Can somebody please explain epsilon and delta arguments to me for calculus. In particular, I'm having trouble understanding what exactly an epsilon-delta argument is, and how you'd use it to, say, show that lim(x->3)(2x - 7) = -1

If x is between 2.9 and 3.1, then 2x - 7 is between -1.2 and -0.8.

More generally, if x is between [tex]3 - \delta[/tex] and [tex]3 + \delta[/tex] (for any [tex]\delta > 0[/tex]), then 2x - 7 is between [tex]-1 - 2\delta[/tex] and [tex]-1 + 2\delta[/tex].

Given an [tex]\epsilon > 0[/tex], choose [tex]\delta = \epsilon / 2[/tex].  Restating the previous paragraph, if x is between [tex]3 - \delta[/tex] and [tex]3 + \delta[/tex], then 2x - 7 is between [tex]-1 - \epsilon[/tex] and [tex]-1 + \epsilon[/tex].

The last paragraph is an epsilon-delta argument that [tex]\lim_{x \rightarrow 3} (2x - 7) = -1[/tex].  Basically, we showed that 2x - 7 is arbitrarily close to -1, provided that we choose x close enough to 3.  The epsilon is how much error we allow in the output.  The delta is how much error we allow in the input.  We choose delta after we are given epsilon.

[quote="A+MATH"]Can somebody please explain epsilon and delta arguments to me for calculus.[/quote]

What Ravi B. said. That is, the idea of epsilon-delta arguments is that you show the error in estimating the limit for an expression can be shrunk done as far as you like, so that we can say the limit is such-and-such quantity, because the expression will tend to a number as close to that quantity as we please, without getting less close.

I have some books at home that show graphic arguments for some limit problems: that is, if you consider a smaller and smaller horizontal expanse on a Cartesian graph (a more and more restricted range of x values) your point on a function graph will be "squeezed" into a narrower and narrower range of y values.

The power of epsilon-delta arguments comes from their being based on undoubtable algebraic manipulation, as in the example Ravi B. showed you. The creativity of epsilon-delta arguments, and why they weren't invented as early as when calculus was developed, comes from figuring WHICH algebraic relationship will relate epsilon to delta in finding the limit for a particular expression. The creativity necessary to prove epsilon-delta arguments convincingly comes from practice.

Have fun with your calculus.


geometry
incenter
symmetry
geometry unsolved

The incircle of triangle $ ABC$ is tangent to the sides $ BC,CA,AB$ at points $ D,E,F$ respectively. Points $ M,N,J$ are centers of incricles of triangles $ AEF,BDF,DEF$. Sohw that points $ F,J$ are symetrical with respect to line $ MN$.

I think there may be some typo in the problem
F,J are not symmetry W.R.T MN

[quote="limes123"]Hint: try to prove, that FMJN is a deltoid.[/quote]

Umm.. that's equivalent to the problem.

[quote="limes123"]Hint: try to prove, that FMJN is a deltoid.[/quote]
what does "that FMJN is a deltoid" mean?

There is a typo. M,N,J are the incenters of triangles AEF, [b]BDF[/b] and DEF. I know that is equivalent to the problem ;) try to use this
[b]Lemma[/b]
let ABC be a triangle, I it's incenter and M the intersection point of bisector A with the circumscribed circle. Then MC=MI=MB

http://en.wikipedia.org/wiki/Kite_(geometry)

notice M,N are on the incircle of triangle ABC
D,J,M    E,J,N are collinear
MJ=MF=ME      NF=NJ=ND


quadratics
Vieta
AMC
AIME
algebra
polynomial
AMC 12

Hi
i was doing a problem on the relationships between the roots of quadratic equations, then i started to wonder, is there such a relationship for cubic roots too? i think i've seen it somewhere but i really cant recall now

does any one know any specail relationships between the roots of a cubic equation?

i googled it and only found a really complex one called Cardano's formula, but i was really thinking about somethign simpler, like the x1+x2=-b/2a, and x1x2=c/a for quadratic equations,

does the relationship exist for cubic roots?

thank you!=)

[url]http://mathworld.wolfram.com/VietasFormulas.html[/url]

it will work for cubic, quartic, quintic, etc.

Yea, Vieta's completely pwnz. It's really useful on the AIME with polynomial questions.

thank you very much!

by the way, do you guys happen to know what the potential topics for amc12 are?
can i find them here at AoPS?

thx!=P

You can find records of past AMC questions at the top in the "Contests" section.

thanks
is there any official lists of topics for amc12?

No there is no official topic list. Just look for the patterns in the problems on previous competitions. They will teach you what to expect. And what you can't expect, you can try to improvise on using usual methods with some insightful applications (but that's kind of hard to do every time...).


calculus
integration
limit
logarithms
calculus computations

For a positive number $ n$, find $ \lim_{n\to\infty} n^2\int_{e^{\minus{}\frac{1}{n}}}^1 x^n \ln x\ dx$.

Okay, Kunny.
The answer is $ \frac{2}{e} \minus{}1$, right?

I agree with akech, but if kunny says it's wrong, may be he could post the correct answer.

How about $ \lim_{n \to \infty} n^2 \int_{e^{\minus{}n}}^{1}x^n \ln x \,dx$ ?

[quote="skylarth"][quote="misan"]
$ \equal{} \lim_{n\to\infty} n^2\int_{e^{ \minus{} \frac {1}{n}}}^1 \;\frac {d}{dn} \; x^n\; dx$

by $ MCT$,

$ \equal{} \lim_{n\to\infty} n^2\; \;\frac {d}{dn} \int_{e^{ \minus{} \frac {1}{n}}}^1 x^n \; dx$[/quote]

This step is wrong, is it not? The limits depend on $ n$. You cant say eg $ \frac {d}{dx} \int_1^x f(y,x)dy \equal{} \int_1^x \frac {\partial f} {\partial x}(x,y)dy$[/quote]

yes, it is !  :blush:  i was trying to be too smart.  :(  i should've said, by $ MCT$,

$ \equal{} \lim_{n\to\infty} n^2\; \left(  \;\frac {d}{dn} \int_{e^{ \minus{} \frac {1}{n}}}^1 x^n \; dx \underbrace{ \plus{}\left( e^{ \minus{} \frac {1}{n}}\right)^n\; \frac{d}{dn} e^{ \minus{} \frac {1}{n}}}_{this \;\;term\;\; was\;\;missing} \right)$

$ \left[ \text{using} \quad \frac{d}{dy} \int_{u_1}^{u_2} f(x,y)\; dx \equal{} \int_{u_1}^{u_2} \frac{\partial f}{\partial  y} \; dx \plus{} f(u_2, y) \frac{du_2}{dy} \minus{} f(u_1, y) \frac{du_1}{dy} \right]$

the [i]extra[/i] term will add

$ \equal{} \lim_{n\to\infty} n^2\; \left( e^{ \minus{} \frac {1}{n}}\right)^n\; \frac{d}{dn} e^{ \minus{} \frac {1}{n}}$

$ \equal{}\;\frac{1}{e} \lim_{n\to\infty} n^2 \; e^{ \minus{} \frac {1}{n}} \;\frac{1}{n^2}$

$ \equal{}\;\frac{1}{e} \lim_{n\to\infty}e^{ \minus{} \frac {1}{n}} \; \equal{} \;\frac{1}{e}$

hence, overall limit is $ \frac{1}{e}\minus{}1 \underbrace{\quad\plus{}  \;\frac{1}{e}\quad }_{i\;\; missed\;\; this \;\;earlier} \equal{}\boxed{\frac{2}{e}\minus{}1 }$

[quote="Euclid's Turtle"][quote="akech"]Okay, Kunny.
The answer is $ \frac {2}{e} \minus{} 1$, right?[/quote]

This is the right answer, I checked it quickly on Maple 11.[/quote]

I'm very sorry, [color=blue]akech's answer was right[/color]. :blush:

kunny

[quote="misan"]
$ \equal{} \lim_{n\to\infty} n^2\int_{e^{ \minus{} \frac {1}{n}}}^1 \;\frac {d}{dn} \; x^n\; dx$

by $ MCT$,

$ \equal{} \lim_{n\to\infty} n^2\; \;\frac {d}{dn} \int_{e^{ \minus{} \frac {1}{n}}}^1 x^n \; dx$[/quote]

What is $ MCT$?


Prove that $ \sqrt{8.000...01}$ is irrational

Do you mean :
$ 8.10^n\plus{}1$ is not  a perfect square for $ n>2$
or $ 8,0..01$ is not a perfect square.
It is quite easy problem.

no my friend.. this is not true.... because it is also allowed that that may be a square of the rational number... so you have to prove for rational numbers that is not the square of rational number... not that this is the not a perfect square.

let (8·10^n +1)/10^n=a/b. Note that a is coprime to b, so the fraction is irreductible. Then both a,b are perfect squares. This is true for b only when n is even. now, let a=x^2. then a-1=(x+1)(x-1)= 8·10^n. Since x-1 and x+1 differ in 2, and one of them must end in zero, then the other ends either in two or eight. If 10 is in both factors, we reach contradiction since x would be congruent modulo 10 to x+2.

Then 10^n is in only one factor. then the other factor must be eight, alone, so the other one must be only 10 so n=1. But n cant be odd, contradiction.


Suppose
$ f(x\plus{}y)\ge yf(f(x)),\forall x>0,y>0$ (*)

1) $ f(x)\le x\plus{}1$.
If $ f(a)\equal{}b>a\plus{}1$, Let $ x\equal{}a$ and $ y\equal{}b\minus{}a$ in (*), we have
$ f(b)\ge (b\minus{}a)f(b)>f(b)$. Contradiction!

2) There exist $ b>0,d>1$ such that $ f(y)\ge d(y\minus{}b),\forall y>b$,
Let $ a\equal{}f(f(1))$, we have
$ f(1\plus{}y)\ge ay,\forall y>0$.
So we can find $ f(b)\equal{}c>1\plus{}\frac1a$
For $ y>0$,
$ f(b\plus{}y)\ge yf(c)\ge y*a*(c\minus{}1)\equal{}dy$, where $ d\equal{}a*(c\minus{}1)>1$.
Hence,
$ f(y)\ge d(y\minus{}b),\forall y>b$.

From 1) 2), we have
$ y\plus{}1\ge d(y\minus{}b),\forall y>b$, which cannot be true.

Q.E.D.

Suppose that it exists a function $f:(0,\infty)\rightarrow (0,\infty)$ with $f(x+y)\ge yf(f(x)),\ (\forall)\ x,y>0$.
Firstly, I'm going to prove that $f(x)\le x+1\ (\forall)\ x>0$. Suppose the contrary, so it exists $x$ such that $f(x)> x+1$.
Then $f(f(x))=f(x+f(x)-x)\ge (f(x)-x)f(f(x))>f(f(x))$ which is a contradiction.
Take $x=y$ and we get $2x+1\ge f(2x)\ge xf(f(x))\Rightarrow 2+\dfrac{1}{x}\ge f(f(x))\Rightarrow 3>f(f(x))\ (\forall)\ x\in \Bbb{N^*}$.

[b]Case 1.[/b] The set $F=\{f(x)\mid x\in \Bbb{N^*}\}$ is bounded, so there is $m\in \Bbb{N^*}$ such that $f(x)\le m\ (\forall)\ x\in \Bbb{N^*}$.
This means that $m\ge f(x+1)\ge xf(f(1))$ for each $x\in \Bbb{N^*}$ which is impossible since $\Bbb{N^*}$ is unbounded.

[b]Case 2.[/b] The set $F=\{f(x)\mid x\in \Bbb{N^*}\}$ is unbounded, so it exists $x$ such that $f(x)$ is very big.
I will prove below that $(\exists)\ k\in \Bbb{N^*}$ such that $(\forall)\ n\ge k, n\in \Bbb{N^*}$ we have $f(n)\ge 3$.
Suppose the contrary, hence we have an infinity of $n$ for which $f(n)<3$, unless we can choose the greatest of them as $k$.
So the set $K=\{n,\in \Bbb{N^*}\mid f(n)<3\}$ is unbounded, therefore it exists a very large $n$ with $n< f(x)$ for such $n$.
But then $3>f(f(x))\ge nf(f(f(x)-n))$ which is a contradiction, so it exists such a $k$.
But now as $f(x)$ is very big, then $3>f(f(x))\ge 3$ which is a contradiction.

Q.E.D.

[quote="Finchy"]Show that for any function $ f: (0,\plus{}\infty)\to (0,\plus{}\infty)$ there exist real numbers $ x>0$ and $ y>0$ such that:

                                                    $ f(x\plus{}y)<yf(f(x)).$

[b]Dan Schwarz[/b][/quote]

Suppose exist a $f$ such that
$P(x,y): f(x\plus{}y)\geq yf(f(x)).$
If there exist $f(a)>a$ then $P(a,f(a)-a)\implies f(a)\leq a+1 \implies f(x)\leq x+1$ for all $x>0$.

So from that we get $f(f(x))\leq 1+\frac {x+1}{y}$ now $y\to\infty \implies RHS \to 1$
$\implies f(f(x))\leq 1$ for all $x>0$

Now suppose there exist $b$ such that $f(b)>1$
$P(f(b)-1,1)\implies f(f(b))\geq (f(b)-1)(f(f(1))\implies f(b)\leq 1+\frac {1}{f(f(1))}=c$
So finally we get $f(x)\leq c$ for all $x>0$ since $c>1$.
So $P(x,y)\implies f(f(x))\leq \frac {c}{y}$ (for a fixed $x$) but for sufficiently large $y$ it's absurd.

This is the first time I ever used unbounded condition in a function.
Suppose otherwise, that there exists a function $f : \mathbb{R}^+ \to \mathbb{R}^+$ such that
\[ f(x+y) \ge yf(f(x)) \ \forall x,y \in \mathbb{R}^+ \]
Let $P(x,y)$ be the assertion to the function above.
[b]Claim 01.[/b] $f$ is unbounded.
[i]Proof.[/i] Fix $x$, take $y \to \infty$.

[b]Claim 02.[/b] $f(x) \le x + 1$ for all $x$.
[i]Proof.[/i] We only consider the case where $f(x) > x$. Then $P(x, f(x) - x)$ (cancellation trick) gives us $f(x) - x \le 1$.

[b]Claim 03.[/b] $f(f(x)) \le 1$ for all $x$.
[i]Proof.[/i] By Claim 02, we know that
\[ yf(f(x)) \le f(x+y) \le x + y + 1 \rightarrow f(f(x)) \le 1 + \frac{x + 1}{y} \]
for a fixed value $x$. Take $y \to \infty$, we get $f(f(x)) \le 1$.

By the unbounded condition, there exists $a$ such that $f(a) > 1$.
$P(1, f(a) - 1)$ gives us
\[f(f(a)) \ge (f(a) - 1)f(f(1)) \]
By Claim 03, we have
\[ (f(a) - 1)f(f(1)) \le f(f(a)) \le 1 \rightarrow f(a) \le 1 + \frac{1}{f(f(1))} \]
This contradicts the fact that $f$ is unbounded as $f$ is bounded by a constant $1 + \frac{1}{f(f(1))}$, which is a contradiction.


After a master-degree it is possible to write a PhD.
Anybody who know how much you get paid when you write a PhD?

Stipends really vary from place to place - you're best off looking at the graduate portion of the websites of the universities you're looking at.

it also depends upon which field you're getting the PhD in...

Try [url=http://www.support4learning.org.uk/money/funds_postgrad.htm]here[/url] for information on funding in the UK.

As an example, at MIT Math, the stipend for a TA is $18,270 / year.
An NSF Graduate Student Fellowship is $19,500 / year (after tuition).

[quote]As an example, at ___ Math, the stipend for a TA is ____ / year.[/quote]

If you want a fairly thorough listing of of this sort of information for the U.S. and Canada, you can download [url=http://www.ams.org/employment/asst.pdf]this rather large .pdf file[/url] from the AMS.


inequalities
algebra
High school olympiad

Let a,b, and c be positive reals. Prove:

$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\ge (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

Expand: \[\sum \frac{a^{2}}{b^{2}}+2\sum \frac{a}{b}\geq  3+\sum \frac{a}{b}+\sum \frac{b}{a}\iff \sum \frac{a^{2}}{b^{2}}+\sum \frac{a-c}{b}\geq 3 \]

No time to post rest right now but I think I can prove the stronger one
\[\sum \frac{(a+b)(a-c)}{b^{2}}\geq 0\] which directly implies this one...

Anything complete or better than just multiplying out?

Wow, this is pretty easy for a MOP problem...

[hide="By employing donuts..."]We have:

$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\ge 3+\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{a}{c}+\frac{c}{a}$

$\Leftrightarrow \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\ge 9$

$\Leftrightarrow \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge 3$

which is clearly true by AM-GM. Note that in the second to third step, $a/b+b/c+c/a \not \le-3$ because $a,b,c > 0.$[/hide]

[quote="Karth"]$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\ge 3+\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{a}{c}+\frac{c}{a}$

$\Leftrightarrow \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\ge 9$

[/quote]

This step is wrong because it's irreversible

[quote="Karth"]Wouldn't it be reversible since $a/b+b/a \ge 2$?[/quote]

You would have to prove

$\Leftrightarrow \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\ge 9 \ge 3+\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{a}{c}+\frac{c}{a}$

which means you would have to prove $2\ge \frac{a}{b}+\frac{b}{a}$

In other words, you have conveniently weakened the bound that you are trying to prove. If we want to prove $a \geq b$ and we know $b \geq c$ and $a \geq c$, well, that does not solve our original problem.

[hide]Let $x=\frac{a}{b}$, $y=\frac{b}{c}$, $z=\frac{c}{a}$; obviously $xyz=1$. Then we have to prove
$\left(x+y+z\right)^{2}\geqslant 3+x+\frac{1}{x}+y+\frac{1}{y}+z+\frac{1}{z}$.
But
$\mathrm{LHS-RHS}= \sum x^{2}+2yz-1-x-\frac{1}{x}= \sum x^{2}+\frac{2}{x}-1-x-\frac{1}{x}=$ $\sum\frac{1}{x}\cdot\left(x+1\right)\cdot\left(x-1\right)^{2}\geqslant 0$
which ends the proof.[/hide]

Multiply by $(abc)^2$, so that we have $$(a^2b+b^2c+c^2a)^2\geq abc(a+b+c)(ab+ac+bc).$$ Because I am not smart, we expand: $$a^4b^2+b^4c^2+c^4a^2+2(a^2b^3c+b^2c^3a+c^2a^3b)\geq (\sum_{sym}a^3b^2c) + 3a^2b^2c^2.$$ Cancelling as many terms as possible, this becomes $$a^4b^2+b^4c^2+c^4a^2+a^2b^3c+b^2c^3a+c^2a^3b\geq a^3b^2c+b^3c^2a+c^3a^2b+3a^2b^2c^2\text{~~~~~(A)}$$ Note that by AM-GM, we have $$\frac{2}{3}(a^4b^2)+\frac{1}{3}ab^2c^3\geq a^3b^2c,$$ $$\frac{1}{3}(a^4b^2)+\frac{2}{3}ab^2c^3\geq a^2b^2c^2.$$ Adding these two together, $$a^4b^2+ab^2c^3\geq a^3b^2c+a^2b^2c^2 \text{~~~~~(C)}.$$ Then, (A) is just the sum of the cyclic variants of (C), so we are done.


For which $n$ do the equations have a solution in integers:  \begin{eqnarray*}x_1 ^2 + x_2 ^2 + 50 &=& 16x_1 + 12x_2 \\ x_2 ^2 + x_3 ^2 + 50 &=& 16x_2 + 12x_3 \\ \cdots \quad \cdots \quad \cdots & \cdots & \cdots \quad \cdots \\ x_{n-1} ^2 + x_n ^2 + 50 &=& 16x_{n-1} + 12x_n \\ x_n ^2 + x_1 ^2 + 50 &=& 16x_n + 12x_1 \end{eqnarray*}

Well, let's just say $n \geq 2$ or it's just weird.  We claim that any even $n$ will work. :)

Note that we can rewrite the equations as $(x_i - 8)^2 + (x_{i+1} - 6)^2 = 50$, for $i = 1,2,\dots , n$ where $x_{n+1} = x_1$.  Because $50 = 7^2 + 1^2 = 5^2 + 5^2$ are the only ways to write $50$ as a sum of two squares, we have one of the following for all $i$.

$\quad x_i - 8 = \pm 7, \quad x_{i+1} - 6 = \pm 1;$
$\quad x_i - 8 = \pm 1, \quad x_{i+1} - 6 = \pm 7;$
$\quad x_i - 8 = \pm 5, \quad x_{i+1} - 6 = \pm 5.$

Thus, $(x_i, x_{i+1})$ must be one of (15, 7), (1, 5), (15, 5), (1, 7) or (9, 13), (7, -1), (9, -1), (7, 13) or (13, 11), (3, 1), (13, 1), (3, 11).  Well, $(x_{i+1}, x_{i+2})$ must be one of these pairs as well, so $x_{i+1} = 1, 7$ for all $i$.  So what do we know?  Each $x_i$ is 1 or 7 and every pair of consecutive $x_i, x_{i+1}$ must be different.  Thus, $x_1, ... , x_n, x_{1}$ has to go $1,7,1,7,\dots , 7, 1$ or $7,1,7,1,\dots ,1,7$ implying that $n$ is even, as desired. :lol:

Even $(1,7)$ works but $(7,1)$ doesn't work, because in your list, there's only $(7,-1)$. In fact if you substitute $(7,1)$ to the original equation, it doesn't satisfy.

Basically, among all the pairs that you listed, you need to find cycles. I found only one cycle:
$(7,13) \to (13,1) \to (1,7)$

Which means $n$ must be divisible by 3.


Square $ ABCD$ is rotated
$ 90^\circ$ clockwise about its center,
and reflected over a diagonal line
determined by the lower left and
upper right vertices. The square
is then reflected over a horizontal
line through the center. What
point now corresponds to the
position originally occupied by $ B$?

[asy]size(75); pair A = (0,0), B = (0,1), C = (1,1), D = (1,0); draw(A--B--C--D--cycle); label("$A$",A,W); label("$B$",B,W); label("$C$",C,E); label("$D$",D,E);[/asy]

OK, when it rotates, the square looks like:

[asy]size(75); pair A = (0,0), B = (0,1), C = (1,1), D = (1,0); draw(A--B--C--D--cycle); label("$D$",A,W); label("$A$",B,W); label("$B$",C,E); label("$C$",D,E);[/asy]

When reflected over the diagonal, it becomes:

[asy]size(75); pair A = (0,0), B = (0,1), C = (1,1), D = (1,0); draw(A--B--C--D--cycle); label("$D$",A,W); label("$C$",B,W); label("$B$",C,E); label("$A$",D,E);[/asy]

Finally when flipped through a horizontal line that passes through the center, it becomes:

[asy]size(75); pair A = (0,0), B = (0,1), C = (1,1), D = (1,0); draw(A--B--C--D--cycle); label("$C$",A,W); label("$D$",B,W); label("$A$",C,E); label("$B$",D,E);[/asy]

Thus, we have the answer of $ \boxed{D}$.

What does it mean if the diagnal is defined by 2 points


wel the problem i have is that when i am logged out mathlinks home page is showing one time.when i log in another time is shown and both these times have a very large difference .
PLS HELP

I think it has to do with the fact that the time zone you see times in is a profile option, so it doesn't apply to guests. However, my time difference is only one hour (due to DST), so I don't know what is going on with your time.

It appears that AoPS is set to DST year-round, if I'm not mistaken. In any case, if you really want to bother, you could always set your time zone back one zone.

For example:

Eastern-->Central,
Central-->Mountain,
Mountain-->Pacific,
Pacific-->Alaskan,
Alaskan-->Hawaiian-Aleutian.

Unfortunately, if you're in awkward time zones like Newfoundland, there's nothing you can do about it.  :(
But then again, it should be really easy to just live with it. No matter how many hours you're behind or ahead, you usually would know the time of day anyway, and the minutes amount will (hopefully) always be right.

[quote="jmerry"]There's a profile option to choose your time zone and whether DST is active. This option must be manually changed each spring/fall, but it's still easier than switching time zones. When not logged in, you see San Diego time, no DST.[/quote]Actually on the mathlinks url you see the Central European time as a guest :)


counting
derangement
combinatorics unsolved
combinatorics

There is given set A of n elements, n is natural number.
How many bijections f ( permutations ) exist on set  A , that don't have fixed points ?
 :maybe:

Such a permutation is called a "derangement," and they have been discussed on the forum many times.  See, for example, http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157629


Hey,

 Are you going to be able to make it over to Walton tomorrow? I'd love it if you can. The MathCounts practice is from 4:30 until 6:30 in my room at Walton. My room is in the back of the building. If you are coming in from Roswell Rd. via Providence., come all the way past the main entrance and go until you are just past the building. Turn right into the school and go all the way past the school. Turn right into the small parking lot next to the school. There is a door into my room and my red Saab will be parked right next to it.

Thanks,
Tom

If you go over to a post by Xantos C. Guin [his username] you can hit the PM button and send it to him without the rest of us knowing.

Thanks. I knew that. I just didn't think about it. I hate to get into trouble!  :)


$ AA_1,BB_2$ are the angle bisectors of triangle $ ABC$. $ A_1B_1$ meets the circumcircle $ (ABC)$ at $ M,N$. $ AA_1$ cuts $ BB_1$ at I. Denote $ R,R'$ are radiuses of the circumcircle $ (ABC),(IMN)$ respectively. Prove that $ R'\equal{}R$

$ \angle BMI \equal{} \angle BMA \equal{} \angle BCA \equal{} \angle C,$
$ \angle BIM \equal{} 180^\circ \minus{} \angle BIA \equal{} 90^\circ \minus{} \frac{\angle C}{2},$
$ \triangle BMI$ is isosceles with $ BM \equal{} MI,$
with $ B, I, N$ collinear, circumcircles of $ \triangle IMN, \triangle BMN$ are congruent.

No! I,A,M [b]are no[/b]t collinear ! We can't solve it by the above proof.

It should be $ R^{\prime}\equal{}2R$ and it was discussed on http://www.mathlinks.ro/viewtopic.php?t=86561 .

  darij

Sorry, I thought $ M, N$ are intersections of $ AA_1, BB_1$ with $ (O)$ (to get $ R' \equal{} R$).

See also [url]http://www.mathlinks.ro/viewtopic.php?t=136301[/url].


In a multiple-choice test (only one answer is correct for every problem), one of the question was illegible, but the choice of answers, given below, was clearly printed. What is the right answer of that particular problem ?

$(a)$ All of the below
$(b)$ None of the below
$(c)$ All of the above
$(d)$ One of the above
$(e)$ None of the above
$(f)$ None of the above

[hide="Answer"]
Answer: (e)

(a) - (d) are not true. (f) is not true because (e) can be true. Thus, the answer is (e).
[/hide]


geometry
circumcircle
incenter
perpendicular bisector
geometry proposed

[quote="N.T.TUAN"]Given a triangle. A variable poin $ D$ is chosen on side $ BC$. Points $ K$ and $ L$ are the incenters of triangles $ ABD$ and $ ACD$, respectively. Prove that the second intersection point of the circumcircles of triangles $ BKD$ and $ CLD$ moves along on a fixed circle (while $ D$ moves along segment $ BC$).[/quote]
[img]10334[/img]
We have $ \angle BKD \equal{} 90\plus{}\frac{\angle BAD}{2}$ and $ \angle CLD \equal{} 90\plus{}\frac{\angle CAD}{2}$
Let $ E$ be the 2nd intersection point of the two circumcircles. Then $ \angle BED \equal{} 180\minus{}\angle BKD$ and $ \angle CED \equal{} 180\minus{}\angle CLD$ [color=black][b](*)[/b][/color]

$ \angle BEC \equal{}\angle BED\plus{}\angle CED$
$ \equal{} 360\minus{}(\angle BKD\plus{}\angle CLD)$
$ \equal{} 360\minus{}180\minus{}\frac{\angle BAD\plus{}\angle CAD}{2}$
$ \equal{} 180\minus{}\frac{\angle BAC}{2}$

So $ E$ belongs to the arc of a circle

The center $ O$ of this circle belongs to the circumcircle of $ \triangle ABC$ and to the perpendicular bisector of $ BC$
And it is with $ A$ on the same side of $ BC$ because we want $ \angle BOC \equal{}\angle BAC$

$ \star$ Note
Since $ BKD>90$ we get that the circumcenter of $ \triangle BKD$ and the point $ A$ is on different sides of $ BC$. We will say that the circumcenter is "below" of $ BC$ Similarly, the circumcenter of $ \triangle CDL$ is also below of $ BC$. Then their midpoint and consequently the point $ E$ are also below of $ BC$. So $ \angle BKD\plus{}\angle BED \equal{} 180$


AMC AIME geometry MATHCOUNTS USA(J)MO USAMO email

Post here if you're taking it (Wednesday) and post what you are doing in the final days to prepare.

I'm finishing up my study of AoPS vol. 2 (I'm on chapter 23) and sleeping, as well as doing as little homework as possible - on the day of the AIME I have the test, Improvisational Comedy class, lunch, and math team practice. :D

Edit: Whoo, 200 posts! :D

I need to find time to do some math problems between now and Wednesday.  Lately, I've been flooded with homework/tests. :(

I'm going for 10-12...

I hope it's not as *interesting* as the AIME I...

i am taking the AIME II. we were on spring break during AIME I. I basically know i am going to fail it, so i'm kinda just trying to learn the higher level math required for it right now. mathcounts really sucked my time up. oh, well. at least i have all of high school to try it.

i'm not taking AIME2, but the forum might be locked down for it tomorrow, so thought i'd get this in today:

good luck everyone! i hope you all do great! :thumbup:

but it's only monday?...i didn't know they closed it down that early *shrugs*. oh well. thanks.

my goal: $\geq 4$
(though it's likely not gonna happen!)

[quote="cognos599"]will the AIME 2 be anywhere as hard as everyone says the AIME 1 was. i hope not because i want to make it to USAMO even though i am only in 7th grade.[/quote]

This kind of question is entirely pointless, because we don't know any better than you do, and the people who wrote the test are not going to tell you.

[quote="Treething"]I'm studying algebra and counting a bit.

I'm just going to ignore every single geometry question on the AIME except the one or two freebie geo questions they give you. I'd rather spend my time eating the donuts they give you.[/quote]

I wish my school gave us donuts. At least this year I think I've convinced them to let us keep our scrap/test booklets. I had to email the AMC Director to convince them.

I am taking it. I have not been preparing for it lately because I have the Canadian Math Olympiad same day and it's more important.
I can probably get 10 if I don't make too many stupid mistakes but I will aim for 12 or 13.

I'm going to be taking it. Pretty much all day during school tomorrow I'll be going through AoPS 2 then after I get home I might work a few problems but I think I'll just take it easy.

I'm aiming for 10+ but I would be happy with making the floor.

[quote="Ignite168"][quote="sapphyre571"]I'm trying to get in some last-minute geometry practice and cram on new geom theorems because geometry is by far my weakest point. I already have a strategy: solve 8 problems, check over them relentlessly for at least a half hour, then continue on, come up with reasonable guesses for the ones I couldn't get, and keep checking.[/quote]

If your strategy is to do 8 problems shouldn't you be going deeper into topics you're good at (maybe counting, combinatorics, NT, I dunno) rather than going wide?[/quote]

I'm hoping that there will be no more than 4 geometry problems, and since the 15 on the AIME I was geometry, I am guessing the 15 on the II will not be, and I've only solved 1 #15 in my life (2003 II).  I'm shooting to get #1, 4 of 2-6, 3 of 7-11, and 1 of 12-15. I just don't want to fall below 8, or worse, below 7.

Good Luck Everyone!

Shoot for the Stars!

 :lol:

[quote="randomdragoon"]
Aim for the moons because even if you miss you will be among the stars...

...unless you fail to reach escape velocity, in which case you're back where you started.[/quote]
Yeah and also the moon is closer than the stars... and if you land in a star you will almost certainly die, whereas on the moon you might live...

I'll be taking it. To be honest, I haven't prepared basically at all for the AIME, other than going through the AIME I from this year. I'd rather focus on USAMO, because the difference between a 10 to 15 on AIME doesn't matter nearly as much as that between a 14 to 21 on the USAMO does.

I'm taking it.

My main goal is to make USAMO.

[quote="diophantient"][quote="randomdragoon"]Good luck everyone!

Aim for the moons because even if you miss you will be among the stars...

...unless you fail to reach escape velocity, in which case you're back where you started.[/quote]
I always heard the converse of that,
"Reach for the stars, even if you miss you'll still be at the moon"

Although that's assuming that stars>moon.
I think that can be proven easily, not sure though.
Applying AM-GM on m, o, o, and n, we see that
$\frac{m+2o+n}{4}\geq \sqrt[4]{moon}$
Now we just need to prove that
$256stars \geq (m+2o+n)^{4}$
Applying Fibonacci's Lemma on s, t, r, s, and a, we see that it telescopes to:
$t^{3}+a^{3}+r^{3}\geq 3tar$ which is just AM-GM on $a^{3}$, $t^{3}$, and $r^{3}$.

Q.E.D.[/quote]

Very nice proof! You have such 1337 math skillz! ;)


search
number theory
relatively prime
number theory unsolved

[quote="srinath.r"]Find all odd integers $ n > 1$ such that for all relatively prime divisors $ a,b$ of $ n$ ,even $ a \plus{} b \minus{} 1$ is a divisor of $ n$ .[/quote]

See :

http://www.mathlinks.ro/viewtopic.php?p=1598693#1598693
http://www.mathlinks.ro/viewtopic.php?search_id=125209069&t=205140

 :wink:


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be an infinite finitely generated nilpotent group. Let $<1> \neq H_{0}\subset H$ be proper subgroups
of $G$ with $[G: H] < \infty.$ Is it possible to find an [b]infinite[/b] sequence of subgroups of [b]finite index [/b]of
$G$, say $H_{1}, H_{2}, ...$, such that $H \supset H_{1}\supset H_{2}\supset ... \supset H_{0}$?

(Here $\subset$ means subset but not equal.)

Not in general... just take $H_{0}=H$ and it is not possible to find even a finite sequence of such groups...
With a similar argument one can see that the condition $|H: H_{0}|=\infty$ is necessary for such a sequence to exist.

Thanks [b]Gockel [/b] for responding.

[quote="Gockel"]Not in general... just take $H_{0}=H$ and it is not possible to find even a finite sequence of such groups...[/quote]
Obviously! It is in the problem that $\subset$ means subset but not equal.

[quote]
With a similar argument one can see that the condition $|H: H_{0}|<\infty$ is necessary for such a sequence to exist.[/quote]
I don't see why it's necessary to have $[H: H_{0}] < \infty$ considering the fact that $[G: H_{0}]$ might be infinity.

Actually, I think he means $[H: H_{0}]=\infty$. The sequence of indices $[H: H_{k}]$ is strictly increasing and bounded by $[H: H_{0}]$.

Yeah of course, that was a typo, I meant = not <


[hide]
The first term is 1
The second term is 1+2
The third term is 1+2+3
And so on.

We want: $\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$
[/hide]

[quote="akimatsu"]Find the equation for the sequence 1, 3, 6, 10, 15.... that gives the n-th number of the sequence[/quote]

[hide]I'll just use x instead and n means the 3 is the second n
x=n+n-1+n-2...when n-y=0[/hide]

[quote="akimatsu"]Find the equation for the sequence 1, 3, 6, 10, 15.... that gives the n-th number of the sequence[/quote]
[hide]They are triangular numbers.

$F_{n}=\frac{n(n+1)}{2}$[/hide]


ARML
geometry
rectangle
perimeter
Pythagorean Theorem

This is a pretty easy problem from ARML (I know that because I was able to solve it after the comptetition  :lol: ).  Here it is:

Two 3x4 rectangles overlap at the corner so that their sides are perpendicular.  The new shape has an area of 22 and a perimeter of 20.  What is the distance between the two corners where the rectangles meet?  Sorry I wasn't able to provide a picture.

[hide="hint"]Try using the Pythagorean Theorem  ;) [/hide]

Does it has a real solution.I can't find it  :wallbash: :wallbash_red:

yep it does.  [hide="hint"]draw a rectangle in the middle and find side lengths[/hide]

correct!  I used the same method with a few slight differences  :)

[hide]Is the answer $2\sqrt{3}$?

I found out the sum of the lengths of that rectangle and then the area so that is how I got my answer by clever manipulation.[/hide]

[quote="fireemblem13"][hide]$2\sqrt{3}?$ let a and b be the side lengths of the middle rectangle. then, finding the perimeter gives $28-2a-2b$ so, $14-a-b=10$ finding the area $24-ab$, $ab=2$ and $a+b=4$. solving gives $a=2+\sqrt{2}$ and $b=2-\sqrt{2}$ using the pythagorean theorem gives the diagonal to be $2\sqrt{3}$[/hide][/quote]
Once you get $ab=2$ and $a+b=4$, you can directly find the value of $\sqrt{a^2+b^2}$ (without solving for $a$ and $b$). Try it ;)


[quote="quykhtn-qa1"]By TRAN MINH VUONG:
Let a,b,c be real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} \frac {4}{3}$.Prove that:
$ \frac {(a \minus{} 1)^2(b \minus{} 1)^2}{(a \plus{} b \minus{} 1)^2} \plus{} \frac {(b \minus{} 1)^2(c \minus{} 1)^2}{(b \plus{} c \minus{} 1)^2} \plus{} \frac {(c \minus{} 1)^2(a \minus{} 1)^2}{(c \plus{} a \minus{} 1)^2}\geq \ \frac {1}{3}$[/quote]
Nobody interested in this inequality? :(  :(


Set $ A$ contains all the two-digit integers that equal the product of their tens and units digits divided by the quotient of their tens and units digits.  What is the square root of the product of all integers in Set $ A$?

Answer [hide]$ 30$[/hide]

I'm afraid I don't quite understand the question :?:

I tried wroting it out algebraically and i got messed up, are you supposed to guess and check?

I didn't find this too difficult. Maybe my problem solving skills have improved :P
[hide]
First, express the value of an element algebraically.
$ \dfrac {10ab}{\frac {10a}{b}} \equal{} 10a \plus{} b$ where a is the tens digit and b is the units digit
simplify that out and you end up with $ b^2 \equal{} 10a \plus{} b$
if you subtract b from both sides $ 10a \equal{} b^2 \minus{} b$
factor the b to get $ 10a \equal{} b(b \minus{} 1)$
now we know that the units digit times one less than itself is a multiple of 10, as a must be an integer.
in the range of 1-9, only two numbers satisfy that requirement, 5 and 6
plug in those values for b and you get $ a \equal{} 2$ when $ b \equal{} 5$ and $ a \equal{} 3$ when $ b \equal{} 6$
so your two numbers are $ 25$ and $ 36$
multiple and square root the result, your answer is $ \boxed{30}$
[/hide]

[quote="AlmightyBob"]I didn't find this too difficult. Maybe my problem solving skills have improved :P
[hide]
First, express the value of an element algebraically.
$ \dfrac {10a \plus{} b}{\frac {10a}{b}} \equal{} 10a \plus{} b$ where a is the tens digit and b is the units digit
simplify that out and you end up with $ b^2 \equal{} 10a \plus{} b$
if you subtract b from both sides $ 10a \equal{} b^2 \minus{} b$
factor the b to get $ 10a \equal{} b(b \minus{} 1)$
now we know that the units digit times one less than itself is a multiple of 10, as a must be an integer.
in the range of 1-9, only two numbers satisfy that requirement, 5 and 6
plug in those values for b and you get $ a \equal{} 2$ when $ b \equal{} 5$ and $ a \equal{} 3$ when $ b \equal{} 6$
so your two numbers are $ 25$ and $ 36$
multiple and square root the result, your answer is $ \boxed{30}$
[/hide][/quote]

I have a question.  How did you get from this step:

$ \dfrac {10a \plus{} b}{\frac {10a}{b}} \equal{} 10a \plus{} b$

to this step:

$ b^2 \equal{} 10a \plus{} b$.

Could  you explain the algebra, please?

Thanks.

Sure. Oops my bad. Made a mistake in my original post. :blush:  Should be $ \dfrac {10ab}{\frac {10a}{b}}$ Sorry for any confusion caused.
[hide]$ \dfrac {10ab}{\frac {10a}{b}} = 10a + b$
change division to multiplication and
$ 10ab\times {\frac b{10a} = 10a + b}$
Cancel out the 10a's and your left with
$ b\times{b} = 10a + b$
thus resulting in $ b^2 = 10a + b$[/hide]


Show that for each prime $p>3$ there exist integers $x,y,k$ satisfying the conditions $0<2k<p$ and $kp+3 = x^{2}+y^{2}$.

I'm sorry about my poor English!!

[mod.: edited]

Why open?
1. If $p=1(mod \ 4)$ exist $1<x\le (p-1)/2$, suth that $x^{2}+2^{2}=kp+3$ and $4k<p$.
2. If $(\frac{2}{p})=1$ exist $1<x\le (p-1)/2$, suth that $x^{2}+1^{2}=kp+3$ and $4k<p$.
3. If $(frac 3p )=1$ exist $1<x\le (p-1)/2$, suth that $x^{2}+0^{2}=kp+3$ and $4k<p$.
Else $p=11(mod \ 24)$. Consider all m, suth that $(\frac{m}{p})=-1$, then $(\frac{3 m}{p})=1$. If $(\frac{3-3m}{p})=1$, then exist $x,y\le (p-1)/2$, suth that $x^{2}=3m(mod \ p),y^{2}=3-3m(mod \ p)$, therefore $x^{2}+y^{2}=3+kp$ and $2k<p$. Else for all case m, suth that $(\frac{m}{p})=-1$ we have $(\frac{1-m}{p})=1$. It give Jacoby summ is $J=\sum_{m}(\frac{m}{p})(\frac{1-m}{p})=1-p$ - contradition.

The problem is from the second round of Polish MO 2003


The teacher whispers positive integer A to Anna, B to Brett, and C to Chris. The students don't know one another's numbers but they do know that the sum of their numbers is 14. Anna says: "I know that Brett and Chris have different numbers." Then Brett says: "I already knew that all three of our numbers were different." Finally, Chris announces: "Now I know all three of our numbers." What is the product ABC?

I honestly don't know what it means, but I'm hoping that you do.

By what Anna knows, she has an odd integer.
By what Brett knows, he has an odd integer $ \geq >  7$.
By what Chris knows, there has to be only one case left for Chris to pick.

Because we concluded that Anna and Brett both have odds, Chris has an even number.

Brett's number is larger than 7, so Chris's is less than 7.  If Chris has 6, then there is no other case other than A=1, and B=7.

Therefore, $ 6 \cdot 1 \cdot 7 \equal{} \boxed{42}$


induction
set theory
real analysis
real analysis theorems

Ok, here is another one of those things that have bugged me throughout my  study of mathematics. In many proofs, one describes a finite inductive process to construct finitely many objects $I_{1}, \ldots, I_{n}$. That is, we know how to construct $I_{1}$, and then given $I_{1}, \ldots, I_{n}$ we can describe how to create $I_{1}, \ldots, I_{n}, I_{n+1}$. Then we say "repeat indefinitely" to get a sequence $(I_{n})_{n=1}^\infty$. Examples of this process can be seen in the proofs I know of for example Heine-Borel, Goursat's theorem and recently the existence of a Hahn decomposition for signed measures.

However, I do not really understand how this repeat indefinitely-process works. And actually, I don't really think it feels natural. I like mathematics to be finite. Before I've always figured it comes from induction but I am too stupid to see it, so I have been afraid to ask anyone. However, at this point I feel I am capable of proving many tough theorems but I still don't see how this follows from the induction (of course my understanding of the foundations of mathematics is poor). Can someone explain what is up?

This is essentially the core of induction. Each case of induction is finite, but the whole isn't. The idea: If it's finite, we'll get there eventually. If our target is $I_{5}$, we run the process 5 times. If our target is the millionth $I$, we run the process a million times. If it's "all of them", that's induction.

Most mathematicians are willing to accept weirder stuff than this- have you ever seen transfinite induction?

I thought induction allowed me to get any finite collection $I_{1}, \ldots, I_{n}$. As I said, I don't know basic mathematics, but as far as I know induction lets me say "For any $n$ I can find a collection $I_{1}, \ldots, I_{n}$ with so-and-so properties". This is not the same as saying you can find a sequence. How exactly do you get the sequence?

No, I haven't seen transfinite induction.

That's different; you can't find a sequence without more to the argument. I thought all of your $I_{1}$ were the same.

There's probably some way that the old stuff is being included in the new stuff in these arguments; we're not just throwing it away.

No, I didn't mean to throw anything away. I think you have understood what I want to do correctly. We begin with an $I_{1}$ and then apply an inductive process to get a sequence. When I said "For any $n$ I can find a collection with so-and-so properties" I meant that this is what I get out of my understanding of "induction". I do not understand how to get the sequence. Could you show me how to apply induction to get a sequence?

In that case, your sequence is the $I_{n}$, and it works. What do you think is bad about the sequence?

It's not just that we can find a suitable collection for any $n$; it's that the sequence we construct in this way has the appropriate properties for every $n$.

The problem is that I don't understand why we can "repeat indefinitely" and get a sequence at all. When I think about it I buy it as just taking the union $\cup \{(n,I_{n})\}$. Intuitively, this should be a well-defined set since for each fixed $n$ we know what we mean by $I_{n}$. On the other hand, I feel that this is just hiding the issue in set notation that I don't understand anyway. I guess this whole thing is a strange thing to choose to doubt when I don't even know any axiomatic set theory. Not understanding mathematics has been starting to bother me too much lately..

[quote="Kalle"]I thought induction allowed me to get any finite collection $I_{1}, \ldots, I_{n}$.[/quote]

The basic induction axiom says the following:

For any formula $F(x)$ in the variable $x$, $\left(F(0) \wedge \forall n\left(F(n) \Longrightarrow F(n+1)\right)\right) \Longrightarrow \forall x F(x)$.

That is, if it's true at 0 and it's truth at $n+1$ follows from its truth at $n$, then it's true for all values.  There is no finiteness limitation in the consequence.

[i]You[/i] don't have to repeat anything indefinitely (although if you were only interested in getting some finite collection, you could repeat some finite number of times to get it).  The "indefinite repetition" is captured by the universal quantifier in $\forall n \left(F(n) \Longrightarrow F(n+1)\right)$.

Explanations are easier if you give a particular example of something that's causing you trouble.

I don't think that giving an example will help, since there is not much more to the problem than what I've described.

The assumptions are that we have some $I_{1}$ and that from every collection $I_{1},...,I_{n}$ we can construct a next element $I_{n+1}$. I do not understand how to formulate the $F$ you describe in such a way that we get the existence of a sequence from the conclusion $\forall x F(x)$. Maybe there is nothing more to the problem than to use the induction to describe that we get an $I_{n}$ for each integer $n$ and then we can consider the union of all of them.

As I said, maybe this is a strange thing to question (not having a basis in set theory or logic or anything), but I don't really understand it. And I don't think anything that has been said so far really deals with what I have an issue with. I do know, in a vague sense, what induction is, but I do not understand how to apply it to get a sequence $(I_{n})_{n=1}^\infty$. So far I don't feel that any reply has addressed this question directly at all.

edit: In trying to read about this I found the axiom of dependent choice. Now, I still don't really know enough to fully understand what this is about, but why would this axiom at all be necessary if there was no problem in just "repeating indefinitely"? Maybe this has something to do with there being choice though. However, this I have also seen many times in "repeating indefinitely". From $I_{1}$ you don't know exactly how to pick $I_{2}$, but you can pick it. Then you can pick $I_{3}$ from $I_{1}$ and $I_{2}$ and so on. Does this picking cause any extra difficulty?


calculus
integration
limit
symmetry
real analysis
real analysis unsolved

P1.
Let $ I_n  = \int\limits_0^1 {\frac{{e^{ - nx} dx}}{{1 + e^{ - x} }}}$, with $ n$ is natural number.
Compute $ {\lim }\limits_{n \to  + \infty } I_n$

P2.
Prove $ \forall n \in Z_ + ^*$, we have: $ I_n  = \int\limits_0^1 {(2x - 1)^{2n + 1} e^{x - x^2 } dx = 0}$

Best,

[quote="L_Euler"]P1.
Let $ I_n = \int\limits_0^1 {\frac {{e^{ - nx} dx}}{{1 + e^{ - x} }}}$, with $ n$ is natural number.
Compute $ {\lim }\limits_{n \to + \infty } I_n$[/quote]

Let $ f_n(x)=\frac {{e^{ - nx}}}{{1 + e^{ - x} }}$

Then $ f_n(x)\le \frac {{e^{ - x}}}{{1 + e^{ - x} }}$, which is integrable over $ [0,1]$

Therefore, by DCT, $ \lim_{n\to \infty}\int\limits_0^1 {\frac {{e^{ - nx}}}{{1 + e^{ - x} }}}dx=\int\limits_0^1 \lim_{n\to \infty}\left(\frac {{e^{ - nx}}}{{1 + e^{ - x} }}\right)dx=0$

Ad 2.: 0.5 is a symmetry point in the interval $ [0,1]$

[quote="L_Euler"]P2. Prove $ \forall n \in Z_ + ^*$, we have: $ I_n = \int\limits_0^1 {(2x - 1)^{2n + 1} e^{x - x^2 } dx = 0}$[/quote]

Here again DCT.

$ |{(2x - 1)^{2n + 1} e^{x - x^2 }| \le e^{x - x^2 }}$, which is integrable over $ [0,1]$

Therefore,  $ \lim_{n\to \infty}\int\limits_0^1 {(2x - 1)^{2n + 1} e^{x - x^2 } dx = \int\limits_0^1 \lim_{n\to \infty}\left((2x - 1)^{2n + 1} e^{x - x^2 }\right) dx=0}$

... we know already that the sequence is zero, thus the limit as well. ;)


Σε ένα νησί βρίσκονται ιππότες και κατάσκοποι   συνολικά  100 άτομα . Οι ιππότες λένε πάντα την αλήθεια και οι κατάσκοποι  λένε είτε  αλήθεια και είτε ψέματα .
Οι ιππότες είναι περισσότεροι από τους κατασκόπους .
Στο νησί βρίσκεται ένας detective ο οποίος  θέλει να μάθει ποιοι είναι ιππότες και ποιοι είναι κατάσκοποι .Αυτός κάνει την ερώτηση σε κάποιον για το τι είναι κάποιος άλλος .(δηλαδή κατάσκοπος ή ιππότης ) .Είναι δυνατόν με 150 τέτοιες ερωτήσεις να βρει πόσοι είνια οι ιππότες και πόσοι οι κατάσκοποι ??

Νίκο  (Ράπανε)  καλά τη θυμάμαι την άσκηση ?? Νομίζω καλά είναι  αλλά επιβεβαίωσε και εσύ.

O detective μπορεί να ρωτήσει κάποιον αν κάποιος άλλος είναι κατάσκοπος ή ιππότης (κατά τη γνώμη του,οπότε αν ο ερωτώμενος είναι ιππότης σίγουρα θα του πει αλήθεια-ενώ αν είναι κατάσκοπος δεν ξέρουμε αν θα πει αλήθεια ή ψέματα).
Ο detective προφανώς δεν συμπεριλαμβάνεται στα 100 άτομα.
Μπορεί να κάνει μόνο 150 ερωτήσεις!

Τώρα είναι εντάξει!!

Mia erwtisi: Enas kataskopos leei stathera alitheia i psemata , i mporei na leei kapoies fores
alitheia kai kapoies fores psemata?

Niko Rapane sigxaritiria gia tin Valkaniada stin Kipro kai kali epitixia stin IMO fetos.

Isws me thimase stin persini Valkaniada gia neous stin Veroia opou eimoun vathmologitis sto 4o thema. Distixws den thimamai pios apo olous ekei eisoun.

Pisteuw oti tha ta poume kai tou xronou stin Valkaniada pou tha diorganwthei an den kanw lathos edw stin Ellada mia kai nomizw oti einai i seira mas ;)

Alexandros Sygelakis

Y.G. Steile mou ena email sto cretanman@gmail.com  na sou steilw photos apo tin perisini valkaniada kai na thimitheis pios apo olous eimoun (an den me thimase).

O kataskopos leei oti goustarei!
Pote alitheia pote psemata!

Λύση:

 Αρχικά πρέπει να παρατηρήσει κάποιος ότι για να βρω πόσοι είναι οι ιππότες και πόσοι οι κατάσκοποι,
πρέπει κάποια στιγμή να βρω ένα ιππότη, γιατί οι κατάσκοποι μπορεί πάντα να λένε την αλήθεια!
Παίρνω τυχαία έναν τον A1 από τους 100. Αρχίζω και ρωτάω τους υπόλοιπους έναν έναν για το τι είναι ο A1.
Όσο το πλήθος αυτών που έχουν πει ότι ο A1 είναι ιππότης είναι μεγαλύτερο ή ίσο από αυτών που έχουν πει
ότι είναι κατάσκοπος , συνεχίζω να ρωτάω. Έτσι στο τέλος θα έχω κάνει 2x1 + 1 ερωτήσεις όπου x1 θα έχουν
απαντήσει ότι είναι ιππότης και x1+1 ότι είναι κατάσκοπος. Είναι εύκολο να δείξει κανείς ότι σε αυτή την ομάδα
των 2x1 + 2 (συμπεριλαμβανομένου και του A1) x1+1 θα είναι τουλάχιστον κατάσκοποι.(το x1 μπορεί να είναι 0)
Έπειτα παίρνω έναν A2 από τους 100-2(x1+1) που απέμειναν και κάνω την ίδια διαδικασία. Κάνοντας αυτή τη διαδικασία παρατηρώ ότι παραμένει το πλήθος των ιπποτών να είναι μεγαλύτερο από αυτό των κατασκόπων(από αυτούς που μένουν). Κάποια στιγμή , μιας και οι ιππότες θα είναι περισσότεροι , θα ρωτάω για τον Ai+1 τους
p = 100 - 2(x1+x2+...+xi + i) που μένουν και θα είναι συνεχώς το πλήθος αυτών που έχουν πει ότι ο Aι+1 είναι ιππότης  μεγαλύτερο ή ίσο από αυτών που έχουν πει ότι είναι κατάσκοπος. Έτσι όταν το πλήθος αυτών φτάσει το
p/2 τότε σίγουρα ο Αi+1 είναι ιππότης(έστω ότι έκανα τότε y ερωτήσεις, άρα y-p/2 είναι οι κατάσκοποι στην i+1 ομάδα).

Έτσι έχω βρει ότι ο Ai+1 είναι ιππότης και έχω κάνει 2(x1+x2+...+xi)+i+y ερωτήσεις. Αλλά επίσης γνωρίζω
ότι στην πρώτη ομάδα υπάρχουν τουλάχιστον x1+1 κατάσκοποι, στη δεύτερη x2+1 ... στην ιοστη xi+1 και άρα με
άλλες i ερωτήσεις θα έχω κάνει 2(x1+...+xi)+2i+y ερωτήσεις και θα ξέρω ότι έχω x1+....+xi+i+y-p/2 κατασκόπους
και τον Αi+1 ιππότη. Άρα μένουν 150 - (2(x1+x2+...+xi)+2i+y) ερωτήσεις που θα κάνω στον Ai+1 για
100 - (x1+x2+...+xi+i+y)-1+p/2 = 150 -2(x1+x2+...+xi+i)-y-1 άτομα (είναι τα άτομα κατά 1 λιγότερα από τις ερωτήσεις που απομένουν). Άρα μπορώ με 150(μπορώ με 149) ερωτήσεις να βρω πόσοι είναι κατάσκοποι και πόσοι είναι ιππότες!

Την άσκηση αυτή την ξαναπροσπάθησα πριν 2 μέρες αλλά δεν είχα αποτέλεσμα. Η κύρια ιδέα μου ήρθε όταν ήμουν
γυμναστήριο...Θέλω να πω ότι οι διαγωνισμοί και η έμπνευση είναι και θέμα ημέρας-ικανότητας συγκέντρωσης εκτός από θέμα προπόνησης και ταλέντου!


If you've changed your profile on the USAMTS website, then prizes will be sent to your new address.  Whatever address is in the profile is the address that we use.

If you're worried, you can PM me your USAMTS ID#, and I'll double check for you.

[quote="mathfanatic"]I recently changed my mailing address (and made the appropriate changes on the profile part of the USAMTS website).  How can I ensure that the USAMTS sends prizes to the new, updated address, rather than an old one that was in the system up until last week?[/quote]

We haven't sent them yet - if you go to the Profile page and it shows your new address, you're fine.


geometry
angle bisector
similar triangles
geometry proposed

In $ \triangle{ABC}$, $ D$ and $ E$ are two points on side $ BC$ with $ E$ between $ B$ and $ D$. $ \angle{DAC}\equal{}\angle{B}$. $ M$ is the midpoint of $ AB$. $ ME$ and $ AD$ meets at $ F$.

Prove that $ AE$ is the angle bisector of $ \angle{BAD}$ if and only if $ AE \parallel CF$.

Let $ D,\ E$ be, the points on the side-segment $ BC$ of the given triangle $ \bigtriangleup ABC,$ such that $ \angle DAC \equal{} \angle B$ and the line segment $ AE$ is the angle bisector of the angle $ \angle BAD.$

We will prove that $ CF\parallel AE,$ where $ F\equiv AD\cap ME,$ and $ M$ is the midpoint of the side-segment $ AB.$

Applying the [b]Menelaos theorem[/b], in the triangle $ \bigtriangleup ABD,$ taken as its transversal the line segment $ ME,$ we have that $ \frac {EB}{ED}\cdot \frac {FD}{FA}\cdot \frac {MA}{MB} \equal{} 1$ $ \Longrightarrow$ $ \frac {EB}{ED} \equal{} \frac {FA}{FD}$ $ ,(1)$

Because of $ AE$ is the angle bisector of $ \angle BAD,$ we have $ \frac {EB}{ED} \equal{} \frac {AB}{AD}$ $ ,(2)$

From the similar triangles $ \bigtriangleup BAC,\ \bigtriangleup ADC,$ we have $ \frac {AB}{AD} \equal{} \frac {AC}{CD} \equal{} \frac {CE}{CD}$ $ ,(3)$

$ ($ because of the isosceles triangle $ \bigtriangleup CAE,$ from $ \angle AEC \equal{} \angle B \plus{} \angle BAE \equal{} \angle DAC \plus{} \angle EAD \equal{} \angle EAC$ $ ).$

From $ (1), (2), (3)$ $ \Longrightarrow$ $ \frac {FA}{FD} \equal{} \frac {CE}{CD}$ $ \Longrightarrow$ $ CF\parallel AE$  and we are done as for the part [b](a)[/b] of the proposed problem.

$ \bullet$ We consider now, the points $ D,\ E$ on $ BC,$ such that $ \angle DAC \equal{} \angle B$ and $ CF\parallel ME,$ where $ M$ is the midpoint of $ AB,$ $ F\equiv AD\cap ME$ and we have to prove that the line segment $ AE,$ is the angle bisector of the angle $ \angle BAD.$

We suppose that $ AE'$ is the angle bisector of $ \angle BAD$ and let be the point $ E',$ between $ B,\ E.$

Let be the point $ F'\equiv AD\cap ME'$ and based on the proof of the part [b](a),[/b] we have that $ CF'\parallel AE'$ $ \Longrightarrow$ $ \angle AE'C \equal{} \angle E'CF'$ $ ,(4)$

But from $ CF\parallel AE$ $ \Longrightarrow$ $ \angle AEC \equal{} \angle ECF$ $ ,(5)$

It is easy to show that $ \angle E'CF' > \angle ECF$ $ ,(6)$ because of the point $ F$ is between $ D,\ F',$ as a consequent from the point $ E$ is between $ D,\ E'.$

So, from $ (4), (5), (6)$ $ \Longrightarrow$ $ \angle AE'C > \angle AEC$ but, this result is [b]atopo[/b] (= contradiction ).

Hence, we conclude that $ E'\equiv E$ and the proof of the proposed problem is completed, because we can prove this result by the same way, also when the point $ E'$ has been considered between $ E,\ D.$

Kostas Vittas.


trigonometry
algebra
system of equations
algebra proposed

If $ x$ , $ y$ and $ z$ are real numbers find all solutions to the following system of equations:

$ {[\cos(x) \cos(y)]}^{2} \equal{} {[\cos(y) \cos(z)]}^{2} \equal{} {[\cos(z) \cos(x)]}^{2}$

$ {[\sin(x) \sin(y)]}^{2} \equal{} {[\sin(y) \sin(z)]}^{2} \equal{} {[\sin(z) \sin(x)]}^{2}$

Put $ a \equal{} \cos^2x,b \equal{} \cos^2y,c \equal{} \cos^2z$ to get

$ ab \equal{} bc \equal{} ca$

$ (1 \minus{} a)(1 \minus{} b) \equal{} (1 \minus{} b)(1 \minus{} c) \equal{} (1 \minus{} c)(1 \minus{} a)$

$ (1 \minus{} a)(1 \minus{} b) \equal{} (1 \minus{} b)(1 \minus{} c)\land ab \equal{} bc\implies a \equal{} c$, hence $ a \equal{} b \equal{} c$

$ \cos^2x \equal{} \cos^2y \equal{} \cos^2z\iff \cos 2x \equal{} \cos 2y \equal{} \cos 2z$, hence

$ 2y \equal{} \pm 2x \plus{} 2k\pi, 2z \equal{} \pm 2x \plus{} 2l\pi$, so the solutions are $ (x,\pm x \plus{} k\pi,\pm x \plus{} l\pi)$ where $ x\in\mathbb{R}$ and $ k,l\in\mathbb{Z}$


[quote="espark52"]how many poitive perfect square factors does the following have?:
2^5*3^5*5^2[/quote]

[hide]$3*3*2=\boxed{18}$[/hide]

[hide]3 ways for the 2, 3 for the 3, and 2 for the 5, so $3 \times 3 \times 2 = \boxed{18}$[/hide]

[quote="kyyuanmathcount"][quote="b-flat"]SplashD, please post your solution.

[hide="my answer"]For a number to be a perfect square, each of its prime factors must have an even exponent (so it can be divided by 2 when the square root is taken).

There are 3 choices for the powers of 2 (0, 2, 4), 3 choices for the powers of 3 (0, 2, 4), and there are 2 choices for the powers of 5 (0, 2).  This makes $3\times3\times2=18$ perfect square factors.[/hide]

espark52:  Surround your equation with dollar signs and use \times for $\times$ to make $\LaTeX$.[/quote]

The star works too, it results in : $1*1$, the times is $1\times1$.
The following link teaches you LaTeX.
[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url][/quote]
Actually, that's the link to download LaTeX.  :wink:  I think this one would be a better link:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5261
Download the file in the first post, which is a .pdf tutorial on how to use LaTeX.

Ha, now I know how to do these problems :D (thanks Neal  :) )
[hide]We want perfect squares, so to get that we want even powers. for $2^{5}$, there is $2^{0}, 2^{2}, 2^{4}$ and etc. for the other numbers, so we have $3\times 3\times 2=\boxed{18}$[/hide]

[quote="espark52"]how many poitive perfect square factors does the following have?:
2^5*3^5*5^2[/quote]

[hide]The perfect squares for this is

$2^{0}$, $2^{2}$, and $2^{4}$
$3^{0}$, $3^{2}$, and $3^{4}$
$5^{0}$, and $5^{2}$.

Any form of $2^{x}3^{y}5^{z}$, is a perfect square as long as it is chosen from the list above.

So the answer is $3*3*2=18$.

$18$[/hide]

[quote="nebula42"][quote="kheg_k"]but isn t $2^{0}$ the same as $3^{0}$ and $5^{0}$?[/quote]
Yes, it is, and multiplying the three of them together ends up in a total product of $1$, one of the perfect square factors of the number in the problem.[/quote]
oh i see


Anyone knows a good conics textbook for creating AIME/AMC problems? I have Volume II for this website but it is very limited. Any one want to suggest any?

Title: Geometry Page.5 to Page. 44

Authors: David A. Brannan, Matthew F. Esplen, Jeremy J. Gray

Publisher: Cambridge University Press

and

[url]http://www.komal.hu/cikkek/dandelin/dandelin.e.shtml[/url]


Let N = 9 + 99 + 999 + ... + 99...99, where the last number is composed of 99 9's.  What is the sum of the digits of N?

i did it for the first few and realized that it stays 9 until we add 99,999,999,999 when it equals to 111,111,111,099 which is 27, it stays 27 as we add the other numbers because the 99 becomes less by one and the number of 1's increases.

ml2 wrote:it stays 27 as we add the other numbers because the 99 becomes less by one and the number of 1's increases.

It might be slightly easier to find the actual sum rather than look for a pattern?

Just a thought: [hide]9 = 10 - 1, 99 = 100 - 1, ... [/hide] and summing them up shouldn't be too difficult.

Edit: [hide]It doesn't stay 27 forever.[/hide]

Not forever, but for at least another 99 numbers.. (until the last two digits reach 0.

ml2, are you saying the answer is 27? That's definitely not what I got.

N = [hide]10 + 100 + ... + 10^99 - 99 = 11...10 - 100 + 1 = 11...011

There are 99 1's, making the sum of the digits 99? [/hide]


Find all real-valued functions $ f$ on the positive reals which satisfy $ f(x \plus{} y) \equal{} f(x^2 \plus{} y^2)$ for all $ x, y$.

1)$ f(x)\equal{}f(\minus{}x)$
2) for $ y\equal{}\minus{}x$ then $ f(2x^2)\equal{}f(0)$
So $ f(x)\equal{}f(0)\equal{}c,\forall x\in R$

another way to show that $ f$ is constant is to prove that for all $ a\in\mathbb{R}^{\plus{}}$ and $ b\in (a\sqrt 2,2a]$ , $ f(b) \equal{} f(a\sqrt 2)$:

let $ x \equal{} \frac {a \minus{} \sqrt {b \minus{} a^2}}{\sqrt 2}$ and $ y \equal{} \frac {a \plus{} \sqrt {b \minus{} a^2}}{\sqrt 2}$ we get that:

$ f(a\sqrt 2) \equal{} f(b)$

QED


Let a,b,c be positive real numbers such that a+b+c>=abc.
Prove that at least two of the following three inequalities hold:
1) 2/a+3/b+6/c>=6
2)2/b+3/c+6/a>=6
3)2/c+3/a+6/b>=6
enjoy.

This one was already posted and solved! I think hxtung posted it!


induction
search
combinatorics proposed
combinatorics

Prove that $ n\geq 3$ points in a plane determine at least $ n$ straight lines.(not all points are collinear)

[hide]I found one approach by induction. Any other possible proofs ?[/hide]

I forgot to mention that not all points are collinear. I have edited it. Thanks.

[quote="dgreenb801"]Well the number of ways to choose two points out of n is nC2=n(n-1)/2, cleary at least n for $ n \ge 3$. But the problem is not true if the points are collinear.[/quote]

You need to address the problem that many of those $ \frac{n(n\minus{}1)}{2}$ lines might coincide.

Induction is quite good. Base case $ n\equal{}3$ is obvious. Assume $ k$ points determine $ k$ straight lines. Adding another point gives us another straight lines, hence we have $ k\plus{}1$ straight lines. ;)

However, we can consider the worst case, that is, $ n\minus{}1$ points are collinear. These collinear give a single straight line. Each of these points can be joined to the lonely point, and hence become $ n\minus{}1$ different lines. So we have $ n$ points. :)

[quote="Johan Gunardi"]Adding another point gives us another straight lines, hence we have $ k \plus{} 1$ straight lines. ;)[/quote]  No: if the first four points are the vertices of a square and the fifth point is the center, no new lines are created.  I believe that this problem has been posted before (and is moderately famous ... solved by Erdos and/or some of his crowd, maybe?), but I wasn't able to find it with the search function.  Perhaps you'll have better luck :)

Induction works, but it's not so trivial. I will use the following famous problem (I'm sure it has been posted before) proved only in the seventies I think, and it is a good example of how an open problem can still have a beautiful, simple solution.

If we have a finite number of points in the plane with the property that each line passing through $ 2$ of them passes through at least $ 3$ of them, then all points lie on a line.

Using this it's clear that we have to have a line which passes through only $ 2$ points. Pick one of these points, and apply induction to the others (if they are not all on a line, but then it's trivial). Then by attaching the chosen point we will necessarily gain a new line.

This problem happens to be [url=http://planetmath.org/encyclopedia/DeBruijnErdHosTheorem.html]Erdos De Bruijn theorem[/url].

I found a proof by induction and another without induction that proves both Erdos De Bruijn theorem and the [url=http://en.wikipedia.org/wiki/Sylvester%E2%80%93Gallai_theorem]Sylvester Gallai theorem[/url]. I am going to publish it elsewhere and provide a link here later.

Erdos's original paper is [url=http://www.renyi.hu/~p_erdos/1948-01.pdf]here[/url].


Often when given an equation or system of equations with some variables (say x, y, z) we can say WLOG that x <= y <= z. I think we can do this because of symmetricness, equality in permutations of the variables, or something like that. But would someone mind posting an official-looking, formal reason why we can do this in some cases?

Well, basically we can do something like that if there is nothing to distinguish the variables from each other. If you change all the xs to ys and vice versa, do you end up with exactly the same problem? If you do, take the minimum of those two, and change it around if you need to until its called x. Then you have x<=y. etc etc. hope that helps slightly.


given directed graph. u have to find out the minimum number of vertices using which u can go from a to b and then back from b to a. that means, u have to select minimum vertices so that there is a loop contaning those vertices!

Any help?

Its quite hard to know if a particular solution you come up with is actually fast enough or not.
I did overhear someone from one team who knew someone from *another* team who had solved it, and said something along the lines of this:
First use Dijkstra both ways to get an upper bound on the answer.
Then start from the start and just do a DFS, somehow checking at each stage if theres no point in going on or not.

Now, thats rather vague. I did come up with the following idea a few days after the competition, which sounds like it may agree with the above..

First use Floyd-Warshall to work out the distance between any two nodes. As above, we can now get an upper bound on the answer as (dist from start to end) + (dist from end to start).

We then do a DFS starting from the start. At each step, we calculate the following value for the current node, which I'll call 'back':
For each node X visited earlier in the current DFS, take the distance from current node to X, plus X.back, -1. Take the minimum of all those values.
Basically that means 'back' is the best way to get back to the start, using some or none of the previous nodes in the current DFS.

Then for the current path, a lower bound for the number of steps required is (current number of nodes visited + distance from final node (b) to one of previous nodes in DFS, plus 'back' for that node), and an upper bound is the same thing, plus the distance from the current node to b. And I'm probably forgetting some + or - 1s somewhere.
So whenever the lower bound gets bigger than the best solution so far, we backtrack, and whenever the upper bound gets smaller than the best solution so far, we update that to be the best.

Probably makes no sense to anybody reading this, but thats the only way I can think of. No idea how fast it is.

If shell, then use tsort. Else preform a topological sort and find a loop. Cut the loop. Then sort the new graph to find a different loop. If it is bigger and a superset of the origional loop it is possible will cut two loops at once, so eliminate the possibility. If it is smaller or not a subset then go again with the smaller one. Eventually the process forms a tree, in which all loops have been cut, so the smallest loop will be found.

Another algorithm requires some parallel processing. Go from a start node down the graph, visiting every possible path. Set a target distance. When the distance is reached and a loop has been hit, compare the sizes of the loops found. If no loop found, raise the distance.

well... watson, i don't know what is shell is is DAG? Directed acyclic graph?

can u plz describe ur topological short method? [i know topological short=tsort] how r u using tsort to find loop? it would be better if u give some example of ur this process...

[ur second idea is time consuming right?]

I agree, I don't get the tsort method at all. I mean, I can see how you can use tsort to get a loop, but I don't get how you 'cut the loop', what 'it is possible will cut two loops at once' means, or how either of the two given vertices are worked into the solution. Can you explain in more detail?

[quote="ar2rd"]What's the input size?[/quote]

Up to 100 nodes.

One solution to your problem would be to apply the Dijkstra algorithm 2 times . The solution that follows will not work or will not give the appropriate result when you graph is a particular tree . What i mean is

   a) all of the edges from the oriented graph should have the same cost let's say X
   b) Apply Dijkstra Algorithm from point A (the base) to point B (the destination) (Path P1)
   c) Remove the  visited edges and in this way you can eliminate a potential cycle that exists in the oriented graph
   d) Check to see if the graph is still conex if not the algorithm will fail because you will not be able to find an appropriate path from B-A (appropriate means a path different from the initial path from A-B). But I think that in the hypothesis is mentioned that there is a cycle in the given graph that contains the vertices A and B.
   e) If the graph is still conex apply again the Dijkstra Algorithm but this time from B to A and you will have a minimum path let’s say path P2 different from P1 but it has the property that it is the minimum path from B to A .
  f) By concatenating the 2 vectors P1 and P2 you will have a complete cycle from A – B – A

I hope this solution will help

tsort is a unix utility for preforming a topological sort. It was once used for ordering files in dependancy order. If no topological sorting is found, it emits a loop. Cutting the loop would mean deleating an edge in the found loop. I didn't read the problem carefully. I thought they ment finding a minimal loop.

[quote="Watson"][quote="TripleM"]That doesn't find the cycle using the minimum number of [b]vertices[/b], though, which is what is wanted.[/quote]
Yeah, but I can't go through a vertex without going on an edge into it and out of it. Remeber V/2=E?[/quote]

No, because you can go in and out of a vertex several times, but it only counts once in this problem.
This is one of the hardest world finals problems - its not going to be that easy!


Consider a parallelogram $ABCD$ with center $O$. Consider an arbitrary point $P$ in the semi-plane determined by the line $AB$ which does not contained $C$. If $M$ is the midpoint of $AP$, and $N$ is the midpoint of $BP$, and we know that $CM$ and $DN$ intersect each other at $Q$, find $\frac{PQ}{PO}$

Observe $\triangle ACP$. Segments $CM$ and $PO$ are its medians, hence their intersection point $Q_{1}$ divides both of them in the ratio of $2: 1$.

Observe $\triangle BDP$. Segments $DN$ and $PO$ are its medians, hence their intersection point $Q_{2}$ divides both of them in the ratio of $2: 1$.

Since both $Q_{1}$ and $Q_{2}$ divide $PO$ in the ratio $2: 1$, they coincide as the point with such property must be unique. Denote both $Q_{1}$ and $Q_{2}$ by $Q_{0}$. From $Q_{0}=Q_{1}\in CM$ and $Q_{0}=Q_{2}\in DN$ it follows that $Q_{0}$ is the intersection point of $CM$ and $DN$, hence $Q_{0}=Q$. Conclusion: $Q$ also belongs to $PO$ and divides it in the ratio $2: 1$, which means that ${PQ\over PO}={2\over 3}$.


Hi All,

I saw this transformer simulator (http://www.ngsir.netfirms.com/englishhtm/Transformer.htm) and it seems to suggest that the voltage ratio (and therefore current) between the primary and secondary windings at any point of time (with sine wave AC) is constant. Is this correct? If so then if the primary winding has rectified (with rectifier bridge and no filtering capacitors or inductors) AC current going through, will the secondary winding have the same output?
Sorry I don't really know proper terms so I may have said stuff wrong.

Thanks

[quote="serialk11r"]I read a bit more into some stuff, and I'm getting that magnetic flux is proportional to current, and emf is proportional to the change in magnetic flux (and therefore current). Is this correct?[/quote]

Yes, it is true. First one s due to relation $ \Phi \equal{} LI$, and second one is due to Faraday's Law $ E \equal{} \minus{} \frac {d\Phi}{dt}$.


I got this from some competition but I can't remember which.

1. Let's say that there exists a function $k(x) = (x^2-7x+9)(x^2-5x+9)(x^2-6x+9)$. Then $k(x)$ intersect the x-axis for how many points?

2. This is really easy one. How many integers n are there such that 100<n<1000 that it is divisible by 7?

P.S. For possible copyrights, I have changed the wording.

1.
[hide]We want to see how many different roots there are for k(x).  The first part is not a square, so it has 2 roots.  The second one cannot be factored so it doesn't have any for it to equal 0, making it have 0 roots.  The third part is a square, so it only has 1 root.  When you add them up, you get that it crosses the x-axis 3 times.[/hide]

2.
[hide]The answer is just [(1000-100-1)/7] = [899/7] = 128.[/hide]

1.[hide]
You want to find how many real solutions there are.  For the first part there are 2 solutions, for the 2nd the discriminant is negative so there are none.  For the last both solutions are the same, so there are [b]3[/b] possible ones.[/hide]

2.[hide]
there are 898 integeres between those, so so 898/7=128.28...and we round down giving us [b]128[/b] possible numbers for n.[/hide]

[quote="eryaman"]
2.[hide]
there are 898 integeres between those, so so 898/7=128.28...and we round down giving us [b]128[/b] possible numbers for n.[/hide][/quote]

Right answer, but not quite the right reasoning.
[hide="An example why"]
How many multiples of 7 are there between 10 and 15? the answer is not 5/7 rounded down to 0.
[/hide]

Hm..
[quote]The answer is just [(1000-100-1)/7] = [899/7] = 128. [/quote]

How do you do that?

why do you subtract 1000-100 by 1?

and why do you divide it?

and why do you divide the 899 with7?

Well, there are 899 numbers to choose from since 100<n<1000, which I got by  doing 1000-100-1.  Then I divided it by 7 since that will say how many numbers are divisible by 7.  The []'s mean that you just take to integer part of the number.  For example, [3.4] = 3, [7.999] = 7.

As I pointed out to eryaman, that reasoning isn't quite true. Look at the counterexample I provided in the above post, and see if you can fix up the reasoning a bit.


A natural numbers $\displaystyle n \geq 2$ is called [i]friendly[/i] if there is a family of non-void subsets $\displaystyle A_1,A_2,\ldots,A_n$ of the set $\displaystyle \left\{ 1,2,\ldots,n \right\}$ such that:

1) $\displaystyle i \not\in A_i$ for all $\displaystyle i = \overline{1,n}$;

2) $\displaystyle i \in A_j$ iff $\displaystyle j \not\in A_i$ for all distinct $\displaystyle i,j \in \left\{ 1,2,\ldots,n \right\}$;

3) $\displaystyle A_i \cap A_j \neq \phi$ for all $\displaystyle i,j \left\{ 1,2,\ldots,n \right\}$.

Prove that:

(a) $7$ is friendly;

(b) $n$ is friendly iff $n \geq 7$.

[i]Valentin Vornicu[/i]

[quote="perfect_radio"]A natural numbers $\displaystyle n \geq 2$ is called [i]friendly[/i] if there is a family of non-void subsets $\displaystyle A_1,A_2,\ldots,A_n$ of the set $\displaystyle \left\{ 1,2,\ldots,n \right\}$ such that:

1) $\displaystyle i \not\in A_i$ for all $\displaystyle i = \overline{1,n}$;

2) $\displaystyle i \in A_j$ iff $\displaystyle j \not\in A_i$ for all distinct $\displaystyle i,j \in \left\{ 1,2,\ldots,n \right\}$;

3) $\displaystyle A_i \cap A_j \neq \phi$ for all $\displaystyle i,j \left\{ 1,2,\ldots,n \right\}$.

Prove that:

(a) $7$ is friendly;

(b) $n$ is friendly iff $n \geq 7$.[i]Valentin Vornicu[/i][/quote]

(a):
$A_1=\{2,3,4\}$
$A_2=\{3.5.7\}$
$A_3=\{4,6,7\}$
$A_4=\{2,5,6\}$
$A_5=\{1,3,6\}$
$A_6=\{1,2,7\}$
$A_7=\{1,4,5\}$
(b): $If n\ge 7$
$A_1=\{2,3,4\}$
$A_2=\{3.5.7\}$
$A_3=\{4,6,7\}$
$A_4=\{2,5,6\}$
$A_5=\{1,3,6\}$
$A_6=\{1,2,7\}$
$A_7=\{1,4,5\}$
$A_i=\{1,2,\ldots,i-1\}$; for all $i\ge 8$
n is friendly

You've left out a pretty important part of the solution. See if you can do it.

I have been taught that the numbers $m$ and $n$ are friendly if $\frac{\sigma(m)}{m}=\frac{\sigma(n)}{n}$. I don't see the equivalence between these two definitions.

Thats a totally different type of friendly ;)


induction
graph theory
combinatorics unsolved
combinatorics

[quote="Matematika"]On sport games there was 1991 participant from which every participant knows at least n other participants(friendship is mutual). Determine the lowest possible n for which we can be sure that there are 6 participants between which any two participants know each other.[/quote]
Remember that we had an old result that :given  a graph having kn+1 vetices,and if the degree of each vertex is larger  than and equal to (k-1)n+1 ,we will find a complete graph K_{k+1}.you can prove that result by induction following n.
To given a graph with each vertex'degree >=(k-1)n and it does'nt contain any complete graph ,we can order all vertices by 1,2,...kn+1 respectively,and connect two vertices i,j of its if (i not $ \equiv$ j mod k)
So my answer is 1593.
Correct my solution if there is any mistake.

We can also solve it using Turan's theorem.
By using it we get that among simple Graphs with $ 1991$ vertices and without any $ K_6$, $ T_{1991,5}$ has the maximum
edges and it is constructed as follows:$ 1$ part with $ 399$ vertices,$ 4$ parts with $ 398$ vertices and all the parts are empty-edges and each vertice is connected to all the vertices of all other parts.by analysing this graph we can understand easily that each vertice's degree is at least $ 1592$.so we can conclude that if $ n \geq 1592$ then we can construct a graph that doesn't have any $ K_6$.But if $ n \geq 1953$ then $ e(G) \geq e(T_{1991,5})$ so  we can conclude that $ G$ must contains $ K_6$.so $ n \equal{} 1593$.

We claim the answer is $n=1593$. We translate to graph theory. First, using turan's graph $T_{1991,5}$ we construct a $5-$partite graph where everyone knows at least $1592$ people. This establishes $n\ge1593$. Hence, if $E$ is the number of edges in the graph, we know $2E=\sum\deg v\ge1593\cdot1991>1592\cdot1992$. Then, since $T_{1991,5}$ has $\frac12\cdot(4\cdot398\cdot1593+399\cdot1592)=\frac12\cdot1593\cdot1991$ edges, by Turan's theorem it follows there must be a $K_6$ subgraph when $n=1593$. $\blacksquare$


This doesn't quite look like a geometry problem to me...I don't know though, maybe my eyes are deceiving me?

Yes, $ S$ is divisible by $ 2007$.  You use a pairing argument: notice $ 2007|x^{2007}\plus{}(2007\minus{}x)^{2007}$.


calculus
derivative
calculus computations

$ y''\minus{}y'\equal{}\frac{e^t\plus{}e^{\minus{}t}}{2}$

$ \therefore e^{\minus{}t}y''\minus{}e^{\minus{}t}y'\equal{}\frac{1\plus{}e^{\minus{}2t}}{2}$

$ \therefore \frac{d}{dt}(e^{\minus{}t}y')\equal{}\frac{1\plus{}e^{\minus{}2t}}{2}$

$ \therefore e^{\minus{}t}y'\equal{}\frac{1}{2}t\minus{}\frac{1}{4}e^{\minus{}2t}\plus{}C_1$

$ t\equal{}0,\ y'(0)\equal{}0\Longrightarrow C_1\equal{}\frac{1}{4}$

$ \therefore y'\equal{}\frac{1}{2}te^t\minus{}\frac{1}{4}e^{\minus{}t}\plus{}\frac{1}{4}e^t$

$ \therefore y(t)\equal{}\frac{1}{2}(t\minus{}1)e^{\minus{}t}\plus{}\frac{1}{4}e^{\minus{}t}\plus{}\frac{1}{4}e^t\plus{}C_2$

$ \therefore y(0)\equal{}1\Longleftrightarrow C_2\equal{}1$

$ \therefore \boxed{y\equal{}\frac{1}{2}\left(t\minus{}\frac{3}{4}\right)e^{\minus{}t}\plus{}\frac{1}{4}e^t\plus{}1}$


probability
limit
Euler
function
number theory
relatively prime
advanced fields

prove that probability of two numbers being relatively prime is 6/pi^2

This is a classical result in number theory, and there are many proofs of it.  A sketch of one proof is as follows:

First show that the probability of two numbers being relatively prime is, in the limit, equivalent to the probability that a given single number is not divisible by the square of any prime.  Having proven that, let $ P(k)$ be the probability that a number is divisible by the square of at least one of the first $ k$ primes.  So, for example, $ P(1) \equal{} \frac {1}{2^2}$, $ P(2) \equal{} P(1) \plus{} \frac {1}{3^2}\left(1 \minus{} P(1)\right)$, and in general
\[ P(k) \equal{} P(k \minus{} 1) \plus{} \frac {1}{p_k^2}\left(1 \minus{} P(k \minus{} 1)\right)
\]
where $ p_k$ denotes the $ k$th prime and $ P(0) \equal{} 0$.  This recurrence is fairly easy to solve using the usual methods (it telescopes rather nicely), and the solution is
\[ \lim_{n\to \infty} P(n) \equal{} 1 \minus{} \prod_{k \equal{} 1}^\infty{\left(1 \minus{} \frac {1}{p_k^2}\right)}
\]
By the Euler identity for the $ \zeta$ function, this is precisely $ 1 \minus{} \frac {1}{\zeta(2)} \equal{} 1 \minus{} \frac {6}{\pi^2}$.  Hence the probability that a number is divisible by the square of some prime is $ 1 \minus{} \frac {6}{\pi^2}$, and therefore the probability that a number is not divisible by the square of any prime is $ \frac {6}{\pi^2}$, which in turn proves the desired result.

Of course the word 'probability' in the above is used rather loosely, and is understood to mean a limit of some appropriate quotients taken in a very particular way.


Let $ a,b,c>0$ and $ a^2\plus{}b^2\plus{}c^2\equal{}3$. Prove that:
$ \frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a} \geq \frac{4}{a^2\plus{}7}\plus{}\frac{4}{b^2\plus{}7}\plus{}\frac{4}{c^2\plus{}7}$

[quote="hetten"]Let $ a,b,c > 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that:
$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \geq \frac {4}{a^2 \plus{} 7} \plus{} \frac {4}{b^2 \plus{} 7} \plus{} \frac {4}{c^2 \plus{} 7}$[/quote]
By the AM-GM Inequality, we have
$ b \plus{} c \le \frac {b^2 \plus{} 1}{2} \plus{} \frac {c^2 \plus{} 1}{2} \equal{} \frac {5 \minus{} a^2}{2}.$
Then, it suffices to prove that
$ \frac {1}{5 \minus{} a^2} \plus{} \frac {1}{5 \minus{} b^2} \plus{} \frac {1}{5 \minus{} c^2} \ge \frac {2}{a^2 \plus{} 7} \plus{} \frac {2}{b^2 \plus{} 7} \plus{} \frac {2}{c^2 \plus{} 7}.$
This inequality can be written as
$ \sum \left[ \frac {1}{5 \minus{} a^2} \minus{} \frac {2}{a^2 \plus{} 7} \minus{} \frac {3}{32}(a^2 \minus{} 1)\right] \ge 0,$
which is true because
$ \frac {1}{5 \minus{} a^2} \minus{} \frac {2}{a^2 \plus{} 7} \minus{} \frac {3}{32}(a^2 \minus{} 1) \equal{} \frac {3(a^2 \plus{} 3)(a^2 \minus{} 1)^2}{32(5 \minus{} a^2)(a^2 \plus{} 7)} \ge 0.$
The proof is completed.

[quote="can_hang2007"]
$ \frac {1}{5 \minus{} a^2} \plus{} \frac {1}{5 \minus{} b^2} \plus{} \frac {1}{5 \minus{} c^2} \ge \frac {2}{a^2 \plus{} 7} \plus{} \frac {2}{b^2 \plus{} 7} \plus{} \frac {2}{c^2 \plus{} 7}.$
[/quote]
I think the inequality above is clearly true by Cauchy-Schwarz
we have: $ \frac {1}{5 \minus{} a^2} \plus{} \frac {1}{5 \minus{} b^2}\ge \frac{4}{10\minus{}a^2\plus{}b^2}\equal{}\frac {4}{c^2 \plus{} 7}$

[quote="leedt26"]
I think the inequality above is clearly true by Cauchy-Schwarz
we have: $ \frac {1}{5 \minus{} a^2} \plus{} \frac {1}{5 \minus{} b^2}\ge \frac {4}{10 \minus{} a^2 \plus{} b^2} \equal{} \frac {4}{c^2 \plus{} 7}$[/quote]
I think, you are wrong. :wink:

I think [b]leedt26[/b] want to say that:
$ \frac {1}{5 \minus{} a^2} \plus{} \frac {1}{5 \minus{} b^2} \ge \frac {4}{10 \minus{} a^2 \minus{} b^2} \equal{} \frac {4}{c^2 \plus{} 7}$ which is true.

P.s.
From what contest is this problem taken???

[quote="leedt26"][quote="can_hang2007"]
$ \frac {1}{5 \minus{} a^2} \plus{} \frac {1}{5 \minus{} b^2} \plus{} \frac {1}{5 \minus{} c^2} \ge \frac {2}{a^2 \plus{} 7} \plus{} \frac {2}{b^2 \plus{} 7} \plus{} \frac {2}{c^2 \plus{} 7}.$
[/quote]
I think the inequality above is clearly true by Cauchy-Schwarz
we have: $ \frac {1}{5 \minus{} a^2} \plus{} \frac {1}{5 \minus{} b^2}\ge \frac {4}{10 \minus{} a^2 \minus{} b^2} \equal{} \frac {4}{c^2 \plus{} 7}$[/quote]
Very nice, leedt26. But what do you think about the following?

If $ a,b,c$ are positive real number such that $ a^2\plus{}b^2\plus{}c^2\equal{}3,$ then
$ \frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\plus{}\frac{1}{a\plus{}b} \plus{}\frac{15}{4} \ge \frac{14}{a^2\plus{}7}\plus{}\frac{14}{b^2\plus{}7}\plus{}\frac{14}{c^2\plus{}7}.$

See my proof at: http://canhang2007.wordpress.com/2009/10/28/inequality-4-v-q-b-can/


probability
geometry
modular arithmetic
Pythagorean Theorem

If a positive three-digit integer that is a multiple of 11 is chosen at random, what is the probability that the sum of its digits is an odd number?  Express your answer as a common fraction.

Answer [hide]4/9[/hide]

A right triangle and an isosceles triangle have equal areas.  The right triangle has sides of length 3, 4, and 5 units.  The isosceles triangle has sides of 3, x, and x units.  What is the value of x?  Express your answer as a decimal to the nearest thousandth.

Answer [hide]4.272[/hide]

still working on the first one  :)

the area of the right triangle is 6, and the area of the isoceles triangle is base*height/2.
the base is 3, and we know that is has the same area as the right triangle, so the height is 4.
to find the length of the other sides, we can use the pythagorean theorem, with a=3/2 because it is half of the base and b=4
solving for c, we get 4.272

[quote="trivea"][hide]
Heres the deal for the first one:
The lowest 3 digit factor of 11 is 110. If you keep finding successive multiples of 11 after that, you realize that the digits of all the multiples add up to an even number. One you get into the 200s, you realize that the first one is 209. This is the only 200 multiple of 11 that has digits that add up to an odd number. The rest are even. For the three hundreds, The first one is 308. That and the next multiple, 319, are the only odd ones. This pattern continues all the way through the 990s. That makes 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36 odd factors. Now, to find the number of three digit factors, you divide 900, the number of 3 digit numbers, by 11 to get 81 9/11. Since you can't have 9/11th of a multiple, there are 81 multiples. The final answer is 36/81 or 4/9
[/hide][/hide][/quote]
[u]Clarified Version[/u].

[hide]There are 81 three-digit numbers that are multiples of 11.

You do case work and find the pattern.
(10 thru 18, 29 thru 38, 49 thru 58, 79 thru 88, 20,40,60,89,90)$ \times 11$= Even sums.
(You can also find the ones that have odd sums which are 19 thru 28, 39 thru 48 and so one)

So $ 81\minus{}45\equal{}36$

36 numbers are what we are trying to find in the set of 81 numbers, so $ \frac{36}{81}\equal{}\boxed{\frac{4}{9}}$[/hide]

[quote="Ocean8"]If a positive three-digit integer that is a multiple of 11 is chosen at random, what is the probability that the sum of its digits is an odd number?  Express your answer as a common fraction.

Answer [hide]4/9[/hide]
 [/quote]
[hide="a start to 1"]a three digit number ABC is divisble by 11. in that case $ A \plus{} C \equal{} B \pmod {11}$ so A+C=B, or A+C=B-1. we want A+B+C $ mod 2$ if A+C=B $ A \plus{} B \plus{} C \equal{} 2B \equal{} 0 \pmod{2}$ we see only the second case works.
the second case only works if $ A \plus{} C > 10$(i dont know how to do greater than/equal to) [/hide]


geometry
3D geometry
factorial
summer program
MathPath
Pascal\u0027s Triangle

1. Who do you think should have won? I think Mark Zhang should have won.

2. What is your opinion of the problem? I thought it was easy. I knew the answers beforehand.

I would say Sergei. Mark Zhang definitely was the best presenter, but Sergei was a good presenter and had what was a seemingly better knowledge of the problem. Mark was definitely a close second though, no doubting that.

The problem seemed easier than last years, but I don't know how I would have taken it under all that pressure!

What was the problem? (I wasn't at nats because of careless mistakes.) And who did win? Sergei?

Sergei did win.

The problem was counting vertices, edges, and faces (not only 2D faces, but 3D, 4D, and 5D faces) on 2D, 3D, 4D, and 5D triangles and squares.

Yeah. Also, I found the guy who wrote the questions rather annoying. When Neal went up, he kept asking questions about the stuff Neal actually got right. It turned out Neal only managed to fix one of his many mistakes.

Being from IN, I think that Nathan should have won.  He will win next year though

No. Sergei, but he was a boring presenter (and i was in the front row!). Mark was the only guy you could hear clearly, but he didnt get the choose thing for the tetrahedrons. I thought that was kinda hard, and I'm glad that I didn't make it!

[quote="Phelpedo"]I don't think Mark was even c lose to sergei. First of all, he didn't even come close to finishing; he barely touched upon the hypercubes. He didn't get the main idea for the n-dimensional triangles either- not only was he wrong, but he failed to notice that the numbers were nC2, etc. and only recognized them as elements of Pascal's triangle when the judge practically pointed it out. Even then, he still didn't explain the correspondence with Pascal's triangle very well.[/quote]

But Mark still found a method for the triangle faces, combinations or no. And there is no explicit answer involving combinations. I say Sergei was the best, but Mark was better than people give him credit for.

I wasn't there, and what was the final order of the Master's round?

how do you count 5D faces? what do 5D figures look like?

They don't. Seeing as our universe is 3-dimensional (and don't argue, because I don't care what Einstein thinks), 5-dimensional figures are impossible to create our imagine. Anyway, 5d cubes are two 4d cubes with lines connecting the corresponding vertices each of which go through 5d space. Same for 4d cubes with 3d cubes. A 4d cube is a "hypercube".

Anyway, the choice numbers are not the [b]only[/b] way to define Pascal's Triangle. The first thing you learn is the recursive rule, which Zhang used. Furthermore, he came about this independently of the choice numbers by correctly pointing out that faces stay at their dimension and increase a dimension when the entire figure increases a dimension. I even thought this was relieving as he was the last presenter. Finally, even if he didn't completely finish, he managed to cover pretty much everything as soon as he learned that he had only a minute left. This in itself is an amazing feat, even if he didn't go in depth.

Well, first of all, we cant really imagine what a 5D figure would look like because that would mean that there are 5 perpendicular directions and we can't visualize that because we see only in 3D. They didn't give the order of the finishers, only the winner. I think Mark talked the clearest, but I think there were a few minor problems:
1. He didn't mention combinatorics
2. He kept saying random things like "woohoo", "yay", "um", "yea", "so", "thats basically it", and things like that that made his presentation a little informal
3. He didn't get to finish and therefore barely touched on half of the question

[quote="Magixmaniak"]
2. He kept saying random things like "woohoo", "yay", "um", "yea", "so", "thats basically it", and things like that that made his presentation a little informal
[/quote]

He said, "woohoo"??? Why?  :rotfl:

yeah really; why? I appear to have missed that.

MY OPINIONS:

Nathan didn't even finish preparing: he just started part III when the 30 minutes were called.

Mark Zhang was the only one who didn't use combinations to explain the properties of n-dimensional triangles. Furthermore, his presentation was very long: he just started part III when the 11 minutes were called.

Neal and Sergei were the only ones that actually finished, and Sergei, as you said, messed up by putting $\binom{n}{...}$, as opposed to $\binom{n+1}{...}$. Plus, Neal has better handwriting.

Thus Neal should have won the Masters Round.

[quote="chess64"]Hey Magixmaniak, what place did you get?[/quote]

Magixmaniac, a.k.a. Dennis Mou, came in 24th place with a score of 29.

Neal Wu, while he had everything else totally under control, was not a good presenter at all. I couldn't even understand him, and I was sitting near the sound system. The beauty of the Masters Round is, you may be a genius, but if you don't know how to talk to an audience you won't win. Mark and Sergei (sort of ) were the only two who could present. And not even Sergei, really.

[quote="biffanddoc"]Magixmaniac a.k.a. Dennis Mou is your name pronounced Dennis Mooooooo like a cow or Mow like mow the lawn? (I bet my teammate $5 on Mooooo)I heard it differently at the countdown/awards banquet[/quote]

It's pronounced like Mao, I think

Who cares if Neal finished? He scribbled random stuff on the paper (a lot of which was completely wrong), and by the time the questioning was over, the annoying judge on the right kept him from correcting a single one of his many mistakes.

I'm new.  I was just in nationals for the first time.
        I thought that Sergei should have won, without a doubt.  I even spoke to him and told him that he was going to win, and everyone else with whom I talked seemed to think the same.  The reason is that he fulfilled two requirements :
        1) He covered the topic completely, and
        2) He was coherent.
        None of the others appeared to have done this.  Neal mumbled; I couldn't understand what he was saying and confess that I started thinking about factorials in general rather than the problem at hand.  I don't remember the other kids' names, but the first one didn't finish and the last didn't get the [i]n[/i] + 1 choose [i]x[/i] format, which, however meritorious his presentation, was a death sentence, as that was what the judges were probably expecting.  I'll admit, however, that I didn't see Sergei's error in saying [i]n[/i] choose [i]x[/i] rather than [i]n[/i] + 1 choose [i]x[/i].
        As for the problem, I think they should have invited me—it was in my domain.  Even if I wouldn't have won, I probably would have given some competition.  But then, I got 105th place, so what do I know?

In all honesty, the Master's Round was boring. I stayed for the first one or two presentations, left out of lack of interest (as well as leaving my other CD in my hotel room), and my coach said she couldn't make it through the third presentation due to boredom.

I like it very much.  This is gewtting more towards high school math.  I watched last year's with Gregory Gauthier and he was amazing and the topic was interesting.  It was the same this year with Sergei.  It's fun.

Ooh. Last year's, I stayed the full two hours or so. It was fun mainly because Ian Frankel and I worked out the Master's Round together in like twenty minutes, and Mike Jin, Ian, and I were comparing answers with the contestants' answers as they presented. I recall making bets on who would win.

Mike Jin? Are you from Missouri? I sat right across from Missouri at the banquet!

[quote="solafidefarms"]Exactly. It was obvious Sergei would win. He was coherent and he brought up how the numbers were n+1 choose x. That was good. Everyone else repeated the same things.

From what I have heard from 2 people who went to MathPath, they did that exact problem last year. :?

Billy[/quote]
For those who didn't attend, each Masters Round competitor was isolated until it was his turn to speak.  However, the presentations were so similar, at least one of the state coaches believed the Masters topic was known in advance.  [It wasn't.]

The actual Masters Round questions were withheld from the audience.  Instead, the audience received a brief lecture outlining the object of the Masters Round.  Several of the students referred directly to the questions, so the audience got a pretty good idea of what the Masters Round prompts must have been.

Emminent geometer Professor Thomas F. Banchoff of Brown University was a guest lecturer at MathPath last summer.  He gave a series of lectures lasting about a week.  Early in the course, the students were assigned homework essentially identical to the Masters Round.  [But without the mandatory presentation and Q&A session.]

It was clear that each of the Masters Round participants understood the problem.  None of the participants came up with a general closed-form solution to the number of m-D objects (vertices, edges, faces, etc) in an n-dimensional cube.  But most participants came up with some closed-form solutions and some recursive formulas to answer some of these specific questions.


I'm not quite sure where to put this topic...

Does anyone have any good resources regarding the Markov Chain?

Thanks in Advance

See here http://en.wikipedia.org/wiki/Markov_chain :wink:


On the board there were written the positive divisors of $30^{19}$. Mike cleaned the one that were divisors of $20^{11}$. How many numbers are still on the board?

[quote="shadysaysurspammed"]180 numbers left?[/quote]

Can you post your solution?

darn,bad mistake.Yes,find the total divisors of the first number and second and subtract the common factors. Stupid me,stupid :blush:

No you have to include all the divisors not just the prime factorization.  I got 7760 numbers remaining.

[quote="shadysaysurspammed"][hide="lets see...."]$30^{19}=2^{19}*3^{19}*5^{19}$

$20^{11}=2^{22}*5^{11}$

So $2^{19}$ and $5^{11}$ get rubbed off leaving on the board

$3^{19}*5^{8}$

So numbers left are $(9)(20)=180$[/hide][/quote]

Wrong. Find yourself your mistake. The answer is much bigger.

[hide]
The prime factorization of $30^{19}$ is $2^{19}\times 3^{19}\times 5^{19}$, so there are a total of $20^3$ numbers on the board to start with.

We're looking for the common factors between $30^{19}$ and $20^{11}$. The prime factorization of $20^{11}$ is $2^{22}\times 5^{11}$. The common factors between the two will then be all the factors of $2^{19}\times 5^{11}$, so there are $240$ common factors.

The number of numbers remaining on the board is then $8000-240=7760$.
[/hide]


is it necessary that when acetylene s passed through $ \text{NaNH}_{2}$ then the primary product is
$ \text{CHCNa}$ and not $ \text{CNaCNa}$

Since $ HC\equiv C^{\minus{}}Na^{\plus{}}$ gives the nucleophile that's needed, I don't see why $ CNaCNa$ would be of any importance in synthesis. Besides, acetylene isn't the strongest of acids, and my best guess is that it can lose just one proton.

That depends on what you want to do: most of the times, the monoanion is used in an nucleophilic addition/substitution step, and for that purpose acetylene and sodium amide should react in nearly equivalent amounts. However, acetylene dianion could be usefull in some synthesis in a double alkylation step. For example:

1. $ C_{2}H_{2}\plus{}2NaNH_{2}\longrightarrow Na_{2}C_{2}H_{2}$

2. $ Na_{2}C_{2}H_{2}\plus{}CH_{3}I\longrightarrow CH_{3}C\equiv CNa$

3. $ CH_{3}C\equiv CNa\plus{}CH_{3}COCH_{3}\longrightarrow CH_{3}C\equiv CC(CH_{3})_{2}OH$.

Dianions derived from alpha,beta-unsaturated carbonyls are often used in similar double alkylation steps.

[quote="Cyprusad"]Besides, acetylene isn't the strongest of acids, and my best guess is that it can lose just one proton.[/quote]

For example, when 5,5-dimethyl-1-hexyne is treated with 2 equivalents of n-butyl litium a usefull dianion is easily formed. Can you find out which?

On second thoughts,
$ Na \plus{} C_{2}H_{2} \longrightarrow NaHC_{2} \longrightarrow Na_{2}C_{2}$ contains a $ [C \equiv C]^{2\minus{}}$ carbide ion.

No I couldn't find out which :huh: Its a terminal alkyne, why is a dianion being formed?

Carbon 3 is deprotonated after the the terminal alkyne. In this way, we can alkylate, not only a terminal alkyne (as we already knew) but also a position alpha to a terminal alkyne, which can be quite usefull in synthesis.

But carbon-3 isn't $ sp$ hybridized, so why does it deprotonate? Does having two methyls on carbon-5 have to do anything with it?

Does a carbon have to be sp in order to be deprotonated? No, the two methyl groups do not interfere here.
The explanation lies in the fact that an sp carbon is electronegative in comparision with a "normal" sp3 carbon, and so an sp3 carbon directly attached to a triple bond is slightly positive: in the presence of a very strong base like butyl litium it can then be deprotonated.