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ceiling function
probability and stats

imagine a ladder with $ n$ steps ($ n$ is very large) and suppose that you have two identical plates. One wants to know the maximum height from where one can let a plate fall without breaking.
- for example, one can use only one plate, start from the step $ 1$, and then try the step $ 2$, etc .. until the plate breaks. In average, one has to do $ \approx \frac{n}{2}$ trials

What would be your strategy ? (minimize the number of trials)

Are we to assume that the critical height is a random variable?  :huh:  What is its distribution?

yes, sorry. I implicitly assumed that the critical height is a random variable uniformly distributed on $ [1,n]$.

There are i think two different problems:

a) Find a strategy wich minimize the number of trials in the worst case
b) Find a strategy wich minimize the average number of trials

Maybe they are the same, maybe not

I'm pretty sure that one cannot do better that a linear growth for the worst case. What about the average case ? (I have something like $ C n^{\frac {1}{4}}$ for the average case, but do not know if one can do better ..)

so we can consider that there is a step limite (number k) and, with j>k, all the plats will be broken, but with j<=k, the plats wont be broken, ok?
so what is the best way to find k? and how does the number of trials depend on k?

It was discussed before (but maybe not conclusively?):
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=774379043&t=117042
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=774379043&t=120466

thanks  :)  but nothing has been proposed as a good strategy with only $ 2$ plates. (indeed, if one has $ Log_2(n)$ plates, it is easy  :lol:  )

In the second thread, weiquan gives what amounts to the following strategy to find the right height in $ x \equal{} \left\lceil \frac{\minus{}1 \plus{} \sqrt{8N \plus{} 1}}{2} \right\rceil$ steps:

Drop the first plate at height $ x \plus{} 1$.  If it breaks, it takes $ x \minus{} 1$ drops of the second plate in the worst case.  If it doesn't break, drop the first plate at height $ 2x \plus{} 1$; if it breaks, it takes $ x \minus{} 2$ drops of the second plate in the worst case.  If it doesn't break, drop the first plate at height $ 3x$; continue in the obvious manner.  Weiquan extended this to show that with $ n$ eggs, the worst-case growth is $ O(\sqrt[n]{N})$ (and it's fairly clear that this is optimal, at least when the problem is reformulated as weiquan wrote it).

I haven't calculated how this strategy does on average.


[quote="Mr.JN2P"]Let a,b,c >0. Prove that :
$ \sum\limits_{cyclic}^{} {a^{b \plus{} c}} \ge 1$[/quote]
This inequality copyright from [b]Mathematics and Youth Magazine[/b]
If you solved, please couldn't post.
Thanks.


AMC
AIME
probability
Functional Analysis

Mine was number 10, the geometric probability one.

What's everyone else's?

The star number-theory problem! It was so easy for so much wording.

14! The unicorn/rope/building question! Even thoug the rope-wrapped-around-a-cylinder type of question is relatively common, i love them!

star problem, definitely.  Unicorn comes in 2nd.[tab]

Number 15, which I had no idea how to do.

I concur. #15 was an excellent combinatorics-disguised-as-a-functional-analysis problem.

Second was #14, because you have to love any problem with a unicorn.


For polynomials,$ {P,Q}$ are such that $ {deg P}=n$,$ {deg Q}=m$,have the same leading coefficient,and $ {P^2(x)=(x^2-1)Q^2(x)+1}$.
Show that $ {P^'(x)}=nQ(x)$ .
I want the solution please and more importantly i want the meaning of this notation$ {P^'(x)}$. as used in the problem.

The tick mark over the P usually notifies of the derivative of a function.
http://en.wikipedia.org/wiki/Derivative

By $ P^2(x)$, do you mean $ P(P(x))$ or $ P(x)^2$?


Angle $\gamma$ in $\triangle ABC$ (with angles $\alpha, \beta, \gamma$) is trisected, with the trisectors meeting $AB$ at $D$ and $E$, with $D$ closer to $A$. If $AB=1$, find the ratio $AD : DE : EB$.

[hide="well i would assume"]law of sines[/hide]

consider angles ACE and DCB (angle bisectors)


function
integration
floor function
inequalities
calculus
algebra
probability

I know how to integrate floor functions, but this....blah

51) Integrate: $ \int_{0}^{\infty}{ \lfloor x \rfloor  e^{-x}dx}$ =

A) $ \frac{e}{e^2-1}$

B) $ \frac{1}{e-1}$

C) $ \frac{e-1}{e}$

D) $ 1$

E) $ +\infty$

Things I tried: I tried to get it into a sum and see if it telescoped or became a nice geometric seriies.  It became one, but not one that I knew a formula for off the top of my head.  It was geometric + arithmetic with no telescoping.  So someone smarter than I might be successful here.

Also tried to just squeeze it with inequality of the same integral just without the floor function from 1 to infinity.

Anyway, any method + as much additional information about anything that you can think of that you think I should know is appreciated.  Cheers.

Here's a hint, think of both series as power series in 1/e.

$ \sum\limits_{n \equal{} 1}^\infty  {nx^n }$has been solved ad nauseam on this forum.

Note that $ \int_k^{k \plus{} 1}e^{ \minus{} x}\,dx \equal{} \left. \minus{} e^{ \minus{} x}\right|_k^{k \plus{} 1} \equal{} e^{ \minus{} k}(1 \minus{} e^{ \minus{} 1}).$

Now, $ \int_0^{\infty}\lfloor x\rfloor e^{ \minus{} x}\,dx \equal{} \sum_{k \equal{} 0}^{\infty}k\int_k^{k \plus{} 1}e^{ \minus{} x}\,dx$

$ \equal{} \left(1 \minus{} \frac1e\right)\sum_{k \equal{} 1}^{\infty}ke^{ \minus{} k}.$

Now, what about $ \sum_{k \equal{} 1}^{\infty}kr^k\ ?$ This is related to the expected value of a geometric random variable, but we can also use calculus means:

$ \frac1{1 \minus{} r} \equal{} \sum_{k \equal{} 0}^{\infty}r^k$ for $ 0 < r < 1.$ Differentiate:

$ \frac1{(1 \minus{} r)^2} \equal{} \sum_{k \equal{} 1}^{\infty}kr^{k \minus{} 1}.$ Multiply by $ r$ to get

$ \frac {r}{(1 \minus{} r)^2} \equal{} \sum_{k \equal{} 1}^{\infty}kr^k.$

$ \sum_{k \equal{} 1}^{\infty}ke^{ \minus{} k} \equal{} \frac {e^{ \minus{} 1}}{(1 \minus{} e^{ \minus{} 1})^2}.$

$ \left(1 \minus{} \frac1e\right)\sum_{k \equal{} 1}^{\infty}ke^{ \minus{} k} \equal{} \frac {e^{ \minus{} 1}}{1 \minus{} e^{ \minus{} 1}} \equal{} \frac1{e \minus{} 1},$ which is answer B).

Kent, when you originally evaluated the integral in series form you left out the floor function

[color=green]Fixed now. It was just that one line. K.M.[/color]

Thanks.  Yea, I didn't think about how easy that sum is and I can just use $ x\equal{}e^{\minus{}1}$.  Ugh, this test makes me feel dumb.

Didn't see Kent's reply just now, but yea, I did same thing.  Grrr...


Starting with WARM, change one letter at a time, each time making a valid word and end with COOL?

[hide]WARM
WARN
BARN
BORN
CORN
COIN
COIL
COOL

I'm sure there is a shorter way...[/hide]

Warm
Worm
Worn
Corn
Coin
Coil
Cool

Only 1 step shorter but that's OK :D

And for the very shortest possible:

  WARM
  WORM
  CORM
  COOM
  COOL

  All real words folks, if you don't believe me consult a dictionary!

WARM
WARD
CARD
CORD
CORK
COOK
COOL

is one.

without reading the above entries, i wanna give it a shot :D

warm
worm
word
cord
cork
cook
cool
 :D

I dunno if this counts...
WARM
WARP
CARP
CORP
COOP
COOL

Let's try longest!
[b]warm[/b]
ward
card
cord
[b]cold[/b]
fold
fond
food
fool
[b]cool[/b]

 :lol:

[quote="perfectnumber628"][quote="pieterminate"]New One!
Change from [b]zebras[/b] to [b]madman[/b]![/quote]

how do you know it's impossible?[/quote]

There is no word one letter different from zebras (as in one letter changed...zebra doesn't count)

I actually made a start with that one, using http://www.onelook.com It found "zebrak"...

[quote="keeyounglee"]come on, proper nouns dont count

wait, it is a proper noun?[/quote]No, it's not a proper noun.

And for the shortest:
 :)

[b]math[/b]
mate
mare
nare
nard
[b]nerd[/b]

Try [b]aops [/b]to [b]math[/b]  :D

[quote="K81o7"][quote="pieterminate"]nare
nard
[/quote]

These are words?[/quote]

Go to:
[url]http://dictionary.reference.com/[/url]   :lol:

...okay, they're bound to be words...but..they're just not that common.

from joke to make

joke
moke
make

moke is a real word

 :D  :D  :rotfl:

Is there a strategy to do this? I have never been able to do these. Any helpful tips would be appreciated.

not really, I just look up words that I haven't heard of that might help make it faster to change

In that case, it seems like to much work for me to be bothered with :rotfl: . If it is guess and check, I am amazed that people can do things like
MIND
RIND
RINK
PINK
PINT
LINT
LIFT
SIFT
SOFT
SORT
PORT
POET

Okay, new one: Get from LEMMA to PROOF. It took me 10 steps.

I bet someone can get llama in there somewhere :rotfl:

jmadsen: how did you do it i can't even start!

I used the wordlist at ftp://gattinger.org/wordlists/english.zip
I haven't heard of most of the words I used.


How many positive integers have their digits in a strictly increasing order?  :rotfl: I have an answer, which is 'correct' though I haven't accounted for the numbers $ 1 \minus{} 9$, which don't seem to have strictly increasing digits to me. I guess they could but I've been unable to find anything on the net that would clear that up.

Add your pennies!

Casework is impractical. It would take far too much time to list them out. [hide="hint"]Start at the end. [hide="Answer"]There are $ 511\equal{}2^9\minus{}1$ such integers if we include the set of integers $ 1\le n_i\le 9$.[/hide][/hide]

I'm more interested in whether or not we consider $ S\equal{}\{1,2,\ldots,9\}$ as having strictly increasing digits. I guess they can be written as $ 0x$, $ x\in S$, which is strictly increasing, but I'd really like to hear some feedback on that. This problem was on a test for some math ed majors and the teacher wouldn't say.

I would consider the statement that the numbers 1,2,3,...9 have their digits (digit) in strictly increasing order to be vacuously true, since we are dealing with a single digit, which cannot be in increasing order.

A somewhat more interesting problem is that of finding all n-digit integers with the property that their digits are in increasing order.


Prove that there is infinitely many dierent maps $ f : N\rightarrow N$ from the set $ N$ of all positive integers into itself, which for all $ n$ satisfy the following condition: $ f(f(n)) \equal{} n^3 \plus{} 1997$ .Can such a map be one-to-one?

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1587013595&t=181199

Pierre.


I recently stumbled upon this:

If a is an odd integer greater than 1, then a, ((a^2-1)/2), and ((a^2+1)/2)) is a Pythagorean triplet.

If a is an even integer greater than 2, then a, (a^2/2-2), and (a^2/2+2) is a Pythagorean tiplet.

(You can see a pattern and come up with this)

I tried it out, and it seems to work. But I am at a loss for deriving it algebraically. o-0. Any ideas?

[hide]i think that this question is a simple question
for odd substitute a = 2k+1
now the given numbers are $2k+1, 2k^{2}+2k, 2k^{2}+2k+1$
k can be any integer
now $(2k^{2}+2k+1)^{2}-(2k+1)^{2}= 2k^{2}(2k^{2}+4k+2)$
      $= 4k^{2}(k+1)^{2}= (2k(k+1))^{2}= (2k^{2}+2k)^{2}$
hence
$(2k^{2}+2k+1)^{2}-(2k+1)^{2}= (2k^{2}+2k)^{2}$
or $(2k^{2}+2k+1)^{2}= (2k+1)^{2}+(2k^{2}+2k)^{2}$

Similarly consider a= 2k for a is even
which convert our numbers to $2k, 2k^{2}-2, 2k^{2}+2$
we will do same for this also or showing that this is pythagorean triplet......[/hide]

If by "derive" you mean "prove," then you don't need to do anything other than substitute.

The general formulae give all primitive triplets (ones that share no common factors) as

$2mn, m^{2}-n^{2}, m^{2}+n^{2}$

Where $m > n, m \not \equiv n \bmod 2$.  Your first observation can be found by letting $n = 1$ and dividing by two.  Your second is incorrect; the correct version can be found by letting $n = 2$ and dividing by two.  You'll also want to substitute $a = 2m$, although I find this unnecessary.

Ah, okay. thanks. That was stupid of me. x.x And I meant derive, as in, where did it come from, not prove.

Some identities don't really "come from" anywhere.  One that comes to mind is the Sophie Germaine identity,

$a^{4}+4b^{4}= (a^{2}+2ab+2b^{2})(a^{2}-2ab+2b^{2})$

Which can't really be said to "come from" anything.

The Pythagorean identities, however, do "come from" somewhere.  Their basis is the fact that $(m+ni)^{2}= (m^{2}-n^{2})+2mni$.


bad day. if we were careful enough, or started warming up a bit earlier,
this year we will have a lot of perfects

I know one kid who got a 78 (Ram Bhaskar)

lol he was sitting RIGHT behind me during the test.

The top scores I know of are 78 for two people my friend knows. That means probably around 5 or 6 with 78+

well COMC didn't mean anything; it only told us who were not rusted with simple math questions in Nov.
I guess most people were hibernating in november. :oops:

Random grade 12 guy in my school, Saroosh, got (from what I can tell) perfect. He came here from Iran last year I believe, and after seeing him for three months in math club, I give him a 60% or greater chance of making the IMO team (not even exagerrating). Part of that 60% accounts for eligibility to make the team itself because of how recently he came, so you better watch out!

Maintenance of Calgary domination becomes that much closer...:ninja:.

Cheers,

Hunter

P.S. I am horrible at probability, so please replace it with whatever percent you see fit.

[quote="jix"][quote="bbi5291"][quote="jix"]I heard Brian Bi got perfect. You know, the IOI guy from last year.[/quote]
You heard this from Cyril, didn't you?[/quote]

Yes, I did. So did you get perfect? (:[/quote]

Yes. But I am not really any better at math than I was last year, and by the standards of this forum community, that translates to thorough mediocrity. I was basically lucky.

btw: You are on my list of people who are more likely to make MIT than I.

[quote="Rofler"]
Maintenance of Calgary domination becomes that much closer...:ninja:.
[/quote]

Ottawa has a similar population yet it lacks basically any math community...

Since my name has been mentioned in these hallowed fora, I might as well stop lurking here for CMOR updates and reveal myself. Bacteria (assuming you are who I assume you are), I told you at camp that I'd join AoPS sooner or later... so here's later!

So, who was the other perfect?

Hmm... in North Carolina (in the states) I know of one person who got a 77 (me) and one that got a 78 (dnkywin)... lol NC > every province other than Ontario.


Given $T$ is a linear transformation mapping from vector space $V_1$ to vector space $V_2$.
If $\tilde{T}=(\ker(T)+x)=T(x)$, then $\tilde{T}$ is an isomorphism from $V_1/\ker(T)$ to $V_2$, the image of $T$.

This seems like a very important proposition, especially now that we are working on the proof of the Cayley-Hamilton Theorem for nilpotent matrices. However, I never entirely grasped it the first time(s) around. Does the above theorem have a name and can someone show me a proof?

Yes, this is the First Isomorphism Theorem.  Analogous results apply to groups, rings and other algebraic structures.  The idea is that you define the space $V_1/K$ of cosets $K+x$ where $K$ is the kernel of the map $T$ and define $\hat T : V_1/K \rightarrow V_2$ by $\hat T : K + x \mapsto T(x)$.  You have to prove that this map is (1) well-defined (2) one-to-one (injective) and (3) maps onto the image of $T$.

(1) and (2) turn out to be the same.  $K + x$ is one of many names for the same coset, so you have to check that if you applied $\hat T$ to $K + y$, and that was a different name for the same coset, you would get the same value.  Now $K + x = K + y$ iff $x-y \in K$ iff $T(x-y)=0$ iff $T(x) = T(y)$ iff $\hat T(K+x) = \hat T(K+y$.  Since each step is if-and-only-if, you've also proved that $\hat T$ is one-to-one.

(3) is easy of course.

As an exercise, transpose this into a theorem for groups, rings, etc.  As a further exercise, compute dimensions and deduce a well-known theorem about rank and nullity.


Determine the positive integer value of $x$ such that $\sqrt{16x^{3}+16x^{3}+16x^{3}+\ldots+16x^{3}}=2006^{2}$, assuming there are $x$ identical addends within the radical.

[hide]Since there are $x$ identical addends, summing them all up would be equivalent to multiplying by $x$.  Therefore, we have:

$\sqrt{(x)16x^{3}}=2006^{2}$
$\sqrt{16x^{4}}=2006^{2}$
$4x^{2}=2006^{2}=(2^{2})(1003^{2})=4(1003^{2})$
$x=\pm1003$

Since we are restricted to positive integers, the negative answer is disregarded and $x=\boxed{1003}$[/hide]


algebra
polynomial
search
linear algebra

Give $ A,B\in M_{n}(k)$. Prove that:

1. $ adj(adjA) \equal{} (detA)^{n \minus{} 2}A$
2. $ adj(AB) \equal{} adj(B)adj(A)$
3. $ adj(A^T) \equal{} (adjA)^T$

2. Observe that

$ AB \,\, adj(B) \,\, adj(A) = A [B \,\,adj(B)] adj(A) = A [(\det B) I_n] \adj(A)$

$ = (\det B) A \,\, adj(A) = (\det B) (\det A) I_n = (\det AB) I_n = (AB) \, adj(AB)$,

and

$ adj(B) \,\, adj(A) \,\, AB = adj(B) [adj(A) \,\, A] B = adj(B) [(\det A)I_n] B$

$ = (\det A) \,\, adj(B) \,\, B = (\det A) (\det B) I_n = (\det AB) I_n = adj(AB) \, (AB)$.

Therefore, $ adj(AB) = adj(B) \,\, adj(A)$. $ \Diamond$

3. Try to find the i,j element of both $ adj(A^T)$ and $ [adj(A)]^T$ and conclude that they coincide.

I think you should explain why you can cancel $ A$'s and $ B$'s from your equations... (For instance, let $ A$ and $ B$ be two matrices whose entries are independent polynomial variables over $ \mathbb{Z}$. Then you have to prove that their determinants are nonzero - thus cancellable - polynomials and therefore they can be cancelled themselves.)

  darij

[quote="Carcul"]1. We first show that $ \det(adj A) \equal{} (\det A)^{n \minus{} 1}$. We have $ A \,\, adj A \equal{} (\det A) I_n$, and taking determinants,

$ \det(A \,\, adj A) \equal{} \det(A) \det(adj A) \equal{} \det[(\det A) I_n] \equal{} (\det A)^n$

$ < \equal{} > \det(adj A) \equal{} (\det A)^{n \minus{} 1}$.

Now, $ adj A \,\, adj(adj A) \equal{} [\det(adj A)] I_n \equal{} (\det A)^{n \minus{} 1} I_n$, and multiplying both sides by $ A$, we have

$ A \,\, adj A \,\, adj(adj A) \equal{} (\det A)^{n \minus{} 1} A$

$ < \equal{} > (\det A) I_n \,\, adj(adj A) \equal{} (\det A)^{n \minus{} 1} A$

$ < \equal{} > adj(adj A) \equal{} (\det A)^{n \minus{} 2} A$. $ \Diamond$[/quote]
I think you are wrong, if $ detA\equal{}0$ ???

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=994667900&t=153199]Link to a previous thread.[/url]

Yes, I should have been more carefull. Darij, to what canceling are you referring to?

For instance, when you conclude $ \mathrm{adj} \left(AB\right) \equal{} \mathrm{adj} B \cdot \mathrm{adj} A$ from $ \mathrm{adj} B\cdot \mathrm{adj} A \cdot AB \equal{} \mathrm{adj} \left(AB\right) \cdot AB$, you have to cancel $ AB$.

  darij


geometry
inequalities
algebra
polynomial
number theory
IMO

I saw a lot of books on this site, but I don't know for what level
of contests are they. So, can you recommend me some good problem
books for preparing for NMO and IMO, Geometry, Number theory,
inequalities, polynomials, anything you have in mind....

http://olympiads.win.tue.nl/imo/books.html
http://www.unl.edu/amc/a-activities/a4-for-students/reading-list.html
These are two good sites you should visit. Also go to
http://www.artofproblemsolving.com/Books/AoPS_B_About.php
which is part of this website. :)
EDIT: read,
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=577660#577660
for free ebooks


trigonometry
inequalities proposed
inequalities

Prove that

$ \frac{r_a\minus{}r}{R}\geq4\sin^2(\frac{A}{2})\cos^2(\frac{B\minus{}C}{2})$

[quote="Ligouras"]Prove that

$ \frac {r_a \minus{} r}{R}\geq4\sin^2(\frac {A}{2})\cos^2(\frac {B \minus{} C}{2})$[/quote]

$ \frac {r_a \minus{} r}{R} \equal{} 4\sin^2(\frac {A}{2})$

To you, based on your Inequality，I get a nice:
$ \frac {9}{2}\frac {r_a \minus{} r}{w_a \plus{} w_b \plus{} w_c} \geq \frac {4r_a}{r_a \plus{} h_a}\sin\frac {A}{2}$
$ < \equal{} > \frac {9}{8}(1 \minus{} \frac {r}{r_a})(r_a \plus{} h_a) \geq \sin\frac {A}{2}(w_a \plus{} w_b \plus{} w_c)$
$ < \equal{} > \frac{9}{8}\tan\frac {A}{2} \geq \sin\frac {A}{2}(w_a \plus{} w_b \plus{} w_c)$
$ < \equal{} > \frac {9}{8}\frac {b \plus{} c}{w_a \plus{} w_b \plus{} w_c} \geq \cos\frac {A}{2}$

So Your a very good question put here.Thank you！

and  You put one,I am able to write a new one.
BQ


I was at MIT's HSSP program this Saturday, taking my Combinatorics class. As I was standing in the Infinite Corridor, I was wondering if there were any AoPSers in the crowd. Are any of you guys taking any classes with the HSSP program?

Hello!
I'm Maria, a teacher at HSSP (doing Explorations in math from 1-3). i love taking part in programs that are run by the MIT Educational Studies Program. if you have any questions about anything related to ESP, feel free to ask me :) maybe we can have an AoPS lunch sometime if enough people are interested!

I used to teach and help run programs at ESP --- it's a great place (hi, Maria).  I still go back and help occasionally, but not with HSSP, since it's too long to fly in from Illinois.  But I'm sure there must be more people around there who are also on here.

I definitely miss teaching for stuff like that, since I haven't found similar programs here.

[quote="Mariag"]:) maybe we can have an AoPS lunch sometime if enough people are interested![/quote]

Yeah that would be cool. It would be great to finally meet some AoPSers, especially at MIT

im taking a cs (intro to programming) course during the last slot (basically its the only one that i can take the commuter rail and get to in time and it also doesnt conflict with arml practices or other "fun" stuff (arml is actually fun not "fun") on saturady mornings)

wow, that's some dedication :)

hope everyone is enjoying hssp  :lol:


trigonometry
geometry
perimeter
inequalities
inequalities unsolved

[quote="arqady"][quote="anthony19891019"]a,b,c>0
$\sum\frac{1}{{a^{2}+1}}$=$2$
Prove: $ab+bc+ac\le\frac{3}{2}$[/quote]
This problem equivalent to follow:
If $\alpha,$ $\beta$ and $\gamma$ are angles of triangle then
$\sin\frac{\alpha}{2}+\sin\frac{\beta}{2}+\sin\frac{\gamma}{2}\leq\frac{3}{2}.$ :wink:[/quote]

How can you change to this step?

[quote="anthony19891019"]
How can you change to this step?[/quote]
Let $\frac{1}{a^{2}+1}=x,$ $\frac{1}{b^{2}+1}=y$ and $\frac{1}{c^{2}+1}=z.$
Then $x+y+z=2$ and $ab+ac+bc\leq\frac{3}{2}\Leftrightarrow\sum_{cyc}\sqrt{\left(\frac{1}{x}-1\right)\left(\frac{1}{y}-1\right)}\leq\frac{3}{2}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\sqrt{\frac{(2-2x)(2-2y)}{xy}}\leq3\Leftrightarrow\sum_{cyc}\sqrt{\frac{(y+z-x)(x+z-y)}{xy}}\leq3\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\sqrt{\frac{z^{2}-x^{2}-y^{2}+2xy}{xy}}\leq3\Leftrightarrow\sum_{cyc}\sqrt{2-2\cdot\frac{x^{2}+y^{2}-z^{2}}{2xy}}\leq3\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\sqrt{2-2\cos\gamma}\leq3\Leftrightarrow\sum_{cyc}\sin{\frac{\gamma}{2}\leq\frac{3}{2},}$ which obviously true. :)

[quote="anthony19891019"]a,b,c>0

$\sum\frac{1}{{a^{2}+1}}$=$2$

Prove: $ab+bc+ac\le\frac{3}{2}$[/quote]

And there is an alternative proof.

We can put $a^{2}= \frac{p}{q+r}, b^{2}= \frac{q}{r+q}, c^{2}=\frac{r}{p+q}$.
(This substitution can be reached by arqady's substitution also, as $x,y,z$ are sides of a triangle whose perimeter is 2.)
So $ab+bc+ca=\sum_{cyc}\sqrt{\frac{pq}{(p+r)(q+r)}}$
$\leq \sqrt{\sum_{cyc}pq \sum_{cyc}\frac{1}{(p+r)(q+r)}}$
(by Cauchy)
$=\sqrt{\frac{\sum_{cyc}pq \sum_{cyc}(p+q)}{(p+q)(q+r)(r+p)}}$
$=\sqrt{\frac{2\sum_{sym}p^{2}q+6pqr}{\sum_{sym}p^{2}q+2pqr}}$
$\leq \sqrt{\frac{\frac{9}{4}\sum_{sym}p^{2}q+\frac{9}{2}pqr}{\sum_{sym}p^{2}q+2pqr}}$
$=\sqrt{\frac{9}{4}}=\frac{3}{2}$.

I thought that this is from Iranian olympiad 2006, (2005?) and very similar problem was proposed in Balkan shortlist 2005.

And this is another proof:

$\sum \frac{1}{a^{2}+1}=2$ is equivalent to
$a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2a^{2}b^{2}c^{2}=1$.
So if $ab=u,bc=v,ca=w$, then $u^{2}+v^{2}+w^{2}+2uvw=1$, which means that
there exists a triangle $ABC$ where $cos A=u, cos B=v, cos C=w$.
Thus $ab+bc+ca=cos A+cos B+cos C \leq \frac{3}{2}.$

Replace $x=\frac{1}{1+a^{2}},y=\frac{1}{1+b^{2}},z=\frac{1}{1+c^{2}}$, then $x,y,z \le 1$ and $x+y+z=2$. So that $a=\sqrt{{1 \over x}-1},b=\sqrt{{1 \over y}-1},c=\sqrt{{1 \over z}-1}$. We must prove: \[\sum_{cyc}\sqrt{{1 \over x}-1}\cdot\sqrt{{1 \over y}-1}\leqslant{3 \over 2}\iff \boxed{\sum_{cyc}\sqrt{y}\sqrt{x+y-z}\sqrt{y+z-x}\le 3\sqrt{xyz}}.\quad (*)\] The square root sign is meaningful iff $x,y,z$ are three sides of a triangle. Applying the cosin law, $y^{2}=z^{2}+x^{2}-2zx\cos B$, \[(*) \iff 3\sqrt{xyz}\ge \sum_{cyc}\sqrt{y}\sqrt{2zx-2zx\cos B}=2\sqrt{xyz}\\ \iff \sin{A \over 2}+\sin{B \over 2}+\sin{C \over 2}\le{3 \over 2}.\] Can anybody prove the inequality in the box by an algebraic inequality?

[quote="anthony19891019"]a,b,c>0

$\sum\frac{1}{{a^{2}+1}}$=$2$

Prove: $ab+bc+ac\le\frac{3}{2}$[/quote]

Just $AM\geq HM$  :wink:

Here is your mistake Fitim:

You used am-hm to yield \[\left(\sum \frac{1}{a^{2}+1}\right)\left(\sum a^{2}+1 \right)\geq 9 \iff \sum a^{2}\geq  \frac32 \]
But we are trying to prove $\frac32 \geq  \sum ab$ which doesn't follow at all from the above...

Solution:
$\sum \frac{1}{a^2+1}=2$
$\Rightarrow 2(abc)^2+\sum (ab)^2=1$
Now using $C-S$,we have:
$1=2(abc)^2+\sum (ab)^2\geq2(abc)^2+\frac{(ab+bc+ac)^2}{3}$
$\Rightarrow (ab+bc+ac)^2\leq 3-6(abc)^2$
So it's enough to prove that:
$\sqrt{3-6(abc)^2}\leq \frac{3}{2}$ $\Leftrightarrow abc\geq \frac{1}{\sqrt8}$
Which is true,since,again by $C-S$ we have:
$2=\sum \frac{1}{a^2+1}\geq \frac{9}{a^2+b^2+c^2+3}$
$\Rightarrow a^2+b^2+c^2\geq 3\sqrt[3]{(abc)^2}\geq \frac{3}{2}$
$\Rightarrow (abc)^2\geq \frac{1}{8}$
$\Rightarrow abc\geq \frac{1}{\sqrt{8}}$
$Q.E.D$


inequalities
logarithms
inequalities unsolved

For the real numbers $ a_1,a_2,...,a_n \in [2,3]$ prove the inequality:

\[ \log _{a_1} (5 \cdot a_2 \minus{}6)\plus{}\log _{a_2} (5 \cdot a_3 \minus{} 6)\plus{}...\plus{}\log _{a_n} (5 \cdot a_1 \minus{}6) \ge 2 \cdot n\]

Really easy!  :)
Since $ a_i\in [2, 3]$ clearly $ 5a_i \minus{} 6\geq a_i^2$ for all $ i$. Now using the monotonicity of logarithm (note that $ a_i\geq 2$ for all $ i$) and AM-GM we have
\[ LHS\geq 2\log_{a_1}{a_2} \plus{} 2\log_{a_2}{a_3} \plus{} ... \plus{} 2\log_{a_n}{a_1}\]
\[ \geq 2n\sqrt [n]{\log_{a_1}{a_2}\cdot\log_{a_2}{a_3}\cdots\log_{a_n}{a_1}} \equal{} 2n.\]


For which $ n$ can we find a permutation $ (a_1, a_2, ... , a_{n\minus{}1})$ of $ (1,2,...,n\minus{}1)$ such that $ 0, a_1, a_1\plus{}a_2, a_1\plus{}a_2\plus{}a_3, ... , a_1\plus{}a_2\plus{}...\plus{}a_{n\minus{}1}$ are all distinct $ \mod n$.

The case when $ n$ is odd is ruled out.
We have $ a_1\plus{}\dots\plus{}a_{n\minus{}1}\equal{}\frac{(n\minus{}1)n}{2}\equal{}0\mod n$

[quote="srikanth"]The case when $ n$ is odd is ruled out.
We have $ a_1 \plus{} \dots \plus{} a_{n \minus{} 1} \equal{} \frac {(n \minus{} 1)n}{2} \equal{} 0\mod n$[/quote]

Right  :)

This is an existence problem for which the associated enumeration problem is open: see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=199166

That link also contains a generalization of this problem:
[quote="JBL"]Generalization: for an arbitrary finite group $ G \equal{} \{g_1, g_2, \ldots, g_n\}$ with identity $ e$, how many permutations $ p \equal{} (p_1, \ldots, p_n)$ of $ G$ are there so that the first differences $ (p_2p_1^{ \minus{} 1}, p_3p_2^{ \minus{} 1}, \ldots, p_np_{n \minus{} 1}^{ \minus{} 1})$ are a permutation of $ G \setminus \{e\}$?

By analogy with [srikanth's argument], the answer is $ 0$ if $ G$ is abelian and the product of the elements of $ G$ is $ e$ (and so in particular if $ G$ has odd order).  This suggests the following, simpler question: find necessary conditions on a group $ G$ for there to be $ 0$ such permutations.[/quote]

Thanks JBL, but I just want an elementary proof to this  :D


Solve the inequation :

\[ x^{\log_4{x\plus{}1}} > 256x ;\] .

[hide="Hint"]Call $ x\equal{}4^t$ we have $ (4^t)^{t\plus{}1}>256.4^t\Leftrightarrow 4^{t^2}>256\Leftrightarrow t^2>4$[/hide]

[quote="tdl"][hide="Hint"]Call $ x \equal{} 4^t$ we have $ (4^t)^{t \plus{} 1} > 256.4^t\Leftrightarrow 4^{t^2} > 256\Leftrightarrow t^2 > 4$[/hide][/quote]

I get a different answer

Taking logs (base 4)

$ \log_4{x} \cdot \log_4{(x \plus{} 1)} > 4 \cdot \log_4{x}$

Therefore $ \log_4{(x \plus{} 1)} > 4$

Or $ x \plus{} 1 > 4^4 \Rightarrow x > 4^4 \minus{} 1$

[quote="SimonM"]Taking logs (base 4)

$ \log_4{x} \cdot \log_4{(x \plus{} 1)} > 4 \cdot \log_4{x}$[/quote]
$ \log(a b) \equal{} \log(a) \plus{} \log(b)$  :roll:

Also, although the notation is ambiguous the exponent on the left is almost certainly intended to be equal to $ 1 \plus{} \log_4 x$, not $ \log_4(x \plus{} 1)$.

To simon:
You have $ 2$ mistake:
1) If $ log_4x<0$ then $ log_4x\plus{}1<4$
2) $ log_4x\plus{}1<4\Leftrightarrow log_4x<3$
Example $ x\equal{}4^{\minus{}3}$ is a root!

[quote="tdl"]To simon:
You have $ 2$ mistake:
1) If $ log_4x < 0$ then $ log_4x \plus{} 1 < 4$
2) $ log_4x \plus{} 1 < 4\Leftrightarrow log_4x < 3$
Example $ x \equal{} 4^{ \minus{} 3}$ is a root![/quote]

Yeah, I realised that just after posting I was in the middle of editing it

[quote="kunny"]Answer: $ 0 < x < \frac {1}{16},\ x > 16.$[/quote] . This is the correct answer . Somebody comes with a solution ? :)


ARML email AMC AIME function USA(J)MO USAMO

You're going to ARML too, right? I just received the following email message today:

"Hello former and/or future ARML mathletes!

You are invited to participate in the 2008 ARML math competition as a member of Iowa ARML. The dates for ARML this year are May 30-31. Please read the enclosed document and let me know if I can be of further help. I look forward to hearing from you.

James Kirpes
Head Coach, Iowa ARML"

It's too bad I have to miss out (like last year...) due to the National Science Olympiad Tournament (May 30-31 as well).
Nevertheless, I wonder what the teams will be and who will be present.

Yeah, I got that email, and will definitely be going.  Did Kirpes send out more than one email, though?  I don't see your email address on the list.

EDIT:  Kirpes also gave me the packet during 7th period today.  Why do I like editing so much?  I don't know.

EDIT:  Did you go to the Iowa State competition?  How'd you do?  (I'm asking because it seems like Ames only goes to that and Iowa Falls).  I failed Iowa State with a 92 on sprint and 50 on target.  Sad.

Mr. Kirpes sent out more than one email; there are only about 20 or so email addresses on the email he sent me.

As to the ISU GPML meet, I did equally miserably. Young probably beat me out in both sprint and target; I got third on each (the same as last year!). We did well at relay though; Lucy Chen was on my team and we only missed one 2-min deadline out of all the problems.

One bit of advice:  watch out for Tyler this year.  He's on like a whole different level or something, and got a 10 on the AIME.  So, naturally, I must get an 11 next year.

As for relay, didn't your team get a perfect like mine?   :lol:

I don't quite remember, actually. I'll go look at the results sheet when I can sort out things in my room this weekend.

And which Tyler are you talking about? (Last name?)

Chenhall.  He's studied a ton.  Yeah.

EDIT:  I just emailed Shi-Ke telling him to get an AoPS account, so hopefully we will soon have another person on this Iowa forum.  (that actually posts, anyway)
I now hate To Kill a Mockingbird.  Giant Essay Test.  Tomorrow.  Agh.

Yeah, I know who you're talking about. It's hard not to know.

Shi-Ke? He's a freshman, so I probably don't know him.

Regarding [i]To Kill a Mockingbird[/i], hate the test, but love the book.

Nah, Shi-Ke's an 8th grader, though he shows up as a freshman in the results for the GPML competitions.  Anyway, I assume you've figured out who I am by now... (my email makes it kinda obvious)

All I know is that you are Matt, a freshman at West High.

Yeah, well, good enough.   :lol:

EDIT:  A bit off topic, but are you taking the ACT on Saturday?

Nope. I probably should, though. I haven't prepared yet for either the SAT or ACT, so it'd be kind of pointless to take either at this point.

Btw, you can really learn a lot just by reading the "Introduce Yourself!" thread (I was the original author of it, but due to bad moderation, I screwed it up and Penguin became the author).

Yeah, I read it a few hours ago.  I now know your first and last name, and all of your AMC/AIME scores from last year...  Mwuahahahahaha

Yeah, this'll be the third time I'll be taking the ACT.  I really have to do better on Reading and Writing.  On Math and Science I should be able to get 35 or 36, but Reading and Writing last year were only 27...  big difference, lol.

EDIT:  My posts have too many "Yeah"s.  weird.  ok...  um....  yeah.

Really? I posted my AMC/AIME scores?   :oops: Now everyone in the world knows...unless I delete it...  :D

No, you didn't.  Have you ever seen the 300+ page AMC summary pdf file?  I can find the AMC scores of every AIME qualifier and more from last year, and AIME scores of anyone who got 5 or more.   :lol:
So... I know you got the same as me on the AIME this year, but how'd you do on the AMCs (this year)?

EDIT:  Yeah, anyone in the world can find your scores, lol.

I'm the AMC 10A 135 male, 10th grade in the state stats.
I got 6 this year, which will probably miss the cutoff. I hope that's not a cutoff of 7, as I missed a cutoff of 6 last year with a score of 5....  :mad:

Did you take the 10/12B also?
For me:
12A:  103.5 (ouch)
12B:  112.5 (yay)
Man, I messed up the AIME really badly.  On the 3x4x6 boxes one I spent 40 minutes multiplying out the generating functions by hand.  Now I could do it in 2 minutes with the Multinomial Theorem, but on March 18...  Oh well...

I took the 10B and got owned...13.5 points lower (even though the test was statistically easier). I don't see any point for underclassmen to take the 12 since that only hurts you in the calculation of your index (given you hit exactly the floor score)...unless you're one of those kids that can get a 140+ in 9th grade.

Anyway, I took the AIME II, which gave me (a little) extra time to cram.

I took the AIME I....
Anyway, I was thinking about taking the 10B, but then reasoned that my brain tends to mess up really easy problems, especially after seeing that Young only got a 124.5 on the 10A.  So, I'm going to try and drastically improve my score on the 12 next year.

Still, you might regret it. Some underclassmen who got 6s on last year's AIME failed to make the USAMO cut simply because they took the 12 and got under a 181 index.

You'll do fine on the 10. At least your score will be higher by a few points, if not more.

If I don't get around a 130 on the 12 next year, I'll probably take the 10B....  or not... I don't know yet...
All of the good underclassmen at my school (well, most of them) have never taken the 10.  Curie didn't, Tyler didn't, you get the idea.  Young took the 10 once, failed miserably, and then failed even more miserably on the 12B (I'll be nice to him and not give out his score - it was bad.  really bad)

EDIT:  Nice, you're about to break 100 posts!

Well, here it is: [b]100TH POST![/b]  :P

Btw, you should stop using "EDIT" even when you don't edit a post (or edit a post just so you can use the letters "EDIT"). I use "btw" or "Note:" all the time. (If you edit a post within 1 min or so, the notice that the post was edited doesn't show. It's just weird when you think a post hasn't been edited because the notice isn't there, but the post says "EDIT.")

Really? You guys mess up the 10 at West High?
Weird; it seems like over here at Ames, we do far better on the 10. Besides Lucy's two 144s, we have gotten at least one 130+ each year on the 10, but since Josh and Qi left, we have never gotten a 120+ on the 12.

I [i]do[/i] actually edit the posts.  I just seem obsessed with the phrase "EDIT:".  Let's see if I can avoid that in this one.

Yeah, our highest scores on the 10 were Young's 124.5 and Tianyi's 132.  I think our next highest score was in the 106 area.  Pretty much all of our "really good people" take the 12.

This is getting kinda off-topic.  Maybe we should create a topic called "Random Discussion of Randomness" where the goal is to be off-topic, or something...

Done.

I guess this is Iowa's mini-Fun Factory.

Yes.  Now if only we could get some more people.  I've told several people I know to be more active in this forum.  They seem lazy.

NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO!!!!!!!!!!
I got the floor but my index wasn't high enough.   :(

Nate [i]did[/i] get in.  I'm Matt.
Had I gotten a 144 on the AMC 10, I would've gotten in...  I think I'll take the 12A next year, and if I score less than 124ish or something, I'll take the 10B.  But I have a whole summer ahead of me to study :D
Anyway, yeah, Curie said she bombed the AIME II, so... yeah...

AIME score: 7
AMC 12 score: 115.5

185.5: I probably should've taken the 10... but I'm really not that lame.

If only I'd known the Multinomial Theorem on March 18...
And maybe taken the 10...

Oh well.  I shall attempt uber pwnage next year.

On #13 on the AIME I, somehow I messed some adding up or something and gotten it to the point where the function was a constant.  Ugh.

Woot!  50 posts!

hillarryous, what'd you get on sprint at state GPML last year?  Just kinda curious, 'cause I took it today.  (argh, 2 stupid mistakes leaving me 106)

ha ha ha ha

I already attained ub3r pw|\|463

On the AIME: I didn't quite like the one about the cars:
you have to take the limit to infinity but if you take it context-wise technically cars won't approach infinity

I completely skipped that problem, which was kinda stupid, because now it seems really easy...
The car's speed approaches infinity, but you can have 375 because you don't have to count the empty space behind the last car.  Grrrr, coulda gotten 7.  Oh well.  Must get 11 next year.

I'm sure of that...

Tyler says the multinomial theorem is useless: it came up in national MATHCOUNTS countdown though.
just thought i'd throw that out there

Tyler's [i]usually[/i] right when he says something is useless, but not always.

right...

some examples of tyler being usually right on uselessness?

[quote=hillarryous]I'm in the same boat.

$ \mbox{AIME II score} \equal{} 6$
$ \mbox{AMC 10A score} \equal{} 135$

$ \mbox{USAMO index} \equal{} 135 \plus{} 6 \cdot 10 \equal{} 195 < 202.5$

If I had only gotten #1 correct...
Missing question #1 has cost me a USAMO invite for the second year in a row. :wallbash_red:

Nate, you should have taken the AMC 10 once--you just might have gotten in.

Congrats to Young, Tyler, and the other qualifiers from West. I'm just a little surprised that Curie didn't make it.[/quote]

Dang that sucks man rip.

Whoa. The forum rises again. I remember when this conversation was first going when I was also in high school. So long ago! :P


modular arithmetic
advanced fields
advanced fields unsolved

If n is a positive integer such that (n − 1)!  congruent to 1 (mod n), prove that n is
prime.

Is it Wilson theorem??
http://en.wikipedia.org/wiki/Wilson_theorem
I think it should be congruent to $ \minus{}1 \pmod n$
 :)

Using Wilson theorem when $ n$ is prime.


inequalities
combinatorics proposed
combinatorics

Ten volleyball teams played a tournament; every two teams met exactly once. The winner of the play gets 1 point, the loser gets 0 (there are no draws in volleyball). If the team that scored $n$-th has $x_{n}$ points ($n=1, \dots, 10$), prove that $x_{1}+2x_{2}+\dots+10x_{10}\geq 165$.

[i]Proposed by D. Teryoshin[/i]

[quote]Looking for the ISL 2006? Check this out![/quote]
 :D

Let $A_{1}$ be the best and $A_{10}$ the worst team. and let $f(A_{1},...,A_{1}0)= x_{1}+2x_{2}+...+10x_{10}$

In a game that $A_{j}$ wins over $A_{i}$, $i < j$ , magicly changing the fate of the game, $f$ decreases.So minimum occures when for all the games $A_{i}$ wins over $A_{j}$ $i < j$. hence
\[f \geq 9+2(8)+3(7)+...+8(2)+9 =165 \]

Isn't this direct consequence of rearrangement inequality?

ok here's a double counting approach
it is obvious that the sum of points of every k persons is at least $\binom{k}2$
then the sum
\[(x_1+x_2+x_3+\cdots+x_{10})+(x_2+x_3+\cdots+x_{10})+\cdots+(x_9+x_{10})+x_{10} \ge \binom{10}2+\binom{9}2+\cdots+\binom{2}2=165\]
\[\color{magenta} \huge{\boxed{\mathbf{Q.E.D}}}\]


geometry
geometric transformation
vector

What you want is called a [url=http://en.wikipedia.org/wiki/Euclidean_vector]vector[/url].  I like the notation $ \left< x_2 \minus{} x_1, y_2 \minus{} y_1 \right>$ for the vector you're looking at.  Then you say "after translating this point by this vector I get this point."

[quote="PowerOfPi"]If you have some point (say $ (x_1, y_1)$), and you are translating it to another (like $ (x_2, y_2)$), how can you express the change? Do you just say it changes by $ (a,b)$? I want a way to say the translation movement without having to know both points (I don't care much about the points).[/quote]

Basically, if you're translating say (x,y) to (x+1, y+6), you are translating the point along the vector <1,6>.


Prove that $n$ is not a perfect number.
$n = 2^{10} (2^{11} - 1)$

$2^{11}-1 = 23 \cdot 89$ is not prime (:blush: ) $n$ can't be a perfect number.

For that to happen 2^11 -1 cannot be a prime number.

Nea is correct, $n$ is not a perfect number.

[quote="tµtµ"]Since $2^{11}-1$ is prime I would rather say that it [b]is[/b] a perfect number  :?:[/quote]
I haven't solvet it yet but it must not be perfect. From theorem it's true, but maybe it is an exception.  :?

A fun  tid  bit:  (easy to prove)
For any  $n>3$  and of the form $4k-1$   (eg 11)
if $n$ is prime and $2n+1$ is prime  ... then $2^n$ is NOT prime.. (It is divisible by $2n+1$)
(Eg:  you right away can dismiss $2^{23}-1$ as prime becuase $2*23+1=47$ is prime. )

[quote="Peter VDD"]Er... when is a number supposed to be perfect? :huh: never heard that name before...[/quote]
If sum of all its proper divisors is equal to the number  it self, ... ((Eg 6(=1+2+3), 28(=1+2+4+7+14),496(=1+2+4+8+16+31+62+124+248))) The number is called
[url=http://mathworld.wolfram.com/PerfectNumber.html]perfect number (see link)[/url]


Το ρωσικό περιοδικό KVANT, που άρχισε να εκδίδεται το 1970, περιλαμβάνει
σε κάθε τεύχος του στήλη διαγωνισμού με προβλήματα μαθηματικών και φυσικής.

Ειδικά, από το τεύχος 11 του 1984 έως και το τεύχος 12 του 1988 περιλάμβανε
παράλληλη αγγλική μετάφραση των προβλημάτων της στήλης διαγωνισμού.

Παρακάτω δίδονται συγκεντρωμένες, σε μορφή djvu,  οι αγγλικές μεταφράσεις των μαθηματικών προβλημάτων της στήλης διαγωνισμού. Συνολικά περιλαμβάνονται 430 προβλήματα (από το πρόβλημα Μ711 εώς το πρόβλημα Μ1140).

[url]http://rapidshare.com/files/60756490/kvant.djvu.html[/url] (~21mb)

Το υλικό αυτό, όπως και όλο το υλικό του περιοδικού KVANT, παρέχεται δωρεάν από
το "Κέντρο δια Βίου Μαθηματικής Εκπαίδευσης Μόσχας", στον σύνδεσμο http://kvant.mccme.ru/  (στα Ρωσικά).

προτείνω, αν κάποιος ασχοληθεί με κάποιο από τα προβλήματα που υπάρχουν στο αρχείο να χρησιμοποιεί τον αντίστοιχο κωδικό του προβλήματος.

το ίδιο ίσως θα μπορούσε να γίνει και στις υπόλοιπες στήλες, πχ τα προβλήματα της γεωμετρίας ή της άλγεβρας να κωδικοποιηθούν με κάποιο σύστημα έτσι ώστε να είναι πιο εύκολο να αναφερόμαστε σε κάποιο από αυτά.


function
limit
calculus
calculus computations

Prove the following limit:

$ \lim_{n\to\plus{}\infty}[n\minus{}\Gamma(1/n)] \equal{}\gamma$

Let $ h\equal{}\frac{1}{n}$ so that $ h\to 0^{\plus{}}$ as $ n\to\plus{}\infty$ and the limit becomes

\[ \lim_{n\to\plus{}\infty}\left[ n\minus{}\Gamma\left(\frac{1}{n}\right)\right] \equal{}\lim_{h\to 0^{\plus{}}}\left[\frac{1}{h}\minus{}\Gamma\left(h\right)\right] \equal{}\minus{}\lim_{h\to 0^{\plus{}}}\frac{h\Gamma\left(h\right)\minus{}1}{h}\]

\[ \equal{}\minus{}\lim_{h\to 0^{\plus{}}}\frac{\Gamma\left(h\plus{}1\right)\minus{}\Gamma(1)}{h}\equal{}\minus{}\Gamma^{\prime}(1) \equal{}\gamma\]

since $ \Gamma(1)\equal{}1$ and $ x\Gamma(x)\equal{}\Gamma(x\plus{}1)$.


[color=blue][b]  [size=150]15th Maths Contest of Viet Nam's Olympic 30-4 for grade 10 organized in Ho Chi Minh city - Duration: 180mins [/size]

1) Solve the equation : $ x^3 \plus{} 3x^2 \minus{} 3\sqrt [3]{3x \plus{} 5} \equal{} 1 \minus{} 3x$

2) Find the prime number $ p$ such as existing all the positive whole numbers $ n,x,y$ that satisfy $ p^n \equal{} x^3 \plus{} y^3$.

3)  Let $ AC$ be a fixed line segment with it's central point $ K$. Set $ B,D$ be $ 2$ changable points that they are symmetric via $ K$. A bisector of $ \hat{BCD}$ intersects the line $ AB$ and the line $ AD$ at $ J,I$ respectively. A circumscribed circle of the triangle $ AIJ$ intersects a circumscribed circle of the triangle $ ABD$ at $ M$. Set $ H$ be projection of $ M$ on the line that is perpendicular to the line $ AC$ at the central point of it. set $ N$ be an intersection point of $ CH$ with $ KM$. Prove that $ N$  always moves on the fixed curved line.

4) Let $ a,b,c > 0$. Prove that :

$ \sqrt {\frac {2a}{a \plus{} b}} \plus{} \sqrt {\frac {2b}{b \plus{} c}} \plus{} \sqrt {\frac {2c}{c \plus{} a}} \le 3$

5) Let $ BC$ be a fixed line segment and $ A$ be anypoint of plane .A straight line that is perpendicular to the line $ BC$ at $ B$ intersects the median $ AI$ of the triangle $ ABC$ at $ K$. Find the locus of $ A$ as the line segment $ IH$ is parallel with the line segment $ CK$ ($ H$ be orthocentric point of the triangle $ ABC$).[/b][/color]

[color=blue][b]May nobody do these problems?  :maybe:  :blush: [/b][/color]


I am starting a movie club at ISI, and want you to suggest two things:
1) A name for it
2) A suitable movie for the inauguration

Why not Hitchhiker's Guide, it's in your signature :P


trigonometry
function
calculus
calculus computations

Find the largest $ a$ and $ b$ so that $ \tan (ax\plus{}b)$ is continuous on $ (0,2\pi)$

Considering graph of tangent function, I got $ \max\left\{a\right\} \equal{} \frac {1}{2}$, $ b \equal{}\frac{1}{2}(\pi \plus{} 2n\pi)$($ n$ is nonnegative integer) for answer.


Let an integer $ n > 3$. Denote the set $ T\equal{}\{1,2, \ldots,n\}.$ A subset S of T is called [i]wanting set[/i] if S has the property: There exists a positive integer $ c$ which is not greater than $ \frac {n}{2}$ such that $ |s_1 \minus{} s_2|\ne c$ for every pairs of arbitrary elements $ s_1,s_2\in S$. How many does a [i]wanting set[/i] have at most are there ?

[quote="mr.danh"]Let an integer $ n > 3$. Denote the set $ T \equal{} \{1,2, \ldots,n\}.$ A subset S of T is called [i]wanting set[/i] if S has the property: There exists a positive integer $ c$ which is not greater than $ \frac {n}{2}$ such that $ |s_1 \minus{} s_2|\ne c$ for every pairs of arbitrary elements $ s_1,s_2\in S$. How many does a [i]wanting set[/i] have at most are there ?[/quote]
The resut is $ [\frac{2n}{3}]$
Am I right

yes Allnames,the result is$ [\frac {2n}{3}]$
the example is {$ {1,...,[\frac {n}{3}],[\frac {2n}{3}] \plus{} 1,....,n}$}
asume set with maximum number is S.we can assume that $ \frac {|S|}{2}$ element of S  is less that $ \frac {n}{2}$.name that subset T.for every $ t\leq\frac {n}{2}$ we have $ S$ and  $ T \plus{} t$ are disjoint so $ \frac {3|S|}{2} \leq n$ so $ |S| \leq \frac {2n}{3}$

*How many wanting set do we have?

[quote="sumita"]
*How many wanting set do we have?[/quote]
Uhm,I think it is very hard,my solution for the first is same to yours

I think the idea of author come from an old problem from USA TST 2003
See and compare two problems :wink:


inequalities
function
inequalities proposed

Given a positive integer $ n$ greater than $ 1$, real numbers $ a_1,a_2,\dots,a_n,b_1,b_2,\dots,b_n$ such that the sequences $ (a_1,a_2,\dots,a_n)$ and $ (b_1,b_2,\dots,b_n)$ are similarly ordered, and for all distinct integers $ i,j$ with $ 1\leq i,j\leq n$, $ |a_i\minus{}a_j|\geq1\leq|b_i\minus{}b_j|$, find the minimum value of
$ \frac{a_1b_1\plus{}a_2b_2\plus{}\dots\plus{}a_nb_n}{n}\minus{}\left(\frac{a_1\plus{}a_2\plus{}\dots\plus{}a_n}{n}\right)\left(\frac{b_1\plus{}b_2\plus{}\dots\plus{}b_n}{n}\right)$

[quote="aadil"]its direct chebyshev:
the answer is
[hide]0[/hide][/quote]You missed the condition  :rotfl:

We can assume wlog $ a_1 < \cdots < a_n$ and $ b_1 < \cdots < b_n$.
We can write the function as

$ f \equal{} \frac1n\sum a_i(b_i \minus{} B)$, where $ B: \equal{} \frac {b_{1} \plus{} \cdots \plus{} b_{n}}{n}$.

Let $ k$ such that $ b_k\le B$ and $ b_{k \plus{} 1} > B$. For fixed $ b_i$, $ f$ is a linear function of the $ a_i$. Thus it decreases with [i]decreasing [/i]$ a_i$ for $ i > k$ and decreases with [i]increasing [/i]$ a_i$ for $ i\le k$. So for a minimum of $ f$, the $ a_i$ must be as close as possible to $ a_k$, i.e. because of the condition, we must have $ a_i \equal{} a_k \plus{} i \minus{} k$ for all $ i$.

Similarly, we have

$ f \equal{} \frac1n\sum b_i(a_i \minus{} A)$, where $ A: \equal{} \frac {a_{1} \plus{} \cdots \plus{} a_{n}}{n}$.

So for a minimum, there is also an $ \ell$ such that $ b_i \equal{} b_\ell \plus{} i \minus{} \ell$.

Let $ a: \equal{} a_k \minus{} k$ and $ b: \equal{} b_\ell \minus{} \ell$. Then
\[ f\ge\sum\frac {(a \plus{} i)(b \plus{} i)}n \minus{} \sum\frac {a \plus{} i}n\sum\frac {b \plus{} i}n \\
 \\
\equal{} \frac {(n \plus{} 1)(2n \plus{} 1)}6 \minus{} \left(\frac {n \plus{} 1}2\right)^2 \equal{} \frac {n^2 \minus{} 1}{12}.\]
How good that $ a$ and $ b$ vanish.  :trampoline:
Equality iff there are $ a,b\in\mathbb R$ such that $ a_i \equal{} a \plus{} i$ and $ b_i \equal{} b \plus{} i$. :lol:

The given expression is exactly $ \frac{\sum_{i > j} (a_i \minus{} a_j)(b_i \minus{} b_j)}{n^2}$.
WLOG we can assume that $ a_1 \ge a_2 \ge \cdots \ge a_n$. Then clearly, the sum is minimized when each $ a_i \minus{} a_j$ attains the minimum, which is $ i \minus{} j$ by given condition. It then suffices to compute $ \frac{\sum_{i > j} (i\minus{}j)^2}{n^2}$.

[quote="Soarer"]The given expression is exactly $ \frac {\sum_{i > j} (a_i \minus{} a_j)(b_i \minus{} b_j)}{n^2}$.
WLOG we can assume that $ a_1 \ge a_2 \ge \cdots \ge a_n$. Then clearly, the sum is minimized when each $ a_i \minus{} a_j$ attains the minimum, which is $ i \minus{} j$ by given condition. It then suffices to compute $ \frac {\sum_{i > j} (i \minus{} j)^2}{n^2}$.[/quote]
Right, that makes it even easier.  :oops: :)

[quote]You missed the condition  [/quote]
sorry  :oops:


The figure shows the graph of the function g whose rule is g(x) = the Pentagon's budget in year x (in billions of 2001 dollars, adjusted for inflation), with x = 0 corresponding to 1975. Use the graph to answer the following questions.

http://img364.imageshack.us/img364/9681/aleak6.jpg

(a) Over what intervals was this function approximately constant?
( ? , ? ) and ( ? , ?)

(b) Over what interval was this function increasing?
( ?  , ? )

(c) When did the Pentagon budget reach a local maximum?
In the year ????

Do your own homework.  This question asks, "Do you know what the definition of these words is?"


I wrote something up in Maple 10. I wanted to convert it to .pdf, so, I went file->export as->.tex on my file, figuring I could build to .pdf in TexNix Center.

The new .tex file opens fine in texnix center. Then, I build. It doesn't. Instead, it gives me this error, and an emergency stop:

! LaTeX Error: File 'maplestd2e.sty' not found

I [i]have[/i] the required file--it installs with Maple. So, I go tools->Import LaTeX command->the required folder in the Maple 10 program file. I save as the default name, and try to re-build: it still doesn't work. In fact, when I look in the "packages.xml" (the thing I saved it as), I see the exact package that Texnix says it doesn't see. What is wrong, and how do I make my generated tex file build the way it should.

It might be worth recapping how LaTeX looks for style files like maplestd2e.sty. It first looks in the same directory as the tex file. If it's not there it looks in a database for where the file is held - this is a lot faster than looking through all the directories.

So what should you do. If you are only compiling one file generated by Maple then just stick the maplestd2e.sty in the same directory as the tex file you are compiling. You may find that compiling shows that other files are also needed.

If you are likely to want to repeat this process often you need to store the style file in the correct place and get its location put into the database. It looks complicated but I assure you it isn't :)

Run MikTeX Options and look on the Roots tab. That will tell you where the latex files are stored.
a) If it tells you that you have a local TEXMF directory (probably called localtexmf) then create a maple directory inside localtexmf\tex to get localtexmf\tex\maple and stick maplestd2e.sty (and any other files you find you need) in there;
b) if you only have a TEXMF directory (maybe called texmf) then create a maple directory inside that to get \texmf\tex\latex\maple and put the files there.
[b]The next step is very important[/b]. In MiKTeX Options, General tab press the Refresh Now button and that will put the paths into the database.

You will need to repeat the process every time LaTeX tells you it can't find some maple file or other.

Note: For anyone else reading this, if a package is listed in MiKTeX Package Manager (which Maple won't be) then use that to install it - it's so much simpler!

I think Import Latex command is just used for auto-completion in the editor.


Does anyone else here feel that (teenage) robotics $ \rightarrow$ drama? I notice this in all programs I see; FLL, FIRST, etc. Now I know that there will always be drama everywhere, but robotics seems to create a lot more than math for example (I think because people cannot agree on what to do).

Really? You should've seen my first high school math team. Epic power-struggle leaking into interpersonal relations = lots of drama.

Interesting. I notice that drama is everywhere, but present in greater amounts in some groups.

Whoa, actually, I think you're right.

- Robotics kids often work for hours on end, sometimes late at night, etc, and it's really hard to be wrapped up in the robotics the whole time. On most math teams, this isn't the case (I hope).

- Also, in robotics, there's a lot of room for disagreement. In math team, people don't have much do disagree about, usually, unless they're solving a problem, in which case they won't end up with many points.

- Finally, robotics is more visibly competitive than mathematics, I think. In mathematics, people just quietly sit down with their papers and pencils and try to solve as many problems as they can. If they loose, they walk out complaining about how they made so many stupid mistakes, or how the guy in front of them kept farting, or whatever. In robotics, people can complain about what specific other teams did, and stuff, among other things.


:P In fact since every term is a perfect square you don't need the positivity. Second of all, this is identical to the inequality in positives
\[\sum a^{2}(a^{2}+1)\geq 6abc \] which, again, doesn't need the positivity condition.

I didn't know this inequality, and in fact, I just copy/past the exercice, I didn't see we don't need positivity  :)


function
topology
real analysis
real analysis solved

If A is closed, B is compact, and A ∩ B = not 0, prove that there is a d > 0 such that |y - x| >= d for all y in A, x in B.

You clearly mean $A\cap B=\emptyset,$ not zero.

We also might as well assume that the context of this problem is that $A$ and $B$ are subsets of an arbitrary metric space.

Hints: for each $x\not\in A,$ consider the function $f(x)=\inf\{d(x,y): y\in A\}.$

Show that $f(x)>0$ for all $x\not\in A.$

Show that $f$ is continuous.

What do you know about continuous functions on compact sets?

And if we merely had that both $A$ and $B$ were closed, but neither one compact, can you find a counterexample to the conclusion?

[quote="Kent Merryfield"]You clearly mean $A\cap B=\emptyset,$ not zero.

We also might as well assume that the context of this problem is that $A$ and $B$ are subsets of an arbitrary metric space.

Hints: for each $x\not\in A,$ consider the function $f(x)=\min\{d(x,y): y\in A\}.$

Show that $f(x)>0$ for all $x\not\in A.$

Show that $f$ is continuous.

What do you know about continous functions on compact sets?[/quote]

Yes, I mean not 0, sorry.

We know that continous functions are bounded on compact sets, correct? Furthermore, I am not sure how to do this, but how do you show that f(x) > 0 given your definition of f(x) being $=\min\{d(x,y): y\in A\}.$

On your first post: "not 0" is not a synonym for "the empty set", or $\emptyset$.

That distance function should be defined as an $\inf$ (which always works for any set); can you show that this takes on nonzero values when $A$ is closed?

If the space also has the property that any closed and bounded set is compact, we can replace that $\inf$ with a $\min$.

[quote="jmerry"]If the space also has the property that any closed and bounded set is compact, we can replace that $\inf$ with a $\min.$[/quote]
We don't need that condition, as long as $A$ is closed. Fix $x\not\in A.$ Define $g(y)=d(x,y).$ $g$ is a continuous function. Now consider $E=\{m|m=d(x,y)\text{ for some }y\in A\}.$ $E$ is a subset of $[0,\infty).$ $E$ is nonempty as long as $A$ is nonempty. There is some $\delta>0.$ such that $[0,\delta)\cap E=\emptyset.$ And, since $A$ is closed and $g$ is continuous, $E=g^{-1}(A)$ is the inverse image of a closed set, hence closed. Any nonempty closed subset of $[0,\infty)$ has a minimum.

No, $E=g(A)$, not $g^{-1}(A)$. Continuity of $G$ isn't good enough.
Here's a bizarre counterexample: Let $X$ be the integers, with metric $d(m,n)=1+\frac1{|n-m|}$. Then $X$ has the discrete topology, so every set is closed. Every distance is greater than 1, and $d(0,X\setminus 0)=1$.

Oh, yeah. $g(A),$ not $g^{-1}(A).$ So I'll edit above to say $\inf$ instead of $\min.$


factorial
ceiling function
floor function
function
algebra

I'm sure most of you have heard of this game.

Basically, you have to give 4 numbers(all with only one digit) and the next person has to use only those numbers once each to make them into 24 by using the 4 simple operations.

First one:

2,3,4,6

[quote="kheg_k"]there's a 17-page long 24 game [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75881]here[/url][/quote]
You shouldn't revive old topics.

And, exponentiation isn't one of the four basic operations ;)

$7(3+3/7)$

$2, 5, 7, 9$

awww.. I was thinking of 4^2+2^3.

1*2*3*4

how about 4,3,0,0?

[quote="mustafa"]4!+3-2-1

5,2,3,3[/quote]

ahh... 3!(5-3+2)

My last one, if you please.

[quote="lingomaniac88"]Yay, floor function!
$\left(\lfloor\frac{8000001}{1000000}\rfloor-5+\frac{5}{5}\right)!$

24, 5, 6, 7[/quote]

$\left(\lfloor\frac{7-6}{5}\rfloor\right)+24$

1,2,3,6

[quote="nutz_for2.718281828"]you can only use each number once. Here is my answer:

[hide]6*2*(3-1)[/hide][/quote]

i did use each of them once...

[quote="nutz_for2.718281828"]used 1 twice. :wink:[/quote]

no i subracted 1 from 2...that is another 1 i can use at my disposal  :lol:

0,6,9,24
u can only use one multiplication


trigonometry
inequalities
geometry
circumcircle
inradius
trig identities
Law of Sines

A triangle $ABC$ has the sidelengths $a$, $b$, $c$ and the angles $A$, $B$, $C$, where $a$ lies opposite to $A$, where $b$ lies opposite to $B$, and $c$ lies opposite to $C$.
If $a\left(1-2 \cos A\right)+b\left(1-2 \cos B\right)+c\left(1-2 \cos C\right) = 0$, then prove that the triangle $ABC$ is equilateral.

By the law of cosine we have $a(1-2\cos A)=a\left(1-2\cdot \frac{b^2+c^2-a^2}{2bc}\right)=a\cdot \frac{a^2-(b^2-bc+c^2)}{bc}$,

in the same way $b(1-2\cos B)=b\cdot \frac{b^2-(c^2-ca+a^2)}{ca}, c(1-2\cos C)=c\cdot \frac{c^2-(a^2-ab+b^2)}{ab}$.

Since $a,b,c>0$, $a(1-2\cos A)+b(1-2\cos B)+c(1-2\cos C)=0$

$\Longleftrightarrow a\cdot \frac{a^2-(b^2-bc+c^2)}{bc}+b\cdot \frac{b^2-(c^2-ca+a^2)}{ca}+c\cdot \frac{c^2-(a^2-ab+b^2)}{ab}=0$

$\Longleftrightarrow a^2\{a^2-(b^2-bc+c^2)\}+b^2\{b^2-(c^2-ca+a^2)\}+c^2\{c^2-(a^2-ab+b^2)\}=0$
$\Longleftrightarrow a^4+b^4+c^4-2(a^2b^2+b^2c^2+c^2a^2)+abc(a+b+c)=0$
$\Longleftrightarrow a^4-2(b^2+c^2)a^2+(b^2-c^2)^2+abc(a+b+c)=0$
$\Longleftrightarrow (a^2)^2-\{(b+c)^2+(b-c)^2\}a^2+(b+c)^2(b-c)^2+abc(a+b+c)=0$
$\Longleftrightarrow \{a^2-(b+c)^2\}\{a^2-(b-c)^2\}+abc(a+b+c)=0$
$\Longleftrightarrow (a+b+c)(a-b-c)(a+b-c)(a-b+c)+abc(a+b+c)=0$
$\Longleftrightarrow (a+b-c)(b+c-a)(c+a-b)=abc\ (\because\ a+b+c>0)$

Let $x=a+b-c,y=b+c-a,z=c+a-b$, solving for $a,b,c$, we have $a=\frac{z+x}{2},b=\frac{x+y}{2},c=\frac{y+z}{2}$,

using $A.M-G.M.$ inequality, $\frac{z+x}{2}\cdot\frac{x+y}{2}\cdot \frac{y+z}{2}\geq \sqrt{zx}\sqrt{xy}\sqrt{yz}=xyz\ (\because\ x,y,z>0)$

$\Longleftrightarrow abc\geq (a+b-c)(b+c-a)(c+a-b)$

The equality holds when $x=y=z\Longleftrightarrow a=b=c$, which completes the proof.

kunny

I will put a strange solution.
Let $t=\frac{a+b+c}{2}$ be the semiperimeter, E the area, R the circumradius and r the inradius of triangle ABC.
we have $a(1 - 2 \cos A) + b(1 - 2 \cos B) + c(1 - 2 \cos C) = 0$, <=>
$a-2a \cos A + b-2b \cos B + c-2c \cos C=0$ (1)
We know that $a \cos A + b \cos B + c \cos C=\frac{2E}{R}$
So (1)<=> $a+b+c=\frac{4E}{R}$
<=>$2tR=4E$ but we know that $ R\geq 2r$
So $4tr\leq 4E$ and the equality holds when the triangle is equilateral. So we are done because this equation tr=E is very famous.

[quote="silouan"]
We know that $a \cos A + b \cos B + c \cos C=\frac{2E}{R}$
[/quote]

I didn't know that, can you prove this?

Silouan, nice solution, but please don't forget to define your notations (e. g. E is the area, t is the semiperimeter etc.) before somebody asks you to do so. Anyway, I have edited your post to contain these definitions.

_______________________________________________________________________________________________

In order to answer Labiliau's question, the identity $a \cos A + b \cos B + c \cos C=\frac{2E}{R}$ can be proven in various ways. Here is one:

Let O be the circumcenter of triangle ABC. We will consider only the case of an acute-angled triangle ABC; if triangle ABC is right-angled or obtuse-angled, the proof will require some modifications.

Since O is the circumcenter of triangle ABC, we have BO = CO = R, where R is the circumradius of triangle ABC, and < BOC = 2 < BAC = 2A by the central angle theorem.

Since the area of a triangle equals $\frac12$ of the product of two sides, multiplied with the sine of the angle between these sides, the area of triangle BOC must equal

$\frac12\cdot BO\cdot CO\cdot\sin\measuredangle BOC=\frac12\cdot R\cdot R\cdot\sin\left(2A\right)$
$=\frac12R^2\sin\left(2A\right)=\frac12R^2\cdot 2\sin A\cos A=R\sin A\cdot R\cos A$.

But by the extended law of sines, a = 2R sin A, so that $R\sin A=\frac{a}{2}$, and thus the area of triangle BOC equals $\frac{a}{2}\cdot R\cos A$. Similarly, the areas of triangles COA and AOB equal $\frac{b}{2}\cdot R\cos B$ and $\frac{c}{2}\cdot R\cos C$. But the sum of the areas of triangles BOC, COA and AOB is obviously the area E of triangle ABC. Hence,

$\frac{a}{2}\cdot R\cos A+\frac{b}{2}\cdot R\cos B+\frac{c}{2}\cdot R\cos C=E$.

Dividing this equation by $\frac{R}{2}$, we get $a \cos A + b \cos B + c \cos C=\frac{2E}{R}$, and the proof is complete.

_______________________________________________________________________________________________

Finally, here is yet another solution of the initial problem:

We will prove that, for any triangle ABC, the inequality

$a\left(1-2\cos A\right)+b\left(1-2\cos B\right)+c\left(1-2\cos C\right)\geq 0$

holds, with equality if and only if the triangle ABC is equilateral.

In fact, WLOG assume $a\geq b\geq c$. Since in a triangle, the longer side lies opposite to the greater angle, this entails $A\geq B\geq C$ and thus $\cos A\leq\cos B\leq\cos C$, so that $1-2\cos A\geq 1-2\cos B\geq 1-2\cos C$. Now, applying the Chebyshev inequality to the equally ordered number arrays $\left(a;\;b;\;c\right)$ and $\left(1-2\cos A;\;1-2\cos B;\;1-2\cos C\right)$, we see that

$a\left(1-2\cos A\right)+b\left(1-2\cos B\right)+c\left(1-2\cos C\right)\geq\frac13\left(a+b+c\right)\left(\left(1-2\cos A\right)+\left(1-2\cos B\right)+\left(1-2\cos C\right)\right)$.

Now, $a+b+c\geq 0$ is obvious, and also $\left(1-2\cos A\right)+\left(1-2\cos B\right)+\left(1-2\cos C\right)\geq 0$ since this is equivalent to $3-2\left(\cos A+\cos B+\cos C\right)\geq 0$, i. e. to $\frac32\geq\cos A+\cos B+\cos C$, but this is a famous inequality (see, for instance, http://www.mathlinks.ro/Forum/viewtopic.php?t=21188 ). So we get

$a\left(1-2\cos A\right)+b\left(1-2\cos B\right)+c\left(1-2\cos C\right)\geq 0$,

and our inequality is proven. As for equality, it is easily seen to occur if and only if the triangle ABC is equilateral (this is already so for the inequality $\frac32\geq\cos A+\cos B+\cos C$). So the problem is solved.

  darij


real analysis
function
calculus
integration
algebra
polynomial
quadratics

A colleague of mine tried to solve a problem using convolution, and as one of the computations needed to be done was to find the Fourier transform of $ t^2 u(t)$. She did manage to get a result (not sure what it was, though), allegedly only by using $ \delta(x)f(x) \equal{} 0$ when $ f(0) \equal{} 0$. Still, I'm not sure whether this transform even exists - it fails the absolute convergence criterion, but I know it still doesn't mean nothing - yet I'm not still convinced in existence of it in manner of 'standard' functions - what I mean is: wouldn't it involve derivations of Dirac delta - those aren't classical functions, they're distributions?

I hope I was prettymuch clear with the question.

Yes, we're talking about distributions, or to be specific, tempered distributions.

(And for that, I'll move this thread out of "Calculus Computations and Tutorials" - a little too high-level a course.)

The function $ f(t) \equal{} t^2u(t)$ does not belong to $ L^1(\mathbb{R})$ or $ L^2(\mathbb{R}),$ and the integral defining the Fourier transform does not converge even conditionally. But it is a locally integrable function that grows no faster than some polynomial, and that makes it a tempered distribution.

You didn't say where the $ 2\pi$ is in your definition of Fourier transform. For the sake of the argument, let's say that you're using $ \widehat{f}(s) \equal{} \frac1{2\pi}\int_{ \minus{} \infty}^{\infty}f(t)e^{ \minus{} ist}\,dt,$ at least for integrable $ f.$

That gives us $ \widehat{f'}(s) \equal{} is\widehat{f}(s).$

For your particular function $ f,$ we have $ f'''(t) \equal{} 2\delta(t).$ (Of course, that's the weak or distributional derivative.) So $ \widehat{f'''}(s) \equal{} \frac {1}{\pi}.$

So then, $ \widehat{f}(s) \equal{} \minus{} \frac {i}{\pi s^3} \plus{} g(s),$ where $ g(s)$ is some distribution supported on the set $ \{0\}.$

Why did I have to invoke $ g(s)?$ Because knowledge of $ f''',$ which is all that we just used, does not give us complete knowledge of $ f.$ We could add any quadratic function to $ f$ and have the same second derivative; that quadratic function has a Fourier transform that is a linear combination of $ \delta,$ $ \delta',$ and $ \delta''.$

So now we know that $ \widehat{f}(s) \equal{} \minus{} \frac {i}{\pi s^3} \plus{} a\delta(s) \plus{} b\delta'(s) \plus{} c\delta''(s).$

(I'll post this much now and be back with more later.)

At this point, I don't think I'm going to bother to sort this out completely. One thing to note: $ \minus{}\frac{i}{\pi s^3}$ is not a locally integrable function, so we can't just write that and call it a tempered distribution. And we can't even make a principal value make sense. No, this has to be a "regularization" of that function. There's a lot going on at zero.

Thank you very much for this explanation - no sorting out completely is necessary, I just needed to check on my reasoning.


invariant
combinatorics unsolved
combinatorics

In a 4*4 square we distribute the digits: 1,0,1,0 en the first, second and third row(in that order); such that there is only a number in a square, in the last row we distribute 1, 0, 1 in that order in the first 3 squares and the last one is empty. A posible change is: in the empty square we write the number of a neighbor, this has an common side(to our election) and we delete the number of our neighbor. Repeting this process. Show that is not posible have the board: 1,0,1,0 an the first, second, and third row; 1,1,0 en the first 3 squares of the last row.

This is almost identical to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157011 -- a version of the "15 puzzle."  The solution is the same as the traditional solution to that puzzle, since the invariant typically used is just a parity invariant.


La cursul de azi de la Institut, domnul profesor Calin Popescu a propus o aplicatie a teoremei de punct fix a lui Brouwer si anume sa demonstram teorema fundamentala a algebrei. :D

 Astept pareri si solutii in legatura cu aceasta chestiune. ;)

E si asta o idee, dar pana atunci astept si alte pareri si de ce nu, solutii. ;)


geometry
incenter
angle bisector
geometry proposed

Let $ \triangle ABC$ be right and $ AH$ be the altitude to the hypotenuse. Prove that Apollonius circles of $ \angle AHB$ and $ \angle AHC$ intersect at the center of Apollonius circle of $ \angle BAC$.

I am not sure for it is a general definition, anyway:
Definition: Let internal and external angle bisectors of $ \angle AXB$ intersect $ AB$ at $ M$ and $ N$. The circle $ (MXN)$ is called the Apollonius circle of $ \angle AXB$.

Let $ [EH]$ be the internal and $ [DH]$ be the external angle bisector of $ \angle AHB$. $ DH \cap AC \equal{} \{F\}$ and $ HE \cap AC \equal{} \{G\}$. So $ [HG]$ is the external and $ [HF]$ is the internal angle bisector of $ \angle AHC$. Let $ [AY]$ be the external and $ [AX]$ be the internal angle bisector of $ \angle BAC$. Assume $ AB < AC$ and $ \angle BAH \equal{} \alpha$. Let $ O$ be the center of the apollonius circle of $ \angle BAC$. So $ OY \equal{} OA$. $ \angle AYB \equal{} 45^\circ \minus{} \alpha \equal{} \angle OAY \Rightarrow \angle OAE \equal{} \angle EAH \equal{} \alpha$. So $ E$ is the incenter of $ \triangle OAH$. If you look carefully $ AF$ is the external angle bisector of $ \angle OAH$, and $ HF$ is the external angle bisector of $ \angle OHA$. So $ F$ is the excenter of $ \triangle HOA$. So $ O,E,F$ are collinear. $ \angle EOH \equal{} \angle EDH \equal{} 45^\circ \minus{} \alpha$. So $ O,E,H,D$ are concyclic. Similarly $ \angle FOH \equal{} \angle FGH \equal{} 45^\circ \minus{} \alpha$. So $ O,G,H,F$ are concyclic. Thus  $ (HGF) \cap (DEH) \equal{} \{H, O\}$. Also you can get  $ H,O,D$ are collinear.


LaTeX
modular arithmetic
number theory
relatively prime
number theory proposed

find all prime number p that: p^(2^(p-1)-1)=a^k, a k is a positive integer (sorry, i cannot typing latex :( )

I'm sorry because I cannot typing latex. case 1 is right  :D

if case 1 is right $ \forall$ $ p$
$ p^{2^{p \minus{} 1} \minus{} 1} \equal{} a^{k}$ :D

sorry for all.  I do this problem for 2 week :">
this is: p*(2^(p-1)-1)=a^k
find all p.

well, obviously we must consider $ k > 1$.
let's suppose that $ k$ is even then
$ (2^{\frac {p \minus{} 1}{2}} \minus{} 1)(2^{\frac {p \minus{} 1}{2}} \plus{} 1) \equal{} pb^2$.
since the two terms on the left side are relatively prime so we have two cases:
1-$ 2^{\frac {p \minus{} 1}{2}} \minus{} 1 \equal{} x^2,2^{\frac {p \minus{} 1}{2}} \plus{} 1 \equal{} py^2$.
now the rest is trivial since we have $ 2^{\frac {p \minus{} 1}{2}} \equal{} x^2 \plus{} 1$ and by checking modulo $ 4$ we are done.
2-$ 2^{\frac {p \minus{} 1}{2}} \plus{} 1 \equal{} x^2,2^{\frac {p \minus{} 1}{2}} \minus{} 1 \equal{} py^2$.
the rest is trivial also by considering the fact that $ x$ is odd and $ x \minus{} 1$ and $ x \plus{} 1$ must both be a power of two.
so let's assume that $ k > 1$ is odd.
now by doing the same we get two cases:
1-$ 2^{\frac {p \minus{} 1}{2}} \plus{} 1 \equal{} x^k,2^{\frac {p \minus{} 1}{2}} \minus{} 1 \equal{} p^{k\minus{}1}y^k$.
now we know that $ x$ is odd and since $ k$ is odd,too and we have $ 2^{\frac {p \minus{} 1}{2}} \equal{} (x \minus{} 1) \frac {x^k \minus{} 1}{x \minus{} 1}$.
but it's obvious that $ \frac {x^k \minus{} 1}{x \minus{} 1} \equiv 1 \pmod {2}$.
but this expression must be a power of two which implies that it must be one.
or in other words we have $ x^k \minus{} 1 \equal{} x \minus{} 1 \implies k \equal{} 1$.
so let's check the second case and finish it...
2-$ 2^{\frac {p \minus{} 1}{2}} \minus{} 1 \equal{} x^k,2^{\frac {p \minus{} 1}{2}} \plus{} 1 \equal{} p^{k\minus{}1}y^k$.
this case is done in the same way
note that $ x$ is odd and $ k$ is odd ,too.
so the expression $ \frac {x^k \plus{} 1}{x \plus{} 1}$ which divides $ 2^{\frac {p \minus{} 1}{2}}$ is odd.
which implies that it must be one or in other words,
$ x^k \plus{} 1 \equal{} x \plus{} 1 \implies k \equal{} 1$.
so we r done!

$ \left\{\begin{array}{l}p^{2^{p\minus{}1}\minus{}1}\equal{}a^k\\ p\minus{}\mbox{ prime number}\end{array}\right.\Rightarrow a\equal{}p\Rightarrow p^{2^{p\minus{}1}\minus{}1}\equal{}p^k\Rightarrow 2^{p\minus{}1}\minus{}1\equal{}k$


Which of the following statements is correct?

A. The internal energy of a system increases when more work is done by the system than heat was flowing into the system.

B. The internal energy of a system decreases when work is done on the system and heat is flowing into the system.

C. The system does work on the surroudings when an ideal gas expands against a constant external pressure.

D. All statements are true.

E. All statements are false.

Could someone please help me out with this question? I have no clue what the answer is.

Recall that internal energy is just the amount of work that a system can do.  If a system loses energy, then its internal energy decreases, and vice-versa.

A. The internal energy of a system increases when more work is done by the system than heat was flowing into the system.

In this case, the system overall lost more energy than it gained (doing work = giving away energy).  Hence this one is false.

B. The internal energy of a system decreases when work is done on the system and heat is flowing into the system.

Both transfer mechanisms in this one put energy in the system.  Hence it is also false

C. The system does work on the surroudings when an ideal gas expands against a constant external pressure.

Think of the surroundings as a new system.  This new system's volume is being decreased, and (assuming P$\neq$0) you can think of it as being forcibly pushed together.  Hence, work is done on this new system in this case, and this answer is correct.

$W_{\mbox{done by the system}} = P \delta V$
where,
$P$ is the pressure of the gas and
$\delta V$ is the change in volume.

When a system expands, $\delta V$ is positive. So, work is done by the system when the ideal gas expands. (C)


MATHCOUNTS
LaTeX
trigonometry
geometry
3D geometry
Pascal\u0027s Triangle
absolute value

I thought I'd make a topic with a new formula, theorom etc. posted daily that would be of use in MathCounts competitions (chapter level and up.)I'll probably forget about this thread a lot so feel free to post your thingy (s). :D

Counting Combination:$\frac{n!}{r!(n-r)!}$ is the number of combinations of r things out of a total of n things.

I may sound like a broken record, but may I ask that you add a little teeny bit of explanation why it's true and maybe an example of it in action?

Nice reasoning, biffandoc. :) Sadly, I could never think of my own proof...
Well sorry I forgot about yesterday. Here's some interesting ones:
1)$(B^2+C^2/2)-(A^2/4)=D^2$
2)Distance formula between two points $(x,y)$ and $(a,b)$: $sqrt(x-a)^2+(y-b)^2$. It's just based on the Pythagorean Theorom.
3)Relationship between $(x+y)^n$ and Pascal's Triangle:
$(x+y)^2=x^2+2xy+y^2$;
$(x+y)^3=x^3+3x^2y+3xy^2+y^3$; and
$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$. Basically, the coefficients of $(x+y)^n$ are just the numbers in the $nth$ row of Pascal's Triangle. That would have saved me ages on this one MathCounts problem I was working on a year ago when I didn't know that. :D Whew, that took awhile to type.

Guess the Distance Formula's already been posted under formulas so here's another cool one:
The distance d between a point $(x1,y1)$ and a line $Ax+By=C$ is $d = |(Ax1 + By1+C)|/sqrt(A^2 + B^2)$.

Just found out that you can use the counting combination in a gridded nxn square when you need to get across from one endpoint of a diagonal to the other endpoint within m moves. :lol:
Sorry for double-posting, by the way.

So that formula for distance between point is
point: $(x_1,y_1)$
line: $Ax+By=C$
$\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$ :?:
I just like my pretty $\LaTeX$ better, its easier to understand
and what does the first formula mean?

The attachment was supposed to go with the first formula. Maybe I should've mentioned that. :blush:

Do you have any idea why these work/proof?

Except distance formula, that's obvious

For the triangle one, it has to do with cosines or something, I'm going to perfect my proof and post it up.
For $(x+y)^n$, that's just based on the binomial theorom.
Here's a proof for the distance between a point and line (finally, I have a decent explanation):
The distance from the point (x1, y1) to the line Ax + By = C is also the distance between the point (0, 0) and the line A(x - x1) + B(y - y1) = C.

This is true because we can translate the line and the point x1 in the negative x-direction and y1 in the negative y-direction.

We have that Ax + By = C + Ax1 + By1.

Note that the origin, the x-intercept of the new line, and the y-intercept form a right triangle such that the lengths of the legs are the absolute values of the intercepts.

Let the distance from the origin to the line be d and the intercepts be x0 and y0. We know that

x0 * y0 = d * √(x0² + y0²).

We can find the intercepts by inserting x = 0 and y = 0 into the line equation.  We have that

x0 = (C + Ax1 + By1)/A and y0 = (C + Ax1 + By1)/B.

Putting these values in, we get

(C + Ax1 + By1)²/AB = d (C + Ax1 + By1)√(1/A² + 1/B²), or

(C + Ax1 + By1) = d√(A² + B²).

So d = |(C + Ax1 + By1)|/√(A² + B²). The absolute value sign is needed as the distance can't be negative. Hope its not to hard to read without tex.

Proof for the triangle thing:(see attachment)
E makes forms the leg of two right triangles, so by the Pythagorean Theorom, $E^2+(A/2+x)^2=C^2$ and $E^2+(A/2-x)^2=B^2.$ Add both of these equations together to get $B^2+C^2=2E^2+2x^2+A^2/2.$ As $E^2+x^2=D^2, B^2+C^2=2D^2+A^2/2$ by substitution. Subtract $A^2/2$ from both sides and divide both sides by two to get $B^2+C^2/2-A^2/4=D^2$.
Yay! I did it without any trig functions.

[quote="biffanddoc"]That one is probably obvious. Let me give an explanation anyway.
First thing: How many ways can you arrange 7 items in a line?
For the first item, we have 7 choices. Then for the second, we have 6. And all of the way down to one. So the answer is 7!.  The formula is $n!$ for $n$ items.
Second thing: What if we chose only 3 items out of the 7 and arranged them in a line?
We can choose any of the 7 items first, then any of the remaining 6, then any of the remaining 5. So the answer is 7*6*5. But how to put that as a formula?
Notice the difference between this and the previous one. On this one, we don't include 4*3*2*1. We can say that we divide the 7! by 4! to get our answer. Thus the formula is $\frac{n!}{(n-r)!}$, if we take the items $r$ at a time.
Third thing: What if we don't care about the order? Say the items are A-G. Before, ABC, ACB, BAC, BCA, CAB, and CBA were considered different. But since they are comprised of the same items, they are really the same, if order doesn't matter. So we overcounted by a factor of $3!$, or $r!$. So we must divide by $r!$. The final formula is $\frac{n!}{(n-r)!r!}$.

Example: How many ways can we choose a set of 4 items from 8 if order doesn't matter?
Solution: It is a combination. $n$ is 8 and $r$ is 4.
Plug into the formula. We get
$\frac{8!}{4!4!}$
You can find the answer here by removing common factors and simplifying.[/quote]

I always get confused on where I should use permutation and where I should use combination.  :blush:

[quote="kidwithshirt"]Heros formula...Triangle area=sqrt[s(sa)(sb)(sc)] where s=(a+b+c)/2[/quote]Actually it's $A=\sqrt{s(s-a)(s-b)(s-c)}$...

In combinations, order doesn't matter. In permutations it does. For example, in permutations 1,0,1 and 1,1,0 would be different. In combinations they are the same.

In heron's formula, a, b, and c are the sides of the triangle. s is the semiperimeter.

basically when they say "group" or "subcommittee" or something like that it is combination.
Otherwise it is a permutation.

If you are unsure, trust your gut whether order matters or not

Sometimes they say to chose something or have X many choices.

To find the number of factors of a number, take the prime factorization, add one to all the exponents of the bases (in the factorization), and multiply.
This method can be adapted to find the number of perfect powers that are factors of the numbers (i.e. number of perfect squares, cubes, etc.) Instead of just adding one to the exponents, find the number of perfect powers in the exponent, add one, then multiply.

There are $(n-1)!$ ways to arrange $n$ objects in a circle.

the 50th digit of pi is 0

just for fun

the 9th decimal digit of root 2 is 2
the 9th decimal digit of root 3 is 8
the 7th decimal digit of e is 8

Sum of first n positive integers: $\frac{n(n+1)} {2}$

Sum of first n squares: $\frac{n(n+1)(2n+1)} {6}$

Sum of first n cubes: $(\frac{n(n+1)} {2})^2$

They've been posted a million times, but in case you need a refresher.

[color=blue]Moderator kyyuanmathcount's edit: You mean positive integers for the first one right?[/color]

EDIT: Yes. Changed it.

[quote="moogra"][quote="JesusFreak197"]You can be more specific than that...

$\pi =3.1415926535897932384626...$
$\sqrt{2}=1.4142135623731...$
$\sqrt{3}=1.7320508075689...$
$e=2.718281828459...$

Although, you don't need to know $e$ for MathCounts, you never need to know $\pi$ past $3.14159$, and probably not even past $3.14$, and you don't need to know $\sqrt{2}$ past $1.414$ and $\sqrt{3}$ past $1.732$, if you even need to know those at all.[/quote]

memorizing pi is fun. once they did ask a pi question (15th digit of pi) in countdown. I know like a little more than average (83)[/quote]

 :what?: What's the average pi memorization??I know like maybe two digits.

Wasn't it sqrt(2)? I remember that problem.

Anyways, its nice to know some of those squareroots.

nah, it was $\sqrt 3$.
Plus the tests that old stresses arithmetic a lot.
One 1992, number 30 was ridiculously easy to  know what to do, but ridiculously hard to do the computation

I noticed that, especially in the handbooks.

You'd be doing brutal calculations on warmup 4, and everything would take you twice as long.

to get back on topic: helpful random thingy of the day
How to do problems like
$(a+b)^2=10$
$ab=3$
What is $a^2+b^2$?

Solution:
Expand the product $(a+b)^2=10$
You get $a^2+2ab+b^2=3$
We know that $ab=3$, so put that into the equation.
Now we have
$a^2+2(3)+b^2=10$
Subtract 6 from both sides to get our desired result.
$a^2+b^2=4$

Now problems like
$(a+b)^3=125$
$a^3+b^3=25$
What is $ab$?\
Solution:
First let's expand $(a+b)^3=125$
We get
$a^3+3a^2b+3ab^2+b^3=125$
Now, factor the middle two terms to get
$a^3+3ab(a+b)+b^3=125$
We can substitute in $a^3+b^3$ and subtract it from both sides to get
$3ab(a+b)=100$
To find out $(a+b)$, we can take the cube root of $(a+b)^3=125$ to get $a+b=5$
Substitute that into the equation to get
$3ab(5)=100$
Divide by 15 to get
$ab=100/15=20/3$

[quote="biffanddoc"]For $\pi$ and square roots, know at least up to 10 first 3 digits.
I ran into a question ON SPRINT ROUND NATIONALS that you needed to evaluate $\pi$ and $\sqrt{3}$ to find the answer which had to be to the nearest hundredth.[/quote]

Omg I did that round just this Wednesday! (That is 2/15/06)

Yes, this is like a formula topic, but nowing like 3 digits of pi and the first 10 roots is a helpful thing in some cases.

As is what I wrote about $a$ and $b$, which isn't a formula

[quote="tz"]Isn't this fourm just like the FORMULAS fourm?[/quote]
I wouldn't say that the forum's like the formulas one, you can post anything you think would be helpful, which isn't just formulas. Getting back to the topic:
An easy way to find a number squared is to first make it a difference of squares, where the product of the difference and sum of the two numbers are easy to calculate. Then, add the number squared that you subtracted from the original number back in.
Ex.:$75^2$ Make this $75^2-25^2$, which can be factored as $(75+25)(75-25). 100\times50$ is easily do able in your head and you just add $25^2$ back in to get $5625$. This method might be of help in Countdown.

i know a problem where you need to know $\sqrt{2}$ is juuuust more than 1.4

The diaz bible has the sq. roots and other important things.

This is ridiculously important, especially for countdown.

Memorize all of the combinations up to 10
$\binom 11=1$
$\binom 21=2 \binom 22=1$
$\binom 31=3 \binom 32=3 \binom 33=1$
$\binom 41=4 \binom 42=6 \binom 43=4 \binom 44=1$
$\binom 51=5 \binom 52=10 \binom 53=10 \binom 54=5 \binom 55=1$
$\binom 61=6 \binom 62=15 \binom 63=20 \binom 64=15 \binom 65=6 \binom 66=1$
$\binom 71=7 \binom 72=21 \binom 73=35 \binom 74=35 \binom 75=21 \binom 76=7 \binom 77=1$
$\binom 81=8 \binom 82=28 \binom 83=56 \binom 84=70 \binom 85=56 \binom 86=28 \binom 87=8 \binom 88=1$
$\binom 91=9 \binom 92=36 \binom 93=84 \binom 94=126 \binom 95=126 \binom 96=84 \binom 97=36 \binom 98=9 \binom 99=1$
$\binom{10}{1}=10 \binom{10}{2}=45 \binom{10}{3}=120 \binom{10}{4}=210 \binom{10}{5}=252 \binom{10}{6}=210 \binom{10}{7}=120 \binom{10}{8}=45 \binom{10}{9}=10 \binom{10}{10}=1$

I guarantee they will be useful on at least one countdown problem

I don't think all of that is nessessary to know(unless you have no life) because one problem is not going to change the world.  Just the ones up to 5 or 6 is fine.

[quote="biffanddoc"]Well, some of us have no life :)

And plus, look at the WINNING question of last years nationals.
$3 \cdot \binom 85$ is the easy part
If you had to evaluare it, you would have to say $8*7*6 \div 6=56$ versus just knowing it.[/quote]

Just cancel out the two 6's and you get 8 * 7 = 56. That isn't so hard  :| .

evaluating all of these is not that hard.
but at countdown at nats, it is so fast, that if you have to compute you will lose

A lot of it is intuitive, but if you're good at intuitive stuff, it's extremely easy. And don't bother actively memorizing them. Do problems involving them, and you'll memorize them over time. Also, remember that the first and last numbers of row $n$ are $1$, the second and second-to-last are $n$, and the third and third-to-last are the $(n-1)^{th}$ triangular number.


Posted in Unsolved Section:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=85236
and Proposed & Own Section:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81889


Find the area of the region where the two circles overlap.

[quote="ch1n353ch3s54a1l"][hide]If you draw triangles of length 5-5-2 and 6-6-2 and then find teh angles of the arcs, then finding the areas of the arcs, subtracting the area of the triangles from the corresponding arcs, and doing that for each triangle, add the lengths up, then you should get your answer...

I got $\approx .23089$[/hide][/quote]

Can you express that in terms of pi?

The reason I did not want to do the problem was because the numbers were so nasty. No one knows $\tan^{-1}\frac{2}{5}$ or anything like that....

[quote="José"]True. There MUST be a nice and elegant solution for this kind of problems. I mean, a solution without using trigonometry.[/quote]
No, actually, in most cases there isn't. Unless the numbers turn out really nicely (such as a 3-4-5 right triangle), we usually have to resort to trigonometry to solve these kinds of problems.

[quote="José"]True. There MUST be a nice and elegant solution for this kind of problems. I mean, a solution without using trigonometry.[/quote]

WHAT IN THE WORLD?  I've seen very few real trig problems that don't use trig...

After doing so many problems with elegant answers, its ok to put a problem in with a not so elegant answer. It may throw some solvers off track. :P

[quote="José"]True. There MUST be a nice and elegant solution for this kind of problems. I mean, a solution without using trigonometry.[/quote]

josé needs to be like an engineer or something when he grows up

"guys are you sure this isn't a 5-12-13"

No, we just use the radii and the length 2 to get 5-5-2 for the top circle and 6-6-2 for the bottom.

I don't see how the problem really requires trigonometry. The closest thing to trig in there is taking $\tan^{-1}$ of something.

Its just that the problem requires a calculator or else you have to do LOTS of simplifying.

[quote="ShoeFactory"]The closest thing to trig in there is taking $\tan^{-1}$ of something.[/quote]

If taking an inverse tangent isn't trigonometry, what is it?

taking inverse tangent = common sense  :wink:

[quote="José"]True. There MUST be a nice and elegant solution for this kind of problems. I mean, a solution without using trigonometry.[/quote]

I doubt there is a non-trig solution more elegant than the trig solution posted previously. There is no other way to find an angle in a triangle given its sides, which was a condition that had to be found, except with the law of cosines.

[quote="José"]We're funny in this forum. Once I made a problem and I was critisiced because I "killed" the problem using non-elegant solutions. Now, you use trigonometry and the answer is an ugly expression...[/quote]
Yeah, it's my first time :lol:
I thought it was a good problem, still do.

i guess if u want it pretty you could use infinite sums...., maybe prettier

[quote="ritchjp"][hide]

Let $A$ be the center of the circle of radius $6$ and let $B$ be the center of the circle of radius $5$. Points $C$ and $D$ will be the end points of the common chord. Point $E$ will intercept the circle $A$ and point $F$ will intercept the circle $B$. Point $G$ is where $AB$ intersects chord $CD$.

The thing we need to do now is look at the right $\triangle ACG$ and the sector $ACE$ in circle $A$. Likewise, look at the right $\triangle BCG$ and the sector $BCF$ in circle $B$.

Next we will call $\angle CAG$ by the name $\theta$ and call $\angle CBG$ by the name $\phi$.

We now need to find the area of sectors $CAE$ and $CBF$.
We also need to find the area of right $\triangle CAG$ and right $\triangle CBG$.

Let's first find the area of sector $CAE$ and $\triangle CAD$.

$AG = \sqrt{35}$ so $\tan \theta = \frac{1}{\sqrt{35}}$ and $\theta = \tan^{-1}\frac{1}{\sqrt{35}}$

We can now find the area of sector $CAE$, but the angle must be in radians to do that. So far, we have the angle expressed in degrees. We then multiply the angle by $\frac{\pi}{180}$

$\theta = \tan^{-1}\frac{1}{\sqrt{35}}(\frac{\pi}{180})$

The area of circle $A$ is $36 \pi$

Now we have the proportion $\frac{36 \pi}{2 \pi}= \frac{CAE}{\tan^{-1}\frac{1}{\sqrt{35}}(\frac{\pi}{180})}$

the area of sector $CAE$ is $3.0141$. The area of sector $CAD$ is $6.0281$

The area of $\triangle CAD = \frac{1}{2}(2)(\sqrt{35}) = 5.9161$

The area of segment $CED$ is $6.0281-5.9161$ or $.112$

Now we find the areas of sectors $CBF$ and $\triangle CBD$

$BG = \sqrt{24}$ so $\tan \phi = \frac{1}{\sqrt{24}}$ and $\phi = \tan^{-1}\frac{1}{\sqrt{24}}$

$\phi = \tan^{-1}\frac{1}{\sqrt{24}}(\frac{\pi}{180})$

The area of circle $B$ is $25 \pi$

Now we have the proportion $\frac{25 \pi}{2 \pi}= \frac{CBF}{\tan^{-1}\frac{1}{\sqrt{24}}(\frac{\pi}{180})}$

the area of sector $CBF$ is $2.5169$. The area of sector $CBD$ is $5.0340$

The area of $\triangle CBD = \frac{1}{2}(2)(\sqrt{24}) = 4.8990$

The area of segment $CFD$ is $5.0340-4.8990$ or $.135$

so the area of the region where the two circles overlap is $.112+.135$ or $.247$

[/hide][/quote]
[quote="ch1n353ch3s54a1l"]If you draw triangles of length 5-5-2 and 6-6-2 and then find teh angles of the arcs, then finding the areas of the arcs, subtracting the area of the triangles from the corresponding arcs, and doing that for each triangle, add the lengths up, then you should get your answer...
I got $\approx .23089$[/quote]
Your solution is very nice ritchjp, but i think you aproximated in many of your steps. ch1n353ch3s54a1l's solution is similar to your's. :)


algebra
polynomial
function
algebra theorems

Do we have a suffice or necessary condition on the coefficients of a real polynomail with the degree $2n$ such that $p(x)\ge 0$
for all real numbers.

I think that $p$ being a sum of squares of polynomials is both necessary and sufficient.

That's not a condition on the coefficients indeed...

[quote="lomos_lupin"]A sufficent and  necessary condition on  a real polynomail with the degree $2n$ such that $p(x)\ge 0$  , $\forall x \in {\mathbb R} .$[/quote]
Indeed as [b] blahblahblah[/b] remark !
Consider $P(x) =\sum\limits_{k=0}^{2n}a_kx^{n-k}$ such that $P(x)\ge 0$ , $\forall x\in {\mathbb R} .$ Moreover let us assume  that \[ P(x_k)=0\hsp ,\hsp x_k=\left\{\begin{array}{lcl} \alpha_k+i\beta_k\; &,&\; k\in \{1,2,...,n\}\\ \alpha_{n+j}-i\beta_{n+j}\; &,&\; j\in \{1,2,...,n\} \end{array}\right.  \]
with real $\alpha_\nu,\beta_{\nu}$ . Therefore  $P(x)=a_0\prod\limits_{k=0}^{2n}(x-x_k)= a_0\prod\limits_{k=0}^{n}\left((x-\alpha_k)^2+\beta_k^2\right) \; \; \; \mbox{with}\; \; a_0>0\; .$ Because  $(A^2+B^2)(C^2+D^2)=(AC+BD)^2+(AD-BC)^2$ it follows  :

[b]Theorem[/b] [i]Every real polynomial function $P(x)$ , of degree $2n$ ,which is non-negative on the real axis ${\mathbb R}$ can be expressed  as [/i]
\[ \begin{array}{|c|} \hline \\ P(x)=A^2(x)+B^2(x)\\ \\ \hline \end{array}\; ,  \]
[i] where [/i] [b]degree[/b]$(A)\le n$,       [b]degree[/b]$(B)\le n\; .$

Well i find this condition which is suffice for a polynomail to be positive for all real numbers.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63239

[quote="behzad"] [i] Let $P(x) = a_nx^n + \ldots + a_1x + a_0$ be a polynomial with real coeficients,  where $n =$ [b]deg[/b]$P$ is even. Suppose that [/i]
[b](i)[/b] $\displaystyle a_0 > 0, a_n > 0$.
[b](ii)[/b] $\displaystyle {a_1}^2+...+{a_{n-1}}^2\ge \displaystyle\frac{4\min({a_0}^2,{a_n}^2)}{n-1}$.
[i]Prove that $P(x) \ge0$ for all real $x$  [/i]
[/quote]
It's  only a [b]sufficient condition[/b] . However nice and of interest .


abstract algebra
superior algebra
superior algebra unsolved

Prove that the kernel of any homomorphism from $C^{*}$ to $(R,+)$ is infinite group.

Isn't this trivial? :? The elements of $\mathbb C^*$ with finite order must be mapped onto $0$ by any homomorphism $: \mathbb C^*\to\mathbb R$.


Please find the sum of factoring of $15x^2+4xy-35y^2+68x+6y+77$

I don't quite understand what you mean by "the sum of factoring"...

We can factorize the given expression as follows:
[hide="Factorization"]
$15x^2+(4y+68)x+(-35y^2+6y+77)$

$=15x^2+(4y+68)x+(5y+7)(-7y+11)$

$=(3x+5y+7)(5x-7y+11)$[/hide]

So the answer is...[hide]$(3x+5y+7)+(5x-7y+11)=8x-2y+18$? (If I understood what you meant correctly...)[/hide]


I haven't received any mail about the USAIO since the initial invitation. Since it's about a week and a half away I am getting worried.

My school's mail server was down for a couple of days, so if anything has been sent out, this could be why I haven't got it. Otherwise, does anyone know what is on Rob's mind?

Thanks

i can do nothing but just hope that u get every thing al right.

anyhow, a request to not only u but also to those who r going to participate at usaco camp, can u plz give some of the materials here at this forum after/during camp?

i would be glad to have those.

I don't think he's emailed anything out after the initial invitation... (though I've talked to him about my flights).  I guess...all will be revealed in due time?

I emailed Kolstad about the camp information last week and he said that he would send out more information in a week or so.  So we should get some more info soon, probably this week.

Thanks for the replies;

yeah, they have not been sent yet, I guess we will be waiting a little longer.

[quote="mahbub"]but no one has said that they will share the materils here! won't u? :([/quote]

camp materials are not really readily "shareable". They have practice contests at camp, but the problems and test cases may be re-used, so they are not released to the participants on a CD or in a book (although this year could be a little different because the Canadian team and a few other internationals are also attending).

If you have not yet completed the USACO training pages, that's probably where you should be concentrating your efforts. The people who attend camp all started with the training pages. Also plan to enter the internest contest being run parallel with the CEOI in July.

is CEOI in online?

i regularly participate in online contest wherevenr it occurs.

i am at 4.5 at usaco training page.

actually which online contest were u talking about? can u given me the URL?

i can't find any info on online version of ceoi2005. when will it occur? if in mid july during imo then  :(

Yea, it's in mid-july.

Does anyone know what language's you can use?

[quote="SnowStorm"]What is CEOI?[/quote]
CEOI = Central European Olympiad in Informatics

See [url=http://ceoi.inf.elte.hu/]http://ceoi.inf.elte.hu/[/url]

No it is not on mid-july. As far i can see it will be on 28 july to 4august. then it must be after IMO. So i must not worry a :lol: bout it!

Check your email now, Hooyoung and kmod.  Bunch of forms and information... finally we got something from Kolstad !!  :)

[quote="Hooyoung"]The packing list alleges that there are three forms, but there are only two.[/quote]

Hmm...  but the email says there are only two.  Maybe the waiver for rafting counts as another one?  :?:


A survey of 100 high school students revealed the following data: 28 eat chicken, 30 eat beef, 20 eat pork.  These numbers include 12 students who eat chicken and beef, 7 who eat chicken and pork, 5 who eat beef and pork, and 3 who eat all three.  How many non-meat eaters are in the sample?

(A)39         (B)40        (C)41             (D)42                  (E)43

Okay, both of your answers are correct.  Good Job.

[hide]From cherry PIE, 28+30+20-12-7-5+3=57 eat meat. So 43 do not eat meat. E.[/hide]


algebra
polynomial
Vieta
Euler
inequalities unsolved
inequalities

a,b are arbitraty positive numbers such that the equation
                            x^3-ax^2+bx-a=0
has three roots greater than 1.Determine a,b so that the expression:
                            ( b^n-3^n)/a^n
 attains its least value and find this value,n is integer

because of vieta, one finds that if p,q,r are the roots of the ecuation. p+q+r=pqr so it ould bnie o try eping p=tanP q=tanQ r=tanR and one gets that P+Q+R=k* \pi  . I dont know if it may help. :)

You shouldn't post that problem, euler.
  It is one of the problems in the latest M and Y magazine.

Troi oi, Euler choi sai luat nha!
Cam on thay Treegoner da nhac nho

Maybe the Mods should consider locking this topic, according to [b]treegoner[/b].