calculus
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he's another proof i know but i've never gotten any feed back on it, im not sure where i learned it, which adds to the uncertainty, but i cant find any flaws.  its for the chain rule.

$y = f(g(x))$

$y = f(u)$
$\frac{dy}{du} = f'(u), du \neq 0$

$u = g(x)$
$\frac{du}{dx} = g'(x)$

$\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}, du \neq 0$

$\frac{dy}{dx} = f'(u)g'(x)$

$\frac{dy}{dx} = f'(g(x))g'(x)$

if du equals zero then $\frac{du}{dx}$ or $g'(x)$ equals 0.  if you applied the rule to the last equation, then the derivitive would equal zero which is most likely false.  also if $\frac{du}{dx} =0$ then the chain rule doesnt apply.

[quote="oooooh ya"]
$\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}$
[/quote]

That's the key. How do you know that ?

the du's cancel or:

$\frac{dy}{dx} = \frac{dy}{dx} \frac{du}{du}$

$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$

eg.
$\frac{5}{7}\frac{4}{3} = \frac{5}{3}\frac{4}{7}$

They are not fractions. The notation is just a way of remembering the chain rule. What you've done is using the chain rule in your proof for the chain rule.

[url=http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/proofs/chainruleproof.html]First result from google for proof of the chain rule[/url]

Kalle is right in saying that leibniz notation uses a fractional form, but each term isn't a fraction and therefore cannot be operated on like that.  The notation is like a mnemonic device.

they're fractions, i dont want to get into a lenghty arguement like the e post so this is all i'll write:

its standard practice for multivariable calculus, and my book has is as an alternative definiton of the chain rule.  math world has it as a definition here: http://mathworld.wolfram.com/ChainRule.html

i've used the definition of the chain rule in my proof, $\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}$, but i could have written out how i arrived to that in my second post.  i dont like when people think i make this stuff up and it happens to work out correctly, i dont, and due to the nature of math if it works out correctly and no rules have been broken then its not made up and its correct.

i dont know anyone who's into math, so i was hoping to get some peer reviews from this forum, but my l'hopital proof got 35 views and not even a single y/n, this forum is turning out to be more aggrivation than its worth!

Do you ever accept that you may be wrong? They are absolutely not fractions, but the notation is clever in that for first order derivatives it seems like they do behave as fractions. Some people think this is beautiful, I think it is deceptive.

edit: The thing is that $\"\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}\"$ is a statement of the chain rule. So, if you prove this, you've proven the chain rule. But I haven't ever seen a proof that actually deals with differentials. I despise writing the chain rule in that way, because it doesn't specify what $y,u,x$ are, and differentials are confusing because $\frac{dy}{dx}$ and $\frac{dy}{du}$ are two formally different things in that context. The notation also lets you solve differential equations without thinking about some important things like local invertability (i.e. the derivative is not allowed to be zero infinitely often in the interval for the "differential notation" to work since we are assuming an inverse exists at least on some interval). I think you should not bother with differentials and instead see the chain rule as

$\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)$ (with the conditions).

Note that the chain rule is not the only reason the notation behaves like fractions. Another theorem that is quite easy to prove from the definition of derivatives is that if $f'(a) \ne 0$ and if $b = f(a)$, then $f^{-1}$ is differentiable in $b$ and $\left(f^{-1}\right)'(b) = \frac{1}{f'(a)}$.

This can be written as $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$. While this again is a rule that can be applied without thinking, I don't like it because you don't think of the necessary conditions.

To exemplify my dislike of the notation:
Suppose $\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}$. Using the theorem I gave this implies that $\frac{dy}{du} = \frac{dy}{dx}\cdot\frac{dx}{du}$. While this will work in any case where $\frac{du}{dx} \ne 0$, do you really have any idea what you just did? Try to sort it out in the other notation and you will have a mess because of the formal differences between $x$ as a variable and $u$ as a function of $x$.

Some technicial/practician math consider derivative as the fraction of two very small number, to help in calculation. But In math we don't define it as a fraction.  Do you ever hear about infintesimal?
$dy$ and $dx$ has been defined as differential,and $\frac{dy}{dx}$ is just the Leibnitz notaion of derivative .
So what we write in chain rule $\frac{dy}{dx}= \frac{dy}{du} \frac{du}{dx}$  is just the way to help us remember it,  just like Kalle said.

The definiton of differential $dy$ and $dx$ in the function $y=f(x)$ is as bellow
\[ dy = f'(x) \, dx  \]
If we have this definition we can now think $f'(x)$ as the fraction of two differential.

the def of a limit is:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

which is the slope, where f(x+h) - f(x) is the rise, and h is the run, $\frac{f(x+h) - f(x)}{h} = \frac{\Delta y}{\Delta x}$

whenever you use a rule for taking a derivitive, its the just a manipulation of the definition, so all derivitives are a limit of a fraction of the rise over run as the run approaches zero.

here's the chain rule in functional notation:

$h(x) = f(g(x))$

$\frac{\Delta h(x)}{\Delta g(x)} = \frac{f(g(x + h)) - f(g(x))}{g(x+h)-g(x)}$

$\frac{\Delta g(x)}{\Delta x} = \frac{g(x + h) - g(x)}{(x + h) - x}$
$\frac{\Delta g(x)}{\Delta x} = \frac{g(x + h) - g(x)}{h}$

$\frac{\Delta h(x)}{\Delta g(x)} \frac{\Delta g(x)}{\Delta x} = \frac{\Delta h(x)}{\Delta x}$

$\frac{\Delta h(x)}{\Delta g(x)} \frac{\Delta g(x)}{\Delta x} = \frac{f(g(x + h)) - f(g(x))}{g(x+h)-h}\frac{g(x + h) - g(x)}{h}$

$\frac{\Delta h(x)}{\Delta x} = \frac{f(g(x + h)) - f(g(x))}{g(x+h)-g(x)}\frac{g(x + h)) - g(x))}{h}$

for h approaches zero:

$\frac{\Delta h(x)}{\Delta x} = f'(g(x))h'(x)$

You mean the definition of the derivative, right?

Anyway, that proof is not correct for some weird functions. The proof doesn't work if you can't choose an interval around 0 where $g(x+h) - g(x) \ne 0$. Of course in most functions you can, but there are better proofs of the chain rule that does not require this.

ooooh ya derivitive :oops:

[quote="Kalle"]that proof is not correct for some weird functions. The proof doesn't work if you can't choose an interval around 0  where $g(x+h) - g(x) \neq 0$[/quote]

im not sure what you mean, a lot of negatives, but the only way you can use the chain rule is if $g(x+h) - g(x)$ as h approaches zero doesnt equal zero, so that $g'(x)$ does not equal zero.

what sorta of wierd function would the proof not work for?

kalle, if your only goal in reading my posts is to try to prove them wrong can you just skip to the next one, we both have better things to do with our time.

For example for the function that is defined as $g(x) = x^2\cos\frac{1}{x^2}$ when $x\ne0$ and $g(x)=0$ when $x=0$. This is a continous and differentiable function everywhere, so $g'(0)$ exists. We have of course that $\lim_{h\to0}g(0+h) = 0$, but since $g$ oscillates infinitely many around the point zero the denominator $g(0+h)-g(0)$ will vanish infinitely many times around $h=0$ in your proof, and therefore your proof of the chain rule will not work for the derivative of $f(g(x))$ in the point 0. Your expression towards the end in the proof will not have a limit as $h$ tends to zero (at least not in the regular sense. there is a thread started by Kent Merryfield which I think might allow this to have a limit).

No, I am not looking for errors, but you are asking whether this is a correct proof, and my lecturer specifically pointed out that this was not a correct proof when he proved the chain rule. He called this a "high school proof". I am not going to write down a correct proof if that's what you're asking, primarily because you could find it in a book, and I also don't remember it.

Frankly, it seems like you are looking for some kind of psuedostringent proofs of things around you, without having the full idea of concepts like limits, continuity and other things. There is no point in trying to prove things until you have the knowledge to do so. It's also quite irritating that I wrote down a long explanation of why differentials behave like fractions, and you then tell me that my only point in replying to your messages is to tell you you're wrong.

kalle this is the last time im responding to one of youre posts

given:
$g(x) = x^2 cos(\frac{1}{x^2})$
$g'(x) = 2xcos(\frac{1}{x^2}) + x^2 \frac{d}{dx}[cos(\frac{1}{x^2})]$
$y = cos(\frac{1}{x^2})$

$y = f(h(x))$
where
$h(x) = \frac{1}{x^2}$ and $f(x) = cos(x)$

thats the only chain rule, the x squared is irrevelent, and neither of this vanish around zero,

i have to go, but dont tell me i have a full idea of concepts, you do.

Ok. Your response doesn't have much to do with what what I said. Read it again. It IS a counterexample to your proof.

I'm not trying to come off as arrogant or rude, but yeah, it does seem like you don't read what I am actually saying, or at least don't trust me at all. Why do you even ask if you're not willing to accept that you are wrong? I am 100 % sure about everything I say this thread unless the sentence starts with "I think" or "I believe".


I've been trying to teach my brothers about negative numbers for a while now. For example, I taught him how to add two negative numbers a while ago. Then, I taught him how to add one positive and one negative, which took A LONG time. As soon as I went back to adding two negative numbers though, I found he had forgotten.

Can anyone give me any suggestions of how to teach him about negative numbers? I'm more comfortable teaching at a higher level than this, because all this basic stuff I just learned as I went through school, since I only started real problem solving in 6th grade. Therefore, I can't remember the way that I learned it, so I can't teach my brother these basic things as well as I can teach higher level concepts.

does he know the basic concept of negative numbers? the idea of negative numbers is initially pretty confusing for most kids just learning it.

the way one of my teachers taught it was the "cha-ching" method.  :D  it goes like:

2+ (-3) = -1 because you "cha" the plus sign to a minus sign and you "ching" the negative sign to a positive sign. it then becomes 2- (+3)= -1. so basically, switch the + and - signs.

or, you can make him think of the negative signs as simply minus signs. that might work better.

[quote="qwertythecucumber"] you can make him think of the negative signs as simply minus signs. that might work better.[/quote]  What I was about to say

I remember that minus is very strong:
It's much stronger than plius  :)

But 2 minuses make a plius  :)

[quote="mewto55555"]Wait but |-x|=|x| so they are equally strong....[/quote]

I think that modulus are too advanced for xpmath's brothers   :roll:

Anyway...

What I used to (and still do) is:

2+(-3)=
3-2 (Switch the digits)
=1 then make negative:
-1

The hard reason it works:

x+(-y)=-(y-x)

He turned 8 on Saturday...

Yeah, I've already tried teaching him the switching numbers way. The main problem is that he can't seem to remember it...I'm guessing I should just teach it in an interesting way...

Mainly, the way I'm teaching him is to disregard the signs of the numbers. If you have an expression involving a positive and a negative, just cross out the minus sign, take the bigger number, and subtract the smaller number. Then, put the sign of the bigger number in front. I can explain better with an example:

Let's say you have -15+6. Don't think of the minus sign, so you just have 15 and 6. 15 is bigger, so first you subtract 6 from 15 to get 9. Then, since 15>6, and the sign in front of it is negative, just add a minus sign to get -9.

If he needs to do something with two negatives, just disregard the minus signs again and add the two numbers. Then, add the minus sign in front. Can anyone help me make my explanation better?

A lot of people can learn well with a visual aid. I suggest making a number line and then showing what it means to add a negative and a negative and a positive and a negative (and vice-versa). So just draw out a number line ranging from some negative number to some positive number. Then you can use circles and arrows to work out an addition or subtraction problem involving a negative (you can teach both with this method).

After getting used to this, they'll probably find it easier to add/subtract without a number line since they know what's going on when they're doing operations on negative numbers.

You can try this link...

I found out its easier to understand....

http://www.homeschoolmath.net/teaching/integers.php

[quote="mewto55555"]He's 8. He just turned 8 on the 9th. His birthdate was 11/9/00[/quote]
He turned 8 on the 15th :P

Thanks for the link. I'll probably use the number line then.

Here are a few tricks that I learned in my Pre-Algebra class last year.

1.) Let the negative=hate and let the positive=love. This trick only works when you are multiplying two integers;

If you are multiplying a positive by a positive, then you could say. I love to love. This is a positive statement. Similarly if you multiply three positive integers by each other, then you could say. I love to love to love. This is also a positive statement. If you are multiplying a negative by a positive, then you could say either one of these;

I hate to love.
I love to hate.

Both of those two statements are negative statements, so a positive number times a negative number results in a negative product. If you are multiplying two negative numbers by each other, then you get;

I hate to hate.

Hating to hate is a good thing, because you should like to hate. So this is a positive statement.

So if you want to compute the following product;

(-2)(9)(-3)(5)(8)(-2)

You could say;

I hate to love to hate to love to love to hate. This might be hard to figure out at first, but there is an easy way to shorten this sentence. You first look at the last two words which say "love to hate". You can just change this to hate to get the sentence;

I hate to love to hate to love to hate. The last two words say "love to hate". This means hate, so you have;

I hate to love to hate to hate. "Hate to hate" = love, so ;

I hate to love to love. "Love to love" = love so;

I hate to love. "Hate to love" = hate, so;

(-2)(9)(-3)(5)(8)(-2) is indeed negative.

When you are subtracting an integer by a negative integer, you are hating to hate, so you are actually adding the absolute value of the number.

Absolute value = The distance between the number and 0 on a number line.

This is the second thing that I learned;

If you are trying to solve the equation 5 + (-4) then I just think.

Wait, (-4) is just (0 - 4), so we substitute (0-4) for the (-4) to get;

5 + 0 - 4
This equals 1.

In other words, if you have the equation (a + -b), then you can substitute (0-b) for (-b) to get;
a + 0 - b. This equals a - b, so (a + -b) = a - b.

You can clarify that this equation is true by adding 'b' to both sides of the equation to get a=a.

Hope that helped.

For multiplying numbers, count up the negative signs.  If it's odd, the answer is negative.  If it's even, then the answer is positive.

edit: Oh wow, mewto.  Not spamming?

[quote="ernie"]

edit: Oh wow, mewto.  Not spamming?[/quote]

Umm ernie;

[quote="mewto55555"] How old is he? [/quote]

It doesn't matter anyway.

I guess me and ernie were both right about this. My explanation of multiplying integers explains mainly how it works, but ernie's explains how to do solve problems once you understand the idea of it.

Here's how I figured it out when I learned negatives.

For me, learning "tricks" doesn't really help - it's confusing and you usually don't understand it. So what I suggest is [i][b]Think About It.[/b][/i] Say you're solving 2+(-3). You've got two apples and you're adding -3. This is the same as 2-3. We have two apples, and lose three. Thus 2+(-3)=-1.

What if we have 4-(-5)? Subtracting a negative means addition(think about it!). So we have 4+5, which is clearly 9.

What about -6-7? The bigger the absolute value of a negative is, the smaller the number. (So -67 is smaller than -42.) Since we're subtracting 7, the number's getting smaller, so the absolute value of the number has to be bigger than the absolute value of -6. This becomes the equivalent of -(6+7)=-13.

As a final example, -9+4. We're adding a number, so the absolute value is going to be smaller than that of -9's. Clearly the answer is -5.

If this doesn't help your brother, here's another method.

Go back to our final example, -9+4. Let that expression equal $ x$.

$ \minus{}9 \plus{} 4 \equal{} x$

We don't like that tricky negative sign. Multiple both sides by a negative.

$ \minus{}1(\minus{}9 \plus{} 4) \equal{} \minus{}x$

$ 9 \minus{} 4 \equal{} \minus{}x$

$ 5 \equal{} \minus{}x$

So obviously x is 5.

This method may or may not work, depending on how comfortable your brother is with equations.

Multiplying by -1 probably won't make sense to him given that he can't add or subtract using negative numbers yet  :wink:

You may want to consider teaching multiplying and dividing negative numbers first. Personally, I have no idea why addition and subtraction are taught first. When I learned this stuff, I found addition extremely confusing, but multiplication was so much easier. I mean, two negatives make positive. Two positives make positive. A negative and positive make negative. Very easy to remember.

okay, does he know the number line.
make him think of it this way. suppose he has a really big numberline that he could stand next to, and the problem is

-7+3

because the first number is -7, he goes to -7. then, because the sign is positive, he faces the positive way and moves up three.

suppose it was -7+(-3)

again, he should go to -7 and face positives. this time, however, he should walk backwards...

now, suppose 7-(-4)

he goes to 7, sees the negative, so faces the negatives, and then walks backwards four, getting 11.

hope that helps...

This might be a good explanation;

If you have five apples, and you are planning on adding negative two it is the opposite of adding positive two, so it can be subtracting 2, or adding negative two, because either way is the opposite. That might be a little confusing.

Here is a site;

http://www.mathleague.com/help/integers/integers.htm

Clarifying (a bit) on what Wicked said:

There are 5 apples on the table.  You take away 2.  How many are left on the table?

The answer is clearly 3.  This can be expressed as 5-2, or 5+(-2).

Do you understand xpmath? Do your brothers understand? Your brothers should get an account on AoPS if they are allowed to.

Explain him the concept of negative number by using the numerical line (or numerical axis, I don't know how you call it)

The numerical line :

   [code]  x'                                                             x
                  --|---|---|---|---||---|---|---|---|---|---|-->
                   -4  -3  -2  -1    0    1   2   3    4   5   6[/code]
In relation with 0
 a) All numbers to the right are positive
 b)All number  to the left are negative

If you want to calculate  (-4)+5

1) Draw an arrow from 0 to (-4)
2) Because of you have to add (+5) , you draw an other arrow with initial point (-4) and move 5 unit numbers to the right
3) After youl do that , you will se that the arrow stoped at  the point which correspond to 1 .

             [code]    (-4)
                 <---------------
                 |---|---|---|---|---|---|---|-->
                -4  -3  -2  -1   0   1    2   3
                 -------------------->
                      +5[/code]

I hope this will help you

And you are going left when you subtract, and right when you add, so when you subtract a negative number, you are going to the left by a negative number, but the only way to do this is by going in the opposite direction by the absolute value of the negative of the number you are subtracting. Does that make sense?

[quote="Wickedestjr"]And you are going left when you subtract, and right when you add[/quote]
I didn't say that

Okay, here's how I like to show it.
Example: $ \minus{}9\plus{}5$
First you subtract the smaller number from the bigger number: $ 9\minus{}5$ to get 4

Look at the numbers from the previous step. Which one is bigger, 9 or 5? (9 is) Look at $ \minus{}9\plus{}5$ now. Since the 9 is bigger, you stick a negative sign onto the answer!!! $ \minus{}4$

hope that helps


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

compute the nth power of the following matrix:

a11=4 a12=1
a21=6 a22=9 (lets say that this is A). compute A^n.

cheers! :D :D

Another way for this problem Lagrangia,

Cayley-Hamilton theorem

A - (TrA)A + det(A)Id=0
the characteristic  polynomial is P(X) = X-13X+30

X^n = P(X)Q(X) + a_nX+b_n

A^n = a_n.A + b_n.Id

find  a reccurence relation on a_n,b_n to get A^n

My Japanese tea is ready

[quote="moubi"]

Cayley-Hamilton [/quote]

Cayley was at Trinity College, Cambridge
http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Cayley.html

Hamilton Trinity College, Dublin
http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Hamilton.html

It is time to take tea with Sir Cayley and Sir Hamilton

Moubinool


Given $ a, b, c$ and $ \frac {ab \plus{} bc \plus{} ac}{\sqrt {abc}}$ are all positive integers, does that imply that $ \sqrt {\frac {ac}{b}}, \sqrt {\frac {ab}{c}}, \sqrt {\frac {bc}{a}}$ must all be integers?

Clearly $ \sqrt {abc} \in \mathbb{N}$ so $ abc \equal{} k^2, k\in \mathbb{N}$

Write $ M \equal{} (a,b,c) \equal{} (\alpha^2 xy, \beta^2yz, \gamma^2zx)$

With $ \gcd(\alpha,\beta) \equal{} \gcd(\beta,\gamma) \equal{} \gcd(\gamma,\alpha) \equal{} 1$

[hide="constructive proof"]
Take $ M \equal{} (a,b,c)$ and let $ \gcd(a,b) \equal{} y \Longrightarrow M \equal{} (a'y,b'y,c)$

Let $ \gcd(a',c) \equal{} x \Longrightarrow M \equal{} (a''yx,b'y,c'x)$

Let $ \gcd(b',c') \equal{} z \Longrightarrow M \equal{} (a''xy,b''yz,c''zx)$

Since $ \gcd(a'',b'') \equal{} \gcd(b'',c'') \equal{} \gcd(c'',a'') \equal{} 1$  it follows that $ a'',b'',c''$ are perfect squares.

$ \therefore M \equal{} (\alpha^2 xy, \beta^2yz, \gamma^2zx)$
[/hide]

This gives

$ \frac {ab \plus{} bc \plus{} ca}{\sqrt {abc}} \equal{} \frac {\sum\alpha^2\beta^2y}{\alpha\beta\gamma}$

Hence $ \alpha | z, \beta | x$ and $ \gamma | y$

Therefore

 $ \sqrt {\frac {ab}{c}} \equal{} \sqrt {\frac {\alpha^2xy \beta^2yz}{\gamma^2zx}} \equal{} \frac {\alpha\beta y}{\gamma} \in \mathbb{N}$ because $ \gamma | y$


trigonometry
geometry
circumcircle
analytic geometry
geometry unsolved

Let $ABC$ a triangle with $A=50$°.Let $BM$ and $CN$($M\in AC,N\in AB$) be 2 secants which intersect in the point $P$ such that $m(PMN)=50$°,$m(PNM)=30$° and $BP=PC$.Find the measure of angle $ABP$.(In fact show that it is equal to 30°)

let me have a try... it ab-uses of trig, but anyway there's my solution

let $\angle ABP=\alpha$, so a simple angle-chase gives $\angle NCA=50-\alpha, \angle PMC=50+\alpha, \angle PNB=100-\alpha$
by the sine law we find that
$PC=\frac{MP\sin{(50+\alpha)}}{\sin{(50-\alpha)}}=\frac{NP\sin{(50+\alpha)}\sin30}{\sin{(50-\alpha)}\sin50}=\frac{PB\sin{(50+\alpha)}\sin\alpha}{2\sin{(50-\alpha)}\sin50\sin{(100-\alpha)}}$, this gives
$\sin{(50+\alpha)}\sin\alpha=2\sin{(50-\alpha)}\cos{(10-\alpha)}\sin50$, and after using the sum-to-product and product-to-sum trig formulas it finally gives
$2\sin{(50+\alpha)}\sin\alpha=\sin80+2\sin50\sin{(60-2\alpha)}$, from here we conclude that if $\alpha>30$ then $LHS>RHS$ and if $\alpha<30$ then $LHS<RHS$, so we must have $\alpha=30$, and we're done!!! :D

Since PB = PC and

$\angle BPC = \angle MPN = 180^\circ - (\angle PMN + \angle PNM) =$

$= 180^\circ - (50^\circ + 30^\circ) = 100^\circ = 2\angle A$

the point P is the circumcenter of the triangle $\triangle ABC$. The barycentric coordinates of the circumcenter are $a \cos \alpha, b \cos \beta, c \cos \gamma$, where a = BC, b - CA, c = AB are the triangle sides and $\alpha = \angle A, \beta = \angle B, \gamma = \angle C$ its internal angles. This is easily seen by using the sine theorem, for example for the triangles $\triangle CNA, \triangle CNB$:

$\frac{CA}{AN} = \frac{\sin[180^\circ - (90^\circ - \beta + \alpha)]}{\sin (90^\circ - \beta)} = \frac{\sin(90^\circ + \beta - \alpha)}{\cos \beta}$

$\frac{BC}{BN} = \frac{\sin[180^\circ - (90^\circ - \alpha + \beta)]}{\sin (90^\circ - \alpha)} = \frac{\sin(90^\circ - \alpha + \beta)}{\cos \alpha}$

$\frac{AN}{BN} = \frac b a \cdot \frac{AN}{CA} \cdot \frac{BC}{BN} = \frac a b \cdot \frac{\cos \alpha}{\cos \beta}$

Putting the masses $m_A = a \cos \alpha, m_B = b \cos \beta, m_C = c \cos \gamma$ at the vertices A, B, C, then piling up $m_A, m_B$ at N or $m_A, m_C$ at M and using the law of moments

$\frac{PM}{PC} = \frac{m_B}{m_A + m_C} = \frac{b \cos \beta}{a \cos \alpha + c \cos \gamma}$

$\frac{PN}{PB} = \frac{m_C}{m_A + m_B} = \frac{c \cos \gamma}{a \cos \alpha + b \cos \beta}$

$\frac{PN}{PM} = \frac{c \cos \gamma }{a \cos \alpha + b \cos \beta} \cdot \frac{a \cos \alpha + c \cos \gamma }{b \cos \beta} =$

$= \frac{\sin \gamma \cos \gamma}{\sin \beta \cos \beta} \cdot \frac{\sin \alpha \cos \alpha + \sin \gamma \cos \gamma}{\sin \alpha \cos \alpha + \sin \beta\cos \beta} = \frac{\sin 2\gamma}{\sin 2\beta} \cdot \frac{\sin 2\alpha + \sin 2\gamma}{\sin 2\alpha + \sin 2\beta}$

Substituting $\alpha = 50^\circ,\ \beta = 180^\circ - (\alpha + \gamma) = 130^\circ - \gamma$,

$\frac{PN}{PM} = \frac{\sin 2\gamma}{\sin (260^\circ - 2\gamma)} \cdot \frac{\sin 100^\circ + \sin 2\gamma}{\sin 100^\circ + \sin (260^\circ - 2\gamma)} = \frac{\sin 2\gamma}{\sin (2\gamma + 100^\circ)} \cdot \frac{\sin 2\gamma + \sin 100^\circ}{\sin (2\gamma + 100^\circ) - \sin 100^\circ}$

Using the sine theorem for the triangle $\triangle PMN$,

$\frac{PN}{PM} = \frac{\sin \widehat{PMM}}{\sin \widehat{PNM}} = \frac{\sin 50^\circ}{\sin 30^\circ} = 2 \sin 50^\circ$

Solving the resulting trigonometric equation for $\gamma$,

$2 \sin 50^\circ = \frac{\sin 2\gamma}{\sin (2\gamma + 100^\circ)} \cdot \frac{\sin(\gamma + 50^\circ) \cos (\gamma - 50^\circ)}{\cos (\gamma + 100^\circ) \sin \gamma} =$

$= \frac{\cos \gamma}{\cos (\gamma + 50^\circ)} \cdot \frac{\cos (\gamma - 50^\circ)}{\cos (\gamma + 100^\circ)} = \frac{2\cos \gamma \cos (\gamma - 50^\circ)}{\cos (2\gamma + 150^\circ) + \cos 50^\circ}$

$\cos \gamma}{\cos (\gamma - 50^\circ) = \sin 50^\circ \cos 50^\circ + \sin 50^\circ \cos (2\gamma + 150^\circ)$

Substituting $\sin 50^\circ \cos 50^\circ = \frac 1 2 \sin 100^\circ = \cos 60^\circ \cos 10^\circ = \frac 1 2 (\cos 70^\circ + \cos 50^\circ)$ and $\cos(2\gamma + 150^\circ) = \cos(2 \gamma - 210^\circ) = \sin(300^\circ - 2\gamma) = \sin(2\gamma - 120^\circ)$,

$\frac 1 2 \left[\cos(2\gamma - 50^\circ) + \cos 50^\circ)\right] = \frac 1 2 (\cos 50^\circ + \cos 70^\circ) + \sin 50^\circ \sin (2\gamma - 120^\circ)$

$\frac 1 2 \left[\cos(2\gamma - 50^\circ) - \cos 70^\circ)\right] = \sin 50^\circ \sin (2\gamma - 120^\circ)$

$\sin(\gamma + 10^\circ) \sin (\gamma - 60^\circ) = 2\sin 50^\circ \sin (\gamma - 60^\circ) \cos (\gamma - 60^\circ)$

Either $\sin (\gamma - 60^\circ) = 0,\ \gamma = 60^\circ$ for $\gamma \in (0^\circ, 180^\circ)$ (actually, for $\gamma \in (0^\circ, 180^\circ - \alpha) \equiv (0^\circ, 130^\circ)$) or

$\sin(\gamma + 10^\circ) = 2\sin 50^\circ \cos (\gamma - 60^\circ)$

$\sin(\gamma + 10^\circ) = \sin(\gamma - 10^\circ) + \sin(110^\circ - \gamma) = \sin(\gamma - 10^\circ) + \cos(\gamma - 20^\circ)$

$\sin(\gamma + 10^\circ) - \sin(\gamma - 10^\circ) = \cos(\gamma - 20^\circ)$

$2\cos \gamma \sin 10^\circ = \cos \gamma \cos 20^\circ + \sin \gamma \sin 20^\circ$

$\tan \gamma = \frac{2 \sin 10^\circ - \cos 20^\circ}{\sin 20^\circ} = \frac{2\cos 80^\circ - 2\cos 60^\circ \cos 20^\circ}{\sin 20^\circ} = \frac{2\cos 80^\circ - (\cos 80^\circ + \cos 40^\circ)}{\sin 20^\circ} =$

$= \frac{\cos 80^\circ - \cos 40^\circ}{\sin 20^\circ} = -\frac{2\sin 60^\circ \sin 20^\circ}{\sin 20^\circ} = -2 \sin 60^\circ = -\sqrt 3,\ \ \gamma = 120^\circ$

Thus our trigonometric equation has 2 possible roots $\gamma = 60^\circ$ or $\gamma = 120^\circ$. But if $\gamma = 120^\circ$, then the triangle $\triangle ABC$ would be obtuse, its circumcenter P would be outside of it on the other side of the line AB than the obtuse angle $\gamma$, which would imply that the points C N, P follow on the line CP in this order and the angle $\angle MPN = 180^\circ - \angle BPC = 180^\circ - 100^\circ = 80^\circ$. This contradicts to the problem condition that the angle $\angle MPN = 180^\circ - (50^\circ + 30^\circ) = 100^\circ$, hence, the root $\gamma = 120^\circ$ is not acceptable and the only possibility is $\gamma = 60^\circ$. Since P is the triangle circumcenter, the desired angle $\angle APB = 90^\circ - \gamma = 30^\circ$.

Please, without trigonometry !
This problem was posted in http://www.mathlinks.ro/Forum/viewtopic.php?t=54312

i'm sorry, i consider that i'm unable to a get a solution for such a kind of problem without trigonometry, maybe darij is able, but certainly i'm not!  :mad:  if you want such a solution, why don't you give it a try?

I tried, but I didn't succed. I am two trigonometric solutions for this problem. I will offer later on this post. You are right, Yetti ! This problem is the best difficulty problem which I came across.


You simply multilply the number of dots on each side by the number of sides, and subtract the overlap. 358*3=1074, and overlaps at each corner, so 1074-3=1071.

Alternately there are $ n\minus{}1$ dots on each side and three sides, so there are $ 3 \times (358\minus{}1)\equal{}3 \times 357\equal{} \boxed{1071}$.
This just removes the confusion about the overlap.


geometric series
algebra
partial fractions
complex analysis
complex analysis unsolved

I feel like this is really easy, so I'll be embarrassed when I see how to do it. How do you shift so that your Laurent series is centered about a point. Example: Find the Laurent series about $z_{0}=1$ of $\frac{z^{2}}{z^{2}-1}$.

Here is what I tried: $\frac{z^{2}}{z^{2}-1}=1+\frac{1}{z^{2}-1}= 1+\frac{1}{(z-1)(z+1)}$, so I need to find the series about $1$ of $\frac{1}{z+1}$ since the other parts are about $1$ already. So I brute forced and said the coefficients are $\frac{1}{2}, \frac{-1}{4}, \frac{1}{8}$ by hand. Then jumped to the conclusion: $\frac{1}{z+1}=\sum_{n=0}^\infty (-1)^{n}\left(\frac{1}{2}\right)^{n+1}(z-1)^{n}$.

Now we have $\frac{z^{2}}{z^{2}-1}=1+\sum_{n=0}^\infty (-1)^{n}\left(\frac{1}{2}\right)^{n+1}(z-1)^{n-1}$. Not sure if this is correct. If it is, there has got to be a more elegant way of shifting where it is centered.

The general principal is that the geometric series (and partial fractions decompositions, if necessary) are all that is needed to find the Laurent series of any rational function.

$\frac1{z+1}=\frac1{2+(z-1)}= \frac12\cdot\frac1{1+\frac{z-1}2}$

$=\frac12\left(1-\frac{z-1}2+\frac{(z-1)^{2}}{2^{2}}+\cdots\right)$

That gives you the series that you already have, at least for $|z-1|<2.$


On the S-I (Letter to USAMO QUALIFIER) it says that we have to send the S-III by no later than April 15th. But the actual S-III says that we have to send it by april 13 (on the top of the S-III form). Which one is right?

I was wondering the same thing, but I'm sending it in by the 13th to be safe.  Chances are that it's the 15th, since the website is often inaccurate and outdated, but you don't want to take chances on whether you get to take the USAMO :P.

[quote="joml88"]I sent mine in last week :P[/quote]

Me too :lol:

You have 2 days...I see no reason why you shouldn't be able to make the deadline. And for that matter, you don't even have to send it in...you can fax it. :-)

Yea... I faxed mine today... :D

BTW, someone else in our school is taking it too, but for some reason he WANTS to bail out...


Let $ V$ be a set of points in space such that $ V = {(x, y, z)| 0 \leq x, y, z \leq 2008, x, y, z \in \mathbb{Z}}$. We wish to color each of the points in $ V$ such that any two points of distance $ 1,\sqrt{2}$, or $ 2$ apart are of different colors. Find the minimum number of colors required to achieve this coloring.


$1)$ Given a circle with center $(4,6)$, find radius $r$ such that the expression $AC-BD$ is maximized where $A$ is the area of the circle in quadrant one; $B$, second; $C$, third; $D$, fourth.
$2)$ Given a circle with radius $10$ find center $(x_1,y_1)$, such that the expression $AC-BD$ is maximized. (Same $AC-BD$)

I dont really understand the question, if the circle is circular :P , then wouldn't the 4 quadrants of the circles be the same?

Here's a pic. For 1), its center is a fixed point whereas the radius is variable, and you want to find the radius that maximizes $A \cdot C-B \cdot D$, where A, B, C, and D are the regions shown in the picture.
[img]http://elnerdo.com/circle.gif[/img]


group theory
abstract algebra
superior algebra
superior algebra solved

Let $(G,\cdot)$ be a group, and let $H_1,H_2$ be proper subgroups s.t. $H_1\cap H_2=\{e\}$, where $e$ is the identity element of $G$. They also have the following properties:

[b]i)[/b] $x\in G\setminus(H_1\cup H_2),y\in H_1\setminus\{e\}\Rightarrow xy\in H_2$
[b]ii)[/b] $x\in G\setminus(H_1\cup H_2),y\in H_2\setminus\{e\}\Rightarrow xy\in H_1$

Prove that:

[b]a)[/b] $|H_1|=|H_2|$
[b]b)[/b] $|G|=|H_1|\cdot |H_2|$

[b]a)[/b]
Fix some $x \in G\setminus(H_1\cup H_2)$  (exists because $|G| > |H_1|+|H_2|$).
Then $x(H_2 \setminus \{ e \}) \subset H_1$ where $e \in H_1$ doesn't occure. So this gives $|H_1| \geq |H_2|$ and by the same way $|H_2| \geq |H_1| \Rightarrow |H_1| = |H_2|$

[b]b)[/b]
For all $a,b \in H_1, c,d \in H_2$ satisfying $ac=bd \Rightarrow b^{-1}a=dc^{-1}$ it follows, since $b^{-1}a \in H_1, dc^{-1} \in H_2$ and because of $H_1\cap H_2=\{e\}$, $a=b,c=d \Rightarrow |H_1| \cdot |H_2| = |H_1 \cdot H_2| \leq G$. But every $x \in G\setminus (H_1\cup H_2)$ can (because of [b]ii)[/b] ) be written as $x=zy^{-1},z \in H_1, y \in H_2$ (and also all $x \in (H_1\cup H_2)$ ), so you can conclude $|G|=|H_1| \cdot |H_2|$

[b]c)[/b]  ;)
Now let $y \in H_1 \setminus \{ e \}  $ be fixed. Then you get from [b]i)[/b] that $y(G\setminus(H_1\cup H_2)) \subset H_1$ (one more time $e$ doesn't occure) and so $|G|-|H_1|-|H_2|+1=(|H_1|-1)^2 \leq |H_1|-1 \Rightarrow |H_1|=2 \Rightarrow |G|=4$ and simple checking gives that $G$ is isomorphic to $\mathbb{Z}_2 ^2$


MATHCOUNTS
modular arithmetic
number theory
least common multiple

I need help with this one problem from Mathcounts National Sprint 2005:

How many integers $ n$ satisfy both of the following conditions:
 (1) $ 100 < n < 200$
 (2) $ n$ has the same remainder whether it is divided by 6 or 8?

Hmm...anyone have an easier way of doing this besides setting up eight systems of congruences?  Although some of them have the same answer...

Well,

n==a(mod 6)
n==a(mod 8)
n==a (mod 48)

Ah, so because $ a \le 5$, we have $ 100$ works plus $ 2(6)$ more which is $ \boxed{13}$?

So then it's just $ 2 \times 6\equal{}\boxed{12}$.  Thanks :)

Right.

In general,

$ n \equiv a \pmod x$
$ n \equiv a \pmod y$

Implies that

$ n \equiv a \pmod xy$.

This is the Chinese Remainder Theorem.

That's only one small part of the Chinese Remainder Theorem. :wink:

$ lcm(6,8) \equal{} 24$, so would it be 6*4=24, since the number can be 0, 1, 2, 3, 4, or 5 (mod 24) and there are 4 possible numbers for each case?

You can get that from the Chinese Remainder Theorem, but it's true for lcm (a,b), not just ab.


Prove that $ \prod_{i\equal{}1}^n \cos x_i \equal{} \sum \frac{\cos(x_1 \pm x_2 \pm ... \pm x_n)}{2^{n\minus{}1}}$,
where the sum is taken over all possible sign combinations.

[hide]
This follows quickly from the product-to-sum identity $ \cos (x_i) \cos (x_j) \equal{} \frac{\cos(x_i\plus{}x_j)\plus{}\cos(x_i\minus{}x_j)}{2}$ and induction.
[/hide]


induction
number theory proposed
number theory

how many solutions of the equation: [tex]\overline{a_1a_2...a_n}.9=\overline{a_n....a_2a_1}[/tex]
NOTE: [tex]a_1\neq{0}[/tex]
goodluck

About this pro: 2 year ago(when i learnt in HANOI),my teacher -Mr PHAM VAN HUNG- gave me this pro: and i solved it, long so not nice but result very cleary:
This equation has exactly [tex]f([\frac{n}{2}]-1)[/tex] solutions, in here [tex]f(o)=1[/tex], [tex]f(1)=1[/tex], [tex]f(n)=f(n-1)+f(n-2)[/tex] is FIBONACI'n
goodluck

why? nobody solve this proooooooooooo :(
[b]HINT[/b]:we can solve this result by induction
NEW PRO: find all solutions of this equation


Yes, today is 3.14, a.k.a. Pi Day.  Anyone doing anything to celebrate?

I'm going to pretend like I knew today was Pi day all along :)

Harvard did "Bad approximation of Pi day" on Friday -- pie eating contest and pi digit competition in three bases (binary, hex, and decimal).  I've decided I'm going to learn some of pi in hex -- it's easier to remember than binary, which gets too difficult after about 30 digits.

EISTEIN"S BIRTHDAY BOO YAH GOLLY GEE!!!!!!!!!!!!!!!!!!!!!

And if you know pi in hexadecimal, then you automatically also know it in binary. We will have a pi day celebration tonight. Yay.

[quote="mathfanatic"]Yes, today is 3.14, a.k.a. Pi Day.[/quote]

Calling today Pi Day assumes a convention of representing dates that is not universal. In some places, where numerical representation of dates precedes numerical representation of months, Pi Day would be the thirty-first day of the fourth month (hmmm, that doesn't work in the Gregorian Calendar), or the third day of the fourteenth month (scratch that), or the 314th day of the year, or . . .

We're not talking about a presidential or gubernatorial proclamation of a holiday here, are we?

Maybe M.I.T. should celebrate Really, Really Bad Approximation of Pi Day on September 9th or something like that.  :wink:

:pi: approximation day has already been taken. It's on July 22nd.

who claims they know the most pi?
jus wondering.

I'll say I do until someone else tells me I'm wrong. Due to extreme laziness, I now *only* know 277 digits. (I used to know over 700 when I was in 5th grade, but I have long since forgotten most of them.)

NOOOOOOOO! I forgot pi day!

I obviously didn't do anything to celebrate this year, but last year I baked a pie with " :pi: " inscribed on top of it.

[quote="JBL"]I know all the digits of pi.  I just don't know them in order ...[/quote]

lol

Boohoo. I only know 100 something. I guess I should memorize more. Class is so boring in my school there just ins't anything else to do.

My school stops math class for a day and everyone brings in pie...

Same here. I brought an ice cream pie, and wrote a big pi on it. We had a party, and got free homework passes and bonus points for giving out random pi facts.


Let $n>2$ be an even integer. We color the squares of an $n\times n$ board with $\frac {n^2}2$ colors, each color being employed exactly twice. Show that we can place $n$ tokens on the board s.t. they are on different colors and there is exactly one token on each column and each row.

Does anybody have a solution? I tried to solve this problem, but I failed each and every single time.

Let a configuration be a set of n squares on n different tows and n different columns. Assume that the statement is not true. Then each configuration is destroyed by a color (we have 2 squares of the same color in that configuration). There are $n!$ such configurations. Now, taking the color that "kills" the most of them, we get that it appears in at least $\frac{n!}{n^{2}/2}$. But it appears in $(n-2)!$ configurations. Therefore, comparing the 2 numbers, we get $n\leq2$, a contradiction.
Is this right? :)

i think that your solution isn't correct, ciprian!  :(

(n-2)! is the number of configurations that have the pair of the color, and 2n!/n² is the minimal number of configurations that have the color (not implies that 'your pair' will be in the configuration!)

this problem isn't easy! i have ideas to solve it, i will try later, and if i solve, i'll post the solution.  :)

BUt I assumed that the pair is in the configuration :maybe:  I said that the pair destroys a good configuration. In order for this to be true I need two squares of the same color. Tell me what I am doing wrong please!


function
algebra
domain
algebra unsolved

Find all functions $ f: \mathbb{R}\rightarrow \mathbb{R}$ such that

(i) $ f(x)\equal{}x^2f(\frac{1}{x})$ for all $ x\neq 0$
(ii) $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ for all reals $ x,y$
(iii) $ f(1)\equal{}1$

i maybe a tad bit stupid here but what does the first line when (the mapping thing)

It means the domain is all reals and the range is all reals.

All functions satisfying $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ are of the form $ f(x) \equal{} kx$ and since $ f(1) \equal{} 1$ ,therefore $ k \equal{} 1$ and hence $ f(x) \equal{} x$ where $ x\neq 0$

But it is nowhere said that $ f$ is continuous so you cannot conclude that $ f(x) \equal{} kx$ for all $ x$. It is only true for all $ x\in\mathbb Q$.

Replace$ x \equal{} y \equal{} 0: f(0) \equal{} 0$
           $ x \equal{} \minus{} y \equal{} 1: f( \minus{} 1) \equal{} \minus{} f(1) \equal{} \minus{} 1$
$ f(\frac {1}{1 \plus{} x}) \plus{} f(\frac {x}{x \plus{} 1}) \equal{} \frac {1}{(1 \plus{} x)^2}f(x \plus{} 1) \plus{} (\frac {x}{x \plus{} 1})^2f(1 \plus{} \frac {1}{x}) \equal{} f(1) \equal{} 1$
$ \Rightarrow\frac {1}{(1 \plus{} x)^2}[f(x) \plus{} 1] \plus{} (\frac {x}{x \plus{} 1})^2[1 \plus{} \frac {f(x)}{x^2}] \equal{} 1\Leftrightarrow\frac {2}{(1 \plus{} x)^2}f(x) \plus{} \frac {x^2 \plus{} 1}{(1 \plus{} x)^2} \equal{} 1 \Rightarrow f(x) \equal{} x$


logarithms
integration
trigonometry
complex numbers
real analysis
real analysis unsolved

Evaluate $ \sum_{n=1}^{\infty}\binom{2n}{n}\frac{1}{n 4^{n}}$

Oh, I didn't expect that there is a closed form for the power series $ \sum_{n=1}^{\infty}\binom{2n}{n}x^{n}$.

This is my approach:

Comparing the real part of $ \ln (1-e^{i\theta})$, one can show that $ \sum_{n=1}^{\infty}\frac{\cos^{n}\theta}{n}\; = \; \ln 2+2 \sum_{n=1}^{\infty}\frac{\cos n\theta}{n}.$

Integrate both side and use the well known fact $ \int_{0}^{\pi}\cos^{n}\theta \; d\theta = \begin{cases}0 & \text{if }n \text{ is odd}\\ \frac{\pi}{4^{k}}\binom{2k}{k}& \text{if }n = 2k\end{cases}$,

$ \frac{\pi}{2}\sum_{k=1}^{\infty}\binom{2k}{k}\frac{1}{k 4^{k}}\; = \; \pi \ln 2$

since $ \int_{0}^{\pi}\cos n\theta \; d\theta$ vanished for all $ n \geq 1$.

[quote="didilica"]The sum equals $ 2\ln 2$.

Use the well-known expansion:

$ \sum_{n=0}^{\infty}\binom{2n}{n}x^{n}=\frac{1}{\sqrt{1-4x}}$.

Divide by $ x$ and integrate and we get that

$ \int_{0}^{x}\left(\frac{1}{t\sqrt{1-4t}}-\frac{1}{t}\right)=\sum_{n=1}^{\infty}\binom{2n}{n}\frac{x^{n}}{n}$.

It follows that

$ \sum_{n=1}^{\infty}\frac{1}{n}\binom{2n}{n}x^{n}=2\ln\left(\frac{1-\sqrt{1-4x}}{2x}\right)$,

and hence, for $ x=\frac{1}{4}$, we get that

$ \sum_{n=1}^{\infty}\frac{1}{n4^{n}}\binom{2n}{n}=2\ln 2$.[/quote]

hey :coolspeak:

how about sums like :D

$ \textbf{(i)}\;\sum_{n=1}^\infty\;\frac{1}{n\cdot \dbinom{2n}{n}}$

$ \textbf{(ii)}\;\sum_{n=1}^\infty\;\frac{n}{\dbinom{2n}{n}}$

$ \textbf{(iii)}\;\sum_{n=1}^\infty\;\frac{1}{n4^{n}\cdot \dbinom{2n}{n}}$

answers hidden below...

[hide]
i get

$ \frac{\pi}{3\sqrt 3}$

$ 2\left(\frac{\pi}{9\sqrt 3}+\frac{1}{3}\right)$

i am still working on the solution for the [b]3rd [/b]one...   :|
seems to me it could be something like $ \left(a+b\cdot\pi+c\cdot\ln 2\right)$
where $ a,b,c\;\in\mathbb R$
[/hide]

All these sums are simple applications of the following   power series:

$ \frac{2x\arcsin(x)}{\sqrt{1-x^{2}}}=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n\binom{2n}{n}}$.

Now let $ x=\frac{1}{2}$ and you get the first sum.

Let $ x=\frac{1}{4}$ and you get the third sum.

$ \frac{d}{dx}\left(x\frac{d}{dx}\left(\frac{2x\arcsin(x)}{\sqrt{1-x^{2}}}\right)\right)\biggm|_{x=\frac{1}{2}}=4\sum_{n=1}^{\infty}\frac{n}{\binom{2n}{n}}$.


$ a,b.c,d > 0$ with $ abcd \equal{} 1$. than we have to prove that : $ (a^4 \plus{} b^3 \plus{} c^2 \plus{} d)(b^4 \plus{} c^3 \plus{} d^2 \plus{} a)(c^4 \plus{} d^3 \plus{} a^2 \plus{} b)(d^4 \plus{} a^3 \plus{} b^2 \plus{} c) \ge 256$

Remember to LaTeX it next time.  :)

[hide="Hint"]
We need to prove that $ (a^4 \plus{} b^3 \plus{} c^2 \plus{} d)(b^4 \plus{} c^3 \plus{} d^2 \plus{} a)(c^4 \plus{} d^3 \plus{} a^2 \plus{} b)(d^4 \plus{} a^3 \plus{} b^2 \plus{} c) \ge 256$. We first divide both sides by 4. Notice that by dividing both sides by a certain factor of 256, we can use the Power Mean Inequality.
[/hide]

Temperal, either I'm misinterpreting your hint or it isn't all that helpful. First, dividing both sides by 4 doesn't really get you anywhere, you have to divide each factor by 4. Also, straight AM-GM is much simpler, I'm not sure how exactly you suggest we apply Power Mean.

Divide all factors on the LHS by 4. By AM-GM on the four factors, we get $ \dfrac{LHS}{256}\ge \sqrt [4]{a^{4}bcd}\cdot\sqrt [4]{ab^{4}cd}\cdot\sqrt [4]{abc^{4}d}\cdot\sqrt [4]{abcd^{4}} \equal{} 1$.

change the condition abcd=1 into abcd=2 makes the problem a little bit harder (yet AM-GM still kills it )
find min
$ P\equal{}(a^4 \plus{} b^3 \plus{} c^2 \plus{} d)(b^4 \plus{} c^3 \plus{} d^2 \plus{} a)(c^4 \plus{} d^3 \plus{} a^2 \plus{} b)(d^4 \plus{} a^3 \plus{} b^2 \plus{} c)$


trigonometry
function
real analysis
real analysis unsolved

Find minimum for $x\in (\alpha, 180^{\circ}-\beta)$ of $f(x)=\frac{ \sin{x}}{\sin{(x-\alpha)}}+\frac{\sin{x}}{\sin{(\beta+x)}}$ if $\alpha+\beta < 90^{\circ}$.

hello, i have simplified your function $ f(x)$ to
$ 2\cos\left(\frac{\alpha\plus{}\beta}{2}\right)\csc(x\minus{}\alpha)\csc(x\plus{}\beta)\sin(x)\sin\left(x\minus{}\frac{\alpha}{2}\plus{}\frac{\beta}{2}\right)$
i hope this will help you.
Sonnhard.


[quote="Virgil Nicula"][color=darkred]Let $ ABCD$ be a cyclical convex quadrilateral. Denote $ E\in AC\cap BD$ and

$ \begin{array}{c} M\in (AC)\ ,\ \widehat {ABM}\equiv\widehat {CBD} \\
 \\
N\in (AC)\ ,\ \widehat {ADN}\equiv\widehat {CDB}\end{array}$ . Prove that $ AM \equal{} CN$ .

[b][u]Remark.[/u][/b] See an aplication of this nice problem at

http://www.mathlinks.ro/viewtopic.php?p=1033914#1033914 .[/color][/quote]
[hide="Solution"]
Notice that $ \angle BAM \equal{} \angle BAC \equal{} \angle BDC$ and $ \angle ABM \equal{} \angle DBC$, so \[ \triangle ABM\sim \triangle DBC\implies \frac {AM}{AB} \equal{} \frac {DC}{DB}\implies AM \equal{} \frac {DC\cdot AB}{DB}\]  Also, we have that \[ \angle ADN \equal{} \angle BDC\implies \angle ADB \plus{} \angle BDN \equal{} \angle BDN \plus{} \angle NDC\implies \angle NDC \equal{} \angle ADB\]  Also, $ \angle ABD \equal{} \angle ACD \equal{} \angle NCD$.  Therefore, \[ \triangle ABD\sim \triangle NCD\implies \frac {NC}{CD} \equal{} \frac {AB}{BD}\implies NC \equal{} \frac {AB\cdot CD}{BD} \equal{} AM\] as desired. [/hide]


In the game Mastermind, one player selects four pegs, each peg having six possible colors, and places them in a line. The second player then tries to guess the sequence of colors. What is the probability of guessing correctly?
(Note: duplicates are allowed, so player one can choose 4 pegs of the same color)

Slightly more difficult:

Suppose the second player knows that the first player always uses at least one red peg. What is the probability of guessing correctly now?

Similarly, if the second player knows that the first player never uses four different colors, what is the probability of guessing correctly?

If the second player knows that the first player always uses at least one red peg and never uses four different colors, what is the probability of guessing correctly?

I'm not sure if these are all right; especially the fourth problem.

[hide="Solution to Problem 1"]The probability of guessing any one peg correctly is $ \frac{1}{6}$. Therefore, the probability of guessing four pegs correctly is $ \boxed{\frac{1}{6^4}\equal{}\frac{1}{1296}}$[/hide]

[hide="Solution to Problem 2"]There are $ 5^4 \equal{} 625$ combinations that do not use any red pegs. There are a total of $ 6^4 \equal{} 1296$ total combinations. Therefore, there are $ 1296\minus{}625 \equal{} 671$ valid boards. Therefore, there is a probability of guessing correctly is now $ \boxed{\frac{1}{671}}$[/hide]

[hide="Solution to Problem 3"]$ 1296$ total combinations, $ 6\times5\times\times4\times3\equal{}360$ combinations are invalid because they use four different colors. Therefore, there are $ 1296\minus{}360 \equal{} 966$ valid combinations. Therefore, there is a probability of guessing correctly is now $ \boxed{\frac{1}{966}}$[/hide]

[hide="Solution to Problem 4"]We use inclusion-exclusion; we subtract $ 625$ and $ 360$ boards, but there are $ 120$ boards that contain a red peg and use four different colors, so there are $ 1296 \minus{} 625 \minus{} 360 \plus{} 120 \equal{} 431$ valid combinations. Therefore, there is a probability of guessing correctly is now $ \boxed{\frac{1}{431}}$[/hide]


Let $\sigma$ be a permutation on the natural numbers and let $\sum a_{k}$ and $\sum a_{\sigma\left(k\right)}$ be convergent series where

\[\sum a_{k}\neq\sum a_{\sigma\left(k\right)}.\]
Prove that the series $\sum a_{k}$ is not absolutely convergent.

Better to state this as the contrapositive: if the series converges absolutely, then the rearranged series converges to the same number as the original series. Stated that way, it gives you something to prove.

Here's something that can be used: Define $F_{n}=\{1,2,3,\dots,n\}$ and $R_{n}=\sigma(F_{n}).$

Show that for any $N,$ there exists $M$ such that for $n\ge M,\ F_{N}\subseteq R_{n}.$

Show also that for any $N,$ there exists $M$ such that for $n\ge M.\ R_{N}\subseteq F_{n}.$


Can someone explain to me stuff about modulus? Like how to use it and notation and things like that?

Thank you.

Fairly recent topic [url=http://artofproblemsolving.com/Forum/viewtopic.php?highlight=modulus&t=29510]here[/url]

There's a "Search" button up there^^^ next to the green "AOPS blogs" button.

you can use mods for sooooo many things.
such as remainder of a division of a very big number and a very small number. and so many many other stuff. actually there was a question  on the this year's USAMO (it was number one) was all about mods. go take a look at it. :lol:

[quote]actually there was a question on the this year's USAMO (it was number one) was all about mods. go take a look at it.[/quote]

actually, it was number 2...


I took time out to snack on a doughnut and a cup of coffee.  I spent the rest of the night trying to figure which one to dunk.

I don't get it... what's the joke here?

It's topology related. Go google it if must.

[quote="bubka"]Equivalent Joke:

A topologist is a person who can't tell a coffee cup from a doughnut.

The point is that a doughnut and a coffee cup are  topologically equivalent: they have the same genus. This is because each has a single hole through it. Given a clay coffee cup, you can turn it into a donut [i]without disturbing the hole[/i]. (I hope my indecent language will be excused.)[/quote]

Well, wobster gave a site where is written the same.

Unless the topologist is on drugs, he can feel the difference :

1.Take a cup of hot coffee.
2. Do a continuous transformation.
3. Look at your hand.
4. If you wanna go to the hospital, it's a donut, if not go back to 2.

:rotfl:  :rotfl:  Heard that one before.

an excuse for not doing homework:

I was doing my homework, but then decided to take a break. I got a cup of coffe and a donut, and spent the rest of the night trying to figure out which one to dunk.

 :rotfl:  :rotfl:

Inside the bottle? okay.

*sticks mucky hand in and takes note out*  :rotfl:

one of the prime reasons for which a klein bottle is noted is that you can't stick a hand, mucky or otherwise, inside it and take notes out.


[b]Welcome to A Time to Quiz Tournament 1![/b]

[u]Description:[/u] Essentially very similar to modern day quizbowl related activities. There will be as many rounds as necessary, with each round eliminating a total of $ \lceil n/2 \rceil$ people. The goal is to progress as far as possible.

[u]Topics:[/u] Almost any topics are liable to be asked.

[u]Policy[/u] Cheating, or looking up information on other websites during the round is absolutely forbidden. All rounds will be Free Response, requiring a user to input in answers into a web form.

[u]Question Format[/u] Nearly all rounds will conform to the following procedure:

I. 5 questions worth 5 points each (25). These are part I and will be free response.

II. 5 questions worth 10 points each (50). These are part II and are worth 50 pts.

III. 1 question worth 25 points (25). This question alone will be worth 25 points.

In the unlikely case of a tie, the following tiebreaker method will be used.

[u]Tiebreakers[/u]
The tiebreakers are like this:
Filter I - Cumulative score.
Filter II - Phase III scores will be used to determine the winner.
Filter III - Phase II score will be used.
Filter V. Problems are numbered as 1,2,...,11, where 1-5 are worth 5 each, 6-10 worth 10 each, and 11 worth 25 points. The following question order will be used: 10, 7, 6, 5, 8, 2.

If all five filters still do not break the tie, then both players will proceed onto the next round.

[u]Final stuff[/u] I need a minimum of 12 people to play, preferably at least 18.
The more the better in this circumstance.
Please post here if you are interested.

I forgot to mention earlier, but I really will need a couple volunteers to help grade/keep score. Volunteers need not make questions :D

Please PM me if you are willing to volunteer

I'll join!
By any topics do you mean any math topics or just any topic at all?

I'll join.  I wonder if this game will be similar to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202390&search_id=1300248327]this one[/url].

P.S.  You only have four filters listed (there is no IV).   :wink:

I know with #H34N1 making this, it'll be good, so I'm in, but I'm willing to volunteer instead also...

In the very unlikely case that this starts before sunday, I just wanna say I'll be gone from thursday to sunday.

Yeah I've tried to run something like this but what happens every time is something occurs and it just blows up.

I'll try my best not to.

That's the reason why I need a couple volunteers (I have one already) since I don't want to like not have time for a few days or so and the whole thing blow up.

Oh yeah get ready for hard problems :D

JOIN, lol

The message is too small. Please make the message longer before submitting.

ill join if it starts after monday and there are pm reminders!

JOIN.

Jus t PM me whenever it starts, I don't check games forum often...

So far in:

mathblitz
crazypianist1116
AIME15
sup3rcrash3r
Brut3Forc3
dragon96
33286
Garyzx
andersonw
SuperMathBoy
james4l
mewto55555
isabella2296
007math

I don't want to state the obvious, but you never posted it. Is it canceled? If not, when can we expect Round 1?

I guess he's too busy to come on games forum. if its not dead can i join?


vector
function
calculus
derivative
real analysis
real analysis unsolved

Let [b][i]r(t)[/i][/b] be a v.v.f -with the first and second derivatives [b][i]r'[/i][/b] and [b][i]r''[/i][/b]. Determine formula for
[b][i]d/dt [r.(r' x r'')][/i][/b] -in terms of [b][i]r[/i][/b]:

I'm thinking that there must be some vector-function identities that I'm missing:

So how do we approach this one?

maybe:
[b][i]d/dt [r.(r' x r'')] = d/dt [r.r' x r.r''] = ((r'.r' + r.r'') x r'.r'') + (r.r' x (r'.r'' + r.r'''))[/i][/b] ??
...and then what? I'm probably looking at this completely wrong --

Is anyone able to give me a starting point -- or starting direction? -- thanks heaps

The product rule for derivatives applies to the dot and cross products as well. (This can be proved by applying the usual product rule to each component separately, or, more abstractly, by noticing that the dot and cross products are nice bilinear maps.)


Let $n\ge5$ be an integer. Find the biggest integer $k$ such that there always exists a $n$-gon with exactly $k$ interior right angles. (Find $k$ in terms of $n$).

This problem originates from ISL 2003. Though it is still unsolved on the Mathlinks. See:
http://www.mathlinks.ro/Forum/viewtopic.php?p=19086#p19086

Sum of exterior angles in $n$-gon is always $360^{o}$, So $k=4$.

4 is not achievable for all $n$.  Also, the problem says nothing about convexity, so more than 4 is achievable for many $n$.  (For example, consider the 12-gon in the shape of a Swiss cross, with 8 internal right angles.)

(Perhaps a moderator could move this to a geometry forum, where it should have been placed originally 7+ years ago.)

Sorry for mistake. I thought that $n$-gon must be convex.


$ ABCD$ is a square and $ \bigtriangleup DCE$ is an equilateral triangle. Given $ FE \equal{} 1$ and $ FE\parallel AD$, find $ DC$
[hide="answer"] Book says its $ 4 \plus{} 2\sqrt {3}$ I got a different answer and I don't understand the solution in the manual  :maybe: [/hide]

EDIT:Nevermind, just realized I forgot to rationalize the denominator. Stupid me. No need to answer this. Is there any way to delete this post?

wait, is there more info given? is F on DC or something?

I get the solution... Could you tell me what you dont get about the solution and then I could explain it...


algebra polynomial Vieta quadratics calculus integration trigonometry

Now I have changed this topic to be a root finding marathon. I will post an equation and people have to find the roots of it. Previous posters in this topic, please edit your posts here.

What are we allowd to use in the formula ;)
(there is none with radicals)
For all others, see http://mathworld.wolfram.com/QuinticEquation.html .

[quote="mathgeniuse^ln(x)"]There is no quintic formula discovered yet, I believe.[/quote]
There are some (thu discovered). But it's impossible to find one with radicals only.

[quote="mathgeniuse^ln(x)"][quote="nutz_for2.718281828"]ok. I meant this to be a learning experience, not a resource-finding topic.

Lets try finding the quintic formula: ax^5+.....+ex+f[/quote]

There is no quintic formula discovered yet, I believe.[/quote]

Indeed, it is proven that there is NO SUCH formula for polynomials of degree 5 or higher.

Here is the first equation.(If you do not understand the meaning of this post, reread the thread.)

Let's start with an easy one.

$x^3+1=0$

[quote="nutz_for2.718281828"]Now I have changed this topic to be a root finding contest. I will post an equation and people have to find the roots of it. Previous posters in this topic, please edit your posts here.[/quote]
Edit our posts¿¿¿
No, thanks ;)
When totally changing the topic/content, you should simply start a new thread...

PS: $-1,\zeta_6 , \zeta_6^5$, $\zeta_n = e^{\frac{2 \pi i}{n}}$ are the roots ;)

I recall reading something about how quintic equations could not be solved with radicals, but with something called ultraradicals.  (I don't know anything else, it's all beyond me right now.)

But the Mathworld page is [url=http://mathworld.wolfram.com/Ultraradical.html]http://mathworld.wolfram.com/Ultraradical.html[/url]

Yes, you can reduce quintics to that by Tschirnhaus transfornations ( http://mathworld.wolfram.com/TschirnhausenTransformation.html ).
The idea is: Let $x_1 , ... x_5$ be the roots of a (monic) quintic polynomial. Then you look for coefficients $c_0, .... , c_3 , c_4$ such that when we set $z_i = c_4 x_i^4 + ... + c_1 x + c_0$ (the highest nonzero of them often $=1$), we have that the $z_i$ are the roots of a polynomial of type $z^5+z-a$.
These $c_i$ can be found by means of solving equations of degree $\leq 4$, so it can be done by normal radicals (but is tricky and uses Vieta's theorem to express some sums).

Special case: when doing the same with $c_n=...=c_3=c_2=0 , c_1=1$ for the polynomial $x^n+a_{n-1} x^{n-1} + ... + a_1 x + a_0$, we can get the substitution $z=x+\frac{a_{n-1}}{n}$, leading to $x=z-\frac{a_{n-1}}{n}$ giving a polynomial of type $z^n+b_{n-2} z^{n-2}+...+b_1 z + b_0$, so with the coefficient of $z^{n-1}$ vanishing. This suffices already to solve quadratic equations and is the first step when solving cubic ones, too.

But see http://www.apmaths.uwo.ca/~djeffrey/Offprints/Adamchik.pdf for full approaches.

1. where did you say "integers"¿
2. $\zeta_n$ is integral  :P
3. when searching for integer answers, the whole thing gets trivial checking.
4. this is not the spam and more spam factory (and such a contest makes less than no sense).
5. at least, I will change not a single post just because you decided to totally change to topic of this thread (or more exactly: decided to make it offtopic).

[quote="nutz_for2.718281828"]I said INTEGERS. Yes, -1 is right. You only need to post the answers.

If it is right, post another question. I'll post one now.

$y^3+y^2-5y+3=0$

No calculators, and integer answers.[/quote]

[hide]
0=$y^3+y^2-5y+3= (y-1)^2 \cdot (y+3)$ so the roots are $\boxed{1 \ , \ -3}$.
[/hide]

Next Problem: Find the roots of $P(x)= x^4-2x^3-10x^2+14x+21$.

All roots are real.

Wait, i don't get it. if all the roots are supposed to be integral or rational for all the problems in this marathon, then you can just use rational root theorem for everything :|

Like the two problems up there^^^

nothing like that.

Try NOT using the rational root theorem.

[quote="nutz_for2.718281828"]
Try NOT using the rational root theorem.[/quote]
Oh...Is being smart cheating?

$x^4-2x^3-10x^2+14x+21=(x+1)(x^3-3x^2-7x+21)=(x+1)(x-3)(x^2-7)$, so the roots are $-1, 3, \pm\sqrt{7}$.

I really don't see the point of this thread except doing busywork...

Post some involving clever factorisations

...i agree that this topic is a rehash of the things we know...the coefficients are symmetric, and that implies that if $r\ne 0$ is a root, then $\frac{1}{r}$ is a root, then either make the substitution

$y=\frac{1}{r}+r$

or

it can factor into $(x^2+ax+1)(x^2+bx+1)$ (this is easily shown by vieta)

there are a few rare problems that are not just application of techniques, here is one that might be a little more difficult, however, it is still pretty easy if you have the knowledge

EDIT: i messed up the problem...it should be

solve:
$8x^3-6x+1=0$

[quote="drunner2007"]$x^4-4x^3+5x^2-4x+1=(x-1)^4-x^2=((x-1)^2-x)((x-1)^2+x)=(x^2-3x+1)(x^2-x+1)$

Using the quadratic formula, the roots are $\frac{3\pm \sqrt{5}}{2}$ and $\frac{1\pm \sqrt{3}i}{2}$

$x^6-6x^4+11x^2-6$

This one has a trick that simplifies things, although the problem in general is not that hard.[/quote]

[hide]This can be factored as

$(x^2-1)(x^2-2)(x^2-3)$.

So the roots of this are

$1$,$-1$,$\sqrt{2}$,$-\sqrt{2}$,$\sqrt{3}$,${-sqrt[3}$.[/hide]

Here's another

$(x^2-3x+2)^2-3(x^2-3x+2)+2=0$

a lot easier than I thought it was going to be

[hide="drunner"]
Quadratic in $x^2-3x+2$.  Let that be $t$.
Then $t^2-3t+2=0\Rightarrow (t-2)(t-1)=0\Rightarrow t=2 \text{ or }1$.
$x^2-3x+1=0$ or $x^2-3x=0$
$x=\frac{-3\pm\sqrt5}2$ or $x=0$ or $x=3$
[/hide]

[hide="Altheman"]
$8x^3-6x+1=0$

Use trig-substitution:
$\cos(3t)=4\cos^3t-3\cos t$
$\left(x=\cos t\right)$
$2\cos(3t)=-1$
$\cos(3t)=-\frac12$
$3t=\frac{2\pi}3+2\pi n$ or $3t=\frac{4\pi}3+2\pi n$
$t=\frac{2\pi}9+\frac23\pi n$ or $t=\frac{4\pi}9+\frac23\pi n$
$x=\cos t=\cos\left(\frac{2\pi}9\right)$ or $\cos\left(\frac{4\pi}9\right)$ or $\cos\left(-\frac{\pi}9\right)$

 :D

[/hide]

Next problem (easy) :
Solve $x^3+6x=20$ for $x$

nobody has solved my question yet...

here is a vague hint, it still gives it away though

[hide="hint"]trig[/hide]

See above.  You posted while I was thinking of a new problem.  :)

[EDIT] Sorry, that's not quite right.  I answered drunner's question, then edited to put Altheman's in when I saw that no one had answered it yet.  I can't believe I wasted that much time on del Ferro's depressed cubic formula!

[quote="Danbert"]
Next problem (easy) :
Solve $x^3+6x=20$ for $x$[/quote]

[hide]$p(2) = 0$ $\Longrightarrow$ $(x - 2)(x^2 + 2x + 10) = 0$ $\Longrightarrow$ $x = 2$, $x = -1 + 3i$, $x = -1 - 3i$.[/hide]

Which is the sum of the real roots of $x^3 - 3x^2 + 3x + 1 = 0$?

[hide]Use Vieta's.  Sum of roots= $-\left(-3\right)=3$.[/hide]
[EDIT]  Oh.  It's just the [i]real[/i] roots.  Oops.
[hide]In that case, change variables to $y=x-1$.  The polynomial becomes $y^3+2=0$.  So $x=y+1=\boxed{1-\sqrt[3]2}$[/hide]
Next:
Find the sum of the [i]squares[/i] of the roots of
$2x^3-18x^2+5x\sqrt3-7\pi$

[quote="Danbert"]
Find the sum of the [i]squares[/i] of the roots of
$2x^3-18x^2+5x\sqrt3-7\pi$[/quote]

[hide]$\alpha_{1}^2 + \alpha_{2}^2 + \alpha_{3}^2 = (\alpha_{1} + \alpha_{2} + \alpha_{3})^2 - 2(\alpha_{1}\alpha_{2} + \alpha_{2}\alpha_{3} + \alpha_{1}\alpha_{3}) = \left(\frac{18}{2}\right)^2 - 2\frac{5\sqrt{3}}{2} = 81 - 5\sqrt{3}$[/hide]

What is the number of different real solutions of the equation $\sqrt[3]{x} + \sqrt[3]{7 - x} = 3$?

[quote="Andreas"]
What is the number of different real solutions of the equation $\sqrt[3]{x} + \sqrt[3]{7 - x} = 3$?[/quote]

a more interesting question is: what is the sum of the real roots of that equation?

(without actually computing the roots)


algorithm
number theory unsolved
number theory

Find all positive integer solutions of the equation \[z=\frac{x}{y}+\frac{y}{x}+\frac{2}{xy}.\]

I come to the conclusion that $gcd(x,y)=1$
if somebody can prove that the system $y/x^2 + 2$ and $x/y^2 + 2$ can not be solved in integers knowing that $gcd(x,y)=1$ then the equation is easy to be solved :D

I am not sure that there are no solutions, actually, I once saw a problem (German olympiad) asking to prove that there exist infinitely many coprime integers $a$ and $b$ such that $a $ divides $b^2 - 5$ and $b$ divides $a^2 - 5$.

The similarity is striking. I will think about this one.

well a very well known idea....
solutions are identified by: $x^2+y^2-zxy+2=0$ see this as a caudratic in $x$  and consider the minimal solution... suppose in the minimal solution $x=y$. then $x^2(z-2)=2$ so $x=1$ and $z= 4$. Now suppose $x>y$. since both roots, $x$ and $x'$ add to a positive integer $yz$ and their product is $y^2+2$, then $x'=\frac{y^2+2}{x}$ we now that $x'>x$ because of the minimality of the solution, and also because $y^2+2$ is not an square so $\frac{y^2+2}{x}>x$ wich implies a contradiction unless $x=y$. So minimal solutions are of the form $x=y$. wich gives $z=4$ and from there we get new solutions, going backwards with our algorithm


trigonometry
geometry
geometric transformation
rotation
analytic geometry

In $ xyz$ space, let $ O(0,\ 0,\ 0),\ A(0,\ 0,\ 1), B(0,\ 2,\ 2\sqrt{2}),\ P(x,\ y,\ z)$ such that $ \cos \angle{OAP}\equal{}\sqrt{\frac{2}{3}},\ \angle{BOP}\equal{}90^\circ$.

Find the maximum value of $ z$.

[hide]
We use the fact that $ a\cdot b \equal{} |a|\cdot|b| \cos \theta$
$ \vec{AO} \equal{} \minus{}k$, $ \vec{AP} \equal{} xi \plus{}yj \plus{} (z\minus{}1)k$, $ \vec{OB} \equal{} 2j \plus{} 2\sqrt{2}k$, $ \vec{OP}\equal{}xi\plus{}yj\plus{}zk$

$ \Rightarrow (1\minus{}z) \equal{} \sqrt{x^2\plus{}y^2 \plus{} (z\minus{}1)^2}\cdot\sqrt{\frac{2}{3}}$

$ \Rightarrow (z\minus{}1)^2 \equal{} 2(x^2\plus{}y^2) \qquad(1)$

And
$ 2y \plus{} 2\sqrt{2}z \equal{} 0 \qquad(2)$

Combining (1) and (2) gives

$ (z\minus{}1)^2 \equal{} 2(x^2 \plus{} 2z^2)$

$ \Rightarrow 3z^2\plus{}2z^2 \minus{} 1 \equal{} \minus{}2x^2 \le 0$

$ \Rightarrow (3z\minus{}1)(z\plus{}1) \le 0$

$ \therefore z \equal{} \boxed{\frac{1}{3}}$  :) [/hide]

Your answer is incorrect. :(

Last Edit:

Your answer is correct.

[hide]

note first of all that BO lies on the zy plane, as does AO.

since $ \angle{BOP} = 90^\circ$, BO acts as the axis of rotation which is fixed in the zy plane, thus z's maximum occurs when $ x = 0$, thus we can completely ignore x for the restt of the question

in $ \triangle{APO}$, $ PO = \sqrt {{p_z}^2 + {p_y}^2}$ and $ PA = \sqrt {{p_y}^2 + (1 - {p_z}^2)}$

since $ \cos\angle{OAP} = \sqrt {\frac {2}{3}}$ then $ \frac {PA}{1} = \sqrt {\frac {2}{3}}$

Mark the point M as $ (0, 2, 0)$
It follows that $ \triangle{APO}$ is similar to $ \triangle{BMO}$, thus

$ \frac {PO}{2} = \frac {PA}{2\sqrt {2}}$
$ 2\sqrt {2}\sqrt {{p_z}^2 + {p_y}^2} = 2\sqrt {{p_y}^2 + 1 - 2{p_z} + {p_z}^2}$
$ {p_y}^2 = 1 - 2{p_z} - {p_z}^2$

$ PA = \sqrt {2 - 4{p_z}}$

therefore
$ \sqrt {\frac {2}{3}} = \sqrt {2 - 4{p_z}}$

$ \boxed{p_z = \frac {1}{3}}$
[/hide]

fixed.. and I get same answer as above but we're wrong :( and I dont even understand ochas answer

I'm very sorry that your answers were correct,  I made a typo.

The coordinate of $ B$ in original problem was $ B(0,\ 2,\ \sqrt{2})$.

kunny

oh wow I didnt think I was in the olympiad program

[hide]
so like in my previous answer we can ignore the x component all together

mark the points $ N(0,0,p_z)$ and $ M(0,2,0)$

It follows that $ \triangle{BMO}$ is similar to $ \triangle{PNO}$
thus
$ \frac {\sqrt {2}}{2} \equal{} \frac {p_y}{p_z}$
$ {p_y}^2 \equal{} \frac {{p_z}^2}{2}$

$ \cos{\angle{OAP}} \equal{} \frac{1 \minus{} p_z}{\sqrt {{p_y}^2 \plus{} (1 \minus{} p_z)^2}} \equal{} \frac{1 \minus{} p_z}{\sqrt{\frac{{p_z}^2}{2} \plus{} 1 \minus{} 2p_z \plus{} {p_z}^2}} \equal{} \sqrt{\frac{2}{3}}$

thus it follows that

$ p_z \equal{} \frac {1}{2}$[/hide]

its also interesting to note then that $ \triangle{ANP} \equal{} \triangle{PNO}$


let the set $ S$ such that:
i) $ 0,1\in S$
ii) $ \forall (a_1,a_2,...,a_n)\in S^n$ such that $ \forall (i,j)\in [|1,n|]^2: i\neq j\iff a_i\neq a_j$
$ \frac {\sum_{k \equal{} 1}^{n}a_k}{n}\in S$

pove that $ [0,1]\cap \mathbb{Q}\subset S$

[hide]
1)we can prove easly that $ \{\frac {k}{2^n}|\ k,n\in\mathbb{N}^*,k\le 2^n\}\subset S$
2)let $ p,q\in\mathbb{N}^*$ such that $ p < q$
we take for lagest $ a$ $ \frac {a_1}{2^a},\frac {a_2}{2^a},...,\frac {a_q}{2^a}$ such that $ a_i\neq a_j$ for $ i\neq j$ and $ \sum_{i \equal{} 1}^qa_i \equal{} p2^a$ (we can pove the existence of the $ (a_i)$)
then $ \frac {p}{q} \equal{} \frac {\sum_{k \equal{} 1}^{n}\frac {a_k}{2^a}}{q}\in S$
[/hide]

can you prove the existence of the $ a_i$

for the existence,we take ,$ a_0 \equal{} p,a_1 \equal{} 2p,a_2 \equal{} 2^2p,.....a_{q \minus{} 1} \equal{} (2^{q \minus{} 1}\plus{}1)p$
we have $ \sum_{i \equal{} 0,q \minus{} 1}a_i \equal{} 2^qp$
so $ \frac {a_0/2^q \plus{} ... \plus{} a_{q \minus{} 1}/2^q}{q} \equal{} \frac {p}{q}$ ,
and all of the $ a_i/2^q$ are in S ;so the conclusion

[quote="drapt@"]for the existence,we take ,$ a_0 \equal{} p,a_1 \equal{} 2p,a_2 \equal{} 2^2p,.....a_{q \minus{} 1} \equal{} 2^{q \minus{} 1}p$
we have $ \sum_{i \equal{} 0,q \minus{} 1}a_i \equal{} 2^qp$
so $ \frac {a_0/2^q \plus{} ... \plus{} a_{q \minus{} 1}/2^q}{q} \equal{} \frac {p}{q}$ ,
and all of the $ a_i/2^q$ are in S ;so the conclusion[/quote]
but here we have $ \sum_{i \equal{} 0}^{q\minus{}1}a_i \equal{} p(2^q\minus{}1)$
nice solution but you must take $ a_0\equal{}2p\minus{}1,a_1\equal{}2p\plus{}1 and \forall i\in\{2,3,...,q\minus{}1\}: \ a_i\equal{}2^ip$
then $ \sum_{i \equal{} 0}^{q\minus{}1}a_i \equal{}4p\plus{}4p(2^{q\minus{}2}\minus{}1)\equal{}2^qp$ :wink:


topology
real analysis
real analysis unsolved

Prove that Grassmanian $ G(n,k)$ (i.e. the set of all k-dimensional subspaces of an n-dimensional linear space) is a smooth manifold of dimension $ k(n \minus{} k)$,

Keyword: homogeneous space. There's a nice action of $ GL_n$ on the space.

A [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=167019]closely related topic[/url].


Let $ C$ be the circumscribe circle of a triangle $ ABC$ and $ I$ the center of the inscribed circle in this triangle.
Consider the circle tangent to the sides $ CA,CB$ in the points $ D,E$ and interior tangent to the cricle $ C$.Prove that $ I$ is the middle of $ DE$

See here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=196786#196786[/url]. Search for "mixtilinear incircles" for other references.


Pat is to select six cookies from a tray containing only chocolate chip, oatmeal, and peanut butter cookies. There are at least six of each of these three kinds of cookies on the tray. How many different assortments of six cookies can be selected?

$ \textbf{(A)}\ 22 \qquad
\textbf{(B)}\ 25 \qquad
\textbf{(C)}\ 27 \qquad
\textbf{(D)}\ 28 \qquad
\textbf{(E)}\ 29$

Let the o's represent the cookies, and the l's are the symbols that divide the cookies into the three different flavors. There are six o's and the two l's. This is a total of eight items. An example of one arrangement is oolooloo, and this means that there are two of each type of cookies. We find the number of arrangements by calculating $ 8!$ and then dividing by $ 6!$ and $ 2!$, and this gives us $ 28$. The answer is D.

I think that this problem can also be solved with generating functions. The generating function would be $(1+x+x^2...)^3=(1-x)^}^{-3}$. Note that the same result would be attained with $(1+x+x^2+x^3+x^4+x^5+x^6)^3=\frac{(x^7-1)^3}{(x-1)^3}$ (why?). Calculating the coefficient of $x^6$ using the former function (since it's easier to calculate) we get ${-3}\choose{6}$ $=\boxed{28}$


calculus
integration
trigonometry
calculus computations

Calculate the following indefinite integrals.

[1] $\int (2x+1)\sqrt{x+2}\ dx$

[2] $\int \frac{1+\cos x}{x+\sin x}\ dx$

[3] $\int \sin ^ 5 x \cos ^ 3 x \ dx$

[4] $\int \frac{(x-3)^2}{x^4}\ dx$

[5] $\int \frac{dx}{\tan x}\ dx$

[quote="djimenez"][hide="Number 3"]Use $u=\sin x$, then

$\int \sin ^ 5 x \cos ^ 3 x \ dx=\int u^5(1-u^2)du=\frac{u^6}{6}-\frac{u^8}{8} = \frac{\sin^6x}{6}-\frac{\sin^8x}{8}$[/hide] Of course, plus the integration constant.[/quote]

Yes, that's right. :)

sorry for reviving this topic, i think 1st one is different.
[hide="1"]
\begin{align*}\int (2x+1)\sqrt{x+2}dx
& =\int (2(x+2)\sqrt{x+2}-3\sqrt{x+2})dx\\ & =\int 2(x+2)^{\frac{3}{2}}dx-3\int \sqrt{x+2}dx\\ & =\frac{4}{5}(x+2)^{\frac{5}{2}}-2(x+2)^{\frac{3}{2}}+C \end{align*}
[/hide]

Yes, you are right.
Your answer is correct and exactly same as mine.  :)

[hide="1"]$I_1=\int (2x+1)\sqrt {x+2}\ \mathrm{dx}\ \stackrel{(x=t^2-2)}{=}\ F\left(\sqrt {x+2}\right)$ , where $F=\int\left(2t^2-3\right)\cdot t\cdot 2t\ \mathrm{dt}=$ $\int\left(4t^4-6t^2\right)\ \mathrm{dt}=$

$\frac {4t^5}{5}-2t^3+\mathbb C$ . Thus, $I_1=\frac 45(x+2)^2\sqrt {x+2}-2(x+2)\sqrt {x+2}+\mathbb C\implies$ $\boxed{I_1=\frac 25(x+2)(2x-1)\sqrt {x+2}+\mathbb C}$ .[/hide]


1) In triangle ABC, angle C measures 60 degrees and angle A>B. The bisector of angle C intersect AB at E. If CE is a mean proportional between AE and EB, find AE/AC.

2) In base fifty the integer x is represented by the numeral CC and $x^3$ is represented by the numeral ABBA. If C>0 express all possible values of B in base 10.

3) External squares are drawn on each side of a rhombus, and the centers of these squares are joined to form convex quadrilateral Q. If the length of each side of the rohombus is 6 and if one of its angles has a degree measure 30, find the the area of the quadrilateral, Q.

[hide="3"]
Let $A$, $B$, $C$ and $D$ be the vertices of the rhombus, and let $E$, $F$, $G$ and $H$ be the centers of the squares that are adjacent to the sides $AB$, $BC$, $CD$ and $DA$, respectively.

$\triangle AEB$, $\triangle BFC$, $\triangle CGD$ and $\triangle DHA$ are isosceles right triangles whose legs are $3\sqrt{2}$.

Also, $\triangle AEH$, $\triangle BEF$, $\triangle CFG$ and $\triangle DGH$ are congruent (by SAS).

Now consider the area $EFGH+\triangle AEH+\triangle CFG$ (a hexagonal region). If we subtract $\triangle BEF$ and $\triangle DHG$, the rest is 4 isosceles right triangles and the rhombus. Since $\triangle AEH+\triangle CFG =\triangle BEF +\triangle DHG$ ($\because$ they are congruent), the sum of the areas of the 4 isosceles triangles (=square) and the rhombus is equal to the area of $EFGH$.

Therefore, $EFGH=6\cdot 6+ 6\cdot 6\sin{30^\circ} =\boxed{54}$
[/hide]

what does it mean

" If CE is a mean proportional between AE and EB, find AE/AC. "

mean proportion is basically the geometric mean between those sides and chess64 i think ur not gonna get too far with problem 2 that way. guaranteed.

Hint for triangle problem: put everything in terms of sine and cosine. for example: if u have sin(150-x) then u would put it all in terms of sine.

[quote="t0rajir0u"][hide="2"]
$x = CC_{50} = (50 + 1)C$
$x^3 = (50+1)^3 C^3 = (50^3 + 1) A + (50^2 + 50) B$
$(50^3 + 1)(C^3 - A) = (50^2 + 50)(B - 3)$

An obvious solution is when $B = 3, C^3 = A$.  Furthermore, taking $\bmod 50$ we know that $C^3 - A \equiv 0 \bmod 50$.

Now, consider the case when $C^3 \neq A$.  Since $C > 0$ and $A < 50$, clearly $C^3 - A > -50$, so the minimum nonzero value of $C^3 - A$ is $50$.  However, since $B < 50$,

$(50^2 + 50)(B - 3) < (50^2 + 50)(47)$

But

$(50^3 + 1)(C^3 - A) \ge (50^3 + 1)(50)$,

Clearly a much larger value.

Hence $C^3 = A, B = \boxed{3}$.
[/hide][/quote]

i think 24 is also a value for B. if you dont get it i'll post my soln. o yea heres a hint: $(50^3+1)=(50+1)(50^2-50+1)$ and try mod 51 also you should see something.

By setting ABBA = (CC)^3 and taking mod 50 and mod 51:

We get that B = 3A and A = C^3, which implies that B = 3C^3

Since B is an integer from 0 to 49, we are looking for integers from 0 to 49 that are 3 times a perfect cube.

Thus, possible answers are:
3*1^3 = 3
3*2^3 = 24

[Oddly enough, those were numbers 2, 4, and 8 on my ARML homework]


integration
vector
function
topology
real analysis
real analysis unsolved

Prove that
\[A: =\{f\in C(0,1)|f(0)=0 \text{ and }\int_{0}^{1}f(t)\,dt=1\}\]
is a closed and convex subset of $C(0,1)$ with no element which minimizes the norm.

Consider the mapping $T: C([0,1])\mapsto\mathbb{R}^{2}$ defined by $T(f)=\left(f(0),\int_{0}^{1}f(t)\,dt\right).$

Then $A=T^{-1}(0,1).$

$T$ is linear, which makes $A$ an affine subset of the vector space $C([0,1]),$ and an affine subset is convex.

$T$ is continuous, which makes $A$ the inverse image of a closed set, hence closed.

Now, about minimizing the norm: If $f$ is continuous, $f(t)\le 1$ for all $t$ and $f(0)=0$, then $\int_{0}^{1}f<1,$ strictly. Hence, any $f\in A$ must be greater than 1 somewhere, hence $\|f\|>1.$ But consider the functions $f_{\delta}$ which are piecewise linear continuous functions with vertices at $(0,0),\left(\frac{2\delta}{1+\delta},1+\delta\right),$ and $(1,1+\delta).$

Each $f_{\delta}$ belongs to $A,$ and $\|f_{\delta}\|=1+\delta.$

Hence $\inf_{A}\|f\|=1$ but no element of $A$ has norm $1.$


[quote="Sunny"][quote="lightrhee"][quote="ebn_mahdi"]Which uni is the best one in canada in economy?[/quote]
What is 'uni'? :?[/quote]

I'm thinking he means university.  I call them 'uni' too, its less to type out.   :P[/quote]
 :o

So many people abbreviate words on this forum. I occasionally do it :blush: , but not on purpose.  Does everyone really want to save one or two seconds that bad?  :?

[quote="TheDreamer"]I'm usually used to "univ" so that one doesn't get confused with "universe" or something ^_^

Neways, back on track ( :rotfl: )best for economy?  Like best for studying economics, or the most economical university to go to?[/quote]
My God
i ask you 1 question but you give me 4 answers(questions)
any way
thank you for your nice answers :lol:
which UNIVERSITY is the best one TO STUDY economy in canad?
is it clear or no!?(i am joking)
thank you

Actually come to think of it it was a stupid question i asked back becasue i woudln't be able to answer either way.  Sorry for troubles : :P

Wow, I guess we canadians like to divert the topics...

Anyways, my friend is going into international finance, and she's considering Ryerson, Wilfrid Laurier and Brescia (Girls-only university, I think). I'm sure she's mentioned some other ones too, but they're the only ones i can think of now. I also think Waterloo (okay, I'm biased, so what?), Toronto, Guelph and Western offers good business programs, but I'm not an expert..

Yes Western is very good, and Waterloo is good at everything :D:D:D

Salaam
thank you for your answers.
one of my friend is studing economy in UBC and i might get to a canadian university.
khoda hafez

[quote="Elyot"]Yes, Laurier is very good for economics.  Apparently there's one insanely amazing economics professor at Waterloo who teaches night courses.  Most of the people who attend his lectures aren't actually enrolled in his class.  They just go, because he is THAT GOOD.  I've been told it's better than watching a Hollywood movie.

Edit: His name's Larry Smith.  search for his record at ratemyprofessors.com[/quote]

Yeah that is correct and he was amazing.  I took his class and his exams are very very challanging, you have to have a good foundation in economics to do well.

He will give you a lot of insight in what strategies to take in day to day life, like how to take a loan or how to invest.

By the way he is a multi-millionaire!  ;)


probability
rotation
geometry
geometric transformation
reflection

how many ways can you put 8 rooks on a regular 8 by 8 chessboard so that they do not attack eachother

[hide]
64*63*62*61*60*59*58*57. Which is the same for any other piece, considering that each piece only takes up one square.

However, if you meant "... a chessboard [u]so that they do not attack each other[/u]", beats me.[/hide]

I have a feeling that that's not what scrambled meant to ask. I will assume that the question was "How many ways can 8 rooks be placed on a chessboard so that they don't attack each other?"

[hide]
There are $64$ possibilities for the 1st rook, and the 2nd rook has to be on a different rank and file. The first rook took away $8+8-1=15$ possibilities for the 2nd, so it has 49 possibilities. The next one has 36, etc. so the answer is $8^2 \cdot 7^2 \cdot 6^2 \cdot 5^2 \cdots 2^2=16257024200.$
[/hide]

[quote="Iversonfan2005"]$64P8$, which is 64x63x62....x57...No, I'm not going to figure that out  :no:[/quote]

That is incorrect because all eight rooks are the same.  We need to divide your answer by $8!$.

Wouldn't it be 2?

I mean I can't think of any other possibility except laying them diagonally from begging to end from bottom left to top right or vice versa.

EDIT*
I mean without attacking each other...

How about

[img]http://home.1asphost.com/WindSlicer/chess.JPG[/img]

Where the blotches are the rooks. There are a good number of combinations

Correct.
You know theyre not in the same row, so theres 8 columns to choose from first, then 7 then 6, because they can't be in the same row, so 8!=40320

Should that answer be divided by 4 (the number of ways to rotate the board) to get rid of repetitions?

Well it would be divided by 8, because you can also reflect it, but then there are those that are the same when rotated and reflected, but for this problem, I don't think you would want that to apply

Why isn't chess's approach correct? That's what I was thinkin.

The first piece has 64 choices, not 8.

Hmm.. does it have to do with the fact that the rooks are identical?

tHis is relatively easy because rooks can only move vertically and horizontally. It's harder with queen, which I think there's a discussion about.

[quote="chess64"][quote="Chigr"]shouldnt it just be 8! becuase teh first one could be placed in the first column then the second one would be placed in the second column, but could only be in one of the 7 left over rows without attacking the other one.
8!=[b]40320 possibilities[/b][/quote]

The 2nd one doesn't have to be in the 2nd column. There are still 7 columns and 7 rows to choose, from!  ;)[/quote]

Yeah, ches but I think you (and I) are looking it at like the chess pieces are not identical - which they are. If each piece were unique then yes your approach would be correct, but since they're not, you have to use Chigr's approach.

At least that's how I see it right now - would be nice if someone could confirm/deny this.

Yes, that's why. If all the rooks were considered distinct, it would be $8^2*7^2*6^2*...*1^2$, but because they aren't disticnt you have to divide by $8!$, the number of ways to "rearrange" the rooks which are really all the same. (Giving you, of course, $8!$.)


Sea $ P$ un punto del plano de $ \triangle ABC$ y sea $ \triangle A'B'C'$  el triángulo ceviano de $ P$ con respecto a $ \triangle ABC.$ Las circunferencias $ \odot(ABB')$ y $ \odot(ACC')$ se cortan en $A$ y $ X.$ Definimos igualmente $ Y,Z$ con respecto a $B$ y $C.$ Mostrar que  $ AX,BY,CZ$ concurren en el complemento del isogonal de $ P$ respecto a  $ \triangle ABC.$

Véase aqui una solución del problema  con coordenadas baricentricas [url]http://www.artofproblemsolving.com/viewtopic.php?t=282732[/url]. También se discuten varias cuestiones interesantes sobre esta configuración.


summer program
MathPath
AwesomeMath
geometry
MATHCOUNTS
LaTeX
percent

The Second round has ended.  The first and second rounds both are posted here, along with a combined answer key.

I'm in. However, I might not have much time, since I will be at Mathpath.

My name is in my signature.

I'll join! I like being in two competitions at once!

Can you give us the format of the competition?

Ditto. When is the competition going to start?

[quote="mad_skillz_aops"]No, I don't need your real names.
And jclarke, yes, mathpath will be a problem, since I was planning on posting the test some time around Friday next week.  Though, I am not sure if you can get internet access there, but if you can, then great.[/quote]

And awesomemath, which will start july 18th.

"Other?" Have you made another competition yet?

Here is a brief overview
Now till July 21
sign up
July 7 till July 21
first round
July 22 till July 30
second round
July 31 till somewhere
brackets
People may participate in both the first round and second round.  The brackets will be based on the total score of the rounds.
The rounds will have 20 no calc problems and 5 calc problems.
I am trusting you not to use a calc on the no calc problems.
A full rules and procedures guide will be posted sometime next week.

EDIT- Only an answer to the problems will be nessecary, no solutions.

If you're doing it today, then anytime from 8:30 est onwards.

[quote="SplashD"][quote="Klebian"]Hey, are we doing the countdown today?
(By the way, I'd like race to 15)[/quote]

race to 10 :!:  and when are you free?[/quote]
Why not race to 15?

because...

can you make it at 8pm est today?

Ack. We've had a ton of guests over for the past two hours!  :(
Shoot. What's your AIM screenname, SplashD?

Well, Klebian didn't sign on before the appointed time and SplashD didn't sign on afterwards, so I guess we're going to do it today.

Are either going to do it or not?  No one's posting when they're on.

i'm free anytime before 8pm est tomorrow or the day after tomorrow

I can do anytime today, not sure about tomorrow though.
I can try to get on at like  7:30 est but I'm not sure i'd be able to. Day after tomorrow I won't be getting home till like 7:30ish est.

[quote="mad_skillz_aops"]Well then, is 7:30 est fine?[/quote]

i'm ok with 7:30 est today.

Well, it's taking longer than imagined, and we're about halfway done.

[size=200]The competition is over![/size]
Klebian won the match by far.  He is the official winner of the mad_skillz competition.  I will post the full list of problems used tommorrow.

mods:  You can lock this after I post that file tommorrow.

Well, here is the file, there may be a few missing, but very very few.  There is also a picture of the bracket match-ups if you want to see the full summary of it.

mods: You may lock this now.

All right, I've locked it.
If for any reason later you need it unlocked, just let me know (by PM).


Let Ai,Bi (i=1,2,..n) real positive such that

A1 \geq (A1+A2)/2  \geq (A1+A2+A3)/3 \geq ... \geq (A1+..+An)/n

B1 \geq (B1+B2)/2  \geq (B1+B2+B3)/3 \geq ... \geq (B1+..+Bn)/n

Prove that

 n( \sum AiBi) \geq ( \sum Ai)( \sum Bi)

I can find a solution only if the signs of the conditions are reversed, that is :

A1 \leq (A1+A2)/2  \leq (A1+A2+A3)/3  \leq ... \leq (A1+....+An)/n

B1 \leq (B1+B2)/2  \leq (B1+B2+B3)/3  \leq ... \leq (B1+....+Bn)/n

so I think this is the correct condition for :

( \sum Ai)( \sum Bi)  \leq n \sum AiBi

I have proved it by induction.

It is clear signs of the conditions are not important. And positiveness of Ai,Bi is unimportant too.

Where is your proof, manlio?

Myth, you are right about the positiveness of Ai,Bi but I don't understand
the unimportance of the signs of the condition.
However, this is my proof. I hope it is correct.

I prove it by induction. I assume true the relation

( \sumAi)( \sum Bi)  \leq n \sum AiBi  for i=1...n

I consider A1....A[n+1] and B1.....B[n+1] satisfying

A[n+1] \geq (A1+...+An)/n  ; B[n+1] \geq (B1+...+Bn)/n

which can be rewritten as

A[n+1] \geq (A1+...+An)/n      B[n+1] \geq (B1+...+Bn)/n

from which I obtain

(A[n+1]-1/n  \sum Ai)(B[n+1]-1/n \sum Bi) \geq 0

A[n+1]B[n+1] - (1/n)B[n+1] \sum Ai -(1/n)A[n+1] \sumBi +

(1/n^2)( \sum Ai)( \sum Bi) \geq 0

A[n+1] \sum Bi + B[n+1] \sum Ai \leq nA[n+1]B[n+1]  +
+(1/n)( \sum Ai)( \sum Bi)

Also

 (\sum (i=1,...,n+1)Ai)( \sum (i=1,...,n+1)Bi)= (\sumAi)( \sum Bi) +

A[n+1]B[n+1] +A[n+1] \sum Bi +B[n+1] \sum Ai \leq

((n+1)/n)( \sum Ai)( \sum Bi) +(n+1)A[n+1]B[n+1] \leq
(n+1) \sum AiBi

by the inductive hypotesis.

Signs are unimportant because if inequality holds for A1,...,An,B1,...,Bn when it holds for -A1,...,-An,-B1,...,-Bn.

Your proof with small modifications can be applied for initial problem.

Myth, you are rigth. Also initial inequality is true. I had made a wrong consideration in my proof.


That works.  I can't think of any other obvious methods.

It's correct.

I'd probably simplify the proof by writing

[tex]\displaystyle
\left(n^4+3n^2+1\right)\left(n^2+1\right)
-
\left(n^3+2n\right)\left(n^3+2n\right)
=
1
[/tex]

and therefore, they are relatively prime.

Of course, you GET the values (n^2+1) and (n^3+2n) to multiply by to get 1 by Euclid's algorithm, so its nothing special. It just looks nice :)


A lottery gives prizes which are powers of 11. Thus the possible prizes are 1, 11, 121, 1331, 14641, and 161051 dollars. They will give no more than 10 awards of any particular amount. The give out exactly 1,111,111 dollars. How many prizes do they award?

[hide="hint"]what is 1,111,111 in base 11[/hide]


Let $ a,b$ be positive real numbers such that $ a^3\plus{}b^2\le 1$. Find the maximum of the following expression:
\[ A\equal{}3\sqrt{a}\plus{}2\sqrt[3]{b}\]

We can make a substitution : $ a \equal{} c^2$ and $ b \equal{} d^3$ for some $ c\leq 1,\; d\leq 1$. We get : $ c^6 \plus{} d^6\leq 1$ and $ A \equal{} 3c \plus{} 2d$.

Now : $ A \equal{} 3c \plus{} 2d\leq 3c \plus{} 2(1 \minus{} c^6)^{\frac {1}{6}}$. Find maximum of the last expression.

$ f(c) \equal{} 3c \plus{} 2(1 \minus{} c^6)^{\frac {1}{6}}$.

$ \frac {df}{dc} \equal{} 3 \minus{} 2\cdot\frac {1}{6}\cdot\frac {6\cdot c^{5}}{(1 \minus{} c^6)^{\frac {5}{6}}} \equal{} 0$ or $ 3^6 \cdot (1 \minus{} c^6)^{5} \equal{} 2^6 c^{30}$ or $ \left(\frac {1 \minus{} c^6}{c^6}\right)^5 \equal{} \left(\frac {2}{3}\right)^6$.

From this we get : $ c_0 \equal{} \frac {1}{\sqrt [6]{1 \plus{} \left(\frac {2}{3}\right)^\frac {6}{5}}}$ and $ f(c_0) \equal{} \frac {3 \plus{} 2\left(\frac {2}{3}\right)^\frac {1}{5}}{\sqrt [6]{1 \plus{} \left(\frac {2}{3}\right)^\frac {6}{5}}}\approx 4.472391146$

Answer : $ \boxed{\max A\equal{}\frac {3 \plus{} 2\left(\frac {2}{3}\right)^\frac {1}{5}}{\sqrt [6]{1 \plus{} \left(\frac {2}{3}\right)^\frac {6}{5}}}}$


Let $ ABC$ be a triangle with $ \measuredangle{BAC} < \measuredangle{ACB}$. Let $ D$, $ E$ be points on the sides $ AC$ and $ AB$, such that the angles $ ACB$ and $ BED$ are congruent. If $ F$ lies in the interior of the quadrilateral $ BCDE$ such that the circumcircle of triangle $ BCF$ is tangent to the circumcircle of $ DEF$ and the circumcircle of $ BEF$ is tangent to the circumcircle of $ CDF$, prove that the points $ A$, $ C$, $ E$, $ F$ are concyclic.

[i]Author: Cosmin Pohoata[/i]

[quote="pohoatza"]Let $ ABC$ be a triangle with $ \measuredangle{BAC} < \measuredangle{ACB}$. Let $ D$, $ E$ be points on the sides $ AC$ and $ AB$, such that the angles $ ACB$ and $ BED$ are congruent. If $ F$ lies in the interior of the quadrilateral $ BCDE$ such that the circumcircle of triangle $ BCF$ is tangent to the circumcircle of $ DEF$ and the circumcircle of $ BEF$ is tangent to the circumcircle of $ CDF$, prove that the points $ A$, $ C$, $ E$, $ F$ are concyclic.[/quote]
[hide="Solution"]
Invert around $ F$, which maps point $ P$ to $ P'$.  Now, since the circumcircles of $ \triangle FBE$ and $ \triangle DFC$ are tangent, we see that $ B'E'\parallel D'C'$.  Similarly, $ E'D'\parallel B'C'$, so $ E'D'C'B'$ is a parallelogram.  This means that $ \angle B'E'D' = \angle D'C'B'$.  Also, the inversion maps $ \angle BED = \angle FEB + \angle FED = \angle FD'E' + \angle FB'E'$.  Similarly,\[ \angle ACB = \angle DCB = \angle FCD + \angle FCB = \angle FB'C' + \angle FD'C'\]so\[ \angle FD'E' + \angle FB'E' = \angle BED = \angle ACB = \angle FB'C' + \angle FD'C'\]Also,\[ \angle B'FD' = 360 - (\angle B'E'D' + \angle FB'E' + \angle FD'E') = 360 - (\angle B'C'D' + \angle FB'C' + \angle FD'C') = 360 - \angle B'FD'\]Which implies that $ \angle B'FD' = 180$, so $ B'$, $ F$, and $ D'$ are collinear.  Now, notice that $ A$ is the intersection between $ BE$ and $ CD$, so $ A'$ is the intersection between the circumcircles of $ \triangle FE'B'$ and $ \triangle FD'C'$.  Now,\[ \angle FA'E' = 180 - \angle FB'E' = 180 - \angle FD'C' = 180 - \angle FA'C'\]so $ \angle FA'C' + \angle FA'E' = 180$, which means that $ E'$, $ A'$, and $ C'$ are collinear, so $ FCAE$ is cyclic. [/hide]

Could anyone solve it without using "inversion"?

From fact that circumcircle of triangle $ BCF$ is tangent to the circumcircle of triangle $ DEF$, we get $ \angle DFE\equal{}\angle FCD\plus{}\angle FBE$ and similarly $ \angle EFB\equal{}\angle FDE\plus{}\angle FCB$. By adding,
$ \angle DFB\equal{}(\angle FCD\plus{}\angle FCB)\plus{}\angle FBE\plus{}\angle FDE\equal{}\angle DCB\plus{}\angle FBE\plus{}\angle FDE\equal{}\angle DEB\plus{}\angle FBE\plus{}\angle FDE$. Hence $ D,F,B$ are collinear. So $ \angle DFE\equal{}\angle FEB\plus{}\angle FBE$. Comparing with $ \angle DFE\equal{}\angle FCD\plus{}\angle FBE$, we get $ \angle FEB\equal{}\angle FCD$ which imply $ A,C,E,F$ are concyclic.

[quote="Leonhard Euler"]$\cdots$ and similarly $ \angle EFB\equal{}\angle FDE\plus{}\angle FCB$$\cdots $  [/quote]
How ?

Perform an [b]inversion[/b] $\mathbf{I}(F,k)$ with arbitrary power $k>0$.(Define $X'$ as the inverse of $X$.)
It's obvious that $E'C'D'B'$ is an parallelogram (because of tangencies)
Let $\angle BEF=\alpha$, $\angle BEF=y$, $\angle ACF=x$.
By angle-chasing we get $\angle E'B'C'=\alpha +y$ and $\angle E'D'C'=2x+y-\alpha$. Since $E'C'D'B'$ is a parallelogram. $\implies$ $x =\alpha$ $\implies$ $\angle BEF=\angle ACF$. So $AEFC$ is cyclic.

By the condition, we know that $\angle BEF+\angle FDC=\angle BFD$, and $\angle FED+\angle CBF=\angle CFD$, so $\angle DEB+\angle CBF+\angle FDC=\angle BFD=360-LHS$, and we must have $\angle BFD=180$. Now, $\angle FCD+\angle FDC=\angle CFB=\angle FDC+\angle FEB$, implying $\angle FCD=\angle FEB=180-\angle FEA$, as desired.

We will be using directed angles.

Invert about $F$. Then in our inverted picture, it's clear that $B^*C^*D^*E^*$ is a parallelogram. Next note that
\[\angle E^*B^*F +\angle FD^*E^*=\angle BEF + \angle FED  =\angle BED = \angle ACB = \angle DCB = \angle DCF + \angle FCB= \angle C^*D^*F + \angle F B^*C^*\]
From this it follows that
\[\angle D^*FB^* =  - \angle B^*E^*D^* - (\angle  FB^*E^* + \angle E^*D^*F ) = -\angle D^*C^*B^* - (\angle FD^*C^* + \angle C^*B^*F) = \angle B^*FD^*\]
Since $\angle D^*FB^* = \angle B^*FD^* = - \angle D^*FB^*$, we have that $D^*,F,B^*$ are collinear. Then, by Reim's, since $B^*E^*\parallel D^*C^*$, it follows that $A^*,C,E^*$ are concurrent.

Inverting back, this means that $A,C,E,F$ are concyclic and we're done. $\blacksquare$.

We invert at $F$ with arbitrary radius.
Then get the following problem.
[quote=Inverted Romanian TST 3 2008, Problem 1]
Let $EDCB$ be a parallelogram. Let $F$ be the point inside $EDCB$ such that $\measuredangle CBF+\measuredangle FDC=\measuredangle FBE+\measuredangle EDF$. Let $A$ be the intersection of $(FBE)$ and $(FCD)$. Prove that $A$ lies on $CE$.
[/quote]
Note that as $\measuredangle CBF+\measuredangle FDC=\measuredangle FBE+\measuredangle EDF$ and $\measuredangle BED=\measuredangle DCB$, thus $B,F,D$ are collinear.
Thus, $$\measuredangle FAC=\measuredangle FDC=\measuredangle FBE=\measuredangle FAE.$$
We are done.


pigeonhole principle
combinatorics unsolved
combinatorics

A chess player plays [u]at least[/u] $ 1$ game [u]a day[/u] and [u]not more than[/u] $ 10$ [u]per week[/u]

Prove that if he does so for a long time, There will be a period of time where he will play exactly $ 21$ games

This problem is from a book I've read the solution but didnt understand a part of it  :(

[hide="Here it is:"]If $ a_i$ is the number of games played [u]till the day[/u] $ i$ with the day $ i$ included, Then

$ \boxed{1\leq a_1 < a_2 < .... < a_k < 10([\frac {k}{7}] \plus{} 1)}$ (according to the text)
Then
$ \boxed{22 \leq a_1 \plus{} 21 < a_2 \plus{} 21 < .... < a_k \plus{} 21 < 10([\frac {k}{7}] \plus{} 1) \plus{} 21}$

For $ k$ big, this ineq is true : $ \boxed{2k \geq 21 \plus{} 10([\frac {k}{7}] \plus{} 1)}$

[color=red][size=150]Then according to pigenhole principle[/size]: there exists $ ai$ and $ aj$ such that $ \boxed{ai \equal{} aj \plus{} 21}$[/color] and done.
[/hide]

The Part in red is the part I didnt understund
ie: I didnt understund how to use pigenhole to get the fact that there exists i and j st $ ai\equal{}aj\plus{}21$

I hope some one can make it clear for me, Thanks a lot  :D

Note that the set $ \{a_1,a_2,...,a_k,a_1+21,a_2+21,...,a_k+21\}$ contains $ 2k$ elements which are integers in the range from $ 1$ to $ 10([\frac {k}{7}] + 1) + 21$ and therefore belong to the set $ 1,2,...,10([\frac {k}{7}] + 1) + 21$ which has $ 10([\frac {k}{7}] + 1) + 21$ distinct elements. Therefore, the set $ \{a_1,a_2,...,a_k,a_1+21,a_2+21,...,a_k+21\}$ with $ 2k$ elements contains at most $ 10([\frac {k}{7}] + 1) + 21$ distinct elements. Therefore, as $ 2k>10([\frac {k}{7}] + 1) + 21$ by the Pigeonhole principle, two of the elements from the set $ \{a_1,a_2,...,a_k,a_1+21,a_2+21,...,a_k+21\}$ have to be the same. However, the set $ \{a_1,a_2,...,a_k\}$ contains all distinct elements and the set $ \{a_1+21,a_2+21,...,a_k\+21\}$ contains all distinct elements, so in order for  two of the elements from the set $ \{a_1,a_2,...,a_k,a_1+21,a_2+21,...,a_k+21\}$ to be the same, we need one element from $ \{a_1,a_2,...,a_k\}$ to be equal to one element from $ \{a_1+21,a_2+21,...,a_k\+21\}$ so there exist $ a_i,a_j$ such that $ a_i=a_j+21$.

I see it now  :o

Thank you Rem  :wink: .