[size=150]Dear All,
I can't speak french.Sorry.
I want your help for the resources of France Team Selection Test 2008-2009.I hope you have the materials.If you have these,Please share with me.Thanks! I want this urgently. You can send these materials to winallthings@yahoo.com
Or,you can upload or post here.
I''m seeking to hearing from you.
 :)
[/size]

Yours.
Steven

J'ai traduit du mieux que je pouvais:

[b]Chers tous,
Je ne parle pas le français. Désolé.
Je veux votre aide pour les ressources de France Sélection Équipe de test 2008-2009.
Je espérons que vous avez la materials. Si vous en avez, S'il vous plaît de partager avec moi. Merci!
Je veux que cette urgence. Vous pouvez envoyer ces documents winallthings@yahoo.com
Ou, vous pouvez télécharger ou de poster ici.
I'm qui cherchent à avoir de vos nouvelles.[/b]
 :)

[b]Votre ami,
Steven[/b]

I hope this helps you get your message across!  :wink:

Have a look here some problems of training 2008-2009
http://www.animath.fr/spip.php?rubrique13

For the courses
http://www.animath.fr/spip.php?rubrique10


I think he means the annoying equations of the form (x-y+k)(x+y+k)=j, j and k constants, x and y integers.

where you have to consider parity, factor j, set one factor equal to one factor of j, set the other, solve simultaneously

A classic example.
Find all solutions to
(1/x)+(1/y)=(1/12)
12y+12x=xy
xy-12x-12y=0
(x-12)(y-12)=144

Though sometimes it can be far worse
Like solve
x^2+6x=9+y^2

Oh ax+by=c...

those aren't so annoying
make sure (a,b)|c.

Then use Euclidean algorithm.

[quote="Fierytycoon"]Please be more specific about the diophantine equation. You mean questions like w+x+y+z=n, find the number of possible solutions?[/quote]

yeah nice combinatorial way for that sorta thing (if you don't know what it is, join a counting clats). but me professor said he didn't know of any nice combinatorial way for something like that with coefficients (besides casework)

[quote]but me professor said he didn't know of any nice combinatorial way for something like that with coefficients [/quote]

You mean like the equation I stated with each variable weighted? Like  :Sigma: (a_i * w_i) = n ?


This problem is from the current BWM.
You shouldn't post it again.


function
real analysis
real analysis solved

Liouville function  \lambda (n) is defined as  \lambda (n)=(-1) a1+a2+...+ak where n=(p1^a1)(p2^a2)...(pk^ak), and w(n) is defined as the number of distinct prime factors of n. Prove  \sum (d|n) (u(d))( \lambda (d))=2^w(n), where u is the Mobius function.

We know that u(d)=0 when d isn't square-free, so we just consider the divisors of n which are square-free. For these ones u(d) =  \lambda (d), so we add 1 for each square-free divisor of n. This gives kC0+kC1+..+kCk=2^k, where n=p1^a1*..*pk^ak.

The notations you use are a bit confusing:

[b]For Charlie:[/b]  \sum (d|n)... means  \sum ... for all d|n try something like  \sum {d|n}... next time.

[b]For grobber:[/b] kCi points to Newton's binomial coefficient, the coefficient of X i in (X+1) k.

Of course they know what they wrote, and most of the readers understand the proof, but I just wanted to make it clear for everybody and of course make another post :D .


The graph of a fifth degree polynomial with 1 as the leading coefficient has five distinct x intercepts, with one at 0,0. Which coefficient cannot be zero?

[hide="I think..."]
Let the polynomial be $f(x)=ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f$.

Since $(0,0)$ satisfies this equation, it follows that $f=0$.

Thus, $f(x)=x(ax^{4}+bx^{3}+cx^{2}+dx+e)$.

Let the other four roots of $f$ be $p$, $q$, $r$, and $s$. Then,

$ax^{4}+bx^{3}+cx^{2}+dx+e=(x-p)(x-q)(x-r)(x-s)$.

Since none of the other roots are 0, $pqrs\neq 0$, so $\boxed{e\neq 0}$. [/hide]

[hide]
Clearly the constant coefficient is zero. The x coefficient cannot be zero.
If it were 0, then you could take out x^2, which means there wouldn't be 5 distinct roots, contradiction..
[/hide]

Yes the coefficient of the x term cannot be zero or there would not be 5 distinct roots, but also, the leading coefficient cannot be zero, or it would be a fourth degree polynomial.

The problem specifies the leading coefficient, so it's assumed that that's not one of the coefficients being considered or otherwise the problem is trivial.


analytic geometry
graphing lines
slope
calculus
Pythagorean Theorem
geometry
AMC

Find the maximum value of the expression $x + 2y + 3z$ subject to the condition that $x - y + z = 1$ and $x^2 + y^2 = 1$.

Please try the problem!

Here's how I thought of it:

Just write it in terms of x and y:
The expression of x+2y+3z will turn into:
5y-2x+3 through substitution of z=y-x+1.

Thus, we will need to maximize 5y-2x.
I assumed that x and y are equal but of opposite parity.
x=-sqrt(2)/2 y=sqrt(2)/2

Answer: 7sqrt(2)/2 +3

your answer is not quite right, but your thinking is on track. What makes you think x and y are equal but of opposite parity?

$y$ should be given the priority over $x$ for the larger value, as $y$ has a larger coefficient. This may not be the largest value, but $y=1$ and $x=0$ gives the value of $8$... is that the answer?

[b]EDIT:[/b] Actually, no... I tested $y=.9$, and got a value above $8$...

a geometric visualization: just draw parallel lines with slope $\frac25$ and the max should occur when one of these lines is tangent to the circle...

Picking up where guangweicao left off:

$f = 5y - 2x$

$x^2 + y^2 = 1$

Since we want to maximize, the values of x and y will be in the second quadrant.  Therefore:

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

$f = 5(1 - x^2)^{\frac{1}{2}} - 2x$

$f' = \frac{5(-2x)}{2(1 - x^2)^{\frac{1}{2}}} - 2 = 0$

$\frac{-5x}{(1 - x^2)^{\frac{1}{2}}} = 2 \rightarrow \frac{-5x}{2} = (1 - x^2)^{\frac{1}{2}}$

$\frac{25x^2}{4} = 1 - x^2 \rightarrow \frac{29x^2}{4} = 1$

$x^2 = \frac{4}{29} \rightarrow x = \pm \frac{2}{\sqrt{29}}$

We want the negative x value to maximize the expression.  So plugging back in:

$x = \frac{-2}{\sqrt{29}}$

$y = \sqrt{1 - \frac{4}{29}} = \frac{5}{\sqrt{29}}$

$f = \frac{25}{\sqrt{29}} + \frac{4}{\sqrt{29}} = \sqrt{29}$

The maximum value is then $\sqrt{29} + 3$.  Finding the tangent point is almost the same as my process and yields the same result.

This problem can, of course, be solved without calculus (through a geometric approach, as mentioned above).

First, let $z = y - x + 1$.  Then we want to find the maximum value of $k$, where

$-2x + 5y + 3 = k$

$ y = \frac{2}{5} x + \frac{k -3}{5}$

Note that $k$ is therefore a vertical shift of the line $y = \frac{2}{5} x$.  From this we can tell that the maximum value of $k$ occurs when the line $ y = \frac{2}{5} x + \frac{k -3}{5}$ is tangent to the circle $x^2 + y^2 = 1$.  Drawing a diagram (not included here), we note that the point of tangency $(x,y)$ will have coordinates of the form $(-2n, 5n)$, where $n$ is some stretch factor such that the point is one unit away from the origin.  Using the Pythagorean theorem and solving, we have

$n = \frac{1}{\sqrt{29}}$

giving the point

$\left(\frac{-2}{\sqrt{29}}, \frac{5}{\sqrt{29}}\right)$

Plugging this into our formula for $k$, we have

$k = -2 \left(\frac{-2}{\sqrt{29}}\right) + 5 \left(\frac{5}{\sqrt{29}}\right) + 3 = \sqrt{29} + 3$


topology
function
calculus
derivative
real analysis
real analysis unsolved

Let $ X$ be Banach space. Assume that $ \phi \in C^1(X, \mathbb R)$ is such that $ \phi ' (0)\equal{}0$. A necessary condition for $ u \in X$ to be a critical point of $ \phi$ is that $ \left\langle {\phi '\left( u \right),u} \right\rangle\equal{}0$. This defines the Nehari manifold
\[ N: \equal{} \left\{ {0 \ne u \in X: \left\langle {\phi '\left( u \right),u} \right\rangle  \equal{} 0} \right\}.
\]
The question is that can we replace $ \phi \in C^1(X, \mathbb R)$ by a weaker condition, e.g., $ \phi \in C(X, \mathbb R)$?

Then how would you define $ \phi'(u)$? :maybe: Of course, a continuous function can have an extremum at a point where its derivative does not exist. You don't need to go into Banach spaces to find such an example.


calculus
derivative
advanced fields
advanced fields unsolved

I'm trying to solve an ordinary differential equation, with non-constant coefficients, using a power series solution.  The question is as follows:

----------------------------------------------------------------

solve the initial value problem:

$ x(2-x)y'' - 6(x-1)y' - 4y = 0$
$ y(1)=1$
$ y'(1) = 0$

hint: since the initial condition is given at $ x_0 = 1$, it is best to write the solution as a series centered at $ x_0 = 1$.

---------------------------------------------------------------

I have attempted the question, but I got stuck in one of the steps.  Here's what I have so far:

assume the solution can be written as a power series centered about x = 1.  Then the solution and its derivatives are:

$ y(t) = \displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n}$

$ y'(t) = \displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^{n-1}}$

$ y''(t) = \displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^{n-2}}$

where the $ a_n$ are constant coefficients.

Substitute y, y', and y'' into the ODE in order to determine the $ a_n$:

$ x(2-x)\displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^{n-2}}-6(x-1)\displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^{n-1}}-4\displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} = 0$

$ (1-(x-1)^2)\displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^{n-2}}-6(x-1)\displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^{n-1}}-4\displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} = 0$

$ \displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n ((x-1)^{n-2}- (x-1)^n)}-6\displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^n}-4\displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} = 0$

$ \displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^{n-2}}- \displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^n} -6\displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^n}-4\displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} = 0$

now I need to shift indices to make all the exponents of (x-1) equal to n, and also to try to make all the summations start at n=0:

$ \displaystyle\sum_{n=0}^{\infty}{(n+2)(n+1) a_{n+2} (x-1)^n}- \displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^n} -6\displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^n}-4\displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} = 0$

I don't know how to proceed from here.  I have 4 summations.  Two of them start at n=0 and has the $ (x-1)^n$ factor.  The other two summations start at n=2 or n=1.  I can't get them to start at n=0 because if I try to do that, then I get:

$ \displaystyle\sum_{n=0}^{\infty}{(n+2) (n+1) a_{n+2} (x-1)^{n+2}}$

and

$ 6\displaystyle\sum_{n=0}^{\infty}{(n+1) a_{n+1} (x-1)^{n+1}}$

As you can see, these have the factors $ (x-1)^{n+2}$ and $ (x-1)^{n+1}$ instead of $ (x-1)^n$.  So I don't know how to combine all this, since the summations do not start at the same number.

--------------------------------------------------------------
I'm told that in the end, the answer to the coefficients should be:

$ a_{2n} = (n+1)a_0$  and  $ a_{2n+1} = \frac{2n+3}{3}a_1$

So I just don't know how to get from the point I'm stuck to this answer.  I'm also quite confused as to how to solve the initial value problem, after obtaining the expected answer.  The initial conditions are: $ y(1)=1$,    $ y'(1) = 0$.  Using the initial condition y(1) = 1, I get:

$ y(1)=  \displaystyle\sum_{n=0}^{\infty}{a_n ((1)-1)^n}$
$ 1 = \displaystyle\sum_{n=0}^{\infty}{a_n (0)}$
$ 1 = 0$

This does not make sense...what am I doing wrong?

$ \displaystyle\sum_{n \equal{} 0}^{\infty}{(n \plus{} 2)(n \plus{} 1) a_{n \plus{} 2} (x \minus{} 1)^n} \equal{} 2a_2 \plus{} 6a_3(x \minus{} 1) \plus{} \sum_{n \equal{} 2}^\infty (n \plus{} 2)(n \plus{} 1) a_{n \plus{} 2} (x \minus{} 1)^n$

I haven't checked to see if up to here is correct, but you can always write out the first two terms when it starts at $ n \equal{} 0$ and the first term when $ n \equal{} 1$ to get them all to match up (except for some stray stuff).

Also, you probably define the first term to be $ a_0$ and not $ (1\minus{}1)^0$ as a convention to avoid that.


linear algebra
matrix
algebra
polynomial

Are the following true or false? If it is true, prove it, if it is false give a counter example (and proof that your counter example is true).

1) If T is a linear transformation such that T: R^2 -> R^3, T can't be one to one.

2) If a matrix has characteristic polynomial f(x) = (x-2)^2(x+4)(x-4), then the matrx is not diagonalizable.

please help. thanks.

Here are some examples. You interpret them.
\[ \begin{bmatrix}1&0\\0&1\\0&0\end{bmatrix}\]

\[ \begin{bmatrix}2&0&0&0\\0&2&0&0\\0&0&4&0\\0&0&0&\minus{}4\end{bmatrix}\]


function
trigonometry
logarithms
algebra
functional equation

find all continuous function $ f : R_{\plus{}} \to R$ such that :
\[ f(x)\equal{}f(x^2)\]

[quote="stvs_f"]find all continuous function $ f : R_{ \plus{} } \to R$ such that :
\[ f(x) \equal{} f(x^2)
\]
[/quote]

Obviously, there are an infinite values of x. Thus, the function must have an infinite degree, which is impossible.

Thus, the function must be constant. Any such function will work.

[quote="BOGTRO"]the function must have an infinite degree[/quote]  What's the degree of the function $ \sin x$?   :roll:

@stvs_f:

Please give your threads less generic names.  "Functional equation" would be an improved title here; "functional equation with square" would be even better.

Let $ g(x) \equal{} f( \ln x)$, so that $ g$ is a function from reals to reals and $ g(x) \equal{} g(2x)$. Suppose $ g(a) \neq g(0)$ for some $ a$. Then $ g(a/2^n) \neq g(0)$, but if $ n$ is high, $ a/2^n$ gets arbitrarily small, and so continuity is contradicted. Only constant functions work.

I can't see an elementary way to rigorize that, so just to make myself feel better: Because $ g$ is continuous, for $ 0 < \epsilon < |g(a) \minus{} g(0)|$ we have a $ \delta$ for which $ |x| < \delta$ implies $ |f(x) \minus{} f(0)| < \epsilon$. Take $ n$ such that $ a/2^n < \delta$. Then $ g(a/2^n) \equal{} g(a)$, but since $ a/2^n < \delta$ we require $ |g(a/2^n) \minus{} g(0)| < \epsilon < |g(a) \minus{} g(0)|$, contradiction.


The vertices of a regular polygon with $ n$ sides are colored with more than one color, such that the points with the same color form regular polygons.

Prove that we can find two equal polygons among the monochromatic ones.

its from all soviet mo 1970.
my sol. uses complex nos. as starting point.
will post sol. in full soon. :D

could u post it now? I'm very grateful if u post all problems in that issue.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=regular&t=23227

Pierre.


An integer $ n$ is called a good number if and only if $ |n|$ is not square of another intger. Find all integers $ m$ such that they can be written in infinitely many ways as sum of three different good numbers and product of these three numbers is square of an odd number.

[quote=andreim]m=a+b+ab/ a,b is not the square of any integer number and a is not equal b[/quote]

andreim, question says, [i]infinitely many ways[/i].


[quote="Myth"]Are you sure that it isn't for Pre-Olympiad? ;)[/quote]

I thought round 1 and 2 is intermediate, round 3 is pre-olympiad and round 4 is for the advanced section.  :)

[quote="Peter Scholze"]orl, i think you got the order of the forums wrong... isn't pre-olympiad the simplest one, intermediate topics something between and advanced the more difficult?[/quote]

Probably you got it wrong. The forums in increasing difficulty are:

Getting Started -> Intermediate -> Pre-Olympiad -> Advanced.  :)

[quote="Myth"]I thought in the same way as Peter.  :?[/quote]

I recommend you to read the sticky threads in each of the categories, especially the one in the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=7255]Pre-Olympiad Forum.[/url] For me this odering was obvious. What do you think ?

[quote="Myth"]This theme is not interested for me.[/quote]

Hmm. I don't understand what you mean. At first we argue whether I posted in the right forum and then I point you to the descriptions of the forums and you say you don't care ? Please explain.  :)

I think we don't need to develop this subject.

[quote="orl"]The number 4018020, the product of the year numbers, 2004 and 2005, of the current contest has some special properties:

(1) it has 7 digits
(2) the final digit is zero
(3) the number composed from the first three digits is half as big as the number composed from the next three digits.

Check whether there are further pairs of consecutive natural numbers satisying these properties. If so, provide them all.[/quote]

:)

jeez this is an interesting problem, ne1 have ne ideas?
the consecutive integers have to lie between:

$1000-1001, 3161-3162$ to satisfy the conditions that it is 7 digits.

There must be a 5 or 0 to satisfy the 0 at the end condition.

how about the three digits half the next three/


Well Miths had said he didn't want a sticky for this topic, but might as well make a normal topic for this.

This is the official topic for signing up to moderate a mafia game.

Since those who have already moderated may not mod until after game 20, game 19 will be dts, and game 20 will be ignite168 (I think, someone correct me if I'm wrong).

After that, it becomes a bit unclear.

I remember that the list was something like..
21. PeterVDD
22. Sunny
23. Melissa

After that, I recall me, samath, a few other people, and then lasagnaman.

Please post here if you had signed up for modding, and if you would like to mod a mafia game.

Sly Si was on there for 22 or 23. Weren't you 25 or 26? That means someone else is still in front of you.

Oh and I'll be away from July 7th-August 4th, this might mean I can't mod Game 20, I'll let people go in front of me and I'll mod Game 21 or 22 or something.

Let's try to keep Mafia on one thread. You guys can discuss mod sign-ups there.

Unfortunately, our mod's discussion is cleared and now some people are unsure of where they were in the line...


trigonometry
algebra
Triangle
triangle inequality
IMO
IMO 1968

Find all triangles whose side lengths are consecutive integers, and one of whose angles is twice another.

Let $ABC$ be a triangle with such conditions, such that $A=2B$. It's very well-known (or it can be deduced from the cosine law and the double-angle formula, or i think it's somewhere in the forum) that $a^2-b^2=bc$
Then, we just have to check the cases $(a,b,c)=(a,a-1,a-2), (a,b,c)=(a,a-2,a-1)$, and $(a,b,c)=(a,a-1,a+1)$.

$(a,b,c)=(a,a-1,a-2)$ gives $a^2-5a+3=0$, but it has no integer roots.
$(a,b,c)=(a,a-1,a+1)$ gives $a^2-2a=0 \Rightarrow a=2 \Rightarrow a+b=3=c \Rightarrow\Leftarrow$
finally, $(a,b,c)=(a,a-2,a-1)$ gives $a^2-7a+6=0$ and since $a=1 \Rightarrow c=0$, then we must have that $a=6$.
then the triangle that satisfies such conditions is $(a,b,c)=(6,4,5)$ with $A=2B$

If <A=2<B, some simple manipulating with the sine rule gives \[ /a^2=b(b+c) /\]. Assuming the lengths of the sides to be \[/ (a-1,a,a+1)/\] and considering all possible cases we get the solutions in (a,b,c) as
\[/(1,2,3) and (4,5,6)/\] and obviously all its permutations.

[quote="sayantanchakraborty"]Campos missed the solution \[(1,2,3)\].[/quote]

That´s NOT a triangle, but a degenerated one!

Best regards,
sunken rock

Well technically a degenerate triangle is a triangle. The problem should be worded more clearly, or both solutions should be accepted as correct.


complex analysis
function
complex analysis unsolved

Let $ f: \Delta(0,0,R)\mapsto \mathbb{C}$ be a holomorphic function and
$ f_{n}: \Delta(0,0,R)\mapsto \mathbb{C}$ where $ f_{n}(z)\equal{}f(\frac{z}{n})$, $ n\equal{}1,2...$
Given that {$ f_{n}$} is normal prove that $ f$ has a removable singularity at $ 0$

Please explain your notation better in the future...

Anyway, take a sequence $ z_k$ converging to 0. Fix an $ 0 < a < R$. Choose a sequence of integers $ n_k$ such that $ n_k|z_k| \in (a/4, a/2)$ (this is possible for large k). Let $ w_k = n_kz_k$. There is a subsequence $ f_{n_{k_j}}$ of $ f_{n_k}$ convering uniformly on compacts. The $ w_k$ lie in a compact, so ${ f_{n_{k_j}}(w_{k_j}) = f(z_{k_j}})$ is bounded.

If $ 0$ was a pole or an essential singularity, then we could choose $ z_k$ so that $ f(z_k) \to \infty$, so $ 0$ must be removable.


Let $ a,\ b,\ c>0$ such that $ a\plus{}b\plus{}c\equal{}1$. Prove that:

$ a\sqrt{1\minus{}bc}\plus{}b\sqrt{1\minus{}ca}\plus{}c\sqrt{1\minus{}ab}\ge\frac{2\sqrt{2}}{3}$ .

[hide="Solution"]
WLOG $ a\geq b\geq c$. Then it is clear that $ \sqrt {1 - bc}\geq\sqrt {1 - ac}\geq\sqrt {1 - ab}$ as well, so by Chebyshev's,
\begin{align*}3\left(a\sqrt {1 - bc} + b\sqrt {1 - ac} + c\sqrt {1 - ab}\right) & \geq(a + b + c)\left(\sqrt {1 - bc} + \sqrt {1 - ac} + \sqrt {1 - ab}\right) \\
& = \sqrt {1 - bc} + \sqrt {1 - ac} + \sqrt {1 - ab},\end{align*}
so it suffices to prove that
\[ \sqrt {1 - bc} + \sqrt {1 - ac} + \sqrt {1 - ab}\geq 2\sqrt 2.\]
First, we have, by AM-GM, the following inequalities:
\[ abc\leq\left(\frac {a + b + c}3\right)^3 = \frac 1{27},\]
and
\[ (1 - a)(1 - b)(1 - c)\leq\left(\frac {3 - a - b - c}3\right)^3 = \frac 8{27},\]
so it follows that
\begin{align*}(1 - ab)(1 - bc)(1 - ac)&=1-ab-bc-ac+a^2bc+ab^2c+abc^2-(abc)^2\\
& = 1 - ab - bc - ac + abc(a + b + c) - (abc)^2 \\
& = 1 - ab - bc - ac + abc - (abc)^2 \\
& = 1 - (1 - a)(1 - b)(1 - c) - (abc)^2 \\
& \geq 1 - \frac 8{27} - \frac 1{729} = \frac {512}{729} = \left(\frac 89\right)^3.\end{align*}
Therefore, by AM-GM again,
\[ \frac {\sqrt {1 - bc} + \sqrt {1 - ac} + \sqrt {1 - ab}}3\geq\sqrt [6]{(1 - ab)(1 - ac)(1 - bc)}\geq \frac {2\sqrt 2}3,\]
or
\[ \sqrt {1 - bc} + \sqrt {1 - ac} + \sqrt {1 - ab}\geq 2\sqrt 2,\]
as desired.[/hide]


function
AMC
AIME
blogs
geometry
geometric transformation
reflection

Hi AoPS Users,

I just thought of a new game. It's something like Scrabble, but you can not repeat any words within the last 20 posts. Using the last letter of the previos word, create a new word. PLEASE no cuss words.  If nobody posts a new word within 2 days, the person who has the last post is the winner.

The first word is
[b]hide[/b]

[b]yak[/b]

p.s. starting now the words are 3-7 letters

I hope I spell this right...  :maybe: [b]xylophone[/b]

gj you guys!!!
LET'S GET GOING WITH THE E STREAK!!!
[b]ebauche![/b]

awesome!!!
yay
e...hmm
[b]equivocate[/b]

[size=84][color=red]If you don't know, don't post! Stop spamming the Fun Factory so much. [/color][/size]


geometry
incenter
circumcircle
inequalities
geometry proposed

Prove or disaprove: (I think it is correct though anyway it's safer to say it this way as I'm not quite sure of my solution)

Denote by $O$ circumcenter and by $I$ incenter of the triangle $ABC$. Let $X_A,X_B,X_C$ be foot of perpendicular bisectors ($X_A$ is opposite to $A$ and so on) and let $Y_A,Y_B,Y_C$ be points in which incircle touches $BC$,$AC$,$AB$. Prove that there exist two indexes $m,n \in \{A,B,C\}$ such that:

$IY_m \geq OX_m$ and $IY_n \geq OX_n$

I don't think it's true (for acute-angled triangles, at least). It's going to be rather hard to give a numerical counterexample, but I drew some sketches on my computer in which the conclusion you wrote fails, and other sketches in which the conclusion you get by reversing the signs fails, so I don't know if you can get anything out of this.

Why did you word it like that, by the way? The "feet of the perpendicular bisectors" are just the midpoints of the sides, aren't they? :) And $IY_m=r,\ \forall m$. So the whole thing is a bit weird to look at :). Maybe I'm misunderstanding the problem because of this? If this is the case, my apologies to you.

Sorry for this strange wording  :D. I'm (almost) convinced to believe you Grobber   :) . Though if anybody finds some numerical example it will be welcomed.

[quote="Megus"]Though if anybody finds some numerical example it will be welcomed.[/quote]

Well, since $IY_A$, $IY_B$, $IY_C$ are the distances from the incenter I to the sides BC, CA, AB of triangle ABC, and $OX_A$, $OX_B$, $OX_C$ are the distances from the circumcenter O to the sides BC, CA, AB of triangle ABC, your conjecture was that every triangle has two sides which are nearer to the circumcenter than to the incenter. The simplest counterexample is an isosceles right triangle: each of the two catets is nearer to the incenter than to the circumcenter, and only the the hypotenuse is nearer to the circumcenter than to the incenter.

The "opposite conjecture", that every triangle has two sides which are nearer to the incenter than to the circumcenter, is equally wrong. As a counterexample, take an isosceles triangle ABC whose base is shorter than its leg.

  Darij


:roll: can anyone find d sum of d reciprocals of d 1st n terms?

If I understand correctly, you're looking for a closed form of the $ n$th harmonic number. Something like that doesn't exist.

http://mathworld.wolfram.com/HarmonicNumber.html


What is defined by problem solving??

I became very interested after my friend solved this problem.'

A girl picks a 2-digit number and reverses the order and adds the result. How many 2 digit numbers, using the process above will yield a perfect square.

About 98% (including me) people solved it using trial and error
12+21=33
13+31=44 etc. etc.

But my friend done like this
[hide]let the 2 digit number be $ 10a+b$ (27=2.10+7 , 31=3.10+1)
reversing, $ 10b+a$

adding , $ 11(a+b)$
for the number to be perfect a and b must equal 11

so 29,92 or 38,83 etc etc. and got 8 soln.[/hide]

Yeah. I'd solve it the way your friend did, but even if you do it by guess and check, you should notice a little pattern: e.g. all of the results are divisible by 11.

[quote="bos1234"]What is defined by problem solving??[/quote]

Well, [url=http://en.wikipedia.org/wiki/Problem_solving]this[/url], for starters.

In the context of this forum, it's good to be more specific; we often mean [url=http://en.wikipedia.org/wiki/Creative_problem_solving]this.[/url]

This website is devoted to mathematical problem solving. There are books written on this to make you much better at this sort of stuff. Look at the AoPs Library.


If a_1>0 and a_(n+1)=a_1^(a_n) I need to show a_n converges iff e^-e<a_1<e^(1/e).

Any pointers on how to get started would be appreciated.

[/code]

[quote="toadsmack7"]If a_1>0 and a_(n+1)=a_1^(a_n) I need to show a_n converges iff e^-e<a_1<e^(1/e).

Any pointers on how to get started would be appreciated.
[/quote]Here's a pointer. Let $a_{1}= a$ for ease of notation. Suppose the sequence converges, then the limit $m$ must satisfy $a^{m}=m$ and thus $a = m^{1/m}$. What is the largest such value?

More pointers (in German) are [url=http://wwwmath1.uni-muenster.de/u/ischebeck/vorkurs2006.pdf]here[/url] on page 31.


induction
number theory unsolved
number theory

[quote="AndrewTom"]If 1/2 is an element of S and if x/(x+1) and 1/(x+1) are elements of S whenever x is , then S consists of  the  rational numbers between 0 and 1.[/quote]

It's quite easy to demonstrate with induction (more precisely, it is easy to show with induction that "If 1/2 is an element of S and if x/(x+1) and 1/(x+1) are elements of S whenever x is , then S [u][b]contains at least all  [/b][/u]the  rational numbers between 0 and 1") :

Let $ P(n)$ be the following assertion : $ \frac{k}{n}\in S$ $ \forall k\in[1,n\minus{}1]$

$ P(2)$ is true.
Suppose now $ P(i)$ is true for any positive integer $ 1\leq i\leq n$
Let then $ x\equal{}\frac{u}{n\plus{}1}$ with $ 0<u<n\plus{}1$
Then :

If $ n\plus{}1\equal{}2u$, $ x\equal{}\frac{1}{2}\in S$

If $ n\plus{}1>2u$, then $ y\equal{}\frac{u}{n\plus{}1\minus{}u}\in S$ since $ P(n\plus{}1\minus{}u)$ is true and so $ x\equal{}\frac{y}{y\plus{}1}\in S$

If $ n\plus{}1<2u$, then $ y\equal{}\frac{n\plus{}1\minus{}u}{u}\in S$ since $ P(u)$ is true and so $ x\equal{}\frac{1}{y\plus{}1}\in S$

So $ x\in S$

So $ P(n\plus{}1)$ is true


[quote="langtuvietnam"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that:
\[ \sum_{cyc}\frac {a^2 \plus{} b^2}{a^2 \plus{} ab \plus{} b^2}\leq\frac {2(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} ac \plus{} bc}\]
[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3.$
Hence, your inequality is equivalent to $ f(w^3)\geq0,$ where $ f$ is linear function.
Hence, it remains to prove this inequality for $ b \equal{} 1,$ $ c \equal{} 0$ and for $ b \equal{} c \equal{} 1,$ which is very easy. :wink:
The minimal $ k$ for which the inequality
\[ \sum_{cyc}\frac {a^2 \plus{} b^2}{a^2 \plus{} ab \plus{} b^2}\leq\frac {k(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} ac \plus{} bc}\plus{}2\minus{}k\]
is true for all non-negatives $ a,$ $ b$ and $ c$ such that $ ab\plus{}ac\plus{}bc\neq0$ is $ k\equal{}\frac{4\sqrt3}{9}\equal{}0.7698...$

It may be true, but can you solve it by using AM-GM. :P

[quote="langtuvietnam"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that:
\[ \sum_{cyc}\frac {a^2 \plus{} b^2}{a^2 \plus{} ab \plus{} b^2}\leq\frac {2(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} ac \plus{} bc}\]
[/quote]
Since
$ \frac {a^2 \plus{} b^2}{a^2 \plus{} ab \plus{} b^2} \equal{} 2 \minus{} \frac {(a \plus{} b)^2}{a^2 \plus{} ab \plus{} b^2},$
the inequality can be written as
$ \frac {2(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} \sum \frac {(a \plus{} b)^2}{a^2 \plus{} ab \plus{} b^2} \ge 6.$
By the Cauchy-Schwarz Inequality, we have
$ \sum \frac {(a \plus{} b)^2}{a^2 \plus{} ab \plus{} b^2} \ge \frac {\left[ \sum (a \plus{} b)\right]^2}{\sum (a^2 \plus{} ab \plus{} b^2)} \equal{} \frac {4(a \plus{} b \plus{} c)^2}{2\sum a^2 \plus{} \sum ab}.$
Therefore, it suffices to prove that
$ \frac {2x}{y} \plus{} \frac {4(x \plus{} 2y)}{2x \plus{} y} \ge 6,$
where $ x \equal{} a^2 \plus{} b^2 \plus{} c^2$ and $ y \equal{} ab \plus{} bc \plus{} ca.$
This inequality is equivalent to
$ \frac {2(2x \minus{} y)(x \minus{} y)}{y(2x \plus{} y)} \ge 0,$
which is true because $ x \ge y.$

it's the best solution  :)
here is my solution http://www.mediafire.com/?i23vytjhdlz
and the solution of the author http://ddbdt.co.cc/forum/ba-t-ae-ng-tha-c-ae-ca-la-oei-gia-i-thcs/56-sa-ka-t-ha-p-cauchy-schwarz-am-gm.html
 :)


Prove that if $ G \equal{} \pi_1(n T^2)$, then $ G/G' \cong \mathbb{Z}^{2n}$, and if $ H \equal{} \pi_1(n \mathbb{R}P^2)$, then $ H/H' \cong \mathbb{Z}^{n\minus{}1} \oplus \mathbb{Z}_2$.

($ G'$ is the subgroup generated by all elements of the form $ aba^{\minus{}1}b^{\minus{}1}$, $ a,b \in G$.)

Telling us the definition of $ nX$ for spaces $ X$ would be nice.

Sorry, $ nX \equal{} \underbrace{X \# \cdots \# X}_{n}$, where $ \#$ is the connected sum.

those factor groups are the first homology group of the respective manifold :)

Draw a picture of the fundamental polygon, maybe use Van Kampen's theorem to read off generators and relations, then just add the relation $ ab\equal{}ba$ to abelianize and see what happens.

If $ X$ and $ Y$ are manifolds, and $ Z \equal{} X \# Y$, then the mayer-vietoris sequence gives a long exact sequence

$ \dots \to H_n(S^1) \to H_n(X \setminus S^1) \oplus H_n(Y \setminus S^1) \to H_n(Z) \to H_{n\minus{}1}(S^1) \to \dots$

If we're just looking for $ H_1$ we have

$ H_2(X \setminus S^1) \oplus H_2(X \setminus S^1) \to H_2(Z) \to H_1(S^1) \to H_1(X \setminus S^1) \to H_1(Y \setminus S^1) \to 0$.

if $ X$ and $ Y$ are both path-connected.

Luckily, all of the above are relatively easy to calculate. :).


Let $ a,b,c>0$ such that $ ab\plus{}bc\plus{}ca\plus{}abc\equal{}4$. Show that:

\[ \frac{a\plus{}b\plus{}c}{ab\plus{}bc\plus{}ca}\le 1\plus{}\frac{(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2}{48}\]

[quote="v235711"][quote="hedeng123"][quote="Pain rinnegan"]Let $ a,b,c > 0$ such that $ ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4$. Show that:
\[ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}\le 1 \plus{} \frac {(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2}{48}\]
[/quote]

substitution  $ a \equal{} \frac {2x}{y \plus{} z},b \equal{} \frac {2y}{z \plus{} x},c \equal{} \frac {2z}{x \plus{} y}$

$ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}\le 1 \plus{} \frac {(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2}{48}$

$ \Longleftrightarrow \sum_{sym} (2x^8y \plus{} 5x^7y^2 \plus{} 3x^7yz \plus{} 3x^6y^3 \plus{} 7x^6y^2z \minus{} 2x^5y^4 \plus{}$ $ x^5y^3z \minus{} x^5y^2z^2 \minus{} 2x^4y^4z \minus{} 12x^4y^3z^2 \minus{} 4x^3y^3z^3)\geq 0$

from Muirhead theorem the inequality holds .

 :lol:[/quote]

how did you think of such a substitution? this is a "classic" one , right?I think i´ve heard before ,but i´m not sure.[/quote]

you are right.

 :lol:


We  can  prove  in any  triangle  that
$4\sqrt{3}A\leq4r(5R-r)\leq ab+bc+ca \leq 4(R+r)^2 \leq a^2+b^2+c^2  $

and also  $4\sqrt{3}A\leq3\sqrt[3](a^2b^2c^2)  $
So my question is : What is  the complete order  betwen  them?

Thanks!

We need only determine the proper relation $?$ in

\[
4r(5R - r)\quad ?\quad 3(abc)^{\frac{2}{3}}
\]

Use the facts $K = \frac{abc}{4R} = rs = \sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}$ to boil this down to

\[
16(a^3 + b^3 + c^3) + 42abc\quad ?\quad 16\left(a^2(b+c) +b^2(c+a)+c^2(a+b)\right) + 3(a+b+c)(abc)^{\frac{2}{3}}
\]

This, however, seems suspicious since there are fewer terms on the left than on the right. Accordingly, setting $a = b = c$ requires that $?$ is $\le$ if anything at all. It is unclear how to proceed. The triangle condition is relieved if we set $a = y+z, b = z+x, c = x+y$ and specify $x, y, z > 0$. (One may be tempted to use AM-GM on $3(a+b+c)(abc)^{\frac{2}{3}}$ and cancel it with $9abc$ but this leads nowhere: the resultant inequality is false when $x = y$ and $z \rightarrow 0$.) Can we show that

\[
10x^2(y+z) + 10y^2(z+x) + 10z^2(x+y) \le 192xyz + 6(x+y+z)\left(x^2(y+z) + y^2(z+x)+z^2(x+y) + 2xyz\right)^{\frac{2}{3}}
\]

[quote="Mildorf"]We need only determine the proper relation $?$ in
\[
4r(5R - r)\quad ?\quad 3(abc)^{\frac{2}{3}}
\]
[/quote]

Unfortunately, there is no relation between the two that would hold for all triangles.

To see an example with $4r(5R - r)\quad <\quad 3(abc)^{\frac{2}{3}}$, [hide="click here"]
Taking $a=b$ and $c\to 0$, we get $r\approx c/2$ and $R\approx a/2$. So, the left hand side behaves like a constant multiple of $c$ near $0$ while the right hand side behaves like that of $c^{2/3}$, which is much bigger.
[/hide]
To see an example with $4r(5R - r)\quad >\quad 3(abc)^{\frac{2}{3}}$, [hide="click here"]
Taking $a=b$ and $c\to 2a$, we get $r\approx h/2$ and $R\approx a^2/2h$ where $h$ is the hight dropped on the side $c$. So, the left hand side tends to $5a^2$ while the right hand side tends to $3\cdot 2^{2/3}a^2$. But $5^3=125>3^3\cdot 4=108$.
[/hide]

Thank  you  very much!  That  was  helpfull! :)


MATHCOUNTS AMC AIME USA(J)MO USAMO geometry Alcumus

I know you're probably tired of answering these kind of posts, but please help me out!

I am a sophomore. I got somewhere between a 110-120 last year on the AMC 10 with not too much practice (just a little bit I remembered from MathCounts and maybe a weekend of studying).

I want to definitely make the AIME this time and hopefully qualify for the USAMO (I know it's tough but it's even tougher to get in as a junior/senior).

I own:
AoPS Volume 1: the Basics
AoPS Volume 2: And Beyond

What chapters should I read and solve all the problems for?, for the AMC 10? For the AIME?
Is there anything else I should be doing?
I can put in atleast an hour in on weekdays and a few hours on weekends.
To be honest, I'm a pretty fast learner so I'll pick up what I need to quickly.

Any help would be greatly appreciated!
Thanks in advance.  :)

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=245551]here[/url].  His guide answers all these questions :)

Thanks, for reminding me about that! (I bookmarked it a while ago and forgot to come back to it).

Yes, it clears up a lot of questions, except for the fact that I only have a month now and I need to know what would be the best allocation of my time for the next month- going through each chapter in AoPS 1 or doing old practice tests or maybe a mix of both?

Are there any more and less-useful chapters in AoPS? (of course I'll read through all of them but I want to know which ones to really study well and do every single problem in).

In the guide, it has the more useful chapters...

Anyway, you might want to take one or two of the old tests NOW, to figure out what subjects you're struggling in, and go to their corresponding chapters in Volume 1 and Volume 2. :)

In general, is using volume 2 useful for the AMC 10?

I suggest doing both --- AMC and practice problems from books. Within a month, what you do not want to do is confuse what you already know. Do enough problems that you learn (almost as if you are cramming for exam) but don't overexert yourself that you do poorly when the real one comes in (analogous to pulling an all-nighter and not waking up in time for the exam).

Everyone learns differently but you should try doing both things and see which one seems to help you more. I definitely suggest taking some practice tests though because you want to get used to the time limit thing definitely.

Volume 2 is useful in some aspects for AMC 10 because with geometry, you can solve some of them with trigonometry. But beyond that and some counting/probability, I suggest just doing old exams and sticking to the fundamentals. I'm not saying that you should close Volume II but if all your goal is to do well on AMC 10, I recommend focusing on previous exams and geometry chapters of Volume I.

My immediate goal is to score as high as I can on the AMC 10.

Next, it's to do well on the AIME, so the combo can possibly get me in USAMO.

So you think it's a good idea to do a bunch of AMC 10s and brush up on geometry and counting for now? Maybe I'll start doing volume 2 after the AMC 10. :)

I think you should get Silver Falcon's Geometry guide, which really did help me on geometry. Also I think if you are weak in geometry you should go through some old AMC's and AHSME geometry problems. I think Alcumus really helps on counting and probability, but I am not sure if you have enough time to just do Alcumus.

For the AMC 10, you don't need any trig...

For the "manipulations" that you asked, you just need experience in spotting the patterns.

My advice is try out everything when you're stuck! This especially helps in geometry and algebra.

Trig isn't needed, but is extremely helpful in problems, such as last year's #25 on the 10...

The ability to recognize possible manipulations comes from experience and practice. It would be best to do a lot of problems.

Trig will be more remembering than learning since I learned the Law of Sines and Cosines last year and we're starting to go over it this year too right now. Hopefully, it will help me do some geometry problems faster.

So I guess I'll just study my geometry, and do all the AMC 10s from 2001-now. (I've only done 2000 so far).

Thanks for helping guys. I hope I can get a 130+ on the AMC 10 after following this advice and working hard!  :)

[quote=".:Logic:."]Alcumus is helpful, it helping me a remember a lot of counting and probability methods that I've forgotten.
I tried taking a past AMC 10 yesterday and I got a 122.5 (with 2.5 for blanks). So if I had taken that this year (with 1.5 for blanks), I wouldn't have made it. Atleast I have a few weeks to bring up my score, to hopefully 130+ with the current scoring system.
For some reason I got 21-25 right, it was 14-20 that gave me trouble for some odd reason. My timing was off too some I'm definitely going to a lot more of those.

I found my main weaknesses to be progressions (easy I'll just read the Aops1 chapter and remember the formulas), geometry (read aops1 and go over trig from school), and [b]algebra[/b]. :|

By algebra, I mean the manipulations. Like when you're given a function which usually can't be solved due to too many variables and you have to manipulate it to get the answer instead. I know the tricks like A^2 - b^2=(A+B)(A-B) but I don;t have the foresight to use them like they're used in some of the harder amc problems. Any tips on improving on that aspect? Or is it small enough that I should just make sure I'm getting the other stuff right?[/quote]

Depending on the type of problem, certain ideas may come to your head. Try asking yourself-Can I Factor? Can i Cancel factors from num. and denominator? can i square this expression? can i multiply it by something useful? can i cleverly add 0? can i cleverly multiply by 1?

to clarify: by (cleverly)adding 0, i mean like

$ x^2 \plus{} 4x \equal{} x^2 \plus{} 4x \plus{} (4 \minus{} 4) \equal{} (x\plus{}2)^2 \minus{} 4$ or
$ A^4 \plus{} 4B^4 \equal{} A^4 \plus{} 4B^4 \plus{} (4A^2B^2 \minus{} 4A^2B^2) \equal{} (A^2 \plus{} 2B^2)^2 \minus{} (2AB)^2 \equal{} (A^2 \plus{} 2B^2 \plus{} 2AB)(A^2 \plus{} 2B^2 \minus{} 2AB)$, which admittedly is tricky to see.

And make sure to bring everything you can. I was told a ruler and protrator weren't allowed (after I brought them) and it would have trivialized #24 on the AMC 10A this year.

OMG! I just noticed.... calculators not allowed? :( There goes my chances of solving problems through graphing and trig. More importantly time might be even more of an issue now. Shouldn't be too bad though....

The problems don't really need a calculator ESPECIALLY the AMC-$ 10$.  As long as you try to simplify things by cancelling out, and not multiplying every single little thing out, a calculator doesn't improve your time by that much at all.  Keep your head out for what you are looking for and an intelligent way to approach it, and you'll be fine.

not having a calculator encourages people to look for clever methods, rather than brute forcing it. the clever methods are generally faster, so i think the calculator won't hurt you too much.

[quote] I was told a ruler and protrator weren't allowed (after I brought them) [/quote]

Please carefully read item 5 on the front cover of the AMC 10 A, and AMC 10 B.
A copy of the front cover of the 2009 AMC 10 A is on page 20 of the Teachers' Manual, you may see it here:
http://www.unl.edu/amc/d-publication/d1-pubarchive/2008-9pub/TM-AMC_1012A,2009.pdf
A copy of the front cover of the 2009 AMC 12 A is on page 21.

This rule has been on the cover of the contest in the same place since the AMC 10 began in 2000.
This rule is in the hands of every contestant at the time of the contest.
In Section II, item 4, page 5 of the Teachers' Manual, the teacher is instructed "As you hand out the papers, tell the students not to open the contest booklet.  They should then read the entire front cover.  Give them 5 minutes to do this."

Steve Dunbar
MAA Director, American Mathematics Competitions

No I know they are allowed, but they were confiscated before I started. I thought there must have been a rule change.


geometry
trigonometry
geometric transformation
rotation
ratio

$P$ is a point inside equilateral triangle $ABC$ such that $\angle BPC=120^\circ$. If the area of $\triangle ABP$ is $18$ and the area of $\triangle ACP$ is $8$, find the area of $\triangle ABC$.

i am unsure how to do it without trig exactly, but rotating 60 about C maping B to A gives a whole bunch of nice results...i will give it more thought

Yeah, rotation would work nicely. (You might need to use it twice) :)

This is sort of like #23 on this years AMC 12, right, except we had to rotate 90 degrees about the right angle vertex.

Anyone?

[hide="Need more hint?"]
Obviously $\angle PBC+\angle PCB=60^\circ$.
What does this imply?

When you rotate $\triangle PBC$, you'll see a pair of parallel lines.
Then you can use area ratios of triangles.

(As I said, you need to rotate $\triangle PBC$ twice.)
[/hide]

amm what is rorating triangles? :?
that what I have found  :P  http://www.dayton.wednet.edu/hs/faculty/dannweb/Rotating%20Triangles.htm


Ei bine, sunt singurul curios in privinta subiectelor de azi? Trebuie sa recunosc ca imi lipseste promptitudinea cu care erau postate acestea atunci cand Valentin era la fata locului. Tin minte ca la IMO 2003 (Japonia), subiectele au fost postate "calde", la aproximativ o ora dupa concurs. Au trecut mai multe ore si totusi inca nimic.

Un nostalgic.

[quote="icx"]Io am asteptat cuminte, fara sa ma gandesc sa fac un topic, iar cand l-am vazut pe al tau credeam ca le-ai si postat. Ah :mad: Vrem subiecte. Apoi rezultate :)[/quote]

eu postasem la doar vreo 2 ore, insa pe un alt thread :wink: clasa a X-a:

problema 1: dem ca ecuatia z^n + z +1=0 are o solutie complexa de modul 1 d.n.d. n restul impartirii lui n la 3 este 2.

problema 2: rezolvati ecuatia: 2^(x patrat +x) + log in baza 2 din x = 2^(x+1)

problema 3: pt ce numere naturale n mai mari sau egale cu 2 numarul a indice n = (n-1)^n^(n+1) + (n+1)^n^(n-1) este divizibil cu n^n?

unul din subiectele de la clasa a XII-a stiu ca zicea sa se demonstreze inegalitatea lui Cauchy cu integrale, sau ceva de genu asta :D

Cel mai bine postati subiectele fiecare intr-un topic separat (in the Olympiad Forums - la rubricile corespunzatoare) si apoi trimiteti-mi si mie linkul va rog.

[quote="icx"]Organizare de tot rasul... au trecut 30+ de ore si nici urma de subiecte* (sau rezultate) :| Vorbesc de site-ul oficial al ON de la Pitesti.

*nu punem la socoteala efortul lui zamfirmihai :D Btw, doar 3 subiecte la clasa a X-a? :maybe:[/quote]

am luat gazeta, acum am ajuns acasa. o sa incerc sa scriu niste probleme de la liceu.

[b]clasa a XI-a[/b]: [u]pr1[/u]: fie A,B apartinand lui M indice 2 (R) a.i. A^2 + B^2 = AB. sa se arate ca (AB-BA)^2 = 0 indice 2. Sper ca intelegeti ca nu am facut inca clasa a Xi-a. sper sa o fac :lol:

[u]pr2[/u]: daca nr reale a si b (a<b) sunt in imaginea unei functii continue  f:R->R, aratati ca exista un interval I inclus in R a.i. f(I) = [a,b]

celelalte 2 sunt greu de scris fara latex

[b]cls a XII-a[/b]
[u]
pr 3[/u]: fie n natural nenul. determinati toate inelele (A,+,*) cu proprietatea:

x^(2^n+1) =1 , oricare ar fi x apartinand lui A-{0}

[u]pr 4[/u]: Fie S indice n grupul permutarilor multimii {1,...,n} , nmai mare sau egal cu 3 si G un subgrup al sau generat de n-2 transpozitii. Aratati ca pentru fiecare i apartinand {1,...,n} multimea { fi(i) : fi apartinand lui G} are gel mult n-1 elemente.

"fi" este semnul acela de permutare, stiti voi  :D

o sa mai postez mai incolo putin si la clasa a IX-a si gimnaziu

[quote="alex__ib"]As aprecia daca ai posta subiectele de gimnaziu, te rog! Multumesc  :) ![/quote]

VII sau VIII? ca sa nu sciru de pomana ca mi-e lene:D

[quote="alex__ib"]VII de preferat  :)
Apropos, tu stai acasa in loc sa stai unde se presupune ca trebuie sa fiti cazati?  :?:[/quote]

da...

clasa a VII-a

pr1:a,b,c lat unui triunghi. a+b-c=2 si 2ab-c^2=4. sa se arate ca triunghiu este echilateral.

pr4: m ,n numere naturale. m>1 si 2^(2m+1) - n^2 mai mare sau egal cu 0. sa se arate ca 2^(2m+1) - n^2 e mai mare sau egal cu 7

Te rog sa postezi si la clasa a IX-a (cu prioritate) si la a VIII-a. :D

Incerc sa le scriu cu latex pe cele de la clasa a X-a ca sa fie mai clare.

[b] [size=134]Clasa a X-a[/size][/b]

[b]Sb1.[/b] Demonstrati ca ecuatia $z^{n}+z+1=0$ are o solutie complexa de modul 1 daca si numai daca restul impartirii lui $n$ la $3$ este $2$.

[b]Sb2.[/b] Rezolvati ecuatia: $2^{x^{2}+x}+\log_{2}x = 2^{x+1}$

[b]Sb3.[/b] Pentru ce numere naturale $n \ge 2$ numarul $a_{n}= (n+1)^{n^{n+1}}+(n+1)^{n^{n-1}}$  $\vdots$ $n^{n}$ ?

[quote="icx"][quote="Valentin Vornicu"]Oricum nu aici se posteaza. Asta e forumul: http://www.mathlinks.ro/Forum/index.php?f=217

si apoi in resources section. Dar o sa mai astept versiunile oficiale.[/quote]

Sorry, noi nu vream decat sa vedem subiectele, nu sa le si comentam/rezolvam. :maybe:[/quote]Stiu, tocmai de asta n-am zis nimic, dar nu e nevoie sa le pui in $\LaTeX$ (cine nu intelege postul initial, ma indoiesc ca o sa aiba vreo treaba cu rezolvarea lor oricum :D).

[quote="pureblack2003"]la a 7-a cine e YE KAI ala de a luat 28 pt ? ne-au invadat chinezii ? this is waaaaar  :!:  :maybe:[/quote]

ye kay e roman cum ar veni. e de la tudor vianu

Ar fi interesant sa se califice la OIM peste cativa ani   :D

Errr,.....de ce nu sunt postate subiectele pe site?

[quote="Zamfirmihai"]ye kay e roman cum ar veni[/quote]

da, si eu sunt japonez

Nu rade pureblack. In liceu un coleg de clasa il chema Yang Yang :D Era din Beijing :)

Could someone be kind to put the file of problems RNMO 2007
grade X
grade XI
grade XII

thx

http://www.mathlinks.ro/Forum/resources.php?c=142&cid=25&year=2007

Mai e nevoie de inca 5 probleme :)

le`am pus aici [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=144026[/url]

felicitari maky ! esti adevarat ! si original . e greu sa intretii parul ala ?  :lol: si mereu te-am vazut c-o fata sau doua . bafta mai departe. daca tot pui de toate p-aci, poate pui si o lista cu rezultatele la baraje ... daca faci rost cumva

ar fi binevenite rezultatele de la baraj ,dar eu as vrea sa vad si subiectele si baremele de la baraj...dak le aveti careva v-as fi recunoscator... :D

Subiectele de la Barajele (de seniori) sunt aici: http://www.mathlinks.ro/Forum/resources.php?c=142&cid=26&year=2007

btw, ordinea problemelor la primul baraj era
1. cea cu poligonul convex si dreptele $PA_{i}$.
2. geometria
3. functia
4. inegalitatea

am gasit un fisier cu rezultatele de la baraj

nu prea sunt surprize... cam astia erau ...


geometry
3D geometry
combinatorics proposed
combinatorics

n*n*n cube partition into n^3 1*1*1 cube. Find out, and prove, the smallest number k such that we can choose k cubes satisfy for any cube (C) of  n^3 cubes, there exists a cube (C') chosen and CC' parallel to one of three segments of original n*n*n cube.

Notice: (C) is the cube with center C.
-------------------------------

Inspired by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=363300#p363300]this problem[/url], which is essentially the same :)..

We regard the cube as being placed on a table, horizontally. In each of the squares in the $n\times n$ table along which the cube touches the table we put a number equal to the number of chosen cubes above that square. A moment's thought will show that the situation on this $n\times n$ square is the following: there are some positive integers in its unit squares, and given a unit square with number $0$, the sum of the numbers in the same row and column with it is at least $n$. The question is: what is the minimal sum of the numbers in the $n\times n$ square for such an arrangement to be possible.

This is essentially the same as the problem in the link above. The rows of the $n\times n$ square represent the sets $A_i$, while the columns represent the sets $B_j$, both families $(A_i)_i,\ (B_j)_j$ being partitions of the set of $1$'s we have placed in the table. The intersection of line $i$ and column $j$ bears the number $|A_i\cap B_j|$. That problem tells us that the minimal sum in such a situation (and the sum coincides with the number of chosen little cubes) is $\frac{n^2}2$, i.e. we must have at least $\frac{n^2}2$ chosen cubes if $n$ is even, and at least $\frac{n^2+1}2$ chosen cubes if $n$ is odd. If we prove that this is also enough, we're done. I'll just treat the case when $n$ is even. The other one can be solved in a very similar way.

First color the small cubes of the first layer of cubes chessboard-style. We number the rows and columns of the first layer of cubes (which is an $n\times n$ square formed of small cubes). The subsquare formed by the rows and columns with odd index have all squares with the same color (black-say). Partition this subsquare into $\frac n2$ groups of black squares, s.t. each group contains exactly one black square on each row and each column with odd index. Now lift each of these groups to a different layer of the $n\times n$ cube exactly as it is (i.e. we change only its vertical coordinate). Repeat for the rows and columns with even index, this time lifting the groups to the $\frac n2$ layers which still have no black cubes. In this way, we obtain the desired group of $\frac{n^2}2$ chosen cubes: the black cubes which we have placed throughout the $n\times n$ cube.


Robin is standing on  the top of a 40-foot flagpole at 1 p.m.  At the same time, a 4-foot child on the ground casts a shadow of length 0.8 feet. If Robin is 6 feet tall, how much longer is the shadow of the flagpole and Robin together than the shadow of the flagpole alone? Specify your answer as a decimal to the nearest tenth.

ratio of kid to shadow is 5/1

with robin = shadow would be 46/5 or 9.2
w/o robin = shadow would be 40/5 or 8

ans = 1.2

this concludes my 40th post (i think).

Or you can just find the length of Robin's shadow. :D

SO THATS THE OTHER PPLZ GET IT SO QUICKLY

 :rotfl:  :rotfl:  :rotfl:


function
real analysis
real analysis unsolved

Kent Merryfield's post in [url=http://www.mathlinks.ro/viewtopic.php?t=323552]this topic[/url] reminded me of a question I was asked recently.

Let $ \{f_n\}$ be a sequence of functions on $ [0,\infty)$ defined by $ f_n(x) \equal{} \frac{x}{1 \plus{} n^2 x^2}$.

We now look at $ \sum\limits_{n\equal{}1}^\infty f_n(x)$.

It is straightforward to show pointwise convergence on $ [0,\infty)$ and uniform convergence on intervals of the form $ [\delta,\infty)$ with $ \delta > 0$.

The question now is to show that we do [b]not[/b] have uniform convergence on $ (0,\infty)$.

From looking at the denominator we suspect that $ x \equal{} 1/n$ might play a role. Since $ f_n(1/n) \equal{} \frac {1}{2n}$ converges to zero, we don't have a cheap counterexample. But $ 1/(2n)$ looks like the harmonic series, and you know what we did with that one... So, we consider the sum $ \sum_{k \equal{} n}^{2n}f_k(1/n)$... and estimate it from below.


algebra
polynomial
inequalities
triangle inequality

Suppose that the coefficients of the equation $x^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 = 0$ are real and satisfy $0 < a_0 \leq a_1 \leq \ldots \leq a_{n-1} \leq 1$. Let $z$ be a complex root of the equation with $|z| \geq 1$. Show that $z^{n+1}=1$.

I think I have a valid solution to this problem, but the answer in the book is different. Here is my solution:

[hide]
If $z^{n+1}=1$,

$z^n = \frac{1}{z}$

Sinze $z$ is a root of the polynomial given,

$z^n + a_{n-1}z^{n-1} + \ldots + a_1z + a_0 = 0$ .......... $(1)$

Replacing $z^n$ with $\frac{1}{z}$,

$\frac{1}{z} + a_{n-1}\frac{1}{z^2} + \ldots + a_1\frac{1}{z^n} + a_0\frac{1}{z^{n+1}} = 0$

Multiplying both sides by $z^{n+1}$,

$z^n + a_{n-1}z^{n-1} + \ldots + a_1z + a_0 = 0$,

which was true as satisfied earlier by $(1)$. Thus, our proof is complete.[/hide]

Yes, it's circular. You should be very suspicious of the fact that you made no use of the condition $|z|\ge 1$ or the very particular conditions on the coefficients $a_k.$

You used the fact that $z^{n+1}=1$ to prove your case. In reality, your proof should proceed backwards and in that case your steps are not reversible so your proof is invalid (circular reasoning).

Let's construct a genuine argument. What follows still feels a little awkward, so I assume there is room for improvement of this proof.

Let $P(z)=z^n+a_{n-1}z^{n-1}+\cdots+a_1x+a_0,$ as above.

Define $Q(z)=(z-1)P(z)=z^{n+1}-b_nz^2-b_{n-1}b^{n-1}-\cdots-b_1z-b_0.$

Here, $b_0=a_0,\,b_1=a_1-a_0,\dots,b_{n-1}=a_{n-1}-a_{n-2}, b_n=1-a_n.$

We want to note that $b_k\ge0$ for all $k$ and that $\sum_{k=0}^nb_k=1.$

Now suppose $|z|\ge1.$ Then

$|Q(z)|\ge|z|^{n+1}-\sum_{k=0}^nb_k|z|^k$

$\ge|z|^{n+1}-\sum_{k=0}^nb_k|z|^n\ge|z|^n\left(|z|- \sum_{k=0}^nb_k\right).$

Since $\sum_{k=0}^nb_k=1,$ this can only possibly be zero if $|z|=1.$ That means that $Q(z)$ (hence also $P(z)$) can have no roots with $|z|>1.$

Now suppose that $|z|=1$ and $Q(z)=0.$ Divide by $z^{n+1}.$

$1-b_nz^{-1}-b_{n-1}z^{-2}-\cdots-b_1z^{-n}-b_0z^{-(n+1)}=0.$

But $|z^{-1}|=|z^{-2}|=\cdots=|z^{-(n+1)}|=1.$

We have that $\sum_{k=0}^nb_kz^{k-n-1}= \sum_{k=0}^nb_k|z^{k-n-1}|=1.$

By the equality case in the triangle inequality, the only way for this to happen would be for $b_kz^{k-n-1}$ to be a nonnegative real number for each $k.$ There are two ways for this to happen: $b_k=0$ or $z^{k-n-1}=1.$

But $a_0>0$ and hence $b_0>0.$ So we must conclude that $z^{-n-1}=1$ and hence $z^{n+1}=1.$


Hi folks,

Here is a thread that we can use to report on things from GHP. I'm currently sitting on the server for the Math Computer Lab. I'm sure that I'm not setting any weird policies for anyone. (Like Allen) I wouldn't do that. Would I?  :lol:

If you are coming, please try to be here at the time specified for you. It's apparently a very delicate dance getting everyone in and unloaded with the limited parking.

Tom

You're teaching at GHP this year!? That would be interesting. So, what classes are you teaching?

Huh whaaaaaat? The Math Computer Lab would definitely not be a place where I would spend my time, unlike other people. (Like Fulton) I wouldn't do that. Would I?  :lol:

Haha, I'm going to take most of your classes. We're already used to each other, aren't we?  :P (In case anyone wants to know, Mr. Fulton and I are both at Walton and I'm going to have Mr. Fulton as a math teacher for 3.5 of my 4 years of high school >_>)

I will be teaching problem solving, Delphi, a class on n-dimensional geometry (with some group theory I hope), and a class on poker math. I plan to use that class (and Jonathan) to fund most of my off campus food! :-)

Allen, classes start at 8am and you actually have to be in class at 8am each day (Monday - Saturday). At 8:01 we actually send people to start looking for you and that is NOT good. You might want start shifting your sleep schedule!

Governor's Honors Program I'll let others give the details

[quote="gtf"]I will be teaching problem solving, Delphi, a class on n-dimensional geometry (with some group theory I hope), and a class on poker math. I plan to use that class (and Jonathan) to fund most of my off campus food! :-)[/quote]

Something I've wondered, are these classes for complete beginners or would you be expected to have some experience in the subject?

[quote="gtf"]Allen, classes start at 8am and you actually have to be in class at 8am each day (Monday - Saturday). At 8:01 we actually send people to start looking for you and that is NOT good. You might want start shifting your sleep schedule![/quote]

Oh that's not good.

[quote="Goldey"]Wait what's GHP?[/quote]

[quote="GHP Website"]The Governor's Honors Program is a six-week summer instructional program designed to provide intellectually gifted and artistically talented high school students challenging and enriching educational opportunities not usually available during the regular school year. Activities are designed to provide each participant with opportunities to acquire the skills, knowledge and attitudes to become independent, life-long learners.[/quote]

Read more at the [url=http://admin.doe.k12.ga.us/gadoe/blogs/ghpblog.nsf/d6plinks/CBYE-6AWNBU]GHP information website.[/url]

[quote="Math Geek"]Two questions:

1). Was that Allen or Edward?

2). From a semi-reliable source, I heard that they're not doing GHP next year cuz of state budget constraints.  Is this true?  Oh, and if you really want to know, that semi-reliable source was a kid doing GHP for piano this year.[/quote]

1. It's Edward's account, though Alan sometimes uses it.
2. wut o.o

And also, there's a course on poker math?? :O
I hope it'll be there next year
Then I'll take it. ^__^

[quote="1077400"][quote="gtf"]I will be teaching problem solving, Delphi, a class on n-dimensional geometry (with some group theory I hope), and a class on poker math. I plan to use that class (and Jonathan) to fund most of my off campus food! :-)[/quote]

Something I've wondered, are these classes for complete beginners or would you be expected to have some experience in the subject?[/quote]

It depends on the class. Dephi is a programming class that all of the math majors will take. (I think.)

[quote="Math Geek"]2). From a semi-reliable source, I heard that they're not doing GHP next year cuz of state budget constraints. Is this true? Oh, and if you really want to know, that semi-reliable source was a kid doing GHP for piano this year.[/quote]

That rumor is absolutely false. That goes around all of the time, but it would probably be political suicide for a governor to kill it. (My opinion only.)

[quote="Arvind_sn"]And also, there's a course on poker math?? :O
I hope it'll be there next year
Then I'll take it. ^__^
[/quote]

Do you mean, if you are chosen? :-)

So Fulton, leave Jonathan with some money, please? I need the chance to take some from him when school starts back.

Have fun at GHP everyone!

Chuck Garner
Georgia ARML
Rockdale Magnet School, Conyers GA

Since I have gone to GHP, and I am somewhat bored, I'll answer some general questions.

1. What is GHP?

GHP (which stands for Govenor's Honors Program) is a 6 week long program at Valdosta State University. It is designed for the "top 2%" of students (10th and 11th graders in the year they apply (rising 11th and 12th graders when they get to GHP)) to get better in a subject that they excel in.

There is a very rigorous interview process, which can include multiple interviews at your school, a county interview, and a statewide interview, to pick the students who will attend.

The number of people in each major vary. I believe that Math is the largest with a little over 100 ppl. (108 in GHP2007). Communicative Arts, Social Studies, and Science are fairly large, but not as large. Some majors, such as Art and Technology, only get like 16. The total number of students is around 700.

2. What is a typical day at GHP like?

My schedule weekday schedule resembled something like this:
6:45 - wake up (yes this early)
7:20 - 7:45 eat breakfast
8:00 - 12:00 Major subject (Mine was Math, duh)
12:05 - 1:15 Lunch
1:30 - 4:00 Minors (I did Science)
4:00 - 10:55 Free time (The fun part about GHP)
11:00 -  Be in halls
11:05 - 11:20 take shower
11:30 - Lights Out be in dorms

Saturdays are like that except free time starts after Majors, there are no minors that day.
Sundays are a complete free day.

3. Are these classes for complete beginners or would you be expected to have some experience in the subject?

People who make it into GHP are already somewhat experienced in their subject and have a strong desire to learn. (If you are on the ARML team, your chances of making it to GHP are pretty good, but NOT guaranteed.)
Anyway, the classes are taught on the basis that you only have a normal high school math education: up to, but not including Calculus. There are very few requirements (at least when I was there). Everyone must take Delphi for the 1st 3 weeks (you can switch classes at this point). Also, the beginning problem solving class is reserved for people who haven't been on their high school's math team before.

@gtf:
Are you replacing one of the teachers, or does Math have one additional teacher now?
Is the problem solving class that you are teaching the most advanced one?
And, are you going by George, or Tom? (For those of you that don't know, GHP instructors go by their first names). EDIT: your forum signature answers this question fairly well.

We have several new instructors this year. Formally, I'll be Dennis for the 1st 3 weeks and he will come in the 2nd 3 weeks. The other instructors that have been here before are Jeff (the new department head), Carol and Vinton. I think the rest of us are all new.

And yes, it's Tom


function
limit
calculus
calculus computations

Using the figures, estimate:

$ \lim_{x \rightarrow 1^ \minus{} }(f(x) \plus{} g(x))$

$ \lim_{x \rightarrow 1^ \plus{} }(f(x) \plus{} 2g(x))$

$ \lim_{x \rightarrow 1^ \minus{} }f(x)g(x)$

$ \lim_{x \rightarrow 1^ \plus{} }\frac {f(x)}{g(x)}$

Stop posting these in the physics forums.  These should be in the Calculus-Computations and Tutorials forum.

Sums, products,... of limits of functions can be applied to each function first and then the operations performed.  So don't try to actually add or multiply the functions together.

[quote="JRav"]Sums, products,... of limits of functions can be applied to each function first and then the operations performed.[/quote]  This is, of course, completely wrong: it only holds if the limits of both pieces [i]exist[/i] (and also requires continuity of the "...").  It happens [i]here[/i] that all the functions involved have one-sided limits, but this hardly excuses giving such bad commentary on the general situation!

Obviously if the limits don't exist then it doesn't hold.  I'm not applying this to the general situation.  I thought that was implied.

If my statement was misconstrued, sorry.


bring that in canonic forme
$x^{2}+1+x/$x^3 -1


When you've private message me with all the problems you solved, you'll get the replying mail with your score. If you've been qualified, your name would appear on this thread. I'll try to grade as many as I can. The name and score would appear:

shinwoo (9)
RC-7th (9)
Farazy (8)
GoBraves (7)
DarkNight (7)

[quote="Silverfalcon"]When you've private message me with all the problems you solved, you'll get the replying mail with your score. If you've been qualified, your name would appear on this thread. I'll try to grade as many as I can. The name and score would appear:

shinwoo (9)[/quote]

Ok. Nice job shinwoo.  :P  Hope i can get 7.

I probably should of gotten an 8, made a typo on one of my explanation.... :( . After round one, can somebody explain number 10 for me, i dont really understand the problem.

how many people are taking the COPSD round 1?

I think that's unknown, since the deadline was extended.

[quote="shinwoo"]howlong?

and y was it extended?[/quote]

Look at the COP topic in getting started for why it extended.

so... today is the last day right?

huh??

I'm so confused....

and when do we take the COPSD 2nd round?

The extended date says the answers are due next weekend.

The reason was due to participants on Mathcounts.

[quote="Farazy"]I made a dumb mistake on the graph one...FORGOT TO SAY THE AREA!!!![/quote]

I made a dumb mistake and used a wrong formula and didn't check my answer on number 4. :(

Sad...  I missed on of the easiset ones.


Knowing:

$\frac{d-1}{a}+\frac{d}{b}+\frac{d+1}{c}=3d$

This problem was given at a Romanian olympiad and it was proposed by [i]Laurentiu Panaitopol[/i]

Well, merry Christmas, mathlinks!  :lol:

By inspection:

if $a = b = c$ then
$\frac{d-1+d+d+1}{a}=3d$
$\frac{3d}{a}=3d$
Thus $a = b = c = 1$ and $d$ can be any natural number...

Are we looking for solutions where
$a \neq b \neq c$

[hide="is it just me or is this really trivial?"]
unless $a=b=c=1$, $\frac{d-1}{a}+\frac{d}{b}+\frac{d+1}{c}<(d-1)+d+(d+1)=3d$ which is BAD!

So $a=b=c=1$ and then $d$ can be any natural number.
[/hide]

What how did I not see that... I have to pay more attention to these things... :roll:

Where did you find this problem? Maybe there are some other restrictions.


floor function
ratio
factorial
advanced fields
advanced fields theorems

Could somebody give me any reference for the following queston:
Fix $\lambda \in (0,1)$, and write $\lfloor z\rfloor$ for the integer part of the real number $z$. What is the asymptotic behavior of
\[\frac{1}{2^{n}}\sum_{i=0}^{\lfloor \lambda n\rfloor }\binom{n}{i}\]

The only easy value for me is $\lambda =\frac{1}{2}$, for which the limit is $\frac{1}{2}$

It is 0 from any $\lambda < \frac{1}{2}$ and 1 for any $\lambda > \frac{1}{2}$, I believe.

Let $\lambda<\frac{1}{2}$. Observe first that for $k$ close to $\lambda n$ the ratio $\binom{n}{k}/\binom{n}{k-1}$ is about $\frac{1-\lambda}{\lambda}$. So the sum in the expression is close to $\binom{n}{\lfloor \lambda n\rfloor}\sum_{-\infty}^{0}\left(\frac{1-\lambda}\lambda\right)^{k}= \frac{1-\lambda}{1-2\lambda}\binom{n}{\lfloor \lambda n\rfloor}$. To find the asymptotics of one binomial coefficient just use Stirling's formula for the factorials involved.

Maybe the ideas described [url=www.asymmetry.gr/images/asymmetry/Asymmetry_V5_March_2014.pdf#page=27]here[/url] could turn out to be useful.


calculus
integration
algebra proposed
algebra

Let Fn be the Fibonacci sequence defined by F0 = F1 = 1 and Fn+2 = Fn+1 + Fn for all n ≥ 0. Find all pairs (a, b) of real numbers such that for each n, aFn + bFn+1 is a member of the sequence. Find all pairs (u, v) of real numbers such that for each n, u(Fn) 2  + v(Fn+1) 2  is a member of the sequence.

The first half seems not very hard -- am I missing something?

$f_m=a f_n+b f_{n+1}$
$f_l=a f_{n+1}+b f_{n+2}$
so
$f_m+f_l=a f_{n+2} + b f_{n+3}=f_k$
Two positive Fibonacci numbers only add to a third fibonacci number when they are consecutive.  Thus we have that, in general,
$a f_n + b f_{n+1} = f_{n + k}$, k an integer.  Then we may simply substitute n = 0  and n = 1 to give us two equations to solve, getting $a = f_{k-2}, b=f_{k-1}$ for any $k \geq 1$ where we define $f_{-1}=0$.  (We could have taken any integral k, in fact, but because we are constrained to having the resulting sums be positive Fibonacci numbers, we must constrain k to be positive.)

I guess the second half is the tougher part -- I'll think on it.

Both questions are easy with a little help from analysis.

What kinf of analisys? can you post a solution so i can understand how you use analisys in this one?

I don't get it, why do you write real numbers???? Or is here a connection to the golden number that I don't see???? :?


calculus
derivative
function
vector
calculus computations

Hey,

[b]1. The problem statement, all variables and given/known data.[/b]
If we consider variables $ {(x, y)}$ we see that the Jacobian is defined as,
$ {{\frac {\partial}{{\partial}{(u, v)}}}{\bigg[(x, y)\bigg]}} \equal{} {{\frac {{{\partial}^{}}{x}}{{\partial}{{u}^{}}}}{\frac {{{\partial}^{}}{y}}{{\partial}{{v}^{}}}} \minus{} {\frac {{{\partial}^{}}{y}}{{\partial}{{u}^{}}}}{\frac {{{\partial}^{}}{x}}{{\partial}{{v}^{}}}}}$

However, we see that we're taking the partial derivative here as opposed to a normal derivative. In addition, we're taking the partial derivative of the variables $ {x}$ and $ {y}$ as opposed to a [b]function[/b] of $ {x}$ and $ {y}$ (ie. $ {f(x, y)}$). The latter is which I wanted to know how to do, in other words how would one define,
$ {{\frac {{d}^{}}{{d}{{(x, y)}^{}}}}{\bigg[{f(x, y)}\bigg]}} \equal{} {\text{???}}$

Where it is understood that we want the derivative of $ {f(x, y)}$ with respect to $ {x}$ and $ {y}$ simultaneously.

[b]2. Relevant equations.[/b]
Knowledge of Differential Equations.

[b]3. The attempt at a solution.[/b]
[hide="My Attempt at a Solution."]
If we refer back to the limit definition of the derivative, we see that for the case of,
$ {y} \equal{} {f(x)}$

We show that the limit exists if the limit is the same when approached from either side of the interval of $ {x}$.
However, when considering the limit of,
$ {z} \equal{} {f(x, y)}$
We find that the limit exists only if the limit is the same when approached from all possible "paths" of $ {(x, y)}$.
I want to say that the above information helps me answer the question I posed, however I am not sure how to put it together or what I am missing.
[/hide]
Thanks,

-[u][b][url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url][/b][/u]

One definition that could work is the is the [url=http://en.wikipedia.org/wiki/Gradient#Interpretations_of_the_gradient]gradient[/url], which is a vector field whose components are the partial derivatives with respect to both variables.  Another is the second derivative $ \frac{\partial}{\partial x \partial y}$ taken with respect to $ x$ and then $ y$ - for sufficiently nice functions, it doesn't matter what order you take the derivatives in.

I don't understand your discussion.  It looks like you're to describe the [url=http://en.wikipedia.org/wiki/Directional_derivative]directional derivative[/url], which is a specialization of the gradient, but you appear to be confused about the definition of the derivative.


a similar question to this was on this years canadian Physics olympiad, i believe that ended this monday.

question:

A pendumum with Mass 1 kg is attached to a string with lenght of 1 meter.  the pendulum is held horizantly and then let go.  assuming no friction in the system, express the tension in the cord as a function of time.

It's not hard to express the tension in terms of $\theta$ as the angular displacement of the mass:

$v=\sqrt{2gR\sin\theta}$

$F=\frac{Mv^2}{R}+Mg\sin\theta = 3Mg\sin\theta$

However, at least from what I know, it is not easy (not possible) to express the displacement of the pendulum as a function of time. (i.e. that differential equation can be solved only numerically.)

exactly, its very easy to write it in terms of angular displacement.

the problem with the time is that the net force acting on the pendulum is constantly changing!
same for the acceleration, that is acceleration is downwards when pendulum is in horizantal position, and its upwards at the very bottom position.


\frac{a^{2}}{a+b}+\frac{b^{2}}{b+c}+\frac{c^{2}}{c+a}\geq\frac{a+b+c}{2}
a,b,c positive reals

[quote="wauwau"]Do not understand the first equality sign. If you mean wolg a>=b>=c the >= would be OK, but than it can only be applied to the first two terms of the inequality[/quote]

Woops I forgot $\sum$.

$\sum_{cyc}\frac{a^{2}}{a+b}=\sum_{cyc}\frac{b^{2}}{a+b}$
$\iff \sum_{cyc}\frac{a^{2}-b^{2}}{a+b}=0$ which is true as $LHS=\sum_{cyc}a-b$.

$\sum_{cyc}\frac{a^{2}}{a+b}=\sum_{cyc}\frac{b^{2}}{a+b}=\sum_{cyc}\frac{a^{2}+b^{2}}{2(a+b)}\geq \sum_{cyc}\frac{a+b}{4}=\frac{a+b+c}{2}$.

OK I understand and the generalization

a,b,c positive reals, n positive Integer

$\frac{a^{n+1}}{a^{n}+a^{n-1}b+...+ab^{n}}+\frac{b^{n+1}}{b^{n}+b^{n-1}c+...+bc^{n}}+\frac{c^{n+1}}{c^{n}+c^{n-1}a+...+ca^{n}}\geq\frac{a+b+c}{n+1}$

I do not understand why the last inequality holds true because of AM GM?? Pls explain. Tahnks

Thanks a lot !
In your last to inequalities on the RHS there is a factor of n+1 missing

Using $\;ab\leqslant\frac{a^{2}+b^{2}}{2}$????
It's easy?

[quote="LoveStorm"]Prove that: \[\frac{a^{3}}{a^{2}+b^{2}}+\frac{b^{3}}{b^{2}+c^{2}}+\frac{c^{3}}{c^{2}+a^{2}}\geqslant\frac{a+b+c}{2}\] [/quote]
hmm...  a better look, http://www.mathlinks.ro/Forum/viewtopic.php?highlight=hungkhtn&t=23111

[quote="Lovasz"]
hmm...  a better look, http://www.mathlinks.ro/Forum/viewtopic.php?highlight=hungkhtn&t=23111[/quote]
Lovasz, it is here
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22937 :)

[quote="arqady"][quote="LoveStorm"]Prove that: \[\frac{a^{3}}{a^{2}+b^{2}}+\frac{b^{3}}{b^{2}+c^{2}}+\frac{c^{3}}{c^{2}+a^{2}}\geqslant\frac{a+b+c}{2}\] [/quote]
It is still easy. But the following vasc's inequality is nice:
Prove that $\frac{a^{4}}{a^{3}+b^{3}}+\frac{b^{4}}{b^{3}+c^{3}}+\frac{c^{4}}{c^{3}+a^{3}}\geqslant\frac{a+b+c}{2}$
for positive $a,$ $b$ and $c.$ :wink:[/quote]

It doesn't belong to VASC, I think.


calculus
integration
probability and stats

It is known (this is the Ito-Clark theorem) that every r.v. $ \xi\in L_2$ measurable w.r.t. $ \sigma(W)$ ($ W\equal{}\{W_t, t\in[0,1]\}$ is a Wiener process) has unique representation $ \xi\equal{}E\xi\plus{}\int_0^1g_t\, dW_t$ with Ito integral in the r.h.s.
 It is useful to get this representation in particular cases. For instance, if $ \xi\equal{}f(W(1))$ with $ f\in C_b^1(\Re)$ then $ g_t\equal{}G(t,W_t)$, where $ G(t,x)\equal{}Ef'(x\plus{}W(1\minus{}t))$.
I know how to prove this fact  with some`scary' technique based on Malliavin derivatives. Can you give a proof using `elementary' tools like the Ito formula?

did you have a look at chapter V.6.41 of Williams-Rogers "Diffusions, Markov processes, and Martingale", entitled "explicit Brownian martingale representation". It is 2 or 3 pages long, but quite relevant.

[b]alekk[/b], thanks, I will take a look


I'm very interested in getting some speed math books or math tricks.  Anything info will be greatly appreciated.  There's such a large selection and I don't feel like wasting money on stuff that isn't worth it or things I already know.  I'm mainly looking for intermediate based texts but I'll check out anything.

Thanks for any replies.

A good way to learn math tricks is to read posts on this forum.  It's free, too  :P

Many solutions in MathContests have speed tricks so look at those.

And as torajirou said, look at the forums.

[quote="mathgeniuse^ln(x)"]Many solutions in MathContests have speed tricks so look at those.[/quote]

I'm pretty sure he means MathCounts. :P


:) I'm going to make an assignment about "area".Could anyone  suggest me some problem about area or using area to solve?Please,include the source or the author's name and the hint.It's very nice if you post the problem that you create. :)

well, in terms of area, if you are/have learned about infinite/geometric series, the koch star(is that how it's spelled?) is quite interesting. it has finite area, with infinite perimeter! showing this is a relatively simple exercise with geometric series


Euler
function
LaTeX
modular arithmetic
number theory
totient function
number theory proposed

Show that for any positive integer $m$, there exists a positive integer $n$ so that in the decimal representations of the numbers $5^{m}$ and $5^{n}$, the representation of $5^{n}$ ends in the representation of $5^{m}$.

It is equvalent $5^{n-m}=1(mod \ 2^{k}), k= [mlog_{10}5]+1$ and obviosl.

Kind of the same.

It suffices to prove that there exists $ n$ such that $ 5^n\equiv 5^m\mod 10^m$, wich is equivalent to $ 5^{n\minus{}m}\equiv 1\mod 2^m$. Which is clearly true.

:)

It could also be finished using Eulers totient function.

By Euler-Fermat: 5^f(2^m) = 1 (mod 2^m) as 5 and 2^m are coprime. [f(x) is Euler's totient function]

But f(2^m) = 2^(m-1)

Thus 5^(2^(m-1)) = 1 (mod 2^m)

So set n = 2^(m-1) + m and you are done

Please, try to use $\LaTeX$, or at least a better syntax. Is it not nicer and clearer like this?

By Euler's, $5^{\varphi(2^m)} \equiv 1 \pmod{2^m}$, as $5$ and $2^m$ are coprime ($\varphi(x)$ is Euler's totient function).
But $\varphi(2^m) = 2^{m-1}$. Thus $5^{2^{m-1}} \equiv 1 \pmod{2^m}$. So set $n = 2^{m-1} + m$, and you are done.

Sure, thanks for asking! :)

[quote="mavropnevma"]By Euler's, $5^{\varphi(2^m)} \equiv 1 \pmod{2^m}$, as $5$ and $2^m$ are coprime ($\varphi(x)$ is Euler's totient function).
But $\varphi(2^m) = 2^{m-1}$. Thus $5^{2^{m-1}} \equiv 1 \pmod{2^m}$. So set $n = 2^{m-1} + m$, and you are done.[/quote]

[hide = Solution]
We will show for all $m$, there exists $n > m$ such that the decimal representation of $5^n$ ends in $5^m$. Notice that $5^m$ contains $\lceil \log{5^m} \rceil$ digits. It suffices to find $5^n$ such that $5^n - 5^m$ contains at least $\lceil \log{5^m} \rceil$ zeroes at its end. Notice that
$$5^n-5^m = 5^m(5^{n-m}-1)$$
We have $m \ge \lceil \log{5^m} \rceil$ for $m \ge 1$, so it suffices to find $n$ such that $5^{n-m}-1$ contains at least $\lceil \log{5^m} \rceil$ factors of $2$. This $n$ satisfies
$$5^{n-m} \equiv 1 \pmod{2^{\lceil \log{5^m} \rceil}}$$
We can choose $n$ such that $n-m$ is a positive multiple of $\phi(2^{\lceil \log{5^m} \rceil}) = 2^{\lceil \log{5^m} \rceil-1}$, and we can choose a sufficiently large multiple for $n$ so that $n > m$.
[/hide]


AMC
AIME
MATHCOUNTS
AMC 10
AMC 10 B
LaTeX

Welcome to the TTTT!

One of the biggest problems many of us have is: [u][b]speed[/b][/u].
Lack of speed causes some of to  lose contests, CD's, etc.
I really we should share our tricks, techniques, and/or faster methods.

So in this thread/marathon/whatever u call it, you have to post a problem, and a specific method you used to solve the problem, and explain clearly. Also, only post problems that use have a special or faster way of solving it.

[u]For Example[/u]:

Bob has $ 3$ pennies, $ 3$ nickels, and $ 3$ dimes. How many different values can he make if he must use at least one coin?

Technique: There are $ 4$ values for each of the coins,(0,1,2, or 3 coins), so the total number of values $ \Longrightarrow$ $ (4^3)\minus{}(1)\equal{}63$

Instead of counting the number of values, that technique helped us solve the problem in a second!

[b][u]Rules[/u][/b]

1.Only post problems with tecniques.
2.You can post any type of porblem (Geo, Algebra, C&P,NT, etc.)
3.I want this to flow smoothly
4.Learn!

I think this is why we have three marathons going on, all that have problems that can be solved using techniques and tricks. No need for a fourth. :)

Yeah, true I dont like too many marathons going at once...really I think one marathon is enough. :blush:
But, I dont really want this to be a marathon. In this the same person that posts the problem, posts a "technique" to solve that problem. :)
But if this isnt a good idea you can lock it. :)

Ok i have one...
There are 2 teams: Team A and Team B. How many ways can a seven game tournement bettween the both end if Team A wins. (When a Team wins 4 games the tournement endsx)

Solution
For this type of problem think of the Wins and Loses as letters. If you arrange them they come out like this:

$ wwwwlll$

For the tournement to win Team A must win 4 games and lose 3. (a game not played is counted as a loss)

So it is just how many ways you can arrange this:

$ \frac{7!}{4!\times3!} \equal{} \boxed{35}$

That's wrong - you must have a win at the end.  Otherwise, you count WWWWLLL as a 7-game series, even though it ends after 4.

Thus, we need to arrange WWWLLL and put a W at the end.  There are $ \binom{6}{3}\equal{}\boxed{20}$ ways to do this, [b]not[/b] 35.

[quote="SuperNerd123"][quote="SuperNerd123"]Ok i have one...
There are 2 teams: Team A and Team B. How many ways can a seven game tournement bettween the both end if Team A wins. (When a Team wins 4 games the tournement endsx)

Solution
For this type of problem think of the Wins and Loses as letters. If you arrange them they come out like this:

$ wwwwlll$

For the tournement to win Team A must win 4 games and lose 3. [b](a game not played is counted as a loss)[/b]

So it is just how many ways you can arrange this:

$ \frac {7!}{4!\times3!} \equal{} \boxed{35}$[/quote][/quote]

You can't just count like that, though - for example, the Lakers destroyed the Orlando Magic with 2 games left, but they didn't play those, and it was a 5-game series, not a 7-game series.

By the way, to mathboy101, you typoed: I think you meant he must have at least 7 socks in the problem statement.

[quote="AIME15"]That's wrong - you must have a win at the end.  Otherwise, you count WWWWLLL as a 7-game series, even though it ends after 4.

Thus, we need to arrange WWWLLL and put a W at the end.  There are $ \binom{6}{3} \equal{} \boxed{20}$ ways to do this, [b]not[/b] 35.[/quote]

Hm, I dont get this. I know that 6 choose 3 covers the situation WWWW, but if you try to arrange the letters WWWLLL into 6 spaces, you get 6!/3!3!=6 choose 3. However, arranging the letters WWWLLL doesnt account for situations like WWWW or WWLWW, does it? What am I doing wrong?

Oh, I'm sorry.  I misread the question as "exactly 7 games long", not a best of 7 series.  Sorry :blush:

Is there a technique for these type of problems?

-What is the maximum number of points of intersection of a line, a circle and a pentagon?

-What is the maximum number of points of intersection of a line, and two circles?

I kinda mess up sometimes when I do these types of questions :blush:

First I like your TTTT idea.
It deserves a seperate thread.

Now your math:
[quote="The Pink Panther"]
Technique: There are $ 4$ values for each of the coins,(0,1,2, or 3 coins), so the total number of values $ \Longrightarrow$ $ (4^3) \minus{} (1) \equal{} 63$
[/quote]

Your counts listing different coin combinations.
It didn't answer correctly about the different values of your selection!

For example 2 Nickles and 0 dime counts as different than 0 Nickle 1 Dime.
But they have the same value.
You have overcount in your trick.

[quote="AIME15"]That's wrong - you must have a win at the end.  Otherwise, you count WWWWLLL as a 7-game series, even though it ends after 4.
Thus, we need to arrange WWWLLL and put a W at the end.  There are $ \binom{6}{3} \equal{} \boxed{20}$ ways to do this, [b]not[/b] 35.[/quote]

Your count has flaws too. Not every win end within 6 or 7 games. It can end with 4 games.
 So  $ \binom{7}{4} or$\binom{6}{3}
are all wrong.

So far, all the quick tricks are not really quick tricks. They leads people with the wrong answers.

I had misread the question as the series ends in [b]exactly[/b] 7 games, not at most.

[quote="mathboy101"]Next Problem:
Bob has 13 blue socks, 213 green socks, and 92343249832749 yellow socks. What is the maximum number of socks he has to pick to be sure that he has at least 8 socks of one color.
Solution: Well, for the maximum, you should pick the most amount of socks that will not satisfy the question. So let's pick 6 socks of each color. That's 18 socks. Then, the only option is to pick the 7th sock of a color. So if you pick 19 socks, you will be sure that you picked at least 7 socks of a color. BTW, you didn't even need the big numbers. So the answer is $ \boxed{19}$.[/quote]

Should be the answer be: 22?

Your question asked for at least 8 socks while your answer answered for at least 7 socks.

I'm pretty sure this works. For any positive integer $ \ b$ then
$ \ 100_b$=$ \ 121_{b\minus{}1}$

That only works when $ b\geq3$. For more information consult Intro. to Number Theory....

This is because:
$ n^2 \equal{} (n\minus{}1)^2 \plus{} 2(n\minus{}1) \plus{} 1$

$ n^2 \equal{} n^2 \minus{} 2n \plus{} 1 \plus{} 2n \minus{} 2 \plus{} 1$
Canceling like terms
$ n^2 \equal{} n^2$

Er, no.
Read Into to NT.
$ \ (b\plus{}1)^{2}\equal{}b^{2}\plus{}2b\plus{}1.$

Er, yes.

Perhaps you should read what he posts before saying he's wrong :wink:

[quote="cgyao15"]Er, no.
Read Into to NT.
$ \ (b \plus{} 1)^{2} \equal{} b^{2} \plus{} 2b \plus{} 1.$[/quote]
If we use this in the problem that is saying b is 6 and b+1 = 7
[quote="I"]
This is because:
 $ n^{2}\equal{} (n\minus{}1)^{2}\plus{}2(n\minus{}1)\plus{}1$

 $ n^{2}\equal{} n^{2}\minus{}2n\plus{}1\plus{}2n\minus{}2\plus{}1$
Canceling like terms
$ n^{2}\equal{} n^{2}$
[/quote]
If we use this in the problem that is saying that n is 7 and n-1 is 6

[quote="Mewto55555"]For quick multiplication, factoring tricks are nice.
Ex: 14x15x16
When n=15, (n-1)(n)(n+1)=n^3-n=3375-15=3360=win[/quote]

How do you do 15^3 quickly?
To me 15^3 and 14x15x16 involved the same amount of calculation.

[quote="The Pink Panther"][u]For Example[/u]:

Bob has $ 3$ pennies, $ 3$ nickels, and $ 3$ dimes. How many different values can he make if he must use at least one coin?

Technique: There are $ 4$ values for each of the coins,(0,1,2, or 3 coins), so the total number of values $ \Longrightarrow$ $ (4^3) \minus{} (1) \equal{} 63$

Instead of counting the number of values, that technique helped us solve the problem in a second![/quote]

Doesn't work. 2 nickels are worth as much as 1 dime. You have to go though casework. There's no easy solution.

[quote="HiDN428"]New Trick:
I'll start by asking a problem. What is the average (mean) of all 5-digit numbers that can be formed by using
each of the digits 1, 3, 5, 7, and 8 exactly once? -2005 AMC10B #20
When i got to this problem i could answer it in a matter of seconds because of this trick. Ok, so i start of this problem looking at the digits we can use. Now, take the average of all the digits we can use, $ (1 \plus{} 3 \plus{} 5 \plus{} 7 \plus{} 8)$ all divided by $ 5$, which get us an average of $ 4.8$. Now multiply this average by 11111, and get the answer. You're done.
[/quote]

Here is the proof for your trick:

Let uvxyz be a such number where u, v, x, y, z can be any number choosing from 1, 3, 5, 7, 8.
We have
SUM_over_all_u,v,x,y,z (uvxyz) = SUM_over_all_u,v,x,y,z (10^4 u+ 10^3v+10^2x+10y+z)
=10^4 SUM_over_all_u (u) + 10^3 SUM_over_all_v (v) +10^2 SUM_over_all_x (x) +10 SUM_over_all_y (y) + SUM_over_all_z (z)
=10^4 *4! ( 1 + 3 + 5 + 7 + 8) + 10^3* 4! ( 1 + 3 + 5 + 7 + 8) +10^2*4! ( 1 + 3 + 5 + 7 + 8) + 10*4! ( 1 + 3 + 5 + 7 + 8) +  4! ( 1 + 3 + 5 + 7 + 8)
= 11111 * 4!  ( 1 + 3 + 5 + 7 + 8)
Therefore:
average = SUM_over_all_u,v,x,y,z (uvxyz)/5!
= 11111  ( 1 + 3 + 5 + 7 + 8)/5

[quote="generalmotors"]
Yeah, thats true because:

For any three digit number

$ \ 10^{3}\plus{}10^{2}\plus{}10\equal{}111$[/quote]
I'm pretty sure 1000+100+10=1110? or was that a typo?

[quote]I'm pretty sure 1000+100+10=1110? or was that a typo?[/quote]
He probably meant to say $ 10^2\plus{}10^1\plus{}10^0$  :wink:

[quote="bull98"]How do you do 15^3 quickly?
To me 15^3 and 14x15x16 involved the same amount of calculation.[/quote]
[quote="5849206328x"]@bull98 - He almost definitely just had $ 15^3$ memorized.  If we were to run this with $ 96\times 97\times 98$, there's isn't much of an advantage in doing that.[/quote]
When squaring any number ending in 5 (like 8.5 or 15), there is a trick. Say your number was $ 10^a \cdot (10n\plus{}5)$. The square of this is $ 10^{2a} \cdot (10(n(n\plus{}1)\plus{}2.5)$.


algebra
system of equations
geometry unsolved
geometry

Can someone show me how to do this,

I'm given three lines L1, L2, and L3 meeting at a point and I'm also given a point X on one of the lines. Construct (with ruler and compass) the triangle XYZ with the lines L1, L2, and L3 as angular bisectors.

Thank you for your help.

Draw $ l_1, l_2$, and $ l_3$, with the given $ X$ on $ l_1$. Let the angle between $ l_1$ and $ l_2$ be $ \theta_1$, the angle between $ l_2$ and $ l_3$ be $ \theta_2$, and the angle between $ l_1$ and $ l_3$ be $ \theta_3$. Call the point of concurrency $ O$. Let angle $ OXY \equal{} \alpha$, $ OYZ \equal{} \beta$, and $ OZX \equal{} \gamma$. Using triangles OXZ, OXY, and OYZ, we can construct the following system of equations:

$ 180\minus{}\theta_2\minus{}\theta_3 \equal{} \alpha \plus{}\beta$
$ 180\minus{}\theta_1\minus{}\theta_3 \equal{} \beta \plus{}\gamma$
$ 180\minus{}\theta_1\minus{}\theta_2 \equal{} \alpha\plus{}\gamma$

But since $ \alpha \plus{}\beta \plus{} \gamma \equal{} 90$, $ \gamma \equal{} \theta_2 \plus{}\theta_3 \minus{}90$, $ \alpha \equal{} \theta_1\plus{}\theta_3\minus{}90$, and $ \beta \equal{} \theta_1\plus{}\theta_2 \minus{}90$, although we really only need $ \alpha$ for our construction.

So, we take angle $ \theta_1\plus{}\theta_3$ and draw a perpendicular to subtract 90. The angle that remains we copy above and below line $ l_1$, and we extend these rays until they meet the other two lines on the other side of point $ O$ (each ray will intersect both $ l_2$ and $ l_3$, but point $ Y$ and $ Z$ are marked off at the intersection that is furthest from $ X$.


quadratics
algebra
calculus
calculus computations

I'm stuck at the basics (as usual). I have an equation x^2-4x+8=0
Using the quadratic I come up with (4(+/-) sqrt of -16)/2

My question is can you just pull out -i to make the sqrt positive so you get
(4(+ or -) sqrt 16*-i)/2?

Hi! davits. :)

$x=\frac{4\pm \sqrt{-16}}{2}=2\pm \sqrt{4}=2\pm \sqrt{4}i=2\pm 2i.$

You can use the following the reduction formula for quadratic equation as follows.

For the quadratic equation $ax^2+2b'x+c=0\ (a\neq 0),$ the roots are

\[ \boxed{x=\frac{-b'\pm \sqrt{b'^2-ac}}{a}} \]

Alternative solutions: You can use the completing square.

$x^2-4x+8=0\Longleftrightarrow (x-2)^2=-4\Longleftrightarrow x-2=\pm \sqrt{-4}=\pm 2i,$ yielding $x=2\pm 2i.$

The short answer is yes:  it is true that $\pm \sqrt{-r} = \pm i \sqrt r$.

Thanks! I was hoping that was the answer :-)


In any $\triangle ABC$, $E$ and $D$ are interior points of $\overline{AC}$ and $\overline{BC}$, respectively. $\overline{AF}$ bisects $\angle CAD$, and $\overline{BF}$ bisects $\angle CBE$. Prove $m\angle AEB+m\angle ADB=2m\angle AFB$.

The solution was like
[hide]label lots of angles [/hide]
but is there any other way? It seems like this problem would have some other sort of solution...

A rough picture of the problem attached...REALLY rough picture...

OK angles is probably the simplest way but if you want another solution this one is pretty cool.(it still ends up using angles but it's cooler :rotfl: ):
Wolog $\angle{EBD}>\angle{EAD}$
Take point $M$ on extension $AD$ through $A$ so $\angle{EMD}=\angle{DBE}$ Then $MFDB$ is cyclic so $\angle{MFB}=\angle{ADB}$
Take point $N$ on on line segment $EB$ closer to $B$ so that $\angle{FNE}=\angle{EAF}$ Then $AEFN$ is cyclic so $\angle{AFN}=\angle{AEB}$
Now, look at triangle $MAF$ and triangle $FNB$. Have $\angle{MAF}=180^{o}-\angle{FAD}=180^{o}-\angle{EAF}=180^{o}-\angle{FNB}=\angle{FNB}$
Also, $\angle{FMA}=\angle{FDB}$
So triangles $MAF$ and $BNF$ are similar so we got $\angle{MFA}=\angle{NFB}$
Then:
$2\angle{AFB}=\angle{AFN}+\angle{MFB}=\angle{AEB}+\angle{ADB}$
QED

when you have a simple way...just use it...angle chasing is fundamental to olympiad level geometry, in this case, it is just trivially labeling angles and using the sum of the degrees in a triangle is 180

Ok.

I was just thinking that the angle on the RHS was the average of the ones on the LHS and we had angle bisectors...but I couldn't really figure that out

Thanks for the solution