A sequence $(x_{n})_{n \in n*}$ of rational numbers is called regular if $|x_{m}-x_{n}| \le m^{-1}+n^{-1}, m, n \in n*$. Let $t=(t_{n})_{n \in n*}$ a regular sequency, it is defined that $k_{t}=\lfloor|t_{1}|+2\rfloor$. Let $x$ and $y$ regular sequences and $K=max[K_{x},K_{y}]$. It is defined $z=(z_{n})_{n \in n*}$ such $z_{n}=x_{2k_{n}}.y_{2k_{n}}, n \in n*$. Prove that $(z_{n})_{n \in n*}$ is a regular sequence.

Ps: n* means the set of natural numbers except zero (0).

Sorry about my english... its clear?

[quote="santosguzella"]Sorry about my english... its clear?[/quote]
No, not at all.
What is $k_{n}$ for $n\in\mathbb{N}$?


Έχουμε ότι $ a\plus{}b\plus{}c\equal{}1,  a,b,c>0$  . Να αποδείξετε ότι

$ 10(a^3\plus{}b^3\plus{}c^3)\minus{}9(a^5\plus{}b^5\plus{}c^5)\geq1$

Nope...
Δεν αφαιρούμε τις ανισότητες κατά μέλη!

Γιατι θα εχεις $ 9(a^5 \plus{} b^5 \plus{} c^5) \geq \frac {1}{9}$

$ \minus{}9(a^5 \plus{} b^5 \plus{} c^5) \leq \minus{} \frac {1}{9}$

Λύση :lol: :
(Αφιερωμένη στον Kritiko)
Θα χρησιμοποιήσουμε τη θεωρία των πολλαπλασιαστών Lagrange.

Η συναρτηση $ f(a,b,c) \equal{} 10(a^3 \plus{} b^3 \plus{} c^3) \minus{} 9(a^5 \plus{} b^5 \plus{} c^5)$ και η $ g(a,b,c) \equal{} a \plus{} b \plus{} c \equal{} 1$ ως πολυωνυμικές έχουν συνεχεις πρώτες μερικές παραγώγους και $ \nabla g\neq 0$ και τα ακρότατα θα εμφανίζονται εκεί όπου  $ \nabla f \equal{} \lambda \nabla g$ για κάποιο $ \lambda$.

Είναι $ \boxed{\nabla f \equal{} < 30a^2 \minus{} 45a^4,30b^2 \minus{} 45b^4,30c^2 \minus{} 45c^4 >}$ και $ \nabla g \equal{} < 1,1,1 >$ και άρα

$ \boxed{\lambda \equal{} 30a^2 \minus{} 45a^4 \equal{} 30b^2 \minus{} 45b^4 \equal{} 30c^2 \minus{} 45c^4}$ και $ g(a,b,c) \equal{} a \plus{} b \plus{} c \equal{} 1$.

Λύνοντας τα συστηματα έχουμε ότι τα τοπικά ακρότατα λαμβάνονται όταν $ a \equal{} b \equal{} c$  ή $ a \equal{} \minus{} b \equal{} c$ , και επίσης λόγω συμμετρίας θα πρέπει να λάβουμε υπ'όψιν και όλες τους τίς μεταθέσεις.

Έχουμε έτσι μετα από πραξούλες ότι $ f\left(\frac {1}{3},\frac {1}{3},\frac {1}{3}\right) \equal{} \frac {29}{27}$   και   $ \boxed{f(1, \minus{} 1,1) \equal{} 1}$

Άρα η ελάχιστη τιμή της συνάρτησης θα είναι το $ 1$ :D Q.E.D.

Κρίμα να σκοτώνεις το κουνουπάκι με οπλοπολυβόλο  :P  (αν και έχει πλάκα  :wink: ) και με τους πολλαπλασιαστές θέλει πολλή προσοχή διότι πρέπει να εξετασεις κάθε περίπτωση και επιπλέον οι πολλαπλασιαστές σου δίνουν πιθανά εσωτερικά σημεία ακροτάτων (πρέπει να εξετάσεις και τα άκρα κάτι το οποίο δεν το κάνεις αλλά εδώ μικρή σημασία έχει )

Τέλος πάντων να μια elementary solution
Η ανισότητα μπορεί να γραφεί ως $ 10(1 \minus{} a^3 \minus{} b^3 \minus{} c^3)\leq 9(1 \minus{} a^5 \minus{} b^5 \minus{} c^5)$

Τώρα απλά παρατηρούμε ότι $ 1 \minus{} a^3 \minus{} b^3 \minus{} c^3 \equal{} (a \plus{} b \plus{} c)^3 \minus{} a^3 \minus{} b^3 \minus{} c^3 \equal{} 3(a \plus{} b)(b \plus{} c)(c \plus{} a)$

και

$ 1 \minus{} a^5 \minus{} b^5 \minus{} c^5 \equal{} (a \plus{} b \plus{} c)^5 \minus{} a^5 \minus{} b^5 \minus{} c^5 \equal{} 5(a \plus{} b)(b \plus{} c)(c \plus{} a)(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)$

Οπότε αρκεί νδο $ 15(a \plus{} b)(b \plus{} c)(c \plus{} a)(3a^2 \plus{} 3b^2 \plus{} 3c^2 \plus{} 3ab \plus{} 3bc \plus{} 3ca \minus{} 2)\geq 0$

δηλαδή αρκεί $ 3a^2 \plus{} 3b^2 \plus{} 3c^2 \plus{} 3ab \plus{} 3bc \plus{} 3ca\geq 2 \equal{} 2(a \plus{} b \plus{} c)^2$ που ισοδυναμεί με
$ a^2 \plus{} b^2 \plus{} c^2\geq ab \plus{} bc \plus{} ca$ που δυστυχώς ισχύει so QED  :)

Όντως ρε Σιλουανέ εχεις δίκιο :lol:  Αλλά τη γουσταρα πολύ αυτή τη λύση.... Εν γνώσει μου δεν εξέτασα τα ακρα γιατί το διαστημα που ορίζονται οι $ a,b,c$ δεν είναι κλειστό. Πάντως δεν ξέρω αν έχεις παίξει ποτέ το παιχνίδι worms, αλλά είναι σα να σκοτώνεις σκουληκάκι με βόμβα υδρογόνου :rotfl:


$ A$ and $ B$ together can do a job in $ 2$ days; $ B$ and $ C$ can do it in four days; and $ A$ and $ C$ in $ 2\frac{2}{5}$ days.  The number of days required for $ A$ to do the job alone is:

$ \textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 3 \qquad
\textbf{(C)}\ 6 \qquad
\textbf{(D)}\ 12 \qquad
\textbf{(E)}\ 2.8$

[hide="Click for solution"]
Denote by $ a$, $ b$, and $ c$ the respective number of days it would take $ A$, $ B$, and $ C$ to complete the job if he were working alone.  Then $ \frac{1}{a}$, $ \frac{1}{b}$ and $ \frac{1}{c}$ denote the fractions of the job done by each in one day.  Since it takes $ A$ and $ B$ together two days to do the whole job, they do half the job in one day, so $ \frac{1}{a}\plus{}\frac{1}{b}\equal{}\frac{1}{2}$.  Similarly, $ \frac{1}{b}\plus{}\frac{1}{c}\equal{}\frac{1}{4}$ and $ \frac{1}{a}\plus{}\frac{1}{c}\equal{}\frac{5}{12}$.  Thus, $ \frac{1}{a}\minus{}\frac{1}{b}\equal{}\frac{1}{6} \implies \frac{2}{a}\equal{}\frac{2}{3} \implies a\equal{}3$, or $ \boxed{\textbf{(B)}}$.
[/hide]


How do you simplify $ 2\sqrt{3(2\minus{}\sqrt{3})}$?

I don't know if this is completely simplified, but what I got was $ 2\sqrt{6\minus{}3\sqrt3}$ because first it would be $ 2\sqrt{3\times2\minus{}3\times\sqrt3}$ and then it would be $ 2\sqrt{6\minus{}3\sqrt3}$. Is that right?

I don't think there is any way to simplify that any more.

Probably this is considered simpler:  $ 3\sqrt{2} \minus{} \sqrt{6}$

[quote="Kouichi Nakagawa"][quote="PowerOfPi"]How did you get it?[/quote]
\begin{eqnarray*} 2\sqrt {3(2 - \sqrt {3})} & = & \sqrt {24 - 12\sqrt {3}} \\
& = & \sqrt {24 - 2\sqrt {108}} \\
& = & \sqrt {(18 + 6) - 2\sqrt {18\times6}} \\
& = & \sqrt {18} - \sqrt {6} \\
& = & \boxed{3\sqrt {2} - \sqrt {6}} \end{eqnarray*}
[/quote]
How did you get from $ \sqrt {(18 + 6) - 2\sqrt {18\times6}}$ to $ \sqrt {18} - \sqrt {6}$

hello, the radicand is $ \left(\sqrt{18}\minus{}\sqrt{6}\right)^2$.
Sonnhard.

The radicand is the expression under the radical sign.

more generic way is to notice that the perfect square form of the radicand expanded has only one irrational term, and you can go from there e.g

from $ \sqrt{24\minus{}2\sqrt{108}}\equal{}\sqrt{24\minus{}(2)\sqrt{108}}$
Since you want the bottom to be of form $ a\minus{}2*\sqrt{a*b}\plus{}b$ (negative since irrational term is negative) you can get equation $ a*b\equal{}108$ and $ a\plus{}b\equal{}24$ and easily solvable into $ a\equal{}18 b\equal{}6$ and you get $ \sqrt{18}\minus{}\sqrt{6}$


Let $ P(x)$  be a polynomial of degree $ n$ satisfying the condition $ P(i)\equal{}2^i$ for $ i\equal{}0,1,2,...,n.$
Find $ P(n\plus{}1)$.

Since $ P(x)$ has a degree of $ n$, and $ P(i) \equal{} 2^i$ for $ n \plus{} 1$ values, we know that $ P(x) \equal{} 2^x$ for all $ x$ (Identity Theorem). So $ P(n \plus{} 1) \equal{} 2^{n \plus{} 1}$.

EDIT: Oh...ok fail

[quote="modularmarc101"]Since $ P(x)$ has a degree of $ n$, and $ P(i) \equal{} 2^i$ for $ n \plus{} 1$ values, we know that $ P(x) \equal{} 2^x$ for all $ x$ (Identity Theorem). So $ P(n \plus{} 1) \equal{} 2^{n \plus{} 1}$.[/quote]
Are you sure that $ P(x) \equal{} 2^x$ for all $ x$?? I didn't get $ 2^{n\plus{}1}$. What is the identity theorem?

Identity theorem- Let $ p(x)$ and $ q(x)$ be two polynomials of degree of at most $ n$. If $ p(x) \equal{} q(x)$ for $ n\plus{}1$ values of $ x$, then $ p(x) \equal{} q(x)$ for all $ x$.

What did you get?

but $ P(x)\equal{}2^x$ is not a polynomial

I think that can be solved by divisibility
because x / P(x) - ao, and P(0) = 1, now ao=1
and x / P(x) - 1
and (a-b)/P(a)-P(b)
next n+1-1 / P(n+1) - 2
But i dont know how to continue
can give a hint? :P

[quote="mahanmath"][quote="Mathias_DK"]I get $ 2^{n \plus{} 1} \minus{} 1$. I guessed $ P(x)$:
$ P(x) \equal{} 1 \cdot \frac {1}{0!} \plus{} x \cdot \frac {1}{1!} \plus{} x(x \minus{} 1) \cdot \frac {1}{2!} \plus{} \cdots \plus{} x(x \minus{} 1) \cdots (x \minus{} n) \cdot \frac {1}{n!}$

Then $ P(m) \equal{} \binom{m}{0} \plus{} \binom{m}{1} \plus{} \cdots \plus{} \binom{m}{n}$.
And using $ \binom{k}{0} \plus{} ... \plus{} \binom{k}{k} \equal{} 2^k$ it is obvious that $ P(i) \equal{} 2^i, i \equal{} 0,1,2,...,n$
Then $ P(n \plus{} 1) \equal{} \binom{n \plus{} 1}{0} \plus{} ... \plus{} \binom{n \plus{} 1}{n}$ $ \equal{} \left ( \binom{n \plus{} 1}{0} \plus{} ... \plus{} \binom{n \plus{} 1}{n \plus{} 1} \right )$ $ \minus{} \binom{n \plus{} 1}{n \plus{} 1} \equal{} 2^{n \plus{} 1} \minus{} 1$[/quote]
You`re right.Lagrange interpolation works ! :P  :P[/quote]
I didn't know Lagrange interpolation, but I looked it up and it seems like you would give a very different function(by the looks of it) using that method?


Eric is catering a party for 152 people. He wants to seat the same number of people at each table. He also wants more than 2 people but fewer than 10 people at a table. How many people can he seat at each table?

Is there an equation for such a question?

well you could guess and check so your choices are 3,4,5,6,7,8, or 9 at each table. you find which ones are divisible by 152... i think so he can seat 4 and 8 at each table.

ch1n353ch3s54a1l, how can you possibly remember that username?? :P

That's exactly what I got for an answer.

Thanks.

[quote="ch1n353ch3s54a1l"]lol
it's leet
but i type it all the time so i remember it anyways  :)[/quote]

  ive been able to decode "chinesechessa1"...... is that anywere near correct?

[hide]152's factors from 3-9 are just (divisibility rules  :D ) are 4 and 8, the answer. [/hide]

That is the wierdest name I've heard, ch1n353ch3s54a1l (copy and pastes), no offense of course  :P

chinese chess for all, the 4 not being an "A."  :lol:


ratio
geometric sequence
algebra unsolved
algebra

Hi all, I'm new and I have a Geometric sequence problem I just can't figure out..

It goes as such: t(7)=5x+2 and t(10)=x-23. The common ratio is 2, so find the numerical value of t(10).

I just can't figure it out and any help, or a point in the right direction would really help. Thanks.

[hide]Notice how t(10)/t(7) is 8. Set the ratio of the terms equal to 8 and solve for x.[/hide]

[hide]t(7)=5x+2. Since this is a geometric sequence with common ratio 2, multiply by 2 to get t(8). 2t(8)=t(9), and 2t(9)==t(10). Thus, 8t(7)=t(10). 40x+16=x-23, 39x=-39, x=-1[/hide]

This seems too easy for the olympiad section. Maybe High School Basics?


geometry
AMC
AIME
AwesomeMath
summer program
email
number theory

im not exactly which subjects i should pic

so far, ive determined that all the subjects will be intermediate level and i will either choose
 comp.geo&counting strategies
or
comp.geo& Modular Arithmetic

im not exactly sure which one to pick because i like combinatorics quite alot and im decent at it but ive never really been exposed to too many intermediate level problems.

as for number theory,its also pretty interesting but im not as good at number theory as i am with combinatorics.  Im also not sure how often it shows  up on the AIME/USAMO  but im guessing that it would be pretty important in general.

do you have any suggestions?

is it a good idea to choos the subjects that are most interesting or is it better to choos subjects that are ur weaknesses?

So, most people take intermediate, and the classes you are taking are good, I hear. Especially comp. geo., but I hear it is the most challenging intermediate class out of the 4.

About either int. comb. or int. number theory, I would really take number theory, especially if you are already kind of good at combinatorics. I would rather go to this camp to improve on fields of mathematics that I am not really good at compared to other fields, than to take courses that you already like and are good at.

In my opinion, I suggest you take the intermediate number theory course, which is.. modular arithmetic. Just because I think the whole reason for this camp is to IMPROVE and GET EXPOSED to "new" fields of math. (By new, I mean, the fields that you don't like, you're not good at, etc.)

Clearly, you should take either geometry or counting, because you hate counting.

[quote="lifeisacircle"]What should I do if I'm sort of in between the levels? For example, I think I fit somewhat in between the intermediate and advanced levels of their number theory courses. Of course, I could be underestimating the difficulty or the overestimating my abilities.[/quote]

if you score 4-10 on aime, you should take the intermediate course
but if ur still not sure, then i would suggest you take the harder one and then switch if it seems toooo hard.

choose something thats ur weakness. you're there to learn. [thats why i chose comp. geo lol]

iYOA: sign up for something dont worry too much because if after a few days u dont feel like its ur level, u can always switch out.

I'm going for Int. Geo and Alg. because I fail them.

@lifeisacircle, ak909090, leafiness0: What are you taking?

I'm not sure yet. Maybe a geometry course, a num theory course (cuz i suck at num theory)...or maybe algebra 2.5.

I think I'm at the intermediate level...

But here are two things what pulls me back from taking comp. geo, although I'm taking Modular Arithmetic. (int. number theory)

1. I hate geometry. I suck at geometry.

2. Everyone is like.. OMG COMP GEO IS SO SO SO HARD!!! OMG OMG SO HARD!!

So.. what should I do.. any advices?

Take the geo class.  I also hate geometry..  I also fail at geometry.  That's why I'm taking it.

Yea.. I guess I should really take geo if I HATE it.. to learn it..

But I'm just wondering if I should like take the entry-level geo class instead of comp. geo.. Because I epic-phail at geo.. Wow.

[quote="AIME15"]I heard that it's really easy to transfer if a class is too easy/difficult.[/quote]

it is.

How do u know if you're good enough to take a certain course? I mean, besides the qualification like.. AIME score between 4~8.. Blah.. Stuff like that. Do they give you a test or something, like a placement test? Or do you just pick.. Plus are you supposed to let them know before you get there? If so, how? Because it's not in the packet or anything..

But but but but taking something I'm interested in is more important to me than taking something to learn something.

Currently, I'm think about the advanced number theory (which will probably be too hard) computational geometry and algebra 2.5

Unfortunately, ALL those classes are afternoon stuff, meaning I'll have to change my choices.

[quote="iYOA"]when did you get teh packet?
i havnt gotten it yet.[/quote]

Um.. I kind of got it a long time ago... But don't panic or anything. You got in (99.9%) but they haven't sent it to you yet.. BTW, where you do you live? Because that might be a factor. Plus even if you didn't get in, which we can rule out that possibility, I mean really.. I can't emphasize this enough.. You got it, so don't panic.. They will send you the packet. If not, e-mail the director.. you can find his e-mail on the awesomemath site.

no placement test.
just pick. when you get there, you'll take the class if you dont' feel its right switch out.

really ? its supposed to be in the packet im pretty sure. a sheet of paper for you to fill out and send back w/ course selection..

lifeisacircle: then take what's important to you ;]
comp. geo is a great class.

Um.. are you sure with the course selection sheet?

Because I'll e-mail the director if you're sure that there is supposed to be some course selection sheet...

Oh well.. I'll email Dr. Andreescu anyways.. Just to get his recommendations on which courses to take, etc.,.


algebra
polynomial
limit
complex analysis
complex analysis unsolved

Let $ P: C\rightarrow C$ be a polynomial of degree $ m\geq 2$ and $ Q_n(z) \equal{} P(P(\cdots P(z))\cdots)$ be $ n^{th}$ iterate of $ P$. Show that $ |\{z: \lim_{n\to\infty}Q_n(z) \equal{} z\}| \equal{} \infty$.

Hint: If $ \lim_nQ_n(z)\equal{}z$, then $ P(z)\equal{}z$. This implies that your set is [b]finite[/b].

[quote="$ math$"]Ok, thanks. But I guess that $ |\{z: Q_{n}(z) \equal{} z\}| \equal{} \infty$ is true. But, why? :blush:[/quote]The way you wrote it here (without the limes), this is [b]not[/b] true because $ Q_n(z)\minus{}z$ is a polynomial for every n and as such it can have only finite number of zeros.


What is the greatest integer value of $ n$ such that 635,040 is divisible by $ 2^n$?

[quote="Butler"]I know a more elegant solution!

You can start by finding the prime factorization of this number...

$635,040= 2^5 \times 3^4 \times 5 \times 7^2$
So, our answer is $ 2^{\boxed{5}}$[/quote]

I'm not sure that your solution is more elegant. Find the prime factorization of that giant number isn't easy (unless you have a factoring program  :) (me)). Here is a different solution.

If the last digit is divisible by 2, then the whole number is divisible by 2. ($\frac{0}{2} = 0$)
If the last two digits are divisible by 4, then the whole number is divisible by 4. ($\frac{40}{4} = 10$)
If the last three digits are divisible by 8, then the whole number is divisible by 8. ($\frac{040}{8} = 5$)
If the last four digits are divisible by 16, then the whole number is divisible by 16. ($\frac{5040}{16} = 315$)
If the last five digits are divisible by 32, then the whole number is divisible by 32. ($\frac{35040}{32} = 1095$)
If the last six digits are divisible by 64, then the whole number is divisible by 64. ($\frac{635040}{64} = 9922 \frac{1}{2}$). So the largest possibly value for $n$ is $\boxed{5}$. However, this will not work for every number. For example, the number 64. The last digit is divisible by 2, so the whole number is divisible by 2, the last two digits are divisible by 4, then the whole number is divisible by 4, but there is no third digit, so you would just divide 64 by 8, and so on.


What is the longest possible sequence of distinct positive integers with first member $1$ and last member $31^a5^b1990^c$ such that each member is divisible by its predecessor?
How many such sequences have this length?

first let us change the problem a bit:

[b]for a given number $ x$,a proper sequence of $ x$ with length $ d$ is a sequence $ x_0,x_1,x_2,\ldots ,x_d$ for which $ 1 \equal{} x_0 < x_1 < x_2 < \ldots < x_d \equal{} x$ and also for $ 1\leq i\leq d \minus{} 1$ we have $ x_i\mid x_{i \plus{} 1}$.
find the longest proper sequence and the number of such proper sequences for $ x \equal{} 31^a5^b1990^c$.[/b]

let $ d$ be the length of the longest proper sequence of $ x$,in such a sequence,note that for $ 1\leq i\leq d\minus{}1$ the number $ \frac {x_{i \plus{} 1}}{x_i}$ must be a prime number,because otherwise we can plug in another number between $ x_i,x_{i \plus{} 1}$.
now let $ x \equal{} p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_r^{\alpha_r}$ where $ p_i$ is a prime number ($ 1\leq i\leq r$).
it follows that:

$ \boxed{d \equal{} \alpha_1 \plus{} \alpha_2 \plus{} \ldots \plus{} \alpha_r}$

also the number of proper sequences of length $ d$ is equal to:

$ \boxed{\frac {(\alpha_1 \plus{} \alpha_2 \plus{} \ldots \plus{} \alpha_r)!}{\alpha_1 !\alpha_2 !\ldots\alpha_r !}}$

now note that in this problem we have:

$ x \equal{} 31^a5^b1990^c \equal{} 2^c5^{a \plus{} c}31^b199^c$

hence the length of the longest proper sequence of $ x$ is $ a \plus{} b \plus{} 3c$ and the number of such proper sequences is $ \frac {(a \plus{} b \plus{} 3c)!}{(a \plus{} c)!b!(c!)^2}$.


A bike rider approaches a hill at a speed of 3.5 m/s. The mass of the bike and rider together is 77 kg. What is the initial kinetic energy of the system?

How do I find the initial kinetic energy of a system in general?

Thanks in advance!

I interpret the problem to mean that the bike/rider are starting at 3.5 m/s and initial kinetic energy refers to when the bike first touches the hill.  So it's just your standard $ \frac{1}{2}mv^{2}$.


function
abstract algebra
trigonometry
algebra
functional equation
algebra proposed

Find all continuous functions $ f: \mathbb{R} \to \mathbb{R}$ such that
\[ f(x^2 \plus{} 1) \equal{} 2f(x)
\]
[hide="my (way over-complicated?) thoughts"]
Let $ u_{n \plus{} 1} \equal{} u_n^2 \plus{} 1$
Define a relation $ r: x \sim y \Leftrightarrow (\exists m,n \in \mathbb{N}, u_0 \in \mathbb{R})\ u_n \equal{} x \wedge u_m \equal{} y)$. Then $ r$ is an equivalence relation and each equivalence class can be represented by the generating $ u_0$. So, for each generating $ u_0$,
\[ f(u_n) \equal{} 2^n f(u_0)
\]
The thing is, I have no idea how to choose the $ f(u_0)$'s such that $ f$ is continuous.
[/hide]

[hide="solution?"]
Clearly, $ 2f(x)\equal{}f(x^2\plus{}1)\equal{}2f(\minus{}x)$. Letting $ x\to0$ and using continuity, $ f(0)\equal{}0$.
If you define a sequence $ a_{n\plus{}1,m}\equal{}a_{n,m}^2\plus{}1$ $ a_0\equal{}m$ then clearly $ f(\pm a_{i,0})\equal{}0$.
Now consider an $ m$ between 0 and 1. Then members of $ \{a_{i,m}\}$ will hit arbitrarily close to members of $ \{a_{i,0}\}$ (I'm missing a  theorem here to back me up) so by continuity, $ f(x)\equiv 0$.
[/hide]

There are nonzero solutions, and lots of them; the set of solutions to this functional equation restricted to $ (0,1)$ is dense in $ C\left((0,1)\right)$.  There are even some $ C^\infty$ solutions on $ \mathbb{R}$.

Let $ f_0: [0,1]\to \mathbb{R}$ be an arbitrary continuous function with $ f_0(1) \equal{} 2f_0(0)$ and let $ \{a_k\}$ be the sequence satisfying $ a_k \equal{} a_{k \minus{} 1}^2 \plus{} 1$ with $ a_0 \equal{} 0$.  Then define $ f_k: [a_k,a_{k \plus{} 1}]\to \mathbb{R}$ for $ k\in \mathbb{N}$ such that
\[ f_k(x) \equal{} 2f_{k \minus{} 1}\left(\sqrt {x \minus{} 1}\right)
\]
Now define $ f: \mathbb{R}\to \mathbb{R}$
\[ f(x) \equal{} f_k(|x|) \ \ \ \ \ \text{for} \ \ a_{k}\leq x < a_{k \plus{} 1}
\]
then by construction $ f$ is a continuous function satisfying the functional equation.  It's also not hard to show that [i]all[/i] solutions are of this form.
[hide="Mathematica script, because seeing is believing"][code]f0[x_] := Sin[2*Pi*x];
f[x0_] :=
 Module[{x = Abs[x0], k=0},
  While[x > 1, x = Sqrt[x - 1]; k = k + 1];
  f0[x]2^k];
Plot[{f[x^2 + 1], 2 f[x]}, {x, -5, 5}]
[/code]Try replacing $ \sin 2\pi x$ by any other function satisfying $ f_0(1) \equal{} 2f_0(0)$.  For example $ f_0(x) \equal{} e^{\frac {1}{x(x \minus{} 1)}}$ is an interesting choice, as in this case $ f$ is infinitely differentiable.[/hide]


USAMTS
inequalities
ARML
special factorizations

[quote="s0mp"]Assume that x, y, and z are real positive numbers that satisfy the equations given on the right.

x+y+xy=8
y+z+yz=15
x+z+xz=35

Determine the value of x+y+z+xyz.[/quote]
Adding 1 to both sides of each equation gives us (x+1)(y+1)=9. (y+1)(z+1)=16. (z+1)(x+1)=36. Multiplying them all together and taking the square root gives (x+1)(y+1)(z+1)=72. Dividing by each in turn gives (z+1)=8, (x+1)=4.5, (y+1)=2. Substituting in our values for x y and z gives that expression to be 36.

One must wonder why they ask for x+y+z+xyz. Is there possibly a way to find x+y+z+xyz without finding x, y, and z individually? I haven't succeeded yet. (I only found the solution that MysticTerminator found.)

[quote="MysticTerminator"]That's what I thought too. however, if we try to get that from (x+1)(y+1)(z+1), we have an extra xy+yz+zx term. We could take care of that with the original equations, but that would give some extra x y and z terms. So, if there [i]is[/i] some clever factorization, it's more trouble than it's worth to find it. I suspect they threw that in just to get people "barking up the wrong tree". I like that expression.[/quote]

I also spent awhile trying to find a "clever factorization" w/o success...

I was thinking about taking (x+1)(y+1)(z+1) and then subtracting the given equations, but then we get a value for xyz-x-y-z unfortunately.

Just a question for mysticterminator, what insight led you to add 1 to both sides?

[color=cyan]It's actually a useful technique, very similar to completing the square -- you want to look for things that will factor.  x + y + xy doesn't factor, but 1 + x + y + xy does.  Then, you get lots of symmetries, and can carry on.  Factorizations are always something to look for.  For another more complicated example, there is a problem called "Inequality" on the Advanced forum where the discussion involved looking for nice factorizations, essentially.[/color]

I solved #2 at ARML in 2003 based on this technique. The problem said something like `'A triangle with sides a, b, and c is said to be log-right if log(a^2)+log(b^2)=log(c^2) and a <= b <= c. What is the largest possible value for a in a triangle that is both right and log-right?'' Try it!


factorial
number theory
prime factorization

[quote="warut_suk"]What's the least value of $n$ such that $n!$ is divisible by $414$?[/quote]

[hide]The prime factorization of 414 is $2*3^2*23$

The lowest factorial that includes $23*9*2$ is $23!$[/hide]


function
number theory proposed
number theory

Bow to  natural solution the number of the equation $ [\frac{x^3}{103x\minus{}1}]\equal{}[\frac{x^3}{103x\plus{}1}]\plus{}1$  :roll:


Does any one have a solution to this one?


analytic geometry
graphing lines
slope
trigonometry
calculus
calculus computations

(a.) A particle moving with a nonzero velocity along the polar curve given by r = 3 + 2 costheta has position (x(t), y(t)) at theta = 0 when t = 0. This particle moves along the curve so that dr/dt = dr/dtheta. Find the value of dr/dt at theta = pi/3 and interpret your answer in terms of the motion of the particle.

(b.) For the particle described above, dy/dt = dy/dtheta. Find the value of dy/dt at theta = pi/3. Interpret your answer in terms of the motion of the particle.

I found the above problems to be kinda confusing. For the first part I answered "0.115 rads/time counterclockwise" and for the second part, "0.5 rads/time vertically." I've no idea if I did the problem right. I think I'm confused because the slope of polar curves are supposed to be the same thing as rate but what about when the slope is exactly vertical? Is the vertical velocity infinitely fast then? If that's the case then the particle would jump off the polar graph instead of tracing along the graph, barely changing its distance from the origin.

I'd appreciate some clarification very much...

we have:

 \[ r(\theta)\equal{}3\plus{}2\cos\theta\]

We also have:

\[ \frac{dr}{dt}\equal{}\frac{dr}{d\theta}\frac{d\theta}{dt}\]

so differentiating the polar equation wrt t:

\[ \frac{dr}{dt}\equal{}\minus{}2\sin\theta \frac{d\theta}{dt}\]

But it was given that:

\[ \frac{dr}{dt}\equal{}\frac{dr}{d\theta}\]

Hence we have accoring to the second equation:

\[ \frac{d\theta}{dt}\equal{}1\]

So:

\[ \frac{dr}{dt}\equal{}\minus{}2\sin\theta\]

at $ \theta\equal{}\frac{\pi}{3}$ it will be:

\[ \frac{dr}{dt}\equal{}\minus{}\sqrt{3}\]

similarly b, can be done...


inequalities
parameterization
calculus
inequalities proposed

I  have checked and proved the inequality for $y\le x$ or $y\le z$. It remains to consider ${y=\max\{x,y,z}$. Let me think a bit more.

it's standart inequality: instead $z$ substitute $3-x-y$ in you inequality. then find max for $x$  where parametr is $y$. then substitute max and find max for $x$ .there is not nothing hard only few calculation  :wink:

[quote="Vasc"]Let $x,y,z$ be nonnegative numbers such that $x+y+z=3$.
Prove or improve that

$x^{2}y+y^{2}z+\frac 3{2}xyz \le 4$.[/quote]
Nice, VasC. But it's also my result, I have also proposed it to Math and Youth magazine before, It will be published in some months so I wish we shouldn't disscus about it anymore, I will send my solution for you through PM. :)

[quote="toanhocmuonmau"][quote="Extremal"]Hm... Why you don't understand it's not hard! I said it's standart inequality: I have not so much free time to solve standart inequality but  if you don't believe it here is my sulution without PM :P
after substitution instead $z$ we get
$-x^{2}y+6y^{2}-5y^{2}x-2y^{3}+9xy\leq 8$
this is true else  if $D \leq 0$ but its hold when  $0\leq y \leq \frac{73-\sqrt{3153}}{34}$ and $1 \leq y\leq 3$      and for all $x$ inequality is true. therefore we only must consider it when $D>0$ it can be achieved when $\frac{73-\sqrt{3153}}{34}<y<1$ . (NOte that $0\leq y \leq 3$) we find $x=\frac{-5y^{2}+9y+(+-)\sqrt{(5y^{2}-9y)^{2}-4y(2y^{3}-6y+8)}}{2y}$
if our inewuality is true then this system  must be wrong:
$\frac{73-\sqrt{3153}}{34}<y<1$
$\frac{-5y^{2}+9y-\sqrt{(5y^{2}-9y)^{2}-4y(2y^{3}-6y+8)}}{2y}>0$
$y+\frac{-5y^{2}+9y+\sqrt{(5y^{2}-9y)^{2}-4y(2y^{3}-6y+8)}}{2y}\leq 3$
Everyone can chek it. yes really
if
$y+\frac{-5y^{2}+9y+\sqrt{(5y^{2}-9y)^{2}-4y(2y^{3}-6y+8)}}{2y}\leq 3$ ( wich mean $x+y\leq 3$
then it's can  be when $y \geq 1$ but we know $\frac{73-\sqrt{3153}}{34}<y<1$
inequality achieved when $z=0$ $x=2$ $y=1$[/quote]
I haven't checked your solution, but I think you don't understand me, I said I had sent it to Math and Youth magazine for the monthly contest, so I think we shouldn't show the solution here! Anyway, the solution for it is very simple, not lots of calculus as yours  :P Moreover, the equality holds for $x=2,y=1,z=0$ and $x=0,y=2,z=1$ :P I think you have a mistake :P[/quote]
Yes, but sulsution my is correct, I only don't calculate full : for wich $x$ and $y$ there hold equality.


calculus
analytic geometry
graphing lines
slope
derivative
algebra
polynomial

Okay, I think this goes here, but I'm not sure.

[quote="Kalle"]I think [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=403036#p403036url][u]this thread[/u][/url] is a good illustration why teaching limits in high school by intuitive reasoning from graphs, and then introducing L'Hopitals rule as some sort of magical way to compute limits should be banned. L'Hopitals rule is just a confusion factor if you don't know what a limit really is.[/quote] So, then where does that leave me? I have been taught limits "by intuitive reasoning from graphs", and then was introduced to L'Hopitals rule "as some sort of magical way to compute limits". I've understood for a while now that I don't know squat about limits. So the question is, where do I go from here? Can anybody recommend a textbook, or papers, or anything that would help clarify what [i]exactly[/i] a limit is.

Here's some basic questions about limits that I have:
1. Why in the world would I want to evaluate the limit of something? What do I learn from knowing that a graph approaches a number from both sides?

2. Why would a graph approach a number from both sides and not actually hit it?

3. Where did the knowledge of limits come from and how does it integrate in with the mathematics I already know?

4. What "real world" (ie physics) application does a limit have?

And that's all I can think of at the moment, but it's a start. Any and all help is greatly appreciated!

This deserves a response, and I intend to respond, eventually. I'm still formulating exactly what I want to say - it could take a couple of days.

Thank you very much, Professor. Your time is much appreciated. I look forward to your reply. :)

Zak

basically limits are nuts, i'd say newton or liebnez came up with the idea, though matematicians back then would communicate and build off eachother.

before calculus, there was just algebra, and a number would either be equal or unequal, and thats it, but not with calc.  limits are like a third case.

the whole idea is what happens when you get as close as you can to a value, "infinitly" close to the value.  the whole idea of a limit was hard for the matematical community at the time to buy into.  its very abstract.  i cant really think of a way to explain it,

i dont really bother with taking limits just by themselves, i just use them for derivitives, which is slope.  in previous math m is constant, but with calculus, you can find the slope of a line with a varying slope.  like $x^2$ just say.  $m = \frac{rise}{run} = \frac{(x + \Delta x)^2 - x^2}{\Delta x}$.  if you imagine a graph, as $\Delta x$ gets smaller and smaller it will become a better and better approximation of what the slope is.  but if you take the limit of \Delta x approaches 0, than you have the slope of the line, but its not an approximation, its exact! which i think is one of the beauties of limits.  the whole idea with infinity is its not a large number or in this case not a very very close value, infinity an idea, unlimited i think is a good synonym, its close to an unlimited extent, as in its impossible to get any closer.

just say you have a car that has a velocity of $v = t^2$, and you need to know the force due to acceleration.  acceleration is the instanteanous change of velocity per change of time:

$lim_{\Delta t\to 0}\frac{(t + \Delta t)^2 - t^2}{\Delta t} = 2t$
[hide="shown"]
$lim_{\Delta t\to 0}\frac{t^2 + 2t\Delta t + (\Delta t)^2 - t^2}{\Delta t}$
$lim_{\Delta t\to 0}\frac{2t\Delta t + \Delta t^2 }{\Delta t}$
$lim_{\Delta t\to 0}2t + \Delta t$
[/hide]

so the accleration equals 2t, and the force equals ma, so the forces equals 2 times the mass times the amount of time since the car started driving.

i wont try #2, kent will do a much better job on that one.  also im still undergrad, so if kent disagrees with anything i say listen to him, he's right.

oooooh ya has his history wrong, but he does spotlight one aspect - the definition of the derivative and the notion of instantaneous velocity or instantaneous rate of change.

Algebra - in the sense of a flexible, powerful system of notation and a clear understanding of equations, especially polynomial equations - was a young, vigorous discipline when calculus was first developed. (And what we call "algebra" they called "analysis.") Most of the notation we're familiar with didn't accrete into place until after 1600. The impulse to apply this new tool to geometry problems (as exemplified by the work of Descartes) didn't really get going until the middle of the 17th century. And once that happened - powerful, flexible algebraic technique applied to geometric problems - the differential and integral calculus was going to follow, and soon.

That's the first historical lesson: calculus [i]is[/i] algebra. That's where I part company with some teaching approaches which try to downplay algebra and encourage calculator-based visualization. Your calculator will mislead you, given a chance. That's not a problem if you can visualize the meaning in algebraic expressions, because then you'll know what you're looking for.

But calculus then developed for over a century without any particularly prominent role for the notion of limit. (What, then does "derivative" - excuse me, that wasn't the language, try "differential" - mean? This is perhaps one of those cases where it doesn't pay to look too closely.) It took until the time of Cauchy, in the early 19th century, for the to be a sense that there was a need for a formal definition of limit, and another 50 years to produce Weierstrass's $\epsilon$ and $\delta.$

So what would I emphasize? I would emphasize the [i]behavior[/i] of functions. Limits are one kind of behavior, but far from the only kind.

Here's an example: how does $x^2-x$ behave for large $x?$ You should be able to look at that and see: if $x$ is very large, the $x^2$ is very much larger than $x$ and this whole thing gets large positive. You should be able to visualize a graph curving upward (like $x^2$) and disappearing from sight. We could express quite a bit of this by saying that $\lim_{x\to\infty}x^2-x=\infty,$ but the limit language is an afterthought.

How does $\sqrt{x^2+1}$ behave as $x\to\infty?$ Why, it has to look quite a lot like $x,$ doesn't it? But now, how does $\sqrt{x^2+1}-x$ behave as $x\to\infty?$ A warning light should now blink on: if we subtract two things that are very nearly the same size, we lose accuracy in the difference unless we take great care. I would see that as a challenge to apply some of the things I know - perhaps an algebraic "rationalize the numerator" trick, perhaps the binomial series - to find out the truth. In fact, $\sqrt{x^2+1}-x$ behaves like $\frac1{2x}$ for large $x.$

Here's an example: suppose $V(r)=\frac{A}{r^6}-\frac{B}r$ for $r>0$ and positive constants $A,B,$ What could that be? Perhaps it's something like the potential energy of two ions, with $r$ the distance between them. And maybe that first exponent should be something other than 6, but it won't make much difference to the analysis.

All I told you about $A$ and $B$ is that they're positive constants (probably emprically derived, and whose actual values would need to be expressed in scientific notation.) That makes it a little awkward for you to use that graphing calculator, doesn't it?

But the essential features can be found rather quickly by considering the extreme cases. What does $V(r)$ do as $r\to0^+?$ You should know that $\frac{A}{r^6}$ becomes much larger than $\frac Br$ so this becomes very large positive - or as we could say it, $\lim_{r\to0^+}V(r)=+\infty.$ Visualize that as a vertical asymptote, as the curve going upward steeply just to the right of the axis.

How does this behave as $r\to\infty.$ "That's easy," you tell me, "it goes to zero." But I come back as say I want more information: it goes to zero like what? For large $r$, the $-\frac Br$ term is much the more important term, so that's the answer: as $r\to\infty,\,V(r)$ looks like $-\frac Br.$ It has a horizonatal asymptote, but it approaches that horizontal asymptote from below.

With just the two things we've established so far, we can draw the graph (do so!) The curve clearly crosses the horizontal axis; with some simple algebra, we could write a formula for where. The curve clearly has a minimum (somehwere to the right of that zero). Locating that minimum, and determining how deep, would be a straightforward calculus problem.

None of this talk about the behavior of functions and none of these examples would be at all strange to a pre-Cauchy practitioner of calculus.

There's a deeper answer to "why limits:" it's how we define things. It's how we define [i]new[/i] things.

What is the area under the curve $y=e^{-x^2}$ for $x$ between $0$ and $1?$ We can draw the object in question easily enough. That ought to have an area, right? We can even write it as an integral, but that isn't solving the problem, it's just changing the notation:

$\int_0^1e^{-x^2}\,dx.$

We know the "names" of a lot of numbers: $\frac3{29},\frac{\pi^2}6,\sqrt[8]{13},$ and so on. This one isn't any of those - it's not a number whose name we know. So how do we know it's anything at all? Why should there be such a number?

The answer is that we [i]define[/i] the integral, and we define it as a limit - a limit of Riemann sums (or as a sup/inf of lower/upper sums, which is very closely related to the idea of limits).

We need a number of possible theorems. One is that for a "nice enough" function (and a continuous function on a closed bounded interval is more than "nice enough") the limit procedures we employ to define the integral behave like limiting procedures that should converge.

And then we need the principal that if we have a limiting procedure that should converge (for instance, a Cauchy sequence), then there is a real number there to [i]be[/i] the limit. This is known as the [b]completeness of the real numbers[/b].

And it's that - the completeness of the real numbers - that lets us define, express, and talk about numbers that we don't have names for. For instance, the number that is the integral I started this with. I know there is such a real number; knowing that, I can fairly quickly find a 12-digit decimal approximation for that number. (In practice, I'd use a power series, which is yet another kind of thing defined as a limit, and to which everything I've just said applies.)

just say you have:
$y = \frac{x^2+x-6}{x-2}$
at x equals two, its undefined b/c of division by zero, but
$lim_{x\to 2}\frac{x^2+x-6}{x-2} = 5$
you can factor out a x-2 and you get y=x+3

then for $y = \frac{1}{x}$
$lim_{x\to 0^+} = \infty$ but
$lim_{x\to 0^-} = -\infty$


function
calculus
integration
Functional Analysis
real analysis
real analysis theorems

I have a problem follow,but I yet solve it ,please help me
Have E is a Banach space,$f: \left[{0,T}\right] \times E \to E$ is continous and bounded on
$\left[{0,T}\right] \times \overline{B\left({x_{0},r}\right)}$ :$\left\|{f\left({t,x}\right)}\right\| \le M$
futhermore f is local Lipschitz on $\left[{0,T}\right] \times E$ (i.e

$\forall \left({t_{0},x_{0}}\right) \in \left[{0,T}\right] \times E\;;$ have a neighbouring of $\left({t_{0},x_{0}}\right)$ that f is Lipschitz.
Prove that the problem : x'(t)=f(t,x(t)) ; $x\left( 0 \right) = x_{0}$ have a solution on $\left[{0,\frac{r}{M}}\right]$

This is a piece of the standard existience-uniqueness theorem. Convert it to an integral equation, and set up an iteration that converges to a solution; it should be a local contraction.

Let prove it :rotfl: ,It in't simple....... :)


MATHCOUNTS algebra function domain Support symmetry blogs

A man's life is ruled by two gods, Fortunately and Unfortunately. Fortunately makes good things happen to the man. Unfortunately makes bad things happen to the man. These gods are supreme rivals, so they frequently try to counteract each other's actions.

Each of the gods alternate doing their bidding on the man.

Ex.
Fortunately, the man finds a treasure chest with million dollars in it
Unfortunately, the treasure chest is locked...
Fortunately, the man has the key!
Unfortunately, he left it at home...

Rules:
1) Gods cannot blatently negate the other god's action. If one god decree that the man fall off the airplane, the other god cannot decree that he [i]not[/i] fall off the airplane.
2) Unfortunately cannot outright kill the man. Fortunately cannot outright exhalt the man (or whatever the opposite of kill is)
3) No god shall have his turn twice in a row.

Fortunately, the man was enjoying himself on his Caribbean cruise.

Unfortunately, the boat he is on, has a big hole in the deck

Fortunately, the boat is still on the ground  :D

Unfortunately, he doesn't get his money back for the boat although the defect was obviously covered in the warranty.

Fortunately, instead of getting his money back, they just upgraded him to a different cruise ship that is twice as big.

unfortunately, that upgraded cruise is for a ship that hasnt been built yet, it's known that its twice as big only from the blueprints

Fortunately, while it may not be built yet, it is only 2 hours away from completion and its maiden voyage is tomorrow.

unfortunately, there was no bottle of champagne to christen it... so it can never go out to sea

Fortunately, there is a champagne store right next door

Unfortunately, it's been closed for 20 years.

Fortunately, it will be under new ownership in five minutes, and the champagne store will open again.

Unfortunately, it is reopening as a furniture shop, not a champagne store.

fortunately, a couch that will be offered in it was acquired secondhand from a drunk and has a bottle of champagne stashed inside

Fortunately, the new boat is also able to transform into a giant mecha robot (which don't have to be christianed) and the man goes on his second choice vacation of rampaging through Japan (because as far as I know, that's the only place where mechas are legal)

Fortunately, the man is a llama.(and has no trace of human left in him.

Unfortuently, the llama is also evil and has super powers

Fortunately, its only super power makes it not evil.

unfortunately, it is a volcanic planet, and the original man is on that planet.

fortunately, he is transported to a safe world by a fairy

unfortunately, he is assigned the job of killing a 40 foot monster who is ravaging the countrysides.

fortuently, the 40 foot monster dies of natural causes

unfortunately, the monster is resurrected by a dark wizard

Fortunately, Mr. Evil Evil's llamas used the SEG to kill both the wizard and the monster.

unfortuently, Mr Evil Evil's llamas take over the world

fortunately, a colossus stomps all the llamas to death

Unfortunately.

Unfortunately takes a break because he has violated the rule on the OP "Thou shalt not kill"

fortunately a fellow employee yells at him to get back to work


Just what is exactly means by:1)Ramanujan Prime;2)Ramanujan Sum;3)Ramanujan Theta Function.Please give example.

                                                                  Respectfully,
                                                                   A.Eddington

These are all questions Google can easily answer.


A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circles and inside the square. What is the radius of the circle?

This was an AMC 8 problem.[hide] Anyway, all of this implies that the circle and the square have the same area, since there are the same number of regions outside the square as outside the circle, and they all have the same area. The square has an area of 4, so $\pir^{2}=4$, implying that $r=\frac{2}{\pi}$.[/hide]

It couldn't have been both. They are held at different times, and one can take both if they are in middle school.

I remember this question, it was 2005 AMC 8, and mathgeniuse^ln(x) missed a perfect score because he tried using calculus on this.


Let $p$ be an odd prime. Without using Dirichlet's theorem, show that there are infinitely  many primes of the form $2pk+1$.

Assume that there are only finitely many primes $\equiv 1 \mod p$ and let $P$ be their product ($P=1$ if there are none).

Let $q$ be any prime divisor of $(P+1)^p-P^p$. We have $q$ coprime to $P$.
Set $Q=\frac{P+1}{P} \mod q$, then $Q^p = \left( \frac{P+1}{P} \right)^p = 1 \mod q$. Now $Q$ has order $1$ or $p$ seen $\mod q$, and order $1$ is impossible, so it has order $p$.
By this, $p|q-1$, thus $q \equiv 1 \mod p$, contradiction (we found such a prime definitely not dividing $P$).

Since $p$ is odd, every prime of type $pn+1$ is of type $2pm+1$.

Assume $ q_i, 1\leq i \leq n$ are the only such primes. Let $ P\equal{}q_1q_2 \dots q_n$, $ N \equal{} P^p \minus{} 1$. Let q be a prime divisor of N. $ d \equal{} \text{ord}_q(P)|p \implies d\equal{}p$, by FLT p|q-1 and since p is odd, $ q\equal{}2kp\plus{}1$. Obviously $ q>q_i$. Repeating the argument there is another such prime and hence infinitely many.

[quote="KevinB"]Assume $ q_i, 1\leq i \leq n$ are the only such primes. Let $ P \equal{} q_1q_2 \dots q_n$, $ N \equal{} P^p \minus{} 1$. Let q be a prime divisor of N. $ d \equal{} \text{ord}_q(P)|p \implies d \equal{} p$, by FLT p|q-1 and since p is odd, $ q \equal{} 2kp \plus{} 1$. Obviously $ q > q_i$. Repeating the argument there is another such prime and hence infinitely many.[/quote]

But for FLT you need to show that $ q \neq 2$ and that $ d \neq 1$. I think it's better to look at $ (P\minus{}1)^p \minus{} 1$ instead, since it's odd and also not divisible by $ p$. For your proof you also need $ P \neq 1$, but that can be shown by looking at $ 2^p \minus{} 1$ for example.


geometry
geometric transformation
reflection
articles
\/closed

I know this is sort of a minor issue, but would you mind adjusting the clock for DST?  Because now, in order to get correct times, I'd have to put GMT - 4 hours instead of GMT - 5 hours because this clock does not adjust for DST, and I cringe at the thought of having to change my profile every 6 months.  (jk)  :lol:

I changed all the people in US timezones to reflect DST.  To all users: If you had already manually changed it (and aren't in the Eastern timezone), you're now one hour ahead of the game and will have to change it back.

DST should be illegal.

There oughta be a law!

Wait...

Lol.  :lol:

Actually, DST is very important.  By setting our clocks forward in the spring and back in the fall, we are forced to stay awake during the most daylight hours possible, conserving energy.  (And I'd assume that mathematicians care about the environment).

In fact, DST also helps safety.  By making people stay awake during daylight hours, it makes people leave the house when it is lighter outside.  This helps every kind of driver and children who need to catch a bus to get to school.

But I'm sure u didn't mean that, MCrawford, so w/e.

I disagree.  DST costs our economy an enormous amount of money.  DST was created to help farmers when our economy was more heavily based in agriculture.

That's interesting about the farmers.   I didn't know that.  I guess you're right.

However, I still think it's useful today.

Could you explain how it costs our economy a lot of money? :?

I don't want to spend a lot of time finding an article similar to one that I read a long time ago, but the sheer number of human production value lost to coordinating such time changes is immense.  Keeping the computers straight alone is no small task.

DST was created by the railroads back in the day so that their time schedules would all match up. I heard that DST saves electricity because people wouldn't have to turn on their lights as early at night.

DST is no longer very useful since a great number of people don't base their waking hours on the position of the sun. (I, however, do.)


hmmm. Don't you do $ n(n\plus{}1)/2$ here, so wouldn't you do: $ 39(40)/2$?

I don't believe so, since you have the $ p \plus{} 2$. The real equation would be $ \sum _{p \equal{} 2} ^{41} p$ (Do you see why?). Then you can apply the formula.

The method alluded to by 7h3.D3m0n can be used, though your formula would have to be tweaked to take into account the fact that you are not starting at $ 1$.

You can also divide the sum into two sums, one of which can use the formula you mentioned:

$ \sum_{p\equal{}0}^{39}(p\plus{}2)\equal{}\sum_{p\equal{}0}^{39}p\plus{}\sum_{p\equal{}0}^{39}2\equal{}\frac{39\cdot40}{2}\plus{}40\cdot2$

[quote="tridecalogism_lover"][quote="jvenezuela716"]The method alluded to by 7h3.D3m0n can be used, though your formula would have to be tweaked to take into account the fact that you are not starting at $ 1$.

You can also divide the sum into two sums, one of which can use the formula you mentioned:

$ \sum_{p \equal{} 0}^{39}(p \plus{} 2) \equal{} \sum_{p \equal{} 0}^{39}p \plus{} \sum_{p \equal{} 0}^{39}2 \equal{} \frac {39\cdot40}{2} \plus{} 40\cdot2$[/quote]

Ah. I see what I did wrong. But how did you get 40 * 2 for that last bit? I'm still confused about p=0. I mean, wouldn't you do $ n* the constant$ ? So wouldn't it be 39 * 2[/quote]

You are adding "2" 40 times. Even if p is 0, value of the function is 2. So in that case, you'll have a 2 as well.

[quote="tridecalogism_lover"]Ah. I see what I did wrong. But how did you get 40 * 2 for that last bit? I'm still confused about p=0. I mean, wouldn't you do $ n* the constant$ ? So wouldn't it be 39 * 2[/quote]
If you started at $ p\equal{}1$, that would be the case. However, you've added one term by making it start at $ p\equal{}0$, so there must be one more term of 2 included.
In general, if you have $ \sum_{k\equal{}a}^b$, there will be $ b\minus{}a\plus{}1$ terms. For the familiar $ \sum_{k\equal{}1}^b$, there are $ b\minus{}1\plus{}1\equal{}b$ terms, and for $ \sum_{k\equal{}0}^b$, there are $ b\minus{}0\plus{}1\equal{}b\plus{}1$ terms.

Use the general formula:
$ 1st$ term: $ 2$
$ 40th$ term: $ 41$
$ \frac{(2\plus{}41)*40}{2}\equal{}860$


Let T: R3 -> R be linear. Show that there exists scalars a, b and c such that T(x,y,z) = ax + by + cz for all (x,y,z) in R3. Generalize this to T: Fn -> F and then show T:Fn -> Fm.

  It looks like all I have to do is say b and c are equal to zero so T(x,y,z) = ax.. this could easily be shown to be linear. and then the other parts are pretty much the same thing.. am I doing something wrong here?thanks

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=311387]this recent thread[/url].

What's wrong with what you said? You don't get to choose the map. You have to show that such coefficients exist for [i]any[/i] linear map $ T$.

[color=green]Not again... Stop with the quotes already![/color]

thanks for the reply and the link.

edit: okay, I think I understand the question.. it's asking me to show that ANY T: R3 -> R  T(x,y,z) can be expressed as ax+by+cz
but I have no idea what to do, I didn't get much from the link that could help me :S


The innkeeper has it because 3*8 doesn't equal 25, but 1 less. I think this is correct, right?

I have no idea how you reached your solution. Where did the 8 come from?

$ 25 can't be divisible by 3 people, but by$8 and with one left over.

That's just a random guess. You can't justify the 8 in the equation.

[quote="fishythefish"]Three travelers walk into an inn and decide to spend the night. The innkeeper charges them $ \$$30. Each traveler pays $ \$$10. Suddenly, the innkeeper realizes that he a made mistake, and that they only owe him $ \$$25. So he gives $ \$$5 to a porter to take to the travelers. However, the money cannot be evenly distributed. So the porter gives each traveler $ \$$1, and keeps $ \$$2 for himself.

So each traveler paid $ \$$9. $ \$$9(3)=$ \$$27.

$ \$$27+$ \$$2 (in the porter's pocket)=$ \$$29.

SO WHERE IS THE REMAINING DOLLAR?

I repeat: those who have seen this (it's a common problem) don't answer and ruin it for others.[/quote]

Say each traveler has $ x$ amount of money.
When they pay, each now has $ x\minus{}10$ and the innkeeper has 30
The innkeeper then give each traveler 1 dollar, so they now each have $ x\minus{}9$
The porter has 2.
The innkeeper has 25.
$ 3(x\minus{}9)\plus{}2\plus{}25\equal{}3x\minus{}27\plus{}27\equal{}3x$
Therefore, they still have the total amount of money they started out with.

Even though this doesn't actually solve the problem...

Law of Conersvation of Mass - The matter that makes (or once made up) the dollar is still present. Tehrefor the dollar is still somewhere.  :D

[quote="fishythefish"]Three travelers walk into an inn and decide to spend the night. The innkeeper charges them $ \$30$. Each traveler pays $ \$10$. Suddenly, the innkeeper realizes that he a made mistake, and that they only owe him $ \$25$. So he gives $ \$5$ to a porter to take to the travelers. However, the money cannot be evenly distributed. So the porter gives each traveler $ \$1$, and keeps $ \$2$ for himself.

So each traveler paid $ \$9$. $ \$9(3)\equal{}\$27$.

$ \$27$+$ \$2$ (in the porter's pocket)=$ \$29$.

SO WHERE IS THE REMAINING DOLLAR?

I repeat: those who have seen this (it's a common problem) don't answer and ruin it for others.[/quote]

Is it the original 25 dollars + the 5?
Or 27+5-2.

ZhangPeijin is correct--where x=10, so just follow his steps (or try to)

Most of you are close, so I'm posting the solution.

(@ryuk: Prove it wasn't in a nuclear reaction... :lol:

They changed the law to include energy.)

[hide="Solution"]

The simplest solution is as follows:

The total price was $ \$$25. Each of the travelers paid $ \$$9 each, totaling $ \$$27. But the money in the porter's pocket should be [i]subtracted[/i], resulting in $ \$$25, the total charge on the rooms. This is a tricky problem, until you realize that you are actually trying to get to $ \$$25, not $ \$$30. [/hide]

This hasn't been responded to for almost a week, so if anyone else has some good riddle problems, post them here. I AM NOT TALKING ABOUT HOMEWORK OR CONTEST PROBLEMS. I'm talking about some nice ones you've heard.

umm. this was  in some forum that i was readin.. i dont think it was this one...

a man has 17 horses and 3 sons. in his will he says to give the oldest son 1/2 of his horses, the middle son 1/3 of his horses, and the youngest son 1/9 of his horses.

the sons realize that 17 isnt divisible by 2,3, or 9! at this point a wise man who has one horse comes along with one horse and is very generous and gives it to the sons, who now have 18 horses. so the first son get 9, the second gets 6, and the third gets 2, and 9+6+2=17. so the wise man takes back his horse and everyone is happy.

first of all, who shouldnt be happy with their share of the horses, and
secondly, why did this division work, i mean, 17 horses were split into halves, thirds, and ninths!

I think I know why.

They couldn't divide 17 horses by two, three, or nine, but they could when the wise man brought an extra horse, making 18, since 18 is divisible by 2, 3, and 9.

However, 1/2 + 1/3 + 1/9 = 17/18, and that is why one horse was left over.

And that problem has already been posted elsewhere.

I have one. I posted it somewhere else, so you should just look at it rather than repeat it here. It's subject is "Logic Problem! Try It! It's not so easy!" or something like that.  :)

A HITMAN STOLE IT! jk, maybe the innkeeper?


linear algebra
matrix
Vieta
trigonometry
algebra proposed
algebra

Solve the system $ \{\begin{matrix}20(x+1/x)=11(y+1/y)=2007(z+1/z)\\xy+yz+zx=1\end{matrix}.$

$ x\plus{}{1\over x}\equal{}{a\over 20}$

$ y\plus{}{1\over y}\equal{}{a\over 11}$

$ z\plus{}{1\over z}\equal{}{a\over 2007}$

$ xy\plus{}yz\plus{}zx \equal{} 1$

From there there are several ways to manipulate the equations to extract $ a$ (e.g. substitution, or exploiting Vieta's), but the numbers are extremely ugly.

Notice the $ 6$ trivial solutions the permutations of $ (\minus{}i,i,i)$ or $ (i,\minus{}i,\minus{}i)$
the other two solutions are ugly as mentioned before.

With this problem, in Vietnam , we mean that $ x,y,z\in\mathbb{R}$.

Not to hard N.T.TUAN  :D

[hide="hint"]Let $ x\equal{}\tan A,y\equal{}\tan B,z\equal{}\tan C,A\plus{}B\plus{}C\equal{}\pi$[/hide]

[quote="quangpbc"]Not to hard N.T.TUAN  :D

[hide="hint"]Let $ x \equal{}\tan A,y \equal{}\tan B,z \equal{}\tan C,A\plus{}B\plus{}C \equal{}\pi$[/hide][/quote]
It is not true. If $ xy\plus{}yz\plus{}zx\equal{}1$ and $ a(x\plus{}\frac{1}{x})\equal{}b(y\plus{}\frac{1}{y})\equal{}c(z\plus{}\frac{1}{z})$ with positive a,b,c, then $ sign(x)\equal{}sign(y)\equal{}sign(z)$. If $ (x,y,z)$ is solution, $ (\minus{}x,\minus{}y,\minus{}z)$ is solution too, therefore we can consider only positive solutions. In these case two of (x,y,z) must be less, than 1. Let they are $ x\equal{}tg\frac{A}{2}, y\equal{}tg\frac{B}{2}$, then $ z\equal{}ctg\frac{A\plus{}B}{2}\equal{}tg\frac{C}{2}, A\plus{}B\plus{}C\equal{}\pi$.
Then $ x\plus{}\frac{1}{x}\equal{}\frac{2}{sin A},y\plus{}\frac{1}{y}\equal{}\frac{2}{sin B},z\plus{}\frac{1}{z}\equal{}\frac{2}{sin C}$, therefore exist triangle with sids (a,b,c). In these case $ (a,b,c)\equal{}(20,11,2007)$ and $ 2007>20\plus{}11$, therefore these equation had not real solution.

I am sorry, i mistyped  :blush: . It must be :

Let $ x\equal{}\tan\frac{A}{2},y\equal{}\tan\frac{B}{2},z\equal{}\tan\frac{C}{2},A\plus{}B\plus{}C\equal{}\pi$

Solution from M&Y:

[hide]First, signs of $ x,y,z$ are same, we have $ 20(x\plus{}1/x)\equal{}\frac{20(x\plus{}y)(x\plus{}z)}{x},11(y\plus{}1/y)\equal{}\frac{11(x\plus{}y)(y\plus{}z)}{y},2007(z\plus{}1/z)\equal{}\frac{2007(z\plus{}y)(x\plus{}z)}{z}$. From here we have a system of $ xy,yz,zx$ and $ xy\equal{}\minus{}\frac{988}{1019}$, contradiction.[/hide]


if you leave in pennslyvannia post something
if you dont, the DONT post anything. sorry, im desperate.

no, you live in PA. your on my MC team.

[hide=":rotfl:"]...no im not impressed[/hide]

I live in PA. Or somewhere around there, I'm not 100% sure. Or rather, I "leave" in PA.

I live in west chester.................if that helps............

Hey all!  I'm up in northeastern PA, so I guess i'm the first from that area  :lol:

i dont live in PA but i used to...ne1 from lancaster or near-lancaster??

Hey guys my name is Ray.

I live in PA.

I don't know what else to say.

Except that I eat curdles and whey.

[quote="VenomX"]Hi guys I live in PA![/quote]
I live in PA. I know you VenomX. You're the guy who called me names and tried to get me banned in FTW.

Hey I'm the AIME12345 I live in Pennsylvania.

I'm a sophomore at Conestoga High School with my bois [b]richuw[/b] and [b]noodlemaster[/b], also known as the "Geo Trio". I'm also honored to be joined this year by National CD participant [b]FATRaichu[/b] and the Lelow Valley Champion [b]Christina26[/b]. We will beat State College High School this year in MathLeague and all the State Competition, because we the best champion in Pennsylvania.

[quote=unimpossible]i dont live in PA. so i guess i shouldn't post this.  :D[/quote]


My watch is 1 second fast per hour and Vasya's is $ 1\frac{1}{2}$ second slow per hour. Right now they show the same time. When will they show the same time again ? When will they show the same correct time again ?

[size=75]
from book "The Moscow puzzles" by Boris A. Kordemsky

Please hide Your solution . Write inside Hide ... /Hide
[/size]

What is up with the watch problems?

[hide="Solutions"]The watches will show the same time when the gain of mine plus the lag of Vasya's is equal to 12 hours. In x hours my watch will be x seconds fast and Vasya's watch will be $ \frac{3}{2}x$ seconds slow. Then $ x\plus{}\frac{3}{2}x\equal{}432000$, and $ x\equal{}720$ days. To show the same correct time again will be even longer. Okay....... This thing will happen to my watch every 1800 days, and to Vasya's every 1200 days. The lcm of these is 3600 days, which is other answer.[/hide]


search
number theory proposed
number theory

Prove that every positive rational number can be written in the form $ \frac{a^3\plus{}b^3}{c^3\plus{}d^3}$, where $ a,b,c,d$ are positive integers.

a little search would have been nice:

[url]http://www.mathlinks.ro/viewtopic.php?p=131812#131812[/url]


:idea: it is said that it takes as long to boil one egg as it takes to boil two eggs. comment on the statement indicating the conditions under which it is valid?
can someone give a mathematical analysis :?:

How do you boil an egg, or several eggs? You bring a pot of water to a boil, then place eggs into the boiling water. The eggs cook becuase heat is transferred from the water into the eggs. The water is at $100^{\circ}\text{C},$ and you will add enough heat continuously to keep the water temperature there. As long as the eggs aren't interacting with each other too strongly - in particular, as long as $100^{\circ}$ water circulates freely between them - then it shouldn't matter how many eggs there are.

The situation in which we placed eggs into cold water in the pot and then put it on the stove might potentially be a little different and should be investigated more fully.

If we tried to cook the eggs by placing them in a microwave oven (not recommend for safety reasons), then it would take longer to cook two eggs than one. The issue there is that one egg would be absorbing the lion's share of the microwave energy bouncing around inside the oven, while two eggs would be splitting the energy absorption.

:blush: Let us take a single pot let us boil eggs there.Then some analysis in entropy of the egg has to be done as the proteins in the yolk become uncoiled.LET us take into account the specific heat capacities of the egg and water too and then let us formulate the cases.After making logical guesses we can provide the equations.This is the way i think this should be attempted as this is a genuine olympiad level question.{it is worth marks, of course!} :lol:
anyways thanks for answering the melting ice problem.i have taken these from british physics olympiads.Will you try out my problem on superconductors and magnets posted in the advanced physics forum?(hey it would be great if you can form the equations of motion in that :rotfl: )


Find all x such that :

$ 8 cosx\equal{}\frac{\sqrt{3}}{sinx}\plus{}\frac{1}{cosx}$

hello, write your equation in the following form
$ \csc(x)\sec(x)\left(2\cos\left(\frac{\pi}{6}\plus{}x\right)\minus{}2\sin(3x)\right)\equal{}0$
Sonnhard.


is it true that if ab-1 is a square then a and b are both
a sum of 2 squares?

let ab-1=x^2
hence ab=x^2+1
suppoe p|ab
then x^2=(-1) mod p, and x^4=1 mod p
but also x^(p-1)=1 mod p, fermat's little theorem
hence p-1=4k
p=4k+1
all primes of x^2+1 is in the form 4k+1
since all primes of that form is expressible as sum of two squares, then
observe the identity (uu+vv)(u'u'+v'v')=uuu'u'+vvv'v'+uuv'v'+u'u'vv=(uu'+vv')^2+(uv'+vu')^2
hence a, and b, being a product of primes in form 4k+1, is also expressible using sum of two squares


geometry
cyclic quadrilateral
geometry proposed

Given a cyclic quadrilateral ABCD. Let E,F on the rays BC,DC resprctively, such that BE=AD, DF=AB. Denote M be the midpoint of EF. Prove that the angle BMC is right.

I think that you've meant the angle $ BMD$ instead of $ BMC$.


I've noticed that Mr. Rusczyk and Mr. Crawford have posted mathcounts-level problems on the mathcounts forum.  Is it possible to post single AMC-AIME-level problems on the American Mathematics Competitions thread in a similar fashion? We students could even do this...given your permission, of course.

We'd be happy to have you put up some problems of AMC/AIME-level difficulty in the AMC thread!


I got a problem with this question. Could anyone please help me, thanks in advance.

Assume that there are 18 board members: 13 females, and 5 males including Carl. There are 4 tasks to be assigned. Note that assigning the same people different tasks constitutes a different assignment. Find the probability that Carl and at least one female are given tasks.?

Well, complement probability is, no females plus case of no arthur but at least one female. I will leave it to you to calculate those (ps you know about binomial coefficients right?)

I still do not know how to slove this prob, can you show me your calculation?


group theory
abstract algebra
superior algebra
superior algebra solved

Let $G$ be a commutative group s.t. there exist $a\neq e$, s.t. $a$ is contained in any subgroup, $\neq \{e\}$, of $G$. Prove that for any two subgroups $K$ and $H$ we have $K\subset H$ or $H\subset K$ or $H=K$.

Under the given hypothesis, the group must be cyclic of order a prime power or eventually a Pruefer group. The result follows.

Could you clarify? I don't understand your proof.

Suppose $x \in K \backslash H$ and $y \in H \backslash K$. Consider the subgroup generated by $x$ and $y$, and we're almost done.

First, $a$ must have prime order $p$, so that $a$ is a power of $a^k$. Second, every element  has order a power of $p$; otherwise, we could construct an element of order $q$ for some other prime $q$. Third, every finitely generated subgroup must be cyclic (by the structure theorem, a non-cyclic abelian group has too many elements of minimal order), and any two elements are commensurable. This actually prescribes the structure; $G$ is of the form $\bigcup_{k=0}^n \frac1{p^k}\mathbb{Z}/\mathbb{Z}$, where $\frac1{p^k}\mathbb{Z}/\mathbb{Z}$ is cyclic of order $p^k$ and $n$ may be either an integer or $\infty$. In either case, every proper subgroup is one of these cyclic groups.


quadratics
linear algebra
matrix
superior algebra
superior algebra solved

I don't know how to say it in English, so I use [] instead of the verb.

A square matrix $A$ is said to be [] with square matrix $B$ if there exists a non-singular matrix $C$ such that $A=C^TBC$.

(e.g. A positive definite matrix is [] with an identity matrix)

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

(1)
Prove that an invertible complex symmetric matrix of size $n\times n$ is [] with the following matrices:

\[
\left(
\begin{array}{cc}
O & E_v\\
E_v & O
\end{array}
\right)
\hbox{if } n = 2v;\quad
\left(
\begin{array}{ccc}
O & E_v & O\\
E_v & O & O\\
O & O & 1
\end{array}
\right)
\hbox{if } n = 2v+1;\quad
\]
======================
2)
Prove that an invertible real symmetric matrix of size $n\times n$ is [] with one of the following matrices:

\[
\left(
\begin{array}{ccc}
O & E_v & O\\
E_v & O & O\\
O & O & E_{n-2v}
\end{array}
\right),\hbox{ or }
\left(
\begin{array}{ccc}
O & E_v & O\\
E_v & O & O\\
O & O & -E_{n-2v}
\end{array}
\right)
\]

(1) Every invertible complex matrix is [] with $E_n$.
(2) Every invertible real matrix is [] with a matrix of the form
$\left( \begin {array}{cc} E_m& O_{m,n}\\ O_{n,m}&-E_n\end{array}\right)$

The matrix
$\left( \begin {array}{cc} O_m & E_m\\ E_m & O_m\end{array}\right)$ is [] with
$\left( \begin {array}{cc} E_m & O_m\\-E_m & O_m\end{array}\right)$

Take v=n and put the sign "+" on $E_{n-2v}$ if n<m and "-" if m<n.


when i make a bulleted list in \begin{enumerate} such as:

\begin{enumerate}
\item N in \ce{NH3} \Rightarrow \textbf{-3} \\
\item C in \ce{CO3^{-2}} \Rightarrow \textbf{+4} \\
\item Cl in \ce{ClO3^{-1}} \Rightarrow \textbf{+5} \\
\end{enumerate}

the bullets dont come aligned. The 2nd and 3rd one come out too close to the left. Could someone tell me how to align them?

Bulleted list? enumerate gives a [i]numbered[/i] list not bullets. But with both \begin{enumerate} ...\end{enumerate} or \begin{itemize}...\end{itemize} I get correctly aligned output. Of course, I presume you have put the dollar signs correctly as in [code]\begin{itemize}
\item N in $\ce{NH3} \Rightarrow$ \textbf{-3} \\
\item C in $\ce{CO3^{-2}} \Rightarrow$ \textbf{+4} \\
\item Cl in $\ce{ClO3^{-1}} \Rightarrow$ \textbf{+5} \\
\end{itemize} [/code]otherwise you'd get lots of errors.

Are there any errors? If you still get a problem please give a complete small document in case something else is causing a problem.


geometry
geometric transformation
reflection
homothety
ratio
circumcircle
trigonometry

Let $ABC$ be a triangle and let $P$ be a variable point. We denote by $A',B',C'$ the symmetric points of $P$ across the triangle's sides and by $P^*$ the centroid of triangle $A'B'C'$. Find $P$ such that $P=P^*$.

You ask for finding a point P such that this point P is the centroid of triangle A'B'C'. Well, if X, Y, Z are the orthogonal projections of the point P on the sides BC, CA, AB of triangle ABC, then, of course, the reflections A', B', C' of the point P in the sides BC, CA, AB are the images of these projections X, Y, Z under the homothety with center P and factor 2. Also, the point P is clearly its own image under this homothety. Hence, the point P is the centroid of triangle A'B'C' if and only if the point P is the centroid of triangle XYZ. Now, the triangle XYZ is the pedal triangle of the point P with respect to triangle ABC, and it is quite well-known that the point P is the centroid of triangle XYZ if and only if the point P is the symmedian point of triangle ABC (see, e. g., the note [i]New Proof of the Symmedian Point to be the centroid of its pedal triangle, and the Converse[/i] on [url=http://de.geocities.com/darij_grinberg/]my geometry website[/url]). Thus, the answer to your problem is the symmedian point of triangle ABC.

On the other hand, it appears strange to me that an IMO shortlist problem has the symmedian point as an answer...

  Darij

Maybe he meant construct. Then we no need all big theory about symmedians, because we know ratio $PX:PY:PZ$.

[quote="enescu"]Thank tou!
Indeed, it was not in the IMO shortlist, but in the Balkan Math Olympiad shortlist, some fifteen years ago...[/quote]

Hi enescu,

may I ask you whether you can tell me whether there any sources for BMO shortlist problems or at least kindly to ask you to post some more of these problems.  :)

I'm sorry, I don't know a source to all BMO shortlisted problems (in fact, for the BMO "listed" is more appropriate than "shortlisted"). I found some problems in "Gazeta Matematica", some were communicated by BMO leaders. Here are few examples:

1. A line passing through the center $O$ of an equilateral
triangle $ABC$ intersects the circumcircles of the triangles
$OAB,OBC,$ and $OCA$ at $K,L,$ and $M.$ Prove that
\[
OK^{2}+OL^{2}+OM^{2}=2AB^{2}.
\]

2. Let $M$ be a point inside the equilateral triangle $ABC$ and let $%
A^{\prime },$ $B^{\prime },$ and $C^{\prime }$ be its projections
on the sides $BC,$ $CA,$ and $AB,$ respectively. Denote by $%
r_{1},r_{2},r_{3},r_{1}^{\prime },r_{2}^{\prime },$ and
$r_{3}^{\prime }$ the inradii of the triangles $MAC^{\prime },$
$MBA^{\prime },$ $MCB^{\prime },$ $MAB^{\prime },$ $MBC^{\prime
},$ and $MCA^{\prime }.$ Prove that
\[
r_{1}+r_{2}+r_{3}=r_{1}^{\prime }+r_{2}^{\prime }+r_{3}^{\prime }.
\]

3. Let $ABCD$ be a convex quadrilateral and let $M,N$ be points on
the sides $AB$ and $BC.$ The line segments $DM$ and $DN$ intersect
$AC$ at $K$ and $L$ and the lines $BK$ and $BL$ intersect the
sides $AD$ and $CD$ at $R$ and $S,$ respectively. Suppose that
$AK=KL=LC$ and
\[
\left[ ADM\right] =\left[ CDN\right] =\frac{1}{4}\left[
ABCD\right] .
\]
Prove that
\[
\left[ ABR\right] =\left[ BCS\right] =\frac{1}{4}\left[
ABCD\right] .
\]

4. On each side of the triangle $ABC$ a regular $n-$gon is
constructed in the exterior on the triangle and sharing the side
with the triangle. Find the values of $n$ for which the centers of
the three $n-$gons are the vertices of an equilateral triangle.

5. Let $O$ be a point in the interior of the triangle $ABC.$ The lines $%
OA,OB,$ and $OC$ intersect the sides $BC,CA,$ and $AB$ at the
points $D,E,$ and $F,$ respectively. The same lines intersect
$EF,FD,$ and $DE$ at the points $K,L,$ and $M.$ Prove that
\[
\frac{KA}{KD}\cdot \frac{LB}{LE}\cdot \frac{MC}{MF}\geq 1.
\]

6. Find the maximum number of points that can be chosen in the
interior of a regular hexagon with side length $1$ such that all
mutual distances between the points are at least $\sqrt{2}.$

7. The side lengths of the obtuse triangle $ABC$ are three consecutive odd
integers and $11\cos \angle B=13\cos \angle C.$ Prove that $OI=\frac{2}{3}\sqrt{ab}$ ($O$ is the circumcenter and $I$ is the incenter of the
triangle).

8. Let $A_{1}A_{2}\ldots A_{n}$ be a regular polygon. Find all
points $P$ in the polygon's plane with the property: the squares
of distances from $P$ to the polygon's vertices are consecutive
terms of an arithmetic sequence.

If you know other examples, please let me know. Thank you!

[quote="enescu"]I'm sorry, I don't know a source to all BMO shortlisted problems (in fact, for the BMO "listed" is more appropriate than "shortlisted"). I found some problems in "Gazeta Matematica", some were communicated by BMO leaders. Here are few examples:

If you know other examples, please let me know. Thank you![/quote]

Thank you for your examples.  :)  Unfortunately I don't have other examples.  :? If you don't mind I post these problems in their own threads.  ;)

BTW I need some help regarding another project. If you have a look at these IMO ShortList/Longlist files[url=http://www.mathlinks.ro/Forum/index.php?f=177]there[/url] you will realize that there are quite some gaps. And I thought as times passes by and many people retire from being a IMO (deputy) leader or just do not keep it, and these materials are very interesting to keep the IMO culture and history with their problems. And Romania is the country who gave birth to the idea to establish an international competition among talented math students.  :)

BTW I figured out that most of the IMO (deputy) leader just have very few years and it might be also in their interest to make up a complete collection...   ;)

Some help would be appreciated.  :)


Let $n\geq2$, $a_i>0$ for $i = 1..n$ and $b>0$. Consider a inequation:
$\sum_1^n a_ix_i\leq b$ where $x_i$ is natural (maybe equal 0) ($i = 1..n$)
$N(a_1,...,a_n,b)$ is the number of solutions. Is there a nice formule for $N(a_1,...,a_n,b)$?
I really need it.

[quote="tµtµ"]I'm sorry I won't answer but I doubt it's possible to give an explicit answer.

Usually this kind of question is answered with generating functions : transform your equation in

$\sum_1^n a_i \cdot x_i + X = b - \sum a_i$ with $0 \leq x_i,X$

Then the number of solutions is the coefficient of $x^{b - \sum a_i}$ in the serie $\prod (1/(1-x)^ {a_i}) * 1/((x-1)$

HTH (at least a little)[/quote]
Thanks. In fact, I need a formule in the case where $a_i$ and $b$ are real, but I think it doesn't exsite.

[quote="vu thanh tung"]in the case where $a_i$ and $b$ are real[/quote]

Are you sure the number of solutions is finite in this case ?


The number $2000=2^{4}\cdot 5^{3}$ is the product of seven not necessarily distinct prime factors. Let $x$ be the smallest integer greater than 2000 with this property and let $y$ be the largest integer less than 2000 with this property. Find $x-y$

[quote="M4RI0"]$x=2^{7}$, still thinking about $y$[/quote]
$2^{7}= 128$ (too small).

On the other hand, $3^{7}=2187.$

Oops, i didn't notice it had to be greater than 2000 :blush:

The question remains, how do we methodically tackle the problem? Trial and error doesn't look too good.

The numbers pointed out by [b]sen[/b] look very good!

Strategy:

$2000 = 2^{4}.5^{3}$

Take all $x=p_{1}p_{2}..p_{n}$ ($7 > n \ge 0$) where $p_{i}= 2, 5$. There's 1 empty subset, 2 of size 1, 3 of size 2, 4 of size 3, 4 of size 4, 3 of size 5, 2 of size 6.

Then take the smallest integer $y$ larger than $x$ with $n$ prime factors for each $x$. You have 18 numbers $\frac{2000}{x}.y$ and you find the smallest of these to find the smallest integer greater than 2000 with 7 prime factors. You have to bash 18 but it beats bashing.. 79.. or trying one by one, I suppose.

Similar process for smaller.


$ B$ and $ C$ trisect $ \overline{AD}$ and $ M$ is the midpoint of $ \overline{AD}$. $ MC \equal{} 8$. How many units are in the length of $ \overline{AD}$?
[asy]size(100); defaultpen(linewidth(0.8));
yaxis(0,1,invisible);
pair A = origin, B = (2,0), C = (4,0), D = (6,0), M = (3,0);
draw(A--D); dot(Label("$A$",align=S),A); dot(Label("$B$",align=S),B); dot(Label("$C$",align=S),C); dot(Label("$M$",align=S),M); dot(Label("$D$",align=S),D);[/asy]

If $ M$ is the midpoint of $ \overline{BC}$ and $ \overline{MC} \equal{} 8$, then we know that $ \overline{BC} \equal{} 2 \cdot \overline{MC} \equal{} 2 \cdot 8 \equal{} 16$.  Since that is $ \frac{1}{3}$ of the total distance $ \overline{AD}$, then our final answer is $ 16 \cdot 3 \equal{} \boxed{48}$.


trigonometry
AMC
AIME
ratio
AIME II
geometry
angle bisector

In triangle $ABC,AB=28,BC=21,CA=14$. Points $D$ and $E$ are on $AB$ with $AD=7$ and $\angle ACD=\angle BCE$. Find the length of BE.

Is it possible to draw a diagram :blush: I just wanna make sure I know exactly where everything was.

First we can find angle $A$ and $B$ by law of cosine, then find $\angle{ADC}$ by law of sine, then find $EB$ by using law of sine and cosine.
I am still working to find a pure geometrical solution.

This is reminiscent of AIME II 2005 #14.  I mean like nearly exactly the same. There are some good methods posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=193340&highlight=#193340]HERE!!!!!![/url]. One of them is mine :lol:. So, extrapolate them to this here problem. Hope that helps. This is my 100th post. Just a quick question who says "a hundred" and who says "one hundred"?

[quote="pontios"]The triangles $ABC,ACD$ are similar because $\frac{AB}{AC}=\frac{AC}{AD}$ and $\angle BAC = \angle CAD$.
So $\angle ABC = \angle ACD$, hence the triangle $EBC$ is isosceles.

The angle bisector of $\angle AEC$ intersects $AC$ at $F$.
$\angle FED = \angle FCD \Rightarrow FCED$ cyclic and $F$ is the midpoint of the arc $CD$. So the perpendicular bisector of $CD$ pass through $F$. Since $\triangle BCD$ is isosceles $(BC=BD=3)$ we get that $BF$ is the angle bisector of $\angle DBC$. So we have

$\frac{AE}{EC} = \frac{AB}{BC} =\frac{4}{3} \Rightarrow \frac{AE}{EB} = \frac{4}{3} \Rightarrow \boxed{AE = 16, BE = 12}$

We used that $\angle FED = \angle FCD \Rightarrow FCED$ cyclic
But we don't know yet if $C,E$ are on the same half-plane of $FD$. Since $F$ is between $A$ and $C$, it's enough to prove that the point $D$ is between $A,E$.
If $CH$ is the internal angle bisector of $\triangle ABC,$ then $H$ divides the side $AB$ in a ratio $AC: CB = 2: 3$
But the point $D$ divides $AB$ in a ratio $1: 4$
$1: 4 < 2: 3\Rightarrow \hat{ ACD} < \frac{1}{2} \hat {ACB}$
So the ray $CD$ is in the interior of the angle $ACH \Rightarrow D$  between $A,H$ and $E$ is more far away[/quote]

here is another way : DB=AB-AD=28-7=21=BC, so DCB is isoceles triangle. Then we have: <DCB=<BDC, <DCB=<DCE+<ECB, <CDB=<ACD+<CAD, since <ECB=<ACD, so <DCE=<CAD, then DCE is similar to ACE, so DE/EC = EC/AE. Since we already have proved EC=EB, so DE/EB = EB/AE, EB=AB-AD-DE= 21-DE, AE=AD+DE=7+DE, sub them into the ratio, then calculate we can get DE=9, so BE=AB-AE=28-16=12


Let $f(x) = ax^7+bx^3+cx-5$, where $a,b$ and $c$ are constants. If $f(-7) = 7$, the $f(7)$ equals

$\textbf {(A) } -17 \qquad \textbf {(B) } -7 \qquad \textbf {(C) } 14 \qquad \textbf {(D) } 21\qquad \textbf {(E) } \text{not uniquely determined}$

[quote="4everwise"]Let $f(x) = ax^7+bx^3+cx-5$, where $a,b$ and $c$ are constants. If $f(-7) = 7$, the $f(7)$ equals

$\text {(A)} -17 \qquad \text {(B)} -7 \qquad \text {(C)} 14 \qquad \text {(D)} 21\qquad \text {(E)} \text{not uniquely determined}$[/quote]

[hide]This question kind of annoyed me... I thought it was going to be a two second problem.

Okay so if $f(-7)=7$, then $(-7)^7a+(-7)^3b+(-7)c-5=7$.
Thus  $-(7^7a+7^3b+7c)=12$.
Thus $f(7)=-12-5=\boxed{-17}$
$\boxed{A}$[/hide]


[quote="236factorial"]The sum of N positive integers is 17. What is the largest possible product these N integers could have?[/quote]

[hide]  $3^5 \cdot 2 = 486$????

we want the smallest integers lager than 1, without actually using 1 (i.e $2^8 \cdot 1$ would not work becuase we need 1)[/hide]

The best number for it value is 3, and the 2nd best is 2.  You want as many 3's as possible without having to use 1's.  That makes the largest product $3^5\times2=486$

The actual best number is e, but it asks for integers.  (I was really really bored one day and found this out :D )


geometry
rectangle
analytic geometry
perimeter

Prove that Pick's Theorem holds for a rectangle with sides of length a and b witch are parallel to the coordinate axes.

A poster to whom this proof is new is likely not to know what Pick's Theorem is, so I think it would help if you provided a statement.

[b]Pick's Theorem:[/b]  Given a polygon whose vertices are lattice points, let $ I$ denote the number of lattice points in its interior, $ B$ the number of lattice points on its perimeter (or boundary), and $ K$ denote its area.  Then $ K \equal{} I \plus{} \frac{B}{2} \minus{} 1$.

You probably want to specify that $ a, b$ are integers.

(I'm going to use $ A$ instead of $ K$).  Lets first recall that $ A\equal{}ab$.  Now the number of points inside this triangle is $ I \equal{} (a\minus{}1)(b\minus{}1) \equal{} ab\minus{}a\minus{}b\plus{}1$.

I probably need to clarify the derivation of this formula: If we write a 1 along every unit of length for a side with length $ a$, and then imagine shifting the numbers all in one direction to a lattice point, we notice that all of the lattice points on the side and one of the corners of the rectangle gets counted, so we have to subtract the corner out, hence the $ a\minus{}1$.  Doing the same to $ b$, we know we have an $ a\minus{}1$ by $ b\minus{}1$ array of lattice points inside the rectangle, hence $ I\equal{}(a\minus{}1)(b\minus{}1)$.

The number of points on the perimeter is the same as the perimeter (think of shifting every unit of length on the perimeter to the point counterclockwise on the perimeter - then every point is accounted for), so we have $ B \equal{} 2a\plus{}2b$.  Substituting, $ A \equal{}ab \minus{}a \minus{}b \plus{}1 \plus{}(2a\plus{}2b)/2 \minus{} 1 \equal{} ab$, which is the area of the rectangle.


The number ${25}^{64}\times{64}^{25}$ is the square of a positiv einteger $N$. In decimal representation, the sum of the digits of $N$ is:

[quote="alkjash"][hide]is it 25?[/hide][/quote]

nope.

the choices are 7,14,21,28, and 35.

I didn't know you could multiply the 5 and the 2 like that and combine.

Wow.

No wonder I didn't get it.

seems pretty cool

That's usually the trick with these types of problems.

[quote="DiscreetFourierTransform"]$n^{2}= 5^{{2}^{64}}\times 2^{{6}^{25}}$
$n^{2}= 5^{128}\times 2^{150}$
$n^{2}= 5^{128}\times 2^{128}\times 2^{22}$
$n^{2}= 10^{128}\times 2^{22}$

That 128 multiples of 10 times 2 to the twentysecond power.

The square root of this is

$n = 10^{64}\times 2^{11}$

10 to the sixtyfourth is just a one with 64 zeroes after it.
2 to the eleventh is 2048

Clearly, we're going to have a number 2048 with 60 zeroes after it.

The sum of the digits of N is just then $\boxed{2+4+8 = 14}$[/quote]
That was an awesome solution. I didn't think about that, but I knew you had to square root it somewhere...

oh oops. i didnt see the square root thing. arg. cant even read a problem right

Thank you! Just thought of that right now!
I took the 12, but didn't do so well, so I'm going over all the problems I can find in HSB and I'll work from there.

Thought that this was a particularly nice problem, because it looks like you'd need a calc, but really you can do it without one.

just a reference that AMC10 problems usually do not require a calculator at all.


When will the results come back by email, roughly? I'm not 100% sure of my score and I'd like to know.

See page 6 of the [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2006-7pub/07-AMC1012A-TM/2007AMC1012TM.pdf]AMC 10/12 Manual[/url]:
[quote]The AMC office will send results by email (if available) and first class mail as soon as the answer forms are scored. If you have not received your results from our office within 30 days after the AMC 10/AMC 12 please contact us to verify that your answer forms were in fact received.[/quote]

I read that. Does anyone have a rough idea of how long it generally takes to score the exams?

It takes about 3 weeks.

Steve Dunbar
AMC Director

[quote="worthawholebean"]Thanks. Too bad it's too late to register for the AMC B if necessary.[/quote]
You can still register today...


AMC
AIME
geometry
USA(J)MO
USAMO
search
number theory
\/closed

[quote="brady00n"]Right now, I can do 90% of the problems in AoPS #1, and I'm starting on AoPS #2, on the triangles chapter.  I'm gonna be a junior next year.  I'm devoting about 7-8 hours/day on math, think it's possible to master AoPS #2 in about a month (master as in have a general competence)?  What level of mastery is needed, say, to get an AIME score of about 10?  It seems like the material in AoPS goes up to the IMO level.  How is it expected that complete mastery can take place?  Seems like I'm making progress at a good rate; what I've done now is virtually entirely self-study (In Volume 1, the transformations chapter is tough, and the problems at the end of the number theory and counting/probability are a little tough, number theory ones are probably the most difficult.

Geometry, I can get all of the non thread and needle, and I generally have the right idea for the needle and thread ones, and I have the right idea 75%, get em right a lil less than that, the number theory, it seems like it's not in depth enough, only one chapter...  same with counting/probability, there's not enough chapters...

It seems to be that AoPS #2 has really tough problems...  Should I get Art and Craft of Problem Solving concurrently, or after getting about 90% of AOPS 2, about half?  Think it would be appropriate to enroll in WOOT or AIME problem series starting this summer?

I probably sound a little excited, I guess I am...  Maybe it's an addiction, but I'm spending all day in aops, i'm neglecting to talk to my friends, and my social life is slowly going down the drain.  But I'm still happy. =) I would like to qualify for USAMO next year and maybe even MOP.  I've heard of people who improved drastically, so they gave me some hope.

So pretty much, these are my questions:
1) What is the range of difficulty of the AoPS #2 problems?  Assume the fairly talented, but not exceptional problem solver who reads the examples and chapters very thoroughly...how much should i be able to solve?
2) The books recommended by my school friends @ Troy were ACOPS and Geometry Revisited.  When should I get these?  Are the books Mathematics of Choice good?  What are really good introductory number theory books other than the aops one?
3) Does WOOT help a lot?
4) What are the best tools to fast improvement?  What are the jumps in ability necessary to go from getting most AMC 10 problems right, 5 or so on the AIME on a good day to getting a 10 on the AIME, to getting 12+ on the aime. getting a 140+ on AMC12.  Then after than, doing well (20+ on the USAMO).  Any exceptionally good books/classes, any advice on how to attack problems in general?

Thanks a lot for reading through this mess, and thanks even more for bothering to respond! =)[/quote]

yikes 7-8 hours! at least in my opinion, you need time to digest the information, and cramming does not really let the information sink in, of course you may be different, but that is my experience, as for improvement, just do problems...there is no secret: PRACTICE, as for aops difficulty there are some IMO problems, but those are very easy ones, geometry revisited is a great book for geometry, in my opinion it works to just sit around and do problems from old contests or something...and if you master in aops, you should do very well on the aime, but when i say master i don't mean do all the problems, i mean understand the techniques that you use and when you should try them, as for how long you approach a problem, i never look at solutions until i get my own answer...this may not be for everyone, but i believe it is better to attack a problem that you cannot do for a while (of course there are some technical problems that you would avoid, but the aops problems are good for this)...then also you want to check your solution to either verify your solution, or see another technique...also to check your aime score, just go to the resources section and take one (using real test conditions) or you can take mine (lol...it displays many ideas you want to learn, but it is very difficult)  also for approaching the book, i often just sit and read through and look at the problems with nothing to think a bit, then i actually take pencil and paper to write out solutions

Man brady, you're going WAY overboard.... (but I guess that's the only way you can catch up)... Now's it's time to get MASSIVE problems like Alex and get a bunch of practices, because at your cramming rate a lot of the things will be forgotten if you don't practice enough.... And the most important thing: relax, I study a lot because next year is my last year, but if what you really said is true, you cram 2 times more than me... :o

Well back to the resources, maybe you can take some online classes, like the trig/complex numbers  (that sounds hard, but parcially because I'm not taking it but I wanted to :P ), and search for a bunch of problems online.  the mock AMC/AIME make by AoPSers sounds good for you now (they're a lot harder than the real one, for example Altheman's...I got a 5?? :oops: )

And book-wise, if you study AoPS II and later get ACoPS you don't need mathematics of choice - it's WAY too elementary (I borrow it from Alex and felt cheated...I trade ACoPS with him :o ), and also Problem Solving Strategy should be good for you now by august (If you really cram that fast...).  Another important thing, which ever area you're weak on, find book that's specifically on that area and study it, just by doing those cover-all-subjects book won't be enough.

So yeah, keep it up :) !! (Not time-wise, but the spirit...)

Its the amount of knowlege you retain matters more than the amount of time you put in(though if u want to do well u do need to put in time, but 6-8 hrs!?  during the summer or during school?  when i can't solve a problem(after thinking for a long time) I go and do something eles, like play a sport or somethign, then I come back to it, and a lot times, I figure out the answer in like 3 min.  I think i was just too tired. hope this helps  :)


calculus
derivative
function
algebra
polynomial
calculus computations

What is the number of real roots for $f=x^5+2x^4+5x^3+5x^2+4x-2$. I've heard that I can use the derivative in order to find the number of real roots,  how ?

yes that is right. For let g(x) be the derivative of f(x). Then if for example g(a) < 0 and g(b) > 0 for some a,b in reals then f(x) has a zero in that interval. Find all intervals like that and find th enr of solutions. Hope it helps

I've seen some theorems in Algebra book about the number of roots of polynoms like Shutma ,Dekart,...
But i dunno them by heart
I think u can find them easly

[quote]any idea ?[/quote]

yes I think that you mean the methode of newton-rapson

here two links

http://www.as.ysu.edu/~burden/Algo23.htm

http://www.ipt.ntnu.no/~kleppe/TPG4155/newton.pdf

Greets

I remember reading some were that a poly can not have more roots then $\textsl{n}$ where $\textsl{n}$ the order of the poly. So in this case $\textbf{x}\leq5$. Hope this helps and correct me if im wrong.

By Descartes's rule of signs, $f(x)$ has exactly one positive root. Applying the same rule to $-f(-x),$ we see that $f(x)$ may have up to four negative roots.

Now consider $f'(x)=5x^4+8x^3+15x^2+10x+4.$ Descartes says that this has no positive roots.

$f'(-x)=5x^4-8x^3+15x^2-10x+4.$ This has three sign changes, so $f'(x)=0$ has up to four negative roots. This puts no particular restriction on the number of negative roots $f(x)$ can have.

By Rolle's Theorem, between any two roots of the function, there must be a root of the derivative. But with so many possible roots of the derivative, this doesn't really restrict us.

So we now know that $f(x)=0$ has one positive root and up to four negative roots, for up to five real roots.

Newton's method is a way of finding the roots. We make some intitial guess $x_0$ and then form the sequence $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.$ This should converge to a root; different initial guess may lead to different roots.

Trying that for this polynomial, we find the following root: $0.32597348914722.$

The function does appear to be negative for all negative $x$, hence there is exactly one real root.


If a square is partitioned into $ n$ convex polygons, determine the maximum possible number of edges in the obtained figure.

(You may wish to use the following theorem of Euler: If a polygon is partitioned into $ n$ polygons with $ v$ vertices and $ e$ edges in the resulting figure, then $ v\minus{}e\plus{}n\equal{}1$.)

Sorry, but to clarify, does number of edges refer to the sum of the number of edges of each of the $ n$ polygons, or number of edges drawn between vertices in the square?

The answer is $3n+1$. For a construction, draw $n-1$ parallel lines through the interior of the square.

To prove this bound, consider the planar graph interpretation. Obviously, there are $n+1$ faces. We proceed by double counting the total degree of the vertices: on one hand, it equals $2E$; on the other hand, because every vertex except for the four corner vertices must lie on at least three edges due to the convexity condition, $$2E = \mathrm{totdeg} \geq 3V-4.$$ Thus, substituting into Euler's formula, $$V+n-1 = E \geq \frac 12(3V-4) \implies V \leq 2n-2.$$ Then, $$E = V+n-1 \leq 3n+1,$$ so we are done.

The answer is $3n+1$. We can get this by drawing $n-1$ parallel lines to form $n$ rectangles.

Now, by Euler's formula, we have $$E+2 = F + V.$$ We have $n+1$ faces with $n$ polygons, so $$E + 2 = n+1 + V.$$ Now we compare $E$ and $V$. First, ignore the four vertices and edges of the original square, then because we require polygons to be convex, each extra vertex must have at least three edges to it, so we have $V-4 \leq \frac{2}{3}(E - 4)$ so $$E + 2 = n+1+v \leq n + 1 + \frac{2}{3}E - \frac{8}{3} + 4$$ $$E \leq 3n+1.$$ As desired.

The answer is $3n+1$. To achieve this, have $n$ rectangles stacked on top of each other.\\

Interpret the configuration as a graph with the vertices of the polygons as vertices and basic segments as edges.\\

Claim: Each non-corner vertex has a degree of at least 2. An edge vertex cannot only have a degree of 2 because then there will be a degenerate polygon and thus it will not be a vertex anymore, so it has a degree of at least 3. An interior vertex must have a degree of at least 3 since the total angle is 360 degrees but each angle has to be strictly less than 180 degrees.\\

This also shows that $$2e\geq 3v-4$$ since there are only 4 corner vertices. However, since this is a connected planar graph, we also have $$e=v+f-2.$$ Clearly, $f=n+1$ ($n$ interior faces and 1 exterior face), so $$e=v+n-1\rightarrow 2e=2v+2n-2.$$ Thus, $$2v+2n-2\geq 3v-4$$ $$v\leq 2n+2,$$ so $$e\leq 3n+1,$$ as desired.\\

Remark: this solution was motivated by drawing lots of pictures and trying to find what the equality cases have in common. I noticed that they all had the same number of vertices $2n+2$, which leads to thinking in terms of number of vertices rather than edges which then of course motivates the graph interpretation and Euler's formula.

Local Approach:

We can just induct. Assume we have drawn $k$ line segments, giving $3k+1$ polygons. Consider adding a segment to increase the number of polygons by $1$. Clearly we will be splitting some polygon into $2$. It is easily seen that splitting a polygon through a segment passing through one of its corners is suboptimal. To see this consider splitting a polygon along a corner. Then the total number of basic lines remains constant. The only other way we can split a polygon is to draw a segment passing through two of its sides, resulting in a total of $3$ additional basic lines.

Thus our induction is complete. For a construction take $n - 1$ parallel lines cutting the square into $n$ rectangles stacked next to each other.

Global Approach:

This is basically the global approach worded differently. Note that Euler's formula gives $V- E + F = 1$, as we are only concerned with polygon bounded by the square. Noting that due to convexity each vertex has at least $3$ edges, except for the four corners of the original polygon, which each contain only $2$. Thus double-counting the degree of this graph we find, $2E  \geq 3V - 4 \iff \frac{1}{3}(2E - 4) \geq V$. Substituting into Euler's we find the desired.


analytic geometry
geometry
calculus
calculus computations

In arch $ OA\frown$ of the line in the plane , with equation $ f(x)\equal{}x^3$ where$ O(0,0)$ and $ A(a,a^3)$ for $ a>0$. Find the coordinates of the point $ M(X0,Y0)$ in order that $ \triangle OAM$ has the biggest area , than find the equation of the tangent and normal line in point$ M$

Let point E be x-coordinate of point M. (Normal from M to x-axis intersects x-axis at E)
Let point F be x-coordinate of point A. (Normal from A to x-axis intersects x-axis at F)
Let the area $ \triangle OAM$ be P.  $ P(x_0, y_0) \equal{} A(AOF) \minus{} A(MOE) \minus{} A(MEFA)$

We know $ y_0 \equal{} x_0^3$, So, at the end you should get:

$ P(x_0) \equal{} (ax_0/2) (a^2\minus{}x_0^2)$

Differentiate it and so forth...  You should get the answer:

$ X_0^* \equal{} a/\sqrt{3}$.  And you should be able to compute the rest.

-polar fox
http://www.8foxes.com