a)
[hide] Since the fourth tail has to be preceeded by $h$ number of heads, there are three tails with the $h$ number of heads preceeding the fourth. The number of ways to arrange that is then $\frac {(h+3)!}{h!3!}$. The probability of getting any one arrangement is $\frac {1}{2}^{h+3}$. Then the probability that $h$ number of heads preceeds the fourth toss is $(\frac {1}{2})^{h+3} * \frac {(h+3)!}{h!3!}$. So the probability that the fourth toss is preceeded by exactly three heads is $(\frac {1}{2})^6 * \frac {6!}{3!3!} = \frac {5}{16}$.[/hide]

[quote="allrighty"]A fair coin is tossed. Let X be the number of tosses for four tails. Find the probability function of X. What is the probability that the fourth tail is preceded by

a) exactly 3 heads?
b) at least 3 heads?[/quote]

The probability fn. of X can be determined as follows.

If a total of X tosses are reqd for 4 tails, there has to be exactly 3 tails in the first X-1 tosses, the remaining toss results being heads.

So P(X) = C(X-1,3)*(1/2)^X

a>
For case(a), X has to be 3 (heads)+ 3 (intermdeiate tails)+1( last tail) =7

So P(E) = C(7-1,3)*(1/2)^7 = C(6,3)/128

b> Here let's calculate the probability of the complement event , i.e. the no. of preceding heads is at most 2.
 The probability of the complement is

then  P(X=4) + P(X=5) +P(X=6) corresponding to the no. of heads being 0,1and 2 respectively

Then P(E) = 1 - ( P(X=4) + P(X=5) +P(X=6))

The values P(X=i) can be calculated from the expression for P(X) substituting the values 4,5,6


There is nothing more annoying than studying math, so Peter is trying to minimize his study time for
passing MAT100. The overall mark will be composed by 15% from the problem sets, 45% from the term
tests and 40% from the final exam. Peter expects his mark P for the problem sets to be proportional to the
time x he spends on them (let’s say, P = x), his mark for the term tests to be T = (2x)/3 + y, where y is the time
he spends studying specifically for the term tests and his mark F for the final to be F = x/2 + y/2 + z, where z
is the time he studies specifically for the final.

What is the minimum amount of time for achieving the pass mark of 50 points, and how should he break down this time?

(The above model is not realistic, because in reality the marks P; T and F cannot go higher than 100
each. Comment on how this would influence your answer.)

$ P(x,y,z) \equal{} x$, $ T(x,y,z) \equal{} \frac {2}{3} x \plus{}y$, $ F(x,y,z) \equal{} \frac {1}{2}x \plus{} \frac {1}{2}y \plus{} z$

want $ S(x,y,z) \equal{} .15P \plus{} .45T \plus{} .4F \equal{} .65x \plus{} .65y \plus{} .4z \equal{} 50$

to be our restriction, and we want to minimize

$ g(x,y,z) \equal{} x \plus{} y \plus{} z$

so.. it would just be lagrange multipliers?

the more interesting question is how you would implement the max score concept..

maybe something like $ 100(1 \minus{} e^{ \minus{} x})$ but this means youll never reach 100.. which isnt realistic..

lagrange multipliers dont work. is there another method to solving this ?


Find all $ n\in\mathbb{N}$ such that polynomial
\[ P(x) = (x - 1)(x - 2) \cdots (x -n)\]
can be represented as $Q(R(x))$, for some polynomials $ Q(x)$ and $R(x)$ with degree greater than $1$.

[quote="Erken"]Find all $ n\in\mathbb{N}$,such that polynomial
$ P(x) \equal{} (x \minus{} 1)(x \minus{} 2)\dots (x \minus{} n)$
can be represented as $ Q(R(x))$,for some non-constant polynomials $ Q(x),R(x)$.[/quote]

It seems I misunderstood something because this problem seems trivial :

Any non constant polynomial $ P(x)$ can be represented as $ Q(R(x))$,for some non-constant polynomials $ Q(x),R(x)$ :

Take $ R(x)\equal{}x\plus{}a$ and $ Q(x)\equal{}P(x\minus{}a)$

[quote="pco"][quote="Erken"]Find all $ n\in\mathbb{N}$,such that polynomial
$ P(x) \equal{} (x \minus{} 1)(x \minus{} 2)\dots (x \minus{} n)$
can be represented as $ Q(R(x))$,for some non-constant polynomials $ Q(x),R(x)$.[/quote]

It seems I misunderstood something because this problem seems trivial :

Any non constant polynomial $ P(x)$ can be represented as $ Q(R(x))$,for some non-constant polynomials $ Q(x),R(x)$ :

Take $ R(x) \equal{} x \plus{} a$ and $ Q(x) \equal{} P(x \minus{} a)$[/quote]
Sorry,i forgot to write,that
$ deg(R),deg(Q)\geq 2$.

[hide="hint"]
Consider two cases,when $ n$ is odd,and when it is an even number :wink:
[/hide]

If $ n > 2$ is even, then
$ P(x) \equal{} \prod_{k \equal{} 1}^{n/2}(x \minus{} k)(x \minus{} (n \plus{} 1 \minus{} k)) \\
\hspace*{0.3in} \equal{} \prod_{k \equal{} 1}^{n/2}((x \minus{} 1)(x \minus{} n) \plus{} k(n \plus{} 1 \minus{} k) \minus{} n) \equal{} Q(R(x))$
with $ R(x) \equal{} a(x \minus{} 1)(x \minus{} n) \plus{} b\in\mathbb{R}[x]$ non-linear
and $ Q(x) \equal{} \prod_{k \equal{} 1}^{n/2}\left({x \minus{} b\over a} \plus{} k(n \plus{} 1 \minus{} k) \minus{} n\right)\in\mathbb{R}[x]$ non-linear
for any $ a,b\in\mathbb{R},a\not \equal{} 0$.

I'll show, that these are the only $ Q(x),R(x)\in\mathbb{R}[x]$ non-linear with $ P(x) \equal{} Q(R(x))$.

Say $ q: \equal{} \deg Q\ge 2$ and $ r: \equal{} \deg R\ge 2$. Then $ n \equal{} qr$.
We can replace $ R(x)$ with $ {1\over a}R(x)$ and $ Q(x)$ with $ Q(ax)$ for any $ a\in\mathbb{R},a\not \equal{} 0$.
Similar we can replace $ R(x)$ with $ R(x) \minus{} R(1)$ and $ Q(x)$ with $ Q(x \plus{} R(1))$.
So we can assume $ R(x)$ has leading coefficient $ 1$ and $ R(1) \equal{} 0$.

From [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=733142&search_id=2134256501#733142]here[/url] we see, that the strictly monotonic sequences
$ R(1 \plus{} kq),R(2 \plus{} kq),\ldots,R((k \plus{} 1)q)$ and
$ R(1 \plus{} (k \plus{} 1)q),R(2 \plus{} (k \plus{} 1)q),\ldots,R((k \plus{} 2)q)$
run through all the roots of $ Q(x)$ in converse order for all $ k \equal{} 0,1,\ldots ,r \minus{} 2$.

This gives
$ (*)$  $ R(k) \equal{} R(2q \plus{} 1 \minus{} k)$ for all $ k \equal{} 1,2,\ldots 2q$.
and
$ (**)$ $ R(k) \equal{} R(k \plus{} 2q)$ for all $ k \equal{} 1,\ldots ,n \minus{} 2q$.

For $ r\ge 3$ and $ (q,r)\not \equal{} (2,3)$ we have $ 1 < (q \minus{} 1)(r \minus{} 2)$ and $ r \minus{} 1 < qr \minus{} 2q \equal{} n \minus{} 2q$.
Then by $ (**)$ $ R(x) \minus{} R(x \plus{} 2q)$ has $ n \minus{} 2q$ different roots and degree $ \le\deg R \minus{} 1 < n \minus{} 2q$.
So $ R(x) \equal{} R(x \plus{} 2q)$ and $ R(1) \equal{} 0$ gives infinitely many roots $ 1 \plus{} 2qk$ of $ R(x)$ for $ k\in\mathbb{N}$,
and therefore $ R(x)\equiv 0$. Contradiction!

For $ (q,r) \equal{} (2,3)$ we get $ 0 \equal{} R(1) \equal{} R(4) \equal{} R(5)$ using $ (*)$ and $ (**)$.
This gives $ R(x) \equal{} (x \minus{} 1)(x \minus{} 4)(x \minus{} 5)$ and $ R(2) \equal{} 6,R(3) \equal{} 4$, which contradicts $ (*)$.

So $ r \equal{} 2$ and $ n \equal{} 2q > 2$ even and $ (*)$ gives $ 0 \equal{} R(1) \equal{} R(n)$.
Then $ R(x) \equal{} (x \minus{} 1)(x \minus{} n)$ and $ P(x) \equal{} Q(R(x)) \equal{} \tilde{Q}(R(x))$
with $ \tilde{Q}(x) \equal{} \prod_{k \equal{} 1}^{n/2}(x \plus{} k(n \plus{} 1 \minus{} k) \minus{} n)\in\mathbb{R}[x]$.
This gives $ Q(x) \equal{} \tilde{Q}(x)$, which completes the proof.
($ Q(x) \minus{} \tilde{Q}(x)$ has infinitely many roots $ x\equal{}R(y),y\in\mathbb{R}$)


Find all integer solutions to the system of equations:
$x^{3}+y^{3}+z^{3}=3=x+y+z$

[b]Solution.[/b] For $x,y,z\geq 0$ by Arithmetic-Cubic Mean Inequality,  $\sqrt[3]{\frac{x^{3}+y^{3}+z^{3}}{3}}=1 \geq \frac{x+y+z}{3}=1$, where equality holds if and only if $x=y=z$. Since equality holds, $3x^{3}=3\implies x=y=z=1$.  As a result, the only solution is $(x,y,z)=(1,1,1)$ for this case.  If we are also considering negative integers, using the well known fact that the only integral solutions to $x^{3}+y^{3}+z^{3}=3$, where at least one of $x,y,z$ is less than 0, then $(x,y,z)=(4,4,-5)$, including all its permutations.

Therefore, all solutions are $(x,y,z,)=(1,1,1),(4,4,-5),(4,-5,4), (-5,4,4)$.$\blacksquare$


I hope this doesn't qualify as Calculus (it shouldn't ...)

For the convergent sequence with $\displaystyle a_1=1$ and $\displaystyle a_{n+1}=\sqrt{3+2a_n}$, determine $\displaystyle \lim_{n\to \infty}{a_n}$

[hide="Might not be right"]I'm not too sure this is right, but as n gets bigger, there are larger and larger strings of $\sqrt{3+2\sqrt{3+2\sqrt{3+2\sqrt{3+ \cdots 2\sqrt{5}}}}}$ Let this equal x, then $x^2$ gets closer to $3+2x$ as the series of sqrts gets longer, solving $x^2=3+2x$ the two solutions are $x=-1, x=3$ and because it can't be negative, then therefore it's 3. This is the only way I can think of how to do such a problem.[/hide]

Before you employ that trick, you first must prove that the limit exists.

[hide="Hint"]Prove that the sequence is strictly increasing and bounded from above.[/hide]

it already says that it converges..."for the convergent series"

when it converges, for large $n$, we can say $a_{n+1}=a_n$

$x^2=3+2x \implies x=-1,3$ and of course the negative is not the answer since we square roots only have a positive range, so $\boxed{3}$


induction
quadratics
Diophantine equation
number theory unsolved
number theory

The sequence $ (x_n)_{n\geq1}$ is defined by:
$ x_1\equal{}1,x_2\equal{}2,$
$ x_{n\plus{}2}\equal{}4x_{n\plus{}1}\minus{}x_n$ $ \forall n\geq1$
Prove that $ x_{2008}$ is the sum of $ 2$ perfect squares.

[quote="KDS"]The sequence $ (x_n)_{n\geq1}$ is defined by:
$ x_1 \equal{} 1,x_2 \equal{} 2,$
$ x_{n \plus{} 2} \equal{} 4x_{n \plus{} 1} \minus{} x_n$ $ \forall n\geq1$
Prove that $ x_{2008}$ is the sum of $ 2$ perfect squares.[/quote]

Let $ u\equal{}2\minus{}\sqrt 3$ and $ v\equal{}\frac 1u\equal{}2\plus{}\sqrt 3$ the two roots of $ x^2\minus{}4x\plus{}1\equal{}0$. We have $ x_n\equal{}\frac{u^{n\minus{}1}\plus{}v^{n\minus{}1}}2$

Then, $ x_{2p}\minus{}1\equal{}\frac{u^{2p\minus{}1}\minus{}2\plus{}v^{2p\minus{}1}}2$ $ \equal{}(\frac{\sqrt 3\plus{}1}2u^p)^2\minus{}1\plus{}(\frac{\sqrt 3\minus{}1}2v^p)^2$ $ \equal{}(\frac{\sqrt 3\plus{}1}2u^p\minus{}\frac{\sqrt 3\minus{}1}2v^p)^2$

And so $ x_{2p}\equal{}y_p^2\plus{}1$ where the sequence $ y_p$ is $ y_n\equal{}\frac{\sqrt 3\minus{}1}2v^n\minus{}\frac{\sqrt 3\plus{}1}2u^n$

And so the sequence $ y_n$ may be defined as :
$ y_1\equal{}1$
$ y_2\equal{}5$
$ y_{n\plus{}2}\equal{}4y_{n\plus{}1}\minus{}y_n$

So $ y_n\in\mathbb Z$ and, since $ x_{2p}\equal{}y_p^2\plus{}1$, we get that any member of rank even of the sequence $ x_n$ is the sum of two perfect squares.

Hence the result.

in fact, $ x_{n}\equal{}X_{n\minus{}1},n \geq 2$, for which $ X_{i}$ are the answers of the pell equation:$ X^2\minus{}3Y^2\equal{}1$(we can prove this  by either induction or (what pco did) finding the formula of $ x_{n}$).the answers are easily obtained by:
$ X_{n\plus{}1}\equal{}2X_{n}\plus{}3Y_{n},Y_{n\plus{}1}\equal{}X_{n}\plus{}2Y_{n}$.we can easily see that $ X_{2k\minus{}1}$ is even. Since
$ (X_{n}\minus{}1)(X_{n}\plus{}1)\equal{}3Y_{2k\minus{}1}^2,\gcd(X_{n}\minus{}1,X_{n}\plus{}1)\equal{}1 (\mbox{for odd n}) \implies X_{n}\plus{}1\equal{}3q^2,X_{n}\minus{}1\equal{}s^2,(s,q)\equal{}1,sq\equal{}Y_{2k\minus{}1},  \implies X_{n}\equal{}s^2\plus{}1$
but we had $ x_{2k}\equal{}X_{2k\minus{}1}$.So we r DONE!(note that $ 2$ is a quadratic non-residue modulo $ 3$ so the case $ X_{n}\plus{}1\equal{}s^2,X_{n}\minus{}1\equal{}3q^2$ is ommitted.)
u can prove that if $ p$ is a prime and $ p \equiv \pm 3 \mod {8}$ and the fondamental answer of the pell equation $ X^2\minus{}pY^2\equal{}1$  is $ (0,1)$ modulo $ 2$, then the rank of  odd answers of this pell equation have the same property.


ratio
conics
function
geometry proposed
geometry

Let A be be a point outside the circle C, and AB and AC be the two tangents from A to this circle C. Let L be an arbitrary tangent to C that cuts AB and AC in P and Q. A line through P parallel to AC cuts BC in R. Prove that while L varies, QR passes through a fixed point. :)

Well done  ;) . A very nice solution.

See my solution( its a little bit longer):
I will prove that the pole of QR is on a fixed line. The pole of QR is the intersection of the polars of Q and R. Let L be tangent to C in X. So the polar of Q is XC . we name the line through P parallel to AC: d. So the polar of R is the line through A and the pole of d .The pole of d (which is Y) is on the diameter of C perpendicular to d( which is the diameter through C which we name CD) .it lies also on BX. So we have found Y=( CD, BX). Let Z be the intersection of CX and AY( so Z is the pole of QR).
Now to finish: its obvious that Z is on BD which is constant ( because AY ,the polar of R, passes through Z so the polar of Z passes through R so Z is on BD). :D

Here's what I did: we must show that the application from line $AC$ to line $BC$ which maps $Q\to R$ is a perspectivity. In order to do this we must show that it's a homographic (preserving the cross-ratio) transformation and it maps $C\to C$.

It's easy to see that it's homographic because it's the composition of $Q\to P$ (from line $AC$ to line $AB$; it's wel-known that this sort of transformation is homographic (even if $(C)$ is a conic, not only a circle)) and $P\to R$ (from line $AB$ to line $BC$).

We must now show that $C\to C$. In this case $Q=C\Rightarrow L=AC\Rightarrow P=A\Rightarrow PR=AC\Rightarrow R=AC\cap BC=C$, and we're done.

grobber, i didn't understand your solution properly could u explain more?

Assume you have two lines $d_1,\ d_2$ and a point $X$ (which doesn't belong to $d_1$ or $d_2$). Take a variable line $d$ through $X$ which cuts $d_i$ in $A_i$. The function $f:d_1\to d_2,\ f(A_1)=A_2$ is called perspectivity and it'sa a particular case of function from $d_1$ to $d_2$ which preserves the cross-ratio of four points.

If you have $g:d_1\to d_2$ s.t. $g$ preserves the cross-ratio, it can be shown (I won't show it here; try searching the Net for stuff like "homographic transformation", "perspectivity" etc.) that this is a perspectivity iff $g(O)=O$, where $O=d_1\cap d_2$. This is exactly what I did here.

oh! Grobber I know homographic transformation I had not understood the last part of your proof. :)  :(  :blush:  ;)

Oh.. Sorry for lecturing you :D Is it Ok now?

What I'm trying to say is that according to the theorem I meantioned in one of my previous messages on this topic, in order to show that the transformation $Q\to R$ is a perspectivity from line $AC$ to line $BC$ (i.e. $QR$ passes through a fixed point), we must show that by this transformation the intersection point of the two lines $AC$ and $BC$ turns into itself. This means we want to prove that when $Q$ coincides with $C$, by constructing $R$, we also find it to coincide with $C$. That's why the last line begins with $Q=C$ and ends with $R=C$ (what we want to show is the implication $Q=C\Rightarrow R=C$).

[quote="Omid Hatami"]Suppose $T$ is intersection of $QR$ and $AB$ If the tangent from $T$ to circle (not $AB$ the other) is' not parallel to $AC$ by Brianchon theorem $PR$ , $AC$ and tangent from $T$ to circle are concurrent.
But $PR$ is parallel to $AC$ contradiction. So $T$ is constant.[/quote]
I don't understand why we have $PR$ , $AC$ and tangent from $T$ to circle are concurrent by Brianchon theorem


The volume of a sphere of radius r is V = (4/3)(pi)(r^3); the surface area S of this sphere is S = 4(pi)(r^2).

(a) Express the volume V as a function of the surface area S.

(b) If the surface area doubles, how does the volume change?

${V(r)=\frac{4 \pi r^{3}}{3}}$
${S(r)=4 \pi r^{2}}$

It' s easy to see

${V'(r)=S(r)}$

Or

${V(r)=\frac{S(r)^{3/2}}{6 \sqrt{\pi }}}$

To double the suface need

${S\left(\sqrt{2}r\right)=2 S(r)}$

So Volume

${V\left(\sqrt{2}r\right)=2 \sqrt{2}V(r)}$

a good tip.

if the side length of a polygon is multiplied by a factor of $x$, the area is multipled by a factor of $x^{2}$.

same with area-volume


Let the sequence $(x_n)$ define by:

$x_1=1$
$x_{m+n}=\frac{1}{2}.(x_{2m}+x_{2n}), \forall m,n \in \mathbb{N}^*$

Find $x_{2007}$.


1. How do you make a space between two paragraphs?

2. Why is it that when I type

$\framebox{a) \ 90^{\circ}}$

it doesn't work. I want to get it so that the box is completely around my answer a) 90 degrees

\framebox{} doesn't work as well as \boxed{}.

[code]\boxed{90^{\circ}}[/code]

$\boxed{90^{\circ}}$

how do you skip a line between paragraphs? right now, everything is just lumped together

You can insert extra space by using
\bigskip \\
or
\medskip \\
or, if you want to vary the space, by using
\vspace{2ex} \\

You can use any number before ex (1ex=height of an x in the current font) and can use other units such as pt, in, mm etc.

But please remember, the style you are using may look best with no extra line between paragraphs. If you pick up a book  this is the way most of them are printed.

this is exactly the result I wanted.

however, the \medskip\\ apparently adds to an "error" whenever I compile. is there a reason for this?

If you let us know the error and a simple document in which you get that error we can try and sort out what is happening.

whenever I add a \medskip\\ an error shows up

for example:

\documentclass{article}
\begin{document}

This document has an error.

\medskip\\

I don't know why.

\end{document}

This works though:
[code]\documentclass{article}
\begin{document}

This document has an error.
\medskip\\

I don't know why.

\end{document}[/code]The problem was the line between [i]This document has an error[/i] and \medskip\\ . You'd get the same error [i]There's no line here to end [/i] even with \\ because LaTeX thinks you want a new paragraph rather than a new line.

I made the change you suggested but now instead of "error" I have "bad box"

Whoops! Missed that as you can ignore bad boxes if the printout looks OK. But to get rid of that warning I should have got rid of the other blank line after \medskip \\ :
[code]\documentclass{article}
\begin{document}

This document has an error.
\medskip \\
I don't know why.

\end{document}[/code]This should have no errors or warnings.

I use the command:

\vskip 2in

You can play with this to make it perfect.  Examples:

\vskip .5in
\vskip 3cm

\vskip is a TeX primitive so may need care in its use.  It is usually best to stick to LaTeX commands which are less likely to have side effects. In this case, the equivalent LaTeX command is \vspace{2in} or whatever.

For example, I found this on a newsgroup and it gives an interesting error not flagged up by LaTeX.
[code]\documentclass{article}
\begin{document}
Beware TeX primitives.
\vskip2cm
Plus, in some cases errors are hard to find.
\end{document}[/code]

That's true, but sometimes TeX primitives let you get around built-in limitations.  Of course, those limitations are probably built in for a reason... I haven't experimented with this much, but does \vskip2cm make a difference if you do \vskip 2cm (with a space) instead?

I think TeX swallows the space so it doesn't matter. I agree that if a TeX primitive works then use it; I have used \def where I should have used \newcommand. Kopka & Daly put it this way
[quote]The primitive TeX commands form the bedrock of any format, and anything defined with them will always do exactly what the programmer expected. However, the equivalent LaTeX command could actually do more as time goes on. The \newcommand checks for name clashes with existing commands, for example. It might even be possible that a debugging device that keeps track of all redefinitions will be added later; any commands defined with \def would be excluded from such a scheme. Even now in LaTeX2e, there is something like this to keep track of all files input with middle and high-level commands.[/quote]
They then go on to discuss the fragility of some low-level commands and say that while one should always try to use high-level commands there will be occasions where this is not possible.
You may find the discussion about \vskip interesting in [url=http://groups.google.co.uk/group/comp.text.tex/browse_frm/thread/7f9388fd02753030]comp.text.tex[/url] where I found the example I gave earlier.


prove that in any triangle  the following inequality holds:

(cyc sum(bc cosA))/(cyc sum (a sin A))>=2r.

I really love these kind of inequalities! :D:D

solution:

try to prove the following identity:
(cyc sum(bc cosA))/(cyc sum (a sin A))=R
and with Euler you get R>=2r.

tschuess!

Proof of the identity which Lagrangia uses in the solution
    We know that bc cosA=(1/2)*(b*b+c*c-a*a)
    That means that  \sum bc cosA=(1/2)(a*a+b*b+c*c) and using the fact that a*a=4R*R*sinA*sinA So we have  \sum bc cosA=2R*R( \sum sinA*sinA)
     We also know that a*sinA=2RsinA*sinA That means that \sum a*sinA=2R( \sum sinA*sinA)
     So  \sum bc cosA/ \sum a*sinA=R which is from Euler Theorem>=2r

     I make that proof because I think that the solution of Lagrangia is too "empty" without it.

     Hope it is corect!

indeed it is kinda "empty".. but I gave him a hint! :D :D

well, the hint that Lagrangia gave me was enough! thanks for the proof RNicula!

I was looking for a more direct proof, without using that identity. I guess that this proof is pretty neat!

tschuess!
:D


trigonometry
search
geometry proposed
geometry

The similar isosceles triangles $ \Delta AC_1 B,\Delta BA_1 C$ and $ \Delta CB_1 A$ with bases $ AB,BC$ and $ AC$ respectively are constructed externally on the sides of non-isosceles triangle $ \Delta ABC$. Prove that if  $ A_1 B_1  \equal{} B_1 C_1$, then $ \angle BAC_1  \equal{} 30^\circ$.

[quote="Edward_Tur"] [color=darkred]The similar isosceles triangles $ \Delta AC_1 B,\Delta BA_1 C$ and $ \Delta CB_1 A$ with bases $ AB,BC$ and $ AC$ respectively are constructed externally on the sides of non-isosceles triangle $ \Delta ABC$. Prove that if  $ A_1 B_1 = B_1 C_1$, then $ \angle BAC_1 = 30^\circ$.[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote $ m(\widehat {BCA_1}) = \phi$ and $ \frac {1}{2\cos\phi} = \rho$ .  Thus, $ \|\begin{array}{c} A_1B = A_1C = \rho a \\
\ B_1C = B_1A = \rho b \\
\ C_1A = C_1B = \rho c\end{array}$

and $ \|\begin{array}{c} m(\widehat {A_1BC_1}) = B + 2\phi \\
\ m(\widehat {A_1CB_1}) = C + 2\phi\end{array}$ . Apply the [b]generalized Pythagoras' theorem[/b] :

$ \|\begin{array}{ccc} \triangle A_1CB_1\ : \ \ A_1B_1^2 = \rho^2[a^2 + b^2 - 2ab\cos (C + 2\phi )] \\
 \\
\triangle A_1BC_1\ : \ \ A_1C_1^2 = \rho^2[a^2 + c^2 - 2ac\cos (B + 2\phi )]\end{array}\ \ \Longrightarrow$

$ \|\begin{array}{c} A_1B_1^2 = \rho^2[a^2 + b^2 - 2ab(\cos C\cos 2\phi - \sin C\sin 2\phi)] \\
 \\
A_1C_1^2 = \rho^2[a^2 + c^2 - 2ac(\cos B\cos 2\phi - \sin B\sin 2\phi)]\end{array}$ . Thus

$ A_1B_1 = A_1C_1\ \ \Longleftrightarrow\ \ b^2 - 2ab\cos C\cos 2\phi = c^2 - 2ac\cos B\cos 2\phi$ , i.e.

$ \cos 2\phi = \frac {b^2 - c^2}{2ab\cos C - 2ac\cos B}$ $ \Longleftrightarrow$ $ \cos 2\phi = \frac 12$ $ \Longleftrightarrow$ $ \boxed {\ \phi = 30^{\circ}\ }$ .

I use the relations $ ab\sin C = ac\sin B$ and $ ab\cos C - ac\cos B = b^2 - c^2$ .[/color]

See and http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1642104637&t=45678


$a_n>0 \forall n\ge 1$, $\lim_{n \rightarrow +\infty} \frac{\sum_1^n a_i}{\Pi_1^n(1+a_i)}=0$.Prove that : $\sum_1^{+\infty} a_n=+\infty$

[hide]
Suppose $\sum_{n=1}^\infty a_i = L$ where $L$ is a finite positive real.

Then $\frac{\sum a_i}{\prod (1+a_i)} \ge \frac{\sum a_i}{\left(\frac{n+\sum a_i}{n}\right)^n}$ by AM-GM.

But $\lim_{n \to \infty} \left(\frac{n+\sum a_i}{n}\right)^n = 1$ because $L$ is finite.

So $\lim_{n \to \infty} \frac{\sum a_i}{\prod (1+a_i)} \ge \lim_{n \to \infty} \frac{\sum a_i}{\left(\frac{n+\sum a_i}{n}\right)^n} = \lim_{n \to \infty} \sum a_i = L > 0$. Contradiction. Hence we must have

$\sum_{n=1}^\infty a_i = \infty$.
[/hide]


calculus
logarithms
function
inequalities
complex analysis
advanced fields
advanced fields unsolved

Let p denote partitions and f fibonacci seqence. For each positive intger k, prove that f(n)>p(n+k) hold for all +ve integers n but finitely many values.

Bomb

Partitions grow as $\exp(c\sqrt{n})$, and Fibonacci sequence exponentially.

Of course, but the problem is, I guess, to prove this special case using combinatorics only.
I think Erdos had a nice upper bound for $p(n)$ with a combinatorial proof.

What do you call "combinatorial"? I may prove that $\ln p(n)=c\sqrt{n}+o(\sqrt{n})$, where $c=2\sqrt{\zeta(2)}$ without using generating functions and complex analysis at all, so any schoolboy, if he wants, may understand it. Maybe, remainder term may be  improved, and we may even show that $np(n)e^{-c\sqrt{n}}$ has a limit (known to be equal to $\frac1{4\sqrt{3}}$).

Fedor, that is extremely interesting.  Can you explain the idea of this elementary method for $p(n)$ asymptotics?

By "combinatorial" proof I meant that inequalities are derived by constructing functions between finite sets.   Erdos argument was (something like) $p(1+2+...+n) < 4^n$  by constructing an injection from partitions to subsets of a set of size $2n$.

The idea is to use an identity $np(n)=\sum\limits_{k=0}^{n-1} p(k)\sigma(n-k)$, where $\sigma(k)$ means the number of divisors. Then, use Abel transform:

\[
\sum_{k=0}^{n-1} p(k)\sigma(n-k)=\sum_{k=0}^{n} p(k)(F(n-k)-F(n-k-1))=\sum_{k=0}^{n-1} (p(k)-p(k-1))F(n-k),
\]

where $F(k)=\sum_{i=0}^k \sigma(k),\,F(0)=0$. Note that $p(n)$ increases (it is obvious), so we may replace $F(n-k)$ to its lower and upper bound to get lower and upper estimates of $RHS$. It is easy to show that $F(k)=\zeta(2)(1+2+\dots+k)+o(k^2)$ (people say that $o(k^2)$ is at fact $O(\ln\ln k)$. But I do not know whether it is elementary). So, do this and you get estimates $np(n)\ge c'\sum_{k=0}^{n-1} p(k)(n-k)$ and $np(n)\le c''\sum_{k=0}^{n-1} p(k)(n-k)$ for aribitrary $c'<c<c''$, where $c=\zeta(2)$. Now we may prove some estimates directly by induction.


Let 0<a<b prove that

There exist $c\in [a,b]$

$ \left(\prod_{k=1}^{n}\frac{e^{kb}-e^{ka}}{b-a}\right)=n!e^{\frac{n(n+1)c}{2}}$

Hmm..

Given that e^x is differentiable we can apply the mean value theorem to get that [tex]\frac{e^{bk}-e^{ak}}{b-a}=k(\frac{1}{k})\frac{e^{bk}-e^{ak}}{b-a}=ke^{c_k}[/tex]
for some [tex]c_k\in(ka,kb)[/tex].  Thus [tex]\prod_{k=1}^{n}\frac{e^{bk}-e^{ak}}{b-a}=n!e^{\sum_{k=1}^{n}c_k}[/tex]

But [tex]c_k\in(ka,kb)\Rightarrow\frac{n(n+1)a}{2}<\sum_{k=1}^{n}c_k<\frac{n(n+1)b}{2}
\Rightarrow{a}<\frac{2\sum_{k=1}^{n}c_k}{n(n+1)}<{b}[/tex].
Now clearly letting [tex]c=\frac{2\sum_{k=1}^{n}c_k}{n(n+1)}[/tex] we get the desired result.

Hah! I can't believe I missed such a nice problem by Moubi!!! :blush:  :D

Ok you have a chance to solve the generalisation 3X.lich

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22991


Consider an isosceles  triangle $ ABC$ such that $ AB\equal{}AC$. Let $ M$ be the midpoint of $ AB$, take a point $ N$ on the extended line of $ AB$ such that $ AN: NB\equal{}2: 1$. Prove that $ \angle{BCM}\equal{}\angle{BCN}$. Note that $ N$ is not on $ AB$.

Let $ D$ be midpoint of $ AC$. Since triangle $ ABC$ is isoceles such $ AB \equal{} AC$, $ \angle BCM \equal{} \angle CBD$. So it is suffice to show $ BD\parallel CN$. $ AN: NB \equal{} 2: 1$ mean $ B$ is midpoint of $ AN$ and $ D$ is midpoint of $ AC$. Hence $ BD\parallel CN$.

[quote="kunny"]Consider an isosceles  triangle $ ABC$ such that $ AB \equal{} AC$. Let $ M$ be the midpoint of $ AB$, take a point $ N$ on the extended line of $ AB$ such that $ AN: NB \equal{} 2: 1$. Prove that $ \angle{BCM} \equal{} \angle{BCN}$. Note that $ N$ is not on $ AB$.[/quote]

[b]ΗΙΝΤ[/b]

[i] (I hope that this hint is not useless)[/i]

 Triangles $ AMC , ACN$ are similar(by the ratio-angle kriterio). Taking the ratios and the inverse of the bisector theorem , we arrive at the result.

 Babis


linear algebra
matrix
linear algebra unsolved

Suppose that $ A \in {\mathbb {C}}^{n \times n}$ satisfies $ AAA^{*}\equal{}AA^{*}A$. Prove that $ A$ is normal (i.e. $ AA^{*}\equal{}A^{*}A$).

$ AA^{*}$ is a positive hermitian matrix, then we may assume that $ AA^{*}\equal{}diag(\lambda_1I_{i_1},\cdots,\lambda_kI_{i_k})$ with $ {i_1}\plus{}\cdots\plus{}{i_k}\equal{}n$, $ \lambda_j\geq{0}$ and the $ \lambda_j$ are pairwise distinct. $ A$ and $ AA^*$ commutate, then $ A\equal{}diag(A_1,\cdots,A_k)$ where $ A_j\in\mathcal{M}_{i_j}$. $ A_{j}A_{j}^*\equal{}\lambda_{j}I_{i_j}$; thus if $ \lambda_j>{0}$ then $ A_j^{*}\equal{}\lambda_{j}A_{j}^{\minus{}1}$ and if $ \lambda_{j}\equal{}0$ then $ A_{j}\equal{}0$; therefore $ A_{j}A_{j}^*\equal{}A_{j}^{*}A_j$.  QED.


videos
calculus
derivative
function
parameterization
real analysis
real analysis unsolved

I always get excited when a video game gives me the opportunity to do a little math.  It usually doesn't help me much, but at least I can feel like I'm exercising my mind :)

I came up with this equation and I'm completely stumped.  I'm trying to solve it for [i]x[/i] in terms of [i]a[/i]:

x=(50/39)*[1-(1-a)^(15/x+6)]+(5/3)*(1-a)^(15/x+6)

I don't even know where to begin.  I've had up to Calc III, so I feel like I should know how to solve this, but it's been a while.  Anyone have any ideas?

Thanks.

Seems pretty transcendental to me.. We could approximate something for a given $ a$, but in situation like this I wouldn't say that anything useful can be done (correct me if I'm wrong, I might have overseen something simple). I'd like to know more about the context: what was the game?

Okay, I did a bit of simplifying (just multiplying through the parentheses) and came up with

x=50/39+(5/13)(1-a)^(15/x+6).

The game is World of Warcraft, and here's the scenario.  I'll try to explain it in term that can be understood by someone who doesn't know anything about the game.

I have a kitty cat that likes to kill monsters.  It does this by "attacking" them and "biting" them.  A bite happens once every 2.5 seconds, and this rate never changes.  An attack normally happens every 5/3 seconds, but sometimes it happens 30% faster (once every 50/39 seconds).

Before I explain how the attack speed changes, I'll throw out there that my ultimate goal is to find the average attack speed ([i]x[/i]).

The attack speed increase is in effect while my cat has a "buff", which, when applied, lasts for fifteen seconds.  The buff is acquired by attacking or biting.  Whenever she attacks or bites, she there is a chance ([i]a[/i]) that she will gain the buff for the next fifteen seconds.  This happens whether or not she already has it.  (If she already has the buff, its duration is refreshed to fifteen seconds; she cannot have more than one buff at a time).

Now here's where the math comes in. First, I solved for the chance she'll have the buff at any given time:

c=1-(1-a)^n, where n is the number of attacks or bites that have happened in the last fifteen seconds.

n was just the number of attacks per fifteen seconds (15/x) plus the number of bites per fifteen seconds (15/2.5).  Thus:

n=15/x+6

and the chance that she'll have the buff at any given time is:

c=1-(1-a)^(15/x+6)

Now, to find the average attack speed [i]x[/i], I had to do a time-weighted average of the two attack speeds.  That would be (attack speed [b]with[/b] the buff)*(fraction of time spent [b]with[/b] the buff)+(attack speed [b]without[/b] the buff)*(fraction of time spent [b]without[/b] the buff):

x=(50/39)c+(5/3)(1-c)

Substitute and simplify to get

x=50/39+(5/13)(1-a)^(15/x+6).

And that's where I am now :)

With an equation like this, there's no reason to expect an exact analytical solution. That leaves numerical methods, which can give a very close approximation to the solution. Here's Newton's method:

Step 1: Rearrange the equation into the form $ f(x)\equal{}0$. Find a formula for the derivative $ f'$.
Step 2: Guess an initial value $ x_0$.
Step 3: Set $ x_{n\plus{}1}\equal{}x_n\minus{}\frac{f(x_n)}{f'(x_n)}$. Repeat as $ n$ increases, until the numbers stop noticeably changing. The final $ x_n$ is a good approximation to the true root $ x$.

If you use a bad initial guess, it may not work. If this happens, try again with a new starting point. The idea behind this method is that we approximate $ f$ with a linear function and solve the resulting linear equation. It converges very rapidly if you start close enough, with $ |x_{n\plus{}1}\minus{}x|$ being proportional to $ |x_n\minus{}x|^2$.

Your example isn't quite in a usable form; we would need a numerical value for the parameter $ a$ in order to use this.


integration
real analysis
real analysis unsolved

Find

 $ \int( x^{x^{x}} \plus{} x^{x} )dx$  :lol:

Actually, I beg to differ.  This can't be evaluated in terms of elementary functions.


calculus
integration
trigonometry
logarithms
calculus computations

[hide]For $ x \in \left]\minus{} \frac{\pi}{2}, \frac{\pi}{2}\right[$, we have

$ P \equal{} \int (\minus{} \tan x) \ln(\cos x)\,dx \equal{} \frac12 \ln^2(\cos x) \plus{} C$.[/hide]

Let $ I$ be the proposed integral.
[hide]if $ u = \sec x \Rightarrow I = \Int {{du\over u}.ln u$
Integrating by parts with $ dv = {{du}\over u}$ and $ w = \ln u$  we get to $ I = {{\ln^2\sec x}\over 2} + C$ [/hide]


function
real analysis
real analysis unsolved

Define a sequence $ X_{0}$, $ X_{1}$, ... of rational numbers by $ X_{0}=2$ and $ X_{n+1}=X_{n}-\frac{1}{X_{n}}$ for $ n\geq 0$. Is the sequence bounded?

It was first asked [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31823]here[/url]. Unfortunately, I still cannot add much to what I wrote then :(.

No convergence.
The recurrence is the formula for finding the zeros of the function $ f(x)=e^{\frac{x^{2}}{2}}$ by the Newton-Raphson method.
This function has no zeros, so the series does not converge.

That much was clear enough from the beginning: the limit would satisfy $ x=x-\frac{1}{x}$, i.e., $ \frac{1}{x}=0$, which is impossible. The question is whether the sequence stays bounded or, if you prefer, whether $ 0$ is not a limit point of this sequence.


search
arithmetic sequence
number theory
relatively prime

The game is as this follows:

I'll think on an integer number between 1 and 100000. You will ask me for characteristics of the number (just yes/no questions), and I will answer all the questions. If you make posts like:

is it 1?
is it 2?
is it 3?
...

I will NOT answer.
If you guess the number, you esrn 12 points. I will deduce points if:

a) You make too many questions
b) You post too many spam
c) You cheat
d) any fail at the search
e) You late too many time.

WE BEGIN RIGHT NOW!!!

Is it a prime? [b]NO[/b]
Is it even? [b]NO[/b]
Is it greater than or equal to 50,000? [b]NO[/b]
Is it a primitive root mod 226? [b]What does it mean? :P [/b]

Is it greater than or equal to 25,000? [b]NO[/b]

Is it greater than or equal to 12,500? [b]YES[/b]

[hide="Observation"] The restriction that the answer be either yes or no means that we can (at best) guarantee to divide our search space in half with each question (which is equivalent to asking for another bit in the binary representation of the desired number), so for those of you wondering whether there is a better chain of questioning than the one that is currently being pursued, there isn't. [/hide]

YES!!!The message is too small. Please make the message longer before submitting.

is the number one of the digits in your username?

YESThe message is too small. Please make the message longer before submitting.

oh yes :winner_second:

You made too many questions by yourself so -3

cf249: 9
dynamo729: 9
panjia123: 9

new number

Is it less than 100?

(ps how much is too much?  [because I wont do it again, I just didnt know sorry!])

and, too much is: I count the number of pages wasted. Each one is a point less. But you made too many questions yourself, so I penalized you.


Let $r_1=2$ and $r_n = \prod^{n-1}_{k=1} r_i + 1$, $n \geq 2.$ Prove that among all sets of positive integers such that $\sum^{n}_{k=1} \frac{1}{a_i} < 1,$ the partial sequences $r_1,r_2, ... , r_n$ are the one that gets nearer to 1.

Proof:Lemma:\[ 1-(\frac{1}{r_1}+\frac{1}{r_2}+\ldots+\frac{1}{r_n})=\frac{1}{r_1r_2\ldots r_n} \]
The proof of Lemma:When $n=1$,it's easy to see it's correct.
Suppose it's correct for $n-1(n\geq2)$,
then \[ 1-(\frac{1}{r_1}+\frac{1}{r_2}+\ldots+\frac{1}{r_n})=\frac{1}{r_1r_2\ldots r_{n-1}}-\frac{1}{r_n}=\frac{1}{r_n-1}-\frac{1}{r_n}=\frac{1}{(r_n-1)r_n}=\frac{1}{r_1r_2\ldots r_n} \]
That means the lemma is correct.

We only need to show that
$\forall a_1\leq a_2\leq\ldots\leq a_n\in\mathbb{N},\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}<1$,
then we must have \[ \frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}\leq\frac{1}{r_1}+\frac{1}{r_2}+\ldots+\frac{1}{r_n} \]
We prove it by induction:
When $n=1$,$\frac{1}{a_1}<1$,$a_1\geq2=r_1$.So $\frac{1}{a_1}\leq\frac{1}{r_1}$.
Suppose the conclusion is correct for all positive integer smaller than $n$,and there exist $a_1,a_2,\ldots,a_n$ such that
\[ \frac{1}{r_1}+\frac{1}{r_2}+\ldots+\frac{1}{r_n}<\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}<1 \eqno(*) \]
By Lemma\[ 1-(\frac{1}{r_1}+\frac{1}{r_2}+\ldots+\frac{1}{r_n})=\frac{1}{r_1r_2\ldots r_n} \]
And it's obvious that \[ 1-(\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n})\leq\frac{1}{a_1a_2\ldots a_n} \]
Because $\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}=\frac{x}{a_1a_2\ldots a_n}<1,x\in\mathbb{Z}$
So $x<a_1a_2\ldots a_n$,and $x\leq a_1a_2\ldots a_n-1$,so
\[ 1-(\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n})=1-\frac{x}{a_1a_2\ldots a_n}\leq\frac{1}{a_1a_2\ldots a_n} \]
So by (*) \[ a_1a_2\ldots a_n\geq r_1r_2\ldots r_n \eqno(**) \]
And by the transform of Abel
\[ \begin{eqnarray*} \sum_{i=1}^n\frac{a_i}{r_i}&=&a_n\sum_{i=1}^n\frac{1}{r_i}+\sum_{k=1}^{n-1}(\sum_{i=1}^k\frac{1}{r_i})(a_k-a_{k+1})\\ &<&a_n\sum_{i=1}^n\frac{1}{a_i}+\sum_{k=1}^{n-1}(\sum_{i=1}^k\frac{1}{a_i})(a_k-a_{k+1})\\ &=&\sum_{i=1}^n\frac{a_i}{a_i}=n \]\end{eqnarray*}
By AM-GM
\[ n>\sum_{i=1}^n\frac{a_i}{r_i}\geq n\sqrt[n]{\frac{a_1a_2\ldots a_n}{r_1r_2\ldots r_n}} \]
So \[ a_1a_2\ldots a_n<r_1r_2\ldots r_n \]
And it's a contradiction with (**).
So there doesn't exist $a_1,a_2,\ldots,a_n$ satisfied (*),that means the conclusion is correct.

[quote="srulikbd"]isn't it obvious? we take the smallest integer such that it doesn't exceed or equal 1..and it's easy to show that it always works..?[/quote]

well, that not necessary give the largest sum as a whole :)

Pretty much as a problem from Bulgaria RMM TST 2023.


geometry
parallelogram
analytic geometry
geometry proposed

[color=darkred]Let $ \triangle\ ABC$ and a point $ P$ for which $ \left|\begin{array}{c}
 D\in AP\ \cap BC\\\
 E\in BP\ \cap\ CA\\\
 F\in CP\ \cap\ AB\end{array}\right|$ . Construct the parallelograms

$ PEXF$ , $ PFYD$ , $ PDZE$ . Prove that the lines  $ AX$ , $ BY$ si $ CZ$ are concurrently.[/color]

[hide]I think this problem is coming from the past and if I am not mistaken, I will post here later the proof I have in mind, based on the [b]Ceva theorem[/b] and the [b]Cevian Nests theorem[/b].[/hide]
Kostas Vittas.

My solution (similar to vittasko's hint)
Denote the points like my figure.
We have:
$ \frac{AM}{ME}\equal{}\frac{AF}{FB}$
$ \frac{AN}{NF}\equal{}\frac{AE}{EC}$
Apply Ceva's Theorem to triangles AFE and ABC,we have:
$ \frac{MA}{ME}.\frac{NF}{NA}.\frac{GE}{GF}\equal{}1$
$ \frac{FA}{FB}.\frac{EC}{EA}.\frac{DB}{DC}\equal{}1$
So:$ \frac{GE}{GF}\equal{}\frac{DB}{DC}$
Similar:
$ \frac{HF}{HD}\equal{}\frac{EC}{EA}$
$ \frac{ID}{IE}\equal{}\frac{FA}{FB}$

So:
$ \frac{GE}{GF}.\frac{HF}{HD}.\frac{ID}{IE}\equal{}1$
According to Ceva's Theorem,we see that DG,EH,FI are concurrent.
Because AD,BE,CF are concurrent,then apply the Ceva Nest Theorem,we see that:$ AG,BH,CI$ are concurrent.
Q.E.D

[img]http://img8.imageshack.us/img8/5634/geo.png[/img][/img]

Let us use barycentric coordinates WRT $\triangle ABC.$ Then $P \equiv (u:v:w)$

$ D \equiv (0: v: w) \ , \ E \equiv (u: 0: w) \ , \ F \equiv (u: v: 0)$

$ AD \equiv  vz \minus{} wy \equal{} 0 \ , \ BE \equiv  wx \minus{} uz \equal{} 0 \ , \ CF \equiv  uy \minus{} vx \equal{} 0$

Line $\ell_1$ passing through $ E$ and parallel to $ CF$ has equation

$ \ell_1 \equiv   uvz \plus{} (uw \plus{} u(u \plus{} v))y \minus{} vwx \equal{} 0$

Line $ \ell_2$ passing through $ F$ and parallel to $ BE$ has equation

$\ell_2 \equiv  uwy \plus{} (uv \plus{} u(u \plus{} w))z \minus{} vwx \equal{} 0$

$ \ell_1$ and $ \ell_2$ meet at the point $ X( ***: uvw(u \plus{} w): uvw(u \plus{} v))$

Hence the foot $X_0$ of the A-cevian of $X$ is $ X_0 \equiv (0: u \plus{} v: u \plus{} v)$

Likewise,  $ Y_0 \equiv (v \plus{} w: 0: v \plus{} u) \ , \   Z_0 \equiv (w \plus{} v: w \plus{} u: 0)$

$AX,BY,CZ$ concur at the complement $Q \equiv (w \plus{} v: u \plus{} w: u \plus{} v)$ of $P.$

[color=darkblue][b]Thanks to all ![/b] Here is a [u]similar nice problem[/u] (more easily !) [b]with very many applications[/b] :[/color]

[quote="Virgil Nicula"][color=darkred]Let $ \triangle\ ABC$ and a point $ P$ for which denote the points $ \left|\begin{array}{ccc} D\in BC & ; & PD\perp BC\\
\ E\in \ CA & ; & PE\perp CA\\
\ F\in AB & ; & PF\perp AB\end{array}\right|$ . Construct

the parallelograms $ PEXF$ , $ PFYD$ , $ PDZE$ . Prove that the lines  $ AX$ , $ BY$ si $ CZ$ are concurrently.[/color][/quote]

[quote][color=darkred]Let $ \triangle\ ABC$ and a point $ P$ for which denote the points $ \left|\begin{array}{ccc} D\in BC & ; & PD\perp BC \\
\ E\in \ CA & ; & PE\perp CA \\
\ F\in AB & ; & PF\perp AB\end{array}\right|$ . Construct

the parallelograms $ PEXF$ , $ PFYD$ , $ PDZE$ . Prove that the lines  $ AX$ , $ BY$ si $ CZ$ are concurrently.[/color][/quote]
 AX,BY,CZ concur at the isogonal conjugate of  P WRT ABC.


Let ABC be triangle ,<C=2<A .CD is the inner angled bisector.S is center of cirlce which touches opposite ray AC and touches externally the circles (ACD),(BCD).Prove that CS perpendicular to AB
Hint :By using the inversion

We use inversion with the center of c and arbitrary power.
The line AB maps to the circle ,crossing the c. assume that  the inversion of the A,B ,D is A',B',D'.as result of  AC,DC,BC cross the center of inversion hence  the map to themselves. The angle between inversion of AC   and AB(A'C and tangent to circle that cross A',B',D')equals to the angle between AC , AB. So A'CD'=ACD,  B'CD'=BCD , and we have BAC=A'CD'=D'CB',   on the other hand we have XA'C=A'CD'=D'CB' hence A'C=A'D'=B'D'  . now the inversion of circle with center S  is a circle that is tangent to the A'C , A'D' ,B'D'.as we know this three chord are equal , the center of this circle coincide the center of the circle crossing A',B',D'.  we name this point O , on the other hand  we name the inversion of S, S' .O lie on the CS' so CS' is perpendicular the circumcircle of A'B'D' so are those inversion ,and we are done

Very nice sol(Because it is same as mine!! :D )
How about sol without inversion?


function
number theory
least common multiple

Here some probs from my book i dont get can you guys help me heres the link to img i did on photoshop [url=http://i6.photobucket.com/albums/y208/outoflimitz/math99.jpg]CLICK HERE[/url] i just added probs there i need help with

thanks for that but i need to know how to do them

1)$\frac{a-b}{c}=\frac{a}{c}-\frac{b}{c}$.
2)Find $x$ in the first equation and substitute this value in the 2º equation (the same for all systems).
3)$\frac{\frac{4}{9}}{\frac{14}{3}}$.
4)see 1).
5)$\frac{(\frac{-2}{3})}{2}$.
6)Just a normal equation.
7)$c$ is the only letter that is common for all the expressions.
10)trivial...

ok ok how you do them? like can you show your work?

They all involve different Algebra 1/2 topics.

1) Use [b]properties of exponents[/b]
2) Use [b]subsitution [/b]or [b]linear combinations[/b]
3) uhh... define [b]fraction[/b]
4) Again with the [b]properties of exponents[/b]
5) Simple [b]function[/b].
6) Define [b]radical[/b], or [b]squaring both sides of an equation[/b]
7) [b]Least Common Multiple[/b] [then just apply to variables]
8) Again, with the [b]subsitution[/b]
9) ...
10) [b]Multiplying and dividing fractions[/b]

Google all the terms in bold [and find a good math site], if you're still confused. I recommend picking up a good Algebra I book, so you can learn essential algebraic properties properly.

i wish you could of help me and showed me

Jose, I think you could be a little nicer; you have no idea how much math Host knows at this point, and your comment, frankly, will add nothing to his understanding of the problems and their solutions.  In fact, your comment will only be detrimental to his desire to ask questions and participate in this community, and therefore your comment merely impedes rather than advances his mathematical pursuits.  Nobody has really explained the problems at a level that an Algebra II student could easily understand them.

You're right, sorry... I have to go now, but then I'll post more details...

[quote="José"]Come on Host!!! is just trivial math! what don't you understand???[/quote]Remember that this is the [b]Getting Started[/b] forum, and many users turn here for guidance rather than answers. If the questions were trivial to him, he would have been posting other questions in higher forums ;)

bro, i just need members of this forum show me how to do them thats all  :(

Host, look
1. $\frac{6r^2st^3-9rs^2t^2}{3rst} =$
$\frac{6r^2st^3}{3rst}-\frac{9rs^2t^2}{3rst} =$
$2rt^2-3st = t(2rt-3s)$

2. Multiply the first equation by $-2$, you will get:
$-4x-6y=22$. Now add both equations and the x's will cancel out and you will have $-8y=24$ which is the same as $y=-3$. Now substitute -3 for y into any of the two equations and solve for x. The solution is (-1,-3)

3. $4\frac{2}{3} =\frac{14}{3}$. We need to know what fraction of $\frac{14}{3}$ is $\frac{4}{9}$. So we set the equation:
$\frac{14}{3}*\frac{x}{1} =\frac{4}{9}$ Now, lets solve for x:
$\frac{14x}{3}=\frac{4}{9}$
$(\frac{3}{14})\frac{14x}{3}=\frac{4}{9}(\frac{3}{14})$
$x=\frac{12}{126}=\frac{2}{21}$

4. Do the same thing that I did on problem #1

5. substitute $-\frac{2}{3}$ into the x and you will have,
$-\frac{\frac{2}{3}}{\frac{2}{1}}$
$-\frac{2*1}{3*2}=$
$-\frac{1}{3}$

6. Subtract 12 from both sides and you will get
$\sqrt{x+5}=-3$, square both sides and you will get+
$x+5=9$, $x=3,-3$, substitute both into the equation independently and you will see there is only one of those that is a solution.

7. I know it is $c^3d^2$ but I don't know how to explain it to you, i don't remember, sorry.

8. Solve for x in the second equationto get $x=9+2y$ substitute that into the first equation and solve for $y$, then substite y for it's volue on the second eqaution and solve for $x$, the correct answer is $(5,-2)$

9. Multiply the fisrt equation by -2 and the second equation by 5. Add both together to get $7x=0$, $x=o$ substitute 0 into the either equation and solve for y. The correct answer is $(0,-2)$

10. $\frac{18}{3} * \frac{5}{4} / \frac{3}{4} =$
$\frac{18}{3} * \frac{5}{4} * \frac{4}{3} =$
$\frac{90}{9}=$
$10$

If you don't understand anything, let me know.

7. $LCM\ of\ cd^2,\ c^2d,\ c^3$

an easy way to do LCM's is to carry over the greatest factor amount in each term into the LCM. In your example, you basically have

$(c)(d)(d), (c)(c)(d), (c)(c)(c)$. You are only working with 2 factors here, and the greatest amount of c's is 3 and d's is 2. The LCM will have 3 c factors and 2 d factors, $cccdd = c^3d^2$

hope that helps


Hi. I'm hosting the Mole 5. You'll have to bear with me :)

If you wish to apply, please PM me the following (slightly hijacked):

Application:
What is your birthday? MM/DD/YY please.

What is your favorite food? Be as specific as possible, as applicable.

How many countries have you visited, besides the one in which you live?

List any foreign languages you know or take in school.

Have you seen the Mole? If so, did you like it?

What is your strategy to win?

Do you want to be The Mole? Elaborate.

Have you played a Mole game before? If so, what was/were the result(s)?

Scenario: You are trapped on an island with 5 others. Do you

1. Take charge
2. Allow someone else to take charge
?

Are you good with codes? Making codes?

On a scale of 1-10, how :

1. Good are you at Making questions?
2. Good are you at Solving AIME level questions?
3. Active are you?
4. high is the chance I will have to replace you?

Do you enjoy:

1. Finding stupid errors (like typos)
2. Reading?

If you had to bring ONE thing with you to a desert island, what would you bring?

Nice job on the submissions. 2 submissions in 25 minutes!

If I PM you back with more questions it means you are being considered.

Update: Only like 3 spots left.

DEADLINE TO APPLY IS 9:30 ET FRIDAY.

Only 2 spots remaining! Sign up quick!

[url=http://www.youtube.com/watch?v=oHg5SJYRHA0]@AIME15[/url]

[quote="BOGTRO"]Application:
What is your birthday? MM/DD/YY please.

What is your favorite food? Be as specific as possible, as applicable.

How many countries have you visited, besides the one in which you live?

List any foreign languages you know or take in school.

Have you seen the Mole? If so, did you like it?

What is your strategy to win?

Do you want to be The Mole? Elaborate.

Have you played a Mole game before? If so, what was/were the result(s)?[/quote]

[quote="I, in Mole 2 Apps,"]
[b]What is your birthday? MM/DD/YY please.

What is your favorite food? Be as specific as possible, as applicable.

How many countries have you visited, besides the one in which you live?[/b]

Can you juggle?

On a scale of 1-10, how good are you at crossword puzzles?

Do you have access to Photoshop, or an equivalent?
[b]
List any foreign languages you know or take in school.

Have you seen the Mole? If so, did you like it?

What is your strategy to win?

Do you want to be The Mole? Elaborate.[/b]

Will I have to replace you? Explain.[/quote]

Just kidding, I don't really care, I'm just glad something I started on a whim many many months ago is still thriving!

@ the rickroll: you can hover your mouse over the link...and it tells you it's a youtube (aka rickroll).  Also, on a mac, hide tags and links show up in different colors :P

4 peopel tried to rickroll me today...hm?


algebra
polynomial
quadratics
topology
number theory
set theory
superior algebra

Let $g(x)\in\mathbb Q[x]$ be an irreducible polynomial of odd degree, and $\rho\in\mathbb C$ be a root of $g$, and $K=\mathbb Q(\rho)$. Let $f$ be a homogeneous polynomial of degree 2 in $\mathbb Q[x_{1},x_{2},\dots,x_{n}]$. Prove that if $f(\alpha)=0$ has a solution in $\mathbb P^{n-1}(K)$, then it has a solution in $\mathbb P^{n-1}(\mathbb Q)$.

I've already posted this, and there was no answer. This is Springer's theorem. If there's no solution in a week, I will post the one I know.

Just to remark: it can be shown (as easy corollary of the Hasse-Minkowski theorem) that any homogeneous quadratic form over $\mathbb{Q}$ in $n \geq 5$ variables has a zero in $\mathbb{P}^{n-1}(\mathbb{Q})$, provided only that it has a zero in $\mathbb{P}^{n-1}(\mathbb{R})$. (Because, by the Hasse-Minkowski theorem, it suffices to find a zero in each $\mathbb{Q}_{p}$. which is an easy job with Hensel liftings). This results subsumes Springer's for $n \geq 5$ (since $[K: \mathbb{Q}]$ is odd, it is easy to see that solvability in $\mathbb{P}^{n-1}(K)$ implies solvability in $\mathbb{P}^{n-1}(\mathbb{R})$), so it suffices to give a proof for $n \leq 4$ variables. The cases $n \leq 2$ are immediate, so the most interesting case in Springer's theorem is $n \in \{ 3,4 \}$.

A propos, it suffices (by the Hasse-Minkowski theorem) to prove the local version: if $K/\mathbb{Q}_{p}$ is a finite extension of odd degree and $f$ a homogeneous quadratic form over $\mathbb{Q}_{p}$ in $n$ variables which has a zero in $\mathbb{P}^{n-1}(K)$, then $f$ has already a zero in $\mathbb{P}^{n-1}(\mathbb{Q}_{p})$.

[Edit: OK, I am sorry if my preceding message was a spam  :roll: . This is not an olympiad problem, although it has an elementary solution; moreover, it isn't discussed in the olympiad section. I find it a bit strange that people expect only a self-contained proof as in an olympiad book and don't appreciate discussions on closely related results and ideas in such exciting topics. I could write down the proof, but it's not mine; that wouldn't be quite interesting. Instead, I chose to mention the above-stated result and the (stronger, in view of Hasse-Minkowski!) local version of Springer's theorem. But, fine, I won't be spamming anymore. ]

Dear Vess, nobody is against such cultural remarks, don't worry: they're always welcomed (at least for me). And don't worry, I (and usually us) don't expect only olympiad type solutions, I certainly prefer conceptual solutions. However, I totally disagree with some "trends" of trivializing everything by dropping atomic bombs on everything: there's a difference between knowing and solving. It is always a pleasure to see a beautiful approach, completely different from the classical one (an excellent example: look at Fedja's fantastic approach for counting matrices in $SL(2,Z)$!) and this is why I sometimes ask for elementary approaches. And really, what's the difference between not posting a proof of Springer's theorem (even though it's not yours) and posting a fact that is not yours also (don't get me wrong: I want to say that anyone knowing Hasse local principle immediately thinks about this approach) and that is also known by quite many people?

OMG...

Yes, it was me who rated Vess' post a 2 or 3 (I don't remember precisely). I certainly have not assumed that a mouseclick of mine would lead to feedback from 3 different people in different threads, but given this feedback I should explain...

First of all, I want to stress that by rating a post I am neither rating its author, nor rating the other posts of that author. Rating a post means rating its form, but also rating its mathematical content, and for this reason I claim the right to rate ugly problems low, even if they are posted "for the resources section" because they were given on a contest.

The reason why the above post was the only one by Vess I have rated until now is just because it is one of the very few posts from Vess that I can halfways understand (I am pretty behind on analysis and am not quite interested in advanced number theory; at least I know what Hasse-Minkowski states, though I don't have the slightest idea how to prove it). And, though this approach yields a solution, I find it ugly because
- it uses theory requiring dozens of pages;
- it seems to use the AC (I am not sure, but most local-global arguments use it);
[Hmm, as this is likely to trigger just another flame... I am not a definitive opponent of the AC. In fields like topology, analysis, set theory, the AC has its significance. I just find it ugly and unconceptual to apply the AC in algebraic contexts like finite field extensions or modules. Probably I am one of the very few people having this opinion...]
- it leaves some cases to be checked by hand which are all but trivial and probably require computation.
Yes, these are subjective criteria, but this is what ratings are for. If there would be enough objective criteria for the quality of posts, probably only the admins would be allowed to rate...

But I must say the problem also deserves a lower rating, because it is just a stupid particular case of
http://www.mathlinks.ro/Forum/viewtopic.php?t=130345
Maybe Omid didn't know this, but again - it's not him but the problem that I am rating.

I hope I did not hurt too many feelings... Let's better try solving the general problem (though I am not going to read solutions until I have solved it on my own).

  Darij


geometry
circumcircle
inequalities
geometry proposed

[color=darkred]Let $ ABC$ be an acute triangle with the incircle $ C(I,r)$ and the circumcircle $ C(O,R)$ .

Prove that there is the inequality $ \boxed {\frac {IA^2 + IB^2 + IC^2}{IA + IB + IC}\le \frac {4(R^2 - Rr + r^2)}{3\sqrt {2Rr}}}\ .$[/color]
[hide="Help."]  [color=darkblue]You can use a strong inequality (acute triangle !) $ \sum\sqrt {\frac {b + c - a}{a}}\ge 3$ ,

the identities $ IA^2 = \frac {bc(p - a)}{p}$ , $ \sum IA^2 = 3\cdot IG^2 + \frac 13\cdot(a^2 + b^2 + c^2)$ ,

$ 9\cdot IG^2 = p^2 + 5r^2 - 16Rr$ and $ p^2\le 4R^2 + 4Rr + 3r^2$ ([b][u]Gerretsen[/u][/b]).[/color] [/hide]
[color=darkblue][b]Remark.[/b] Prove easily that $ R\le \frac {4(R^2 - Rr + r^2)}{3\sqrt {2Rr}}\ .$

[u][b]Unsolved question[/b][/u] : in any triangle $ ABC$ ,  $ \frac {IA^2 + IB^2 + IC^2}{IA + IB + IC}\le R\ ?$

[b]Example.[/b] For $ \{\begin{array}{c} a = 5 \\
 \\
b = 4 \\
 \\
c = 3\end{array}\|$ $ \implies$ $ \{\begin{array}{c} R = \frac 52 \\
 \\
r = 1\end{array}$ and $ \{\begin{array}{c} IA^2 = 2 \\
 \\
IB^2 = 5 \\
 \\
IC^2 = 10\end{array}\|$ $ \implies$

$ \frac {IA^2 + IB^2 + IC^2}{IA + IB + IC}\le R$ because $ 34 < 5(\sqrt 2 + \sqrt 5 + \sqrt {10})\ .$[/color]

[quote="Virgil Nicula"][u][b]Unsolved question[/b][/u] : in any triangle $ ABC$ ,  $ \frac {IA^2 + IB^2 + IC^2}{IA + IB + IC}\le R\ ?$[/quote]

For $ \{\begin{array}{c} a = 13 \\
 \\
b = 12 \\
 \\
c = 5\end{array}\|$ $ \implies$ $ \{\begin{array}{c} R = \frac{13}{2} \\
 \\
r = 2\end{array}$ and $ \{\begin{array}{c} IA^2 = 8 \\
 \\
IB^2 = 13 \\
 \\
IC^2 = 104\end{array}\|$ $ \implies$ $ \frac {IA^2 + IB^2 + IC^2}{IA + IB + IC} > R$ because $ 250 > 13 (\sqrt 8 + \sqrt {13} + \sqrt {104})\ .$


Set up a coordinate plane and place 1 ohm resistors between ever pair of adjacent lattice points. What is the resistance between any 2 adjacent lattice points? (This is not a trick question!)

Isn't the resistance simply $ 0.5$ Ohms??

yes, it's $ 0.5$ Ohms. I thought [b]dominicleejun[/b] was asking for the resistance between any two points :)

Adjacent, as far as I doesn't mean any two points  :lol:. jk

what is the resistance between the points (0,0) and (x,y)?

sorry for revive..


function
algebra
system of equations
inequalities proposed
inequalities

For $ a,b,c$ positive real numbers such that $ a\plus{}b\plus{}c\equal{}1$, determine the minimum and maximum value of $ 2a^2\plus{}2b^2\plus{}2c^2\plus{}9abc$.

sorry if it's a silly problem for you  :(

[quote="Raja Oktovin"]For $ a,b,c$ positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$, determine the minimum and maximum value of $ 2a^2 \plus{} 2b^2 \plus{} 2c^2 \plus{} 9abc$.

sorry if it's a silly problem for you  :([/quote]
$ A\equal{}2a^2 \plus{} 2b^2 \plus{} 2c^2 \plus{} 9abc\equal{}(2a^2 \plus{} 2b^2 \plus{} 2c^2)(a\plus{}b\plus{}c)\plus{}9abc\ge (a\plus{}b\plus{}c)^3\equal{}1$ its equivalent
$ a^3\plus{}b^3\plus{}c^3\plus{}3abc\ge a^2(b\plus{}c)\plus{}b^2(a\plus{}c)\plus{}c^2(a\plus{}b)$ and its true by Schur. A=1 when a=b=c=1/3 And its obvious that $ A\equal{}(2a^2 \plus{} 2b^2 \plus{} 2c^2)(a\plus{}b\plus{}c)\plus{}9abc\le 2(a\plus{}b\plus{}c)^3$ (just open brackets) A=2 a=1,b,c=0 :wink:

Do you want to mean that it lies within $ [1,2]$??

[quote="Raja Oktovin"]For $ a,b,c$ positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$, determine the minimum and maximum value of $ 2a^2 \plus{} 2b^2 \plus{} 2c^2 \plus{} 9abc$.

sorry if it's a silly problem for you  :([/quote]
Try to use function $ F(a,b,c) \equal{} 2a^2 \plus{} 2b^2 \plus{} 2c^2 \plus{} \alpha (a \plus{} b \plus{} c \minus{} 1)$.

So, in the extremum $ \frac {df}{da} \equal{} 0$, $ \frac {df}{db} \equal{} 0$, $ \frac {df}{dc} \equal{} 0$, and $ a \plus{} b \plus{} c \equal{} 1$.

After solving this system of equations (with unknown $ a,b,c,\alpha$), you will have that $ a \equal{} b \equal{} c \equal{} 1/3$ - only one extremum-point.

So, $ F(1/3,1/3,1/3) \equal{} 1$.

Now, if one of $ a,b,c$ is 0 (for example, $ c$), then you will have $ a \plus{} b \equal{} 1$, and $ F(a,b,0) \equal{} 2a^2 \plus{} 2b^2 \equal{} 2(a^2 \plus{} (1 \minus{} a)^2) \equal{} 2(2a^2 \minus{} 2a \plus{} 1)$ - the minimum value of $ F(a,b,0)$ if and only if $ a \equal{} b \equal{} 1/2$.

So, $ F(1/2,1/2,0) \equal{} 1$.

If both $ b \equal{} c \equal{} 0$, then $ a \equal{} 1$, and $ F(1,0,0) \equal{} 2$.

So, the minimum value is 1, and the maximum is 2.

[quote="Raja Oktovin"]halo...

i don't understand your solution.. sorry..

please explain  :(

Raja[/quote] Whats you  don't understand?


What is the YTL? Is that some type currency?

sorry it's turkish liras :) it isn't important.  :)

[quote="Shahzada"]A businessman has some money in his pocket.For the first two days he spends (1/3)+100YTL of it.And for the last two days spends (1/5)+100 of available money.After 4 days there is 140YTL in his pocket.How much money did he have at start?[/quote]
[hide="Work Backwards"]
Say that $x-(\frac{x}{5}+100)=140$.  Solving, we find x to be 300.  Again, say that $y-(\frac{y}{5}+100)=300$.  Thus, we have that y=500.  Now, say that $a-(\frac{a}{3}+100)=500$.  Thus, a=900.  Now, we have that $b-(\frac{b}{3}+100)=900$.  Thus, $b=\boxed{1500}$, which is our answer. [/hide]

Yes it's true.This is another simple way.
[hide][img]http://img128.imageshack.us/img128/8992/dfgoc4.gif[/img][/hide]

that not needs more than on simple equation
but its beneficial for mental calculus[/hide]


OK. The second revolution has started!

Rules:

There are 10 ppl, 1 king, 1 queen, 2 knights (one known to public and one unknown), 1 renegade, and 5 citizens in each country, bringing the total to 20 ppl.

King tries to kill the queen and 3 citizens without losing any knights or kill all citizens and queen to win the game.

Queen must kill King firsthand to win.

Renegades try to have lowest number of points when the kingdom falls.

Citizens have same objective as queen.

Knights can cooperate (have same goal) as the king, or kill queen, other knight, and the king.

King starts w/ 20 pts, known knight 20 pts, and others 10 pts.

All type of people except renegade can only win if the kingdom wins.

When queen or knight (2nd objective for knight) completes their objective before kingdom falls, the person becomes the king.

Kingdom falls when the king is dead; but obviously it doesn't fall if the other kingdom has already fallen.

Everyone will be private except King and the known knight.

Daily, king will receive 3 pts, known knight 2 pts, and 1 pt for rest.

A person may attack once per day, removing same number of points from both sides and removal can only be up to 2 pts, and attack 1 person only.

King can execute someone by giving up 15 pts every 3 days.

Also, king may not be attacked unless both knights are dead. Also, anyone can give points (as much as wanted, once per day) to the king, but only king can give points to knights.

I will PM who's who if all 10 ppl are signed up; and king and knight will be announced.

Mob Lynch: during the whole game, a Rest (those who are not known) has 1 vote to mob lynch someone. Whenever majority is met, the person will be lynched (cannot be known knight or the king). Majority does not include abstained votes.

There are 9 events (one of them happens twice, so total 10) that the player may choose in order. I have decided them, but I may change them so I am not posting yet. Every day, 1st person to post what actions to take (doesn't matter with the kingdom, same effect will go on to same kingdom regardless of the poster. The poster also may choose to not to use any of events.) i will have events posted later.

Current signups:

1.(^_^)
2.packman2812
3.chenhsi

17 or 7 spots still available (if i can't get 20 people, i will do this by having only one kingdom).

I'll join. I think others after me should join with PM to make this topic not clog up. (Hehe, hypocrisy!)

Updated current signups:

1.(^_^)
2.packman2812
3.chenhsi
4.lotsofmath
5.budi713
6.Brut3Forc3
7.mnmath

Comeon people, it'll be more fun when we have 20 people!

I could join, but right now, I'm vacationing in Boston, so I might not be able to get internet access everyday.  I hope everyone forgives me for that.

/in

Updated signups:

1.(^_^)
2.packman2812
3.chenhsi
4.lotsofmath
5.budi713
6.Brut3Forc3
7.mnmath
8.142857
9.alexhhmun
10.BOGTRO

do you want to start or get more people?

Looks like I missed it.

If you guys need more people, I'm here!

please wait. There are more than 20 active people in this forum. Continue waiting.

me want to join :)

Updated signups:

1.(^_^)
2.packman2812
3.chenhsi
4.lotsofmath
5.budi713
6.Brut3Forc3
7.mnmath
8.142857
9.alexhhmun
10.BOGTRO
11.142857
12.iLord
13.Garyzx
14.Crazychinesemaniac

Unless 6 more people join anyone below BOGTRO will not play I guess, which includes me, so come on people...I wanna play!!!

C'mon people, we need signups!!!  The game has to start quickly, or many of us will have to leave!

Yeah...... I have to leave in 10 days or so and I won't be back for long......

I PMed Revolution I players, only dynamo and iLord responded.

Maybe we can cut down some:

1 king, 1 knight, 1 queen, and 4 citizens each country.

If we don't get any more people by this Friday, I will start like this.

Let's hope some people join. (especially since MOSP is over, i think).

1.(^_^)
2.packman2812
3.chenhsi
4.lotsofmath
5.budi713
6.Brut3Forc3
7.mnmath
8.142857
9.alexhhmun
10.BOGTRO
11.142857
12.iLord
13.Garyzx
14.Crazychinesemaniac

PM people you can reach so we can get more people.

Anyways, I am posting 10 events scheduled and the first person each day may choose which (or not choose at all)

1. Everyone must give up 1 point. If this is not done by the end of the day, the person loses 5 points.
2. A person from each kingdom must give 2 points to respective king.IFINDBTEOTD (if this is not done by the end of the day) king can execute someone without 15 points.
3. Everyone except king receives 1 point. Knight receives 2 extra points.
4. People may PM me a number by subtracting 1 point. The person to PM the largest number loses that much, and 2nd largest receives that many. You can only bid up to how much you have.
5. People may PM me a number by subtracting 1 point. The person to PM the lowest unique positive integer receives 5 points.
6. Knights of both kingdom must give up points how many each king wants to.
7. King may give up 10 points to revive a dead knight (knight starting with 10 points) or give up 15 points to bribe a citizen into a knight (point become the number of points left of that citizen)
8. Declare War; a Rest (unknowns) with lowest points on both sides die. ( if multiple, at most 2 will die. Choosing it will be by getting last people on the list. This may not be used for first 3 days) X 2
9. Swap a Rest with highest points from each kingdom (if multiple, all will swap.)

[quote="dannyhamtx"]The problem is, that i have to go (so does budi, alexhmun, (^_^) and 1=2 at least) will leave for AMP pretty soon (or July 13 :P )

So I have no option unless I want to mod it during school, which will be quite painful.

I have to hurry up with this.

Please put this on your sigs if you want:

[code]Start [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=212218]Revolution: The War![/url]
 :help:[/code][/quote]

You misspelled my name...  *sniffles*

Oh, and maybe we change that into
[code]Start [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=212218]Revolution 2: The War![/url]

This game should be freezed now. I hope mods will not consider this "dead."

It MAY start again when we come back from AMP which is August 3rd.

See you till then and i hope more ppl still sign up anyways.

1.(^_^)
2.packman2812
3.chenhsi
4.lotsofmath
5.budi713
6.Brut3Forc3
7.mnmath
8.142857
9.alexhhmun
10.BOGTRO
11.142857
12.iLord
13.Garyzx
14.Crazychinesemaniac
15.mathlete9999
16.!!!!!

We just need 4 more people now

I hope the new people join now *sigh*

Finally, I almost forgot about this game :D

Is this considered dead? just curious. Sorry for posting if it is  :D .

[quote="dannyhamtx"]1.(^_^)
2.pac[color=red]k[/color]man2812
3.chenhsi
4.lotsofmath
5.budi713
6.Brut3Forc3
7.mnmath
8.142857
9.alexhhmun
10.BOGTRO
11.142857
12.iLord
13.Garyzx
14.Crazychinesemaniac
15.mathlete9999
16.!!!!!

We just need 4 more people now

I hope the new people join now *sigh*[/quote]

packman $ \neq$ pacman :mad:

And yeah, this is possibly dead.


algebra
polynomial
inequalities
vector
function
algebra unsolved

Find all polynomials $ P(x)$ of degree $ \le n$ with nonnegative real coefficients

such that the inequality $ P(x)P(1/x) \le (P(1))^2$ holds for all positive real

numbers x.

By Cauchy-Schwarz, \[ P(x)P(x^{\minus{}1})\equal{}\left(\sum x^ka_k\right)\left(\sum x^{\minus{}k}a_k\right)\geq\left(\sum a_k\right)^2\equal{}P(1)^2\](valid only for positive real $ x$).
Therefore, $ P(x)P(x^{\minus{}1})\equal{}P(1)^2$, and equality holds above. The equality case for the above inequality is when the two "vectors" are parallel... i.e. each "component" $ x^ka_k\equal{}N\cdot x^{\minus{}k}a_k$, for some constant $ N$. When $ k\equal{}0$, $ N$ drops out to be $ 1$. We conclude that $ x^k\in\pm1$ for all $ k$. Since $ x$ is an arbitrary positive real, $ k$ can't take on the value one - that is, \[ \boxed{P(x)\text{ is a constant function}}\]

Let $ P(x)$ be such a polynomial, and let its degree be $ k$. Since all of the coefficients of $ P(x)$ are nonnegative, $ P(x)P\left(\frac{1}{x}\right)\ge 0$ for all $ x\ge 0$. We consider the behavior of $ P(x)P\left(\frac{1}{x}\right)$ as $ x\to\infty$. Clearly, $ P(x)$ goes to infinity as $ x^k$. It follows, since $ P(x)P\left(\frac{1}{x}\right)$ is bounded, that $ P\left(\frac{1}{x}\right)$ goes to zero as fast or faster than $ x^{\minus{}k}$, or equivalently, $ P(x)$ has a zero of multiplicity not less than $ k$ at zero. But is $ P(x)$ is a polynomial of degree $ k$, this implies that it has exactly one zero of multiplicity $ k$ at zero. The only polynomials satisfying this requirement are those of the form $ Cx^k$. It is easy to see that, if $ C\ge 0$ and $ 0\le k\le n$, then $ Cx^k$ all the required conditions.


Does there exist an real number $ c < \frac{1}{2}$ and a positive integer $ m$ such that for every positive integer $ n$ and family $ \mathcal F$ of subsets of $ \{1, \ldots, n\}$, we have: If $ |A \cap B \cap C| \geq m$ for all $ A, B, C \in \mathcal F$, then $ |F| \leq c \cdot 2^n$.

That seems to be a nice question! Before proceeding, how about the case of two-set-intersection, i.e. replacing the $ |A \cap B|$ with $ A \cap B \cap C|$ in the condition? Is there such a $ c$ in this case? This simpler version seems to be interesting too ...

The similar problem with $ A \cap B$ is not that difficult: There is no such $ c$. For a proof, consider an arbitrary positive integer $ m$. For every $ n \in \mathbb N$, define $ \mathcal F_n \equal{} \{A \subset \{1, 2, \ldots, n\} : |A| \geq \frac{n\plus{}m}{2}\}$. We have $ A,B \in \mathcal F \Rightarrow |A \cap B| \geq m$ and $ \lim_{n \rightarrow \infty} \frac{|\mathcal F_n|}{2^n} \equal{} \frac{1}{2}$.


[hide="Solution"]If he had $x$ barrels, then:

1. After the first city he was left with ${2\over 3}x$ barrels;

2. After the second city he was left with ${3\over 4}\cdot{2\over 3}x={1\over 2}x$ barrels;

3. After the third city he was left with ${4\over 5}\cdot{1\over 2}x={2\over 5}x$ barrels.

${2\over 5}x=72\iff x=180$ barrels.[/hide]

[hide]He ended up with 72 barrels after 1/5 of them were taken, so $72=4/5*x=90.$ $90=3/4*y=120$, and $120=2/3*z=180$
[/hide]

[/hide]4/5*3/4*2/3=2/5. This represents the fraction of wine that was taken from him. 72 is 2/5ths of [b]180[/b][hide][/hide]

[quote="moldlee"]A wine dealer, who carries his wine in identical barrels, travels through

three cities. Each city charges a different tax which is collected upon

entrance into the city. The first city takes $\frac{1}{3}$  of his wine

as a tax, the second city takes $\frac{1}{4}$  of his wine, and

 the third city takes $\frac{1}{5}$  of his wine.

After paying all these taxes he sold all of his remaining 72  barrels

of wine. How many barrels of wine did he have originally?[/quote]
$\frac{4}{5}\cdot{3}{4}\cdot{2}{3}\cdot 72=\frac{2}{5}\cdot 72=\boxed{180}$


limit
function
calculus
derivative
geometry
3D geometry
calculus computations

My book states the following:

If the alternating series

$\sum_{n=1}^{\infty} (-1)^{n-1} b_n = b_1 - b_2 + b_3 - b_4 + ... (b_n > 0)$
satisfies
(i) $b_{n+1} \leq b_n$ for all $n$

(ii) $\lim_{n \rightarrow \infty} b_n = 0$

then the series is convergent.

That much I understand....I read the proof and it makes sense to me.

One of the examples says to test the following series for convergence or divergence.
$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n^2}{n^3 + 1}$

It says to find out if $b_n = \frac{n^2}{n^3 + 1}$ is decreasing we take the function $f(x) = \frac{x^2}{x^3 + 1}$
And since the derivative is $f'(x) = \frac{x(2-x^3)}{{x^3 +1}^2}$ is decreasing on the interval (cube root 2, $\infty$) which means that $b_{n+1} < b_n$ when $n \geq 2$

It goes on to show the limit as n approaches infinity is zero, so the given series converges....

But in the definition of the test, it says it should be $b_{n+1} < b_n$ for all $n$, yet this example considers only $n$ greater or equal to 2. Doesn't it really matter if $b_n$ is ultimately decreasing, or is it for all $n$?

It seems that this example conflicts with the definition....

If you remove  a finite number of terms from a series, the nature of the series remains the same. If the series you get from removing the first $2$ terms converges then the initial series converges as well.

Okay, thanks

That's what I thought because a definite number of terms would converge, so it shouldn't matter to add that to a convergent series...


Given the function $ f : \mathbb{R} \rightarrow \mathbb{R}$ such that $ f(x^3 \plus{} y^3) \equal{} (x\plus{}y)(f(x)^2 \minus{} f(x)f(y) \plus{} f(y)^2)$, $ \forall x,y \in \mathbb{R}$. Prove that  $ f(2010x)\equal{}2010f(x)$.

[quote="rajelaz"]There's function $ f(x) \mathbb{R} \rightarrow \mathbb{R}$ such that $ f(x^3 \plus{} y^3) \equal{} (x \plus{} y)(f(x)^2 \minus{} f(x)f(y) \plus{} f(y)^2)$ , $ \forall x,y \in \mathbb{R}$ .. prove that  $ f(2010x) \equal{} 2010f(x)$[/quote]
Let $ P(x,y)$ be the assertion $ f(x^3\plus{}y^3)\equal{}(x\plus{}y)(f(x)^2\minus{}f(x)f(y)\plus{}f(y)^2)$
Let $ \mathbb S\equal{}\{x\in\mathbb R$ such that $ f(xy)\equal{}xf(y)$ $ \forall y\in\mathbb R\}$

$ 1\in\mathbb S$
$ P(0,0)$ $ \implies$ $ f(0)\equal{}0$ and $ 0\in\mathbb S$
$ P(\sqrt[3]x,0)$ $ \implies$ $ f(x)\equal{}\sqrt[3]xf(\sqrt[3]x)^2$ and so $ f(x)$ and $ x$ have same signs.
If $ p,q\in\mathbb S$, then $ P(p\sqrt[3]x,q\sqrt[3]x)$ $ \implies$ $ p^3\plus{}q^3\in\mathbb S$ (1)

If $ p\in\mathbb S$, then :
$ P(x,0)$ $ \implies$ $ f(x^3)\equal{}xf(x)^2$
$ P(\sqrt[3]px,0)$ $ \implies$ $ pf(x^3)\equal{}f(px^3)\equal{}\sqrt[3]pxf(\sqrt[3]px)^2$
Comparing these two lines, we get, for $ x\neq 0$, and using the fact that $ f(x)$ and $ x$ have same signs  : $ f(\sqrt[3]px)\equal{}\sqrt[3]pf(x)$ and so $ \sqrt[3]p\in\mathbb S$ (2)

Now :
$ p\in\mathbb S$ $ \implies$ $ \sqrt[3]p\in\mathbb S$ (using (2))
$ \sqrt[3]p\in\mathbb S$ and $ 1\in\mathbb S$ $ \implies$ $ p\plus{}1\in\mathbb S$ (using (1))

So $ p\in\mathbb S$ $ \implies$ $ p\plus{}1\in\mathbb S$, and, since $ 1\in\mathbb S$, $ \mathbb N\subseteq\mathbb S$ and so $ 2010\in\mathbb S$
Q.E.D.


"Most fractions look better in \displaystyle (remember, you don't need the \displaystyle declaration if you are in \[...\ or \$\$...\$\$ mode.)"
it doesn't work! \[(\frac{2-x}{2})^{2}=2x\] but
$ \left(\frac{2-x}{2}\right)^{2}=2x$
why??

Parentheses will never get bigger or smaller on their own (as far as I know) -- you always have to use \left and \right if you want them bigger.

To answer the question: on the forum we have displaystyle on automatically.


geometry
3D geometry
analytic geometry
greatest common divisor

A 150X324X375  rectangular solid is made by gluing together 1X1X1  cubes. An internal diagonal of this solid passes through the interiors of how many of the 1X1X1  cubes?

[quote="Kamil Witek"]This is just like the chessboard and a line problem, just in three dimensions, so to get the answer wouldn't you just add the dimentions of the figure?[/quote]
[hide="I am pretty sure the answer is:"]
\[150+324+375-\left[ (150,324)+(324,375)+(375,150) \right]+(150,324,375) = \boxed{768}\] where $(a,b)$ is the greatest common factor of $a$ and $b$. We count the number of times the line must pass through a lattice plane (integer constant), then subtract the number of times that it goes through an edge of two cubes (i.e. two of three coordinates are integers); but we overcount, so by the Principle of Inclusion-Exclusion we have to add back the number of times the diagonal goes through a vertex of a cube. [/hide]


limit
induction
real analysis
real analysis unsolved

Let $I=(0,1)$, and let $C$ be a countable subset of $I$. Also let $(a_{n})_{n\ge 1}$ be a sequence of positive reals with $\sum a_{n}\le 1$.

Prove that we can place disjoint compact intervals $I_{n}, n\ge 1$ inside $I$ such that the length of $I_{n}$ is $a_{n}$ for all $n\ge 1$ and every point in $C$ belongs to the interior of some $I_{n}$.

We consider $(x_{n})$ a sequence s.t. every number in $C$ appears exactly once in the sequence.
Firstly, we consider the case $\sum_{n\geq 1}a_{n}=a\in(0,1)$
$I_{n}: =[c_{n},d_{n}], \forall n\geq 1$
Let  $(b_{n})_{n\geq 1}$ be a sequence of positive reals s.t. $\sum_{n\geq 1}b_{n}=1-a$
We take $I_{1}=[b_{1},b_{1}+a_{1}]$ and $J_{n}=\bigcup_{k=1}^{n}I_{k}$.

Let $M=\{c_{n}\mid n\in \mathbb{N}\}$
We assume we have fixed the intervals $I_{1},...,I_{k-1}$
Let $t$ be the smallest integer for which $x_{t}\notin J_{k-1}$

1)If $x_{t}< \max M$ and $a_{k}<l$ then we can position $I_{k}$ in that open interval (of length $l$) s.t. $x_{t}\in I_{k}$ and $I_{k}\bigcap J_{k-1}=\emptyset$.

2) $x_{t}\ge \max M$ or $a_{k}\ge l$
$I_{k}=[ d_{1}+b_{2}+a_{2}+b_{3}+a_{3}+b_{4}+a_{4}+...+a_{k-1}+b_{k-1}+b_{k}, d_{1}+b_{2}+a_{2}+b_{3}+a_{3}+b_{4}+a_{4}+...+a_{k-1}+b_{k-1}+b_{k}+a_{k}]$

$(I_{2}=[d_{1}+b_{2}, d_{1}+b_{2}+a_{2}], I_{3}=[d_{1}+b_{2}+a_{2}+b_{3}, d_{1}+b_{2}+a_{2}+b_{3}+a_{3}], I_{4}=[ d_{1}+b_{2}+a_{2}+b_{3}+a_{3}+b_{4}, d_{1}+b_{2}+a_{2}+b_{3}+a_{3}+b_{4}+a_{4}])$

Now we prove that this sequence verifies the conditions of the problem.
Let $n\in \mathbb{N}$. We prove that there exists $k\in \mathbb{N}$ s.t. $x_{n}\in I_{k}$.
Assume that $x_{n}\notin I_{k}, \forall k\in \mathbb{N}$ and $n$ is the smallest with this property.
We have now 2 cases:
I) $\sup\{c_{n}\mid n\in \mathbb{N}\}=1$
Let $k\in \mathbb{N^{*}}$ s.t. $x_{n}\le c_{k}$. $\Rightarrow x_{n}$ is in an uncovered interval of length $t_{n}$. $\exists a_{i}\le t_{n}, i\ge k$, because we have $\lim_{n\rightarrow \infty}a_{n}=0$. Then because of the way we constructed the sequence, we have $x_{n}\in I_{i}$, contradiction.
II) $\sup\{c_{n}\mid n\in \mathbb{N}\}<1$
Then there exists the integer $n$ s.t. $c_{k}\le c_{n}, \forall k\ge n$
It follows that $\forall k\ge n$ we have the first case from the induction, so $C\subset \bigcup_{n\geq 1}I_{n}$

In all cases we proved that the sequence $(I_{n})$verifies the conditions of the problem.

I do not have a proof for the second case ($\sum_{n\geq 1}a_{n}=1$)
I hope this is correct even if it doesn’t look very clear.

The case $\sum_{n\ge 1}a_{n}=1$ is interesting even if you forget about the countable set and just try to fit disjoint closed intervals of length $a_{n}$ into $(0,1)$.

I have managed to find a way of fitting disjoint closed intervals of length $a_{n}$ in $(0,1)$ in the case $\sum_{n\geq 1}a_{n}=1$, but it doesn’t fulfil the countable set conditions. I will probably post it if I do not succeed in solving this case.

[quote="ovidius"]
1)	$x_{1}<c_{1}$ and $a_{2}< l$, where $l$ is the length of the interval not covered by $J_{1}$ in which is $x_{1}$.
In this case, we take $I_{2}$ in this interval s.t. $x_{1}\in I_{2}$ and $I_{2}\bigcup J_{1}=\emptyset$ and  the next interval from $I_{1}$ to the right (if there will be such an interval denoted by $I_{k}$) will be fixed such that $c_{k}-d_{1}=b_{2}+t$, where t is the sum of the lengths of the intervals positioned left from the rightest interval.

.[/quote]
Hi,can you explain how you constructed $I_{2}$ in this case?

PS:I think the original problem call for $x_{n}$ is the $interior$ of some $I_{n}$.

I have edited it. I hope it is more clear now  :)

I know how you construct in case(2).
But what you edited still not explain for case (1).

Any way,is x_k should must be the interior in your constructing...? :?:

In case (1), $x_{t}$ is in an uncovered open interval (of length $l$). And we position $I_{k}$ such that $x_{t}\in I_{k}$ and $I_{k}\bigcap J_{k-1}=\emptyset$.

Considering the problem of fitting disjoint compact intervals in $(0,1)$ and ignoring the countable set, this is the idea for the case $\sum a_{n}=1$:
We split the sequence $(a_{n})$ into 2 subsequences: $a_{2},a_{4},a_{6},. . .$ and $a_{3},a_{5},a_{7},. . .$, with$u=\sum_{k\geq 1}a_{2k}$ and $v=\sum_{k\geq 1}a_{2k+1}$.
Now we fix $I_{1}=[u,1-v]$, and we have the intervals $(0,u)$ and $(1-v,1)$ and the subsequences above.We repeat the way we have fixed $I_{1}$ in $(0,1)$ for the interval $(0,u), I_{2}$ and $a_{2},a_{4},a_{6},. . .$  and for $(1-v,1), I_{3}$ and $a_{3},a_{5},a_{7},. ..$We proceed by induction  and we have $I_{n}$ positioned in $(0,1) , \forall n\geq 1$, with $I_{k}\bigcap I_{j}=\emptyset , \forall k,j\geq 1, k\not =j$.

It gets more complicated when we must have $C\subset \bigcup_{n\geq 1}I_{n}$  :(


Hello everyone,

How do you determine with proofs which of the following infinite sets are countable and which is uncountable:

A) The set of all triples (x, y, z) where x, y, and z are rationals
B) The set of all subsets of N (N = Natural numbers)
C) The set of all finite subsets of N (N = Natural numbers)

The book only tells me if they are countable or not like some sort of rule but I like to see the proof for all these, can anyone help me please ?

Basic facts about countability and cardinality:

- $ \mathbb{N}$ is countable.
- A countable union of countable sets is countable.  (In particular, $ \mathbb{Q}$ is countable, so A) is countable, as is, I think, C).)
- The power set of a set $ S$ has strictly greater cardinality than $ S$.  (In particular, B) is uncountable.)

We have an easy bijection $ \mathbb{N}^2\to\mathbb{N},(n,m)\mapsto 2^{n\minus{}1}(2m\minus{}1)$.
By induction we find a bijection $ f_n: \mathbb{N}\to\mathbb{N}^n$ for all $ n\in\mathbb{N}$.

Let $ M_n$ be the set of all subsets of $ \mathbb{N}$ with at most $ n$ elements for all $ n\in\mathbb{N}$
and $ M: \equal{}\cup_{n\equal{}0}^\infty M_n$ the set of all finite subsets of $ \mathbb{N}$.
We have a surjection $ g_n: \mathbb{N}^n\to M_n,(k_1,\ldots,k_n)\mapsto \{k_1,\ldots,k_n\}$ for all $ n\in\mathbb{N}$.
This gives a surjection $ h: \mathbb{N}^2\to M,(n,m)\mapsto g_n(f_n(m))$,
and a surjection $ h\circ f_2: \mathbb{N}\to M$.
This gives C).

Another bijection for (c) is the following:
$ (a_{1}, a_{2}, a_{3} ... a_{k}) \mapsto 2^{a_{1}} + 2^{a_{2}} + ... + 2^{a_{k}}$
which is clearly injective, if we look at things in base 2.
The slightly stronger statement is true as well:
All sets, constituting of elements from $ \mathbb{Z}$ (or any countable set), but where an element can occur multiple times (but only finitely many times), is also countable (so these are like subsets, except elements can occur more than once).

Firstly, I would like to say thank you to the people who have made inputs to my questions.

I still have some things to clarify.

For A) It is countable, because the reason is "A countable union of countable sets is countable. In particular Q". Can someone write out the proof to help me to clarify it ?

For B) It is uncountable. The reason is "The power set of a set S has strictly greater cardinality than S". Can someone write out the proof to help me to clarify it please ? Also, is it possible to use Cantor Diagonalisation to show it is uncountable, if so how ?

For C) I understand why it is countable now, thanks.


The slope of the line $ \frac{x}{3} \plus{} \frac{y}{2} \equal{} 1$ is:

$ \textbf{(A)} \minus{}\frac32 \qquad \textbf{(B)} \minus{}\frac23 \qquad \textbf{(C)} \frac13 \qquad \textbf{(D)} \frac23 \qquad \textbf{(E)} \frac32$

[hide="Click for solution"]$ \frac {x}{3} \plus{} \frac {y}{2} \equal{} 1 \implies 2x \plus{} 3y \equal{} 6 \implies 3y \equal{} \minus{} 2x \plus{} 6 \implies y \equal{} \frac23 x \plus{} 2 \implies \boxed{\textbf{(B)}\ \minus{}\frac23}$[/hide]


You are looking for Two numbers. Here are the hints
Both integers are even
The sum of the integers is negative
The product of the integers is -96
Neither integer is a square
Neither integer is a factor of the other
Neither integer is a cube.
THE END
Solve the problem :D
Good luck! (Its fairly simple)
EDIT- Ahem... not necessarily 2 digits... my bad... Thank you for pointing that out JRav

How can you have 2 two-digit integers whose product is -96?  The minimum two-digit number is 10, and $ 10\times 10\equal{}100$, so that is impossible.

Again, I'm still not sure there's a solution.  Every pair of factors of -96 violates one of the conditions.

[quote="joyofpi"]I think that THIS is the answer...
-12, and 8.[/quote]

The number 8 is a cube of 2.

$ \minus{} 96 \equal{} \minus{} 1*2^5*3$

since neither integer is factor of one another, and since their sum is a negative number, its -32 and 3.

answer : -32 and 3

EDIT:

oh... didnt see that both numbers are even

its [hide] before you look [hide="here"]-16,6[/hide], note that numbers like -1, -4, and -9 are not perfect squares[/hide]

fantasy's anser:
the #s must be even

Congratz!
I have mislead you all to wrong answers...

[quote="QuincyCello"]isn't it just 6, -16? guess and check...[/quote]

Please do not post the exact same answer.

96/4=26 so two numbers must multiply to 24 one of which is negative.

1,24
2,12
3,8
4,6

Multiply by 2 to make all even,

2,48<---2 is factor of 48
4,24<-----FACTOR!
6.16
8.12<----PERFECT CUBE!

In order for 16 not to be a square, it must be negative so as posted above, 6,-16.

Is -2 a factor of 48?

Also Quincy, i herd you liek mudkipz?

[quote="joyofpi"]You are looking for Two numbers. Here are the hints
1) Both integers are even
2) The sum of the integers is negative
3) The product of the integers is -96
4) Neither integer is a square
5) Neither integer is a factor of the other
6) Neither integer is a cube.
THE END
Solve the problem :D
Good luck! (Its fairly simple)
EDIT- Ahem... not necessarily 2 digits... my bad... Thank you for pointing that out JRav[/quote]

[hide="Proof"]Suppose that our two integers are a and b.
By (3) one of the integers is negative.
Now suppose that the negative number is a.
By (2), |a|>|b|
By (1), we can make the substitution: a=2c, b=2d
Where cd=-24
d cannot equal to 2 (or else it violates (4))
|c|,|d| cannot equal to 4 (or else it violates (6))
Remember that c is negative.
The only integer pairs left (c,d) are:
(-1,24),(-2,12),(-3,8),(-4,6),(-8,3),(-24,1)
We can now remove those that are factors of each other:
(-3,8),(-4,6),(-8,3)
Also remember from (2). The only (c,d) pair that satisfies all criteria is (-8,3)
which leads us that the two numbers are -16 and 6.

QED[/hide]


prove that the equation $ x^2\plus{}y^2\plus{}1\equal{}z^2$ has infinitely many solution

[quote="aadil"]prove that the equation $ x^2 \plus{} y^2 \plus{} 1 \equal{} z^2$ has infinitely many solution[/quote]

$ (2k^2)^2\plus{}(2k)^2\plus{}1\equal{}(2k^2\plus{}1)^2$

So $ (x,y,z)\equal{}(2k^2,2k,2k^2\plus{}1)$ is solution for all $ k$, done.

[hide="a much less intelligent approach..."]Since $ y^2 \plus{} 1 \equal{} z^2$ has infinitely many solutions,

$ (0,k,\sqrt {k^2 \plus{} 1})$ works for $ x^2 \plus{} y^2 \plus{} 1 \equal{} z^2$.[/hide]

EDIT: Sorry, too careless  :blush:

I suppose he meant infinitely many solutions in integers, which invalidates chrischris's proof but stephencheng's proof remains correct.

(Otherwise, you can just set $ z\equal{}\sqrt{x^2\plus{}y^2\plus{}1}$ and then [i]any[/i] ordered pair $ (x,y)$ works.)

[quote="Yongyi781"]I suppose he meant infinitely many solutions in integers, which invalidates chrischris's proof but stephencheng's proof remains correct.

(Otherwise, you can just set $ z \equal{} \sqrt {x^2 \plus{} y^2 \plus{} 1}$ and then [i]any[/i] ordered pair $ (x,y)$ works.)[/quote]

0 is a integer. You meant naturals?

$ (0,0,1)$ is the only solution that contains 0's. :maybe:

[hide="Discussion"] One way to solve this problem is to write it as $ x^2 \plus{} 1 \equal{} z^2 \minus{} y^2 \equal{} (z \minus{} y)(z \plus{} y)$.  Then it's reasonable to set $ z \minus{} y \equal{} 1, z \plus{} y \equal{} x^2 \plus{} 1$ for $ x \equal{} 2k$ even, which gives stephencheng's parameterization.

You can use similar techniques to deduce the parameterization of the [url=http://en.wikipedia.org/wiki/Pythagorean_quadruple]Pythagorean quadruples[/url], and then it remains merely to specialize. [/hide]


function
algebra
polynomial
integration
real analysis
real analysis unsolved

Let $a,b\in\mathbb{R}, a<b$. Let $f(x)$ be a continuous function on $]a,b[$. Prove that for every $\epsilon>0$ there exists a polynomial $P$ such that
\[\int_{a}^{b}|f(x)-P(x)|<\epsilon. \]

does it work with $f(x)=\frac{1}{x}$ on $]0;1[$ ?

alekk has a point. You'd better add a condition on $f$ beyond merely being continuous on the open interval. You need some kind of control of behavior at the endpoints.

Now, it turns out that a sufficient condition would be for $f$ to be absolutely integrable on $(a,b)$ - that is, that

$\int_{a}^{b}|f(x)|\,dx<\infty.$

But once you put in this condition, you'll find that you don't really need for $f$ to be continuous - the continuity doesn't even help.


When the mean, median, and mode of the list
10; 2; 5; 2; 4; 2; x
are arranged in increasing order, they form a non-constant arithmetic progres-
sion. What is the sum of all possible real value of x ?

(A) 3 (B) 6 (C) 9 (D) 17 (E) 20

I was really confused by non-constant arithmetic progression  :!:

It was actually #23
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=438929#438929


geometry
circumcircle
search
perpendicular bisector

I need help with these 5 geometry problems:

1. A semicircle Γ is drawn on one side of a straight line l. C and D are points on Γ. The tangents to Γ at C and D meet l at B and A respectively, with the center of the semicircle between them. Let E be the point of intersection of AC and BD, and F the point on l such that EF is perpendicular to l. Prove that EF bisects angle CFD.

2. (1995 IMO) Let A, B, C, and D be distinct points on a line, in that order. The circles with diameters AC and BD intersect at X and Y. O is an arbitrary point on the line XY but not on AD. CO intersects the circle with diameter AC again at M, and BO intersects the other circle again at N. Prove that the lines AM, DN, and XY are concurrent.

3. (2007 IMO) In triangle ABC the bisector of angle BCA intersects the circumcircle again at R, the perpendicular bisector of BC at P, and the perpendicular bisector of AC at Q. The midpoint of BC is K and the midpoint of AC is L. Prove that the triangles RPK and RQL have the same area.

4. In the plane we are given two circles intersecting at X and Y. Prove that there exist four points P, Q, R, S with the following property: For every circle touching the two given circles at A and B, and meeting the lines XY at C and D, each of the lines AC, AD, BC, BD passes through one of these points.

5. (2000 IMO) Two circles G1 and G2 intersect at M and N. Let AB be the line tangent to these circles at A and B, respectively, such that M lies closer to AB than N. Let CD be the line parallel to AB and passing through M, with C on G1 and D on G2. Lines AC and BD meet at E; lines AN and CD meet at P; lines BN and CD meet at Q. Show that EP = EQ.

Thanks!

 :)  :)  :)

Still, can anyone solve 1, its driving me crazy  :help:  :bomb:  :wallbash:

[quote="SHSMCMXCIV"]1. A semicircle Γ is drawn on one side of a straight line l. C and D are points on Γ. The tangents to Γ at C and D meet l at B and A respectively, with the center of the semicircle between them. Let E be the point of intersection of AC and BD, and F the point on l such that EF is perpendicular to l. Prove that EF bisects angle CFD.
[/quote]
Look here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=352892#p352892


algebra
polynomial
floor function
ceiling function
logarithms
search
number theory unsolved

Does there exist a positive integer $ n$ such that $ n!$ starts with $ 2005$?

Yes. the sequence {n!}, n=1,2,3,... is what one calls a left-normal sequence, i.e. it can start with any string of digits in any base.

Other left normal sequences:
{p_n}, the nth prime
{F_n}, the nth Fibonacci number
{P(n)}, where P is a nonconstant polynomial in Z[x]

Suhaimi

I have never heard anything like that before. :blush:

Can you prove it? And do you have any reference about left-normal sequence?

The first four decimal digits of $ n!$ can be written as
\[ d \equal{} \left\lfloor \frac {n!}{10^{\left\lceil\log_{10} n!\right\rceil \minus{} 4}}\right\rfloor
\]
Replace $ n!$ by Stirling's approximation (which agrees to more than four digits for sufficiently large $ n$), and then use the fact that $ \lbrace an\minus{}\lfloor an \rfloor \rbrace_{n\in\mathbb{N}}$ is dense in $ (0,1)$ for any irrational $ a$.

The first few values are $ 4968!\approx 2.0059\cdot 10^{16207}$, $ 6230!\approx 2.0056\cdot 10^{20936}$, $ 6527!\approx 2.0050\cdot 10^{22066}$.  There are infinitely many such values, and the argument works not just for $ 2005$ but any integer.

I get this problem from a competition in Singapore for secondary school student (about 15 years old). I think there must be a simpler solution, correct?

Have a look here for a more general statement
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=237215850&t=23417

Pierre.


Find all positive intergers n, satisfying the following condition:
for all pairs of the positive intergers p, q, if p+q is a square number , then p^n + q^n is also a square.

I suppose the only solution to the problem is n=1, but I haven't got the proof, I'll appreciate that if anyone could help me.

$ 1 \plus{} (k^2 \minus{} 1)$ is a square for all $ k$, but $ 1^n \plus{} (k^2 \minus{} 1)^n$ can't be a square for sufficiently large $ k$; it's the standard bounding trick.  Alternately, the Mason-Stothers theorem might work, since if the above is a square for all $ k$ it must be the square of a polynomial.

I've tried that idea, but I still can't get the exact proof. Could you explain it more clearly?

Okay, the bounding trick is a little annoying to work out for all $ n$, so I'll just use the Mason-Stothers theorem.  It's a well-known result (posted on the board many times) that if a polynomial such as $ P(k) \equal{} 1 \plus{} (k^2 \minus{} 1)^n$ is square for all positive integers $ k$ then it must in fact be the square of a polynomial $ Q(k)$, hence

$ 1 \plus{} (k^2 \minus{} 1)^n \equal{} Q(k)^2$

as a polynomial identity.  By the [url=http://mathworld.wolfram.com/MasonsTheorem.html]Mason-Stothers theorem[/url], $ (k^2 \minus{} 1)^n Q(k)^2$ must have at least $ 2n \plus{} 1$ distinct roots.  But since $ Q(k)$ has degree $ n$, this polynomial can have at most $ n \plus{} 2$ distinct roots.  It follows that $ n \le 1$.


function
integration
calculus
derivative
calculus computations

Can anyone tell me how to solve a ordinary differential equation?

I know this is kind of easy, but it's rather common in calculus.

 :lol:  :blush:

You provided no examples. There are assorted techniques. You must [b]classify[/b] your problem.

First classifying question: What is the order of the equation? First order? Or second or higher order? (Note that there's some room for creativity here: $y''+k(y')^{2}=g$ is ostensibly a second order equation but we can view it as a first order equation for the function $y'.$)

If it is first order: now ask whether it is linear. That is, can it be written as $y'+p(x)y=q(x)?$ If so, then put it into that standard form and use the multiplying factor technique. (Multiply both sides by $\mu(x)=e^{\int p(x)\,dx}$ and then recognize the left hand side as the derivative of a product.)

If it is not linear, then ask if it is separable. That is, can it be factored as $y'=f(x)g(y).$ If so, then employ the technique of separation of variables.

Equations that are both linear and separable pose no problem; both methods work (or at least they "work" if we can do the integrals involved).

Many first order differential equations are neither linear nor separable. There are a number of other possible narrower classifications, each bringing its own specialized technique. We will not be able to usefully classify all of them.

Now, what if the equation is second or higher order? We start by asking if it's linear - hoping the answer is "yes" because there are few available techniques for nonlinear equations. If linear we ask if it is constant coefficient, and we ask whether it is homogenous or nonhomogeneous. Our answers to each of these classifying questions bring various techniques into play.


geometry
3D geometry
tetrahedron
geometric transformation
reflection
AMC

Can someone restate the problem for the tetrahedron(i dont have my booklet) because i got 18, and i just want to know how i got the extra 3

heres what i have
4 faces same color(like BBBB) 3 choices
3 same(like BBBW), 1 different 6 choices
2 same 2 same (so like BBWW) 3 choices
2 same 2 different (so like BBWR)-is this suppose to be 3 choices, because i get BBRW RRWB WWBR, but cant you also have BBWR or does it say the tetrahedrons can be reflections of eachother?

Yeah for 3 colors, there are 3 possibilities.

Think of it this way. If you have BBWR. Lay the tetrahedron so that the edge between the two B faces is on the ground horizontally. Then the face on the left top is W and the right top is R. Buti f you spin it 180 degrees about the y axis, R is on the left and W is on the left, which is BBRW, so BBWR=BBRW

dang.. if i ddint think about it i wouldve guessed it right.. i guessed A first because i have 1 other A up to that point in the test..


IMO Shortlist
inequalities unsolved
inequalities

It's IMO Shortlist 2001 algebra problem 3. See http://www.mathlinks.ro/Forum/viewtopic.php?t=17449 , http://www.kalva.demon.co.uk/short/soln/sh01a3.html , or http://www.mathlinks.ro/Forum/viewtopic.php?t=15624 .

  darij


On the number line, point $ A$ is located at 0, point $ B$ is located at 4, and point $ C$ is located at 6.
[asy]defaultpen(1);
draw((0,0)--(6,0));
for (int i=0;i<7;++i){
  draw((i,-.1)--(i,.1));
  label(string(i),(i,-.1),(0,-1));
}
label("\(A\)",(0,0),(0,1));
label("\(B\)",(4,0),(0,1));
label("\(C\)",(6,0),(0,1));[/asy]
A dart randomly lands somewhere on the number line between $ A$ and $ C$.  What is the probability that it lands closer to $ B$ than it does to $ A$ and $ C$?

The distance between the midpoint of A and B, and the midpoint of B and C yields where the dart would be closer to B than the other two.

This is from 2 to 5, or 3 units. There is a total of 6 units where the dart could land, so the probability is $ \frac{3}{6} \equal{} \frac{1}{2}$