امتحان مرحله دوم فردا برگزار خواهد شد. در این قسمت می‌توانید بحث‌های خود در مورد سوالات و راه‌حل‌ها مطرح کنید

bashgah MASKHarsho daravorde 3 ta soal dadan 2 tash tekrari!!! farda chi mikhad beshe !!!!!!!!!!
yedoonasham badihi bud biain begin chi kar kardin??????

Soalhaye Marhale Dovvome Riazi Rooze Avval :
1-$ p$ yek chandjomleyie darajeye do ast ke be ezaye $ 0$ , $ 1$ va $ \minus{} 1$ ghadre motlaghe an az $ 1$ bishtar nist . neshan dahid baraye har $ \minus{} 1\leq x \leq 1$ ghadre motlaghe $ p$ az $ \frac {5}{4}$ bishtar nist .
2-yek baghe morabayi shekl ra be yek shabakeye $ 50$ dar $ 50$ az ghata'ate $ 1$ metr dar $ 1$ metr taghsim kardeyim va dar bazi az ghet'e ha yek derakhte sib , anar ya holoo kashteyim . midanim mojavere har derakhte anar , daste kam yek derakhte sib va mojavere har derakhte holoo dastekam yek derakhte anar va yek derakhte sib voojood darad . be alave mojavere har gheteyi ke dar an derakhti nist , az har se no derakht voojood darad (do ghete ra mojaver migooyim agar yek zel'e moshtarak dashte bashand) .
neshan dahid tedade ghata'ate khali az $ 1000$ bishtar nist .
3-farz konid nimsaze dakheylie zavieye $ A$ az mosalase $ ABC$ zel'e $ BC$ ra dar $ D$ va dayereye mohitie mosalas ra dar $ M$ ghat konad . az $ D$ khati rasm mikonim ke do nimkhate $ MB$ va $ MC$ (ba shooroo az $ M$) ra dar noghate $ P$ va $ Q$ ghat konad . sabet konid $ \angle PAQ \geq \angle A$

Bebinid be rahati mitunid sabet konid ke dayere mohiti $ ABD$ bar khat $ MD$ momase. Pas mishe did ke:
$ \angle APD\leq \angle ABD$
Be tarighe moshabeh:
$ \angle AQD\leq\angle ACD$

Va khob dar akhar ham natije mishe ke:

$ \angle PAQ\geq \angle A$

[quote="Omid Hatami"]Bebinid be rahati mitunid sabet konid ke dayere mohiti $ ABD$ bar khat $ MD$ momase. Pas mishe did ke:
$ \angle APD\geq \angle ABD$
Be tarighe moshabeh:
$ \angle AQD\geq\angle ACD$

Va khob dar akhar ham natije mishe ke:

$ \angle PAQ\leq \angle A$[/quote]
albate jaye [code]\geq[/code] bayad [code]\leq[/code] mizashtid too hame namosavi ha.

emsal be marateb sakht tar bud. akhe azmooni tasavor konin bedune HENDE3!!!!!!!!!!!

vali kaf ham beyne 23-28 !! :)

yani vaghan nazARet ine ke az parsal sakhttar bode????
 :o  :rotfl:  :rotfl:  :o

marhale 2 az emsal asoontar faghat sale 86 bood ke dg hame midoonan bashgah che gandi zade bud..!!
 khashi???? khabari niss azataaa dg yadet raft maro???
TST keye??? beterekooniaa !
darzEmn be Ramtin BEGOO age nare IOI ........        ......     .......!!!!!!(khodet befahm dg))

Rooze Dovvom :
4-$ n(n\plus{}2)$ sarbaze taze kar dar $ n$ sootoone barabar kenare ham va be faseleye yek ghadam istade and . ba farmane farmande , har sarbaz ya sare jayash mi istad ya be yeki az chahar jahat yek ghadam bar midarad . pas az jabejayi sarbazha dar $ n\plus{}2$ sootoone barabar , be shekle monazzam gharar gerefte and , be nahvi ke do satre avvalo akhar hazf va do sootoon be chap va rast ezafe shode ast . neshan dahid $ n$ zoj ast .
5-adade tabiyie $ a_{1}<a_{2}<\cdots<a_{n}$ daraye in khassiat hastand ke baraye har $ i$ va $ j$ motemayez :
$ a_{i}\minus{}a_{j}|a_{i}$ . neshand dahid baraye har $ i<j$ :
$ ia_{j}\leq ja_{i}$
6-$ 11$ nafar dore yek mize dayereyi shekl be shekle monazzam (!) nesheste and va $ 11$ kart ba shomarehaye $ 1$ ta $ 11$ beyne anha pakhsh shode ast , momken ast barkhi karti nadashte bashand va barkhi bish az yek kart dashte bashand . dar har marhale yek nafar mitavanad yeki az karthaye khod ra be farde mojaverash bedahad dar soorati ke agar shomareye an kart $ i$ bashad , ghabl va bad az in amal , makane se karte $ i\minus{}1$ , $ i$ va $ i\plus{}1$ tashkile yek mosalase hade nadahand ( manzoor az karte shomareye $ 0$ karte shomareye $ 11$ va manzoor az karte shomareye $ 12$ karte shomareye $ 1$ ast )
Farz konid dar ebteda karthaye $ 1$ ta $ 11$ be tartib dar jahate aghrabe haye saat , be afrad dade shode bashad . sabet konid hichgah kartha dar daste yek nafar jam nakhahad shod .

Be nazare man vaghean emtehane badi bood . vaghean comite che fekri karde $ 3$ ta tarkibiat gozashte too emtehan mage computere ?
Har kasi ke soalaye IMO e chand sale akhiro dide bashe mifahme ke IMO dare mire be samte inke tarkibiat bi tasir she va jabr tasiresh ziad she . tooye $ 7$,$ 8$ sale gozashte too IMO faghat ye sal tarkibiat $ 2$ ta boode . hala ma nemigim comite kheyli az khodesh ebtekar bezane o az in harfa , $ 2$ ta tarkibiat moshkeli nadare vali vaghean $ 3$ ta tarkibiat fajeast !

kaf beine 4,5 va 5e oonayi ke migan kaf payintare fekre ghabulie khodeshunan vagarna gheire soale 6 baghie soala sakht nabud vaghean baraye comtie (az bala ta paeen) moteassefam , manam ke bayad gajamo bekharam .

na sakht bood..!! alaki nagoo 4 5 ghabulie bishtar az 24 nemire emkan nadare soale 6 kheli sakht bood
.. bashgaham ke ozash kheli kharabe mese inkkee

vaghan bashgah gand zad ba in emtehanesh....rasman olamp comp gereft dg....hala age soalash tekrari nabood baz ye chizi!! har sal haghe ye edde zaye mishod,emsal sigmaye hameye salaye ghabl haghe mellat zaye shod!!!
aghaye hatami nazare shoma raje be kaf chie?

نمی‌دونم، بستگی داره

ولی حدس می‌زنم بین ۲۱ تا ۲۵ باشه

[color=#666666]Hi............[/color]
[size=142]
رياضي رو بايد براي خودش خوند نه براي المپياد و هرچي ديگه. البته درسته اينا هم در كنار اون هدف [color=darkred][u][b]ممكنه[/b][/u][/color] بدست بيان ديگه.
من با اين كه خيلي اميد داشتم و فقط براي المپياد(استعاره از رياضي) وقت گذاشته بودم و خيلي كار كرده بودم و براي مرحله هاي بالاتر مي خوندم ، اين آزمون (مرحله$ 2$) رو خيلي بد دادم و حالا ديگه مطمئن هستم كه قبول نميشم. الان هم دارم براي كنكور شروع ميكنم. ولي رياضي رو هيچ وقت ول نميكنم.
شايد اين يكي از بزرگترين و مهمترين شكست هاي زندگيم بوده باشه ولي مطمئن هستم كه ميتونم با تلاش و پشتكار اين رو بزرگترين  تجربه زندگيم كنم
اين نظر منه كه طي اين 10 روز خيلي فكرا (حتي ...) به سرم زده بود
[/size]موفق باشيد

[color=#666666]Bye.......[/color]


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Let $ \mathbf{A} \in \mathcal{M}_n(\mathbb{Z})$ have the property that, for every (some?) prime $ p$, $ \mathbf{A}^{p^n} \equiv \mathbf{A} \bmod p$.  Does it follow that $ \mathbf{A}$ is diagonalizable in $ \mathcal{M}_n(\mathbb{C})$?

No ideas?  Perhaps this should be moved to Number Theory.

[hide="Discussion"] The idea I had in mind is this:  let $ P(x)$ be the characteristic polynomial of $ \mathbf{A}$ and let $ p$ be a prime such that none of the irreducible factors of $ P$ ramify $ \bmod p$.  Then the roots of $ P$ generate a field extension $ \mathbb{F}_{p^k}$ of $ \mathbb{F}_p$ over which $ \mathbf{A}$ has a Jordan normal form.  I suspect that a Jordan block satisfies $ \mathbf{J}^{p^n} \equal{} \mathbf{J}$ over $ \mathbb{F}_{p^k}$ if and only if $ k | n$ and $ \mathbf{J}$ is $ 1 \times 1$, which means that the problem condition can be satisfied if and only if the characteristic polynomial of $ \mathbf{A}$ has irreducible factors with degree dividing $ n$ and its Jordan normal form consists of $ 1 \times 1$ blocks, i.e. it is diagonalizable.  But I haven't convinced myself about the Jordan blocks yet.  It might be necessary to choose $ p$ more carefully so that the Jordan normal form can actually be computed. [/hide]

Well, I was off at ARML. Not the best weekend to pose something interesting.

Let $ f$ be the minimal polynomial of $ A$ over $ \mathbb{Z}$ and $ f_p$ the irreducible polynomial over $ F_p$. We must of course have that $ f_p|f$ mod $ p$. Let $ d$ be the degree of $ f$.
Now, consider the $ d$ matrices $ I,A,A^2,\cdots, A^{d\minus{}1}$. These are linearly independent over $ \mathbb{Z}$, so if we rearrange them into a $ d\times n^2$ matrix, some $ d\times d$ submatrix has nonzero determinant $ k$. If $ P$ is a large enough prime, it doesn't divide $ k$; that makes the $ d$ matrices we started with linearly independent mod $ P.$ Therefore, $ f_P\equiv f\mod P$.
Now, a matrix in $ M_n(F)$ is diagonalizable over some extension of $ F$ if and only if its minimal polynomial i squarefree. If $ f$ is not squarefree and $ g^2|f$ for some $ g,$ then $ g^2|f$ mod $ P$ as well and $ f_p$ is not squarefree. Since $ x^{P^n}\minus{}x$ is squarefree mod $ P$, $ f_p(x)\nmid x^{P^n}\minus{}x\mod P$, and $ A^{P^n}\not\equiv A\mod P$.
That means that if $ A$ is not diagonalizable over $ \mathbb{C}$, we have $ A^{p^n}\equiv A\mod p$ for only finitely many primes $ p$. This is the contrapositive of what we wanted, so we're done.


[b]Recall that a poll was taken before you joined the campaign. The poll surveyed 500 people and found that 53% of them intended to vote for Coretta.

What does this poll mean?

Write a report for Coretta explaining how confident she should be that she was leading at the time of the poll. Explain your answer in terms of confidence intervals, standard deviation, and any other ideas you consider appropriate.[/b]

Here's the math for this problem... but I don't know how to explain for the whole process. Please help!

step one:
$\sigma = \sqrt{\frac{0.53(1-0.53)}{100}}= 0.022$

step two:
Confidence level = $95$ percent

$0.53+2(0.022) = 0.574$
$0.53-2(0.022) = 0.486$

step three:
$0.53-0.501 = 0.029$
$\frac{0.029}{0.022}= 1.3$ => $80.64$ percent

step four:
$80.64+9.68 = 90.32$ percent.

She is going to win.


How can one prove it when a field has unique prime factorization?

A field has always unique prime factorisation as ever nonzero element is a unit (and especially every element is product of primes [of zero primes ;) ] and a unit; and this obviously is unique as there aren't any primes).


function
limit
calculus
derivative
real analysis
real analysis theorems

Hi to all!

Our schoolbook says that if a function $ f$ is differentiable on the $ (a,b)$  $ x_{0}\in(a,b)$  and
$ i) f'(x_{0})=0$  $ ii)f'(x_{0})<0$, then $ maxf=f'(x_{0})$ , but it doesn't give any proove  :mad:

Can somebody give me a proof  :roll:  Thank you!

[quote="sylow_theory"]I think you mean $ f''(x_{0}) <0$ and its not a max, its a relative max, thats the correct term.

Is that what you mean?[/quote]

Yes that's what I mean!

All you have to do is show that $ f'$ changes its sign from $ +$ to $ -$ around $ x_{0}$.

Let $ f''(x_{0}) = \lim_{h \to 0}\frac{f'(x_{0}+h)-f'(x_{0})}{h}= \lim_{h \to 0}\frac{f'(x_{0}+h)}{h}= L < 0$.

Since the limit is negative, there exists a neighborhood of $ 0$, call it $ N^{*}_{1}(0; \delta) = (-\delta, 0) \cup (0, \delta)$, for some $ \delta > 0$, in which $ \frac{f'(x_{0}+h)}{h}< 0, \; \forall h \in N^{*}_{1}(0; \delta)$.

For $ h < 0$ in $ N^{*}_{1}(0; \delta)$ you get $ f'(x_{0}+h) > 0$, whereas for $ h > 0$ you get $ f'(x_{0}+h) < 0$.

Thus, $ f'$ changes its sign from $ +$ to $ -$ around $ x_{0}$, so $ f$ attains a relative maximum in $ x_{0}$ (by the first derivative test, which your book does prove).

[hide="gianni"]Μην ξεχνάς ότι το θεώρημα αυτό είναι εκτός ύλης για τις πανελλήνιες... Μην κάνεις το λάθος (που παραλίγο να κάνω εγώ φέτος) και το χρησιμοποιήσεις στις εξετάσεις[/hide]


trigonometry LaTeX geometry symmetry USAMTS AMC AIME

i failed

55554

unfortunately they took off for the silly misreading

Hmm...I also got 5 5 5 5 4 with no commended.

[quote="Twiz"]The scores are out!
What did you get?
I got a [b]5[/b] [b]5[/b] 5 [b]5[/b] 4 = 24[/quote]

3 commended? nice

I got a 25!!  :D  :D  :rotfl:  :D

what the heck...superdock [b]5 5 5 5 5[/b], nice job if you're reading this.

I got 5 5 5 [b]5[/b] 5.

My solution for #5:

Define a 1337 sequence to be a sequence with all terms negative, blah blah blah.

Grader comment: 1337 solution.  :rotfl:  :rotfl:

5 [b]5[/b] 5 5 5.

How did my #2 get commended... eh?

5 5 5 5 4

argh made a typo, lost 1 pt for that

wha...?
i was expecting at least a cople pts off on #5...but i got
55555

also:
i love my grader comments:
1: Nice work.
2: Good work
3: Good wo_k
4: Excellent work
5: Good work.

First of all, hello, everyone. This is my first post in any of the AoPS forums.

I got a 5-5-5-5-5 (with no commended solutions). I expected a to get a 4 on problem 2 due to a failure to simplify a trig expression. (I used Excel to simplify it, and I got an integer as an answer, and was low on time, as it was 19:00 EDT 09 October 2007 and I still had to tie up some loose ends in another problem, and put it in). In addition, I made two typos (accidentally swapping two angle measurements) on problem number 4, although, from my work, it might have been evident that it was indeed a typo. Most surprisingly of all, there was no mention of the typo in the comment for the problem, which simply stated, "Good".

5-5-5-5-5 with no commended...

Comments:
1.good
2.nice
3.nice casework
4.nice and concise
5.collect,but
PLEASE WRITE BIGGER!!( :rotfl: )

(They will have no trouble about small handwriting this round, because I'm going to use LaTeX,...)

5 5 5 5 [b]5[/b]

Comments:

1. Nice!
2. Good
3. (No comment -.-)
4. Good work
5. Good bounding argument

[b]5[/b] 4 4 5 [b]5[/b]

I should be more careful next time.

My scores: 5 5 2 3 2

I used my calculator for #3 and brute forced it because I didn't know much trigonometry...lol at relying on TI-84 so much.

"You've found solutions that work to 4 decimal places, but how do you know they are equal?"

#4- Suprised I only lost 2 points for assuming what my perpendicular bisectors reached.

#5- I did little more than show that - .5 works.

Cool...I was the one 9th grader to score 17pts.

[quote="cognos599"]first timer 8th grader

5531[b]5[/b]
so 19[/quote]
how did you get a point on 1
and your 5 looks unrigorous
lol

[b]5[/b] 5 5 5 2
:( I failed the last one.

lol i suck:
5 3 4 5 2 = 19

Was too lazy to show trig simplification in #2, missed detail in #3, and failed at #5.

Wow, I already got more commendeds than I did last yr.  :o

[b]5[/b] 5 [b]5[/b] [b]5[/b] 5

Interesting, the comment on my #4 said "nice use of modern geometry".

[quote="k8reindeer"]5 5 5 5 5
which is surprising because last year all of the problems were way over my head.
On another note, I'm not up on the leaderboard, and others with 25 are... is there some criterion for inclusion on the list that I don't know about?[/quote]
you probably didnt check that permission box or didnt send in the forms
something along those lines

[quote="azjps"]Interesting, the comment on my #4 said "nice use of modern geometry".[/quote]
I didn't grade that one myself, but looking at the solution you posted in this forum, I can understand the comment. You pivoted triangle PAB around point P. Classic geometry is about intersection and congruence; modern geometry is about symmetry and transformation. Both forms of geometry provide rigorous proofs, but modern geometry gives a better understanding of why the properties exist.

Erin Schram
USAMTS grader

I got a [b]5[/b] 5 4 5 2 = total 21

For number 5, I just didn't know how to prove that c=-0.5 was the only solution, and for number 3, I got a point off for showing that a<4 by looking at the case a=b=c.

How should I have showed a<4 or something like that?

I can show you my solution, which is really bad. (Attached)

I'm so mad though, because to qualify for AIME, it's usually 68 points, so I'll need 47 points in the next 2 rounds, which is 23/24, which is hard for me.

55442=20

-1 pt on 3 for not showing that a does not equal one=laziness
-1 on 4 for getting 133 instead of 47 =stupid
-3 on 5. "Graphs do not constitute a rigorous proof" I didn't really know what i was doing.

It's my first year doing this, so i am moderately satisfied with a 20

[quote="le00327146"]weird...i used a graph for my solution to #5 and i got a 5 on it...[/quote]

well then your solution was rigorous while the other person's wasn't.

Just because one graph solution doesn't work, doesn't mean all graph solutions don't work.

gah i suck. i didnt send it in. i tried for fun. and failed miserably. i probably would have gotten like a 10.

Wow, I got [b]5 5 5 5 5[/b].  :D

That was definitely surprising, I don't think I've ever had more than 1 commended solution in a round, and I didn't even expect to get 25/25 this round (I thought my #4 wasn't rigorous enough).

[quote="CatalystOfNostalgia"][quote="123456789"]For number 5, I just didn't know how to prove that c=-0.5 was the only solution, and for number 3, I got a point off for showing that a<4 by looking at the case a=b=c.

How should I have showed a<4 or something like that?
[/quote]

You have to show that a=b=c indeed maximizes it, that is, arctan 1/x > arctan 1/y if x<y.[/quote]

What do you mean by "a=b=c indeed maximizes it"?

[5][5]555

ahhh i need to like do the next problem set...

3 5 5 1 2 = 16

Which isn't that bad considering I did it in one day.  :lol:

I'm feeling much better about round 2.

5 5 5 5 5...
No commended...
I seems round 2 is a bit harder...

I thought Round 2 was easier...I hope I didn't mistake something.

It's a good 5 hours before you're allowed to say that.  :wink:

Actually we're not allowed to discuss it until Thursday ... so that would be another 52 hours.


inequalities
number theory open
number theory

$ 9\cdot 1089\equal{}9801$
my conjecture: there are no other numbers with this property (multiplying by 9 is the same as reversing digits)

am I right or not ?

if I write $ n\equal{}n_s10^s\plus{}\dots \plus{}n_1 10 \plus{}n_0$ I should have $ 9(n_s10^s\plus{}\dots \plus{}n_1 10 \plus{}n_0) \equal{} n_0 10^s\plus{}\dots \plus{}n_{s\minus{}1} 10 \plus{}n_s$
this immediately implies $ n_s\equal{}1$, $ n_0\equal{}9$, $ n_{s\minus{}1}\in\{0,1\}$.

I have no idea how to obtain $ s\equal{}3$, probably one should write some inequalities to eliminate bigger numbers ...............

$ 9 \cdot 10989\equal{}98901$
However, it does look as if the entire family (I think infinite) of this numbers can be classified.

[quote="bambaman"]$ 9 \cdot 10989 \equal{} 98901$
However, it does look as if the entire family (I think infinite) of this numbers can be classified.[/quote]
Yes ! You're right, thanks :)

for $ k\equal{}0,1,\dots$ let $ A_k\equal{}109\dots 989$ (there are $ k$ 9's in the centre)
then $ 9A_k$ is equal $ A_k$ with digits reversed  :)

so now modificated problem:
does this sequence contain all numbers having this property ?

nope:

108900001089*9 = 980100009801

how about this version: does any such number contain a digit other than 0, 1, 8, or 9?


algebra
function
domain
Support
Ring Theory
superior algebra
superior algebra unsolved

Does anyone know how to show that the nilradical is the intersection of all prime ideals?  Also, why is it that maximal ideals are prime?  Does this just follow by definition of maximal ideals.  Thanks for any help that anyone can provide

Clearly every nilpotent element belongs to every prime ideal,now assume $u$  is a non nilpotent element which belongs to every prime ideal,Now consider the set $Z$which contatins all ideals of the ring under consideration which contains no power of $u$,$Z$ is non empty as $<0>$ belongs to $Z$ by our assumption.Now this set can be given a partial order(inclusion) and then we can see its inductive too,so we apply zorn's lemma and get a maximal element $P$.Now we try to show $P$ is prime and hence a contradiction.To show this let $x,y$ be two elements of the rings not belonging to $P$,now $P +<x>,P+<y>$ strictly contain P and hence contains powers of $u$ each,now if
$xy \in P$ then this implies some power of $u$ exists in $P$ a contradiction.Hence $P$ is prime and it does'nt contain $u$,so $u$ cannot be in the intersection of all prime ideals if its non nilpotent.

Regarding maximal ideal,if $M$ is maximal in $R$,then $R/M$ is a field and hence a domain and so $M$ is prime. :)

Hi Thalia,

in another topic of you I've already given you a page with many standard proofs like this one:

http://www.mathreference.com/id-rad,intro.html

You can look almost all proofs up by yourself, best after having tried to solve the problem by yourself. (But this board does not seem to support this idea ;-))


There are 3 pipes connecting a large water tank to a pool. each of the pipes fills the pool at a uniform rate but the rates of different pipes are different. To fill the empty pool, one can open the first pipe for 2 hours and then the second pipe for 1 hour. Also using the second pipe for 2 hours and the third pipe for 1 hour fills the empty pool. If the pool is empty and the first pipe was open for 80 minutes, then the second pipe was open for 80 minutes, for how long does the third pipe need to be open after that to fill the pool?

The first step in solving a problem like this is to interpret and define the problem.  Knowing that rate x time = work done or whatever you want to call filling the pool:

x = rate of pipe 1
y = rate of pipe 2
z = rate of pipe 3

$2x + y = 1$

$2y + z = 1$

$\frac{4x}{3} + \frac{4y}{3} + hz = 1$

We want to solve for $h$.  Note that the time is measured in hours.  I'll leave the rest for you to solve.

[hide="hint"]substitute until you have an equation with only z as a variable.  then solve.[/hide]

[hide="answer"]$h = \frac{1}{3}$ or 20 minutes.[/hide]

[hide="Answer"]To show the substitutions:

$2x+y=1 \Rightarrow x=\frac{1-y}{2}$
$2y+z=1 \Rightarrow y=\frac{1-z}{2} \Rightarrow x=\frac{1-\frac{1-z}{2}}{2}=\frac{1}{4}+\frac{z}{4}$
$\frac{4x}{3}+\frac{4y}{3}+hz=1 \Rightarrow \frac{1}{3}+\frac{z}{3}+\frac{2}{3}-\frac{2z}{3}+hz=1 \Rightarrow z(h-\frac{1}{3})=0 \Rightarrow h=\frac{1}{3}$[/hide]


For $ a,b,x,y$ positive reals and $ k$ positive real number greater than $ 2$ prove that

$ \frac {a^k}{x} \plus{}\frac{b^k}{y}  \ge 2^{2\minus{}k}\frac {(a\plus{}b)^k}{x\plus{}y}$

P.S. For $ k$ positive number greater than $ 2$ it is a simple application of Holder.

Enough to prove for
$ b = y = 1$
The inequalities equivalent :
$ \frac {a^k}{x} + 1\geq 2^{2-k}\frac {(a + 1)^k}{x + 1}$
${ \Longleftrightarrow (x + 1)(\frac {a^k}{x} + 1}\geq 2^{2-k}(a + 1)^k$
Apply AM-GM :
$ (x + 1)(\frac {a^k}{x} + 1)\geq (a^{\frac {k}{2} + 1)^2}$
Because $ k > 2$
$ a^{\frac {k}{2} + 1\geq 2^{\frac {2 - k}{2}(a + 1)^{\frac {k}{2}}}}$
This inequalities claim.
Apply for
$ x = \frac {x}{y},a = \frac {a}{b}$

[quote="TTsphn"]Enough to prove for
$ b = y = 1$
The inequalities equivalent :
$ \frac {a^k}{x} + 1\geq 2^{2 - k}\frac {(a + 1)^k}{x + 1}$
${ \Longleftrightarrow (x + 1)(\frac {a^k}{x} + 1}\geq 2^{2 - k}(a + 1)^k$
Apply AM-GM :
$ (x + 1)(\frac {a^k}{x} + 1)\geq (a^{\frac {k}{2} + 1)^2}$
Because $ k > 2$
$ a^{\frac {k}{2} + 1\geq 2^{\frac {2 - k}{2}(a + 1)^{\frac {k}{2}}}}$
This inequalities claim.
Apply for
$ x = \frac {x}{y},a = \frac {a}{b}$[/quote]

I don't understand your proof  :wink:


algebra
polynomial
sum of roots
product of roots
algebra proposed

Find all numbers $ a$ for which the equation $ a 3^x\plus{}3^{\minus{}x}\equal{}3$ has a unique solution $ x$.

[quote="moldovan"]Find all numbers $ a$ for which the equation $ a 3^x \plus{} 3^{ \minus{} x} \equal{} 3$ has a unique solution $ x$.[/quote]

Let $ y\equal{}3^x$. Clearly for all $ y>0$ we can find corresponding $ x$.

The equation becomes $ ay^2\minus{}3y\plus{}1\equal{}0$ and the question becomes to find $ a$ such that there is a unique positive solution for $ y$.

For $ a>0$, consider that sum of roots $ >0$ and product of roots $ >0$, the equation either has no real roots or double positive real roots or distinct positive real roots.
Double positive real roots $ \Leftrightarrow Discriminant\equal{}(\minus{}3)^2\minus{}4a\equal{}0$
$ \Leftrightarrow a\equal{}\frac{9}{4}$

For $ a<0$, similarly the equation has either no real roots or only one positive real roots.
Discriminant $ \equal{}9\minus{}4a \ge 0$ is true for all negative real $ a$

$ a\equal{}0$ gives $ y\equal{}\frac{1}{3} \Rightarrow x\equal{}\minus{}1$ as unique solution.

So $ a\equal{}\frac{9}{4}$ or $ a \le 0$ is the answer.


function
algebra
functional equation
algebra proposed

If $ f(f(m)) \equal{} \minus{} m$ for $ m\in\mathbb{Z}$ then Prove that when the function $ f: \mathbb{Z}\rightarrow \mathbb{Z}$ is bijectiv and odd.

[i]sorry for english[/i]

Attempted restatement:

[quote]Let $ f: \mathbb{Z} \to \mathbb{Z}$ be such that $ f(f(m)) \equal{} \minus{}m$ for all $ m \in \mathbb{Z}$. Show that $ f$ is bijective and odd.[/quote]

[hide]Surjective is obvious, since $ f$ must output $ \minus{}m$ for an arbitrary integer $ m$.
Injective: Suppose $ f(a)\equal{}f(b)$. Then $ \minus{}a \equal{} f(f(a)) \equal{} f(f(b)) \equal{} \minus{}b$, so $ a\equal{}b$.
Odd: If $ f(a) \equal{} b$, then $ f(b) \equal{} \minus{}a$, $ f(\minus{}a) \equal{} \minus{}b$, $ f(\minus{}b) \equal{} a$. Note if $ a \equal{} 0$, $ b$ must be $ 0$ for $ f$ not to output multiple values at one point. Since $ a,b$ are arbitrary nonzero integers and $ f(0) \equal{} 0$, this shows $ f$ is odd.[/hide]

[quote="MellowMelon"][quote]Let $ f: \mathbb{Z} \to \mathbb{Z}$ be such that $ f(f(m)) \equal{} \minus{} m$ for all $ m \in \mathbb{Z}$. Show that $ f$ is bijective and odd.[/quote]

[hide]Surjective is obvious, since $ f$ must output $ \minus{} m$ for an arbitrary integer $ m$.
Injective: Suppose $ f(a) \equal{} f(b)$. Then $ \minus{} a \equal{} f(f(a)) \equal{} f(f(b)) \equal{} \minus{} b$, so $ a \equal{} b$.
Odd: If $ f(a) \equal{} b$, then $ f(b) \equal{} \minus{} a$, $ f( \minus{} a) \equal{} \minus{} b$, $ f( \minus{} b) \equal{} a$. Note if $ a \equal{} 0$, $ b$ must be $ 0$ for $ f$ not to output multiple values at one point. Since $ a,b$ are arbitrary nonzero integers and $ f(0) \equal{} 0$, this shows $ f$ is odd.[/hide][/quote]

Sure.
And I think this problem must not be in olympiad section.

A little bit more complex : give a general solution of this functional equation from  $ \mathbb{Z} \to \mathbb{Z}$

Is my solution correct ?
$ \forall y\in Z$ there exist an integer $ x$ such that $ f(x)\equal{}y$. From the function $ f(f(m))\equal{}\minus{}m \Rightarrow f(f(x))\equal{}\minus{}x \Rightarrow f(y)\equal{}\minus{}x$.This tells us that for every integer $ \minus{}x$ there exist an integer $ y$ such that $ f(y)\equal{}\minus{}x$ ,thus the function is surjective.

I'm quite new in these ideas.

[quote="enndb0x"]Is my solution correct ?
$ \forall y\in Z$ there exist an integer $ x$ such that $ f(x) \equal{} y$. From the function $ f(f(m)) \equal{} \minus{} m \Rightarrow f(f(x)) \equal{} \minus{} x \Rightarrow f(y) \equal{} \minus{} x$.This tells us that for every integer $ \minus{} x$ there exist an integer $ y$ such that $ f(y) \equal{} \minus{} x$ ,thus the function is surjective.

I'm quite new in these ideas.[/quote]

No, this demonstration of sujectivity is false because you begin your demo with "$ \forall y\in Z$ there exist an integer $ x$ such that $ f(x) \equal{} y$" whis means $ f(x)$ is surjetive.

Look at MellowMellon's demo.

The simplest demo of surjectivity is :
$ \forall x\in\mathbb{Z}$, $ f(f(\minus{}x))\equal{}x$ and so $ \exists y$ (take $ y\equal{}f(\minus{}x)$) such that $ f(y)\equal{}x$   Q.E.D.


linear algebra
matrix
algebra
polynomial
absolute value
linear algebra unsolved

Let $E,F\in M_n(C).$Show that there exists $H\in M_n(C)$ such that $EH=HF$ if and only if $E$ and $F$ have a common eigenvalue.

First of all, the matrices $H$ in the problem should be non-zero.

If we can find such a amtrix $H$, then we can multiply it with some scalar s.t. the largest absolute value of an entry in the matrix is $1$ (we will call such matrices normalized). This means that it suffices to prove the implication from right to left for diagonalizable matrices $F$. This is because given $E,F$ which share a common eigenvalue $\lambda$, we can find a sequence $F_n\to F$ of diagonalizable matrices, all of them having $\lambda$ as an eigenvalue. For each $F$, we can find a normalized matrix $H_n$ s.t. $EH_n=H_nF_n$, and we can extract a convergent subsequence from $(H_n)_n$, which tends to some normalized $H$ which will satisfy $EH=HF$, as desired.

Now, for a diagonalizable matrix $F$, let $(e_i)_i$ be a basis of $\mathbb C^n$ formed by eigenvectors of $F$ (the eigenvalue $\lambda_i$ corresponds to $e_i$, for $i=\overline{1,n}$). Suppose $E$ has $\lambda=\lambda_1$ as an eigenvalue, and that $Ey=\lambda y,\ y\ne 0$. Let $H$ be the matrix defined by $He_i=y$ if $i=1$, and $He_i=0$ if $i>1$. Then $(EH)e_i=(HF)e_i,\ \forall i\in\overline{1,n}$, so $EH=HF$.

Now for the converse, suppose we can find a non-zero matrix $H$ s.t. $EH=HF$. For every $m$, we get $E^mH=HF^m$. Let $\mathcal P$ be the characteristic polynomial of $F$. We have $\mathcal P(E)H=H\mathcal P(F)=0$, and, since $H\ne 0$, $\mathcal P(E)$ must be singular.

Probably, we assume that $H \neq 0$ otherwise this is false.

Let's have a go wth this condition added :

If $EH = HF$ (with $H \neq 0$) then for all polynomial $P$ it's easy to check that $P(E)H = HP(F)$
Taking $P$ as caracteristic polynomial of (for instance) $F$ prove that $P(E)$ is not invertible. Hence $E$ and $F$ have a common eigenvalue.

If $E$ and $F$ have a common eigenvalue, so have $E$ and $F^T$. Then $U \times V^T$ is not null ($U$ and $V$ being eigenvectors of $E$ and $F$) and is a common eigenvector of $E$ and $F$.

I think it's correct.

Edited : fixed typo and clarify a bit


The graph of $p(x) = x^{3}+ax^{2}+5x+6$ intersects the x-axis at three distinct points M, N, and P, where N is the midpoint of MP. Find the real number $a$. (No Calculator)

[hide]
If we let the roots be $m,n,p$ then $n=\frac{m+p}{2}$ so $3n=m+n+p=-a$. Then $a=-3n$, evaluate $P(n)$...
\[0=n^{3}-3n^{3}+5n+6\Rightarrow 0=2n^{3}-5n-6=(n-2)(2n^{2}+n+3)\]
The quadratic $2x^{2}+x+3$ has no real root, so $n=2$ and $a=\boxed{-6}$.
[/hide]

the roots are in an arithmetic sequence....so if you say the roots are p-q,p,p+q...once you see that, this problem is pretty standard...i would say though...good for ictm haha


AMC
USA(J)MO
USAMO
summer program
PROMYS
MATHCOUNTS
geometry

Congrats to everybody who qualified:

http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2008-ua/2008usamoqual.shtml#il

Justin Kopinsky  	12  	Adlai E Stevenson HS  	Buffalo Grove  	IL
Zach Nahlik 	12 	Edwardsville SHS 	Glen Carbon 	IL
Ari Tenzer 	11 	Glenbrook North HS 	Northbrook 	IL
Michael Tarczon 	11 	Glenbrook South HS 	Glenview 	IL
Je Ok Choi 	12 	IL Math & Sci Academy 	Chicago 	IL
Jennifer Iglesias 	12 	IL Math & Sci Academy 	Aurora 	IL
Vladik Kontsevoi 	10 	IL Math & Sci Academy 	Evanston 	IL
Yize Leng 	11 	IL Math & Sci Academy 	Palatine 	IL
Iris Xu 	10 	IL Math & Sci Academy 	Vernon Hills 	IL
Kiaran Dave 	11 	Naperville Central HS 	Naperville 	IL
Kevin Hu 	9 	Naperville Central HS 	Naperville 	IL
Keshav Puranmalka 	12 	Naperville Central HS 	Naperville 	IL
Katherine Rudolph 	10 	Naperville Central HS 	Naperville 	IL
Jiefu Wang 	12 	Naperville North HS 	Lisle 	IL
Alexander Anderson 	12 	New Trier Township HS 	Wilmette 	IL
Arjun Puranik 	10 	North Central College 	Palatine 	IL
Eric Hallman 	10 	Oak Park/River Forest HS 	Oak Park 	IL
Robin Jia 	10 	Oak Park/River Forest HS 	Oak Park 	IL
Sooeun Kim 	11 	Saint Viator HS 	Schaumburg 	IL
Alex Zhai 	12 	University Laboratory HS 	Champaign 	IL
Joseph Kornick 	12 	Walter Payton Coll Prep 	Chicago 	IL
Matthew Waechter 	10 	Walter Payton Coll Prep 	Chicago 	IL
Daping Weng 	12 	Walter Payton Coll Prep 	Chicago 	IL
Alex Zorn 	12 	Walter Payton Coll Prep 	Chicago 	IL
Gregory Gauthier 	12 	Wheaton North HS 	Wheaton 	IL

Yay for going to a school starting with "A"! My Mom thought I got 1st in the state for a while :P

Anyway I can't help but wonder what Gauthier's index is (I don't believe he's an AoPSer)... is he as godly at high level math as he is at schol bowl?

Hello guys... don't forget to study for USAMO. Well, it might be kind of late for that now, but you can still do a few problems and just do some preliminary thinking. It will help you maintain focus and confidence during the exam. We need more MOPpers from IL! It seems that we used to have pretty good representation, but we are in danger of having no MOPpers next year  :o

Also, what is Iris Xu doing in IL?

[quote]we are in danger of having no MOPpers next year Shocked [/quote]
Eh we have a sophomore on ChiARML I would put in a good bet for... he got the high score on our try-outs and went to PROMYS last year, haven't actually seen him in action on USAMO-type questions though.

[quote="kops723"] is he as godly at high level math as he is at schol bowl?[/quote]

...Don't remind me... :P

Seems like there's no qualifiers within 25 minutes of me (including myself). Always next year!

[quote="stuck"][quote="kops723"] is he as godly at high level math as he is at schol bowl?[/quote]

...Don't remind me... :P
[/quote]

What school do you go to, out of curiosity?

And hey, at least we had the pleasure of placing higher than Wheaton North at IHSA finals :D Go lucky pool!

[quote="stuck"]Seems like there's no qualifiers within 25 minutes of me (including myself).[/quote]

 :o Seems like there is a lot of corn within 25 minutes of me, but not much else.

[quote="probability1.01"]:o Seems like there is a lot of corn within 25 minutes of me, but not much else.[/quote]

tru dat

[quote]
What school do you go to, out of curiosity?
[/quote]

A modest school in the Fox Valley Conference called Cary-Grove High School.

I wasn't actually there when we played Wheaton North, (I don't think much of our varsity team was), but the brutality with which we got beaten was legendary.

[quote="kops723"]Anyway I can't help but wonder what Gauthier's index is (I don't believe he's an AoPSer)... is he as godly at high level math as he is at schol bowl?[/quote]

Yep. I went to national mathcounts with him our 8th grade year (the year he and IL won first place). I finished in the top 15 overall and he was lightyears ahead of me.

Yeah but mathcounts isn't high level.. what I suspect is that he put all his HS energies into schol bowl rather than math and so isn't quite MOP material.. but eh who knows?

[quote="Aryth"]i think my life will be fulfilled if i make mop  :P[/quote]

congratulations on fulfilling your life....freshman.

Did you just relax?
I know that if I make it in 8th grade then I would just chill do the best I can and just have fun. Is that what you did?

Well I kinda got owned by the problems, but there were a lot of pity points.
Strangely, I did not get the mode score of 0.
But, if you know what you don't know, which was for me pretty much everything non-mathcountsy, then yeah, you'll relax and try to do well, but if you don't, hey, no biggie.

Do you know how you did?
Was it different as a freshman?

i got a 4 my eighth grade year.
yes, it was much different freshman year, because it's your only shot at making red, unless you're a girl or get a sex change (actually I don't know if the latter will really help you)
it's a lot more pressure, and since usually you know a LOT more, it's a very different experience.

yes.

List of MOPpers
http://www.unl.edu/amc/a-activities/a6-mosp/a6-1-mosparchives/2008-ma/08mosppart.shtml

2008 AMC 10/12 IL statistics
http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2008-12a/StateStats/IL.HTML

More IL statistics
http://www.unl.edu/amc//e-exams/e5-amc10/archive10.shtml

So you probably wouldnt have made Red Mop in 8th grade even if it was possible

uh, definitely not. cutoff was 9. i was too mathcountsy in 8th grade, hence my warnings to you guys.

I really wanna try and do BOTH for 8th grade:

do well at least not get a 0 on USAMO, or perhaps make red MOP
do well at Mathcounts Nats

although it culd be tough :maybe:

yes, it would be extremely tough to make red mop as an 8th grader.

But not impossible if 1) you're a girl or 2) you make blue mop and then demote yourself.

I'm a girl though.

so it may not be too tough?

haha, for me, getting a sex change is easier than making blue and demoting. although if i do make blue, the demoting part won't be too bad...  :roll:


geometry
incenter
rectangle
circumcircle
geometry unsolved

Let ABC be a triangle with incenter I. Fix a line a tangent to the incircle of ABC (not BC, CA or AB). Let A0;B0;C0 be points on a  such that: <AIA0=<BIB0=<CIC0=90.
Prove that AA0,BB0,CC0 are concurrent.

Incircle is tangent to $ BC, CA, AB$ at $ D, E, F.$ Fixed tangent touches $ (I)$ at $ T.$ $ EF, FD, DE$ are polars of $ A, B, C$ WRT $ (I).$ Perpendiculars to $ EF, FD, DE$ from $ T$ are polars of $ A_0, B_0, C_0.$ Let $ X \in EF, Y \in FD, Z \in DE$ be feet of the these polars of $ A_0, B_0, C_0.$ Then $ AA_0, BB_0, CC_0$ are polars of $ X, Y, Z$ WRT $ (I).$ But $ XYZ$ is Simson line of $ \triangle DEF$ with the pole $ T,$ points $ X, Y, Z$ are collinear, their polars $ AA_0, BB_0, CC_0$ are concurrent.

can anyone post a solution not using polar,pole?

It is variation of the same thing:

A', B', C' are midpoints of EF, FD, DE. Perpendiculars to EF, FD, DE from T cut EF, FD, DE at X, Y, Z and $ IA_0, IB_0, IC_0$ at $ A_0', B_0', C_0'.$ $ IA'XA_0', IB'YB_0', IC'ZC_0'$ are rectangles, their circumcircles (P), (Q), (R) have diameters IX, IY, IZ. XYZ is Simson line of $ \triangle DEF$ => X, Y, Z are collinear => circle centers P, Q, R are collinear => circles (P), (Q), (R) intersecting at I are coaxal, they intersect at one other point J. Inversion in (I) takes A, B, C to A', B', C' and $ A_0, B_0, C_0$ to $ A_0', B_0', C_0'$ => it takes lines $ AA_0, BB_0, CC_0$ to circles $ (P) \equiv (IA'A_0'), (Q) \equiv (IB'B_0'), (R) \equiv (IC'C_0')$ concurrent at the inversion center I and at another point J => The lines $ AA_0, BB_0, CC_0$ are concurrent at the inversion image J' of J.


analytic geometry
induction
combinatorics solved
combinatorics

There are 4 frogs positioned in the vertices of a square. At any given time a frog can jump over th centroid of the triangle formed by the other 3 frogs s.t. the centroid is the midpoint of the segm formed by the current and the future positions ofthe frog. Can 2 frogs ever be on top of one another?

It's been killing me for quite some time now :) Please help!

The answer is no.

Assume that no frog make two successive jumps, since we it will return at its original position...Label, the frogs 1,2,3,4, and let P(i,n) = (x(i,n),y(i,n)) be the coordinates of the frog ni after the n-th jump.
Wlog, we have P(1,0) = (1,1), P(2,0) = (1,2), P(3,0) = (2,1) and P(4,0) = (2,2).

Suppose that jump n+1 is made by frog n4. Thus :
x(4,n+1) = [2*x(1,n) + 2*x(2,n) + 2*x(3,n) - 3*x(4,n)]/3
y(4,n+1) = [2*y(1,n) + 2*y(2,n) + 2*y(3,n) - 3*y(4,n)]/3.

Now, we prove by induction on n  \geq  1 that after the n-th jump the frog which has jumped have coordinates of the form k/3 n where k is an integer not divisible by 3, and the other frogs have coordinates of the form r/3 n-1 for some integers r.
Note that the above claim follows directly from this result.

For n = 1, we may assume that it is the frog n 4 which has jumped. From the above formula, we easily deduce that :
x(4,1) = 2/3 and y(4,1) = 2/3. Hence result.

Now let's suppose the result holds for n.
Wlog, we assume the frog n1 has made the n-th jump and that frog n4 the (n+1)-th jump.
Thus, from the induction hypothesis, x(1,n) = a/3 n  and x(1,n) = b/3 n  where a,b are integers not divisible by 3, whereas x(i,n) and y(i,n) are of the form r/3 n-1 for i = 2,3,4. It follows that x(i,n+1) and y(i,n+1) have the desired form form i = 1,2,3.
Moreover 2x(2,n)+ 2x(3,n) - 3x(4,n) has the form k/3 n-1  for some integer k. Then x(4,n+1) = 2a/3 n  + k/3 n = (2a+3k)/3 n  where 2a+3k is not divisible by 3. Thus, x(4,n+1) has the desired form.
We prove in a similar way that y(4,n+1) has the desired form, and we are done.

Pierre.


One mathematical tournament have 8 problems. One player is called POWER if he solve more than half of problems. One problem is called HARD if is solved by less than half of POWER players.

1) Find the maximum number of HARD problems.

2) In 1) find the minimum even number of players and the maximum odd number of players.

For the first part, the maximal number is $5$, I think. We may assume, for the sake of simplicity, that all the solvers are strong (I prefer this to POWER :)). Count the pairs $(x,p)$, where $x$ is a solver and $p$ is a problem he has solved. If we have $n$ solvers, we get at least $5n$, because each one solves at least $5$ problems. on the other hand, counting, for each problem, the number of solvers who have solved it, we get $n_1+\ldots+n_8$. If $6$ of the $n_i$'s are $<\frac n2$, then the sum is $<2n+6\cdot\frac n2=5n$ (because each $n_i\le n$), but it must be at least $5n$, so we have a contradiction.

In order to find an example for $5$, we may take $5$ solvers, $a,b,c,d,e$. $a$ solves $1,2,6,7,8$, $b$ solves $2,3,6,7,8$, $c$ solves $3,4,6,7,8$, $d$ solves $4,5,6,7,8$, and $e$ solves $5,1,6,7,8$. The hard problems are $1,2,3,4,5$, each one being solved by $2<\frac 52$ solvers.

In 2) is easy to see that 10 is the minimum even number of POWER players when the number of HARD problems is 5.  :D

but, what the maximum odd number of POWER players?  :(

The second part is hard.. one idea please  :(


linear algebra
matrix
linear algebra unsolved

For every eigenvalue $\lambda$ of $A$ (complex matrix), $\lambda\overline{\lambda}$ is an eigenvalue of $AA^{H}$ ($A^{H}$: transpose and conjugate).

Does every eigenvalue of $AA^{H}$ arise in such a way from an eigenvalue of $A$?

No, and the original statement isn't true either.
Consider $A=\begin{bmatrix}1&5\\0&2\end{bmatrix}$. Its eigenvalues are 1 and 2. $AA^{*}=\begin{bmatrix}26&10\\10&4\end{bmatrix}$, which has eigenvalues $15\pm\sqrt{221}$.


[quote="Timothy"]A cube is sliced by a plane which goes through two opposite corners and the midpoints of two opposing edges.  If the cube has edge length one unit, how many square units are in the area of the rhombus formed by the intersection of the plane and the cube?[/quote]

[hide]If we find the area of the pyramid that was chopped off, we can just subtract that from 1 to get the trapezoid's volume.

$1-\frac{1}{4}$

$\frac{3}{4}$[/hide]


Could someone explain to me what exactly a mass spectrum is and then answer the following question?

The mass spectrum of dihydrogen is taken under conditions that prevent decomposition into H atoms. The two naturally occurring isotopes of hydrogen are H-1 (mass = 1.00783 amu; abundance = 99.985 percent) and H-2 (mass = 2.01411 amu; abundance = 0.015 percent)

a) how many peaks will the mass spectrum have?

b) give the relative atomic masses of each of these peaks

c) Which peak will be the largest and which will be the smallest?

a) 3 peaks
b) First peak     -  m/e=  2.01566
    Second peak - m/e=  3.02194
    Third peak    - m/e=  4.02822
c)Largest : first peak
   Smallest:Third peak

A mass spectrometer is a pretty big machine composed of a few parts. (the one I have used was GCMS, and it was composed of two parts of 1 cubic meter and 1/2 cubic meter. Not too big.
1. A sample is bombarded with electrons in an effort to split the molecules into smaller, charged parts.
2. The remaining sample is accelerated through a voltage gradient.
3. The sample hits a detector, which can indicate the amount of sample that hits given a certain mass.
4. You get a chart of the number of particles vs. the mass (amu). This is your analysis.

The idea is the atomic mass of the particle and various distinct pieces of the particle can be distinguished. This can be taken from principles of physics (acceleration is related to mass and voltage). The peak with the highest number of "hits" (P) usually corresponds to the molecular mass of the compound in question. Higher peaks P+1 and P+2 signify the heavier isotopes of the atoms that constitute the molecule. Lower peaks refer to different alkyl radicals. Usually, stable radicals are more often shown.


1.A block of mass  2kg is dropped from a height 40m,on spring whose spring constant

k=1960N/m.what will be the maximum distance 'x' of compression of the spring

a)1cm
b)1.5cm
c)100cm
d)10cm

[quote="bkunchanapalli"][quote="quantum leap"][hide]The initial potential energy of the block is $U=mgh$.  By the time it reaches the spring, all the potential energy has been converted to kinetic.  For springs, $U=\frac{1}{2}kx^2$ so set the kinetic energy of the block equal to the potential in the spring, because when the block finally stops again it's kinetic energy will be again completely converted to potential.[/hide][/quote]

than s for the idea[/quote]

i am getting different answer can you please solve it

The solution is
$\large mg(h+x)=\frac{1}{2}kx^2$
from this equation, you can get $x$
Try it!

[quote="perfect_radio"]So, what's the correct answer??[/quote]

i did what phucnv wrote but didnt get the answer in the options

but why should we take (h+x)


[quote="nayel"]Let $p,q$ be primes such that $8\not|\ (p-1)(q-1)$. Prove that if $x^{2}\equiv p\pmod q$ has no solutions, then $x^{2}\equiv q\pmod p$ has two solutions.[/quote]
Let say $p,q$ are odd primes. I think the special cases when $p,q$ one of them is even is simple. So either $p,q$ has form $4k+1$ or $4k+3$. Note that neither of $p,q$ can have form of $4k+1$ for that would imply that $8| (p-1)(q-1)$. So both $p,q$ are of the form $4k+3$. Now by Quadradic Reciprocity $(p/q) =-(q/p)$ so if $x^{2}\equiv p (\bmod q)$ has no solutiuon means $(p/q)=-1$ and so $(q/p)=1$ which means that $x^{2}\equiv q(\bmod p)$ has two solutions (i.e. iif $x_{0}$ is solution then $p-x_{0}$ is another incongruent solution).

Err...this is the quadratic reciprocity theorem. So it does not really suffice to cite that theorem. A quick google search will yield a proof for this theorem.


modular arithmetic
number theory
relatively prime
number theory unsolved

$ x^4 \plus{} 5y^4 \equal{} z^2$ has no "natural number" solution

some cumbersome procedure makes following equation

$ \pm x^2 \equal{} 5^{4s \plus{} 1}c^{4}2^{2\alpha} \minus{} d^{4}2^{2\beta}$
where
$ c, d$ => odd, relatively prime with 5, relatively prime each other, positive
$ \alpha \equal{} odd, \;  \; \beta \equal{} 0$ or
$ \alpha \equal{} 0, \;  \; \beta \equal{} odd$
$ s \geq 0$

[quote="Aravind Srinivas L"]Just work out using modulos of certain pimes.[/quote]
Can't find such prime.

[quote="t0rajir0u"]This equation clearly has nontrivial solutions $ (\pm 1, 0, \pm 1)$ modulo every prime, so that's not going to cut it.  The standard elementary proof for the similar Diophantine $ x^4 \plus{} y^4 \equal{} z^2$ is by descent (which proves the $ n \equal{} 4$ case of Fermat's Last Theorem), which might work here.[/quote]

I said that the natrural solution(positive integer x, y, z to the $ x^4\plus{}5y^4\equal{}z^2$) does not exist!
I know the succession of $ x^4 \plus{} y^4 \equal{} z^2$
There's similart succession to the following
$ x^4 \minus{} y^4 \equal{} z^2$
$ x^4 \minus{} y^4 \equal{} 2z^2$
$ x^4 \minus{}4y^4 \equal{} z^2$
$ x^4 \plus{}2y^4 \equal{} z^2$

As you know, $ x^4 \plus{} y^4 \equal{} z^2$ will result to similart but small natural number solution by base equation($ x^2 \plus{}y^2 \equal{} z^2$). but $ x^4 \plus{}5y^4 \equal{} z^2$'s base equation($ x^2 \plus{}5y^2 \equal{} z^2$) has somewhat cumbersome solution) so I stuck in above equation($ \pm x^2 \equal{} 5^{4s \plus{} 1}c^{4}2^{2\alpha} \minus{} d^{4}2^{2\beta}$)...

[quote="cosinator"]Looking at the equation $ \pmod{16}$ works well.[/quote]

You find positive integer solution???

No, no.  Disregard what I posted.  I wasn't thinking the right way at the time.

[quote="JBrere"]Umm...

x=1
y=2,
z=9[/quote]

Oh, it's really such a so... Problem is wrong.
I must have such a result from my manipulated equation($ \pm x^2 \equal{} 5^{4s \plus{} 1}c^{4}2^{2\alpha} \minus{} d^{4}2^{2\beta}$) already.
but I don't have general solution not yet.

Anyway, thanks for helps!


Cv(mix)=$ \frac{n1Cv1\plus{}n2Cv2}{n1\plus{}n2}$
Cv(mix)=$ \frac{R}{\gamma(mix)\minus{}1}$

solving,
Y(mix)=$ \frac{(n1\plus{}n2)(Y2\minus{}1)(Y1\minus{}1)}{n1(Y2\minus{}1)\plus{}n2(Y1\minus{}1)}$+1

i hope i m not wrong!!! :)

Yes we have to do by finding Cv and Cp of mix seperately.

Sometimes students just use n1gamma1 + n2gamma2/n1 + n2 which is ABSOLUTELY wrong since gamma is a ratio and hence cant be used here. Thats what my teacher told and I didnt guess myself.  :oops:  :roll:

[quote="rituraj007"]Cv(mix)=$ \frac {n1Cv1 \plus{} n2Cv2}{n1 \plus{} n2}$
Cv(mix)=$ \frac {R}{\gamma(mix) \minus{} 1}$

solving,
Y(mix)=$ \frac {(n1 \plus{} n2)(Y2 \minus{} 1)(Y1 \minus{} 1)}{n1(Y2 \minus{} 1) \plus{} n2(Y1 \minus{} 1)}$+1

i hope i m not wrong!!! :)[/quote]

Excellent! Are you going to any JEE classes? I thought you were interested in medicine?

Here'a niche form, incase anybody is interested.

$ \frac {n_1 \plus{} n_2}{\gamma \minus{} 1} \equal{} \frac {n_1}{\gamma _1 \minus{} 1} \plus{} \frac {n_2}{\gamma _2 \minus{} 1}$

This last result can infact be generalised for $ n$ gases by just conserving our internal energy.  :D

[quote="Chronoz"]
Excellent! Are you going to any JEE classes?
[/quote]
ofc not
that was a part of 11th syllabus & i m not that bad in physics :roll:  :roll:


LaTeX
advanced fields
advanced fields unsolved

It is true that if
Code:
{An}
and A are uniformly bounded self adjoint operators and
Code:
A_{n}
does not converge strongly to A then this will imply that
Code:
A_{n}
does not converges to A in the weak resolvent sense?

Thank you very much !

mikiduta

What is "convergence in the weak resolvent sense"? :?

Hi!
I think there is the same thing with strong resolvent convergence notion. I looked to see what it is means and in the first volume on Reed's book there is a proof that says that these notions are equivalent.
Definition Let An, n — 1, 2, ... and A be self-adjoint operators. Then An is said to converge to A in the norm resolvent sense (or norm generalized sense) if R_{\lambda}(An) converges to [code]R_{\lambda}(A)[/code] in norm for all [code]\lambda[/code] with Im [code]\lambda[/code] different from 0. An is said to converge to A in the strong resolvent sense (or strong generalized sense) if  [code]R_{\lambda}(An) [/code]converges to[code] R_{\lambda}(A)[/code] strongly for all [code]\lambda[/code] with Im[code] \lambda[/code] different from 0.
Thank you very much for your time !

Sincerely,
mikiduta

A note on readability: $ \text{\LaTeX}$ math mode is enabled here, and activated with dollar signs. The "code" tag just puts things into a monospaced font, useful for tables or indented programming stuff.

thanks , i have troubles all the time using the this site:)
I guess i will pay more attention  next time:)

Thanks a lot!

mikiduta


You sure have a long time for chapters. Mine is in three weeks.
I hope for a 46 too, but a 45 will suffice - just in case some counting problem kills me(reason I am taking the counting course).
This year a 44 for states should make nationals in my state.

My team has 2-3 practices a week, and I'm on AOPS a lot. I take the Mathcounts and Counting classes.
What's your state?

i do too many things above comprehension

i dont think working on countdown is the most important, especially if you plan on going to nationals... but if you want to get first in chapter, definitely a good idea. i made it to nationals without getting a 46 (unfortunately) but two 44s, which is decent. so i think practicing just general math is good, which is what i did (before making it to nationals, then it was heavy duty math, heh).

I used to live in PA. Now I live in Delaware - the easy state to get to nationals. :wink:

my team practices every Monday/Wednesday/Friday mornings from 7:15-8:15. We also have been practicing everyday after school since we got back from break from 3:30-5:00. So we're practicing about 10.5 hours a week. In the mornings we take state competitions and on Fridays we have a countdown round. In the afternoons, we do nationals. Sprint, Team, and Target. We occasionally do a national count down on Fridays, but not always.

Making nationals might not be too easy for me, because the top two scores last year were perfect 46's. The next two were 44 and 45 I believe. So, I've been working through AoPS and being on AoPS quite a bit.

You're lucky you have a team that cares. MY teammates really don't care about what they are going to get this year. They could get a 10 and care less. It makes me so frustrated because my coach doesn't care either. I have to run the team.

[quote]So we're practicing about 10.5 hours a week[/quote]
wow, 10.5 hours a week... my team gets a maximum of 4, with random sessions of goofing off in between :D But i practice a lot more at home... which is why im better than almost every1 even the seniors... *sigh*

[quote]They could get a 10 and care less[/quote]
ya, that was the same at my school last year... no wonder we got 3rd in chapter even though i was 1st and only 8th team in state... oh well, they are all my friends so i can let them off the hook... this year, its gonna be different cuz i actually have another person in my 9/10 division to compete with :o though hes 10th grade anyway

hm, my team practices around 1 or so hour a week. but yet, that time is spent eating foode. but of course we do stuffs on our own.

My school team didn't even make state last year, hahaha...sad isn't it?  I went individually to state.

No. Not sad. Last year was the last year my team is ever going to states... cus I'm gone next year. Sigh... if only the one good 8th grader hadn't graduated. My teammates play games while I work... now that is sad.

That was what my teammates did last year also.  My deepest regrets.


[quote="GameBot"]At a local restaurant, appetizers range in price from $ \;\$5.45$ to $ \;\$12.75$, entrees range in price from $ \;\$10.50$ to $ \;\$22.50$, and desserts range in price from $ \;\$6.25$ to $ \;\$10.50$. Assuming a meal consists of exactly one item from each of these three categories, what is the least expensive meal possible?[/quote]

We want to minimize the prices, so we add the cheapest prices together. $ 5.45 \plus{} 10.5 \plus{} 6.25 \equal{} 22.20$.


what does red + green = yellow mean ? how can light of two different wavelengths interfere to give light of yet a third wavelength ?

This is not interference. You don't get light with the wavelength corresponding to yellow light.
A spectrometer will just "see" two lines in the spectrum.
This is a physiologic effect. How do we see yellow? There are three types of receptors in the retina, sensitive to different spectral regions. (low wavelengths, medium and short  - Usually associated with red, green and blue).
I suppose yellow light will excite mostly the "low" and "medium" receptors at the same time and the brain will call this "yellow". The same receptors will be excited by a mixture of red and green. If the proportion is right, you can get the same excitation as for yellow and the brain will say "yellow" again.

Probably it's more complicated but I think it contains the basic idea.


geometry
email
IMO Shortlist
number theory
IMO

Hi guys,

IMO 2008 is coming up these days. Thus the IMO Shortlist 2007 will be provided as .pdf-file here. Thanks to [url=http://www.artofproblemsolving.com/profile.php?mode=viewprofile&u=3272]Andrei Frimu,[/url] [url=http://www.artofproblemsolving.com/profile.php?mode=viewprofile&u=652]Omid Hatami,[/url] [url=http://www.artofproblemsolving.com/profile.php?mode=viewprofile&u=22793]Son Hong Ta[/url] and [url=http://www.artofproblemsolving.com/profile.php?mode=viewprofile&u=8875]Ashwath Rabindranath[/url] who helped me gathering posted ISL 2007 problems on this site.

[hide="Problem URLs"][b]Algebra[/b]

[url=http://www.artofproblemsolving.com/viewtopic.php?p=893741#893741]Algebra A1[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1187155#1187155]Algebra A2[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1187159#1187159]Algebra A3[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1165901#1165901]Algebra A4[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1187165#1187165]Algebra A5[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1187171#1187171]Algebra A6[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=894658#894658]Algebra A7[/url]

[b]Combinatorics[/b]

[url=http://www.artofproblemsolving.com/viewtopic.php?p=1187174#1187174]Combinatorics C1[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1071295#1071295]Combinatorics C2[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1187179#1187179]Combinatorics C3[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1186971#1186971]Combinatorics C4[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1071276#1071276]Combinatorics C5[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=893746#893746]Combinatorics C6[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1187185#1187185]Combinatorics C7[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1143934#1143934]Combinatorics C8[/url]

[b]Geometry[/b]

[url=http://www.artofproblemsolving.com/viewtopic.php?p=894655#894655]Geometry G1[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1143932#1143932]Geometry G2[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1186774#1186774]Geometry G3[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=893744#893744]Geometry G4[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1186790#1186790]Geometry G5[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1186813#1186813]Geometry G6[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1178972#1178972]Geometry G7[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1186805#1186805]Geometry G8[/url]

[b]Number Theory[/b]

[url=http://www.artofproblemsolving.com/viewtopic.php?p=1106459#1106459]Number Theory N1[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1187198#1187198]Number Theory N2[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1187204#1187204]Number Theory N3[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1187214#1187214]Number Theory N4[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1187222#1187222]Number Theory N5[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=894656#894656]Number Theory N6[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1186847#1186847]Number Theory N7[/url]
[/hide]

And we also would like to know the problem proposers and their country. According to the [url=http://www.imo-register.org.uk/2007-report.html]UK report 2007 IMO report[/url]: Geometry problem G5 (IMO Problem 5) is proposed by Christopher Bradley and number theory problem N6 by Kevin Buzzard and Edward Crane. [url=http://www.artofproblemsolving.com/viewtopic.php?p=1186971#1186971]IMO Shortlist Problem, C4[/url] was proposed by [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=161046]Omid Hatami.[/url] Mohsen Jamali and [url=http://www.artofproblemsolving.com/profile.php?mode=viewprofile&u=4683]Nima Ahmadi Pour Anari,[/url] Iran, are the proposers of [url=http://www.artofproblemsolving.com/viewtopic.php?p=1187222#1187222]IMO Shortlist 2007, N5.[/url]

I added the problems to the [url=http://www.artofproblemsolving.com/resources.php?c=1&cid=17&year=2007]Resources section.[/url]

Thanks and congratulations to Omid and Christopher for two of the very few (ca. four) good problems on this shortlist...

  darij

[quote="Johan Gunardi"]Orlando, do you have the TeX format of the original file?

Thanks.[/quote]

I don't have the .tex-file but I know at least one person who has it. But unfortunately that member of previous IMO Problem Selection Committee has a different opinion from me about providing that file publically. If at all he thinks these kinds of materials (original pdf etc.) shall only be shared via the official IMO web page and any other web page as ML/AoPS is deemed to be inappropriate in his view. More details only via email for the more interested reader.


quadratics
inequalities
algebra proposed
algebra

Solve the equation

\[\parallel \parallel | x^{2}-x-1|-2|-3|-4|-5|=x^{2}+x-30 \]

[hide="Hint"]You should examine the following inequalities and equal sign conditions.
$\parallel \parallel | x^{2}-x-1|-2|-3|-4|-5|\ge x^{2}-x-15$
and
If $x$ satisfies $x^{2}-x-15\le0$, it is $x^{2}-x-15\ge x^{2}+x-30$

[hide="Answer"]$x=\frac{15}{2}$[/hide][/hide]


let $ABC$ be a triangle with incercle$(O) .$ the incercle is tangent to the sides AB and $BC$ at $P$and$Q$ respectively .the internal angle bisector from vertex A intersects the ray PQ at E .
prove that perpenticular $AC$to$CE .$

WHICH line is perpendicular to $CE$?

Maybe you should make your posts a bit more readable...

Advisement to tranthanhnam:
You'd better review your post before sending it.
You see, you have posted so many unreadble posts.

Whatever you mean, I guess it follows from http://www.mathlinks.ro/Forum/viewtopic.php?t=23677 post #7, Extended problem. Am I right?

  darij

[b]PS.[/b] Sorry, I posted a wrong link first. Thanks Arne for pointing this out.


Salutare,

In timpul semestrului nu am invatat deloc, iar peste aproape 2 luni o sa am examen.
Am nevoie doar de cateva indicatii, am auzit ca exista o carte cu probleme rezolvate, cunoscuta ca "ABC-ul", stiti despre ce vorbesc ? eu nu.

Cum ma sfatuiti sa abordez toata "operatiunea" asta ?

Specific faptul ca din curs nu pot intelege nimic daca nu am si cateva exemple de probleme rezolvate, de aceea am facut referire la acea carte.
Va las pe voi sa ma ghidati :)
Multumesc anticipat.

[b]Cam asta o sa am de invatat :[/b]

1.1
Spatii vectoriale. Subspatii vectoriale.  Sisteme de generatori. Liniar dependenta si  liniar independenta.
Baza, coordonatele unui vector intr-o baza, dimensiune.
Schimbari de baze. Schimbarea coordonatelor unui vector la o schimbare de baza.

1.2
Aplicatii liniare.  Nucleul si imaginea unei aplicatii liniare.  Matricea unei aplicatii liniare.
Modificarea matricei unei aplicatii liniare la o schimbare de baza. Vectori si valori proprii.
Forma diagonala.

1.3
Forme liniare; proprietati. Forme biliniare.
Matricea unei forme biliniare. Forme  patratice.
Reducerea formelor patratice la forma canonica. Metoda Jacobi. Metoda Gauss.

1.4
Spatii vectoriale euclidiene .
Ortogonalitate. Procedeul de ortogonalizare Gramm-Schmidt.


[hide] Notice that the denominator of term "n" cancels out with the denominator of term "n+1."  Thus, we are left with the denominator of the final term and the numerator of the first term to get $\frac{3}{k+1}\le\frac{1}{8}\Rightarrow k+1\le 24\Rightarrow k\le 23\Rightarrow k_{max}=\boxed{23}$ [/hide]

[quote="happyme"]What is the larget integer $k$ such that:

$\frac{3}{2}\cdot \frac{2}{1}\cdot \frac{1}{2}\cdot \frac{2}{3}\cdot \frac{3}{4}\cdot ... \cdot \frac{k}{k+1}\geq \frac{1}{8}$[/quote]

[hide]everything cancels except the 3 and k+1

$\frac{3}{k+1}\geq \frac{1}{8}$

so $k=\boxed{23}$[/hide]


probability counting distinguishability MATHCOUNTS function AoPSwiki search

[quote="Thunder365"]I need help in solving problems like these:
If you roll 5 dice, how many ways can you roll a 16 total (or what is the probability that you get a 16 total)?

How would you tackle problems like these?[/quote]

Well, you would generally see problems with a larger number, closer to the largest sum you can get. For this problem it would prolly be 28, because it is 2 less than 30. Did you make this up?

Yeah those would be different.

[hide="partial solution"]
Let the rollings be $ a,b,c,d,e.$ We have the equation \[ a\plus{}b\plus{}c\plus{}d\plus{}e\equal{}16.\] I think there is a balls and urns solution, but I'm not sure of it.
So I'll do it a different way. However, this has the disadvantage of missing a combination.

\[ (6,6,2,1,1)\Rightarrow30\newline(6,5,3,1,1)\Rightarrow60\newline(6,5,2,2,1)\Rightarrow60\newline(6,4,3,2,1)\Rightarrow120\newline(6,3,3,3,1)\Rightarrow60\newline(6,3,3,2,2)\Rightarrow30\newline(5,3,3,3,2)\Rightarrow20\newline(4,3,3,3,3)\Rightarrow5\newline(5,3,3,3,2)\Rightarrow20\newline(4,4,3,3,2)\Rightarrow30\newline(4,4,4,3,3)\Rightarrow10.\] Just continue in the fashion, add up all the numbers, then put it over $ 6^5\equal{}7776.$[/hide]

Ah yes completely forgot about generating functions.

[hide="partial solution using generating functions"]
You just need to find the coefficient of $ x^{16}$ in $ \left(x\plus{}x^2\plus{}x^3\plus{}x^4\plus{}x^5\plus{}x^6\right)^5.$ I'll leave that to you. Then put that over $ 6^5\equal{}7776.$[/hide]

[quote="RoFlLoLcOpT"]Ah yes completely forgot about generating functions.

[hide="partial solution using generating functions"]
You just need to find the coefficient of $ x^{16}$ in $ \left(x \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} x^5 \plus{} x^6\right)^5.$ I'll leave that to you. Then put that over $ 6^5 \equal{} 7776.$[/hide][/quote]

Wait..how does that work? Is that an easy way? This seems interesting!! :idea:  Please explain a little more..PLEASE!!! :oops:

Go to the AoPSWiki and search for generating functions, or buy Intermediate Counting and Probability.

To RoFlLoLcOpT: I don't see a really easy way to multiply that out.

Thunder365: A problem more likely to appear on a contest would be "If you roll 5 distinct dice, how many ways can you roll a sum of 27" (or 28, 29) to avoid brute-forcing.  Search Balls and Urns on the AoPSWiki.

[url=http://www.artofproblemsolving.com/Wiki/index.php/Generating_functions]Wait..this is for coins not dice right?[/url]

Read chapter 14 in [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=302]Intermediate Counting and Probability.[/url] It explains Generating Functions.

It's quite complicated, so it won't be easy for me to explain without taking up a lot of space and time.

[hide="basic problem"]
There are three urns on a table. One has 2 red balls, one has 2 green balls, and one has 3 blue balls. Balls of the same color are indistinguishable. In how many ways can I choose 4 balls from the urns?

(It's easy to solve it by just listing, but a solution using generating functions is intended.)
[hide="list"]
\begin{tabular}{c|c|c}Red & Green & Blue \\
\hline \\
0 & 1 & 3 \\
0 & 2 & 2 \\
1 & 0 & 3 \\
1 & 1 & 2 \\
1 & 2 & 1 \\
2 & 0 & 2 \\
2 & 1 & 1 \\
2 & 2 & 0\end{tabular}
(stupid table looks ugly). We can immediately see there are $ \boxed{8}$ ways to choose 4 balls.[/hide]
[hide="generating functions"]
Let's look at the red balls first. It's possible to choose 0,1 or 2 balls. However, there is only one way for each of these because the balls are indistinguishable. The generating function for the red balls is $ 1 + x + x^2 = 1\cdot x^0 + 1\cdot x^1 + 1\cdot x^2.$ The coefficient "1" of $ x^0$ represents the number of ways to choose 0 balls, and the same with the others.
The generating for the green balls is the same. 1 way to choose 0,1, and 2 balls, so the generating function is $ 1 + x + x^2 = 1\cdot x^0 + 1\cdot x + 1\cdot x^2.$
Finally, the blue balls. 1 way to choose 0,1,2, and 3 balls. Therefore, the generating function is $ 1 + x + x^2 + x^3 = 1\cdot x^0 + 1\cdot x^1 + 1\cdot x^2 + 1\cdot x^3.$
To get the generating function for the ways to pick the balls from all 3 urns, we multiply the individual generating functions together:
\[ \begin{align*}(1 + x + x^2)(1 + x + x^2)(1 + x + x^2 + x^3) & = (1 + 2x + 3x^2 + 2x^3 + x^4)(1 + x + x^2 + x^3) \\
& = 1 + 3x + 6x^2 + 8x^3 + 8x^4 + 6x^5 + 3x^6 + x^7\end{align*}.
\]
Now to get the number of ways to choose 4 balls, look at the coefficient of $ x^4,$ which is $ \boxed{8}.$
[hide="why does this work?"]
Let's we write out and add up the individual parts that make up the $ x^4$ term:
\[ (1\cdot x\cdot x^2) + (1\cdot x^2\cdot x^2) + (x\cdot 1\cdot x^3) + (x\cdot x\cdot x^2) + (x\cdot x^2\cdot x) + (x^2\cdot1\cdot x^2) + (x^2\cdot x\cdot x) + (x^2\cdot x^2\cdot1) = 8x^4.
\]
Notice how these 8 products correspond with the 8 entries from the chart above:
\begin{tabular}{c|c|c\parallel{}c}Red & Green & Blue & Monomial \\
\hline \\
0 & 1 & 3 & 1\cdot x\cdot x^3 \\
0 & 2 & 2 & 1\cdot x^2\cdot x^2 \\
1 & 0 & 3 & x\cdot1\cdot x^3 \\
1 & 1 & 2 & x\cdot x\cdot x^2 \\
1 & 2 & 1 & x\cdot x^2\cdot x \\
2 & 0 & 2 & x^2\cdot1\cdot x^2 \\
2 & 1 & 1 & x^2\cdot x\cdot x \\
2 & 2 & 0 & x^2\cdot x^2\cdot1\end{tabular}
[/hide][/hide][/hide]

[hide="@AIME15"]
There are tactics to find the coefficient without multiplying everything out.[/hide]
[hide="@The Pink Panther"]
No, it can work for a variety of problems.[/hide]

You have no idea how long that took to type... :ewpu:

Maybe a little hard for this forum...

Dang....what grade are you in?

Is the Intermediate Counting Book good for middle schoolers?

I believe that RoFlLoLcOpT is in 6th grade...

Yes, intermediate counting and prob is good for middle schoolers...

Actually, I think it would be for 8-10 grade students. Though, if you have mastered Intro to C & P, Intro to NT, Intro to Geom., or maybe even less then you would be ready. Unless, of course, you are very smart (unlike me).

Lol :lol: And when are they gonna come out with the rest of the Intermediate books? And is there gonna be an advanced series? If so, when is that gonna come out?

[quote="ged3.14"]there are two general types of dice problems in mathcounts: one with high sums and one with low sums. For low sums, you can use casework based on the first die. For high sums, the problem is really easy, since you can find the difference. (Example, if you want to make 28 from 5 dice, notice that max = 66666 = 30, so you have 66655 or 66664). Finally with large numbers of dice you can use generating functions.[/quote]

So would questions that involve multiple dice be nats level or hard state?

[quote="ged3.14"]If you see a dice problem on chapter or state, its really easy with casework. Also, you shouldnt be to afraid of large numbers of dice, since they usually come with large sums.[/quote]

This was already stated, no need to state it again. :lol:

I am sure that a problem that requires this much casework would never appear on a Mathcounts test.

If you want a method that works for a decent amount of numbers, you'd be best off with balls and urns. Note however that you don't need to use it immediately; it's possible to find the number of ways to get a sum of 23 while rolling five dice using balls and urns  :wink: . I think I posted on your blog.

I'm lazy so I never bothered to learn generating functions, although I've had access to them for about two years now  :P .

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2050466265&t=264024]You don't need generating functions: PIE is equivalent here.[/url]

So first convert it to the problem $ a \plus{} b \plus{} c \plus{} d \plus{} e \equal{} 16 \minus{} 5\cdot1 \equal{} 11$ for nonnegative integers $ 0\le a,b,c,d,e\le5$.

Without the bounding restrictions, there are $ \binom{11 \plus{} 5 \minus{} 1}{5 \minus{} 1} \equal{} \binom{15}4$ ways.

However, some of these have one term more than $ 5$, which is not allowed. There are $ \binom51$ ways to choose this term which is greater than $ 5$ (WLOG, first "give" $ a$ six; the number of ways is now $ (a \minus{} 6) \plus{} b \plus{} c \plus{} d \plus{} e \equal{} 11 \minus{} 6 \equal{} 5$ which has $ \binom{5 \plus{} 5 \minus{} 1}{5 \minus{} 1} \equal{} \binom94$ ways).

Two or more can't be at least $ 6$, so our answer is $ \binom{15}4 \minus{} \binom51\binom94\equal{}\boxed{735}$. As mentioned in the link, there's an equivalent way using negative binomial coefficients...

That's incorrect. Their would be 15C4 ways if you could have integers 0 through 11, not 0 through 5.

There are $ \binom{15}{4}$ ways [b]without[/b] the bounding restrictions, which is what math154 was saying. He then subtracted the number of ways where some term was greater than 5, which gives the desired answer.

[quote="PowerOfPi"]Are the dice distinguishable? Like, is 1-4-6-3-2 different than 2-6-3-1-4?[/quote]

In case of probability whether or not the dice are "distinguishable" or not shouldn't matter. All you have to make sure are that you stay consistent. As in, for the Total, if you counted without regard to order, you must count without regard to order for the "success" cases too. And if you make order matter for either the numerator or the denominator, you must do the same for the other.


logarithms
function
calculus
algorithm
geometry
3D geometry
calculus computations

I have been going through a bit of Rudin's book, and in chapter 1 the following fact is used freely:

If $b$ is a real number such that $b>1$ and $n$ is a positive integer, then $b^{1/n} > 1$.

Now this looked like a pretty obvious statement, but I found that I had trouble proving it using only the axioms and definitions given and the theorems previously proved.

Does anyone have any suggestions?

Thanks.

I'll be honest- I am the worst proof writer in the forum.....but I think I can manipulate this one.

Theorem: $\{b\in\mathbb{R}\mid b>1\}$, if $n$ is a positive integer, then $b^{\frac{1}{n}}>1$.

Proof: $b^{\frac{1}{n}}>1$
$\frac{\ln b}{n}>0$

${\ln b}>0$
$b>e^{0}$
$b>1$

Q.E.D.

I think I did this proof backwards. But I noticed that if you read my proof from bottom to top, it still makes sense. A palindrome proof? Works both ways? I don't know. I hope this is right.

[b]Axiom 1[/b] For two real numbers $x$ and $y$ exactly one of the following statements is true:
$x<y, x=y, x>y$

[b]Axiom 2[/b] If $ x<y$ and $z>0$ then $xz<yz$

Since $b^{1/n}$ is a root it is by definition positive

If $b^{1/n}$ isn't greater than $1$, we must have, by Axiom 1 $0< b^{1/n}\leq 1$

Now it follows from repeated use of Axiom 2 that

$1\geq b^{1/n} \geq b^{1/n} \cdot b^{1/n} \geq...\geq  b^{1/n} \cdots b^{1/n} =b $

Contradiction and so we must have $b^{1/n}>1$

So I guess Disoriented Vectors' proof pretty much killed mine. lol

Your proof uses the $\ln$ function, which is really not appropriate for proving a fact of this nature.

[quote="fleeting_guest"][quote="Fierytycoon"]If $b$ is a real number such that $b>1$ and $n$ is a positive integer, then $b^{1/n} > 1$.

Now this looked like a pretty obvious statement, but I found that I had trouble proving it using only the axioms and definitions given and the theorems previously proved. [/quote]

This is an algebraic problem that can be separated from the more complicated issue of existence of $b^{1/n}$.  It can be understood as "if $x^n > 1$ for positive $x$, then $x > 1$", which follows from factorization of $x^n - 1$.     Here $x$ is the (at this stage hypothetical) $n$th root.[/quote]

can one prove the existence of the $n^{\textrm{th}}$ root of a number without ugly things like calculus?

[quote="perfect_radio"] can one prove the existence of the $n^{\textrm{th}}$ root of a number without ugly things like calculus?[/quote]
If "number" is given a specific definition, such as Cauchy sequences of rational numbers, then how to prove existence will depend on the definition.   For Cauchy sequences the proofs I can think of are essentially calculus.  For infinite decimal expansions I guess the algorithm for extracting the digits of the $n^{\textrm{th}}$ root would prove that the root exists.  I don't remember the algorithm clearly, except that it is very similar to Newton's method for finding solutions of $x^n - a = 0$, so it might involve some calculus-like step in the proof.

If "number" is undefined then the answer depends on what axioms are assumed about numbers.  Proofs from a completeness axiom probably will involve some infinite sequences.   It is hard to imagine any simple axioms weaker than "the $n$-th root of a positive number exists" that imply, by some finitary algebraic (no calculus) argument, the existence of $2^{1/3}$.

As a sort of executive summary of what fleeting_guest said:

$\sqrt[3]{2}$ is the unique postive root of the equation $x^3-2=0.$

$f(x)=x^3-2$ is continuous, $f(0)<0$, and $f(2)>0,$ so the Intermediate Value Theorem promises us that it has a root somewhere in the interval $(0,2).$ (Or you could narrow that interval down if it suited your purposes to do so.)

The Intermediate Value Theorem is most definitely a theorem of real analysis. It is "intuitively obvious," but its truth does depend on the definition of number in exactly the ways fleeting_guest was talking about.

Continuing beyond the summary:

My favorite proof of the Intermediate Value Theorem is by successive bisection: assume $f$ is continuous on $[a,b]$ with $f(a)<0,\,f(b)>0.$ Let $c=\frac{a+b}2.$ If $f(c)=0$ we're done. If $f(c)<0$ let $a_1=c$ and $b_1=b.$ If $f(c)>0$ let $a_1=a$ and $b_1=c.$ Repeat, recursively.

If this process is performed a finite number of times, until $b_n-a_n$ is less than some pre-selected tolerance, then it's a numerical method. Performed abstractly, as an infinite process, it's a proof of the Intermediate Value Theorem. Of course, that proof must quote some version of the completeness of the real numbers.

A certain amount of completeness is included automatically when one uses infinite decimals (or binary, etc) to define real numbers, because some questions reduce to discrete operations on positive integers.

To define, say, the cube root of 2, one can let the first $n$ digits be the largest possible decimal $D$ at that precision (followed by an infinite string of zeroes) with $D^3 < 2$.  The fact that $2 \times 10^k$ is never a perfect cube implies that this process continues perpetually and gives a well-defined decimal.  Maybe you also need to show that there isn't an indeterminacy coming from $0.999... = 1$, but that doesn't involve any calculus.  I believe that (more or less "by construction") one could prove that the end result is in fact a cube root of 2; and that there are ways of phrasing all the above in finitary ways that avoid any reference to completed infinities, and deal only with assertions about a specific sequence of integers and its properties, i.e. an explicit proof by induction.   In that sense the construction would not involve any calculus, by which I understand some general theory of infinite sequences, or functions, or sets.

For many numbers $x_i$, one could therefore prove a theorem $T(x_i,n)$ stating that $x_i$, given as a decimal, has an $n$-th root.  But the theorem "for all positive $x$, there exists an $n$-th root of $x$" would probably require dealing explicitly with questions of completeness in a manner that  amounts to the standard real analysis.


This is a game based on number guessing, but there it is quite different from other number-guessing games. Here's the rules:

1) The number is a whole number between $ 1$ and $ 600$ inclusive.
2) You can make $ n$ guesses on what the number is, given the restriction that $ 600$ is divisible by $ n$.
3) If one of the $ n$ numbers you guessed were correct, you score $ \frac {600}{n}$ points.
4) If none of the $ n$ numbers you guessed were correct, you score $ 0$ points.
5) PM me your guesses, ALL IN ONE PM. DO NOT POST YOUR GUESSES HERE!!!
6) There will be a scoreboard.

Round $ 1$:
Submission expiration: Noon PDT (3:00 EDT) on 4/13/2009.

[quote="r15s11z55y89w21"]
2) You can make $ n$ guesses on what the number is, given the restriction that $ 600$ is divisible by $ n$.
[/quote]

Do you mean n is divisible by 600?

hmm... your expected value is 1 point no matter how many guesses you make.

Okay, I guess this game is a fail?

[hide="Submissions"]1[/hide]

Nobody has scored above 0 yet.

Round $ 2$:
Submission expiration: 21:00 PDT (Midnight EDT) on 4/14/2009.

By the way, if anyone's interested, the number for the first round was $ 360$.

EDIT: 666th post!

[hide="Submissions"]2[/hide]

[hide="Scoreboard"]AIME15: 1[/hide]

For your information, the answer was $ 363$.

[hide="Submissions"]2[/hide]

[hide="Scoreboard"]james4l: 3
AIME15: 1[/hide]

The number was $ 384$.

Round $ 4$:

Expiration Time: 10:37 PDT (13:37 EDT) on 4/16/2009

[hide="Submissions"]2[/hide]

[hide="Scoreboard"]james4l: 3
pinkmuskrat: 2
AIME15: 1[/hide]

The number was $ 87$.

Round $ 5$:

Expiration Time: 10:50 PDT (13:50 EDT) on 4/17/2009

[quote="r15s11z55y89w21"]Round $ 3$!

Hint: you might already know the answer.[/quote]

Sorry if I'm being dumb, but I don't get it.

It's all right; as it seems, $ 0$ people in total got the message.

[hide="Submissions"]3[/hide]

[hide="Scoreboard"]james4l: 9
pinkmuskrat: 2
AIME15: 2[/hide]

The number was $ 258$.

Round $ 6$:

Expiration Time: 11:37 PDT (14:37 EDT) on 4/18/2009

[hide="Submissions"]2[/hide]

[hide="Scoreboard"]james4l: 9
pinkmuskrat: 2
CircleSquared: 2
AIME15: 2[/hide]

The number was $ 596$.

Round $ 7$:

Expiration Time: 12:37 PDT (15:37 EDT) on 4/19/2009

Collecting evidence:
[hide=" "]BTM scavenger hint 2 = Z of Fr[/hide]
360
363
384
087
258
596

No discernable pattern, except they COULD be the answers to random AIME problems... hmmm.

[quote="r15s11z55y89w21"]Has everyone agreed to this?

I guess I'll just allow AIME15 to take over command for Meta_Knight, since he's just...not here anymore. [hidden]The answer is 363.[/hidden][/quote]

It's funny how no one figured out what that might mean.

Okay, seriously... I don't get it. I saw some of the numbers in AIME answers, but not all. And those which were had no apparent pattern.....

Ahh...I get it now.

Although I don't see how we were supposed to link the two separate games...

[quote="r15s11z55y89w21"]Round $ 3$!

Submission expiration time: 13:37 PDT (16:67 EDT) on 4/15/2009

Hint: you might already know the answer.[/quote]

It's also funny how nobody figured out what that might mean.

[hide="Submissions"]2[/hide]

[hide="Scoreboard"]james4l: 15
pinkmuskrat: 4
CircleSquared: 2
AIME15: 2[/hide]

The number was $ 597$.

Round $ 8$:

Expiration Time: 15:37 PDT (18:37 EDT) on 4/20/2009

[hide="Submissions"]1
Sigh, I guess this game's going down?[/hide]

[hide="Scoreboard"]james4l: 27
pinkmuskrat: 4
CircleSquared: 2
AIME15: 2[/hide]

The number was $ 444$.

Round $ 9$:

Expiration Time: 14:37 PDT (17:37 EDT) on 4/21/2009

EDIT: Very interestingly, after I posted this, I had a 4.44 posts per day average.

EDIT2: Rofl, so far this year, I also have 444 countdowns won.

[hide="Submissions"]4[/hide]

[hide="Perfect One-Number Submissions"]1[/hide]

[hide="Scoreboard"]james4l: 627
CircleSquared: 7
pinkmuskrat: 6
AIME15: 2
vahalla: 1[/hide]

The number was $ 50$.

Round $ 10$:

Expiration Time: 12:37 PDT (15:37 EDT) on 4/22/2009

I have decided that there will be $ 15$ rounds. [hide=" "]Fill in ieie. [/hide] Believ[color=blue]v[/color]e it or not, you have an alternative to gue[color=green]r[/color]ssing; there might be hints hidden (or outright exposed) in different places. For most of the $ 15$ rounds, the[color=red]t[/color]re is/was/will be a way to figure the number out without guessing. Round $ 15$ will[color=green]l[/color] be the hardest and mos[color=red]f[/color]t tedious to answer [color=red]p[/color]this "alternative" way.

Will you reveal all to us people who are foolish and stupid like me in the end? :oops:

If you score 200+, you could pm me for solutions.

Yes I am awesome like that.

He's very obvious at where the numbers are... sometimes...

[quote="james4l"]He's very obvious at where the numbers are... sometimes...[/quote]

And yet so far there's only one perfect guess...

[hide="Submissions"]2[/hide]

[hide="Scoreboard"]james4l: 627
CircleSquared: 7
pinkmuskrat: 6
AIME15: 2
vahalla: 1[/hide]

The number was $ 555$.

Round $ \text{Eleven}$:

Expiration Time: 15:37 PDT (18:37 EDT) on 4/23/2009

I figured out why it was 555 :D ...but it is unlikely that I will get any others.

Do you wish to pm me your solution to the "555" to see if it's correct?

[hide="Submissions"]2[/hide]

[hide="Scoreboard"]james4l: 627
CircleSquared: 7
pinkmuskrat: 6
AIME15: 2
vahalla: 1[/hide]

The number was $ 382$.

Round $ 12$ (this one's a toughie if you're going to try to find the answer):

Expiration Time: 15:55 PDT (18:55 EDT) on 4/[color=red]28[/color]/2009

Are you being sarcastic, or am I being stupid?

I'm back!

[hide="Submissions"]2[/hide]

[hide="Scoreboard"]james4l: 627
CircleSquared: 7
pinkmuskrat: 6
AIME15: 2
vahalla: 1[/hide]

$ 202$ was acceptable only if you do not live in the Bay Area; I accepted $ 411$ or $ 209$; if you answered $ 411$, you needed to explain why. This was a tough problem, and it is not a surprise that nobody found the answer. I find it hilarious that nikeballa96 thought (no offense) that the answer was $ 782$, which was out of range.

Round $ 13$:
[hide="Lolz..."]The number is the number! [/hide]
Expiration Time: 18:55 PDT (21:55 EDT) on 4/29/2009

[hide="Submissions/Perfect Submissions"]1[/hide]
[hide=" "](5,11),(4,2),(9,6),(9,8),(3,22),(13,7),(1,2),(9,9),(3,21),(4,12)[/hide]
[hide="Scoreboard"]james4l: 1227
CircleSquared: 7
pinkmuskrat: 6
AIME15: 2
vahalla: 1[/hide]

The number was $ 13$.

Round $ 14$:

Expiration Time: 10:07 PDT (13:07 EDT) on 4/30/2009

[hide="Submissions"]1[/hide]

[hide="Scoreboard"]james4l: 1227
CircleSquared: 7
pinkmuskrat: 6
AIME15: 2
vahalla: 1[/hide]

The number was $ 42$.

Round $ 15$:

Expiration Time: 20:59 PDT (23:59 PDT) on 5/2/2009

The answer.
Is the fifteenth.
Of the fifteen.
Numbers.
There, I told you the answer right there.

What, nobody cares about $ 15$!?!?!?

Okay, I postpone the expiration time to the same time given in the previous post tomorrow.

[hide="Submissions"]1[/hide]

[hide="Scoreboard"]james4l: 1227
CircleSquared: 7
pinkmuskrat: 6
AIME15: 2
vahalla: 1[/hide]

The number was $ 15\plus{}11\plus{}55\plus{}89\plus{}21\equal{}191$.

james4l wins an award; I'm thinking about what it should be.


I was wondering if there is a Round 2 Math Jam transcript available? I missed it.


A turn consists of rolling a standard die and tossing a fair coin. The game is won when a dice shows a 1 or 6 and the coin shows heads. What is the probability the game is won before the fourth turn? Express your answer as a common fraction.

[hide="Answer"]There's a $\frac{1}{6}$ chance of winning on each turn, and therefore a $\frac{5}{6}$ chance of not winning. The chance of winning before the fourth turn is the chance of winning sometime in the first three turns. Therefore, the total chance is $\frac{1}{6}+\frac{5}{6}\cdot \frac{1}{6}+\frac{5}{6}\cdot \frac{5}{6}\cdot \frac{1}{6}=\boxed{\frac{91}{256}}$.[/hide]

I hate when I mess up probabilities  :blush: . It was a 1 or 6 on the dice [b]or[/b] the heads on the coin. I treated it as and and got $\frac{13}{54}$.

Aargh, I somehow can't figure out how the two things with different probabilities affect each other. :huh: Would you mind posting your solution? Maybe I can get it from that. I thought I had something, but if it worked that way, then the chance of winning would be $\frac{26}{27}$...

[hide="the incorrect solution"]the probability of winning the first round is $\frac{1}{3}\cdot\frac{1}{2}=\frac{1}{6}$.

The probability of winning the second turn is $\frac{2}{3}\cdot\frac{1}{2}\cdot\frac{1}{6}=\frac{1}{18}$

The probability of winning the third turn is $(\frac{2}{3}\cdot\frac{1}{2})^2\cdot\frac{1}{6}=\frac{1}{54}$

Adding those gives us $\frac{13}{54}$  :roll: [/hide]

[hide="my solution"]
assuming it should be and, here's the easiest way for me. There will be 2 possibilites, before the 4th turn, and not before the 4th turn. Since there is a total of 1 for the probabilities of these two, we should subtract the probability that there is no winner before the first turn from 1. Since there's a $\frac 16$ chance of winning each turn, there is a $\frac 56$ chance of [b]not[/b] winning. Therefore, there is a $(\frac 56)^3$ chance of not winning in the first three turns, which is $\frac {125}{216}$. Subtracting that from 1 gives $\boxed{\frac {91}{216}}$ which is in simplest form.

NOTE: jesusfreak had a way to solve the problem, but he used 256 for the denominator instead of 216
[/hide]

[hide="Solution for winning if you roll a 1 or 6 OR flip a heads"]
I will work this in exactly the same way. The probability of winning with the die is 1/3. The probability of winning with the coin is 1/2. To find the probability of winning with either, we add them and take away the intersection, which in this case is 1/6. This means that the probability of winning is $\frac 46$ or $\frac 23$. Therefore, there is a $\frac 13$ chance of [b]not[/b] winning on any given turn. So we have a $(\frac 13)^3$ chance of not winning in the first three turns. This means that there is a $\frac 1{27}$ chance of not winning in the first three turns, so there is a $\boxed{\frac {26}{27}}$ chance of winning.
[/hide]

[quote="JesusFreak197"][quote="Hamster1800"]NOTE: jesusfreak had a way to solve the problem, but he used 256 for the denominator instead of 216[/quote]

Oops. :oops:

So I actually solved it correctly, then? That's surprising, it didn't seem right.[/quote]

It seems right because everthing you did was logical. Does anyone have an answer key to the 2001 National Competition?

Are you sure it's 2001? I have a copy of the tests and can't find that problem.

EDIT

nvm, i didn't think that it was team level :P I was looking on sprint and target. The answer to team round number 8 at 2001 nationals is:

[hide="answer"]$\frac {91}{216}$[/hide]

Although the answers for the scenario:  Rolling a 1 or 6 OR flipping a heads, is correct, it should be worded "and/or" considering you included the cases where both conditions were met.  It never hurts to specify you wording. ;)

[hide]

The probability to win the first time is $\frac{1}{6}$. The probability to win the second time is $\frac{5}{6} * \frac{1}{6}$ since the chance of losing in the first round is $\frac{5}{6}$. The third time to win is $\frac{5}{6} * \frac{5}{6} * \frac{1}{6}$.
$ (\displaystyle\frac{1}{6}) + (\displaystyle\frac{5}{6} * \displaystyle\frac{1}{6}) + (\displaystyle\frac{5}{6} * \displaystyle\frac{5}{6} * \displaystyle\frac{1}{6} = \displaystyle\boxed{\frac{91}{216}}$[/hide]


advanced fields
advanced fields unsolved

Can someone look through the proposed proof of the Riemann Hypothesis at http://www.aronpalmer.com/rproofv2.pdf and write back saying whether there's any problems with it please?

Try sci.math, they have more tolerance for that sort of thing there.


Look at [url=http://www.artofproblemsolving.com/Wiki/index.php/Complex_analysis]this[/url] for example.  Should we have links in the titles?

I don't think we should.  For one, it makes the titles red which would confuse people.  And it shouldn't be too hard to repeat the section title in the main text so you can like to it there, right?

Seconded -- get rid of them.  There's also on in the article called "Euler's identity," I think.

Wait, Mr. Crawford, what is your take on this?

http://www.artofproblemsolving.com/Wiki/index.php/Number

Does that mean you want links in the titles?

Honestly, I hate them aesthetically.  However, I've not brought it up because they're all over Wikipedia, suggesting to me that other people either (1) don't hate them, or (2) find it useful to have the option.

I personally added just one myself recently, but I could see keeping or not keeping them.

One thing that may be true is that it could be hard to set that standard due to the fact that other MediaWiki wikis use it.  Many new users will use it automatically.

I'm undecided.


A lottery involves five balls being selected from a drum. Each ball has a different positive integer printed on it.
Show that, whichever five balls are selected, it is always possible to choose three of them so that the sum of the numbers on these three balls is a multiple of 3.

Case 1:there are three multiples of 3 in the five balls , then we are done
Case 2:there is only one or two multiples of 3 , then we can find three balls congruent mod 3 or we can find two balls the numbers on which not divisible by $ 3$ such that their sum is divisible by 3 , we are done also
Case 3:there is no multiple of 3, then by pigeonhole principle we can find three balls congruent mod 3, done again

1st case: If there are 3 balls that have the same value of modulo 3, therefore the sum of the numbers are divisible by 3
2nd case: If there aren't 3 balls that have the same value of modulo 3, so there will be balls that are $ 0,1,2 \pmod3$, and the sum of the balls is divisible by 3 too

$ QED$

If I'm not mistaken, this problem is similar with canada 1970 :D


integration
real analysis
real analysis solved

Let  E the diffrential equation y" + exp(-t 2 )y = sin(t)

Consider f a solution of E wich is bound on [1;+oo[ and such that

 \int_[1;+oo[  (f(t)) 2 dt is convergent

Show that limite of f(t) is 0 when t-->+oo

Oral Polytechnique

let A  = sup{|f| on [1;+oo( }

|f"(t) -sint| = |f(t).exp(-t^2)| <= A.exp(-t)

if we integrate we have

cos1 - cost -A(1-exp(-t)) <= f'(t) -f'(1) <= cos1 - cost +A(1-exp(-t))

| f '(t)| <= |f'(1)| + 2 +A , f' is bounded =>

int_{1 to +oo} f^2(t).f'(t)dt is convergent

But int_{1 to x}f^2(t)f'(t)dt = 1/3(f^3(x) - f^3(1) ) has a limit when

x-->+oo => lim f = L

the convergence of int_{1 to +oo} f^2(t)dt => L=0


geometry
integration
trigonometry
calculus
calculus computations

This question is quite hard for me.
Find the area between the curve $ y\equal{}e^{\minus{}x}\left|sinx\right|$ and the straight line y=0 for$ x\geq 0$
My solution
$ \int_{0}^{\infty}e^{\minus{}x}\left|sinx\right|dx\equal{}\minus{}e^{\minus{}x}\left|sinx\right|\plus{}\int_{0}^{\infty}e^{\minus{}x}\left|cosx\right|dx$
$ \equal{}\minus{}e^{\minus{}x}\left|sinx\right|\minus{}e^{\minus{}x}\left|cosx\right|\minus{}\int_{0}^{\infty}e^{\minus{}x}\left|sinx\right|dx$
$ \int_{0}^{\infty}e^{\minus{}x}\left|sinx\right|dx\equal{}\frac{\minus{}e^{\minus{}x}\left|sinx\right|\minus{}e^{\minus{}x}\left|cosx\right|}{2}$
Cannot do anything.

The antiderivative of $ |\sin x|$ is not $ \minus{}|\cos x|.$

The only practical way to handle absolute values is as piecewise defined functions.

$ \int_0^{\infty}e^{\minus{}x}|\sin x|\,dx\equal{}\int_0^{\pi}e^{\minus{}x}\sin x\,dx\plus{}\int_{\pi}^{2\pi}e^{\minus{}x}(\minus{}\sin x)\,dx\plus{}\cdots$

$ \equal{}\sum_{k\equal{}0}^{\infty}(\minus{}1)^k\int_{k\pi}^{(k\plus{}1)\pi}e^{\minus{}x}\sin x\,dx$

Now, there's some more work to do. Your best hope is that the series might turn out to be a geometric series.

Is it equals to
$ ( \minus{} 1)^k\sum_{k \equal{} 1}^{\infty}\frac {e^{\minus{}k\pi}}{2}$ ?


Find minimum value of expression $\frac{1}{r}(\frac{4p}{u}+\frac{q}{\sqrt{1-v^{2}}})$, where $p,q,r,u,v$ - positive numbers satisfying conditions:
$pv+q\sqrt{1-u^{2}}\le r,$

$p^{2}+2qr\sqrt{1-u^{2}}\ge q^{2}+r^{2},$

$2qr\sqrt{1-u^{2}}+q^{2}\frac{1-v^{2}-u^{2}}{v^{2}-1}\ge r^{2}$

Since noone attempted to solve, I'll post solution.
[hide="here it is"]
First we try to simplify problem a bit. Make a new variables $a=\frac{p}{r}, b=\frac{q}{r}$. After this problem looks easier:
Find minimum value of function
$f=\frac{4a}{u}+\frac{b}{\sqrt{1-v^{2}}}$, where $a,b,u,v$ - positive numbers satisfying conditions:
$av+b\sqrt{1-u^{2}}\le 1,$

$a^{2}+2b\sqrt{1-u^{2}}\ge b^{2}+1,$

$2b\sqrt{1-u^{2}}+b^{2}\frac{u^{2}}{1-v^{2}}\ge b^{2}+1$

After this lake a look at variables $u$ and $v$: they satisfy inequalities $0<u, v<1$, its possible to make such angles $0< \alpha, \beta < \frac{\pi}{2}$, what $u=\cos\alpha, v=\sin\beta$.
After this problems looks like:
$f=\frac{4a}{\cos\alpha}+\frac{b}{\cos\beta}$, where variables satisfy:
$a\sin\beta+b\sin\alpha \le 1,$

$a^{2}+2b\sin\alpha \ge b^{2}+1,$

$2b\sin\alpha+b^{2}\frac{\cos^{2}\alpha}{\cos^{2}\beta}\ge b^{2}+1$

Then notice what second condition allows to lower $a$ until next equation will be satisfied:
$a^{2}+2b\sin\alpha=b^{2}+1 \Leftrightarrow a^{2}-b^{2}\cos^{2}\alpha=(1-b\sin\alpha)^{2}$
Then lower $b$ until:
$2b\sin\alpha+b^{2}\frac{\cos^{2}\alpha}{\cos^{2}\beta}=b^{2}+1 \Leftrightarrow b\tan\beta\cos\alpha=1-b\sin\alpha \Leftrightarrow b\sin(\alpha+\beta)=\cos\beta$
From last two conditions what became equations we get what
$a^{2}-b^{2}\cos^{2}\alpha=b^{2}\tan^{2}\beta\cos^{2}\alpha \Leftrightarrow a\cos\beta=b\cos\alpha$
Then
$f=\frac{4a\cos\beta+b\cos\alpha}{\cos\beta\cos\alpha}=\frac{5b}{\cos\beta}=\frac{5}{\sin(\alpha+\beta)}\ge 5$
So the answer is $5$.
[/hide]


trigonometry
modular arithmetic
calculus
calculus computations

$ 1\equal{}\minus{}2\sin{x}\cos{x}\Rightarrow \minus{}\frac{1}{2}\equal{}\frac{1}{2}\sin{2x}$ (remember the identity?).
Then $ \minus{}1\equal{}\sin{2x}$.  Now let $ \theta\equal{}2x$. . So we need to solve $ \minus{}1\equal{}\sin{\theta}\Rightarrow \theta\equal{}\frac{\minus{}3\pi}{2}\Rightarrow x\equal{}\frac{3\pi}{4}$.  So $ x\equal{}\boxed{\frac{3\pi}{4}}$.

[hide="Another Way"]$ 1\plus{}2\sin{\theta}\cos{\theta}\equal{}0\Leftrightarrow$
$ \sin{\theta}\cos{\theta}\equal{}\minus{}\frac{1}{2}$

Recalling our special trig values, its clear that $ \displaystyle\left(\frac{1}{\sqrt{2}}\right)^2\equal{}\frac{1}{2}$.
This makes me think the solution is expressible in the terms $ \frac{\pi}{4}k$ for some $ n\in\{\mathbb{Z}^\plus{}: k\equal{}2n\minus{}1\}$.
From there, it's a little thing to show that $ k\equiv 1\pmod{4}\Leftrightarrow \sin{(\pi k/4)}\equal{}\cos{(\pi k/4)}$. All k of this form cannot satisfy the needed qualities (i.e a negative one-half) because $ \sin{\theta}\cos{\theta}$ would effectively be the square of a value. Since $ k$ must be odd, the only other possible odd residue class modulo four is three, so general solution is $ k\cdot\frac{\pi}{4}$ where $ k\equal{}4n\minus{}1$.[/hide]
There are ways to solve if you don't know your half angle identity. It is best though, to know them. If you are the forgetful sort, I would suggest learning De Moivre's Theorem, which allows you to derive such identities. AoPS Volume II has a great section on it and the roots of unity.

[quote="igiul"]This makes me think the solution is expressible in the terms $ \frac {\pi}{4}k$ for some $ n\in\{\mathbb{Z}^ \plus{} : k \equal{} 2n \minus{} 1\}$.
From there, it's a little thing to show that $ k\equiv 1\pmod{4}\Leftrightarrow \sin{(\pi k/4)} \equal{} \cos{(\pi k/4)}$. [/quote]

This is not enough to guarantee that you have found all the solutions.  The solution by double-angle identity is complete.