algorithm
AMC
USA(J)MO
USAMO
number theory
Euclidean algorithm

(9th graders only, please)

What score did everyone think they got?

Maybe I'm pessimistic, but probably 2 (1/0/0/1/0/0).  :stretcher:

That contradicts your sig, which claims you got 0. On the other hand, my sig says 2=1, so you actually probably got a 1.

Kinda off-topic, but how many freshmen took this?

hoping for a 15.... :D but my first time....and I wrote my 3problems with at most 4papers....so like 10? :blush:
I counted the ninth grader and I think somwhere between 50-60...

I think 70 ninth graders took it, and ~25 are invited to Red MOSP.

Shouldn't it be "1 out of 3"? :D  Well, it's a lot harder than it sounds (unless, you already think it's hard)...There are some pretty beastly 9th graders...

u see...that's y I have no confidence....~~my English sucks....I even wonder if the teachers can read my solution without guessing  :(

I'm pretty sure I solved 1 & 4, for a total score of probably 7/0/0/7/0/0 =14.  I think I could have solved 5, but I spent over 3 hours on #4 and then misread #5 (so I thought that the Euclidean algorithm wouldn't work with integers; I had x & y switched in the replacement).

[quote="CatalystOfNostalgia"]I believe I got 7/0-1/0/6/0-1/0. So 13-15. I probably got 1 on #2 and 0 on #5, but we shall see.[/quote]

No way -_-... That's like... What I think I would get... but the USAMO graders are harsh. So. Dead.

Is it like for freshman they won't give out 7s? :(

[quote="Carbon57"]Of course not...it's just that freshmen usually aren't that experience, since most of them are first time USAMOers.[/quote]
yeah I doubt I get any thing close to 6 or 7.... :(

Expecting about 19....

Got Questions 1 and 4,

Solved 5, but in a way different from the official solution - dividing cases by rationals and irrationals.

Well...since it's not like the official one, I'm not expecting a 7, but I HOPE for a 5... :lol:

Phew....first ever USAMO, my goal was Red MOSP, but I think I almost made it...Very HAPPY  :D

[b]P.S. Taking it in South Korea was one "ULTIMATE" experience....1:30 Am~ 6:00 AM, and it was mid-term period. :P [/b]

[quote="physisnu"]
Solved 5, but in a way different from the official solution - dividing cases by rationals and irrationals.

Well...since it's not like the official one, I'm not expecting a 7, but I HOPE for a 5... :lol:
[/quote]

Um... the general consensus we seem to be getting is that such a solution is impossible. Can you please post your [b]full[/b] solution. Unless your solution contains something [i]significant[/i] that was not in your post in the other thread, I'm predicting that you're going to get a zero on that problem.

Sean Gu: Of course they will give 7's to freshmen, the graders have no way of determining who wrote the solutions, unless for some reason they read and memorized the thread were everyone was giving away their USAMO numbers (or are very familiar with the handwriting from MOP grading or whatever). The thing is that first time USAMOers, which just so happen to sometimes be freshmen, overestimate the rigor and correctness of their solutions, and thus their scores.

physisnu: I thought only Americans can go to MOP...?

[quote="diophantient"]My chances at red mop look slim :(
I suppose the most painful thing is knowing that last year my score was good enough for red mop, but I was only in 8th grade so I couldn't go...[/quote]

Pretty much the same thing here.   :(

I died! Spent 4 hours on number 2 and then 30 minutes on number 1, so i ran out of time. otherwise i could have gotten 7 on number 1, i'm pretty sure. This is what I think i got:

1-2/2/0/2/2/0

7-8? I died. :(

might make it to red mop if graders are nice and give me a 6 for number 4. It would give me 11-12. So, i died. (I'm saying that a lot, I know)

I've talked to at least one of Kevin Chen and Wenyu Cao, but I think that if they want how they did to be known, they would talk about it publically themselves.

I don't know why I haven't posted this...

I'm very unsure, anywhere from 9 to ~14. I would be shocked with <7 on 1, as I essentially had official #1. Imay have gotten a point on one of 2 and 5, but it seems unlikely. On 4, I was missing a case, so I'm not sure how much I'm going to lose.

I thought I had a 14ish for a long time, but yesterday I realized that my #4 was holey :( :( :( .

New projected score is 7/0/0/3/0/0. Hopefully that'll make MOP...

I'm in the same boat as everyone else who has recently posted.

Good luck to you guys, man this is stressssfullll

I really wish I have  a 10+....that would be encouraging~~ :D

Ok seems like im the stupidest here
on number 1, i got the x^2-x+1, etc. for a specific x
and I was like
hmmm
now how do you prove coprimeness?
afterwards Quatto's like......Eclidean alg?
gah

so thats 1 point

on number 4, I thought I had a good solution
but
nopes, I have a small hole
gah

so thats another point

I wrote absolute bs for 5, so 0

a total of 2
which ties my last year's score!
yay, no mop  :) *sarcasm*

*ponders what to do 4 summer*
max: 3/0/0/6/0/0
min: 1/0/0/1/0/0

You don't have to be to take the USAMO. Most foreign students don't, however, since they'd have to follow US time.

[quote="serialk11r"][quote="Anaxerzia"]Any one know how Kevin Chen or Wenyu Cao did?

@Catalyst: Same. Maybe he has a U.S. passport?[/quote]
Um, to take it in Korea I think he needs a US passport? (he's stated this in other threads, he's an American citizen living in Korea)[/quote]

I think you're mistaken.....there are only 2 students who took the USAMO in South Korea, Michael Jang from KMLA

and James Lim from Daecheong MS. (Includes me, YAY :D )

I'm quite sure those names sound "Chinese"....but then, Mark Chen is the only student who took in Taiwan.

(So, only 3 students who took it as an  "international"...)


Well the units digit would clearly be 0.  :wink:

Perhaps the tens digit, which would be $ 4$? (I would guess it would come from a [url=http://www.artofproblemsolving.com/Wiki/index.php/2006_AMC_10B_Problems/Problem_11]2006 AMC[/url]?)

I do believe that this user wants you to actually sum the series... might be a spammer. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=184071

Yea, I saw that post too, after I posted here.

Haha, looks like I did it from 4! instead of 7!  :P

Why is this relevant to math competitions and just out of curiosity, how did you find the sum?

(i) It's not relevant, and (ii) I used Mathematica.


Let $n$ be a positive integer and let $a_k = \dfrac{1}{\binom{n}{k}}, b_k = 2^{k-n},\ (k=1..n)$.

Show that $\sum_{k=1}^n \dfrac{a_k-b_k}{k} = 0$.

Who can imagine such formula ??

A few known (and always usefull :)) facts :

${n\choose k} = n/k \times {n-1 \choose k-1}$
${1/{n\choose k}+1/{n\choose k+1} = \dfrac{n+1}{n} \times 1/{n-1\choose k}}$

Using both formulae successively (grouping 2 successive binomials) :

$2n \times \sum_{k=1}^{n} \dfrac{{n \choose k}}{k} = (n-1) \times \sum_{k=1}^{n-1} \dfrac{{n-1 \choose k}}{k} + 1/{n}$

Then we can finish by induction or by noting that $1/2^n \times \sum 2^k/k$ has same induction relation.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=371496&sid=e0319f030d85bf1390a8fb335fd87c9d#p371496]Just in another form lol.[/url]

Let's split the sum first so we have $\sum\limits_{k=1}^n \frac{a_k-b_k}{k}=\sum\limits_{k=1}^n \frac{a_k}{k}-\sum\limits_{k=1}^n \frac{b_k}{k}$. By a combinatorial identity $\left(k\binom{n}{k}=n\binom{n-1}{k-1}\right)$, the first sum is now $\frac{1}{n}\sum\limits_{k=1}^n \frac{1}{\binom{n-1}{k-1}}$. This looks very familiar now if we just take the inverse of both sides of the binomial formula with $x=1,\;y=1,\;n-1$, we know the sum evaluates to $\frac{1}{2^{n-1}}$. As for the right sum, we have $\sum\limits_{k=1}^n \frac{2^k}{2^n}=\frac{1}{2^n}\left[2^n\left(2^{-{n-1}}+2^{-(n-2)}+\cdots + 1\right)\right]=\frac{1}{2^{n-1}}$ as it's a geometric series. Thus with both partial sums the same, we have proven that the entire expression sums to 0.


Which of the fractions $ \frac{13}{40}$, $ \frac{17}{52}$, $ \frac{21}{64}$, $ \frac{25}{76}$ is the smallest?

$ a_n \equal{} \frac {9 \plus{} 4n}{28 \plus{} 12n}$. (for $ n \equal{} \{n \in \mathbb{N}|N > 0\}$)
Let, $ f(x) \equal{} \frac {9 \plus{} 4x}{28 \plus{} 12x}$. (for $ x \equal{} \{x \in \mathbb{R}|x > 0\}$)
So $ f'(x) \equal{} \frac {1}{4(7 \plus{} 3x)^2} > 0$.

Thus, $ \boxed{\frac {13}{40}}$ is the smallest.


[quote="agolsme"]I have a few questions:
1. Which is more difficult of the two? Is Woot more difficult?
2. How much overlap is there between the two classes?
3. Is there a pretest for WOOT?(as there is for olympiad problem solving)
4. How much time is a student expected to spend on each class?[/quote]

The Olympiad Problem Solving class is very different from the WOOT program.

There are three levels of WOOT difficulty ranging from students just learning how to advance from medium-level AIME problem to learning how to write proofs...all the way up to students who can already medal at the IMO.

There is certainly some overlap as the topics covered in the Olympiad Problem Series are a subset of the possible topics covered in WOOT.  However, not every idea covered will be the same wince WOOT instructors will cover a plethora of topics both in and out of the curriculum of Olympiad Problem Solving.

There is no pretest for WOOT.  WOOT is for students who are capable of learning how to write a proof all the way up to students who have been doing so for several years already.  If you can consistently score a 5 or higher on the AIME, we think you will get a lot out of the program.

WOOT is different from our other programs and the amount of time we recommend spending is variable.  Some weeks you might want to spend 2 to 4 hours and on others you might want to spend 10 or 12.  This is mostly for a couple of reasons: the Olympiad-style tests will be administered every other week and class difficulty will vary greatly meaning that students might choose to attend and focus on some lessons more than on others.

Does that answer all your questions?

[quote="K81o7"]So is WOOT a lot more rigorous?[/quote]

More rigorous than what?

WOOT is designed to focus on both rigor and problem solving advancement.  Proofs on the tests will be judged as they are on national or international Olympiads and feedback will be designed to help students reach that understand of both the mathematics and their writing.


What is the remainder when x^27 minus 3x^24 plus x^13 plus 2x^7 minus x plus 1, is divided by x + 1.

The Remainder Theorem states that the remainder when a polynomial $ P(x)$ is divided by the binomial $ (x \minus{} k)$, then the remainder is $ P(k)$.

With that, I think you can solve this.  :)

If I understood all this correctly, the answer would be p^27 - 3p^24 + p^13 + 2p^7 - p + 1, or 1.

What are you doing? If we let $ P(x)\equal{}x^{27}\minus{}3x^{24}\plus{}x^{13}\plus{}2x^7\minus{}x\plus{}1$, then $ P(\minus{}1)$ is the remainder when we are dividing by $ x\minus{}(\minus{}1)$, as described above by tinytim.

:fool: Yeah, you're right. Sorry about that.


geometry
3D geometry
sphere
tetrahedron
analytic geometry

A sphere is inscribed in a regular tetrahedron with edges measuring 12 cm. What is the radius of a smaller sphere that can be inscribed touching the original sphere and three faces of the tetrahedron?

[hide="Hint"]If you prefer an analytic geometry approach, keep in mind that $ (0,0,0)$, $ (1,1,0)$, $ (1,0,1)$, $ (0,1,1)$ are the vertices of a regular tetrahedron. You should very easily be able to find the centroid of the figure, the side-length, and the centroids of the faces from these co-ordinates.[/hide]

Cool story, bro.

Could you... prove that your answer is right?

Could you also... hide your answer using a hide tag?

[hide="Like... this?"][code][hide=Like... this?]Please?[/hide][/code]Please?[/hide]


limit
real analysis
real analysis unsolved

Assume $ a_n\geq0,b_n\geq0,c_n > 0$ for $ n \equal{} 1,2,3,\ldots$ and holds following:
$ i)\displaystyle\sum_{n \equal{} 1}^{\infty}c_n \equal{} \plus{} \infty;$
$ ii)\displaystyle\lim_{n\to\infty}\frac {b_n}{c_n} \equal{} 0;$
$ iii)a_{n \plus{} 1}\leq b_n \plus{} \sqrt {a_n}(\sqrt {a_n} \minus{} c_n);$
then prove that $ \displaystyle\lim_{n\to\infty}a_n \equal{} 0$.


Am compus urmatoarea problema:
  Fie un triunghi echilateral $\ ABC$ si $\ M$ un punct interior astfel incat $\ MA=\sqrt{n+1}, MB=\sqrt{n+2}$ si
$\ MC=\sqrt{2n+3}$.Aflati $\ S_{ABC}$.
    Exista o metoda de rezolvare a problemei la nivelul clasei a VII-a, cam semestrul I?Asa e banala folosind o rotire
de $\ 60$ de grade in jurul lui $\ A$. :lol:

cred ca merge ca idee sa duci paralele prin M la laturi...se obtin 3 trapeze isoscele avand fiecare una din diagonale ..MA
MB respectiv MC si cum diag. sunt congruente in tr isoscele ...se obtine un triunghi dreptunghic cu laturile MA , MB , MC inscris in tr ABC

Nu jtiu daka merge mai departe, ca nu am facut pe foaie :)

Da.Ar fi o idee, pentru ca si in rezolvarea mea cu rotirea ajungi la un triunghi dreptunghic, caci numerele
$\sqrt{n+1},\sqrt{n+2},\sqrt{2n+3}$ sunt pitagoreice.


geometry
3D geometry
tetrahedron
geometry proposed

Let $ABCD$ be a tetrahedron and let $d$ be the sum of squares of its edges' lengths. Prove that the tetrahedron can be included in a region bounded by two parallel planes, the distances between the planes being at most $\frac{\sqrt{d}}{2\sqrt{3}}$

Let E, F and M be the midpoints of AC, AB and DC, respectively; also let the edge’s length of the tetrahedron be l. From D and M draw the altitudes to plane ABC to meet it at H and I, respectively. We will prove that the tetrahedron fits into the parallel planes with DC and AB on either plane.

The sum of squares of six lengths is 	6l² = d	or l =  sqrt(d/6) .
Consider equilateral triangle DAC,  DE²  = l² - l²/4 	or  DE  =  l sqrt(3) /2.
We also have BE = DE and since H is also the centroid of triangle ABC,
HE = BE/3  = DE/3 = l sqrt(3) /6.

Now consider right triangle DHE where 	DH²  =  DE² - HE² = 3l²/4 - 3l²/36 = 2l²/3  or  DH = l sqrt(2/3) =  sqrt(d) /3
Now FM²  =  MI² + FI² = (DH/2)² + (2HE)² = 1/2 sqrt(d/3), but as we can see FM is orthogonal to AB (in triangle BMA) and it’s also orthogonal to DC (in triangle DFC), therefore, the plane containing AB and the plane containing DC that are both orthogonal to FM are parallel to each other. The  tetrahedron, therefore, fits into the two planes being at most 1/2 sqrt(d/3) apart.

By Vo Duc Dien

Vo Duc Dien, ABCD is not regular tetrahedron.


summer program
PROMYS
blogs
algorithm
number theory
relatively prime
prime factorization

Show that if $ GCD(n,m)=1$ and $ a\equiv b\bmod n$ and $ a\equiv b\bmod m$ then $ a\equiv b\bmod nm$

hmmm, seems too easy, am i doing something wrong?
[hide]
by CRT, there is exactly one solution mod mn and the two given congruences satisfy the claimed one so we're done?[/hide]

well i am working through a NT book and this is in a section before the CRT so i assume that you can't use it.

fair enough, then we can just use the standard method of solving systems of linear congruences.[/hide]

[quote="Altheman"][quote="cincodemayo5590"][hide="Solution"]
Since $ (m,n)=1$, we have that $ m|a-b \text{ and }n|a-b \implies mn |a-b$.[/hide][/quote]

You didn't prove anything. You have to explain why $ (mn)|(a-b)^{2}\implies (mn)|(a-b)$ when $ \gcd(m,n)=1$.[/quote]
Um? I'm not multiplying them. $ m$ and $ n$ are distinct divisors of $ a-b$. Since $ m$ and $ n$ are relatively prime, their product is also a divisor of $ a-b$... e.g. $ 12=(2 \cdot 3) \cdot 2 = 6 \cdot 2$.

[quote="cincodemayo5590"]
Um? I'm not multiplying them. $ m$ and $ n$ are distinct divisors of $ a-b$. Since $ m$ and $ n$ are relatively prime, their product is also a divisor of $ a-b$... e.g. $ 12=(2 \cdot 3) \cdot 2 = 6 \cdot 2$.[/quote]

Okay then. I guess I added logic that wasn't there. So then your original post just restated the problem with different symbols.

I will admit that the problem is sort of trivial, but citing the the theorem as obvious [i.e. "citing" in your case would be writing an implies symbol].

I suppose a proof would go something along the lines of:

If $ a=b$, it is trivial.

Otherwise, consider the prime factorization of $ a-b$. We can factor $ a-b$ into primes that are also in $ m$, primes that are also in $ n$, and the rest of the primes. Since we have unique prime factorization in integers, a primes to a power of $ m$ must divide the corresponding prime to a power in $ a-b$. Likewise for $ n$. The sets of primes of $ m$ and $ n$ are disjoint since $ m$ and $ n$ are relatively prime. Hence $ m$ and $ n$ divide different factors of $ a-b$, so the result follows.

[quote="t0rajir0u"]I'm not sure if you can even assume prime factorization.  In any case, it's not necessary since a version of the CRT holds over general rings.

[b]Restatement:[/b]  $ gcd(m, n) = 1, m | k, n | k \implies mn | k$

[hide="Hint"] $ gcd(m, n) = 1 \Leftrightarrow \exists x, y : mx+ny = 1$ [/hide][/quote]

Obviously it is clear that my proof "proves" something. I suppose it is arguable of whether we can cite integers as a UFD. Consequently, it is reasonable for me to assume that we are explaining for that type of audience.

Anyway, I think we should consider our audience. We are doing this from a problem solving perspective. Both the AoPS books and ACoPS assume that integers are a UFD.

If we were studying this in college, that might be a concern.

Also, isn't that bezout's lemma. Perhaps the proposed problem and CRT are both direct results of that? I will assume that the proof for that doesn't assume UFD. Does the proof use the awesome extreme principle proof that normally works? [that you have learned at PROMYS, and posted like 4 times on the forums and on your blog?] :lol:

[quote="Altheman"]Perhaps the proposed problem and CRT are both direct results of that? I will assume that the proof for that doesn't assume UFD. Does the proof use the awesome extreme principle proof that normally works? [that you have learned at PROMYS, and posted like 4 times on the forums and on your blog?] :lol:[/quote]

CRT can be proven directly by a counting argument (I think) or in a few other ways, and no, it doesn't require unique factorization at all.  The division algorithm is the most important part.


find all reals $ (a,b,c)$ that makes $ \lbrace a,b,c \rbrace$ the real roots for $ x^3\plus{}ax^2\plus{}bx\plus{}c\equal{}0$

Apply viete law
$ a\plus{}b\plus{}c\equal{}\minus{}a$(1)
$ ab\plus{}bc\plus{}ca\equal{}b$(2)
$ abc\equal{}\minus{}c$(3)
from (3) we have $ c(ab\plus{}1)\equal{}0\Leftrightarrow c\equal{}0 \text{or} ab\plus{}1\equal{}0$
Case 1:$ c\equal{}0$ we have
$ 2a\plus{}b\equal{}0$
$ ab\equal{}b$
So $ (a,b)\equal{}(0,0),(1,\frac{\minus{}1}{2})$
Case 2:$ ab\equal{}\minus{}1$
$ a\plus{}b\plus{}c\equal{}\minus{}a$
$ bc\plus{}ca\minus{}1\equal{}b$
So $ \frac{\minus{}2}{b}\plus{}b\plus{}c\equal{}0$
$ bc\minus{}\frac{c}{b}\minus{}1\equal{}b$
Let $ c\equal{}\frac{2}{b}\minus{}b$ (4)
Replace to  (4) then solve equation.

[quote="waelalghamdi"]find all reals $ (a,b,c)$ that makes $ \lbrace a,b,c \rbrace$ the real roots for $ x^3 \plus{} ax^2 \plus{} bx \plus{} c \equal{} 0$[/quote]

So $ (x\minus{}a)(x\minus{}b)(x\minus{}c)\equal{}x^3 \plus{} ax^2 \plus{} bx \plus{} c$ And so :
$ a\plus{}b\plus{}c\equal{}\minus{}a$
$ ab\plus{}bc\plus{}ca\equal{}b$
$ abc\equal{}\minus{}c$

S1) : $ c\equal{}0$ and so $ 2a\plus{}b\equal{}0$ and $ (a\minus{}1)b\equal{}0$ and the two solutions $ (0,0,0)$ and $ (1,\minus{}2,0)$
S2) : $ c\neq 0$ and so $ ab\equal{}\minus{}1$, $ 2a\plus{}b\plus{}c\equal{}0$ and $ bc\plus{}ca\equal{}b\plus{}1$
$ c\equal{}\minus{}2a\minus{}b$ implies $ \minus{}(a\plus{}b)(2a\plus{}b)\equal{}b\plus{}1$ and so $ 2a^2\plus{}b^2\plus{}b\equal{}2$ and so (multiplying by $ b^2$) $ b^4\plus{}b^3\minus{}2b^2\plus{}2\equal{}0\equal{}(b\plus{}1)(b^3\minus{}2b\plus{}2)$

So we have two solutions for $ b$ : $ b\equal{}\minus{}1$ and $ b\equal{}u$, unique real root of $ b^3\minus{}2b\plus{}2\equal{}0$ : $ u\equal{}\sqrt[3]{\sqrt{\frac{19}{27}}\minus{}1}\minus{}\sqrt[3]{\sqrt{\frac{19}{27}}\plus{}1}$

And all the four solutions for $ (a,b,c)$ :

$ (0,0,0)$
$ (1,\minus{}2,0)$
$ (1,\minus{}1,\minus{}1)$
$ (\minus{}\frac{1}{u},u,\frac{2}{u}\minus{}{u})\equal{}(0.565...,\minus{}1,769...,0,639...)$ with $ u\equal{}\sqrt[3]{\sqrt{\frac{19}{27}}\minus{}1}\minus{}\sqrt[3]{\sqrt{\frac{19}{27}}\plus{}1}$


Functional Analysis
real analysis
real analysis unsolved

Suppose $ \sum a_{n}x_{n}$ is convergent for all $ x_{n}\to 0$. Prove that $ \sum |a_{n}|<\infty.$

In functional analysis terms, the dual of $ c_{0}$ is $ \ell^{1}.$

Suppose $ \sum|a_{k}|=\infty$ then there always exists a series $ \sum b_{n}=\infty$ which diverges more slowly, i.e. $ x_{n}: =\frac{b_{n}}{|a_{n}|}\to{0}$ and we have contradiction to the assumption that  $ \sum a_{n}x_{n}<\infty$ for all $ x_{n}$ (you can always choose signs of $ x_{n}$ or arguments for complex $ x_{n}$ so that $ \sum a_{n}x_{n}=\sum |a_{n}x_{n}|$).

A nice specific way to choose $ x_{n}$ was given by Abel:
$ x_{n}=\frac{1}{\sum_{k=1}^{n}|a_{n}|}$

In general it is proved in functional analysis textbooks that the space $ l_{1}$ (comprising complex sequences with $ \|a\|=\sum|a_{k}|<\infty$) is dual to the space $ c_{0}$ comprising complex sequences $ x_{n}\to{0}$ with norm $ \|x\|=\sup|x_{n}|$.  This means that every bounded linear functional on $ c_{0}$ is given by some sequence from $ l_{1}$ and its action is $ a(x)=\sum{a_{n}x_{n}}$ which must be finite because $ a$ is bounded.  Conversely every sequence from $ l_{1}$ defines such bounded linear functional.

Thank you both. This is given as an exercise to the students of analysis, so it should have a elementary proof (or at least in elementary terms), does anybody have one?


I'm currently working through volume 1 of AoPs, and I'm trying to work through all the problems, without using the solutions manual. However, many times, after 20 minutes even, I make no progress with the answer to certain problems. How long should I spend banging my head against the problem before I give up and look it up?

As long as you still want to find the solution on your own.

Also, it's never bad to skip a question and go back to it later.


How can I track my child's progress as a parent?

The parent should create an Art of Problem Solving Account.

The child should log in with his or her username. Click Alcumus on the left menu navigation, then click on the Friends tab within Alcumus. In the friends area, the child should add the parent as a friend.

The parent can then log in to their own account, click the Friends tab in Alcumus, and click on their child's username. This will bring up the child's overall statistics.


Let  f(x) = 5 x^13 + 13 x^5 + 9ax

Find the least positive integer  a  such that 65 divides f(x) for every integer x.

uhh

[hide]

we know $x^4 \equiv 1 \mod 5$ and $x^{12} \equiv 1 \mod 13$ from fermats

then we have:

$f(x) = x(5x^{12} + 13x^4 + 9a)$, at which point i took this into two cases

in order for this to be divisble by $65$, $f(x) \equiv 0 \mod 13$ and $f(x) \equiv 0 \mod 5$

so we have:

\[ f(x) \mod 5 \equiv x(0 + 3 \mod 5 + 9a \mod 5)  \]
\[ f(x) \mod 13 \equiv x(5 \mod 13 + 0 + 9a \mod 13)  \]

In which case we know $f(x)$ will be $0\mod 5$ and $0\mod 13$ if $9a \equiv 2 \mod 5$, and if $9a \equiv 8 \mod 13$

then we have \[ 9a \equiv 2 \mod 5 \equiv 27 \mod 5 \implies a \equiv 3 \mod \displaystyle\frac{5}{\gcd(9,5)}  \]

\[ 9a \equiv 8 \mod 13 \equiv 99 \mod 13 \implies a \equiv 11 \mod \displaystyle\frac{13}{\gcd(13,9)}  \]

and we see the smallest value of $a$ for which these are true is $a = 63$

edited
[/hide]

I think you're wrong, take $x=1: 5+13+9(68) = 630 = 2\cdot3^2\cdot5\cdot7$.

[hide]

since it must be divisible by 65, it must be divisible by 5 and 13, so looking in mod 5 and 13, we get that

$3+4a \mod{5} \equiv 0$ testing a bit, we get

$a\equiv 3 \mod{5}$

and $5+9a \mod{13} \equiv 0$ after a bit of value testing, we get

$a \equiv 11 \mod{13}$

then combining those with CRT, we get
$a\equiv 63 \mod{65}$, so the least value of a is 63
so we get

$f(x)=5x^{13}+13x^5+567x$ where $a=63$

[/hide]

[quote="Altheman"][hide]

since it must be divisible by 65, it must be divisible by 5 and 13, so looking in mod 5 and 13, we get that

$3+4a \mod{5} \equiv 0$ testing a bit, we get

$a\equiv 3 \mod{5}$

and $5+9a \mod{13} \equiv 0$ after a bit of value testing, we get

$a \equiv 11 \mod{13}$

then combining those with CRT, we get
$a\equiv 63 \mod{65}$, so the least value of a is 63
so we get

$f(x)=5x^13+12x^5+567x$ where $a=63$

[/hide][/quote]

it's $13x^5$, not $12x^5$.

sorry, i was a little hasty with typing my answer...edited

aw man... i had like all of that... except i did my math wrong at the end  :(


just consider a right angle triangle with the hipotenuse 5.


it is a consequence of ptolemy theorem: in an equilateral triangle $ABC$, if we take $M \in \widehat{BC}$, from the circumcircle, then $AM = BM+CM$.

[quote="Zamfirmihai"]How is this theorem? Anybody knows?[/quote]

mersi....stiam teorema asta, dar nu stiam ca este a lui Schooten si imi trebuie pt nationala


Source: Massachusetts Mathematics League March 2003

The sides of a right triangle are $x+4, x+1,$ and $2x-1$. Find the area of the smallest possible right triangle having these sides.

And can it be even smaller if the hypotenuse is $x+1$ or $x+4$?

[hide]Yes, it can, for $x={7\over 2}$, and the area is ${27\over 2}$[/hide]

This is how I did it.
For any triangle with sides $a,b,$ $and$ $c$ this must be true:
$a+b>c, b+c>a+c>b$ So
$x+4+x+1>2x-1=2x+5>2x-1=5>-1$
$2x-1+x+1>x+4=3x>x+4=2x>4=x>2$
$x+4+2x-1>x+1=3x+3>x+1=2x>-2=x>-1$
So in order for this to be defined, the only restriction we have is that x>2 and we could evaluate any number after 2.
Ok, now the question asks to find the area of the smallest possible triangle but there is no smallest possible length for the sides,meaning that we can always get closer and closer to two, so from here i know not what to do. Any suggestions?

My bad... Never mind guys
I did not see it for a right triangle. Sorry

[quote="mysmartmouth"]Can't you just keep cutting it down to smaller fractions?  What limits you?[/quote]

Aren't there only 3 different sizes? [since the permutations of the legs won't change area]

Oh.  I get it now.  Because you can boil this down to a quadratic, you have a limited number of solutions.  Gotcha!


Simone has a collection of coins consisting of 1 quarter, 1 nickel and 1 penny. Using at least two coins from the collection, how many different sums of money are possible?

There are 4 possible sums:
1Q+1N=30 cents
1Q+1P=26
1N+1P=6
1Q+1N+1P=31

Tip: When faced with these kinds of problems, just list out all the possibilities. Write them on paper, so you can verify you didn't repeat any.

Or you could do $ \binom{3}{2} \plus{} \binom{3}{3}$, which is $ 3 \plus{} 1$, or $ \boxed{4}$.


geometry
circumcircle
perpendicular bisector

Let E be the point of intersection between the diagonals of convex quadrilateral ABCD, and let P, Q, R, and S be the centers of the circles circumscribing triangles ABE, BCE< CDE, and ADE respectively. Prove that PQRS is a parallelogram.

The circumcenters $P, Q$ of the triangles $\triangle ABE, \triangle BCE$ both lie on the perpendicular bisector of their common side $BE$, a segment of the diagonal $BD$. The circumcenters $R, S$ of the triangles $\triangle CDE, \triangle DAE$ both lie on the perpendicular bisector of their common side $DE$, a segment of the diagonal $BD$. These bisectors, both perpendicular to the diagonal $BD$, are parallel. Hence, $PQ \parallel RS$.

The circumcenters $Q, R$ of the triangles $\triangle BCE, \triangle CDE$ both lie on the perpendicular bisector of their common side $CE$, a segment of the diagonal $AC$. The circumcenters $S, P$ of the triangles $\triangle DAE, \triangle ABE$ both lie on the perpendicular bisector of their common side $AE$, a segment of the diagonal $AC$. These bisectors, both perpendicular to the diagonal $AC$, are parallel. Hence, $QR \parallel SP$.

Consequently, the quadrilateral $PQRS$ is a parallelogram.


Consider the sequence $ (x_n),n\geq 1$ of real numbers so that $ x_1 \equal{} 1; x_2 \equal{} 0; x_3 \equal{} \frac13$ and for any $ n\geq 2$ we have $ (n \plus{} 2)x_{n\plus{}2} \plus{} (2n \plus{} 1)x_{n\plus{}1} \plus{} (n\minus{}1)x_n \equal{} 0$. Find $ x_n$.

[quote="tdl"]Consider the sequence $ (x_n),n\geq 1$ of real numbers so that $ x_1 \equal{} 1; x_2 \equal{} 0; x_3 \equal{} \frac13$ and for any $ n\geq 2$ we have $ (n \plus{} 2)x_{n \plus{} 2} \plus{} (2n \plus{} 1)x_{n \plus{} 1} \plus{} (n \minus{} 1)x_n \equal{} 0$. Find $ x_n$.[/quote]

[color=green]Hi [b]tdl[/b], it was posted before but i can't find a link for you. It is a problem in a book of Adrei too.

Ps. Do you know [b]let[/b] in mathsope.org ?  :lol:[/color]


Given $ a, b, c > 0$. Prove that:
$ \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \plus{} \frac{32abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \geq\ 7$

$ k\equal{}32$ is the best constant ?  :maybe:

Let $ a\equal{}4,b\equal{}3,c\equal{}5,$then

  \[ \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \plus{} \frac{32abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} <{7}!\]


number theory
greatest common divisor
geometry
geometric transformation

Hello  :)

Which are the $6$ numbers of $2$ digits each such that the $\text{m.d.c}$ is the biggest possible ?

gcd....

lalalalalalallalalallalaalalala just so I can have enough words to make this post valid.

O.K. so it is asking "What are the 6 two-digit numbers such that the GCD is as large as possible?"

These assume that they must be different.  Otherwise, it would just be 99, 99, 99, 99, 99, 99.  :D

[hide="if so"]Well, we need a number that has 6 multiples less than 100.  $\frac{100}{6}\approx16.6$, so 16 would be the GCD of 6 numbers.  The 6 multiples of 16 are 16, 32, 48, 64, 80, and 96.[/hide]

Well, I thought english speakers used the same acronym.

I'm not really sure about the translation, and I apologize for that, but I think it's like: maximum common divisor  :blush:

I should have thought of that  :wallbash:

Well, English is a very different language:  we like to change everything and make it complicated.

Now, I learned something new, so I am glad you posted it as MCD  :lol:

Just a small correction  :blush:

I didn't post it as mcd, but as mdc  :lol:

Well, that doesn't really matter, as long as people know what it means.

Would you like me to change the title to MDC (GCD)?  This might help other people who don't read all of the thread.


p is a prime number, p>3; k=[2p/3].
Prove p2 | C(p,1)+C(p,2)+...+C(p,k), where C(n,i) is a binomial coefficient.

Some unexpected changes happened when I pasted the solution from a Word document. So one must replace ? in the previous solution by "..." .
I appologise for the inconvenience.

no problem - try to use notepad for pasting .... word has some text formattings tags and stuff and they don't comply when pasted

you can also hit the edit button located in the right part of the page - above your posting text


Some biological questions have been bothering me. What is considered alive? Is being alive and being an organism the same thing?

To be more specific, I heard that a hair follicle, if put in a nutritional medium, will continue to grow. Suppose it were able to mitotically divide. Would a human hair follicle be considered alive then? Would it be considered an organism?

Here are some more examples: headless chickens can survive for days; suppose their wounds healed over so that they could continue to exist with being fed by their owners. Would they be alive? Would they be organisms?

Suppose a human finger was cut off but a continuous supply of nutrient rich blood is provided so that its cells do not die. Is it alive?

Obviously these are slightly far-fetched situations.
This question might also pertain to abortion-related topics, but that is not the original intent.

There is, in fact, a very specific biological definition of life. From Wikipedia:
[quote]
   1. Homeostasis: Regulation of the internal environment to maintain a constant state; for example, electrolyte concentration or sweating to reduce temperature.
   2. Organization: Being structurally composed of one or more cells, which are the basic units of life.
   3. Metabolism: Transformation of energy by converting chemicals and energy into cellular components (anabolism) and decomposing organic matter (catabolism). Living things require energy to maintain internal organization (homeostasis) and to produce the other phenomena associated with life.
   4. Growth: Maintenance of a higher rate of anabolism than catabolism. A growing organism increases in size in all of its parts, rather than simply accumulating matter.
   5. Adaptation: The ability to change over a period of time in response to the environment. This ability is fundamental to the process of evolution and is determined by the organism's heredity as well as the composition of metabolized substances, and external factors present.
   6. Response to stimuli: A response can take many forms, from the contraction of a unicellular organism to external chemicals, to complex reactions involving all the senses of higher animals. A response is often expressed by motion, for example, the leaves of a plant turning toward the sun (phototropism) and by chemotaxis.
   7. Reproduction: The ability to produce new individual organisms either asexually, from a single parent organism, or sexually, from at least two parent organisms.
[/quote]

[color=green]Moderator says: there is never a good reason to quote the entire post immediately preceding yours.[/color]

You're thinking of the biological definition of an [i]organism[/i]. Being "alive" is different altogether.

Terms that refer to real stuff evade clear-cut definitions; that's one reason we (at least I) do math.

[quote="sdkudrgn88"]You're thinking of the biological definition of an [i]organism[/i]. Being "alive" is different altogether.[/quote]
I don't understand what you mean by this.  What's the difference?

I also don't think this is a meaningful question as stated.  Using one word to talk about the subject of biology is just too restrictive.  For example, many people don't consider viruses to be alive because they aren't capable of [b]independent[/b] replication, but they do have DNA (or RNA) and do undergo natural selection.  Biologists study viruses because they are important to understanding our own health and because they are a biologically interesting phenomenon in their own right, regardless of whether they're "alive."

The only philosophically meaningful distinction between what is "alive" and what isn't that I can think of occurs if you think life is sacred.  There are at least three reasons why someone would think this:

- The [b]social[/b] reason:  it is immoral to kill other humans, and it is also immoral to kill, for example, pets.  By extension, it is immoral to kill animals with which we can have a meaningful social relationship.

- The [b]environmental[/b] reason:  biological life is an integral part of the proper functioning of the world at large.  Biology is also a fantastic source of important pharmaceutical chemicals and other products important to industry.

- The [b]spiritual[/b] reason:  life is embued with some kind of special property that nonliving matter is not and deserves unconditional respect.

Neither of the first two reasons depends on the definition of life; the former because it is more restrictive, and the second because it is less.  I don't find the third reason meaningful; for all practical purposes, it decomposes as a combination of the first two.


The Monroe Monkeys scored in each of the odd-numbered innings of a 9-inning baseball game.  The Freeport Ferrets scored in each of the even-numbered innings of the game.  The two teams never scored in the same inning.  The Monkeys scored consecutive even numbers of runs starting with two runs in the first inning.  The Ferrets scored consecutive odd numbers of runs starting with one run in the second inning.  By how many runs did the winning team win?

After thinking random stuff we get $ (2 \plus{} 4 \plus{} 6 \plus{} 8 \plus{} 10) \minus{} (1 \plus{} 3 \plus{} 5 \plus{} 7) \equal{} \boxed{14}$.

Edit: Thanks? :P

[quote="math154"]After thinking random stuff we get $ (2 \plus{} 4 \plus{} 6 \plus{} 8 \plus{} 10) \boxed{\minus{}} (1 \plus{} 3 \plus{} 5 \plus{} 7) \equal{} \boxed{14}$.[/quote]

 :wink:


The beginning point is $(1,0)$ and the end point is $(2006,2005)$.
In each turn you can move one up(increase $y$ in $1$) or to move one right (increase $x$ in $1$).

A) How many ways do you have from the beginning point to the end point?
B) How many ways do you have from the beginning point to the end point while each way of your point doesn't contain a point with coordinat $x=y$.

If I am true the first number of the ways is $\binom{4010}{2005}$. The second number is much more complicated.

Naphthalin

Well, this is some version of the Catalan-problem, so giving the Catalan numbers (I will consider paths from $(0,0)$ to $(n,n)$ now):

Let's count the number of paths $p$ intersecting the line $l$ given by $y=x-1$ at the point $P$ first.
By mirrowing the rest of the path (from $P$ to $(n,n)$) on the line $y=x-1$ we get now a path from $(0,0)$ to $(n+1,n)$ 'crossing' the line at $P$ first. Inversly, every path to $(n+1,n-1)$ can be mirrowed back (there is one first point of intersection for sure), so the total number of paths crossing $l$ equals the total number of paths from $(0,0)$ to $n+1,n-1)$.

So as result for the number of paths not crossing this line we get:
\[ \binom{2n}{n} - \binom{2n}{n-1} = \frac{(2n)!}{n!(n-1)!} \cdot \left( \frac{1}{n} - \frac{1}{n+1} \right) = \frac{1}{n+1} \binom{2n}{n}  \]

For part B, as ZetaX has shown, the value is the $n$th Catalan number if you go from $(0,0)$ to $(n,n)$. This is in fact one of the ways the Catalan Numbers are defined.

$C_n$ can be defined as the number of sequences $\{s_1, s_2, \cdots, s_{2n}\}$ containing only $1$'s and $-1$'s such that $s_1+s_2+\cdots+s_{2n} = 0$ and $\sum_{i=1}^j s_i > 0$ for $j = 1,2,\dots,2n-1$. Taking a move to the right to be $1$ and the move up to be $-1$, we know that the number of moves to the right has be be greater than the number of moves up (so it doesn't touch or cross $y=x$) so their sum is greater than zero, exactly as the Catalan numbers are defined. Furthermore, the end position is $(n,n)$ so the total number of moves to the right is equal to the total number of moves up which means the sum of the $s_i$'s is zero.

Hence $C_n = \frac{1}{n+1}\binom{2n}{n}$ is the number of paths.

A similar problem is if you considered starting at $(0,0)$ and only moved left and right along the $x$-axis, but you had to stay to the right of the origin and your final position is the origin.


A triangle and a trapezoid are equal in area. They also have the same altitude. If the base of the triangle is $ 18$ inches, the median of the trapezoid is:
$ \textbf{(A)}\ 36\text{ inches} \qquad\textbf{(B)}\ 9\text{ inches} \qquad\textbf{(C)}\ 18\text{ inches}\\
\textbf{(D)}\ \text{not obtainable from these data} \qquad\textbf{(E)}\ \text{none of these}$

$ mh\equal{}\frac{ah}2\Rightarrow m\equal{}\frac{a}2\equal{}9$

B


topology
advanced fields
advanced fields unsolved

[b]
Hey guys , I'm trying to prove some of theorem about $ T_2 \ Space$" But I stuck in some part of the theorem :

For a topological Space $ X$ ,the F.A.E :

a) $ X$ is a $ T_2 \ Space$

b) $ \forall p\ne q$ in $ X$ , there exist an open set U containing p s.t. q not In $ \bar{U}$.

c) the diagonal set $ D \equal{} \{ (x,x) \ : x \in X \}$ is Closed in $ X \times X$.

d) for each $ p \in X$ , then $ \{p\} \equal{} \bigcap\{\bar{U} : p\in U , U \ is \ open \ in X \}$.

I just need to prove it as (a) $ \Rightarrow$ (b). ...etc .

BUT I NEED ONLY THE PART (c) $ \Rightarrow$ (d). how could I prove this part ?  :mad: ...

I do all of them but this no...

Thanks[/b]

every equivalence here is trivial, you don't have to do a circular argument.

$ D$ is closed
$ \Leftrightarrow$ for all $ (p,q) \notin D$ there are neighboorhoods $ U$ of $ p$ and $ V$ of $ q$ such that $ U \times V$ does not intersect $ D$
$ \Leftrightarrow$ all $ p \neq q$ may be seperated by disjoint open neighborhoods, i.e. $ X$ is T2
$ \Leftrightarrow$ for all $ p \neq q$, there is a neighborhood $ U$ of $ p$ such that there is a neighborhood of $ q$ which does not intersect $ U$
$ \Leftrightarrow$ for all $ p \neq q$, there is a neighborhood $ U$ of $ p$ such that $ q \notin \overline{U}$.
$ \Leftrightarrow$ for all $ p$, we have $ \{p\} \equal{} \bigcap \{\overline{U} : U \text{ is open neighborhood of } p\}$.

[b]Thx -oo-    :)

I deleted the stupid statements from me.

Yes you deserve to laugh  :blush:

I don't read the proof .....

Sorrrrrrrrrrrrrrrrrrrrrrry , I just read the 1st statement only.

Sorry.


Let $p$ be a prime number, $p \ge 5$, and $k$ be a digit in the $p$-adic representation of positive integers. Find the maximal length of a non constant arithmetic progression whose terms do not contain the digit $k$ in their $p$-adic representation.

Let $d$ be the common difference of the sequence. If $p|d$, the units' digit never changes and we could strike it away, resulting in a sequence of common difference $\frac dp$ still having the same property.
By this, we may assume $\gcd(d,p)=1$. Now $d$ is invertible $\mod p$, let $c$ be an inverse. If the sequence is $dn+e$, we find that $n\equiv c(k-e) \mod p$ gives $dn+e \equiv cd(k-e)+e \equiv k-e+e =k \mod p$, thus if we choose a nonnegative representant of $c(k-e) < p$, then we see that one of the first $p$ terms always contains the digit $k$.
This shows that the length is at most $p-1$, archieved by $1,2,...,p-1$ itself.

PS: why $p \neq 2,3$

[quote="ZetaX"]This shows that the length is at most $p-1$, archieved by $1,2,...,p-1$ itself.[/quote]I am probably asking something extremely stupid, but why is $1,2,...,p-1$ an example of $p-1$? $k$ appears in this sequence for any $k$, doesn't it? :blush: (never worked with p-adic numbers before so I'm completely unsure)

As for $p=2,3$, I think they only call it an arithmetic progression if it has at least $3$ terms.

You are right, I mistakenly considered $k=0$ only (this problem by the way is one purely in $\mathbb Z$, so it's not really $p$-adical).

For all other cases:
If $k \neq 1$ consider the sequence $k+1 , k+2,  k+3, ... p-1 , p+0, p+1, ... , p+(k-1)$.
If $k=1$, use $2p+2,2p+3,...,3p+0$.

[quote="ZetaX"]If $k=1$, use $2p+2,2p+3,...,3p+0$.[/quote]Not if $p=3$, then $3p=100$ if I'm not mistaken. But of course there are enough trivial examples for $p=3$.


If $ m \equal{} \frac {cab}{a \minus{} b}$, then $ b$ equals:

$ \textbf{(A)}\ \frac {m(a \minus{} b)}{ca} \qquad\textbf{(B)}\ \frac {cab \minus{} ma}{ \minus{} m} \qquad\textbf{(C)}\ \frac {1}{1 \plus{} c} \qquad\textbf{(D)}\ \frac {ma}{m \plus{} ca}$
$ \textbf{(E)}\ \frac {m \plus{} ca}{ma}$

[hide="Click for solution"]
$ ma\minus{}mb\equal{}cab \implies ma\equal{}b(m\plus{}ca) \implies b\equal{}\frac{ma}{m\plus{}ca}$, or $ \boxed{\textbf{(D)}}$.
[/hide]


Is it true?
[hide]Use De Moivre identity for $x^4+y^4+z^4$[/hide]

[quote="Soarer"][quote="hardsoul"][quote="Marius Damian"]Let $x,y,z>0$, with $xyz=3$.
Prove that:
\[\frac{x^4+y^4+z^4}{4}\geq\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}+\frac{x+y+z}{9}.\][/quote]
$\frac{1}{4}\sum_{cyc}\frac{x^2}{y}\geq\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$
Now using the fact that $\frac{1}{4}\sum_{cyc}\frac{x^2}{y}=\frac{1}{12}\sum_{cyc}x^3y\leq \frac{1}{6}\sum x^4$ and from $\frac{1}{27}(x^4+y^4+z^4)\ge \frac{1}{27}xyz(x+y+z)=\frac{x+y+z}{9}$ we get that

RHS$\leq \frac{11}{54}(x^4+y^4+z^4)<\frac{x^4+y^4+z^4}{4}$[/quote]

Do you mean symmetric or cyclic?[/quote]

I meant $\sum_{cyc}\frac{x^2}{y}=\frac{x^2}{y}+\frac{x^2}{z}+\frac{y^2}{z}+\frac{y^2}{x}+\frac{z^2}{y}+\frac{z^2}{x}$  I think it must  be symmetric .

I think $\sum_{sym}$ should be the correct expression. $\sum_{cyc} \frac{x^2}{y} = \frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x}$


Let $ N \equal{} 1111...111$ (with $ 2008$,  $ 1's$)

What is the $ 1005^{th}$ digit after ther decimal place in the expansion of $ \sqrt {N}$?

No Calculator!!  :)

[hide]Let $ x \equal{} 1 0 \ldots 0$, with 1004 zeros there.

We approximate $ \sqrt{N} \equal{} \sqrt{\frac{x^2\minus{}1}{9}} \approx \sqrt{\frac{\left (x \minus{} \frac{1}{2 x} \right)^2}{9}} \equal{} \frac{x \minus{} \frac{1}{2 x}}{3} \equal{} \frac{x}{3} \minus{} \frac{1}{6 x}$

The 1005th digit after the decimal of this equals the 1st digit after the decimal of $ \frac{x^2}{3} \minus{} \frac{1}{6}\equal{} \frac{x^2\minus{}1}{3} \plus{} \frac{1}{6} \equal{} 3N \plus{} \frac{1}{6}$, so the answer is $ \boxed{1}$[/hide]


analytic geometry
trigonometry
geometry open
geometry

I'm tranquoc
Let $ ABC$ be a triangle with $ \angle{A} \equal{} 120^\circ$; $ AB>AC$ and altitudes $ BM,CN.$ Denote by $ K,L,O$ the midpoints of $ BM,CN, BC,$ respectively; $ A \in \Delta \perp KL$; $ O \in d \perp BC$; $ \Delta \cap d \equiv I$. Find $ \angle{BIC}.$

Ugly solution with coordinates and trigonometry: B(-1,0),C(1,0) and after a lot of computations and several simplifications one get BIC equals to two times the angle which has tangent (3^1/2)/5.

[quote="tran300491"]I'm tranquoc
Let $ ABC$ be a triangle with $ \angle{A} \equal{} 120^\circ$; $ AB > AC$ and altitudes $ BM,CN.$ Denote by $ K,L,O$ the midpoints of $ BM,CN, BC,$ respectively; $ A \in \Delta \perp KL$; $ O \in d \perp BC$; $ \Delta \cap d \equiv I$. Find $ \angle{BIC}.$[/quote]
$ A \in \Delta \perp KL$; $ O \in d \perp BC$; $ \Delta \cap d \equiv I$ could you tell me in language


[quote="chemrock"]What happens when acid is added to CrO42-?(easy)
In the group V confirmatory test the medium is acidic yet Ba2+ is precipiated as BaCrO4.Why?

[hide="click here to reveal the hidden content"]For those of u who I told the answer pls don't answer[/hide][/quote]

1) Cr gets converted to +7 oxdn state.

2) Is it because the acid used is too weak......  we use only organic acid.

Ok rakesh ,i know u are referrin to me.....

@ shreyas:ya partly correct but then there is one more answer for that.

you could have asked better da chemrock  :)

Oops I did not mean +7, I meant Cr2O7.... Serious TYPO.

Cr2O7 2- is in equilibrium with CrO4 2- in ph=4 if my pradeep's nv is right.. Below that you have an the concentration of Cr2O7 2- forming.. That is why you use mineral acids I gess as their ph>4 and hence will precipitate only chromate salt.. Why only barium precipitates even then has been dealt wit quite adeptly by Varunarasu..[/url]

[quote="VARUNARASU"][quote="Cyber-Shreyas"]

1) Cr gets converted to +7 oxdn state.

2) Is it because the acid used is too weak......  we use only organic acid.[/quote]

cr does not exist in +7 oxdn state. Cro42- is converted to Cr2O72- in acid medium.

and shreyas is rite, we use only acetic acid, as the ppt is soluble in mineral acids.
the solubility product of bacro4 is quite low(1.6*10^-10) whereas the ksp of other chromates is high (strontium- 3.5*10^-5 and calcium even higher) so only bacro4 is ppted.[/quote]
Yes perfectly correct.Good :10:

dei plz dont insult chappli by posting ques that are there in pradeeps da  :D  Chappli has karachikudichified pradeeps u know?  :o   :P

I didn't want to reveal Cr2O7 2- so that's why I said that the first was damn easy(even 12th std. normal person will know)

i mean chappli even knws the corresponding pH da for the 2nd one :o  :o  :o


Given that
\[ x^4 \plus{} px^2 \plus{} qx \plus{} r \equal{} (x^2 \minus{} ax \plus{} b)(x^2 \plus{} ax \plus{} c),
\] express $ p$, $ q$ and $ r$ in terms of $ a$, $ b$ and $ c$.

Show also that $ a^2$ is a root of the cubic equation
\[ u^3 \plus{} 2pu^2 \plus{} (p^2 \minus{} 4r)u \minus{} q^2 \equal{} 0.
\] Explain why this equation always has a non-negative square root, and verify that $ u \equal{} 9$ is a root in the case $ p \equal{} \minus{}1$, $ q \equal{} \minus{}6$, $ r \equal{} 15$.

Hence, or otherwise, express
\[ y^4 \minus{} 8y^3 \plus{} 23y^2 \minus{} 34y \plus{} 39
\] as a product of two quadratic factors.

hello, if we expand the product $ (x^2\minus{}ax\plus{}b)(x^2\plus{}ax\plus{}c)$ we get $ x^4\plus{}x^2(b\plus{}c\minus{}a^2)\plus{}x(ab\minus{}ac)\plus{}bc$ comparing this with $ x^4\plus{}px^2\plus{}qx\plus{}r$ we have to solve the system
$ p\equal{}b\plus{}c\minus{}a^2$
$ q\equal{}ab\minus{}ac$
$ r\equal{}bc$
with respect to $ a,b,c$.
Sonnhard.

[hide="Last one"]

Let $ y^4\minus{}8y^3\plus{}23y^2\minus{}34y\plus{}39\equal{}(y^2\plus{}ay\plus{}b)(y^2\plus{}cy\plus{}d)$

Then: $ \minus{}8\equal{}a\plus{}c,23\equal{}b\plus{}d\plus{}ac,\minus{}34\equal{}ad\plus{}bc,39\equal{}bd$

One of the 2 solutions: $ a\equal{}\minus{}7,b\equal{}3,c\equal{}\minus{}1,d\equal{}13$

$ y^4\minus{}8y^3\plus{}23y^2\minus{}34y\plus{}39\equal{}(y^2\minus{}7y\plus{}3)(y^2\minus{}y\plus{}13)$

[/hide]

[quote="aidan"]How about the middle part of the problem? That's where I'm having a problem - how to prove the non-negative root  :)[/quote]

I don't know yet...

BTW, is it too hard to prove that $ u\equal{}9$ is a root for the given values...

Wait, can't you substitute $ u \equal{} 9$? If it satisfies the equation then it's a root. So there remains the problem of how to prove that that cubic equation always has a non-negative root.

I think the middle part of the question should read :Show that this equation always has a non-negative root....

Using the equations derived in post#2 above
 $ p = b + c - a^2$
$ q = ab - ac$
 $ r = bc$ , we can eliminate b and c to find an equation connecting a,p,q,r by means of the equation

$ (b + c)^2 - (b - c)^2$  =4bc (*)

 Substituting for b+c, b-c and bc from the first three equations  into equation (*) gives

$ (p + a^2)^2$  - $ \frac{q^2}{a^2}$ =4r which reduces to

$ a^6 + 2pa^4 + (p^2 - 4r)a^2 - q^2 = 0$ which shows that $ a^2( > 0)$ is positive root of the equation

$ u^3 + 2pu^2 + (p^2 - 4r)u - q^2$ =0

Edit:Notice that the product of the roots of this equation is $ q^2$ which is positive.
This means that either all three of the roots are positve or one root is positive and the other two are negative.
We cannot have all three roots negative or one root negative and the other two positive.
We can therefore conclude that the equation always has a positive root.


Al and his friend Jack decided to invest in shares of a certain stock.  They pooled their resources and bought $ \$900$ worth of this stock. Six months later, the price of each share had increased by $ \$5.00$. They kept two shares and sold all the rest for $ \$800.00$. How many shares did they buy originally?

$ mn \equal{} 900$

OK, you can probably deduce that $ 60 < m < 90$ and $ 10 < n < 15$.

Then, we see that $ (75,12)$ works.  So they originally bought $ \boxed{12}$ stocks.

I tried using algebra, but it got a little messy.

Algebra solution:
Let $x$ be the number of stocks they bought. So the initial price of the stock $900/x$. We express the amount of money they have at the end in two ways:
$$ 900 + 5x = 800 + 2(900/x+5)$$
Simplifying and rearranging,
$$900x + 5x^2 = 810x+1800$$
$$x^2 + 18x - 360 = 0$$
$$ (x-12)(x+30) = 0$$
Thus $x = \boxed{12}$.


The circumference of a circle is divided into eight arcs by a convex quadrilateral $ ABCD$ with four arcs lying inside the quadrilateral and the remaining four lying outside it. The lengths of the arcs lying inside the quadrilateral are denoted by $ p,q,r,s$ in counter-clockwise direction. Suppose $ p\plus{}r\equal{}q\plus{}s$. Prove that $ ABCD$ is cyclic.


[quote="mdk"]Can a square be divided into several pairwise distinct squares[/quote]
i'm very bad in english,
if i inderstood you
$ (n!)^{2}=\prod_{i=1}^{n}i^{2}$

can a cube be divided into several pairwise distinct cubes.

@mdk:  Can you please clarify your question?  Also, try to use correct punctuation/capitalization.

[b]Question 1:[/b]  Can a square in the plane be cut into pairwise distinct squares?

[b]Question 2:[/b]  Can a cube in space be cut into pairwise distinct cubes?


If two marbles are removed at random from a bag containing only black and
white marbles, the chance that they are both white is one-third. If three are removed at random, the chance that they are all white is one-sixth. How many white marbles are there in the bag?

Is there anyway to to solve the problem ( except by trial and error )?

[quote="Suou"][hide]$ \frac {\binom{n}{2}}{\binom{n \plus{} b}{2}} \equal{} 1/3$ and $ \frac {\binom{n}{3}}{\binom{n \plus{} b}{3}} \equal{} 1/6$
Be $ n$ the number of white balls.
I got $ n \equal{} 6$ and $ b \equal{} 4$.[/hide][/quote]

how did you get 6 & 4? By trial and error?

Is there any real number in the result of the problem?

[quote="Suou"]Did you know what is $ \binom{n}{k}$? I expanded it.[/quote]

What equations did you write to solve for n and b variables ?

I think there are many results of the problem


Does someone know who is the perfect scorer from the Moldavia team?

As far as I know it is Mathlinks member Iura (Boreico)! Congratulations to him!

[quote="Valentin Vornicu"]It is true. Iura's track record is 1 Silver, 1 Gold, 1 Perfect Score and 2 more IMOs to go :) Congrats once again![/quote]
So it means that he went to his first IMO around 14 years old?

[quote="manuel"]valetin, can you post his solution for #3?[/quote]It is posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=282691#p282691]here[/url].


There are three digits $ A$, $ B$, and $ C$ such that the product of two-digit numbers $ \underline{AB}$ and $ \underline{CA}$ is 1995. Find $ A\plus{}B\plus{}C$.

1995 is obviously a multiple of 5.
1995=5*399
399 is obviously a multiple of 3.
1995=5*3*133
133 is not-so-obviously a multiple of 7 by the divisibility rule
1995=5*3*7*19
After a few trials and errors, two 2-digit numbers that multiply to 1995 are 57 and 35. The tens of 57 and units of 35 are same, so AB is 57 and C is 3. A+B+C=15


What is the easiest table or chart to set up that can help me solve mixture questions?  I really do not understand mixture word problems at all.

Here is a sample:

Steven wants to market a new blend of coffee that will cost 3 dollars and 90 cents per pound by mixing two coffees that sell for 2 dollars and 75 cents and 5 dollars per pound, respectively. What amounts of each coffee should be blended to obtain the desired mixture?

HINT GIVEN:

Assume that the total weight of the desired blend is 100 pounds.

[hide="Solution"]
For these sort of problems, you need to visualize them by dividing what's being asked into two questions.

1) How much stuff do I want?
2) What concentration or price is it per unit?

For example:
Your goal is 100 pounds of coffee at 3.90 per pound. Which comes out to 390 dollars.
So we have two equations we can set up: one based on the final weight, and one based on the final price.

Let $x$ be the amount (in pounds) of cheap coffee, and $y$ be the amount of expensive coffee.

Then we have
$x+y = 100$ (that's because the total amount between the two must be 100 pounds)
$2.75x+5y = 390$ (that's because the total price of the two is 390 dollars)
Then you can solve this system for $x$ and $y$.

That's the general solution for this kind of problem.
You get two equations based with two different types of information, and then solve the system.

I'll let you solve this one.
[/hide]

I can solve this system of equation in two variables.

However, there is an easier method using charts or tables setting up information leading to an equation(s) where we need to solve for a variable(s).

How do I set this given data into a chart or table?

Thanks

P.S. I understood your explanation and can solve the equations but I am interested in charts and/or tables that can help me better understand NOT ONLY mixture questions but word problems in general.

I tried the purplemath lady but her method is not too clear.

Her method can be found here:

http://purplemath.com/modules/mixture.htm

In general it is a good strategy on word problems to pull out the crucial information.
Because the spectrum of information can be so large, it is hard to give advice for general word problems.
Here's what I do when I see one:
1) Pull out all the numbers, and decide which are important and which are not.
2) Look for any proportions/constants that I should know.
3) Look for any relations between values or variables that seem key.
4) Set up equations involving these constants and relations, and solve them in terms of a variable that I find convenient/necessary.
5) Express all information in terms of as few variables as possible, and then extract necessary data for the solution to the problem.

That's my meta-strategy.
You can organize this however you want, i.e. a table or a graph or whatever. Sometimes a quick sketch of the problem can help as well.

I guess daily practice with word problems can help, right?

How much time should I spend playing with word problems daily?

It's really up to your style of math practice.
I spend no more than 45 minutes a day on math, and roughly 25 mins of that are old competition problems. The other 20 mins are devoted to learning new things.

That time split seems to work well for me, as I can do a couple problems and learn a new thing roughly every day.

It's really up to you and how you choose to devote your time.

I love math.

Do you think one hour of daily word problems practice and another hour of pratice with other math topics (like equations, functions, graphing, trig, etc) will suffice?

That's plenty.
If you keep working at it, and ask for advice from people who are better than you (and believe me, you can find all of them on AoPS, then you'll get better quickly.

A few good senior members can help you out:
Altheman,
T0rajir0u,
JBL,
MysticTerminator (though he tends to spam a lot, he's won Gold at IMO),
JSteinhardt

and a few others whose name I forget.


A. In a test of qualitative control, a router of network (router),
undertakes not distinguished network "parcels" to four different channels.
With how many ways it can become a test routing of 300 parcels in one
second if the first channel has "inexperienced" capacity,
second and third 50 parcels/second while fourth 40 parcels/second
should be dispatched ten at least parcels in each channel?
(write the generator interelation and solved it so that is found
the factor of term that gives the right answer).

B. In a half test, the router blames moreover different (distinguished)
heading in each one from the parcels front him it undertakes. ¶With how
many ways it can become the routing of 60 such parcels in one second if
we doesn't interest the line with which they will pass from the each
channel? (write the generator interelation and determine
the term of the factor that gives the right answer.(Is not required
the calculation of factor).With how much ways it can become the routing
of 60 such parcels in one second if we are interest for the
line with which they will pass from the each channel?
(write the generator interelation and determine the term
of the factor tha gives the right answer.(Is not required the calculation of factor).

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=172156

So are you the same person, or are you cheating on the same work, or what?

I felt that I might run into some trouble :)
No, he pm me (wrong person for advanced math) asking for help with the problem and said that he posted but "there is no answer" so I thought that he is having trouble with his AoPS account, so I posted the problem for him. I should have use the search button. I am really sorry for this.

he pmed me too but from a different username than that. that name only posted in the greek forums...not sure what that signifies, but w/e.


calculus
integration
calculus computations

Evaluiate the folowing integrals:

$ \int_{0}^{4}\int_{0}^{\sqrt{16\minus{}x^2}}xy^3dydx$

$ \int_{\minus{}1}^{1}\int_{0}^{\sqrt{1\minus{}y^2}}e^{\minus{}x^2\minus{}y^2}dxdy$

First one's easy.  Integrate first with respect to $ y$.
$ 128x^{2}\minus{}\frac{32x^{3}}{3}\plus{}\frac{x^{6}}{6}$ from $ x\equal{}0$ to $ x\equal{}4$, whatever it is.  Here, all I did was that I integrated with respect to $ y$ to get $ \frac{xy^{4}}{4}$ and then plug in the two boundaries.  The $ y\equal{}0$ makes it disappear so only the top one matters.  Plug it in, square it, integrate it, and game over.

For the second one, it actually can't be integrated the way it is, so you're going to have to convert to polar coordinates.  Don't forget that extra $ r$ from the Jacobian transformation.


ratio
LaTeX
geometric series
geometric sequence

I was bored during math class today so I came up with this formula for infinite sums of geometric sequence.
$ x\equal{}n^1\plus{}n^2\plus{}n^3\plus{}n^4\plus{}...\plus{}n^\infty$
$ x\equal{}n(n^0\plus{}n^1\plus{}n^2\plus{}n^3\plus{}...\plus{}n^\infty)$
$ x\equal{}n(1\plus{}x)$
$ x\equal{}n\plus{}nx$
$ x\minus{}nx\equal{}n$
$ (1\minus{}n)x\equal{}n$
$ \boxed {x\equal{}\frac{1\minus{}n}{n}}$
Then I experimented with some numbers and found interesting results:
$ n\equal{}\frac{1}{2}$
$ \frac{1\minus{}\frac{1}{2}}{\frac{1}{2}}\equal{}1$

$ n\equal{}\frac{1}{3}$
$ \frac{1\minus{}\frac{1}{3}}{\frac{1}{3}}\equal{}\frac{1}{2}$

$ n\equal{}1$
$ \frac{1\minus{}1}{1}\equal{}0$

$ n\equal{}2$
$ \frac{1\minus{}2}{2}\equal{}\minus{}2$

$ n\equal{}3$
$ \frac{1\minus{}3}{3}\equal{}\minus{}\frac{2}{3}$

I think the results for fractions are very interesting. What happened? Why are they negative for integers? I know they are infinity, but why isn't it giving $ \frac {1}{0}$?

Isn't it supposed to be $ \frac{n}{1\minus{}n}$? But I don't think that changes anything.

Like the one theorem with proving .99999....=1, the repeating part is messed up because if you take one of infinity, there really is no exact number like infinity-1 because infinity-1=infinity. That's part of the problem i think.

You're right. I messed up. For fractions it doesn't matter, but for 1 it is $ \infty$ and for 3 it is $ \minus{}\frac {3}{2}$.

[quote="jonathanchou711"]Isn't it supposed to be $ \frac {n}{1 \minus{} n}$? But I don't think that changes anything.

Like the one theorem with proving .99999....=1, the repeating part is messed up because if you take one of infinity, there really is no exact number like infinity-1 because infinity-1=infinity. That's part of the problem i think.[/quote]
:) $ 0.9999999999... \equal{} 3\times 0.33333333... \equal{} 3\times \frac13 \equal{} 1$ :)

you have to specify that the common ratio has to be less than the absolute value of 1

[hide="long post"]I'm pretty sure the problem arises when validity of (real number) axioms is assumed while dealing with the counterintuitive subject of infinity. In your derivation of the formula, it seems that you assume that x and n are real numbers, which is not necessarily true (what if the sum diverges?) You have to be [b]very[/b] careful when multiplying and dividing by infinity. For example, with n=2, this is what is happening(maybe slightly differently, but it's pretty much the same_:

1+2+4+8+...=x
1+2(1+2+4+...)=x
1+2x=x
1=-x so x=-1. There are several places where this could have a problem (i'm not sure which one in particular is the problem though)

1) In line 2, where the distributive law was applied. Can the distributive law be applied to infinite sums? Try to think about the example $ \minus{}1\equal{}\minus{}1\plus{}0\plus{}0\plus{}0\plus{}0\plus{}...\equal{}\minus{}1\plus{}(1\minus{}1)\plus{}(1\minus{}1)\plus{}(1\minus{}1)\plus{}...\equal{}\minus{}1\plus{}1\minus{}1\plus{}1\minus{}1\plus{}1...\equal{}(\minus{}1\plus{}1)\plus{}(\minus{}1\plus{}1)\plus{}(\minus{}1\plus{}1)\plus{}....\equal{}0\plus{}0\plus{}0\plus{}....\equal{}0$ to see why it's not entirely obvious whether or not these real number axioms can be applied. This one illustrates problems with the associative law(being used infinitely many times).
2)In line 3, where a substitution was made. are we allowed to substitute infinity?

3)In lines 3-4, we subtracted 2x (which is infinite). What happens when you subtract infinity from both sides of an equation?

On the other hand, if it is specified that x is a real number (because the infinite sum converges), then this formula is correct. (except i think you wrote x as the reciprocal of what it should be; check your algebra in the last few steps)

'Another' way to try to do the same thing is the following:
1+2+4+8+....=x
2+4+8+16+....=2x
....2+4+8+....=2x
1+2+4+8+....=x

The third and fourth lines are rewrites of equations two and one, respectively, to make it clearer. Now subtract the last two equations to get -1=x.

I think that most of these contradictions (if not all) rely on a subtle way to bend the axioms, whether it is dividing by 0, or subtracting infinity. Heck, I even used to have a proof that 1=-1 which relied on a sketchy understanding of limits. If i remember correctly, i think the problem was that i said that (I don't know the latex for limits)

Lim(x->2)  $ i^x\equal{}i^2\equal{}\minus{}1$, where i is the imaginary blah blah blah. Even if this were true(I'm pretty sure it's not) I still had all these other holes, such as performing an operation infinitely many times, and undoing it later, and blah blah blah this is such a long post...LOL[/hide]

[hide="proving 0.9~=1 a few ways"]$ 0.\overline{9} \equal{} \frac {9}{10^1} \plus{} \frac {9}{10^2} \plus{} \frac {9}{10^3} \plus{} \ldots$. From the formula for an infinite geometric series sum, we have $ \frac {\frac {9}{10}}{1 \minus{} \frac {1}{10}} \equal{} \frac {\frac {9}{10}}{\frac {9}{10}} \equal{} 1$.
$ Q.E.D.$
_________________________________________________________________
Let $ x \equal{} \text{number corresponding to }0.\overline{9}$. This is basically
\[ x \equal{} 0.\overline{9}.
\]
Multiplying both sides by $ 10$ gives
\[ 10x \equal{} 9.\overline{9}.
\]
Taking the difference between both equations gives us
\[ 9x \equal{} 9\implies x \equal{} 1.
\]
$ Q.E.D.$ (of course, the following $ 9$'s all disappear)[/hide]

[hide="formula for infinite geometric series sum, and proof"] Let $ a \equal{} \text{first number}, r \equal{} \text{common ratio where } |r| < 1, S \equal{} \text{sum of all the terms}$.
\[ S \equal{} a \plus{} ar \plus{} ar^2 \plus{} ar^3 \plus{} \ldots
\]

\[ rS \equal{} ar \plus{} ar^2 \plus{} ar^3 \plus{} ar^4 \plus{} \ldots
\]
Taking the difference between these two sums gives
\[ S \minus{} rS \equal{} a.
\]
How do all of the other terms disappear? This is because $ ar^{\infty} \equal{} ar^{\infty \plus{} 1}.$ Simplifying, we get $ S(1 \minus{} r) \equal{} a\implies S \equal{} \frac {a}{1 \minus{} r}.$
If $ |r|\ge1,$ the denominator is either negative or $ 0$. Don't let this fool you into thinking the sum of an infinite geometric series when $ |r|\ge1$ is negative!  It actually obviously diverges, since the terms just get larger and larger and larger and larger... so the sum is [i]infinite[/i][/hide]

Maybe this should be moved to classroom math? :maybe:

It should be moved to classroom. But proving that .99999=1 doesn't matter! That's not the point!!!!!!

Our objective is to see what PowerofPi has done wrong.

But you see PowerofPi: In the step that you converted $ x=n^1+n^2+n^3+.....n^\infty$ to $ x=n(n^0+n^1+....n^\infty)$, the $ n^\infty$ should be $ n^{\infty-1}$.

As I said in my last post, there really is no limit to infinity. Infinite sums equations always have some flaw in them. Like the .99999...=1 proof.

$ x=.99999....$
$ 10x=9.9999....$
$ 9x=9$
$ x=1$

In this case, the 10x=9.9999 is wrong. If we were to say that x=.999 not repeating, then 10x=9.99. Which means x is not exactly equal to 1. But no matter how far you go, you will always have infinity - 1. But infinity is forever. Math is confusing.  :huh:  :ninja:

But we're not saying $ x\equal{}0.999$. We're saying that $ x\equal{}0.\overline{9}!$ There is a difference. We have a string  of an infinite number of $ 9$'s. When we multiply by $ 10,$ there is still an infinite number of $ 9$'s, so all of them cancel out when we subtract.

What does geometric sums have to do with the proof of $ .\overline{9}\equal{}1$? The concept is similar, but the different proofs has nothing to do with this thread.

heee heee haaa haaa how is it not a geometric sum?
$ 9\times\left(\frac{1}{10}\right)^1\plus{}9\times\left(\frac{1}{10}\right)^2\plus{}9\times\left(\frac{1}{10}\right)^3\plus{}9\times\left(\frac{1}{10}\right)^4\plus{}\dots$

[quote="CaptainAwesome"]heee heee haaa haaa how is it not a geometric sum?
$ 9\times\left(\frac {1}{10}\right)^1 \plus{} 9\times\left(\frac {1}{10}\right)^2 \plus{} 9\times\left(\frac {1}{10}\right)^3 \plus{} 9\times\left(\frac {1}{10}\right)^4 \plus{} \dots$[/quote]
It isn't funny and we're talking about the general formula. We don't care about the proof of a single one.

Sorry for laughing i didn't know you hated laughter so much
You should make a change to your formula then since if $ x>1$ then $ x^\infty \equal{} \infty$ and put the condition $ \minus{}1 \le x\le 1$
If $ x<\minus{}1$, then you will find that it will be extremely screwed up since no one knows whether $ \infty$ is even or odd, and that will greatly affect the result.

When $ n>1$, the difference between $ n^x$ and $ n^{\left(x\plus{}1\right)}$ approaches $ \infty$ as $ x$ approaches $ \infty$ so you can't really factor out that $ n$ i think. But $ \infty$ is a complicated thing, so don't blame me if I made some wrong statements.

[quote="CaptainAwesome"]When $ n > 1$, the difference between $ n^x$ and $ n^{\left(x \plus{} 1\right)}$ approaches $ \infty$ as $ x$ approaches $ \infty$ so you can't really factor out that $ n$ i think. But $ \infty$ is a complicated thing, so don't blame me if I made some wrong statements.[/quote]
Wouldn't the difference approach 0 because $ n^\infty\equal{}n^{\left(\infty \plus{} 1\right)}$?

ahhhh no you got it wrong :) Just plug in some numbers in your calc and you'll see. $ e$ has something to do with it but that's special

Don't use infinity as a number.

Consider the sequence 1,3,5,7,9,...

Clearly the difference between adjacent terms is 2.

However, you might reason like this:

The nth term is 2n+1.

When n is infinity, 2n+1 and 2(n+1)+1 are the same number: infinity.

Therefore when n gets large, the difference between adjacent terms approaches 0.

This is clearly false...

[quote="PowerOfPi"]I like laughing and sorry. Thanks, but [u][i][b]why [/b][/i][/u]doesn't it work for numbers greater than 1 and less than -1? (I know it is undefined, but why does it give defined answers?)[/quote]

Someone correct me if I'm wrong, but I think this is because otherwise, the series wouldn't converge, and would sum to infinity.

[hide="geometric sequences"]A geometric sequence is a sequence where the $ n\plus{}1$st term is $ r$ times the $ n$th term. So if the first term is $ a$, then the second term would be $ ar$, the third term would be $ arr$, the fourth term would be $ arrr$, etc. Thus a geometric sequence is in the form

$ a, ar, ar^2, \cdots$

$ r$ is defined to be the common ratio.

Some people are more concerned about the sum of the entries in the sequence other than the sequence itself, so I shall cover that. I shall assume that $ a$ and $ r$ are nonzero real numbers.

If $ a>0$ and $ r>1$, the terms just keep getting larger and larger without bound, hence the sum is too big to contemplate. The same for $ a<0$ and $ r>1$, just that the terms get smaller and smaller without bound. If $ r\equal{}1$, then we have so many $ a$'s, and their sum is infinitely large/small. If $ 0<r<1$, then the terms get closer and closer to 0, so we might have a shot at evaluating the sum this way. So we shall assume that $ 0<r<1$.

Let $ S\equal{}a\plus{}ar\plus{}ar^2\plus{}\cdots$. We have $ rS\equal{}ar\plus{}ar^2\plus{}\cdots \equal{}S\minus{}a$. Therefore $ a\equal{}S(1\minus{}r)$, or $ S\equal{}\frac{a}{1\minus{}r}$.

The same formula holds true for $ \minus{}1<r<0$. For $ r\equal{}\minus{}1$, we do not know what the sum will be; it could either be $ a$ or $ 0$, from common sense. If $ r<\minus{}1$, then the terms get farther and farther away from 0, and it becomes all the more difficult to evaluate the sum. Thus for $ \minus{}1<r<1$ and $ r\neq 0$, we have

\[ a\plus{}ar\plus{}ar^2\plus{}\cdots \equal{}\frac{a}{1\minus{}r}.\][/hide]

Here's a sequence to find the sum of:

$ \frac{1}{7}\plus{}\frac{2}{49}\plus{}\frac{3}{343}\plus{}\cdots \plus{}\frac{n}{7^n}\plus{}\cdots$

[hide="hint"]How would you compute $ \frac{1}{7}\plus{}\frac{1}{49}\plus{}\cdots$? Could a similar method be used here?[/hide]

[quote="jonathanchou711"]You cannot exactly post an infinite number sum of a whole number. Only fractions. If it was the infinite sum, it would be equal to infinity.[/quote]

Not true: 0+0+0+0+0+...

The numbers don't have to be fractions:

2^(-1/2) + 2^(-3/2) + 2^(-5/2)+... converges, but non of it's terms are fractions.

Infinite sums: Some sequences converge, some sequences don't.

Geometric sequences aren't the only sequences that converge:

The following sequence converges:

1/1+1/4+1/9+1/16+1/25+...


algebra
polynomial
Vieta
number theory
relatively prime
number theory unsolved

Find the minimal value k such that in every k element subset of {1, 2, ... 50} there exists a pair (a,b,) such that a+b divides ab.

$ a \neq b$?
If a can equal b then its every non-empty subset or $ 2^{50} \minus{} 1$

What if a and b are odd? I don't think 3+3 divides 3*3.

Also, the answer is at most 50, I think you may have misread the problem statement.

Edit: It would be nice if you could state whether a could equal b, though.

yes i did. I think you use vieta jumping to get a condition on when $ a\plus{}b|(ab)$

The answer is $ k \equal{} 26$.
No two elements from the subset $ \{1,3,5,...,49\}$ satisfies $ a \plus{} b \mid ab$, since $ a \plus{} b$ is even while $ ab$ is odd for any $ a,b \in \{1,3,5,..,49\}$.
If there is any $ k$ element subset of $ \{1,3,..,49\}$ with $ k \ge 26$, then there exists an even number $ 2n$ from the set. So the pair $ 2n,2n \in \{1,3,...,49\}$ satisfies $ 2n \plus{} 2n \mid (2n)^2 \iff 4n \mid 4n^2$, which is clearly true.

I claim that the answer is 39.  I am going to assume that a and b are distinct as kalva has that as distinct.

Let a and b be in S such that a+b divides ab.  Then, let c=gcd(a,b), and put $ a\equal{}ca_{1}$, and $ b\equal{}cb_{1}$, so that $ a_{1}$ and $ b_{1}$ are relatively prime.  Then, $ c(a_{1}\plus{}b_{1})$ divies $ ca_{1}b_{1}$.  Since $ a_{1}$ and $ b_{1}$ are relatively prime, so are $ a_{1}$ and $ a_{1}\plus{}b_{1}$ and $ b_{1}$ and $ a_{1}\plus{}b_{1}$.  So, $ a_{1}\plus{}b_{1}$ divides c.

Now, a+b is at most 99, so $ c(a_{1}\plus{}b_{1}) \le 99$, which mean that $ a_{1}\plus{}b_{1} \le 9$, but $ a_{1}\plus{}b_{1} \ge 3$.  Now, we can go case work to see sumwise that if $ a_{1}\plus{}b_{1}\equal{}3$, the pairs are (6,3), (12,6), (18,9), (24, 12), (30, 15), (36, 18), (42, 21), (48, 24)

for 4, (12, 4), (24, 8), (36, 12), (48, 16)
for 5, (20, 5), (40,10), (15, 10), (30, 20), (45, 3),
for 6, (30, 6)
for 7, (42, 7), (35, 14), (28,21)
for 8, (40, 24)
for 9, (45, 36)

Let A={6,12,15,18,20,21,24,35,40,42,45,48} and B={1,2,...,50}-A.
Since each pair listed above contains an element of A, B does not have the desired property.  Hence we must take at least $ k \ge |B|\plus{}1\equal{}39$.

Now the following twelve pairs from the 23 mentioned above are mutually disjoint (we choose an element from A and B).
(6,3), (12,4), (20,5), (42,7), (24,8), (18,9), (40,10), (35,14), (30,15), (48,16), (28,21), (45,36).

Any 39-element subset must contain both elemnts of one of these pairs.  We get the desired minimal number is k=39.


geometry
circumcircle
trigonometry
geometry proposed

Let $ \Delta ABC$ be a triangle and let $ BE$ and $ CF$ be its altitudes.
we consider a point $ P$ on the circle with diameter $ BC$ and draw the perpendiculars from it to $ AB$ and $ AC$ and let them intersect $ AB$ and $ AC$ at $ C'$ and $ B'$ respectively.
let $ H'$ be the orthocenter of the triangle $ AB'C'$
find the locus of the point $ H'$ when $ P$ varies on the circle $ BFEC$
(sorry if it has been posted before :oops: )

Denote $ O$ the midpoint of $ AP$ then $ O$ is the circumcenter of $ AC'PB'$.
Since $ H'$ is the altitude of $ \Delta AB'C'$ then $ \angle FAH'\equal{}\angle PAC$.
On the other side, denote $ M$ the midpoint of $ B'C'$.
We have $ \frac{AH'}{AP}\equal{}\frac{OM}{OB'}\equal{}\cos A\equal{}\frac{AF}{AC}$
Therefore $ \Delta FAH'\sim \Delta CAP \Rightarrow \angle AFH'\equal{}\angle ACP$. But $ \angle AFE\equal{}\angle ACB \Rightarrow \angle H'FE\equal{}\angle PCB$.
Similarly, $ \angle H'EF\equal{}\angle PBC$.
$ \Rightarrow \angle FH'E\equal{}\angle BPC\equal{}90^o$. So $ H'$ lies on the circle with diameter $ EF$.

my solution was to consider the point $ H''$ which is constructed as follows:
Let $ PB' \cap (BC)\equal{} \{P,B'' \}$ and similarly for $ PC'$
then let $ H''\equal{} FB'' \cap EC''$
then we can prove that $ H''\equal{}H'$ also the equality $ \angle EH''F\equal{}90^\circ$ can be obtained easily
we can do all of this using just angle we just use the fact that $ BE \parallel{} PB'$ and $ CF \parallel{} PC'$


geometry
geometric transformation
reflection
circumcircle
inradius
analytic geometry
inequalities

An interesting problem I think was proposed by me for Mathematical Reflections (http://www.awesomemath.org).

Let $O$, $I$ and $r$ be the circumcircle, incircle and the inradius of a triangle $ABC$.
Let $M$ be a point inside the triangle $ABC$ and let $d_{a}, d_{b}, d_{c}$ be the distances to the sides $ABC$.
Prove that if  $d_{a}\cdot d_{b}\cdot d_{c}\geq r^{3}$ then $M$ lies inside the circle with center $O$ and radius $OI$.

A large solution with barycentric coordinates

Hi prowler your peoblem is really nice problem,I have found a similar problem:
Let $O,G$ and $R$ be the circumcircle, centroid and the circumradius of a $\triangle ABC$.Let $M$ be a point inside the triangle $ABC$ and let $d_{a}, d_{b}, d_{c}$ be the distances to the sides $ABC$.Prove that if $d_{a}\cdot d_{b}\cdot d_{c}\geq \frac{R^{3}}{8}$ then $M$lies inside the circle with center $O$ and radius $OG$.
Hope you are interested in!  :P

You are very clever gemath :)

We have $2S=ad_{a}+bd_{b}+cd_{c}\geq 3\sqrt[3]{abc}\sqrt[3]{d_{a}d_{b}d_{c}}$

Hence $d_{a}d_{b}d_{c}\leq \frac{8S^{3}}{27abc}\leq \frac{R^{3}}{8}$.

Last inequality is obviously true $27R^{4}\geq 16S^{2}$ so this is true iff
$P$ is center of gravity in the equilateral triangle.

Actually I have a nice solution for my problem in about five lines, which I will post tomorrow if nobody will post his one.

Hi prowler! Thank you for your praise! but  I made this inequality weakly really it is Prove that if $d_{a}d_{b}d_{c}\geq \frac{Rr^{2}}{2}$ then $M$ lies inside the circle with center $O$ and radius $OG$.
Hope you are interested!
I have a solution for your problem with barycentric but really It isn't large becasue I wait for you post your solution! :)
Another idea for extend this problem but I see it isn't nice:
Prove that if $\frac{d_{a}}{h_{a}}(s-a)^{2}+\frac{d_{b}}{h_{b}}(s-b)^{2}+\frac{d_{c}}{h_{c}}(s-c)^{2}\le \frac{2R}{r^{2}}d_{a}d_{b}d_{c}-9r^{2}$ then $M$ lies inside the circle with center $I$ and radius $IG$ with $I$ incenter ,$G$ centroid, $s$ semiperimeter.
Hope you are interested in!  :P  :P

I have a solution to prowler's problem.

Notice that $\sum \frac{d_{b}d_{c}}{bc}= \frac{S_{pedal}}{S}= \frac{1}{4}|1-\frac{OP^{2}}{R^{2}}|$

On the other hand, $\sum \frac{d_{b}d_{c}}{bc}= \frac{d_{a}d_{b}d_{c}}{abc}(\sum \frac{a}{d_{a}}) \geq \frac{r^{3}}{2Sabc}(\sum ad_{a})(\sum \frac{a}{d_{a}}) \geq \frac{4p^{2}r^{3}}{8S^{2}R}= \frac{r}{2R}$.

Thus $|R^{2}-OP^{2}| \geq 2Rr$. Since $P$ is inside the triangle, it follows that $OP^{2}\leq R^{2}-2Rr = OI^{2}$. And the result follows.

A very nice problem, prowler.

To gemath, it seems that your last problem is quite intimidating. However, I need to ask you if you have some typo or not. About the other problem, in fact, you always have the inequality $d_{a}d_{b}d_{c}\leq \frac{Rr^{2}}{2}$, (here $M$ lies inside the triangle) by the same argument presented above. The equality happens when $M$ is the circumcenter of equilateral triangle $ABC$.

ok treegoner you are right I don't know that $d_{a}d_{b}d_{c}\leq \frac{Rr^{2}}{2}$ but really it is $d_{a}d_{b}d_{c}\ge \frac{r^{2}(a^{2}+b^{2}+c^{2})}{18R}$ I hope there isn't inequality:$d_{a}d_{b}d_{c}\le \frac{r^{2}(a^{2}+b^{2}+c^{2})}{18R}$ about  problem with incenter I think it is true but really it isn't nice  :)  :)

My solution is almost the same as treegoner's one.
Well done !

[quote="Summerburn"][quote="treegoner"]I have a solution to prowler's problem.

Notice that $\frac{S_{pedal}}{S}= \frac{1}{4}|1-\frac{OP^{2}}{R^{2}}|$

[/quote]

May I ask here, how this come?[/quote]

This is an equivalent form of http://www.mathlinks.ro/Forum/viewtopic.php?t=38164 .

  darij


A box contains red, green, blue, and white balls, 111 balls in all. If you take out 100 balls without looking, then there will always be 4 balls of different colors among them. What is the smallest number of balls you must take out without looking to guarantee that among them there will always be balls of at least 3 different colors?

[quote="Xixas"]A box contains red, green, blue, and white balls, 111 balls in all. If you take out 100 balls without looking, then there will always be 4 balls of different colors among them.[/quote]
This condition means that thre are at least 12 balls of each color.
Therefore, if we take 88 balls then it is impossible that we have at most two colors.
On the other hand, if there are 12 red, 12 blue, 12 green and 75 white balls then we can choose 87 balls of two colors and for any 100 balls we have all four colors.

There must be at least 12 balls of each colour, otherwise if there are fewer than 12 balls of a particular colour, you can select 100 balls of the other 3 colours.  Hence it suffices to take out at most $111-2 \times 12+1 = 88$ balls.  To show that you must take out at least 88 balls, consider the case where there are 12 red, green and blue balls each.


AMC
USA(J)MO
USAMO
geometry
ratio
analytic geometry

I will try it... I guess. Wait until the weekend to post your solution. However, I do have MOCK problems, of which I have only solved the geometry one.

[quote="n^4+4"]This was a #3 from some IMO wasn't it?  Maybe I am not remembering properly.  I will post a solution soon (and another problem) if no one else jumps on it.  Come on Bryce and Damien... you want to make MOP don't you? (and Vitek, Jane, etc. if you read this)[/quote]

lol #3? It's an IMO problem, but a relatively easy one. And a #1.

It is a little difficult for me to make sense out of this problem (probably because English is not my native language), but could you post the link where you've got this problem?  Thanks.

I agree, and English is my native language.

By sequence of moves, do you mean one point hopping over another? Or what? Define it please.

Also, I couldn't find this wording looking through the IMOs.

EDIT: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=354112#354112

It was indeed a #3.

[quote="jb05"]Given n not-necessarily-distinct points on a line and a constant λ, a move consists of jumping a point A to the left of a point B over B to a point λ times as far away from B as is A from B. Find, in terms of n, all possible λ such that any original configuration of n points (not all coincident) has a sequence of moves moving the leftmost point arbitrarily far right.[/quote]
 :huuh: what does that mean :?:

umm so try to read it?

or look up the original formulation in the contests section

it is like you have a bunch of fleas and they jump over each other with ratio lambda

badly worded imo (even though i couldnt do any better)

I think it's perfectly clear. If it's a little wordy and formal, that's for the sake of mathematical precision.

How about this for the people who can't understand the problem as stated:

Let $ n$ be an positive integer with $ n \ge 2$.  Let $ \vec{X} \equal{} \{x_1,x_2,\cdots,x_n\} \in \mathbb{R}^n$ and let $ \lambda \in \mathbb{R}$.  A move consists of choosing some integers $ 1 \le i \neq j \le n$ such that $ x_j > x_i$ and replacing $ x_i$ with $ x_j \plus{} \lambda(x_j \minus{} x_i)$.  For given $ \vec{X}$, we define $ f(\vec{X})$ to be the maximum possible value of $ \max{\{x_i\}}$ as the number of moves grows arbitrarily large.  Let $ L$ be the set of all values of $ \lambda$ such that we can vary all of the $ x_i$ (without having them all equal) and still have $ f(\vec{X}) \equal{} \infty$.  Find $ L$ in terms of $ n$.

[hide="My suspicions"]
I assume that $ \lambda$ is taken to be positive, as jb05's formulation does not make sense if $ \lambda < 0$.  Obviously if $ \lambda \equal{} 0$ then we are finished.

Now if $ \lambda \ge 1$, then we are pretty obviously done.  For given $ \vec{X}$, let $ d(\vec{X}) \equal{} \max{\{x_i\}} \minus{} \min{\{x_i\}}$.  Then we simply take the rightmost point in $ \vec{X}$ and jump it over the leftmost point.  It is fairly obvious that if $ \lambda \ge 1$, $ d(\vec{X})$ will never decrease.  Hence if we look at the sequence $ r_i$ of the coordinates of the rightmost fleas after they have jumped the leftmost fleas, we will see that in each case $ r_{i \plus{} 1} \minus{} r_i \ge d$.  Hence $ \sum_{i \equal{} 1}^{\infty} r_i \equal{} \infty$.  So if $ \lambda \ge 1$ then we can always get a flea as far away as we want.

As for $ 0 < \lambda < 1$, I suspect that the set of such values is empty, but I'm not sure.[/hide]

bleargh hello so let the minimal distance between any two be $ k$

so if $ \lambda$ is at least $ \frac{1}{n\minus{}1}$, then make the furthest-left one jump over the furthest-right one

then the new furthest-right flea is at least distance $ k$ from the old furthest-right flea

so the minimal distance is still $ \ge k$

blah I forgot the other direction though


calculus
derivative
geometry
calculus computations

If f(1)=1+f(0) and f'(1)=f'(0)=0,then prove that |f''(x)|>=4 for at least one real x belongs to (0,1)

I know I posted essentially this several years ago in two places: once in the calculus/analysis section, and once more in the physics section. At the moment, I'm too lazy to look up the links.

If $ \left|f''\left(x\right)\right|\leq4$ for all $ x\in\left(0,1\right)$, then the positive area under the  graph of $ f'\left(x\right)$ should lie entirely within the triangle whose veritices are $ \left(0,0\right),\left(\frac{1}{2},2\right),\left(1,0\right)$ which is less than 1


Our school's girls volleyball team has 14 players, including a set of 3 triplets: Alicia, Amanda, and Anna.  In how many ways can we choose 6 starters if exactly two of the triplets are in the starting lineup?

Choose 2 of the triplets. There are $ \binom{3}{2}\equal{}3$ ways to do this. Now choose 4 of the remaining 11 players. There are $ \binom{11}{4}\equal{}330$ ways. The final answer is $ 3\cdot330\equal{}990$.

Why don't you divide by two? Because you are counting: you first choose the triplets, and THEN the rest of the people AND you first choose the rest of the people and THEN the triplets, so you are overcounting by double, right?

No, the 3 clearly only consists of choosing the triplets and the 330 consists only of the possibilities of choosing from the remainder. The two groups are distinctly separated.


basically, use numbers from previous posts to make the next number in the sequence. you may use the past 2 numbers, past 3 numbers, or however many you would like

for example
$ 2$
$ 3$
$ 2^3 \equal{} 8$
$ \frac {8} {2} \plus{} 3 \equal{} 7$

use whatever math functions you would like, but you many only use previous numbers in the sequence, and you cannot skip back like the next number could not be $ \log_2 8 \equal{} 3$

ONLY POST THE PAST 10 NUMBERS IN THE SEQUENCE. otherwise this topic will become huge and get locked.

first 2 numbers in the sequence
$ 5$
$ 25$

[quote="abacadea"]you cannot skip back[/quote]

*cough* *cough*


modular arithmetic
superior algebra
superior algebra unsolved

HINT Scale by $ \frac{1}{5}\pmod{108}$, computed via extended Euclidean algorithm.

OR directly, mod 5 we see $ 5x \plus{} 108y \equal{} 34 \implies \minus{}2y \equiv 4\pmod{5}$

So $ y \equal{} \minus{}2\plus{}5n\implies x \equal{}\frac{1}{5} (34\minus{}108 (\minus{}2\plus{}5n)) \equal{} 50 \minus{} 108 n$

NOTE:  I edited the above to simplify it slightly.


number theory
greatest common divisor
relatively prime
prime numbers
prime factorization

A set of integers are said to be [b]relatively prime[/b] if any two of them have no common factors other than 1.
In other words, any two of the numbers have their GCD ( or HCF ) being 1.

[i][size=150]Question:
Taking out 10 numbers from 1,2,....,100 so that these 10 numbers are relatively prime and the product of these numbers is [color=violet]AS LARGE AS POSSIBLE[/color].[/size][/i]

[quote="i_like_pie"][hide]$ 100, 99, 97, 91, 89, 87, 83, 79, 77, 73$[/hide][/quote]

luimichael quoted that these 10 numbers are relatively prime, so wouldn't they be
$ 97,89,83,79,73,67,61,59,53$?

I'm pretty sure my 10 numbers are relatively prime. I took the 9 largest primes < 100 and 100 itself, and changed the set a little. For example, it is possible to add 77 because it is still relatively prime to the other primes. Even if there is a little error in my list, the largest 9 primes and 100 definitely have a larger product than your 10 numbers.

[quote="i_like_pie"]$ 100, 99, 97, 91, 89, 87, 83, 79, 77, 73$[/quote]

A good trial.  :)
But please note that 99 and 77 are not relatively prime.
These two numbers contain a common factor of 11.

We say that the set of numbers are relatively prime provided that ANY TWO OF THEM ARE RELATIVELY PRIME.
Maybe I have to put some emphasis on the definition by saying :
the set of numbers are [b]pairwisely relatively prime[/b]......

http://primes.utm.edu/glossary/page.php?sort=RelativelyPrime

Kong,

You have chosen the numbers to be [b]all prime numbers[/b].
This fufils the criterion of "pairwisely relatively prime" but the product of these numbers is [b]not the largest possible[/b].

For example, you may [u]replace the smallest candidate 53 by 100[/u] and the newly formed set of numbers , however, are still pairwisely relatively prime.
Don't mix up the concepts of "being prime" and "being relatively prime."

Example:
8 and 5 are relatively prime but 8 itself is not prime.
8 and 9 are relatively prime but both of these two numbers are not prime.

My suggestion is that if you want to know whether some numbers are relatively prime or not, you may  look into the [b]prime factorization[/b] of the numbers.

[quote="luimichael"]Don't mix up the concepts of "being prime" and "being relatively prime."
[/quote]
Oh.. Okay..

Very good trial!
What a pity that $ 99\equal{}3^2*11$ and $ 87\equal{}3*29$, which means that your sequence of numbers does not satisfy the required condition. :roll:

Better and better now!
Your answer is [b]very close [/b]to the largest possible value.

Note that both 98 and 91 contain [b]a factor of 7[/b].
You [b]almost[/b] hit the correct answer! :)

Congratulations!
Fraenkel got the correct answer. :idea:

The product should be 99*97*95*94*91*89*83*79*73*71.

cant you replace 95 with 100 to make the largest product?

and what does it mean by pairwisely relatively prime?

That means that every number in the set shares no common factors other than 1.
And no, because then 94 and 100 are even.


inequalities
combinatorics unsolved
combinatorics
Probabilistic Method

Let $ F$ be a finite collection of binary strings of finite lengths and assume no member of $ F$ is a prefix of another one. Let $ N_i$ denote the number of strings length $ i$ in $ F$. Prove that: $ \sum_{i} \frac {N_i}{2^i} \le 1$

Well, we can do it probabilistically or we can clear denominators and do it combinatorially.  I like the second way (though of course the two methods are identical except for some linguistic differences): choose some large $ n$; we need to show that $ \sum_i N_i 2^{n \minus{} i} \leq 2^n$.  The left side counts the number of strings of length $ n$ that have a prefix in $ F$; by the given condition, there is no string double-counted in this expression and so the result is no larger than the number of all strings of length $ n$, which is the right side.

Not a different solution [i]per se[/i], but a visual one nonetheless.

[b]Lemma:[/b] Given a (rooted) binary tree with levels indexed from $0$ (so that the root is the zeroth level). Let $N_i$ be the number of leaves at the $i^{\text{th}}$ level. Then $\sum \frac {N_i}{2^i} \leq 1$

[b]Proof:[/b] Let $S$ be the desired sum. Define a 'collapse' to be the process by which we select a (non-leaf) node whose children are all leaves, and remove said leaves. Then $S$ is easily verified to not decrease under this process. After some finite iterations of this process, we are left with the root only, and here $S=1$, so we conclude $\Box$

Now consider the natural binary tree generated by $F$. Specifically, a string is in $F$ iff it is a leaf. Done $\blacksquare$

Edit: It has [url=https://drive.google.com/file/d/0B_QyOg5Kp1eHOGNlWndxMWt1VEk/view?usp=sharing]come to my attention[/url] that the problem has a corollary: the expected length of an element in $F$ is at least $\log_2(n)$. Refer back to the notation above, and let $H_i$ be the $i^{\text{th}}$ level of the universal binary tree encompassing $F$. Let $S=\frac{1}{|F|}\sum i\cdot N_i$. In particular, suppose that $P=N_0\cup N_1\cup\cdots \cup N_{\left\lfloor\log_2(n)\right\rfloor-1}$ is nonempty. By "projecting" the nodes in $P$ onto $H_{\left\lfloor\log_2(n)\right\rfloor}$, we see that $Q=N_{\left\lfloor\log_2(n)\right\rfloor+1}\cup N_{\left\lfloor\log_2(n)\right\rfloor+1}\cup\cdots$ is nonempty as well. Now choose arbitrary $p\in P,\ q\in Q$, and let the former's children be $p',p''$. Reform set $F'$ by replacing $p$ with $p'$ and $q$ with $p''$. Clearly the leaf-only property of $|F|$ is invariant under this process, and $S$ can only decrease. So iterate until $P=\emptyset$, and then project the remaining $Q$ all the way down onto $H_{\left\lfloor\log_2(n)\right\rfloor+1}$, which can only decrease $S$ as well. The final result, $S\geq \log_2(n)$, follows quickly from some algebra. $\Box$

I like that this is an analog of Sperner's(more closely [url=https://en.wikipedia.org/wiki/Lubell%E2%80%93Yamamoto%E2%80%93Meshalkin_inequality]LYM[/url] inequality for binary strings).

Pick the largest sequence say $s_i$ and let it have length $L$ and let $\mathcal{F}=\{s_1,s_2,\dots s_k\}$.Now define a uniform distribution on binary sequences of length $L$.Let $X_i$ be the event of $s_i$ being a prefix of a binary string.Notice that by the condition $X_i,X_j$ are always disjoint and hence by union bound we have :
$$\sum N_i\cdot 2^{-i}\leq 1$$
Where $P(X_i)$ is obtained just by fixing the first $|s_k|$ indices and picking the rest randomly.