summer program
MathPath
MATHCOUNTS
email
AMC
AMC 8
function

Could I just get an idea of who from AoPS is/wants to going/go to Mathpath?

Thanks,
Jorian Hoover

You could just ask this in the MathPath forum since people who are interested in MathPath are bound to go in there....

I saw their whole application thingy, and boy it wasn't too hard (especially with a few weeks!).

Colorado?

I don't think my dad would even consider it.

you have more than a couple of weeks

ea is end of march though :lol:

jorian

and please say you're going again... :)

jorian

MATHPATH IS AWESOME IM TOTALLY GOING SEE YA THERE
i sent in my application early so i already got in

Yep, MathPath is wonderful (I went last year, but am too old this year).

Good luck to everyone applying :) .

[quote="daermon"][quote="Walk Around The River"]To anyone whos gone before...Am I too stupid to go?  like how smart are the kids there?  i mean, im good at math, but im not some super genius.  i heard, from my coach who taught there last summer, it was pretty intense stuff.  and no, im not like ubemaya who thinks hes stupid, yet hes a genius.[/quote]
well, try the qualifying quiz
also, maybe email somebody who went last year or something to see how much is expected at minimum
if you decide to go, see ya there![/quote]
notice that the number of questions on the quiz changed.

oh they sent me an invitation. I don't think I am going though. My priorities are philmont, cty, debate camp, rowing camp, guitar camp, and etc so I really don't have the time
hope you have fun(?) in the math camp.

haha, maybe they don't like massachusetts (spelled right?)...

e-mail me at jhredsox@cox.net if you're not already in alan, eric, my, alina, valyrie, etc.'s group

[quote="guyinPA"]When you say Mathpath sent you invitations or certificates, what does that MEAN???? [b]Does that mean you can just get in without having to take and Pass the qualifying test?[/b] Please do tell.  I know several people who went to Nats this year (and ones who ended up in top 57 - including USAMO qualifiers, and I asked them - None of the people I asked received anything at all from Mathpath unless they personally requested it!!  Should they feel gyped? Maybe they only send to certain states??[/quote]
If you are a first time participant, regardless of whether you received any invitation or certificate, you still have to do the qualifying test.

Thanks, does anyone know if you take the qualifying test one year, and pass, but don't go to camp, do you have to retake the test the following year?  Just wondered.. :?:

WRT the certificate/coupons: My son made nationals (DC team) both last year and this year.  This year he received a certificate entitling him to \$250 off mathpath tuition, last year he didn't.  The certificate does not allow you to bypass the entrance exam, though.

My guess is that these certificates are intended for all national mathcounts competitors, and that some are getting lost on their way to the students (maybe the state coordinator, the state coach, the school teacher didn't pass it along--my son got his from his school coach).  I think that the folks at mathpath would gladly extend the \$250 discount to any student who made it to mathcounts nationals, with or without the certificate.

It's a pretty intense sounding camp, though!  My son and his friends had a ball poking fun of the brochure with its sun-up to sun-down schedule of math talks/problem solving sessions, etc., although I think I would have had a really good time there, had I been able to go. :-)

Hope that helps, guyinPA!

[quote]not sure why they don't like to try to balance some more "social" activities in between - for ex: soccer, frisbee, tennis, swimming - anything - to break up the day[/quote]

From what I recall, there was about 2 hours after lunch and before the problem solving sessions for activities such as those you mentioned.  Last year soccer was the big sport, but a variety of things were offered.  After dinner there is also plenty of time for recreation.

yeah come on look at the schedule carefully, there's like 3 hours of free time after lunch and two hours after RSSM, come on.

the following activities were readily available pretty much whenever you wanted (during free time) at uc santa cruz

-foosball (well they broke the table in the first week)
-ping pong (they have a tournament, in which i choked in the semifinals -_-)
-soccer (we didn't have a real field though...)
-basketball
-tennis

so you're completely wrong to say they don't balance social activities. also, they're big on chess (if that's fun for you), they had something like 15 boards readily available last year

and the math isn't as tiring as it seems. yeah, some of the lectures can be pretty boring (coughberndtcough, but he's not coming back this year), but breakouts are really fun (if you pick the right ones), and with a good lecturer, the lectures are interesting. mathematical writing seems pretty boring, but it's quite helpful and definitely helped me. history of math is boring as a subject (to me, at least), but the history teachers there are great. i mean come on, drucker sings the french national anthem and sandifer has like the most uber beard ever.

[quote="guyinPA"]Yeah, it seemed pretty intense to me too! Not sure why they don't like to try to balance some more "social" activities in between - for ex: soccer, frisbee, tennis, swimming - anything - to break up the day -- I mean, it IS summer, after all.  I Love math - yes, I can admit that - but I think of myself being a bit more well rounded, and enjoy other aspects of "fun" (if math is fun to many on here) as well.  But thanks for the info![/quote]
The mathematics are intense.  But not too intense for a middle school student who loves math.  Absolutely not too intense for most students who frequently visit this Forum.  I think the program is very well balanced.  And *all* of these things you say are missing (soccer, frisbee, tennis, and swimming) were present when my daughter went.  The founder (Dr. Thomas) will probably eat you alive at soccer, despite the fact that he's elibible for senior citizen discounts at some restaurants.  Camp director Kip Sumner supervised the swimming.  Some students bring musical instruments, and other sports like basketball and volleyball have also been pretty popular.

guyinPA:  If you've already decided you wouldn't like it, then you shouldn't go.  But please don't spread false information to others, who still might be considering the camp.

jhredsox:  The first year my daughter went, I went along on the walking tour of campus the first day of camp.  They pointed out the tennis courts and the swimming pool, and I know they purchased a new ping-pong table.  They also held a ping-pong tournament.  If you are wondering about a specific activity, you should contact MathPath, directly.

Further discussion of this topic should probably take place in the MathPath Forum.

Uh, first - I was ABOUT to thank everyone on here who set the record straight for me and others about the social activities (and I love all the ones mentioned) -- -- and so, first, Thanks Everyone!  I mean I hadn't realized there was ample time for them after reading the brochure they sent ... I guess I should have read the timing of activities more carefully ...  :blush:  my bad ...

To rcv -- my intent was clearly (well, hopefully clearly) not to deliberately mislead anyone nor to discourage anyone - in fact, I heard it's a very, very good program -- I was merely expressing my own opinion and thoughts -- thought that's what this forum was for -- but I can see your point - so I apologize if anyone on here was misled!  I can't go so far as to say "I stand corrected" until that time if and when I ever experience the camp but since others on here who have gone say differently than what I thought, I guess they should know better than me!  Hope this sets the record straight

[quote="rcv"]The founder (Dr. Thomas) will probably eat you alive at soccer, despite the fact that he's elibible for senior citizen discounts at some restaurants.

jhredsox:  The first year my daughter went, I went along on the walking tour of campus the first day of camp.  They pointed out the tennis courts and the swimming pool, and I know they purchased a new ping-pong table.  They also held a ping-pong tournament.  If you are wondering about a specific activity, you should contact MathPath, directly.

Further discussion of this topic should probably take place in the MathPath Forum.[/quote]

I don't think he plays premier soccer though :) (oh by the way rcv, i changed my mind and i'm going :D ) Did your daughter go 2 years ago, because if I am thinking correctly, it was held at Cololorado College that year, as well.

[quote="13375P34K43V312"][quote="jhredsox"]Should I bring my good tennis racket/ping pong paddle?

The soccer will be lopsided and unfair with me there (like seriously) :wink: .

1337...which breakouts should I choose, and what was wrong with Berndt?

Any like actual good speakers? (that have practiced speeches before)[/quote]

um practiced speeches? well first, conway is an internationally reknowned mathematician who has made huge contributions in several branches of mathematics (number theory, knot theory, cellular automata, for example), and saying he's a beast at computation is an understatement. he's probably given maybe 5000 speeches to all audiences in his lifetime, and he's spoken at mathpath like the last five years I believe
berndt was mad boring...i mean his stuff on generating functions was somewhat interesting (not that i was listening...), but he spent two days on like the most boring parts of ramanujan's life
hartshorne was decent, his construction workshops were pretty fun but it looks like he's not doing that this year, last year he focused only on euclidean geometry. he's written several books, he's well-known (i think he's retired now though)
never heard of the other two[/quote]

thanks for all this! i just didn't know cause it's not like you hear about these guys on the front page of the new york times every day...

to anyone that's been to mathpath before: what was the most useful one *small* thing that has helped you out in math (come on, there must be one)?

thanks,
jorian

[quote="jhredsox"]

I don't think he plays premier soccer though :) (oh by the way rcv, i changed my mind and i'm going :D ) Did your daughter go 2 years ago, because if I am thinking correctly, it was held at Cololorado College that year, as well. [/quote]

no dude he's insane
i have never seen anyone run so fast
let alone a sixty-year-old man

also with a decent goalie in net
the goalie like stepped out of the way to avoid getting hit by the ball

ok, this is getting a little off-topic but what kind of "decent" goaley does that. but w/e it's math camp

umm...can't wait, i hope it's fun there


algebra
polynomial
limit
LaTeX
real analysis
real analysis unsolved

let $ p\in C[X] $ a polynomial such that $ p^{-1}(R)\subset R. $ Prove that p=az+b with a,b reals.

$p^{-1}(0)\subset \mathbb{R}$ means that all roots of p are real. This means, by Viete, that $p \in \mathbb{R}[X]$. Also all roots of p' are real.

Assume degp>1.

WLOG, assume that the leading coefficient of p is 1.

If degp is even then p has a minimum m. So, for m'<m, $p^{-1}(m') \subset \mathbb{C} \setminus \mathbb{R}$. This is a contradiction.

So n=degp is odd. Then $\displaystyle \lim{x\to\infty}p(x)=\infty$, so there is an M such that for all y>M, $p^{-1}(y) \cap \mathbb{R}$ is reduced to one point.
But the equation p(x)=y has n complex solutions, so the only possibility is that all n solutions are equal and real for all y>M.
Define $f : (M,\infty) \to \mathbb{R}$ by $f(y)=p^{-1}(y)$. As we seen abolve, f is well defined for y>M. It is easy to see that f is injective.

For all y>M, we have $p(X)-y=(X-f(y))^n$. This means that $p'(f(y))=0\ \forall y\in (M,\infty)$. Because f is injective, this means that p' has infinitely many roots. We get a contradiction.

So degp=1.

[b]Moderator edit: Edited for readability (a problem with ":" in $\LaTeX$)[/b]


probability
counting
distinguishability
advanced fields
advanced fields unsolved

Can we find a probability space and three events on this space (with pairwise distinct probability) which are pairwise independent but not independent ?

The answer has to be yes, you can find such an example.

Roll two fair, distiguishable, six-sided dice. Let $A$ be the event that the first die shows $1,2$, or $3.$. Let $B$ be the event that the second die shows. $1$ or $2.$ Let $C=\{(1,1),(1,2),(1,2),(5,6),(6,5),(6,6)\}.$

Check the details of that.

Here's another example.  Flip five coins.  Let $A$ be the event that the first two coins both come up heads.  Let $B$ be the event that the last three coins all come up heads.  Let $C$ be the event that an odd number of the five coins come up heads.


When is it?  Who is eligible?  How do you register?

I believe its this friday. (the day after tomorrow)

[quote="VINDOZO"][b]YES[/b].  ASK ME MORE. :!:[/quote]

It took me 20 seconds to figure out who you are.
What a shame (for me).

It's very very nice to see you here!


[quote="watchcatno1"]If $ a,b,c\in\mathbb{R}^{\plus{}}$ let
\[ X\equal{}\text{max}\{ a^{2}b\plus{}b^{2}c\plus{}c^{2}a , ab^{2}\plus{}bc^{2}\plus{}ca^{2}\}\]

\[ Y\equal{}\text{min}\{a^{2}b\plus{}b^{2}c\plus{}c^{2}a , ab^{2}\plus{}bc^{2}\plus{}ca^{2}\}\]
Prove that
\[ 13Y\minus{}5X\geq 24abc\]
[/quote]
I think, it's wrong. Try $ a\equal{}0.1,$ $ b\equal{}1$ and $ c\equal{}3.$

hey are you sure you're right? Cause I saw this stated in another form -
                       \[ 4\sum_{sym}a^{2}b\plus{}9|(a\minus{}b)(b\minus{}c)(c\minus{}a)|\geq 24abc\]
and I don't think this is wrong. Can anyone try and find a solution?

u have already got a counter example.so make sure that u r statement is correct.
May be it is true when $ a,b,c>1$
(But i haven't proved it.....)

suppose$ a\geq b\geq c$, and  let  $ a\equal{}x\plus{}y\plus{}z,b\equal{}y\plus{}z,c\equal{}z, x\geq 0, y\geq 0, z\geq 0$;

then

$ 4(a^{2}(b\plus{}c)\plus{}b^{2}(c\plus{}a)\plus{}c^{2}(a\plus{}b))\plus{}9|(a\minus{}b)(b\minus{}c)(c\minus{}a)|\minus{}24abc$
$ \equal{}4((x\plus{}y\plus{}z)^{2}(y\plus{}2z)\plus{}(y\plus{}z)^{2}(2z\plus{}y\plus{}x)\plus{}z^{2}(x\plus{}2y\plus{}2z))\plus{}9|xy(\minus{}x\minus{}y)|\minus{}24(x\plus{}y\plus{}z)(y\plus{}z)z$
$ \equal{}8xyz\plus{}13x^{2}y\plus{}8y^{2}z\plus{}8y^{3}\plus{}21xy^{2}\plus{}8zx^{2}\geq 0$

Suppose
\[ a\geq b,b\geq c,\]
then
\[ X\equal{}a^{2}b\plus{}b^{2}c\plus{}c^{2}a,Y\equal{}a^{2}c\plus{}c^{2}b\plus{}b^{2}a,\]

\[ 13Y\minus{}5X\minus{}24abc\equal{}13a^{2}c\plus{}13b^{2}a\plus{}13c^{2}b\minus{}5a^{2}b\minus{}5cb^{2}\minus{}5ac^{2}\minus{}24abc,\]
but  let
\[ a\equal{}16,b\equal{}4,c\equal{}1 ,\]
then
\[ 13Y\minus{}5X\minus{}24abc\equal{}\minus{}108<0,\]
The inequality
\[ 13Y\minus{}5X\geq 24abc\]
does not hold!

notice :

if $ a\geq b\geq c$, then

$ 4a^{2}(b\plus{}c)\plus{}4b^{2}(c\plus{}a)\plus{}4c^{2}(a\plus{}b)\plus{}9|(a\minus{}b)(b\minus{}c)(c\minus{}a)|\minus{}24abc$
$ \equal{}13X\minus{}5Y\minus{}24abc$
$ \neq13Y\minus{}5X\minus{}24abc$

$ 13X\minus{}5Y\minus{}24abc\geq 0$ is true,but  $ 13Y\minus{}5X\minus{}24abc\geq 0$ is not true.


geometry
calculus
integration
calculus computations

Prove that if two solids contained between parallel plates [b]P[/b] and [b]Q[/b] possess the property that on being cut by any plane [b]R[/b] parallel to these planes equivalent figures are obtained in their section, then the volumes of these solids are equal.

[quote="LkNsngth"]Heuristically, if we let $ f(x)$ be the area bounded by the plane figures, the total volume of the solid is the integral of f(x) along an interval that goes between P and Q. If the plane figures are the same then their integrals are the same.[/quote]
Ok. Good. Thanks....


inequalities
induction
inequalities unsolved

[hide="hint"]Put $P_{k}(x)=\sum_{i=0}^{k}(-1)^{i}\cdot\frac{x^{i}}{i!}(k=1,2,...)$. Induction on $k$ that
$P_{2k}(x)$ hasn't got real root, and $P_{2k+1}(x)$ has only one real root.[/hide]


What percent of the set of even natural numbers end in zero?

even-> ends in $ 2,4,6,8,0$
and equal number of each of the cases above listed in the set of even natural numbers, so $ \frac15$ in percentages, this is 20


Circle $ O$ has radius 12 cm and angle 1 has a measure of 60 degrees. Find the number of square centimeters in the area of the shaded region. Express your answer to the nearest tenth of a square centimeter.
[asy]size(4cm,4cm);
import markers;
pair B,C,a,O;

draw(unitcircle);

O=(0,0);
B=(1,0);
C=(0.5,0.86);

draw(C--O--B--C);

a=(0.85,0.53);

markangle(n=1,radius=15,C,O,B,marker(stickframe(n=0),true));

fill((B--C..a..cycle),black);

label("O",O,SW);

label(rotate(57)*"\textbf{$12$}",shift(-0.18,0.18)*(O--C));

label("$1$", (0.31,0.15));[/asy]

I asked this question on forum because I was wondering if I should report error because the answer rounded to the nearest tenth is 13.0, which is the same as rounding to the nearest whole number.

I'm pretty sure that both $ 13.0$ and $ 13$ are acceptable in most actual competitions, but I don't know what FTW wants.


Μπορει καποιος να με βοηθησει σε αυτο το προβλημα?

Θεωρουμε ενα πεπερασμενο συνολο ανθρωπων (μη κενο) οπου καθε ενας γνωριζει τουλαχιστον εναν απο τους αλλους.Δειξτε, χρησιμοποιωντας την αρχη του περιστεριωνα οτι υπαρχουν τουλαχιστον 2 ανθρωποι του συνολου που εχουν τον ιδιο αριθμο γνωριμιων μεσα στο συνολο.(προφανως γνωριμια ενος ανθρωπου Α ειναι διαφορετικος απο τον Α)
 Ευχαριστω

Ρε συ, είναι απλή εφαρμογή! Σε ποιο πανεπιστήμιο σπουδάζεις?

ΑΣΟΕΕ!!!και παραδιδω εργασια αυριο...αλλα δεν ειναι αυτο το αντικειμενο μου...γι αυτο εχω μια "μικρη" δυσκολια...βοηθα λιγο ρε king!!

Κάθε άνθρωπος έχει από 1 μέχρι ν-1 γνωστούς (ν-1 ομάδες) , ενώ υπάρχουν ν άνθρωποι
(θες να χωρέσεις ν ανθρώπους σε ν-1 ομάδες)
Άρα τουλάχιστον 2 άνθρωποι θα έχουν τον ίδιο αριθμό γνωστών

[b]O ENGEL  τη λύνει έτσι :[/b]

 << Το άτομο  $a_i$ μπαίνει στο κουτί  $i$  , αν έχει  $i$  γνωστούς, όπου $0, 1 , 2 , ... , n-1$ είναι τα κουτιά.Αλλά τα κουτιά

   $0 , n-1$  δεν μπορεί να είναι και τα δύο γεμάτα . Έχουμε όμως  $n$   άτομα  κλπ. >>

 Μπάμπης

Εύστοχο... άρα το πρόβλημα έχει λύση ακόμα και αν δε δίνεται ότι
[b]"καθε ενας γνωριζει τουλαχιστον εναν απο τους αλλους"[/b]


function
calculus
integration
topology
number theory
prime numbers
real analysis

Is there $ (a_n) \in \mathbb{R}^{\mathbb{N^*}}$ other than the trivial one, s.t. $ \forall \ell \in \mathbb{N^*}, \sum_{k \geqslant 1} a_{k\ell} \equal{}0$?

Plenty of them. Just take any open set containing all "rational" points on the unit circumference of small measure and consider the Fourier series of any integrable function $ f$ supported on its complement with integral $ 0$. If $ f\asymp \sum_{k\ne 0}c_k z^k$, then $ a_k\equal{}c_{\minus{}k}\plus{}c_k$ is the series with the desired property. Most of the time, we'll get a non-trivial sequence this way.

Now the interesting part  :wink: : what if we add the constraint that $ \sum_{k\geqslant 1} |a_k|< \infty$?

I think this is simplest  part of the problem  :lol: ,
all $ a_{n}$ will be equal to zero.
I don't know how to explain it in normal (reasonable language )  but anyway I'll try. this is idea :
let's show that $ a_{n} \equal{} 0$ suppose $ n \equal{} p_{1}^{k_{1}}p_{2}^{k_{2}} ... p_{m}^{l_{m}}$
fix some sufficiently large number $ J$ such that $ \sum_{i\geq J} |a_{i}| < \epsilon$
 and consider all prime numbers  $ p_{j}$ such that $ p_{j} < J$.
here we will use http://en.wikipedia.org/wiki/Inclusion-exclusion_principle
thus from  the seire $ \sum a_{j}$ we can minus  all such $ a_{j}$ such as $ j\neq n$ and $ j < J$ for example by this way :
let $ j \equal{} c_{1}^{l_{1}}c_{2}^{l_{2}} ... c_{m}^{l_{m}}$
then consider set of numbers $ \{c_{f}^{l_{f}}p\}$ where $ f$ is natural numbers. then by the sufficiently large operation (for example $ g$ operaton ) via intersection and union of such sets we get serie $ \sum a_{i} \approx \sum _{i\neq j} a_{i}$ and now we need to add and minus   $ g$ times such serie as $ \sum_{k} a_{c_{1}^{l_{1}}c_{2}^{l_{2}} ... c_{m_{u}}^{l_{m_{u}}}n k }$  " and everything will be ok ".  thus finnaly we get that $ a_{n}$ is at most $ \sum_{s\geq J}|a_{s}|$ wich tends to zero as $ J \to \infty$

I agree that there exist function for which  (in the case : conditional convergence of serie) when there is nontrivial $ a_{n}$ and fedja has already proved this.
Yes I agree such serie exist ...  but can you post any example, I mean write  here such function ? :)

[quote="fedja"]Plenty of them. Just take any open set containing all "rational" points on the unit circumference of small measure and consider the Fourier series of any integrable function $ f$ supported on its complement with integral $ 0$. If $ f\asymp \sum_{k\ne 0}c_k z^k$, then $ a_k \equal{} c_{ \minus{} k} \plus{} c_k$ is the series with the desired property. Most of the time, we'll get a non-trivial sequence this way.[/quote]

Ok, this might be a dumb question.

If we evaluate the fourier series at $ e^{{2\pi i}/l}$ (for some $ l \in \mathbb{N}$) where the function is (if I understood correctly) equal to zero we get $ \sum_{k \in \mathbb{Z}, k \ne 0}c_{kl} \plus{} \sum_{k \ne {jl}, j \in \mathbb{Z}}c_k e^{{2\pi i k}/l} \equal{} 0$, and in order to get the desired property, we need the first sum to be zero. But how do we get that, I mean, what about the second sum, how do we get rid of it? Is what I'm asking making any sense?:lol:

$ \sum_{k\in\mathbb Z}c_{k\ell}\equal{}\frac 1\ell \sum_{j\equal{}0}^{\ell\minus{}1}f(e^{2\pi ij/\ell})$.

[quote="fedja"]$ \sum_{k\in\mathbb Z}c_{k\ell} \equal{} \frac 1\ell \sum_{j \equal{} 0}^{\ell \minus{} 1}f(e^{2\pi ij/\ell})$.[/quote]

Ah, I see. Neat trick, wasn't quite obvious to me. Thanks for pointing it out! :)


[asy]size(150);
draw((0,0)--(1,0)--(1,1.5)--(0,1.5)--cycle);
draw((2,0)--(3,0)--(3,1.5)--(2,1.5)--cycle);
draw((4,0)--(5,0)--(5,1.5)--(4,1.5)--cycle);
draw((2,2.5)--(3,2.5)--(3,4)--(2,4)--cycle);
draw((4,2.5)--(5,2.5)--(5,4)--(4,4)--cycle);
label("3",(0.5,0.5),N);
label("4",(2.5,0.5),N);
label("6",(4.5,0.5),N);
label("P",(2.5,3),N);
label("Q",(4.5,3),N);[/asy]

Five cards are lying on a table as shown.  Each card has a letter on one side and a whole number on the other side.  Jane said, "If a vowel is on one side of any card, then an even number is on the other side."  Mary showed Jane was wrong by turning over one card.  Which card did Mary turn over?

\[ \textbf{(A)}\ 3 \qquad
\textbf{(B)}\ 4 \qquad
\textbf{(C)}\ 6 \qquad
\textbf{(D)}\ \text{P} \qquad
\textbf{(E)}\ \text{Q} \qquad
\]

[hide="Solution 1"]
If Jane is wrong, then there is a card with a vowel on one side and an odd number on the other side.  Such a card cannot have a consonant or an even number on either side.  Thus the only card which could prove Jane wrong is the one with a "$ 3$" on one side, or $ \boxed{\textbf{(A)}}$.
[/hide]

[hide="Solution 2"]
The easiest way for Mary to show Jane was wrong is for her to turn over a card showing a vowel, but there is no such card.  But if Jane were correct, then so is the contrapositive: "If an odd number is on one side of a card, then a consonant is on the other side."  Thus, Mary showed Jane was wrong by turning over the card marked with a "$ 3$", or $ \boxed{\textbf{(A)}}$.
[/hide]

[hide="Solution 3"]
Note that the only things that make cards "different" from each other are the parity of the numbers and consonant/vowel.  Because there are $ 2$ even cards and $ 2$ consonants, they cannot be the correct choice or there would be $ 2$ correct choices.  Thus, she turned over the last card, $ \boxed{\textbf{(A)}}$.
[/hide]

$x\implies y=0$ iff $x=1, y=0$.

Here, $x$ is vowel or not and $y$ is odd or even.

Because $x=1$ and $y=0$ the card must either have a vowel or odd number to be able to prove the implication false.

the only vowel or odd number is $\boxed{(A) 3}$.

We can't see any vowels on the table, so if there are any, they're on the other side of a number. The only way to prove the rule wrong is to have a card that has a vowel on one side and an odd number on the other. The only card that could possibly have both of those is [b](A) 3[/b], since P and Q are consonants, and 4 and 6 are even numbers.

This is an example of the [url=https://en.wikipedia.org/wiki/Wason_selection_task]Wason selection task[/url].

the fact that cards $\text{P}$ and $\text{Q}$ are not centered infuriates me ):<

[quote=titanlord]this post was made before I was born...[/quote]

Not me but wow this is old.

This problem is certainly very easy and is clearly not meant for a #25.

[quote=aopsonline2020888]this problem reminds me of that one problem on mc state[/quote]

yeah the one with like the circles are on the other side of the other thingy with 12 thingies

[hide="Statement Solution"]
===============================

If the statement:  If $a$ then $b$, is true then the statement: If not $b$ then not $a$ is true.
Using the fact above and the statement given we can produce another statement:
If there is an odd number then there is a consonant on the other side.
So we find an odd number and $ \boxed{3} $ is the only odd number.
Thus the answer is $\boxed{A) \ 3}$

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


inequalities
trigonometry
trig identities
Law of Cosines
inequalities solved

In  the  triangle  ABC , we have  2b^2 = a^2 + c^2 . Prove  that
(cotB)^2 >=cotA * cotB .
Thanks  !!

[i]unknown source[/i]

yes  you  are  rirght!!  Sorry for  my  mistake  :?

2b2 = a2 + c2
<==>
sin2 B  = (sin2 A+ sin2 C)/2.

cot2 B \geq cot A cot C
<==>
cos2 B/sin 2 B \geq cos A cos C/(sin A sin C)
<==>
1/sin2 B - 1 \geq cos A cos C/(sin A sin C)
<==>
1/sin2 B \geq cos A cos C/(sin A sin C) + 1 = cos (A-C)/(sin A sin C)
<==>
2/(sin2 A+ sin2 C) \geq cos(A-C)/(sin A sin C).

Does this hold?

OK.

We need to prove cos B cos B / sin B sin B \geq cos A cos C / sin A sin C.

All the sines are nonnegative, so we may multiply:

<=> cos B cos B sin A sin C \geq cos A cos C sin B sin B.

<=> cos B cosB a c \geq cos A cos C b b

<=> cos B cos B \geq cos A cos C b^2 / ac *, where a, b, c are sides of the triangle.

By the Law of Cosines, b^2 = a^2 + c^2 - 2ac cos B.

But 2b^2 = a^2 + c^2, so cos B = b^2 / 2ac.

Substituting in *, we get

<=> b^2 / 4ac \geq cos A cos C.

We use the Law of Cosines again:

cos A = (b^2 + c^2 - a^2) / 2bc
cos C = (b^2 - (c^2 - c^2)) / 2ab.

So we must prove b^4 \geq b^4 - (c^2 - a^2)^2, which is trivial since

c^2 - a^2 \geq 0 with equality iff a = c.

We also can use nice relation (so called cotang rule - this rule follows from cosin rule and formula S = absinC/2)
cotB = (a^2+c^2-b^2)/4S = b^2/4S, cotA = (b^2+c^2-a^2)/4S, cotC = (a^2+b^2-c^2)/4S
So we have to show that  b^4 >= b^4 - (c^2-a^2)^2 which is trivial.

Namdung


function
integration
algebra
linear equation
calculus
calculus computations

Solve $ \frac{dy}{dx}\equal{}\frac{y^{2}}{xy\plus{}\sqrt [3]{xy^{2}}}$.

[hide="Solution"]We have here a Bernoulli's differential equation:

$ \frac{dx}{dy}\equal{}\frac{xy\plus{}x^{1/3}y^{2/3}}{y^{2}}\equal{}\frac{x}{y}\plus{}x^{1/3}y^{\minus{}4/3}$

$ \Rightarrow x'\minus{}\frac{1}{y}x \equal{} x^{1/3}y^{\minus{}4/3}$.

Two methods that can be used to solve a Bernoulli's d.eq. $ y'\plus{}A(x)y \equal{} y^{k}B(x)$ are:

1) make the substitution $ z \equal{} y^{1\minus{}k}$, and after some manipulation we obtain the linear equation $ \frac{1}{1\minus{}k}z'\plus{}A(x)z \equal{} B(x)$;

2) make the substitution $ z \equal{} u\cdot v$, where u and v are functions of x to be determined.

Let's use the first: in our case (the x plays the role of y), k = 1/3 and so we make $ z \equal{} x^{2/3}$, from which we get

$ z'\minus{}\frac{2}{3y}z \equal{}\frac{2}{3}y^{\minus{}4/3}$

whose solution is

$ z \equal{} e^{\frac{2}{3}\int\frac{dy}{y}}\left(\int\frac{2}{3}y^{\minus{}4/3}e^{\minus{}\frac{2}{3}\int\frac{dy}{y}}\,dy\plus{}C\right) \equal{} y^{2/3}\left(\frac{2}{3}\int y^{\minus{}2}\,dy\plus{}C\right)$

$ \Rightarrow z \equal{} y^{2/3}\left(C\minus{}\frac{2}{3y}\right)$

and so the general solution is $ x^{2/3}\equal{} y^{2/3}\left(C\minus{}\frac{2}{3y}\right)$.

[/hide]

Hi carcul;
1) $ x\in]0,\infty[\rightarrow{0}$ and $ x\in]\minus{}\infty,0[\rightarrow{0}$ are other solutions.
2) Moreover your ODE shows that you seek functions $ y: x\rightarrow{y(x)}$ and these functions aren't defined over whole $ \mathbb{R}$ but only over a subinterval. For example let $ C\equal{}1$:
if $ y(2)\approx2.94$ then $ def(y)\equal{}]0,\infty[$.
if $ y(2)\approx\minus{}0.13$ then $ def(y)\equal{}]\approx1.74,\infty[$.
if $ y(2)\approx\minus{}0.8$ then $ def(y)\equal{}]\approx1.74,\infty[$.

Hi loup blanc. Could you explain your first point? Thanks.

Hi carcul;
in fact your ODE is $ y'\equal{}\dfrac{y^{4/3}}{xy^{1/3}\plus{}x^{1/3}}$.
You may apply the Cauchy-Lipschitz theorem; the points $ (x,0)$ with $ x\not\equal{}0$ aren't singular. The singular points are those of the curve $ y\equal{}\minus{}1/x^{2}$ or the $ y$-axis points; when a solution runs to a neibourhood of a such point (except($ (0,0)$) then the tangent becomes vertical. In addition to the solutions $ y\equal{}0$, the only solutions which haven't a vertical tangent are obtained for $ C\equal{}0$: the 2 branches of the curve $ y\equal{}\minus{}8/(27x^2)$.


Έστω $ x_{1},x_{2},...,x_{n}$ πραγματικοί αριθμοί τέτοιοι ώστε $ 0<x_{j}\leq \frac{1}{2}$ για $ 1\leq j\leq n$. Αποδείξτε ότι
$ \frac{\prod^{n}_{j\equal{}1}x_{j}}{\left(\sum^{n}_{j\equal{}1}x_{j}\right)^{n}}\leq \frac{\prod^{n}_{j\equal{}1}\left(1\minus{}x_{j}\right)}{\left(\sum^{n}_{j\equal{}1}\left(1\minus{}x_{j}\right)\right)^{n}}$.

Αρχικά αντιστρέφουμε την ανισότητα:

$ \frac {\left(\sum^{n}_{j \equal{} 1}x_{j}\right)^{n}} {\prod^{n}_{j \equal{} 1}x_{j}} \geq \frac {\left(\sum^{n}_{j \equal{} 1}\left(1 \minus{} x_{j}\right)\right)^{n}} {\prod^{n}_{j \equal{} 1}\left(1 \minus{} x_{j}\right)}$

Εφαρμόζουμε την ln και στα δύο μέλη, οπότε παίρνουμε:

$ n \cdot ln(\sum^{n}_{j \equal{} 1}x_{j}) \minus{} \sum^{n}_{j \equal{} 1}ln(x_{j}) \geq n \cdot ln(\sum^{n}_{j \equal{} 1}(1 \minus{} x_{j})) \minus{} \sum^{n}_{j \equal{} 1}ln(1 \minus{} x_{j})$

$ n \cdot ln \frac {\sum^{n}_{j \equal{} 1}x_{j}} {\sum^{n}_{j \equal{} 1}(1 \minus{} x_{j})} \geq \sum^{n}_{j \equal{} 1} ln \frac {x_{j}} {1 \minus{} x_{j}}$

Θεωρούμε τώρα την:
$ f(x) \equal{} ln \frac {x} {1 \minus{} x}, 0 \leq x \leq \frac {1} {2}$

Έχουμε:
$ f''(x) \equal{} \frac {2x\minus{}1}{[x(1 \minus{} x)]^2} \leq 0$ αφού $ x \leq \frac {1} {2}$

Άρα η f είναι κοίλη και άρα με βάση την ανισότητα Jensen:

$ \frac {\sum^{n}_{j \equal{} 1}f(x_{j})} {n} \leq f(\frac {\sum^{n}_{j \equal{} 1}f(x_{j})} {n})$

Άρα:

$ \sum^{n}_{j \equal{} 1} ln( \frac {x_{j}} {1 \minus{} x_{j}}) \leq n \cdot ln(\frac {\frac {\sum^{n}_{j \equal{} 1} (x_{j})} {n}} {1 \minus{} \frac {\sum^{n}_{j \equal{} 1} (x_{j})} {n}}) \equal{} n \cdot ln \frac {\sum^{n}_{j \equal{} 1}x_{j}} {\sum^{n}_{j \equal{} 1}(1 \minus{} x_{j})}$

Και άρα αποδείξαμε αυτό που θέλαμε...


number theory
greatest common divisor
relatively prime
combinatorics

[quote="discredit"]A natural number is written on the blackboard. Whenever number $ x$ is written, one can write any of the numbers $ 2x \plus{} 1$ and $ \frac {x}{x \plus{} 2}$. At some moment the number $ 2008$ appears on the blackboard. Show that it was there from the very beginning.[/quote]

Notice that beggining with a natural, all the numbers that can appear on the board are rational.
Assume $ x\equal{}\frac {a}{b}$ is on the board. Define $ f(x)\equal{}a\plus{}b$  (sum of numerator and denominator).
Then the new number is either:
1) $ 2x\plus{}1\equal{}\frac {2a}{b}\plus{}1\equal{}\frac {2a\plus{}b}{b}$ and then $ f(x')\equal{}2f(x)$
Or:
2) $ \frac {x}{x \plus{} 2}\equal{}\frac {\frac {a}{b}}{\frac {a}{b} \plus{} 2}\equal{}\frac {a}{a\plus{}2b}$ and then $ f(x')\equal{}2f(x)$
We begin with sum natural value of $ f$, so $ f$ stays natural, and each turn is multiplied by $ 2$.
But $ f(2008)\equal{}2009$, odd, so $ 2008$ isn't a result of a series of such manipulations.
Therefore it was there from the beggining.
Right?

By the way, what is ARO?

Be careful that you are assuming that all the fractions are irreducibles.

How about $ x\equal{}6$, $ f(x)\equal{}7$ then $ \frac{x}{x\plus{}2}\equal{}\frac{6}{8}\equal{}\frac{3}{4}$ with $ f(x')\equal{}7$ ?

Lol, we start with the fraction 2008/1. We reduce the fractions to lowest form. From a different natural number, we need 2008k/k after a certain number of these operations.

2x+1 takes a/b and gives (2a+b)/b
x/(x+2) takes a/b and gives a/(a+2b)

Note that the sum of numerator and denominator is going to be 2/k of the previous one for some k, where k is the gcd of the new numerator and denominator.

The only other thing you need to solve this problem is to notice that a,b are initially relatively prime, so by linear combination, we know that the gcd is not bigger than 2, and will continue not being bigger than 2 throughout. So 2/k is either 2 or 1, so the original sum had to be 2009, since you can't have a sum of 1004.5. Therefore 2008 must have started out on the board, since the numerator and denominator had to sum to 2009, so the problem is now trivial.

[quote="rofler"]

Note that the sum of numerator and denominator is going to be 2/k of the previous one for some k, where k is the gcd of the new numerator and denominator.
[/quote]

Why is this?

This problem has been coincidented by APMO 2015..

Nice!

Note that sum of numerator and denominator of a rational number $x$ doubles after the operation. As this sum for $2007$ is $2009$, an odd number, it cannot occur in the sequence unless it was already there from the beginning.

Consider a number $\tfrac{a}{b}$ where $\gcd(a,b)=1$. I claim that of the 2 numbers written from it, the numerator plus the denominator must be either equal to $a+b$ or $2(a+b)$. It is easy to see that from this number, only the numbers $\tfrac{2a+b}{b}$ and $\tfrac{a}{a+2b}$ can be written. First assume that $a$ is even and $b$ is odd. Hence, $\gcd(2a+b, b)=\gcd(2a, b)=1,$ and $\tfrac{2a+b}{b}$ is in simplest form. The sum of the numerator and denominator is $2(a+b)$. In the other case, $\gcd(a, a+2b)=\gcd(a, 2b)=2,$ so $\tfrac{a}{a+2b}$ in lowest form is $\tfrac{a/2}{a/2+b}$, and the sum of the numerator and denominator is $a+b$. In the case where $a$ is odd and $b$ is even, the proof is similar, with the sum of the numerator and denominator being $2(a+b)$ or $a+b$ as well. From this, it can be established that from a number $\tfrac{a}{b}$, all the numbers formed after it must have the sum of the numerator and denominator be in the form $2^n(a+b)$.

Note that $2008$ is $\frac{2008}{1}$, so the sum of the numerator and denominator is 2009. Let the initial number on the blackboard be $N$ (the sum of the numerator and denominator is $N+1$). Hence, $2009$ must in the form $2^n(N+1)$. This is clearly absurd for $n \geq 1,$ so $n=0$. This means that $N$ must be 2008, and we are done.


superior algebra
superior algebra unsolved

Let $ Q$ be a $ P$ -primary ideal of commutative notherian ring prove that there is an $ n$ such   that $ P^n \subset Q$.

Atiyah, Macdonald "Introduction to commutative algebra"
Corollary 7.16


lol. My handwriting and mouse controlling skills on paint :P...rofl I was bored...:D

Sine cose, tan  theta..ya it's in the geometry album too..you know :P

These are in the geometry album too...random circles in the middle of nowhere :P

fail...lol the I is a little big..not in a straight line, otherwise it's fine :P

Yea, I made it..I made a lot..more. For her.

I'll post both. this is the 1st :D Oh. wait..I can't post th eother one!


number theory
prime numbers
number theory unsolved

Given numbers $a_1, a_2, ..., a_n \in \mathbb{Z_+}$ ; $1 < a_i < (2n-1)^2$ ; $gcd(a_i,a_j) = 1$ for all $i \neq j$. Prove that there exists $i$ such that $a_i$ is prime.

Consider, that all $a_i$ are not primes.
Then each $a_i$ have at least 2 prime divisors, one of them $p_i\leq \sqrt{a_i}<2n-1$. But all $p_i$ are different :arrow: there is $n$ different prime numbers is $(1,2n-1)$.
Variants $n=1,2$ are evident.
If $n\geq 3$, then number of prime in $(1,2n-1)$ $\leq (2n-3)-(n-2)=n-1$(since $4,6,...,2n-2$ are not prime), but it's $<n$.Contradiction


how can we get general formula with having the recurence relation.
example: the sequence 5, 6, 7, 8, 9,.......

General formula $U_{n}=n+4$
Recurence relation $U_{n+1}=U_{n}+1$

how can we find the general formula when we have the recurence relation or itterative method.(for any sequence)

 from this $U_{n+1}=U_{n}+1$ to this $U_{n}=n+4$

I dont that you could finde the general formula using the recurrence relation, at least in this case.

For example:

the sequence 2,3,4,5,... has the same recurrence that the sequence 5,6,7,8,... but differente general formula.

Maybe if the recurrence is very specific you could then use to find the general formula.

For all recurrence formula, one must know both the formula [b]and[/b] the initial case.

One technique commonly used to get a general formula from a recursive formula is generating functions. The generating function of a sequence $a_{0}, a_{1}, a_{2}, a_{3}, ...$ is
$F(x) = a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+...$
It basically represents the values of a sequence as the coefficients of a polynomial. The point of using generating functions is that there are all sorts of useful ways to work with functions that are not available for sequences, from simple multiplication to things like geometric series.
If you have the AOPS books, the second volume has a section on generating functions. [url=http://www.math.wisc.edu/~propp/491/gfologyLinked.pdf]generatingfunctionology[/url] is a book released free online about generating functions, but it is more at the college level.


Prove that for $ a,b,c \ge 0$,

$ (ab\plus{}ac\plus{}bc)(a\plus{}b\plus{}c)^4 \le 27 (a^3\plus{}b^3\plus{}c^3)^2$

Use:

1. $ 9(a^3 \plus{} b^3 \plus{} c^2)\ge (a \plus{} b \plus{} c)^3$

2. $ 3(a^3 \plus{} b^3 \plus{} c^3)\ge (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)\ge (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$.

EDIT: A harder problem: show that for $ a,b,c\ge 0$ we have
\[ (a\plus{}b\plus{}c)^5\ge 27(a^3\plus{}b^3\plus{}c^3)(ab\plus{}bc\plus{}ca)\]
Have fun!

[quote="nayel"]
EDIT: A harder problem: show that for $ a,b,c\ge 0$ we have
\[ (a \plus{} b \plus{} c)^5\ge 27(a^3 \plus{} b^3 \plus{} c^3)(ab \plus{} bc \plus{} ca)
\]
Have fun![/quote]

Try (1,1,0) ;)

[quote="nayel"]Use:

1. $ 9(a^3 \plus{} b^3 \plus{} c^2)\ge (a \plus{} b \plus{} c)^3$

2. $ 3(a^3 \plus{} b^3 \plus{} c^3)\ge (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)\ge (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$.

EDIT: A harder problem: show that for $ a,b,c\ge 0$ we have
\[ (a \plus{} b \plus{} c)^5\ge 27(a^3 \plus{} b^3 \plus{} c^3)(ab \plus{} bc \plus{} ca)
\]
Have fun![/quote]

how do you get inequality #2?  I know #1 is Power Mean Inequality.

[quote="Zellex"][quote="nayel"]2. $ 3(a^3 \plus{} b^3 \plus{} c^3)\ge (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)\ge (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$.[/quote]
how do you get inequality #2? [/quote]

wlog let $ a \geq b \geq c$

[b]First half:[/b] [i]Lines 2 and 3 follow from the rearrangement inequality[/i]
$ a \cdot a^2 \plus{} b \cdot b^2 \plus{} c \cdot c^2 \equal{} a \cdot a^2 \plus{} b \cdot b^2 \plus{} c \cdot c^2$
$ a \cdot a^2 \plus{} b \cdot b^2 \plus{} c \cdot c^2 \geq b \cdot a^2 \plus{} c \cdot b^2 \plus{} a \cdot c^2$
$ a \cdot a^2 \plus{} b \cdot b^2 \plus{} c \cdot c^2 \geq c \cdot a^2 \plus{} a \cdot b^2 \plus{} b \cdot c^2$
Summing, $ 3(a^3 \plus{} b^3 \plus{} c^3) \geq (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)$

[b]Second half:[/b] [i]By the rearrangement inequality,[/i]
$ a \cdot a \plus{} b \cdot b \plus{} c \cdot c \geq b \cdot a \plus{} c \cdot b \plus{} a \cdot c$
$ \implies a^2 \plus{} b^2 \plus{} c^2 \geq ab \plus{} bc \plus{} ca$
$ \implies (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2) \geq (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$

[b]Note:[/b] JavaMan identified this as an instance of Chebyshev's sum inequality. The general case of that inequality follows from rearrangement in the same manner as the proof above.

For the second half, you could also note that $ \frac {a^2 \plus{} b^2}{2} \ge \sqrt {a^2b^2}$ and etc. for the other variables.  Sum it up to get $ a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} ac$

EDIT: I looked up rearrangement inequality on AoPS wiki.  So, does it basically mean that for $ a_{1} > a_{2} > a_{3}$, and $ b_{1} > b_{2} > b_{3}$, then, $ a_{3}b_{3}\plus{}a_{2}b_{2} \plus{} a_{1}b_{1}$ is the maximum of all such permutations.

However, I don't quite understand the minimum from AoPS wiki.

So, in this case, I guess Zuton Force has $ a_1 \equal{} a, a_2 \equal{} b, a_3\equal{} c$  Then, $ b_1 \equal{} a^2, b_2 \equal{} b^2, b_3 \equal{} c^2$

So, if $ a > b > c$, then $ a^2 > b^2 > c^2$.  So, the maximum of the possible permutations is $ a^3 \plus{} b^3 \plus{}c^3$.  Thus, lines two and three follow from this, right?

That's exactly what I used.  :) And your interpretation of the upper bound from rearrangement is right

Rearrangement says that if you have two positive lists of equal length, that the sum of pairwise products is maximized when the lists are both ordered in the same way (both increasing or both decreasing), and is minimized when the lists are oppositely ordered (one is increasing and one is decreasing).

For example, if you had lists (1, 2, 3) and (5, 6, 4), then 1*4+2*5+3*6 >= 1*5+2*6+3*4 >= 1*6+2*5+3*4, where the first part represents like ordering, the third represents opposite ordering, and the second is some other random ordering. Indeed, in this case, it evaluates to 32 >= 29 >= 28.

One can also interpret the first inequality as a case of Holder's.


geometry
trapezoid
ratio
calculus
factorial
function
arithmetic sequence

[color=cyan]1. Three integers a, b, and c are three consecutive terms of an arithmetic sequence, and a*b*c is a positive prime number.
Find (a, b, c).

2. If f(x+1) = -(x+1)f(x) for x >= 0 and f(0) = 1, compute f(10)/f(8).

3. 12! = 47A001600 for some digit A.  Find A.

4. Given a trapezoid ABCD with AB || CD.  If AC and BD intersect at E, and the area of ABE = 18 and the area of CDE = 32, what is the area of triangle BCE?

If I've given any of these before, just tell me and I'll replace them with others.[/color]

For #1, does abc refer to the three-digit number with digits of a, b, and c (i.e. 100a + 10b + c)?

It refers to a*b*c. Prove that you can't get any primes of the form 100a+10b+c where a, b, c are from 0 to 9 and in arithmetic progression.

proof for simon's

write a,b,c as a, a+d, a+2d.  So 100a + 10b + c = 100a + 10(a+d) + (a+2d), which simplifies to 111a + 12 d, which is divisible by 3.

Wait, so how is that possible? a|abc, b|abc, c|abc, and so unless a=b=c=1, which is not a prime, abc would have more than 2 factors.

Be careful Fierytycoon. You missed some cases.

[quote]3. 12! = 47A001600 for some digit A. Find A. [/quote]

If this were on the AMC, I'd do it out.

He doesn't want you to multiply (at least I don't think he does). He wants you to be clever with the factors of 12!.

JBL wrote:1. Three integers a, b, and c are three consecutive terms of an arithmetic sequence, and a*b*c is a positive prime number.
Find (a, b, c).

2. If f(x+1) = -(x+1)f(x) for x >= 0 and f(0) = 1, compute f(10)/f(8).

3. 12! = 47A001600 for some digit A.  Find A.

4. Given a trapezoid ABCD with AB || CD.  If AC and BD intersect at E, and the area of ABE = 18 and the area of CDE = 32, what is the area of triangle BCE?

If I've given any of these before, just tell me and I'll replace them with others.

[hide]#1. either a and b are negative and c is positive, or all three are positive. if all three are positive, then only a can be 1, therefore abc will have more than two factors =>not prime. Therefore a and b are negative, while c is positive.  two factors have to be :pm:1, else we will have more than two positive factors, therefore c=1, b=-1, and by continuing the arithmetic sequence, we find a=-3, making our prime 3.

#2. the general formula for f(x) is (-1)^x*x!, which can be proved by induction. f(10)/f(8)=10!/8!=90

#3. 9 is a factor of 12! therefore we use the div. by 9 rule to find that A=0 or 9. 11 is also a factor of 12! we use the div. by 11 rule to find A=9 (come to think of it, we didn't need to use div. by 9 oh well)

#4. The area of BCE will be the geometric mean of the areas of ABE and CDE. We use the area formula absin:theta:/2 and the fact that ABE is similar to CDE. If we let the ratio of similarity between ABE and CDE be x, and we let AE=a and BE=b, we see that [AEB]=absin:theta:/2 [CED]=xaxbsin:theta:/2 and [BCE]=bxasin(180-:theta:)/2. Therefore [BCE] is the geometric mean of [AEB] and [CED] which is 48.[/hide]

[quote]The area of BCE will be the geometric mean of the areas of ABE and CDE.[/quote]

What? Never heard about this. How did you derive it?

that was a claim that I subsequently showed to be true.

Hehe..sorry about that Mystic, I had seen your assertion and then panicked that I hadn't known a simple theorem about trapezoids--I saw the following calculations but didn't bother to read them before understanding your original assertion.  8-)  Im tired, its been a long week...

[hide]

3. Use the divisibility rules for 9 and 11. The answer is 9.

[/hide]

[color=cyan]Nice work.  I really liked that prime one.[/color]

[color=cyan]Honestly, I'm not entirely sure.  These may be the product of my predecessor as math team captain or they might be mine.  They are very familiar, however.  I'm sure darkquantum can tell us, but I don't really know.  This particular one was stolen from somewhere else, though, I think.[/color]

aghh
gasp

i don't know about these problems, but number 4 on the trapezoids is definitely a very popular topic

It's time for my stupid question....

What's the div by 11 rule? :\

Also, for those of you into calculus theory (like I)... notice the relationship of question #2 with f(n) = Int(ln(x)^n,x,0,1)... very neat.

Now, I am very bad at series stuff, especially induction... can someone please show the techniques required to show that the definition for #2 is equal to (-1)^x*x!????

I just can't see it.

[color=violet]I'm not entirely sure why those were three seperate posts, but taking life as it is . . .

1)  Consider a number with digits a(0), a(1), . . . a(n) from right to left.  (That is, a(0) is the units digit, etc.).  This number equals
a(0) + 10a(1) + 100a(2) + . . . + 10^n * a(n).

Now, just take that expression mod 11 and look at what happens to powers of 10.  You should get the result
a(0) - a(1) + a(2) - . . . + (-1)^n * a(n).  In other words, you add and subtract alternate digits.  Now, these are congruent mod 11.  So the first is divisible by 11 only if the second is.  That's the divisibility rule for 11.  Ask if you want clarification.

2) Pretty cool.

3) Check it out.  Just do the first few values.  Then see if you can generalize it, using the definition of the factorial function:
0! = 1
n! = n * (n-1)! for n >= 1.[/color]

they were separate posts because i thought about them after i had posted the previous.

[color=indigo]When you've posted something, you can click on "Edit" up in the upper right and add anything you want to it.  Thus, you can combine other things into the same post.[/color]

[quote="Spoon"]

What's the div by 11 rule? :\[/quote]

meh, its like stubtracting and adding consecutive digits. like to test if 1331 is divisible by 11, start from the left, so 1-3+3-1=0 so if that is 0, its divisible by 11.

i think theres more, but i have to go to my violin lesson. darnit! im late


Name 3 angles that can't be constructed.

How do you do that?

If you want to get a satisfactory answer, you should study Galois Theory, which is far beyond the level of this forum.

Should I bother learning that? I am  in Junior High.

I think this also goes for regular polygons and whether they are construcible or not.

what's a junior high doing with a galois theory lvl problem...?

this isn't schoolwork...is it?

You could just cheat and use angles that have no relation to geometric figures, i.e. $e, e^{2}, e^{3}$  :P

That's a complicated result.  In fact, it turns out that as far as angles with integer degrees, precisely the multiples of $3^{\circ}$ can be constructed.  The construction of a $1^{\circ}$ angle is equivalent to trisecting a $3^{\circ}$ angle, which is a classically impossible construction.

Well there's a very well known proof that $20^{\circ}$ cannot be constructed, which basically proves that you can't trisect an angle.

So if we know 20 is impossible, it's trivial that 40 and 80 are too, because if 40 was possible, we could just bisect it to get 20.


modular arithmetic
number theory
relatively prime

This probably belongs in Intermediate topics, I guess?

[hide="Anyway"]
$ 2008^{2008} \equiv 8^{2008} \equiv 2^{6024} \pmod{1000}$.

Decomposing, we have $ 2^{6024} \equiv 0 \pmod{8}$ and (since $ \phi(125) \equal{} 100$) $ 2^{6024} \equiv 2^{24} \equiv 128^3 \cdot 8 \equiv 27 \cdot 8 \equiv 91 \pmod{125}$. Then by CRT, we get that $ 2008^{2008} \equiv \boxed{216} \pmod{1000}$

[/hide]

How do you use the Chinese Remainder Theorem?

Chinese Remainder Theorem.

EDIT Refer below, since 123456789 editted the post, this one does not make sense.  :P

uh AstroPhys that doesn't exactly answer his question.

The Chinese Remainder Theorem states that if we have

$ x\equiv a_{1}\pmod{m_{1}}$
$ x\equiv a_{2}\pmod{m_{2}}$

for [b]relatively prime[/b] moduli $ m_{1},m_{2}$, there exactly once solution for $ x$ modulo $ m_{1}m_{2}$. Now, how do we find this? I'll show the example in this problem, then you can apply it to other problems.

Let $ x\equal{}2008^{2008}$. We have

$ x\equiv0\pmod{8}$;
$ x\equiv91\pmod{125}$.

Since 8 and 125 are relatively prime, there is exactly one solution modulo 1000, and this will give the last three digits. Now, since $ x\equiv0\pmod{8}$, we write $ x\equal{}8a$ for some positive integer $ a$. Now, we have

$ 8a\equiv91\pmod{125}$. Thus, $ 8a\minus{}91\equal{}125b$ for some positive integer $ b$. Rearranging, $ 8a\minus{}125b\equal{}91$. Now, taking the equation modulo 8, we get $ 3b\equiv3\pmod{8}$, so $ b\equal{}1,a\equal{}27$, and $ x\equal{}8\cdot27\equal{}216$.[/hide]


limit
integration
function
calculus
calculus computations

Where is the Error?!

$ \lim_{n\to \infty } \, \frac{\sum _{i\equal{}0}^n \alpha _i}{\sum _{i\equal{}0}^n 1}\equal{}\lim_{n\to \infty } \, \left(\sum
   _{i\equal{}0}^n \frac{\alpha _i}{n\plus{}1}\right)\equal{}\lim_{n\to \infty } \, \alpha _n$

$ \lim_{n\to \infty } \, \left(\sum _{i\equal{}0}^n \frac{f\left(x_i\right)}{n\plus{}1}\right)\equal{}\lim_{n\to \infty } \,
   f\left(x_n\right)$

$ \left(x_i\equal{}a\plus{}i \Delta x \right) \land \left(\Delta x \equal{} \frac{b\minus{}a}{n} \right)$

$ \frac{f\left(x_i\right)}{n\plus{}1}\equal{}\frac{f\left(x_i\right)}{\frac{b\minus{}a}{\Delta
   x}\plus{}1}\equal{}\frac{f\left(x_i\right) \Delta x}{(b\minus{}a)\plus{}\Delta x}$

$ \lim_{n\to \infty } \, \left(\sum _{i\equal{}0}^n \frac{f\left(x_i\right)}{n\plus{}1}\right)\equal{}\lim_{\Delta x\to
   0} \, \left(\sum _{i\equal{}0}^n \frac{f\left(x_i\right) \Delta x}{(b\minus{}a)\plus{}\Delta
   x}\right)\equal{}\left(\lim_{\Delta x\to 0} \, \left(\sum _{i\equal{}0}^n f\left(x_i\right) \Delta
   x\right)\right) \left(\lim_{\Delta x\to 0} \, \frac{1}{(b\minus{}a)\plus{}\Delta x}\right)$

$ \lim_{n\to \infty } \, \left(\sum _{i\equal{}0}^n \frac{f\left(x_i\right)}{n\plus{}1}\right)\equal{}\frac{1}{b\minus{}a} \int_a^b
   f(t) \, dt\equal{}\lim_{n\to \infty } \, f\left(x_n\right)\equal{}b$

In the first formula $ \alpha_i$ are supposed to be independent of $ n$.

there is no dependence.

$ \lim_{n \to \infty}  \frac{1}{n} \sum_{i\equal{}1}^n a_i  \equal{} \lim_{n \to \infty} a_i$

if it begin to count in $ i\equal{}0$ we add a extra term so it need to be dividided by $ n\plus{}1$

this formula is completely right

$ x_i$ depends on $ n$ as well as $ i$- that's the problem here. There is no single sequence $ a_i$.

$ x_i$ is independent of $ n$, only of $ i$

$ S_n$ is dependent of $ n$ but not $ i$

http://en.wikipedia.org/wiki/Ces%C3%A0ro_mean

this is the same think that say that $ f(x)$ is a function that depend of its value at infinity

nonsense

if you are right. cesaro is wrong, since he is right, you are wrong QED

Look at the definition in your third line. What is $ x_5$? See the $ n$?

When you think you've contradicted well-established results, look for your mistake. Don't waste our time claiming you're right.

The contradiction is not between Cesaro and [b]jmerry[/b], it's between [b]EulerIntegral [/b]and [b]EulerIntegral[/b].
[quote="EulerIntegral"]$ \left(x_i\equal{}a\plus{}i \Delta x \right) \land \left(\Delta x \equal{} \frac{b\minus{}a}{n} \right)$[/quote]
[quote="EulerIntegral"]$ x_i$ is independent of $ n$, only of $ i$[/quote]

the first formulae is right.
and the second too.
the error is on the third line. thats the question, ho ho

but why they cannot be dependent?


number theory
greatest common divisor
least common multiple
function
algorithm
calculus
Euclidean algorithm

How do you prove...

gcd(a, lcm(b, c)) = lcm(gcd(a, b), gcd(a, c))

and

lcm(a, gcd(b, c)) = gcd(lcm(a, b), lcm(a, c))

???

[quote="s.tringali"]1) Let $p \in \mathbb{N}$ be any prime and $\alpha = v_{p}(a)$, $\beta = v_{p}(b)$ and $\gamma = v_{p}(c)$, where $v_{p}(\cdot)$ a $p$-adic valuation. Wlog, we can admit $\beta \ge \gamma$. Then

$v_{p}(\mbox{LHS}) = \min(v_{p}(a), v_{p}(\mbox{lcm}(b,c))) = \min(\alpha, \max(\beta, \gamma)) = \min(\alpha, \beta)$

and as well in a similar way

$v_{p}(\mbox{RHS}) =\max(v_{p}(\gcd(a,b)), v_{p}(\gcd(a,c))) = \max(\min(\alpha, \beta), \min(\alpha, \gamma))$

Indeed, if $\alpha \ge \beta$, we get

$\min(\alpha, \beta) = \beta = \max(\beta, \gamma) = \max(\min(\alpha, \beta), \min(\alpha, \gamma))$.

Otherwise

$\min(\alpha, \beta) = \alpha = \max(\alpha, \min(\alpha, \gamma)) = \max(\min(\alpha, \beta), \min(\alpha, \gamma))$.

In both cases, $v_{p}(\mbox{LHS}) = v_{p}(\mbox{RHS})$. Since this is true for [i]any[/i] prime $p \in \mathbb{N}$, the claim is finally proved.

2) Mutatis mutandis, it's just the same as before.[/quote]

Don't mean to sound dumb, but I have no idea what most of that notation means. Just to clarify, I don't have a college education in math. Most of the GCD and LCM properties I proved using Euclid's division lemma and also using fractional part and integer part function properties. For instance,

I proved that GCD(a+bm,b) = GCD(a,b) by doing the following.

If <> denotes an integer part function and {} denotes a fractional part function,

a/b = <a/b> + {a/b}

a = b<a/b> + b{a/b}

Since we know that <a/b> is an integer, we can substitute <a/b> by an integer q. In addition, {a/b} = r/b where c<a

a= bq + b(r/b)

thus we get the familiar a= bq + r where a is the dividend, b is the divisor, q is the quotient and r is the remainder after division of a and b.

From this form, we can eventually solve the GCD(a,b) using a Euclidean algorithm or a continued fraction.

Now, (a+bm)/b = <(a+bm)/b> + {(a+bm)/b}

a+bm = b<(a+bm)/b> + b{(a+bm)/b}

a+bm = b<m+(a/b)> + b{m+(a/b)}

The next steps involve a few lemmas.

1) If m is an integer, <m> = m and {m}= 0. Also, if m is a decimal, {m} = m and <m>= 0
2) For a belonging to all reals

a = <a> + {a} meaning any number can be defined as the sum of its integer and fractional parts

3) If a<b, then {a/b} = a/b
4) Given m is an integer

{m+ a/b} = {<m> + a/b} since an integer has no effect on the calculation of a fractional part, we get:

{m + a/b} = {a/b}

5) if m is an integer and n belongs to all reals, then:

<m + n> = <m + <n> + {n}> = < <m> + <n> > = <m> + <n>

Thus <m+n> = <m> + <n>

Using these properties, I continued to find:

a+bm = b<m> + b<a/b> + b{a/b}

a+bm = bm + bq +r

Since bm cancels out, we return to a = bq + r, meaning

GCD(a+bm, b) = GCD(a,b)

Note that the fpart and ipart properties listed aren't the only properties I've found.

Sorry if I seemed to be digressing there. It's just that I am not accustomed to proofs using calculus notation. In this case, I've been able to prove most of the GCD and LCM properties I've encountered, all with non-calculus methods. Perhaps if someone gives me a word explanation of s.tringali's post, I'll be able to follow.

Indeed, I didn't appeal to any methods from calculus. Probably, you need just to know that, for any integer $n \ne 0$ and any prime $p \in \mathbb{Z}$, $v_{p}(n)$ represents the greatest $k \in \mathbb{N}$ such that $p^{k}$ divides $n$.

Re-edit: Is there any proof that doesn't involve prime factorization?


On the classes page, it says that Intermediate Trigonometry/Complex Numbers will be offered in fall 2005. Has this already started? And if so, when is it going to be offered again?

Thank you!

[quote="MCrawford"]The class has already started.  If you wish to sign up now, you will need to call the office (619) 659-1612.

We have not yet scheduled that class again, but it is very possible that the next offering will be Summer 2006.[/quote]
I couldn't find this anywhere, but when did the class start and on what day is it held?

Thanks again. (:

It started September 22 and runs Thursday nights from 8:00 to 9:30 PM ET.


A river flows 1.2m/s south. Starting on the east bank of the river,  a girl swims at 1.4 m/s [W 30degrees N] relative to the water. The girl reaches the other bank in a minute. What is the width of the river?

[quote="sludgethrower"]Solution

No component of the river's velocity acts west, so the velocity of the girl in the west direction is: $ v \equal{} 1.4\cos{30^{o}}\equal{} 0.7\sqrt{3}\,ms^{\minus{}1}$

The distance between the banks is:

$ s \equal{} vt \equal{} 0.7\sqrt{3}\times 60 \equal{} 42\sqrt{3}\,m$[/quote]

Shouldn't the westward component of the girl's velocity be calculated by $ \sin{30}$? And since she travels at an angle of $ 30^{o}$ shouldn't the width of the river be calculated by halving the distance she travels since we can draw a 30-60-90 trinagle? ( Sorry I am new at physics so its taking me some time to understand the basics).

[quote="sludgethrower"]Well if she is travelling $ 30^{o}$ from north, then certainly you would use $ \sin{30^{0}}$ or $ \cos{60^{o}}$, but I assumed she is swimming $ 30^{o}$ north of west. Just remember if from $ \cos{\theta}\equal{}\frac{adjacent}{hypotenuse}$. The adjacent will give the component of velocity west, as $ {\theta}$ is measured from west.[/quote]

Lol I am quite aware of the definition of cosine  :P . I was just confused of how we determine the "width" of the river.

I agree with sludgethrower's argument -- the river flow does not contribute west-ward velocity, thus the only west-ward $ v\equal{}1.4cos30^{\circ}$

note that, the girl actually traveled more distance than the river width (due to the south-ward combined velocity), but it does not matter in our case.


A prime number less than twenty is randomly selected. What is the probability, expressed as a common fraction, that the units digit of its square is one?

Only occurs when last digit is $ 1$ or $ 9$.  Two primes are $ 11$ and $ 19$ out of $ 8$ total, answer is $ \frac{1}{4}$.

EDIT: beaten


BUENO LO QUE PONGO, ES LA PREGUNTA DE UN AMIGO, BUENO SERIA SU CONJETURA,  POR QUE NINGUNO ANTES VIMOS LA SIGUIENTE EXPRESION.
SI $(n,p)=1$, donde $n, p \in Z^{+}$ y $p$ primo
$\Rightarrow  \exists k , P_{1}/ nk+p = P_{1}$, donde $k, P_{1}\in Z^{+}$ y $P_{1}$ es primo

[quote="jasamper88"]El teorema de Dirichlet no solo nos dice eso sino que nos dice que hay infinitos primos de esa forma. Además la condición de $p$ primo sobra. Basta con que $(a,p)=1$[/quote]
Bueno, gracias amigo, porfavor no si podrian postear la demostracion.

La demostración del teorema de dirichlet no es elemental.

$Tipe$


You have nine coins: a collection of pennies, nickels, dimes, and
quarters having a total value of $ \$1.02$, with at least one coin of
each type. How many dimes must you have?

There are either 2 pennies, or 7 pennies or else you couldn't have 2 cents.  But there are nine coins and at least one of each type so there must be 2 pennies.  Now if you have 2 quarters, you would need exactly 5 dimes to reach $ \$1.02$.  But then there are no nickels so there must be 3 quarters.  So that leaves 4 coins which must give 25 cents.  This can be done with 3 nickels, and [b]1 dime[/b].


floor function
modular arithmetic
blogs
Congruences

If $p$ is an odd prime, prove that \[{k \choose p}\equiv \left\lfloor \frac{k}{p}\right\rfloor \pmod{p}.\]

It's a special case of Lucas' theorem (I'm not sure if it was already posted here):

If $a_{0}, a_{1}, ... , a_{n}$ and $b_{1}, b_{2}, ... , b_{n}$ are the digits of $a$ and $b$ respectively in $p$-adic representation, then
\[\binom ab \equiv \binom{a_{0}}{b_{0}}\binom{a_{1}}{b_{1}}... \binom{a_{n}}{b_{n}}\mod p.\]

Let $ k\equal{}pt\plus{}u$ with $ 0\le u<p$. Consider the element of $ S_k$ of order $ p$ specified by $ (1,2,3,\ldots,p)(p\plus{}1,p\plus{}2,\ldots,2p)\ldots((t\minus{}1)p\plus{}1,(t\minus{}1)p\plus{}2,\ldots,tp)$, and consider the action of this element on the subsets of $ \{1, \ldots, k\}$ of size $ p$.

Clearly all of the orbits are of size 1 or $ p$, and hence the total number of subsets, namely $ C(k,p)$ must be equivalent (modulo $ p$) to the number of fixed points, which is $ t$.

Luke
See my puzzle blog at [url]http://bozzball.blogspot.com[/url]


I recently came across the following problem:

The convex quadrilateral ABCD of area 2t has no parallel sides. Locate the point P1 on CD such that P1 is on the same side of AB as C and the area of the triangle ABP_1 is t. If P_2, P_3 and P_4 are similarly defined for the sides DA, AB and BC, respectively, prove that P_1, P_2, P_3 and P_4 are collinear.

I don't understand, because if ABCD is "approximately" a square, then P_1 will be around the mid-point of CD, P_2 will be around the mid-point of AB, etc., and these points are not collinear.  Can anyone clarify?

A line thru midpoints of diagonals is your line.   Triangles with vertices on
    this line and opposite sides being the quad sides add up to 2T in area, and each
    opposite pair adds to T.  Each point P_i  sends area of one triangle in the pair
    to zero.

    It is Leon Anne theorem, of course.

    Pestich.

Son of a gun, it works.  Thank you, pestich!


Given triangle $ABC$, $D$ is a point on $BC$ such that $AD \perp BC$. Construct a circle with diameter $AD$ to intersect $AB$ and $AC$ at $M$ and $N$. Let $l_{A}$ be a line through $A$ and perpendicular to $MN$. Similarly we define $l_{B}$ and $l_{C}$. Show that $l_{A}$, $l_{B}$ and $l_{C}$ are concurrent.

They meet at the circumcenter $O$. It is easy to be proven by calculating the angle.


Let the function ${f\text{:}D\rightarrow \mathbb{R}, f(x)=\log_{5}(\log_{\frac{1}{3}}{(\log_{8}{(2x+1)}})})$  where $D$ is the maximum domain of $f(x)$. If $S$ represents the sum of the absolute values of all integers from $D$, find $S$.

Domain: 0 < x < 3.5
The integers in this interval are 1,2 and 3.  $1+2+3=\boxed{6}$.


quadratics
geometry
geometric transformation
reflection

How many distinct $ x$-intercepts does the graph of $ y \equal{} (x\minus{}5)(x^2\plus{}5x\plus{}6)$ have?

before we solve this problem, i'm going to ask you how we find the $ x$-intercept
the answer: if $ y$ is $ 0$, then we have the $ x$-intercept

with that in mind, we move to our equation
$ y \equal{} (x\minus{}5)(x^2\plus{}5x\plus{}6)$

we see that we can simplify our second expression for $ x$ a little more
since $ x^2\plus{}5x\plus{}6\equal{}(x\plus{}2)(x\plus{}3)$, we can rewrite the equation as:
$ y \equal{} (x\minus{}5)(x\plus{}2)(x\plus{}3)$

from here we can see that in order for $ y$ to be $ 0$, the one or more of the expressions $ (x\minus{}5), (x\plus{}2), (x\plus{}3)$ have to be $ 0$

so, we can rewrite this as:
$ (x\minus{}5)\equal{}0$
$ (x\plus{}2)\equal{}0$
$ (x\plus{}3)\equal{}0$
if any of these equations are true, then our original equation is true (make sure you see why this is)

solving those equations, we get $ x\equal{}5, \minus{}2, \text{or} \minus{}3$

therefore, we have $ 3$ distinct $ x$-intercepts of the graph $ y\equal{}(x\minus{}5)(x^2\plus{}5x\plus{}6)$

First we note that, since we are looking for the $x$ intercept, $y$ must be $0$. Thus, we have the quadratic:
\[0=(x-5)(x^2+5x+6)\]
We could also note that in the expansion of:
\[(x-5)(x^2+5x+6)=x^3+5x^2+6x-5x^2-25x-30=x^3-19x-30\]
the exponent of $x$ is $3$.

Since we are solving a quadratic in which the quadratic portion's $x$ is raised to the third power, we note that there a total of $3$ solutions, and therefore:
\[\boxed{3}\]
distince $x$ intercepts.

(See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=300&t=412078&p=2316963#p2316963]here[/url] for an explanation of why the degree of a quadratic reflects the number of solutions it has).

While it is true that an expression raised to the $n$ power has $n$ number of solutions, please be careful when translating this to mean that it has $n$ number of x-intercepts. It is quite possible that the solutions are complex and thus do not cross the x-axis


In a sequence of coin tosses, one can keep a record of instances in which a tail is immediately followed by a head, a head is immediately followed by a head, and etc. We denote these by TH, HH, and etc. For example, in the sequence HHTTHHHHTHHTTTT of 15 coin tosses we observe that there are two HH, three HT, four TH, and five TT subsequences. How many different sequences of 15 coin tosses will contain exactly two HH, three HT, four TH, and five TT subsequences?

[hide]
Let us look at the sequence: (HTHTHT)
There are 3 HTs and 2 THs.
All the possibilities of the sequences that satisfy the given conditions can be met by inserting 5 H's next to another H and 4 T's next to another T.
For example, (HHHHHHTHTHTTTTT) has 5 H's added at the front and 4 T's added at the end of the base sequence. Likewise, inserting an H next to any H will produce a (HH) sequence.

All we need to do is count how many ways there are to partition 5 H's into 3 spots and 4 T's into 3 spots. (as inserting a H to the left of another H is the same thing as inserting it to the right)
There are:
21 ways to put 2 dividers between HHHHH
15 ways to put 2 dividers between TTTT

Therefore the answer is $21(15)=\boxed{315}$.

[/hide]

[hide="Rubiks Cube"]We start with the HT and TH sequences. We note that there is only one possible way to arrange these sequences: HTHTHT. From there, we are allowed to add 5 Hs as long as they are adjacent to another H and 4 Ts as long as they are adjacent to another T. In other words, our 15 flip sequences goes:

[H block][T block][H block][T block][H block][T block]

Now, we have to determine how many are in each block. Starting with the Hs, we see there are 8 Hs: HHHHHHHH. Now, we must insert two blocks of Ts in between two different pairs of Hs. So, we can insert them anywhere that there is an underscore: H_H_H_H_H_H_H_H. There are seven underscores, and thus $\binom{7}{2} = 21$ possible H block combinations.

Now for the Ts, there are 7: TTTTTTT. We have to insert two H blocks between two different pairs of Ts. We can insert them anywhere where there is an underscore: T_T_T_T_T_T_T. There are six underscores, and thus $\binom{6}{2} = 15$ possible T block combinations.

Now, to count all ways for us to have our sequence, we multiply. $21 \times 15 = 315$. And we're done.[/hide]

Wait a sec...
I'm getting [hide]560[/hide] as my answer.  :huh:

Edit: Nevermind... Kalva says I'm right. :P

[quote="chesspro"]Actually, 4everwise, I don't think so. :P

[quote="4everwise"]
...there are five HH, three HT, two TH, and four TT subsequences
[/quote]

[quote="kalva"]2 are HH, 3 are HT, 4 are TH, 5 are TT.[/quote]

So you probably have the correct answer to the answer that shows up on the AIME, but you posted a different problem than that.  ;)[/quote]Yeah... I was solving the problems posted on Kalva...  :?

That's really messed up. Different content? It's supposed to be unchanged. :mad:

[quote="chesspro"]So you probably have the correct answer to the answer that shows up on the AIME, but you posted a different problem than that.  ;)[/quote]I just checked and I don't think I typed the wrong problem in...  :huh:

That's really weird.

Well...one of them isn't the original AIME question. But at least we know why we have two different solutions now.  :)

Ok, this problem is now FIXED.  :)

I'm not sure what happened when I was typing it in. I probably dozed off. :D

The problem doesn't match that on the [url=http://www.artofproblemsolving.com/Wiki/index.php/1986_AIME_Problems/Problem_13]AoPS Wiki[/url]

4everwise says the problem is fixed, but it still looks like something is wrong to me.

What is the motivation for viewing this problem as blocks of types of flips; how do you think of that?

Ok, so this is pretty old but just some insight on this problem, as I saw it for the first time today and solved it not too long ago.

To answer the question in the post above:
$1$) You want to look at the most restrictive condition.
$2$) You want to find things that don't change.

  The most restrictive condition happens to be the condition that there can only be $2$ $HH$'s. You can do casework based off the number of $H$'s in a block. Say that there exists a block of $H$'s with three $H$'s, well then there cannot exist another slot with two consecutive $H$'s. So we say that this sequence starts off with $HHH$, well if there are $x$ groups of $T$'s in this sequence, then you can see that you reach a contradiction, as you cannot keep both the conditions that there are three $HT$'s, and four $TH$'s. Thus, you have to have any sequence that satisfies the problem statement starting off with $T$'s. Now do casework based off the largest number of $T$'s in a group, this is very fast. Primarily because you have to keep a fixed number of groups of heads and tails.

All-in-all, you want to look for fixed things in this problem, once you find the number of groups of $H$'s and the number of groups of $T$'s, you're basically done. You then do casework on the largest number of consecutive $H$'s and $T$'s in the groups of $H$'s and $T$'s, and find how many ways you can arrange those groups within their "allowed" region.


let P is a point  in ABC triangle.that $ <PAB\equal{}10 , <PAB\equal{}20 ,<PCA\equal{}30 , <PAC\equal{}40$
prove AB=BC

[quote="stvs_f"]let P is a point  in ABC triangle.that $ < PAB \equal{} 10 , < PAB \equal{} 20 , < PCA \equal{} 30 , < PAC \equal{} 40$
prove AB=BC[/quote]

$ \angle PAB \equal{} 10$ or $ 20$??

It can be proven with a regular 18 gon: [url]http://www.cut-the-knot.org/triangle/TrigCeva.shtml[/url]
The given triangle is a 40-40-100 triangle inscribed in the 18 gon, then draw the red cevians as you gave, they concur at a point. This is easily verified by the identity $ sin 20 sin 20 sin 30\equal{} sin 20 sin 80 sin 10$, which is true because $ sin80 sin 10\equal{}sin 10 cos 10\equal{}\frac{1}{2} \cdot sin20 \equal{}sin20 sin 30$.

There is a typo: it should be $\angle ABP=20^\circ$. Take $D$, the symmetrical of $A$ w.r.t. $PB$, see that $\triangle ABD$ is equilateral, i.e. $D$ is the circumcenter of $\triangle ACD\implies CD=AD$. Because $\angle DAC=\angle PAD-\angle PAC=20^\circ$, and $\angle ADB=70^\circ$ we infer $BD\bot AC\implies ABCD$ a kite, or $BC=AB$, done.

Best regards,
sunken rock


$f: \mathbb R\longrightarrow\mathbb R^{+}$ is a non-decreasing function. Prove that there is a point $a\in\mathbb R$ that \[f(a+\frac1{f(a)})<2f(a)\]


Let $ a,b,c>0$ such that $ a\plus{}b\plus{}c\le3$. Find the maximum value of expression
$ P\equal{}\frac{a^2}{a^2\plus{}1}\plus{}\frac{b^2}{b^2\plus{}1}\plus{}\frac{c^2}{c^2\plus{}4}\minus{}\frac1{ab}\minus{}\frac1{bc}\minus{} \frac1{ca}$

[quote="bunhiacovski"][quote="akai"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c\le3$. Find the maximum value of expression
$ P \equal{} \frac {a^2}{a^2 \plus{} 1} \plus{} \frac {b^2}{b^2 \plus{} 1} \plus{} \frac {c^2}{c^2 \plus{} 4} \minus{} \frac1{ab} \minus{} \frac1{bc} \minus{} \frac1{ca}$[/quote]
I'm think
Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c\le3$. Find the maximum value of expression
$ P \equal{} \frac {a^2}{a^2 \plus{} 1} \plus{} \frac {b^2}{b^2 \plus{} 1} \plus{} \frac {c^2}{c^2 \plus{} 1} \minus{} \frac1{ab} \minus{} \frac1{bc} \minus{} \frac1{ca}$
$ \frac {c^2}{c^2 \plus{} 1}$ not $ \frac {c^2}{c^2 \plus{} 4}$[/quote]

Oh no,this problem's [b]not sym[/b]. Because if it's sym ~> too easy  :lol:

Oh,It's hard problem. Because it's not sym,then equality don't hold for $ a\equal{}b\equal{}c$. I predict that Max $ P\equal{}\sim  \minus{}1.8$  :(
Can some body help me?


Find all pairs of non-negative integer $(x,y)$ which satisfy\[p^x-y^p=1\]where $p$ is an odd prime number

isn´t this a 2004 romanian tst??? i think it is  :P

First note that $(0, 0)$ always works. Also, clearly if one of them is $0$ the other must be too, so assume $x, y$ are positive.

Then we have $p^x \mid y^p+1$. This implies $p \mid y^p+1$. Since $p \mid y^p-y$ by Fermat, $p \mid y^p+1-y^p+y=y+1$. Write $y=pk-1$. Then we have $p^x=(pk-1)^p+1=pk((pk-1)^{p-1}-...-(pk-1)+1)$. This implies $k \mid p^x$, so $k=p^{n-1}$ for some $n$. Thus $y=p^n-1$.

Our equation can be rearranged to $(p^n-1)^p=p^x-1$. Now, from this we conclude that $p^n-1 \mid p^x-1$, so by a well-known lemma $n \mid x, x=na$. Now it seems to be easy...

EDIT: I had made a gross mistake.  :blush: Here goes again.

A special case is $p=2, n=1$ giving the solution $p=2, x=1, y=1$. Otherwise, suppose $p^{na}=(p^n-1)^p+1$. Then $p^{na}=p^{np}-...-\binom{p}{p-2}p^{2n}+\binom{p}{p-1}p^n$. Since the LHS is a power of $p$, the RHS is too. However, for $p>2, n>0$ all the terms in the RHS are divisible by $p^{2n+1}$ except the last, which is divisible by just $p^{n+1}$, so that the result is false. So there are no other solutions.


I only decided to go to the camp on Monday so I have to do late application. (Stupid mail coming so close to the deadline.)

Do I submit my application now and then submit the test when I'm done? I see that it isn't supposed to be posted until May. Do I wait to apply until then?

Help would be much appreciated!

I'm not 100% sure about their policy on this, but its probably best you submit your application along with the admissions test. Plus, there's no point in rushing through your application to get it in this early if the test isn't done anyway. I'd advise you to keep looking over your application and keep revising it as you wait for test C to be posted, as effort put into the app does matter.

send your application and solutions to the test together.
no rush.

Thanks for the help.

I'll follow your advice.


Here's a nice card trick I learned:
From a standard deck of cards (without the jokers), remove the aces, kings, queens, jacks, and tens. Then shuffle the cards. Next, make sure you have an audience ( :lol: ) and ask one of them to pick a card randomly from the deck; they shouldn't tell you what it is, just memorize it and keep it with them. This is where the fun begins, ask them to double the value of the card, add 5, and then multiply by 5. When they are done, ask them to take another random card and add it the the value they calculated. ask them to tell you this value. Finally in your head, subtract 25 from that total. Each of the digits of the 2-digit number you get represent the 2 cards the person picked!
-------------------------------------------------
Now the mathematical part:
a) Prove or disprove: This trick will always work for our custom-built deck (without the jokers, aces, kings, queens, jacks, and tens).
b) Why would the trick breakdown if we were to add the tens?

[hide="A"]Let the number on the original card be $ x$. The volunteer doubles the value, adds 5, and then multiples by 5, making it $ 10x+25$. Then, adding the number on the second card makes it $ 10x+y+25$. Subtracting 25 gives you $ 10x+y$, which is a 2-digit number with the tens digit being the first card and the ones being the second card.[/hide]
[hide="B"]If $ x>9$, then you'll just have a 3-digit number with the first two digits being the first card and the last digit being the second card. If, however, $ y>9$, then you'll get the cards wrong.[/hide]

Well you'll still be able to find the values of the cards, cuz if you get 80 as the number then you know he got a 7 and a 10.


AMC
AIME
email
LaTeX
analytic geometry
graphing lines
slope

Hey, I have an idea. How about the AoPS forum members (anyone who wants) compile two nice exams: one with short answer (AIME-type) problems, and the other with olympiad proof-type problems. We would have someone be the designated compiler, and anyone who wants to participate would send 2 problems to this compiler--1 of each type. The compiler would then copy all problems of each type into two files, which he/she would upload onto the site (or simply post the contents of in a post). Then, since each person would be choosing two of the nicest problems to submit, we would have a very nice small collection of both short-answer and proof-type problems.

Fierytycoon

Sounds like a plan... how you email them all to this email.. (don't worry just send me the problems...)

idiotwithoutavillage@hotmail.com

Please place in the subject line, AoPS math problems.. or something with math problems in the subject line.. I'll compile it and send 'em out or post 'em on a webpage or somethingl like that.. please send me the problems...

$10 says this starts a flame war as people fight over who gets to be the compiler.

Why don't we set a deadline for the sending of the problem? How about, say, by next saturday at midnight (which is technically sunday)?

Fierytycoon

I have a bunch of problems I wrote for a competition that never happened. I can send them on if you think it's a good idea.

By next Saturday, I meant the night of August 9th, not this Saturday. And to Rep123max, the problems will just be posted on this forum or something, so that everyone can see them.

Fierytycoon

Idiot, if you'd like me to LaTeX all the problems once you've compiled them, I'd be happy to do that.

So what exactly is short answer.. is it multichoice or is it just that only an answer is required? And what about difficulty level.. I've got hundreds of problems thanks to training over the last few years..

Well, short-answer as in an actual number (like on the AIME). And I was sorta hoping that the short-answer questions would be about the difficulty of a moderate-tough AIME problem, and the proof-type ones to be tough USAMO/moderate IMO ones, so that the problems would appeal to a wide audience.

Fierytycoon

Short answer means that it isn't multiple choice, and only the answer matters. For example, if I asked what is the slope of x^2 at the point (4, 16), then you would have to write 8 if you want credit.

Short answer is, in math, where you input an answer but don't have to show any work.  If you know what the AIME is, that answering scheme is "short answer".

In other classes, such as English, short answer is writing a sentence or two.  Or three or four, if you feel like elaborating - but it's a lot shorter than an essay.

Yeah, it was the AIME bit that I wasn't sure about :)

IW.OAV, did you look at the problems I sent you? The short-answer problems are probably AIMEish level. The proof questions are somewhat simple, but I do not know if I can come up with good USAMO-level questions. I'll play around with some inequalities later tonight to see what I can do. I think I might be able to create some USAMO-level inequalities problems, but that's about it.

I don't seem to be getting the point of this?

The point is to make a list of questions that people on AoPS made up themselves.  It's recreational, in the same way that browing the forum is recreational.  Anyway, if you don't understand what we're talking about you should read the beginning of the thread, also.  Because if you just read the most recent post, it will make no sense whatsoever.  :wink:

I did read the first post, but didn't catch the "created by AoPs members" part, if there was one. Now I get it.

[quote]but didn't catch the "created by AoPs members" part, if there was one. [/quote]

Subject: Compiled Exam of AoPS Forum Members

aww...I feel so sad...you don't think me important enough to even read in the subject that I type in... *tears*

Fierytycoon[/quote]

My oh my, you get tearful a lot, Fierytycoon.  You need a backpat.  I'll give you one since rippledance isn't here: *patpat*

How hard should these be? Because i came up with one that i think is moderate in my opinion.

Awww...just submit whatever problems you come up with.  If one's too hard, well, it's too hard...

Hey Idiot_without_a_village, could you post the compiled problems in a post now? thx


abstract algebra
invariant
topology
calculus
integration
geometry
3D geometry

$ \text{A Prize\minus{}winner partly nocturnal}$
$ \text{Sends variants from Weil to a journal}$
$ \text{I'm willing to wager}$
$ \text{This paper is major}$
$ \text{(Rank one perturbation of kernel)}$

Post other ones you know. Or make up your own  :wink: .

random one i did for ELA last year:

π (And Other Math Nonsense)
                  By Travis Yeh

3.1415926535897932384…………………
Ugh! What a bore!
That transcendental number, pi.
It just looks displeasing to the eye.

There’s also Euler’s Number, that deceivingly simple e.
2.7182818…………
It’s just too much for me.
If I memorize it all, then my brain just might deflate.

And imaginary numbers,
Why are they there at all?
While trying to explain why z = x + iy,
I think that I might die.

Knot Theory? Oh, please no!
That will make my cranium blow.
With knot invariants and Floer homology,
I’d rather “knot” do this and take up biology.

Then, there’s topology, that endlessly complicated math,
With confusing homeomorphisms,
(No, I did not just sneeze)
I think I’ll go down a different path, if you please.

Champernowne constant? Do you mean C10?
Now that seems like it’s more for me.
0.123456789….
This irrational number seems just fine.

 Hmmm… these constants aren’t that bad!
What about Chaitin’s constant, it seems rad!
0.0078749969978123844…..
Never mind, I can’t memorize it for sure.

Mathematics is really quite strange,
The only thing I can remember is mode, median, mean, and range!
Why didn’t I listen to my mother
And take Engineering like the others?

dont get me wrong, i love math.

wow, nice travis!

how 'bout this one I made up?

one plus one is equal to two
ten plus twelve is twenty-two
forty-five and thirty is not seventy-two
eighty-six plus sixteen is a hundred and two!

see? it all rhymes.

Well, there's this limerick (not mine):
A dozen, a gross and a score,
plus three times the square root of four,
divided by seven, plus five times eleven,
is nine squared and not a bit more.

To prove Pythagoras was right
With new-found ease and much delight, --
Let's say hypotenuse is $ c$,
With one leg $ a$ and 'tother $ b$.

On c we draw a square with pride,
Then triangles on every side.
The square of $ (a \plus{} b)$, you see,
Is $ c$ squared plus two times $ ab$.

Equate and cancel $ 2ab$
--- And we have reached the Q.E.D.

Said Mrs. Isosceles Tri,
"That I'm sharp I've no wish to deny;
But I do not dare
To be perfectly square --
I'm sure if I did I should die!"

Integral v-squared dv
From one to the cube root of three
Times the cosine of three pi over nine
Equals log of the cube root of e.


sum of 3 #s is 84. the second # is twice the first., and third is 4 more than the second. find the #s

The first number is $ x$.

The second number is $ 2x$.

The third number is $ 2x \plus{} 4$

Thus, $ 5x \plus{} 4 \equal{} 84$, and $ x$ is clearly $ 16$.

The numbers are $ \boxed{16,32,36}$.

Hmmm, is this like your homework???

no i am just making these up i like these kind of problems

Then do what the mods are telling you and stop posting so many.

Don't flood the forum.


Let 2006 positive real number $a_1,a_2,...,a_{2006}$.Prove that there exists 2006 real number$b_1,b_2,...,b_{2006}$:
(i)$a_i<b_i\forall i=1,2,..,2006$
(ii)$\frac{1}{2}<\frac{b_i}{b_{i+1}}<2\forall i=1,2,..,2005$
(iii)$\sum_{i=1}^{2006}b_i<3\sum_{i=1}^{2006}a_i$
  Prove that 3 is the best number when we replace 2006 by n.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88917]View all problem for VN IMO team 2006[/url]

Let $a_k$ is maximal terms. If sequence $b_k=a_k+\epsilon,b_{k+i}=\frac 12 b_{k+i-1},i>0,b_{k-i}=\frac 12 b_{k-i+1},i>0$ satisfyed
(1)$b_i>a_i$,
then $\sum_i b_i <3(a_k+\epsilon )$, and 3 is minimal possible coefficient. If this sequence is not satisfyed (1) it may be modificied aprokcimated sequence $x_i=max_{j\le i} a_j$ if i<k, and $x_i=max_{j\ge i} a_j$ if i>k.


A certain pitcher has a $\frac 1 2$ chance of pitching a strike.  If he pitches a strike, the batter will hit the ball $\frac 3 4$ of the time, ending the at bat.  Also, if he pitches 3 strikes, the batter strikes out, ending the at bat.  Assuming there are no foul balls, what is the probability the batter gets a walk?  A walk is 4 balls, or non-strikes.

Hm, interesting solution... there is a more direct solution, involving the balls and dividers trick, which helps with the extended question:

Generalize the probability for the probability of getting a strike $s$ and the probability of getting a hit $h$.

Let's say you wanted to find the number of solutions to x + y = 10 for positive integers x and y. You could take a group of 10 balls:

o o o o o o o o o o

With these ten balls, let's say we took a divider and placed it in between any two balls, for example

o o o|o o o o o o o

We could say that the number of balls to the left of the divider represents x and the number of balls to the right of the divider represents y. We can therefore represent each solution (x, y) graphically as balls/dividers. We see that there are 9 locations for our 1 divider, so there would be 9 choose 1, or 9 solutions. This can be generalized for any number of balls or dividers. (You can also generalize it to include zero, etc.)

Regarding how that relates to this problem, as mentioned by deej21, you can do this:

[hide]
We can either can 4 balls, 4 balls and 1 strike, or 4 balls and 2 strikes. We can represent each ball with a B and each strike with an S. (Note: Regarding balls/dividers, this would mean each B represents a ball and each S represents a divider.)

0 strikes:
We have to find all arrangements of BBBB that end with B. This is of course 1. Since there is a probability of 1/2 for each B, there are thus a $\left(\frac{1}{2}\right)^4$ probability for this case.

1 strike:
We have to find all arrangements of BBBBS that end with a B (since we cannot end with a strike). We note that this is the same as rearranging BBBS and then sticking a B at the end, so there are 4 possibilities. Since a strike has probability 1/8, we have $4\cdot \frac{1}{8}\cdot\left(\frac{1}{2}\right)^4$

2 strikes:
We have to arrange BBBBSS such that our arrangement ends with a B. We note this is the same as rearranging BBBSS and then tacking on a B, so there are 10 arrangements. The probability for this case is thus $10\cdot\left(\frac{1}{8}\right)^2\cdot\left(\frac{1}{2}\right)^4$

If we add all these probabilities, we end up with $\frac{53}{512}$, the same as before.[/hide]

Nice solution!  :D  Yeah, I thought there should be a faster way.


invariant
combinatorics proposed
combinatorics

Given a nxn table, where n is odd. There is either 1 or -1 in its every
field. A product of the numbers in the column is written under every
column.  A product of the numbers in the row is written to the right of
every row.
Prove that the sum of 2n products doesn't equal to 0.

It can only be $0$ when $n$ of the numbers are $-1$. Since $n$ is odd, an odd number of these $-1$s will come from rows, and an even number from columns, or vice-versa. This gives two different values for the product of all the numbers in the table (we compute this product in two different ways, by taking the product of all the rows, and then the product of all the columns).

My solution is an invariant, change a -1 by a 1, then the sum remains invariant mod 4.

The Union of Soviet Socialist Republics.
What age are you?

He,he
:))))
I mean about USSR. Only Iura knows it is USSR, not URSS :) . You should know a little history besides math ...

It is easy to know the history but hard to remember the names.

Actually, I know it as CCCP. :)

So many acronyms to remember...

No. In france too we said URSS :
Union des Rpubliques Socialistes Sovitiques.

Pierre.


Pls could someone help me out with the function of a semiconductor diodes and their uses.

These links might be helpful:
[url=http://en.wikipedia.org/wiki/Semiconductor_diode]Wikipedia[/url]
[url=http://hyperphysics.phy-astr.gsu.edu/hbase/solids/diod.html]HyperPhysics[/url]

.

If you have some specific questions, dilemmas about diodes, electronic circuits, etc, post them, there are a few Electrical & Electronic Engineering students here, willing to help (including myself).  :)


Ελπίζω να μην την ξέρετε ώστε να έχει ενδιαφέρον.

Λοιπόν αν $a,b,c,d>0$ με $abcd=1$  να δείξετε ότι

$\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2}\geq 1$

Λοιπόν  Σιλουανέ  πολυ  καλή  η  ασκηση  αυτή  και  μπορώ  να  σου  πω  οτι  ενώ  στην  αρχή  μου  φάνηκε  εύκολη  τελικά  χρειάσθηκε  αρκετό χρόνο για  να  την  λύσω.

Λοιπόν  από  γνωστή  ανισότητα  έχουμε    ότι  $\frac{1}{(1+a)^{2}} + \frac{1}{(1+b)^{2}} \geq \frac{1}{1+ab}$  (1) ενώ  ομοίως  $\frac{1}{(1+c)^{2}} + \frac{1}{(1+d)^{2}} \geq \frac{1}{1+cd}$ (2)  έτσι
προσθέτοντας  κατα  μέλη  τις  (1)  και  (2)  έχουμε  οτι

$\frac{1}{(1+a)^{2}} + \frac{1}{(1+b)^{2}} + \frac{1}{(1+c)^{2}} + \frac{1}{(1+d)^{2}} \geq\frac{1}{1 +ab} + \frac{1}{1+cd}$  όμως
$\frac{1}{1 +ab} + \frac{1}{1+cd} = {\frac{2 + ab + cd }{1 + ab + cd +abcd} = \frac{ 2 + ab + cd}{ 2 + ab + cd } = 1}$   άρα  $\frac{1}{(1+a)^{2}} + \frac{1}{(1+b)^{2}} + \frac{1}{(1+c)^{2}} + \frac{1}{(1+d)^{2}} \geq 1$

Ακριβώς. Αυτή είναι και η λύση που είχα και εγώ. Λοιπόν θα τα πούμε.


Hi.

I got the function that goes like this: f(t)=11+4,9*sin((pi/12)(t-8))
0 < or = t
and
t < or = 24

Now what I need to do, is to find the place where t, is higest and where t is lowest. I have been able to find where it's higest, by diffrentiate f(t) and them let f'(t)=0 and then solve after t. t is then 14, and y is 15,9. But I dont know how to find the lowest one, can some one please help me?

Our minimum occurs where $ \sin[\frac{\pi}{12}\cdot (t\minus{}8)]\equal{}\minus{}1$. This occurs when $ t\equal{}2$, and our min is 6.1.

Yeah, the extreme values of a sine or cosine function are $ \pm 1$, so whenever the sine or cosine is $ \pm 1$, with a linear transformation, you'll have an extreme on the transformed function.


geometry
calculus
AMC
AMC 10
trigonometry

i just made this problem up.  two circles have radius r.
their centers are r apart.
what is the area of the two circle's intersecion?

I don't think you need calculus for this problem.  Just need the area of circles and triangles.

Haha... Calculus!

[hide]So..... We can draw 2 equilateral triangles in there.   Then if we draw in the radii and examine the top portion of the (-)... Thats supposed to be the shape...  We will have 60/360 :pi: r^2 for one of the arcs... If we do the same thing for the other arc, we get 1/3 :pi: r^2.... We have overlapped the area of the equilateral triangle whose side is r...   :rt3: /4r^2.... Subtract that from the 1/3 :pi:  and you get the top half.  Multiply this by 2 and you get the area of the () shape.[/hide]

this problem sounds really familiar. are you sure it isn't an amc10 or other contest problem? just wondering.

This problem is perhaps one of the most popular contest problems, after "the average of 40 and 60 is 48."  Using that trick to find the area of the lens shape appears all over the place, including the AMC and many other contests.

belenos wrote:Are you sure it's a "getting stated" problem ?

[hide]
Using Pythagoras and a little bit of calculus, it is :

2 * int (x=r/2 .. r)  :sqrt: (r^2 - x^2)

Now it's calculus gymnastics, int (sqrt (1-x^2)) = x*sqrt(1-x^2)/2 + arcsin(x)/2, change of variable y=x/r, ...

I found (it makes sense because it is  :sim: 20% of the area of one circle, but I didn't checked it carefully) :

r*(:pi:/3- :rt3:/4)[/hide]]

calculus?  What the ****?

this needs only a simple trig or knowledge of 306090 triangle.

i made this up during school.

Well, well, obviously I gave a perfect example of the use of the wrong tool !! :D


calculus
integration
function
calculus computations

I posted this in the wrong forum before but here it is

I am trying to find the integral of (1/((1 + (x^4))^.5)) between 0 and 1 but as of now have had no luck. Does anyone have a suggestion?

thanks for replying, but that was actually the first place I tried and it didn't give a good result, and I ended being confused.Any more suggestions?

Indeed, Mathematica's answer is confusing. Maple returns $ \frac12\mathcal{K}(1/\sqrt{2})\approx 0.927$

thanks for replying, now is there any way could you tell me that answer with more digits? It's just that I need it for finding the error to approxmiations.

Will 50 be enough? 0.92703733865068595921692517359763002310879941176088
You know, you could download SAGE (http://www.sagemath.org) and get this kind of stuff done in seconds.

Could it be that you were asked to [b]estimate [/b]the error of approximation?

I've been using MATLAB for the last while as that is what my prof recommended, but after I had trouble with this, maybe I'll give that a try.

No the thing is I have to use the trapezodial and simpsons rule to approxmiate it and then use the actual value to find the absolute error, and for that I needed the actual value. Which was the thing I couldn't find.

 Thanks again for the help

With MATLAB:
[quote]>> syms x

>> a=int(1/sqrt(1+x^4),0,1)

a = 1/2*EllipticK(1/2*2^(1/2))

>> vpa(a,50)

ans = .92703733865068595921692517359763002310879941176085
[/quote]

I just saw that you posted this question twice. Wouldn't bother replying if I noticed this earlier.

If you have a TI-83 calculator or similar, it has a numerical integration function you should take advantage of.

Anyway, as the MATLAB result suggests, this is an [url=http://mathworld.wolfram.com/CompleteEllipticIntegraloftheFirstKind.html]elliptic integral[/url].


Find all $f: R^{+}\rightarrow R$ such that

$f(x)f(y)=f(xy)+2005(\frac{1}{x}+\frac{1}{y}+2004)$


Let $ a, b$ and $ c$ are integer, such that \[ \frac ab\plus{}\frac bc\plus{}\frac ca \equal{}3.\] Show that $ abc\equal{}d^3$, for some $ d$ is integer.

more generally, if $ a,b,c$ are integers such that $ \frac ab \plus{}\frac bc \plus{}\frac ca$ is an integer, then $ abc$ is a perfect cube.


Let $ \phi(n)$ be the number of factors of n. What is the maximum value of
$ \frac {\phi(n)}{n}$ for $ n \ge 100$?

I don't know the answer, I was just curious, is this a well-known problem? Maybe it's  when n=120 or something...
Also, is there an asymptotic function for this?

No, that is infinitely high for an infinitely high prime number. Unless you want to find the minimum value of that, in that case it's the same as my problem. I meant maybe it's maximum when n=120.

But how can the number of factors of $ n$ be greater than $ n$?

Sorry, I meant maybe it's maximum when n=120, in which case the value is $ \frac{2}{15}$

The usual notation is $ \sigma_0(n)$ or $ d(n)$; $ \phi(n)$ usually denotes the [url=http://en.wikipedia.org/wiki/Euler%27s_totient_function]totient function[/url].  See [url=http://en.wikipedia.org/wiki/Divisor_function#Approximate_growth_rate]here[/url] for asymptotics.

As for the actual maximum, that seems like a computational question.  It's enough to compute the value for a few of the [url=http://en.wikipedia.org/wiki/Highly_composite_number]highly composite numbers[/url] and then use the trivial bound $ \frac{\sigma_0(n)}{n} \le \frac{2}{ \sqrt{n} }$.

I was interested in this problem a few years back, before I joined AoPS... I had forgotten about it...

YAY it's solved!


group theory
abstract algebra
superior algebra
superior algebra unsolved

One more question: we have a finite group G and for each $ p$ we choose a p-Sylow subgroup of G. The problem is to prove that G is generated by this subgroups.
It is obvious if $ |G|\equal{}p_1^{\alpha_1}p_2^{\alpha_2}$, but then i'm stuck.

The intersection between the two Sylow subgroups must be (e). Think about why that is and the counting formula for the product of two subgroups of a group. If you can't do from that, you should probably drop the class.....

Thanks MOZART1756, i have already found the solution.


let $ a_1, a_2, ..., a_9$ are permutation of $ 1,2,...,9$. Find the maximum and the minimum value of $ a_1 a_2 a_3 \plus{}a_4 a_5 a_6 \plus{}a_7 a_8 a_9\plus{} a_1 a_4 a_7\plus{}a_2a_5 a_8 \plus{} a_3 a_6 a_9$

[quote="Pluraln"]let $ a_1, a_2, ..., a_9$ are permutation of $ 1,2,...,9$. Find the maximum and the minimum value of $ a_1 a_2 a_3 \plus{} a_4 a_5 a_6 \plus{} a_7 a_8 a_9 \plus{} a_1 a_4 a_7 \plus{} a_2a_5 a_8 \plus{} a_3 a_6 a_9$[/quote]
Use computer.


A 2.42-g sample of a mixture of KCl and MgCl2 is treated with excess
>AgNO3(aq). If 5.46 grams of AgCl(s) are obtained, then what is the
>percentage by mass of KCl in the original mixture?
>
>Molar masses (in g/mol):
>
>KCl, 74.55
>
>MgCl2, 95.21
>
>AgCl, 143.35

So far  what I know is that two reactions are occuring independent of one anohter which would be:

   KCl +  AgNO3 -->  AgCl(s)  + K+(aq)  + NO3-(aq)        and
MgCl2 + 2 AgNO3 --> 2 AgCl(s) + Mg{2+}(aq) + 2 NO3-(aq)

I don't know how I would go about next...in particular i am confused when they say the word excess..any help would be appreciated.

Try substituting a variable in for the mass of KCl.  You can then describe the mass of MgCl2 in terms of this variable, and can set up some equations.

Excess just means that AgNO3 will not be the limiting reagent.

When setting variables for the mass of each substance, try not to confuse yourself by setting to many unknowns. Eventually you should come down to solve some simultaneous linear algebraic equations. If you still cannot find the solution then just post another message and I will post a solution.