Inequality
function
three variable inequality
rearrangement inequality
inequalities
crux

Now this inequality is symmetric in a,b,c. So let a \geq b \geq c>0.
From here we have:
1/(b+c) \geq 1/(a+c) \geq 1/(a+b).
So by Rearrangement inequality we have:
a/(b+c)*(y+z)+b/(c+a)*(x+z)+c/(a+b)*(x+y)>=
1/2*(b+c)/(b+c)*(y+z) + 1/2*(c+a)/(c+a)*(x+z)+1/2*(a+b)/(a+b)*(x+y) = 1/2(2x+2y+2z)=x+y+z.
But x 2 +y 2 +z 2  \geq xy+yz+zx. Thus
(x+y+z) 2  \geq 3(xy+yz+zx). Which finishes the proof.

It is wrong applying of Rearrangement inequality and inequality LHS \geq x+y+z is wrong for a = b = 0.1, c = 1, x = y = 0.01, z = 1.

Hi all,

It is quite interesting and hard inequality. I have some idea but no time to realize it (the aproach is not so nice). Anyone has solution or idea?

Namdung

Hi all,

I think this nice inequality has a nice solution. But, while waiting this nice solution, I post my solution. It's not very nice but I think it solves the problem.

1. Firstly, we put m = a/(b+c), n = b/(a+c), p = c/(a+b) then m, n, p > 0 and  mn + np + pm + 2mnp = 1. We have to show that
        m(y+z) + n(z+x) + p(x+y) >= 3(xy+yz+zx)/(x+y+z)
<=> (n+p)x^2 + ((m+n+2p-3)y+(m+2n+p-3)z)x + (m+p)y^2 + (m+n)z^2 + (2m+n+p-3)yz => 0  (1)
Consider LSH (1) as trinom of x, we have
        D = ((m-n)^2-6p + 9 - 6(m+n+p))y^2 +
  2(m^2-mn-mp-6m+np-3n-3p+9)yz+ ((m-p)^2-6n + 9 - 6(m+n+p))z^2
If we can show that  D <= 0 then we get (1).
2. To prove  D <=0 WLOG we assume that p is maximum among m, n, p. Then, from the known fact m+n+p >=3/2 we have  (m-n)^2-6p + 9 - 6(m+n+p) < 0. So, is sufficient to show that
D = (m^2-mn-mp-6m+np-3n-3p+9)^2 - ((m-n)^2-6p + 9 - 6(m+n+p)) ((m-p)^2-6n + 9 - 6(m+n+p)) <=0
It equivalent to (p+n)(m^2+n^2+p^2-mn-np-mp+9-6m-6n-6p) <= 0 or
	F = m^2+n^2+p^2-mn-np-mp+9-6m-6n-6p <= 0  (2)
3. The (2) is generally wrong, even with the condition mn + np + pm + 2mnp = 1 (For example m=0, n=1/5, p=5). Fortunately, we can restrict the value of m, n, p in such a way that (2) holds. Note that
              3(xy+yz+zx)/(x+y+z) <= 3min{x+y,y+z,z+x}
we can suppose that  m, n, p <= 3.
Now if we can prove (2) holds for m, n, p in [0, 3], mn + np + pm + 2mnp = 1 then we have done.
4. This last problem can be solved by Lagrange method: Take function
               F_ \lambda  = m^2+n^2+p^2-mn-np-mp+9-6m-6n-6p + \lambda (mn + np + pm + 2mnp  1)
The system  dF_\lambda /dm = 0, dF_\lambda /dm=0, dF_\lambda/dp = 0, dF_\lambda/d\lambda  = 0 has only roots m = n = p = 1/2. So, the minimum can reach only at this point (F=0) or in the boundary.
              + If  for example m = 0 then  np = 1 and  F = (n+1/n)^2  6(n+1/n) + 6 < 0  because 2<= n+1/n < 4  ( 1/3 < n <3!)
              + If  for example p = 3 then m, n < 1 and  F = m^2+n^2-mn-9n-9m < 0.

The problem is done. I really dont like it, but the worst existing thing is better than the best_thing_that_we_dont_have.

Have fun,

Namdung

I have much appreciated your effort to find a solution.
It is not neat but it is correct.Congratulation.

P.S.I hope it will be possible to find a neat solution and the solver of it will reach a very nice winning post and a deserved celebrity in the forum.

It seems that the title was a good choice: nice inequality comes back and back and back.... But it is too difficult for me.

This beautiful inequality has been proved a lot of times (see also http://www.mathlinks.ro/Forum/viewtopic.php?t=50006 ). Let me add a different proof for a slightly more general version:

[color=blue][b]Extended Janous inequality.[/b] If u, v, w, x, y, z are six reals such that the reals y + z, z + x, x + y are all positive and such that the reals v + w, w + u, u + v and vw + wu + uv are all nonnegative, then

$\frac{x}{y+z}\cdot\left(v+w\right)+\frac{y}{z+x}\cdot\left(w+u\right)+\frac{z}{x+y}\cdot\left(u+v\right)\geq\sqrt{3\left(vw+wu+uv\right)}$
$\geq 3\frac{vw+wu+uv}{u+v+w}$.[/color]

[i]Proof.[/i] According to the Conway substitution theorem ([url=http://www.mathlinks.ro/Forum/viewtopic.php?p=204393#p204393]http://www.mathlinks.ro/Forum/viewtopic.php?t=2958 post #3[/url]), since the reals v + w, w + u, u + v and vw + wu + uv are all nonnegative, there exists a triangle ABC with sidelengths $a=BC=\sqrt{v+w}$, $b=CA=\sqrt{w+u}$, $c=AB=\sqrt{u+v}$ and area $S=\frac12\sqrt{vw+wu+uv}$. Applying the Extended Nesbitt-Pedoe inequality ([url=http://www.mathlinks.ro/Forum/viewtopic.php?p=535443#p535443]http://www.mathlinks.ro/Forum/viewtopic.php?t=90999 post #3[/url]) to this triangle ABC and to the reals x, y, z satisfying the condition that the reals y + z, z + x, x + y are all positive, we obtain

$\frac{x}{y+z}\cdot a^2+\frac{y}{z+x}\cdot b^2+\frac{z}{x+y}\cdot c^2\geq 2\sqrt3 S$
$\Longleftrightarrow\ \ \ \ \ \frac{x}{y+z}\cdot\left(\sqrt{v+w}\right)^2+\frac{y}{z+x}\cdot\left(\sqrt{w+u}\right)^2+\frac{z}{x+y}\cdot\left(\sqrt{u+v}\right)^2$
$\geq 2\sqrt3\cdot\frac12\sqrt{vw+wu+uv}$
$\Longleftrightarrow\ \ \ \ \ \frac{x}{y+z}\cdot\left(v+w\right)+\frac{y}{z+x}\cdot\left(w+u\right)+\frac{z}{x+y}\cdot\left(u+v\right)\geq\sqrt{3\left(vw+wu+uv\right)}$.

Remains only to show that $\sqrt{3\left(vw+wu+uv\right)}\geq 3\frac{vw+wu+uv}{u+v+w}$. Before we do this, we make sure that both terms vw + wu + uv and u + v + w are nonnegative. For vw + wu + uv, this was a condition of our theorem; for u + v + w, this is clear since $u+v+w=\frac{\left(v+w\right)+\left(w+u\right)+\left(u+v\right)}{2}$, and the reals v + w, w + u, u + v are all nonnegative. Thus,

$\sqrt{3\left(vw+wu+uv\right)}\geq 3\frac{vw+wu+uv}{u+v+w}$
$\Longleftrightarrow\ \ \ \ \ 3\left(vw+wu+uv\right)\geq\left(3\frac{vw+wu+uv}{u+v+w}\right)^2$
$\Longleftrightarrow\ \ \ \ \ \left(u+v+w\right)^2\geq 3\left(vw+wu+uv\right)$,

but this inequality is obvious, since

$\left(u+v+w\right)^2-3\left(vw+wu+uv\right)=\frac12\left(\left(v-w\right)^2+\left(w-u\right)^2+\left(u-v\right)^2\right)\geq 0$.

This completes the proof of the Extended Janous inequality.

  Darij

[quote=Shanghai][quote=sqing]If $ x, y, z $ are positive reals ,$u, v, w$ are  reals . Then
$$\frac{y+z}{x}u^2+\frac{z+x}{y} v^2+\frac{x+y}{z} w^2\geq 2\left(vw+wu+uv\right)$$[/quote]

this i can do :D ,simple AM_GM :-D[/quote]
Yeah.
$$\frac{y}{x}u^2+\frac{x}{y} v^2\geq 2uv$$
Thanks.

[quote=harazi]This time it's a very nice inequality proposed by Walther Janous in Crux:
Prove that if $ a$, $ b$, $ c$, $ x$, $ y$, $ z$ are positive reals, then
$F=\frac{a}{b\plus{}c}\cdot\left(y\plus{}z\right)\plus{}\frac{b}{c\plus{}a}\cdot\left(z\plus{}x\right)\plus{}\frac{c}{a\plus{}b}\cdot\left(x\plus{}y\right)\geq\frac{3\left(yz\plus{}zx\plus{}xy\right)}{x\plus{}y\plus{}z}$.[/quote]

$F\geq \sqrt{3(xy+yz+zx)}\geq\frac{3(xy+yz+zx)}{x+y+z}$
$RHS$ equivalent to $(x+y+z)^2\geq 3(xy+yz+zx)$
You can look to this [url=https://keoserey.files.wordpress.com/2012/07/zdravko-cvetkovski-inequalities-theorems_-techniques-and-selected-problems.pdfbook]book[/url]$($on page $32$ and $33)$ for $LHS's$ proof. :)

[quote=harazi]This time it's a very nice inequality proposed by Walther Janous in Crux:
Prove that if $ a$, $ b$, $ c$, $ x$, $ y$, $ z$ are positive reals, then
$ \frac{a}{b\plus{}c}\cdot\left(y\plus{}z\right)\plus{}\frac{b}{c\plus{}a}\cdot\left(z\plus{}x\right)\plus{}\frac{c}{a\plus{}b}\cdot\left(x\plus{}y\right)\geq\frac{3\left(yz\plus{}zx\plus{}xy\right)}{x\plus{}y\plus{}z}$.[/quote]
Prove that if $ a$, $ b$, $ c$, $ x$, $ y$, $ z$ are positive reals, then
$$\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}\geq\frac{3\left(a+b+c\right)}{ax\plus{}by\plus{}cz}$$
(Wanhuihua)

Prove that if $ a$, $ b$, $ c$, $ x$, $ y$, $ z$ are non-negative numbers with $x\ge a,z\le c$, then
$$\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}+\frac{y^2}{x}+\frac{z^2}{y}+\frac{x^2}{z}
\geq \sqrt{6(a^2+b^2+c^2+x^2+y^2+z^2)}$$(Wanhuihua)

[quote=arqady][quote=sqing]Prove that if $ a$, $ b$, $ c$, $ x$, $ y$, $ z$ are positive reals, then
$$\frac{ax}{a+x}+\frac{by}{b+y}+\frac{cz}{c+z}\leq\frac{(a+b+c)(x+y+z)}{a+b+c+x+y+z}$$
Old ? Where ?[/quote]
It's just C-S.[/quote]
Yeah.
Thank arqady.

[quote=harazi]This time it's a very nice inequality proposed by Walther Janous in Crux:
Prove that if $ a$, $ b$, $ c$, $ x$, $ y$, $ z$ are positive reals, then
$ \frac{a}{b\plus{}c}\cdot\left(y\plus{}z\right)\plus{}\frac{b}{c\plus{}a}\cdot\left(z\plus{}x\right)\plus{}\frac{c}{a\plus{}b}\cdot\left(x\plus{}y\right)\geq\frac{3\left(yz\plus{}zx\plus{}xy\right)}{x\plus{}y\plus{}z}$.[/quote]

Amazing problem .My solution,with references here:
https://www.cut-the-knot.org/m/Algebra/InequalityWithTwoTriples.shtml

[quote=harazi]This time it's a very nice inequality proposed by Walther Janous in Crux:
Prove that if $ a$, $ b$, $ c$, $ x$, $ y$, $ z$ are positive reals, then
$$ \frac{a}{b\plus{}c}\cdot\left(y\plus{}z\right)\plus{}\frac{b}{c\plus{}a}\cdot\left(z\plus{}x\right)\plus{}\frac{c}{a\plus{}b}\cdot\left(x\plus{}y\right)\geq\frac{3\left(yz\plus{}zx\plus{}xy\right)}{x\plus{}y\plus{}z}$$.[/quote]
[url=https://artofproblemsolving.com/community/c6h1141511p5364028]For positive reals prove that[/url] $$\frac{a}{b+c}(y+z)+\frac{b}{c+a}(z+x)+\frac{c}{a+b}(x+y)\geq\sqrt{3( xy+yz+zx)}$$$$\geq\frac{3\left(yz\plus{}zx\plus{}xy\right)}{x\plus{}y\plus{}z}$$
For #8

Prove that if $ a$, $ b$, $ c$, $ x$, $ y$, $ z$ are positive reals such that $\frac{a^2}{yz}+\frac{b^2}{zx}+\frac{c^2}{xy}+\frac{abc}{xyz}\leq 4.$   Prove that
$$a+b+c\leq x+y+z.$$
Let $ a$, $ b$, $ c$, $ x$, $ y$, $ z$ are positive reals.Prove that
$$\frac{by+cz}{a(y+z)}+\frac{cz+ax}{b(z+x)}+\frac{ax+by}{c(x+y)}\geq 2.$$
[url=https://artofproblemsolving.com/community/c6h374237p2066057]here[/url]

Let $ a$, $ b$, $ c$, $ x$, $ y$, $ z$ are reals such that $x+y+z=0.$ [url=http://kuing.orzweb.net/viewthread.php?tid=8077&extra=page%3D1%26amp%3Bfilter%3Dtype%26amp%3Btypeid%3D1&page=2]Prove that[/url]
$$x^2bc+y^2ca+z^2ab\leq\frac{1}{4}(a+b+c)^2 \max\{x^2,y^2,z^2\}.$$
Let $ a$, $ b$, $ c$  are positive reals and   $ x$, $ y$, $ z$ are  reals.Prove that
$$a^3(y^2+z^2)+b^3(z^2+x^2)+c^3(x^2+y^2)\geq 2abc(xy+yz+zx).$$[quote=sqing]
Let $ a$, $ b$, $ c$, $ x$, $ y$, $ z$ are positive reals.Prove that
$$\frac{by+cz}{a(y+z)}+\frac{cz+ax}{b(z+x)}+\frac{ax+by}{c(x+y)}\geq 2.$$
[url=https://artofproblemsolving.com/community/c6h374237p2066057]here[/url][/quote]


Your product equals:

$ \pi^{\sum_{k\equal{}2}^{\infty}\frac{1}{k!}}\equal{}\pi^{e\minus{}2}$.


topology
function
real analysis
real analysis unsolved

X is a compct metric space and f an function on X  to X such that d(f(x),f(y)) >OR EQUAL f(x,y)
Show that f is an isometry.


Let $[mn]$ be a diameter of the circle $C$ and $[AB]$ a chord with given length on this circle. $[AB]$ neither coincides nor is perpendicular to $[MN]$.
Let $C,D$ be the orthogonal projections of $A$ and $B$ on $[MN]$ and $P$ the midpoint of $[AB]$.
Prove that $\angle CPD$ does not depend on the chord $[AB]$.

Let $O$ be circle center, it is more straight forward to prove $\triangle PCD \sim \triangle OAB$.
First, because $OA=OB$, $\triangle OAB$ is an isosceles triangle.
Because $P$ is mid point of $AB$, then we have $PA=PD$, which indicates that $\triangle PCD$ is also an isosceles triangle.
Because $\angle ACO = 90 degree = \angle OPA$, $OPAC$ is on a circle, from which we have $\angle PAO = \angle PCO$.
Now that we have $\angle BAO = \angle PCO$, we can say  $\triangle PCD \sim \triangle OAB$.
Then we have $\angle CPD = \angle AOB$, which must be constant.

[quote=Singular]Let O be center of circle.  APOC, PBDO are cyclic.  Now APB is constant, so the sum CPA + BPD is constant, so APC + BPD is constant, and since APB is constant, it follows CPD is constant.[/quote]

I don't think your reasoning is valid when you say [quote]so the sum CPA + BPD is constant[/quote].


linear algebra
matrix
geometry
3D geometry

Show, by row operations, that the determinant of the matrix

$ \begin{bmatrix}x&y&z\\ z&x&y\\ y&z&x\\ \end{bmatrix}$

has factors $ x\plus{}y\plus{}z$,  $ x\plus{}ay\plus{}a^2z$,  $ x\plus{}a^2y\plus{}az$, where a is a complex cube root of unity


If $ a \plus{} b \plus{} c \equal{} 0, a \plus{} b \equal{} \minus{}\frac32, b \plus{} c \equal{} 4$ and $ a \plus{} c \equal{} \minus{}\frac52$, find $ abc$.

Subtracting the second equation from the first yields $ c\equal{}\frac{3}{2} \implies b\equal{}\frac{5}{2} \implies a\equal{}\minus{}4$, so $ abc\equal{}\boxed{\minus{}15}$.


what is the lopngest word you can think of. PM me your answer. also count the letters because i dont want to have to count all of them. and please dont not use google. ill post the results in 2 weeks or whenever i feel like it

dont post the word! now everyone will know it. if you post a word, nobody is allowed to copy it.

i know a 1185-letter word so i can't vote in the poll

it's a protein.

Please tell me that organic compound names don't count...

[quote="unimpossible"]wait i know a 1909 letter word too[/quote]
i doubt you know how to spell it, let alone take the time  to remeber all 1909 letters of it and type them. theres something know as copy and paste. im not stupid

Wait, I can't vote.  The longest word I know is < 0.

[quote="unimpossible"]well, i know the word. it counts[/quote]

have you read my post. im not stupid. theres something know as copy and paste. you dont know the word by heart because it would take apx 5 minutes to recite it. unless your that guy on tv who recited the first 2 mmillion digits of pi in 2 hours.

and miyomiyo, i wont even ask.

[quote="Lawrence Wu"]no u don't look up the chemical name of titin. and i know your 1,185 word too.[/quote]

same. it starts w/ an m right?

[quote="abacadaea"]YEAH ANY WORD THAT IS LONGER THAN 100 LETTERS DOESNT COUNT BECAUSE I KNOW YOUR USING GOOGLE!!! geez. i went a bit over but u get the point[/quote]

no, i'm using a dictionary :lol:

[quote="abacadaea"]thts cheasting 2[/quote]

i know lol :lol:

[quote="abacadaea"]YEAH ANY WORD THAT IS LONGER THAN 100 LETTERS DOESNT COUNT BECAUSE I KNOW YOUR USING GOOGLE!!! geez. i went a bit over but u get the point[/quote]

Chill man. This is the fun and games section. It's not like you get accurate results on fun and games polls anyway.

[quote="miyomiyo"]If they guessed pretty much anything, they'd survive.[/quote]

not z or x or q

You know what I mean.

How long did it take you to figure that out?

not very long. just copy-paste it on word, then call up the find key


linear algebra
matrix
function
linear algebra unsolved

The formula $\det(AB)=\det(A)\det(B)$ is well known. Do we have a formula for $\det(A+B)=?$  :oops:
On the other hand ${\rm Tr}(A+B)={\rm Tr}(A)+{\rm Tr}(B)$
How about ${\rm Tr}(AB)=?$  :oops:

In the following link, I gave a formula for the determinant of a sum, and used it to evaluate a determinant:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=591700#591700[/url].

There's no non-grubby way to get at the determinant of the sum- we have to break it down into rows or columns.

If we're willing to go a little farther, there are some nice ways to get at the determinant of a sum of many matrices:
Suppose $A_{i},$ $1\le i\le n+1$ are $n\times n$ matrices. Then $\det\left(\sum_{i=1}^{n+1}A_{i}\right)= \sum_{S}(-1)^{n-|S|}\det\left(\sum_{i\in S}A_{i}\right)$, where $S$ ranges over all subsets of $\{1,2,\dots,n+1\}$ with at most $n$ elements.
There are similar formulas for the determinant of a sum of $n+k$  $n\times n$ matrices, but we would need coefficients. In all cases, we can reduce the determinant to a sum over subsets with at most $n$ elements.

As for the trace of a product- there's a very nice elementwise formula. I don't know anything else you can do with it.

There are some nice special cases: $\det (A+tB)$ is an affine function of $t$ when $B$ has rank $1$. This sometimes makes it possible to find the determinant by interpolation, as [b]Ravi B[/b] did.


MATHCOUNTS
AMC
AMC 12
AMC 10
algebra
polynomial
geometric sequence

MINGMO

Due by March 30.

**
Rules:
1.No calculators.
2.No one else can do the problems for you (non-intellegent resources are fine, books, sites, etc.).
3.Send the answers to me via pm, do not discuss openly.
4.All contests must be submitted within 1 week.
5.All questions will be graded out of 5 points, a correct answer will earn 1 point. So a full solution is neccesary.
6.Solutions *may* gain bonus points if they're really, really good.
7.The score you earn per round will be equal to your number of points in the round (5 questions, 5 points max) times you level.
Example: A level 2 scores a 20, that person earns 40 points and 40 is added to their total.
8.No late submissions.
9.If you are not enlisted, do not send me a solution, instead, wait for the next round.
**

The second MINGMO
1. A real root to $\displaystyle x^3 + x^2 +x +1=0$ always equals an integer. Find that integer.
2. Two tangent circles with centers $\displaystyle A$ and $\displaystyle B$ have radii $\displaystyle 5$ and $\displaystyle 7$, respecticely. A line is drawn which is tangent to both circle $\displaystyle A$ and circle $\displaystyle B$ at points $\displaystyle C$ and $\displaystyle D$, respectively. A line is drawn from $\displaystyle A$ which passes $\displaystyle BD$ at $\displaystyle E$. If $\displaystyle \angle ABE$ is $\displaystyle 90^{\circ} (\frac{\pi}{2})$, determine the length of $\displaystyle AE$.
3. Find the set of k in which $\displaystyle x^3 + 3x^2 + 2x + k$ is divisible by $\displaystyle 3$. $\displaystyle x \in Z$.
4. Find the sum of the geometric sequence ${\displaystyle -3 - 1 - \frac{1}{3}....\frac{1}{6561}}$
5. A triangle has side lengths $\displaystyle 5$, $\displaystyle \frac{5\sqrt2}{2}$, and $\displaystyle \frac{5(\sqrt3+1)}{2}$. The three angles are denoted $\displaystyle\theta_1$, $\displaystyle\theta_2$, and $\displaystyle \theta_3$. Given $\displaystyle f(n)=\frac{n}{\theta_n}$. What is the maximum value of $\displaystyle f(1)+f(2)+f(3)$.

EDIT: Sorry, question 3 had a fault, it is changed. Sorry.
EDIT: Changed back.  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:

Since some other moderator or admin made this "sticky", I will defer to their judgement.  [I might have bumped this topic.]

It was stated in the Basics and MATHCOUNTS Forums that this contest would be "easy".  It is definitely *not* easy.

[hide="to hiitsme"] [size=75]I see from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=188339#p188339]this thread[/url] that you were concerned about the difficulty.  If you get fewer responses to this contest than you expect, please consider trying again with easier questions.

The mathematics behind each of the individual questions is probably within the MATHCOUNTS level.  And I'm sure you will get some good responses.  However, the overall contest is above the level of a National MATHCOUNTS competition.  Partly, this is because the problems are *hard*.  Partly, this is because answers must be justified (proven?).  And, partly, this is because some notations are unfamiliar to many middle school students.[/size][/hide]

Oh, it seems that I overestimated. But I have seen people with extrodanary skills around the forums, skill that I feel are far superior to mine. I wrote these quetions, they're definately not hard, they're probably easier than most amc12 quetions and probably around amc10 level, if I remember corrrectly.

If this contest is too difficult, speak up. I might lower the difficulty of level 1 questions and keep this difficulty for a further level.

That is odd though, I've seen some questions in this forums and I feel they are harder than mine  ;) .

EDIT: This may be because I never wrote mathcounts, sorry if I do not estimate the difficulty correctly. But I think nationals should be harder than my contest is it not?

even if we didnt enlist, can we just pm u to see how we do?

I have answers to all of them I'm just making sure they are correct... and I'm rather unsure of how I'm interpreting one of the questions...

sry im not quite i understand number 2, based on info given cant AE be many diff. possibiliites? need more info?

[quote="hiitsme"]You have a very good point. Since nobody has pmed me so far, I'm gonna make exceptions. You count as an enlisted contestant now sheepwarrior.

EDIT: I'm still a little shocked at how little people submitted, currently, it's 0.............. People seemed interested while I was advertising it, what happened now?[/quote] It is somewhat harder than advertised. :(

Even though the contest is not over, I want to congratulate tarquin on achieving 26/25 on this first round. (Remember the bonus marks idea?). Your solution-writing skills are very amazing.

BTW: 2 is a fault question, anyone who submits a reasonable solution will recieve full marks, tarquin has a misunderstanding of that question, but he did the rest very well, so he recieved full marks.

I don't see how the total is out of 30, you have 5 problems, each marked out of 5.  :?

I've moved this to Getting Started because it is way beyond MATHCOUNTS.  If the next one looks like this, I'll move it to Intermediate.  While the content of some of these questions is not Intermediate, the way they are asked is way beyond Getting Started.  For example, #3 should read something like

Given f(x) = ( polynomial ), describe all k for which f(n) is divisible by 3 for all integers n.  (The notation you use for 'all integers' is definitely not known by many Getting Started level students)

I duno either, must have thought I had 6 questions. *sigh BTW: why don't you submit as well? The points add up after a while, like cnml  :lol: .

Mind if I enter? I thought this was for the middle school, therefore I didn't sign up. And if it's still for middle school, ignore the post

May I still enter?
I have nothing else to do tomorrow, I'm on spring break...

Everyone feel free to enter as you will, only 3 people did, so results will be up very soon.

argh, didnt get to look at this thread until now heh.  I wanted to enter.


geometry perimeter trigonometry calculus derivative function

[quote="diophantient"][quote="The QuattoMaster 6000"][quote="mayhemkerim"][url=http://imageshack.us][img]http://img228.imageshack.us/img228/6282/triny2.gif[/img][/url][/quote]
[hide] AB and AC must be just over 12.  This makes angle ACB go to 90.  Thus, angle BCD goes to 90, thus BD is just above 25 (which is the minimum value of BD).  So, the minimum perimeter is 12+12+7+25=$ \boxed{56}$ [/hide][/quote]
But equality is unattainable.
How would 12-12-24 make ACB go to 90?[/quote]
I think the degenerate thing makes C go to 0...

Quick question: If we are asked "find the min perimeter of a triangle with one side equal to 3 and the other two sides equal to each other", would we able to say 6? But we cannot attain that...  What would a good answer be? Or would we never be asked such a question?

[quote="archimedes1"][/quote][quote="diophantient"][quote="The QuattoMaster 6000"][quote="mayhemkerim"][url=http://imageshack.us][img]http://img228.imageshack.us/img228/6282/triny2.gif[/img][/url][/quote]
[hide] AB and AC must be just over 12.  This makes angle ACB go to 90.  Thus, angle BCD goes to 90, thus BD is just above 25 (which is the minimum value of BD).  So, the minimum perimeter is 12+12+7+25=$ \boxed{56}$ [/hide][/quote]
But equality is unattainable.
How would 12-12-24 make ACB go to 90?[/quote][quote="archimedes1"]
I think the degenerate thing makes C go to 0...

Quick question: If we are asked "find the min perimeter of a triangle with one side equal to 3 and the other two sides equal to each other", would we able to say 6? But we cannot attain that...  What would a good answer be? Or would we never be asked such a question?[/quote]
There is no minimum perimeter in that case...

What? How are A,C,D collinear if BD=25, AC=AB=12?

Let $ \angle BCA\equal{}x$. Then $ AC\equal{}AB\equal{}12\sec x$. $ BD^2\equal{}24^2\plus{}7^2\minus{}2*24*7\cos x$.

It suffices to minimize
\[ 7\plus{}24t\plus{}\sqrt{25^2\minus{}2*24*7*\frac{1}{t}}\] where $ t\in (1,\infty)$. If you look at the derivative of that function, it is strictly positive. Hence the function is increasing, so the minimum occurs when $ t$ approaches the left bound. The maximum value does not occur...

:wallbash_red:  :wallbash_red:  :wallbash_red:
Sorry, I messed up.

[quote="Altheman"]What? How are A,C,D collinear if BD=25, AC=AB=12?

Let $ \angle BCA \equal{} x$. Then $ AC \equal{} AB \equal{} 12\sec x$. $ BD^2 \equal{} 24^2 \plus{} 7^2 \minus{} 2*24*7\cos x$.

It suffices to minimize
\[ 7 \plus{} 24t \plus{} \sqrt {25^2 \minus{} 2*24*7*\frac {1}{t}}
\]
where $ t\in (1,\infty)$. If you look at the derivative of that function, it is strictly positive. Hence the function is increasing, so the minimum occurs when $ t$ approaches the left bound. The maximum value does not occur...[/quote]
Wouldn't it be $ 7 \plus{} 24t \plus{} \sqrt {25^2 \plus{} 2*24*7*\frac {1}{t}}$ since $ \cos (\pi \minus{} x)\equal{}\minus{}\cos x$?
Also, aren't we looking at $ t^{2} \in (1,\infty)$ since t can be negative?


geometry
rectangle
circumcircle
blogs
geometry unsolved

there is a rectangle ( FMOH )  with sides |HO|=11 and |OM|= 5.
H is the intersection points of heights in a triangle ABC;
O is the center of circumcenter of ABC;
M is the middle point of |BC| and finally F is the foot of the heigt which blongs to the point A.

|BC|= ?

you problem is too easy
it's known that the center of gravity is on the line $ (OH)$ such that
 $ \vec{OH}\equal{} 3\vec{OG}$ so $ AH \equal{} 15$ we deduce $ AO\equal{}\sqrt{AH^{2}\plus{}OH^{2}}$
then we get $ CM \equal{}\sqrt{MO^{2}\plus{}OC^{2}}$ ....
done

[quote="castigioni"]THE ANSWER IS BC=28[/quote]

right but ı couldn't draw it :(


Find the value of $ \cos 36^\circ \minus{} \cos 72^\circ$.

[hide="Hint"]$ \cos 0 \plus{} \cos 72 \plus{} \cos 144 \plus{} \cos 216 \plus{} \cos 288 \equal{} 0$ by considering 5th roots of unity.[/hide]


find the ratio of $0$'s to $1$'s in binary expantion of $\frac{2^i-1}{2^j-1}$, where $\{i>j\}\in\mathbb{Z^+}$

I would like to (non-accusationally) point out that this question is currently the second problem of the SIMUW summer program application, so it would be nice to have this thread locked and perhaps the hints deleted, in hopes of discouraging people from seeking help on the problem. You can find it [url=http://www.math.washington.edu/~simuw/2006/pdf/prob_simuw.pdf]here[/url].


analytic geometry
graphing lines
slope
geometry
perpendicular bisector

[b]12. Find the standard form of the equation of the perpendicular bisector of (2,1) and (-1, -2)[/b]

CANT REMEMBER HOW TO DO THESE!!!  :mad:   :blush:

Find the slope of the line that goes through those two points. The slope of the perpendicular line is the negative reciprocal of the slope of the first line.
Then, use the midpoint formula to find the midpoint of the first line. Combine the two to get your equation.

The answer should be $ y\plus{}x\equal{}0$ <----Standard Form...right? or is it y=x+b

[quote="Nexmus"]The answer should be $ y \plus{} x \equal{} 0$ <----Standard Form...right? [/quote]

$ Ax \plus{} By \equal{} C$ is standard form, and yes, $ x \plus{} y \equal{} 0$ is the correct answer.


$ a,b,c,d$ -  integers, $ n$ - positive integers
solve equation
$ a\plus{}b^2\plus{}c^3\plus{}d^4\equal{}n$

Perhaps I am confused about the intent of the question but why would we not take any $ b,c,d$ and then $ a \equal{} n \minus{} b^2 \minus{} c^3 \minus{} d^4$ to get all solutions ?

Did you mean that $ a,b,c,d$ must be positive?

This problem you have shown , has infinite number of solutions and it seems to be senseless.
May i ask you where did you get this "problem" ?


All right since the other one is almost done or abandoned we will have a new Beat the Mod where you will be battling an actual mod. This game will be shorter than the last one. I plan to have maybe 10 rounds where 2 of them it is possible to kill the mod. Come to think of it ill play too as a whisperer to give you hints like pythag did in the other game. I don't know who the mod will be but sign-ups for regular players begin now.

Round ten will be a modkill (literally) round and so will another one.

Signups:

hmmm we could have brute as the mod again or we could get someone else. It doesnt have to be a mod. I guess we should have mod signups too.

My role was much more than a whisperer.

I recommend you not start until the last game ends...

[quote="mathblitz"]oh by the way you can remove Karamata from your signature pythag. Well than what else were you?[/quote]

I can't say yet.

First I sign up for the mod part. If brut gets it or someone else I will join as a competitor.

Seeing as I can't edit anymore here are the sign-ups:

dragon96
FantasyLover
SuperMathBoy
alekim
rt_08

mod signups: jjx1

"whisperer":Mathblitz

theres not enough people yet but if no more join then Ill figure it out.

idk how to play, but i guess ill join

PMme when it starts

Nothing appealing from the others, but there is stuff unappealing bout mew :wink:

Meh.  Whatever.  I'll mod.

I have a setup already written  :wink: . Meaning we could start now. But I'm waiting until BTM 1 finishes.

I really impatient to start this
 :wallbash_red: HURRY UP GUYS! :furious:

[quote="Mewto55555"]I was under the impression that Brut3, the original BTM Mod, gets to pick his successor?[/quote]

He was originally planning to choose me.

Ok I'll be player. And what are you talking about Aime15. I am very active. I go on ftw every day for at least 5 hours or so. I just don't go on games forum very much. I spend all my time on ftw :lol:

mathblitz, you cheater!!! i said i was in on the first page!!!

[quote="mathblitz"]yes I'm conferencing with Brute to figure out who is the mod and then I will see whether or not to start.[/quote]
Sorry, but I really, really don't have an opinion on this.

it seems this thread has been killed because of my absence. use mewtos thread but leave the setup the same plz. im not a whisperer im too busy right now to participate too much


1. What is the least number of objects required to way any integral number of pounds up to 63 pounds if one is allowed to put weights in only one pan of a balance.

2. Prove that each nonzero integer may be uniquely represented in the form

\[n=\sum_{j=0}^s c_j3^j,\]

where $c_s \ne 0$, and each $c_j$ is equal to $-1$, $0$, or $1$.

3. Using the result of problem 2, determine the least number of weights required to weigh any integral number of pounds up to 80 pounds if one is allowed to put weights in both pans of a balance.

(Number Theory: Andrews, George E.)

[hide]1. 6: the numbers 1, 2, 4, 8, 16, 32
2. We must prove that we can construct any number with a sequence of this form.  Notice that for every negative number, we can take the opposite, the positive number, and simply reverse all of the signs in its representation.  Thus, we must show this for all positive integers.  Note that we can express any number $2(3^r)$ as $3^{r+1}-3^r$.  Therefore, the question is the same as proving that any number can be expressed as a sume of $c_j3^j$ where $c_j$ is $0$, $1$ or $2$.  Then we can get exactly $3^{s+1}$ unique non-negative integers through a summation like this.  Therefore, if we let $s=\infty$, then we can construct any number.[/hide]

How did you figure out #1? And can you prove that your answer is indeed the smallest possible?

[hide="1."]
You obviously need $1$. Then you obviously need $2$. Now you can make $3$, but you need $4$. You can make anything up to $7$, so you need $8$. Continue this until you have $1,2,4,8,16,32$ and you can make anything up to $63$. Good enough?[/hide]

[hide="2."]
Write the number in base $3$ (i.e. $1(3^0)+2(3^1)+0(3^2)+1(3^3)+...$). Whenever there is a 2, change it to a -1 and increase the next power of 3 by 1. We have the original number still in terms of only $-1, 0, 1$ powers of $3$.[/hide]

[hide="3."]
I think you can pretty much do it the same was as #1, from which we get a result of $5$ (the powers of $3$ up to $81$). I'm not completely sure, though, since you can actually make anything up to $121$ with this...[/hide]

why can't 1 be 5, with weights 1, 3, 9, 27, and 81 (although there is a lot of waste).

How can you weigh something that's 2 pounds? (236!)

Sorry, only one pan  :roll: i mis read it


arithmetic sequence
number theory proposed
number theory

Prove that in every arithmetic sequence a, a + d, a + 2d, ..., a + nd, ..., where a and d are natural numbers, we can find infinitely many numbers whose canonic representations consist of the same prime factors.

[An example of two numbers whose canonic representations consist of the same prime factors: $8232=2^{3}\cdot 3\cdot 7^{3}$ and $2016=2^{5}\cdot 3^{2}\cdot 7$.]

  Darij

For all integers $k \geq 0$, we have $(1+d)^k = 1 + m_k \cdot d$ for some nonegative integer $m_k$.
Clearly, the $m_k$'s are pairwise distinct.
Now, let $n_k = a\cdot m_k$.
Then :
$a+n_k \cdot d = a(1+d)^k$, which easily leads to the fact that all the terms $a + n_k \cdot d$ have xactly th same prime divisors.

Pierre.


Express $ 0.\overline{3}\plus{}0.\overline{2}$ as a common fraction.

\begin{align*}
0.\overline{3}+0.\overline{2} &= \frac{3}{9}+\frac{2}{9} \\
&= \boxed{\frac59 }
\end{align*}


geometry inequalities AMC USA(J)MO USAMO USAJMO Putnam

Yes pohoatza,I too noticed it just now.
That makes problem IMO 2009/2 actually useless infact. There is nothing invovled int he problem. I shall be surprised if some contestant doesnt solve this problem.

[quote="Valentin Vornicu"]Watch this. At most 50% will get 7 points. And this is a large margin that I'm giving :)[/quote]

Hmm... I remember that on IMO 2002, the entire Korean team got 6's for #2 (Geometry) because they forgot to explain how the diagram was formulated a certain way... [url]http://imo-official.org/country_individual_r.aspx?code=KOR[/url]

Is this a simillar case?

[quote="Agr_94_Math"]Why do you say so Mr. Valentin?
DOn't you think the problem is tooeasy to lose marks?
Where can someone possibly go worng??[/quote]

[quote="Valentin Vornicu"]Watch this. At most 50% will get 7 points. And this is a large margin that I'm giving :)[/quote]

I would also say that many won't have 7. First, for such a problem there are lot of misleading approaches, and even if on a right track, the contestants may not do the problem, and of course there are mistakes. No, not 50% of 7 points...

[quote="Valentin Vornicu"][quote="kevinatcausa"]A better question might be: What percent of competitors do you predict to get at least $ 5$ points?[/quote]A maximum of 66.67% :)[/quote]

Problem number 2 from IMO 2002 was maybe even easier that this one, and than about 250 students (which was about 50%) got 5 or more points. Problem number 1 on that olympiad was maybe a little hard than problem number 1 from this year, so I think that this percentage is going to be around 50.

You take longer drawing it than solving it.
By angles, is obviously prove that AQP similar to KLM, then AQ/AP=ML/MK=2ML/2MK=PC/QB then P(Q)=AQxQB=APxPC=P(P)
I think 50%, is very vulnerable this problem.
I cant understand why the imo jury discriminates the geometry in the last 4 years

[quote="Agr_94_Math"]The IMO is having an easy geometry problem for the 3rd consecutive year.  Infact, in 2007, I think on Day 2, there was aproblem relating to an acute nagled triangle $ ABC$. Even there, you can do the problem without any construction.
Last year, there was Problem 1.People who knew radical axes could have solved it in maximum 5 minutes. I am sure all IMO contestants know Radical Axes.
And this year this one. Probably the idea might not have struck the people in the stress over there very soon, but whether they tried this for an hour or not, they would have defintely solved the problem.
Anyway, I want an inequality on Day 2 along with a nice number theory.[/quote]

I guess not all of them knows radical axis... You're just putting your own standard on them...

[quote="Concyclicboy"]I cant understand why the imo jury discriminates the geometry in the last 4 years[/quote]We only have 6 problems at the IMO. The jury is not formed with the best countries, but with [b]all[/b] countries. Some leaders want easier problems so that they students don't feel discouraged and hope they will go back home [b]at least[/b] with a Honorable Mention. So the appearance of 3-4 easy problems out of 6 is no coincidence, or fault, for that matter.

Either we have a secondary IMO with like top 15 countries qualifying for it, with two days and 4 questions each day, or simply add another day of competition to the IMO. This would surely help in boosting the competition standards.

[quote="Valentin Vornicu"][quote="Agr_94_Math"]That makes the problem actually useless infact. There is nothing invovled int he problem. I shall be surprised if some contestant doesnt solve this problem.[/quote]I will bet you that more than 50% of the contestants won't get 7 points on this problem :)[/quote]

Yes, let's wait for the marking scheme  :wink:

Anyway, it is quite boring that, for each problem in each IMO since Mathlinks exists, there are always guys who are complaining that the problem is too easy for an IMO.  :wink:

Pierre.

[quote="huang"]Did they really consider this hard enough for a problem 6??[/quote]

As the coordinations take place, it turns to be the most difficult problem of the IMO for the last 50 years in terms of full solutions (maybe only 2, surely less than 5), and in terms of nonzero scores (probably less than 10 contestants will receive a nonzero score).
So, please stop with all these complaining about the level of difficulties of the IMO problems (each problem and each year), it becomes ridiculous.

Pierre.

[quote="kdano"]

Can you provide us some partial results? :P[/quote]

I have no complete results but, according to one of the coordinators, it seems that there would have 4 complete solutions and near 20 non-zero contestants. This is not over, but the coordinations are sufficientely advanced to think that considering this problem as easy, even for a #6, is stupid.
That a simple solution exists does not means that the problem is easy... :wink:  And I think it is quite insulting for all the contestants...

Pierre.

[quote="Erken"]
P.S another option is to make somewhat a qualifying round for the IMO. Like qualifying rounds of FIFA World Cup, so only strongest countries will get to the last round of the IMO. But this would be too expensive, I guess...[/quote]The problem here is the same, this will harm strong students from weak countries. I think 3 days IMO is perfect.

[quote="Bugi"]Maybe if the students themselves choose the category they want to compete... But that would mean throwing out country rankings. I think it's not a good idea...[/quote]

That is a good idea. You just could add some weighting such there is trade-off by getting more points on the easier IMO test but a lower weighting factor as it enters the inofficial cumulative team score.

They also could provide sample tests to see what each IMO level aims at in terms of difficulty level.

Next thing you have to think on how to distribute medals with multiple IMO difficuly levels!

[quote="mszew"]My opinion in order to be fare there are two main factors:
The time limit of the competition vs. the whole Imo experience.
In this case either solution of 2 tests of 5/6 hs or 3 test of 4,5hs is possible, however base on my experience I will prefer 2 tests of 6hs and 4/5 problems each.

The other factor is the clasification of the students based on the result, in this case there are several approaches, identifying the top student (no ties with 42 points, the special prize is an option) / having clear medal cuts (For example: max(the lowest gold - the best silver), nowadays is only 1/42 points.) or finally at the extreme having no ties in the score ?!?!?!?

To sum up probably the best solution is as it is or
at most to add one more problem each day or
another possibility somehow difficult to implement is the special prize concept or bonus points for interesting generalizations of the given problems...there are several examples on the respective subforums of different generalization of 2009 problems.[/quote]

I think if you have like 2 days with 4 problems a day it would require an additional day of coordination and marking the exam papers. But three days of competition would already extend the IMO by two days, and increase the required budget for the IMO. For example some country want to host the IMO but cannot due to insufficient funds. This also excludes the idea of Erken to have qualifying exams to select those countries which may be able to attend IMO. But anyway for the sake of experience I do not like idea either, I think all the countries should have the unique chance to meet so many math people every year.

Can you please do elaborate on your second part of the post which I do not quite understand...

Weighing would have to be carefully done... But someone would always complain why it is 1.62 and not 1.65 when that would bring their team to higher position... Although it's doable.

And medals? Same for all categories, because students who know less shouldn't be discriminated by not winning golds at any rate. But, then some advanced students could go to lower category and smash for a sure gold. Maybe then, some country students would be prohibited from going into lower category (China, Russia, USA, ... ).

And back on the time&problems:

4 problems, for an IMO is too much, except if the 4th added would be difficult as 1/4 or easy 2/5. And 3 days - why not? It's implemented at some competitions. Maybe some days of pause in between (IJSO is problem day-pause-problem day-pause-problem day).

Actually I would prefer having 2 days with 4 problems each, rather than 3 days. Anyway don't worry I doubt anything is going to change anytime soon :D

In IMC they give 6 problems per day :D

Btw, I know that my country tries to avoid combinatorical problems because we have no experience in them and I think that out team leader votes for combinatorical problem to be the 3rd or 6th problem (we have never get more then 1 point in p3 and p6).

Pretty much nobody solved IMO 6 completely, so why are we complaining about the test not being hard enough?

I thought the discussion now was about changing the format of the competition. Some complained about A/G being dominant to NT/C.

If you set 4 problems a day like how much time to attempt the problems should students have? The "Romanian Master in Mathematics" has five hours for four problems, I cannot quite recall but the 5th (final) round in Russia has also five hours. Though there are two competition days with eight problems in total.

I guess the question from above also depends on how hard you make those questions in a four problem test. The USA TST has quite a few hard problems but this may a bit not the goal for the more diverse IMO participant group.

The US undergraduate competition Putnam for example sets twelve problems a day where six of them are posed in an afternoon session from 10am to 1pm and another six problems in the afternoon session from 3pm to 6pm.

[url=http://www.imo-official.org/year_statistics.aspx?year=2009]Here[/url] you can find the score statistics of the IMO 2009. Posing eight or nine problems you could cover a wider range of topics. For eight problems I suggest the following aim for the target mean score [i](5.5, 5, 4, 3, 2, 1.5 ,0.8 ,0.3)[/i].

For the last 5 years, mean on problems was like rounded on 2 decimals:

1. 4.28
2. 2.74
3. 0.74
4. 4.35
5. 1.96
6. 0.42

4 is easier than 1!!! 2007 and 2005 influenced much on this...

On a 4 problem 2 days, I would go with 1,2,5,3 and 4,2,5,6 setup (problems from IMO now), so maybe means:
4.5, 2.5, 2.2, 0.7
4, 3, 1.8, 0.4

Somehow that 5,5 and 5 would make very easy problems  :wink:

I already suggested to give IMO papers at different difficulty level because the countries are quite diverse with regard to their average problem-solving capabilities. I do not think one problem of targeted mean score 5, 5.5 or even 6 is too easy. Check the [url=http://www.imo-official.org/year_individual_r.aspx?year=2009]IMO 2009 individual results[/url] and you will see that of those 565 participants 137 students (24.2%, group 1) scored 5 points or less, and 187 (33.1%, group 2) could not not get full score, i.e. 7 points, on any of the six problems.

If you want to increase the number of problems (via more days and/or more problems per day) I guess it is even a less positive experience for those participants who fall into that score range of group 1 or 2. I think less than 5-10% should fall into group 2. And 1/3 as in IMO 2009 is far too many people in my view. If you look at the top participants they will eat those very easy problems for breakfast, no matter whether the mean score is 5 or 6.5.

Think about which targets you may have using more than eight problems? Like a more equal distribution fo the 4 broad topics (algebra, combinatorics, geometry, number theory) covered? You may not only want to differentiate participants in the very top range but also in the medium and lower range? You may want to make sure that the big majority can solve at least one problem fully?

Thinking a little bit more, I agree. IMO with 4x2 could be very nice for all contestants, 1 and 2 easier, 3 and 4 harder, everybody satisfied. And there are lots of competitions which are 4 problems or more:

BMO
APMO - 5
JBMO
RMM

These all except APMO are 4 problems (although it's one day).

And, 5 hours would be enough I think.

To have some comparisons: the [url=http://www.imc-math.org.uk/]International Mathematics Competition[/url] poses five problems in each of the two five hour sessions. And yes the [url=http://www.rmm.lbi.ro/]Romanian Masters of Mathematics (RMM)[/url] sets one test with four problems that needs to completed in five hours and is roughly at the difficulty level of the IMO problems 2/5 and 3/6 respectively.


Let $ C$ be a point on a fixed circle whose diameter is $ AB \equal{} 2R$ ($ C$ is different from $ A$ and $ B$). The incircle of $ ABC$ touches $ AB$ and $ AC$ at $ M$ and $ N$, respectively Find the maximum value of the length of $ MN$ when $ C$ moves on the given fixed circle.

This is [color=red]T8/275[/color].

Sholud be locked!


[quote="worthawholebean"]([b]3[/b]) A $ 3\times3\times3$ cube composed of $ 27$ unit cubes rests on a horizontal plane. Determine the number of ways of selecting two distinct unit cubes from a $ 3\times3\times1$ block (the order is irrelevant) with the property that the line joining the centers of the two cubes makes a $ 45^\circ$ angle with the horizontal plane.[/quote]

[hide]
Easy bruteforce gives 10 per block
there are 6 ways to choose a 3x3x1 block for which this works
so 60?

[/hide]


trigonometry
calculus
calculus computations

[quote="Leonard"]there are explicit formulas for coefficients in power expansion of $ (\arcsin x)^n$ for $ n \equal{} 1,2,3,4$

for $ n\ge 5$ this formulas are much more complicated

in general, I would like to know why ?
what causes that for $ n\ge 5$ this formulas become much more complicated[/quote]

well, it really depends on how [u]you define[/u] complicated.  :|

from trigonometry, you can show

$ \sin\,(n\theta)\;\;\equal{}\;\;n\sin\,\theta\minus{}\frac{n(n^2\minus{}1)}{3!}\sin^3\,\theta\plus{}\frac{n(n^2\minus{}1)(n^2\minus{}9)}{5!}\sin^5\,\theta\minus{}\frac{n(n^2\minus{}1)(n^2\minus{}9)(n^2\minus{}25)}{7!}\sin^7\,\theta\plus{}\;\;\minus{}\;\;\ldots$

now comparing this (co-efficients of $ n$, $ n^3$, $ n^5$, etc...) with
$ \sin\,(n\theta)\;\;\equal{}\;\;n\theta\minus{}\frac{n^3\theta^3}{3!}\plus{}\frac{n^5\theta^5}{5!}\minus{}\frac{n^7\theta^7}{7!}\plus{}\;\;\minus{}\;\;\ldots$

you will get the series expansion of $ (\sin^{\minus{}1}\,\theta)^n$ for [b]odd [/b]$ n$, i.e.

$ 1.\;\;\;\quad\sin^{\minus{}1}\,\theta\;\;\equal{}\;\;\theta\plus{}\frac{1}{2}\;\frac{\theta^3}{3}\plus{}\frac{1\cdot 3}{2\cdot 4}\;\frac{\theta^5}{5}\plus{}\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\;\frac{\theta^7}{7}\plus{}\plus{}\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}\;\frac{\theta^9}{9}\plus{}\;\;\ldots$

$ 2.\;\;\;\quad\frac{1}{3!}\left(\sin^{\minus{}1}\,\theta\right)^3\;\;\equal{}\;\;(1)\frac{1}{2}\;\frac{\theta^3}{3}\plus{}\left(1\plus{}\frac{1}{9}\right)\frac{1\cdot 3}{2\cdot 4}\;\frac{\theta^5}{5}\plus{}\left(1\plus{}\frac{1}{9}\plus{}\frac{1}{25}\right)\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\;\frac{\theta^7}{7}\plus{}\left(1\plus{}\frac{1}{9}\plus{}\frac{1}{25}\plus{}\frac{1}{49}\right)\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}\;\frac{\theta^9}{9}\plus{}\;\;\ldots$

and to get the series expansion of $ (\sin^{\minus{}1}\,\theta)^n$ for [b]even [/b]$ n$, you need to compare
co-efficients of $ n^2$, $ n^4$, $ n^6$, etc...) between

$ \cos\,(n\theta)\;\;\equal{}\;\;1\minus{}\frac{n^2}{2!}\sin^2\,\theta\plus{}\frac{n^2(n^2\minus{}4)}{4!}\sin^4\,\theta\minus{}\frac{n^2(n^2\minus{}4)(n^2\minus{}16)}{6!}\sin^6\,\theta\plus{}\frac{n^2(n^2\minus{}4)(n^2\minus{}16)(n^2\minus{}36)}{8!}\sin^8\,\theta\minus{}\;\;\plus{}\;\;\ldots$

and

$ \cos\,(n\theta)\;\;\equal{}\;\;1\minus{}\frac{n^2}{2!}\theta^2\plus{}\frac{n^4}{4!}\theta^4\minus{}\frac{n^6}{6!}\theta^6\plus{}\frac{n^8}{8!}\theta^8\minus{}\;\;\plus{}\;\;\ldots$

and you will get

$ 3.\;\;\;\quad\frac{1}{2!}\left(\sin^{\minus{}1}\,\theta\right)^2\;\;\equal{}\;\;\frac{\theta^2}{2}\plus{}\frac{2}{3}\;\frac{\theta^4}{4}\plus{}\frac{2\cdot 4}{3\cdot 5}\;\frac{\theta^6}{6}\plus{}\frac{2\cdot 4\cdot 6}{3\cdot 5\cdot 7}\;\frac{\theta^8}{8}\plus{}\frac{2\cdot 4\cdot 6\cdot 8}{3\cdot 5\cdot 7\cdot 9}\;\frac{\theta^{10}}{10}\plus{}\;\;\ldots$

$ 4.\;\;\;\quad\frac{1}{4!}\left(\sin^{\minus{}1}\,\theta\right)^4\;\;\equal{}\;\;\left(\frac{1}{4}\right)\frac{2}{3}\;\frac{\theta^4}{4}\plus{}\left(\frac{1}{4}\plus{}\frac{1}{16}\right)\frac{2\cdot 4}{3\cdot 5}\;\frac{\theta^6}{6}\plus{}\left(\frac{1}{4}\plus{}\frac{1}{16}\plus{}\frac{1}{36}\right)\frac{2\cdot 4\cdot 6}{3\cdot 5\cdot 7}\;\frac{\theta^8}{8}\plus{}\left(\frac{1}{4}\plus{}\frac{1}{16}\plus{}\frac{1}{36}\plus{}\frac{1}{64}\right)\frac{2\cdot 4\cdot 6\cdot 8}{3\cdot 5\cdot 7\cdot 9}\;\frac{\theta^{10}}{10}\plus{}\;\;\ldots$

i will leave it to you to get the series for $ n\equal{}5$ and others (see if you spot a pattern) and then judge whether you think it is complicated or not... :maybe:


calculus
integration
vector
function
real analysis
real analysis theorems

Could somehow help explain to me why equations (2) and (3) on the following webpage are true (i.e. how to prove them):

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

I am having some trouble easily grasping some of the basics of surface integrals.

Thanks for any help you can provide.

Those equations aren't theorems, rather they constitute a definition. What is being defined is the  integral

$\int_S F\cdot da$

The only things to be proved or demonstrated are that

(1) The various forms given there are equivalent, and

(2) That the concept makes some kind of heuristic physical sense, as the net flux of the vector field across the surface.

The big theorems associated with this concept are Stokes's Theorem and Gauss's Theorem (also known as the Divergence Theorem.)

Right. So, if the proofs/justifications are fairly short, would it be possible for you (or someone else) to post a quick explanation of why the three forms listed are equivalent?

Or perhaps someone knows of a place online that provides a proper justification (my attempts with Google were not very successful)...


AMC
AMC 12
probability
AIME
geometry
ratio
AMC 12 A

Hi..I'm wondering if this is a problem.
Last year, as a 7th grader, I got 104.5 on the AMC 12.
This year, on the AMC 12 A, I got 82.5
I felt this year's was much harder, not only because of the omit changed form 2.5 to 1.5 but also the question content..Is this something I should be worrying about? I'm doing one past AMC12 every two days, Should I be studying more and scanning these forums more? I'm good at solving problems, but I'm up to pre calc and lots of the concepts I dont know (especially in the 20's and late 10's)
Thanks guys. Also I have trouble with unrelated probability, how is that calculated? Such as the question on AMC 2006A/B? test about the dice and if they remove one dot randomly from a side whats the chance of it being odd on the top if they roll it?
Thanks a LOT!

It sounds like you would have done better on the AMC 10- you haven't gotten to everything that's on the AMC 12 yet.
In any case, 82.5 is still a fine score. Last year's contest was one of the easiest ever to qualify for the AIME on, and this year's contest was significantly harder. If you account for the difficulty and the scoring change, there's probably only a question worth of difference, and anyone can have a bad day.

You still have four years to make your mark. You will do better in years to come, because you'll know more background and techniques.

Why are you taking the AMC12?

Go take the AMC10B.

Was this year's a ton harder? My score improved over 20 points...

Anyway, really pay attention to basic geometry. The algebra stuff is pretty easy up to the teens.

Geometry intuition - perpendicular lines, power of a point, area ratios, point to line, similar triangles... etc are extremely helpful and basic stuff can generally always solve the AMC geometry problems.

The difficulty depends on what range of scores you're looking at. From the little information I have, this year's AMC 12 seemed easier than usual to get very high (130-150) scores on, but significantly harder to get 100 on. The middle pages were tough, and the back page wasn't evil.

I heard it was the same for the 10. Easier to get a perfect or really high score in the 140s, but harder to qualify. I don't really follow why people have stated this though.

Thanks, guys, also how can I prepare for the AIME if i do make it?
Last year, I got 20 on the AIME (2 questions right) by brute-forcing numbers 1 and 2..Is there a way to "study" for the AIME? other than taking mock exams?
Thanks  :lol:

Well, first of all, you got a 2 on the AIME, not 20. You don't multiply by ten.
To prepare: Besides taking mock and past exams? Art of Problem Solving Vol. II helps, I suppose.

[quote="white_horse_king88"]Was this year's a ton harder? My score improved over 20 points...
[/quote]

Lol my score improved like that too.

28 points.

My 138 on the AMC 10 last year went down to a 112.5 or 118.5 on the AMC 12 this year. Not too bad, is it?

[quote="bpms"][quote="mathgeniuse^ln(x)"][quote="bpms"]My score went down 36 points  :huh:[/quote]

Hah, you're funny.

Not.[/quote]
I wasn't meaning to be funny, apparantly I can't subtract...[/quote]
ooh did u make addition/subtraction errors too?
one of those cost me a perfect score on the AMC 8
not sure if i messed up on them on the AMC 10 though


Hmm...I am hoping that I would be able to attend Mathcamp next year...but I doubt I'd make it though.  Since on the qualifying quiz this year, I think I could only do like 3 problems or something around that.  Anyhow, has anyone here went there, was it fun, and how well did ya do on the qualifying quiz?  Anything I should be particularly good at to complete the quiz?  (Maybe I'll do a challenge thread and have all the Mathcamp questions...hmmm :-P)

Go [url=http://www.artofproblemsolving.com/Forum/search.php]here[/url].  In the first box, type the word "mathcamp."

In the pulldown box that says "Forum," select "Other US Contests & Programs."  Press the button labeled "Search."

This is crazy!  We mods need to put instructions on how to search in our respective forums.  That way new members will know because it isn't their fault that they don't know about how to search the forums.

haha...i'll use that now.  Sorry about that, and thanks.  (I would delete this thread I knew how....)


algorithm
number theory
Euclidean algorithm
superior algebra
superior algebra unsolved

The smallest ideal which contains
$ 2009^{2009}\minus{}1>$ and $ <2009^{4510}\minus{}1>$
is ...

Assuming that you are working aver $ \mathbb Z$ what does a generator of the smallest ideal correspond to? Can you produce a generator using the Euclidean algorithm in $ \mathbb Z$?


[quote="arqady"]If $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} 1$, where $ x_i$ are non-negative real numbers
and $ 2\leq k < n$, $ k\in\mathbb \mathbb{N}$, then prove or disprove that
$ x_1x_2...x_k \plus{} x_2x_3...x_{k \plus{} 1} \plus{} ... \plus{} x_nx_1...x_{k \minus{} 1}\leq \max\left\{\frac {1}{k^k},\frac {1}{n^{k \minus{} 1}}\right\}$.[/quote]
Maybe anybody knows, what known about it?
I know that the inequality is true for cases:
$ (2,n),n\geq3, (3,n),n\geq4,(4,n),n\geq8,n \equal{} 6,n \equal{} 5,(n \minus{} 1,n),n\geq5$
and $ (k,km),(k,km \plus{} 1),$ where $ k\geq5,m\geq2.$
Thanks.

[quote="arqady"]If $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} 1$, where $ x_i$ are non-negative real numbers
and $ 2\leq k < n$, $ k\in\mathbb \mathbb{N}$, then prove or disprove that
$ x_1x_2...x_k \plus{} x_2x_3...x_{k \plus{} 1} \plus{} ... \plus{} x_nx_1...x_{k \minus{} 1}\leq \max\left\{\frac {1}{k^k},\frac {1}{n^{k \minus{} 1}}\right\}$.[/quote]

I think you know this, but for $ k \equal{} 2$ and $ n\geq 4$, your inequality becomes

$ x_1x_2 \plus{} x_2x_3 \plus{} ... \plus{} x_nx_1\leq\max\left\{\frac {1}{2^2},\frac {1}{n^1}\right\}$,

what rewrites as

$ x_1x_2 \plus{} x_2x_3 \plus{} ... \plus{} x_nx_1\leq\frac14$

(because $ \max\left\{\frac {1}{2^2},\frac {1}{n^1}\right\} \equal{} \frac {1}{2^2} \equal{} \frac14$), what was proven in the homogeneous form $ \left(x_1 \plus{} x_2 \plus{} ... \plus{} x_n\right)^2\geq 4\left(x_1x_2 \plus{} x_2x_3 \plus{} ... \plus{} x_{n \minus{} 1}x_n \plus{} x_nx_1\right)$ in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=190966#190966]http://www.mathlinks.ro/Forum/viewtopic.php?t=30787 post #4[/url].

In [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=251155#p251155]the same topic, post #9[/url], I also showed that

$ x_1x_2...x_k \plus{} x_2x_3...x_{k \plus{} 1} \plus{} ... \plus{} x_{n \minus{} k \plus{} 1}x_{n \minus{} k \plus{} 2}...x_n\le \frac {\left(x_1 \plus{} x_2 \plus{} ... \plus{} x_n\right)^k}{k^k}$

for all $ k$ (not just for $ k \equal{} 2$), so that with the condition $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} 1$ we obtain

$ x_1x_2...x_k \plus{} x_2x_3...x_{k \plus{} 1} \plus{} ... \plus{} x_{n \minus{} k \plus{} 1}x_{n \minus{} k \plus{} 2}...x_n\leq\frac {1}{k^k}$.

But this does [i]not[/i] prove the Janous conjecture

$ x_1x_2...x_k \plus{} x_2x_3...x_{k \plus{} 1} \plus{} ... \plus{} x_nx_1...x_{k \minus{} 1}\leq\max\left\{\frac {1}{k^k},\frac {1}{n^{k \minus{} 1}}\right\}$,

because there are additional terms on the left hand side of the Janous conjecture.

Neither is

$ x_1x_2...x_k \plus{} x_2x_3...x_{k \plus{} 1} \plus{} ... \plus{} x_nx_1...x_{k \minus{} 1}\leq\frac {1}{n^{k \minus{} 1}}$

always true (n = 5, k = 2 being a random counterexample), nor is

$ x_1x_2...x_k \plus{} x_2x_3...x_{k \plus{} 1} \plus{} ... \plus{} x_nx_1...x_{k \minus{} 1}\leq\frac {1}{k^k}$

always true (here, a counterexample is n = 5, k = 3).

[Thanks to Arqady for several corrections in this post.]

  Darij

Particular case of

$ x_1x_2...x_k \plus{} x_2x_3...x_{k \plus{} 1} \plus{} ... \plus{} x_nx_1...x_{k \minus{} 1}\leq\max\left\{\frac {1}{k^k},\frac {1}{n^{k \minus{} 1}}\right\}$

for $ n \equal{} 5,k \equal{} 3$:

If $ \sum_{i \equal{} 1}^5 x_i \equal{} 1$ and $ x_i$ are non-negative numbers, then
$ x_1x_2x_3 \plus{} x_2x_3x_4 \plus{} x_3x_4x_5 \plus{} x_4x_5x_1 \plus{} x_5x_1x_2\leq\frac {1}{25}.$ ;)

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=112980 for this inequality.

Particular case for $ n\equal{}6,k\equal{}3$: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=192763 .


geometry
geometric transformation
rotation
congruent triangles

1. ABCDEF is a convex hexagon with all its sides equal. Also A + C + E = B + D + F. Show that A = D, B = E and C = F.
(54th Kürschák Competition 1953/3)

2. The hexagon ABCDEF is convex and opposite sides are parallel. Show that the triangles ACE and BDF have equal area.
(58th Kürschák Competition 1958/3)

 I would like some help with these problems please :lol:

Are these angles, i'm assuming?

the capital letters.

Hint:

[hide]Think of Tesselations with the hexagons, and keep in mind of the conditions given (because points will coincide).

Then try to look for congruent triangles, collinearity, parallel lines, and then use rotation to try to get to the fact that angles are equal.[/hide]

If you need more, i will elaborate.

I have 8 different solution for second problem, unfortunately only in [b]hungarian[/b].

Solution for Problem 2.: Olympiad Section >> Geometry >>Geometry Proposed & Own Problems: [b]gemath/equal of area // Message 2.[/b] (Darij Grinberg)


function
invariant
geometry
geometric transformation
real analysis
real analysis unsolved

Find  all   continuous  function $ f(x)$ on $ \mathbb{R}$ such  that   $ f(x_1)\neq f(x_2)$ if  $ x_1 \neq x_2$  and
\[ f(2x \minus{} f(x)) \equal{} x \forall x \in \mathbb{R}\]

[hide="Hint"]
First note that an injective function that is continuous must be monotonic increasing or monotonic decreasing.
Case I : The function $ f$ is monotonic decreasing. We will prove that no function exist. If any function exist, take any $ x<y$ then $ f(x)>f(y)$ implies $ \minus{}f(x)<\minus{}f(y)$ then $ 2x\minus{}f(x)<2y\minus{}f(y)$, $ x\equal{}f(2x\minus{}f(x))>f(2y\minus{}f(y))\equal{}y$, a contradiction
Case II : Please help me to continue :(
[/hide]

I have a claim. Consider the case $ f(x)\equal{}x$ . This is injective and an strictly increasing function.
 :maybe: Won't this work out...

[quote="DineshManohar"]I have a claim. Consider the case $ f(x) \equal{} x$ . This is injective and an strictly increasing function.
 :maybe: Won't this work out...[/quote]

obviously it works. but the question is, which other functions work, if any? and if no others work, why not?

Let me give some serious hint for this problem.
let $ g(x) \equal{} x \minus{} f(x)$. Then the problem reduces to $ g(g(x) \plus{} x) \equal{} g(x)$. Now one can see that $ g(x \plus{} ng(x)) \equal{} g(x)$(1) hold for every natural n.
If g isn't monotone then we can pick $ x < y < z$ such that $ max\{g(x),g(z)\} < g(y)$ but then $ max\{g(x \plus{} ng(x)),g(z \plus{} ng(z))\} < g(y \plus{} ng(y))$ for "big" $ n$. Something impossible since  $ g(x) \plus{} nx$ is increasing.
Now it easy to take cases and see that $ g$ is constant because of its monotone behaviour and (1).  :wink:

I always get this functional equations wrong, but still, here it goes. We can define $ f(x) \equal{} g(x) \plus{} x$. First note that since $ f$ is one-to-one, so is $ g$. Now, we have:

$ f(2x\minus{}f(x)) \equal{} x$

$ f(2x\minus{}g(x)\minus{}x) \equal{} x$

$ f(x\minus{}g(x)) \equal{} x$

$ g(x\minus{}g(x))\plus{}x\minus{}g(x) \equal{} x$

$ g(x\minus{}g(x))\equal{}g(x)$

So, since $ g$ is one-to-one:

$ x\minus{}g(x) \equal{}x$

$ g(x) \equal{} 0$

so

$ f(x) \equal{} x \plus{} g(x) \equal{} x$.

Would this do?

[quote="milin"]First note that since $ f$ is one-to-one, so is $ g$.[/quote]

this isn't right, e.g. for $ g(x)\equiv 0$. ironically enough, after saying $ g$ was one-to-one, you concluded that $ g(x)\equiv 0$.

Yes, well.. I warned you I will probably do something stupid..  :blush:

By applying: $ x : \equal{} 2x \minus{} f(x)$ accordingly many times, we have:
$ f^n(x) \equal{} x \minus{} n (x\minus{}f(x))$  for all whole $ n$  ($ n \in Z$)
so, must be $ f(x) \equal{} x \plus{} c$ for $ x \in  A \subset R$ where set $ A$ is invariant under translation by c in $ R$

thus we define function $ f$ in this way for  given distribution $ R$ on disjoint and translation-invariant subsets.

for example we can define: $ f(x) : \equal{} x\plus{}1$ for $ x \in Q$ and $ f(x) : \equal{} x\plus{}2$ for other $ x \in R$

sorry for not perfect latex.

[quote="kelsar"]By applying: $ x : \equal{} 2x \minus{} f(x)$ accordingly many times, we have:
$ f^n(x) \equal{} x \minus{} n (x \minus{} f(x))$  for all whole $ n$  ($ n \in Z$)
so, must be $ f(x) \equal{} x \plus{} c$ for $ x \in A \subset R$ where set $ A$ is invariant under translation by c in $ R$

thus we define function $ f$ in this way for  given distribution $ R$ on disjoint and translation-invariant subsets.

for example we can define: $ f(x) : \equal{} x \plus{} 1$ for $ x \in Q$ and $ f(x) : \equal{} x \plus{} 2$ for other $ x \in R$

sorry for not perfect latex.[/quote]

it said that $ f$ should be continuous


Find b and c such that x= (sqrt 19)+ i(sqrt 39) is a root of x^4+bx^2+c.

hello, we get
$ f(\sqrt{19}\plus{}i\sqrt{19})\equal{}38ib\plus{}c\minus{}1444\equal{}0$ and from here $ c\equal{}1444$ and $ b\equal{}0$.
Sonnhard.

Dr Sonnhard Graubner, $ x\equal{}\sqrt{19}\plus{}\sqrt{39}i$.

I got $ x^4\plus{}116x^2\plus{}3440\equal{}0$.

hello kunny, yes this was a typo of mine, but i have got
$ f(x)\equal{}x^4\plus{}40x^2\plus{}3364$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello kunny, yes this was a typo of mine, but i have got
$ f(x) \equal{} x^4 \plus{} 40x^2 \plus{} 3364$
Sonnhard.[/quote]

You are right.

$ x\equal{}\sqrt{19}\plus{}\sqrt{39}i\Longrightarrow (x\minus{}\sqrt{19})^2\equal{}(\sqrt{39}i)^2$

$ \Longleftrightarrow x^2\plus{}58\equal{}2\sqrt{19}x$, Squaring both sides gives $ x^4\plus{}40x^2\plus{}3364\equal{}0$.


Suppse that $X=\{1,2, \ldots, 2^{1996}-1\}$, prove that there exist a subset $A$ that satisfies these conditions:

a) $1\in A$ and $2^{1996}-1\in A$;

b) Every element of $A$ except $1$ is equal to the sum of two (possibly equal) elements from $A$;

c) The maximum number of elements of $A$ is $2012$.

[i]Romania[/i]

[quote="Omid Hatami"]This is my solution. I hope it's clear.
We will write at most $2012$ numbers, starting with $1$ and ending with $2^{P}{1996}$, such that every number written is the sum of two numbers previously written. If $2^{n}-1$ is already written then we can  obtain $2^{n}(2^{n}-1)$ by doubling $n$ times. Also $2^{2n}-1$ can be obtained in one more step. So we can obtain $2^{2}-1,2^{4}-1,\dots,2^{256}-1$ in $263=(1+2)+(1+4)+\dots+(1+128)$ steps. Doubling $243$ times we obtain $2^{499}-2^{243}$. Then in $6$ we obtain $2^{4}99-1$, then $2^{998}-1$ in $500$ steps and finally $2^{1996}-1$ in $499$ steps. So we count $263+243+6+500+999+1=2012$ numbers written. (Including the initial $1$.)[/quote]
Thank you very much ,Omid Hatami.

[quote=Omid Hatami]This is my solution. I hope it's clear.
We will write at most $2012$ numbers, starting with $1$ and ending with $2^{P}{1996}$, such that every number written is the sum of two numbers previously written. If $2^{n}-1$ is already written then we can  obtain $2^{n}(2^{n}-1)$ by doubling $n$ times. Also $2^{2n}-1$ can be obtained in one more step. So we can obtain $2^{2}-1,2^{4}-1,\dots,2^{256}-1$ in $263=(1+2)+(1+4)+\dots+(1+128)$ steps. Doubling $243$ times we obtain $2^{499}-2^{243}$. Then in $6$ we obtain $2^{4}99-1$, then $2^{998}-1$ in $500$ steps and finally $2^{1996}-1$ in $499$ steps. So we count $263+243+6+500+999+1=2012$ numbers written. (Including the initial $1$.)[/quote]

Excuses me. How can you obtain $2^{499}-1$ in $6$


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^2\plus{}b^2\plus{}c^2\equal{}3.$ Prove that:
\[ 8(2\minus{}a)(2\minus{}b)(2\minus{}c)\geq(a\plus{}bc)(b\plus{}ac)(c\plus{}ab)\]
I have a proof, but my solution is very ugly.
Maybe there is a nice proof.

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3.$ Prove that:
\[ 8(2 \minus{} a)(2 \minus{} b)(2 \minus{} c)\geq(a \plus{} bc)(b \plus{} ac)(c \plus{} ab)
\]
I have a proof, but my solution is very ugly.
Maybe there is a nice proof.[/quote]
Yes, of course. :)

You can find a beautiful proof for it in the follow attachment. It's also MIC problem for the 1st round. :)

Indeed, the follow is the MIC 1st round problems (Vietnamese version). :)


algebra
polynomial
calculus
calculus computations

Given a polynomial $ f(x) \equal{} x^3 \minus{} 6x \plus{} p$, where $ p$ is an arbitrary constant,
(b) find the value / all values of $ p$ where $ f(x)$ has three distinct roots.

This was off of an old AP AB Calc test we took today, and I'm recalling this from memory (I don't remember if p was one value or more than one), but does anyone have a tip on how to solve this?

(please move this post if not in the right forum)

then look at $ h(x) \equal{} x^3 \minus{} 6x$ has local min/max at $ x \equal{} \pm \sqrt 2$
which are $ \mp 4\sqrt 2$
so if you shift up or down more than that, you will have only one real root, meaning
you want $ |p| < 4\sqrt 2$ to ensure three distinct real roots..
or $ |p|\ne 4\sqrt 2$ if you want to include three distinct complex roots


[quote="MR.129"]Given x,y,z are real numbers such that $ x \ge 0,y \ge 0,z \ge 0$ and
$ x \plus{} y \plus{} z \le 3$.Prove that:
$ \frac {x}{{1 \plus{} x^2 }} \plus{} \frac {y}{{1 \plus{} y^2 }} \plus{} \frac {z}{{1 \plus{} z^2 }} \le \frac {1}{{1 \plus{} x}} \plus{} \frac {1}{{1 \plus{} y}} \plus{} \frac {1}{{1 \plus{} z}}$[/quote]
we have $ \frac {1}{1 \plus{} x} \plus{} \frac {1}{1 \plus{} y} \plus{} \frac {1}{1 \plus{} z} \minus{} \frac {x}{1 \plus{} x^2} \minus{} \frac {y}{1 \plus{} y^2} \minus{} \frac {z}{1 \plus{} z^2}$
             $ \equal{} \frac {1 \minus{} x}{(1 \plus{} x)(1 \plus{} x^2)} \plus{} \frac {1 \minus{} y}{(1 \plus{} y)(1 \plus{} y^2)} \plus{} \frac {1 \minus{} z}{(1 \plus{} z)(1 \plus{} z^2)}$
Suppose $ x\geq\ y \geq\ z \geq\ 0$,this inequaliti is true By chebyshev's inequaliti   :)


Hey,

this is my first time on mathlinks.
I have a nice problem for you:

you have a 7*7 square.  you have to cut it to 5 pieces
and then rearrange them to make a 2*2 square, a 3*3 square and a 6*6 square.

good luck

 :lol:

"you have a 7*7 square. you have to cut it to 5 pieces
and then rearrange them to make a 2*2 square, a 3*3 square and a 6*6 square. "

The challenge is to take a 7 by 7 square and divide the square up into 5 pieces using 4 lines.  Then, take those 5 pieces and put them together to form a 2*2, 3*3 and 6*6 square (all separate squares I think).  If you notice, 2^2 + 3^2 + 6^2 = 7^2 so all pieces should be used.

That's what I interpret the question to mean.  I could be wrong though.

you have a 7*7 square and the problem is to cut it with scissors
to 5 pieces (using broken lines, not just lines) and then rearrange them in a position so that you get a 2*2 , 3*3 and 6*6 square at the same time.

Could someone upload and hide a diagram?

[quote=Treething wrote]I just chugged a 16 ounce bottle of Dr. Pepper, so I'm feeling slightly stoned and I kinda feel like throwing up, but I believe I can do that in four cuts. Three if you can fold the paper.
[/quote]

whoa there!!!!

never drink Dr.Pepper!!!!
the worst drink ever!!!!!!!

i've seen this problem before, somewhere in my quiz book or something


For some reason there is no solution for this one . . .  :maybe:

A toroidal chess board is like an ordinary chess board except that pieces can move off the edges and appear on the other side.  A knight can attack by moving 2 squares in one direction and one square in a perpendicular direction.  Four knights are placed at random on a toroidal chess board.  What is the probability that none of them can attack each other?

This is a stab:
Placing the first knight, you have a choice of 64 un-attacked squares.

Each knight attacks eight squares simultaneously, and occupy one.
The next knight you place has only $(64-9)$ safe squares, or 55 possible safe choices.

The next knight only has 46 possible safe squares, and the next only 38.
This should be equal to $\frac{64}{64}\cdot\frac{55}{64}\cdot\frac{46}{64}\cdot\frac{38}{64}\cdot$,
or $\frac{46805}{131072}$

However, this doesn't take into account the fact that more than one knight may attack the same square. I think that screws my solution up a bit.

When you've already placed 1 knight, there's only 63 places for the next..   :)

True. I'll change the first post to match that, but we're still not accounting for more than one knight attacking the same square.


probability
number theory
relatively prime

[quote="4everwise"]Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is $25.$ One marble is taken out of each box randomly. The probability that both marbles are black is $27/50,$ and the probability that both marbles are white is $m/n,$ where $m$ and $n$ are relatively prime positive integers. What is $m+n?$[/quote]

[hide="Very bad, but I think it works"]If $a$ is the amount in one box and $b$ the other, then $a+b=25$ and $50|ab$. The only possible values are $(a,b)=(20,5),(15,10),(10,15), (5,20)$. In each case they work out to be the same. So then for $(a,b)=(20,5)$ the only possible combination is 18 black marbles in the box with 20 marbles and 3 black marbles in the box with 5 marbles. So then the probability of drawing both white marbles is $\frac{2}{20}\cdot \frac{2}{5}=\frac{1}{25}$, so $m+n=\boxed{26}$.[/hide]

Guess and check actually works awesomly there are no cases or anythink..its straight forward.

Sorry for revive, but I have a solution.  It's pretty bashy, but it worked.

We will denote the marbles from the first box $b_1$ and $w_1$, and the marbles from the second box $b_2$ and $w_2$.

We know that $b_1+w_1+b_2+w_2=25$.  We also know that $\frac{b_1}{b_1+w_1} \cdot \frac{b_2}{b_2+w_2} = \frac{27}{50}$, because the probability that both marbles are black is $27/50$.

We can tell that $b_1b_2$ will be equal to either $27$ or a multiple of $27$.  We can also tell that $(b_1+w_1)(b_2+w_2)$ will be equal to either $50$ or a multiple of $50$.

We use all of the factors of $27 (1, 3, 9, 27)$ and all of the factors of $50 (1, 2, 5, 10 ,25, 50)$, but are unable to find any values that satisfy both $\frac{b_1}{b_1+w_1} \cdot \frac{b_2}{b_2+w_2} = \frac{27}{50}$ and $b_1+w_1+b_2+w_2=25$. We next look at $54$ and $100$, the next multiples of $27$ and $50$.

Looking through all the combinations of factors, we know only $(3, 18)$ and $(6, 9)$ will work for $54$, for they are the only factors that add to a sum less than $25$, and we know only $(10, 10)$ and $(5, 20)$ will work for $100$, for the same reason.

We find that $(3, 18)$ and $(5, 20)$ successfully combine to make the values $3, 18, 2, 2$ for $b_1, b_2, w_1,$ and $w_2$, respectively.  These both add to $25$ and satisfy $\frac{b_1}{b_1+w_1} \cdot \frac{b_2}{b_2+w_2} = \frac{27}{50}$.

We plug these values in to $\frac{w_1}{b_1+w_1} \cdot \frac{w_2}{b_2+w_2} = \frac{m}{n}$ to find our final answer and end up with $m=1$ and $n=25$.

Finishing it up, $m+n=1+25=\boxed{026}$, because this is the AIME.

Let there be $a$ black marbles in the first box, $x-a$ white marbles in the first box, $b$ black marbles in the second box, and $25-x-b$ white marbles in the second box.

Then $frac{a}{x} \cdot \frac{b}{25-x} = \frac{27}{50}$, so $x(25-x)$ has a factor of $50$, meaning that $x = 5, 10, 15, 20$.

If $x=5$, then $ab = 54$, so the first box has 3 black, 2 white and the second has 18 black, 2 white. This checks out, so the answer is $\frac{2}{5} \cdot \frac{2}{20} = \frac{1}{25} \implies \boxed{26}$.

Let the number of black marbles in the first box be $b_1,$ the number of black marbles in the second box be $b_2,$ the number of white marbles in the first box be $w_1,$ and the number of white marbles in the first box be $w_2.$ We know that $$\frac{w_1}{b_1+w_1} \cdot \frac{w_2}{b_2+w_2} = \frac{27}{50}$$ and $$b_1+w_1+b_2+w_2=25.$$ We immediately realize that $b_1+w_=20$ and $b_2+w_2=5$ is the only case that works. Hence, $b_1=18,b_2=3 \implies w_1=18, w_2=2.$ Thus, the probability that both marbles are white is is $\frac{2}{20} \cdot \frac{2}{5} = \frac{1}{25}.$ So the answer is $m+n=\boxed{26}.$


I need this theorem to do something, but can't seem to figure out the proof:

Let f be continuous on on [0, infinity). Let lim x->infinity f(x) = A in R. Then f is uniformly continuous on [0, infinity)

Thanks!

Given $\epsilon>0,$ find $M$ such that if $x>M,\,|f(x)-A|<\frac{\epsilon}2$

Now consider $f$ on the interval $[0,M+1].$ Since $f$ is continuous on this compact set, it is uniformly continuous on that set. Find $\delta>0$ and while we're at it, insist that $\delta<1$ such that for $x,y\in[0,M+1]$ such that $|x-y|<\delta$, then $|f(x)-f(y)|<\epsilon.$

The situation for $x,y>M$ and $|x-y|<\delta$ is the same as for any $x,y>M.$ In particular, for such $x,y,\,|f(x)-f(y)|\le|f(x)-A|+|A-f(y)|<\frac{\epsilon}2+\frac{\epsilon}2 =\epsilon.$

The reason for the $M+1$ and the insistance that $\delta<1$ was to assure that for any $|x-y|<\delta,$ at least one of $x,y\in[0,M+1]$ or $x,y\in[M,\infty)$ would be true.


AMC
AIME
MATHCOUNTS
AMC 10
number theory

I hope to qualify for the AIME this year through the AMC 10 and/or AMC 12.  In terms of the concepts I know and how well I can utilize them to get a decent score on the AMC 10, I'm probably in the bottom 10% of this site.  (Last year, as a 7th grader I scored a 93.)  I see this, but I'm determined to make an improvement.  I spend and anywhere from about 2-3 hours a day on weekdays and 4-6 on weekends preparing for these competitions.  Still, I don't think these long hours are as productive as they could be.  On the AMC 10, generally I can get 15 of the first 20 questions and if I'm lucky I'll be able to get one of the last 5.  I have AoPS Volume 1 and 2, and AoPS Introduction to Number Theory at the moment, and I'm going to be buying new books.  My question is, how should I prepare for these exams?  The problem with look at past tests is me not knowing how to solve t he difficult ones.  Is it a good practice to look at the solution and see how they solved it if I cannot solve it myself?  Or should I not look at the solutions at all and just let it be?

How should I prepare for both of these exams?

EDIT:  Also, what books should I go through?

Well......
first of all go through all the AMC 10/12 to famiriliaze yourself with the questions and [b]understand all the solutions[/b].

The books on AOPS are great.
However, if you're looking for more practice I would recommend (books):

A Primer for Mathematics Competitions- great practice for Diophantine equations/number theory/trig./geometry/counting/probability

Competition math for middle schools- more for MATHCOUNTS but could help for the AMC 10

Mathematics Circles Russian Experience- proof and numerical answers oriented- perfect for numerous concepts especially ones on logic, parity, graphs, combinametrics, number theory, and problems

First Steps for Math Olympians- great practice problems and solutions from the AMC competitions

If you aren't sure which book is best, please visit google books to see a snippet of each book.

Books:
For all of AMC 10 and the first half of AMC 12, do AOPS Volume 1.
I recommend that you don't move on to AOPS volume 2 just yet, because mastering 1 will get you to the AIME and maybe more. By mastering, I mean reading the chapters (active reading by taking notes and completing the inner steps to the solutions in the examples), doing the exercises, and doing the problems. Try to do all the problems by February, and I should be able to make the AIME. Obviously, there will be problems that you cannot solve, so after spending at least 15 minutes on them, look at the AOPS solutions book. Learn how they solved the problem, what parts of the problem motivated each part of the solution, and what you can take away from the problem. Lastly, you should redo the problem by yourself. This will make you much better.

Contests:
In the month of December (hey that's tomorrow!), you should be taking AMC 10's and doing them without time limits (make sure to not do all of them). Then, from January up until the contest, you should do timed AMC 10's.

Bottom line: finish volume 1 and do contests and you should be fine.


inequalities
invariant
linear algebra
linear algebra unsolved

Let $ A\in M_{n\times n}(R)$, $ \|A\otimes I \minus{} I\otimes A\|^2\leq 2\|A\|_F^2$,
 where $ \|\cdot\|$ denotes the spectral norm, $ \|\cdot\|_F$ is the
 Frobenius norm, and $ \otimes$ is the usual Kronecker product.
  Note that $ A\otimes I$ commutes with $ I\otimes A$.

[quote="loup blanc"]Hi miwalin, you repeat yourselves
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=258803
Do your spectral norm refer to $ \parallel{}.\parallel{}_2$ ?
If yes then I have no proof.

[color=green][The other topic has been deleted. Let this be the only topic for this problem. - moderator][/color][/quote]

Yes, you are right.
I wish you all pay attention to this inequality.

I think the result is true for all $ n$ (the computer says OK).
If $ n\equal{}2$ it is correct because    $ \|A\otimes I \minus{} I\otimes A\|_2^2\plus{}trace(A)^2\equal{} 2\|A\|_F^2$
That we must prove: let us recall that $ \parallel{}A\parallel{}_F^2\equal{}trace(A^TA)$.
Let $ \phi\equal{}A\otimes I \minus{} I\otimes A: M\rightarrow{A^TM\minus{}MA}$.
The question is: show that if $ trace(M^TM)\equal{}1$ then
$ \parallel{}\phi(M)\parallel{}^2\equal{}trace((\phi(M))^T\phi(M))\leq{2trace(A^TA)}$
or $ trace(M^TAA^TM\plus{}A^TM^TMA\minus{}M^TAMA\minus{}A^TM^TA^TM)\leq{2trace(A^TA)}$.
Remark 1: $ trace(M^TAA^TM)\equal{}\parallel{}A^TM\parallel{}_F^2\leq\parallel{}A^T\parallel{}_F^2\parallel{}\parallel{}M\parallel{}_F^2\equal{}\parallel{}A\parallel{}_F^2$ because $ \parallel{}.\parallel{}$ is submultiplicative.
By the same way $ trace(A^TM^TMA)\leq\parallel{}A\parallel{}_F^2$ .
Unfortunately $ trace(M^TAMA\plus{}A^TM^TA^TM)$ may be negative.
Remark 2:the norms are unitarily invariant; then we may assume $ A$ is an upper triangular matrix.


I have trouble with the following: when writting in the double dollar environment it's seems that I can't break the formulas into a new line. I tried to use \newline command, but it doesn't seem to work.
I would be grateful for any suggestions.

Double dollars are obsolete and have been replaced by \ [ ... \ ] (without the spaces). See [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=dolldoll]Why use \ [ ...\ ] in place of double dollar signs?[/url].
However, this also only works for a single line so a new line won't do anything. For that situation you ned the align environment:
\begin{align}
x &=1 \\
y &= 2
\end{align}
where you use \\ for a new line and & to line up the following symbol and align* instead of align if you don't want equation numbers.

Thanks for the quick reply Steve.  :lol:

I used your suggestion, and tried it right away but it seems I don't understand fully the use of &. An example:

\begin{align*} & \psi_{\tilde{N}}(f_2) = \mathbb{E}[\exp\{ - \tilde{N}(f_2)\}] = \mathbb{E}[\exp\{ - \int_{E_2}f_2(x_2)d(N\circ T^{ - 1})(\omega,x_2)\}] = \\
\mathbb{E}[\exp\{ - \int_{E_1}f(Tx_1)dN(\omega,x_1)\}]\end{align*}.

As you can se it doesn't break the way it should be. Where should I put the & sign? Thanks in advance.

Edit: As a reference I'm using "Not so short introduction to Latex2_{\varepsilon}"

The symbols following & will be lined up. So using & before = will line up the equals signs. In your example doing this would give you
\begin{align*}\psi_{\tilde{N}}(f_{2})&=\mathbb{E}[\exp\{-\tilde{N}(f_{2})\}]=\mathbb{E}[\exp\{-\int_{E_{2}}f_{2}(x_{2})d(N\circ T^{-1})(\omega,x_{2})\}]\\
&= \mathbb{E}[\exp\{-\int_{E_{1}}f(Tx_{1})dN(\omega,x_{1})\}]\end{align*}which is I think what you want.
& = occurs where the two = line up so you have
&=\mathbb{E} ...
on the two lines
Also the \\ should be before not after the equals sign.

[quote="Mr.Doe"]Edit: As a reference I'm using "Not so short introduction to Latex2_{\varepsilon}"[/quote]The align environment is covered in section 3.4.1 of the latest edition.

Again thanks for the quick reply.

Thats not exactly what I wanted, I wanted this:

\[ \begin{align*}&\psi_{\tilde{N}}(f_{2})=\mathbb{E}[\exp\{-\tilde{N}(f_{2})\}]=\mathbb{E}[\exp\{-\int_{E_{2}}f_{2}(x_{2})d(N\circ T^{-1})(\omega,x_{2})\}]\\ &=\mathbb{E}[\exp\{-\int_{E_{1}}f(Tx_{1})dN(\omega,x_{1})\}]\end{align*}, \]

but your example made it perfectly clear how to do it.

Obviously I don't know what you want this equation for, but I thought I would note that in almost any context your reader will find it easier to understand if you were to write this same equation as
\begin{align*} \psi_{\tilde{N}}(f_{2}) & = \mathbb{E}[\exp\{ - \tilde{N}(f_{2})\}] \\
& = \mathbb{E}[\exp\{ - \int_{E_{2}}f_{2}(x_{2})d(N\circ T^{ - 1})(\omega,x_{2})\}] \\
& = \mathbb{E}[\exp\{ - \int_{E_{1}}f(Tx_{1})dN(\omega,x_{1})\}].\end{align*}

(Having multiple equalities on the same line often makes everything too dense to follow.)


Are there any relations between $a,b,c$ and $x,y,z$?

[quote="Gibbenergy"][quote="Mathsniper"]for any x, y, z $\in \mathbb{R}$, prove that
\[ (x^2+yz)(y^2+zx)(z^2+xy) \ge 2xyz(x^2+y^2+z^2-1)((xy)^2+(yz)^2+(zx)^2-(xyz)^2)  \][/quote]

Let say $z=0$
$RHS=0$,  $LHS= x^2y^2xy=(xy)^3$ and you say," for any x, y, z $\in \mathbb{R}$" :D[/quote]   sorry for some condition missing :blush:

maybe something wrong when I generated the inequality. Try this one.  :)
I will check my wrong inequality again.


limit
logarithms
function
calculus
derivative
ratio
real analysis

Determine whether there exists a sequence $x_{n}$ of positive real numbers such that
\[x_{n+1}=\left(1+\frac{1}{n}\right)^{x_{n}},\]
and
\[\lim_{n\to +\infty}\frac{n}{x_{n}}=\lim_{n\to
+\infty}\frac{x_{n}}{n^{2}}=0.\]

Enjoy!

There exists a unique sequence $A_n$ satisfying the recurrence with $n<A_n<n^2$ for all sufficiently large $n$. This sequence has, approximately, $A_1=2.42411597$, $A_2=5.36700031$, $A_{25}=124.153798$, $A_{100}=652.488129$, $A_{1000}=9124.38704$, $A_{10000}=116678.722$, $A_{1000000}=16626518.9$, and $\displaystyle \lim_{n \to \infty} \frac{A_n}{n \ln n} = 1$.
The proof relies on the nature of the real numbers, defining these by a limit.

Let $\displaystyle f_n(x) = \frac{\ln(x)}{\ln(1+\frac{1}{n})}$ and $g_n(x)=(1+\frac{1}{n})^x$

Define sequences $b_{n,m}$ for $n \le m$ by $b_{m,m}=m$ and $b_{n,m}=f_n(b_{n+1,m})$.
Then $b_{n+1,m}=(1+\frac{1}{n})^{b_n}$.
We will show that for each $n$, the sequence $b_{n,m}$ is increasing in $m$ and has a limit; let $\displaystyle B_n=\lim_{m \to \infty} b_{n,m}$ such that $B_{n+1}=g_n(B_n)$

Also define $c_{n,m}$ for $n \le m$ by $c_{m,m}=m( \log_2 (m+1)+ \log_2(m+3))$ and $c_{n,m}=f_n(c_{n+1,m})$.
Then $c_{n+1,m}=(1+\frac{1}{n})^{c_n}$ and for each $n$, the sequence $c_{n,m}$ is decreasing in $m$ and has a limit; let $\displaystyle C_n=\lim_{m \to \infty} c_{n,m}$.

First, $f_m(x)$ and its inverse $g_m(x)$ are monotonically increasing functions of $x$ for all $m$. We have $g_m(m)=\frac{(m+1)^m}{m^m}$; since $(m+1)^{m-1} \cdot 1 \le m^m$ by AM-GM (equality only if $m=1$), $g_m(m) \le m+1$ and we must have $b_{m,m+1} \ge b_{m,m}$, implying $b_{n,m+1} \ge b_{n,m}$.
Similarly, $\displaystyle g_m(m( \log_2 (m+1)+ \log_2(m+3))) = ((1+\frac{1}{m})^m)^{(\log_2 (m+1)+ \log_2(m+3))} \ge 2^{(\log_2 (m+1)+ \log_2(m+3))} = (m+1)(m+3)$ by Bernoulli's inequality.
We have $m+3>\log_2 (m+2)+ \log_2(m+4)$, so $g_m(m( \log_2 (m+1)+ \log_2(m+3))) > (m+1)( \log_2 (m+2)+ \log_2(m+4))$ and $c_{m,m+1} \le c_{m,m}$, implying $c_{n,m+1} \le c_{n,m}$.

Since $c_{m,m} > b_{m,m}$, $c_{n,n} \ge c_{n,m} > b_{n,m} \ge b_{n,n}$ for all $n,m$.
This gives an upper bound for each sequence of the $b_{n,m}$ and a lower bound for each sequence of the $c_{n,m}$. Since monotone bounded sequences converge, $B_n$ and $C_n$ exist.

Clearly, $\displaystyle \lim_{n \to \infty} c_{n,n}/n^2 = 0$, so $\displaystyle \lim_{n \to \infty} C_n/n^2 = \lim_{n \to \infty} B_n/n^2=0$.
Also, since $\ln \frac{n+1}{n} < \frac{1}{n}$, $b_{n,n+1} > \frac{\ln (n+1)}{\frac{1}{n}} = n \ln (n+1)$. This gives  $\displaystyle \lim_{n \to \infty} n/b_{n,n+1}= 0$ and $\displaystyle \lim_{n \to \infty} n/B_n= \lim_{n \to \infty} n/C_n=0$.

The sequences $B_n$ and $C_n$ therefore satisfy the conditions. Moreover, if a sequence $a_n$ does, we have $B_n \le a_n \le C_n$ for all $n$; if $a_n < b_{n,m}$ for some $m,n$, $a_m < b_{m,m} = m$ and if $a_n > c_{n,m}$ for some $m,n$, $a_{m+1} > g_m(c_{m,m}) \ge (m+1)(m+3)$, contradicting the inequalities.

Now the derivative $g'_n(x)$ is $\displaystyle \frac{1}{x \ln (1+\frac{1}{n})} < \frac{n}{x}$. If $x,y > c \cdot n$, $|g_n(x) - g_n(y)| < \frac{x-y}{c}$. This gives $\displaystyle c_{n,m}-b_{n,m} < \frac{c_{n+1,m}-b_{n+1,m}}{\ln n}$ for $n<m$ and $\displaystyle c_{n,m}-b_{n,m} < m(\log_2(m+1)+\log_2(m+3)-1) \prod_{j=n}^{m-2} \frac{1}{\ln j}$.
The product goes to zero exponentially in $m$, so $\displaystyle \lim_{m \to \infty} (b_{n,m}-c_{n,m})=0$ and $B_n=C_n$.

The approximations of $A_n=B_n=C_n$ above were derived by iterating $f_n$. I actually used $b_{n,n+20}$ for these, which were accurate to at least ten digits in each case. We can approximate $A_n$ by $n \ln n$; the ratio $\frac{A_n}{\ln n}$ tends to 1.


I would really appreciate it if anyone could get me a link for national mathconts competitions or state competitions pre-1996.  Thanks for your consideration

actually, my email is really messing up, so can you just pm me the link?


geometry
circumcircle
LaTeX
ratio
geometric transformation
reflection
perpendicular bisector

Dear all,
ABC a triangle, 0 the circumcircle of ABC, O the center of 0, P and Q two points on 0, Q1 the symmetric of Q wrt BC, Q2 the symmetric of Q wrt CA, S1 the point of intersection of PQ1 and BC, S2 the point of intersection of PQ2 and CA, 1 the circumcircle of triangle PQS1, O1 the center of 1, and U the second point of intersection of S1S2 with 1.
Prouve that AU and OO1 intersect on 1.
Any synthetic proof?
Jean-Louis

Just want to rewrite it:
[quote="jayme"]
   Let $ \triangle ABC$ be a triangle with circumcircle $ \omega(O,R)$.Consider two arbitrary points on $ \omega$,denote them as $ P$ and $ Q$.Let $ Q_1,Q_2$ be the symmetric points to $ Q$ w.r.t to $ BC$ and $ AC$ respectively.Denote $ S_1\in PQ_1\cap BC$,$ S_2\in PQ_2\cap CA$.Let $ \zeta(O_1,R_1)$ be the circumcircle of $ PQS_1$ and $ U$ be the second point of intersection of $ S_1S_2$ with $ \zeta$.
   Prove that $ AU$ and $ OO_1$ intersect on $ \zeta$.
[/quote]
P.S:Use $ \text{\LaTeX}$ please.

[quote="Domi"][quote="Erken"]P.S:Use $ \text{\LaTeX}$ please.[/quote]
The Jayme's version was not so bad  :maybe:

Domi[/quote]
It is not bad it is become easier to read.Please jayme correct the statement.

Dear all,
I am not easy with latex. I am waiting for a synthetical proof.
Sincerely
Jean-Louis

[b]Lemma:[/b] Let $ \mathcal O_1, \mathcal O_2$ be two circles (different from each other, but otherwise arbitrary - intersectiong or not, possibly even concentric.) Let their center line $ o$ meet $ \mathcal O_1$ at $ K, M$ and $ \mathcal O_2$ at $ L, N.$ (If concentric, $ o$ is an arbitrary line through their common center.) Let $ s$ be an arbitrary line different from $ o$ intersecting both circles, $ \mathcal O_1$ at $ S, U$ and $ \mathcal O_2$ at $ T, V.$ (Order of the points $ K, L, M, N$ on the center line $ o$ and order of the points $ S, T, U, V$ on the arbitrary line $ s$ does not matter.) $ KU, MS$ are two secants of $ \mathcal O_1$ and $ LV, NT$ are two secants of $ \mathcal O_2.$ Let $ D \equiv KU \cap LV,$ $ A \equiv KU \cap NT,$ $ B \equiv MS \cap LV,$ $ C \equiv MS \cap NT$ be the four intersections of a secant of $ \mathcal O_1$ with a secant of $ \mathcal O_2.$ The points $ D, A, B, C$ lie on a circle $ \mathcal O$ coaxal with the circles $ \mathcal O_1, \mathcal O_2.$

Using directed angles modulo $ 180^\circ.$ The $ \triangle DUV \sim \triangle CMN$ are similar, because

$ \measuredangle DUV \equal{} \measuredangle KUS \equal{} \measuredangle KMS \equal{} \measuredangle NMC \mod 180^\circ$

$ \measuredangle UVD \equal{} \measuredangle UVL \equal{} \measuredangle TVL \equal{} \measuredangle TNL \equal{} \measuredangle CNL \equal{} \measuredangle CNM \mod 180^\circ$

Consequently, $ \frac {\overline{DU}}{\overline{DV}} \equal{} \frac {\overline{CM}}{\overline{CN}}.$ For likewise reasons, the $ \triangle DKL \sim \triangle CST$ are similar and $ \frac {\overline{DK}}{\overline{DL}} \equal{} \frac {\overline{CS}}{\overline{CT}}.$ Combining,

$ \frac {\overline{DU} \cdot \overline{DK}}{\overline{DV} \cdot \overline{DL}} \equal{} \frac {\overline{CM} \cdot \overline{CS}}{\overline{CN} \cdot \overline{CT}}.$

Thus, the powers of the points $ C, D$ to the circles $ \mathcal O_1, \mathcal O_2$ are in the same ratio, which means that they lie on a circle $ \mathcal O$ coaxal with the circles $ \mathcal O_1, \mathcal O_2$ (see solution of the problem [url=http://www.mathlinks.ro/viewtopic.php?t=121062]Constant value[/url]). In exactly the same way, we can show that the point pairs $ (D, A)$; $ (A, B)$; $ (B, C)$ lie on circles coaxal with $ \mathcal O_1, \mathcal O_2.$ Since exactly one circle coaxal with $ \mathcal O_1, \mathcal O_2$ passes through every point of the plane (with the exception of their possible intersection(s), two at most), the four points $ D, A, B, C$ all lie on the same circle $ \mathcal O$ coaxal with $ \mathcal O_1, \mathcal O_2.$ This proves the lemma.

[b]Converse of lemma:[/b] Let $ \mathcal O_1, \mathcal O_2, \mathcal O$ be three coaxal circles. Let $ \mathcal O_1$ cut their common center line $ o$ at $ K, M$ and let $ \mathcal O_2$ cut $ o$ at $ N, L.$ Let $ C \in \mathcal O$ be arbitrary (not on the center line $ o$) and let $ CM, CN$ meet $ \mathcal O$ again at $ A, B.$ By solution of the problem [url=http://www.mathlinks.ro/viewtopic.php?t=190473]Concurrency on circumcircle[/url], the intersection $ D \equiv AK \cap BL$ also lies on $ \mathcal O.$ Let $ CM, DK$ meet $ \mathcal O_1$ again at $ S, U$ and let $ CN, DL$ meet $ \mathcal O_2$ again at $ T, V.$ The four points points $ S, T, U, V$ are then collinear.

Let the line $ TV$ meet $ CM, DK$ at $ S', U'.$ The $ \triangle U'VD \sim \triangle MNC$ are similar, having equal angles

$ \measuredangle VDU' \equal{} \measuredangle BDA \equal{} \measuredangle BCA \equal{} \measuredangle MCN \mod 180^\circ$

$ \measuredangle U'VD \equal{} \measuredangle U'VL \equal{} \measuredangle TVL \equal{} \measuredangle TNL \equal{} \measuredangle CNL \equal{} \measuredangle CNM \mod 180^\circ$

Consequently, $ \frac {\overline{U'V}}{\overline{U'D}} \equal{} \frac {\overline{MN}}{\overline{MC}}.$ For likewise reasons, the $ \triangle U'TA \sim \triangle MLB$ are similar and $ \frac {\overline{U'T}}{\overline{U'A}} \equal{} \frac {\overline{ML}}{\overline{MB}}.$  Combining,

$ \frac {\overline{U'V} \cdot \overline{U'T}}{\overline{U'D} \cdot \overline{U'A}} \equal{} \frac {\overline{MN} \cdot \overline{ML}}{\overline{MC} \cdot \overline{MB}}.$

Thus, the powers of the points $ U', M$ to the circles $ \mathcal O_2, \mathcal O$ are in the same ratio, which means that $ U'$ lies on the circle $ \mathcal O_1$ coaxal with the circles $ \mathcal O_2, \mathcal O$ and passing through $ M.$ In exactly the same way, we can show that the point $ S'$ lie the circle $ \mathcal O_1$ coaxal with the circles $ \mathcal O_2, \mathcal O$ and passing through $ K.$ Therefore, $ U' \equiv U$ and $ S' \equiv S$ are identical. But $ S', T, U', V$ are collinear by definition of $ S', U'.$ This proves converse of the lemma.

[b]Problem:[/b] $ P, Q$ are two arbitrary points on the circumcircle $ \mathcal O$ of the $ \triangle ABC.$ $ Q_1$ is a reflection of $ Q$ in $ BC$ and $ PQ_1$ cuts $ BC$ at $ S_1.$ The perpendicular bisector of the segment $ PQ$ is the center line $ o \equiv OO_1$ of the circumcircles $ \mathcal O, \mathcal O_1$ of the $ \triangle ABC, \triangle PQS_1.$ $ S_1Q_1$ is a reflection of $ S_1Q$ in $ BC$ $ \Longrightarrow$ the line $ BC$ bisects the angle $ \angle QS_1Q_1,$ internally, when $ P, Q$ are on the same side of $ BC$ and externally, when $ P, Q$ are on opposite sides of $ BC.$ Perpendicular to $ BC$ at $ S_1$ bisects the suppplemetary angle $ \angle PS_1Q,$ internally, when $ P, Q$ are on the same side of $ BC$ and externally, when $ P, Q$ are on opposite sides of $ BC.$ Thus this perpendicular passes through the midpoint $ K \in o$ of the arc $ PQ$ of $ \mathcal O_1$ on the opposite side of $ BC$ than $ S_1$ resp. on the same side of $ BC$ as $ S_1.$ Let $ \mathcal O_1$ meet $ BC$ again at $ M.$ Since $ \angle KS_1M \equal{} 90^\circ$ is right, $ KM$ is a diameter of $ \mathcal O_1$ and $ M \equiv o \cap BC.$ Likewise, $ Q_2$ is a reflection of $ Q$ in $ CA$ and $ PQ_2$ cuts $ CA$ at $ S_2.$ The circumcircle $ \mathcal O_2$ of the $ \triangle PQS_2$ cuts the perpendicular bisector $ o \equiv OO_2 \equiv OO_1$ of the segment $ PQ$ at $ L,$ so that $ S_2L \perp CA$ and again at $ N \equiv o \cap CA.$

By solution of the problem [url=http://www.mathlinks.ro/viewtopic.php?t=190473]Concurrency on circumcircle[/url], the intersection $ D \equiv AK \cap BL$ lies on the circumcircle $ \mathcal O.$ Let $ AK$ meet $ \mathcal O_1$ again at $ U$ and let $ BL$ meet $ \mathcal O_2$ again at $ V.$ We arrived to the configuration of the above lemma: $ KU, MS_1$ are two secants of $ \mathcal O_1$ and $ LV, NS_2$ are two secants of $ \mathcal O_2.$ The four points $ D \equiv KU \cap LV,$ $ A \equiv KU \cap NS_2,$ $ B \equiv MS_1 \cap LV,$ $ C \equiv MS_1 \cap NS_2,$ are pairwise intersections of a secant of $ \mathcal O_1$ with a secant of $ \mathcal O_2.$ They lie on the circumcircle $ \mathcal O$ coaxal with $ \mathcal O_1, O_2.$ By converse of the above lemma, the points $ S_1, S_2, U, V$ are collinear.


Gay marriage, prayer in schools, abortion, or any "moral value".

I know! How about the current LD topic? It's in the other forum.

Which part of the sandwich is more important, the peanut butter, or the jelly?

If teflon never sticks to anything, what makes teflon stick to the pan??? LOL.

Don't bother PMing me about this. It was my old signature, and a bunch of people came arguing that blahblahblah, and I don't mean blahblahblah himself, just characterizing other people blabbering about the reasons and stuff.  :?

abortion! i did that debate in english this year and it was fascinating...err...that didn't sound right.  :)

1) how ppl who are prejudiced shouldnt be blamed

2)war with iraq. (....its so commonly spoken about, but why not?)

3)is it better or worse for children to be exposed to swear language?

4)which line of barbie dolls should be banned (or all of them)

5)

Whether vegetarians should animal crackers or not.

[quote]Can someone suggest a debatable topic for my L.A. speech? I can't think of anything that can create controversy.[/quote]

For Mrs. Altemann?

[quote="filletwho"]Can someone suggest a debatable topic for my L.A. speech? I can't think of anything that can create controversy.[/quote]

How about some more details?

Are you looking for a strictly two-sided AFF v NEG debate topic (e.g. the existence of free enterprise) or like something that could have more than two viewpoints (e.g. what form of government is most beneficial for the economic growth of a country)?

How many people?

Time limits?

This speech is for Altemann. Thx for the ideas. As for the NFL or AGE thingymajigs, i really don't kno wat it is: I didn't take debate yet. The class sounds fun though... ne ways, i think it'll be better two sided since it makes my job easier.  :)