practice Session[/b]
                                                                                Saturday May 28, 2005
                                                                                12 PM –

Hello everyone, this is Geehoon.
We are planning to run another online relay practice session before ARML.
This is totally voluntary (well, A team members are expected to come), but you are highly recommended to come!

Ok, here is a brief description on what to do.

If you decide to come, you have to form a three-member-team.
This does not have to be the same team as your ARML team, but feel free to do so.
Once a team is formed, please send me your teammates’ names and screen names prior to Saturday. (and please show up if you decided to come)

On Saturday, I will open three chat rooms.
In the first room, the first person of all teams will be invited, and the questions for them will be presented as well as time warning. In the second room, questions for the second person of each team will be presented and so forth. People in the third position have to give me the answers before the signal is called. In sum, the first person of each team will be getting questions from Window 1 and will pass his/ her answer to their second person; second person of the team will get questions from Window 2 and will pass his/ her answer to your third person, and the third person of the team will get questions from Window 3, and will pass the final answer to me.

So, If you’re the second person on your team, just IM me (yeahhhh16) and say “Please invite me onto Window 2”

It will be continued as long as you guys want,
After the session is over, I will post up questions / solutions / results on http://www.artofproblemsolving.com, if anyone is interested.
Please ask Ravi B for the access for NYC forum.

If you have any questions, please contact me at Geehoon@mit.edu
Thank you.

14 days until ARML.

Maybe the captains should remind their teams...

or if they dont, you can just send an email to the whole team...i mean, thats what the mailing list is for, right?

yes, that is what im going to do, after coaches and I decide who the proctor will be.

[quote="Geehoon"]Also, it seems like there will be a practice session with power questions after the award ceremony at Stuyvesant.[/quote]

then you should send an email to everyone asap.  you can't really run a practice if many people don't show up. ;)

Seems? Is the practice definite? If so, I need to get the rest of the team out of class. Not all of the city math team is going to the ceremony, from my school, at least.

exactly.  i know that hunter won't be attending the award ceremony because we do our own thing, but we need to know whether the practice is official.  there's really no point if only some people will show up... also, our math department is complaining about having to excuse so many people from classes in the second semester, so please make sure it is actually going to happen

The idea of the practice session was brought up by A team members, after poor performance on the practice session at CCNY.

It will be a power round practice, and I dont think this is official practice for all city math team people, but A team members should come.

=======================================================
From me to Mr. Linker

Are you going to be at the Award Ceremony?
Stan from A team asked that if it is possible for us to organize another practice session at the award ceremony.

We would like to practice another power round if possible.
We're going to practice relays online this Saturday, as planned, but many people seem not to like the practice results.

Thank you,
Geehoon Hong

=======================================================
From Mr. Linker to me

I will be there.  I can provide a power question and I am sure that we can get you a room following the ceremony.  Are all team members attending the ceremony?

DL

=======================================================
From me to Mr. Linker again

14 of A Team members are confirmed to attend the ceremony, and I believe Tai Ho will come as well.

Thanks,
Geehoon Hong

=======================================================

Tai Ho's coming. I really have to get him to join this forum.
He tells me a bunch of people on A team fooled around after doing the first 5 questions. Is that true? :P

so we are having the practice tomorrow, right?!

unless mr.linker forgets to bring power questions.
I guess I'm going to bring something as Plan B then..

Ohk, so who's coming tomorrow's $LAST$ online practice?

I am expecting at least 5 teams?

i think there will be more than 5 teams.
because this practice is for all ANYCR teams(more than 20 relay teams) :)

you forgot to factor in apathy ;)

but i'll be there :?

ill be there...but i wouldnt have high hopes in that every person will show up...

Xiao Ling	loser1986330
Francisco Cai	ocsicnarf16
Yang Bai	thats my shirt

Barry Weng	ManicBarry
Yitian Liu	YiddyBiddy015
Zoe Chiang	comeflyflywithme

Tao Yang	dark plutos
Max Engelstein	Maxeng88
Olga Zverovich	ZZZ48

Brady Tang	darkporpoise
Yuri Lin	mayoiko
Mark Spindler	kewlman8207

Edward Sung	Celihque
Yi Wu		randomname13579

Well, these are the sns of my team. I'm not saying they will all show up, but if some of them do, this will help things run more smoothly, I suppose.


Calculate a0, a1, a2, and a3 (give four decimal places)

Am I missing something, or is this problem just a plug in to your calculator type? As in just plug in the value for n and then yea...


integration
real analysis
function
real analysis unsolved

Let $ g\in L(0;1)$ such that for every $ f\in C^{\infty}_c(0;1)$ we have $ \int_0^1 g(t)f'(t)dt \equal{} 0$. Show that $ g(t) \equal{} \text{const}$ for almost everywhere in $ (0;1)$.

And it can be proved in essentially the same way. Suppose there are two Lebesgue points  of $ g$ such that $ g(a)\ne g(b)$. Define $ f$ so that $ f'$ is the difference of a bump function around $ a$ and another bump function around $ b$. If the bumps are narrow enough, $ \int g f'\ne 0$.

What is [url=http://en.wikipedia.org/wiki/Bump_function]Wikipedia[/url]?  :maybe:


function
algebra
domain
limit
topology
real analysis
calculus

Is the function

$ f(x)\equal{}\begin{cases}1,&x<\sqrt{2}\\ \minus{}1,&x>\sqrt{2}\end{cases},\,\,\,\,\, x\in\mathbb{Q}$

continuous?

What does it means continous? For every sequence $ \{ x_{n}\}$ converging to $ x$ we have $ \{ f(x_{n})\}$ converges to $ f(x)$. So is it true that any sequence $ \{x_{n}\}$ of rational numbers converging to a rational number $ x$ we have $ \{ f(x_{n})\}$ converges to a rational number $ f(x)$? I think so.

[quote="LeFromage"]Is the function

$ f(x) \equal{}\begin{cases}1, & x <\sqrt{2}\\ \minus{}1, & x >\sqrt{2}\end{cases},\,\,\,\,\, x\in\mathbb{Q}$

continuous?[/quote]

I wonder what is a continous function defined in $ \mathbb{Q}$  :?:

$ \mathbb{Q}$ is a metric space, and it makes perfect sense to talk about a function being continuous on it. We can define continuity in either of two ways, which are completely equivalent.

Epsilon-delta definition: $ f$ is continuous on a domain $ D$ if for every $ y\in D$ and for every $ \epsilon>0,$ there exists $ \delta>0,$ $ \delta$ depending on both $ \epsilon$ and $ y,$ such that  if $ d(x,y)<\delta,$ then $ d(f(x),f(y))<\epsilon.$

[Of course, we have the usual distance on $ \mathbb{Q}$ and can write $ |x\minus{}y|$ instead of $ d(x,y)$ and $ |f(x)\minus{}f(y)|$ instead of $ d(f(x),f(y)).$]

Sequential definition: $ f$ is continuous on a domain $ D$ if for every $ y\in D$ and for every sequence $ (x_{n})$ in $ D$ such that $ \lim_{n\to\infty}x_{n}\equal{}y,$ then $ \lim_{n\to\infty}f(x_{n})\equal{}f(y).$

--

The function defined above is continuous on $ \mathbb{Q}.$ Let's look at the sequential definition. Choose a $ y.$ There are two cases - either $ y>\sqrt{2}$ or $ y<\sqrt{2}.$ Suppose for now that $ y>\sqrt{2}.$ Take any sequence of rational numbers $ (x_{n})$ such that $ x_{n}\to y.$ Then for all large enough $ n,$ we have that $ x_{n}>\sqrt{2},$ and thus $ \lim f(x_{n})\equal{}1\equal{}f(\sqrt{2}).$ The case $ y<\sqrt{2}$ is similar.

If your mind rebels at that and still insist that something must be wrong with this function: we can find something "wrong." The function is not uniformly continuous. The sequential version of uniform continuity is that a uniformly continuous function takes Cauchy sequences to Cauchy sequences. We could find a sequence of rational numbers that alternates between $ >\sqrt{2}$ and $ <\sqrt{2}$ while tending to $ \sqrt{2}.$ Such a sequence would not be a convergent sequence if our universe consisted only of $ \mathbb{Q},$ but it is a Cauchy sequence. And $ f(x_{n})$ alternates and thus diverges.

Could you explain this in more "common sense" terms? I don't see how the function can be continuous, since there is a jump as you go from rationals less than $ \sqrt{2}$ to rationals greater than $ \sqrt{2}$.

Also, does it make sense to ask whether the function is continuous at $ x\equal{}\sqrt{2}$?

It doesn't make any sense to ask whether the function is continuous at $ \sqrt{2}$ because that point does not lie in the domain of the function. But note: the function is also not continuous at $ \sqrt{3}$ [i]for exactly the same reason![/i] And the function is continuous at every rational number, which is everywhere in its domain.

As for the "common sense" language, the best I can do is this: the function is continuous at $ y$ if you can make $ f(x)$ arbitrarily close to $ f(y)$ by making $ x$ close enough to $ y.$ Now, that's just a paraphrase of of the technical definition - it's the best I can do because I can't afford to lose any of the meaning of that technical definition.

The problem is that you reach a point in mathematics at which you cannot proceed based on squishy, intuitive notions of what words like "continuous" ought to mean. I guarantee that whatever you would write, informally, for what you think that word ought to mean, I could find an example to break it. That is, unless you approached the care and rigor of the accepted definition of continuity.

Thanks for the explanation.

As far as I know, a function is continuous at $ x \equal{} a$ if $ \lim_{x\to a}f(x) \equal{} f(a)$.

I guess it is intuitively clear that the function is continuous at x not near $ \sqrt{2}$, since, for example, you only "see" 1's as you hop from one rational number to the next.

But it's the "jump" near $ \sqrt{2}$ that is a little confusing. Are you saying that despite this "jump," the function is still continuous?

there is no jump

no matter which number you pick, all sufficiently nearby values of the function are the same

[quote="LeFromage"]
As far as I know, a function is continuous at $ x \equal{} a$ if $ \lim_{x\to a}f(x) \equal{} f(a)$.
[/quote]
I used to think the same thing too until I learned some analysis and basic topology. This definition isnt exactly right. Dr. Kent gave a perfect explanation of what it means continous.

[quote="LeFromage"]As far as I know, a function is continuous at $ x \equal{} a$ if $ \lim_{x\to a}f(x) \equal{} f(a)$.[/quote]

Contra sylow_theory, this is a perfectly good definition of continuous if applied correctly.  Here, the key is that the limit is through the rationals.  For any rational, the rationals "nearest to it" are all on the same side of $ \sqrt{2}$, so the limit exists everywhere (and is equal to the value of the function) -- rationals don't "see" the jump at $ \sqrt{2}$.

[quote="LeFromage"]As far as I know, a function is continuous at $ x \equal{} a$ if $ \lim_{x\to a}f(x) \equal{} f(a)$.[/quote]
In all the ways that matter here, that definition [b]is[/b] correct. (The only reasons to quibble with it have to do with points in the domain that aren't limit points of the domain, and there are none of those here.)

The reason we don't say the function is discontinuous at $ \sqrt{2}$ is that $ \sqrt{2}$ doesn't exist. (Not as point in the domain of the function, anyway.) In other words, exactly one of the following statements is true:

(1) $ f$ is continuous at $ a.$
(2) $ f$ is discontinuous at $ a.$
(3) There is no such point as $ a,$ as far as the domain of $ f$ is concerned.

And for this function, there are no points that fall into category (2).

-----

You're trying to "see" a picture, and you are thinking of $ \mathbb{Q}$ as a dense subset of $ \mathbb{R}.$ Since $ \sqrt{2}\in\mathbb{R}\setminus{Q},$ trying to ask about what happens "at" $ \sqrt{2}$ is not a question about $ f$ itself but about a hypothetical possible extension of the definition of $ f$ to $ \mathbb{R}.$

Is $ f$ continuous on $ \mathbb{Q}?$ Yes, it is.
Can $ f$ be extended to be a continuous function on $ \mathbb{R}?$ No, it cannot. The particular objection is that we cannot define it at $ \sqrt{2}$ in such a way as to make it continuous.

Suppose $ D$ is a subset of $ \mathbb{R},$ suppose $ f$ is a continuous function on $ D,$ and suppose let $ \overline{D}$ be the closure of $ D$ in $ \mathbb{R}.$ (Note that $ \overline{\mathbb{Q}}\equal{}\mathbb{R}.$)

Under what circumstances is it possible to find a continuous function $ g$ defined on $ \overline{D}$ such that $ g(x) \equal{} f(x)$ for all $ x\in D?$ A sufficient condition would be that $ f$ is uniformly continuous on $ D.$ (To make it a necessary and sufficient condition, require that $ f$ be uniformly continuous on each bounded subset of $ D.$)

The function $ f$ on this problem cannot be extended to a continuous function on $ \mathbb{R}.$ Clearly, it is not uniformly continuous on $ \mathbb{Q}.$

----

Definition of uniform continuity: a function $ f$ is uniformly continuous on $ D$ if and only if for every $ \epsilon > 0,$ there exists $ \delta > 0$ ($ \delta$ depending only on $ \epsilon$ and not on $ x$ or $ y$) such that if $ |x\minus{}y| <\delta,$ then $ |f(x)\minus{}f(y)| <\epsilon.$

Sequential version of the same definition: $ f$ is uniformly continuous on $ D$ if and only if for every Cauchy sequence $ (x_{n})$ in $ D,$ then $ (f(x_{n}))$ is a Cauchy sequence.

[color=green][I was working on this when JBL posted. Obviously, I agree with JBL on this point.][/color]

Here is the way continuity is defined in $ \mathbb{R}$ and then I will explain when your definition works and fails.

[b]Definition:[/b] Let $ S$ be a non-empty subset of $ \mathbb{R}$ with the function $ f: S\mapsto\mathbb{R}$. We say $ f$ is continous at $ x_{0}\in S$ when for any $ \epsilon > 0$ there exists $ \delta > 0$ so $ |x\minus{}x_{0}| <\delta$ and $ x\in S$ implies $ |f(x)\minus{}f(x_{0})| <\epsilon$.

So for example $ f:\{ 0\}\mapsto\mathbb{R}$ defined as $ f(0) \equal{} 0$ is continous. Kinda strange but that's the definition. However $ \lim_{x\to 0}f(x) \equal{} f(0)$ just doesn't make sense because limits only make sense for function defined on open intervals (except possibly at that point).

Here is the limit version $ \lim_{x\to x_{0}}f(x) \equal{} f(x_{0})$.
And this means that,
For any $ \epsilon > 0$ there exists a $ \delta > 0$ so that $ |x\minus{}x_{0}| <\delta$ implies $ x\in S$ and $ |f(x)\minus{}f(x_{0})| <\epsilon$.

There is a minor difference between this limit version and the one I posted first. Look at were the statement $ x\in S$ is place. In one version is goes [i]before[/i] the word "implies" in the other version it goes [i]after[/i] the word implies. Minor difference but sometimes it leads to problems, i.e. in this example (first post on this thread) limits dont work anymore. However, JBL is right, these two version are exactly the same [b]if the function is defined on an open interval[/b].

Hope that helps.

EDIT: There is a way to define [i]limits on sets[/i] and using this notion your understanding if fully correct. Maybe if I have time in the future I post that. By the way, I'm not disagreeagring with the other posters.


Hi Guys,
If you know the solution of this question then please post the answer here.
Mary makes mugs in Miami. She makes two sizes of mugs: a small mug that she sells for 2.50 doller and a large mug that she sells for 5.75 doller. Yesterday Mary made 56 doller. Before she opened her shop in the morning, she had 200 mugs in her inventory. At the end of the day she has 188. How many mugs of each price did she sell?

Mary must have sold 200-188=12 mugs yesterday. Let x be the number of \$2.50 mugs and y be the number of \$5.75 mugs. From here we can set up an equation between both the total number of mugs and the price:

$ \mbox{(1) } x\plus{}y\equal{}12$
$ \mbox{(2) } 2.5x\plus{}5.75y\equal{}56$

From statement 1, I multiply by 2.5:
$ 2.5x\plus{}2.5y\equal{}30$

Next, subtract the two equations from each other:
$ \mbox{(2)}\minus{}\mbox{2.5(1) }\Longrightarrow5.75y\minus{}2.5y\plus{}2.5x\minus{}2.5x\equal{} 56\minus{}30 \Longrightarrow y\equal{}8$

Knowing y, we can easily solve for x to get a final answer of [b]4 \$2.50 mugs and 8 \$5.75 mugs.[/b]

[hide="Answer"]She sold 4 small and 8 large. This is a pretty simple systems of equations problem basically you gather as many equations as you can and then see what you can do with them:

Equations:
1. x + y = 12
2. 2.5x + 5.75y = 56

Then you can solve by substituting the first equation into y of the second:
2.5x + 5.75(12 - x) = 12

Solve this to find x = 4, then substitute 4 into either equation to find the y = 8[/hide]

Edit: didn't even notice this was already solved haha


What is the fastest way to solve the problem?

Check that A^9 = (3^16)*A where

A=
      5 -2 -4
     -2  8 -2
     -4 -2  5

D =
      0 0 0
      0 9 0
      0 0 9

in D = inv(T)*A*T

what I do is the following:

if A^9 = (3^16)*A then [T*D*inv(T)]^9 = (3^16)*[T*D*inv(T)] which is T*(D^9)*inv(T) = (3^16)*[T*D*inv(T)]
so I take out twice 9^8 out of D^9 and I get (9^8)*(9^8)*[T*D*inv(T)] =  (3^16)*[T*D*inv(T)]
But then 9^16 is not 3^16.....Do I make a mistake anywhere or it seems that the problem is wrong?

silly me....I apologize for wasting space here :)
I take 9^8 out in front only once...must have been blind


The Art of Problem Solving Books have really good sections on number theory. If you haven't looked at those yet I really recommend them.

For books devoted specifically to number theory, I think the book Number Theory and It's History by Oystein Ore is a good one to start with. It's pretty old, but still very popular. You can find it in just about any book store. It's pretty cheap too, like less than 10 dollars.

The other book I would recommend is The Higher Arithmatic by H. Davenport. Some of it starts to touch on deeper topics, like quadratic reciprocity, but it doesn't go too far. It's also a kind of old book, so some things, like the chapter on computers may be a slightly outdated.

Someone mentioned G.E. Andrew's book "Number Theory" (published by Dover).  I have this book, and I've found it pretty interesting, although I haven't gotten through much of it.  I'd recommend this book - maybe - but I'm not sure how good it is.  Gauss202?  I believe you mentioned it in a ComplexZeta problem post, no?

Yes, I have been recommending Andrews book. I think it is very good. But probably a little bit more advanced than the books I mentioned above. I would probably recommend Andrews' book for upperclass high school students.

Here is a post I made about Number Theory Books earlier - http://www.artofproblemsolving.com/wforum/viewtopic.php?p=948&highlight=andrews#948

Hardys An Introduction to the Theory of Numbers, after you've mastered that one go to Nivens An Introduction to the Theory of Numbers. The first book might be a little difficult for you to read at the beginning, but you will come over that I think.

I think that Hardy book is too difficult to start out as an 8th grade student. It's an excellent book, but it is much too deep.

In addition to the books I mentioned above I would also recommend Beiler's Recreations in the Theory of Numbers: The Queen of Mathematics Entertains

I agree with gauss202's last post.  Beiler's [i]Recreations in the Theory of Numbers: The Queen of Mathematics Entertains[/i] is a good introduction to Number Theory.  It doesn't teach you all the methods, but it gives you a good taste of number theory and tells you about a lot of special little patterns and "tricks" that you might even apply to your math competitions.  Besides, it's a fairly common book (I found it at my local book store, which isn't exactly math-nerd-friendly) and isn't too expensive (12.00 or so)

As an 8th grader a really nice book to read is A friendly introduction to number theory. It's written very much like AoPS books where they have a conversation with you rather than format the book like this, Thm 1. ~~~ Proof of Thm 1 ~~~~ Thm. 2 ~~~ Proof of Thm 2. ~~~~ Where the ~~~ represent mathspeak.

Enjoy,
IW.OAV


AMC
AIME
calculus
integration
geometry
3D geometry
tetrahedron

By a proper divisor of a natural number we mean a positive integral divisor other than 1 and the number itself. A natural number greater than 1 will be called "nice" if it is equal to the product of its distinct proper divisors. What is the sum of the first ten nice numbers?

Haha - nice. Large sentences simple concept.

[hide]You know that such a number which is the product of its proper divisors has to be in the form a * b, where a and b are prime and distinct. So,

$2*3=6 \\
2*5=10 \\
2*7=14 \\
2*11=22\\
2*13=26\\
2^3=8\\
3*5=15\\
3*7=21\\
3*11=33\\
3^3=27$

First 10.

So, $6+10+14+22+26+8+33+15+21+27=182$?

I can't count, but that's what I got.[/hide]

What about 8?

I think you must miscounted something :-D

Do it again. In AIME, you should be sure of getting these "easy" ones. :P

white_horse_king88 forgot the case of $p^3$ where $p$ is prime. its proper divisors are $1,p,p^2$ which multiply out to $p^3$.

Agh... noooo. Oh well. Thanks guys. Like tetrahedron said - it's good to get it out of your system.  :blush:

[quote="eryaman"]To whitehorse: you added wrong, anyways  :D[/quote]

However painful, I admit my addition skills in my head tend to stink... eh. Haha... I need to take 2nd grade math again.

[hide]For a number to be "nice," it must either be the product of two primes or the cube of a prime. For example, if $p$ is prime, $p^3$ has proper divisors $p$ and $p^2$ and is "nice." The first ten nice numbers are 6, 8, 10, 14, 15, 21, 22, 26, 27, and 33, so our answer is $182$. [/hide]

why can't the product be two non relatively prime number?

If the two numbers a and b are not relatively prime, say a gcd of $ a \equal{} d*a'$ and $ b \equal{} d*b'$, then the number could be written as $ ab \equal{} adb' \equal{} da'b$. Now if you multiply the proper divisors it won't give you $ ab$.

The condition here is not relative primality, it's absolute primality.

Let $ N \equal{} \prod_{n \equal{} 1}^{\infty} p_n^{a_n}$ be the prime number decomposition of a number $ N$, where $ a_n \in \mathbb{Z}^{\geq 0}$

The product of its proper factors must have exactly $ a_j$ factors of $ p_j$ for each $ j$.

[b]Case 0:[/b] If there are no positive $ a_j$, then the product of proper factors is the empty product, or $ 1$, which does not equal any $ N > 1$, for sure.

[b]Case 1:[/b] If there is exactly one positive $ a_j$, say at $ j \equal{} r$, then the proper factors are $ p_r^1$, $ p_r^2$, ... $ p_r ^{a_r \minus{} 1}$.

Their product is $ p_r^{1 \plus{} 2 \plus{} \ldots (a_r \minus{} 1)}$, which must equal $ p_r ^{a_r}$, so $ 1 \plus{} \ldots \plus{} (a_r \minus{} 1) \equal{} a_r$, or $ a_r (a_r \minus{} 1)/2 \equal{} a_r$, or $ a_r \equal{} 3$. Then, $ N \equal{} p_r ^3$ for some $ r$.

[b]Case 2:[/b] If there are more than one positive $ a_j$, then they must all be 1. If there were, let $ i$ be the lowest integer such that $ a_i > 1$. Then, $ p_i^{a_i}$ and $ p_i$ are distinct proper factors, and the product of the proper factors will have at least $ a_i \plus{} 1$ factors of $ p_i$, a contradiction.

There cannot be more than two positive $ a_j$. If there could be, let three of them have indexes $ r,s,t$. Then, each of the pairwise products of $ p_r ^{a_r}$, $ p_s ^{a_s}$, $ p_t ^{a_t}$ is a proper factor of $ N$. Multiplying these three, we realize that the product of proper factors of $ N$ would have at least $ 2a_r > a_r$ factors of $ p_r$, a contradiction.

Thus, if there is more than one positive $ a_j$, there must be exactly two, both equal to 1; so $ N \equal{} p_r p_s$ for some $ r,s$.

[quote="Ihatepie"]Isn't it basically all numbers with 4 factors?[/quote]
yep, because 2 of the factors are not proper and the other 2 multiply to the original


Let $s(n)$ denote the sum of the decimal digits of the positive integer $n$. Find the set of all positive integers $m$ such that $s(km) = s(m)$ for $k = 1, 2, ... , m.$

Rust, are you sure? And why 99, 999,... don't work?

[quote="Rust"]$9|m$, because $s(n)=n(mod \ 9)$. If $[log_{10}m]=k$, then $m>10^{k}$ (because $9|m$), but $s((10^{k}+1)m)\not =s(m)$, therefore only $m=9$.
Interesting question, when exist $k\not =10^{l}$, suth that $s(km)=s(m)$.[/quote]
m=99...9  satisfies the question :)

I am sorry, my method prove only all digits are 9.
If $s((10^{k}+1)m)=s(m)$, were $m=a_{k}10^{k}+....+a_{0}, \ 0\le a_{i}\le 9$, then $s(m+a_{k})=a_{k}$. Therefore $a_{k}=a_{k-1}=...=a_{1}=9$, and $a_{0}=9$ because $a_{0}+a_{k}\ge 10$ and $9|m$. It is easy to chek, that $m=10^{k+1}-1$ work.


summer program
PROMYS
Ross Mathematics Program
Mathcamp
email
Galois Theory
number theory

I just got into PROMYS 2009!! Has anybody else found out yet? I am so excited  :D
I have some q's:
1) are research groups yourself or 2-3 other people
2) what about learning Galois Theory etc. - was that  a seminar t0ra or a minicourse
3) I have basic NT knowledge already- does that mean that some things will be review???

2)  It was a second-year course, taught by a professor.  They change every year.

3)  This is unavoidable when you learn a subject from multiple sources.  Rest assured that the way the program starts out will get you to think about the subject in a different way.

I got accepted, too. I'm not sure if i'm going yet though.

[quote="mathemonster"]did you get in as a counselor?[/quote]
As it turns out, no. :|

Hm, I have to decide between Ross and PROMYS so far. I haven't heard back from HCCSiM yet.
how about you, mathemonster?

first of all congratulations. I wanted to go to PROMYS over Ross. @t0ra: ?????!!!????  :o  :(  :mad: what happened??

thanks. They're both really good... what makes you want to go to PROMYS?

And yeah, t0raji0u, that's really surprising that you didn't make it...

Oh, I see. Boston's a nice place.

Ross is a bit more challenging though, right? And more structured, I hear.

I'm not actually sure. I chose PROMYS though.

Okay, that's cool. I'm still deciding. I guess I'll leave this thread to "meet and greet"-ers now. :)

[quote="t0rajir0u"]It's quite likely that most people haven't found out yet, considering the deadline isn't even for almost another month.

In any case, one of the counselors this year is a good friend of mine, so I'll be hanging around PROMYS talking about math anyway :)[/quote]
cool. I still don't know why u didn't get in. did u rush the app?

how long do you usually wait before you get either an acceptance or rejection letter?

Is it a letter, email, or phone call?

i got an email first.
then letter.
but if you havent gotten anything back yet, you should probably contact them.
because all the other forms are due the 12th.......
two of my friends got their notifications yesterday & today. one hasnt heard back , but she's going to call them tomorrow.

did your friend get in?

i should definitely call

yea b/c also there cud be a mix up- i didnt know i got into ross till last week b/c they misread my email address.

Hi! I also was accepted to PROMYS and will be attending it over Ross...
Does anyone know what the average age at PROMYS is? I just recently turned 16 and will be a senior in a few months (having skipped a grade).

There isn't really an average age.  We get every high-school grade.  Keep in mind that some students choose to attend PROMYS a second time, whereupon they take harder classes and get harder problem sets and tests.

@ smartguy: my friend got through and they said she had a pretty strong app and were trying to see whether they had spots for her [and several similar cases] ... so i guess they're still deciding..

on a different note: is there wifi? pool? free time? tennis courts?

[quote="shelly32494"]is there wifi?[/quote]
Not when I was there, although this may have changed.  There are ethernet ports in the rooms and a place to buy cables not too far away.

[quote="shelly32494"]pool?[/quote]
Not close to the dorms, at least that I know of.

[quote="shelly32494"]free time?[/quote]
That all depends on how much of the problem sets you finish :P  If by "free" you mean "unstructured," there's plenty of that.

[quote="shelly32494"]tennis courts?[/quote]
Again, not close to the dorms that I know of.  Keep in mind that Warren Towers is in the middle of downtown.

are there any fun activities planned? how many problems are in the problem sets. what happens if u don't finish?

[quote="mathemonster"]are there any fun activities planned?[/quote]
There's lots of "mandatory fun" planned, and you'll find out what that is once you get there.  You can also join the frisbee team to train for the annual RSI-PROMYS game.

[quote="mathemonster"]how many problems are in the problem sets.[/quote]
The problem sets are split up into sections with varying numbers of problems in each section; there are usually maybe 30 per set.  Most aren't too bad, but the last few are "exploration" questions which aren't intended to be answered quickly.

[quote="mathemonster"]what happens if u don't finish?[/quote]
Nothing.  How much of the problem sets you finish is totally up to you.  You aren't expected to finish any of them; they're not supposed to be easy.

I was wondering...are there basketball courts nearby?
And is there time on weekends for those who might want to attend religous services?

Uh of course there'd be tennis courts, basketball courts, and a pool near the dorms.. It's all at FitRec (the Fitness and Recreation Center)

Does anybody know how the midterm exam and final exam ("exam") work at PROMYS? I am currently attending the program...

Both consist of essentially writing down proofs from your problem sets.  If you've done your problem sets, you shouldn't have a problem with either.


Hello there,

I was wondering how can I write a document to form it as a PDF file in $ \text{\LaTeX}$?
Do you have any idea about it?

Start at [[LaTeX:Downloads]] if you are using Windows.

Hi Hasan,

perhaps have a look at [url=http://texblog.net/latex-link-archive/distribution-editor-viewer/]Essential Software for LaTeX[/url], choose your operating system there and you will find some recommendable software like distributions, editors and viewer.
For your first steps you could also have a look at the [url=http://texblog.net/latex-link-archive/introduction-guide/]LaTeX Introductions and Guides[/url].

Stefan

--
[url=http://texblog.net]TeXblog.net[/url]


integration
advanced fields
advanced fields unsolved

Let $g(x)=\begin{cases}x^{-1/2}&\text{if }x\in(0,1)\\0&\text{otherwise}\end{cases}$ and let $a_{k}>0$ such that $\sum_{k=1}^\infty a_{k}<\infty$ and let $\{r_{k}: k\in\mathbb{N}\}$ is a set of rational numbers. Define \[f(x): =\sum_{k=1}^\infty a_{k}g(x-r_{k})\] Show that $f\not\in L^{2}([a,b])$ for every $a<b$.

I'm assuming you meant to say "the set of rational numbers" rather than "a set of rational numbers." That is, I'm assuming that $\{r_{k}\}$ is an enumeration of all of the rational numbers. We have to assume that $\{r_{k}\}$ covers at least a dense set or the argument doesn't work.

Given $a<b,$ there exists a rational number $r_{m}\in(a,b).$

Then $f(x)>a_{m}g(x-r_{m})\ge0$ so that

$\int_{x}^{b}f(x)^{2}\,dx\ge a_{m}^{2}\int_{r_{m}}^{b}\frac{dx}{x-r_{m}}=\infty.$

However, $f\in L^{1}(\mathbb{R})$ so in particular it is finite almost everywhere and integrable on any interval.

By the Monotone Convergence Theorem

$\int_{\mathbb{R}}f(x)\,dx=\int_{\mathbb{R}}\sum_{k}a_{k}g(x-r_{k})\,dx$

$=\sum_{k}a_{k}\int_{\mathbb{R}}g(x-r_{k})\,dx =\sum_{k}a_{k}\int_{0}^{1}g(x)\,dx = 2\sum_{k}a_{k}$

and that sum is finite by hypothesis.


George and Henry started a race from opposite ends of the pool. After a minute and a half, they passed each other in the center of the pool. If they lost no time in turning and maintained their respective speeds, how many minutes after starting did they pass each other the second time?

$ \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 4\frac {1}{2} \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 7\frac {1}{2} \qquad\textbf{(E)}\ 9$

We know that it takes 1.5 minutes for each swimmer to reach the center. Then it will take another 1.5 minutes for them to reach opposite ends of the pool again. From there, they will turn around and meet in the center again in 1.5 minutes, so after starting, it will take 1.5 * 3 = 4.5 minutes for them to meet twice.


If $x$ is a real number, at most how many integers can lie between the numbers $x-\frac{2004}{2005}$ and $x+\frac{2004}{2005}$?

[hide]The range looks like 4006/2005, so the answer must be 1.[/hide]

[hide]I came up with two. If you make x a fraction like 5012/2005, then

5012/2005 - 2004/2005 = 3008/2005

5012/2005 + 2004/2005 = 7016/2005

You end up having the integers 2 and 3 in between the two rational numbers. [/hide]

Let me know if anyone comes up with more. I'd be interested in seeing how you did it.

need I repeat what has been said already?
~-1<x<~1
>>2<<

The range is more than 1 and less than 2, so it must be 2.


Bad, but I did my best. We went before the countdown round began since I wasn't one of the top ten :( . I thought there would be more Dickerson, Westminster, and Hightower Trail kids up for countdown though. You?

Wasn't there :). [I'm in 10th grade anyway.]

Heh. I did make Nats 2 years ago though. Aren't you a 7th grader? You've still got another year :).

Nah, 8th. I was in a disadvantage since I'm only in Geometry.

Well.... It was interesting to say the least.

Mandatory reminder: Do [b][u]NOT[/u][/b] talk about any problems for now. There are still states that will be taking it for the next week or so. I know that you all know this, but it can't hurt to mention it anyway. If you have a real question about it and it just can't wait another week or so, talk to your coach or PM me.

I was pretty disappointed when they announced the top 10 (really 12 since 2 weren't there.) There wasn't ANYONE from Dickerson! I thought that Minh and Claus at least had a great shot at top 10. Then there were 3 Westminster kids there and I figured, "There goes the team championship!" Once they start announcing the team scores, Westminster was 3rd or 4th! Turns out that one of the Westminster kids in the top 10 was there as an individual and not on the team. Wasn't the first time that happened and won't be the last. It's REALLY hard to predict performance on a particular day. (You should see us trying to choose the last few positions on the ARML team.) Then they announced that Dickerson had WON! I was completely shocked. The team was extremely consistent and did very well in the team round. Taylor Sands from Hightower was one of the top 4 and will be going to Nationals. (Me TOO!  :D )

Arvind and Jenn, it was nice meeting you today. (Again for Jenn.) Make sure to come up and say hi when we cross paths down the road.

Take care,
Fulton

Wow! That's pretty amazing. Congrats to you and your team. I'm glad someone from Cobb won.

Congratulations! Have fun!

Who is on the Nats team? Did Arvind make it?

The state team is:

John Whelchel: Westminster
Alex Commandey: Hopewell Middle
Sitan Chen: River Trail Middle
Taylor Sands: Hightower Trail Middle

Any suggestions about how to prepare for Nationals? The best that I can think of is to work old test and do things like we do for ARML.

Cool, someone from Hightower made it. That makes 3 total (I think) . BTW, how did they do as a team?

I suggest holding meetings once a weekend for about 3 hours and practicing old national tests. This is what Mr. Hite did for our team 2 years ago.

I'm not sure how Hightower did as a team. To be honest, I was so disgusted, thinking that we had been completely shut out, that I wasn't paying that much attention! I don't think they were in the top 4.

Unless I'm much mistaken, Hightower got either fifth or sixth. (Neither of the top 2 teams had any students in the top 10, which attests to the closeness of the state competition this year!)

I think that in general, if you are an individual who [i]could[/i] qualify for state individually, then you are more likely to do it that way than by riding along on your school team.  (I would know—I qualified for state three times, but my school never made it as a team :( .)  Consequently, I don't think it's that surprising that many of the top schools didn't have individuals in the top ten.  In my day, the first two teams didn't have any individuals in the top four, so they waived coaching to Mr. Hite, the coach of the third-place team.

I'm curious why you would say this. The individual rounds are first, the tests are exactly the same. The only possibility that I can think of is that you wouldn't practice for the team round. But I consider the team round problems just more of the same type.

Do you have a reason? Or is this just based on personal experience? (Which is valuable in itself.)

It's much easier for an individual to advance than for a team to do so, since 1/4 of all individuals in a chapter competition advance as individuals, whereas only 1/10 of all teams do (according to [url=http://web.archive.org/web/20060209055432/http://gspe.org/mathcounts.htm]last year's guidelines[/url], anyhow).  In my three years at middle school, two of us advanced to state each year, but we never made it as a team—including the year when we were #2 and #3 in chapter on written individual.  Furthermore, if you couldn't get to state on your own, you might be able to go if you are part of a good enough team.  Thus not only are school qualifiers in the minority (approx. 5:2 ratio), but they also reach to a lower-performing segment of the population, so I would not be surprised if three or fewer of the top ten students qualified with a team.  (After all, you'd [i]think[/i] that the third place state student and the second place chapter student would be enough to pull a four-person team up to second in the county.)

So, in brief, I am using reason coupled with personal experience. :)

woot

sitan (stan is his american name) made nationals

that's like the first person from river trail

that's quite a turkey

who is coaching the national team this year then?

terrible :mad:  i didn't make countdown...hmm...aah well, still have jv state to look to!

I'm the coach! Any suggestions on how to work with the team are appreciated. I'm planning on running it pretty much the same way we run ARML practices. BTW, any of you MATHCOUNTS nationals veterans are both invited and encouraged to come to a practice or two. (As soon as we get a practice schedule.)

[quote="gtf"]Let me restate your premise and see if you agree. Maybe it's not easier to make it into the top 4 if you qualify as an individual rather than a team, but there is a higher percentage of individual qualifiers that are actually competitive as opposed to team qualifiers. The Dickerson kids were 2nd, 4th, 5th, and 11th at chapter. So all 4 of them would have probably qualified as individuals also. (I assume that at least one of the top 10 would have been on a team. But maybe it was top 8.)[/quote]
I agree :) .

Hm, my idea for MATHCOUNTS practice would be, of course, to practice with things that are harder than nationals tests.  But that's just me.  I might be able to come to a few practices, depending on when they are—I'd certainly like to.

i got 22nd.... but with one more question (i 4got 2 round on that one) i should have been 11th. because 2 peeps weren't there, i would have been in countdown! i seriously got almost every question before that stan guy. did u guys hear me yelling "Yes!" when they showed us the answer to that decagon question and it was 210? well, that's cuz i got it! but peeps were looking at me so i stopped doing that.
but we were 2nd as a team! we actually tied w/ first, you know. also, hyungi kang made it 2 countdown because 2 people left early who were in top 10. she got 4th in it! but im sad cuz i was captain, and that one thing which i was REALLY good at was countdown. like, hyungi and i were pretty much tied in everything else. my coach and my teammates r still mad at me. one question cost me all of that :mad: !
EDIT:well, at least i have another year. and we got 3rd in the state in uncle remus! (i know its not math but it has math it it, right? and i got every math question first. we have the best academic team of any public school!)

[quote="funcia"]no wonder. no offense meant but he seemed like an idiot savant...the questions he got right and wrong. :maybe:
i hope he makes national countdown[/quote]

hah he said he was messing around throughout the entire thing. He's really good... if he really tried he probably could've gotten most of them he attempted right.

[quote="Karth"]Heheh, I talked to Sitan afterwards, and he said that he pretty much saw numbers and did random stuff with them. He originally asked whether the countdown round mattered, and when they said no, he pretty much went all out and did random stuff, which still made him do well  :D[/quote]

Yup that's Sitan all right (he goes to my school)...and btw for anyone who duzznt know this its not pronounced (sit-an), its pronounced (stan)... :D