trigonometry
analytic geometry
logarithms
geometry
calculus
calculus computations

The curve C has equation

$ y \equal{} a^{\sin (\pi e^{x})}$,

where $ a>1$.

(i) Find the coordinates of the stationary points on C.

(ii) Use the approximations $ e^{t} \approx 1\plus{}t$ and $ \sin t \approx t$ (both valid for small values of $ t$) to show that

$ y \approx 1 \minus{} \pi x \ln a$

for small values of $ x$.

(iii) Sketch C.

(iv) By approximating C by means of straight lines joining consecutive stationary points, show that the area between C and the $ x$-axis between the $ k$th and $ (k\plus{}1)$th maxima is approximately

$ \left(\frac{a^{2} \plus{}1}{2a} \right)\ln\left(1\plus{}\left(k\minus{}\frac{3}{4}\right)^{\minus{}1}\right)$.

$ y \equal{} a^{\sin{(\pi e^x)}}\text{ , }a > 1$
#1
Finding all $ (x,y)$ such that $ f^{\prime}(x) \equal{} 0$,
$ \ln{y} \equal{} \sin{(\pi e^x)}\ln{a}\Rightarrow y^{\prime} \equal{} y \pi e^x \cos{(\pi e^x)}\ln{a}$
& so, $ y^{\prime} \equal{} \pi \ln{a} \cdot e^x a^{\sin{(\pi e^x)}}\cos{(\pi e^x)}$
To solve,
$ a > 0\Leftrightarrow a^u > 0$, thus we seek the roots of $ \cos{(\pi e^x)} \equal{} 0$, which have solutions at $ \pi /2 \plus{} \pi n$
then the solution in $ x$ is $ \ln{(1/2\pm n)}$, which gives
\[ \displaystyle p_n \equal{} \left(\ln{\left(\frac {1}{2} \plus{} n\right)}, a^{( \minus{} 1)^n}\right)
\]
#2
This is just a first order approximation based on $ x \equal{} 0$.

Thanks igiul. Unfortunately I can't access hidden text on my computer.

[quote="igiul"]$ y \equal{} a^{\sin{(\pi e^x)}}\text{ , }a > 1$
Finding all $ (x,y)$ such that $ f^{\prime}(x) \equal{} 0$,
$ \ln{y} \equal{} \sin{(\pi e^x)}\ln{a}\Rightarrow y^{\prime} \equal{} y \pi e^x \cos{(\pi e^x)}\ln{a}$
& so, $ y^{\prime} \equal{} \pi \ln{a} \cdot e^x a^{\sin{(\pi e^x)}}\cos{(\pi e^x)}$
To solve,
$ a > 0\Leftrightarrow a^u > 0$, thus we seek the roots of $ \cos{(\pi e^x)} \equal{} 0$, which have solutions at $ \pi /2 \plus{} \pi n$
then the solution in $ x$ is $ \ln{(1/2\pm n)}$, which gives
\[ \displaystyle p_n \equal{} \left(\ln{\left(\frac {1}{2} \plus{} n\right)}, a^{( \minus{} 1)^n}\right)
\]

This is just a first order approximation based on $ x \equal{} 0$.[/quote]

For part (ii) , Maclaurin's theorem gives the same result as the method specified.

Any ideas for part (iv)? I make the relevant stationary points $ (\ln (\frac {4k \plus{} 1}{2}), a), (\ln (\frac {4k \plus{} 3}{2}, \frac {1}{a})$ and $ (\ln (\frac {4k \plus{} 5}{2}), a)$.

Then we just have to add the areas of two adjacent trapezia.

Unfortunately, I have $ (k \plus{} \frac {1}{4})^{ \minus{} 1}$, instead of $ (k \minus{} \frac {3}{4})^{ \minus{} 1}$.

I would appreciate if someone could  work through this so I can could see where I've gone wrong.  I expect I've done something silly as usual.

Thanks, in anticipation.

Ignore this. I should have had $ (\ln(\frac{4k\minus{}3}{2}),a), (\ln(\frac{4k\plus{}1}{2}, \frac{1}{a})$ and $ (\ln(\frac{4k\plus{}3}{2}), a)$ which then gives $ (\frac{a^{2}\plus{}1}{2a})\ln(1\plus{} (k\minus{}\frac{3}{4})^{\minus{}1})$.

I forgot about the maximum when $ x\equal{}\ln 0.5$.

Thanks to anyone who tried it.


probability LaTeX logarithms trigonometry

A point is randomly chosen in an equilateral triangle of side length 2. Find the probability it is within $\sqrt{2}$ units of exactly two of the triangles vertices.

[hide]\[\sqrt{3}*2-3\][/hide]
Is that right?

Yes. Please post a solution so that other people can see what to do to solve it.

[quote="PathofNeo"]I would...but I can't do LaTeX!  It took me 10 minutes just to figure out how to display the solution.[/quote]

Put dollar signs around everything, some common commands are:

1234567890+=-*/  ->
$1234567890+=-*/$

5^3=125 ->
$5^{3}=125$

\pi -> $\pi$

\frac{ top}{bottom} ->

$\frac{ top}{bottom}$

a_1, a_2, a_3...a_i ->

$a_{1}, a_{2}, a_{3}...a_{i}$

\log_{e}{5} -> $\log_{e}{5}$

\sum^{\infty}_{k=0}
->
$\sum^{\infty}_{k=0}$
\sin, \cos, \tan etc...

More Symbols can be found on this page:
http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php
Commands can be found here: http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php


Sketch: $a$ must be an integer so we are left with checking some cases: namely depending on whether $2,3,5$ divides $a$ or not. Using that $[\frac{a}{b}]=\frac{a-x}{b}$ for some $x \in [0,b)$ we can deal with it. It's nothing nice but it works.

Leaning on megus's reasoning:

$a$ must be integer, hence we can write it as $a=30k+r$ where $0\leq r\leq 29$.

Then $\left[{a\over 2}\right]=15k+[{r\over 2}], \left[{a\over 3}\right]=10k+[{r\over 3}], \left[{a\over 5}\right]=6k+[{r\over 5}]$, so the equation becomes

$31k+[{r\over 2}]+[{r\over 3}]+[{r\over 5}]=30k+r\iff k=r-[{r\over 2}]-[{r\over 3}]-[{r\over 5}]$

Hence the value for $k$ is uniquely determined by $r$, therefore we have $30$ values for $a=30k+r.$ (Obviously, no two values for $a$ can be equal, since they are pairwise mutually incongruent modulo $30$.)

Note that a is an integer. Now, by rewriting the floor function as
$\frac{a-x}{2} + \frac{a-y}{3} + \frac{a-z}{5} = a$              for some integers $x, y, z$
Multiplying by 30 we obtain,
$15a - 15x + 10a -10y + 6a - 6z = 30a$
$a = 15x + 10y + 6z$
So the solutions of the equation are all possible values of this form for any given x, y, z integers.
$a = {0, \pm6, \pm10, \pm15 ...}$

Oriol

[quote=Canada 1998 P1]Determine the number of real solutions $a$ to the equation:
\[ \left[\,\frac{1}{2}\;a\,\right]+\left[\,\frac{1}{3}\;a\,\right]+\left[\,\frac{1}{5}\;a\,\right] = a. \]
Here, if $x$ is a real number, then $[\,x\,]$ denotes the greatest integer that is less than or equal to $x$.[/quote]
[b][color=#000]Solution:[/color][/b] First of all, Clearly, $a$ $\in$ $\mathbb{Z}$ $$a= \left \lfloor \frac{a}{2} \right \rfloor + \left \lfloor \frac{a}{3} \right \rfloor + \left \lfloor \frac{a}{5} \right \rfloor=\frac{31a}{30} - \left( \left \{\frac{a}{2} \right \} + \left \{\frac{a}{3} \right \} + \left \{\frac{a}{5} \right \} \right) \implies \left \{\frac{a}{2} \right \} + \left \{\frac{a}{3} \right \} + \left \{\frac{a}{5} \right \}=\frac{a}{30} $$
Denote: $a \equiv a_n \pmod{n} ~ $ Hence, we have
$$\left \{\frac{a}{2} \right \} + \left \{\frac{a}{3} \right \} + \left \{\frac{a}{5} \right \}=\frac{a}{30} \implies a=15a_2 + 10a_3 + 6a_5$$
Which (after manually checking) has $30$ solutions, namely: $$\{0,6,10,12,15,16,18,20,21,22,24,25,26,27,28,31,32,33,34,35,37,38,39,41,43,44,47,49,53,59\} \qquad \blacksquare$$


Nice proof.  In fact, we need only the condition $ x_{i}\ge\minus{}1$, not $ x_{i}\le 1$.  If $ x_{i}\ge\minus{}1$, we have
\[ 4 x_{i}^{3}\minus{}3 x_{i}\plus{}1 \equal{} (2x_{i}\minus{}1)^{2}(x_{i}\plus{}1)\ge 0.\]
Summing over all $ i$ gives
\[ \minus{}3\sum_{i\equal{}1}^{n}x_{i}\plus{}n\ge 0.\]
That's equivalent to what we want.


Show that there exist positive integer $N$ such that the decimal represantation of $2000^{N}$ starts with the digits
$200120012001$


geometry
circumcircle
trigonometry
geometry proposed

[color=darkred]Let $ABC$ be a  triangle for which $B>C$. Let $M,N$ be the points for which
$\{\begin{array}{c}MB\perp AB\ ,\ (CM)\cap AB=\emptyset\\\ NC\perp AC\ ,\ (BN)\cap AC\ne\emptyset\end{array}$ so that $m(\widehat{MAB})=m(\widehat{NAC})=B-C$.
Prove that the lines $BN$, $CM$ and the tangent in the point $A$ to the circumcircle of the triangle $ABC$ are concurrently.[/color]

[hide="Hint"][color=darkblue][b][u]I"ll study more generally this problem.[/u][/b] Let $ABC$ be an any triangle. Let $M,N$ be the points for which
$\{\begin{array}{c}MB\perp AB\ ,\ (CM)\cap AB=\emptyset\\\ NC\perp AC\ ,\ (BN)\cap AC\ne\emptyset\end{array}$. Denote $\{\begin{array}{c}m(\widehat{BAM})=x\\\ m(\widehat{CAN})=y\end{array}$ and $\{\begin{array}{c}P\in BN\cap CM\\\ D\in AP\cap BC\end{array}$. Then $\boxed{\frac{DB}{DC}=\frac{c\cdot\cos B}{b\cdot\cos C}\cdot\frac{\sin(A+y)\cdot\sin x}{\sin (A-x)\cdot\sin y}}$.[/color][/hide]

let $ \angle NBC =\angle B_{1}\ ,\angle NBA =\angle B_{2}\ ,\angle ACM =\angle C_{1}\ ,\angle BCM =\angle C_{2}\ ,\angle BAL =\angle A_{1}\ ,\angle CAL =\angle A_{2}$ we have:
$ \frac{\sin B_{1}}{\sin (90^{\circ}+\angle C)}=\frac{NC}{BN}\ ,\frac{\sin B_{2}}{\sin(A+B-C)}=\frac{AN}{BN}\Longrightarrow\frac{\sin B_{1}}{\sin B_{2}}=\frac{NC}{AN}.\frac{\sin (90^{\circ}+\angle C)}{\sin(A+B-C)}=\sin (B-C).\frac{\cos C}{\sin (2C)}=\frac{\sin (B-C)}{2\sin C}$
similary we have $ \frac{\sin C_{1}}{\sin C_{2}}=\frac{2.\sin B}{\sin (B-C)}\ ,\ \frac{\sin A_{1}}{\sin A_{2}}=\frac{\sin C}{\sin B}$
hence:
$ \frac{\sin C_{1}}{\sin C_{2}}.\frac{\sin B_{1}}{\sin B_{2}}.\frac{\sin A_{1}}{\sin A_{2}}= 1$


function
trigonometry
real analysis
real analysis unsolved

[u]Ascertain[/u] the [b]range[/b] of the function $ f(x,y) \equal{} \frac {\sin x \minus{} \sin y}{\cos x \plus{} \cos y \minus{} 1}$ , where $ \left\|\begin{array}{c} x > 0\ ,\ y > 0 \\
 \\
x \plus{} y < \pi \\
 \\
\cos x \plus{} \cos y\ne 1\end{array}\right\|$ .

Maybe someone presents and a geometrical interpretation. Thank you very much !

$ \minus{}\frac 12<f(x,\ y)\leq \frac{1}{\sqrt{3}}$ is answer, right?

We have $ f(x,y) \equal{} \minus{} f(y,x)$, and for $ y\in(0,\pi/2)$, $ f(\pi/2,y) \equal{} \frac {1 \minus{} \sin y}{\cos y \minus{} 1}$ takes on all values in $ ( \minus{} \infty,0)$. Clearly $ 0$ is in the range. So the range is all of $ R$.


function
topology
linear algebra
matrix
vector
real analysis
real analysis unsolved

:?: I had got the conjecture two years ago as soon as I had successfully constructed continuous functions from R onto R^n (including the care n=∞ and taking product topology when necessary), which I couldn't prove or disprove over the past two years. The case n =2 is crucial to construct continuous functions from R onto R^n which I had used the Peano curve and the pasting lemma to obtain by induction.

:what?: What do you mean, jmerry? Both are $ c$ :? What you probably intended to say is that the image of $ [\minus{}n,n]$ is compact, hence contains no inner points and we run into contradiction with Baire category theorem :)

[quote="jmerry"]It's impossible, since the cardinality of $ \ell^2$ is larger than that of $ \mathbb{R}$.[/quote] I don't think so, because we can prove that card(R)=card(R^∞). The cardinal number of  the set of all the sequences formed by natural numbers is c, given an element of R^∞ we can correspond to a Z*Z matrix of natural numbers, then the matrix corresponds to a sequence which corresponds to an element of R, thus we obtain an injective mapping from R^∞ to R. By Cantor-Bernstein theorem, the result is obvious.

[quote="fedja"]:what?: What do you mean, jmerry? Both are $ c$ :? What you probably intended to say is that the image of $ [ \minus{} n,n]$ is compact, hence contains no inner points and we run into contradiction with Baire category theorem :)[/quote] What do you mean, fedja? It seems that Baire category theorem doesn't work  to this problem.

What did I mean? I was wrong. The dimension of the vector space is $ c$, but that's not enough.

The Baire category argument is next on the list of ways to show that $ \mathbb{R}$ is too small to map onto $ \ell^2$, and it works. Note that we're applying Baire in $ \ell^2$ to show that it isn't $ \sigma$-compact.

Well, I just can't understand why a compact subset of l^2 contains no inner points.

Any ball contains an infinite discrete set (the distances between points are bounded below by a positive number). For the unit ball, one example of such a set is $ \frac12e_1,\frac12e_2,\dots$, where $ e_i$ are the standard basis vectors. You can't fit in infinite discrete set in a compact set, so no compact set contains a ball.

Thank jmerry and fedja for all the help, I look back and wonder why I didn't manage to solve it.


x^2+ax+1=0
x^2+x+a=0
IF at least one of the roots are equal,find all possible values of a

Cant believe i took more than 10MIN to do this crap!!!

I think 10 mins is acceptable.

[hide]

The key is to use quadratic formula. First of all, let me rephrase the question to avoid symbol conflicts:

find y such that

1. x^2 + yx +1 = 0
2. x^2 + x + y = 0
3. at least one of the roots are equal.

recall: Quad. form. (Q.F.):

x = [-b+ (b^2 - 4ac) ^ (0.5)] / 2a

apply Q.F. to 1, then we must have y^2 - 4 >= 0
apply Q.F. to 2, then we must have 1 - 4*y >=0

now at least one of the roots are equal means one of them must exactly = 0. (too lazy to cont.)

[/hide][/code]

quad formula?
[hide="Sol"]
$x^2+ax+1=x^2+a+x$
$a(x-1)=(x-1)1$
$a=1$
and when we do it differently we get $a=0$ 2 values?
[/hide]

[hide="diff sol"]
$x=\frac{-a+-\sqrt{a^2-4}}{2a}$
this is the quadratic root of 1st one
then $2$nd one is
$x=\frac{-1+-\sqrt{1-4a}}{2a}$
Atleast one of them is equal,
So, try out all options
Too tired to continue zzzzzz good night...
[/hide]

[hide="Not hard click for solution"]2 possible cases:

Suppose both equations have one root in common
$x^2+ax+1=(x-b)(x-c)$
$x^2+x+a=(x-c)(x-d)$
Put $x=c$ and get $ac+1=c+a\Rightarrow a(c-1)=c-1$

If $c=1$, so far no restriction on $a$. But $c$ is a root of the equations. If you put in $c=x=1$, you get $a=-2$

If $c\neq 1$, $a=\frac{c-1}{c-1}=1$

Time to think and solve: 2 minutes

Time to write it into AoPS: 10 minutes[/hide]

I didn't solve it yet, but i'll bet i'll take MORE than 10 minutes!

[quote="eminem"]hey if  one root  is 0 then the equations arent gettin satisfied . just try  them

i did the same as riddler's second soln and got the answers as 1, -2[/quote]
If one root is 0, only second equation can have that root, and this means $a=0$. But if $a=0$, $x^2+1=0$ and $x^2+x=0$ do not have a common root. Your comment above is not correct. Show your work please.

hey 10000th user thats what i meant ,  that 0 is NOT an answer, but someone above has said it IS..... :?:


Does anyone know where I can find all the Mock AMCs and AIMEs?
or could someone send them to me?

thanks


parameterization
calculus
quadratics
algebra
system of equations
Parametric Equations

find the exact solutions for the system
$ \frac{1}{x^3}\plus{}\frac{1}{y^3}\plus{}\frac{2}{xy}\equal{}\frac{8}{27}$
$ \frac{1}{x^2}\plus{}\frac{1}{y^2}\minus{}\frac{1}{xy}\equal{}\frac{3(m\plus{}1)}{a^3\plus{}mb^3}$ (where m is real parameter)

[quote="hoangvn.fix"]find the exact solutions for the system
$ \frac {1}{x^3} \plus{} \frac {1}{y^3} \plus{} \frac {2}{xy} \equal{} \frac {8}{27}$
$ \frac {1}{x^2} \plus{} \frac {1}{y^2} \minus{} \frac {1}{xy} \equal{} \frac {3(m \plus{} 1)}{a^3 \plus{} mb^3}$ (where m is real parameter)[/quote]

As often, a crazy calculus college exercise far from olympiad training :(

Let $ u \equal{} \frac 1x \plus{} \frac 1y$ and $ v \equal{} \frac 1{xy}$. Let $ \frac 8{27} \equal{} c$ and $ \frac {3(m \plus{} 1)}{a^3 \plus{} mb^3} \equal{} d$ The system becomes :

$ u^3 \minus{} 3uv \plus{} 2v \equal{} c$
$ u^2 \minus{} 3v \equal{} d$

The second equation gives $ u^3 \minus{} 3uv \equal{} du$. Plugging this in the first, we get $ du \plus{} 2v \equal{} c$, so $ v \equal{} \frac {c \minus{} du}2$. Plugging in the second :

$ 2u^2 \plus{} 3du \minus{} 3c \minus{} 2d \equal{} 0$, whose discriminant is $ 9d^2 \plus{} 8(3c \plus{} 2d) \equal{} 9d^2 \plus{} 16d \plus{} 24c$ $ \equal{} 9d^2 \plus{} 16d \plus{} \frac {64}9 \equal{} (3d \plus{} \frac 83)^2$. So two possibilitues for $ u$ :

1) $ u \equal{} \frac 23$ and then $ v \equal{} \frac {c \minus{} du}2 \equal{} \frac {4 \minus{} 9d}{27}$
Then $ \frac 1x$ and $ \frac 1y$ are roots of $ X^2 \minus{} ux \plus{} v \equal{} 0$ and so $ x,y$ are roots of $ vX^2 \minus{} uX \plus{} 1 \equal{} 0$
So $ \frac {4 \minus{} 9d}{27}X^2 \minus{} \frac 23X \plus{} 1 \equal{} 0$

1.1) $ 4 \minus{} 9d \equal{} 0$ would here imply $ v \equal{} \frac 1{xy} \equal{} 0$, which is impossible

1.2) $ 4 \minus{} 9d\ne 0$ lead to a quadratic $ (4 \minus{} 9d)X^2 \minus{} 18X \plus{} 27 \equal{} 0$ whose discriminant is $ 108(9d \minus{} 1)$

And so $ (x,y)$ or $ (y,x) \equal{} (\frac {9 \minus{} \sqrt {27(9d \minus{} 1)}}{4 \minus{} 9d},\frac {9 \plus{} \sqrt {27(9d \minus{} 1)}}{4 \minus{} 9d})$ with the condition $ 9d \minus{} 1\ge 0$

2) $ u \equal{} \minus{} \frac {9d \plus{} 4}6$ and then $ v \equal{} \frac {c \minus{} du}2 \equal{} \frac {3d^2}4 \plus{} \frac d3 \plus{} \frac 4{27}$
Then $ \frac 1x$ and $ \frac 1y$ are roots of $ X^2 \minus{} ux \plus{} v \equal{} 0$ and so $ x,y$ are roots of $ vX^2 \minus{} uX \plus{} 1 \equal{} 0$

So $ (\frac {3d^2}4 \plus{} \frac d3 \plus{} \frac 4{27})X^2 \plus{} \frac {9d \plus{} 4}6X \plus{} 1 \equal{} 0$ whose discriminant is $ \left(\frac {9d \plus{} 4}6\right)^2 \minus{} 4(\frac {3d^2}4 \plus{} \frac d3 \plus{} \frac 4{27})$ $ \equal{} \minus{} \left(\frac {\sqrt 3d}2 \minus{} \frac {2}{3\sqrt 3}\right)^2$

And so solution only if $ d \equal{} \frac 49$. Then $ u \equal{} \minus{} \frac 43$ and $ v \equal{} \frac {4}{9}$ and the equation is $ 4X^2 \plus{} 12X \plus{} 9 \equal{} 0$ and so $ x \equal{} y \equal{} \minus{} \frac 32$

Synthesis of solutions :
======================
If $ d \equal{} \frac {3(m \plus{} 1)}{a^3 \plus{} mb^3} < \frac 19$ : no solution

If $ d \equal{} \frac {3(m \plus{} 1)}{a^3 \plus{} mb^3}\ge\frac 19$ and $ \ne \frac 49$ : $ (x,y)$ or $ (y,x) \equal{} (\frac {9 \minus{} \sqrt {27(9d \minus{} 1)}}{4 \minus{} 9d},\frac {9 \plus{} \sqrt {27(9d \minus{} 1)}}{4 \minus{} 9d})$

If $ d \equal{} \frac {3(m \plus{} 1)}{a^3 \plus{} mb^3} \equal{} \frac 49$ : $ (x,y) \equal{} ( \minus{} \frac 32, \minus{} \frac 32)$

dear mr pco i am terribly sorry for my mistakes in typing the second equation of this system must be
$ \frac{1}{x^2}\plus{}\frac{1}{y^2}\minus{}\frac{1}{xy}\equal{}\frac{3(m\plus{}1)}{x^3\plus{}my^3}$ (where m is real parameter)

[quote="hoangvn.fix"]Dear mr pco i am terribly sorry for my mistakes in typing the second equation of this system must be
$ \frac {1}{x^2} \plus{} \frac {1}{y^2} \minus{} \frac {1}{xy} \equal{} \frac {3(m \plus{} 1)}{x^3 \plus{} my^3}$ (where m is real parameter)[/quote]

My condolences.

Factoring the first equation gives us $ (\frac {1}{x} \plus{} \frac {1}{y} \minus{} \frac {2}{3})(\frac {1}{x^2} \plus{} \frac {1}{y^2} \plus{} \frac {4}{9} \minus{} \frac {1}{xy} \plus{} \frac {2}{3x} \plus{} \frac {2}{3y}) \equal{} 0$

Either $ (x \equal{} \minus{} \frac {3}{2}, y \equal{} \minus{} \frac {3}{2})$ or $ \frac {1}{x} \plus{} \frac {1}{y} \equal{} \frac {2}{3}$.

The second case gives $ y \equal{} \frac {3x}{2x \minus{} 3}$. Our second equation becomes

$ \frac {4}{27} \equal{} \frac {1}{xy} \plus{} \frac {m \plus{} 1}{x^3 \plus{} my^3}$

Multiplying, plugging in $ y$ (not very fun), and factoring (not fun either) gives us $ \frac {3x^2(x \minus{} 3)((2x \minus{} 3)^3 \minus{} 27m)}{(2x \minus{} 3)^2} \equal{} 0$

Thus, for the second case we get $ x \equal{} 3$ and $ x \equal{} \frac {3(\sqrt [3]{m} \plus{} 1)}{2}$. The second solution only applies when $ m \neq 1$. Finding $ y$ reveals that another restriction is $ m \neq 0$.

So our final solutions are $ ( \minus{} \frac {3}{2}, \minus{} \frac {3}{2}), (3, 3), (\frac {3(\sqrt [3]{m} \plus{} 1)}{2}, \frac {3\sqrt [3]{m \plus{} 1}}{2\sqrt [3]{m \plus{} 1} \minus{} 1})$, with conditions on the last one. There are some imaginary roots too but I won't bother finding them.


Im just curious as to what people think about this (if they do):

What do you think are the top 5 hardest aime problems?

I'd guess the game-like problems in the harder years.

-interesting_move

In a math jam earlier this year, Mr. Rusczyk said that one of the hardest AIME problems he had ever seen was problem 15 from the 2004 AIME A.


analytic geometry
vector
geometry
number theory
geometry solved

Is it possible to have a regular pentagon whose vertices are lattice point coordinates? I remember seeing something on ML that said that this is impossible as well, but I am looking for a proof of that. I know that the equilateral triangle can't have lattice point coordinates, but I was wondering if anyone had a proof that a regular pentagon can't have lattice point coordinates in the plane?

Thanks

Assume that such a pentagon does exist. Among all the lattice pentagon, choose one with minimal length sid, say $P_1P_2P_3P_4P_5$.
It is easy to verify that the convex pentagon $B_1B_2B_3B_4B_5$ is regular too, where $B_1$ is the common point of the diagonals $A_2A_4$ and $A_3A_5$ the other points $B_i$ are similarly defined)
Moreover, in terms of vectors $A_1A_5 = A_2B_1$ since $A_1A_5B_1A_2$ is a parallelogramm. Since the three other points are integer points, we deduce that $B_1$ is an integer point too.
The same is true for the other $B_i$.
Thus $B_1B_2B_3B_4B_5$ is a convex regular pentagon whose vertices are integer points and with side length less than the side length of $A_1A_2A_3A_4A_5$, which contradicts its minimality.

Pierre.

I also remember the general statement being proved on ML, by using $\mathbb Z[i]$: if we can find a regular lattice polygon with $n$ sides, then $n=4$. I'm not sure right now, but I think I posted a proof.

[quote="pbornsztein"]Assume that such a pentagon does exist. Among all the lattice pentagon, choose one with minimal length side, say $P_1P_2P_3P_4P_5$.
It is easy to verify that the convex pentagon $B_1B_2B_3B_4B_5$ is regular too, where $B_1$ is the common point of the diagonals $A_2A_4$ and $A_3A_5$ the other points $B_i$ are similarly defined)
Moreover, in terms of vectors $A_1A_5 = A_2B_1$ since $A_1A_5B_1A_2$ is a parallelogramm. Since the three other points are integer points, we deduce that $B_1$ is an integer point too.
The same is true for the other $B_i$.
Thus $B_1B_2B_3B_4B_5$ is a convex regular pentagon whose vertices are integer points and with side length less than the side length of $A_1A_2A_3A_4A_5$, which contradicts its minimality.

Pierre.[/quote]

Sorry if I don't understand so quickly, but what exactly do you mean by "its minimality"? You say that "Thus $B_1B_2B_3B_4B_5$ is a convex regular pentagon whose vertices are integer points and with side length less than the side length of $A_1A_2A_3A_4A_5$, which contradicts its minimality.", does this mean that the pentagon does or does not exsist? (i.e doe this mean that you can find a regular pentagon with lattice point vertices in the plane?)

I should note that Fanaticsm sent me this problem yesterday on PM, and I must admit I hadn't find any time yet to reply, since I am lately being bombardized with lots of preparation problems for the German IMO training in Rostock. Just to note that Grobber indeed solved the problem before, namely on http://www.mathlinks.ro/Forum/viewtopic.php?t=10569 , but I myself don't understand his solution because of the advanced algebra he uses. Actually, there is a elementary solution, but it is hard to explain (if you have Engel's "Problem Solving Strategies", you will find it there - in the chapter about infinite descent, I believe).

[b]EDIT:[/b] Now I see that the elementary solution was posted on MathLinks long ago - see http://www.mathlinks.ro/Forum/viewtopic.php?t=4249 .

And, sorry, Fanaticsm - I have no idea about your first number theory problem. Probably I am not the right one to contact about number theory.

  Apologies again,
  Darij

Not a problem darij, thanks for replying, but as you said, I can't seem to understand this problem. Is it possible that anyone else would have a more elementary solution (or maybe Pierre could clarify his solution a little bit?)

Actually, thre is very simple solution for $n\geq 7$.
If $P_1P_2...P_n$ is a regular $n$-gon with minimal side length then consider  a regular $n$-gon $Q_1...Q_n$ s.t. $\vec{OQ_i}=P_iP_{i+1}$. Since $n\geq 7$ we have $|Q_iQ_{i+1}|<|P_iP_{i+1}|$.
For $n=6$ and $n=3$ problem can be easily  solved using rationality of area of lattice polygons.

[quote="fanaticsm"]does this mean that the pentagon does not exsist?[/quote]

Yes.

Pierre.

In fact there are many related results :
In $\mathbb {Z}^p$ for $p \geq 3$ we may find a planar regular $n$-gon if and only if $n=3,4,6$. And if $n=3,6$ such a polygon does exist even for $p=3$.

Moreover, my proof for the pentagon and Myth's one for the case $n \geq 7$ can easily be generalised word for word in the case of an arbitrary lattice (defined as the intersection points of two infinite families of regularly spaced parallel lines). That is the only regular convex $n$-gon which can be inscibed into a lattice are the ones for $n=3,4,6$. It is easy to find lattices into which we can inscribed a square or an equilateral triangle (and then a regular hexagon).
But, there is no lattice into which we can inscribed simultaneously a square and an equilateral triangle.

And, it has been proved that in $\mathbb {Z}^n$ we can inscribed a regular $n$-simplex if and only if $n+1$ is the sum of $1,2,3$ or $8$ odd squares.

Pierre.


Was he doing the correct thing? I mean, is it correct to steal reach people, or just people, because you're poor or you have nothing to eat?

He wasn't stealing only because he was poor and had nothing to eat. He did it for the poor in general.

Of course he was doing the right thing! I mean, where is the confusion? Private property should be abolished right?

i've never really thought much about it being wrong the but more i think about it, its kind of messed up.

let's say that you work really hard to become wealthy and someone steals lots of your money and gives it freely to the poor. doesn't seem right.
but it's a little different when he's stealing from a king, because a king usually inherits money as opposed to working really hard for it.

I think the point of the story isn't just that he steals from the rich and gives to the poor, but also that the poor are poor because the rich are evil and cheat them and keep them oppressed. Remember, Robin Hood is from the 13th century or something.

I think the main point of the story was to help the poor and those in need.  I don't think it matters how he did it as long as he helped the poor.

Please, remember that I'm not asking about the story, I talk about stealing rich people to help poor ones

[quote="bubka"]You seem to have doubt as to the nobility and righteousness of stealing from the rich and giving to the poor. May I know the source of these doubts?[/quote]

ehhhm.... my head

i don't think one can just say in general that it's "right" or "wrong."  it kind of depends on the situation.  the rich should be able to keep their wealth, but not to the extent that the poor are hungry and homeless, in which case, they should be provided enough money to survive given that they work for it.  not everyone should be able to afford mansions or new clothes every weekend, but every person that works hard should has a right to be able to afford food and shelter.  in robin hood, nothing's as complicated as life really is because it's a fictional novel.  life isn't all in terms of integers.

^ WTF?

I seem to recall somewhere that with the historical robin-hood and king whoever it was that is portrayed as evil, the king person was actually a pretty responsible leader, and the "good king" that replaced him just gave out lots of money sort of haphazardly and ruined the treasury. But that may just be heresay.

[quote="bubka"][quote="K81o7"]Clearly, Robin Hood was communist.  You know, he was just trying to share his ideas, and perhaps restructure the development of western politics...
Did I mention he was communist?  A red just like this smiley  :blush: .  And therefore he was evil!  So yes, he WAS wrong.

[hide]
[color=red]Just to clear things up, I did not mean to offend anyone by the above post...I'm not anti-communist nor anti-red.[/color][/hide][/quote]

Sure thing: you are only anti Robin Hood!  :wink:[/quote]

No Robin Hood was as bad as....as Hitler...okay no but almost.

Robin Hood helped the rich too. The rich saw poor people walking around, and didn't do anything about it. They would have burned in hell if Robin didn't help them out by handling their charity for them.

Also, it isn't like he made the rich people poor, he just made the poor people not so poor.

hes stealing afrom the bad guys. of course it is rgood

Are you sure that they were the bad guys though? You only see robin hood's perspective.... :huh:

[quote="JSteinhardt"]king whoever it was that is portrayed as evil[/quote]
That would be King John "the only". He was so unpopular among the royals that nobody ever took the name again.

The main reason for unpopularity is probably the Magna Carta- the nobles forced John to give up some power.

[quote="Treething"][quote="bubka"][quote="Treething"]one thing's for sure, he definitely wasn't straight

and that is all that matters.[/quote]

And what does straight mean here?[/quote]

heterosexual[/quote]

And that is [i]all[/i] that matters?

My god, you seem to have had some really bad experience with guys who are not straight! Ouch - that musta hurt!


$a,b,c,x,y,z$ are positive real numbers such that $a+b+c=x+y+z.$
Prove that $a^2x+b^2y+c^2z+abc\geq4xyz.$

Thank you, zhaobin. :lol:
This is super nice inequality!


Is there anybody here whose Chapter advances All-Star team(s) in addition to the winning school team(s) to the State Event?

I'll take it that you mean the top 4 as a team when you say All-Star team.

My chapter didn't do that. We just had the normal teams and individual thing.

I wouldn't think that that would work very well, because what if a person was in the top four, yet their team won. Then, they would be placed on two teams, which isn't allowed.

Sorry, let me clarify what I meant by All-Stars.

California's chapters used to allow 1-3 ALL-STAR teams of 4 high scoring mathletes per chapter who were NOT on any school teams already qualified for the State. These students could have been competing as individuals or as members of any non-qualified teams. Those All-Star teams even competed for State title!

But the idea was mainly to allow strong mathletes who did not belong to a (good enough) team to still have a chance to qualify for State and National.

For some reason they discontinued that in 2004.

Was California alone in doing that in the past, or is California alone in not doing that in the present?

That's just like the individuals system, except just a tad different. Each school can bring 8 students to the competition to compete in the actualy competition (not including alternates). 4 of them make up the school's team, the other 4 participate as individuals in the competition. Those four can still make it to States or Nationals, if they score high enough. If they're in the top 60 or so after Chapters, they can go to States to also participate as an individual. So it's basically what you explained, except they're not placed on a team.

[quote="MathChauffeur"][quote="yif man12"]... the other 4 participate as individuals in the competition... [/quote]

Individuals cannot advance to the State events in California as far as I know. So I take that your state allows it. Then I wonder how many states allow individuals to go to the State?[/quote]

I'm pretty sure individuals can advance to state (the All-Star team from San Diego won Southern CA in 2003, and one of them made it to nat's).  While I don't think they are allowed to proceed as a team anymore, the individuals still are allowed to continue, at least from San Diego and the Bay Area (whatever chapter Miller is in).

Yeah, individuals can still go to state, in the Bay Area at least. One from my school did--I was so jealous!--though I don't know what happens afterwards. Do those who go as a team have a team round like during the chapter, or are they a team in name only? What about in nationals?

[quote="rrusczyk"]I'm pretty sure individuals can advance to state (the All-Star team from San Diego won Southern CA in 2003, and one of them made it to nat's).  While I don't think they are allowed to proceed as a team anymore, the individuals still are allowed to continue, at least from San Diego and the Bay Area (whatever chapter Miller is in).[/quote]
You're right as usual  ;). Our chapter did select 2 individuals to the State.

[quote="LynnelleYe"]...Do those who go as a team have a team round like during the chapter, or are they a team in name only? What about in nationals?[/quote]
They only compete as individuals at the State. But if they end up representing the State, they will be part of the State team at the National.

In California, the bummer for 2004 and beyond is that in the past a lot more individuals were allowed to go to the State as members of the "All-Star Teams". In our chapter, that number had gone down from 8 to 2.

Now the coaches better pick the "right" team members. It could look rather bad/sad if a school team qualifies but some or all of the team members have lower score than another schoolmate who barely misses out for one of the rare individual spots. That scenario almost happened to us in February.

I guess all Putnam coaches have the same dilemma  :D

I am still curious about the ratio of mathletes advancing to the State as team members vs. as individuals for each state ...

In my chapter, the top 3 teams advance to states as well as the 4 highest scores who are not actually on a team.  We only have 6 chapters in NH, so we still only get 100 kids at state.  Last year one of my kids who wasn't even on my team finished 3rd overall at chapter, so he made it.  Actually 3 kids who didn't make my team finished in the top 8 at chapter so they all made it to state!! (my 4 team members were all in top 7 at state).

Every state is different though, we're just lucky enough to be in a small state :P

Josh Frost
Coach
New Hampshire

Yeah, I was an individual last year (i.e. if I didn't come in first, second, third, or fourth on Chapter, I wouldn't advance). I came in first in Chapter, so I advanced to States, where I came in fourth. I came in 36th in Nationals.

[quote]
Every state is different though, we're just lucky enough to be in a small state
[/quote]

I'm in a small state, New Jersey, yet the competition is pretty tough, as all of the NJ team members came in the top 100 (17th, 36th, 64th, 81st)...

[quote="frost13"]by "small state" I meant participation.  We had less that 60 schools participating, and we also have NO large cities which can often produce large competition within a school.  NJ is always great at MC and is very very competitive, so in my mind it is a BIG Mathcounts state.

If you live in NJ you aren't "lucky" in terms of making it to Nationals even though geographically NJ is small.[/quote]

Very true. My state is relatively easy to make it into nationals as well. The last couple of years, the 4th place score has been 39 and 38, and the year before last 33.

Well, I got 41 and made it into fourth place...

Also, I don't know how well NJ did in the previous years, as I'm currently a seventh grader.

41 in California would have put you in about 7th last year...

Dunno... that's why I'm practicing :) .

Today I just got a 29 on a state Sprint Round :) .

[quote="MathChauffeur"]So it seems some states have decreased the number of participants in the State event. Just for the sake of a provocative discussion, I wonder if the following strategy may work for a school that is gung ho about its position in its chapter and wants to send as many students to the State as possible.

After ranking within the school is determined, instead of putting the top 4 students on the official team, how about put the 3rd to 6th ranked students on the team and have the top 2 students compete as individuals. Obviously the school has to assume that the resulting team still has an excellent chance of advancing.

If all goes well the school will advance 6 students. If it does not go well in certain fashion the coach might have to skip town  :D[/quote]

Good plan, but the only problem with it is coaches want their team to get a high score. Doing this will make their team round score worse, since the best students aren't on the team.

If the first to sixth are about equal (or all are far better than the competition), then it wouldn't quite make a difference...

[quote="MathChauffeur"]

After ranking within the school is determined, instead of putting the top 4 students on the official team, how about put the 3rd to 6th ranked students on the team and have the top 2 students compete as individuals. Obviously the school has to assume that the resulting team still has an excellent chance of advancing.

If all goes well the school will advance 6 students. If it does not go well in certain fashion the coach might have to skip town  :D[/quote]

Last year my team won Chapter by like 16 points, so I could have taken my 4th-8th kids and made them the team and we still would have won Chapter, problem is coaches are not supposed to rearrange their team between chapter and state, so at State even though we had won by 10 points, my 4th-8th kids probably would have finished 2nd or 3rd at state.

I talked to my state coordinator when they started having Individuals and made sure they set the advancement rules up so that some individuals would make it to State, otherwise why bother paying for them to register.

In my state now, the top 4 kids who aren't on an advancing team get to advance as well wether they are a strong kid from a weak team, or a strong kid from a school with a really strong team they couldn't make.  I really like this method since I don't have to debate with myself about leaving top kids off the team, last year 7 of my 8 kids went to state even though I put the top 4 on the team.

I wish more kids could advance since we always have kids who end up actually ranking higher at state then they did at chapter.  last year, my 9th at chapter took 6th at state and my 4th at chapter took 2nd at state(2nd time that's happened.

[quote="yifman12"] Good plan, but the only problem with it is coaches want their team to get a high score. Doing this will make their team round score worse, since the best students aren't on the team. [/quote]
That depends on whether we wish to send more students to the State, or to try a best showing at the State. My choice (not necessarily shared by my fellow coaches) would be to send more students if possible.

[quote="Treething"] If the first to sixth are about equal (or all are far better than the competition), then it wouldn't quite make a difference. [/quote]There is usually a slight drop-off after the top 1 or 2, luckily we usually also have even depth after that. Competition at our Chapter is somewhat predictable so we hope we can afford to be slightly creative.

At the State its far less predictable. One year our school won with only one team member in the top 10. Another year a team member placed 3rd but the team was barely in the top 10. Another year our team was around top 25% but no team member was in the top 25% individually. So team performance seems to count more than individual member performance.

Assuming they dont hang math coaches, selecting team members with the objective to maximize the number of students advancing to State will be tricky but fun.

[quote="frost13"] Last year my team won Chapter by like 16 points  problem is coaches are not supposed to rearrange their team between chapter and state  my 9th at chapter took 6th at state and my 4th at chapter took 2nd at state (2nd time that's happened.[/quote] Well done coach, you prepare your students well!

As we cant predict our competition at the State, not being able to change team composition between Chapter and State may not even matter. Besides, our 3rd-6th students could out-perform our 1st-2nd  :)

If our chapter only sends as many students to the State as they used to, we dont even have to think about being creative  :D

[quote]
There is usually a slight drop-off after the top 1 or 2, luckily we usually also have even depth after that. Competition at our Chapter is somewhat predictable so we hope we can afford to be slightly creative.
[/quote]

Yeah, in my school, there's me and Jason. We both get high 20s (27s, 28s, 29s, maybe even a 30 on a good day) on previous State rounds, but the third place, fourth place, fifth place, and sixth place people get from the upper teens to the mid-twenties on State rounds...

At Chapter, there's two schools that routinely place first and second in the district - they're the two middle schools in the school district I live in. 1st and 2nd were a quarter point off this year at Chapter... luckily, our team was a wild-card team (the best-scoring team after all of the first place teams in Chapter also go to States), so me and the school I go to's team all went to States.

At States, three people in our school placed in the top five - 2nd, 4th, and 5th. An eight grader, a sixth grader, and a seventh grader.

here's the way it works in my chapter.... depending on number of teams present (last year one chapter had students taking the SAT on the same day as chapter.... so they joined ours):
normally 4 to  5  top individuals NOT on  team advancing to state. so say you are on a school team but your team does not advance to state you have a chance to if you get in the top 4 or 5 individual scores

and  4 to  5 teams go as well

This will be our first year competing in MC, but this is what I was told:

[i]We average 8-10 schools competing at the chapter level and
the top two schools from each chapter go to the state competition and the remaining teams are determined by score.

		We also make up a team of the top 4 individuals that are not part of one of the teams listed above.[/i]

So, if I am reading this correctly, at least 2 teams from each chapter go to state, plus some unspecified number of wildcard teams.  In addition the top 4 remaining individuals from each chapter also go to states.  (I am reading that as 4 individuals per chapter, but I suppose it might be only 4 individuals in the whole state.)

Any hints that coaches of highly successful MathCounts teams want to share would be most welcome.  These kids are new to math competition this year, and I don't want to overwhelm them, but it would be nice to send a couple of kids to state.

mostly... it depends upon how many students and schools sign up for the chapter contest... eing a math contest, the chapter and state cooridinators use proptions to see how many teams and individuals go to the state contest... so more teams there are, more people go.... less teams, less people go.

There is considerable variation from state to state in what rules are used to decide how many people advance from chapter to state. In my state, there are two chapter with really strong teams (= teams from schools in which many parents have advanced degrees in mathematics) and it is HARD to advance to state from those chapters. Minnesota's team to Nats nearly always includes students from both of those chapters. But there are other chapters that have the occasional surprising lone individual, or even the surprising team, who can do very well at state. My goal as a coach this year is just to get one or two of my guys on to state from our chapter, which is probably the toughest chapter in the state.


See that PBCQ is cyclic from the right angles <PBC and <PQC ---> <CQB=<BPC , but <BPC=<APD=<BCQ ---> <BCQ=<CQB , done

[geogebra]fb936f8f750443add055fb9dcb47c598262048f3[/geogebra]

[quote="AndrewTom"]$ ABCD$ is a rectangle, $ P$ is the mid-point of $ AB$ and $ Q$ is the point on $ PD$ such that $ CQ$ is perpendicular to $ PD$.

Prove that the triangle $ BCQ$ is isosceles.[/quote]
Another solution.
Draw $ BM \perp QC$. We have $ \triangle BMC \sim \triangle CQD \sim \triangle DAP$. So, $ \frac{BM}{AD}\equal{}\frac{MC}{AP}$, Also, $ \frac{BM}{BC}\equal{}\frac{QC}{DC}\equal{}\frac{QC}{2AP}$. So, $ 2MC\equal{}QC$. That is, $ M$ is the midpoint of $ QC$, thus $ BCQ$ isosceles.


Asymptote geometry parameterization incenter

Is there a way I can mark points of tangency in asymptote? For example, I want to mark the points of tangency of the circle and triangle in the following diagram:
[asy]pair A,B,C,I;
A=(0,0);
B=(0,3);
C=(4,0);
draw(A--B);
draw(A--C);
draw(C--B);
draw(incircle(A,B,C));[/asy]

Hi,

With the package [url=http://piprim.tuxfamily.org/asymptote/travaux/index.html]geometry_dev[/url]:
[code]import geometry_dev;
size(7cm);

triangle t=triangleabc(5,3,4);
draw(t);
label(t);
draw(incircle(t));

triangle it=intouch(t);
dot(it);
label( "$A^{\prime}$", "$B^{\prime}$", "$B^{\prime}$", it);[/code]

Let's see if we can reproduce some of it here:
[asy]import geometry_dev;

triangle t=triangleabc(2,3,4);
draw(t);
label(t);
draw(incircle(t));

triangle it=intouch(t);
dot(it);[/asy]

[quote="fedja"]Or, for the same result with cse5:
[code]
pair A=MP("A",(0,0),SW), B=MP("B",(4,0),E), C=MP("C",(0,3),N);
path P=D(A--B--C--cycle), Q=D(incircle(A,B,C));
D("C'",IP(A--B,Q));
D("A'",IP(B--C,Q),NE);
D("B'",IP(A--C,Q),W);
[/code]
[asy]size(200pt);
pair A=MP("A",(0,0),SW), B=MP("B",(4,0),E), C=MP("C",(0,3),N);
path P=D(A--B--C--cycle), Q=D(incircle(A,B,C));
D("C'",IP(A--B,Q));
D("A'",IP(B--C,Q),NE);
D("B'",IP(A--C,Q),W);[/asy]
In the ideal world, Asymptote would allow you to use intersectionpoint() for finding the tangency point of a line and a circle, but it is not so now. You have to either choose the fuzziness parameter carefully to make it work (which is what was done in IP()) or to define the point as the projection from the center to the line (so, if you are using just the Olympiad package, you have to write
[code]
pair D=incenter(A,B,C);
label("$A'$",foot(D,B,C),NE);
[/code]
to get the tangency point on BC marked).

Geometry_dev seems to be a great thing for triangles: the code in the previous post is very short and elegant :).[/quote]
How do I use cse5 in Texnic Center?
I get an error when I say.
\usepackage{cse5}

[quote="ZhangPeijin"]Directory?
Is that another way to say file?[/quote]
[url=http://tinyurl.com/5zmsxn]This link[/url] does it help ?

[quote="Ph. Ivaldi"]Hi,
[quote="kokan"]I'm interested in geometry_dev package, but I don't know where to get it from and how to set it on my unix system. Thanks in advance![/quote]
Place the file [url=http://piprim.tuxfamily.org/asymptote/travaux/modules/geometry_dev.asy]geometry_dev.asy[/url] in the directory ~/.asy (create it if it does not exist) simply.[/quote]

Directory is the same as folder. ~ is your home folder on a Unixlike system. ~/.asy/ is a folder. So you could do:

[code]
mkdir -p ~/.asy && cd ~/.asy && wget http://piprim.tuxfamily.org/asymptote/travaux/modules/geometry_dev.asy;
[/code]

and all would be well.

Directories with names that begin with . are hidden.  On the command line,
[code]ls -a[/code]
will show them.  In a graphical file browser, there is probably a "show hidden files"-like option.

on a Mac, you can either use command line inputs like those mentioned above, or download a system tweak utility like Onyx (freeware). Be careful with those hidden files and system tweaks, though - they're disabled for good reason, and you risk damaging your system (possibly permanently) if you mess with the wrong ones. Don't touch them if you don't have to.

in Windows Explorer, you can go to Tools > Folder Options and there should be something in one of the tabs that lets you show hidden system files (though it's pretty useless for the average user when working on Windows). Again, same precautions.

you probably won't have a folder with that particular name if you're on Windows (.asy is UNIX format)

erm I'm not too sure, but you might try looking in C:\Program Files\Asymptote

[quote="ZhangPeijin"]Hmm, I have a folder called .asy, is that right?
Anyways, how do I find the directory on a Windows?[/quote]
You can put it in [b]/home/<name>/.asy[/b], or you can put it directly in [b]/usr/local/share/asymptote[/b]. I use the latter because it keeps all the files in one place. :)

Unfortunately, I don't know about Asymptote in Windows, but I would guess that the folder you want is the one in which you originally installed Asymptote.

[quote="ZhangPeijin"]Hmm, I have a folder called .asy, is that right?
Anyways, how do I find the directory on a Windows?[/quote]
It's at C:\Documents and Settings\[i]your_user_name[/i]\.asy

Yeah, that's the folder I have.
However, the cse5 file is an internet file, is it supposed to be like that or am I supposed to copy the information on a notepad document and save it in the folder?

In IE, right-click on the link to the file, Save Target As and save in the correct directory.

In firefox it's very similar: right click on the link, and do "Save link as...".


Can someone explain the closure property to me in simple terms?

I searched the internet but found complexed definitions for this algebra property.

Can you also provide an example of the closure property?

Thanks!

The term "closure" has several meanings in mathematics but here you surely mean the term "algebraic closure". Take a field $K$. Then for some polynomial $P(x)=k_{0}+k_{1}x+\cdots+k_{n}x^{n}$ with coefficients in $K$ has some roots $x_{1},\cdots,x_{m}$ in $K$ where $m\leq n$. If $m<n$ then we can find (by Zorn's lemma) a field $K'\supset K$ such that the rest $n-m$ roots of $P(x)$ lies in $K'$. Now again by Zorn's lemma the smallest $K'$ such that every $P(x)$ splits into linear factors exists. That $K'$ is called by an algebraic extension of $K$.

Example: $x^{2}+1=0$ has no roots in $\mathbb{R}$ but it has roots in $\mathbb{C}$. It turns out that no proper subfield of $\mathbb{C}$ contains every root of some real coefficient polynomial and that every polynomial with real coefficients splits into factor of first degree polynomials in $\mathbb{C}$ (the fundamental theorem of algebra). Therefore $\mathbb{C}$ is the algebraic closure of $\mathbb{R}$.

Your data is great but beyond simplicity.

I am talking about the closure property in terms of algebra 1 or basic algebra.

Do you have a more simple definition?


(a) for any integer $i$ such that $0\leq i\leq 2p_k-k-1$ $\implies k\leq i+k\leq 2p_k-1$ $\implies p_k<i+k<2p_k$...(*)

If $n|(n-k)!$ so does $p_k|(2p_k-k)!$ or $p_k|(2p_k-k-i)$ . But this is impposible by (*)

(b) havent solved :P

Since k > p, we have p > 2p-k and hence p does not divide any of 1, 2, 3, ... , 2p-k. But p is prime, so it does not divide (2p - k)! So 2p does not either.

Now consider composite n > 2p. Note that k > 14, so p ≥ 13, so n > 26. Take q to be the largest prime divisor of n and put n = qr. We now have three cases.

(1) q > r ≥ 3. Then n > 2p ≥ 3k/2, so 2n/3 > k. Hence n-k > n-2n/3 = n/3 ≥ n/r = q > r. So q and r are distinct integers < n-k. Hence n = qr divides (n-k)!.

(2) r = 2. Then n = 2q > 2p. But p is the largest prime < k. Hence q ≥ k, so n-k ≥ n-q = q. Obviously q > 2 (since n > 26), so q and 2 are distinct integers < n-k. Hence n = 2q divides (n-k)! .

(3) The final case is n = q2. We show that 2q ≤ n-k. Suppose not. Then 2q ≥ n-k+1 ≥ (2p+1)-k+1 ≥ 3k/2 + 1 - k + 1 = k/2 + 2. So q ≥ k/4 + 1. Also since 2q ≥ n-k+1, we have k ≥ n-2q+1 = (q-1)2 ≥ k2/16. If k > 16, this is a contradiction. If k = 15 or 16, then p = 13 and k ≥ (q-1)2 gives q ≤ 5, so n = q2 ≤ 25. But n > 2p = 26. Contradiction. So we have established that 2q ≤ n-k. But that means that q and 2q are distinct integers ≤ n-k and so their product 2n divides (n-k)!.

I got this, but I forgot the case where $n=q^2$ for part b. How many do you think I would get if ever?

Thanks!

[quote=jin](a)$n=2p_k$, since $k>p_k$,$p_k>2p_k-k=n-k$.
Since $p_k$ is a prime number,$p_k$ does not divide $(n-k)!$,$n$ does not divide $(n-k)!$.
(b)$n>2p_k$, since $n$ is not prime,suppose $n=ab(2\leq a\leq b)$.
(I)$a\geq3$,$n>2p_k\geq\frac{3k}{2}$,$\frac{2}{3}n>k,n-k>\frac{n}{3}$.
(i)$a\neq b$.By $b\leq\frac{n}{3}$,$1<a<b<n-k$,$n=ab|(n-k)!$.
(ii)$a=b$,$n=a^2,n-k>\frac{n}{3}=\frac{a^2}{3}$.By $k\geq14,p_k\geq13,n>26,a\geq6$.
$\frac{a^2}{3}\geq2a,n-k>2a$,$\therefore a\cdot 2a|(n-k)^2,n|(n-k)^2$.
(II)$a=2$,since $n>26$,
(i)$b$ isn't prime,$b=b_1b_2(2\leq b_1\leq b_2)$.
Since $b>13,b_2\geq4,ab_1\geq4,n=(ab_1)(b_2)$,both of $ab_1,b_2$ are larger than 3,
so by (I),$n|(n-k)!$.
(ii)$b$ is prime,$b=\frac{n}{2}>p_k$,since $p_k$be the largest prime number which is strictly less than k,
$b\geq k$,$n-k=2b-k\geq b$,so $b|(n-k)!$,$n|(n-k)!$.
So if $n>2p_k$,$n|(n-k)!$.[/quote]

I like your solution, but I think you meant to write $ a\cdot 2a|(n-k)!,n|(n-k)!$ instead of $a\cdot 2a|(n-k)^2,n|(n-k)^2$ on line seven.

Bashy problem :(

[quote=shobber]Let $k\ge 14$ be an integer, and let $p_k$ be the largest prime number which is strictly less than $k$. You may assume that $p_k\ge 3k/4$. Let $n$ be a composite integer. Prove:
(a) if $n=2p_k$, then $n$ does not divide $(n-k)!$;
(b) if $n>2p_k$, then $n$ divides $(n-k)!$.[/quote]

[b]Solution:[/b]

(a) We have \[n - k < 2p_k - p_k < p_k\] so $p_k \nmid (n-k)!$.

(b) Suppose there is a prime $q$ such that $q\mid n$, but $q\nmid (n-k)!$. Therefore $q<n$ and $q>n-k$.

If $n = 2q$, then $n=2q>2k$. Therefore $2q > 2n -2k > n = 2q$ which is a contradiction.
If $n \neq 2q$, then as $n$ is composite, $n\geq 3q$. Therefore $n\geq 3q>3(n-k)\implies k>\frac{2n}{3}$ which is a contradiction.

Therefore there is no such prime $q$.

Suppose there is a prime $q$ such that $a=v_q(n)>v_q((n-k)!)$. Also, $a \geq 2$. We have  $n\geq q^a$ and \[a>\lfloor \frac{q^{a-1}}{3} \rfloor > \frac{q^{a-1}}{3}-1\implies 3a+3> q^{a-1}.\] As $a\geq 2$, we need to check for primes $q=2,3,5,7$.

A simple case bash for $q=2,3,5,7$ using the bounds we are done $~~\square$

Part A: Since $2p_k-k<p_k$ as $p_k<k$, $p_k$ does not divide $(2p_k-k)!$, so $2p_k$ also does not divide it.\\

Part B: Case 1: $n$ can be expressed as $ab$ for $a,b\geq 3$. Note that this means $a,b\leq n/3$. Then, $$k\leq \frac{4}{3}p_k<\frac{2}{3}n,$$ so $$n-k\geq \frac{1}{3}n.$$ Thus, the factorial will be divisible by $ab$.\\

Case 2: $n$ cannot be expressed in such a way. Then, we must have $n=2p$ for a prime $p$. Then, we wish to show $$2p\mid (2p-k)!.$$ However, $p>p_k$, and since by definition $p_k$ is the largest prime under $k$, we must have $p\geq k$, so $2p-k\geq p.$ Thus, the factorial contains the factor of $p$ (and obviously a factor of 2 as well), so we are done.


Two players A and В alternately take chips from two piles with $ a$ and $ b$ chips, respectively. Initially $ a > b$. A move consists in taking a multiple of the other pile from a pile. The winner is the one who takes the last chip in one of the piles. For what $ \alpha$ can A force a win, if initially $ a > \alpha b$.


Hi, I'm trying to do a reality check on a feature of the forum that seems to have disappeared. I thought it used to be that when I read one of my old posts, there was a control (which looked like a little X on the screen) that allowed deletion of the post. I thought every user could delete his or her own posts after posting. Now I see a control (which looks like a pair of scissors cutting a piece of paper) for which the tooltip says "Edit/Delete this post," but I don't quite get how to use that control to delete the post. I edit posts from time to time, and wish that more forum users would edit posts after posting, but how do I delete a post if need be, using that control? Today I tried setting all the text of a message to nothing being there (that is, deleting the formally posted text in the edit window), but then I got an error message saying that a post has to contain text. I suppose this is a feature, not a bug  :P  but in any event I'm not sure how to delete posts (of my own) now that the former control is gone.

Meanwhile, I think I can still delete any post in the forum I moderate, although I can't say that I've ever tried it.

It used to be that you could only delete a post if nobody had made a post in the same topic. Because then things could get a bit mucked up, I suppose.

The same is still true now. The x is there as long as its the last post in that thread. As for why it says edit/delete when you hold your mouse over it, I don't know, perhaps thats a bug.

I think TripleM is basically right.  If you did want to delete a post in the middle of a thread, I suppose you could just replace it with ' . . . '.  It would leave a little 'placeholder' of the three dots where your post would have been.


function calculus derivative trigonometry algebra polynomial

Determine the function $f: \mathbb{Z} \to \mathbb{Z}$ satisfying:
$f(x+y)+f(x-y)=2f(x)f(y)$
Same question for condition:
$f(x+y)+(f(x-y))^2=2f(x)f(y)$

The first problem (I know it uses calculus, but it's still an interesting solution).
[hide]If y=0, we see that f(0)=1 or 0. Letting y=x, 1+f(2x)=2f(x)f(x). Now letting y=-x, 1+f(2x)=2f(x)f(-x), so f(-x)=f(x) and f is an even function. If we hold y constant, we note that the derivative of f has the same property. But since f is an even function, its derivative is odd. Since the derivative has the same property, it must be both even and odd. Therefore, f'(x)=0 and f(x)=c for some constant c. Since f(0)=1 or 0, the only functions with this property are f(x)=1 and f(x)=0.[/hide]

A lesson I learned from stergiu is that nowhere from the problem says that $f$ has a derivative. And we are working with integers...

[quote="JohnConstantine"]what about $\cos \alpha$?[/quote]Let me restate it

\[ f: \mathbb{Z}\to\mathbb{Z} \]

John Constantine is headed in the right direction, however.  [hide]Let $y\to 0$ and we have $2f(x) = 2f(0)f(x)$ so $f(0) = 0$ or $f(0) = 1$.  From $f(0) = 0$ it follows immediately from that $f\equiv 0$.  Thus, assume $f(0) = 1$.  Let $f(1) = a$.  Then plugging in a few values one realizes that there is a solution $f(n) = P_n(a)$ where $P_n$ is the polynomial such that $\cos(nx) = P_n(\cos x)$.  That these are all solutions follows from the properties of the cosine.  That they are the only solutions follows from the fact that $f(n)$ can be solved for recursively by taking $x\to n-1, y\to 1$.[/hide]

I haven't thought about the second question yet.

Well, let me give it a little start:
[hide]Take $y\to 0$ so that $f(x) + f(x)^2 = 2f(x)f(0)$.  Thus either $f(x) = 0$ or $f(x) = 2f(0) - 1$, for each $x$.  In particular, either $f(0) = 0$ or $f(0) = 1$, and so in fact the function either has range $\{0, -1\}$ or $\{0, 1\}$ and the problem is reduced in each of these two cases to determining the values of $x$ for which $f(x) = 0$.[/hide]


linear algebra
matrix
linear algebra unsolved

The assertion is false.  Choose, for example $ y_k = x_k^4$, then the columns look like
\[{{ {{\bf{\alpha }}_{\bf{k}}} = \left({\begin{array}{*{20}{c}}{x_{k}^{3}} \\{
{x_{k}^{6}}} \\
{{x_{k}^9}} \\
{x_{k}^{12}} \\
{x_{k}^{2}} \\{
{{x_{k}^5}}} \\
{x_{k}^{8}} \\
{{x_{k}}} \\
{{x_{k}^4}} \\
1 \\
\end{array}}\right)\]
But then the columns of the matrix are $ 9$ rows of a Vandermonde matrix which, as long as $ \lbrace x_k\rbrace$ are distinct, is known to have full rank.

A way of seeing that this assertion must be false is to note the relationship between this matrix and two dimensional Lagrange interpolation.  Since 'typically' interpolants are unique, we should expect this matrix to be rank deficient for only very specially selected  $ \lbrace (x_k,y_k)\rbrace$.


inequalities
quadratics
Pythagorean Theorem
geometry

If someone finds a new proof of something, say the Pythagorean Theorem, what should he/she do? How should he/she find out if it has already been discovered? How should he/she publicize it under his/her name? For example, I know Mr. Vornicu added something to the Schur inequality, and now it is sometimes called the Vornicu-Schur inequality.

Find as many other proofs as you can.  For example, people often publish new proofs of old theorems in the AMM.  You would be surprised at the sheer number of different proofs - there are some four hundred or more proofs of the Pythagorean theorem and at least two hundred of quadratic reciprocity.  You should be extremely skeptical that you've come up with something original until you have considerably evidence to the contrary.

Once you've convinced yourself you have something interesting, you should try to get it published somewhere reasonably reputable; AMM is not a bad choice for such things.


[quote="236factorial"]If xy=18, yz=24, xz=12, x,y,z are positive integers, what is xyz?[/quote]

[hide]If we can just find out ONE term, then we can find $xyz$

$xy=18$

$yz=24$

$xz=12$

$\frac{yz}{xz}=\frac{24}{12}$

$\frac{y}{x}=2$

$y=2x$

$xy=18$

$x(2x)=18$

$2x^2=18$

$x^2=9$

$x=3$

Since $yz=24$, $xyz=24*3=72$[/hide]

[quote="236factorial"]There's a much easier way to do this  ;)  you don't have to find anything...[/quote]

D'oh!  :blush:
[hide]
$x^2y^2z^2=18*12*24$

$(xyz)^2=5184$

$xyz=72$[/hide]


A polygon can be dissected into 100 rectangles, but not into 99. Prove that it cannot be dissected into 100 triangles. :D

Hmm... Do know a nice solution?
Though my solution is simple enough, but all explanations doesn't look nice.

I have prooved that such rectangle necessarily have 99 angles of $270^\circ$. So the sum of angles is too big for only 100 triangles.


Perpendiculars $BE$ and $CF$  are dropped onto $AD$ frpm the points $B$ and $C$ of a semicircle with diameter $AD$ .The straight lines $AB$ and $DC$ intersect in $P$ and the segments $EC$ and $BF$ intersect in $Q$ . Prove that $PQ\bot\ AD$

[quote="silouan"][color=red]Perpendiculars $BE$ and $CF$  are dropped onto $AD$ frpm the points $B$ and $C$ of a semicircle with diameter $AD$
The straight lines $AB$ and $DC$ intersect in $P$ and the segments $EC$ and $BF$ intersect in $Q$ . Prove that $PQ\bot\ AD\ .$[/color][/quote]
[color=darkred][b]An equivalent enunciation.[/b] Let $ABC$ be an [u]acute[/u] triangle with the orthocenter $H\ .$ Denote : the feet $E$ and $F$ of the altitudes $BH$ and $CH\ ;$
the projections $M$ and $N$ to the line $BC$ of the points $E$ and $F$ respectively. Prove that the lines $AH\ ,\ NE\ ,\ MF$ are concurrently.
[hide="Hint."]For the point $D\in AH\cap BC$ prove easily that $\triangle DME\sim\triangle DNF$, i.e. $\frac{DN}{DM}=\frac{FN}{EM}$ and apply this relation in the trapezoid $MNFE\ .$
[/hide][/color][color=darkblue][b]Extension.[/b] Let $ABC$ be an [u]acute[/u] triangle. Denote the projection $D$ of the vertex $A$ on the line $BC\ .$  For a point $P\in [AD]$ define the points $E\in BP\cap AC$ , $F\in CP\cap AB$ , the projections $M\ ,\ N$ of the points $E\ ,\ F$ respectively on the line $BC$ and the point $L\in NE\cap MF\ .$ Prove that $L\in AD\ .$[/color]


quadratics
modular arithmetic
number theory proposed
number theory

Find the smallest nomial of this sequence that $a_1=1993^{1994^{1995}}$ and
\[ a_{n+1}=\begin{cases}\frac{a_n}{2}&\text{if $n$ is even}\\a_n+7 &\text{if $n$ is odd.} \end{cases} \]

First of all, I'm pretty sure $a_{n+1}$ is supposed to be $\frac{a_n}2$ when $a_n$ is even, not $n$ (and $a_n+7$ when $a_n$ is odd). Second of all, I think $a_1=1993^{1994^{1995}}$, that is, the topmost number is $1995$, not $1994$.

Every number $>7$ is turned into one $<7$ in at most $2$ steps, so the minimum is $\le 7$. Since our initial term is not divisible by $7$ and it's clear that the rules can't produce terms divisible by $7$ when there are none, it means that the minimum is $\le 6$. $6,5,3$ go to $3$, while $4,2,1$ go to $1$, so there are two possibilities: the minimum is either $3$ or $1$.

$2$ is a quadratic residue modulo $7$, so either all the terms of the sequence are quadratic residues modulo $7$, or none are. Since the initial term is $4\pmod 7$, it means that all terms are quadratic residues of $7$, so the minimum can't be $3$, meaning that it must be $1$.


find the maximum the value of (a+1)/(ab+a+1)+(b+1)/(bc+c+1)+(c+1)/(ca+a+1) such abc=k>0 and
a,b and c>0;it is abvious that if abc=1  l'expression equals 2.

if $a=k$, $bc=1$, $b\rightarrow \infty$ then $\frac{b+1}{bc+c+1}\rightarrow\infty$

you should check your inequality.


inequalities
calculus
function
inequalities unsolved

how to prove that a^3+b^3+c^3+3abc>=a^2(b+c)+b^2(a+c)+c^2(a+b)

This is exactly Schur's Inequality.
For non-negative $x, y, z, r$, $a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b)\geq 0$
When $r=1$, we get
$a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)\geq 0$
Expanding everything,
$a^3+b^3+c^3+3abc\geq \sum_{sym}a^2b$

Let $f(a,b,c)=a^3+b^3+c^3-ab(a+b)-bc(b+c)-ca(c+a)$.

Our inequality is sym. , so $f(a,b,c)=...=f(c,b,a)$.

\[f(a,b,0)=a^3+b^3-ab(a+b)=(a+b)(a-b)^2\geq 0\]

\[\frac{\partial f}{\partial a}+\frac{\partial f}{\partial b}+\frac{\partial f}{\partial c}=a^2+b^2+c^2-ab-bc-ca\geq 0, \mbox{Q.E.D.}\]

Or observe that $a^3+b^3+c^3+3abc-ab(a+b)-bc(b+c)-ca(a+c)=abc-(a+b-c)(b+c-a)(c+a-b)$, substitute $x=a+b-c$, $y=a-b+c$, $z=b+c-a$ and use AM-GM.

i've seen a very simple proof of this in Kiran Kedlaya's "A>B".

Since the inequality is symmetric on a,b,c we may suppose that
$a \geq b \geq c$, so we have that
$a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b) \geq 0
\\\Leftrightarrow (a-b)[a^r(a-c)-b^r(b-c)]+c^r(c-a)(c-b) \geq 0$,
but this is obvious since $a^r \geq b^r$, $a-c \geq b-c \geq 0$, and $a-b \geq 0$ imply that $(a-b)[a^r(a-c)-b^r(b-c)]$ and $c^r(c-a)(c-b) \geq 0$

hope you find this helpful.

[quote="lotharek"]

\[f(a,b,0)=a^3+b^3-ab(a+b)=(a+b)(a-b)^2\geq 0\]

\[\frac{\partial f}{\partial a}+\frac{\partial f}{\partial b}+\frac{\partial f}{\partial c}=a^2+b^2+c^2-ab-bc-ca\geq 0, \mbox{Q.E.D.}\]

[/quote]
I don't catch your prove
 especially the derivation calculus

[quote]This is exactly Schur's Inequality.
For non-negative a, b, c, r, $a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b)\geq 0 $[/quote]

It's true for any convex function, so u can write $f(a)$ instead $a^r$ etc.

[quote]
I don't catch your prove
especially the derivation calculus[/quote]

[i]Theorem[/i]:

$x_1,x_2,\ldots , x_n\in \mathbb{R}_+\cup \{0\}$. Let $f(x_1,x_2,\ldots , x_n)$ be function that satisfies the following conditions:

- $f(x_1,x_2,\ldots , x_{k-1}, 0, x_{k+1}, \ldots , x_n)\geq 0$,
- $\frac{\partial f}{\partial x_1} + \frac{\partial f}{\partial x_2} + \ldots +\frac{\partial f}{\partial x_n} \geq 0$,
then $f(x_1,x_2,\ldots , x_n)\geq 0$.

This theorem has easy proof.


geometry
parallelogram
perimeter
Pythagorean Theorem

A parallelogram has three of its vertices at $ (\minus{}1,0)$, $ (2,4)$ and $ (2,\minus{}4)$.  What is the positive difference between the greatest possible perimeter and the least possible perimeter of the parallelogram?

There are only 2 possible perimeters given three points of a parallelogram (think about it).

One example is found if the final point is $ (5, 0)$.  Using the Pythagorean theorem, the sides all have length $ 5$ (a square).  This means that the perimeter is $ 20$.

The other perimeter is found for point $ (\minus{}1, 8)$
Using the Pythagorean theorem, we can find that the perimeter is equal to
$ 5 \plus{} 8 \plus{} 5 \plus{} 8 \equal{} 26$.

The difference in these perimeters is $ 26 \minus{} 20 \equal{} \boxed{6}$

Did someone forget (-1,-8)? That also has a perimeter of 26, so the answer 6 is correct.

lol, same answer, so therefore, no one "forgot" it.

All thing are obvious From the following figure
[img]http://photobase.ro/di-F31L.png[/img]


number theory
relatively prime
number theory proposed

[quote="N.T.TUAN"][quote="zweig"]
[b][u]Problem 2[/u][/b] : Let be $ (a, b, c)\in\mathbb{N}^{*}$. Show that $ a^{3}+b^{5}=c^{2}$ admit infinite solutions.
[/quote]
  Let $ a, b, c$ be positive integers such that $ a$ and $ b$ are relatively prime and $ c$ is relatively prime to $ a$ or $ b$. Show that there are infinitely many solutions to $ m^{a}+n^{b}= k^{c}$, where $ m, n, k$ are distinct positive integers.

Can you?  :wink:[/quote]

If $ ab$ and $ c$ are coprime, we can find $ \alpha$ and $ \beta$ such that $ \alpha ab+1=\beta c$. Then :

$ (x(x^{a}+y^{b})^{\alpha b})^{a}+(y(x^{a}+y^{b})^{\alpha a})^{b}=((x^{a}+y^{b})^{\beta})^{c}$

[quote="N.T.TUAN"]Nice, but we haven't $ \gcd (c,ab)=1$, an example $ a=6,b=5,c=4$.[/quote]

OK, I'm sorry, I didn't read carefully the conditions. But the answer is nearly the same :

Say WLOG $ a$ and $ c$ are coprime, then $ a$ and $ bc$ are coprime, so it exist $ \alpha$ and $ \beta$ such that $ \alpha bc+1=\beta a$

Then $ ((x^{c}-y^{b})^{\beta})^{a}+(y(x^{c}-y^{b})^{\alpha c})^{b}=(x(x^{c}-y^{b})^{\alpha b})^{c}$

You are right, this is one in ''38th IMO 1997 shortlisted problems'' . But i don't know why our site Resources hasn't it!  :maybe:


let $ x$ ,$ y$ and $ z$ are strictly reals positives  numbers  prove that :
for all quadrilateral which their lenghts are $ a$,$ b$,$ c$, and $ d$ such that:$ a^2x\plus{}b^2y\plus{}c^2z>d^2 \Leftrightarrow \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z} \leq 1$

it seems no one would help me so this implication is hard or what.

[quote="ghostchungho"]let $ x$ ,$ y$ and $ z$ are strictly reals positives  numbers  prove that :
for all quadrilateral which their lenghts are $ a$,$ b$,$ c$, and $ d$ such that:$ a^2x \plus{} b^2y \plus{} c^2z > d^2 \Leftrightarrow \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \leq 1$[/quote]
$ x\equal{}\frac{1}{2}$, y=z >>, $ a^2x \plus{} b^2y \plus{} c^2z > d^2$, but $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}>2$  :?:


inequalities function calculus derivative integration logarithms algorithm

[quote="flip2004"][color=green] [b] Find  $A, A > 0,$such that  $\; \; \; \; \frac{x^2+6x+12}{x^2-6x+12} \le A^x \le\frac{2+x}{2-x} \; , \; \; \forall x \in [0,2).$   [/b] [/color][/quote]
Let $fx) = \frac{x^2+6x+12}{x^2-6x+12}$ and $g(x) = \frac{2+x}{2-x}$. One checks that $f(0) = g(0) = f'(0) = g'(0) = 1$, thus $A = e$ is the only choice.

To show that the inequality is actually correct for $A = e$, compute
$\frac{d^2}{dx^2}\frac{e^x}{f(x)} = e^x x^3 \frac{x^3 + 6x^2 + 24x + 48}{\left(x^2 + 6x + 12\right)^3}$,
a positive term for $x > 0$, and similary
$\frac{d^2}{dx^2}\frac{g(x)}{e^x} = x e^{-x}\frac{x^2 -2x + 4}{\left(2-x\right)^3}$,
which is also positive for $0 < x < 2$.

The function $h(x) = \frac{e^x}{f(x)}$ therefore satisfies $h(0) = 1$, $h'(0) = 0$, $h''(x) > 0$ for $0 < x < 2$. Thus $\frac{e^x}{f(x)} \ge 1$ on this interval. By the same argument, $\frac{g(x)}{e^x} \ge 1$ on this interval. [i]Mathcad helped on this problem[/i] :lol:

[quote="Kalle"]This is a VERY annoying problem if you can't use a computer for the differentiation. Is there a more elegant approach?[/quote]
You can also try to show that $\frac{f'(x)}{f(x)} \le 1 \le \frac{g'(x)}{g(x)}$ for $0 \le x < 2$. This requires only one (logarithmic) differentiation, but more algebraic work.

But tell me - why is it more elegant to walk when you can use a bicycle?

[quote="Kalle"] Is ....there a more elegant approach ....?[/quote]
[b]Hint by proposer.[/b] Instead of [color=green] [b] differentiation[/b] [/color] try to use  [color=blue][b]integration ! [/b][/color]

[b]Solution.[/b] [color=green][b]  For $n\in \{1,2\}$ consider the integrals ${\mathbf I_n}=\int\limits_{0}^{x}t^n(x-t)^nA^t\;dt$ and integrate by parts  . [/b][/color]

[size=75]Details:  By denoting $\lambda=\ln{A}$  and integrating by parts one finds ([color=red][b]please verify[/b][/color])
\[ (*)\; \; \; \; A^x\sum\limits_{k=0}^{n}{n\choose k}\frac{(2n-k)!}{n !}(-\lambda^kx^k) =\sum\limits_{k=0}^{n}{n\choose k}\frac{(2n-k)!}{n !}(\lambda^kx^k) + \frac{(-1)^n}{(2n)!}\lambda^{2n+1}{\mathbf I_n}\; .  \]
Observe that   ${\mathbf I_n } >0 \;,\; \forall x \in (0,\infty).$ [b] Further consider cases  [/b]   $n\in \{1,2\} .$ If you wish ,  put $A=e$ in (*) , that is    $\lambda=1 .$
 [color=green][b] If  $Q_n(z): =\sum\limits_{k=0}^{n}{n\choose k}\frac{(2n-k)!}{n !}z^k$ observe that (*)  enables us to write  $\displaystyle (**)\; \; \; \; \; (-1)^n \left( A^xQ_n(-\lambda x)-Q_n(\lambda x) \right) > 0 \; \; ,\; x> 0 .$
Because  $\; Q_1(z)=z+2 ,\; \; Q_2(z)=z^2+6z+12$ and  $Q_1(-x) >0$  for   $x \in [0,2)$ , the solution is complete.
[/b][/color]
[/size]

$(x^2+6x+12)\leq (x^2-6x+12)A^x$ use asymptotic expansion when x near 0

 gives $A\geq e$

Similarly asymptotic expansion near 0 for $A^{x}(2-x)\leq (2+x)$
 gives $A\leq e$

A=e

[quote="Kalle"]Is there a more elegant approach?[/quote]
Here's a mechanical approach (I wouldn't quite call it an algorithm yet) to produce any number of inequalities of the form $\frac{p(x)}{q(x)} \le e^x$ or their reverse, on some interval containing $x=0$.

Start with a known inequality, say $p(x) \le q(x)e^x$ on $[0,x_0)$, where $p$ and $q$ are polynomials. Find polynomials $P$ and $Q$ such that $Q'(x) + Q(x) = q(x)$ and $P'(x) = p(x), \, P(0) = Q(0)$. $Q$ can be found by comparing coefficients, $P$ by direct integration. Then $P(x) \le Q(x) e^x$ on the same interval, by integration. Repeat as often as needed. :)

[b]Example 1:[/b] Start with $0 \ge -xe^x$. Then $Q(x) = 1-x, \, P(x) = 1$, and $1 \ge (1-x)e^x$. Repeat, then $Q(x) = 2-x, \, P(x) = 2+x$, and $2+x \ge (2-x)e^x$, one of the inequalities stated in the problem. Of course we may continue and obtain $Q(x) = 3-x, \, P(x) = 3 + 2x + \frac{1}{2}x^2$, resulting in $3 + 2x + \frac{1}{2}x^2 \ge (3-x)e^x$. And so on.

[b]Example 2:[/b] Start with $0 \le x^2 e^x$ and do three steps, the result is $x^2 + 6x + 12 \le (x^2 - 6x + 12)e^x$, the second inequality.

[b]Example 0:[/b] Start with $0 \le e^x$ and do $n+1$ steps, arrive at $\sum_{k=0}^n\frac{x^k}{k!} \le e^x$ :)

There's probably a connection here with Laguerre polynomials and Pade approximants to the exponential functions.

[quote="hpe"] ... a connection here with ... Pade approximants to the exponential functions.[/quote]
Indeed,proposer


[quote="Marius Mainea"]If $ x,y,z > 0$ then
\[ \frac {x^3}{x^2 \plus{} y^2} \plus{} \frac {y^3}{y^2 \plus{} z^2} \plus{} \frac {z^3}{z^2 \plus{} x^2}\geq \frac {x \plus{} y \plus{} z}{2}
\]
[/quote]
$ \frac {x^3}{x^2 \plus{} y^2}\equal{}x\minus{}\frac{xy^2}{x^2\plus{}y^2}\ge x\minus{}\frac{y}{2}$
$ \Rightarrow \sum\frac {x^3}{x^2 \plus{} y^2}\ge x\plus{}y\plus{}z\minus{}\frac{x}{2}\minus{}\frac{y}{2}\minus{}\frac{z}{2}\equal{}\frac {x \plus{} y \plus{} z}{2}$ :wink:

[quote="Marius Mainea"]If $ x,y,z > 0$ then
\[ \frac {x^3}{x^2 \plus{} y^2} \plus{} \frac {y^3}{y^2 \plus{} z^2} \plus{} \frac {z^3}{z^2 \plus{} x^2}\geq \frac {x \plus{} y \plus{} z}{2}
\]
[/quote]
$ \frac {x^3}{x^2 \plus{} y^2} \plus{} \frac {y^3}{y^2 \plus{} z^2} \plus{} \frac {z^3}{z^2 \plus{} x^2}\geq \frac {x \plus{} y \plus{} z}{2}$
$ \leftrightarrow \frac {y}{x^2 \plus{} y^2}(x\minus{}y)^2\plus{}\frac {z}{y^2 \plus{} z^2}(y\minus{}z)^2\plus{}\frac {x}{z^2 \plus{} x^2}(z\minus{}x)^2 \ge\ 0$
We have done!


Incorrect; there is a solution, but I don't know what it is.

hmmm... i can't seem to find an answer. I factor the problem to (x+y)(x+y+3) = 24, but now what???  Are you SuperSure that there is an answer???

There are definitely no real roots. Subtract the equations to get $x^2-xy+y^2=0$, and note that $(0,0)$ is not a solution of either equation.

An intersection occurs when $x$ is -1 in the second equation.  $y$ is undefined though.

No, points at infinity don't make sense here; the first equation is a circle, so its graph is bounded.

No, there are no values that satisfy the two equations (I might be overlooking something though, this solution seems too easy):

[hide]
Subtracting the second equation from the first equation:
$
x^2-xy+y^2=0
$
Completing the square, we get:
$
(x-\frac{1}{2}y)^2+\frac{3}{4}y^2=0
$
In order for this to equal 0, both of the terms must equal zero, which implies that x=1/2y and y=0 (therefore x=0)
But substituting back, x=0, y=0, we get 0 equals 8, which does not make sense.
Therefore there are no solutions.
[/hide]


When riding on a train a person often listens to the click of the rails as the wheels pass over their joints. If the rails are each 30 feet long, for how many seconds must a person count clicks in order that the number of clicks counted will exactly equal the measure of the speed of the train in miles per hour.

There are 5280 feet in a mile.

5280/30 = 176

Thus there are 176 rails per mile and you would need to count for 176 seconds.


geometry perimeter parallelogram rhombus

i think he meant intersection point of $ AC$ and $ BD$

Again WLOG we can assume $ AO\geq CO$ and $ BO/geq DO$. We consider points C',D' symmetric to C,D with respect to O => ABCD must be a parallelogram + condition (it's cicrumscribed quadrilateral) => ABCD is a rhombus QED

[quote="Virgil Nicula"] [color=darkred]Let $ ABCD$ be a convex quadrilateral for which denote $ O\in AC\cap BD$ .

Suppose that $ AB \plus{} CD \equal{} AD \plus{} BC$ , the triangles  $ AOB$ , $ COD$ have same perimeter

and the triangles $ BOC$ , $ AOD$ have same perimeter. Prove that  $ ABCD$ is a rhombus. [/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote the perimeter $ p(XYZ)$ of $ \triangle XYZ$ . Observe that the quadrilateral $ ABCD$ is tangential

and denote its tangent points $ M\in (AB)$ , $ N\in (BC)$ , $ P\in (CD)$ , $ R\in (DA)$ with its incircle.

Is well-known that $ O\in MP\cap NR$ and observe that $ \left\|\begin{array}{ccc} AR \equal{} AM & , & BM \equal{} BN \\
\ CN \equal{} CP & , & DP \equal{} DR\end{array}\right\|$ . Therefore,

$ \left\|\begin{array}{c} p(AOB) \equal{} p(COD) \\
\ p(BOC) \equal{} p(DOA)\end{array}\right\|$ $ \implies$ $ \left\|\begin{array}{c} p(AOB) \plus{} p(BOC) \equal{} p(COD) \plus{} p(DOA) \\
\ p(AOB) \minus{} p(BOC) \equal{} p(COD) \minus{} p(DOA)\end{array}\right\|$ $ \implies$

$ \left\|\begin{array}{ccc} OB \plus{} BN \equal{} OD \plus{} DR & \stackrel {(*)}{\implies} & OB \equal{} OD \\
\ OA \plus{} AR \equal{} OC \plus{} CN & \stackrel {(*)}{\implies} & OA \equal{} OC\end{array}\right\|$ $ \implies$ $ ABCD$ is a rhombus.

For the understanding the implication $ (*)$ you must read the proof of my proposed problem from[/color] [url=http://www.mathlinks.ro/viewtopic.php?t=238903][color=red][b]here[/b][/color][/url] .


Has the 2006 Nationals Test been posted anywhere online? I would like to see it. :D

I would too, although I was looking for it and I could not find it... :?

Nationals do not get posted.  If you were there, the coach gets a copy, otherwise, you must buy it.

Yeah... I'll have to ask my coach about it. I'd like to see what kind of mistakes I made after getting only about six hours of sleep. :(

My coach scanned and sent us pdfs by email. I already gave it to my brother, who got a 36.

[quote="jclarkemathL314159"]ohhh you scared me. i thought your brother would be in like...4th grade.

Oh and by the way, does anyone know what neal got? (and everyone else in the countdown)

George (MA), got 38, and he came in 8th.[/quote]

42, Daniel Li got 41, it should be on the distribution sheet.

We still haven't gotten anything from our coach yet.  She said she would copy it, but she hasn't.  We only know our scores and places.

Mathcounts has made 13 doubly unlucky now :D.
Who were the top 10 7th graders?

I thought alot more people got a 35, but only 3 people did, this is just weird.

[quote="Fanatic"]Kevin Chen is it I think (3rd)[/quote]
I was wondering about top ten. I want to know about competition for next year. :D

I will just post the places to ensure privacy among people.

Top 7th Graders
1. 4th Place
2. 15th Place
3. 16th Place
4. 23rd Place
5. 24th Place
6. 27th Place
7. 35th Place
8. 40th Place
9. 46th Place
10. 48th Place
11. 53rd Place
Then it goes from there

I am pretty sure this is correct.
Correct me if I am mistaken.

Yeah, that's right. I had counted 10, but had overlooked 53rd place.

Why not post names though? It's not like the Rankings for this year aren't publicly viewable...

Yeah, but some people may not want their real name to be associated with their username...

Username?
I'm not talking about usernames, I'm talking about each person's ranking this year...


inequalities
induction
inequalities proposed

Hmm... I've got straightforward solution, I hope there is some niecer ;)
We decompose:
$(\sum a_i)(\sum b_i)\geq (\sum a_i+b_i)(\sum\frac{a_ib_i}{a_i+b_i}) $
into sum of many "smaller" inequalities which will turn to be true:

For each $i$ this ineq is true:
$(a_ib_i)\geq (a_i+b_i)\frac{a_ib_i}{a_i+b_i}$
cause it's equivalent to:
$a_ib_i\geq a_ib_i$
which is true.

For each $i<j$ we have:
$a_ib_j+a_jb_i\geq (a_i+b_i)\frac{a_jb_j}{a_j+b_j}+(a_j+b_j)\frac{a_ib_i}{a_i+b_i}$
It's true because it's equivalent to:
$\frac{(a_ib_j-a_jb_i)^2}{(a_i+b_i)(a_j+b_j)}\geq 0$
And it's certainly true.

Now we have:
$(\sum a_i)(\sum b_i)=(\sum a_ib_i)+(\sum_{i<j} a_ib_j+a_jb_i)\geq \\ \geq (\sum (a_i+b_i)\frac{a_ib_i}{a_i+b_i})+(\sum_{i<j} (a_i+b_i)\frac{a_jb_j}{a_j+b_j}+(a_j+b_j)\frac{a_ib_i}{a_i+b_i})=\\=(\sum a_i+b_i)(\sum\frac{a_ib_i}{a_i+b_i}) $

[quote="Pascual2005"]1. Prove the following inequality for positive reals $a_1,a_2...,a_n$ and $b_1,b_2...,b_n$:

$(\sum a_i)(\sum b_i)\geq (\sum a_i+b_i)(\sum\frac{a_ib_i}{a_i+b_i})$[/quote]

In http://www.mathlinks.ro/Forum/viewtopic.php?t=16582 post #9, I proved the inequality

$\sum_{i=1}^{n}A_{i} \cdot \sum_{i=1}^{n}B_{i}\geq \sum_{i=1}^{n}\frac{A_{i}B_{i}}{A_{i}+B_{i}}\cdot \left( \sum_{i=1}^{n}A_{i}+\sum_{i=1}^{n}B_{i}\right) $

for positive reals $A_1$, $A_2$, ..., $A_n$, $B_1$, $B_2$, ..., $B_n$. Of course, since $\sum_{i=1}^{n}A_{i}+\sum_{i=1}^{n}B_{i}=\sum_{i=1}^{n}\left(A_{i}+B_{i}\right)$, this becomes

$\sum_{i=1}^{n}A_{i} \cdot \sum_{i=1}^{n}B_{i}\geq \sum_{i=1}^{n}\frac{A_{i}B_{i}}{A_{i}+B_{i}}\cdot \sum_{i=1}^{n}\left(A_{i}+B_{i}\right) $,

what is exactly your inequality.

  darij

My solution is to prove the case $n=2$ and then aply induction putting $a_n'=a_n+a_{n+1}$ and using case $n=2$

I see that this problem has allready been solved (in many ways ), but i'll post my solution cause i liked it.
Let $S_a = \sum a_i$ , $S_b= \sum b_i$ and without loss of generallity let $S_a + S_b = 1$.
Moreover let us assume that $(\sum a_i)(\sum b_i) < (\sum a_i+b_i)(\sum\frac{a_i\cdot b_i}{a_i+b_i})$.
Then easily follows that $\sum \frac{(a_i)^2}{a_i +b_i} < (S_a)^2$ and $\sum \frac{(b_i)^2}{a_i +b_i} < (S_b)^2$. So by multiplying these two inequalities and apply C-S we get :
$(\sum \frac{a_i\cdot b_i}{a_i+b_i})^2 \leq (\sum \frac{(a_i)^2}{a_i +b_i})(\sum \frac{(b_i)^2}{a_i +b_i})<(S_a\cdot S_b)^2$. This is clearly a contradiction.
The rest follows.

[quote="Anto"]
Then easily follows that $\sum \frac{(a_i)^2}{a_i +b_i} < (S_a)^2$ and $\sum \frac{(b_i)^2}{a_i +b_i} < (S_b)^2$[/quote]
How ???

Since $ S_a\cdot S_b < \sum \frac{a_i\cdot b_i}{a_i+b_i} $ we get that $ S_a\cdot (1-S_a)  < \sum \frac{a_i\cdot b_i}{a_i+b_i} $ or $ \sum (a_i - \frac{a_i\cdot b_i}{a_i+b_i}) < (S_a)^2$ e.t.c.

thanx, now I see  :blush: , really nice solution :)

This inequality is not so hard,it could be proved by mathematical induction for i.
      I tried that one and I get long but very simple solution  :lol:


A space shuttle travels from Earth towards the moon at a speed of
30,000 kilometers per hour for 12 hours. The total distance to the
moon is 400,000 kilometers. At the current velocity, how
many additional minutes are needed to reach the moon?

The shuttle has already traveled $ 30000(12)\equal{}360000$ km. Thus, it needs to travel another $ 40000$ km. Therefore, it needs to travel an additional $ 60(\frac43)\equal{}\boxed{80}$ minutes.


[quote="Silverfalcon"]If $ AB \equal{} 2, BC \equal{} 6, AE \equal{} 6, BF \equal{} 8, CE \equal{} 7,$ and $ CF \equal{} 7,$ compute the ratio of the area of quadrilateral $ ABDE$ to the area of $ \triangle CDF$.[/quote]

Whee, more geometry!

[hide]$ \triangle ACE \equiv \triangle BFC$ so $ \angle CAE \equal{} \angle CBF$ and then $ BF \parallel{} AE$. then $ \triangle CBD$ is similar to $ \triangle CAE$ and then with some calculation get $ BD \equal{} \frac {9}{2}$, $ DF \equal{} \frac {7}{2}$, $ CD \equal{} \frac {21}{4}$, $ DE \equal{} \frac {7}{4}$.

Then we have $ \frac {[CBD]}{[CDF]} \equal{} \frac {9}{7}$.

We also have $ \frac {[CAE]}{[CBD]} \equal{} \left(\frac {4}{3}\right)^2 \equal{} \frac {16}{9} \equal{} \frac {[CBD] \plus{} [ABDE]}{[CBD]} \equal{} 1 \plus{} \frac {[ABDE]}{[CBD]}$ or $ \frac {[ABDE]}{[CBD]} \equal{} \frac {7}{9}$

Then multiply the two results and get $ \frac {[ABDE]}{[CDF]} \equal{} \boxed{1}$? Hmm...[/hide]

[hide]Could you just see that triangles ACE and BFC are congruent?[/hide]

I appreciate first replier's answer but second poster's answer is way much more fast.

This problem is interesting because it's not obvious from just reading the problem. But, once you list out the sides and their values, it becomes evident.  :rotfl:


Here's what i'm gettin...
$15bs-15s^2-b=0$

Where b is speed of boat and s is speed of current.

[quote="symmetry"]There are two bridges running parallel over river Kauvery, and they are 1KM apart.  After a man started rowing the boat upstream under the first bridge, his hat fell into river while he was exactly under second bridge. He realized that his hat was missing after 15 minutes, and started rowing in the opposite direction. He eventually caught his hat under First Bridge. What is the speed of river Kauvery? What is the speed of boat?

MY WORK:

U:SPEED OF THE BOAT PER HOUR
V:SPEED OF THE RIVER PER HOUR
I CAN FORM ONLY ONE EQUATION:
1/V hr.=1/(U+V)hr. +1/4hr.

What do I do next?  Help....[/quote]
I get
$rs=2(r-s)$
Now you can't solve it from here. :?

I'll try both equations to see which one works best.


I just sent my solution set.  I did one of the problems yesterday and wrote and cleaned up another problem today.  Just letting others know you aren't alone  ;) .

Also, seeing as the deadline is today, may we discuss the questions starting tomorrow or should we wait until grading is finished?

Edit: I see that the question was already answered.  That's what I get for hitting the new post button hours ago.

Please wait until next week to start discussing.

And no, you are definitely not alone among procrastinators.  Our fax machine hasn't stopped all day except during paper jams.

They're due on 12 PM Pacific, right? Since the address is to California. Otherwiswe, i'm doomed.

7: 30 PM here and still need to writ ehte solutions for the last three problems. ;)

At this point, it would probably be easier to scan it and email it.

[quote="yanaru"]I am having trouble faxing my work.  Is there any advice you can give?

Should I try to find another fax machine or a scanner and email it?[/quote]

Agreed, scan and email it if you can.  The fax machine is working fine on this end, but if you're having trouble, you're better off just scanning and emailing.

I sent mine in panickingly at 11:59pm EST. =(

Ouch.  I'm glad that wasn't me; I noticed 2 glaringerrors in question 5 after I typed it up (things were named incorrectly).

GAh I bet I missed the deadline our fax machine broke and my parents won't let me email and I didn't get them into the post office till saturday and I don't know if they made the outgoing mail....I'm dead I managed to solve a few too!

[quote="Magnara"]I started the problems the day after they came out, and had solved all the problems within 2 days.  I waited until Sunday to type up my solutions.[/quote]

Yeah... I did the same thing.

When exactly *did* round1 problems appear on the usamts website?
I'm curious to know, because I got them about 2-3 days before they were due.

i finished everything the day before the due date and faxed it the next day and i was so happy when i received email saying they received my solutions. however, i did submit the solutions which i wrote with pencil, so can anyone tell me if they'll grade it? :maybe:

The problems were up 6-7 weeks before the deadline.

darksword -> we'll let you know if we can't read it.  Try not to do that in the future though.  Use pen or mail it.


Hi, I have two things here that I really can't solve.

1) Prove that if the equation
ax^4 + bx^3 + cx^2 - bx + a = 0 (a != 0)
has a solution x = r, then x = -1/r is also a solution to the equation.

2) Solve the equation
6x^4 - 35x^3 + 62x^2 - 35x + 6 = 0
Start by dividing everything by x^2.
Bring together all terms with the same coefficient and assume that
x + 1/x = y.

What I don't get in the second one is the how to do the second step.

 :o

[quote="KentBrostrom"]

2) Solve the equation
$6x^4 - 35x^3 + 62x^2 - 35x + 6 = 0$
Start by dividing everything by $x^2$.
Bring together all terms with the same coefficient and assume that $x + \frac{1}{x} = y$.

What I don't get in the second one is the how to do the second step.

:o[/quote]

After dividing you get: $6x^2-35x+62-\frac{35}{x}+\frac{6}{x^2}=0$ By substituting $x+\frac{1}{x}$ with $y$ the eq becomes: $6y^2-12-35y+62=0\Longleftrightarrow 6y^2-35y+50=0$ with $y_{1}=\frac{10}{3}$ and $y_{2}=\frac{5}{2}$

Now we have to solve $x^2-xy_{1}+1=0$ and $x^2-xy_{2}+1=0$ and we get the solutions $x_{1}=3,x_{2}=\frac{1}{3},x_{3}=2$ and $x_{4}=\frac{1}{2}$ ;)

Thanks! I have another one though. I have 'till the 1st of April to do it but anyways it would be great if anyone had any idea.

Solve the eq: x^6 -3x^5 - x^4 + 6x^3 - x^2 - 3x + 1=0

[quote="JBL"]Well, based on the other two questions, what do you think the first steps will be?  These are all excercizes in using a single technique, so use it![/quote]

Well, to break out variables... but I don't get it, as I said.:\

[quote="JBL"]Later edit: Jimmy posted before I did, but you should ignore his post because the point of this excercize is to solve this by a different method than just guessing the factors.  Also, I'm going to use the word "symmetric polynomial" to refer to what we've got here.  This is non-standard terminology.

You have a symmetric polynomial, yes?  What did you do in the last question where you had a symmetric polynomial?

The key idea here, by the way, is to consider the powers of $y = x + \frac{1}{x}$.  When you expand out $y^n$, you will always get expressions in which the coefficient of $y^i$ and $y^{-i}$ are the same.  You can then multiply these by an appropriate power of $x$ to get a symmetric polynomial.  The interesting thing is that we can go in reverse: given this symmetric polynomial, we can express it in terms of $y$, a new polynomial of smaller degree.  Small degrees are good, because we can solve polynomials of small degree much more easily.[/quote]

I'm terriby sorry but I don't follow your logic. Maybe I'm to tired or I think it's because of the mathematical lingo. ;) A clarification would save my day (night)!


complex analysis
algebra
polynomial
complex analysis unsolved

The problem is to show that the map $ F: \mathbb{P}^{1}\rightarrow\mathbb{P}^{1}$ defined by $ F([z: w]) = [f(z,w): g(z,w)]$ is holomorphic, where $ f$ and $ g$ are homogeneous polynomials of the same degree with no common factor (and both not identically zero).

I know that for each $ p = [\alpha ,\beta]\in\mathbb{P}^{1}$, we need to find complex charts $ \phi_{1}$ defined at $ p$ and $ \phi_{2}$ defined at $ F(p)$ such that
\[ \phi_{2}\circ F\circ {\phi_{1}}^{ - 1}
\]
is holomorphic at $ \phi_{1}(p)$.  For $ U_{1} = {[z: w]\ : \ z\neq 0\}}$ and $ U_{2} = \{[z: w]\ : \ w\neq 0\}$, can we somehow use the following charts?:

$ \varphi_{1}: U_{1}\rightarrow\mathbb{C}$ defined by $ \varphi_{1}([z: w]) = \frac {w}{z}$;

$ \varphi_{2}: U_{2}\rightarrow\mathbb{C}$ defined by $ \varphi_{2}([z: w]) = \frac {z}{w}$.

I am having trouble seeing how we prove the composition is holomorphic since the choice of $ \phi_{2}$ depends on if $ f(z,w)$ or $ g(z,w)$ is zero at $ (z,w)$.

Check on the open sets $ U_j\cap F^{\minus{}1}(U_i)$ ($ i,j\equal{}1,2$) covering $ \mathbb P_1$.

Thanks for the reply.  How do we show that the composition is holomorphic though?

What does $ F$ look like under the charts subordinated to the open covering defined in my first reply? [hide="Solution"]They are rational maps, i.e. a quotient of polynomials in one variable.[/hide]


This was from COP 4th.. First question does not have answer (it does but I'm not sure it's right) and the second one does but I'm looking for elegant solution.

#6: There are 12 points distributed around the circle so that all $\binom{12}{2}$ pairs of points are connected by segments, no three of them being concurrent. Denote the number of intersection of points inside the circle as $\phi$. What is $\phi$ in base 100?

#14: Let's say there are three guys with different functions. Abe has function $\cos^2 x$. Bob has function $\cos^2 {2x}$.  Cyn has function $\cos^2 {3x}$. Given that the period of $x$ is $[0, \pi)$, find all values of $x$ such that when Abe's function, Bob's function, and Cyn's function is added, it equals the smallest positive integer. List them.

[hide="6"]
Consider one segment. It intersects with another iff the other segment as one point on either side of the original segment.

So call the points $p_1, p_2,...,p_{12}$ arranged around the circle.

Consider the line $p_1p_2$. Since there are no points on one side of it, no lines intersect it.
Consider $p_1p_3$. There is $1$ point on one side and $9$ on the other, there are $(1)(9)=9$ lines intersecting it.
$p_1p_4$: $(2)(8) = 16$
$p_1p_5$: $(3)(7) = 21$
$p_1p_6$: $(4)(6) = 24$
$p_1p_7$: $(5)(5) = 25$

After that, we have the intersections with $p_1p_{14-a}$ equal to those with $p_1p_a$. So from $p_1$ we have $9+16+21+24+25+24+21+16+9 = 165$ intersections.

Do that with each point and we get $12(165)$ intersections. But since each intersection point is counted twice, we divide by $2$ to get $6(165) = 980$ intersections.

I don't know why you would want to put that into base $100$ but yeah...[/hide]


Beat the Mod
AMC
AIME
probability
symmetry
Mafia
blogs

Same rules apply, that is, no rules (well, really - rules will come later).

You may need to use a calculator for some of these things... just a tip.

16 signups, no more, no less.

Signups
1. pythag011
2. nt811581147
*3. isabella2296
4. fun[b]TWO[/b]
5. Joe10112
*6. lotsofmath
*7. AIME15
8. budi713
9. jjx1
*10. alekim
11. Brut3Forc3
*12. zapi2007
*13. dragon96
*14. Flame
15. cf249
*16. BOGTRO

aadil, I checked your previous posts, it seems that you are new to the Games Forum. You can be a replacement.

[hide=" "]omg hidden text! Is this going to be crucial 59 563 5173 0 557[/hide]

Also, type all irrational numbers to three decimal places (3.445, 5591.228, etc.)

[quote="isabella2296"]james4l makes cool games(better than me :P .) I'm in![/quote]

I agree. My games are like... just have random stuff happen.

i agree. i dont even host games.
i join.

[quote="james4l"]I've got to say this, but I will use any randomness ONLY FOR SELECTION PURPOSES (will not directly affect your aliveness). Anyone who catches me using randomness for other purposes can offer to become a martyr and tell me where I randomized. If they are correct I die, if they are wrong they die. Deal?[/quote]

No randomness yay!

Randomness = bad.

[quote="jjx1"]I will join.

NO math! :([/quote]

No math? In a math forum? Don't worry there will be very little math (slight computation, such as what is rt(2)+rt(3) may happen, but remember your calculator!)

btw, how is a multi determined?

was this task predetermined before the game or was however it is just chosen as a random person?

what time is it?
it's arml time!

Multi is randomly chosen.
pythag011 is being annoying.
There is a deadline.

Does my multi know that she is my multi?

*sigh* Is anyone going to try to crack my code?

[quote="zapi2007"][quote="dragon96"]Does my multi know that [b]she[/b] is my multi?[/quote]

in addition, does dragon know who the multi is?[/quote]

It goes both ways.

I'm telling you, kill eli! He's not my multi.

I'm not sure if I'm still in this game, but if I am, I would just like to say that I have not received a PM concerning this round, so I can only assume that I am not the multi.

yes, all who don't receive pms are townies

This IS dead.

BOGTRO was the multi.

dragon96 dies for not convincing anyone to vote.

Next round.

Pick:

[b]Math
English
Science[/b]

ANAGRAMS!

Solve one, you save one person who chose False and one person who chose True.

1. pythag011 - False
*7. AIME15 - True
*10. alekim - True
*12. zapi2007 - True
15. cf249 - False
*16. BOGTRO – False

(6) MATH: aaaaaccceimmmmmmoopssu
(12) FORUM: bbbcdddeeeeeeggghiiiiillmnooooqrrsttuuuuuxy
(5) BTM: aaaeflmmmmnnopprsttttw

Solve 3, you win!


inequalities
logarithms
induction
strong induction

Prove that:

$ In2*In3*In4* ... *In(n\plus{}1) \le \frac{(n\plus{}1)!}{e^n} ; n\in N$

hello, do you mean the natural logarithm $ \ln(x)$ the base $ e$?
Sonnhard.

[hide="Solution"]We will use induction in order to prove the inequality above.

i) $ n\equal{}1$
$ \ln 2\le \frac{2!}{e}$, which is true.

ii) $ n\equal{}k$
Assume that $ \ln 2\cdot \ln 3\cdots \ln k\cdot \ln (k\plus{}1)\le \frac{(k\plus{}1)!}{e^k}$.

iii) $ n\equal{}k\plus{}1$
Now we are to prove that $ \ln 2\cdot \ln3\cdots \ln k\cdot\ln (k\plus{}1)\cdot\ln (k\plus{}2)\le \frac{(k\plus{}2)!}{e^{k\plus{}1}}$.

$ \ln 2\cdot \ln3\cdots \ln k\cdot\ln (k\plus{}1)\cdot\ln (k\plus{}2)\le \ln(k\plus{}2)\cdot\frac{(k\plus{}1)!}{e^k}\le\frac{(k\plus{}2)!}{e^{k\plus{}1}}$.

We now have to prove that $ \ln(k\plus{}2)\le \frac{k\plus{}2}{e}\implies \ln x\le \frac{x}{e}\implies \ln x^e\le \ln e^x$, which is true.

Hence, the induction is complete.[/hide]

How might it be possible to prove this by [b]strong induction[/b]?[/b]

I have found that it is possible to use strong induction to prove this problem, but will not post it here since the solution is very similar to that of FantasyLover.


algebra
polynomial
number theory proposed
number theory

Find all polinomials from Z[X] for wich p prime implies f(p) prime.

May I ask how you solve it without Dirichlet theorem?

Well, the solution uses Dirichlet's theorem.

Maybe this somewhat belongs to the advanced section???....

can somebody tell me what's Dirichlet theorem?? maybe somebody should put it in the Theorem's section.
thanks

Dirichlet theorem:
If p,q are primes then the arithmetic sequence p+kq, k in N contains infinitely many prime numbers.

You mean 'p and q are coprimes' that is gcd(p,q) = 1.
The particular case where p and q are both primes is obviously true, if you add 'p and q are distinct primes'.

Pierre.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=91236 for something more general.