logarithms
AMC
AIME
algebra
system of equations
AIME I
binomial theorem

k i get this but for some reason my values are complex :huh:
[hide="Partial"]
first let:
$a=\log_{10}x,b=\log_{10}y,c=\log_{10}z$
Notice:
$\log_{10}2000=\log_{10}1000+\log_{10}2=3+\log_{1}0{2}$

So, Equation One:
$3+\log2+a+b-ab=4$
$\log2+a+b-ab=1$

$\log2+b+c-bc=log2+a+b-ab$
$(c-a)-(cb+ab)=0$
$(c-a)-b(c-a)=0$
$(1-b)(c-a)=0$
$\therefore{b=1\text{ or }a=c}$

if $b=1$ then substituting it back in reveals no solutions..so then $a=c$

So sub $a=c$ into Equation three and you eventually get..
$\log2+2a-a^{2}=-3$
$\Rightarrow{a^{2}-2a-(3+\log2)=0}$
$\Rightarrow{a^{2}-2a-\log2000=0}$
$a=\frac{2\pm\sqrt{4+4\log2000}}{2}=1\pm\sqrt{\log20000}$

We want to find the two possible b's..so
sub $a=1+\sqrt{\log20000}$ into 1

$\log2+(1+\sqrt{4+\log2})+b-(1+\sqrt{4+\log2})b=1$
$\Rightarrow{b_{1}=\frac{\log2+\sqrt{4+\log2}}{\sqrt{4+\log2}}}$

sub $a=1-\sqrt{4+\log2}$ into 1

$\log2+(1-\sqrt{4+\log2})+b-(1-\sqrt{4+\log2})b=1$
$\Rightarrow{b_{2}=\frac{-\log2+\sqrt{4+\log2}}{\sqrt{4+\log2}}}$

$y_{1}=10^{b_{1}}=10^{\frac{\log2+\sqrt{4+\log2}}{\sqrt{4+\log2}}}$

$y_{2}=10^{b_{2}}=10^{\frac{-\log2+\sqrt{4+\log2}}{\sqrt{4+\log2}}}$

$y_{1}+y_{2}=10^{\frac{\log2+\sqrt{4+\log2}}{\sqrt{4+\log2}}}+10^{\frac{-\log2+\sqrt{4+\log2}}{\sqrt{4+\log2}}}$

if you can simplify this even more please do, I think there is a much better approach to this question
[/hide]
EDIT: yes i did have a sign error..

[quote="SM4RT"]k i get this but for some reason my values are complex :huh:
[hide="Partial"]
first let:
$a=\log_{10}x,b=\log_{10}y,c=\log_{10}z$
Notice:
$\log_{10}2000=\log_{10}1000+\log_{10}2=3+\log_{1}0{2}$

So, Equation One:
$3+\log2+a+b-ab=4$
$\log2+a+b-ab=1$

$\log2+b+c-bc=log2+a+b-ab$
$(c-a)-(cb+ab)=0$
$(c-a)-b(c-a)=0$
$(1-b)(c-a)=0$
$\therefore{b=1\text{ or }a=c}$

if $b=1$ then substituting it back in reveals no solutions..so then $a=c$

So sub $a=c$ into Equation three and you eventually get..
$\log2+2a-a^{2}=3$
$\Rightarrow{a^{2}-2a+(3-\log2)=0}$
$\Rightarrow{a^{2}-2a+\log500=0}$
however the discriminant is$4-4\log500<{0}$

Why am i getting this??[/hide][/quote]

[hide="SM4RT, maybe because of this?"]Wait, $\log(2000zx)-(\log_{10}z)(\log_{10}x) = 0$, using your substitutions, should simplify as follows:

$\log2+\log(1000)+\log_{10}z+\log_{10}x-(\log_{10}z)(\log_{10}x) = 0$

$\log2+3+a+c-ac = 0$

I think you switched the sign on the $3$.

To continue, Using your a = c result,

$(\log2+3)+2a-a^{2}= 0 a^{2}-2a-(\log2+3) = 0$

The discriminant would then be $4+4\log(2000)$, perfectly valid, but very messy. As a result, I'm stuck... Could someone check my work and expand upon this please?[/hide]

[quote="4everwise"]The system of equations
\begin{eqnarray*}\log_{10}(2000xy)-(\log_{10}x)(\log_{10}y) &=& 4\\ \log_{10}(2yz)-(\log_{10}y)(\log_{10}z) &=& 1\\ \log_{10}(2000zx)-(\log_{10}z)(\log_{10}x) &=& 0\\ \end{eqnarray*}
has two solutions $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2}).$ Find $y_{1}+y_{2}.$[/quote]

This problem is wrong. I kept getting imaginary values for $\log_{x}$ and $\log_{z}$ until I checked and realized that the third equation should be $\log_{10}(zx)-(\log_{10}z)(\log_{10}x) = 0$

[hide="Solution"]
[b]Solution.[/b]
\begin{eqnarray*}\log_{10}(2000xy)-(\log_{10}x)(\log_{10}y) &=& 4\\ \log_{10}(2yz)-(\log_{10}y)(\log_{10}z) &=& 1\\ \log_{10}(2000zx)-(\log_{10}z)(\log_{10}x) &=& 0\\ \end{eqnarray*}
The above system is equivalent to
\begin{eqnarray*}(\log x-1)(\log y-1) &=& \log 2\\ (\log y-1)(\log z-1) &=& \log 2\\ (\log x-1)(\log z-1) &=& 4+\log 2\\ \end{eqnarray*}
Let $\log x-1=u,\log y-1=v,\log z-1=w$.  Hence,
\begin{eqnarray}uv &=& \log 2\\ vw &=& \log 2\\ uw &=& 4+\log 2\end{eqnarray}
Thus, from (1), $u=\frac{\log 2}{v}$. Using (3), $\frac{w\log 2}{v}=4+\log 2$.  Using (2), $w=\pm\sqrt{4+\log 2}$.  Then it follows that
\[(u,v,w)=\left(\pm\sqrt{4+\log 2},\pm\frac{\log 2}{\sqrt{4+\log 2}},\pm\sqrt{4+\log 2}\right). \]
This implies that
\[(x,y,z)=\left(10^{\pm\sqrt{4+\log 2}+1},10^{\pm\frac{\log 2}{\sqrt{4+\log 2}}+1},10^{\pm\sqrt{4+\log 2}+1}\right) \]
And the conclusion follows.$\blacksquare$[/hide]

[quote="usaha"]I was doing over this problem..and I cant seem to find aninteger solutions for y1 + y2..

Is this problem written wrong???[/quote]
It never said it had to be an integer.

[quote="usaha"]I was doing over this problem..and I cant seem to find aninteger solutions for y1 + y2..

Is this problem written wrong???[/quote]

If you're thinking about the 2000 AIME, the problem on there had the last equation be

$\log zx-(\log z)(\log x)=0$

[quote="usaha"]wow..that wouldnve made it a lot easier, the y1 and y2 are 5 adn 20 so 25...

you get the same thing, but you find that log x=log z and also log x=0 or 2.. plug them in and you get log y...[/quote]

Is there an official solution for 2000's problems?

[hide="Full Solution to 2000 AIME I #9"]Let $ \log(\omega) \equal{} \log_{10}(\omega)$, and by this shorthand let $ a \equal{} \log(x)$, $ b \equal{} \log(y)$ and $ c \equal{} \log(z)$.  Because $ \log(pq) \equal{} \log(p) \plus{} \log(q)$, the system becomes

$ \log(2) \plus{} 3 \plus{} a \plus{} b \minus{} ab \equal{} 4\Rightarrow ab \minus{} a \minus{} b \plus{} 1 \equal{} \minus{} \log(2)\Rightarrow (a \minus{} 1)(b \minus{} 1) \equal{} \log(2)$
$ \log(2) \plus{} b \plus{} c \minus{} bc \equal{} 1\Rightarrow bc \minus{} b \minus{} c \plus{} 1 \equal{} \log(2)\Rightarrow (b \minus{} 1)(c \minus{} 1) \equal{} \log(2)$
$ a \plus{} c \minus{} ac \equal{} 0\Rightarrow ac \minus{} a \minus{} c \plus{} 1 \equal{} 1\Rightarrow (a \minus{} 1)(c \minus{} 1) \equal{} 1$

Now we substitute $ \alpha \equal{} a \minus{} 1$, $ \beta \equal{} b \minus{} 1$ and $ \gamma \equal{} c \minus{} 1$ to get

$ \alpha\beta \equal{} \log(2)$ (1)
$ \beta\gamma \equal{} \log(2)$ (2)
$ \alpha\gamma \equal{} 1$ (3)

We take (1) divided by (2) and multiplied by (3):

$ \alpha^2 \equal{} 1$,

which gives us $ \alpha \equal{} \pm 1$.  Then $ \beta \equal{} \pm\log(2)$, so $ b \equal{} 1\pm\log(2)$.  Then the two solutions for $ b$ are

$ b \equal{} 1 \plus{} \log(2) \equal{} \log(10) \plus{} \log(2) \equal{} \log(20)$ and
$ b \equal{} 1 \minus{} \log(2) \equal{} \log(10) \minus{} \log(2) \equal{} \log(5)$.

Thus the two solutions for $ y$ are $ y_1 \equal{} 20$ and $ y_2 \equal{} 5$, so our answer is

$ y_1 \plus{} y_2 \equal{} \fbox{025}$.[/hide]

Have they made a hard logs question yet?

[hide]
\[\log_{10}(2000xy)-(\log_{10}x)(\log_{10}y) = 4. \]
Therefore,
\[\log_{10}(2xy)-(\log_{10}x)(\log_{10}y) = 1. \]
By comparing the first two equations, it becomes clear that
\[x=z \].
Solving the third equation gives
\[x=z=1,100 \]
Plugging this back into the equations gives
\[y=5,20 \]
\[20+5 = 025. \]

[/hide]

[hide]First equation gives $3\log_{10}2+\log_{10}(x)+\log_{10}(y)-\log_{10}x\log_{10}y=4$.

Wait first off let's let $\log_{10}x=a, \log_{10}y=b, \log_{10}z=c$.

Therefore $3\log_{10}2+a+b-ab=4$ or therefore $3\log_{10}2-(a-1)(b-1)=3$.  This is the same as $(a-1)(b-1)=3\log_{10}2-3$.

The second equation gives $\log_{10}2+y+z-yz=1$.  Therefore, $\log_{10}2-(y-1)(z-1)=0$ or $\log_{10}2=(b-1)(c-1)$.

Lastly, the third equation gives $a+c=ac$ or therefore $(a-1)(c-1)=1$.

Multiply all the equations to give us $((a-1)(b-1)(c-1))^2=(3\log_{10}2-3)(\log_{10}2)$.

We care most about the variable $b$, so we go from teh equation $(a-1)(c-1)=1$ to give us $(b-1)^2=(3\log_{10}2-3)(\log_{10}2)$.  First off, simplify this, to give us $\log_{10}(2000)-\log_{10}(1000)=\log_{10}2$.  Therefore $b-1=\pm \log_{10}2$.  Therefore, substitute $b=\log_{10}y$ to give us $\log_{10}y-1=\log_{10}2, -\log_{10}2$.

The first gives us $\log_{10}y=\log_{10}(20)$ and second gives us $\log_{10}y=\log_{10}(5)$ so their sum is $\boxed{25}$. [/hide]

[hide="fast solution"]
let a=log x
b=log y
c= log z

log 2000xy-logxlogy=3+log 2+a+b-ab=4: (a-1)(b-1)=log2
do similar stuff to get (b-1)(c-1)=log 2 and (a-1)(c-1)=1
Thus b-1=plus or minus log 2 while both the others are plus or minus log 1
thus y1+y2=10^(1+log2)+10^(1-log2)=10*10^(log2)+10*10^(-log2)=20+5=

025

[/hide]

[hide="bro"]
Let us denote the equations as 1,2,3 in descending order respectively.
Subtracting $2$ from $1$ yields,
$\log 1000 + \log 2 + \log x + \log y - \log x\log y - \log 2 - \log y - \log z + \log y\log z = 3$
We get this from the log property $ \log xy = \log x + \log y$
From here it is easy to see,
$ \log x - \log z + \log y\log z - \log x\log y = (1 - \log y) (\log x - \log z) = 0$
$\rightarrow$ $x=z$ or $y=10$
substituting $x=z$ we get $y= 20$ or $y=5$ and by checking if $y=10$ works, we find $y=5,20$ are the only solutions so,
$y_1 + y_2 = 25$
QED

[/hide]

@Binomial-theorem: How is $ \log_{10} (2000) = 3 \log_{10} (2) $. I am getting $ 3 + \log_{10} (2) $. But, with what you did, everything turns out beautifully. I did exactly what you did but I used $ 3 + \log_{10} (2) $ and got a bunch of nice but non-working dead ends. Hm.

[hide="Can someone tell me where my reasoning fails, if x, y, and z are positive real numbers?"]

Let $a=\log_{10}\left(x\right)$, $b=\log_{10}\left(y\right)$, and $c=\log_{10}\left(z\right)$.  The last equation gives us

$a+c-a\cdot c=0$

Thus either $a+c=0$ or $\dfrac{a^3+c^3}{a+c}=\dfrac{\left(a+c-a\cdot c\right)\cdot\left(a+c\right)}{a+c}=0$.  Let's prove $a+c=0$ with a proof by contradiction.  If $a+c\ne0$, then $\dfrac{a^3+c^3}{a+c}=0$, so $a^3=-c^3$, so $a=-c$, a contradiction.

Thus, with that and $a^2+c^2-a\cdot c=0$, $a=c=0$.  Which doesn't look good.[/hide]

Ah, sorry, messed up a little, \Edits

Though that doesn't destroy my "proof."

EDIT: Oh.  It does.  Nvmd, the LHS isn't $a^2+c^2-a\cdot c$.

[quote=SM4RT]k i get this but for some reason my values are complex :huh:
[hide="Partial"]
first let:
$a=\log_{10}x,b=\log_{10}y,c=\log_{10}z$
Notice:
$\log_{10}2000=\log_{10}1000+\log_{10}2=3+\log_{1}0{2}$

So, Equation One:
$3+\log2+a+b-ab=4$
$\log2+a+b-ab=1$

$\log2+b+c-bc=log2+a+b-ab$
$(c-a)-(cb+ab)=0$
$(c-a)-b(c-a)=0$
$(1-b)(c-a)=0$
$\therefore{b=1\text{ or }a=c}$

if $b=1$ then substituting it back in reveals no solutions..so then $a=c$

[/quote]

By substituting a = c into the third equation, a = logx = 0 or 2. Plugging these values for equation 1, logy = log 5 for a = 0 and logy = log20 for a = 2. Didn't feel like it was too hard

I just tried this and only got one solution, because I got to $(\log y)^2-\log y^2=(\log 5)^2-\log 5^2.$ I looked at a solution and it showed that the other value is $20$ but used a different method. How do I find $y=20$ from here?

EDIT: Oh I see! We complete the squares: $(b-1)^2=(\log 5-1)^2\Longleftrightarrow b=1\pm(\log 5-1)$ and thus the other solution is $b=-\log5+\log100=\log20$.

p5: Denote $\log_{10}{x} = a, \log_{10}{y} = b, \log_{10}{z} = c$. Then $3 + \log_{10}{2} + a + b - ab = 4, \log_{10}{2} + b + c - bc = 1, a + c - ac = 0$. Simplifying gives $\log_{10}{2} + a + b - ab = 1, \log_{10}{2} + b + c - bc = 1, a + c - ac = 0$. Subtracting the first two equations gives $a - c - ab + bc = (a-c) - b(a-c) = (1-b)(a-c) = 0$. If $b = 1$, then the second equation gives $4+\log_{10}{2} = 4$, which is obviously not possible. If $a = c$, then the third equation gives that either $a = c = 0$ or $a = c = 2$. In the first case, we get $b = 1 - \log_{10}{2} = \log_{10}{5} \implies y = 5$. In the second case, we get $b = 1 + \log_{10}{2} = \log_{10}{5} \implies y = 20$. The problems tells us that there are two solutions, so we don't have to worry about extraneous solutions. The answer is $5+20=\boxed{25}$.

[quote=RJ5303707]In [url=https://youtu.be/sOyLnGJjVvc?t=372]this video[/url], at 6:12 he converts $\log_{10} y_1=1-\log 2$ to $\frac{10}{10^{\log_2}}=y_1$. He also converted $\log_{10} y_2=1+\log 2$ into $y_2=(10)(2)$. How exactly did he do this? I don't really understand which properties he used.[/quote]

I don't know why he would do that for the first one (my youtube is blocked so I can't check if there's a typo) but here's what you do to solve it; write $1-\log 2 = \log 10 - \log 2 = \log \frac{10}{2} = \log 5$ and for the second one write $1+\log 2 = \log 10 + \log 2 = \log 10 \cdot 2 = \log 20$.

[b]Solution:[/b] Let $a = \log_{10}x, b = \log_{10}y, c=\log_{10}z$. Then, by simple log manipulations we have that $\log(2) + 3 + a + b - ab =4, \log(2) + b + c - bc = 1, a + c - ac = 0$. Then, subtracting the first two equations, we have that $a - c - ab + bc = 0 \Rightarrow (1-b)(a-c) = 0$. Clearly, $b=1$ fails. It follows that $a=c$. Using our third equation, we now see the only possible values are $(2, y_1, 2), (0, y_2, 0)$. One can use the system of equations once again to calculate these values as $10^{1 + \log(2)} + 10^{1 - \log(2)} = 20 + 5 = \boxed{025}$.


calculus
integration
logarithms
real analysis
real analysis solved

Compute $\int_{0}^{+\infty}ln(thx)dx$ where
$thx = \frac{shx}{chx}$

American textbooks would use the notation tanh(x), sinh(x), and cosh(x) rather than thx, shx, or chx.

tanh.

The answer is [hide]-(pi^2)/8.[/hide] Intermediate steps tomorrow.

The intermediate steps, as promised.

Let $u =$ tanh$(x)$. Then $du = $1 - tanh$^2(x) dx = (1-u^2)dx.$ We rewrite this as $\displaystyle dx = \frac{du}{1-u^2}.$ As $x$ goes from 1 to $\infty$, then $u$ goes from 0 to 1.

Make that substitution, and the integral becomes:

$\displaystyle \int_0^1 \frac{\ln u}{1-u^2}du = \int_0^1 (\sum_{k=0}^{\infty}u^{2k}\ln u) du = \sum_{k=0}^{\infty}\int_0^1 (u^{2k}\ln u) du = -\sum_{k=0}^{\infty}\frac1{(2k+1)^2}.$

This sum is the negative of 3/4 of the sum $\displaystyle \sum_{n=1}^{\infty}\frac1{n^2}$, which is well known to be $\displaystyle \frac{\pi^2}6.$ Therefore, the answer to the problem is $\displaystyle -\frac{\pi^2}8.$

There are approaches to calculating $\zeta (2)$, such as using the Fourier series of the integral of the square wave, that first calculate the sum of reciprocal odd squares - the sum we have here.

There are several possible justification for the interchange of sum and integral. The Monotone Convergence Theorem certainly suffices.


Let $ p > 2$ be a prime number. Find the least positive number $ a$ which can be represented as
\[ a \equal{} (X \minus{} 1)f(X) \plus{} (X^{p \minus{} 1} \plus{} X^{p \minus{} 2} \plus{} \cdots \plus{} X \plus{} 1)g(X),
\]
where $ f(X)$ and $ g(X)$ are integer polynomials.

[i]Mircea Becheanu[/i].

The remainder of $x^{p-1}+\ldots+x+1$ when divided by $x-1$ is $p$, so $a=p$ works. It's also the smallest one, since plugging $1$ into $(x-1)f(x)+(x^{p-1}+\ldots+x+1)g(x)=a$ gives $p|a$.

Why do we need $p$ to be prime? :? The remainder when we divide $\frac{x^n-1}{x-1}$ to $x-1$ is always $n$ (since $x^{n-1}+\ldots+x+1=(x-1)(x^{n-2}+2x^{n-3}+\ldots+(n-2)x+(n-1))+n$), and the exact same argument as above shows that $n|a$ in the general case.. Sorry if I'm missing something obvious.

[quote="Fermat's Little Turtle"]Are we to assume the degree of $f(x),g(x)$ are $\geq 1$?[/quote]
It doesn't matter.


geometry circumcircle trigonometry ratio

let ABC be a triangle with(A=90).and let (C) be its circumcircle .we draw AM wich is bisector of the diameter BC.

we draw CE ,wich is the bisector of the segment AM,and let it intersect (C) is point P.

we draw the altitude of triangle ,from point A.and let it intersect (C) in point Q.

prove that :

                   PQ bisects the segment AB

Let the median of $ AM $ meets $ AM $ at $ E $ and $ AB $ at $ F $.

$ PQ $ bisect $ AB $ if and only if $ \triangle{APQ}=\triangle{BPQ} $

Since $ \angle{PAQ}+\angle{PBQ}=180^o $, so we have: $ \frac{\triangle{APQ}}{\triangle{BPQ}}=\frac{AP \cdot AQ}{BP \cdot BQ} $

Since $ M $ and $ E $ are mid-points, so use Menelaus theorem on triangle $ ABM $ and line $ CEF $:
\[\frac{AF\cdot BC \cdot EA}{FB \cdot MC \cdot ME}=1\]
\[\Longleftrightarrow \frac{AF}{FB}=\frac12\]
Because $ \frac{AF}{FB}=\frac{AP}{BP}\frac{\sin{\angle{APC}}}{\sin{\angle{BPC}}}=\frac{AP}{BP}\sin{\angle{ABC}} $

Hence: $ \frac{AP}{BP}=\frac{1}{2\sin{\angle{ABC}}} $
Put it back to the original area ratio equation, we will have:
\[\frac{\triangle{APQ}}{\triangle{BPQ}}=\frac{1}{2\sin{\angle{ABC}}}\frac{AQ}{BQ}=\frac{AD}{\sin{\angle{ABC}}}\cdot \frac{1}{AB}=1\]

Therefore, $ PQ $ bisect segment $ AB $.

:yup:  :first: very very good.ure a good solver ;)

angle ABC= angle CPQ(guess why?).

angleBAM= angleABH=angleMBA. and it means that the quadrilaterall

PDEA is cyclic.(D is the intersection of PQ and AB) and (E is the mid point of MA).

if PDEA is cyclic we will have : angle PAD=angle PED.

but:angle PAD=anglePCB.   and it means  that angle PED=anglePCB.

then ED and and BC will be paralell. the problem is solved by looking at the two simillar triangle AED and AMB,we

can find that :   AD=BD


Let $ ABCD$ be a convex quadrilateral inscribed in the circle $ C(O)$ . Denote $ E\in AC\cap BD$ .

Suppose $ EB=3\cdot ED$ and define $ \{\begin{array}{c}
M\in (BC)\ ,\ MB=MC\\\\
N\in (BE)\ ,\ NE=ED\\\\\
P\in MN\cap AB\end{array}$ . Prove that $ EP\perp EO$ .

We draw the line through the point $ M$ and parallel to $ AC,$ which intersects the line segments $ BD,\ AB,$ at points $ K,\ F$ respectively and so, we have $ BK \equal{} KE$ and $ BF \equal{} FA.$

Because of now, the points $ K,\ N,$ are the midpoints of the segments   $ BE,\ BD$ and because of $ BE \equal{} 3\cdot EB,$ we conclude that ( easy to prove ) $ BK \equal{} 3\cdot KN$ $ ,(1)$

It is easy to show that the points $ B,\ M,\ O,\ N,\ F$ are concyclic, with diameter of their circumcircle $ (O'),$ the segment $ BO.$

The point $ P,$ as the intersection point of the line segments $ BF,\ MN,$ lies on the polar of $ K,$ with respect to the circle $ (O')$ and let $ S$ be the orthogonal projection of $ P,$ on the line segment $ O'K.$

It is clear that $ PS\perp OE,$ because of $ O'K\parallel OE,$ where $ O'$ is the center of $ (O').$

We denote the point $ E'\equiv PS\cap BD$ and it is enough to prove that $ E'\equiv E.$

Because of the point $ E',$ is the intersection point of the line segment $ BN$ passing through the point $ K,$ from the polar of $ K$ with respect to the circle $ (O'),$ we conclude that the points $ E',\ N,\ K,\ B,$ are in harmonic conjugation.

So, we have $ \frac {E'B}{E'N} \equal{} \frac {KB}{KN}$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ \frac {E'B}{E'N} \equal{} 3 \equal{} \frac {EB}{ED} \equal{} \frac {EB}{EN}$ $ ,(3)$

From $ (3)$ $ \Longrightarrow$ $ E'\equiv E$ and the proof is completed.

$ \bullet$ This proof is dedicated to [b][size=100]Anastasios Vafeidis[/size][/b].

Kostas Vittas.

[quote="Virgil Nicula"] [color=darkred]Let $ ABCD$ be a convex quadrilateral inscribed in the circle $ w = C(O)$ . Denote $ E\in AC\cap BD$ .

Suppose $ EB = 3\cdot ED$ and define $ \{\begin{array}{c} M\in (BC)\ ,\ MB = MC \\
 \\
N\in (BE)\ ,\ NE = ED \\
 \\
\ P\in MN\cap AB\end{array}$ . Prove that $ EP\perp EO$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote $ \{\begin{array}{c} R\in MN\cap BC \\
\ S\in PE\cap DC\end{array}$ . Apply the [b]Menelaus' theorem[/b] to the transversal $ \overline {MNR}$ and the triangle $ BEC$ :

$ \frac {RE}{RC}\cdot\frac {MC}{MB}\cdot\frac {NB}{NE} = 1$ $ \implies$ $ RC = 2\cdot RE$ $ \implies$ $ ER = EC$ . Thus, $ \{\begin{array}{c} ER = EC \\
\ EN = ED\end{array}\|$ $ \implies$ $ \overline {MNPR}\parallel \overline {DSC}$ $ \implies$ $ EP = ES$ .

Apply the [b]property of butterfly[/b] to the interior point $ E\in BD\cap AC$ of the circle $ w$ : $ \boxed {\ EP = ES\ \Longleftrightarrow\ EP\perp EO\ }$ .[/color]

[quote="Virgil Nicula"] [color=darkred][b][u]An easy extension.[/u][/b] Let $ ABCD$ be a convex quadrilateral inscribed in the circle $ w = C(O)$ . Define the points :

$ E\in AC\cap BD$ ; $ \{\begin{array}{c}
N\in (BE)\\\
EN=ED\end{array}$ ; $ \{\begin{array}{c}
R\in AC\\\
A\in (RC)\\\
ER=EC\end{array}$ ; $ P\in NR\cap AB$ . Prove that $ EP\perp EO$ .[/color] [/quote]

Dear Mathlinkers,
this problem comes also from
Duman  A. N., Problem 1669, Mathematics Magazine vol. 76, 2 (2003) 151
and can be solved with the butterfly theorem
see
http://perso.orange.fr/jl.ayme  vol. 7  A new metamorphosis of the butterfly theorem p. 70
Sincerely
Jean-Louis
Sincerely
Jean-Louis

To avoid using butterfly theorem: let $PE$ cut the circumcircle at $K, H$ and cut $DC$ at $G$. It is obvious that $PE = EG$, if we can prove that $KP = GH$ then we will be done.
Using sinus rule we have $GC/sinE_{1} = EG/sinC_{1}$ and $AP/sinE_{1} = PE/sinA_{1}$, hence $GC/AP = sinA_{1}/sinC_{1}$. Similarly $BP/DG = sin D_{1}/sinB_{1}$. Thus $GC/AP = BP/DG$ or  $GC.GD = PA.PB$ or $GH.GK = PK. PH$, it follows that $GH = KP$ (q.e.d)


analytic geometry
FTW
email
ARML
probability
expected value

What scores do people think they got? What did people get as answers?

I got 134, that is, 22 right, 2 wrong, 1 omit. I got 15 and 24 wrong, and I omitted 23.

I don't remember all the answers, but here are a few, checked with others at my school:

#25 is 26
#24 is D
#23 is 272
#21 is E

the giant problem is 30
#15 is 13

And a certain freshman at my school got a very good score...he'll tell you more later...

although i got more problems correct this year vs. last year,
i thought it was harder

(also consider that ive gotten better at math)

I thought last year's was a little harder.
There was that tricky expected value one and also the weird mass points problem.
This year's was pretty straightforward; only 23 required a lot of work.

I will not make any claims about my score just in case.

Hint: He got a 150.

I did not take it becuase my school doesn't do it this year (for some odd reason...)

Yongyi, why is your school so amazing but so stupid at the same time?
They let you take calc in 8th grade but not MAML...

I have no idea :(

:noo: :noo:

On Tuesday, I knew MAML started then and I asked my coach about it, and he said there's no math competition today (meaning Tuesday).

I got an 88, same as last year, partially because I thought east was a different direction........ seemingly its along the positive x axis if you r at the origin. who knew?

I got 144, switched the people on #19. Oh well.

grr...made 2 careless mistakes

132 this year... :(

i thought the jack didn't move when the giant was chasing him...

didn't we all learn that when a big animal is chasing you, you stand still? (jurassic park any1?)

[quote="alkjash"]Yea...
That would make so much sense
An AMC level question in which there is only one thing moving...[/quote]

well..mmo is so ez that i assumed that there was only one thing that was moving... :maybe:

like seriously, the first 10 questions are like free points

[quote="Pineappleperson"]like seriously, the first 10 questions are like free points[/quote]

lol it took me like 20 minutes to get through like the first 5 questions...then I happened to look at the clock and I realized I needed to work faster to get through the questions

132: 21 right, 1 wrong, 3 omit

I got 11 wrong (Somehow I managed to say: $ \frac {5}{3} \pi \left(\frac {r}{2}\right)^2 \equal{} \frac {5}{4} \pi r^2 \equal{} \cdots \equal{} 54\pi$. Because obviously $ \frac {5}{3} \pi \left(\frac {r}{2}\right)^2 \equal{} \frac {5}{4} \pi r^2$.  :wallbash_red: ). I omitted 23 (too lazy to coordinate bash/trig bash it), 24 (thought the calculations were really long...only after the contest did I read the word "integer"), and 25 (is the answer 26 or 50? in other words, could "coincide" include the cases are in opposite directions but on the same line?)

[quote="CatalystOfNostalgia"]Is the meeting being re-postponed because of snow?[/quote]
It just was.

I'm 99% sure the finalist cutoff was settled by email (I participated in the discussion but it's hard to "vote" by email, and not everyone contributed to the conversation).  The finalist cutoff was needed so L2 site numbers could be determined and the sites could be contacted last week.  The semifinalist cutoff is probably set, just by tradition.

I just sent an email to the board proposing we release the results at noon today using the agreed-upon finalist and traditional semifinalist cutoffs.  If no one objects, I'll send an email here when the information is online.  If the information isn't here by, say, 1 PM, send me a PM and I'll tell you your score and rank.

[quote="Phelpedo"]Why doesn't the MAML board just use a set formula, like setting the cutoff to be the highest possible score that will make at least 100 people be finalists?[/quote]
Basically we do.  We discuss the L1 cutoffs at the meeting mainly for the formality.  (The meeting has to exist anyway, to discuss details of the state championship math meet, the L2 Olympiad, ARML funding, Math Day, etc.).  In case there is something like a 400 way tie for 99th place, of course, a formula like you mention would break down.  e.g., this year the choice was between 96 finalists and 120 finalists, and we chose the latter.

I'll have the results up on the WOCOMAL website shortly.

Here are the finalists and semifinalists:  http://www.wocomal.org/statistics/olympiad/200809/200809o_level1.html

As always, the names you see are the names that the Scantron picked up, so if there were miscodings (or incomplete codings) of names there will be errors.  Let me know any errors that you see, and I'll fix them.  Thanks.

Do only the finalists take the next round?

Mr. Yanco, would it be possible to also put up an unofficial list that includes Phillips Academy participants?

Thanks for putting this up!

Lexington name corrections:
"Elkin Ro" should be Welkin Uttaro. Welkin has failed to bubble his name correctly on pretty much all of his AMC tests...
"Kristine Im" should be Kristine Kim. She is a junior.
Ha-Young as in Ha-Young Kang is hyphenated. Not a big deal.
"JaeYoon Lee" should be Jaeyoon Lee, without a capital Y.
"Ju Zheng" should be Julia Zheng (but in this case feel free to leave it as Ju purely for comedic value).

[quote="rjyanco"][quote="Phelpedo"]would it be possible to also put up an unofficial list that includes Phillips Academy participants?[/quote]  All of the PA participants who scored high enough to be a semifinalist or better are listed at the bottom of the page (as is the DA postgraduate and the Newton eighth grader).  Were you asking for them to be sorted into the list?[/quote]

Ah, I didn't look at the bottom. It's great as is, thanks.

I just fixed all of the name errors I was told about (five Lexington, one AB).  Let me know if you see anything else.

This has been corrected on the main MAML site, but Sherrie Ng of Sharon should be Sherrie Wang.  Thanks.

Thanks also for keeping us all up to date on just about everything.

MM

[quote="rjyanco"]Thanks for pointing that out -- the official results are now online at http://www.maml.org.  (Yesterday the complete results, of all ~2500 students, were available there behind a very weak password.  I didn't want to direct you there until that got straightened out.)[/quote]

I was wondering what that site was going to be used for; I guess now we know.

Just to let you know, some (if not all) of the mistakes that were corrected on the wocomal site were not corrected on the maml site. For instance, "Im Kristine" of Lexington High School should be "Kim Kristine", and she's in 11th grade; "Ro Elkin" is really "Uttaro Welkin"; and so on.

[quote="calc rulz"]Just curious: how come 8th graders aren't allowed to advanced?[/quote]
The obvious answer is that a majority of those on the MAML board voted that way.  I believe the majority argument was that talented middle school students would get their chance to advance when they reached high school, so they shouldn't displace current (albeit less talented) high school students.


inequalities
calculus
derivative
Cauchy Inequality
three variable inequality

I posted some time ago but there was noone interested in: http://www.mathlinks.ro/Forum/viewtopic.php?t=119604

[quote="toanhocmuonmau"]Let $a,b,c$ be positive real numbers. Prove that
\[\underset{cyc}{\sum}\frac{a}{b}\ge \underset{cyc}{\sum}\sqrt{\frac{a^{2}+c^{2}}{b^{2}+c^{2}}}\]
:)[/quote]
Actually, the following inequality holds for any non-zero real $x$:
\[\underset{cyc}{\sum}\frac{a}{b}\ge \underset{cyc}{\sum}{(\frac{a^{x}+c^{x}}{b^{x}+c^{x}}})^\frac 1{x}\]

[quote="Vasc"]
Actually, the following inequality holds for any non-zero real $x$:
\[\underset{cyc}{\sum}\frac{a}{b}\ge \underset{cyc}{\sum}{(\frac{a^{x}+c^{x}}{b^{x}+c^{x}}})^\frac 1{x}\]
[/quote]
How two prove it?

I am guessing that the only way is derivatives

Vasc, does\[\sum_{cyc}\frac{a_{1}}{a_{2}}\geq \sum_{cyc}\left(\frac{a_{1}^{x}+a_{3}^{x}}{a_{2}^{x}+a_{3}^{x}}\right)^{\frac{1}{x}}\]hold?

[quote="me@home"]I am guessing that the only way is derivatives

Vasc, does
\[\sum_{cyc}\frac{a_{1}}{a_{2}}\geq \sum_{cyc}\left(\frac{a_{1}^{x}+a_{3}^{x}}{a_{2}^{x}+a_{3}^{x}}\right)^{\frac{1}{x}}\]
hold?[/quote]
Without derivatives.
I think it is not true for $n>3$.

To prove $\underset{cyc}{\sum}\frac{a}{b}\ge \underset{cyc}{\sum}{(\frac{a^{x}+c^{x}}{b^{x}+c^{x}}})^\frac 1{x}$, we can use the known statement:

Let $a,b,c$ and $x,y,z$ be positive numbers such that
$abc=xyz$, $max\{a,b,c\}\geq max \{x,y,z\}$ and $min\{a,b,c\}\leq min \{x,y,z\}$.
Then $a+b+c \geq x+y+z$. :wink:

Dear toanhocmuonmau I am very curious to see your elegant solution as I cannot see it.

Thank you very much.

[quote="HTA"][quote="cefer"]Let's prove that $\frac{a^{2}}{b^{2}}+2\frac{b}{a}\geq \frac{a^{2}+c^{2}}{b^{2}+c^{2}}+2\sqrt{\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$
[/quote]
wonderful step :D , could you plz tell me how did you find out this ? :D[/quote]

As you see I solved this problem in December, so I dont remember why I did this step,but may be I did it for a symmetry.

[quote="Vasc"]To prove $ \underset{cyc}{\sum}\frac {a}{b}\ge \underset{cyc}{\sum}{(\frac {a^{x} + c^{x}}{b^{x} + c^{x}}})^\frac 1{x}$, we can use the known statement:

Let $ a,b,c$ and $ x,y,z$ be positive numbers such that
$ abc = xyz$, $ max\{a,b,c\}\geq max \{x,y,z\}$ and $ min\{a,b,c\}\leq min \{x,y,z\}$.
Then $ a + b + c \geq x + y + z$. :wink:[/quote]

WLOG    $ a\ge b\ge c$  and $ x\ge y\ge z$  then $ a\ge x$ , $ c\le z$ and $ ab\ge xy$

From Abel formula and AM-GM inequality we have

${ a + b + c - x - y - z = (x - y)(\frac {a}{x} - 1) + (y - z)(\frac {a}{x} + \frac {b}{y} - 2) + z(\frac {a}{x} + \frac {b}{y} + \frac {c}{z} - 3) \ge(x - y)(\frac {a}{x} - 1) + (y - z)(2\sqrt {\frac {ab}{xy}} - 2) + z(3\sqrt [3]{\frac {abc}{xyz} }- 3})\ge0$
Hence

$ a + b + c\ge x + y + z$

Now

${{{ \frac {a}{b}.\frac {b}{c}.\frac {c}{a} = (\frac {a^{x} + c^{x}}{b^{x} + c^{x}}})^\frac 1{x}. (\frac {b^{x} + a^{x}}{c^{x} + a^{x}}})^\frac 1{x}. (\frac {c^{x} + b^{x}}{a^{x} + b^{x}}})^\frac 1{x} = 1$

Can help me to prove

${{{{ \max\{{\frac {a}{b},\frac {b}{c},\frac {c}{a}}\}\ge\max\{(\frac {a^{x} + c^{x}}{b^{x} + c^{x}}})^\frac 1{x},(\frac {b^{x} + a^{x}}{c^{x} + a^{x}}})^\frac 1{x},(\frac {c^{x} + b^{x}}{a^{x} + b^{x}}})^\frac 1{x}}\}$

and

${{{{{ \min\{\frac {a}{b},\frac {b}{c},\frac {c}{a}}\}\leq \min\{(\frac {a^{x} + c^{x}}{b^{x} + c^{x}}})^\frac 1{x},(\frac {b^{x} + a^{x}}{c^{x} + a^{x}}})^\frac 1{x},(\frac {c^{x} + b^{x}}{a^{x} + b^{x}}})^\frac 1{x}}\}$

[quote="manlio"]Dear can_hang2007, can you please post your impressive solution by contradiction method?

Thank you very much.

I am very curious to see it.[/quote]
Here is it :)

Thank you very much, can_hang2007.

Very nice inequality!!!


algebra polynomial MATHCOUNTS calculus integration inequalities induction

N is a natural number such that $2^x>x^8$ for all x>N. What is the minimum value for N?

(Source: Mathcounts 2000-2001 Workout 2)

Another way to do it:

Taking log (base 2) of both sides, we have
2^x>x^8 iff
x>8*log_2(x)

Testing powers of 2s, we find that x is between 32 and 64. After some guessing and checking, we find that
43<8*log_2(43) and
44>8*log_2(44),
so N=43.

But what's the non-calculus way to prove that
2^x>x^8
for all x>43? I couldn't find a rigorous way.

Well, you know that N has to be bigger x when $2^x = x^8$, so you could just solve for $2^x = x^8$.  But I really don't know where to start.

[quote="EFuzzy"]N is a natural number such that $2^x>x^8$ for all x>N. What is the minimum value for N?[/quote]The answer to that actually depends on whether $x$ is defined as a real variable or an integral variable. I'll assume $x\in\mathbf{N}$.[quote="SnowStorm"][hide]
44
I couldn't think of any better way than to just do:
$\root 8 \of {2^{x}} > x$
And then just guess and check:
$\root 8 \of {2^{44}} = 45.25$
[/hide][/quote]If you're going to use a calculator you don't need to do any manipulation of the inequality beforehand. Also, I doubt that $\root 8 \of {2^{44}}$ is rational, since $\root 8 \of {2^{44}}=\root 8 \of {(2^8)^5\cdot2^4}=\root 8 \of {(2^5)^8\cdot2^4}=2^5\cdot\root 8 \of {16}$. The eighth root of sixteen is not rational. Say $\frac{m}{n}$ is the eighth root of sixteen and $m,n$ are two relatively prime integers. Intuitively, if $n\nmid{m}$ then $n^8\nmid{m^8}$, so $\frac{m^8}{n^8}\not\in\mathbf{Z}\Rightarrow\frac{m^8}{n^8}\ne16$.[quote="Wumbate"]Here's a way.
First, assume $2^x>x^8$ for some x (43 for example).  Also, $2<(1+\frac{1}{x})^x<3$.  So
$2^x*(1+\frac{1}{x})^x>x^8*(1+\frac{1}{x})^x$
And $(1+\frac{1}{x})^x>(1+\frac{1}{x})^8$ for x>8, which it is in this case.
Therefore $2^x*(1+\frac{1}{x})^x>x^8*(1+\frac{1}{x})^8=(x+1)^8$
Now since $2^x>3>(1+\frac{1}{x})^x$ for x>1, the left side can become

Edit: I made a mistake.  Can anyone fix this?[/quote]I've seen induction proofs that use $2\le\left(1+\frac{1}{n}\right)^n<3$ but I can only see how to use this to show that $m^n>n^k$ for all $n\ge{N}$ for some given $N$ when $m\ge3$, but unfortunately $m=2$..[quote="h_s_potter2002"]you could just solve for $2^x = x^8$[/quote]I'll assume you meant to solve for $x$. Still, the problem will require some sort of reasoning rather than just solving an equation. I'm not sure that equation can be solved algebraically anyway.


There is a right triangle, ABC, where C is the right angle. What is the value of (Cos A squared) plus (Sin A squared)?

It's a well-known trig identity. [hide="Answer"] $ \sin^{2} \plus{} \cos^{2}x \equal{} 1$.[/hide]

Or you can prove it by relating to pythagorian theorem..

@aznotaku13=:lol: Nice problem, my friend. Just out of curiousity................
Isn't this the problem you show this to me in Chinese class?

yea, it was, but i was bored and wanted to see who could figure it out besides you since you already know.


So are these questions that we thought of that can become research questions, or can this also be ideas for new approaches to already stated problems?

[quote="solafidefarms"]And ... why would someone post an idea here instead of doing it with themselves or their friends?[/quote]
To share it with others here.  There was a request in the round table for this forum; many wanted it, including me.  Why would you keep an idea to yourself?

Check http://www.mathlinks.ro/viewtopic.php?t=194500 for the meaning of this forum, before Valentin or anyone else has given an official description. And feel free to contribute ideas.

  darij


algebra
polynomial
calculus
calculus computations

Find the sequence Yn that satisfies the following difference equation
Y(n+1) - 2Y(n-1) = n  , where n=1,2,3,...

$ Y_{n\plus{}1} \minus{} 2Y_{n\minus{}1} \equal{} n ,\;  n\in \mathbb{N}$.

OK, this shouldn't be a problem. First, solve the homogeneous analogue of the equation. Its solution is $ 2^{n \minus{} 1}C$, right? (make the characteristic polynomial equation). Then, guess a particular solution. One possibility is to look for it in a form of $ c_1 n \plus{} c_2$, as the RHS of the equation is $ n$. After plugging it in your equation, you get $ c_1 \equal{} \minus{} 1, c_2 \equal{} \minus{} 1$, so your solution is:

$ Y_n \equal{} 2^{n \minus{} 1}C \minus{} n \minus{} 1$. Hope this is ok, as I haven't done recurrence relations for a long time.

hsiljak, you missed that the equation is 2nd order, not first order. The characteristic polynomial is $ x^2\minus{}2,$ so the solutions of the homogeneous equation are of the form $ c_12^{n/2}\plus{}c_2(\minus{}1)^n2^{n/2}.$

Also, I get a particular solution of $ \minus{}n\minus{}3$ rather than $ \minus{}n\minus{}1.$ So the general solution is

$ Y_n\equal{}c_12^{n/2}\plus{}c_2(\minus{}1)^n2^{n/2}\minus{}n\minus{}3$

where $ c_1$ and $ c_2$ are determined by initial conditions (which the original poster didn't give).

$ Y_{n \plus{} 1} \equal{} 2Y_{n \minus{} 1} \plus{} n\Longleftrightarrow Y_{n \plus{} 1} \plus{} n \plus{} 1 \plus{} 3 \equal{} 2(Y_{n \minus{} 1} \plus{} n \minus{} 1 \plus{} 3)$. Let $ Z_n \equal{} Y_n \plus{} n \plus{} 3$, we have $ Z_{n \plus{} 1} \equal{} 2Z_{n \minus{} 1}$, yielding $ Z_n \equal{} 2Z_{n \minus{} 2}\ (n\geq 2)$.

Thus For $ n \equal{} 2k\ (k\equal{}0,\ 1,\ 2,\ \cdots)$, we have $ Z_{2k} \equal{} 2Z_{2(k \minus{} 1)}$, yielding $ Z_{2k} \equal{} Z_0\cdot 2^k\ \therefore Z_n \equal{} Z_0\cdot2^{\frac {n}{2}}$,

For $ n \equal{} 2k \minus{} 1\ (k\equal{}1,\ 2,\ 3,\ \cdots)$, we have $ Z_{2k \minus{} 1} \equal{} 2Z_{2(k \minus{} 1) \minus{} 1}$, yielding $ Z_{2k \minus{} 1} \equal{} Z_1\cdot 2^{k \minus{} 1}\ \therefore Z_n \equal{} Z_1\cdot 2^{\frac {n \minus{} 1}{2}}$.

Answer: $ Y_n \equal{} C\cdot 2^{\left[\frac {n}{2}\right]} \minus{} n \minus{} 3\ (n\geq 1)$.

Again, the difference equation is second order, not first order. kunny hasn't captured all solutions.

Using the same basic idea as kunny, but varying the presentation a little:

Define $ Z_n \equal{} Y_n \plus{} n \plus{} 3$ so that $ Y_n \equal{} Z_n \minus{} n \minus{} 3.$

Upon substituting that into the original equation and simplifying, we get

$ Z_{n \plus{} 1} \equal{} 2Z_{n \minus{} 1}.$

Both $ Z_n \equal{} \left(\sqrt {2}\right)^n$ and $ Z_n \equal{} \left( \minus{} \sqrt {2}\right)^n$ are solutions to this, as are also all linear combinations of those two. kunny only wrote the first of those two possibilities.

I believe my answer is right, because I classified the parity of $ n$.

Yes, you did. It's only the very last line of your post, which looks like a single formula with a single constant, that looks misleading. There are two degrees of freedom here, and two pieces of information (such as $ Y_0$ and $ Y_1$) would be needed to pin down the solution.

I'm sorry. I've been solving the equation $ Y_{n\plus{}1} \minus{} 2Y_{n} \equal{} n ,\; n\in \mathbb{N}$, and not the OP's one... Typo...

Oh, that gets easier than first problem.

We have at least 3 solutions. :lol:

For the second difference equation, $ y_{n\plus{}1} \minus{} 2 y_n \equal{} n$,

[hide="Solution"]Applying the z-transform to both sides we get

$ \mathcal{Z}(y_{n\plus{}1}) \minus{} 2 \mathcal{Z}(y_n) \equal{} \mathcal{Z}(n)$

$ z[Y(z) \minus{} y_0] \minus{} 2 Y(z) \equal{} \frac{z}{(z\minus{}1)^2}$

and solving for Y(z) we get

$ Y(z) \equal{} y_0 \plus{} \frac{2y_0z^2 \plus{} (1\minus{}4y_0)z \plus{} 2y_0}{(z\minus{}2)(z\minus{}1)^2} \equal{} y_0 \plus{} \frac{2\plus{}2y_0}{z\minus{}2} \minus{} \frac{1}{(z\minus{}1)^2} \minus{} \frac{2}{z\minus{}1}$.

Now let's find the inverse z-transform of each term:

$ y_n \equal{} \mathcal{Z} ^{\minus{}1}(y_0) \plus{} (2\plus{}2y_0) \mathcal{Z} ^{\minus{}1}\left(\frac{1}{z\minus{}2}\right) \minus{} \mathcal{Z} ^{\minus{}1} \left(\frac{1}{(z\minus{}1)^2}\right) \minus{} 2 \mathcal{Z} ^{\minus{}1}\left(\frac{1}{z\minus{}1}\right)$

$ \equal{}> y_n \equal{} y_0 \delta_n \plus{} (2\plus{}2y_0) 2^{n\minus{}1} u_{n\minus{}1} \minus{} (n\minus{}1) u_{n\minus{}1} \minus{} 2 u_{n\minus{}1}$.

For n=0, we get $ y_n \equal{} y_0$, and for $ n \geq 1$ we have

$ y_n \equal{} (1\plus{}y_0)2^n \minus{} n \plus{} 1 \minus{} 2 \equal{}> y_n \equal{} (1\plus{}y_0)2^n \minus{} n \minus{}1$.
[/hide]

For the first difference equation, $ y_{n\plus{}1} \minus{} 2 y_{n\minus{}1} \equal{} n$:

[hide="Solution"]Start by shifting the index to get $ y_{n\plus{}2} \minus{} 2 y_{n} \equal{} n \plus{} 1$. Now let0s apply the z-transform to both sides:

$ \mathcal{Z} (y_{n\plus{}2}) \minus{} 2 \mathcal{Z}(y_n) \equal{} \mathcal{Z}(n) \plus{} \mathcal{Z}(1)$

$ \equal{}> z^2[Y(z) \minus{} a \minus{} bz^{\minus{}1}] \minus{} 2Y(z) \equal{} \frac{z}{(z\minus{}1)^2} \plus{} \frac{z}{z\minus{}1}$

(where $ a \equal{} y_0$ and $ b \equal{} y_1$)

$ \equal{}> z^2Y(z) \minus{} az^2 \minus{} bz \minus{} 2Y(z) \equal{} \frac{z^2}{(z\minus{}1)^2}$,

and solving for Y(z) we get

$ Y(z) \equal{} a \plus{} \frac{A}{z\minus{}\sqrt{2}} \plus{} \frac{B}{z \plus{} \sqrt{2}} \plus{} \frac{C}{(z\minus{}1)^2} \plus{} \frac{D}{z\minus{}1}$,

where A, B, C, and D are constants that are expressed in terms of a and b. Now applying the inverse z-transform to this equality yields

$ y_n \equal{} \mathcal{Z} ^{\minus{}1}(a) \plus{} A \mathcal{Z} ^{\minus{}1} \left(\frac{A}{z\minus{}\sqrt{2}}\right) \plus{} B \mathcal{Z} ^{\minus{}1} \left(\frac{B}{z \plus{} \sqrt{2}}\right) \plus{} C \mathcal{Z} ^{\minus{}1} \left(\frac{C}{(z\minus{}1)^2}\right) \plus{} D \mathcal{Z} ^{\minus{}1} \left(\frac{D}{z\minus{}1}\right)$

$ \equal{}> y_n \equal{} a \delta_n \plus{} A \cdot (\sqrt{2})^{n\minus{}1} u_{n\minus{}1} \plus{} B \cdot (\minus{}\sqrt{2})^{n\minus{}1} u_{n\minus{}1} \plus{} C(n\minus{}1) u_{n\minus{}1} \plus{} D u_{n\minus{}1}$.

For n = 0 we get $ y_0 \equal{} a$; for n = 1, we have $ y_1 \equal{} A \plus{} B \plus{} D$. For n > 1, the expression for $ y_n$ reduces to

$ y_n \equal{} \frac{A}{\sqrt{2}} \cdot 2^{n/2} \plus{} \frac{B}{\minus{} \sqrt{2}} \cdot (\minus{}1)^n 2^{n/2} \plus{} Cn \minus{} C \plus{} D$.

I have derived the (complicated) expressions for A and B, but for C and D there was no more patience. Of course, this would be much more easy if the original poster has given the initial conditions a and b.
[/hide]

[quote="Carcul"]For the second difference equation, $ y_{n \plus{} 1} \minus{} 2 y_n \equal{} n$,

[hide="Solution"]Applying the z-transform to both sides we get

$ \mathcal{Z}(y_{n \plus{} 1}) \minus{} 2 \mathcal{Z}(y_n) \equal{} \mathcal{Z}(n)$

$ z[Y(z) \minus{} y_0] \minus{} 2 Y(z) \equal{} \frac {z}{(z \minus{} 1)^2}$

and solving for Y(z) we get

$ Y(z) \equal{} y_0 \plus{} \frac {2y_0z^2 \plus{} (1 \minus{} 4y_0)z \plus{} 2y_0}{(z \minus{} 2)(z \minus{} 1)^2} \equal{} y_0 \plus{} \frac {2 \plus{} 2y_0}{z \minus{} 2} \minus{} \frac {1}{(z \minus{} 1)^2} \minus{} \frac {2}{z \minus{} 1}$.

Now let's find the inverse z-transform of each term:

$ y_n \equal{} \mathcal{Z} ^{ \minus{} 1}(y_0) \plus{} (2 \plus{} 2y_0) \mathcal{Z} ^{ \minus{} 1}\left(\frac {1}{z \minus{} 2}\right) \minus{} \mathcal{Z} ^{ \minus{} 1} \left(\frac {1}{(z \minus{} 1)^2}\right) \minus{} 2 \mathcal{Z} ^{ \minus{} 1}\left(\frac {1}{z \minus{} 1}\right)$

$ \equal{} > y_n \equal{} y_0 \delta_n \plus{} (2 \plus{} 2y_0) 2^{n \minus{} 1} u_{n \minus{} 1} \minus{} (n \minus{} 1) u_{n \minus{} 1} \minus{} 2 u_{n \minus{} 1}$.

For n=0, we get $ y_n \equal{} y_0$, and for $ n \geq 1$ we have

$ y_n \equal{} (1 \plus{} y_0)2^n \minus{} n \plus{} 1 \minus{} 2 \equal{} > y_n \equal{} (1 \plus{} y_0)2^n \minus{} n \minus{} 1$.
[/hide][/quote]

Hum, your solution is overdone for the problem.

Here is my solution.

Solution 1:

$ y_{n \plus{} 1} \equal{} 2(y_n \plus{} n)\Longleftrightarrow y_{n \plus{} 1} \plus{} (n \plus{} 1) \plus{} 1 \equal{} 2(y_n \plus{} n \plus{} 1)$, yielding $ y_n \plus{} n \plus{} 1 \equal{} 2^n(y_0 \plus{} 0 \plus{} 1)$.

$ \therefore y_n \equal{} (y_0 \plus{} 1)2^n \minus{} n \minus{} 1\ (n\geq 0)$.

Solution 2:

$ y_{n \plus{} 2} \equal{} 2y_{n \plus{} 1} \plus{} (n \plus{} 1)$
$ y_{n \plus{} 1} \equal{} 2y_n \plus{} n$

Subtracting both sides gives $ y_{n \plus{} 2} \minus{} y_{n \plus{} 1} \equal{} 2(y_{n \plus{} 1} \minus{} y_n) \plus{} 1\Longleftrightarrow y_{n \plus{} 2} \minus{} y_{n \plus{} 1} \plus{} 1 \equal{} 2(y_{n \plus{} 1} \minus{} y_n \plus{} 1)$, yielding

$ y_{n \plus{} 1} \minus{} y_n \plus{} 1 \equal{} 2^n(y_1 \minus{} y_0 \plus{} 1)$, since $ y_1 \equal{} 2y_0 \plus{} 0$, we have $ (2y_n \plus{} n) \minus{} y_n \plus{} 1 \equal{} 2^n(y_0 \plus{} 1)$.

$ \therefore y_n \equal{} (y_0 \plus{} 1)2^n \minus{} n \minus{} 1\ (n\geq 0)$.


[quote="rrusczyk"]The requesting, posting, and sharing of copyrighted files of books on this forum must stop.  From now own, we will be banning anyone we see doing this.  Sharing publicly available files is ok, but asking for books that are clearly under copyright is not.  Authors work very hard to share their knowledge, at a very small price compared to the value of that knowledge.  Your theft of this work makes it harder for people to be able to afford producing any more new materials (or to support free materials, like this site in general).  The authors are not getting rich off this material - believe me, they could make much more money elsewhere.  But they and their families do need to eat.[/quote]

Ok! We are agree  :wink:

Again I repeat, do not share files on copyright on the forum without the author's permission. If you wanna do that in private that's your business (although still wrong), but don't do it over the forums please!

:blush: I don't think I am wrong, sorry!

Sorry i edited my post this will not happen again

[quote="rrusczyk"]The requesting, posting, and sharing of copyrighted files of books on this forum must stop.  From now own, we will be banning anyone we see doing this.  Sharing publicly available files is ok, but asking for books that are clearly under copyright is not.  Authors work very hard to share their knowledge, at a very small price compared to the value of that knowledge.  Your theft of this work makes it harder for people to be able to afford producing any more new materials (or to support free materials, like this site in general).  The authors are not getting rich off this material - believe me, they could make much more money elsewhere.  But they and their families do need to eat.[/quote]
You're right.

[quote="rrusczyk"]The requesting, posting, and sharing of copyrighted files of books on this forum must stop.  From now own, we will be banning anyone we see doing this.  Sharing publicly available files is ok, but asking for books that are clearly under copyright is not.  Authors work very hard to share their knowledge, at a very small price compared to the value of that knowledge.  Your theft of this work makes it harder for people to be able to afford producing any more new materials (or to support free materials, like this site in general).  The authors are not getting rich off this material - believe me, they could make much more money elsewhere.  But they and their families do need to eat.[/quote]

Good decision.


trigonometry
factorial
search
algebra
polynomial
function
geometric sequence

I'm writing next year's Math Team Selection Test for our school and I need to know the difficulty of the two variations of the following problem. I also want to know if I should have it as a multiple choice or an open answer. Any help would be appreciated. Also, please post how long it took you to do it. Thanks!

- tZf

NO CALCULATORS:
Variation 1: Compute $\lfloor100 sin(\frac{\pi}{31})\rfloor$

Variation 2: Compute $\lfloor1000 sin(\frac{\pi}{31})\rfloor$

well, you could do multiple choice and part of the problem would be looking at the answers and eliminating the wrong answer, i don't see how you could reasonably be able to figure out that trig value without a calculator

well if you had to and had way too much time on your hands, you could find an approximation using the half angle identities.

zuton force, how were you planning to solve that problem on your own?

It's actually somewhat trivial
For small angles, sinx is approximately equal to x

$\pi/31$ is about 0.1013
$sin(\pi/31)$ is about 0.1012

So the answer would be 10 for variation 1 and 101 for variation 2

Meh. I guess I'll trash it

that is not really a rigirous answer, you cannot know that it will not deviate enough for the hundredths digit to remain constant

change the problem slightly to be the sum of a gemoetric series or something, at least try to find a trick..

[quote="t0rajir0u"]I can't think of a smart way to do this and Taylor approximation without a calculator would be horrid.  :(  I think you should focus on problems with elegant and clever solutions.[/quote]

Well actually if you think about it, you only need the first two terms. The third term is about $\frac{0.1^5}{5!}$, which is obviously too insignificant to affect anything in the first three terms to the right of the decimal. (Even the second term is completely insignificant, but that's going back to saying sinx is nearly x.)

Of course, I wouldn't expect people to use taylor series expansions anyway, so I guess I'll scrap it, especially since I've accumulated a large pool of non-trivial questions :|

Thanks for the feedback everyone!  :)

You really need the Lagrange Remainder term of the taylor expansion to say anything about significant figures here. Just because you feel intuitively that all the other terms in the series won't matter...

This is basically a pointless problem that doesn't measure anything, especially since it's not even clear that you want an approximate value of the function.

Your school has an entrance exam for the math team? Anyone can go to ours (TJ's), and at my base school, I imagine they would be begging people to join due to lack of intrest.

[b]Lord[/b] - Unlike TJ, public schools are filled with people that don't know anything that join random clubs for the sake of adding extra-curricular activities to their resumees.

I don't know about what it's like down there in VA, but in our school is filled with resumee-freaks that purposely burn away to achieve nothing but a 10000 page resumee.

Now, (arguably thanks to Bush) we're facing severe shortages in money and the math team is basically now running on it's own money. We can't afford to pay for people that don't have their head in it, and we don't want to start charging obscene ammounts of money to people that join for the math.

[b]Kalle[/b] - I know what you're driving at. However, intuition is a very important skill, especially since that's all that some of the people come in with. At most American public schools, math is taught extremely mechanically, so even for people that have covered trig and such material, many of them have not had much practice with creative mathematical thinking.

About the "Lagrange Remainder term" - I don't really know what this is. But the terms are obviously decreasing faster than a geometric progression. Pretend that after the second term, the terms decrease geometrically. If we draw out an infinite geometric progression based on terms 2 and 3, it's still insignificant. So in the real expansion with the factorials, it would be even more insignificant.

I am starting a search for the problem with the most creative solution. Take it to mean what you want, buty PM me with the question/answer if you think you have a good problem.

[quote="The Zuton Force"]
About the "Lagrange Remainder term" - I don't really know what this is. But the terms are obviously decreasing faster than a geometric progression. Pretend that after the second term, the terms decrease geometrically. If we draw out an infinite geometric progression based on terms 2 and 3, it's still insignificant. So in the real expansion with the factorials, it would be even more insignificant.[/quote]

The Lagrange Remainder for the $k^{th}$ order Taylor polynomial $T_k(x)$ of an infinitely differentiable function $f(x)$ is given by $\frac{ f^{(k+1)} (z) x^{k+1}}{(k+1)!}$, where $f^{(k+1)} (z)$ is the maximum value of $f^{(k+1)} (x)$ (and $f^{(k+1)} (x) = \frac{d}{dx} f^{k} (x), f^{(0)} (x) = f(x)$.

Now, for $\sin x$ the term simply becomes $\frac{x^{k+1}}{(k+1)!}$, which, for $x \approx 0.1$ only definitely has error less than $\frac{1}{1000}$ for $k \ge 3$.  Hence it is only necessary to take $\sin x \approx x - \frac{x^3}{3!}$ to calculate an approximation.

The point is that Lagrange is a way to say, [i]rigorously[/i], that any new terms are insignificant.  You can't just say it.

Well, I have it, uh... learned.  I guess I have an advantage over the other people in my Calc class in that I taught myself Taylor series sometime in 8th grade due to ridiculous boredom, so I didn't have to deal with how poorly the book explains it.  :)

(Oh, I forgot to add that the Lagrange Remainder Term gives $\epsilon$ such that $| f(x) - T_k (x) | \le \epsilon$.)

[quote="me@home"]how did you teach it to yourself? sat down with dad's books at home? :huh:[/quote]

Internet.  Actually, I didn't clearly remember what I read about Taylor series on the internet, so I basically ended up re-deriving the concept myself... minus all those qualms about convergence :)


geometry
perimeter
inradius
vector
linear algebra

Greetings,

Got a question.  Does anybody how to [b]prove that the perimeter of a right triangle is equal to twice the area[/b]?

This was asked as an extra credit in a brain teasers contest here and the organizer didn't have the solution, and nobody got it either...

Todd

Hmm..try a right triangle with sides 1,1, $\sqrt{2}.$  Perimeter is $2+\sqrt{2}$ and area is $\frac 12.$  Obviously this is a contradiction.

yes, yes, that's an example.

 But the question is how to PROVE it, much like
proving the pythagorean (sp?) theorem.

I think it starts with writing the perimeter as a+b+c and then
substituting in a^2+b^2 for c, then a bit of magic occurs,
and you end up with it equaling ab/2.

I see the contradiction, so maybe I'm not remembering the qualifiers on what kind of triangle completely.  But isn't there a perimeter to area relationship for triangles?

Uh, no, that was a counterexample.  The perimeter is $2+\sqrt{2}$ whereas twice the area is 1.  $2+\sqrt{2}\neq 1.$  Therefore [b]the perimeter of every right triangle is not necessarily equal to its twice its area.[/b]

[edit: ok, i just saw your post (i was writing this while you posted)]

Perhaps it was $A=rs$ where $r$ is the radius of the inscribed circle and $s$ is half the perimeter?

If that's it I can write up a proof for you.

Maybe it has to have whole number sides... but then 20 21 29 wouldn't work because you have the perimeter=90 and the area=210... *thinks about it* *5 seconds later* *gets bored* *gives up*

I dont really get what this thread is talking about, but oh well...

A=rs=rp/2=2p => r=4. So the perimeter is only twice the area when the inradius is 4. Can't find a specific example though...

Treething, to disprove something a simple counterexample is all that is needed.  I already provided one so I don't see why you are doing that.

The problem could be to prove that there is only one right triangle with integer sides whose perimeter is equal to twice its area.

Well, everyone, this problem is posted in the [b]Game And Fun Factory[/b]
So i think we could try to find some interesting ways to prove that it is [b]true[/b] for [b]Every right triangles[/b]!

Use Heron's. Direct disproval.

$2\sqrt{(\frac{a+b+c}{2})(\frac{-a+b+c}{2})(\frac{a-b+c}{2})(\frac{a+b-c}{2})}$ is not equal to $ab$.

Solafide, those two expressions [i]are[/i] equal for any right triangle!  Just because it isn't obvious that two things are equal doesn't mean they aren't.  (I get this very much when grading linear algebra problems ... "Well, these vectors are not necessarily the same, so their span is different."  If you want to show something doesn't work, simply give a counter-example.

[quote="Treething"]Time to disprove it.

I'm assuming it's true and working from there. x is the length of one leg, and y is the length of the other one.

x + y + sqrt(x :^2: + y :^2: ) = xy/2
sqrt(x :^2: + y :^2: ) = xy/2 - x - y
sqrt(x :^2: + y :^2: ) = xy - 2x - 2y
x :^2: + y :^2: = (xy - 2x - 2y) :^2:
x :^2: + y :^2: = x :^2: y :^2: - 2x :^2: y - 2xy :^2: - 2x :^2: y + 4x :^2: + 4xy - 2xy :^2: + 4xy + 4y :^2:
x :^2: + y :^2: = x :^2: y :^2: - 4x :^2: y - 4xy :^2: + 8xy + 4x :^2: + 4y :^2:
x :^2: y :^2: - 4x :^2: y - 4xy :^2: + 8xy + 3x :^2: + 3y :^2: = 0
x :^2: y :^2: + 8xy + 3x :^2: + 3y :^2: = 4x :^2: y + 4xy :^2:

(unfinished)[/quote]
You made a mistake between lines two and three of your proof.

There are infinitely many such triangles.  If one leg is [b]a[/b], then the other leg is $2(\frac{a-1}{a-2})$.
and


vector
analytic geometry
trigonometry
geometry unsolved
geometry

1.  Find two unit vectors in $ R^2$ that are perpendicular to $ ( \minus{} 3,4)$
2. If $ \vec{a}$ and $ \vec{b}$ are vectors in 3-space, where $ \vec{a} \equal{} t\vec{b}$, show that $ (\frac {\vec{a}\cdot\vec{b}}{\vec{b}\cdot\vec{b}})\vec{b} \equal{} \vec{a}$
3. Calculate the angles that $ (3,4,1)$ makes with the coordinate axes.
4. In triangle ABC, E is the midpoint of AC and BE is extended to D so that ED=BE. Prove that $ \vec{AB} \equal{} \vec{DC}$.
5. In the quadrilateral ABCD, the midpoints of the sides AB, BC, CD, and DA are P, Q, R, and S, respectively. Prove that PR and QS bisect each other. Do the points A, B, C, D have to lie in a plane for the result to hold?
6. A man weighing 70 kg lies in a hammock whose ropes make angles of 30 and 45 degrees with the horizontal. What is the tension in each rope?
7. Two tugs are towing a ship. The smaller tug is 10 degrees off the port bow and the larger tug is 20 degrees off the starboard bow. The larger tug pulls twice as hard as the smaller tug. In what direction will the ship move?
8. Find the work done by the force $ \vec{F}$ exerted on an object moving along the vector $ \vec{s}$. Assume that the force is measured in newtons and the distance is in meters. ( $ \vec{F} \equal{} (3,2)$, $ \vec{s} \equal{} (4,1)$ )
9. Find a vector perpendicular to both $ \vec{u}$ and $ \vec{v}$. ( $ \vec{u} \equal{} ( \minus{} 2,3,1)$, $ \vec{v} \equal{} ( \minus{} 1,4,0)$ )

[quote="smellypoopy"]1.  Find two unit vectors in $ R^2$ that are perpendicular to $ ( - 3,4)$
2. If $ \vec{a}$ and $ \vec{b}$ are vectors in 3-space, where $ \vec{a} = t\vec{b}$, show that $ (\frac {\vec{a}\cdot\vec{b}}{\vec{b}\cdot\vec{b}})\vec{b} = \vec{a}$
3. Calculate the angles that $ (3,4,1)$ makes with the coordinate axes.
4. In triangle ABC, E is the midpoint of AC and BE is extended to D so that ED=BE. Prove that $ \vec{AB} = \vec{DC}$.
5. In the quadrilateral ABCD, the midpoints of the sides AB, BC, CD, and DA are P, Q, R, and S, respectively. Prove that PR and QS bisect each other. Do the points A, B, C, D have to lie in a plane for the result to hold?
6. A man weighing 70 kg lies in a hammock whose ropes make angles of 30 and 45 degrees with the horizontal. What is the tension in each rope?
7. Two tugs are towing a ship. The smaller tug is 10 degrees off the port bow and the larger tug is 20 degrees off the starboard bow. The larger tug pulls twice as hard as the smaller tug. In what direction will the ship move?
8. Find the work done by the force $ \vec{F}$ exerted on an object moving along the vector $ \vec{s}$. Assume that the force is measured in newtons and the distance is in meters. ( $ \vec{F} = (3,2)$, $ \vec{s} = (4,1)$ )
9. Find a vector perpendicular to both $ \vec{u}$ and $ \vec{v}$. ( $ \vec{u} = ( - 2,3,1)$, $ \vec{v} = ( - 1,4,0)$ )[/quote]

1. $ (\frac{4}{5}, \frac{3}{5}), (-\frac{4}{5}, -\frac{3}{5})$
2. $ (\frac {\vec{a}\cdot\vec{b}}{\vec{b}\cdot\vec{b}})\vec{b} = \frac{t \cdot \vec{a} \cdot \vec{a}}{t^2 \cdot \vec{a} \cdot \vec{a}} \cdot t \cdot \vec{a} = \vex{a}$. (That is when $ \vec{a}^2 \neq 0 \iff \vec{a} = (0,0)$)
3. What do you mean?
4. $ E$ is the midpoint of $ A$ and $ C$ so $ E = \frac{A+C}{2}$. $ E$ is the midpoint of $ B$ and $ D$ so $ E = \frac{B+D}{2} \iff D = 2E - B = A+C-B$. Now $ AB = B-A$ and $ DC = C-D = C-A-C+B = B-A$ so $ AB = DC$.
5. $ P = \frac{A+B}{2}$ and so on. Note that the midpoint of $ PR$ is $ \frac{P+R}{2} = \frac{A+B+C+D}{4}$. The same with $ QS$ and we are done. No they do not.

The rest seems like physics, which I'll not waste my time with.

3.it is actually to find [url=http://mathworld.wolfram.com/DirectionCosine.html]direction cosines[/url] of vector $ \vec{u}\equal{}(3,4,1)$
formula is:
If $ \alpha$ is angle made by vector$ (a,b,c)$ with X axis,then:
$ cos {\alpha}\equal{}\frac{a}{\sqrt{a^2\plus{}b^2\plus{}c^2}}$

6 and 7:Please put in physics forum(Hint:just break each force in horizontal and vertical components and add)

8.It also belongs to physics forum.
$ W\equal{}\vec{F}\cdot \vec{s}$
so,$ W\equal{}3\cdot 4\plus{}2\cdot 1 \equal{}14N$

9.A vector perpendicular to both $ \vec{u}$ and $ \vec{v}$ is given by $ \vec{u} \times \vec{v}$
Please check [url=http://en.wikipedia.org/wiki/Cross_product]cross products[/url]


[quote="ch1n353ch3s54a1l"]Suppose that S is a set of three positive integers. At most, how many of the following four properties could the set S have?

I. S has mean 10
II. S has median 8
III. S has sum 45
IV. S contains the number 35[/quote]

[hide]3 properties.[/hide]

[hide]3 because if the third 1 is true the first one can't be, but if the first one is true the third and fourth one can't and the second is possible no matter what.[/hide]

[hide]3  :P

Number one was used to trick all the mathcounts noobs that want to blow their brains out....

Sorry  :D [/hide]

3   :jump:  :harhar:   1st one isn't right

[hide]
At best, either 1 or 3 isnt true.
so $3$[/hide]

[hide]3 II, III, and IV can all be true.[/hide]


[quote="eryaman"]Do the properties for trig functions hold for inverse trig function?  ie, ${\sin^{-1}x=\cos^{-1}({\frac{\pi}{2}}-x})$[/quote]

No, it should be $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$


Find all natural pairs $ (x,y)$, such that $ x$ and $ y$ are relative prime and satisfy equality: $ 2x^2 \plus{} 5xy \plus{} 3y^2 \equal{} 41x \plus{} 62y \plus{} 21$.

I tried to translate, but may be I made mistake. I meant (x,y)=1. Can you say how it is in English? And there are not 2, but about 12 solutions.

[quote=bambaman]$ 2x^2 \plus{} 5xy \plus{} 3y^2 \equal{} (x \plus{} y)(2x \plus{} 3y)$
$ 41x \plus{} 62y \plus{} 21 \equal{} 21(2x \plus{} 3y) \minus{} (x \plus{} y) \plus{} 21$
$ 2x^2 \plus{} 5xy \plus{} 3y^2 \equal{} 41x \plus{} 62y \plus{} 21$ iff $ (x \plus{} y \minus{} 21)(2x \plus{} 3y \plus{} 1) \equal{} 0$. x, y>0, hence we must have x+y=21. If "reciprocal prime" is a mistake (what can it possible mean?...), then the possible solutions are only (2,19) and (19,2).[/quote]

Nice sol! I did the same thing, but the solutions are (1, 20), (2, 19), (4, 17), (5, 16), (8, 13), (10, 11), and it’s permutations, which gives a total of 12 solutions.


probability
induction
modular arithmetic
floor function
limit
strong induction

Nathan and Abi are playing a game. Abi always goes first. The players take turns changing a positive integer to a smaller one and then passing the smaller number back to their opponent. On each move, a player may either subtract one from the integer or halve it, rounding down if necessary.  Thus, from $ 28$ the legal moves are to $ 27$ or to $ 14$; from $ 27$, the legal moves are to $ 26$ or to $ 13$. The game ends when the integer reaches $ 0$. The player who makes the last move wins. For example, if the starting integer is $ 15$, Abi might move to $ 7$, Nathan to $ 6$, Abi to $ 3$, Nathan to $ 2$, Abi to $ 1$, and now Nathan moves to $ 0$ and wins. (However, in this sample game Abi could have played better!)

[list]a) Assuming both Nathan and Abi play according to the best possible strategy, who will win if the starting integer is $ 1000$? $ 2000$?  [i]Prove your answer.[/i]

b) As you might expect, for some starting integers Abi will win and for others Nathan will win. If we pick a starting integer at random from all the integers from $ 1$ to $ n$ inclusive, we can consider the probability of Nathan winning. This probability will fluctuate as $ n$ increases, but what is its limit as $ n$ tends to infinity?  [i]Prove your answer.[/i][/list]

[hide="a"]Let the starting integer be $ n\equal{}2^{k}\cdot a$, where $ a$ is an odd positive integer and $ k$ is a
nonnegative integer. We claim that Abi wins if and only if $ k$ is even, and Nathan if and only if $ k$ is odd. Thus, since $ 1000\equal{}2^{3}\cdot125$ and $ 2000\equal{}2^{4}\cdot125$, Nathan wins when we start with 1000, and Abi wins when we start with 2000.

We prove this by strong induction on $ n$. The claim is true for $ n\equal{}1$: we have $ 1\equal{}2^{0}\cdot1$, and Abi wins here because she can begin by moving from 1 to 0.

Now, assume the claim holds for all $ n\le x\minus{}1$ for some integer $ x\ge2$. We prove it for $ n\equal{}x$. We have a few cases to consider. First, $ k\equal{}0$, that is, $ x$ is odd. If $ n\equiv1\pmod{4}$, let $ x\equal{}4i\plus{}1$. We want to show that Abi wins. She has two choices for her first move: she can either move to $ 4i$ or $ 2i$. We can write these in the form $ 2^{k}\cdot a$; note that exactly one of these has $ k$ odd. Then, if Abi moves to the one where $ k$ is odd, she has reduced the game to starting with this new number, either $ 4i$ or $ 2i$, which by the inductive hypothesis Nathan wins. However, this game is reversed, since Nathan is "starting". Thus, Abi will win. Now, if $ n\equiv3\pmod{4}$, let $ x\equal{}4i\plus{}3$. Again, we want to show that Abi wins. She can either move to $ 4i\plus{}2$ or $ 2i\plus{}1$. If she does the former, she reduces the game to one in which we start with $ 4i\plus{}2$, but reversed because Nathan starts. But Nathan wins this game, because this is the $ k\equal{}1$ case (as 2 divides $ 4i\plus{}2$, but 4 doesn't). Thus, in the reversed version, Abi wins.

Now, consider $ k>0$ and even (note that $ k\ge2$). We want to show that Abi wins. In her first move, we can have Abi divide the number by 2, in which we get a number with $ k$ odd, a game that Nathan would win. However, the game is now reversed, so Abi wins. Finally, consider $ k>0$ and odd. We want to show that no matter what move Abi moves, she will set up Nathan in a winning position. If she subtracts 1, we will end up with an odd integer (since we start from an even), so $ k\equal{}0$. This gives Nathan a position in which Abi would win, but the game is now reversed, so Nathan will win. Now, if she divides by 2, we will end up with an integer with $ k$ even, in which Nathan also wins (because Abi wins in the non-reversed version). We have thus exhausted all cases, completing the proof.[/hide]

[hide="b"]The answer is $ \dfrac{1}{3}$.

As proven in part (a), Nathan wins if and only if $ k$ is odd. An equivalent statement is that Nathan wins with starting integer $ z$ if and only if $ z\equiv2^{i}\pmod{2^{i\plus{}1}}$ for some odd positive integer $ i$. As $ n$ goes to infinity, the probability that $ z\equiv2^{i}\pmod{2^{i\plus{}1}}$ is $ \dfrac{1}{2^{i\plus{}1}}$, since the residues modulo $ 2^{i\plus{}1}$ are periodic with period $ 2^{i\plus{}1}$. Thus, we want to sum this over all odd positive integers $ i$; we can make the transformation $ i\equal{}2k\minus{}1$, so that we are summing over all positive integers $ k$. Thus, the probability is

$ \displaystyle\sum_{k\equal{}1}^{\infty}(\dfrac{1}{2})^{2k}\equal{}\displaystyle\sum_{k\equal{}1}^{\infty}(\dfrac{1}{4})^{k}\equal{}\dfrac{\dfrac{1}{4}}{1\minus{}\dfrac{1}{4}}\equal{}\dfrac{1}{3}$,

as desired.[/hide]

[hide]Define the sequence $ \{a_k\}_{k \ge 1}$ such that for every positive integer $ i$, if a player can always win if he starts with the number $ i$, then $ a_i = 1$, and if the player must always lose if he starts with $ i$, then $ a_i = 0$.  In other words, if having $ i$ is a winning position, then $ a_i = 1$, and if $ i$ is a losing position, $ a_i = 0$.

Note that $ i$ is a winning position if either $ \lfloor i/2 \rfloor$ or $ i-1$ is a losing position, because then the opponent can be given a losing position.  In addition, $ i$ is a losing position if both $ \lfloor i/2 \rfloor$ and $ i-1$ are winning positions, because then the opponent always gets a winning position.  Thus, using the language of Boolean logic, \[ a_i = (\neg a_{\lfloor i/2 \rfloor }) \vee (\neg a_{i-1}).\] Trivially, $ a_1 = 1$, because either operation will give 0.

Lemma: $ a_{2n-1} = 1$ and $ a_{2n} = \neg a_n$ for all positive integers $ n$.

Proof:  We proceed by induction.  The base case is clearly satisfied: $ a_1 = 1$ and $ a_2 = 0$.  Assume that that the inductive hypothesis is true for $ n=k$.  Then $ a_{2(k+1) - 1} = a_{2k+1} = (\neg a_k) \vee (\neg a_{2k}) = a_{2k} \vee (\neg a_{2k}) = 1$, and $ a_{2(k+1)} = a_{2k+2} = (\neg a_{k+1}) \vee (\neg  a_{2k+1}) = (\neg a_{k+1}) \vee 0 = \neg a_{k + 1}$, as desired, so the induction is complete.

Corollary:  For any positive integer $ n$, let $ 2^p$ be the highest power of $ 2$ dividing $ n$.  Then if $ p$ is odd, $ a_n = 0$, and if $ p$ is even, $ a_n = 1$.

PART A
From our lemma, it is clear that $ a_{1000} = \neg a_{500} = a_{250} = \neg a_{125} = \neg 1 = 0$, so since Abi goes first, he will lose.  Finally, $ a_{2000} = \neg a_{1000} = 1$, so Abi will win.

PART B Let $ \{S_k \}_{k \ge 1}$ be a sequence of strings of 1s and 0s such that for every positive integer $ i$, we have $ S_i = a_1 a_2 a_3 \dots a_{2^{i-1}}$, so $ S_1 = 1$, $ S_2 = 10$, $ S_3 = 1011$, $ S_4 = 10111010$, and so on.

Then by our lemma, it is clear that $ S_{n+1}$ is obtained by replacing each 1 in $ S_n$ with the string 10, and replacing each 0 in $ S_n$ with the string $ 11$: if $ a_n=1$, then $ a_{2n-1}$ and $ a_{2n}$ are 1 and 0, and if $ a_n = 0$, then $ a_{2n-1}$ and $ a_{2n}$ are 1 and 1.

Thus, if we let $ p(n)$ be the probability that a randomly chosen digit from $ S_n$ is a 1, then we have $ p(n) = \frac{1}{2} \cdot p(n-1) + 1 \cdot (1 - p(n-1))$, or \[ p(n) = 1 - \frac{1}{2} p(n-1).\]

We will prove that $ p(n) = -\frac{2}{3} (-\frac{1}{2})^n + \frac{2}{3}.$  The base case $ p(1) = 1$ is clear.  If this formula works for $ n=k$, then
\begin{eqnarray*}
p(k+1) &=& 1 - \frac{1}{2} p(k) \\
&=& 1 - \frac{1}{2} \left(-\frac{2}{3} \left(-\frac{1}{2} \right)^{k} + \frac{2}{3} \right)\\
&=& 1 + \frac{1}{3} \left( -\frac{1}{2} \right)^{k} - \frac{1}{3} \\
&=& \frac{1}{3} \left( -\frac{1}{2} \right)^{k} + \frac{2}{3} \\
&=& -\frac{2}{3} \left(-\frac{1}{2} \right)^{k+1} + \frac{2}{3},
\end{eqnarray*}
so the formula works for $ n=k+1$ and the induction is complete.  Thus, we have

\[ \lim_{n \rightarrow \infty} p(n) = \lim_{n \rightarrow \infty} \left( -\frac{2}{3} \left(-\frac{1}{2} \right)^n + \frac{2}{3} \right) = \frac{2}{3}.\]

Since Abi always goes first and Nathan goes second, the number we want (the probability that Nathan wins) is $ 1 - \frac{2}{3} = \boxed{\frac{1}{3}}$[/hide]


topology
real analysis
real analysis unsolved

Let K be a planar, compact and convex set in $\mathbb{R}^n$. Prove that the boundary countains four points which are the verticies of a arbitrary square.

This seems pretty badly stated. If "planar" really means "2-manifold with boundary", we shouldn't bother with $\mathbb{R}^n$; convexity means we can restrict to a plane anyway. If it means something else, it should still obviously reduce to a plane problem- and then you have to do something to exclude a line segment.
Also, the topological boundary of a "2-dimensional" set in $\mathbb{R}^n$ for $n> 2$ is the whole set.

In $\mathbb{R}^2$, the statement is much simpler: compact, convex, nonempty interior.

adapt the solution of this topic : http://www.mathlinks.ro/Forum/viewtopic.php?highlight=area&t=39780
(post No 2). It works fine here.


AMC USA(J)MO USAMO inequalities calculus search function

Is there a list of theorems that you are allowed to use without proof on the USAMO? (yeah, it would be pretty long) I know there are like 12 inequalities that you are allowed to use

Also, if you give a calculus solution that's correct, will you get full credit?

You can pretty much use any theorem, as long as it doesn't trivialize a problem, and as long as you cite it properly. The more bizarre or brutish a theorem you use, the more and more scrutiny it is put under for whether it is cited properly. If you give a calculus solution that's correct, it gets full credit, but it's generally frowned upon, and like citing bizarre or brutish theorems, or generally using any brute force or dumb(you know the last three letters) solution such as calculus on an olympiad problem, the smallest mistake can knock off all of your points. (this question's been asked a lot before, so I'd use the search function next time).

[quote="JSteinhardt"]Give the name, conductions, and result of the theorem.[/quote]

Is that for all theorems? Even for like Pythagorean Theorem?

I thought I read somewhere that for very-well known theorems you only have to state the theorem, I could be wrong though.

You can generally gloss over stuff like the Pythagorean theorem if it's pretty obvious.  The main point of citing theorems is to let the graders know that you know what you are doing.  It is true, however, that some theorems are more equal than others.  A good example of this is Nesbitt's inequality, which is not a fundamental inequality per se, but is nevertheless encountered by vast numbers of students, as it is a good instructional problem.

Anyhow, better than any advice :  look at examples.  Try working problems, and then look at the official solutions to see where you could have been more direct, what different approaches you could have used, and whether you are being needlessly verbose or abstemiously stingy.

Are there any olympiads with problems of the difficulty of AIME problems ranging from #6-15 (medium to hard AIME problems)


algebra
polynomial
number theory
Rational Root Theorem
number theory solved

Solution

Let $ x \equal{}\sqrt{2}\plus{}\sqrt{3}$

$ \Rightarrow x^{2}\equal{} 2\plus{}2\sqrt{6}\plus{}3 \equal{} 5\plus{}2\sqrt{6}$

$ \sqrt{6}$ is irrational so this makes $ x^{2}$ is irrational. Hence $ x$ is irrational. If this were not the case then $ x$ would be rational, and can be expressed as $ x \equal{}\frac{a}{b}$ where $ a$ and $ b$ are integers. This would make $ x^{2}\equal{}\frac{a^{2}}{b^{2}}$ which is rational and this contradicts the fact that $ x^{2}$ is irrational.

Rationality/irrationality is more number theory than it is analysis. sludgethrower's proof does qualify as number theory. I'm moving the topic.

Although it is quite simple to prove the assertion by very elementary tools, you can also prove that the number $ \sqrt{2}\plus{}\sqrt{3}$ is irrational, like this:
 Suppose, by the way of contradiction, that $ \sqrt{2}\plus{}\sqrt{3}$ is rational. Since $ \sqrt{2}$ and $ \sqrt{3}$ are algebraic integers (because for example $ \sqrt{2}$ is root of a monic polynomial with integer coefficients, in our case $ x^{2}\minus{}2$). So, $ \sqrt{2}\plus{}\sqrt{3}$ is algebraic integer (the sum of two algebraic integers is an algebraic integer). It follows that the number $ \sqrt{2}\plus{}\sqrt{3}$ is actually an integer (because a rational number which is algebraic integer, is an integer number). But, it is easy to see that $ 3 <\sqrt{2}\plus{}\sqrt{3}< 4$, contradiction, because there is no integer in the interval $ (3,4)$.

[quote="Cezar Lupu"]Since $ \sqrt{2}$ and $ \sqrt{3}$ are algebraic integers [/quote]

What do you mean please by [size=150][b]algebraic integer[/b][/size]??

An algebraic number is the root of a polynomial with integer coefficients.  An algebraic integer is the root of a [i]monic[/i] polynomial with integer coefficients.  Here $ \sqrt{2}$ is the root of the polynomial $ x^{2}\minus{}2$, and $ \sqrt{3}$ is the root of the polynomial $ x^{2}\minus{}3$.

The algebraic numbers are an interesting and important set because they are the [i]algebraic closure[/i] of the rationals: in other words,

- every rational number is an algebraic number,
- the sum of any two algebraic numbers is an algebraic number,
- the product of any two algebraic numbers is an algebraic number,
- the root of any polynomial with algebraic coefficients is an algebraic number.

It turns out that the algebraic integers are also closed under addition and multiplication.  Hence, Cezar Lupu's reasoning is as follows:  based on some results in algebraic number theory, because $ \sqrt{2},\sqrt{3}$ are roots of monic polynomials with integer coefficients, $ \sqrt{2}\plus{}\sqrt{3}$ is also the root of a monic polynomial with integer coefficients.  This is not hard to show explicitly:

$ x \equal{}\sqrt{2}\plus{}\sqrt{3}\implies$
$ x^{2}\equal{} 5\plus{}2\sqrt{6}\implies$
$ x^{2}\minus{}5 \equal{} 2\sqrt{6}\implies$
$ (x^{2}\minus{}5)^{2}\equal{} 24$

which is a monic quartic polynomial with integer coefficients.  Now by the Rational Root Theorem we conclude that if $ x$ is rational it must be an integer (moreover, an integer that divides $ 25\minus{}24 \equal{} 1$), absurd.

$ \sqrt 2 \equal{}1.41..,\ \sqrt 3 \equal{}1.73...$, therefore $ \sqrt 2\plus{}\sqrt 3\equal{} 3.14...\equal{}\pi$ is irrational.  :D

[quote="t0rajir0u"]An algebraic number ...[/quote]

Thanks a lot for your explanation

Im in class 12 but I've never heard about algeabric numbers, it does'nt exist in Morrocan system  :mad:  :lol:

[quote="Rust"]$ \sqrt 2 \equal{} 1.41..,\ \sqrt 3 \equal{} 1.73...$, therefore $ \sqrt 2\plus{}\sqrt 3 \equal{} 3.14... \equal{}\pi$ is irrational.  :D[/quote]
I don't think that because $ \pi$ is a advance number but $ \sqrt2\plus{}\sqrt3$ is an algebraic number
So $ \sqrt2\plus{}\sqrt3\equal{}\pi$ not true.
Have a same problem
$ Let a,b\in Z$ satisfy
$ \sqrt[3]{3}a\plus{}\sqrt[3]{9}b$ is a rational .
Prove that
$ a\equal{}b\equal{}0$
b)Find the polynomial $ P[x]\in Z[x]$has least degree such that
$ P(\sqrt2\plus{}\sqrt3)\equal{}0$

[quote="TTsphn"][quote="Rust"]$ \sqrt 2 \equal{} 1.41..,\ \sqrt 3 \equal{} 1.73...$, therefore $ \sqrt 2\plus{}\sqrt 3 \equal{} 3.14... \equal{}\pi$ is irrational.  :D[/quote]
I don't think that because $ \pi$ is a advance number but $ \sqrt2\plus{}\sqrt3$ is an algebraic number
So $ \sqrt2\plus{}\sqrt3 \equal{}\pi$ not true.
[/quote]

He was just joking  :lol:

[quote="t0rajir0u"]Rational Root Theorem[/quote]

[b]Nobody knows the name of this theorem in French please.[/b]

[quote="TTsphn"] $ \pi$ is a [color=red][b]advance[/b][/color] number but $ \sqrt2\plus{}\sqrt3$ is an algebraic number
So $ \sqrt2\plus{}\sqrt3 \equal{}\pi$ not true.
[/quote]

What is advance number please?

I think, he mean ''Transcendental Number ''.  :wink:

[quote="t0rajir0u"][b]Theorem:[/b]  If $ r$ is a rational root of a polynomial $ P(x)$ of degree $ n$ with integer coefficients with leading term $ a_{n}$ and constant term $ a_{0}$, then $ r \equal{}\frac{p}{q}$ where $ gcd(p, q) \equal{} 1, p | a_{0}, q | a_{n}$.

In other words, $ (qx\minus{}p) | P(x)$.  For monic polynomials, $ q \equal{} 1$, hence $ r$ is an integer.[/quote]

Ok, now I see, Its a very interresting theorem.

Thanks  :)

[quote="TTsphn"]Yes  prove it quite easy
A problem is same from Viet Nam Olympiad
$ \frac{x^{3}\plus{}1}{y\plus{}1}\plus{}\frac{y^{3}\plus{}1}{x\plus{}1}$ is a integer
Prove that
$ \frac{x^{2007}\plus{}1}{y\plus{}1}$ is an integer.[/quote]

Your problem isn't same to allrighty'problem , TTsphn  :(

No I want to apply theory
If $ x_{0}$ is a rational root a monic polynomial then $ x_{0}\in Z$

[quote="Rust"]$ \sqrt 2 \equal{} 1.41\ldots\ ,\ \sqrt 3 \equal{} 1.73\ldots$. Therefore $ \left(\sqrt 2 \plus{} \sqrt 3\right)\approx\pi\ .$[/quote]
[color=darkblue][b]Fine remark and a nice joke,  Rust ![/b]

[b]Exercise.[/b] What is greater between $ \left(\sqrt 2 \plus{} \sqrt 3\right)$ and $ \pi$ (without computer) ?[/color]

[quote="allrighty"]Show that sqrt(2) + sqrt(3) is irrational.[/quote]
This a good link http://www.mathlinks.ro/Forum/viewtopic.php?t=27089  :wink:

Let $ x\equal{}\sqrt{2}\plus{}\sqrt{3}.$
Suppose,by the way of contradiction,that $ x$ is rational. Then $ x^2$ is rational. But $ x^2\equal{}5\plus{}2\sqrt{6}$ implies that $ \sqrt{6}\equal{}\dfrac{x^2\minus{}5}{2}$ is rational, contradiction ($ \sqrt{6}$ is irrational!). :ninja:

[quote="TTsphn"]Yes  prove it quite easy
A problem is same from Viet Nam Olympiad
$ \frac {x^{3} \plus{} 1}{y \plus{} 1} \plus{} \frac {y^{3} \plus{} 1}{x \plus{} 1}$ is a integer
Prove that
$ \frac {x^{2007} \plus{} 1}{y \plus{} 1}$ is an integer.[/quote]

Whats the year of this problem?
Are $ x,y$ integers??

Thank you TUAN  :wink:

It seems to be an interresting problem, Ill try to solve it  :roll:


In the picture, let $ V_{A}\minus{}V_{b}\equal{}5$ volts, $ E\equal{}10$ volts, $ C1 \equal{} 1\mu F,C2 \equal{} 2\mu F$. What's the voltage over $ C1$ and $ C2$ respectively?

Solution

I assume $ A$ is on the left.

Let the voltages across the capacitors $ C_{1}$ and $ C_{2}$ be $ V_{1}$ and $ V_{2}$ respectively. Voltages are measured from the subtracting the potential on the left from the potential on the right. The total potential difference, $ \minus{}5\,V$, is equal to the sum of the potential differences of the various components.

$ \Rightarrow V_{1}\plus{}V_{E}\plus{}V_{2}\equal{}\minus{}5\Rightarrow V_{1}\plus{}10\plus{}V_{2}\equal{}\minus{}5\Rightarrow V_{1}\plus{}V_{2}\equal{}\minus{}15$

Note that the sign of $ V_{E}$ is positive since the positive terminal is on the right.

The charge across a capacitor is given by $ Q \equal{} CV$. The current in this portion $ AB$ is the same at at any point. If the capacitors initally have the same charge then the charge on the capcitors will always be the same. This follows from:

$ Q(t) \equal{}\int I(t)\,dt$

Thus: $ C_{1}V_{1}\equal{} C_{2}V_{2}$

$ \Rightarrow V_{1}\equal{} 2V_{2}$

Substituting this into the other equation:

$ 2V_{2}\plus{}V_{2}\equal{}\minus{}15\Rightarrow 3V_{2}\equal{}\minus{}15\Rightarrow V_{2}\equal{}\minus{}5\,V$

$ \Rightarrow V_{1}\equal{}\minus{}10\,V$


hello hello hello.

There are no stategies in FTW, just read as fast as possible, and get the answer as fast as possible.  Or mem.

:P:P:P

this should be moved.

[size=75][color=red]Moved. [/color][/size]

most of the time....

Whenever long questions appear, I just skip to the last sentence.

Have paper and pen or pencil by you at all times during a game.

please don't purposefully try to mem just to gain rating...
it takes out the learning part of FTW and makes it a memming game

[quote="agentcx"]Here are some strategies if you are low on rating:
1. play a lot of 2/5 games
[/quote]
This tends to be 1. very annoying and 2. you don't learn anything because you don't have enough time to work the problem through.
[quote="agentcx"]3. Try to mem (this is the "foolish" player's option)[/quote]
Please don't mem. This makes the game very unfair for other players, and, once again, you don't learn anything. (as vallon22 said)

I would recommend making sure that you are familiar with formulas. Practice. Take a course if you need to improve on any one subject. AoPS has many courses available. Try to pinpoint what you are having trouble with (geo, algebra, etc), and work to get better at these subjects. :)

Do you mean like strategies to get higher rating or strategies for answering questions faster?

If you want to get higher rating, there's really no strategy for it. You just have to get better.

And for answering questions faster, here's my strategy.

I skim the passage and then read the last sentence thoroughly because that's usually the sentence that tells you what its asking for. (It kind of makes me confused sometimes when the last sentence isn't the question.)
Skim the passage as fast as you can but comprehending it at the same time.

[quote="AIME15"][quote="agentcx"]Here are some strategies if you are low on rating:
1. play a lot of 2/5 games
2. bomb your rating (pretending that you're bad at FTW) so that when you play high-rated people, you gain rating epicly with such low rating..
3. Try to mem (this is the "foolish" player's option)[/quote]

1. People won't join after awhile, except memmers who will take away your rating.
2. Uhm, this is pretty dumb...losing a game then winning a game will give you less rating than just winning a game without losing one first :wink:
3. Good luck[size=0]kids[/size] with that.[/quote]

strategy 2 is good for dumb people!

Well thats why I said it was the foolish players option. I don't try to purposefully mem, and I don't want anyone else to try to mem on purpose, so I labeled it the foolish players option. The player is foolish because memming is strongly discouraged (even though it's not cheating).
Well if you've come to the point where you've started memming most of the problems, then I suggest you go to other resources for learning (like Alcumus).

The main reason that memming is "foolish" is because memming will not help you with the real CD. FTW was developed to help users do better at the real CD.  :)

[cue="to moderators"]Hi, I think this entire thread has became quite foolish, don't you?[/cue]