linear algebra
matrix
linear algebra unsolved

Prove or disprove that diagonally dominant matrix is always invertible. Thanks in advance.

Hint:  Disprove.

[hide]How about a diagonal matrix with a 0 on the diagonal?[/hide]

And $ \begin{bmatrix}1&1\\1&1\end{bmatrix}$ isn't invertible, either.

The correct statement is that a [i]strictly[/i] diagonally dominant matrix is invertible.

Strictly diagonally dominant means that for all $ k,$ $ |a_{kk}|>\sum_{j\ne k}|a_{jk}|.$


calculus
function
integration
derivative
calculus computations

Find all [u][b]positive[/b][/u] real-valued functions $ f(x)$ such that $ f(x)$ satisfies the following equation for all $ x$.

$ (f(x))^{2} \equal{} \int_0^x [(f(t))^2 \plus{} (f'(t))^2] dt \plus{} 2009$

By the way, $ f'(t)$ is the derivative of $ f(t)$.

Thanks!

How do you get from the first step to the second? It seems like differentiation but how do you differentiate a definite integral?

And what's the proof to show that when $ f(x) \equal{} f'(x)$, then $ f(x) \equal{} f(0) e^x$? I heard of it before but i forgotten it. thanks!

(a) It is a differentiation and $ \int_0^x [f(t)^2 \plus{} f'(t)^2] dt$  it isn't a definite integral because the upper bound $ x$ is variable.
The rule is $ \left( {\int_a^x f(t)\,dt}\right)^{\prime} \equal{} f(x)$.
(b) $ f^{\prime}(x) \minus{} f(x) \equal{} 0 \ \Leftrightarrow \ \displaystyle\frac {f^{\prime}(x)\,e^{x} \minus{} f(x)\,(e^{x})^{\prime}}{e^{2x}} \equal{} 0\ \Leftrightarrow\ \left({\frac {f(x)}{e^{x}}}\right)^{\prime} \equal{} 0\ \Leftrightarrow$
$ \frac {f(x)}{e^{x}} \equal{} c\ \Leftrightarrow \ f(x) \equal{} c\,e^{x}$ and  $ f(0) \equal{} c\,e^{0}\equal{}c$.
So $ \ f(x) \equal{} f(0)\,e^{x}$.

Ok I think i got it thanks!

But just one question.
For $ \left( {\int_a^x f(t)\,dt}\right)^{\prime}\equal{}f(x)$, does it hold true for all constant $ a$? Or only 0?

It's true for any constant a

[And if the lower limit is a function of (x), it's not true.]

This is one way to phrase the fundamental theorem of calculus.


Um so here it is.

I tried googling for a fancy l symbol but couldn't find anything that would help insert that nasty l.  $ \mathit{l}$
$ \mathnormal{l}$
$ \mathcal{l}$
and
$ Bbb{l}$

all don't work. So...

\ell as in $ \mathbb{R}_\ell$
See [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=comprehensive]The Comprehensive LaTeX Symbol List[/url] page 51.


The numbers 1, 3, 5, 7, and 9 are used to create 5-digit integers. What's the sum of all the 5-digit integers that can be created with all 5 digits?

[quote="jonathanchou711"]I don't think it's you who made that up. I have seen this problem before. And it's actually pretty simple. For the first part, let's find out how many numbers can be made with 1 _ _ _ _ So we are going to have 24 numbers beginning with 1 right? So 24*10000 = 240000. If we start with 3 and then have the remaining four numbers as _ _ _ _, then we will have 24* 30000 = 720000! So if we continue this process until every number given has been in the ten-thousands spot, then we add up the total (240000+720000+....) which totals to 6000000! So you can expect that we are going to take the thousands digit spot now. ITS THE SAME PROCESS!!! It's just taking down one more 0 from the last process. And so on for the hundreds and tens and ones. So if you add this all up, you get a total of 6666660! And that's the answer... Can someone rate this post? It took me a long time to type this... And this is the first time I've actually answered a smart guy's post...   :(  :blush:[/quote]
I'm just saying that I made this up because I thought of those numbers naturally in my head.

[hide="Well known solution"]
nth number from the front+nth number from the last is constant.

That 'constant' is $ 13579\plus{}97531\equal{}13597\plus{}97513\equal{}\cdots\equal{}111110$

There are total of $ 5!\equal{}120$ numbers, so the sum is $ 111110*120\div2\equal{}6666600$

Answer: 6666600
[/hide]

[quote="jonathanchou711"]total of 6666660! And that's the answer... Can someone rate this post? It took me a long time to type this... And this is the first time I've actually answered a smart [b]guy's [/b]post...   :(  :blush:[/quote]

I think you mean "gal" (or whatever the terminology they use for a female)

Guy, gal. Doesn't really matter for me. I don't really care. But I'm sorry. And I think i might've typoed... the other guy/gal who answered this question had 66666600 or something like that... HM.............................. But did you guys vote? :D

i put 5/5 on both of ur posts jonathan, lol...now rate mine :roll:

THANKS HIDDEN!! YOU'RE AWESOME. I WILL! IF YOU ANSWER SOME QUESTIONS ON THE FORUMS! :)


i misplaced my 1991 AIME A solutions (i have the problems), could someone post them in spoiler, please?

um, that would be asking a lot for someone who actually had the official solutions.. but kalva has solutions to the problems, not necessarily official though.

http://www.kalva.demon.co.uk/aime.html

my bad i didn't mean solutions i meant answers, klava site has different questions, i want to know what the answers to this were, if someone can post them.

All of the questions on Kalva are the same - except in the obvious ways. If you look at the answers there you can tell if you got the question right or not.

there are quiet a few errors on kalva...  its scary to know that nobody pointed them out before I did.


Let be $ a,b,c\in (0,\infty )$ such that $ ab\plus{}bc\plus{}ca\equal{}1$ . Show that :

$ 9a^2b^2c^2\plus{}abc(\sqrt{1\plus{}a^2}\plus{}\sqrt{1\plus{}b^2}\plus{}\sqrt{1\plus{}c^2})\le 1$

[quote="alex2008"]Let be $ a,b,c\in (0,\infty )$ such that $ ab \plus{} bc \plus{} ca \equal{} 1$ . Show that :

$ 9a^2b^2c^2 \plus{} abc(\sqrt {1 \plus{} a^2} \plus{} \sqrt {1 \plus{} b^2} \plus{} \sqrt {1 \plus{} c^2})\le 1$[/quote]

Since $ 1\plus{}a^2\equal{}(a\plus{}b)(a\plus{}c)$, by AM-GM we have
\[ LSH\le 9ab.bc.ca\plus{}abc(a\plus{}b\plus{}b\plus{}c\plus{}c\plus{}a)\le \frac1{3}(ab\plus{}bc\plus{}ca)^3\plus{}\frac2{3}(ab\plus{}bc\plus{}ca)^2\equal{}1
\]
:)


Find all the $ x,y\in \mathbb{R}_\plus{}$ such that $ \frac{2}{y}\plus{}\frac{3}{x}\equal{}12$ and $ 2x\plus{}3y\le 2$

We have $ 12xy \equal{} 2x \plus{} 3y \le 2$, so $ xy \le \frac {1}{6}$

Now by AM-GM, $ 12 \equal{} \frac {2}{y} \plus{} \frac {3}{x} \ge 2 \times \sqrt {\frac {6}{xy}} \ge 12$

Equality holds so $ \frac {2}{y} \equal{} \frac {3}{x}$
$ \frac {2}{y} \equal{} 6$ and $ \frac {3}{x} \equal{} 6$
$ y \equal{} \frac {1}{3}$ and $ x \equal{} \frac {1}{2}$

Typo corrected now, keyree10 .

[quote="stephencheng"]
Now by AM-GM, 12 =$ \frac{2}{y}+\frac{3}{x} \ge 2 \times \sqrt{\frac{6}{xy} \ge 12}$
[/quote]
@stephencheng: you seem to have made a typo.

hello, i have also found $ x\equal{}\frac{1}{2},y\equal{}\frac{1}{3}$.
Sonnhard.


number theory
prime numbers
combinatorics unsolved
combinatorics

Set $ A\subseteq \{1,2,...,100\}$, for any three different elements $ a,b,c$ in $ A$, $ bc$ is not multiple of $ a$ . Find the maximum $ |A|$.

The answer is $ 27$ :D
The construction is not difficult,but it cost much time on argument(seems boring)

[quote="sansuunoousama"]hmmm....I get $ 26$ numbers,
the basic argument hxy09 is talking about is that if you don't want $ bc$ to be a multiple of $ a$, then $ a$, $ b$, $ c$ must be relatively prime....experiment and see for yourself......the set $ A$ becomes all the prime numbers and $ 1$[/quote]

Try $ a\equal{}6$, $ b\equal{}8$, and $ c\equal{}10$.

@sans
That is exactly what I was thinking but I didn't want to write anything because I didn't figure out how to prove it was maxiaml...I figured it wasn't. Too easy ^_^
kevin : can you post some solution?


According to a nursery rhyme recipe, 24 blackbirds were to be baked in a pie. But due to a blackbird shortage, the baker cut the recipe to make the pie filling three-fourths the normal volume. How many fewer blackbirds were needed?

Set up porportions. This gives

\[ \frac{3}{4}\equal{}\frac{x}{24}\implies24\cdot\frac{3}{4}\equal{}x\rightarrow x\equal{}18
\]

Thus, the total amount you need want is $ 24\minus{}18\equal{}\boxed{6}$

Or use that $ x - \frac{3x}{4} gives \frac{x}{4}$
$\frac{x}{4}$ for $x = 24$ gives \boxed{6}$


Find all polynomials $f(x);g(x)\in \mathbb{R}[x]$ sastifying the relation:
$f(x)-f(y)=(x+y)g(x-y)$ for all $x,y \in \mathbb{R}$

+$f(x)=f(0)+xg(x)$
So: $xg(x)-yg(y)=(x+y)g(x-y)$
Then:$g(0)=0$, so $g(x)=xt(x)$
We have: $x^{2}t(x)-y^{2}t(y)=(x+y)(x-y)t(x-y)$
Easily show that: $t(x)=t(-x)$, then $t(x-y)=t(x+y)$, $t \equiv const$
Finally $f(x)=ax^{2}+b, g(x)=ax$

@hieu:you know who I am?  :D

$x=y \implies g(0)=0,~y=0\implies f(x)=xg(x)+f(0)=xg(x)+c$
$y \to x, x\to y$ and adding the original equallity $\implies g(x-y)=-g(y-x)\implies g(-z)=-g(z)\implies f(-x)=f(x)$
$\implies \frac{f(x)-f(y)}{x-y}=g(x+y), x\ne y\implies f'(y)=g(2y)\implies yg'(y)+g(y)=g(2y)$
But g is a polinomal so its coefficients must satisfie:$a_{k}+ka_{k}=2^{k}a_{k}\implies k=0$ or $k=1$ But g cannot be constant couse $g(-x)=-g(x)$ so $g(x)=ax+b,g(-x)=-g(x)\implies b=0$ So $f(x)=xg(x)+c=ax^{2}+c$


A committee is to be made up from a group of 10 candidates. The committee must consist of a chairman and some (not necessarily nonzero) number of members-at-large. How many ways can the committee be chosen?

There are $ 10$ ways to pick a chairman, then $ 2^9$ ways to determine the status of everyone else, giving $ 10\cdot2^9\equal{}5120$ ways total.


Wow.....I have no clue....I would like to know the answer to that myself!!!!!

[quote="Big M"]A 2 by 2003 rectangle consists of unit square, the middle unit square of each row(only two rows) is shaded. if the rectangle is chosen at random, waht is the probability that the rectangle does not include a shaded square?[/quote]
[hide]The total number of rectangles in the given $2 \times 2003$ rectangle is $\binom{3}{2}\binom{2004}{2}$.

The total number of rectangles that do not not include a shaded square = The total number of rectangles on the left of the shaded column + the total number of rectangles on the right of the shaded column. This is equal to,
\[2 \times \binom{3}{2}\binom{1002}{2}.\]
So the desired probability is,
\[\frac{2\times \binom{3}{2}\binom{1002}{2}}{\binom{3}{2}\binom{2004}{2}}= \boxed{\frac{1001}{2003}}.\][/hide]

[quote="Quevvy"][quote="mthiyaga"][quote="Big M"]A 2 by 2003 rectangle consists of unit square, the middle unit square of each row(only two rows) is shaded. if the rectangle is chosen at random, waht is the probability that the rectangle does not include a shaded square?[/quote]
[hide]The total number of rectangles in the given $2 \times 2003$ rectangle is $\binom{3}{2}\binom{2004}{2}$.

The total number of rectangles that do not not include a shaded square = The total number of rectangles on the left of the shaded column + the total number of rectangles on the right of the shaded column. This is equal to,
\[2 \times \binom{3}{2}\binom{1002}{2}. \]
So the desired probability is,
\[\frac{2\times \binom{3}{2}\binom{1002}{2}}{\binom{3}{2}\binom{2004}{2}}= \boxed{\frac{1001}{2003}}. \]
[/hide][/quote]
How did you get those equations? Where did all the 3 choose 2 things come from?[/quote]
When you have a $2 \times 2003$ rectangle, each column has 3 vertices and each row has 2004 vertices. You need to choose 2 vertices from a row and 2 vertices from a column to make a rectangle. So this can be done in a total of $\binom{3}{2}\binom{2004}{2}$ ways.


Polynomial of $ 5$th degree $ P(x)$ is such that $ P(x)\plus{}1$ is dividable by $ (x\minus{}1)^3$ and $ P(x)\minus{}1$ is dividable by $ (x\plus{}1)^3$.
Find $ P(x)$ .

[quote="enndb0x"]Polynomial of $ 5$th degree $ P(x)$ is such that $ P(x) \plus{} 1$ is dividable by $ (x \minus{} 1)^3$ and $ P(x) \minus{} 1$ is dividable by $ (x \plus{} 1)^3$.
Find $ P(x)$ .[/quote]

$ (ax^2\plus{}bx\plus{}c)(x\minus{}1)^3\minus{}1\equal{}(ux^2\plus{}vx\plus{}w)(x\plus{}1)^3\plus{}1$

Changing $ x\to\\minus{}x$ gives $ (ax^2\minus{}bx\plus{}c)(x\plus{}1)^3\plus{}1\equal{}(ux^2\minus{}vx\plus{}w)(x\minus{}1)^3\minus{}1$ and so $ (u,v,w)\equal{}(a,\minus{}b,c)$ and the equation is :

$ (ax^2\plus{}bx\plus{}c)(x\minus{}1)^3\minus{}1\equal{}(ax^2\minus{}bx\plus{}c)(x\plus{}1)^3\plus{}1$

Setting $ x\equal{}0$ gives $ c\equal{}\minus{}1$ and the equation becomes $ (ax^2\plus{}bx\minus{}1)(x\minus{}1)^3\minus{}1\equal{}(ax^2\minus{}bx\minus{}1)(x\plus{}1)^3\plus{}1$

So $ (ax^2\plus{}bx\minus{}1)(x^3\minus{}3x^2\plus{}3x\minus{}1)\minus{}1\equal{}(ax^2\minus{}bx\minus{}1)(x^3\plus{}3x^2\plus{}3x\plus{}1)\plus{}1$

So $ ax^5\plus{}(b\minus{}3a)x^4\plus{}(3a\minus{}3b\minus{}1)x^3\plus{}(\minus{}a\plus{}3b\plus{}3)x^2\plus{}(\minus{}b\minus{}3)x$ $ \equal{}ax^5\plus{}(3a\minus{}b)x^4\plus{}(3a\minus{}3b\minus{}1)x^3\plus{}(a\minus{}3b\minus{}3)x^2\plus{}(\minus{}b\minus{}3)x$

So $ b\equal{}3a$ and $ a\minus{}3b\minus{}3\equal{}0$ and so $ a\equal{}\minus{}\frac 38$ and $ b\equal{}\minus{}\frac 98$

And $ P(x)\equal{}\minus{}\frac 38x^5\plus{}\frac 54x^3\minus{}\frac {15}8x$


At least it can be solved with pretty basic algebra since difference is :

$(a_2 \cdot b_1 - b_2\cdot a_1)^2/((a_1+b_1)\cdot (a_2+b_2)\cdot (a_1+a_2+b_1+b_2))$


superior algebra
superior algebra unsolved

Let $\left( A,+, \cdot \right)$ be a ring that verifies the following properties:
(i) it has a unit, $1$, and its order is $p$, a prime number;
(ii) there is $B \subset A, \, |B| = p$, such that: for all $x,y \in A$, there is $b \in B$ such that $xy = byx$.

Prove that $A$ is commutative.

[i]Ion Savu[/i]

[quote="perfect_radio"]
(i) it has a unit, $1$, [color=blue]and its order is $p$[/color], a prime number;
[/quote]
i.e order of $1$ in $(A,+)$ is $p$ ? Am i wrong?  :oops:

Indeed, that's what I meant $\left( \underbrace{1+1+\ldots+1}_{p \, \text{times}}= 0 \right).$

The best way to say that: $R$ has characteristic $p$.

it took me one hour :-P

first note that we can identify $\mathbb{F}_{p}$ with a subset of $A$, and the elements $\neq 0$ of this subset lie in the center and the unit group of $A$. besides, $1 \in B$ (there is some $b \in B$ with $1*1 = b*1*1$).

now assume that there are $x,y \in A$ with $xy \neq yx$. then also $(x+i)y \neq y(x+i)$ for $i \in \mathbb{F}_{p}$. by the assumption, there are $b_{i}\in B$ so that $(x+i)y = b_{i}y (x+i)$, i.e. $xy+iy = b_{i}y x+b_{i}y i$. the trick is to see that these $b_{i}$ are distinct: if $b_{i}= b_{j}$ and $i \neq j$, we get by subtracting the according equations $(i-j)y = b_{i}y (i-j)$, hence $y = b_{i}y$. but this implies the contradiction $(x+i) y = y (x+i)$. so, the $b_{i}$ are distinct. they are $p$ elements, and since $B$ has only $p$ elements, they already fill $B$ out :-). especially, $b_{i}=1$ for some $i$, which again yields the contradiction $(x+i) y = y (x+i)$.

hence $xy = yx$ for all $x,y \in A$, i.e. $A$ is commutative.


geometry
circumcircle
rectangle
parallelogram
geometric transformation
reflection
radical axis

k is a circle centered at Q, and it contains points B and C. It also intersects sides AB and AC of triangle ABC in K and L. Circumcircle of AKL intersects circumcircle of ABC in A and N. Prove that $ \angle QNA\equal{}90^\circ$

[hide]Let $ O$ be the circumcenter of triangle $ ABC$. Extend $ NQ$ along ray $ NQ$ to point $ P$ on the circumcircle of $ ABC$. Draw a line $ l$ through $ O$ parallel to line $ AN$. Let $ l$ intersect $ NQ$ at $ T$. Let $ R$ be the point on $ AN$ such that $ AN \equal{} 2AR \equal{} 2RN$; $ OR$ is perpendicular to $ AN$, so $ \angle NRO \equal{} \angle ROT \equal{} 90^\circ$. Extend $ AO$ along ray $ AO$ to point $ P'$ on the circumcircle of triangle $ ABC$.
We need to show that $ P$ and $ P'$ are the same point.
Since $ AN$ and $ OT$ are parallel, $ \angle RAO \equal{} \angle TOP'$
Since $ AO$ and $ OP'$ are radii of the same circle, $ AP \equal{} 2AO \equal{} 2OP'$.
By SAS similarity triangle $ OAR$ is similar to triangle $ P'AN$.
Thus $ \angle AOR \equal{} \angle AP'N$.
By ASA, triangle $ OAR$ is congruent to triangle $ P'OT$. $ OT \equal{} AR \equal{} RN$, and since $ \angle NRO \equal{} \angle ROT \equal{} 90^\circ$, $ ROTN$ is a rectangle. $ \angle OTP \equal{} \angle OTP' \equal{} 90^\circ$. By ASA triangles $ OTP$ and $ OTP'$ are congruent, so $ P$ and $ P'$ are the same point.
Therefore, since $ ROTN$ is a rectangle, $ \angle QNA \equal{} \angle TNR \equal{} 90^\circ$, as desired.[/hide]

[quote="Bugi"]k is a circle centered at Q, and it contains points B and C. It also intersects sides AB and AC of triangle ABC in K and L. Circumcircle of AKL intersects circumcircle of ABC in A and N. Prove that $ \angle QNA \equal{} 90^\circ$[/quote]

I do not know if this helps, but it is an IMO - 1985 problem .

 There are more than two nice solutions to this problem.One uses radical center and powers. It was also posed in an Australian olympiad.

 Babis

I didn't know it was an IMO problem. Thanks!

P is circumcenter of $ \triangle ALK.$ BCLK is cyclic $ \Longrightarrow$ $ \triangle ALK \sim \triangle ABC$ are oppositely similar $ \Longrightarrow$ $ AP \perp BC, AO \perp LK.$ $ QO \perp BC$ is the perpendicular bisector of BC, $ QP \perp LK$ is center line of the circles (Q), (P) perpendicular to their radical axis LK $ \Longrightarrow$ AOQP is a parallelogram, its diagonals cutting each other in half at a point S. (S)  is a circle with center S and diameter AQ. Centers of the circles (O), (S), (P) are collinear and they all go through A $ \Longrightarrow$ they are coaxal and intersect again at the reflection N of A in their center line OSP. The angle $ \angle ANQ$ is right, because $ N \in (S)$ and AQ is a diameter of (S).

[quote="Michelle13"]Could you please reveal the secret what was the book in which you found that problem? :) Thank you :)[/quote]

It was on a paper (not in English, Serbian).


geometry
parallelogram
geometry proposed

http://www.mathlinks.ro/viewtopic.php?t=286222
They are similar.

thank you Mr.Nguyen Van Linh...but is there any other solution?

If $ U \equiv AB \cap CD,\ V \equiv BC \cap DA,$ project circle $ (O)$ to a circle $ (O')$ and the line $ UV$ to infinity. $ ABCD$ is projected to a tangential parallelogram $ A'B'C'D',$ a rhombus. The proposition is then true by symmetry.

Can you Solve it by Pascal`s Theorem??? :maybe:


In a convex quadrilateral,five in eight straight connect a top with midpoints of opposite side is equal.
Prove that all eight is equal

[quote="christ"]Do you sure about this topic ?I could prove that only in a special case by means of isosceles triangles.[/quote]
Yes,I sure it is true! :wink:


invariant
linear algebra
matrix
vector
algebra
linear equation
linear algebra unsolved

Consider the matrix \[A= \begin{pmatrix}\lambda & 1 & 0\\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{pmatrix}\] What are the A-invariant subspaces of $\mathbb{C}^{3}$ (besides $0$ and $\mathbb{C}^{3}$)?

The only eigenvalue of A is $\lambda$ and the corresponding eigenspace is $\{ (a,0,c): a,c \in \mathbb{C}\}$. The 1-dimensional A-invariant subspaces must be generated by eigenvectors so these are $\{ k(1,0,0): k \in \mathbb{C}\}$, $\{ k(0,0,1): k \in \mathbb{C}\}$, $\{ k(1,0,1): k \in \mathbb{C}\}$, etc.

$\{ (a,0,c): a,c \in \mathbb{C}\}$ and $\{ (a,b,0): a,b \in \mathbb{C}\}$ are 2-dimensional A-invariant subspaces. Are there any others?

Are you saying that there are only three 1dimensional invariant subspaces?  Because you have a twodimensional subspace of eigenvectors of $\lambda$, all 1dimensional subspaces in there are invariant

For the second, why don't you pull the same trick I did last time?

Those 1-dimensional subspaces that I listed are just meant as examples. I don't mean there are only 3.

As for the second part, I still can't quite figure out what you did the last time. Can you please explain why you have $W = \{ x| v^{t}x=0 \}$?

[quote="chungyc"]
As for the second part, I still can't quite figure out what you did the last time. Can you please explain why you have $W = \{ x| v^{t}x=0 \}$?[/quote]
That is a general theorem about hyperplanes (spaces of dimensions n-1 in a n dimensional space)

Let $v_{1},\dots,v_{n-1}$ be linearly independent column vectors spanning a space $A$
A vector $v$ will be an element if and only if this determinant is zero :

$|v_{1}\cdots v_{n-1}v|$

Expand this with the last column and you will find your linear equation!


induction
LaTeX
number theory unsolved
number theory

For any $h = 2^{r}$ ($r$ is a non-negative integer), find all $k \in \mathbb{N}$ which satisfy the following condition: There exists an odd natural number $m > 1$ and $n \in \mathbb{N}$, such that $k \mid m^{h} - 1, m \mid n^{\frac{m^{h}-1}{k}} + 1$.

Let $2^a||m^h-1$ and $2^b||k$. Then $2^{a-b}||\frac{m^h-1}{k}$, so since $m$ divides $n^{\frac{m^h-1}{k}}+1$ there is a residue modulo $m$ of order $2^{a-b+1}$, it easily follows that $2^{a-b+1}$ divides $p-1$ for every prime divisor of $m$, and hence $2^{a-b+2}|m^2-1$.

By induction on $r$, we can prove that  $2^t||m^h-1$ iff and only iff $2^t||(m^2-1)2^{r-1}$ so we get that:
$a-b+2+r-1$ is at most $a$, so from here it follows that  $b\geq r+1$. We prove this is a sufficient condition.

Let $k=2^{r+1+t}h$ with $h$ odd. Iff t is not $0$, then choose an $m$ such that $2^{t+2}$ is the max power of 2 dividing $m^2-1$ and also such that $h|m-1$, and put $n=m+1$, in this case everything works.

iff $t=0$, let $p$ be a prime of the form $8j+5$, and such that $h|p-1$, then there exist an $n$, such that $n^2+1$ is a multiple of $p$, so this $n$ works!

Pascual2005 haven't considered the case when $ k$ is odd, but of course every odd $ k$ works, since we can take $ m=k+1$ and $ n=m-1$.

[quote="orl"]For any $ h \equal{} 2^{r}$ ($ r$ is a non-negative integer), find all $ k \in \mathbb{N}$ which satisfy the following condition: There exists an odd natural number $ m > 1$ and $ n \in \mathbb{N}$, such that $ k \mid m^{h} \minus{} 1, m \mid n^{\frac {m^{h} \minus{} 1}{k}} \plus{} 1$.[/quote]
The proof is rather long and complicated,but it is not that difficult  :D

[quote="orl"]How about writing it in English and maybe $ \text{\LaTeX}$ here?  :)[/quote]
Thank you for your advice :)
But the above documents were taken from my notebook(I wrote it down last year not today)If anyone had any questions,I would like to explain.

This is my solution. I hope it's right.  :oops:   :)
Firstly, setting $ k=2^{b}q$ with$ gcd(2;q)=1.$
Case 1: $ b \ge r + 2$
Applying Chinese remainder theorem, exist $ m \in N$; $ m$ is odd  such that $ q| {m - 1} $and $ 2^{b - r} \| {m - 1} $
From that, we have $ v_2 (m^h  - 1) = r - 1 + v_2 (m^2  - 1) = r - 1 + b - r + 1 = b = v_2 (k)$
So, $ \frac{{m^h  - 1}}{k}$ is odd and we only choose $ n\in N$ such that $ m| {n+1}.$
Case 2:$ 1\le b \le r + 1.$
Setting$ m^h-1=k.N$. Suppose that $ N=2^u.M ( gcd(2;M)=1).$
We have $ u + b = r - 1 + v_2 (m^2  - 1).$ If exist $ n$ such that $ m| {n^N+1}$ then $ m| {(n{}^2{}^{u})^M+1}$. So$ v_p (m) \le v_p (M) + v_p (n{}^{2{}^u }  + 1) = v_p (n{}^{2{}^u }  + 1) \Rightarrow m| {n{}^{2{}^u }  + 1} $. From that, we can easily see that$ m \equiv 1(\bmod 2^{u + 1} )$. So $ v_2 (m^2  - 1) \ge u + 2 \Rightarrow b \ge r + 1$.
Thus, $ k=2^{r+1}q$. We choose $ m \in N$; $ m$ is odd  such that $ q| {m - 1} $ and $ 2^{b - r} \| {m - 1} $ Then $ v_2 (\frac{{m^h  - 1}}{k}) = 2.$ Setting $ \frac{{m^h  - 1}}{k}=2.N$ with$ N$ is odd. Because $ (-1/m)=1$ so exist $ n$ such that
$ m| {n^2  + 1} | {n^{2N}  + 1} $

We claim that $k$ works if and only if $2^{r+1}\mid k$. We divide the proof into two parts.
-----
[b]Necessity:[/b] Let $A=\dfrac{m^{2^r}-1}{k}$ The main claim is

[color=red][b]Claim:[/b][/color] $\nu_2(p-1)\geq\nu_2(A)+1$ for any prime divisor $p$ of $m$.

[i]Proof:[/i] Let $u=\mathrm{ord}_p(n)$. Then notes that $u\mid 2A$ but $u\nmid A$ (notice that $m$ odd). Therefore $\nu_2(u)=\nu_2(A)+1$. But $u\mid p-1$ so we are done. $\blacksquare$

Let $t=\nu_2(A)$. The claim implies $m\equiv 1\pmod{2^{t+1}}$. However, by LTE,
\begin{align*}
t &= \nu_2(m^{2^r}-1)-\nu_2(k) \\
&= \nu_2(m-1)+\nu_2(m+1)+r-1-\nu_2(k) \\
&\geq (t+1)+1+r-1-\nu_2(k)
\end{align*}
thus $\nu_2(k)\geq r+1$ as claimed.
-----
[b]Sufficiency:[/b] By Dirichlet, take prime $p\equiv 1\pmod{4k}$ and $m=p$ so the first divisibility is clearly satisfied. Let $t=\nu_2((p^{2^r}-1)/k)$. Since $p+1\equiv 2\pmod 4$, we get
\begin{align*}t &=\nu_2(p-1)+\nu_2(p+1)+r-1-\nu_2(k) \\
&= \nu_2(p-1)+(r-\nu_2(k))\end{align*}
Thus $t+1\leq\nu_2(p-1)$. Since groups $\mathbb{Z}_p^{\times}$ and $\mathbb{Z}_{p-1}$ are isomorphic, there exists an element $n$ of $\mathbb{Z}_p^{\times}$ that has order $2^{t+1}$. This means
$$n^{2^t}\equiv -1\pmod p\implies n^{(p^{2^r}-1)/k} = n^{2^t\cdot\text{odd}}\equiv -1\pmod p$$
so this $m,n$ works.


Find all functions $f: N -> N$ such that
$f(m^{2}+f(n)) = f(m)^{2}+n$, for all $m,n \in N$.

Change $n$ to $m^2+f(n)$. So $f(m^2+f(m)^2+n)=f(m^2+f(m^2+f(n)))=f(m)^2+m^2+f(n)$. Let $m=1$ and $C=1+f(1)$, so $f(C+n)=C+f(n)$ $ \forall n \in \mathbb{N}$. Therefore, $f(kC+n)=kC+f(n)$ for all non-negative integer $k$ takes place by simple induction. Put $k=2m+C$ into previous equality and change $n$ to $kC+n$ in the main equality. On one hand $f(m)^2+2mC+C^2+n=f(m^2+f(2mC+C^2+n))=f(m^2+2mC+C^2+f(n))$ but on the other $f((m+C)^2+f(n))=f(m+C)^2+n=f(m)^2+2f(m)C+C^2+n$. So, only one function, namely $f(m)=m$, fulfils the condition.


In all triangles have the same area, find the infimum of radius of their Lester circles.

See Lester circle [url=http://mathworld.wolfram.com/LesterCircle.html]here[/url].


logarithms
calculus
derivative
real analysis
real analysis unsolved

Study the convergence of the sequence \[ a_n = \log_2 3 + \log_3 4 + \ldots + \log_n (n+1) - (n + \ln \ln n), \, \, n \geq 2 \, .  \]

[i] Problem 105 of the Romanian MO Shortlist 2005, 11th grade[/i]

From Taylor series, for $x<1$ we get: $x-\frac{x^2}{2}\le \ln (1+x)<x$ (A proof using derivative can also be done). We have that $f(x)=x-\ln (x+1)$ is positive for every $x>0$ and $f(x)\le \frac{x^2}{2}$. (well-kown).
$a_{n+1}-a_n=f(\frac{\ln (n+2)-\ln (n+1)}{\ln (n+1)} )>0$, hence $a_n$ is increasing.
And $a_{n+1}-a_n\le \frac{\left(\frac{\ln (n+2)-\ln (n+1)}{\ln (n+1)} \right)^2}{2}$
But $\ln (n+2)-\ln (n+1)-\frac 1 {n+1}=f(\frac 1 {n+1})<0$
From the last two relations, we get that $a_{n+1}-a_n \le \frac 1 {2(n\ln n)^2}$, Hence $a_n \le \sum \frac{1}{2(n\ln n)^2}<\sum \frac{1}{2 n^2}<1$.

$a_n$ being , monotonic and bounded is convergent.

I get:
\begin{eqnarray*} a_{n+1} - a_n &=& \log_{n+1} (n+2) + \left( n + \ln \ln n \right) - \left( n+1 + \ln \ln (n+1) \right) \\ \ &=& \frac{\ln (n+2)}{\ln (n+1)} - 1 + \ln \frac{\ln n}{\ln (n+1)} \end{eqnarray*}

But this doesn't look like $f (\ldots)$...

Am I missing something, as I always do?

It's me that I missed smth now :D

Another proof (correct proof) :D
The monotonicity:

\[ a_{n+1} - a_n =\left[\frac{\ln (n+2)-\ln (n+1)}{\ln (n+1)}\right] - [\ln\ln (n+1)-\ln\ln n] \le \frac 1 {(n+1)\ln (n+1)}-\frac 1 {n+1\ln n+1}= 0 \]

I used Lagrange theorem for both terms to prove this. Hence $a_n$ is decreasing.

The boundness:

\[ a_{n+1}-a_n=\left[\frac{\ln (n+2)-\ln (n+1)}{\ln (n+1)}\right] - [\ln\ln (n+1)-\ln\ln n]\ge \frac 1 {(n+2)\ln (n+1)}-\frac{1}{n \ln n}\Rightarrow \]
\[ a_m\le sum_{n=1}^m \frac 1 {(n+2)\ln (n+1)}-\frac{1}{n \ln n}=k+\sum_{i=1}^{m} \left[ \frac 1 {(n+1)\ln n}-\frac 1 {n\ln n}\right]=k-\sum_{i=1}^{m}\frac 1 {n(n+1)\ln n}\le k-\sum_{i=2}^{m}\frac 1 {n^2\ln n} \]

Since $\sum_{i=1}^{m}\frac 1 {n^2\ln n}$ converge (a nice exercise for condensation test) the boundness follows.

Since $a_n$ is monotone and bounded, it is convergent. Hope that it's ok ;)

[quote="perfect_radio"]The proof is very nice. (and correct, also)

I have just one thing to add: instead of

[quote="xirti"]Since $\sum_{i=1}^{m}\frac 1 {n^2\ln n}$ converge (a nice exercise for condensation test) the boundness follows.[/quote]

we could say $\displaystyle \frac{1}{n^2 \ln n} < \frac{1}{n^2}$, and the latter leads to a convergent series.

(I'm saying this because I want to keep the ending proof at 11th grade level)[/quote]

you can also compare this with $\sum \frac 1 {k(k+1)}=1-\frac 1 n \to 1$ and is simpler.


how long (character-wise) does a post have to be? (the minimum)

Huh: Apparently I can post a one letter message. Perhaps this has something to do with mod powers?

1234567891

Yeah, I got 10 as well, which isn't really a lot, considering you can uses spaces and whatnot.


[u][b]Clasa 9-a:[/b][/u]
[b]P3).[/b] Demonstrati ca: $ (\forall)a;b;c > 0\Rightarrow \frac {a^2}{(a + c)(a + b)} + \frac {b^2}{(a + b)(b + c)} + \frac {c^2}{(a + c)(b + c)}\ge\frac {3}{4}.$
[b]P4).[/b] Cercurile $ S_1$ si $ S_2$ avand centrele in punctele $ O_1$ si $ O_2$ sunt secante in punctele $ A$ si $ B.$ Cercul circumscris $ \triangle {O_1O_2A}$ intersecteaza a doua oara cercurile $ S_1$ si $ S_2$ in mod respectiv in punctele D si E, iar dreapta AB in punctul $ C.$ Demonstrati ca: $ |CD| = |CB| = CE|.$
[b]P6). [/b] Doua cercuri avand razele $ R$ si $ r$ sunt tangente la dreapta $ l$ in punctele $ A$ si $ B$ si sunt secante in punctele $ C$ si $ D$(punctul $ D$ se gaseste mai aproape de dreapta $ l$, ca punctul $ C$). Demonstrati, ca cercul circumscris triunghiului $ ABC$ nu depinde de lungimea segmentului $ [AB].$

[u][b]Clasa 10-a:[/b][/u]
[b]P1).[/b] Fiecare dintre polinoamele $ P_1(x) = x^2 + px + q$ si $ P_2(x) = x^2 + qx + p$ are radacini. Demonstrati ca fiecare dintre polinoamele $ Q_1(x) = x^2 + (p - 2)x + 1$ si $ Q_2(x) = x^2 + (q - 2)x + 1$ are radacini.
[b]P2).[/b] Se da functia: $ f(x) = \frac {1}{\sqrt [3]{1 - x^3}}.$ Definim sirul de functii: $ f_1(x) = f(x); f_{n + 1}(x) = f(f_n(x));(\forall);n\ge 2$. Gasiti $ f_{95}(19).$
[b]P4).[/b] Fie $ ABCD$ un patrulater in care: $ |AB| = |AD|$ si $ m(\angle{ABC}) = m(\angle{ADC}) = 90^0.$ Pe laturile $ [BC]$ si $ [CD]$ luam punctele $ F$ si respectiv $ E$, a.i. $ DF\perp AE.$
Demonstrati ca: $ AF\perp BE.$
[b]P5).[/b] In triunghiul ascutitunghic $ ABC$; cercul $ S$ avand drept diametru inaltimea $ [BK]$ intersecteaza laturile $ (AB)$ si $ (BC)$ in mod respectiv in punctele $ E$ si $ F$.
Demonstrati, ca tangentele in punctele $ E$ si $ F$ la cercul $ S$ se intersecteaza intr-un punct al dreaptei suport a medianei duse din varful B al triunghiului $ ABC.$

[b][i]Clasa 11-a:[/b][/i]
[b]P1).[/b] Se da polinomul $ P(t) = t^2 - 4t.$ Demonstrati, ca: $ P(x^2 + y^2)\ge P(2xy);(\forall) x;y\ge 1.$
[b]P2).[/b] Unghiurile $ \alpha, \ \beta, \ \gamma$ indeplinesc conditia: $ sin {\alpha} + sin{\beta} + sin{\gamma}\ge2.$ Demonstrati, ca: $ coss {\alpha} + cos{\beta} + cos{\gamma}\le \sqrt {5}.$
[b]P4).[/b] Sirul $ (a_n)_{n\ge1}$, are proprietatea ca: $ a_{m + n} + a_{m - n} = \frac {1}{2}.(a_{2m} + a_{2n}); \ (\forall)m\ge n.$ Gasiti $ a_{1995}$ in cazul in care $ a_1 = 1.$
[b]P6).[/b] Cercurile $ S_1$ si $ S_2$ avand centrele in punctele $ O_1$ si $ O_2$ sunt secante in punctele $ A$ si $ B.$ Semidreapta $ [O_1B$ intersecteaza cercuk $ S_2$ in punctul $ F$, iar semidreapta $ [O_2B$ intersecteaza cercul $ S_1$ in punctul $ E.$ Paralela dusa prin punctul $ B$ la dreapta $ EF$, intersecteaza a doua oara cercurile $ S_1$ si $ S_2$ in mod respectiv in punctele $ M$ si $ N.$ Demonstrati, ca: $ |MN| = |AE| + |AF|.$

Clasa a 9-a, Problema 3

[hide] Cu Lema Cauchy-Schwarz, se obtine $LHS \ge \frac{(a+b+c)^2}{(a+b)(a+c)+(a+b)(b+c)+(a+c)(b+c)} $.

Prin cateva calcule,
$ \frac{(a+b+c)^2}{(a+b)(a+c)+(a+b)(b+c)+(a+c)(b+c)}\ge \frac{3}{4} \Leftrightarrow a^2+b^2+c^2\ge ab+bc+ac $, evident adevarat.[/hide]

Clasa a 11-a, Problema 1

[hide]WLOG, $x\ge y$.
$ P(x^{2}+y^{2})\ge P(2xy) \Leftrightarrow \frac{x^2-y^2}{x-y}\ge 2 \Leftrightarrow (x-2)^2\ge (y-2)^2$, adevarat.[/hide]

Pentru Problema 3, clasa a IX-a, altă soluție:

[hide]\[\frac{a^2}{(a+c)(a+b)}+\frac{b^2}{(a+b)(b+c)}+\frac{c^2}{(a+c)(b+c)}\geq\frac{3}{4}\]
\[\Leftrightarrow\]
\[\frac{a^2(b+c)+b^2(c+a)+c^2(a+b)}{(a+c)(b+c)(c+a)}\geq\frac{3}{4}\]
\[\Leftrightarrow\]
\[4(a^2b+a^2c+b^2c+b^2a+c^2a+c^2b)\geq3(a+b)(b+c)(c+a)\]
\[\Leftrightarrow\]
\[4(a^2b+a^2c+b^2c+b^2a+c^2a+c^2b)\geq3(a^2b+a^2c+b^2c+b^2a+c^2a+c^2b+2abc)\]
\[\Leftrightarrow\]
\[a^2b+a^2c+b^2c+b^2a+c^2a+c^2b\geq6abc, ~ceea ~ ce ~ este ~ evident ~ din ~ Inegalitatea ~ mediilor ~ pentru ~ 6 ~ numere.~\][/hide]

[quote=mihai miculita][u][b]Clasa 9-a:[/b][/u]
[b]P3).[/b] Demonstrati ca: $ (\forall)a;b;c > 0\Rightarrow \frac {a^2}{(a + c)(a + b)} + \frac {b^2}{(a + b)(b + c)} + \frac {c^2}{(a + c)(b + c)}\ge\frac {3}{4}.$[/quote]
[url=http://artofproblemsolving.com/community/c6h1160188p5607389][b]Croatia 2004[/b][/url]

Clasa a 9-a, Problema 1
Din Ineg. T.A => LHS>=(a+b+c)^2/s, unde s=a^2+b^2+c^2+3ab+3bc+3ca
Notand x=a^2+b^2+c^2, y=ab+bc+ca este suficient sa aratam ca x+2y/x+3y>=3/4 <=> 4x+8y>=3x+9y <=> x>=y care este bine-cunoscuta


[quote="Γιώργος"]Μήπως ξέρει κανείς σας  απο ποια σελίδα μπορώ να αγοράσω το βιβλίο ''101 problems in algebra from the training of the USA IMO team'' δίοτι ούτε το amazon το έχει αλλά ούτε και ο Παπασωτηρίου. :blush:[/quote]

Για προσπάθησε εδώ :

 http://www.amtt.com.au/ProductList.php?pager=1&startpage=1

 Να βάλεις στο search  : 101 problems   και θα στο βγάλει.

 Αλλιώς να στο φτιάξω βιβλιαράκι και να σου το στείλλω !
 Θα βάλω τον Αμπράζη να στο κάνει δώρο :)  !

 Μπάμπης

Το βρήκα το βιβλίο και το παρήγγειλα. Σας ευχαριστώ πάντως για τη προθυμία σας να μου το τυπώσετε. :wink:


[quote]SQL requests not achieved

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SQL Request: SELECT fav_id FROM phpbb_favorites WHERE user_id = 28419 AND topic_id = 175340 ORDER BY topic_id ASC LIMIT 1
Line : 561
File : weblog_entry.php[/quote]

While posting a blog entry.

XP, ML, FF2.

All these fill (hopefully) be fixed sometime next week.


Hi,
I have a question about the Ti-89 Titanium..
Since this calc doesn't have the reciprocal button...
How do you find the reciprocal?
And is there a way to solve equations and letting it show procedures?
Thanks

1)  Yes.  Just do 1/(statement).  That's the equivalent of (statement)^-1.

2)  No, unless you can find some 3rd party software program that will do this.  It's a pretty complex task, and I imagine the AMS solves equations in a rather different "style" than a human would.  Since it's a computer, it will probably favor certain reductions that we wouldn't prefer, or do large steps in one.  When solving the roots of polynomials of high degree, often it will give the answer in decimal format (not exact form) because it doesn't use a systematic approach to solving the equation; it's more of an estimation method (though it's still very accurate).  For this, it couldn't show steps because its steps are more or less a computational "trial and error."

I hope this helps.


AMC
AIME
geometry
probability
search
number theory

I have done a couple amc 10s (I'm going into 8th grade) and both times I would have qualified for the AIME. Unfortunately, on AIMEs I can only get the first few (sad I know). Most of Vol. 2 seems too advanced although I get some of it. What preperation advice do you all have?

The most important thing you can do is review or learn from AoPS Volume 1, specifically the geometry chapters (9-20 I think?).  The key to AMC is not making mistakes.  One or two can really kill your score if you're having trouble with it.  Make sure you're really solid on your probability and combinatorics, which is usually very key on AMCs, so read both the Volume 1 and 2 Chapters on it.

And needless to say, do problems in the getting started and intermediate forums of the forum! Taking practice tests will improve your score to an extent.  I think practice tests are more useful for the sake of getting used to what the tests feel like, tend to ask, and figuring out what topics you really need to know.

Always first search for the specifically looking for help on.  Then post and ask for a solution. You'll be fine if you continue to do math on a consistent basis.  Don't worry too much about it.

As for the AIME, taking classes on AoPS is nice.  Definitely doing old practice tests will help.  AoPS Volume 2 isn't a text that is meant to be read through and through in 2-3 days type of thing.  You could spend a week or two on a chapter and still not understand all of it.  Its just a matter of asking the questions you don't understand. That you can't find the text that is.

I'll say first you want to learn all the basic/intermediate concepts in all the areas of math AIME covers (number theory, geometry, algebra, combinatorics, trig/complex numbers).  Then when you see the easier problems (1-7) you can immediately see the purpose of the problem (as in what the grader wants you to use to solve the problem) then using the corresponding theorems usually is trivial.  Then the harder problems requires ingenuity, which oyu get by solving hard problems.

For the AIME, my best advice would just be to go through lots of old tests.  (Use Kalva or the resources or wiki on this site)

Last year I improved from an ability around 3 AIME problems to more around 9 problems.  The key is, find the solutions to the problems you don't know, (if you're getting only a few problems right, focus on figuring out how to do problems 6-10)and make sure you learn all the reasons, techniques, etc. behind them.

AoPS Volume 2 is helpful as well, especially if you aren't well acquanted with stuff like trig.


Bob has B blue cards, O orange cards and W white cards. He scores points as he play each cards.
   If he plays a blue card, his score is the number of white cards remaining in his hand. If he plays an orange card, his score is three times the number of blue cards remaining in his hand. If he plays a white card, his score is twice the number of orange cards remaining in his hand. Find
   a) the lowest possible score of Bob.
   b) how many ways he can achieve that score.

Note: I use R for the no. of orange cards to avoid confusion with zero.
Let (i) be diacarding a blue card, (ii) and (iii) discarding orange and white.

a) Note that if we run out of cards of one colour, we can easily clear up with no extra score (there is only one way to do this). So we want to reduce BRW to 0. (i) reduces BRW by RW, with score W, so we reduce by R per point scored. Similarly (ii) and (iii) reduce BRW by W/3 and B/2 per point scored, respectively. This shows that a mixed strategy cannot work strictly better than one consisting entirely of one action, e.g. if we do (i), B decreases, so (iii) is a worse choice, but (i) is just as good. So the lowest score is min(BW, 3BR, 2RW).

b) If one of R, W/3, B/2 is strictly greater than the other two, then we have only one choice: keep doing the best action. If R = W/3 and B/2 is smaller, then (ii) remains just as good as we do (i), but as soon as we do (ii), (i) becomes inefficient. So we can do (i) from 0 to B times and then finish off with (ii). We have B+1 choices. Similarly, if W/3 = B/2, then we have R+1 choices, and if R = B/2, we have W+1. If R = W/3 = B/2, we have free choice of our first move, and then we are in one of the situations above with the relevant variable reduced by 1. So we have B+R+W choices.

The exception to all of the above is when one of B, R, W is zero, when of course we have only one choice.


If you could run a class specifically for the AMC/AIME/USAMO, what would it be like?  (You don't have to pay attention to your typical school standards and such if you don't want to)

Or if you were a student in this class, what would you like taught?

It'd be doing problems, nonstop, with 5 minutes of teaching tops.

Also, the teacher should have done some competitive math when they were in school. The teacher should administer several tests and mock tests, and be able to teach good problem solving techniques.

[quote="SplashD"]Isn't that sort of like the online school here?[/quote]Yes, except you have a lot more time in school. :)

I think it's easier to learn and see things in person  :oops:

If I had this class in my school, I'd definitely want AoPS Vol 2 as our "textbook" (mostly because I'm still at the level of hard AIMEs/easy USAMOs) and the teacher should give short lectures each day (10-15 min) and maybe 1 long lecture per week.  The rest of the time for problems

I'd rather not do problem during class, but rather on my own time. I'd like for the teacher to assign problems/administer mock tests and be able answer questions/give hints on practically any AMC/AIME problem (USAMO is pushing it).

To deal with the wide range of skills, we can have a pre-test to make sure the range of skills is not so big. We can also have a post-test for students who might be to advanced for the class. (Kind of like AoPS classes).

My Class:
Given a 60 minute class period. I would lecture the class for 20 minutes on a topic in AoPS Vol. 2. I would use the Excercise problems as classwork problems, and then I would use the end-of chapter problems as Homework. Once a week, I would have my students take a practice AMC, AIME, etc. The final exam at the end of the year would be a collection of problems from all levels of competition that apply what the students have learned from the year.

A subject-focused problem series, with every problem reworded to include penguins.

[quote="rzsolt"]A forseeable problem:
How do you deal with the wide range of skill levels?
This could potentialy ruin the class for most involved. Being too fast for the kids just starting, and being to slow for the people who are really interested.

A subject class seems more apropriate, such as 'intermediate geo' or somesuch.

A class just around the amc12, looks hard to do, in my oppinion.[/quote]

Many universities offer problem solving classes and have to deal with this problem.  Usually grading is done on how much effort you put into the class.

Here are some course descriptions from college problem solving seminars.  It doesn't seem like it would be too hard to get a good grade in one of these classes, you also have to keep in mind that you will usually only get one credit hour for it.  Additionally, I imagine that most people who take these courses do so because they love math and problem solving.

[quote="MIT"]Problem sets will be graded and returned. The class is Pass-Fail only, and anyone making a reasonable effort will pass. There will be problems at all levels of difficulty, so no student should be unable to solve close to six problems each week. Some class participation is required of every student; in particular, every student must occasionally present a solution to the rest of the class.[/quote]

[quote="Harvey Mudd"]
All HMC students are welcome to attend the seminar. You may register for Math 191 as a one credit course. To receive credit you must attend every meeting of the seminar (with at most one unexcused absence) and submit a written solution to a Putnam problem by the end of the semester.[/quote]

[quote="St Olaf"]
Students in the Seminar will work on problems which challenge them. These won't be the same problems for everybody. To get your grade for the course, you must satisfy the following requirements.

    * Buy the books and carry them which you everywhere you go.
          o Problem Solving Through Problems, Loren Larson.
          o How to Solve It, George Pólya.

    * Show up every week.

    * Participate in discussions on math-282 and math-probsolv mailing lists.

    * Keep a notebook or journal for the course. The graph-paper pads with the hard backs are nice.
          o Anything which you feel fits in with your progress in problem solving belongs in this notebook. That is, the notebook need not - and probably should not be limited to problems from this course.
          o Making up problems is also excellent practice for solving problems.
          o The notebook can be decorated with stickers from Tacoasis.

    * I would like you to take the Putnam exam, which is traditionally held the Saturday of Christmasfest. I am not comfortable requiring you to take a six-hour exam on a Saturday right before finals, although as an undergraduate I always found it to be a nice way to relieve stress. You are however required to prepare as though you were preparing to take the Putnam exam. Putnam participants will receive tori.

    * St. Olaf's Carlson Contest will take place October 7th and 8th. This is a team competition (up to three per team); you don't have to work with people from the Seminar.

    * The North Central Section of the Mathematical Association of America (NCS-MAA) holds its annual problem solving contest on Saturday, November 8. Teams of three again.

    * Hand in some stuff once in a while. What and how much will depend on your level of preparation coming in. But I would like at a bare minimum by the end of the semester to be comfortable saying that everyone in the Seminar knows how to use mathematical induction and the pigeonhole principle.

    * Brag to everyone about how freakin' smart you are.
[/quote]

I would check their website/talk to the head of the math department there.

[quote="rzsolt"]Interesting. I'm going to uwaterloo in september, I would love to attend such a class if they offer it. Any one know if there is?[/quote]

On [url=http://www.ucalendar.uwaterloo.ca/COURSE/course-CM.html]this[/url] page I found:

[quote]
CO 380 LEC 0.50	Course ID: 003901

Mathematical Discovery and Invention
A course in problem solving. 100 problems are studied. Problems are taken mainly from the elementary parts of algebra, geometry, number theory, combinatorics and probability.

[Note: Offered in spring term, even-numbered years.]
Prereq: MATH 135/145, 136/146, 138/148; Level at least 3A; Not open to General Mathematics students[/quote]

I would still check with someone in the mathematics department there to make sure this is the class you want to take.

These online classes are especially hard because it is hard to visually explain problems. Also, when in class it would be much easier to ask questions. Basically communication would be enhanced when learning through a class instead of an online "chat room." As to what I would like to be taught, I would want to learn how to increase my speed when working on problems. The topics should be pretty much the same as here.


geometry
incenter
trigonometry
inequalities
geometric inequality
inequalities solved

let $I$ be the incenter of a triangle $ABC$,prove that $IA^{2}+IB^{2}+IC^{2}\ge\frac{BC^{2}+CA^{2}+AB^{2}}{3}$

[quote="ac1072"]let $ I$ be the incenter of a triangle $ ABC$,prove that $ IA^{2} \plus{} IB^{2} \plus{} IC^{2}\ge\frac {BC^{2} \plus{} CA^{2} \plus{} AB^{2}}{3}$[/quote]
$ IA \equal{} \frac {r}{\sin{\frac {\alpha}{2}}} \equal{} \frac {s}{p\sin{\frac {\alpha}{2}}} \equal{} \sqrt {\frac {(p \minus{} a)(p \minus{} b)(p \minus{} c)}{p}}\cdot\sqrt\frac {2}{1 \minus{} \frac {b^{2} \plus{} c^{2} \minus{} a^{2}}{2bc}} \equal{} \sqrt {\frac {bc(p \minus{} a)}{p}}.$
Hence, $ IA^{2} \plus{} IB^{2} \plus{} IC^{2}\ge\frac {BC^{2} \plus{} CA^{2} \plus{} AB^{2}}{3}\Leftrightarrow$
$ \Leftrightarrow3\cdot\sum_{cyc}bc(b \plus{} c \minus{} a)\geq(a \plus{} b \plus{} c)(a^{2} \plus{} b^{2} \plus{} c^{2})\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^{3} \minus{} 2a^{2}b \minus{} 2a^{2}c \plus{} 3abc)\leq0.$
Let $ a \equal{} y \plus{} z,$ $ b \equal{} x \plus{} z$ and $ c \equal{} x \plus{} y.$ Then $ \sum_{cyc}(a^{3} \minus{} 2a^{2}b \minus{} 2a^{2}c \plus{} 3abc)\leq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^{3} \minus{} x^{2}y \minus{} x^{2}z \plus{} xyz)\geq0,$ which Schur.

According to Leibniz, MA2+MB2+MC2=TA2+TB2+TC2+3*MT2 where Tis the centroid and M is any point,so putting I instead of M gives the result.

For [i]any[/i] point M in the plane of triangle ABC, we have $ MA^{2} \plus{} MB^{2} \plus{} MC^{2}\geq\frac {BC^{2} \plus{} CA^{2} \plus{} AB^{2}}{3}$. In fact, substituting x = y = z = 1 into the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15558]Fairly General Geometric Inequality[/url], we obtain $ \left(1 \plus{} 1 \plus{} 1\right)\left(1\cdot MA^{2} \plus{} 1\cdot MB^{2} \plus{} 1\cdot MC^{2}\right)\geq 1\cdot 1\cdot a^{2} \plus{} 1\cdot 1\cdot b^{2} \plus{} 1\cdot 1\cdot c^{2}$, what simplifies to $ 3\left(MA^{2} \plus{} MB^{2} \plus{} MC^{2}\right)\geq a^{2} \plus{} b^{2} \plus{} c^{2}$, so that $ MA^{2} \plus{} MB^{2} \plus{} MC^{2}\geq\frac {a^{2} \plus{} b^{2} \plus{} c^{2}}{3}$, what is exactly what we wanted to prove.

  darij


In the Flower-city there are $ n$ squares and $ m$ streets, where $ m \geq n \plus{} 1$. Each street connects two squares and does not pass through other squares.
According to a tradition in the city, each street is named either blue or red. Every year, a square is selected and the names of all streets emanating from that square are changed. Show that the streets can be initially named in such a way that, no matter how the names will be changed, the streets will never all have the same name.

[hide]

Consider the number of street namings for which it is possible to get the namings from an initial naming where all the streets are named the same thing. We will show this number is less than the total number of street namings.

For each square, consider the number of times the square has been selected. If it is odd, call the square ON. if it is even, call the square OFF.

Note that the state of each square automattically determines the naming given the initial naming. This is because, if we assign numbers to every street so that streets are red if and only if their integer is odd, changing the state of a square is really just adding 1 to every street next to it, changing the state twice will not change any of the streets.

Therefore, the number of namings starting from an initial position (and being able to get back to the initial position) is equal to 2*2^n (The two is because we have two possible initial positions: all red and all blue.)

The total number of namings is (n+1)!, which is greater than 2*2^n when n is greater than or equal to 3.

If n is 1, then there can be no streets, and 0 < 1+1

If n is 2, then there can only be 1 street, and 1 < 2+1.

[/hide]

[quote="pythag011"]

The total number of namings is (n+1)!, [/quote]

I don't agree with this. Each street is named either "blue" or "red"...  :)

Oops. Change it to 2^(n+1)

But the number of street namings for which it is possible to get the namings from an initial naming where all the streets are named the same thing is less than this. I claim it is less than or equal to

2*(2^n-1)

this is because when everything is on, it is the same naming as when everything is off.


Rules are simple , every body should form a word with the letters of the word from the previous post.

Ya, repetitions of the letters are allowed!
All letters need not be utilized.
But no extra letter can be included.

Suppose if the previous word is "[i]milkman[/i]",
then you can form "[i]milk[/i]" or "[i]man[/i]" or "[i]link[/i]" or  "[i]nail[/i]" but not "[i]king[/i]" since "[i]g[/i]" is not there in "[i]milkman[/i]"

[i][b]In other words,  the set of the letters of a word must be a subset of the set made up of the letters contained in the previous word.[/b]
[/i]

I begin with

[b]Pneumonoultramicroscopicsilicovolcanoconiosis[/b] :!:

[hide]Ya the above word exists, see[url]http://en.wikipedia.org/wiki/Longest_word_in_English[/url][/hide]

wait... we can only make one word?? or more than one?? bcuz right now it's impossible to make just one word using the letters from Pneumonoultramicroscopicsilicovolcanoconiosis...

[b]One word[/b].
you just have to take each of a kind , not the same number of times.

I mean to say the set of the letters of a word must be a subset of the letters making the word in the previous post

Falcon1: volcanoes...

crops
canoes
digits

[b]primoris[/b]

[b]vocal[/b]

[b]indigestion[/b]

Well this thread is  dead for good,

According to the rules of FTW, I can't change the rules of the game,once it has started and if I don't change the rules this game's gonna end very soon.
 :P


trigonometry
function
limit
calculus
calculus computations

Evaluate the limit:

limit of (1-cosx) / (x) as x approaches pi/4

*non-calculator*

Just substitute $ x\equal{} \frac{\pi}{4}$.

So i get it simplified to [1- sqrt(2)/2] / (pi/4).

Can this be simplified anymore?

the problem could be a "little" more interesting if x approaches 0.

hello, i have
$ \lim_{x \to \frac{\pi}{4}}\frac{1\minus{}\cos(x)}{x}\equal{}\frac{4\minus{}2\sqrt{2}}{\pi}$
as the answer.


I am trying to use hyperref to create a hyperlink to a website with a tilde (~) in the url.  When I hold the cursor over the link, the url is displayed correctly in the pop-up box, but when I click to follow the link, the tilde changes into the string %7E.  Why does this happen, and is there a workaround?

I can't reproduce that behaviour so could you give a minimal working example where it happens? Is it the browser changing ~ to %7E? Which browsers have you tried? Some browsers are happy to use either ~ or %7E so there's not really a problem.

Strictly speaking %7E shoud be used instead of ~ [url=http://www.cs.tut.fi/~jkorpela/tilde.html]Why tilde (~) should not be used in Web addresses (URLs)[/url] but nobody takes any notice of that these days :)

[code]
\documentclass[12pt]{article}
\usepackage[pdftex,letterpaper]{hyperref}

\begin{document}
\href{http://thomas.loc.gov/cgi-bin/query/D?c107:4:./temp/~c107i5W8q3::}%
{http://thomas.loc.gov/cgi-bin/query/D?c107:4:./temp/\~{}c107Ao1cpz::}
\end{document}
[/code]

I'm using Firefox.

It appears to be a problem between the pdf reader and the browser. If you have a URL which does not have ~ after a ? then it works. So http://mysite.abc/Dxyz/temp/~mypage works fine.

I also tried making a pdf with your URL without involving LaTeX (I used a print to pdf program from Notepad) and I get the same problem so it is definitely not LaTeX but an Adobe Reader problem. In that case it might be best to ask about what to use in an Adobe forum. Give an example URL which causes the problem, since, as I say, it is only when ~ occurs in a query that you get the problem.

Update: Acrobat changes the text whether or not the tilde is in a query (in case it makes a difference, I am using Reader v8.1.0).


group theory
superior algebra
superior algebra unsolved

let G be finite simple group of order m and n be any natural no. s.t.(m,n)=1.
 show each element of G is product of n th powers of some elements of G.

We do not need that the group is simple, nor that we can multiply powers:

$x \mapsto x^{n}$ is a surjective map (in fact bijective).


Let $ ABC$ be a triangle, and $ D, E, F$ be the midpoints of $ BC, CA, AB$, respectively.

Prove that $ \angle DAC \equal{}\angle ABE$ if, and only if, $ \angle AFC \equal{} \angle ADB$.

Let $ G$ be the centroid of $ ABC$.
If $ \angle DAC \equal{} \angle ABE \equal{} \alpha$, then $ \angle FDA \equal{} \alpha$ since $ FD$ is parallel to $ AC$. Hence, $ \angle GDF \equal{} \angle ABE \equal{} \angle GBF$ making $ GDBF$ cyclic. Hence $ \angle GDB \equal{} \angle GFA$.
If instead $ \angle AFC \equal{} \angle ADB$ then $ GDBF$ is cyclic and so $ \angle ABE \equal{} \angle FBG \equal{} \angle FDG \equal{} \angle DAC$ ($ FD$ parallel to $ AC$), and we are done.


Given $ f(x) \equal{} x^2 \minus{} x \minus{} 1$ find $ f(\minus{}x)$.

Plugging in $ \minus{}x$ for $ x$, we get $ f(x) \equal{} x^2 \plus{} x \plus{} 1$


$ \frac {2^{n \plus{} 4} \minus{} 2(2^n)}{2(2^{n \plus{} 3})}$ when simplified is:

$ \textbf{(A)}\ 2^{n \plus{} 1} \minus{} \frac {1}{8} \qquad\textbf{(B)}\ \minus{} 2^{n \plus{} 1} \qquad\textbf{(C)}\ 1 \minus{} 2^n \qquad\textbf{(D)}\ \frac {7}{8} \qquad\textbf{(E)}\ \frac {7}{4}$

[hide]

This is equal to $ \frac{2^{n\plus{}4}\minus{}2^{n\plus{}1}}{2^{n\plus{}4}}\equal{}\frac{2^{n\plus{}1}(8\minus{}1)}{2^{n\plus{}4}}\equal{}\frac{7}{8}$ or $ \mathrm{ (D) }$

[/hide]


modular arithmetic
algebra
binomial theorem
number theory unsolved
number theory

prove that for every positive integers n ,there is an positive integers K that:

2^n | (19^k    - 97)

I don't understand you demonstration... :maybe:   can you explain me,please?

[quote="Rust"]If $ n\le 5$ work $ k \equal{} 0$. Let $ n > 5$ and exist $ k(n \minus{} 1)$, suth that $ 2^{n \minus{} 1}|19^{k(n \minus{} 1} \minus{} 97$.
If $ 2^n|19^{k(n \minus{} 1)} \minus{} 97$ we take $ k(n) \equal{} k(n \minus{} 1)$, else $ k(n) \equal{} k(n \minus{} 1) \plus{} 2^{n \minus{} 3}$.
\[ 19^{k(n \minus{} 1) \plus{} 2^{n \minus{} 3}} \minus{} 97 \equal{} (2^{n \minus{} 1} \plus{} 97)*(19^{2^{n \minus{} 3}}) \equal{} 97\mod 2^n \equal{} 0\mod 2^n
\]
.[/quote]

Sorry, but I dont understand why $ 19^{2^{n\minus{}2}}\equal{}1\pmod{2^n}$ (I know that $ 19^{2^{n\minus{}1}}\equal{}1\pmod{2^n})$

It's clearly true, so the demo is OK, but I dont understand why.

It is because the order $ k$ modulo $ 2^n$ (with $ k$ odd) is no greater than $ 2^{n \minus{} 2}$

The proof is actually fairly simple:  apply the binomial theorem to $ (1 \plus{} 2k)^{2^{n\minus{}2}}$.  But we can be more precise: it is in fact well-known that $ (\mathbb{Z}/2^n\mathbb{Z})^{*}$ is isomorphic to $ \mathbb{Z}/2^{n\minus{}2} \mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.

[quote="t0rajir0u"]The proof is actually fairly simple:  apply the binomial theorem to $ (1 \plus{} 2k)^{2^{n \minus{} 2}}$[/quote]

Ahh yes, thanks.  :blush:


find out all k in R for which we have:
a^4+b^4+c^4+abc(a+b+c)>=k(ab+bc+ca)^2!!!!!

bye bye

This one is again cute! :D

because the ineq must take place for any k in R. lets take a=b=c=1 and we get that k<=2/3 and in this case we get that 2/3(ab+bc+ca)^2>=k(ab+bc+ca)^2.
Now we have to prove that:

a^4+b^4+c^4+abc(a+b+c)>=2/3(ab+bc+ca)^2 for any a,b,c in R
this ineq is equivalent to:
3(a^4+b^4+c^4)>=2(a^2b^2+b^2c^2+c^2a^2)+abc(a+b+c) and it is true because a^4+b^4+c^4=sum((a^2)^2)>=sum(a^2b^2)=sum((ab)^2)>= sum((ab)(bc))=abc(a+b+c).
that's all! :D :D

tschuess! :D :D


function
trigonometry
complex analysis
complex analysis unsolved

I put this in matlab $ acos(1.2)$ Then it returned to me $ 0 \plus{} 0.6224i$ why? Shouldn't it return to me a message of error?
$ \minus{}1 \le cos x \le 1$

Arccosine is a multi-valued function defined to be
$ \cos^{\minus{}1}(z)\equal{}\minus{}i\log(z\plus{}\sqrt{z^{2}\minus{}1})$
Inserting $ z\equal{}1.2$, and choosing a suitable branch, we get
$ \cos^{\minus{}1}(1.2)\equal{}\minus{}i\log(1.863)\equal{}\minus{}.6224i\equal{}.6224i$ (Remember, this is an even function).


logarithms
calculus
number theory
relatively prime
calculus computations

Problem:

Prove that $ \log_2 5$ is irrational.

[hide="My Solution(non-calculus)"]
Let $ \log_2 5\equal{}x$ where $ x$ is a rational number.

Then by the property of a rational number, let $ x\equal{}\frac{a}{b}$ where $ a,b$ are relatively prime integers.

Then, $ \log_2 5\equal{}\frac{a}{b} \implies 2^{\frac{a}{b}}\equal{}5\implies 2^a\equal{}5^b$ which is impossible.[/hide]

This problem was in a calculus textbook.

How do you prove that $ \log_2 5$ is irrational using calculus?

Thanks. :)

This isn't a calculus question.  Your solution is fine.

My solution is exactly same as yours, FantasyLover.

THe problem appeares in Japanese university entrance examination every year.


i have a few math problems that i'm having trouble with, can anyone help me solve them
1. Rewrite the expression in terms of the first power of the cosine
sin^2xcos^4x
answer: 1/32(2+cos2x-2cos4x-cos6x)
2.Find the exact zeros of the function in the interval 0,2pie
g(x)=sin6x+sin2x
3. Verify the identity algebraically
sec2θ=(sec^2θ)/(2-sec^2θ)
4. Verify the identity algebraically
sin4β=4sinβcosβ(1-2sin^2β)

If I am not wrong.I think it should be moved to pre-olympiad. :wink:
[color=red]Moved (from Algebra Unsolved problems) to Intermediate Topics. -mathmanman[/color]

Do not double post. http://www.mathlinks.ro/Forum/viewtopic.php?p=849097#849097


logarithms
function
algebra unsolved
algebra

[quote="gunes"]with your help, i will learn the answer,too.
$a^{b}=b^{a}$ and $a=3^{b}$
$ab=?$[/quote]
This equation have not solution in natural numbers b ($a=3^{b}$).

yes, but b may be a real number, and I'm asking a*b=?

For it you must solve equation $b^{2}=3^{b}\log_{3}b$ and calculate $ab=b*3^{b}=\frac{b^{3}}{\log_{3}b }$.

I realized that  :) , but I can't solve it, I asked here if somebody solve it or find a good way to calculate a*b=?

I'm asking whether we express a*b in some form that can be calculated by using only pencil or not.
 :huh:
for example $a*b=3-\sqrt{3}$

It had only one solution ab~9.77, and can not expessed by elementary functions.

at least let me know why it can not be expressed by elemantary functions?
 :idea:
smth. like a proof.


1. Sa se rezolve in $M_3(N)$ ecuatia $A^3=I_3.$
2.Sa se arate ca daca $f$ este o funcite cu proprietatea lui Darboux si $g$ o functie reala astfel incat $\lim_{h\to 0}\frac{f(x+h)-g(x)}{h}$ exista si este finita oricare ar fi $x\in R$ atunci $f=g$.
3.Daca $A,B\in M_n(R)$ si $AB=A,BA=B$ atunci $A+B-I_n$ inversabila.
4.Fie $m,n$ impare . Sa se arate ca $[(a+b+c)^n-a^n-b^n-c^n][(a+b+c)^m-a^m-b^m-c^m]\geq 0, \forall a,b,c \in R$.

Da..... a 2a problema e de pe shortlist si a fost propusa de Mihai Piticari. Am fost si eu la concurs si am facut 27p  :D

Spix ,matematica te-a adus in 'halul ' asta,sa te dai cu capul de masa ?Daca te vad asa suparat,parca imi piere optimismul pt. mate.(sper ca sti de gluma )Iti doresc succese mai departe si...cat mai putine batai de cap !

Te asigur eu ca avatarul nu are nici o legatura cu matematica sau cu mine in general, reprezinta o stare in care te afli cand te blochezi mai ales la programare PhP, de asta nici nu am continuat cu PhP.

1. N-o sa-mi iasa in vecii vecilor. Probabil Spix l-a facut cu valori proprii.

2. $0 = \left( \lim_{h \to 0} h \right) \cdot \left( \lim_{h \to 0} \frac{f(x+h)-g(x)}{h} \right) = \lim_{h \to 0} ( f(x+h) - g(x) )$, deci $\lim_{h \to 0} f(x+h) = g(x)$. Daca $f(x) \neq g(x)$, atunci $f$ nu prea ar avea cum sa aiba IVP.

(E gresit? Pare prea scurta solutia pentru o problema din lista scurta.)

3. $AB = A | \cdot A \, \, \Longrightarrow \, \, ABA = A^2 \, \, \Longrightarrow \, \, AB = A^2 \, \, \Longrightarrow \, \, A = A^2$ si $B^2 = B$, deci $(A+B-I)^2 = I$.

4. Nici asta.

[quote="pureblack2003"]a doua a fost pe lista scurta de anul acesta de la ONM . ai fost si tu la concurs spix ? le-ai facut pe toate nu ? :) mie mi se par grele la prima vedere ...  :blush:[/quote]

Esti mai ironic din fire ?

radio, la care parte te referi ? a doua chiar a fost pe lista scurta de anu asta . si nu, nu i-ai dat o rezolvare completa . e ceva mai complex de atat, nici n-ai folosit faptul ca  $f$ are proprietatea lui Darboux. si la partea cu mi se par grele la prima vedere, nu glumesc :) una le gandesti la concurs, si alta le gandesti cu ditai berea in fatza si vorbind pe net cu domnisoare :P

intr-adevar, la 2 avem asa
$\lim_{h\to 0}f(x+h)-g(x)=0$, deci $\lim_{h\to 0}f(x+h) = g(x)$ , ceea ce inseamna ca f are limite laterale finite in orice punct . deci nu vom gasi discontinuitati de speta a II-a . dar mai stim din ipoteza ca f are proprietatea lui Darboux, deci f nu are nici discontinuitati de speta I . aha, deci nu are discontinuitati, deci este o functie continua. atunci , $\lim_{h\to 0}f(x+h) = f(x) = g(x)$
si cam asta e ... totul era sa observi ca $\lim_{h\to 0}f(x+h)-g(x)=0$ , si asta o deduceai ori cum ai zis tu, si la varianta ta nu m-as fi gandit, ori eventual prin reducere la absurd, pentru ca daca era diferita de 0, atunci aveai un caz de [finit/0], care oricum o sa dea plus sau minus infinit. asa , trebuie neaparat sa ai caz de nedeterminare [0/0] , pentru ca limita aceea sa aibe sansa sa exista si sa fie finita .

[quote="perfect_radio"]Daca $f(x) \neq g(x)$, atunci $f$ nu prea ar avea cum sa aiba IVP.[/quote]

Daca, de exemplu, $f(x) > g(x)$, atunci $f(x) > g(x) + \epsilon$, unde $\epsilon > 0$, pe o vecinatate a lui $x$ (fara $x$). Iei un punct de acolo si il iei si pe $x$; aici se foloseste proprietatea lui Darboux pt. obtinerea contradictiei.

Asta nu prea imi pare complex.

Poti sa fii mai explicit te rog? Adica daca am $f(x)>g(x)$ ,$f$ nu poate avea proprietatea lui Darboux?

Fie $x_0$ un număr fixat. Atunci $\lim_{x \to x_0} f(x) = g(x_0)$ (sunt două demonstraţii mai sus... pe coperta manualului meu de mate e demonstratia faptului ca $f$ derivabilă implică $f$ continuă - seamănă foarte mult cu ce-am scris în primul post). Dacă $g(x_0) > f(x_0)$, atunci există o vecinătate $V$ a lui $x_0$ astfel încât $f(x) > \frac{g(x_0) + f(x_0)}{2}$, $\forall x \in V \setminus \{ x_0 \}$. Fie $y \in V \setminus \{ x_0 \}$, $y<x_0$. Valoarea $t = \frac{\frac{g(x_0) + f(x_0)}{2} + f(x_0)}{2}$ nu este atinsă pe $[y,x_0]$, deşi $f(y)>t$ & $f(x)<t$.

Cam ce-a zis şi pureblack, dar mai mult mă bagă în ceaţă discontinuităţile astea, cu speţele lor, de zici că-s avocaţi.

E clar/corect acuma? Tu ai rezolvat-o altfel?

[quote="Marius Damian"][quote]1. Sa se rezolve in $M_3(N)$  ecuatia $A^3=I_3$.[/quote]
Din $A^3=I_3$, deducem ca $\det A=1$ si ca $\textrm{Tr} A=3$.[/quote]

M-ai pierdut de la $\textrm{Tr} (A) = 3$   :? Poti dezvolta?

nup nup nup . $Tr(A)$ poa sa fie si 0 . valorile proprii ale lui $I_3$ sunt 1,1,1 . deci A va avea valorile proprii (1,1,1) , sau ( intr-un alt subcaz ) $(1, \epsilon , \epsilon^2)$ $Tr(A)$ e suma valorilor proprii deci obtinem cazu cand $Tr(A)=0$ si cazul $Tr(A)=3$

[quote="Marius Damian"][quote]1. Sa se rezolve in $M_3(N)$  ecuatia $A^3=I_3$.[/quote]
Din $A^3=I_3$, deducem ca $det A=1$ si ca $Tr A=3$.
Conform teoremei Cayley-Hamilton, $A^3-3A^2+nA-I_3=O_3$.
Tinand cont din nou de $A^3=I_3$, obtinem ca $n=3$ si imediat, $A^2=A$.
Prin urmare, $A^3=A^2=A=I_3$, deci singura solutie a ecuatiei este $A=I_3$.[/quote]

[quote="perfect_radio"]M-ai pierdut de la $\textrm{Tr} (A) = 3$   :? Poti dezvolta?[/quote]

Ai dreptate, rationamentul nu e corect!
Am revizuit demonstratia si iata ce a iesit:

Din $A^3=I_3$, deducem ca $det A=1$.
Conform teoremei Cayley-Hamilton, $A^3-mA^2+nA-I_3=O_3$, unde $m=Tr(A)$.
Tinand cont din nou de $A^3=I_3$, obtinem ca $m=n$.
Apar acum doua cazuri:
I. $m$ nenul si atunci $A^3=A^2=A=I_3$, deci o solutie a ecuatiei este $A_1=I_3$.
II. $m=n=0$ si in acest caz matricea $A$ are una si numai una din formele:
\[ A_2=\left(\begin{array}{ccc}0&1&a\\b&0&1\\1&c&0\end{array}\right), A_3=\left(\begin{array}{ccc}0&p&1\\1&0&q\\r&1&0\end{array}\right).  \]
De aici, prin ridicare la cub si egalarea cu $I_3$, gasim $a=b=c=p=q=r=0$, deci ecuatia data mai are si solutiile:
\[ A_2=\left(\begin{array}{ccc}0&1&0\\0&0&1\\1&0&0\end{array}\right), A_3=\left(\begin{array}{ccc}0&0&1\\1&0&0\\0&1&0\end{array}\right).  \]
Sper ca acum nu mi-a scapat nimic.


Show that there are infinitely many primes congruent to $1\mod{3}$.

gcd(3,1)=1
then according to direchlet theorem

there are infintly many primes in the form
1+3n

hence
there are infinitely many primes congruent to $\ 1\ mod3$

Consider divisors of $x^{2}+x+1$, all exceot $3$ itself will work.

I assume countdownmath wants something that could be called "elementary" whatever this means...


analytic geometry
calculus
calculus computations

Dear friends,

I got a small doubt. Let $ (x_0,y_0)$ be the centre of a square of length $ r$. Let a point $ (x_1,y_1)$ be chosen on the square such that it makes an angle $ \theta$ with $ (x_0,y_0)$. How do we express $ (x_1,y_1)$ in terms of  $ (x_0,y_0)$ and $ \theta$?

Any help is appreciated.

Regards,

bhupala.

You have not described the problem with sufficient precision.  The following information must also be supplied:

Are the sides of the square parallel to the coordinate axes?

How do you measure the angle $ \theta$?  The line segment from $ (x_0, y_0)$ to $ (x_1, y_1)$ makes an angle $ \theta$ relative to what?  The positive $ x$-axis?

Yeah,

Sorry for that. Yes it is the line joining $ (x_0,y_0)$ and $ (x_1,y_1)$ also the sides of the square are parallel to co-ordinate axes. Centre of the square is $ (x_0,y_0)$.

bhupala.


Solve $ x^2\minus{}2x\minus{}\frac{\sqrt{3}}{2}\plus{}\frac{5}{3}\equal{}0$ for $ x$.

[quote="PhireKaLk6781"]Is the quadratic formula the only method of approach possible?[/quote]

Pretty much, yeah...

hello, alternatively you can write
$ x^2\minus{}2x\plus{}1\plus{}\frac{5}{3}\minus{}1\minus{}\frac{\sqrt{3}}{2}\equal{}0$
$ (x\minus{}1)^2\equal{}\frac{\sqrt{2}}{2}\minus{}\frac{2}{3}$
and therefore we get
$ |x\minus{}1|\equal{}\sqrt{\frac{\sqrt{3}}{2}\minus{}\frac{2}{3}}$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, alternatively you can write
$ x^2 \minus{} 2x \plus{} 1 \plus{} \frac {5}{3} \minus{} 1 \minus{} \frac {\sqrt {3}}{2} \equal{} 0$
$ (x \minus{} 1)^2 \equal{} \frac {\sqrt {2}}{2} \minus{} \frac {2}{3}$
and therefore we get
$ |x \minus{} 1| \equal{} \sqrt {\frac {\sqrt {3}}{2} \minus{} \frac {2}{3}}$
Sonnhard.[/quote]

That's the derivation of the quadratic formula. It comes from completing the square on $ x$ and solving. It's not really different...


Duke
college
geometry
MATHCOUNTS
probability
AoPS Books

Is anyone taking a DUKE TIP camp this summer?  I'm in seventh, going on eighth, grade.  Should I take Geometry or Algebra II.  My primary goal is MathCounts, and I will be taking Geometry next year anyway.  These are full summer courses.  So should I take this higher-thinkimg Geometry class, or take an Algebra II class that I could only take after MathCounts in 8th grade is over.

PS - I think the BEST way to study for MC is to pore through these forums!  I have learned more from this Fourm than from all of the AoPS books combined!

Edit:  If you're taking a DUKE TIP camp. let me know when.  It would be cool to be in the same class.....

DUKE TIP...
huh?

As for Geometry of Algebra II...
Geometry. I think that more points could be gained by taking geometry, but thats me.

On PS: I wholeheartedly agree

Duke Talent Identification Program is one of several talent searches for the different regions of the country.  Talented 7th graders are able to take the SAT through their program and later take enriching summer classes based on their scores. The classes are offered at Duke and at other colleges throught the country.

[quote="mathgeek2006"]I would suggest taking Geometry.  The probability that Geometry problems will show up on MathCounts is higher than for Algebra II.[/quote]

But this is an actual course, and I will take geometry next year anyway.  Is it still worth it to take it at Duke?  My way of looking at is that their might be two Algebra II questions on all of the 2006 MathCounts (chapter/state) that I would be able to because of Algebra II.  Geometry at Duke won't help AT ALL, because I will take it again, and our teacher teaches for MathCounts, meaning MathCounts starts after we finish our Geometry course. (We finish in late February!)

I'm in eighth grade. I took Algebra II at Duke last summer, and I am taking geometry right now. Algebra II never shows up on Mathcounts, but you can sure learn a whole lot more by taking it.

[quote="shake9991"]my best bet would be to take geometry. Reasoning: so that you learn the geometry stuff right away and have an easier time in school... hence more time for mathcounts. I'm attending Duke TIP but am doing Intro. to medical sciences... what campus are you going to and what term are you in?[/quote]

Davidson - Term 1

um... this is kind of irrevelant... but...
I took the TIP SAT this year. I'm also thinking of taking a camp, probably won't...
Did you take the SAT? What did you get?

SAT was 1430 (old version), bad, i know, but i'm only in 7th grade.


If Greg rolls four dice, what is the probability that he rolls more 1's than 6's?

Since the probability is symmetrical, we'll have to subtract the probability that the number of 1's and 6's are the same, and halve it.

[i]Case 1: No 1's and 6's are rolled.[/i]

This is simply $ \left(\frac23\right)^4$, or $ \frac{16}{81}$.

[i]Case 2: One 1 and one 6 are rolled; the other two dice rolled are the same.[/i]

This is $ \left(\frac{1}{6}\right)^3 \times \frac{2}{3} \times \frac{4!}{2}$, or $ \frac{1}{27}$.

[i]Case 3: One 1 and one 6 are rolled; the other two dice are different.[/i]

This is $ \left(\frac{1}{6}\right)^2 \times \frac{2}{3} \times \frac{1}{2} \times 4!$, or $ \frac49$.

[i]Case 4: Two 1's and two 6's are rolled.[/i]

This is $ \left(\frac16\right)^4 \times \frac{4!}{4}$ or $ \frac{1}{216}$.

The answer seems to be a little low?  Do I need Case 3?  Gah, I fail C & P.

[quote="ernie"]
[i]Case 3: One 1 and one 6 are rolled; the other two dice are different.[/i]

This is $ \left(\frac{1}{6}\right)^2 \times \frac{2}{3} \times \frac{1}{2} \times 4!$, or $ \frac49$.[/quote]

@ernie, Case 3 is incorrect.

You overcounted $4$ times.  Look at it this way:

There are $\binom{4}{1}=4$ ways to place the $6$, and then $\binom{3}{1}=3$ ways to place the $1$.  Then there are $\binom{4}{2}=6$ ways you can choose the two different numbers that are rolled and $2!=2$ ways to order the two other numbers.  So the number of ways one $1$, one $6$, and the other two dice are different can be rolled is $4\times3\times6\times2=144$.  So therefore the probability is:

\begin{align*}\frac{144}{6^4}&=\frac{144}{1296}\\&=\frac{1}{9}\end{align*}

The rest of your cases are correct.  Since the probability that we get more $1$'s than $6$'s is the same as getting more $6$'s than $1$'s, then the probability that we get the same amount of $6$'s and $1$'s added to the probability $2p$ that we get more of $1$'s or $6$'s equals $1$:

\[2p+\frac{16}{81}+\frac{1}{27}+\frac{1}{9}+\frac{1}{216}=1\]
\[2p=1-\frac{227}{648}\]
\[p=\frac{1}{2}\times{\frac{421}{648}}\]
\[p=\boxed{\frac{421}{1296}}\]

One thing I don't understand in the Alcumus explanation was that it mentioned how to choose [u]which[/u] dice to have the number be on. However, aren't all the dice the same, so the order or which number appears on which not matter?

Solving it with order mattering and solving it with order not mattering still results in the same answer.  Think about it.

In the [hide=alcumus solution]
We notice that the probability that he rolls more 1's than 6's must equal the probability that he rolls more 6's than 1's. So, we can find the probability that Greg rolls the same number of 1's and 6's, subtract it from 1, and divide by 2 to find the probability that Greg rolls more 1's than 6's. There are three ways Greg can roll the same number of 1's and 6's: he can roll two of each, one of each, or none of each. If he rolls two of each, there are $\binom{4}{2}=6$ ways to choose which two dice roll the 1's. If he rolls one of each, there are $\binom{4}{1}\binom{3}{1}=12$ ways to choose which dice are the 6 and the 1, and for each of those ways there are $4\cdot4=16$ ways to choose the values of the other dice. If Greg rolls no 1's or 6's, there are $4^4=256$ possible values for the dice. In total, there are $6+12\cdot16+256=454$ ways Greg can roll the same number of 1's and 6's. There are $6^4=1296$ total ways the four dice can roll, so the probability that Greg rolls more 1's than 6's is $\dfrac{1}{2} \left(1-\dfrac{454}{1296}\right)=\boxed{\dfrac{421}{1296}}$.

[/hide], I couldn't understand this part: "there are $4\cdot4=16$ ways to choose the values of the other dice" . Isn't there overcounting here? Doesn't this count rolls such as 164[size=50]1[/size]4[size=50]2[/size] and 164[size=50]2[/size]4[size=50]1[/size] twice? But doing this the [hide=casework way]
$\text{Case 1:}$ The two numbers shown are different
There are $4$ ways to pick the first one, and $3$ ways for the last; so there are $4\cdot3 = 12$ ways.

$\text{Case 2:}$ The two numbers shown are the same
There are $\binom41 = 4$ ways to pick which number is repeated and only one way to rearrange them, since they are indistinguishable.

Adding these two cases we get $12+4=\boxed{16}$
[/hide] results in the same answer: 16. So maybe we aren't overcounting there...

Would it be possible for someone to help me figure out why we aren't overcounting? Is it because (going back to the previous example with 1644) the 3rd 4 can't be the 4th 4? (If you understand that)

Thanks in advance,
Stephy

I missed this problem because I couldn't figure out if to multiply one of the numbers 1, 16, 256 or not. How can I easily remember which numbers to multiply?

we are counting the opposite of what we want, casework is messy. complementary does not give u any overcounting methods in this case'

[quote=stephy2003]I was so excited and thought someone was finally going to answer my question :dry:[/quote]

Same here :dry:

I felt the same thing, I check if my answer is right, it was.

Try to do the problem a different way.  See if you get the same thing.

P.S. Do you mean you got the problem wrong after checking your answer?

No, sometimes I do, but this time that was not the case, I was just suspicious if I got it wrong.

That's a great strategy, Richard always says that, but sometimes it isn't possible to do so.

[quote=totally_not_a_hacker]Re Stephy:
No it doesn't.  If you think about it, the first four means you're choosing a value for the first die, and the second four means you're choosing a value for the second die.  You'll never count $4_{1}4_{2}$ twice. because the only way for that to happen is when both dice say the same number, and when you think about the cycle: 1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 etc. it won't happen[/quote]

Why did u revive?

[hide=Wrong solution]We approach with casework:
Case 1: Four 1s and no 6s: There's obviously only $1$ way this can happen.
Case 2: Three 1s and one 6: There are $\dbinom{4}{1}=4$ ways this can happen.
Case 3: Three 1s and no 6s: There are $\dbinom{4}{1}=4$ ways to choose where the 6 goes, and 4 choices for the last dice, so we have $4\cdot 4=16$ ways this can happen.
Case 4: Two 1s and no 6s: There are $\dbinom{4}{2}=6$ ways to choose where the 1s go, and $4\cdot 4$ choices for the other two dice, so we have $6\cdot 4\cdot 4=96$ ways this can happen.
Case 5: One 1 and no 6s: There are $\dbinom{4}{1}=4$ ways to choose where the 1 go, and $4\cdot 4\cdot 4$ choices for the other three dice, so we have $4\cdot 4\cdot 4\cdot 4=256$ ways this can happen.
Adding up all these together and dividing by $6^4$, we have $\dfrac{373}{1296}$ as our answer.
[/hide]

Sadly, I missed the "Two 2s and one 6" case, but how could I have counted that?

The way I did it was very similar to your way:

Regular casework solution:

[hide]
Case 1: 1 one and zero 6’s.
There are (4 choose 1) = 4 places to put the 1 and 4^3=64 ways to place the remaining numbers (since each number can be anything but a 1 and a 6). Thus the total number of possibilities for this case is 4*(4^3)=4^4=256.

Case 2: 2 ones and zero or 1 6’s.
There are (4 choose 2)=6 places to put the two ones. From here, we can have zero or one sixes, so the number of ways to place the remaining numbers is (5^2)-1 = 24 (because there are 5 possibilities for each number (cannot be one) and we have to get rid of the case where there are two sixes). Thus, the total number of possibilites for this case is 6*24= 144.

Case 3: 3 ones and 1 or 0 6’s.
There are (4 choose 3)=4 places to put the three ones and then there are 5 possibilites for the last number because having just one six will still be less than the three ones. Thus, there are 4*5=20 possibilities for this case.

Case 4: 4 ones and zero sixes.
There is only (4 choose 4)=1 way that this can happen.

The total number of possibilities is obviously 6^4=1296, so our answer is (256+144+20+1)/1296 = 421/1296.
[/hide]

[quote=OlympusHero]Exactly @above, that's the easiest way.[/quote]

It was also the official solution. I'm happy I didn't waste my time bashing here :P. There was a problem like this on the AMC 10, I'm going to try to see if I can find it...

[url=https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_16]2010 AMC 12A/16[/url]

Just a comment: I think Adv. Probability w/ Combinations is the best topic for sure. The problems are really fun! :)

[quote=Valkoorsky18][quote=crosby87]you do not need to post 4 times in a row.[/quote]

it's likely that the other posts got deleted; i don't think he's just having a conversation with himself

(chances that you'll see this are impossible, though)[/quote]

(S)he was responding to the person above, but (s)he just didn't edit his posts.

[quote=618173][hide=hint]do 1/2(1-P(tie))[/hide][/quote]

Yeah clever solution


I remember last year from dissection of a frog that frogs have similar organs as humans.

Other than apes, are there any other animals that have similar organs as humans? I'm just wondering.  :)

Pigs. Usually if a pig is infected with a disease, it's likely a human will be soon too. (or so I read on business week)

Yeah I remember pigs now.

I think advanced biology class is dissecting pigs this year (my general biology class last year did frog ---> already died and among popular species of frog so no worries  :D )

I really like this question. The answer is...YES! There are lots of awesome creatures that have organs and organ systems similar to humans. Frogs, apes, pigs, cows all have body parts that are analogous to those of Homo sapiens. This is essentially how you can intuitively tell how close we are to other organisms.

Get a phylogenetic tree of organisms; it doesn't have to be detailed. Find humans and look at the different clades that surround them. You can expect the organisms in class Mammalia to be much more like humans than the rest of the vertebrates. You can expect the rest of the vertebrates to be closer to us than the rest of the chordates. You can expect the rest of the chordates to be more like us than, for example, the organisms in Platyhelminthes. This is the organization of life and it is really, really neat!

Of course, each organism is very special and different. Frogs, for example, appear to have the same exact organs that humans do. Take a closer look, say, at the heart, and you will notice that it has only three chambers! Why does it have only three chambers? Think about it a while. Each organism is highly specialized.

I think this is part of what makes biology the coolest science ever!

Thanks!

I'm glad my question was pretty interesting.  :)

Why does a frog heart have only three chambers..?

I think that depends on whether animal is cold-blooded or warm-blooded (that's not right term.. someone please correct this). I think warm-blooded animal has more chambers because they need to pump the blood and clean it and then move to the body.

I'm not sure on where these blood go and get cleaned.. But I'm pretty sure it has to do with cold-blooded or not.

Good thinking, silverfalcon. The reason lies here:

Mammals have four-chambered hearts. This keeps deoxygenated and oxygenated blood completely separate during circulation and is, therefore, more efficient for human respiration and lifestyle. It enhances the concentration of oxygen gas that gets delivered to tissues. This 4 chambered heart is a shared derived character to the mammals (although I believe crocodiles are one exception.) And, as silverfalcon noticed, mammals are endothermic.

A step closer to frogs, we will examine reptiles. Reptiles generally have a three chambered heart, but the heart is partially divided by an incomplete septum. This separates the oxygen-poor and oxygen-rich blood a little more than in frogs but a little less than in humans. Now think about the lifestyle of a reptile. Think about how they usually eat and how active they are. They are ectothermic- their heat from respiration alone cannot sustain a constant body temperature. Lizards and snakes and most reptiles in general are pretty lazy except when hunting or escaping something. You can kind of now see how, due to the lifestyle of reptiles, a 4 chambered heart would not necessarily be beneficially adaptive.

And now to the frog. The frog has two respiratory surfaces- skin and lungs. It is ectothermic from what I know and true frogs spend much of their time near water. The oxygen rich and oxygen poor blood don't completely mix even in the frog heart (that would be really dumb and inefficient) but they are together in a single ventricle. Having a three-chambered heart just makes sense for amphibians. After all, when is the last time you saw a frog run 30 mph for its dinner? If frogs suddenly went into an environment where activity and efficiency was of utmost importance, then natural selecter would favor a more efficient heart.

As a final example, take a look at birds. Birds are arguably the most active animals and it comes at a price (hey, no one said flying is free.) They also have a SUPER efficient respiratory system, more efficient even when compared to ours, and you guessed it, they have a four chambered heart. Interesting stuff, eh?

Yay.. I'm glad I got it right.  :D

By the way, great explanation. That's more than I learned last year in school AND internet.

 :clap2:  :clap2:  :clap2:  :clap2:  :10:  :winner_first:

I'll just start by saying that I am dissapointed with the mention of the phylogenetic tree and the apparent relationship of organisms that has been mentioned.  I'm sure you all believe in evolution as an answer to the age old question of where we came from, but I'm here to try to open a few eyes and catch a few ears.
     A lot of organisms in the world all have very similar organs within them.  All mammals, but not restricted to mammals, have lungs for exchange of gases in the blood.  But this doesn't prove common descent among organisms.  I noticed the comparison of hearts among mammals, crocodiles/alligators, and birds.  These organisms all have four chambered hearts.  It's a fact.  The four chambered heart is more effective for these animals than a three chambered one.  But does a frog have genes for a four chambered heart?  I highly doubt it.  No matter how long the frogs live on earth, they will never have a four chambered heart.  Natural selection doesn't add genetic information.  It simply weeds out the weaklings.  Natural selection actually decreases the gene pool by removing unwanted traits.  Once a gene or a trait is lost, it's gone forever.  If the entire population of frogs, that being all the frogs in the entire world, lost the trait or genes that express the red color in eyes, then no matter how many times the frogs mate, in no matter what combination, and no matter how many billion years go by, there will never be another frog with red eyes.

what about mutations?

That adds information.

[quote="camaroracer214"]Ok, mutations...show me a mutation that doesn't end in complete disaster.  One you have probably heard about is probably sickle-cell anemia, a mutation that changes the shape of blood cells from a nice round shape to a bent, sickle shape.  This mutations makes carriers incapable of contracting malaria, a life threatening disease that is carried by mosquitos.  But having sickle-cell anemia is also very deadly.  Many people die each year from sickle-cell anemia.  But like sickle-cell anemia, all mutations end up being very harmful and even deadly.  Evolution teaches that billions of years ago non-living dirt became ingredients to life.  From here, single cell organisms arose and evolved into the life we see around us.  So how many mutations would have to occur before a single celled prokaryoted evolved into a human being?  Beneficial mutations, if they actually exist, are so rare that it would take a lot longer than the proposed 4+ billion years for life to arise and evolve into what it is today.[/quote]

hmmm, well, my only claim was that mutations add information. I don't have an example of mutations adding good information outside of my experiments with genetic algorithms in computer science(which introduce randomness given constraints meaning you eventually end up with something that works well); I am guessing they work the same in biology which is what genetic algorithms are based off of. Though I think everything technically started out as a mutation and now that I think of it, polydactylism could be beneficial, you never know.

However, your example of sickle cell anemia and your subsequent extrapolation that "but like sickle-cell anemia, all mutations end up being very harmful and even deadly" is a logical fallacy. Just because sickle-cell anemia is harmful doesn't mean all mutations are harmful. Your later bit about beneficial mutations has no substance because I don't think you take into account the speed at which bacteria divides. Nor is there any external criteria to calculate whether something is "harmful" or "beneficial" other than to have it happen and see. So, I don't see how you get that the probability of something being harmful is greather than the probability of something being beneficial.

Not only that, but you picked a genetic mutation which completely moots your point. There's a reason sickle cell is so prevalent (.. prevalant? sp sorry) in African Americans- it actually IS a beneficial mutation! When combined with a normal RBC gene, the sickle-cell gene actually helps its carrier to survive maleria. Also, it is completely untrue that "no new genetic information is added to the gene pool," as anyone who has taken college genetics can tell you. And just so you know, my pet peeve is uneducated people who try to discuss topics without the knowledge to understand what they are talking about. It makes conversation extremely difficult because, to get my point across, I am forced to educate the other person, which should really not be my responsibility. It is your responsibility to educate yourself on the facts before you form an opinion.

Blumoose, you shouldn't be so hard on others.  If you are like me, you didn't know anything about everything until someone told you.  Why would you resent sharing with others what was shared with you?

I'm not sure that anyone has shown that sickle cell anemia is a mutation, so this example doesn't do what it is said, regardless of whether one believes that it is a benefit or detriment.

To discuss this subject efficiently, one must distinguish between data and information, which was the point made earlier.  One can easily see how gamma radiation can strike a one celled human, eliminating the fingernail gene (or the like) and produce a human that doesn't have fingernails.  One can even see how a mutation can restore a turned off gene and have expression of a gene that one's species hasn't used for generations.

The prior message asked for an example where a random change added information that was useful, not just data.  The example of microcondrea is not sufficient, as one can argue that this is not new information added to a species, it is just a symbotic relationship that orginated from an infection that passed through the egg cell to the offspring.

I am looking for a new specises or a new function that is inherieted.  I believe that the extent of human knowledge in this area is exactly zero.  I would like to be educated if someone is aware of any such finding . . . . . . . anywhere.

However, logically one must not leap that lack of proof of a theory like evolution is the same as proof of the lack of applicability of a theory like evolution.


induction
recurrence relation
algebra
Sequence
IMO Shortlist
IMO Longlist

For any positive integer $ x$ define $ g(x)$ as greatest odd divisor of $ x,$ and

\[ f(x) \equal{} \begin{cases} \frac {x}{2} \plus{} \frac {x}{g(x)} & \text{if \ \(x\) is even}, \\
 2^{\frac {x \plus{} 1}{2}} & \text{if \ \(x\) is odd}. \end{cases}
\]

Construct the sequence $ x_1 \equal{} 1, x_{n \plus{} 1} \equal{} f(x_n).$ Show that the number 1992 appears in this sequence, determine  the least $ n$ such that $ x_n \equal{} 1992,$ and determine whether $ n$ is unique.

[hide="approach"]prove by induction (or perhaps otherwise?) that if $ n$ is odd then $ x_{\frac {(n \plus{} 3)(n \plus{} 1)}{8}} \equal{} n$, and that it is the only time $ n$ appears in the sequence.  After that it follows immediately that 1992 appears at position $ \frac {258\cdot 256}{8} \minus{} 3 \equal{} 66045$, and that it is the only time it appears.  It also follows that every number appears in the sequence exactly once, so $ \eta(k) \equal{} f^k(1) \equal{} x_k$ is a bijection on $ \mathbb{N}\to \mathbb{N}$.[/hide]

[hide]Observation 1:
Say $ x_n \equal{} (2k\plus{}1)2^p$ where $ p$ is positive.
Then, $ x_{n\plus{}1} \equal{} (2k\plus{}1)2^{p\minus{}1} \plus{} 2^p \equal{} (2k\plus{}1)2^{p\minus{}1} \plus{} 2(2^{p\minus{}1}) \equal{} (2k\plus{}3)2^{p\minus{}1}$

Observation 2:
From repeated use of Observation 1, we can see that \[ x_n \equal{} 2^p \implies x_{n\plus{}k} \equal{} (2k\plus{}1)2^{p\minus{}k}\] given $ p \ge k$.

Observation 3:
It follows that if there exist $ n_1$ such that $ x_{n_1} \equal{} 2^p$, then there exist $ n_2 \equal{} n_1\plus{}p$ such that $ x_{n_2} \equal{} (2p\plus{}1)$.
Since $ 2p\plus{}1$ is odd, $ x_{n_2\plus{}1} \equal{} 2^\frac{2p\plus{}1\plus{}1}{2} \equal{} 2^{p\plus{}1}$.
Thus \[ x_{n} \equal{} 2^p \implies x_{n\plus{}p\plus{}1} \equal{} 2^{p\plus{}1}\]

Observation 4:
Since $ x_1 \equal{} 1 \equal{} 2^0$ is in the sequence, it follows $ 2^p$ is in the sequence for any $ p$, which in turn implies $ (2k\plus{}1)2^{p\minus{}k}$ is in the sequence for any given $ p \ge k$.

In fact, $ 1992 \equal{} (2*124 \plus{} 1)2^{127\minus{}124}$, so $ x_1 \equal{} 2^0$ so, $ x_{1\plus{}(1)\plus{}(2)\plus{}...(127)}\equal{}2^{127}$ so, $ x_{1\plus{}(1)\plus{}...(127)\plus{}124} \equal{} (2*124 \plus{} 1)2^{127\minus{}124} \equal{} 1992$

It's not hard to see that $ n \equal{} 1 \plus{} 1... \plus{} 124 \equal{} 8253$ is the smallest and the only value such that $ x_n \equal{} 1992$[/hide]

Any comments on correctness or how I could word it better are welcome.

Nice pattern finding problem. The following two claims are the key to the problem:

Claim: For all nonnegative integers $n,$ we have that $x_{\frac{n^2+n+2}{2}} = 2^n$.

Claim: Call a [i]block[/i] the numbers from $x_i$ to $x_k$, including $x_i$ and excluding $x_k,$ such that $x_i$ and $x_k$ are the consecutive powers of two of the sequence which have the form given in the first claim. We claim that between two adjacent blocks, the only thing that happens is the numbers in the previous block get multiplied by $2$, and there is one extra odd number at the end, which is just the next odd number after the previous odd number in the first block.

Both of these claims are easily provable by induction by just listing out the terms.

From here it is easy to find the index at which 1992 appears, and because 1992 has a unique prime factorization, it is also unique. :)


If circular arcs AC and BC have centers at B and A respectively, then there exists a circle tangent to both AC and BC and to line AB. If the length of arc BC is 12, what is the circumference of the circle?

A. 24
B. 25
C. 26
D. 27
E. 28

Sorry for not including a picture...you can find one here:
http://www.math.uncc.edu/~hbreiter/

Just scroll down to the AMC 12 2000. There are also a couple of other AMCs on that site and some old ASHMEs.

[hide]Draw the center of the small circle in, call it O; connect the line between A and O. It'll go through the tangent point on BC (common tangent, both lines perpendicular to same line at same point). Call that tangent point P. We have AO+OP= AP = AB
Drop the perpendicular from O to AB, call it D.

You have: (r is radius of small circle around O)
r = OP = OD
R= AB = AO+OP = sqrt( OD^2 + (1/2 AB) ^2  )  + OP^2

sqrt ( r^2 + (1/2 R)^2  ) = R

you can solve for R using the arc info of BC   ( BC is 1/6th of a circumference since AC=AB=BC makes ABC an equilateral)

and then you'd get r also; i got 27[/hide]

[quote="krustyteklown"][hide]connect the line between A and O. It'll go through the tangent point on BC (common tangent, both lines perpendicular to same line at same point). [/quote]
Could you please explain this part...I don't really understand why A,O and P are collinear.

It's a geometric principle.

If you have two lines perpendicular to another line at the same point, those two lines coincide...

So all points on either one of them are collinear. Hope that made some sense.


I just got my placement letter for Calgary today, and I was wondering if I'm going to see any other AoPSers there.

oo thats awesome, I had heard of it but didnt think I would get in so I didnt bother to apply.. cant afford it anyway  :P

I have heard good things about Shad Valley program. How hard is it to get into Shad Valley? What does your average have to be?

It's not very difficult to get in. My average was ~mid 90s, but my friend with mid 80s got in.

Marks are not the only thing Shad is looking at; there's a ton of other criteria.

I know that mark's are not the only thing- they have like six to seven different critierias (a very long application). I just wanna know if you have to be super smart or just smart to get into shad.


projective geometry
calculus
integration
geometry proposed
geometry

Find center of quadrant without using calculus.

[asy]pair O=(0,0), A=(3,0), B=(0,3), G=(4/pi, 4/pi);
draw(O--A, black);
draw(O--B, black);
draw(Arc(O, 3, 0, 90), black);
dot(G);
label("(0, 0)", O, SW);
label("(a, 0)", A, SE);
label("(0, a)", B, NW);
label("(0, a)", B, NW);
label("G", G, SW);[/asy]

[hide="Answer"]$ G\equal{}\left( \frac{4a}{3\pi}, \frac{4a}{3\pi} \right)$[/hide]

I think that we cannot solve this problem if we cannot use calculus...

[b]Gouldin's theorem[/b] : $ V\equal{}2\pi x\cdot S\ \implies\ \frac {2\pi a^3}{3}\equal{}2\pi x\cdot \frac {\pi a^2}{4}\ \implies\ x\equal{}\frac {4a}{3\pi}$

Proof of Pappus-Guldinus theorem needs calculus...
So this solution is inadequacy.

Whatever "center of quadrant" means if you cannot use integrals to define a center...

  darij

I think this problem　cannot　be solved　algebraic.

Because $ \Bbb{Q}\left(\frac {4}{3\pi}\right) \equal{} \Bbb{Q}(\pi)$ is transcendental extention of $ \Bbb{Q}$.