inequalities
induction
combinatorics solved
combinatorics
graph theory

In a group of kn persons, each person knows more than (k - 1)n others.
(Here k and n are positive integers.)
Prove that one can choose k + 1 persons from this group,
such that every chosen person knows all the others chosen.

(A knows B if and only if B knows A, as usual)

Myth, I'm wondering ... Is this really that easy for you ?

indeed is not such a simple induction. This is actually a particular case of [b]Zarankiewcz's[/b] theorem:

[i] If a graph G does not contain any k-complete sub-graphs, then the set of its vertices' degrees verifies the following inequality

min { d(x) | x\in G } \leq [ n(k-2) / (k-1) ]

where by [x] we have denoted the integer part of the real x.[/i]

One can see that the problem above is an application of this theorem for  n -> kn , k -> k+1.

The solution I know is based on using the cardinal set inclusion and exclusion theorem.

Is it the same with yours, Myth?

Nope I didn't solve it ... :( I'm not so strong you know :(

And I almost don't know what a graph is ...
Yeah of course I heard of it but I never used it.
And we didn't learn about graphs in our IMO preparation session.

So Myth, what is your simple induction ?

If each persons knows greater than (k-1)n+1 other persons then we delete some acquaintances.

Let P be person with (k-1)n+1 acquaintances. Let M be a set of all friends of P, |M|=(k-1)n+1. Each person in M has in M at least ((k-1)n+1)-(n-1)=(k-2)n+2 friends. Now we delete arbitrary element of M ==> |M|=(k-1)n and each person in M has at least (k-2)n+1 friends in M ==> there are k persons in M who know each other ==> these k persons + person P know each other.

Am I right?


search
number theory unsolved
number theory

find all $ x,y \in R$ such that:
$ (x^2\minus{}x)(x^2\minus{}2x\plus{}2)\equal{}y^2\minus{}1$

I thin we should search for solution ins $ \mathbb{N}$ or $ \mathbb{Z}$ but not in $ \mathbb{R}$. However wherever you look either in $ \mathbb{R}$ or in $ \mathbb{N}$ or $ \mathbb{Z}$ solving towards $ x$ we get $ x \equal{} \{\frac {1 \minus{} \sqrt {4y \plus{} 5}}{2},\frac {1 \plus{} \sqrt {4y \plus{} 5}}{2},\frac {1 \minus{} \sqrt {5 \minus{} 4y}}{2},\frac {1 \plus{} \sqrt {5 \minus{} 4y}}{2}\}$. From these solutiom we can make infinitely many solutions in every set.

why $ x\equal{}{\frac{1\minus{}sqrt{4y\plus{}5}}{2},...}$?
Could you discribe?


trigonometry calculus integration algebra polynomial AMC

[quote="bpms"]Is there anyway to "simplify" the following:
$\sin^{-1}\cos(a)$
$\cos^{-1}\sin(a)$[/quote]

well $\cos a=\sin \frac{\pi}{2}-a$, so your expression both equal to $\frac{\pi}{2}-a$

[quote="bpms"]Is there anyway to "simplify" $\sin(x) \sin(y)$.
To splash D, I realized that moments after posting so I changed it.[/quote]
$\sin(x) \sin(y)=\frac{\cos(x-y)-\cos(x+y)}{2}$
$\cos(x)\cos(y)=\frac{\cos(x-y)+\cos(x+y)}{2}$
$\sin(x)\cos(y)=\frac{\sin(x+y)+\sin(x-y)}{2}$

[quote="bpms"]How about $\sin(\cos^{-1}x)$? I got this far.
[hide]$\cos(90-\cos^{-1}x)$
$\sin(90)\sin(\cos^{-1}x)+x\cos{90}$
$\sin(\cos^{-1}x)$
$\sin(90-\cos^{-1}x)$
$\sin(90)x-\cos(90)\sin(\cos^{-1}x)$
$x$[/hide]
Which is obviously not true... I made a mistake somewhere... Could someone point it out...[/quote]

dude draw a triangle

[quote="bpms"]How about $\sin(\cos^{-1}x)$? I got this far.
[hide]$\cos(90-\cos^{-1}x)$
$\sin(90)\sin(\cos^{-1}x)+x\cos{90}$
$\sin(\cos^{-1}x)$
$\sin(90-\cos^{-1}x)$
$\sin(90)x-\cos(90)\sin(\cos^{-1}x)$
$x$[/hide]
Which is obviously not true... I made a mistake somewhere... Could someone point it out...[/quote]
You just said $\sin(\cos^{-1}x)=\cos(90-\cos^{-1}x)$ and then $\sin(\cos^{-1}x)=\sin(90-\cos^{-1}x)$. The first one is correct, but that doesn't help.
From Iversonfan's hint: [hide] $\sin(\cos^{-1}x)=\sqrt{1-x^{2}}$ [/hide]

[quote="bpms"]What about $\sin(xy)$?[/quote]
That's unlikely to have a simpler form. Consider the trouble of getting a formula for integral $x$: $\sin(xy)=\sqrt{S_{x}(\sin^{2}y)}$, where $S_{x}$ is the x-th degree spread polynomial described by the explicit formula $S_{x}(n) = n\sum_{k=0}^{x-1}{x \over x-k}{2x-1-k \choose k}(-4n)^{x-1-k}$.
(all found on Wikipedia :D)

you could use complex numbers for sin(xy)

More specifically, $\sin(xy) = \Im{(\cos x+i \sin x)^{y}}.$

Then do some complicated work involving non-integer factorials.


geometry
parallelogram
ratio
cyclic quadrilateral

When doing certain geometry problems, mainly the olympiad ones, what tactics do you guys use?

When you think you have to join lines up (as you usually do with olympiad geometry) to prove and relate certain things, how do you know where to join, produce or connects lines?

When you start an olympiad geometry problem, do you just look at and think about possible ways? Or do you just consider a way to do it and rush into it? Do you think about what lines you could draw to help before you start?

--I really want to know how you guys think as i want to be able to do that. When i start problems, its usually train wreaks, i will randomly connect lines and hopes it works. I also think of one way to start it and just stick with it. Leonardo Da Vinci believed that the first way he looked at a problem was to bias.

So how can i think like the way yous guys think? What did yous do?

I would really appreciate any input!
-Kurt

This is someone that's probably as equally stunted in geometry as you are, but I have heard some things from people that are very good at geometry (sometimes in a failed attempt to help me get better :D ), so hopefully this will help.

Most of the time when these extra lines or points are constructed, it's not a random process. There's a goal in mind; maybe making that extra line gives you a parallelogram, or that extra point creates a cyclic quadrilateral, or you want a right angle.

In the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1247506#1247506]only geometry problem I ever solved in contest time[/url], I recall needing to prove something about ratios of sides as my first step. It seemed intuitively obvious to me from all of the parallel lines, but the only way I found to rigorously do it was to extend those parallel lines to form several parallelograms and use a whole bunch of equal sides. From this experience, my guess would be that people good at geometry have an idea of what they're trying to do when they make these extra constructions, the same way I knew exactly what I was going for.

The way you would get to the point of noticing what extra things can be made by extending lines of stuff, I presume, is by doing lots of problems and thus finding what patterns and situations give an opportunity for constructing them.

Thanks, thats really good info.

Anyone else?


Let $p$ be an odd prime congruent to 2 modulo 3. Prove that at most $p-1$ members of the set $\{m^2 - n^3 - 1 \mid 0 < m,\ n < p\}$ are divisible by $p$.

The equation $x^3 \equiv a \mod p$ has exactly one solution for given $a$.
Proof: Take some $s,t$ with $3s-(p-1)t=1$, since this shows $(x^3)^s \equiv x \cdot x^{t\cdot (p-1)} \equiv x \mod p$, thus $x \to x^3$ is injective, thus surjective.
Now for any $m$, there is exactly one $n$ when $n=0$ is also allowed, thus at most $p-1$ solutions (indeed there are $p-3$ of them for $p \geq 5$: $m \equiv \pm 1 \mod p \iff n\equiv 0 \mod p$ fall out, too).

Sorry for reviving the old topic ,

I feel ,this problem can be generalized for $m^{j} - n^{3} - k $  where $k$ and $j$ are fixed constants . :maybe: , instead of just  $m^{2} - n^{3} - 1 $ :maybe:

Isn't it true that there are exactly $p-1$ members of the set?

Actually, this is weird. Given that this is an old problem, and (supposedly) known to Eastern European community, I would not expect it, but it seems, there is a typo, and it is asking to prove that such a set has at most $\mathbf{p}$, rather than $\mathbf{p-1}$ elements, and $S=\{m^2-n^3-1:0\leq m,n\leq p-1\}$. (In fact, Andreescu's contest problems book shows the same modification, too).

Assuming this is the problem statement, here is a proof. The key here is that if $p\equiv 2 \pmod{3}$ prime, then $-3$ is not a quadratic residue modulo $p$. Show first that, $x^3+1 \equiv x'^3 +1 \pmod{p} \iff x\equiv x' \pmod{p}$. This is indeed true, since,
$$
p\mid (x'-x)(x'^2 + x'x+x^2) \quad \text{and if } p\mid  x'^2 + x'x+x^2\implies p\mid (2x'+x)^2+3x^2  \implies ((2x'+x)x^{-1})^2 \equiv -3 \pmod{p},
$$
a contradiction. This further proves that, $\{1^3,2^3,\dots,(p-1)^3\}\equiv \{1,2,\dots,p-1\}\pmod{p}$. Therefore, for any given $y$, there is a unique $x$ for which, $x^3 \equiv y^2-1 \pmod{p}$, showing that at most $p$ pairs of $(x,y)$ can exist.

Alternative way to show that $x^3$ forms a CRS $\pmod{p}$.$x^{p}\equiv x\pmod{p}$ and $x^{p-1}\equiv 1\pmod{p}$ multiplying the two yields $x^{2p-1}\equiv x\pmod{p}$ since $p\equiv 2\pmod{3}$, $3|2p-1$ so it must form a CRS.

Notice that third powers are surjective modulo $p$, thus for every $y \in \{0, 1, 2, \dots, p-1\}$, there is exactly one $x$ under modulo $p$ that works. Thus, there are at most $p$ pairs of solutions.

Actually,if p is congrunt to1 modulo 3,{1³,2³,...,（p-1）³}has only（p-1）/3values（modp）

The problem comes down to noticing that $\frac{p-2}{3}$ is integer implies
\[x^3 \equiv y^3 \iff x^{p-2} \equiv y^{p-2} \iff x^{-1} \equiv y^{-1} \iff x \equiv y \pmod p.\]

Therefore $n^3 \pmod p$ is both injective and surjective, so for every value of $m$, there is exactly one corresponding value of $n$ to make $m^2-n^3-1 \equiv 0 \pmod p$. Therefore we have exactly $p$ multiples of $p$.


algebra
polynomial
LaTeX
linear algebra
matrix
linear algebra theorems

Let T:V-->V be any linear map.Suppose ch(X)=(X-c_1)^m_1...(X-c_k)^m_k is the characteristic polynomial such that c_i not equal to c_j, for all i not equal to j.Put W_i =Ker(T-ci.Identity)^m_i, with 1<=i<=k..then prove that dim(W_i)=m_i.

Please learn to type in $ \text{\LaTeX}$, otherwise reading your posts will be a nightmare: in other words (and symbols), if $ \lambda_1, \lambda_2, ..., \lambda_s$ are the $ s$ disctint eigenvalues of any matrix $ A$ that represents $ T$, whose (algebraic) multiplicities are $ m_1, m_2, ..., m_s$, prove that $ \dim N(A - \lambda_iI)^{m_i} = m_i$.

Observe that $ A$ is similar to a Jordan canonical form $ J$: therefore, $ \dim N(A - \lambda_iI)^{m_i} = \dim N(J - \lambda_iI)^{m_i}$. But $ J$ has on its main diagonal the eigenvalue $ \lambda_i$ exactly $ m_i$ times. So, $ J-\lambda_iI$ has exactly $ m_i$ zero elements on the main diagonal, the same happening with $ (J - \lambda_iI)^{m_i}$ (and in fact for $ (J - \lambda_iI)^k, \, \forall \, k \in \mathbb{N}$). $ \Diamond$


I know this is really random, but what's your favorite word? mine's holyschmanitzenwiggle  :D  which is totally Jeffbot5000's word but i love it

Definitions please.  :maybe: My pick: [b]flam·boy·ant[/b]
1: characterized by waving curves suggesting flames
2: marked by or given to strikingly elaborate or colorful display or behavior

holyschmanitzenwiggle: an exclamation of surprise

sesquipedalian(person who uses long words)

[quote="infinity4ever"]how do you pronounce that?!?  :?:[/quote]

don't say the ch

then its easy

crwth is another one i like

-a Celtic instrument

[b]buffalo[/b]

It's an amazing word.
- The animal
- to bully, confuse, or intimidate

Plus, "Buffalo buffalo Buffalo buffalo buffalo buffalo Buffalo buffalo." is a grammatically correct sentence  :)

bubka
it is a real amazing word for reasons which have presently escaped my mind

first of all, if its your favorite word, YOU NEED TO KNO HOW TO SPELL IT

I like all words... no favorite :D

but i do have a favorite number

whats your favorite number? mine are 7,12 and infinity  :D

I've liked 4 ever since i've had a favorite number ...
 the others are 22/7, 42, and 2400... or $ \sqrt{5760000}$  :wink:

Ever since I've learned them, I've particularly liked fustigate and spifflicate.

permutation, lamentation, jubilant, luminescent, phosphorescent, galvanizing, morbid, sequential, impetuous, insipid, frivolous, convivial, temperamental, all fun words. Try using the word deplorable in a sentence with a bunch of lowlife idiots and see how they respond  :D

Bubbling.  Just say it to yourself.  Bubbling.  Bubbling.

aperture = opening that admits light, in cameras for example

Hippopotomonstrosesquippedaliophobia- Fear of long words

transcendentalism...it was one of the last words at a local spelling bee...

aardvark

First word in the dictionary and it's named after some kind of animal.

[quote="1=2"]PNEUMONOULTRAMICROSCOPICSILICOVOLCANOCONIOSIS (also spelled PNEUMONOULTRAMICROSCOPICSILICOVOLCANOKONIOSIS) = a lung disease caused by breathing in particles of siliceous volcanic dust.[/quote]

that's a made up word

my favorite word

is

I

referring to the first person subject pronoun

no, thats a real word 1337whatevercomesnext

ouch. It's the greatest word of all tim. Say it over and over again:

"ouch ouch ouch ouch ouch ouch ouch ouch chow chow chow mmm... chow is good."

rhetorical. reason: one of the only times that r is next to h.


[quote="cnyd"]$ P(x) \equal{} x^{2} \plus{} ax \plus{} b,(a,b\in \mathbb{Z})$
Prove that,there is such an $ m\in\mathbb{Z}$ for  $ \forall n\in \mathbb{Z}$
$ P(n).P(n \plus{} 1) \equal{} P(m)$  sorry for my bad english[/quote]

$ \forall m,n$ :¨$ P(n)P(n\plus{}1)\equal{}P(n^2\plus{}(a\plus{}1)n\plus{}b)$

[quote="pankajsinha"]How?Please show.[/quote]

Are you joking ? That's just basic verification calculus :

$ P(n)\equal{}n^2\plus{}an\plus{}b$
$ P(n\plus{}1)\equal{}(n\plus{}1)^2\plus{}a(n\plus{}1)\plus{}b\equal{}n^2\plus{}(a\plus{}2)n\plus{}a\plus{}b\plus{}1$

$ P(n)P(n\plus{}1)\equal{}(n^2\plus{}an\plus{}b)(n^2\plus{}(a\plus{}2)n\plus{}a\plus{}b\plus{}1)$ $ \equal{}(n^4\plus{}(a\plus{}2)n^3\plus{}(a\plus{}b\plus{}1)n^2)$ $ \plus{}(an^3\plus{}(a^2\plus{}2a)n^2\plus{}(a^2\plus{}ab\plus{}a)n)$ $ \plus{}(bn^2\plus{}(ab\plus{}2b)n\plus{}ab\plus{}b^2\plus{}b)$
and so $ P(n)P(n\plus{}1)\equal{}n^4\plus{}(2a\plus{}2)n^3\plus{}(a^2\plus{}3a\plus{}2b\plus{}1)n^2$ $ \plus{}(a^2\plus{}2ab\plus{}a\plus{}2b)n$ $ \plus{}ab\plus{}b^2\plus{}b$

$ P(n^2\plus{}(a\plus{}1)n\plus{}b)\equal{}(n^2\plus{}(a\plus{}1)n\plus{}b)^2\plus{}a(n^2\plus{}(a\plus{}1)n\plus{}b)\plus{}b$ $ \equal{}n^4\plus{}(a\plus{}1)^2n^2\plus{}b^2\plus{}2(a\plus{}1)n^3\plus{}2bn^2\plus{}2(ab\plus{}b)n$ $ \plus{}an^2\plus{}(a^2\plus{}a)n\plus{}ab\plus{}b$
and so $ P(n^2\plus{}(a\plus{}1)n\plus{}b)\equal{}$ $ n^4\plus{}(2a\plus{}2)n^3\plus{}(a^2\plus{}3a\plus{}2b\plus{}1)n^2\plus{}(a^2\plus{}a\plus{}2ab\plus{}2b)n\plus{}b^2\plus{}ab\plus{}b$

Hence the claim.

Is there a step you want to be explained a little bit more ?

Thank You and Sorry for making you work hard .I thought there would be some method without getting into calculations


In how many ways can a committee of 4 people be chosen from five married couples if no committee is to include a husband-and-wife pair?
A: 105 B:54 C:10 D:80

I'm really bad with any kind of combinatorics or probability... I almost skipped over this problem on the test.  But I tried it out.

I found that first we find ${10} \choose 1$ ways to choose the first person.  Then his/her spouse can't be chosen, so there are $8$ people left.  Then do $8 \choose 1$ for the second, and so on.  I realized of course that my answer would be a lot bigger than the possibilities.  Now this was a problem I felt confident on... I never know when to use combinations or permutations, and I felt I had picked the right time here - obviousy not so.  What am I doing wrong here?

Hi!

I think a differrent way

We have 5 couples, 1,2,3,4, and 5

All 4 members must belong into different couples. Hence, there is just one couple which will not be choosen
We have 5 ways to choose this couple.

After that, we have to choose the Man or Woman from any other couples

e.g. we take out the couple number 5
Then for every of the couples 1,2,3,4 we choose M or W

2 possible choices for every couple, total ways=5*2*2*2*2=80

deej, that's not a bad idea, but you have to check two things:

(1)  Did you count every possibility?
(2)  Did you count any possibility more than once?

In this case, you did (1) fine -- any possible committee can be found by your process.  The problem is in (2) -- how many ways did you count each committee?


I'll just post the required steps, please give your comments about the validity of the apparent fallacy, I'm not really sure.

First we derive $ K_{eq}$ for $ A \plus{} B \minus{}\minus{}> C \plus{} D$

$ R_f \equal{} K_f[A][B]$
$ R_b \equal{} K_b[C][D]$

At Equilibrium, $ R_f \equal{} R_b$
$ \frac{K_f}{K_b} \equal{} K_{eq} \equal{} \frac{[C][D]}{[A][B]}$

And so we get a formula for Keq.

Looking at the 1st step we have written rate of the reaction is proportional to concentration of its reactants raised to the power of its stoichiometric co-efficients. However we know that this is not valid, in Kinetics the order of the reaction is not always equal to the sum of co-efficents/stoi and it is only a experimentally determined quantity.

So is the formula for $ K_{eq}$ valid in case ALL the reactions. Or perhaps are all the reactions we consider under equilibrium only elementary reactions with obvious orders or in the worst case Pseudo first order reactions?

i had already posted this [url=http://www.mathlinks.ro/viewtopic.php?search_id=150332017&t=159089]here[/url]

I don't think you got an answer because Carcul didn't understand your question. So let me wait.

[quote]
$ \Delta G \equal{} \Delta G^{\circ} \plus{} RT \ln{Q}$ where $ Q \equal{} \prod^{N}_{i \equal{} 1} (\frac {f_i}{P^{\circ}})^{v_i}$[/quote]

Just elaborating on this. Say the gas is not ideal. By definition $ P^{\circ} \equal{} 1$. We define the activity for gases as $ \frac {f_i}{f^{\circ}_i} \equal{} a_i$. Since we defined $ f^{\circ} \equal{} 1$ hence $ f_i \equal{} a_i$.

So $ K \equal{} \prod^{N}_{i \equal{} 1} a_i^{v_i}$ defined in terms of activity.

Edit: Oh [censored] I just killed my own post. Anyway, it wasn't a hard derivation. Did everyone see it before I stuffed it up?


calculus
integration
function
trigonometry
calculus computations

\[ \int {\sqrt {\frac{{\sin (x \minus{} a)}}{{\sin (x \plus{} a)}}} } dx
\]

[quote="Xaenir"]
\[ \int {\sqrt {\frac {{\sin (x \minus{} a)}}{{\sin (x \plus{} a)}}} } dx
\]
[/quote]

thanks Xaenir
nice integral this my solution

hello, alternatively you can set $ t=\sqrt{\frac{\sin(x-a)}{\sin(x+a)}}$ then you will get $ x=-a+\arctan\left(\frac{\sin(2a)}{-t^2+\cos(2a)}\right)$ and we have $ dx=-\frac{2t\sin(2a)}{-t^4+2t^2\cos(2a)-1}\,dt$
and our integral is $ \int\frac{2t^2\sin(2a)}{t^4-2t^2\cos(2a)+1}\,dt$.
Sonnhard.


Welcome to the 2nd Pytheagle Mathematical Competition (PMC)!

[b]Purpose:[/b]
This competition was created by me, Pytheagle, because of boredom. However, I spent hours of thought on the problems, so I hope there will be nothing wrong there.

[b]Rules:[/b]
1) Calculators are allowed but for only adding, subtracting, multiplying, dividing, and for exponential work. Acceptable uses of calculators include $ 2^{12}, 134224/433, 342\cdot33,$ etc. Unacceptable uses include graphing, $ \sin x$, and any other "non-basic" operation.

2) Solutions are required; if only an answer is given, no more than 1 point shall be awarded.

3) I grade each solution out of 5 points based on Clarity, Completeness, and Correctness. NO GRADING VOLUNTEERS.

4) NO CHEATING (asking others for help, asking other's answers).

5) Submissions are due by September 12, 2009, however, turning it in by September 9, 2009 will get you 2 more points.

=====

The last PMC ended horribly because of no one submitting, inactive graders, and inexperience from my part. I hope this will be better.

=====

If you are participating, post that in this thread. ONLY PM ME TO SEND YOUR SOLUTIONS OR IF YOU HAVE ANY QUESTION OR PROBLEM.

=====

Award: The winner will get X posts of his rated 6 cubes (X will be atleast 1). X is the difference between the top score and the 2nd top score.

DOWNLOAD PMC HERE!!!

1. ABCAK!
So hopefully this will be rated 6 CUBE!
First post!
 :!:

Ok will there be a finals round?

[quote="pytheage"]
Award: The winner will get X posts of his rated 6 cubes (X will be atleast 1). X is the difference between the top score and the 2nd top score.
[/quote]

You realize this defies AoPS rules?

--

I might join, depending on how much homework I get this week.

It is?
Cmon, cause I really need that :P:P
And how hard will this be?

[quote="abcak"]And how hard will this be?[/quote]
Look at the problems.

Unlimited time? Internet resources allowed?

BTW, I might be in for the same reasons izzy stated.

No it's not against the rules; I asked Mr. Rusczyk and he said it's fine.

And yes, you have unlimited time, but you MAY NOT USE ANY RESOURCES OTHER THAN YOURSELF.

It still defeats the purpose of post rating, but whatever. :D

Oh, yourself counts your brain right...?  :D

I got only ONE submittion. You hear that? ONE submission.

That's why PMC is canceled and thunder365 gets 1 post rated 6 cubes.

What do I need to do to make more people participate? Make the problems easier?

No, make them harder...it's a waste of my time to type up 6 solutions to 6 (relatively) easy problems.


An aluminum sphere with an irrational value as its radius can be melted down to form two new spheres with radii of integer length in two different ways. What is the smallest possible volume of that sphere?


If log base 4 of 6 = p,  express log base 3 of 12 in terms of p.

$ p \equal{} \log_46 \equal{} \log_4(2\cdot3) \equal{} \frac12 \plus{} \log_43\implies p \minus{} \frac12 \equal{} \log_43$

$ \implies \frac2{2p \minus{} 1} \equal{} \log_34\implies \frac1{2p \minus{} 1} \equal{} \log_32$

$ \log_3{12} \equal{} \log_3\left(2^2\cdot3\right) \equal{} 2\log_32 \plus{} 1$

$ \equal{} \frac2{2p \minus{} 1} \plus{} 1 \equal{} \boxed{\frac {2p \plus{} 1}{2p \minus{} 1}}$


I remember this problem, because I actually participated in NCIML this year.  DOes anyone else here participate in it?

I'm going to give you a detailed explanation even if you can't read it  :diablo:

okay:

log2 of (x+2) + 5= 8 + log2 of (x)

log2 of (x+2)-log2 of (x)=8-5

log2 of (x+2)/x=3

Then exponentiate to get (x+2)/x=2^3

From there, just cross multiply and get [b]x=2/7[/b]

[quote="kunny"]Everyone, You all forgot the condition for anti-logarithm. :D

kunny[/quote]

Hm? Can you expand on that statement? Never heard of that before.

[quote="kunny"]Everyone, You all forgot the condition for anti-logarithm. :D

kunny[/quote]

Um.. Not sure what you mean because that's what I got and it's what the site had too.  ;)

Here's what I found out what AntiLogarithm is:

It's the inverse function of a log defined as:

log b of (antilog b of z) = z = antilog b of (log b of z)

The antilog of base b of z is therefore $b^z$

However, I don't see how we needed this, I think the solutions posted above should be allright  :sleep2:  :sleeping:

Can anyone  solve the following logarithmic equation and inequality?

[1] $\log_2(x-2)+\log_2(7-x)=2$

[2] $\log_2(x-2)+\log_2(7-x)\leq2$

I don't get it...for the first one isn't the answer just $6$ or $3$... :?:

[quote="amirhtlusa"]second one is:
$x<3 \cup x>6$ :( [/quote]

Second one: $-x^2+9x-18\leq 4$

So $x\in(-\infty,3]\cup[6,\infty)$

Everyone,you forgot [color=blue]the condition for anti-logarithm[/color], that is to say the condition for which the function of logarithm can be defined. :)

[quote="kunny"]Everyone,you forgot [color=blue]the condition for anti=logarithm[/color], that is to say the condition for which the function of logarithm can be defined. :)[/quote]
But it still doesn't affect Silverfalcon's problem, does it?

Yes, you are right, but in case of logarithmic inequality, you can't skimp. ;)

[quote="amirhtlusa"]errr...
yup kunny is right i forgot the defenition of anit logaritm here is the answer:
$2\le x\le3 \cup 6\le x\le 7$ :?  :( [/quote]

yes it's right its open interval
$x\in (2,3]\cup [6,7)$

Yeah.. thanks kunny for reminding us about anti-logarithm

[quote="kunny"][quote="YooSam"]So everytime we deal with logarithms, we have to think of the anti-logarithms too?

Is that the same with the restriction?

How come in the school, my teacher has never mentioned anything like anti-logarithms before?  :(  :([/quote]

Yes. We should take into consideration the condition for which the logarithm can be defined, that is to say,

[color=blue]anti-logarithm is always positive real number[/color] and moreover [color=red]the base is positive real number and not equal 1[/color].

kunny[/quote]

Oh, that is what we call restriction  :D . I never knew that people call it anti-logarithms  :( . Thanks kunny.


inequalities
geometry
circumcircle
percent
email
national olympiad

I'd like to know who are the members of mathlinks who are going to partecipate at the 8th Junior Balkan in Novi Sad.It would be very nice to meet there. :)

Cheers!

there are also some romanian members going to jbmo 2004, but I assume that they do not visit the website that often ...  :maybe:

And you? Have you ever been at the JBMO?If yes could you please tell me something about it. :) How was that like?The level of the problems, the organisation, everything...

Iris

I have participated in the 2nd JBMO in 1998 (first one at which romania went) and got 39p :( out of 40p. Gold medal and special award of the jury for the best solution of problem 4. The JBMO was my first official contest where I represented Romania, and I just loved it. Plus it was in Athens, and I got to see a lot of stuff there, which I hope to revisit now.

Wow,congratulations :) 39 out of 40!I guess that's the best score ever achieved at the JBMO.I also guess that that gold medal has been the first of a long serie of one.Your solution of problem 4 was the best one.Congratulations for that too,because I've noticed that it was the hardest one there.(it was proposed by Bulgaria,wasn't it)I can't solve it and i've posted it here at mathlinks,but no acceptable solution came :( .It would be very nice of you to send a detailed solution. ;) I've been to Athens too,but not for a competition cause i'm just a freshmen.It's a lovely city.You said you hope to visit it again this year.Are you going at the IMO?

well a lot of people got 40p afterwards, for example Claudiu Raicu, Andrei Negut, Kappa :D so it's not such a great performance :P

and yes I do plan to visit at IMO.

PS in the future you may want to use space " " character after the punctuation "," and "." :D

[quote="Sasha"]BTW last year I got 39p. :)[/quote] we all know that 39p is the best score :P:P;)

iris, ku jeton ne shqiperi?? une jam shqiptare po jetoj ne amerike. nice to hear from a fellow albanian who is interested in math. keep in touch.

[quote="Persu Madalina"]Well, I have participated at JBMO and I am from the roumanian team.I have got 40. :D  :D  ;)[/quote]Hi there ... congrats I must admit that I did not expect such a nice result from the Romanian team (4 maximum scores, one 39 and one 38 :)).

More about Madalina, she is 13, so she's got another 2 more years of potentially making a full-score at the Jbmo. Good luck in that.

So, tell us how was the trip there?

Congrats, Madalina! Bravo!
P.S. Meet you next year in Moldova??? or Greece!!

Hi Madalina!
I had the pleasure to get to know you there at the JBMO and I congratulated you on your perfect score, but I didnt know you were 13.Thats amazing. Good job! Wish you the best in all the competitions youre going to participate in the future. :)

[quote="Valentin Vornicu"]I was wondering, what did you do at the jbmo, iris_aliaj?[/quote]

Well, I got bronze medal and i'm so happy but i wanna thank mathlinks and those active members like: Darij, Pierre, Lagrangia, Grobber, Arne, Zabelman, Mindflyer, Maja, Dante, Myth, Valiowk for all their help and supply during the period i was training for it. :)
And thanks to you Valentin for the solution of that second JBMO problem. :)

Thanks a lot.

P.S: It's Iris Aliaj and not "iris_aliaj" :D

Hi, Iris Aliaj!
I also participated at JBMO and also got bronze. How many points did you get? (I got 24)

strange jbmo ... too easy maybe ... so many full-scores, getting 39 / 40 would mean silver  :maybe: jeez ...

anyway, can you guys post the questions?

Hi freemind!
Congrats on your bronze medal.
I got 25 points :D (of course 1 point doesn't show anything :) )
Problem 1    10p
Problem 2     5p
problem 3     10p
Problem 4      0p :maybe:

Hi Valentin!
well maybe it was easy for the strong teams like Romania and Bulgaria since there were 11 gold medals.
I think determining sth as easy or as difficult is never completely true or false cause it depends on the training you have had and on the level of problems you have most dealed with(e.g i've never dealed with any combinatoric problem though i'm gonna start now).
I'll see to posting the problems here as soon as possible. :)

Cheers! :D

Hi Iris Aliaj!
Actually, i did like you, just at the 2nd problem I got 4 points.
By the way, how old are you? I'm 13.
Keep in touch.

Hi, V.V. I already have the problems. Here you are:
                        8th JUNIOR BALKAN MATHEMATICAL OLYMPIAD
                                NOVI SAD, Serbia & Montenegro, 26.06.2004

PROBLEM 1

Prove that the inequality
			(x+y) / (x² – xy + y²) ≤ 2√2 / √(x² + y²)
Holds for all real numbers x and y, not both equal to 0.

PROBLEM 2

Let ABC be an isosceles triangle with AC=BC, M the midpoint of AC, and Z be the line through C perpendicular to AB. The circle through B, C, M intersects the line Z at points at C and Q. Find the radius of the circumcircle of ABC in terms of m = CQ.

PROBLEM 3

If the positive integers x and y are such that 3x + 4y and 4x + 3y are both perfect squares, prove that both x and y are both divisible with 7.

PROBLEM 4

Consider a convex polygon having n vertices, n≥4. We arbitrarily decompose the polygon into triangles having all the vertices among the vertices of the polygon, such that no two of the triangles have interior points in common. We paint in black the triangles that have two sides that are also sides of the polygon, in red if only one side of the triangle is also a side of the polygon and in white those triangles that have no sides that are sides of the polygon.

Prove that there are two more black triangles that white ones.

					Working time: 4½ hours

[color=red][Moderator edit: See also http://www.mathlinks.ro/Forum/resources.php?c=1&cid=21&year=2004 .][/color]

Hi!!! I am back again!!! :lol: All the problemes were easy for the roumanian team!!!I mean first we all said we did all the problems. Freemind what's up? I hope to see you soon... maibe at our ONM. What's the first strong contest in Roumania?Vranceanu?I am so bored in this holiday... :?

[quote="Persu Madalina"]Hi!!! I am back again!!! :lol: [...] I am so bored in this holiday... :?[/quote] nice of you to join our "little" gathering here on the site. if you are bored, you can always try to solve problems from the site ;) :D

Hello to everyone  :D . I have oalso participate on JBMO 2004 and got bronze medal with 25 points.I am defintly late to post in this topic but i am posting with 2 reasons : 1.Does anybody know in which city in Moldova will be held next JBMO?
    And the second is to show my solution of problem 2(unfortunatly i did not do it on competention pmčy 4 or 5 month after :)))
[solution reposted at http://www.mathlinks.ro/Forum/viewtopic.php?t=14344 ]

       P.S. i almost forgot to ask from which to which date will JBMO 2005 be held

These details are not yet known I'm afraid ... probably in the city of Chisinau (the capital of Moldova)...

Hi..
I thought jbmo will be in greece this year
and in moldova in 2006
and all contestants from moldova, adn the teachers say it will be in greece this year!

[quote="Valentin Vornicu"][quote="vedran6"]I am 99,99 percent sure that the competition is in Moldova but Valentin knows the best  :D  doesnt he ? :)[/quote]Well I didn't say that it will be in Moldova. All I said was that if the olympiad is held there, then it probably will be in the Capital city :)[/quote]
     Than we will have to wait until we got the information about the host of JBMO 2005 :).I would that it will be Greece because it is more atractive than Moldova  :lol:

I went to Greece on my JBMO in 1998 and it was wonderfull :) Also the IMO last year was great, so I assume it'll be great if the JBMO is held again in Greece ...

[quote="ruza"]I hope it will be in moldova[/quote]
    I hope i will be in the first six on BiH Junior Mathematical Olimpiad this week  :D

why does everyone want jbmo to be in moldova?
it would be much nicer to be in greece
:)))

Well if I get into BiH JBMO team(hopefully I will :) ) I think it would be my only chance to visit Moldova in my life.Beside I don`t know much about Moldova and this would be perfect opportunity to learn more about it :lol: .

Moldova is a small and i'd say that it's a poor country compared with romania, bulgary or greece. of course becuse i'm a contestant i don't want jbmo to be in moldova. but anyway if it is not this  year then it will be next year and i will also participate (if i'll qualify :))) )
good luck tu ruza and vedran6 at their tst

[quote="Persu Madalina"]I found out that JBMO this year will be in Greece, between 20 and 26 June! I am going to participate,too! :D See you there and good luck at your tst![/quote]Great! Good luck in Greece Madalina. You have the chance of making for the first time two consecutive maximum scores at the jBMO (and even 3 right? :) ).

Not really...because I miss my third JBMO with only about 2 month  :(  :mad: Anyway, thanks for your wishes! :D

Just to mention that ruza and I have qualifaid for JBMO which will be held in city of Varlea in Greece :).I was first with 100 points  :lol: (i am very lucky man  :lol:).and ruza was fourthSo see you in Greece

[quote="ruza"]It`s not Varlea  it`s Veria :lol:[/quote]
             There is also problem number two on the map of Greece there isnt neither Varlea or Veria there is only something like Vorua

Iris Aliaj, give me you're email please?!


geometry
3D geometry
prism
Pythagorean Theorem

Does anyone know a way to find the space diagonal of a hexagonal prism?

[hide="Assuming the bases are regular:"]
Let the side length be $ s$, and the height be $ h$.  The long diagonal of the base is $ 2s$, and thus, by the Pythagorean Theorem, the space diagonal is $ \sqrt{h^2\plus{}4s^2}$.
[/hide]

Okay, that was not a very good explanation, but hopefully it will suffice.


geometry rectangle inradius incenter ratio trigonometry perimeter

Let $ ABCD$ a cuadrilateral inscribed in a circumference. Suppose that there is a semicircle with its center on $ AB$, that
is tangent to the other three sides of the cuadrilateral.

(i) Show that $ AB \equal{} AD \plus{} BC$.

(ii) Calculate, in term of $ x \equal{} AB$ and $ y \equal{} CD$, the maximal area that can be reached for such quadrilateral.

If $AD \parallel BC,$ then $AD, BC \perp AB$ and ABCD is a rectangle with $AD = BC = \frac{AB}{2}.$ Assume that AC, BC are not parallel and they meet at a point P. Since ABCD is cyclic, $\angle PCD = \angle PAB,\ \angle PDC = \angle PBA$ and the triangles $\triangle ABP \sim \triangle CDP$ are similar. Let $E \in AB$ be the center of the semicircle tangent to AD, BC, CD. (E) is the excircle of the triangle $\triangle CDP$ against the vertex P with exradius $r_p.$ Let (I) be the incircle of the triangle $\triangle ABC$ with inradius r. Denote $p = CD,\ c = DP,\ d = PC$ the sides and $s = \frac{c+ d + p}{2}$ the semiperimeter of the triangle $\triangle CDP.$ Let the bisector $PI \equiv PE$ of the angle $\angle P$ meet the opposite side CD at a point K. The similarity coefficientb of the triangles $\triangle ABC \sim \triangle CDP$ is $\eta = \frac{PE}{PK}.$ The incenter I cuts the angle bisector PK in the ratio $\frac{PI}{IK} = \frac{c + d}{p},$ hence $\frac{PI}{PK} = \frac{c + d}{c + d + p}.$ Also, $\frac{PE}{PI} = \frac{r_p}{r} = \frac{s}{s - p} = \frac{c + d + p}{c + d - p}.$ Thus the similarity coefficient is equal to

$\eta = \frac{PE}{PK} = \frac{PE}{PI} \cdot \frac{PI}{PK} = \frac{c + d + p}{c + d - p} \cdot \frac{c + d}{c + d + p} = \frac{c + d}{c + d - p}$

Consequently,

$AB = \eta CD = \frac{(c + d)p}{c + d - p}$

$AD = PA - PD = \eta PC - PD = \eta d - c$

$BC= PB - PC = \eta PD - PC = \eta c - d$

$AD + BC = (\eta - 1)(c + d) = \frac{p(c + d)}{c + d - p} = AB$

a) Call $ O$ the center of the circle $ \Gamma$ tangent to $ BC$, $ CD$, $ DA$, and $ P$, $ Q$, $ R$ the respective points where $ \Gamma$ touches $ BC$, $ CD$, $ DA$.  Call $ 2\alpha=\angle DAB$, $ 2\beta=\angle CBA$.  Triangles $ OPC$ and $ OQC$ are obviously equal and rectangle at $ P$ and $ Q$ respectively, or $ \angle POC=\angle QOC=\frac{\BAD}{2}=\alpha$,  Similarly, $ \angle QOD=\angle ROD=\beta$.  Calling $ r$ the radius of $ \Gamma$, $ AR=\frac{r}{\tan(2\alpha)}$ and $ CP=r\tan(\alpha)$.  Therefore, since $ \frac{1}{\tan(2\alpha)}+\tan(\alpha)=\frac{\cos(2\alpha)+2\sin^{2}\alpha}{\sin(2\alpha)}=\frac{1}{\sin(2\alpha)}=\frac{OA}{r}$, then $ AR+CP=OA$, and similarly $ BP+DR=OB$, or $ AD+BC=AB$, qed.

b) If $ BC$ and $ AD$ are parallel, so are $ OP$ and $ OR$, or $ BC$ and $ AD$ are perpendicular to $ AB$.  The area of $ ABCD$ is then $ \frac{AB(BC+DA)}{2}=\frac{x^{2}}{2}$ (incidentally, $ ABCD$ is then a rectangle or it could not be cyclic, and $ AB=CD=2BC=2DA$).

If $ BC$ and $ AD$ are not parallel, let $ E$ be the point where they meet.  Since $ ABCD$ is cyclic, $ ABE$ and $ DCE$ are similar, with proportionality constant $ \frac{AB}{CD}=\frac{x}{y}$, or the area of $ ABCD$ is equal to $ 1-\frac{y^{2}}{x^{2}}$ times the area of $ ABE$.  Let us now show that the perimeter of $ ABE$ is determined by $ x$ and $ y$: it is obvious that $ DE+CE=\frac{y}{x}(AE+BE)=AE+BE-(AD+BC)=AE+BE-x$, or $ AE+BE+AB=\frac{x^{2}}{x-y}+x$.  Since $ AB$ is given, the area will reach a maximum when $ ABE$ is isosceles in $ E$, ie, when $ ABCD$ is an isosceles trapezoid with $ AB$ parallel to $ CD$, and $ BC=AD=\frac{x}{2}$.  Then, the altitude from $ C$ or $ D$ onto $ AB$ is given by $ \sqrt{AD^{2}-\left(\frac{x-y}{2}\right)^{2}}=\frac{1}{2}\sqrt{2xy-y^{2}}$.  Multiply this by $ \frac{x+y}{2}$ and the maximum area of $ ABCD$ is found.

Let $ O$ be the center of the circle centered on $ AB$ and tangent to $ AD,DC,CB.$ Rays $ OC,OB$ bisect $ \angle BCD \equal{} \theta$ and $ \angle CDA \equal{} \varphi.$ Pick the point $ M$ on $ AB$ such that $ AD \equal{} AM.$ Then the triangle $ \triangle ADM$ is isosceles, where

$ \angle DMA \equal{} \frac {\pi \minus{} \angle DAB}{2} \equal{} \frac {\pi}{2} \minus{} \frac {\pi \minus{} \theta}{2} \equal{} \frac {\theta}{2}$

$ \angle DCO \equal{} \frac {\theta}{2}$ $ \Longrightarrow$ $ DCOM$ is cyclic $\Longrightarrow$ $ \angle CMB \equal{} \angle ODC \equal{} \frac {\varphi}{2}$

On the other hand, in the $ \triangle CMB,$ we have

 $ \angle MCB \equal{} \pi \minus{} \frac {\varphi}{2} \minus{} (\pi \minus{} \varphi) \equal{} \frac {\varphi}{2} \Longrightarrow \angle CMB \equal{} \angle MCB \equal{} \frac {\varphi}{2}$

$ \triangle CMB$ is then isosceles with apex $ B$ $ \Longrightarrow$ $ AB \equal{} AM \plus{} BM \equal{} AD \plus{} BC.$

[quote=Fouad-Almouine][quote=Erken]Part a) is IMO 1985 :wink: I know one very nice proof of this part(without using trigonometry)  :) If someone want i could post it.[/quote]

Yes it's the circle centered at A with the length AD then we show that the second triangle is isosceles by some angle chasing and we are done ...[/quote]
Yes, I like this solution...
For who wants to see this solution here (first solution) :
https://artofproblemsolving.com/community/c6h60782p366584

Could someone explain why in part b of daniel73 solution it is needed to show that the perimeter of $ ABE$ is determined by $ x$ and $ y$?
Thanks


geometry
parallelogram
geometry proposed

Pint $ M$ is taken in interior of quadrilateral $ ABCD$ such that $ ABMD$ is parallelogram. If $ \angle CBM \equal{} \angle CDM$ prove that $ \angle ACD \equal{} \angle BCM$.

This is my proof.Nice problem! :)
Point N is taken such that DMCN is a parallelogram.
So ABCN is also a parallelogram.
$ \angle ADC \equal{} \angle ADM \plus{} \angle MDC \equal{} \angle ABM \plus{} \angle MBC \equal{} \angle ABC \equal{} \angle ANC$
$ \Rightarrow \angle ADC \equal{} \angle ANC$
$ \Rightarrow \angle ACD \equal{} \angle AND$
We have $ \angle BCM \equal{} \angle AND$
Thus  $ \angle BCM \equal{} \angle ACD$ :lol:


Find the number of positive integers x satisfying the following two conditions:
1. $x<10^{2006}$
2. $x^{2}-x$ is divisible by $10^{2006}$

Funny how I guessed this property from long before the IMC... I never thought there would be an easy proof though. :)

Really easy for IMC :D !!!
There are 3 answer $x=1$ and there exist exactly one $0<x<10^{2006}$ that $x\equiv0\pmod{5^{2006}},\ \ x\equiv1\pmod{2^{2006}}$ and also there exist one $0<x<10^{2006}$ that $x\equiv0\pmod{2^{2006}},\ \ x\equiv1\pmod{5^{2006}}$. So there are 3 such numbers.


1)A man goes into a bank to cash a check. The teller mistakenly reverses the
amounts and gives the man cents for dollars and dollars for cents. (Example: if the check
was for 5.10 dollars, the man was given $ 10.05 dollars.). After spending five cents, the man finds that he
still has twice as much as the original check amount. What was the original check amount?
Find all possible solutions.
2)Three farmers sell chickens at a market. One has 10 chickens, another has 16,
and the third has 26. Each farmer sells at least one, but not all, of his chickens before noon,
all farmers selling at the same price per chicken. Later in the day each sells his remaining
chickens, all again selling at the same reduced price. If each farmer received a total of$35
from the sale of his chickens, what was the selling price before noon and the selling price after
noon?

There is a serious problem with your syntax and as a result, the important part of the problem is not there.  Please fix it.

[quote="bigman"]1)A man goes into a bank to cash a check. The teller mistakenly reverses the
amounts and gives the man cents for dollars and dollars for cents. (Example: if the check
was for 5.10 dollars, the man was given 10.05 dollars.). After spending five cents, the man finds that he still has twice as much as the original check amount. What was the original check amount? Find all possible solutions. 2)Three farmers sell chickens at a market. One has 10 chickens, another has 16, and the third has 26. Each farmer sells at least one, but not all, of his chickens before noon, all farmers selling at the same price per chicken. Later in the day each sells his remaining chickens, all again selling at the same reduced price. If each farmer received a total of 35 moola
from the sale of his chickens, what was the selling price before noon and the selling price after
noon?[/quote]

There we go.


I am studying for the SAT II World History test and I know quite a bit, but is there a way to maximaze your score other than just memorizing discrete facts?

THINK about when and why events happened. Be sure to note the maps, cartoons, and other illustrations in your textbook too.

Try to live in history-trust me it works-i have been in greece for 8 years
other than that research is the best bet...(by that i mean learning MORE than your text) oh and perhaps sparknotes but i don't know if they have them on world history...probably


videos
geometry
geometric transformation

[youtube]ZA1NoOOoaNw[/youtube]

May be slightly inappropriate for the younger audience at times, but worth it nonetheless.
This is an Indian music video, with subtitles of what English words the Indian words sound like.
It is [i]not[/i] a translation.

Favorite lines:
"Have you been high today?"
"I see the nuns are gay."
"Jet pack.  Operation.  Send him the crazy Hindi."
"I'd like to see you pee on us tonight."

Are you sure they are Indian words? They sound like English funniness with an Indian accent.

[quote="lifesapain"]The language they're singing in is an Indian language called Tamil. Parts of it just happen to make sense in English. Only a coincidence. But a hilarious coincidence[/quote]...you mean seem like they make sense.   And yeah, it is in Tamil.

Yeah, it did sound like it's in Tamil... def. not hindi

:rotfl:  I can't stop.  :rotfl:  Help someone!!!!!!!

Oh, and who can forget such classics as
"The puppy had a fee."
"Who put the goat in there?"
"Some day I sell DNA."
"That stuff is pink-colored."

seen it but not with the translation,  :rotfl:

"We know what's in butter rum!"
 :rotfl:

[youtube]yRmqZRPgK1w[/youtube]

Without the lyrics...

[quote="kimmystar94"]you should see the indian thriller that Buffalax did, too..
it's hilarious...[/quote]

There is another one:

[youtube]yR2jjFB_qnc[/youtube]

eh.. i'll post the indian thriller one...

it's funnier. I saw the one that 1=2 posted..
[youtube]TtJRNyPK-lc[/youtube]


Hi guys....its been a long time since I was last here!!!
By the way, if some of you may remember (or notice) , I was trying to maintain Indian MO resources(INMO, IMOTC,RMO etc...) for previous years but just lost touch now.....I request that somebody take over especially by putting this years different papers and also try to get a hand on those things that are not online but maybe in print....

[size=150]So, anyone interested!?!!!!![/size]

Hi Rushil ,
well i am ready and i think  i have almost all of them(too optimistic though  :D  ) well surely would try to get those i don't have but if you meant those in the resources section :
http://www.mathlinks.ro/Forum/resources.php then i don't exactly know how to post them there  :(
well we obviously have people from all over india here so we can get all of them.

thats nice that you have those resources....

the way to get them onto the resources section is to contact Valentin Vornicu (the administrator) and he'll tell you the format in which to send them to him so that he can post them there.....

please coordinate amongst yourselves so that all the resources that anyone has access to can be put online!

Dude start.. I have been waiting since 15 years........

ikr I've been waiting since before I was born


A $ 100$-gon $ P_1$ is drawn in the Cartesian plane.  The sum of the $ x$-coordinates of the $ 100$ vertices equals 2009.  The midpoints of the sides of $ P_1$ form a second $ 100$-gon, $ P_2$.  Finally, the midpoints of the sides of $ P_2$ form a third $ 100$-gon, $ P_3$.  Find the sum of the $ x$-coordinates of the vertices of $ P_3$.

[quote="AceOfDiamonds"]the sum of the x coordinates of the midpoints don't change, so it's still 2009[/quote]
That is true but I don't get it.  Why don't the sum of the $x$-coordinates change?

Two times the x coordinate of the midpoint of a line segment is equal to the sum of the x coordinates of the endpoints of the segment.

Why:

Letting the endpoints be $(x_1,y_1)$ and $(x_2,y_2)$ we have that the midpoint is $(\frac{x_1 + x_2}{2}, \text{blah})$, where blah is something else :) .  So twice the midpoint is $x_1 + x_2$, which is the sum of the x coordinates of the endpoints.  So the sum of the midpoints does not change.


Isn't this trivial? I mean we just observe that $ i\cdot x_i\leq 1+x_1+...+x_n $ and then write that $ (\frac{i\cdot x_i}{1+x_1+...+x_n})^a\geq\frac{i\cdot x_i}{1+...+x_n} $ and then multiply by $\frac{1}{i} $ and then add up.


I have 85000 words in my book. HOw many words would I have left over on the last page if I organized the pages so that 300 words were on each page?

[hide]$\frac{85000}{300} = 283 \text{remainder} 100$
so the answer is 100
[/hide]

Plug it into a calculator, and you get something that ends in .333333333333333333333, and .333333333333333333333333 is about 1/3, and 1/3 of 300 = 100. Yay.


Find all possible solutions to $x^2+y^2+z^2=1996$, where x, y and z are positive integers.

[hide]
$1996\equiv 0 (mod 4)$. The unic quadratic residual $mod 4$ are $0$ and $1$. So it must be $x^2\equiv y^2 \equiv z^2 \equiv 0 (mod 4)$

$x=4k$
$y=4m$
$z=4n$

The equation becomes $4k^2+4m^2+4n^2=499$ The left-hand side is $\equiv 0 (mod 4)$ $\forall k,m,n \in \mathbb{N}$, the right-hand is $\equiv 1 (mod 4)$. So there are no solutions.[/hide]

I think there is at least one solution, here is an example:
[hide]
(x,y,z) = (6, 14, 42 ) :!:
[/hide]

[quote="hydro"]So it must be $x^2\equiv y^2 \equiv z^2 \equiv 0 (mod 4)$

$x=4k$
$y=4m$
$z=4n$[/quote]

$x^2\equiv 0 \pmod {4}\not \implies 4|x$.

[hide="Partial Solution"]So $4|x^{2}\implies 2|x$.  This means $x=2m$, $y=2n$, and $z=2k$.  So then $m^{2}+n^{2}+k^{2}=499$.  The right side $499\equiv 3\pmod{4}$ so $m^{2},n^{2},k^{2}\equiv 1\pmod{4}$.

We have $\left(m-1\right)|4$ or $\left(m+1\right)|4$ (the same holds true for $n$ and $k$).  This shows why $\left(6,14,42\right)$ works.

Someone else will have to finish it from there, I have to go...[/hide]

once we reduce it to

$m^2+n^2+p^2=499$ , by looking at mod 4 , we find that they are all odd number . Also , every odd square number ends with 1,5,9 . So the combination to have a $9$ behind is $[1,9,9]$ or $[5,5,9]$

For the case $[5,5,9]$ by trial and error ( only need to try $5^2,15^2$ ) and we find only $(15,15,7)$ satisfy it .

For the case $[1,9,9]$ , those end with $1$ are $[1,81,121,361,441]$ while those end with $9$ are $[9,49,169,289]$ . By some calculation , only $(21,7,3)$ satisfy it . Hence the two solution is

$(30,30,14)$ and $(42,14,6)$ and their ordered pair satisfy the condition

Congratulations :!:
My solution is almost the same as yours except in the stage of "trial and error".

maybe we can also consider $\mod 3$ and deduce that two of them are divisible by 3. Then taking $\mod 9$ to deduce that the third one must be $7 (\mod 9)$ ( since every square mod 9 can only be 0,1,4,7).

From this , it is obvious that one of them is $7$ ( or else it will be 7+9x2=25 which is too big) .

So now the equation is reduce to

$m^2+n^2=499-49=450=2\cdot 5^2\cdot 3^2$

So we can have $(5^2+5^2)(3^2+0^2)=15^2+15^2$ or $(6^2+3^2)(3^2+1^2)=21^2+3^2$


function
induction
percent
algebra proposed
algebra

Find all functions $f: \mathbb R \to \mathbb R$ which satisfy \[f( x + f ( xy ) )= f( x ) + x f( y ) \] for all reals $x$ and $y$.

[quote="kenan aze"]f:R - R/      x, y are real numbers, Find all functions such that
f[ x + f ( xy ) ] = f( x ) + x f( y ) :)[/quote]

I found this problem rather hard. Thank you for giving us a simpler solution if you got one.

Let $ P(x,y)$ be the assertion $ f(x+f(xy))=f(x)+xf(y)$

$ f(x)=0$ $ \forall x$ is a trivial solution. So we'll now consider non all-zero solutions.

1) $ f(x)=0$ $ \iff$ $ x=0$
========================

Suppose that $ \exists u\neq 0$ such that $ f(u)=0$. then, for $ x\neq 0$, $ P(\frac ux,x)$ $ \implies$ $ \frac uxf(x)=0$ and so :

$ f(x)=0$ $ \forall x\neq 0$ and then $ P(0,0)$ $ \implies$ $ f(f(0))=f(0)$ and so $ f(0)=0$ (else we would have a non zero real $ f(0)$ whose image would be non zero.

So $ f(x)=0$ $ \implies$ $ x=0$.
Then $ P(-1,-1)$ $ \implies$ $ f(f(1)-1)=0$ and so $ f(1)=1$ and $ f(0)=0$
Q.E.D.

2) $ f(n)=n$ $ \forall n\in\mathbb Z$
==================================

$ P(1,x)$ $ \implies$ $ f(f(x)+1)=f(x)+1$
$ P(1,f(x)+1)$ $ \implies$ $ f(f(x)+2)=f(x)+2$
An immediate induction gives $ f(f(x)+n)=f(x)+n$ $ \forall x$, $ \forall n>0\in\mathbb N$

Then $ P(-n,-1)$ $ \implies$ $ f(-n)=nf(-1)$

Then $ P(-1,-2)$ $ \implies$ $ f(1)=f(-1)-f(-2)=-f(-1)$ and so $ f(-1)=-1$ and $ f(n)=n$ $ \forall n\in\mathbb Z$
Q.E.D.

3) $ f(x)=1$ $ \iff$ $ x=1$
=========================

We already know that $ f(1)=1$. Suppose now $ f(a)=1$, then :

$ P(\frac 1x,x)$ $ \implies$ $ f(\frac 1x+1)=f(\frac 1x)+\frac{f(x)}x$

$ P(\frac 1x,ax)$ $ \implies$ $ f(\frac 1x+1)=f(\frac 1x)+\frac{f(ax)}x$

And so $ f(ax)=f(x)$ $ \forall x\neq 0$ and so $ f(ax)=f(x)$ $ \forall x$

So $ f(-a)=-1$ and then $ P(a,-1)$ $ \implies$ $ f(a-1)=1-a$
So $ f(a(a-1))=f(a-1)=1-a$ and then $ P(a,a-1)$ $ \implies$ $ f(a+f(a(a-1)))=f(a)+af(a-1)$ and so $ 1=1+a(1-a)$ and so $ a=1$
Q.E.D

4) $ f(x)$ is injective
=======================

Consider now $ f(y_1)=f(y_2)$ with $ y_1\neq y_2$. Obviously $ y_1,y_2\neq 0$ (since $ f(x)=0$ implies $ x=0$) and let then $ a=\frac{y_1}{y_2}$ :
Comparing  $ P(y_2,a)$ and $ P(y_2,1)$, we get $ f(a)=1$, so $ a=1$ (point 3 above) and $ y_1=y_2$ and $ f(x)$ is injective.
Q.E.D.

5) $ f(x)=x$ $ \forall x$
=======================

Let $ A=f(\mathbb R)$

$ P(1,x)$ $ \implies$ $ f(f(x)+1)=f(x)+1$ and so $ x\in A$ $ \implies$ $ x+1\in A$
$ P(-1,x)$ $ \implies$ $ f(f(-x)-1)=-1-f(x)$ and so $ x\in A$ $ \implies$ $ -1-x\in A$ and so, using line above, $ -x\in A$
So, since $ x\in A$ implies $ x+1\in A$ and $ -x\in A$ we get $ -x+1\in A$ and $ x-1\in A$

Then for a given $ x$, and since $ f(x)-1\in A$, let $ z$ such that $ f(z)=f(x)-1$ :
$ P(1,z)$ $ \implies$ $ f(f(z)+1)=f(z)+1$ and so $ f(f(x))=f(x)$ $ \forall x$ and, since $ f(x)$ is injective, $ f(x)=x$ $ \forall
x$

And it's immediate to check back that this indeed is a solution.

6) synthesis of solutions
==========================

Hence the two solutions of this equation :
$ f(x)=0$ $ \forall x$
$ f(x)=x$ $ \forall x$

congrats you. i think your solution is very good. but  i prove it by finding f(1) and f(-1). i have no time to write solution. i will send my solution in a week when school holiday will begin. again thank you  for your solution. :)

f(x+f(xy))= f(x) + xf(y)

1) let find f(0)=0
let f(0) gets other value than 0. x=y=o  ff(0)=f(0)
x=f(0),y=1 then x=f(0),y=0.  then we get f(0)=f(1)
x=y=1 then x=1, y= 1 + f(0) then x= 1 + f(1) , y =0    we will get f(0)f(1)=0 . which means f(0)=0

2)let find f(1)=1,
if f(1) + 1 = A = 0  then  x = y = 1 then we get f(1) = 0 condraction A is not zero
x= 1 => f( 1 + f ( y )) = f(1) + f ( y ) is B
at B y= 1 + f( y ) => f( A + f(y) ) = f( A ) + f (y)
x=A y= y/A  f ( A + f(y) ) = f( A ) + A f ( y/A )
then we get f(y) = A f ( y / A )
at B y=A  then we get f(1) ( f(1) - 1 ) = 0
if f(1) = 0 x= 1/m y= m then we get  f(m) = 0 and f(0) = 0 which means f is obviously zero
then f(1) = 1

3)let prove that if n is N then f(yn)=nf(y)
it is easy to prove by helping of B by induction if n is NATURAL then  C=> f( n + f(y) ) = n + f(y)
at C y = 1 then xis natural f( x ) = x
at C y = yn f( n + f( yn ) ) = n + f( yn )
x= n, n is N, f ( n + f( yn ) ) = f (n) + nf(y) = n + f (yn)  we get if n is N then f(yn)=nf(y)
n=2 y= -1/2 then f( -1/2 ) = - 1/2

4)let find f( -1 ) = -1
if f(k) = 0 x=k , y= 1 then we get k = 0
x= -1 y= 1 then f( -1 ) = 1 + f ( -1 + f( -1 )) then ff( -1 ) = f(1 + f ( -1 + f( -1 ))) = 1 + f( -1 + f( -1 )) = 1 + f ( -1) -1 we get ff(-1) = f ( -1)
x=1 y= -1 then f(-1) = -1 + f ( 1 + f ( -1 )) then ff( -1 ) = f ( -1 + f ( 1 + f ( -1 ))) =  f(-1) + f(-1 - f( -1 )) then f(-1 - f( -1 ))= 0 => f(-1) = -1

5) let finish problem
at B y = -1 + f( -y)    f( 1 + f( -1 + f(-y)))= 1 + f ( -1 + f(-y) ) = -f(y) then we get D=>f( - f(y)) = -f(y)
at D let give y = -f(y) we get ff(y) = f(y)
y= -1    f(x - f(-x)) = f(x) - x then   f( f(x - f(-x)))= f(f(x) - x) = f ( x + f( -x)) = x + f( -x) = f(x) - x then we get 2x + f(-x) = f(x)
then we get 2x + f(-x) = f(x) , then f(2x + f(-x)) = ff(x) = f(x) = f(2x) + 2xf( -1/2 ) = 2f(x) - x  which means f(x) = x

6) last part is checking
............................................ is obviously true :D

=========================================================================================================================
the land of fire AZERBAIJAN

[quote="kenan aze"]f(x+f(xy))= f(x) + xf(y)

1) let find f(0)=0
let f(0) gets other value than 0. x=y=o  ff(0)=f(0)
x=f(0),y=1 then x=f(0),y=0.  then we gets f(0)=f(1)
x=y=1 then x=1, y= 1 + f(0) then x= 1 + f(1) , y =0    we will get f(0)f(1)=0 . which means f(0)=0
[/quote]

$ x \equal{} y \equal{} 1$ $ \implies$ $ f(1 \plus{} f(1)) \equal{} f(1) \plus{} f(1)$
$ x \equal{} 1, y \equal{} 1 \plus{} f(0)$ $ \implies$ $ f(1 \plus{} f(1 \plus{} f(0))) \equal{} f(1) \plus{} f(1 \plus{} f(0))$

And I dont understand how you conclude from these two lines that $ f(0)f(1) \equal{} 0$
 :blush:

[quote="kenan aze"] x= 1 + f(1) , y =0   f(1 + f(1) + f(0))= f( 1 + f(1)) + (1 + f(1))f(0) = 2f(1) + f(0) + f(1)f(0) = 3f(1) + f(1)f(0)

i think there is no dificculty to calculate some simple operations.... :D
[/quote]

Sure, but you did not give the last step and since you gave the two first, I thought there was no missing step.

And I dont worry finding $ f(0)$. I gave an earlier proof of this problem and was just reading yours in order to help you to improve.

I'm sorry for your reaction and will no longer read your proofs.

Enjoy mathlinks and maths.

i am sorry if you feel worse yourself for my reply. i use it randomly.  big percent you are elder than me. i dont want to make you unhappy.  maybe i make somethink wrongly

If we just assume that $f(x)$ is continuous and differentiable then we can proceed as follows.

If, $r$ be a root of $f(x)$, then we take $(x,y)=(x,\frac{r}{x})$. The given equation then implies, $xf(\frac{r}{x})=0 \rightarrow f(t)=0 \forall t \in \mathbb R$. So,a non trivial solution would imply that there are no non-zero root....$(1)$.

Now, let $f(0)=a > 0$ without loss of generality. So, $f(x+a)=f(x)+ax$, which contains a special periodicity, namely $f'(x+a)=f'(x)+a$. Now, for $4x \in [0,a)$, let $\text{min}\{f(x)\}=-g; \text{max}\{f(x)\}=G$ ($g>0$ as $f(0)=f(a)$). Hence, for all $x>\lceil \frac{g}{a} \rceil ,f'(x)>0$ and for all $x<\lfloor \frac{G}{a} \rfloor, f'(x)<0$. Hence, $\exists x_0 <0$ such that $f(x_0)=0$. But that is not possible as $(1)$ implies. Hence, $f(0)=0$.

Now, let $y=k \to 0, f(xk)=h \to 0$, the given equation gives $\lim \limits_{k \to 0} \frac{f(x+h)-f(x)}{h}=\frac{xf(k)}{f(xk)}=f'(x)$. Using L'Hospital rule in the middle expression of $\frac{0}{0}$ for we get $f'(x)=\lim \limits_{k \to 0} \frac{f'(k)}{f'(xk)}$. this implies that $f'(0) \neq 0$. Because using L,Hospital rule repeatedly if there exists some $k >1$ s,t $f^{(k)}(0) \neq 0$, then as $x \to 0$, the middle equation blows up even though $f'(x)=0$ (what is just assumed). So, either $f^{(k)}(0)=0$ $\forall k \in \mathbb N \rightarrow f(x)=0 \forall x \in \mathbb R$ or $f'(0) \neq 0 \rightarrow f'(x)=1 \rightarrow f(x)=x$


[b]Thank you reply soon![/b]

Let circle (O) with diameter AB. Let line $ \Delta$ is perpendicular with AB at I. Let a point M change on circle (O). MA, MB meet $ \Delta$ order at P and Q. QA meet (O) at M. Prove that MN pass a fixed point.

--------------------------------------------
I love you. Because you a good friend.

[quote="thanhnam2902"][b]Thank you reply soon![/b]

Let circle (O) with diameter AB. Let line $ \Delta$ is perpendicular with AB at I. Let a point M change on circle (O). MA, MB meet $ \Delta$ order at P and Q. QA meet (O) at M. Prove that MN pass a fixed point.

--------------------------------------------[quote]
I thinh QA meet (O) at N
$ MN\cup (AB) \equal{} K$
Then $ (KIAB) \equal{}\minus{}1$ so  K is the fix point.


search
algorithm
advanced fields
advanced fields unsolved

Simple question. My mind isn't working right now. Consider that there is a graph G = (V,E) which is an undirected, connected graph, and let sEV. The breadth-first search obviously constructs a breadth-first tree rooted at s that, for each vEV, represents a shortest path from s to v in G. Of course, there may be many shortest paths from s to v. All such shortest paths are represented in the breadth-first directed acyclic graph (BFDAG) G^D = (V,D), where

D = {(u,v) | d(s,v) = d(s,u) + 1} ----> d represents a lower-case delta symbol.

Basically, I need to modify the BFS algorithm to also compute D. My problem is that I can't seem to break down the definition of D into english terms. If someone could help me to sort out the meaning of the above equation for D then I can easily finish this problem. Like I said, my mind just isn't working right now.

Thanks.

I'm not sure, but I hope $\delta (s,u)$ represents the distance from $s$ to $u$ in the tree built by the DFS algorithm. Let $T$ be this tree.

This $\delta$ allows us to see $T$ as structured in levels.
$s$ is the only vertex on the $0$ level.
On the first level lie all the vertices that are connected to $s$ by an edge.
On the $i+1^{th}$ level lie all the vertices that are connected to a vertex on the $i^{th}$ level and than don't appear in one of the previous levels. This gives the DFS algorithm.

Now that you have this tree, link all the vertices on the level $i$ with all the vertices on the level $i+1$, for all $i$. Like in a complete bipartite graph. This gives the set of edges $D$. You say this is a directed set of edges, but I don't see the reason why we can't see it as undirected.
$D$ is a set of edges that doesn't have much to do with $E$, the edges of $G$.
The idea is that if we have an edge in $(E\cap D)\setminus T$ then we have a new shortest path from $s$ to the highest leveled vertex of this edge. This is easily seen if you draw the picture, but if you need help I can solve it.

If you make the definition of $D$ like this:
$D=\{(u,v)|\delta(s,v)=\delta(s,u)+1\mbox{ and } \{u,v\}\in E\}$, then the set of directed paths from $s$ to a vertex $v$ is precisely the set of shortest paths from $s$ to $v$.


The n'th row, starting at $ n\equal{}2$, has value $ \dfrac{\frac{1}{n^2}}{1\minus{}\frac{1}{n}}\equal{}\frac{1}{n^2\minus{}n}\equal{}\frac{1}{n\minus{}1}\minus{}\frac{1}{n}$. This clearly telescopes to 1.


inequalities
algebra
polynomial
inequalities solved
4-variable inequality
Symmetric inequality
AM-GM

Let $a,b,c,d$ be positve numbers. Prove that
\[\sum_{\rm cyclic}\frac{a^{3}}{(a+b)(a+c)(a+d)}\geq\frac12.\]

Today, I not delete any post :D . Please post at least one solution, That link is from diendantoanhoc.net and NOT English.  :maybe:
@pvthuan: You are wrong when post this problem in Inequalities Solved Problems.

It is a solved problem I received from one of my friends and I put it here in hope of a new solution. Of course the author sent me the solution in diendantoanhoc.

See here ''The problems that were discussed and solved in the topics above are moved here.'' You are wrong, please don't delete my post and someone's in this topic :P

Wonderful! That is what I am looking for. You will be credited for this solution in my book, Naoki Sato (on Crux editorial).

As for the claim, we may write
\[\left(\frac x{x+2}\right)^{3}\geq \frac1{27}+k(x-1). \]
We come up with this idea by taking into account the fact of equality case. We need to find $k$ such that the claim is valid. Write it in the form
\[(x-1)\left[\frac{2x^{2}}{(x+2)^{3}}-k\right]\geq0. \]
We wish to find $k$ such that the polynomial in the square bracket has $x-1$ as an factor. Thus, we can plug $x=1$ into that bracket to get $k=\tfrac2{27}$. Alternatively, we can check the $g'(1)=0$, $g(x)$ be the polynomial in the bracket. Then we can verify the claim as what nsato did.

[quote="pvthuan"]Wonderful! That is what I am looking for. You will be credited for this solution in my book, Naoki Sato (on Crux editorial).
[/quote]

Thank you!  These days, I need all the credit I can get, although I don't work for Crux anymore.

Incidentally, the corresponding inequality for five variables is false.

The following is easy
\[\sum_{\rm cyclic}\frac{a^{4}}{(a+b)(a+c)(a+d)}\geq\frac{a+b+c+d}8.\]
I guess that
\[\sum_{\rm cyclic}\frac{a^{4}}{(a+b)(a+c)(a+d)}\geq\frac{\sqrt{a^{2}+b^{2}+c^{2}+d^{2}}}4.\]


AoPS will host another AMC contest [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]Math Jam[/url] at 7 PM ET (4 PM PT) on Friday, February 17. We will work through solutions for each of the last 5 problems on each the AMC 10 and AMC 12 and then take requests for discussing solutions to other problems on the exams.

sorry to revive an old post, but is there an AMC 10A or 10B math jam this year?

You can find these on the math jam schedule:
[url]http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php[/url]


If two real numbers $ x$ and $ y$ satisfy the equation $ \frac{x}{y} \equal{} x \minus{} y$, then:
$ \textbf{(A)}\ {x \ge 4}\text{ and }{x \le 0}\qquad \\
\textbf{(B)}\ {y}\text{ can equal }{1}\qquad \\
\textbf{(C)}\ \text{both }{x}\text{ and }{y}\text{ must be irrational}\qquad \\
\textbf{(D)}\ {x}\text{ and }{y}\text{ cannot both be integers}\qquad \\
\textbf{(E)}\ \text{both }{x}\text{ and }{y}\text{ must be rational}$

$ x=xy-y^2$
$ x-xy=-y^2$
$ x(1-y)=-y^2$
$ x=\frac{y^2}{y-1}$

Note that $ y^2$ is always nonnegative. $ x$ is always negative if $ y<1$. If we want to find a lower bound for $ x$, set $ y^2%Error. "geky" is a bad command.
-k$, where $ k$ is a real constant. $ y^2-ky+k\ge0$, which means the LHS of that inequality must be a perfect square trinomial. So $ k=4$, which means $ x\ge4$ as well.

Answer: A

I'm not sure if I'm right or if my solution's legit though... :maybe:

Your solution would be what one should do at the actual contest, but the complete solution to the problem is provided as follows:

[hide="My Solution"]
Continuing from where gauss left off and solving for $ y$, we obtain

$ x \equal{} \frac{y^2}{y \minus{} 1}\implies xy \minus{} x \equal{} y^2\implies y^2 \minus{} xy \plus{} x \equal{} 0.$

The quadratic in $ y$ yields roots of $ \frac{x \pm \sqrt{x^2 \minus{} 4x}}{2}.$ Clearly, the quantity under the square root must be nonnegative. Therefore, $ x^2 \minus{} 4x \ge 0\implies x(x \minus{} 4)\ge 0\implies x \in (\minus{}\infty, 0] \cup [4, \infty) \implies \fbox{(A)}.$
[/hide]

[quote="sunehra"]Your solution would be what one should do at the actual contest[/quote]

By that, do you mean in terms of the general gist of the idea one should be thinking, instead of writing/noting the whole solution?

Yes, because a contest is about speed. Especially in a large contest like this one. I'm not really sure how much time they had when taking the AHSME back then.


LaTeX
limit
calculus
calculus computations

Sorry for my bad latex but.. :)

Here is the problem
Find lim(x->0)[1/ln(x+sqrt(1+x^2)) - 1/ln(1+x)]

I think its a L'hopital problem but not sure :)

c'mon guys.. my exam is in 3 hours.. i need this solution

$= \lim_{x \rightarrow 0}\frac{\ln(1+x)-\ln(x+\sqrt{1+x^{2}})}{\ln(x+\sqrt{1+x^{2}})\ \ln(1+x)}$
Then 2 times l'Hôpital.

Let's not use L'Hôpital's rule. Starting from the simplification made by Petruspan we have:

$ln(1+x) = x-\frac{x^{2}}{2}+\frac{x^{3}}{3}$
$ln(x+\sqrt{1+x^{2}}) = argsinhx = x-\frac{x^{3}}{6}$

Now substitute these expansions in your limit (which I will designate by L) and we get:

$L = \frac{\frac{x^{3}}{2}-\frac{x^{2}}{2}}{x^{2}-\frac{x^{3}}{2}}= \frac{x-1}{2-x}=-\frac{1}{2}$


This problem was communicated to me and I really love it.

Given a circle X and three points A, B, C inside it, let us draw three circles :

1) a circle $c_1$ tangent to the circle X, to the line AB, and to the line BC.
2) a circle $c_2$ tangent to the circle X, to the line BC, and to the line CA.
3) a circle $c_3$ tangent to the circle X, to the line CA, and to the line AB.

Let us denote $C_1$, $C_2$, $C_3$ the points of tangency of the circles $c_1$, $c_2$, $c_3$ with X.

Prove that the lines $BC_1$, $CC_2$, $AC_3$ are concurrent.

Let N be the external center of similitude of the circle X and the incircle I of triangle ABC. We are going to prove that the lines $BC_1$, $CC_2$ and $AC_3$ pass through N. (This will, of course, establish the concurrency of these lines.)

The Monge theorem states that the pairwise external centers of similitude of three circles lie on one line. Applying this to the circles X, I and $c_1$, we obtain that the external center of similitude of the circles X and I, the external center of similitude of the circles I and $c_1$, and the external center of similitude of the circles $c_1$ and X lie on one line. In other words, the points N, B and $C_1$ lie one one line (in fact, the point B is the external center of similitude of the circles $C_1$ and I because B lies on the two external common tangents AB and BC of these circles, and the point $C_1$ is the external center of similitude of the circles $c_1$ and X because it is the point of tangency of these two circles). This means that the line $BC_1$ passes through N. Similarly we can prove the same for $CC_2$ and $AC_3$, and we are done.

  Darij

Let K be that intersection. You can also prove I , O, K are collinear. :cool:
   The solution uses ditaliation. I 'll give you later.

I intend to give you my solution next time, but I have time now.
 You can easily prove that the detalitation centers of (c1) and (c2), (c2) and (c3), (c3) and (c1) such that the product of the three propotions is positive are collinear (remember there are two detalitation centers of (c1) and (c2)).
    Now, what you have to do now is to apply it. ( Sorry, I 'm not very good at English geometry languange  :(  ).


I took acting fundamentals as a freshmen.  I need a total of 2 credits in art to graduate. (so far I have 1/2 credits)  acting was alright, i'm deciding weather i should take actingII as a sophmore  or should I take Introduction to pyschology(after taking the class you get get to be a leader/mentor in camps my school organizes, n i've heard it's a fun class) instead.  How much do art(meaning acting,band, ect) classes count for colleges?  I could fullfill my art requirement in my junior n senior year, but I don't want to have a whole bunch of classes on top of college apps.  advice anyone?

Thanks :)

It depends what colleges you are applying to, and what you like to do. If you are applying to University of California campuses, it's a good idea to meet the "a-g" requirements for admission in California

http://www.ucop.edu/a-gGuide/ag/welcome.html

and those include art classes of specified kinds. But study what you like for your electives. PLEASE don't try to microanalyze what a college wants you to do--believe me, within broad limits college admission officers want you to do what YOU want to do. Do something challenging and interesting, and it will all work out.

Do you like it? If you like it, do it. If you love it, do it more.

I heard you need a full year of art to get into some colleges in the west coast.  Is this only for university of California or is it for all colleges on the west coast?  How about stanford?  Acting was alright, it wasn't great but it wasn't horrible either.  I enjoy participating in the Speech and Debate club more than taking the class.

[quote]I heard you need a full year of art to get into some colleges in the west coast. Is this only for university of California or is it for all colleges on the west coast?[/quote]
We in the CSU system had that as a requirement before the UC did. But as far as I know, it's just a UC-CSU requirement, that is, public universities in California. I can't speak for any private universities or for other states. And it's "visual or performing arts," so yes, for all you violin players out there, your orchestra class does count.


I can't figure out how to order the Aops books online.  Is ordering something online the same as ordering by fax?

Go to "bookstore"
Add a book to your cart
Click "view cart"
Then click "complete order"

I'm not exactly sure about these steps...but I think they're correct..


Point $B$ is in the exterior of the regular $n$-sided polygon $A_1A_2\cdots A_n,$ and $A_1A_2B$ is an equilateral triangle.  What is the largest value of $n$ for which $A_n, A_1,$ and $B$ are consecutive vertices of a regular polygon?

[hide="Simon-Favorite Factoring Trick"] We have $m\angle A_nA_1A_2=\frac{(180^\circ)(n-2)}{n}$ and $m\angle A_1A_2B=60^\circ$. Hence, $m\angle A_nA_1B=360^\circ-\frac{(180^\circ)(n-2)}{n}-60^\circ$. For this angle to be one of a regular polygon, it must be equivalent to $\frac{180^\circ(m-2)}{m}$ for some positive integer $m$. \[ 360^\circ-\frac{(180^\circ)(n-2)}{n}-60^\circ=\frac{180^\circ(m-2)}{m},  \] \[ 2-\frac{n-2}{n}-\frac{1}{3}=\frac{m-2}{m},  \] \[ \frac{6n-3(n-2)-n}{3n}=\frac{m-2}{m},  \] \[ \frac{2n+6}{3n}=\frac{m-2}{m},  \] \[ (2n+6)(m)=(3n)(m-2),  \] \[ 0=nm-6m-6n,  \] \[ 36=nm-6m-6n+36,  \] \[ 36=(m-6)(n-6).  \] $n$ must be an integer, so it is clearly maximized when $n-6=36$ and $m-6=1$. Therefore the desired answer is $\boxed{042}$. Go figure. :roll: [/hide]

Shouldn't the order in the problem be $ A_n$, $ A_1$, and then $ B$? The current order doesn't seem to make sense...

You're right.

(well, except for that #1 in your sig .. )  :P


function
quadratics
real analysis
real analysis unsolved

Study if there is a derivable function $ f: R->R$ with:

$ f'(x+f(y))=y+f'(x) \forall x,y \in R$

Integrate to $ x$:
$ f(x \plus{} f(y)) \equal{} yx \plus{} C \plus{} f(x)$
Differentiate to $ y$:
$ f'(x \plus{} f(y))\cdot f'(y) \equal{} x$
Integrate with respect to $ x$:
$ (f(x \plus{} f(y)) \minus{} f(f(y)))\cdot f'(y) \equal{} \frac {x^2}{2}$
So if $ f$ exists, then it is a quadratic function.
Substitution of  a quadratic function shows that there is no solution.

Yes
Another proof:if we take x=0 it results $ f'(f(y))\equal{}y\plus{}a$ (1)
where we put $ a\equal{}f'(0)$
From this it is clear that $ f$ is an injection,and because it is also continuous(it is differentiable) it results that $ f$ is strictly monotonous.
This is obvious a contadiction,because $ f'$ must have the same sign on $ R$ but we can see from (1) $ f'(f(y\minus{}a))\equal{}y$,so $ f'(R)\equal{}R$.

$ x \mapsto f(y)f'(x) \minus{} yx$ is periodic with period $ f(y)$.
Then (differentiate) $ x \mapsto f'(y)f'(x) \minus{} x$ is periodic with period $ f(y)$,
which means $ yf'(y) \equal{} f(y)$, or $ f$ is linear, which is not possible.


Unlike [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=169876]previous[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=164380]examples[/url], this one doesn't require extremely high precision, and is accessible using relatively simple inequalities that can be done by hand.
I still don't see the appeal, but some might like this.

[I used a spreadsheet to test my suspicions, and won't say any more about the answer]

I suggest truncating the problem, meaning actually alter the problem to a close one, but slightly shorter. (Get rid of the 15^(1/8) term; it doesn't fit in with inequalities like the ones below, because it crowds the problem.)

The following does not use calculus or decimal approximations of roots.  It does use inequalities involving roots of perfect powers.

15^(1/2) <  16^(1/2)......= 4
15^(1/3) <  27^(1/3)......= 3
15^(1/4) <  16^(1/4)......= 2
15^(1/5) <  32^(1/5)......= 2
15^(1/6) <  64^(1/6)......= 2
15^(1/7) < 128^(1/7)....= 2
------------ . . . . . . . . . . . . -----
15^(1/2) +...+ 15^(1/7) < 15


This is actually an easy problem but I found it hard to show algebrically and I want to know other solutions. Please help. Thanks in advance.

A, B and C run around a circular track of length 750m at speeds of 3 m/sec, 6 m/sec and 18 m/sec respectively. If all three start from the same point, simultaneously and run in the same direction, when will they meet for the first time after they start the race?

[hide="hint"]
Think mods.
[hide="further hint"]
$3x \equiv 6x \equiv 18x \mod{750}$
[/hide][/hide]

[hide]They meet again for the first time only when A finishes one lap.  Proof: Clearly, B will have to have lapped A exactly once when B meets A again.  B meets A again at the starting point, when B has run 2 laps, and A has run 1 lap.  Well, the first time that happens, C is there as well, having run 6 laps...[/hide]


$ x,y$ integer number . $ y^{3}\equal{}x^{3}\plus{}8x^{2}\minus{}6x\plus{}8$ find $ x,y\equal{}?$

[hide="hint"] we use the well-known idea,we put the expression between two cubes(consecutive one) :wink: [/hide]


The perimeter of a retangle is 100 and diagonal mesure is x. What's the area of the retangle ?

A) 625 – x²
B) 625 – $ \frac{x^{2}}{2}$
C) 1250 – $ \frac{x^{2}}{2}$
D) 250 - $ \frac{x^{2}}{2}$
E) 2500 - $ \frac{x^{2}}{2}$

[hide="Solution"]Let the length of the rectangle by $ y$, and the width $ 50\minus{}y$. $ x^2 \equal{} y^2 \plus{} (50\minus{}y)^2 \equal{} y^2 \plus{} 2500 \minus{} 100y \plus{} y^2 \equal{} 2y^2 \minus{} 100y \plus{} 2500$. We want to compute $ A \equal{} y(50\minus{}y) \equal{} 50y \minus{} y^2$. $ x^2 \equal{} \minus{}2A \plus{} 2500$. $ 2A \equal{} 2500\minus{}x^2$. $ A \equal{} 1250 \minus{} \displaystyle\frac{x^2}{2}$. The answer is $ \boxed{C}$.

[/hide]

[quote="darkdieuguerre"][hide="Solution"]Let the length of the rectangle by $ y$, and the width $ 50 \minus{} y$. $ x^2 \equal{} y^2 \plus{} (50 \minus{} y)^2 \equal{} y^2 \plus{} 2500 \minus{} 100y \plus{} y^2 \equal{} 2y^2 \minus{} 100y \plus{} 2500$. We want to compute $ A \equal{} y(50 \minus{} y) \equal{} 50y \minus{} y^2$. $ x^2 \equal{} \minus{} 2A \plus{} 2500$. $ 2A \equal{} 2500 \minus{} x^2$. $ A \equal{} 1250 \minus{} \displaystyle\frac {x^2}{2}$. The answer is $ \boxed{C}$.

[/hide][/quote]

Thanks, but I don't understand this line:

$ x^2 \equal{} \minus{} 2A \plus{} 2500$. $ 2A \equal{} 2500 \minus{} x^2$. $ A \equal{} 1250 \minus{} \displaystyle\frac {x^2}{2}$. The answer is $ \boxed{C}$.

How do you find that x² = -2A +2500 ?
How to do this ?

\begin{align*}
\left\{\begin{array}{l}
2(a+b)=100 \\
a^2+b^2=x^2
\end{array}\right.
&\iff
\left\{\begin{array}{l}
a+b=50 \\
a^2+b^2=x^2
\end{array}\right. \\
&\iff
\left\{\begin{array}{l}
a+b=50 \\
ab=\frac{(a+b)^2-(a^2+b^2)}{2}=\frac{50^2-x^2}{2}=1250-\frac{x^2}{2}
\end{array}\right. \\
\end{align*}

Thus, $ \boxed{C}$.

Thanks Nakagawa, good method, and now I understand. Thanks ;)


What books would anyone recomend for olympiad level geometry? Im so terrible at it, i need it for the IMO...

For inequalities I recomend "Old and new inequalities" from Gil Publishing House.The authors are T.Andreescu,G.Dospinescu and others.

I recomend two books:

   1) Virgil Nicula-"Geometrie sintetica,vectoriala si analitica" GIL Publishing House
   2) Marian Dinca & Marcel Chirita-"Numere complexe in geometrie"

Errr, i only speak english, are those books in english?

No they are not. But, I think you understand what I have written. Am I right?

I'm sorry. I find the above conversation hilarious. :laugh:

Vector, Analytical, and Classical??? Geometry, and Geometry with Complex Numbers?  LOL :rotfl:

Youve never done geometry with complex numbers??? I had more classical geometry in mind...

Books: FJM 1,2,3 and Titeica and Lalescu.


Find the maximum and minimum of $ f(a,b,c) \equal{} \frac {a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3}{(a^2 \plus{} b^2 \plus{} c^2)^3}$ for $ a,b,c\in\mathbb{R}$.

@bakerbakura: Please check your problems after posting them. $ d$ is missing.  :wink:

[quote="bakerbakura"]Find the maximum of $ f(a,b,c,d) \equal{} \frac {a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3}{(a^2 \plus{} b^2 \plus{} c^2)^3}$ for $ a,b,c,d\in\mathbb{R}$.[/quote]
$ \frac{1}{8}.$

Sorry about the mistake! Also notice that I have included the minimum as well...


quadratics
geometry
3D geometry
algebra
quadratic formula
number theory solved
number theory

Find all integer solutions of:

x^2+1=y^3

Bye

if i remember correctly it is from one of mathlinks contest and if my memory doesnt betray it has been solved by Z[i].

Ok but do you know the smallest solution?

Wouldn't the smallest solution be x=0 and y=1?

Becuase 0^a and 1^b equal 0 and 1 respectively for all numbers a and b.  Since 0 is 1 less than1, and both 2 and 3 are numbers, 0^2+1 = 1^2

A try
x^2=(y-1)(y^2+y+1)
Since y^2+y+1 is between y^2 and (y+1)^2, so y^2+y+1 can never be a perfect square, so (y-1) and (y^2+y+1) must have common factors>1.
then (y-1)(y+2)=y^2+y-2 and y^2+y+1 must have common divisors. The only possible common divisor they have must be 3, thus y=1(mod 3).
y-1=3a^2, y^2+y+1=3b^2. Use quadratic formula $y= \frac{-1+\sqrt{3(4b^2-1)}}{2}$, so 4b^2-1=3k^2 where k is integer. (2b)^2-3k^2=1. Apply Pell's equation, solutions for (b, k) are (1, 1), (13, 15), (181, 209)....
But y=3a^2+1, stuck there....

the soln that i mentioned before was from valentin vornicu.and it is realy nice .u can easily conclude that x is even number.and then in Z[i] u can also conclude that x+i and x-i are coprime so they should be product of unit and cube in Z[i] .so we can say x+i=(a+ib)^3 .and the remain is trivial.


Can we use "let alone" in affirmative sentence?

Thanks in advance

Do you mean something like "It is not true that (weak statement), let alone (stronger statement)." E.g., "I can't even picture higher dimensions, let alone do geometry in them." It's perfectly good English, but is not terribly common. The only other use I can think of is for "leave alone", as in "let that man alone!".

You can look at the free dictionary's idioms section: [url]http://idioms.thefreedictionary.com/let+alone[/url]

[quote="kunny"]Thank you for your reply.

Is the following sentence correct in grammaticaly?

She speaks French let alone English.[/quote]
I'm not an expert on this but this sentence doesn't sound right to me. EDIT: I see what you mean here, I guess you want to use 'let alone' as an inclusion, like 'She speaks French and English', as if that's not enough language! ;)

The affirmative usage would be something like '(general statement), let alone (less general statement)'

For example: kunny is an avid calculus problem solver, let alone doing 'Today's Calculation of Integral'!!! :)

In that sentence construction, "and" is definitely the right word to use. If you want the emphasis and connotations of "let alone", modify the second part with something like "as well".

For something that sounds right: "I can't believe she speaks French, let alone English!"

I am pretty sure the first clause has to be negative to use "let alone".

[quote="kunny"]Thank you for your reply.

Is the following sentence correct in grammaticaly?

She speaks French let alone English.[/quote] I am not sure but it looks wrong


quadratics
algebra
quadratic formula
algebra unsolved

1) $p,q \in Z$ satisfy the equation:
$x^{2}+px+q=0$
Determine the roots of equation.
2) Let $m,n \in N$ satisfy the equation
$x^{2}-m(n+1).x+m+n+1=0$ which has all roots are positive integers. Prove that $mn \leq 4$

Oh, are they very difficult?
Who can solve them? :D

I think people just aren't interested.
[hide="1)"]
$x^{2}+px+q=0=\left(x+\frac{p}2\right)^{2}+q-\frac{p}4$, so we can easily solve for the two roots which are the positive or negative square root (yes that was a derivation of the quadratic formula)
[/hide]
Same Idea for other one


[quote="mayhem"]Find a $ n$ such that the number $ n^2 \minus{} 1$ have $ 10$ positive divisor.
Find a $ n$ such that the number  $ n^2 \minus{} 2$ have $ 10$ positive divisor.
Find a $ n$ such that the number  $ n^2 \minus{} 3$ have $ 10$ positive divisor.

> I edited <[/quote]
Let $ d(n)$ be the number of divisors of $ n$.

$ n^2 \minus{} 1 \equal{} (n \minus{} 1)(n \plus{} 1)$.
Case $ n$ is even:
Then $ d(n^2 \minus{} 1) \equal{} d(n \minus{} 1) \cdot d(n \plus{} 1)$. $ n \equal{} 2$ gives $ n^2 \minus{} 1 \equal{} 3$ which doesn't have $ 10$ divisors. So assume $ n > 2$. Obviously $ d(n) > 1 \forall n > 1$ so either $ d(n \minus{} 1) \equal{} 2$ and $ d(n \plus{} 1) \equal{} 5$ or $ d(n \minus{} 1) \equal{} 5$ and $ d(n \plus{} 1) \equal{} 2$.
Case $ d(n \minus{} 1) \equal{} 2$ and $ d(n \plus{} 1) \equal{} 5$:
Then $ n \minus{} 1 \equal{} p$ and $ n \plus{} 1 \equal{} q^4$ where $ p$ and $ q$ are primes. So $ q^4 \minus{} 2 \equal{} p$. $ q \equal{} 3$ gives $ p \equal{} 79$. $ q \equal{} 7$ gives $ p \equal{} 2399$. $ q \equal{} 11$ gives $ p \equal{} 14639$. $ q \equal{} 13$ gives $ p \equal{} 28559$. $ q \equal{} 29$ gives $ p \equal{} 707279$. $ q \equal{} 41$ gives $ p \equal{} 2825759$. usw...
Case $ d(n \minus{} 1) \equal{} 5$ and $ d(n \plus{} 1) \equal{} 2$:
Then $ n \minus{} 1 \equal{} q^4$ and $ n \plus{} 1 \equal{} p$ where $ p$ and $ q$ are primes. So $ q^4 \plus{} 2 \equal{} p$. If $ q \neq 3$ then $ 3 \mid q^4 \plus{} 2$ so $ q \equal{} 3$. Giving $ p \equal{} 83$. So $ n \equal{} 82$ is a solution. $ 82^2 \minus{} 1 \equal{} 6723 \equal{} 3^4 \cdot 83$.
Case $ n$ is odd:
Then $ 8 \mid n^2 \minus{} 1$. So since $ n^2 \minus{} 1 \equal{} pq^4$ for primes $ p,q$. we get $ q \equal{} 2$. So $ n^2 \minus{} 1 \equal{} 16p$. So $ (n \minus{} 1)(n \plus{} 1) \equal{} 16p$.
Case $ n \minus{} 1 \equal{} 8$, $ n \plus{} 1 \equal{} 2p$:
Then $ 2p \equal{} 10$ or $ p \equal{} 5$. So $ n \equal{} 9$ is a solution.
Case $ n \minus{} 1 \equal{} 2p$, $ n \plus{} 1 \equal{} 8$:
Then $ 6 \equal{} 2p$ so $ p \equal{} 3$, so $ n \equal{} 7$ is a solution.

So a smallest solutions are $ n \in \{7, 9, 80,82,2398,14638,27557,707278,2825758,...\}$


Find positive integers $ a,b$ such that the following equation satisfies: \[ (a\minus{}b)^5\equal{}(a\plus{}b)^3\minus{}a^2b^2\]
 :)

No solutions!
A very obvious one is $ (a,b) \equal{} (5,2)$
 :)
I am not sure whether any more solution exists.... :maybe:

If $ a\equal{}b$,
$ (a\minus{}a)^5 \equal{} (a\plus{}a)^3 \minus{} a^2a^2 \iff 8a^3\equal{}a^4 \iff a\equal{}0, 8$
So, $ ((a, b)\equal{}(0, 0), (8, 8))$

If $ b\equal{}0$,
$ (a\minus{}0)^5 \equal{} (a\plus{}0)^3 \minus{} a^20^2 \iff a^5\equal{}a^3 \iff a\equal{}0, \pm 1$
So, $ (a, b) \equal{} (0, 0), (1, 0)$

I don't try the other possibility.
But, In the range of $ 0 \le a \le 500$, $ 0 \le b \le 500$, , a solution is only $ (a, b)\equal{}(0, 0), (1, 0), (5, 2), (8, 8)$.
(by Mathematica)

In my problem, I mentioned $ a,b>0$, see.
So, only two solutions within $ 500$ is $ (5,2) \& (8,8)$! That's amazing indeed! :lol:
I wonder what surprises lie beyond 500!

[quote="Sunkern_sunflora"]???????

Are you writing Java? I don't know Java.
What does $ \{3455.41 \ \text{Second}, \text{Null}\}$ mean?[/quote]
No, That's Mathematica programing. The computer was instructed to find all possible solutions with $ a,b\le 10000$, and it returned the two solutions and displayed that no other solution found and that total time taken= $ 3455.41$ seconds.

[quote="Sunkern_sunflora"]???????

Are you writing Java? I don't know Java.
What does $ \{3455.41 \ \text{Second}, \text{Null}\}$ mean?[/quote]

cf. [url=http://reference.wolfram.com/mathematica/ref/Timing.html]Timing[/url] - Wolfram Mathematica

But is there a soln? Can this be mathematically deduced?

[hide="does it help?"]
Let $ d\equal{}a\minus{}b$ then $ (a \minus{} b)^5 \equal{} (a \plus{} b)^3 \minus{} a^2b^2$ becomes
$ d^5\equal{}(2b\plus{}d)^3\minus{}(b\plus{}d)^2b^2$
trivial cases for $ d\equal{}0,1,2,3$

[/hide]


function
analytic geometry
graphing lines
slope
calculus
calculus computations

If the line tangent to the graph of the function f at the point (1,7) passes through the point (-2,2) then f'(1) =
Can someone please explain this to me? Thanks

The line passing through those 2 points is defined by $y = 5/3*x+16/3$
$f'(1)$ means: what is the slope of the tangent line at that point and from the equation of the line we know that the slope is $5/3$ so $f'(1)=5/3$

could it possibly be another answer because 5/3 dosen't match any of the choices that i have?
the choices are
a. 1
b. 3
c. -5
d. 7
e. undefined

This is simple. $f'(1)=\frac{7-2}{1-(-2)}=3.$ thus the ansawer is $b.$

Last edit: $f'(1)=\frac{7-2}{1-(-2)}=\frac{5}{3}.$ thus the answer is $e.$

kunny

You may want to check your arithmetic.... Maybe the second point is really supposed to be $(-2, -2)$? Then kunny is right :D

edit: it's been fixed. Yay!

Yeh, oops. :blush: Thanks, Xevarion.

kunny


The perimeter of a square lot is lined with trees, and there are three yards between the centers of adjacent trees. There are eight trees on a side, and a tree is at each corner. What is the number of yards in the perimeter of the lot?

Since there are 8 trees on a side, there are 7 spaces between trees.
Each space is 3 yards, so the length of each side is $ 7*3 \equal{} 21$ yards.
The perimeter is therefore $ 21 * 4 \equal{} 84$ yd.


let O is the center of circumcircle of ABC triangle.M is a point on AB arc.K & N are 2 point on AB such that :
$ MK\bot OA and MN\bot OB$
(MK,AC)=L and (MN,BC)=P.
what is <MLP ?
 :) say it by <A,<B,<C


geometry
trigonometry
circumcircle
function
calculus
calculus computations

Consider the situation such that $ n$ circles with radius $ r$ are contained between 2 concentric circles with radius $ 1,\ 1\minus{}2r$ without intersecting one anoyher.Let $ r$ vary with being fixed $ n\geq 2.$

(1) Show that $ r$ should be satisfied $ 0<r\leq \frac{\sin \frac{\pi}{n}}{1\plus{}\sin \frac{\pi}{n}}.$

(2) Find the values of $ r$ for which the sum of the areas of $ n\plus{}2$ circles is minimized.

OK, we can easily see that the centers of small circles make a cyclic $n$-gon, circumradius $1-r$. If there are no 2 small circles intersecting, then the side which those 2 centers make is at least $2r$. Now, let $2a$ be the side of a regular $n$-gon inscribed in the same circle with radius of $1-r$. It's easy to see that $a=(1-r)x$, where $x=\sin\frac{\pi}{n}$.
Now, an easy argument gives us that $a\ge r$. Which one?
Well, let's start from one of $n$ centers of small circles. Then if each side of the $n$-gon is more than $2a$, we would get that the last side must be smaller than $2a$, so every side is smaller or equal to $2a$, and we've explained before why every side is at least $2r$. That means $a\ge r$, from which 1) is easily derived.

Now, sum of the areas of all the circles is $P=r^2\pi n+\pi +\pi+4r^2\pi-4r\pi$, so we should minimize $A=r^2(4+n)-4r$. It's easy to see that $(r\sqrt{n+4}-\frac{2}{\sqrt{n+4}})^2\ge 0$, so $A\ge \frac{-4}{n+4}$

Equality holds for $r=\frac{2}{n+4}$. Now we must prove that that $r$ satisfies 1). To prove that, we need to prove $nx\ge 2(1-x)$, which always holds, because the RHS is at most 2, and LHS is a increasing function of $n$ with limit $\pi$, and $LHS(2)=2$


[hide]As 2.7=3^3*10^-1, we need to find this a certain power of 10 that is one greater than a perfect cube. The only possibility is 2.7*10^28.[/hide]

[hide]$2.7=3^2*10^-1$ so we need a number that is exactly one greater than a perfect cube to have a solution.  28 is that number.[/hide] [color=green]mod edit: use hide tags[/color]

[quote="kishiman1"]i disagree 2.7 does not equal 3 sqaured x 10 to the -1. that equals .9. so that makes you wrong.[/quote]

wow give him a break you can see what he meant and you can also see that it was a typo

[hide="My answer"]The answer is 2.7 *10^28[/hide]

I don't get it... what's "whole number"??

[quote="fireemblem13"]whole numbers are...
the numbers 0, 1, 2, ...[/quote]

I know now, thanks...


Warning: I saw this mentioned on MathWorld and I thought it was neat. I haven't tried to prove it myself so I don't know how hard (or easy) it is.

Prove that if a, b, and c are three distinct rational numbers, then [tex] \displaystyle {\frac {1}{(a-b)^2} + \frac {1}{(b-c)^2} + \frac {1}{(c-a)^2}}[/tex] is the square of a rational number.


geometry
Siemens Competition
number theory

Is there a way to see the research papers of old Siemens Competition Winners/Applicants?  I want to see what it takes to write a very good research paper, and I want to see what a good research paper requires (especially by a HS student).

I found a link where it gave abstracts of old winners, but I can't find one that gave me entire research papers.

The reason that I am asking is because I am interested in the Siemens Competition, and I will defiently be submitting the project that I am working on this summer.  However, I have never written a formal research paper and do not know what it requires/what it looks like.  It would be very nice to see some old research papers to see how to properly write/cite a research paper.

Also, I have seen research papers of my mentor and also I can find tons of research papers online, but all of them are written by Ph.D.'s, unversity professors, that have worked on their projects for a long time.  Obviously, the work of a high school student should not be at the same level, especially working on it for only 3-5 months.  For people that have done this competition, are the research papers of high school students comparable to real research papers by professors (or am I underestimating)?

Thanks.

You're not going to find entire papers.  I know that Intel, in their semifinalist booklet, gives abridged versions of the top 3 or so papers, but not the whole thing.  They care about not only how well the paper is written, but your project itself, the results, how the results were analyzed, and major implications.

I believe that both t0rajir0u and JBL were Intel Finalists so you may want to ask them for more detail.

actually nvm, I found tons of research papers on RSI's website.

Can you give me the exact website?
Also, do you know if the judges want the contestants to solve the previously unsolved problems or find new approaches to them? or is there something easier? i just want to know if i have the purpose of this competition right....

the future- is yours a math project?
If so i would be interested how you found a mentor? did he suggest a problem etc.. Yea ill probably enter too..

If you go to a program like RSI, you'll get a mentor.  If you're going to enter these competitions on your own, you'll have to find one yourself.  Contact people at local universities and see if anyone is willing to work with you.

When I contact a professor do I have to already have the problem I want to work on in mind, or typically do they assist you in choosing a topic?

I would first look at the faculty list and check out their areas of research.  For example, if you know you want to do research in number theory, it probably wouldn't be a good idea to contact someone who does all their research in PDE and Fourier analysis.  Once you've narrowed down your list, contact them, tell them what you're interested in, what you would like to do in the future (for example, enter a research competition), and ask them for advice, and if they know anyone who could help you.  I wouldn't straight out ask them to help you-let them bring up the suggestion.  If all else fails, then yes, you should probably ask.

Thank you JRav, hopefully contacting them in August won't be too late.

Are you trying to finish your research for this year competition?  I'm working on a project in physics - I just emailed a professor that did research in the field that interested me and asked him if he would be willing to serve as a sort of "mentor".  Luckily, he said yes and had a topic for the research project!  So if you really know what you want to do great, otherwise get some suggestions from some professors.

yes i am planning to try and do something before october this year.

I guess it always possible  :) but I think typical Intel/Siements projects take at least one year to do.

Siemens is entered in either September or October, and Intel in November.  Plus you'll need to complete the entire application, which for Intel, will take you several weeks to make perfect (not to mention you'll need letters of recommendation).

I would say you're really pushing it.

yea, i agree. I was thinking I might* do siemens in october or just work for a year and enter in 2010

You can't enter Intel until you're a senior so if you're a junior or lower, you still have time.  I still wouldn't enter Siemens with something you worked on for a month.

I originally intended to submit my research in the Intel competition (that was before I had ever heard of Siemens), however the competitions seem to be pretty similar so I was thinking I would enter both.  Do a lot of people do this?  Also would there be any disadvantages?

A lot of people do enter both-Siemens usually receives less applicants, one major reason being that Siemens does not allow projects involving humans (i.e., psychology and behavioral sciences).  At my high school, the teachers typically view Siemens as "warm-up" for Intel, although that doesn't really make sense to me.  This is in spite of the fact that a few years ago, we had the number 3 finisher nationally (he took home 50,000 I think).


inequalities
geometry
3D geometry
triangle inequality

[quote][color=darkred]Consider the complex numbers $ a_k\ ,\ k\in\overline {1,n}$ so that for any $ p\in \mathcal N^*$ ,
$ \prod_{k \equal{} 1}^n\left(1 \plus{} a_k^p\right) \equal{} 1$ . Prove that for any $ k\in\overline {1,n}$ , $ a_k \equal{} 0$ .[/color][/quote]

[quote="Singular"]Remove all $ a_i$'s that are zero without affecting the product.

Then $ |1| \equal{} \prod^n |1 \plus{} a_i^p | \ge \prod^n (1 \plus{} |a_i|^p) \ge (1 \plus{} (\prod^n|a_i|)^{\frac {p}{n}} )^n$

implying $ 0 \neq \prod^n|a_i| \le 0$.[/quote]

I think that the triangle inequality step does not point the correct way. I don't see how you used all of the information in the problem anyway.

please give solution!!  im bashing my head on this

Hope this is right.

Let $ arg(x) : \mathbb{C} \rightarrow (\minus{}\pi, \pi]$, the angle of $ x$ in radians.

Say $ x$ is good iff $ arg(x) / \pi$ is rational.  If $ x_i$ is good, we can find $ a_i$ with $ arg( x_i^{a_i} ) \equal{} 0$.  So taking $ A \equal{} lcm(a_1, \dots )$ we have $ arg(x^{Ap}) \equal{} 0$ for any natural $ p$, any 'good' $ x$.

Now if $ x_i$ is not good, for any $ \epsilon$ we can find $ B$ so that all $ |arg(x_i^{AB})| < \epsilon$. [hide="Why?"] Put $ arg(x_i^A) \equal{} 2\pi \alpha_i$ where $ \alpha_i \in (0,1)$ is irrational. (Here we are using $ arg : \mathbb{C} \rightarrow [0,2\pi)$.)  For any $ N$, consider the points in $ [0,1]^n$, $ p_k \equal{} (\{k\alpha_1 \}, \{k\alpha_2\}, \dots, \{k\alpha_n\})$ for integer $ k \in  [1,1 \plus{} N^n]$.   If we split $ [0,1]^n$ into $ N^n$ $ n$-cubes of length $ 1/N$, then two of the points $ p_i$ and $ p_j$ are in the same mini-cube.  Wlog $ i>j$, so $ \forall_{k\equal{}1}^n (i\minus{}j)\alpha_k < \frac{1}{N}$. [/hide]
Now for $ P \equal{} AB$, we have $ | 1 \plus{} x_i^P | \ge 1$ with equality when $ x_i \equal{} 0$.

Note that this solution only used the condition $ 1 \equal{} \prod | 1 \plus{} x_i^p |$.


[color=darkblue]Let $ O$ be the center of the circumscribe circle of the triangle $ ABC$.The lines $ AO,BO,CO$ are meeting the oposite sides in points $ D,E,F$.Prove that :
\[ \frac{1}{AD}\plus{}\frac{1}{BE}\plus{}\frac{1}{CF}\equal{}\frac{2}{R}\][/color]

By relation of Gergonne we obtain:
$ \frac{OD}{AD}\plus{}\frac{OE}{BE}\plus{}\frac{OF}{CF}\equal{}1$
$ \frac{AD\minus{}R}{AD}\plus{}\frac{BE\minus{}R}{BE}\plus{}\frac{CF\minus{}R}{CF}\equal{}1$
$ 3\minus{}(\frac{R}{AD}\plus{}\frac{R}{BE}\plus{}\frac{R}{CF})\equal{}1$
$ R(\frac{1}{AD}\plus{}\frac{1}{BE}\plus{}\frac{1}{CF})\equal{}2$
$ \frac{1}{AD}\plus{}\frac{1}{BE}\plus{}\frac{1}{CF}\equal{}\frac{2}{R}$, q.e.d.

This is the simple and the nicest way to solve this problem.
Here is a Trigonometric proof, if you don;t know Gergonne's relation.
[hide]Let $ AA'$ be the alttude from $ A$.Then with the sin law:
$ AA'%Error. "ADcos" is a bad command.
(C-B)=ACsinC=2RsinBsinC$ therefore $ \frac{2R}{AD}=\frac{cos(C-B)}{sinBsinC}$
For $ BE$ and $ CF$ is the same and by suming we obtain \[ 2R(\frac{1}{AD}+\frac{1}{BE}+\frac{1}{CF})=\frac{cos(C-B)}{sinBsinC}+\frac{cos(A-C)}{sicCsinA}+\frac{cos(B-A)}{sinBsinA}\]
Therefore we have $ 2RsinAsinBsinC(\frac{1}{AD}+\frac{1}{BE}+\frac{1}{CF})=4sinAsinBsinC$[/hide]

This is a repeated post... Problem 6 of 1985 Iberoamerican Olympiad :)

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