function
real analysis
real analysis unsolved

I guess we all know that uniform convergence of a sequence of functions implies pointwise convergence :D

The converse is not true. However, it is true in the special case of a sequence of distribution functions.

But how to prove this?  :huh:

You cannot prove it, since it's false... You need at the very least continuity of the limit distrubution.

[quote="Arne"]
The converse is not true. However, it is true in the special case of a sequence of distribution functions.

But how to prove this?  :huh:[/quote]
I posted sometime ago a theorem that gives the converse.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=147666

Is that what you are asking?

No, I'm not asking for a proof of Dini's theorem... This question is different. I want to show that if a sequence of distribution functions converges pointwise to a distribution function which is continuous, then the convergence is uniform.

Distribution functions have the property that for every $n$, you can split $\mathbb{R}$ up the in $2^{n}$ intervals $f^{-1}(2^{-n}[k,k+1])$. So we only need to prove it for the intervals $f^{-1}([1-2^{-n},1])$ and $f^{-1}([0,2^{-n}])$, since for the compact intervals we can take $n_{k}=\max_{x\in2^{-n}[k,k+1]}\{n_{x}\}$ and then take $\max_{k=1,\ldots,n-2}\{n_{k}\}$. Name our 2 intervals $]-\infty,A]$ and $[B,+\infty[$.

Now we use the fact that $f_{n}$ increases: taking $n_{0}$ for which $|f_{n}(A)-f(A)|<\frac{\varepsilon}{2}$ and $n_{1}$ for which $|f_{n}(B)-f(B)|<\frac{\varepsilon}{2}$, and $n=\max(n_{0},n_{1})$. Then for $x<A$ we have $|f_{n}(x)-f(x)|<\frac{\varepsilon}{2}+2^{-n}$ and for $x>B$ we have $|f_{n}(x)-f(x)|<\frac{\varepsilon}{2}+2^{-n}$.

Take now $n$ also sufficiently large to have $2^{-n}<\frac{\varepsilon}{2}$, and you're done. :)


algebra
polynomial
calculus
calculus computations

How do I solve by hand the equation

$ x^3\minus{}7 x^2\plus{}14 x\minus{}7\equal{}0$

thanks guys

You don't.

[url=http://en.wikipedia.org/wiki/Eisenstein%27s_criterion]Eisenstein's[/url] tells us that this polynomial is irreducible, so all of its roots are irrational.  The only way to find the exact solutions, therefore, is the [url=http://mathworld.wolfram.com/CubicFormula.html]cubic formula[/url].


Ok, i have a $50 bet riding on this, so i need some help.  The problem on the test that said n=a^bc, and a,b,c are all prime, how  many factors of n are there whose divisors are greater than 2 and less then the number.  Thats the problem as best as i can recall.  As is, what is the answer?

I'm fairly certain that if the College Board finds out that you posted this, they'll send their College Board SWAT Team to hunt you down and kill you.

yes, but i live dangerously, come on, help me out here.

Any one here take the sat and remember this problem?

Well, that, and the way you worded the question, I'm fairly certain that it makes no sense.

yea, what about a = 2? Doesn't seem to make much sense

[quote]are there whose divisors are greater than 2 and less then the number. [/quote]

Are you sure the problem was greater than 2 (and not greater than 1 (or greater  or equat to two ) ?

If so the answer would be  (bc-1)

Yhe way you posted the problem, the answer is one  less when  a=2

If you go to the college confidential board, there's a huge post about SAT problems.... I think the question and answer was on there.

This problem should get moved down, no way a SAT question is Pre-Olympiad level.....(be pretty scary if they are)

I think the answer is bc+1, if a is greater than 2.  If a = 2, then the answer is 0.

And like beta said, this definetly isn't a Pre-Olympiad problem.  If it were, I'd be a gold medalist in the IMO.

Wait. Is it even legal to make a bet on confidential problems?

They are not allowed to be distributed until a certain date, like this example.

It's like owning old money.

If it's five years old, it, crud
If it's fifteen years old, it's illegal because of forgery problems
If it's fifty years old it's an antique...

Someone explain how old bills exist!

By the way, explain the problem better; is it just $\&a^{bc}\&$?

where a,b, and c are prime?

I didn't get the rest of the problem...

It doesn't look like a SAT problem; what section did you get this from?

[quote="pakagawa"]Wait. Is it even legal to make a bet on confidential problems?[/quote]

I think underage betting in general is illegal.

I think all betting that isn't reported to the government is illegal.  Including home poker games etc.  Am I right?

[quote="Andretti"]The problem on the test that said n=a^bc, and a,b,c are all prime, how  many factors of n are there whose divisors are greater than 2 and less then the number.[/quote]
I think the original problem was more like this: If $\textstyle n = a^2 b c$, where $a$, $b$, and $c$ are different prime numbers greater than 2, how many positive divisors does $n$ have besides 1 and $n$?

The answer is 10.

Yea.. .it was 10, but i'm sure they stressed that it was NOT 1 and the number

this is an example of a Pre-Olympiad Problem

AIME 2004A  #15
Define f(1) = 1, f(n) = n/10 if n is a multiple of 10 and f(n) = n+1 otherwise. For each positive integer m define the sequence a_1, a_2, a_3, ... by a_1 = m, a_n+1 = f(a_n). Let g(m) be the smallest n such that an = 1. For example, g(100) = 3, g(87) = 7. Let N be the number of positive integers m such that g(m) = 20. How many distinct prime factors does N have?

That's a lot harder than SAT problem

[quote="Ravi B"]Did you see the word "besides"?  The original wording might have been more emphatic, like "not including 1 and n".[/quote]

Fine.  See if I care.   :)


How many pairs of positive integers $ (x,y)$ with $ x \le y$ are there such that $ \gcd (x,y)=5!$ and $ \text{lcm} (x,y)=50!?$

[quote="moldovan"]How many pairs of positive integers $ (x,y)$ with $ x \le y$ are there such that $ \gcd (x,y) = 5!$ and $ \text{lcm} (x,y) = 50!?$[/quote]

Let $ x=(5!)x_1, y=(5!)y_1$, then $ \gcd(x_1,y_1)=1$

$ x_1y_1= \text{lcm} (x_1,y_1)=\frac{50!}{5!}$ has $ 15$ prime factor.

Write $ \frac{50!}{5!}=p_1^{a_1}p_2^{a_2}p_3^{a_3}...p_{15}^{a_{15}}$
Then each of $ p_i^{a_i}$ divides either $ x_1$ or $ y_1$.

So number of pairs $ =2^{15}$


Find all numbers of four digits that are perfect squares, and that perfect square is the result of adding a two digit number made by the first two digits of that square, and another two digit number made by the last two digits of that square. In example   9801 because   $ 9801\equal{}(98\plus{} 01) ^2$

Please explain it clear, and step-by-step!

Probably not much help but $ 0001 \equal{} (00 \plus{} 01)^{2} \equal{} 01^{2}$. Putting this with your example $ 99^{2}$, we have $ 01 \plus{} 99 \equal{} 100 \equal{} 10^{2}$. Looks interesting: I'll try it later if no-one else answers.

P.S. Reversing, $ (10 \plus{} 89)^{2} \equal{} 9801$.

P.P.S. I've found two more: $ 2025 \equal{} (20 \plus{} 25)^{2}$ and $ 30 \plus{} 25 \equal{} (30 \plus{} 25)^{2}$. Still have nothing general.

And $ 81 \equal{} (8 \plus{} 1)^{2} \equal{} 9^{2}$.

Just looked on the internet and found this: http://wapedia.mobi/en/D._R._Kaprekar

Hope this is useful.

[quote="AndrewTom"]Probably not much help but $ 0001 \equal{} (00 \plus{} 01)^{2} \equal{} 01^{2}$. Putting this with your example $ 99^{2}$, we have $ 01 \plus{} 99 \equal{} 100 \equal{} 10^{2}$. Looks interesting: I'll try it later if no-one else answers.

P.S. Reversing, $ (10 \plus{} 89)^{2} \equal{} 9801$.

P.P.S. I've found two more: $ 2025 \equal{} (20 \plus{} 25)^{2}$ and $ 30 \plus{} 25 \equal{} (30 \plus{} 25)^{2}$. Still have nothing general.

And $ 0801 \equal{} (08 \plus{} 01)^{2} \equal{} 9^{2}$.[/quote]

$ 0801\equal{}9^2$?


On the solution on http://www.artofproblemsolving.com/Wiki/index.php/2006_IMO_Problems/Problem_5 shouldn't the last line after Lemma 1 have just $ a_i\plus{}b_i$ instead of $ {a_i\plus{}b_i}/2$

Yes, that must have been a typo, I have now fixed it. But strangely, this is the first time I hear about an IMO solution being posted in the Wiki; I thought the discussion topics were enough.

Can we harmonize the Wiki with the forum? For instance, every solution will first be posted in the forum, then receive some feedback (e. g. typo corrections, improvements) until it becomes "publishable", and then be posted on the Wiki. And I think the Resources Section (e. g. [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16&year=1989]here[/url]) should link to the Wiki as well, not just to the discussion topic in the forum.

  darij

Indeed in the [url=http://www.mathlinks.ro/resources.php]resources section[/url] you can add up to three solutions for each problem. Surely not everybody can edit the forum posts except mods/admins but it also a form of protection against vandalism. So my question is: How often do regular non-mod/admin user edit the problems sections of the wiki by adding problems/solutions, correcting errors/typos etc. in the wiki?

As Darij I feel it is kind of duplication but I also see the advantages that everybody can edit it. So what about the following idea: When we post problems and/or solutions to the resources section it will automatically also be posted to the wiki? But then we need to think how to synchronize it. Wiki entries should not be overwritten by an update of the resources section but merged. If a certain competition existed in the wiki before the resources section there is also the issue to find this mapping. If you consecutively update a competition in the resources section with problems $ P \equal{} \{P_1, \ldots, P_6\}$ then the merging process should not turn into adding multiple instance copies of $ P$ to the wiki. But it may be a good idea for a subset of solutions corresponding to problems of $ P$ have changed a lot. Then those items could be combined manually. To simplify synchronization a solution should always first posted to the forum.

I agree with orl completely. I spend a huge amount of time doing stuff in the Resources section, so it would be *really* helpful if we didn't have the duplication - there are contests which are in the Wiki but not the Resources section and vice versa. Obviously, it would be difficult to resolve conflicts.


Where can anybody but you sit?

Why would a cleaning person prefer to wash a mirror
than a window of the same size, assuming both are easily
accessible?

How can you throw a ball so it will reverse direction and return
to you without the ball bouncing against or touching any
solid object?

Your bed and light switch are 15 feet apart. Without using
any object or mechanical device, no pole, no remote
switch, how can you turn off the light and get into bed
before the room is dark?

Rusty's Montana cabin is almost buried in snow, and the
temperature is 25 below zero. Rusty's eyes move from the
single match he has in his hand to a candle, an oil lamp and
a fireplace with kindling all ready to be lit. Which does he light
first?

What common mechanized transports in New York City
(120,000 of them) carry 3 times as many passengers daily
as all the cities subways, trains, buses, cars, and taxis
combined?

Why didn't William Howard Taft run for a third term as
president?

What is cowhide chiefly used for?

Can you arrange these letters into one long word:
d o o r n o n e g w l

If Betsy Ross were living today, what would she be most
noted for?

I got these from a Reader's Digest.
Please hide your answers so other people will have a chance to figure them out on their own.   :)

Why would a cleaning person prefer to wash a mirror
than a window of the same size, assuming both are easily
accessible?
[hide]A mirror has only one side, whereas a window has two[/hide]

Your bed and light switch are 15 feet apart. Without using
any object or mechanical device, no pole, no remote
switch, how can you turn off the light and get into bed
before the room is dark?
[hide]It's daytime[/hide]

Rusty's Montana cabin is almost buried in snow, and the
temperature is 25 below zero. Rusty's eyes move from the
single match he has in his hand to a candle, an oil lamp and
a fireplace with kindling all ready to be lit. Which does he light
first?
[hide]The match[/hide]

What common mechanized transports in New York City
(120,000 of them) carry 3 times as many passengers daily
as all the cities subways, trains, buses, cars, and taxis
combined?
[hide]Is it like pigeons or some bird like that?[/hide]

What is cowhide chiefly used for?
[hide]To be worn on cows of course!!![/hide]

Where can anybody but you sit?
[hide]On top of my stomach. Or under my legs. Something like that.[/hide]
How can you throw a ball so it will reverse direction and return
to you without the ball bouncing against or touching any
solid object?
[hide]Throw straight up.[/hide]
Why didn't William Howard Taft run for a third term as
president?
[hide]Because the amendment to ban more than two terms was already in effect by then? Just guessing.[/hide]
Can you arrange these letters into one long word:
d o o r n o n e g w l
[hide]one long word.[/hide]
If Betsy Ross were living today, what would she be most
noted for?
[hide]Defying the usual limits on the length of human lives?[/hide]

[quote="LynnelleYe"]
Why didn't William Howard Taft run for a third term as
president?
[hide]Because the amendment to ban more than two terms was already in effect by then? Just guessing.[/hide][/quote]

[hide]I don't think he ever made a second term.[/hide]

Where can anybody but you sit?

[hide]Next to you, perhaps?[/hide]

What common mechanized transports in New York City

(120,000 of them) carry 3 times as many passengers daily

as all the cities subways, trains, buses, cars, and taxis

combined?

[hide]Elevators, presumably.[/hide]

Why didn't William Howard Taft run for a third term as

president?

[hide]Usually one has to become president for two terms before running for a third.[/hide]

Where can anybody but you sit? [hide] on my lap[/hide]

How can you throw a ball so it will reverse direction and return
to you without the ball bouncing against or touching any
solid object? [hide] throw the ball straight up [/hide]

Your bed and light switch are 15 feet apart. Without using
any object or mechanical device, no pole, no remote
switch, how can you turn off the light and get into bed
before the room is dark? [hide] just do it during the day when there is light [/hide]

Rusty's Montana cabin is almost buried in snow, and the
temperature is 25 below zero. Rusty's eyes move from the
single match he has in his hand to a candle, an oil lamp and
a fireplace with kindling all ready to be lit. Which does he light
first? [hide] the match [/hide]

What common mechanized transports in New York City
(120,000 of them) carry 3 times as many passengers daily
as all the cities subways, trains, buses, cars, and taxis
combined? [hide] elevators? [/hide]

Can you arrange these letters into one long word:
d o o r n o n e g w l
[hide] one long word [/hide]

Why would a barber want to cut two out-of-towners than one resident of the town?
[hide]
Because he would get twice the money[/hide]

why did I get a 19-year bump?
[hide]Because I necropost[/hide]


A group of $ N$ students, where $ N < 50$, is on a field trip. If their teacher puts them in groups of 8, the last group has 5 students. If their teacher instead puts them in groups of 6, the last group has 3 students. What is the sum of all possible values of $ N$?

N mod 8 = 5
N mod 6 = 3

So, N+3 = 5+3 = 8 = 0 (mod 8)
and N+3 = 3+3 = 6 = 0 (mod 6)

So, N+3 is divisible by both 6,8. There are two possible ways: 24 and 48.

So, N=21 or 45, so our answer is 66.


A new sports stadium has a seating capacity of 24,000 people. One-fifth of the seats are box seats, and $ \frac58$ of the remaining seats are reserved seats; all other seats are general admission. How many seats are general admission?

After the $ \frac{1}{5}$ of box seats are taken away, $ 1\minus{}\frac{1}{5}\equal{}\frac{4}{5}$ of the original seats are left. When $ \frac{5}{8}$ of the remaining are reserved, only $ 1\minus{}\frac{5}{8}\equal{}\frac{3}{8}$ of the seats that aren't box seats are also not reserved seats. The number of remaining seats, therefore, is $ \frac{3}{8}(\frac{4}{5})(24000)\equal{}\fbox{7200}$.


Η παρακάτω άσκηση είναι πολύ καλή για διαγωνισμούς. Χαρείτε την!

Ενδιαφέρον πρόβλημα και μπορώ να πω όχι εύκολο . Να η λύση μου

Έστω $ F,R$ οι προβολές των κέντρων $ O_1,\ O_2$ στην $ AB$ . Προφανώς είναι $ \angle{DMC}\equal{}90$ οπότε αρκεί να δείξουμε
ότι αν $ O$ το μέσο του $ DC$ και $ K$ to μέσο του $ AB$ τότε $ OK\equal{}OM$ .
Θεωρώ $ P$ το μέσο του $ O_1O_2$ και $ S_1,\ S_2$ οι τομές των $ O_1O_2,\ DM$ και $ O_1F,\ DM$

Θα δείξω αρχικά ότι η $ OP$ περνά από το μέσο  $ L$ του FR δείχνοντας ότι $ OP//O_1F$ .
Έχουμε τώρα ότι η κάθετη από το $ O$ στην $ DM$ περνά από το μέσο του $ DM$ άρα και από το $ O_1$ και όμοια
η $ OO_2$ είναι κάθετη στην $ MC$ άρα θα έχουμε ότι $ \angle{O_1OO_2}\equal{}90$

Τώρα έχουμε $ \angle{O_1S_1S_2}\equal{}90\minus{}\angle{S_1MN}\equal{}90\minus{}\angle{AMD}\equal{}\angle{FS_2M}$ άρα $ O_1S_1S_2$ ισοσκελές
και $ O_1O$ ύψος άρα και διχοτόμος άρα $ \angle{S_2O_1O}\equal{}\angle{OO_1S_1}\equal{}\angle{O_1OP}$ άρα $ OP//O_1F$
και συνεπώς $ OP\bot FR$ άρα $ OP$ περνά από το μέσο $ L$ του FR

Όμως το $ L$ είναι και μέσο του ΚΜ οπότε $ OL$ ύψος και διάμεσος στο ΚΟΜ άρα ΚΟ=ΚΜ , το ζητούμενο

Σιλουανέ ,
 πολύ καλή σκέψη και ωραία λύση :lol: . Θα στείλλω μια παραλλαγή στο αγγλικό φόρουμ , αλλά σε παρακαλώ να μη στείλλεις λύση , μέχρι να μαζέψουμε και καμιά άλλη ιδέα.
 Κάποια στιγμή , αν θελήσετε , θα στείλλω και μια άλλη λύση που έστειλα στο CRUX για το θέμα αυτό. ΑΝ θέλεις γράψε τη λύση σου στην αγγλική με το όνομά σου από κάτω(και το σχήμα) και στείλτο (ή στείλτο σε μένα να το κάνω εγω) στον R.Woodrow του CRUX [2007:411], moldavian olympiad

 Μπάμπης

Έξυπνη λύση, Σιλουανέ!
Καλή χρονιά σε όλους!  :)

Η βασική ιδέα της λύσης που προτείνω είναι ότι το $ A$ έχει διπλάσια δύναμη ως προς τον $ \Gamma_2$ σε σχέση με τη δύναμη ως προς τον κύκλο με διάμετρο $ DC$.

Έστω $ \phi \equal{} \widehat {NMC} \equal{} \widehat {CMB}$. Από το εγγεγραμμένο στον $ \Gamma_1$ τετράπλευρο $ MADN$ προκύπτει $ \widehat {DAN} \equal{} \widehat {MNB} \equal{} 2\phi \ (1)$

Η $ AC$ τέμνει τον κύκλο $ \Gamma_2$ στο σημείο $ L$

Είναι  $ \widehat {NLC} \equal{} \phi \equal{} > \widehat {ALN} \equal{} \pi \minus{} \phi \ \ (2)$
Αφού το $ D$ είναι μέσον του τόξου $ AN$, θα είναι $ DA \equal{} DN$. Έστω $ d$ ο κύκλος με κέντρο το $ D$ και ακτίνα $ DA$.
Από τις $ (1),(2)$ προκύπτει ότι το $ L$ ανήκει στον κύκλο $ d$

[img]11902[/img]
Έστω τώρα $ K$ η προβολή του $ D$ πάνω στην $ AL$. Σύμφωνα με τα προηγούμενα, το $ K$ είναι μέσον του $ AL$.

Ο κύκλος $ k$ με διάμετρο $ DC$ διέρχεται προφανώς από το $ K$ και το $ M$. Έστω ότι τέμνει την ευθεία $ AB$ σε ένα ακόμη σημείο $ P$

Δύναμη του $ A$ ως προς τον κύκλο $ \Gamma_2: \ AM \cdot AB \equal{} AL \cdot AC$

Δύναμη του $ A$ ως προς τον κύκλο $ k: \ AM \cdot AP \equal{} AK \cdot AC$

Διαιρώντας κατά μέλη παίρνουμε $ \frac {AB}{AP} \equal{} 2$

-----------------------------------------------------------------------------------------
[b]Παρατήρηση![/b]

Αν το $ M$ βρίσκεται εκτός του ευθυγράμμου τμήματος $ AB$, φαίνεται ότι δεν ισχύει. Δεν το έψαξα αλλα ορίστε ένα αντιπαράδειγμα (S μέσον του AB):

[img]11901[/img]
[i]Εξάλλου, στην περίπτωση αυτή ο κύκλος d δεν περνάει ούτε καν από το M
Όπως το παρατηρώ όμως έχει κάποιο ενδιαφέρον και ίσως θα ήταν καλό να το δούμε σαν ξεχωριστή άσκηση[/i]

Λουκάς


search
number theory unsolved
number theory

Let $ A\equal{}\{2^{2^n}\plus{}1|n\in\mathbb{N}\}, B\equal{}\{6^{2^n}\plus{}1|n\in\mathbb{N}\}$. Prove that $ A\cup B$ contains infinitely many composite numbers.


group theory
abstract algebra
modular arithmetic
superior algebra
superior algebra unsolved

Let $ G$ be a finite group, en assume that every maximal subgroup $ M \leq G$ has the property that $ [G : M]$ is prime.
Let $ p$ be the biggest prime divisor of $ \left|G\right|$, and let $ P$ be a Sylow $ p$-subgroup of $ G$.

Prove that $ P \trianglelefteq G$.

[hide="Some Progress"]Suppose $ P \ntrianglelefteq G$. Then $ N(P) \neq G$, so let $ M \ge N(P)$ be a maximal subgroup. Then we have that $ [G: M] \equal{} q$, where $ q < p$ is prime.

Letting $ G$ act on the left cosets of $ M$, we get a homomorphism from $ G$ into $ S_q$ with kernel $ K$ equal to the largest normal subgroup of $ G$ contained in $ M$. Since $ p \nmid q!$, we must have $ P \le K$. Now, since $ K$ is normal and contains $ P$, it must contain all $ p$-Sylow subgroups of $ G$, of which there are
$ [G: N(P)]$. This means that $ [K: K \cap N(P)] \equal{} [G: N(P)]$.[/hide]

Ok, you have very good ideas! I don't know if you can complete them to a full solution though, mine was different.
Here's something you might try and which will work: What can you say about the Sylow $ p$-subgroups of $ M$ ?  :)

Let $ N$ be the normaliser of $ P$ in $ G$. Suppose that $ N\neq G$ and let $ M$ be a maximal subgroup of $ G$ containing $ N$. Then the number of Sylow $ p$-subgroups in $ G$ is equal to $ \frac{|G|}{|N|} \equal{} \frac{ q\cdot |M|}{|N|} \equiv 1 \pmod p$ for some prime number $ q$ dividing $ |G|$. Clearly $ q\neq p$.
But $ M$ contains $ N$, so the normaliser of $ P$ in $ M$ is equal to $ N$. It follows that the number of Sylow $ p$-subgroups in $ M$ is equal to $ \frac{|M|}{|N|} \equiv 1\pmod p$.
It follows that $ q\equiv 1\pmod p$, clearly contradicting $ p$ being the biggest prime divisor of $ |G|$.

Exactly what I answered Yimin, so I think and hope this is correct.
You wrote it down short and elegant!


Prove that there are no integers $ x>1, y>1, z>1$ such that $ x!+y!=z!$

Assume WLOG $ x \geq y$ so $ x!+y! \leq 2x! <(x+1)!$ and at the same time $ x!+y! > x!$ so we have $ x!<z!<(x+1)!$. So, no solutions exist for $ x!+y!=z!$ if $ x>1,y>1,z>1$.

Edit: Hope my reasoning is correct. Also is the integer restriction necessary? Because don't $ !$ only apply for non-negative integers anyway?

This reminded me of this problem: Find all solutions to $ x!+y!+z! = t!$.

[quote="BanishedTraitor"]Also is the integer restriction necessary? Because don't $ !$ only apply for non-negative integers anyway?[/quote]

[url]http://en.wikipedia.org/wiki/Factorial#The_gamma_function[/url]


[quote="arqady"][quote="shyong"]for all positive real $a,b,c$ such that $abc=1$ prove that

$\frac{a^{2}-4a+3}{a^{2}+4a+3}+\frac{b^{2}-4b+3}{b^{2}+4b+3}+\frac{c^{2}-4c+3}{c^{2}+4c+3}\geq 0$[/quote]
$\frac{a^{2}-4a+3}{a^{2}+4a+3}+\frac{b^{2}-4b+3}{b^{2}+4b+3}+\frac{c^{2}-4c+3}{c^{2}+4c+3}\geq 0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(9a^{2}b^{2}+12a^{2}b+12a^{2}c+27a^{2}-44ab+20a-36)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(9a^{2}b^{2}+a^{2}b+b^{2}a+6.5a^{2}+6.5b^{2}+10a+10b-44ab)+$
$+\sum_{cyc}(11a^{2}b+11a^{2}c+14a^{2}-36)\geq0,$ which obvious.[/quote]

Wow  :o . you are really an expanding expert  :D

So what do you think with 4 variables $abcd=1$ and

$E(a,b,c,d)\geq 0$ where $E(a)=\frac{a^{2}-4a+3}{a^{2}+4a+3}$

I dont have a proof by I conjecture that it is true .


algebra
polynomial
induction
superior algebra
superior algebra solved
first college math post

seems to me that you fogot to mention what A and B are? I suppose matrixes with real entries ... ?

Yes , A and B are square matrices of size n with complex entries. Sorry , I was a bit in a rush last night :)

So, let's denote Ak-Bk by Ck. Thus, we have C0=0 and we must find a Cr, with r>0, that equals 0.

Let us consider the sequences (Ci,Ci+1,...,Ci+n), for all i \geq 0. Because the Ck's take on finitely many values, therefore there are only finitely many values that the sequences may take on, because n is fixed (being the order of the matrices). Therefore we will be able to find m \geq 0 and r>0 such that:
(Cm,Cm+1,...,Cm+n)= (Cm+r,Cm+r+1,...,Cm+r+n).
Thus we have  Cm=Cm+r,  Cm+1=Cm+r+1, ..., Cm+n=cm+n+r.

Now comes the interesting part.
Let Xn+an-1Xn-1+...+a0 and Let Xn+bn-1Xn-1+...+b0 be the characteristic polynomials of the matrices A and B, respectively. Because A and B are invertible, and (-1)na0 is the determinant of A, and (-1)nb0 that of B, a0 and b0 will be different from 0. According to the Cayley-Hamilton theorem, we have:

An+an-1An-1+...+a0=0 and

Bn+bn-1Bn-1+...+b0=0, so for any k  \geq 0 we have

An+k+an-1An+k-1+...+a0Ak=0       (1) and
Bn+k+bn-1Bn+k-1+...+b0Bk=0, which leads to

Bn+k+an-1Bn+k-1+...+a0Bk = (an-1-bn-1)Bn+k-1+...+(a0-b0)Bk=Bk(-X)       (2),
where -X= (an-1-bn-1)Bn-1+...+(a0-b0).

By substracting (2) from (1) we obtain, for any k \geq 0, the following:

Cn+k+an-1Cn+k-1+...+a0Ck=BkX.

Now let's return to the m and r we found at the beginning of the solution. We have
Cn+m+an-1Cn+m-1+...+a0Cm=BmX   and
Cn+m+r+an-1Cn+m+r-1+...+a0Cm+r=Bm+rX

But for any i with 0 \leq i \leq n we have Cm+i=Cm+r+i.
Therefore by substracting the two equalities above we have Bm(Br-1)X=0.
Because B is invertible we have (Br-1)X=0.

Now, we know that
Cn+m-1+an-1Cn+m-2+...+a0Cm-1=Bm-1X and

Cn+m+r-1+an-1Cn+m+r-2+...+a0Cm+r-1=Bm+r-1X.

By substracting we have
a0(Cm+r-1-Cm-1)=Bm-1(Br-1)X=0, therefore Cm+r-1=Cm-1.
In the same manner we obtain Cm+r-2=Cm-2,..., and by induction, in the end we will have Cr=C0=0, which completes the proof.

______________________________________________________

Funny thing ! It seems like this is the only solution , in the book (the dreaded Caragea:D) there is a solution in the same manner , slightly simplified.

[quote=Valentin Vornicu]nevertheless andrei negut's solution look nice to me ... i would have probably cooked up the same solution :) well done andrei[/quote]

yeah nice job :)

[quote=mathpro12345][quote=iandrei]If A,B are invertible and the set {Ak - Bk | k is a natural number} is finite , then there exists a natural number m such that Am = Bm.[/quote]

what does  mean????[/quote]

its actually [code][sup][/sup][/code] but it is in AoPS wiki formatting :)


geometry
geometric transformation
rotation
real analysis
real analysis unsolved

Can someone please help me with this question? I can do nothing more for this than draw pictures at the moment.

Let $ x, y \in \mathbb{R}^n$ ($ n \geq 3$) and $ \left\Vert x\minus{}y \right\Vert\equal{}d>0$ according to the standard euclidean ("dot product") norm.  Set $ r>0$ and let $ Z\equal{}\{z: \left\Vert z\minus{}x\right\Vert\equal{}\left\Vert z\minus{}y\right\Vert\equal{}r \}$. Then,
$ 2r>d \Rightarrow Z$ is infinite,
$ 2r\equal{}d \Rightarrow |Z|\equal{}1$.

Thanks for any help.

By a rotation and translation, you can assume $ x\equal{}(\minus{}d/2,0,0,\cdots)$, $ y\equal{}(d/2,0,0,\cdots)$.

Ok so then we have (nothing changes for n>3)

$ \left( \left(z_1\plus{}\frac{d}{2} \right) ^2 \plus{} z_2^2 \plus{} z_3^2 \right) ^{1/2} \equal{} \left( \left(z_1\minus{}\frac{d}{2} \right) ^2 \plus{} z_2^2 \plus{} z_3^2 \right) ^{1/2}$
$ \Rightarrow  \left|z_1\plus{}\frac{d}{2} \right| \equal{} \left|z_1 \minus{}\frac{d}{2} \right| \Rightarrow z_1\equal{}0$

Then $ 2r\equal{}d \Rightarrow 4\left( \left(\frac{d}{2} \right)^2 \plus{} z_2^2 \plus{} z_3^2 \right) \equal{} d^2 \Rightarrow z_2\equal{}z_3\equal{}0$
and  $ 2r > d \Rightarrow 4\left( \left(\frac{d}{2} \right)^2 \plus{} z_2^2 \plus{} z_3^2 \right) > d^2 \Rightarrow z_2\plus{}z_3>0$.

And that seems to be all...  Thanks for the tip.


function
calculus
derivative
probability
Functional Analysis
topology
real analysis

give example of function which is continuous at all real pts. but differentiable nowhere

am interested in knowing other example your example is well known.can we in general say something of such class of func.?

I don't know of any compelling reason to be interested in this class, but here is a curious theorem: "Every separable Banach space is isometric to a space of continuous nowhere differentiable functions". This is actually the title of a 1995 paper by L. Rodríguez-Piazza. It is a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=111284]classical theorem[/url] of Banach and Mazur that any separable Banach space is isometric to a linear subspace of $C[0,1]$. Rodríguez-Piazza shows that the linear subspace can be chosen so that all of its elements, except for zero function, are nowhere differentiable.

There are "more" nowhere differentiable continuous functions than there are continuous functions with a derivative at some point.

You can't use cardinality for this - the sets you're interested in have the same cardinality. But there are at least two other senses of "more": Baire category and measure.

Baire category: the set of all contionuous functons on $[a,b]$ that have a derivative at any point can be contained is a countable union of nowhere dense subsets of the metric space $C[a,b].$

Measure (probability): Construct a probability measure on the set of continuous functions with a certain fixed value at $a$ by constructing Brownian motion. The probability that such a continuous function has a derivative at any point is zero.


[quote="leviethai"]Let $ a,b,c$ be positive real numbers such that

$ abc \ge 1$

Prove 2 following inequalities

[quote]$ \dfrac{1}{{3{a^4} \plus{} 4{b^4} \plus{} 5}} \plus{} \dfrac{1}{{3{b^4} \plus{} 4{c^4} \plus{} 5}} \plus{} \dfrac{1}{{3{c^4} \plus{} 4{a^4} \plus{} 5}} \le \dfrac{1}{4}$[/quote]

 [quote]$ \dfrac{1}{{{a^7} \plus{} {b^5} \plus{} 1}} \plus{} \dfrac{1}{{{b^7} \plus{} {c^5} \plus{} 1}} \plus{} \dfrac{1}{{{c^7} \plus{} {a^5} \plus{} 1}} \le 1$[/quote][/quote]
The first inequality follows from the two known inequalities:
$ \frac{1}{a^4\plus{}b^4\plus{}1}\plus{}\frac{1}{b^4\plus{}c^4\plus{}1}\plus{}\frac{1}{c^4\plus{}a^4\plus{}1} \le 1,$
$ \frac{1}{a^4\plus{}2}\plus{}\frac{1}{b^4\plus{}2}\plus{}\frac{1}{c^4\plus{}2} \le 1.$

For the second one, see my proof here: http://can-hang2007.blogspot.com/2010/01/inequality-38-l-v-hai.html

[quote="leviethai"]Let $ a,b,c$ be positive real numbers such that

$ abc \ge 1$

Prove 2 following inequalities

$ \dfrac{1}{{3{a^4} \plus{} 4{b^4} \plus{} 5}} \plus{} \dfrac{1}{{3{b^4} \plus{} 4{c^4} \plus{} 5}} \plus{} \dfrac{1}{{3{c^4} \plus{} 4{a^4} \plus{} 5}} \le \dfrac{1}{4}$

[/quote]

See the general problem here (#28). :)
http://www.maths.vn/forums/showthread.php?t=16412&page=3
Your problem is the case with $ k \equal{} \frac{1}{3}$.


Positive integers $ a$, $ b$, and $ 2009$, with $ a<b<2009$, form a geometric sequence with an integer ratio. What is $ a$?

$ \textbf{(A)}\ 7 \qquad
\textbf{(B)}\ 41 \qquad
\textbf{(C)}\ 49 \qquad
\textbf{(D)}\ 289 \qquad
\textbf{(E)}\ 2009$

[hide="Click for solution"]
Note that $ 2009 \equal{} 7^2 \times 41$.  Thus, the common ratio is $ 7$, so $ a \equal{} 41$, or $ \boxed{\textbf{(B)}}$.
[/hide]

its helpful to memorize the prime factorization of 2009, isn't it?  :P


number theory
prime numbers
number theory unsolved

1, Show that $ 1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n}$ is not an integer

2, Show that $ 1 \plus{} \frac {1}{3} \plus{} \frac {1}{5} \plus{} ... \plus{} \frac {1}{2n \minus{} 1}$ is not an integer

3, Determine the largest of all integers n with the property that n is divisible by all positive integers that are less than $ ^3\sqrt n$.

4, Let $ p_n$ denote the $ n$ th prime number,

(a) Show that $ p_n \leq 2^n$ for all $ n$. When does the equality hold?

(b) Show that $ p_n \leq 2^{2^{n\minus{}1}}$.

(c) Show that for all natural numbers $ n$, there exist $ n \plus{} 1$ primes less than $ 2^{2^n}$.

5, Find all prime numbers p for which $ 2^p \plus{} p^2$ is prime.

6, Prove if a, b, c, d are integers satisfying
\[ (3a \plus{} 5b)(7b \plus{} 11c)(13c \plus{} 17d)(19d \plus{} 23a) \equal{} 2001^{2001}
\]
then a must be even.

7, If $ p$ is any prime other than $ 2$ or $ 5$, prove that $ p$ divides infinitely many of the integers $ 9, 99, 999, 9999,$ ….

[1]  http://www.mathlinks.ro/viewtopic.php?t=212837

[2] $ S(n)\equal{}\sum \frac{1} {k}$ and $ P(n)\equal{}\sum \frac{1} {2k\minus{}1}$

$ S(2n) \equal{}P(n)\plus{}\frac{1} {2}S(n)$  so $ P(n)\equal{}S(2n)\minus{}\frac{1} {2} S(n)$

and we know $ S(k)$ is not an integer that mean $ P(k)$ is not integer

[/hide]

5) for $ p\equal{}2$ we have $ p^2\plus{}2^p\equal{} 8$ !!!!!!!

supposing $ p$ different to $ 3$ so $ p^2\equiv1[3]$ and $ 2\equiv\minus{}1[3]$

so $ 2^p\equiv\minus{}1[3]$ so $ 2^p\plus{}p^2\equiv0[3]$ ==> $ 2^p\plus{}p^2\equal{}3$ !!!!!!!

so $ p\equal{}3$

[quote="rogerlee"]
5, Find all prime numbers p for which $ 2^p \plus{} p^2$ is prime.
[/quote]
If $ p > 3$, then p is odd and $ 2^p \plus{} p^2$ is divisible by 3, but $ 2^p \plus{} p^2 > 3$.
It easy to check, that p=2 not solution, and p=3 - solution.
[quote="rogerlee"]
7, If $ p$ is any prime other than $ 2$ or $ 5$, prove that $ p$ divides infinitely many of the integers $ 9, 99, 999, 9999,$ ….[/quote]
In sequence 9, 99, 999, ... there are exist pair (x, y) that x=y (mod p), because the set {0, 1, ..., p-1} is finite.
So, $ (x \minus{} y)/10^k$ in this sequence and divisible by $ p$ (here $ k$ - the number of zeros in the end of $ x \minus{} y$).

So, we can see, there exist $ z \equal{} 999...9 \equal{} 10^t \minus{} 1$, which is divisible by $ p$.

Thus, every element in sequence $ z_1 \equal{} z \equal{} 10^t \minus{} 1$, $ z_n \equal{} 10^{nt} \minus{} 1$ is divisible by $ p$.

[quote="rogerlee"]
6, Prove if a, b, c, d are integers satisfying
\[ (3a \plus{} 5b)(7b \plus{} 11c)(13c \plus{} 17d)(19d \plus{} 23a) \equal{} 2001^{2001}
\]
then a must be even.
[/quote]
If $ xyzt \equal{} 2001^{2001}$, then $ xyzt\equal{}1$ $ (mod 4)$
So, $ x \plus{} y \plus{} z \plus{} t$ is divisible by 4.
In this problem we have $ x \plus{} y \plus{} z \plus{} t \equal{} 26a \plus{} 12b \plus{} 24c \plus{} 36d$, so $ a$ is even.

[quote="rogerlee"]
3, Determine the largest of all integers n with the property that n is divisible by all positive integers that are less than $ ^3\sqrt n$.
[/quote]
Suppose $ n > 512$.
$ n$ divisible by $ 8*7*5*3 = 840$, so $ n\geq840$.
$ ^3\sqrt 840 > 9$.
So, we see, that $ n$ divisible by $ 8*7*5*9 = 2520$, and $ n$ must be divisible by $ 11$.

Let $ p$ - the largest prime number, less then $ ^3\sqrt n$.
It's well-known, that $ \frac {^3\sqrt n}{2}\leq p < ^3\sqrt n$.
We have $ n\leq 8p^3 (1)$.
Let $ p_1 = 5, p_2 = 7, ..., p_k, p$ - prime numbers, and $ n$ must be divisible by $ p_1, p_2, ..., p_k, p$.

So, $ n \geq 9*(2p_k)*(4p_{k - 1})*p \geq 9*p*p*p > 8p^3$ - if $ p \geq 11$ then $ p_k$ and $ p_{k - 1}$ exists both.
Contradiction with (1).
So, we have $ n\leq 512$.

The answer is $ n = 420$.
If $ n > 420$, then $ n$ divisible by 3*4*5*7=420, which impossible in $ 420 < n\leq512$.

4c)  By Bertrand Postulate, we have that $ \forall n\in\mathbb{N}\ \exists i: \ n\le p_i\le 2n$.
In $ [2,7]$ there are $ 4$ primes, and for every $ n\in\mathbb{N}$ we have \[ 2^3\le p_{a_1}\le 2^4\] \[{ 2^4\le p_{a_2}}\le 2^5\] \[ \ldots\] \[{ 2^{n-1}\le p_{a_{n-3}}\le 2^{n}}\] So there are at least $ n-3+4=n+1$ primes less than $ 2^n<2^{2^n}$

[quote="rogerlee"]
2, Show that $ 1 \plus{} \frac {1}{3} \plus{} \frac {1}{5} \plus{} ... \plus{} \frac {1}{2n \minus{} 1}$ is not an integer
[/quote]

Let $ S_n \equal{} 1 \plus{} \frac {1}{3} \plus{} \frac {1}{5} \plus{} ... \plus{} \frac {1}{2n \minus{} 1} \equal{} \frac {\alpha_n}{\beta_n}$, where denominator is $ \beta_n \equal{} 1*3*5*...*(2n \minus{} 1)$.

Let $ p$ - the largest prime number, less or equal $ 2n \minus{} 1$.
There is only one number in the set {1, 3, 5, ..., 2n-1}, which is divisible by $ p$ (we can use Bertrand Postulate to prove this fact: $ n < p < 2n$, so $ 2k \minus{} 1$ not divisible by $ p$, if $ p\neq 2k \minus{} 1$).

So, the nominator $ \alpha_n$ is a summ of $ n$ elements and $ n \minus{} 1$ of them are divisible by $ p$, and one of them isn't divisible by $ p$.
We see that nominator isn't divisible by $ p$, but denominator is divisible by $ p$, and $ \frac {\alpha_n}{\beta_n}$ is not an integer.

As we can see, some of these problems (1-5) can be solved by using Bertrand Postulate very easy.

[quote="rogerlee"][quote="Ibn Rochde"][1]  http://www.mathlinks.ro/viewtopic.php?t=212837

[2] $ S(n) \equal{} \sum \frac {1} {k}$ and $ P(n) \equal{} \sum \frac {1} {2k \minus{} 1}$

$ S(2n) \equal{} P(n) \plus{} \frac {1} {2}S(n)$  so $ P(n) \equal{} S(2n) \minus{} \frac {1} {2} S(n)$

and we know $ S(k)$ is not an integer that mean $ P(k)$ is not integer

[/hide][/quote][b][/b]and we know $ S(k)$ is not an integer that mean $ P(k)$ is not integer？？？[/quote]

[quote="Kirill"]Ibn Rochde, what means your post?[/quote]
sorry i had some  PC problems .i want to say that his right and we can solve it by using the same method of the first question


Ragingg wrote:Here's a challenge...

ABC is a triangle, D lies on the side BC and E lies on the side AC. AE = 3, EC = 1, CD = 2, DB = 5, AB = 8. AD and BE meet at P. The line parallel to AC through P meets AB at Q, and the line parallel to BC through P meets AB at R. Find area PQR/area ABC.

Spoiler:

[hide]

Let x=[APB]/[ABC] then [APC]/[ABC]=2x/5 and [CPB]/[ABC]=x/3. So

x + 2x/5 + x/3 =1. Solve x = 15/26

Let F be the intersection of CP and AB, then

[PQR]/[ABC] = (PR/CB)^2 = (PF/CF)^2 = ([APB]/[ABC])^2 = x^2 = (15/26)^2 = 225/676

[/hide]


email
search
floor function
geometry
3D geometry
algebra
factorization

I'm quite curious - how did you actually find the numbers that you have?

I did a search for solutions by brute force (using Excel), which I'm sure is not the most efficient thing to do.  I managed to check all the numbers up to 20 million without too much difficulty even at that though, so I'm sure a more refined approach could go a lot further.

Your set of solutions for $n=3,4,5,6$ is complete (as far as I can see), but there are three other solutions for $n=7$, giving a full set:

$1741725=1^7+7^7+4^7+1^7+7^7+2^7+5^7$
$4210818=4^7+2^7+1^7+0^7+8^7+1^7+8^7$
$9800817=9^7+8^7+0^7+0^7+8^7+1^7+7^7$
$9926315=9^7+9^7+2^7+6^7+3^7+1^7+5^7$

I haven't been able to find any solution with $8$ digits.

Incidentally, there are also (trivial) solutions for $n=1$ (i.e. all numbers from $1$ to $9$), but interestingly enough there are no solutions for $n=2$.

It's interesting that the problem (which is intuitively very clear) has an extremely messy formal statement.  If I've formulated it correctly, we're actually looking for solutions to the equation

\[n=\displaystyle\sum_{i=1}^{\lfloor log_{10}n\rfloor +1}\left( \lfloor 10^{-i+1}n \rfloor-10\lfloor 10^{-i}n\rfloor \right)^{\lfloor log_{10}n\rfloor +1}.\]

[quote="Diarmuid"]Or of course, you could just go and look it up...  :blush:

Apparently they're called Armstrong numbers...

[url]http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eismum.cgi[/url][/quote]

That link doesn't work.. But I think Armstrong nubmers are nubmers that are equal to sum of cubes of their digits, so it's not this..

And to topic starter.. There are more of these numbers.. Starting with all one-digit numbers.. And there are those I e-mailed you.. Here are they:

1741725
4210818
9926315
24678050
24678051
88593477

I don't find this fascinating as others probably do.. And you didn't write any question concerning those numbers.. Are there infinitely many of them, are there infinitely many of pairs $(n, n+1)$ of those numbers..

[quote="kayo"]It can be proved if k is a "perfect-digit-number" then k has less than 61 digits. Thus, they must be finite.

(115132219018763992565095597973971522400 and 115132219018763992565095597973971522401, n=39)[/quote]

Yeah, I realised what stupid thing I wrote.. Then.. How many of them are there?

And about those examples with 39 digits.. How can we check it? And more important, how did you find them?

[quote="kayo"]It can be proved if k is a "perfect-digit-number" then k has less than 61 digits.[/quote]

How?

Sorry if the link doesn't work - but just type the first few into the integer sequence search engine and you'll get it.

I'm not sure if this would prove the bound of $61$, but it's easy to see that this sequence can only have finitely many terms, since the sum of the $n^th$ powers of $n$ digits is at most $n9^n$, whereas an $n$ digit number is at least $10^{n-1}$, which is larger for sufficiently large $n$.

According to the Sloane website, the last Armstrong number is the one kayo posted, with only $39$ digits.

[quote="idioteque"]
And about those examples with 39 digits.. How can we check it? And more important, how did you find them?[/quote]

Sloane, right ;)

The sequence mentioned by [b]Diarmuid[/b] is here : http://www.research.att.com/~njas/sequences/A005188 . ;)


algebra
polynomial
Rational Root Theorem
algebra unsolved

Let n >= 3 be an integer and x be a real number such that the number x, x 2, x^n  have the same fractional parts. Prove that x is an integer.

Have a look here :

http://www.animath.fr/

This is pb #83 (solutions are at the end of the book)

Pierre.

sorry,I can read English only. I cannot find the solution.
Would you mind showing the solution?

I didn't read it, but I have a solution.
Suppose that $x^2-x=k$ for some positive integer $k$ and also that $x$ is not an integer. By the Rational Root theorem, $x$ is irrational.
Consider the sequence $a_n$ defined by $a_0=1,a_1=x,a_n=a_{n-1}+ka_{n-2}$. Then $a_n=x^n$. We also have $a_n=b_n+xc_n$, where $b_n$ and $c_n$ are integer sequences defined by the same recursion with $b_0=1,b_1=0,c_0=0,c_1=1.$ Therefore $x^n\equiv xc_n\mod 1.$ We have $c_1=c_2=1$, and $c_n$ increases; $c_n\ge c_3=k+1\ge2$ for all $n\ge3$. Since $x$ is irrational, we can only have $xc_m\equiv xc_n\mod 1$ if $c_m=c_n$. Since this never happens for $n\ge3$, we are done.

([tex]\Longrightarrow )[/tex] We have [tex]x = \frac{a}{b}[/tex], [tex]x^2 = \frac{c}{b}, x^n =
\frac{d}{b}[/tex] for [tex]n \geqslant 3[/tex] and [tex]a, c, d \in N[/tex]

Thus [tex]v_p ( x^2 ) = 2 v_p ( a ) - 2 v_p ( b ) = v_p ( c ) - v_p ( b )
\Leftrightarrow v_p ( c ) = 2 v_p ( a ) - v_p ( b )[/tex]

But because [tex]c \in N[/tex] we have [tex]v_p ( c ) \geqslant 0[/tex] which means that [tex]2 v_p
( a ) \geqslant v_p ( b )[/tex] (i)

And [tex]v_p ( x^n ) = n ( v_p ( a ) - v_p ( b ) ) = v_p ( d ) - v_p ( b )
\Leftrightarrow v_p ( d ) = nv_p ( a ) - ( n - 1 ) v_p ( b )[/tex]

But because [tex]d \in N[/tex] we have [tex]v_p ( d ) \geqslant 0[/tex] which means that [tex]nv_p (
a ) - ( n - 1 ) v_p ( b )[/tex] (ii)

Then substracting (ii)-(i) gives us [tex]( n - 2 ) v_p ( a ) \geqslant ( n - 2 )
v_p ( b )[/tex] thus [tex]v_p ( a ) \geqslant v_p ( b )[/tex] and [tex]b|a[/tex]

And [tex]x[/tex] must be an integer.

[tex]( \Longleftarrow )[/tex] If [tex]x[/tex] is an integer then [tex]x = \frac{x}{1}[/tex] and thus [tex]x^2
= \frac{x^2}{1}[/tex], and [tex]x^n = \frac{x^n}{1}[/tex] have the same fractional part.

but who has told you that $x$ is a rational number??

Right, when he said fractional part , I forgot x is after all a real number


geometry
incenter
trigonometry
Asymptote
geometry proposed

Through the incenter $ I$ of $ \Delta ABC$ we draw the perpendicular to $ AC$ that cuts the sidelines $ BC$ and $ AB$ at $ M,N$
respectively. Show that if $ \frac {1}{IM^2} \plus{} \frac {1}{IN^2} \equal{} \frac {1}{r^2}$ then $ \angle A$ is right

$ r$ is the inradii

[b]Are there any other triangles that satisfy the condition besides the right-angled triangle?[/b]

[hide]The answer is yes.figure it out.[/hide]

[quote="Virgil Nicula"] [color=darkred]Let $ ABC$ be a triangle with the incircle $ w \equal{} C(I,r)$ . Denote $ M\in BC$ , $ N\in AB$ so that $ I\in MN$

and $ \overline{IMN}\perp AC$ . Prove  that $ \frac {1}{IM^2} \plus{} \frac {1}{IN^2} \equal{} \frac {1}{r^2}\ \Longleftrightarrow\ B \equal{} 90^{\circ}\ \ \vee\ \ |A \minus{} C| \equal{} 90^{\circ}$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] $ r \equal{} MI\cdot\cos C \equal{}$ $ NI\cdot\cos A$ . Therefore, $ \frac {1}{IM^2} \plus{} \frac {1}{IN^2} \equal{} \frac {1}{r^2}\ \Longleftrightarrow\ \cos^2C \plus{}$ $ \cos^2A \equal{}$ $ 1\ \Longleftrightarrow$

$ \cos 2A \plus{} \cos 2C \equal{}$ $ 0\ \Longleftrightarrow\ \cos (A \plus{} C)\cos (A \minus{} C) \equal{} 0\ \Longleftrightarrow\ \ldots\ \Longleftrightarrow\ B$ $ \equal{} 90^{\circ}\ \ \vee\ \ |A \minus{} C| \equal{} 90^{\circ}$ .[/color]

Thanks for your fast post[b] Virgil[/b].Your a master at trigonometry, as i can see in all your posts,What do you think of the question?.I found actually that if we fixe the side BC, the locus of A that satisfies the condition is a [b]Descartes Folium[/b].
Well i used cabri to visualize the curve and it seems like a Desartes Folium,Im performing the computations to see if this curve is really a Descartes folium

after simplifications, the hypotesis is equivalent to this following equation

$ b^2 \equal{} \frac {(a^2 \minus{} c^2)^2}{a^2 \plus{} c^2}$

If we fix $ BC \equal{} a$ $ B(0,0)$, $ C(a,0)$, then $ A(x,y)$,  the locus of $ A$ is the curve $ y^2 \equal{} \frac {x(x \minus{} a)^2}{2a \minus{} x}$

this is the locus of all the points A of the other triangles besides the right angle that satisfy the above condition.

this curve has asymptote $ x \equal{} 2a$ and a loop around $ BC$. It really looks like a Descartes folium, but i'm not sure if that
formula represents a Descartes Folium.


Find two convergent series $\Sigma_{k=1}^{\infty}x_{k}$ and $\Sigma_{k=1}^{\infty}y_{k}$ s.t. $\Sigma_{k=1}^{\infty}x_{k}y_{k}$ diverges. Can this happen if one of the series is aboslutely convergent?

1. One of typical example is:

$\sum_{n=1}^{\infty}\frac{(-1)^{n}}{\sqrt{n}}$ converges but $\sum_{n=1}^{\infty}\left( \frac{(-1)^{n}}{\sqrt{n}}\right)^{2}= \sum_{n=1}^{\infty}\frac{1}{n}$ diverges to $\infty$.

2. Suppose $\sum_{n=1}^{\infty}x_{n}$ absolutely converges and $\sum_{n=1}^{\infty}y_{n}$ converges. Then $y_{n}$ is bounded by some positive real number $M$, so $|x_{n}y_{n}| \leq M|x_{n}|$ and by comparison test, $\sum_{n=1}^{\infty}x_{n}y_{n}$ absolutely converges.


probability
number theory
least common multiple

A four-digit number is chosen at random from all four-digit numbers. Express as a common fraction the probability that the number is divisible by 2, 3, 4, and 5.

Is this just $ \frac1{60}$?  Can someone explain?

To be divisible by 2,3,4, and 5, it must be divisible by their LCM, which is 60. The number of 4-digit numbers divisible by 60 is as follows:

$ 9999 \div 60 \approx 166$

$ 1000 \div 60 \approx 17$

$ 166 \minus{} 17 \equal{} 149 \plus{} 1 \equal{} 150$

There are $ 9999 \minus{} 1000 \plus{} 1 \equal{} 9000$ 4-digit numbers in total, so the probability is indeed $ \boxed{\frac{1}{60}}$.


inequalities integration calculus function geometry rectangle factorial

This question may not be too clear:
Can anyone give an upper bound for this

sum of square roots of 1 ... n, i.e.
sqrt1 + sqrt 2 + ... + sqrt n <= x.
Find  suitable x.

Or, can anyone find a lower bound?

By Cauchy-Schwarz,
${(\sqrt{1}\cdot 1+\sqrt{2}\cdot 1+\sqrt{3}\cdot 1+\cdots+\sqrt{n}\cdot 1)}^{2}< (1+2+3+\cdots+n)(1^{2}+\cdots+1^{2})$
$\Rightarrow \sqrt{1}+\sqrt{2}+\cdots+\sqrt{n}< n\sqrt{\frac{n+1}{2}}$

However, it is certain there are stricter upper bounds.

[quote="t0rajir0u"]We can get a nice bound with calculus.

$\int_{1}^{n+1}\sqrt{x}\, dx > \sum_{k=1}^{n}\sqrt{k}> \int_{0}^{n}\sqrt{x}\, dx$

Where $\int \sqrt{x}\, dx = \frac{2}{3}x^{ \frac{3}{2}}$.  Hence our final bound is given by

$\frac{2}{3}(n+1)^{ \frac{3}{2}}-\frac{2}{3}> \sum_{k=1}^{n}\sqrt{k}> \frac{2}{3}n^{ \frac{3}{2}}$

This is better than the one given by Cauchy since the equality case of Cauchy doesn't really lend itself to giving a good approximation of a sum where the terms go to infinity.[/quote]

how do u know that $\sum_{k=1}^{n}\sqrt{k}>\int_{0}^{n}\sqrt{x}dx$?

The sum in question is a right-hand approximation of the right integral, and a left-hand approximation of the left integral...and both are increasing, so...

[quote="maokid7"]how do u know that $\sum_{k=1}^{n}\sqrt{k}>\int_{0}^{n}\sqrt{x}dx$?[/quote]

[img]http://www.ecr.unimelb.edu.au/~xlr8tr/techs/calculus/integration2.png[/img]

This may not be the correct function, but I hope it helps.

Essentially, the sum is the area of the rectangles and the integral is the area under the curve.  We can pick the sum so that we get rectangles that are exactly within the curve, or so that the curve is exactly within the rectangles.

[quote="thealchemist"]This question may not be too clear:
Can anyone give an upper bound for this

sum of square roots of 1 ... n, i.e.
$\sqrt1+\sqrt 2+...+\sqrt n \leq x$.
Find  suitable x.

Or, can anyone find a lower bound?[/quote]

For a lower bound, you can use AM-GM (i'm not sure of a way to get a more strict bound) to get

$\sum_{k=1}^{n}{\sqrt{k}}\geq n\cdot \sqrt[2n]{n!}$.  (for integral $n$.)

Using Stirling's approximation for factorial, that lower bound simplifies to $\frac{n^{3/2}}{\sqrt{e}}$ (I think.. maybe I made a mistake :P). That will be more accurate for larger $n$.

Of course, an integral will give the sharpest approximation again. And if you want to get an even better estimate, you can approximate the difference between the integral and the sum by another integral, and so on. It just gets more tedious and complicated as you go on.

[quote="t0rajir0u"][quote="cincodemayo5590"]For a lower bound, you can use AM-GM (i'm not sure of a way to get a more strict bound) to get

$\sum_{k=1}^{n}{\sqrt{k}}\leq n\cdot \sqrt[2n]{n!}$.  (for integral $n$.)[/quote]

I think you mean $\geq$.

In any case, the bound I gave is both sharper and easier to calculate  :wink:

Edit:  As with Cauchy, the problem with using AM-GM is that the terms tend to infinity, which means that the result will not be close to the equality case (where the terms are equal).[/quote]

Right  :oops:  (It should be better now)


Here we go :)

Show that $(0,1]$ and $[0,1]$ have the same cardinality

Let $S=\{1,\frac{1}{2},\frac{1}{3},\dots,\frac{1}{n},\dots\}$
Then we define
$f(x) = \left\{
\begin{array}{cc}
0, & x =1\\
\frac{1}{n-1}, & x=\frac{1}{n}\\
x, & x\notin S
\end{array}
\right.
$
then $f:(0,1]\to[0,1]$ is a bijection


Let's ignore the two "decorative" squares near the middle for a moment. In the $ 4$-by-$ 4$ matrix of squares, there is the $ 1$ largest square, $ 4$ of the $ 3$-by-$ 3$ squares, $ 9$ of the $ 2$-by-$ 2$s, and $ 16$ small squares. Adding these four quantities up gives $ 16\plus{}9\plus{}4\plus{}1\equal{}30$ squares.

Now for the extra ones we've ignored previously: there are $ 8$ tiny squares and $ 2$ larger squares, making $ 10$. Our grand total is $ 30\plus{}10\equal{}\fbox{40}$ squares.


Varfurile unui poligon regulat cu $\ n$ laturi se coloreaza in culori diferite, astfel incat punctele cu aceeasi culoare sunt varfurile  unor poligoane regulate.Sa se arate ca printre aceste poligoane se gasesc doua egale.

Observam ca pentru un poligon regulat cu $n$ laturi, orice sub-poligon regulat cu $k$ laturi verifica $k | n$.

Deci $n$ nu este prim. Fie $\frac{n}{k}$ numarul laturilor pe care le are un sub-poligon al sau in configuratia data. Daca $k$ este numar prim si $k \| n$ (adica $k|n$ si $k^{2}\nmid n$), atunci orice alt divizor al lui $n$ verifica $(d,k) = 1$, deci vom avea $k$ sub-poligoane regulate, toate egale intre ele.

[i]Observatie.[/i] In pasul precedent, daca exista $k$ a.i. $k$ e prim si indeplineste conditiile, atunci il alegem neaparat. Altfel facem pasul 2 de mai jos.

Daca $k$ nu-i prim sau $k^{2}| n$, atunci fie $p$ prim, $p|k$. Consideram apoi sub-poligonul $A_{1}\ldots A_{\frac{n}{p}}$, luat din $p$ in $p$ varfuri. Facem chestia asta pentru ca daca un sub-poligon al celui initial are un varf in $A_{1}\ldots$, atunci acolo are toate varfurile (sper sa fie corect ce-am zis eu in ultima fraza). Acest sub-poligon n-are cum sa fie monocromatic, deci putem aplica aceasta procedura pentru $A_{1}\ldots$ (de la "Fie $\frac{n}{k}$" incolo).

Daca notam $\prod n_{i}^{\alpha_{1,i}}$ factorizarea lui $n$ in pasul 1, $\prod n_{i}^{\alpha_{2,i}}$ factorizarea noului $n$ (adica $\frac{n}{p}$) in pasul 2 etc. observam ca $\sum \alpha_{1,i}> \sum \alpha_{2,i}> \ldots$, deci chestia asta se va termina pana la urma (mai exact, atunci cand vom da de un $k$ prim & $k | n$).

Nu stiu cat de bine e, dar macar am incercat  :D


[quote="linzk3"][2 5]*[x]=[4]   how would you solve for x and y??
[3 1]  [y]  [7]

"["
 [ is supposed to be 1 bracket...sorry i didnt know how to do that![/img][/quote]

I have no experience with matricies, but...[hide] You could just multiply out the matricies, to get
$2x+5y=4$
$3x+y=7$
And then solve. [/hide]

Another way would be to change the augmented matrix to reduced row-echelon form. (or find the inverse of the matrix, use cramer's rule, etc.)


geometry
projective geometry
geometry proposed

Projective geometry:

let $\Delta ABC$ be a triangle in the projective plane, let $D$ be a fixed point on $AB$ and let $H$ and $K$ be fixed points which do not lie on one of the sides of the triangle. Let $P$ be a variable point on $CD$. Let $Q = AP \cap DH$ and $R = BP \cap DK$. Show that $QR$ intersects $AB$ in a fixed point. .

Let lines  $AC=a$, $BC=b$, $DK=k$, $DH=h$, $DC=d$.
Consider a central proiection which put line $AB$ to infty.Then $h' \parallel d' \parallel k'$ ,angels $(b',d')$ and $(k',d')$ are fixed. All triangels $P'Q'R'$ which could be get when $P'$ goes over $d'$ are equal and making equal angels with $d'$.So all $P'Q'$ are parallel $\Longleftrightarrow$ they passing throug a fixed point on $A'B'$.

Why my posts are rated like a spam?! :mad:
Someone don't know what is 'central projection' or  how I use it in solution or why there are angels in it?
Or you know but think there is mistake.
Please ask questions.


geometry
circumcircle
incenter
geometry unsolved

Let two circles $(O1)$ and $(O2)$ meet at $A,B$ ( $A$ $\neq$ $B$) . Let circumcircle of triangle $O1AO2$ meet $(O1) ,(O2)$,$AB$ at $M1,M2$,$K$ ,respectively. Prove that $KM1$ = $KM2$

[quote="Megus"]Let $N_{1}, N_{2}$ be points where $KM_{1}$, $KM_{2}$ meets O_1, O_2 respectively. It's enough to prove that $KN_{1}=KN_{2}$. We have:

$2 \angle M_{1}N_{1}A=\angle M_{1}O_{1}A=\pi-\angle M_{1}KA= \angle M_{1}N_{1}+\angle N_{1}KA$ so $KN_{1}=AK$. Analogically $KN_{2}=AK$ so $KN_{1}=KN_{2}$ as wanted. QED[/quote]

there's something wrong with your angle calculations... :>

It 's obvious that $B$ is the intersection of $M_{2}O_{1}$ and $M_{1}O_{2}$. Since these are the bisectors of angles $AM_{2}M_{1}$ and $AM_{1}M_{2}$, we obtain $B$ is the incenter of $O_{1}AO_{2}$. Therefore $KM_{1}=KM_{2}$.

[quote="gollywog"]there's something wrong with your angle calculations... :>[/quote]

where exactly?  :huh:


Let $ x\equal{}\frac{1}{cosy}$, then $ \frac{cosy}{4cosy\minus{}\sqrt 3}\equal{}siny$. It equavalent to $ sin(y\plus{}\frac{\pi}{6})\equal{}sin2y$.
It give solution $ x\equal{}\frac{1}{cosy}$, were $ y\equal{}\frac{\pi}{6}$ or $ y\equal{}\frac{5\pi}{18}$.

Yes. Consider $ \pi\minus{}\pi/6\minus{}y\plus{}2k\pi\equal{}2y$. It give $ y\equal{}\frac{5\pi}{18}\plus{}\frac{2k\pi}{3},k\equal{}0,1,2.$


limit
LaTeX
calculus
calculus computations

$ \lim_{x\to 0}\left( \frac{1}{x}\minus{}\frac{1}{\ln(x\plus{}1)} \right)$

[hide="Solution."]

Edit: oops.  Apparently, fraction bars can't magically be transformed into minus signs.

[/hide]

[hide="Solution"][img]http://imgcash5.imageshack.us/img213/9093/lospital1as9.jpg[/img]

Using L'ospital twice you can find  -1/2.[/hide]

If you want more information ask me ( i didn't put more because of my dificult in Latex )

There's something novel going on with exponents in the first solution

Starting where others left off

$ \lim_{x\to{0}}\frac{\ln(1+x)-x}{x\ln(1+x)}$

Now inserting the Maclaurin series gives

${ \lim_[x\to{0}}\frac{x-\frac{x^2}{2}+\cdots-x}{x\left(x-\frac{x^2}{2}+\cdots\right)}=\lim_{x\to{0}}\frac{\frac{-x^2}{2}+\cdots}{x^2-\frac{x^3}{2}+\cdots}=\frac{-1}{2}$

Essentially the same as Mathstud28's solution but organized slightly differently (avoiding the common denominator):

$ \frac1x\minus{}\frac1{\ln(1\plus{}x)}\equal{}\frac1x\minus{}\frac1{x\minus{}\frac12x^2\plus{}O(x^3)}$

$ \equal{}\frac1x\minus{}\frac1x\cdot\frac1{1\minus{}\frac12x\plus{}O(x^2)}$

$ \equal{}\frac1x\minus{}\frac1x\left(1\plus{}\frac12x\plus{}O(x^2)\right)$

$ \equal{}\frac1x\minus{}\frac1x\minus{}\frac12\plus{}O(x)\equal{}\minus{}\frac12\plus{}O(x).$

And the limit is $ \minus{}\frac12.$


The number $ 2^{1993}\plus{}3^{1993}$ is a multiple of 5. What is the units digit of the quotient $ \frac{2^{1993}\plus{}3^{1993}}{5}$?

The units digit of $ 2^{1993}$ is 2 and the units digit of $ 3^{1993}$ is 3.  The the units digit of the sum is 5, so when divided by 5, it has a units digit of 1.

how do you solve this?
i tried
$ {(5 - 3)^{1993} + (5 - 2)^{1993} \pmod{50}}$ with binom. theorem, didn't work :(

If you don't like ugly format, don't read this, or else write it up nicely for someone else (if you confirm the approach).

Factoring  2^1993 + 3^1993  we get
(2 + 3)(2^1992 - 2^1991*3^1 + 2^1990*3^2 - 2^1989*3^3 + . . . + 3^1992)

Dividing by 5 leaves
(2^1992 - 2^1991*3^1 + 2^1990*3^2 - 2^1989*3^3 + . . . + 3^1992)

Patterns for powers of 2 and powers of 3 give us the following numbers with only the units digits showing.
[(_6) – (_8)(_3) + (_4)(_9) – (_2)(_7) + (_6)(_1). . . + (_2)(_7) + (_1)]

Which simplifies to the repeating pattern:
[_6 – _4 + _6 – _4 – . . . + _6 – _4 + _1]

Reducing to:
[ _2 * 996 + _1] = _2 + _1 = _3
[/hide]

$2^1+3^1=5$. This is a multiple of $5$, $5\cdot1$.
$2^3+3^3=35$. This is $5\cdot 7$.
$2^5+3^5=275$. This is $5\cdot 55$ (units digit 5)
$2^7+3^7=128+2187=2315$ This is $5\cdot463$ (units digit 3)

since the numbers are gonna get big now, it's gonna be easier to reduce everything to $\pmod{50}$

$2^9\equiv 12\pmod{50}$
$3^9\equiv3^7*9\equiv87*9\equiv33\pmod{50}$

so

$2^9+3^9\equiv 12+33\equiv 45 \pmod{50}$ and dividing by $5$ gives us $9$.

if you keep doing this, you'll see that the sequence goes $1, 7, 5, 3, 9, 9, 3, 5, 7, 1$, repeating every 10 odd numbers. since $1993+1/2 = 997$th odd number, which means its the $7$th term, $\boxed{3}$

[hide="birdmath's solution"]

$x^a+y^a$ can be factored into
$(x+y)(x^{a-1}-x^{a-2}y^1+x^{a-3}y^2...-x^1y^{a-2}+y^{a-1})$

which is where birdmath got
$(2+3)(2^{1992}-2^{1991}\cdot3...-2\cdot3^{1991}+3^{1992})$

and dividing by five, we're only left with the polynomial.

now we're only looking for the units digit.

this gives us

$(6-8(3)+4(9)...-4+1)$

finding the units digit we get $2$ $\frac{1992}{2}=996$ times with an extra $1$ at the end.

so we have $996*2$ (units digit of 2) plus $1$ so $2+1=3$.

good bump.


real analysis
advanced fields
advanced fields solved

Show that a countable number of lines on [b]R[/b]2 cannot fill the whole plane.
(Please give an elementary proof without using Baire category)

[i]Chinese undergraduate students camp, year 1992[/i]

Let's give a countable number of lines of the plane.
There are an uncountable number of different directions for the lines in the plane. Thus, we may find a line D which is not parallell to any of the given lines. Each of the given lines meet D in exactly one points, thus they cover only a at most countable part of D, and thus do not cover D entirely.

Pierre.

Quite nice. This is exactly the first proof in official solutions.. it's the easiest one to understand..

To be honest, I did not even try to think about any Baire's theorem. :D
Pierre.

another simple proof is to consider a circle on the plane.
Does anyone have other proofs?

The Lebesgue measure of countable lines is zero where the Lebesegue mesure of $\mathbb{R^2}$ is not zero.


A pair of $ \it{emirps}$ consists of two prime numbers such that reversing the digits of one number gives the other. How many pairs of two-digit emirps exist such that each number in the pair is greater than 11?

Zomg, emirp is prime backwards! How clever.

So, basically, you test all primes greater than 11 that could be emirps.  This means that primes starting with an even number are out (when reversed, they'll be divisible by 2), and primes starting with 5 are out (when reversed, they'll be divisible by 5, of course).  Test all others and you get:

(13,31)
(17,71)
(37,73)
(79,97)

Which is 4.


Let $ a,b,c>0$.Prove that:
$ 3\sqrt{3}(\sum \frac{a^3}{2a^2\plus{}b^2}) \geq \sum \sqrt{a^2\plus{}ab\plus{}b^2}$

It seems that the following inequality is true too.
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that
\[ \sum_{cyc}\frac {18a^3}{2a^2 \plus{} b^2}\geq\sum_{cyc}\sqrt {17a^2 \plus{} 2ab \plus{} 17b^2}
\]
For $ k>8.5$ the inequality
$ \sum_{cyc}\frac {a^3}{2a^2 \plus{} b^2}\geq\frac{1}{3\sqrt{2k\plus{}1}}\sum_{cyc}\sqrt {ka^2 \plus{} ab \plus{} kb^2}$ is wrong already.

[quote="No Reason"]Let $ a,b,c > 0$.Prove that:
$ 3\sqrt {3}(\sum \frac {a^3}{2a^2 \plus{} b^2}) \geq \sum \sqrt {a^2 \plus{} ab \plus{} b^2}$[/quote]

See also : http://www.mathlinks.ro/viewtopic.php?t=228984


Each of $ n > 2$ lines on a plane is cut by the others into $ 2$ rays and $ n \minus{} 2$ equal segments.
Prove that $ n \equal{} 3$.


floor function
number theory proposed
number theory

Prove that if $ b^{n}-1$ divides $ a$ then the sum of digits of $ a$ in base $ b$ representation is not less than $ n(b-1).$

Some definitions:

s(n) is the sum of the digits of n in base b.
f(n) is the greatest integer k such that $ b^{k}|n$
$ g(n)=\sum_{i=1}^{n}f(n)$ (and g(0)=0)

Now (by simple double counting): $ g(x)=\sum_{i=1}^{\infty}\left\lfloor\frac{n}{b^{i}}\right\rfloor=\frac{x-s(x)}{b-1}$

Moreover $ g(x)+g(y)\le g(x+y)$ (by the first equality, because $ \left\lfloor\frac{x}{b^{i}}\right\rfloor+\left\lfloor\frac{y}{b^{i}}\right\rfloor\le\left\lfloor\frac{x+y}{b^{i}}\right\rfloor$)

Let $ a=k(b^{n}-1)$ where k is a positive integer

So we have that:

$ g(a+k-1)-g(a)\ge g(k-1)$

$ g(a+k-1)+f(a+k)-g(a)\ge g(k-1)+f(a+k)$

But $ f(a+k)=f(kb^{n})=f(k)+n$, so:

$ g(a+k)-g(a)\ge g(k-1)+f(k)+n$

$ \frac{a+k-s(a+k)}{b-1}-\frac{a-s(a)}{b-1}\ge g(k)+n$

$ \frac{k-s(a+k)}{b-1}+\frac{s(a)}{b-1}\ge \frac{k-s(k)}{b-1}+n$

But $ s(a+k)=s(kb^{n})=s(k)$, so: $ \frac{s(a)}{b-1}\ge n$, that is to say $ s(a)\ge n(b-1)$ as we wished to prove.


how many solutions $(a,b,c)$ to

$a+b+c\le d$

infinite
unless you specify $d$ and restrict $a,b,c$

sorry a, b, c positive integres, d is positive integer

[hide="shazam"]$a-1+b-1+c-1=d-3$
let $(a-1,b-1,c-1)=(u,v,w)$
imagine $d-3$ dots in a line
we need to partition these dots into three sections
this requires two partitions
so we have $d-1$ total objects, so we can place the partitions in ${d-1}\choose2$ ways. now we know $u,v,w$ are nonnegative integers, so $a,b,c$ are positive integers[/hide]
I think
EDIT: no wait
that's for equality
to satisfy the inequality, you get
[hide="this"]
${{d-1}\choose{2}}+{{d-2}\choose{2}}+{{d-3}\choose{2}}+...+{2\choose{2}}$
which, by the hockeystick theorem, equals
${d-1}\choose3$[/hide]

are you sure it is not [b]dC3[/b]?

Try a case: a+b+c <=5, wehre a,b,c are positive integers.

so a+b+c can equal 2,1,or 0

when a+b+c=2, there are 6 solutions
when a+b+c=1, ther are 3 solutions
when a+b+c=0, ther is 1 solution
so 6+3+1=10, which is percisely 5C3,

you can think about it like this: a+1+b+1+c+1+x=d, where x is nonnegative, so get rid of the less than sign,

so a+b+c+x=d-3, now thats just dC3..

Sorry folks, I had thought it was an $\emph{equality}$ as well.

[quote="yeppyyep"]How many solutions $(a,b,c)$ to
$a+b+c\le d$
exist for $(a,b,c) \in \mathbb{N}^{3}$ and $d \in \mathbb{N}$?[/quote]

[hide="Solution"]
First, let's change the restriction on $a,b,c$: put $a=x+1, \ b=y+1, \ c=z-1$ so that $(x,y,z) \in \left( \mathbb{N}^{*}\right)^{3}$.

Now, we have $x+y+z \leq d-3$. Imagine a line of $d-3$ dots, each one representing one unit of value. We can place three dividers either directly before, directly after, or within the line of dots. The number of dots to the left of all dividers will be the value of $x$; the number between the first two dividers will be $y$; the number between the second and third will become the value of $z$; and whatever is left (past the last divider on the right) will be left-over. We allow the left-over since the value of $x+y+z$ will not always be equal to $d-3$.

Consequently, the number of distinct ways to place the dividers, and the number of ordered solutions, is:
\[\dbinom{ (d-3)+3 }{3}= \boxed{ \dbinom{d}{3}}\]
[/hide]

Good call. The so-called "pivot" variable.

[quote="Art of Owna"]you can always make it an equality by adding a fourth variable thats nonnegative.[/quote]
very nice art of owna!

my bad - hockey stick says the answer is $d\choose3$


function
inequalities unsolved
inequalities

[quote="thienthan89"]Prove that:
$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + 3 \geq 2(a+b+c)$
Where a,b,c are positives numbers satisfy condition: abc=1[/quote]
First, substitute $a=e^x,b=e^y,c=e^z$. We now have $x+y+z=0$. WLOG $x\geq y\geq z$.
Now there are two cases: $x\geq y\geq 0\geq z$ and  $x\geq 0\geq y\geq z$. As I'm lazy, I will only show the first case.
We have $-2x\leq -2y\leq 0\leq 0\leq 0\leq -2z$, $z\leq z\leq y\leq y\leq x\leq x$ and
\[ -2x\leq -x-y=z \\-2x-2y=2z\leq z+z \\-2x-2y-0\leq z+z+y \\-2x-2y-0-0\leq z+z+y+y \\-2x-2y-0-0-0\leq z+z+y+y+x \\-2x-2y-0-0-0-2z=z+z+y+y+x+x \]
Thus we can use Karamata on the convex function $e^x$ to show \[ e^{-2x}+e^{-2y}+e^{-2z}+e^{0}+e^{0}+e^{0}\geq e^{x}+e^{x}+e^{y}+e^{y}+e^{z}+e^{z} \] :D  :D  :D


Not sure if you guys can help this guy, but I run the  Q&A site http://www.ExpertBee.com and someone offered to pay to find out " how do you seperate the oxygen and hydrogen molecules in water," and no one has bid to answer.
If you know, help the guy out and make a little money.  here is the posting link:
http://www.expertbee.com/auction-query_i12899.html
Thanks
ExpertBee

I don't know anything about paying for answers.

The question as asked is meaningless -- the question assume a fact that is incorrect: there is no molecules of hydrogen and oxygen to seperate.  If what was meant was how to break down the water into oxygen atoms and hydrogen atoms, which will seperately combine into oxygen molecules and hydrogen molecules that can be seperately collected.  The answer is an extremely easy electrochem. question.  Run a direct current through an electrolyte solution of water, be careful to collect the oxygen and hydrogen before they can recombine in a fire or explosion.

If the question is how to do this comercially, there are questions of the amount of electrolyte (perhaps table salt) voltage and gas seperation techniques that I am not qualified to address, having never done this reaction commercially.  The real question woudl be why would one want to develop hydrogen and oxygen this way.  Normally, hydrogen is evolved as a by-product of a chemical reaction conducted for other purposes (sometimes evolution of chlorine from brine, or salt water).  Normally, oxygen (the by-product waste gas of plants living and making sugars) is made by cooling air and fractionally distilling the resulting liquid into the separate gases as needed (nitrogen, oxygen, etc.).

I hope this helps.

be careful guys, expertbee.com is a well documented scam.

why would somone want to develop hydrogen from water....dah...lots of water around, which could be used to make valuable hydrogen for running hydrogen cars and more oxygen for human consumption, they sell it in japan dont they....there are wonderings if there are oceans of oil underground, to run machines, we already know there are oceans of water above and below ground which could be used minimally to run clean burning machines..that could use both hydrogen and oxygen separately...doing this with plants from gasses is just way too slow and expensive sort of like the current cosly alternative of making ethanol from corn, eventually we will probably learn to make ethanol cheaply from other bio sources, but for now we need people who can do the math...........before asking silly questions...


rotation
combinatorics unsolved
combinatorics

Can we put positive integers $1,2,3, \cdots 64$ into $8 \times 8$ grids such that the sum of the numbers in any $4$ grids that have the form like $T$ ( $3$ on top and $1$ under the middle one on the top, this can be rotate to any direction) can be divided by $5$?

The answer is no.
Suposse it is possible:
Let's enumerate de grids in the following way: in the first row with 1,2,3,4,5,6,7,8; in the second row with 9,10,11,12,13,14,15,16, etc.
Let $A$ the number in the square 2, taking the $T$'s (9,10,11,2) and (9,10,11,18) it follows that the number in the square 18 is congruent to $A$ mod 5, analogous the numbers in the squares 34,50,52,36,20,4,22,6,38,54,56,40,24 are congruent to $A$ mod 5, the trouble is in 8, since the numbers in 7 and 23 are congruent, and taking the $T$'s (6,7,8,15) and (22,23,24,15) and using that the numbers in 6,22,24 are congruent mod 5, it follows that the number in 8 is also congruent the number in 24, i.e. is congruent to $A$.
Analogous the numbers in:
 1,3,5,7,17,19,21,23,33,35,37,39,49,51,53,55 are conguent mod 5, say to $B$.
9,11,13,15,25,27,29,31,41,43,45,47,57,59,61,63 are congruent mod 5, say to $C$.
10,12,14,16,26,28,30,32,42,44,46,48,58,60,62,64 are congruent mod 5, say to $D$.
It's obvious that $A;B;C;D$ are pairwise distinct mod 5 (if not we have 32 numbers in 1,2,3,...,64 mod 5) , so we have only 4 possible residues mod 5, it is a contradiction because for example 50,51,52,53,54 are pairwise distinct mod 5.

$Tipe$


Find an integer $n$ with two digits which satisfies \[\sqrt{n + \sqrt{n + 7}} \in \mathbb{N}.\]

[hide]

 is a square.  We are given that n is two digits.  Because  is an integer, it follows  is.

This gives us only a few possibilities for n+7:  .

So the expression could be 18+5, 29+6, 42+7, 57+8, 74+9, 93+10.  Only one of these is a square, namely n = 42.

[/hide]

[quote="Fermat -Euler"]:10:[/quote]

Please don't do that again, it's a useless post. And I suggest you learn LaTeX (and stop posting attachments) since I have to spend a LOT of time on editing your posts.


For $ 0 \leq x \leq 1$, evaluate $ \lim_{n\to\infty} \sum_{k\equal{}0}^{n} \binom{n\plus{}k}{n} x^n (1\minus{}x)^k$.

hello, it is $ \lim_{n\to\infty} \sum_{k\equal{}0}^{n} \binom{n\plus{}k}{n} x^n (1\minus{}x)^k\equal{}\frac{1}{x}$.
Sonnhard.


geometry Support AMC USA(J)MO USAMO \/closed

I thought you get sponsored if you make it to MOSP.

[quote="h_s_potter2002"]I know that rrusczyk was suggesting to xxreddevilzxx to get a sponsorship to help him (or her) to pay for WOOT.

But I was just wondering if anyone has ever tried this, for WOOT, or for anything else, and how you went about doing it, because I'd like to get sponsored, but I have no idea of who to ask, or how to ask them.[/quote]

xxreddevilzxx actually succeeded with this, getting wealthy individuals in his area to sponsor him; you may want to hunt him down to ask how.

Actually it was just one individual who had interest in where I was going with the AoPS education. I actually met with him after writing the letter and we talked about what AoPS is (types of things done here). Then we got into financial matters, which didn't take long. He paid through most (I paid minimal amount) of the cost for WOOT and I was really happy I got to do it. I am not sure if I could do WOOT again (I am not sure, is it more expensive this year?) but good luck with the sponsorship for those who are interested. I would suggest writing a letter around all the businesses, or a few if time is of essence. Just write about a few things you have done, a little bit about AoPS and WOOT, and then what you would get by participating.

Hope this helped.

Would one be more likely to get a team sponsored or an individual sponsored?

I think it would be more likely to get an individual sponsored, because it wouldn't cost as much money.

Basically you're trying to convince someone that your enrollment in WOOT is worth their money.  If you talk to someone you want to sponsor your team, you can tell them that their money is going to support the (your school here) math team.  That may be a good way to gain support, but remember that you need a few thousand dollars, and I doubt a single person would be willing to fork over that much money. You may need to talk to a lot of different people.  xxreddevilzxx only needed a few hundred dollars, because he is only one person, and because the cost of WOOT was less last year.  Good luck to you and your team!  :)

Er, if xxredddevilzxx is still here, how do you speak to the people such that they get persuaded? Here are my main reasons for taking WOOT:

- My interest in math
- My weakness in proofs
- Making USAMO next year
- Having fun with other top people in the world

How would I phrase it in such a way that it is convincing?


Find $ x, y \in R$ of this system of this equation: $ \begin {cases} 1\plus{}x^3y^3\equal{}19x^3 \\ y\plus{}xy^2\equal{}\minus{}6x^2\end {cases}$

Ok I solved it a bit long, but in a nice way, I hope :D

$ y(xy \plus{} 1) \equal{} \minus{} 6x^2 \implies$ from the second equation,

$ (xy \plus{} 1) \equal{} \frac { \minus{} 6x^2}{y}$

Then I have taken the 3rd power of both sides, which gave me,

$ x^3y^3 \plus{} 3x^2y^2 \plus{} 3xy \plus{} 1 \equal{} \frac { \minus{} 216x^6}{y^3}$

I already know the value for $ x^3y^3 \plus{} 1$.. which is $ 19x^3$

$ 19x^3 \plus{} 3xy(xy \plus{} 1) \equal{} \frac { \minus{} 216x^6}{y^3}$ I also know the value for $ xy \plus{} 1$

$ 19x^3 \minus{} 18x^3 \equal{} \frac { \minus{} 216x^6}{y^3}$

$ x^3 \equal{} \frac { \minus{} 216x^6}{y^3}$

$ y^3 \equal{} \minus{} 216x^3 \implies y \equal{} \minus{} 6x$ --- I used this value at the second equation, it looks easier for me.

$ \minus{} 6x \plus{} x(36x^2) \equal{} \minus{} 6x^2 \implies 36x^3 \equal{} 6x \minus{} 6x^2$

$ 6x^2 \equal{} 1 \minus{} x \implies 6x^2 \plus{} x \minus{} 1 \equal{} 0$

Now from that, I have the roots $ \minus{} 1/2,1/3$ Since I now, know the value of x, why shouldn't I put that in the equation I found $ y \equal{} \minus{} 6x$

I get for $ x \equal{} \minus{} 1/2$ , $ y \equal{} 3$ and for $ x \equal{} 1/3$ , $ y \equal{} \minus{} 2$


trigonometry
ratio
geometry
circumcircle
angle bisector

how do you prove that the interior angles of a triangle are concurrent using ceva's theorem?

You mean the interior angle [b]bisectors[/b]?

Well, if you use the trig version of Ceva, it's very easy.  Using the other form of Ceva, you can use the angle bisector theorem.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=422[/img]

[hide="Using Ceva"]
By Ceva, we wish to show $\frac{BF}{CF}\cdot\frac{AE}{BE}\cdot\frac{CD}{AD}=1.$  By the angle bisector theorem:

\begin{eqnarray*} \frac{BF}{CF} &=& \frac{AB}{AC}\\ \frac{AE}{BE} &=& \frac{AC}{BC}\\ \frac{CD}{AD} &=& \frac{BC}{AC} \end{eqnarray*}

Multiplying these produces the desired result.[/hide]

[hide="Trig Ceva"]
Well,  $\sin \alpha\sin \beta\sin \gamma=\sin \alpha\sin \beta\sin \gamma$ obviously holds...so we're done![/hide]

Question: when you do the trig Ceva, in which direction does one take the ratios of the sin's? More specifically, what is the general form of the trigonometric version of Ceva?

Triangle ABC with Cevians AD, BE, CF. The three cevians are concurrent if and only if:
\[ \frac{\sin{\angle{DAB}}}{\sin{\angle{EBA}}} \cdot \frac{\sin{\angle{EBC}}}{\sin{\angle{FCB}}} \cdot \frac{\sin{\angle{FCA}}}{\sin{\angle{DAC}}} = 1 \]

Joml88, doesn't that assume each of those triangles in which you take the sin of one of its angles has the same circumcircle? I am somewhat confused.

[quote="joml88"]Oops, what I meant to say was use \[ AE=\frac{CE\sin\gamma}{\sin \angle CAB}  \] and similar relations.  Then they will cancel.[/quote]

Cool stuff, thanks for the clarification.  It took me some time to realize you were using the diagram up above, heh.  :blush:


isn't there a solutions guide at the end?

First, you do 6 c 2, which is 15.  Then you subtract two.  THe reason is that there are 3 ways to make a circle with points across from eachother.  However, there are four units between each thing across from eachother (the ones where you choose a unit farthest from it lies on only one circle.  Not three).  you subtract two because it must be distinct.


Given:

$-\frac{1}{\sqrt{a}}\leq x_{i}\leq \sqrt{a}$

and

$x_{1}+x_{2}+\cdots x_{m}= b*\sqrt{a}$

what is the maximum value of:

$x_{1}^{p}+x_{2}^{p}+\cdots x_{m}^{p}$

what about the constrait of $p$?$p\geq 1$?


Use each of the digits 1,2,3,4,5,6,7,8,9 exactly twice to form distinct prime numbers whose sum is as small as possible. What must this minimal sum be? (Note: The five smallest primes are 2,3,5,7, and 11)

Hello,

23,29,43,47,619,857 and 6581. The sum is 8199. It existe a better solution ?

We can't use 2 or 5 as units digits unless we include 2 and 5 themselves.  So our first two numbers are 2 and 5.  However, we still need to use the digits 2 and 5 once more, and the digits 4, 6, and 8 twice more.  None of these can be units digits, so to minimize the sum we can try to use them as tens digits.  To show that this is possible,

$2,5,23,41,47,59,61,67,83,89$

The sum is $477$.


[url=http://usinfo.state.gov/ei/Archive/2005/Apr/19-681391.html]Madagascar[/url]

This program is one of the many programs instituted by major world powers designed to promote the economic growth of LDC, or Third world countries.  Just out of curiousity, what are your opinions on the role of the US and other developed and industrialized nations in aiding those that are not?  What limits, if any, should be placed on such activity, if allowed at all?  Post your thoughts.

It seems to me that LDCs do better as they trade more and that artificial external programs cause at least us much long term harm as good.  When wealth is imported into a nation before the education required to manage and sustain that wealth, it seems to create corruption or even warlordism depending on local stability.

I'm sure there are counterexamples to my point and I would be interested in hearing them for the sake of diagnosing from exactly where the failures of various economic programs stem.

I also take some issue with the word "responsibility" which seems to me highly mutable and easily politicized.  I think it often steers attention away from real economic analysis.  Questions of responsibility can be more meaningfully phrased in terms of the utility of an individual.


[size=150]Find two numbers y so that the distance from point
S(-3, 2) to point T(5, y) = 10.[/size]

Great reply but what does D^2 mean here?


[quote="Bazooka"]Let $f: Z^{+}\to Z^{+}$ such that:
i. $f(n+1)= f(n)-1$ or $4f(n)-1$
ii. $\forall m \in Z^{+}, \exists n \in Z^{+}$ such that $f(n)=m$

Find $f(2002)$[/quote]
May be $f: N\to N$.
If $f(n)=0$, then $f(n+1)=-1$ :lol:

$\forall m\geq 1$ then  $\exists !n$ such that
$f(n)=m$


Here's a really funny video - [url]http://www.bobricci.com/flash/blockedme.htm[/url]

that's hilarious!!! my mom was sitting behind me and was like "what does 'wtf' mean?" i'm just like um....lol. but it was great!

hahahaha yea - but um... at the end he subscribes to um... susn... p***, don't wanna say pg-18 word here, so edit out, so i dunno - rather funny yes - but don't wanna ur parents behind u like Mathfiend... lol... my momdunno what wtf means either

[quote="white_horse_king88"]my momdunno what wtf means either[/quote]this is funnier than the movie itself :D

lol lol lol. funny. it actually has a plot and stuff not lyke those songs that spend five minutes whining about life and not really getting anywhere. good one for chat ppl.

that was HILARIOUS!!!!!!!!!!!! Wonder who bothered to make the movie.....
but it was alsome

[quote="themonster"]That was great.  At a summer camp that I went to one year (by the way, not math related), there was a kid who brought a stereo and played the song that the movie parodies (and the word "frickin'" wasn't used) consecutively time after time for a whole day at full volume.[/quote]You should email him the topic :)

[quote="theone853"]You know that in this generation, our minds have been dirtied (or at least mine has been) when you know what deals with the internet and rhymes with "torn"[/quote]

Hey, what do you have against corn? Corn's good, especially corn on the cob.

:ninja:

it's so yellow, that's why i'm prejudice against corn

  ;)  :alien:  ;)

[quote="Athena"]o yes........did u listen to the "mrs. claus" one? its so funny. just click on the "other songs by bob ricci" link and choose mrs. claus. but then you would need to be familiar with "stacy's mom" by fountains of wayne" but not really, cause theyre both equally funny but.....still innapropriate. id rate it a little less than pg-13[/quote]

LOL, that's awesome.

[quote="peter"]Off topic - GASP!!! the holiday smilies are gone.[/quote]It took you some time to notice that  :D  :D


Solve the system of equations \[ \{\begin{array}{lll} y+xy^2&=&6x^2\\ 1+x^2y^2&=&5x^2\end{array}\]

[quote="April"]Solve the system of equations
\[ \{\begin{array}{lll} y + xy^2 & = & 6x^2 \\
1 + x^2y^2 & = & 5x^2\end{array}
\]
[/quote]
divide both sides of both equation by x^2 we get the system :
\[ \{\begin{array}{lll} \frac{y}{x^2} + \frac{y^2}{x} & = & 6 \\
\frac{1}{x^2} + y^2 & = & 5\end{array}
\]
now let $ \frac{1}{x}=a ; y=b$ the system is equivalent to
\[ \{\begin{array}{lll} a^2b + b^a & = & 6 \\
a^2 + b^2 & = & 5\end{array}
\]
now let S=a+b and P=ab we have
\[ \{\begin{array}{lll} SP & = & 6 \\
\S^2-2P& = & 5\end{array}
\]
so we have $ S^3-5S-12=0$ which means $ (S-3)(S^2+3S+4)=0$ gives $ S=3$

[quote="HTA -- typos fixed"]divide both sides of both equation by x^2 we get the system :
\[ \{\begin{array}{lll} \frac {y}{x^2} + \frac {y^2}{x} & = & 6 \\
\frac {1}{x^2} + y^2 & = & 5\end{array}
\]
now let $ \frac {1}{x} = a ; y = b$ the system is equivalent to
\[ \{\begin{array}{lll} a^2b + \mathbf{\color{red}{b^2 a}} & = & 6 \\
a^2 + b^2 & = & 5\end{array}
\]
now let S=a+b and P=ab we have
\[ \{\begin{array}{lll} SP & = & 6 \\
\mathbf{\color{red}{S^2}} - 2P & = & 5\end{array}
\]
so we have $ S^3 - 5S - 12 = 0$ which means $ (S - 3)(S^2 + 3S + 4) = 0$ gives $ S = 3$[/quote]

Using $ SP = 6$, we get $ P = 2$. Thus, $ a$ and $ b$ are the roots of $ x^2 - 3x + 2$, or $ 1$ and $ 2$.

$ (a,b) = (1,2) \implies \boxed{(x,y) = (1,2)}$
$ (a,b) = (2,1) \implies \boxed{(x,y) = ( \frac {1}{2}, 1 )}$

I believe there are also four very ugly complex solutions, which result from using the two complex values of $ S$ from that cubic in $ S$.


geometry
trigonometry
calculus
integration
analytic geometry
function
trig identities

I would like to find the area under a certain polar curve.  However, I can not solve for $r$.  Is there any general way this can be done?  If not, is there a way to get a calculator or Mathematica to give me a good estimate?

I'm trying to integrate $r^{2}$ from $0$ to somewhere around $\pi/4$ (wouldn't want to spend time finding the exact upper bound if I won't be able to complete it anyway), where $r=\sqrt{r^{2}+1-2r\cos{\theta}}+\sqrt{r^{2}+2-2\sqrt{2}r\cos{(\pi/4-\theta)}}$.

Any input appreciated.

There is a TeX error in the formula: it should contain $\cos(\pi/4-\theta)$. You may be able to do this in Mathematica by applying [url=http://documents.wolfram.com/mathematica/functions/Solve]Solve[/url] to your equation and then using [url=http://documents.wolfram.com/mathematica/functions/NIntegrate]NIntegrate[/url].
But are you sure that the problem is set up correctly? It's not often that one integrates $r(\theta)$ with $r$ representing a polar curve.

heh good point, $r^{2}$ then I suppose.  Problem editted, I'll give it a little more thought with the adjustment and try out Mathematica, but I'll probably need a bit more assistance.

Maybe someone would help if you told us how this problem came about.  :maybe:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=29997

I was trying a different approach using D as the pole because then it would for sure only have one $r$ for each $\theta$ so it seemed nicer.  Because when making my sketch of the curve I used a ruler pivoting at D.  Law of cosines lead to my aforementioned equation.

I think the conversion to polar coordinates does not help much. You could try to define a function u(x,y) = 1 if AP+BP < DP and 0 otherwise. (Here P has coordinates (x,y). Mathematica has a function [url=http://documents.wolfram.com/v5/TheMathematicaBook/PrinciplesOfMathematica/EvaluationOfExpressions/2.6.8.html]If[/url]). Then integrate $u(x,y)$ numerically via NIntegrate.

Edit: after much wailing and gnashing of teeth, Mathematica spat put 0.113882, confirming an earlier computation of [b]JBL[/b].

[size=75]I doubt that the problem has any nice answer or solution.[/size]


I was reading the book AoPS volume II, and in chapter 4 the last question (Page 50) wasn't solved clearly in the solution book. The author said "after lots of algebra we get the disired expression."  :D
Can someone help me with this?
(OMG i tried to write the problem in latex code but i can't, can someone write the question in latex code with ABCD and s in italic?)
Prove that if in AMCD we let a=AB, b=BC, c=CD, and d=DA, we have [ABCD]^2=(s-a)(s-b)(s-c)(s-d)=abcd cos^2((B+D)/2) where s is the semiperimeter of the quadrilateral. What does this expression, called Brahmagupta's formula, yield for a cyclick quadrilateral?

Prove that if in $ABCD$ we let $a=AB$, $b=BC$, $c=CD$, and $d=DA$, we have $[ABCD]^2=(s-a)(s-b)(s-c)(s-d)-abcd\ cos^2((B+D)/2)$ where $s$ is the semiperimeter of the quadrilateral. (its $ABCD$ not $AMCD$, i guess!)

the same (similar) thing is true for a triangle. that how you can easilly find the area of the trianlge, if you have the three sides of the triangle.

[quote="amirhtlusa"]Prove that if in $ABCD$ we let $a=AB$, $b=BC$, $c=CD$, and $d=DA$, we have $[ABCD]^2=(s-a)(s-b)(s-c)(s-d)=abcd cos^2((B+D)/2)$ where $s$ is the semiperimeter of the quadrilateral. What does this expression, called Brahmagupta's formula, yield for a cyclick quadrilateral?
(its $ABCD$ not $AMCD$, i guess!)[/quote]

actually your version of Brahmagupta is not correct. it should be $[ABCD]^2=(s-a)(s-b)(s-c)(s-d) - abcd cos^2((B+D)/2)$

use \cos to make the cos in $\LaTeX$ look nicer: $\cos ABC$ as opposed to $cos ABC$ ;)

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=112006

Proof by darij.

sorry guys I was abit lost I know whats A abd B

Hey I got Heron fromula for this one
But i havnt been able to yet figure out the other one
Is there anyother way apart from Brahmaguptas generallised formula to prove it
Sicne I teach in class 8 and 9 We dont yet teach trigonometry so using that method would be abstract


inequalities
function
real analysis
real analysis unsolved

Prove that :

$ \sum\limits_{i\equal{}1}^{k}{\sum\limits_{j\equal{}1}^{k}}{|x_i \minus{}x_j|} \leq k^2$,

where $ x_t \in [0,2]$ , $ \forall t\equal{}1,2,\cdots k$.

Does the equality hold ? if so when?

 :D  :D  :D

The left-hand side is a piecewise convex function of any given $ x_i$, so it takes on its maximum values at the endpoints, namely $ x_i\equal{}0$, $ 2$, or $ x_j$ for some $ j$.  Since setting some of the variables equal to each other [i]still[/i] yields a piecewise convex function, we can assume without loss of generality that all of the $ x_i$ are equal to $ 0$ or $ 2$.

Assume that $ n$ of the $ x_i$ are equal to $ 0$, and $ k\minus{}n$ are equal to $ 2$.  Then we compute that the sum is equal to $ 4n(k\minus{}n)\leq k^2$, with equality if and only if $ k\equal{}2n$.

Note that if $ k$ is odd then the sum's maximum is $ k^2\minus{}1$, achieved when either $ \frac{k\minus{}1}{2}$ or $ \frac{k\plus{}1}{2}$ of the variables are $ 0$.


An ugly but straitforward solution may be to use Cauchy-Schwarz ($(\sum a_i/b_i) \cdot (\sum a_i b_i ) \geq (\sum a_i)^2$), which ends proving $8 (\sum a^3)^2 \geq 3 (\sum a^3(a+b)^3)$ which is true by rearrangement after expansion.

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=45819 .

  darij


geometry
geometric transformation
reflection
rotation
search
geometry proposed

Let ABC be a triangle.
Reflect the vertex A in the sideline BC; you get a new point A'.
Construct an equilateral triangle CAB' on the side CA of triangle ABC inwardly (i. e. such that the points B and B' lie in the same halfplane with respect to the line CA).
Construct an equilateral triangle ABC' on the side AB of triangle ABC outwardly (i. e. such that the points C and C' lie in different halfplanes with respect to the line CA).
Show that the points A', B', C' are collinear.

Mentioned in

[url=http://forumgeom.fau.edu/FG2003volume3/FG200304index.html]Lawrence S. Evans, [i]Some Configurations of Triangle Centers[/i], Forum Geometricorum 3 (2003) pp. 49-56.[/url]

  Darij

Take a point $A^*$ on $BC$ s.t. $\angle A'AA^*=\frac {\pi}3$ (a directed angle). It's clear that $AA^*=AA'$, and all we have to do now is rotate the line $BCA^*$ around $A$ with an angle of $-\frac {\pi}3$, and we get the line $B'C'A'$.

Cool proof, I didn't think it was so simple! (Actually, when I posted the problem, I thought: it should not be very hard, but I have no time to search for a synthetic proof now, at least I have one using barycentrics...)

  Darij


Prove that $AM \geq HM$ for $\{a_0,a_1,\dots,a_n\}$ where all $a_i > 0$ for $0\leq {i} \leq {n}$. Using either AM-GM or Cauchy and show that there is equality iff all $a_i=1$.

notes:
$AM=\frac{a_0+a_1+a_2+\dots+a_n}{n}$
and
$HM=\frac{n}{\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}}$

[hide="Cauchy-Schwarz inequality Solution"]By the Cauchy-Schwarz inequality, we have

$(a_1 + a_2 + \dots + a_n)(\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n}) \geq (1 + 1 + \dots + 1)^2 = n^2$

Rearranging, we get
$\mbox{AM} = \frac{a_1 + a_2 + \dots + a_n}{n} \geq \frac{n}{ \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n}} = \mbox{HM}$
with equality iff
$a_1^2 = a_2^2 = \dots = a_n^2$
or since all $a_i > 0$,
$a_1 = a_2 = \dots = a_n$
(instead of all $a_i = 1$)[/hide]

Hopefully I didn't mess anything up...  :P
EDIT: I did  :P
EDIT 2: I did again  :P  :rotfl:

nebula beat me to it.  :P

Incidentally, this is one of several inequalities that can be proven very simply with a slight rewriting of Jensen's.

o yeah woops idk what i was thinking when i wrote the $a_i=1$ sorry  :blush:  :oops:

[quote="nat mc"][quote="nebula42"]with equality iff
$\frac{a_0}{\frac{1}{a_0}} = \frac{a_1}{\frac{1}{a_1}} = \frac{a_2}{\frac{1}{a_2}} = ... = \frac{a_n}{\frac{1}{a_n}}$
or
$a_0^2 = a_1^2 = a_2^2 = ... = a_n^2$
(instead of all $a_i = 1$)

Hopefully I didn't mess anything up...  :P[/quote]
Hmm are you sure about this last part? Let $a_1=3,a_2=3,a_3=-3$, then $a_1^2=a_2^2=a_3^2$ But $\frac{3+3-3}3\neq\frac3{\frac13+\frac13-\frac13}$
I think equality holds iff $a_1=a_2=\cdots=a_n$

[quote="maokid7"]$AM=\frac{a_0+a_1+a_2+\dots+a_n}{n}$
and
$HM=\frac{n}{\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}}$[/quote]
I think it's
$AM=\frac{a_1+a_2+\dots+a_n}{n}$
and
$HM=\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}}$

Because by yours the AM of $a$ and $b$ would be $a+b$[/quote]

Oh yeah, I can't believe I didn't notice that  :oops: ... oh well. Is there a stronger inequality that can prove that $a_1 = a_2 = ... = a_n$ is necessary?

[quote="maokid7"]yes i know it should be $a_1,a_2,\dots,a_n$ and not $a_0,a_1,\dots,a_n$

and if you noticed it said $a_i>0$ so $a_1^2=a_2^2,\dots,a_n^2$ works.[/quote]

Oh yeah... I can't believe I didn't see that either...  :blush:  :P

[quote="maokid7"]yes i know it should be $a_1,a_2,\dots,a_n$ and not $a_0,a_1,\dots,a_n$

and if you noticed it said $a_i>0$ so $a_1^2=a_2^2,\dots,a_n^2$ works.[/quote]
oh ok, I didn't see that. Then it's fine  :)


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inequalities proposed

If a,b,c,d,e are positive real numbers such that $ abcde\equal{}1$, prove that
\[ \frac{a\minus{}1}{a^2\plus{}a\plus{}1}\plus{}\frac{b\minus{}1}{b^2\plus{}b\plus{}1}\plus{}\frac{c\minus{}1}{c^2\plus{}c\plus{}1}\plus{}\frac{d\minus{}1}{d^2\plus{}d\plus{}1}\plus{}\frac{e\minus{}1}{e^2\plus{}e\plus{}1}\leq0.\]

by AM-GM we know that:
$ a^2\plus{}a\plus{}1\geq 3a$
thus we get that:
$ \frac{a\minus{}1}{a^2\plus{}a\plus{}1}\leq\frac{a\minus{}1}{3a}$
thus its sufficient to show that:
$ \frac{a\minus{}1}{3a}\plus{}\frac{b\minus{}1}{3b}\plus{}\frac{c\minus{}1}{3c}\plus{}\frac{d\minus{}1}{3d}\plus{}\frac{e\minus{}1}{3e}\leq 0$
$ \Rightarrow \frac{a\minus{}1}{a}\plus{}\frac{b\minus{}1}{b}\plus{}\frac{c\minus{}1}{c}\plus{}\frac{d\minus{}1}{d}\plus{}\frac{e\minus{}1}{e}\leq 0$
which is equivalent to:
$ 5\leq \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{d}\plus{}\frac{1}{e}$
which can be obtained directly from AM-GM...

[quote="BaBaK Ghalebi"]thus we get that:
$ \frac {a \minus{} 1}{a^2 \plus{} a \plus{} 1}\leq\frac {a \minus{} 1}{3a}$
[/quote]

I that here you are wrong because you don't know the sign of $ a\minus{}1$ , you you can't multiply ...

This is actually a problem in Secrets in Inequalities, and I (or zaizai) did post this problem in this forum, too. Try to search.

@ hitek. I think using a fake nickname in Mathlinks at this time is not a good idea.

[quote="hitek"]If a,b,c,d,e are positive real numbers such that $ abcde \equal{} 1$, prove that
\[ \frac {a \minus{} 1}{a^2 \plus{} a \plus{} 1} \plus{} \frac {b \minus{} 1}{b^2 \plus{} b \plus{} 1} \plus{} \frac {c \minus{} 1}{c^2 \plus{} c \plus{} 1} \plus{} \frac {d \minus{} 1}{d^2 \plus{} d \plus{} 1} \plus{} \frac {e \minus{} 1}{e^2 \plus{} e \plus{} 1}\leq0.
\]
[/quote]
$ \sum_{cyc}\frac{a\minus{}1}{a^2\plus{}a\plus{}1}\leq0\Leftrightarrow\sum_{cyc}\left(\frac{a\minus{}1}{a^2\plus{}a\plus{}1}\minus{}\frac{\ln a}{3}\right)\leq0.$
Let $ f(a)\equal{}\frac{a\minus{}1}{a^2\plus{}a\plus{}1}\minus{}\frac{\ln a}{3}.$ Then $ f'(a)\equal{}\frac{(1\minus{}a)(a^3\plus{}3a^2\plus{}9a\plus{}1)}{2(a^2\plus{}a\plus{}1)^2}.$
Thus, $ a_{max}\equal{}1$ and it's done.

[quote="arqady"]
Let $ f(a) \equal{} \frac {a \minus{} 1}{a^2 \plus{} a \plus{} 1} \minus{} \frac {\ln a}{3}.$ Then $ f'(a) \equal{} \frac {(1 \minus{} a)(a^3 \plus{} 3a^2 \plus{} 9a \plus{} 1)}{2(a^2 \plus{} a \plus{} 1)^2}.$
Thus, $ a_{max} \equal{} 1$ and it's done.[/quote]
why is $ a_{max}\equal{}1$
for example $ (a,b,c,d,e)\equal{}(100,200,1,\frac{1}{100},\frac{1}{200})$

[quote="pardesi"][quote="arqady"]
Let $ f(a) \equal{} \frac {a \minus{} 1}{a^2 \plus{} a \plus{} 1} \minus{} \frac {\ln a}{3}.$ Then $ f'(a) \equal{} \frac {(1 \minus{} a)(a^3 \plus{} 3a^2 \plus{} 9a \plus{} 1)}{2(a^2 \plus{} a \plus{} 1)^2}.$
Thus, $ a_{max} \equal{} 1$ and it's done.[/quote]
why is $ a_{max} \equal{} 1$
for example $ (a,b,c,d,e) \equal{} (100,200,1,\frac {1}{100},\frac {1}{200})$[/quote]
$ f(100) \equal{} \frac {99}{10101} \minus{} \frac {\ln100}{3} \equal{} \minus{} 1.525255719... < 0.$
By the way, is proved that $ f(a)\leq f(1) \equal{} 0$ for all $ a > 0.$
Hence, $ f(200) < 0,$ $ f\left(\frac {1}{100}\right) < 0$ and $ f\left(\frac {1}{200}\right) < 0.$
Thus, $ \sum_{cyc}f(a)\leq0.$ :wink:
Maybe you have meant to $ \max a$ $ ?$
But I saw about $ a_{max}.$  :wink:

[quote="arqady"]Let $ f(a) \equal{} \frac {a \minus{} 1}{a^2 \plus{} a \plus{} 1} \minus{} \frac {\ln a}{3}$. Then $ f'(a) \equal{} \frac {(1 \minus{} a)(a^3 \plus{} 3a^2 \plus{} 9a \plus{} 1)}{2(a^2 \plus{} a \plus{} 1)^2}.$
Thus, $ a_{max} \equal{} 1$ and it's done.
[/quote]
u said $ a_{max} \equal{} 1$ which is wrong as a can be any real no.
actually is suppose u meant $ f_{max}$ is at $ a\equal{}1$

[quote="arqady"]
$ \sum_{cyc}\frac {a \minus{} 1}{a^2 \plus{} a \plus{} 1}\leq0\Leftrightarrow\sum_{cyc}\left(\frac {a \minus{} 1}{a^2 \plus{} a \plus{} 1} \minus{} \frac {\ln a}{3}\right)\leq0.$

[/quote]
Is it $ \sum_{cyc}\left(\frac {a \minus{} 1}{a^2 \plus{} a \plus{} 1} \minus{} \frac {\ln a}{3}\right)\leq0.$ instead of

$ \sum_{cyc}\left(\frac {a \minus{} 1}{a^2 \plus{} a \plus{} 1} \plus{} \frac {\ln a}{3}\right)\leq0.$??? :maybe:

[quote="pardesi"][quote="arqady"]Let $ f(a) \equal{} \frac {a \minus{} 1}{a^2 \plus{} a \plus{} 1} \minus{} \frac {\ln a}{3}$. Then $ f'(a) \equal{} \frac {(1 \minus{} a)(a^3 \plus{} 3a^2 \plus{} 9a \plus{} 1)}{2(a^2 \plus{} a \plus{} 1)^2}.$
Thus, $ a_{max} \equal{} 1$ and it's done.
[/quote]
u said $ a_{max} \equal{} 1$ which is wrong as a can be any real no.
actually is suppose u meant $ f_{max}$ is at $ a \equal{} 1$[/quote]
let $ f(x)\equal{}\frac{a\minus{}1}{a^2\plus{}a\plus{}1}\minus{}\frac{\ln a}{3}$ as he said,then we get that for $ a\equal{}1$ the derivative of $ f$ becomse $ 0$ i.e. $ f'(1)\equal{}0$ and also for $ a>1$ the function will become decreasing and for $ a\leq 1$ we get that $ f(a)\leq f(1)$...

yes i already got the proof .but was confused with my earlier interpretation due to the way the proof was presented

I am sorry. My proof is wrong.
$ f'(a)\equal{}\frac{(1\minus{}a)(a^3\plus{}6a^2\plus{}3a\minus{}1)}{3a(a^2\plus{}a\plus{}1)^2}$ and it's nothing.

yes but the general idea is great and can be seen pretty often now in ML.
after Harazi Substitution i think this is a pretty good method to solve inequalities of this type.

Ok. Here is the problem in my book (we prove it, for six numbers instead of five)

Is any idea for it, Arqady?  :lol: You are close to the end of the solution.

Yes, hungkhtn! It's easy enough and true for six numbers. :lol:
The beginning is the same! :wink:

10p of searching, finally, I've gone to

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1021565391&t=152115

Above problem was once chosen for VIF problem of day. I think Zaizai also has the same solution. It is very easy if we know the technique, but very hard if not. Am I true, Arqady?

I agree with you, hungkhtn. The same technique see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45516