trigonometry
analytic geometry
quadratics
geometry
circumcircle
ratio
angle bisector

It's always nice to see the different ways people approach problems and about 4 more solutions to #14 have already been given on another thread so I figured, "Why not open up a thread for everyone to share their insights!?" So here it is, please give at least an outline of your solution.

I'll start things off by bringing over one of those solutions to #14(mine of course.)
Number 14:
Before we start, let it be noted that the infamous 13-14-15 triangle is just a 5-12-13 right triangle and a 9-12-15 right triangle being put together with the sides of 12 put together as the height and the sides of 5 and 9 running together to make the side of 14. Using this fact we can easily find the sines and cosines of the angles of the 13-14-15 triangle.

Say $\angle DAC = \theta$
First we must find $\tan\theta$. Reason will be obvious later.
Given $\Delta ADC$, we drop the altitude from $D$ to a point $R$, so $\tan\theta = \frac{DR}{AR} = \frac{DR}{AC-RC}$. We know $\sin(C) = \frac{4}{5}$ and $\cos(C) = \frac{3}{5}$. These give us two facts: $\sin(C) = \frac{4}{5} = \frac{DR}{6} \Rightarrow DR = \frac{24}{5}$ and $\cos(C) = \frac{3}{5} = \frac{RC}{6} \Rightarrow RC = \frac{18}{5}$ Using these, we get $\tan\theta = \frac{\frac{24}{5}}{14-\frac{18}{5}} = \frac{6}{13}$.
Now to set up a coordinate system.
Points:
$A(0,0),B(5,12),$ and $C(14,0)$ give us
Lines:
$\displaystyle AD \Rightarrow y=\tan(\theta)x = \frac{6}{13}x
\\
AB \Rightarrow y=\tan(A)x=\frac{12}{5}x
\\
AE \Rightarrow y=\tan(A-\theta)x
\\
BC \Rightarrow \frac{y-0}{x-14}=\frac{12-0}{5-14} \Rightarrow y = -\frac{4}{3}(x-14)$
Also, $\displaystyle\tan(A-\theta)=\frac{\tan(A)-\tan(\theta)}{1+\tan(A)\tan(\theta)}=\frac{\frac{12}{5}-\frac{6}{13}}{1+\frac{12}{5}\frac{6}{13}}=\frac{12(13)-6(5)}{13(5)+12(6)}=\frac{126}{137}$
We find point E based on the intersection of lines BC and AE, but we only need find the x-coordinate of E.
\begin{eqnarray*}
y=\frac{126}{137}x&=&-\frac{4}{3}(x-14)\\
\frac{189}{274}x&=&(14-x)\\
\frac{463}{274}x&=&14\\
x&=&14\left(\frac{274}{463}\right)=\frac{3836}{463}\\
\end{eqnarray*}
If we drop the altitude from $B$ to a point $P$, and if we drop the perpendicular from $E$ to $AC$ we get a point $Q$. Since $\angle BPC \cong \angle EQC \cong 90$ and $\angle BCP \cong \angle ECQ$, $\Delta EQC \sim \Delta BPC$. Also, $CQ = AC - AQ = 14 - x$ where $x$ is the x-coordinate of E. By similir triangles,
\begin{eqnarray*}
\frac{EC}{BC}&=&\frac{QC}{PC}\\
EC&=&BC\frac{14-x}{9}\\
BC-BE&=&BC\frac{14-x}{9}\\
BE&=&BC\left(1-\frac{14-x}{9}\right)\\
BE&=&5\left(\frac{x-5}{3}\right)\\
BE&=&5\left(\frac{\frac{3836}{463}-5}{3}\right)\\
BE&=&5\left(\frac{1521}{3(463)}\right)\\
BE&=&5\left(\frac{507}{463}\right)\\
\end{eqnarray*}

OK, I follow your step.

Here is another solution for #14.

Let $F$ be on $BC$ such that $AF$ is the angle bisector of $\angle{BAC}$, then it's also the angle bisector of $\angle{EAD}$.

By the angle bisector theorem:

$\frac{AE}{AD}=\frac{EF}{FD}=\frac{BF-BE}{FC-DC}$

Using the Law of Sines in triangles $\Delta{ABE}$ and $\Delta{ADC}$, we have:

$AE = BE\cdot\frac{\sin{B}}{\sin{\angle{BAE}}}$

$AD = DC\cdot\frac{\sin{C}}{\sin{\angle{CAD}}}$

So $\frac{AE}{AD}=\frac{BE}{DC}\cdot\frac{\sin{B}}{\sin{C}}=\frac{BE}{DC}\cdot\frac{AC}{AB}$

Therefore $\frac{BF-BE}{FC-DC}=\frac{BE}{DC}\cdot\frac{AC}{AB}$

This is a single variable equation in $BE=x$ where $BF=\frac{13\cdot15}{27}$ and $FC=\frac{14\cdot15}{27}$ can be easily calculated by the angle bisector theorem on $\Delta{BAC}$.

The equation:

$\frac{\frac{13\cdot15}{27} - x}{\frac{14\cdot15}{27} - 6} = \frac{x}{6}\cdot\frac{14}{13}$

Solve, $x = \frac{13^2\cdot 15}{463}$.

Here's a really fast way to do #12:
Scale down the figure by 100(so AB=9, EF=4)

Note that $\angle FAO = \angle EBO = 45^{\circ}$, $\angle AOF = \angle OEB = 45^{\circ} +\angle AOE$, therefore $\triangle AOF \sim \triangle BEO$, so

$\frac{AO}{BE} = \frac{AF}{BO}$. Let $BF=x$, since $AO = BO = \frac{9\sqrt{2}}{2}$, we have

$AO(BO) = \frac{81}{2} = BE*(AF) = (4+x)(9-x)$

Simplify we get

$2x^2 - 10x + 9 = 0$, solve the quadratic we can get $x = \frac{5 + \sqrt{7}}{2}$

Multiply the 100 back we have it to be $250 + 50\sqrt{7}$, 250+50+7=307

[quote="fuzzylogic"][quote="beta"]
$\triangle AOF \sim \triangle BEO$
[/quote]

beta, nice observation! Once you discovered this, the rest became straightforward.[/quote]

It makes this problem seems really simple too(similarity and you're done).

Here is another way for #14 (correction and more details):

Extend $AD$ to meet the circumcircle of the triangle $\Delta{ABC}$ at $F$.

By power of a point, we have $DF = \frac{BD\cdot DC}{AD}$, So $AF=AD+DF= \frac{AD^2+BD\cdot DC}{AD}$

Since $\Delta{CFD} \sim \Delta{ABD}$, we have $CF = \frac{AB\cdot CD}{AD}$.

Since $\Delta{ABE} \sim \Delta{AFC}$,  we have $EB = \frac{AB\cdot CF}{AF} = \frac{AB^2\cdot CD}{AD^2+BD\cdot DC}$

By Stewart's theorem: $BC\cdot (AD^2+BD\cdot DC) = AC^2\cdot BD + AB^2\cdot CD$

Therefore, $EB =  \frac{AB^2\cdot CD\cdot BC}{AC^2\cdot BD + AB^2\cdot CD} = \frac{13^2\cdot6\cdot15}{14^2\cdot9+13^2\cdot6}=\frac{13^2\cdot15}{463}$

No trig, no equation, no special right triangles, pure geometry and simple calculation!

Yet another method for #14.  No trig.  Just similar triangles, one unpleasant application of the Pythagorean Theorem, and a little algebra.  [Click the image to view it at a legible size.]

Figure (a) breaks the 13-14-15 triangle down to a 5-12-13 and a 9-12-15.

Figure (b) shows a triangle similar to the 9-12-15.

Figure (c) shows that we can find the legs of a right triangle containing the angle of interest.  The legs are in the ratio of 6:13.  A similar triangle with one leg equal to [i]p[/i] is shown.

Figure (d) shows the altitude to the 13" side.  The altitude is known because we know the area of the triangle is 84.  After a slightly nasty application of the Pythagorean Theorem, we find the right triangle shown is in the proportions 99:168:195.  A smaller, similar triangle, with one leg equal to [i]q[/i] is shown.

Figure (e) combines the two triangles with legs [i]p[/i] and [i]q[/i].  These two triangles have the same altitude and have a combined base equal to 13.  We have two linear equations in two unknowns, which we can solve for [i]q[/i].  [Omitted, but not very tedious.]  Then we compute (195/99)[i]q[/i] to find the denominator of the required length is 463.

Here is yet another approach to #14:

Draw DG parallel to AC with G on AB.  Since triangle GBD is similar to triangle ABC, it follows DG = (BD/BC)(AC)=42/5 and AG=(CD/CB)(AC)=26/5.

Draw EF parallel to AB with F on AC.  Since triangle EFC is similar to triangle BAC, it follows that  AF=(BE/BC)(AC)=(AC/BC)(BE)=(14/15)(BE) and
EF=(EC/BC)(BA)=(BA/BC)(BC-BE)=(13/15)(15-BE).

Next, observe that angle GDA = angle CAD = angle BAE = angle AEF (Why?)
and angle GAD = angle EAF.  Therefore, triangle DAG is similar to triangle EAF, from which it follows that  AG/AF=DG/EF, or equivalently, (DG)(AF)=(AG)(EF).  Substitution into this last equation gives (42/5)(14/15)(BE)=(26/5)(13/15)(15-BE).

Finally, solve to obtain BE= 2535/463, so q = 463.


integration
function
Support
geometry
3D geometry
calculus
real analysis

$f: \mathbb{R}\to \mathbb{R}$ a measurable function. One can define its mean on the ball $B(x,r)$ by $M(f,x,r)=\frac{1}{2r}\int_{[x-r;x+r]}f(t) dt$. Suppose that there exists a constant $C$ such that for all $x \in \mathbb{R}$ and all $r<1$:
\[\int_{[x-r;x+r]}|f(t)-M(f,x,r)|^{2}dt \leq C r^{1+2 \alpha}\]
where $\alpha > 0$. Prove that $f$ is Hölderian with exponent $\alpha$.

Here's a framework I'd put this in - it's related to things I was working on for my thesis long ago.

Let $\varphi$ be an even function with mean value zero. (Even is just a convenience, but it is convenient.) Let it be compactly supported - say, supported on $[-1,1]$ - and assume that it is smooth. Normalize the Fourier transform of $\varphi$ by

$\int_{0}^{\infty}\frac{[\widehat{\varphi}(\xi)]^{2}}{\xi}\,d\xi=1.$

For $t>0,$ define $\varphi_{t}(x)=\frac1t\varphi\left(\frac{x}{t}\right).$

Define $\mathbb{R}_{+}^{2}$ to be $\{(x,t): t>0\}.$

Define $F$ on $\mathbb{R}_{+}^{2}$ by

$F(x,t)=\varphi_{t}*f(x)= \int_{\mathbb{R}}\varphi_{t}(x-y)f(y)\,dy$

For reasonable $f$ (say, $f\in L^{2},$ or, as long as $\varphi$ is compactly supported, $f\in L^{2}(\text{loc})$) we can recover $f$ from $F$:

$\mathbf{[*]}\ \ f(x)=\int\int_{\mathbb{R}_{+}^{2}}\varphi_{t}(x-y) F(x,y)\,\frac{dy\,dt}t$

Your condition implies that $|F(x,t)|\le Ct^{\alpha}.$

That in turn implies this: for the interval $I,$ let the Carleson region $S(I)$ be the square in $\mathbb{R}_{+}^{2}$ whose base is $I.$ Then for each $I,$ then

$\int\int_{S(I)}\frac{|F(x,t)|^{2}}{t}\,dx\,dt\le C|I|^{2\alpha+1}.$

OK, I haven't gotten around to the Hölder condition yet. I'll fill that in in the next post.

These Hölder spaces turn out to be the dual spaces to the (real) Hardy spaces $H^{p}$ with $p<1.$ The limiting case as $\alpha\to0$ would be the space BMO.

Now that I've got this elaborate setup, I might as well use it. Use the line marked $\mathbf{[*]}$ above to write

$f(x)-f(y)=\int\int_{\mathbb{R}_{+}^{2}}F(z,t) [\varphi_{t}(x-z)-\varphi_{t}(y-z)]\,\frac{dz\,dt}t$

Estimating,

$|f(x)-f(y)|\le C\int\int_{\mathbb{R}_{+}^{2}}t^{\alpha-1}|\varphi_{t}(x-z)-\varphi_{t}(y-z)|\,dz\,dt$

The support of the function being integrated is the union of two cones: $\{(z,t): |z-x|\le t\}\cup\{(z,t): |z-y|\le t\}.$ Let's split this up into the part where $t>|x-y|$ and the part where $t<|x-y|.$

Let's start with the upper region. The width of this region is $|x-y|+2t<3t.$ On it, we can use the Mean Value Theorem to estimate that $|\varphi_{t}(x-z)-\varphi_{t}(y-z)|\le C\,\frac{|x-y|}{t^{2}}.$

We can then estimate this part of the integral as no greater than

$C\int_{|x-y|}^{\infty}|x-y|t^{\alpha-2}\,dt=C|x-y|^{\alpha}.$

(Note that we are assuming that $0<\alpha<1.$)

The lower part of the integral, that over $t<|x-y|,$ splits into the integral over two separate cones. (They may overlap a little - that doesn't hurt this argument.) Each such cone has width $2t$ at height $t$ and we use the estimate $|\varphi_{t}(\cdot)|\le \frac{C}{t}.$

Then the contribution of the lower integral is no more than

$C\int_{0}^{|x-y|}t^{\alpha-1}\,dt=C|x-y|^{\alpha}.$

Put these together, and we have $|f(x)-f(y)|\le C|x-y|^{\alpha}.$

I would note that the mean-deviation condition as stated is slightly stronger that I needed. Using Cauchy-Schwarz, I gave away some information to arrive at $|F(x,t)|\le Ct^{\alpha}.$

I will also note that if you simply state that original hypothesis with $\alpha=0$ you are defining the space BMO.

But with $ \alpha \equal{} 0$ for functions from BMO space thesis is false because it implies that functions from BMO are bounded but we have $ log|x|$ in BMO.

The limiting case as $ \alpha\to0$ is not that the function that I called $ F(x,t)$ is bounded but rather that $ \frac{|F(x,t)|^2}{t}\,dx\,dt$ is a Carleson measure in the upper half plane - that is (using my previous notation) that

$ \int\int_{S(I)}\frac{|F(x,t)|^2}{t}\,dx\,dt\le C|I|.$

Also, note the condition that $ \varphi(x)$ have mean value zero. It's entirely possible for $ F(x,t)$ to be bounded even when $ f$ is not bounded - and it doesn't even matter whether $ F$ is bounded as long as the Carleson measure hypothesis is satisfied.

If you go back to the original statement by alekk and put $ \alpha \equal{}0$ in that, that is the very definition of BMO right there (or at least one version of that definition). I guarantee that $ \log|x|$ satisfies alekk's condition with $ \alpha\equal{}0.$

Yes, I use the same definition of BMO on $ \mathbb{R}$.  I wrote it only to point out that thesis of alekk's question: ' Prove that $ f$ is Hölderian with exponent $ \alpha$' is not true for $ \alpha = 0$ because then $ |f(x)| - |f(y)|\leq|f(x) - f(y)|\le C|x - y|^{0}\Rightarrow |f(x)| < M$ for all $ x$.


Let $t \in (0,1)$.Determine all functions $f: \mathbb{R}\rightarrow \mathbb{R}$ which are continous in point 0 and satisfy: for all $x \in \mathbb{R}$  $f(x)-2f(tx)+f(t^{2}x)=x^{2}$


Ok... I decided that a Tournament would be fun, so if you are

interested, post here [b]Clearly[/b] That you want to participate.

The pairings will be based on rating. If you want to compete, post

your name, and your rating. Then, I will take the players and pair

them. Format is Single-Elimination.

I will decide based on how many players whether to do 2-player

countdowns or 3+player normal games.

Your post should look something like this

I'm interested

Name: BOGTRO

Rating: 1294

<Whatever else you want to say, please keep it related.>

Thanks!

-BOGTRO

P.S. Tell other people who don't come to this forum about this!

EDIT: Other people beat me to this, so if you are registered for something else forget this. Date is TBD. I'm making this mainly so that tourneys don't get too big...

I am very much interested.

Verizonwires:

1275,

I have a lot of enemys and friends on FTW, so im perfect this is perfect to get revenge on:

iin77 and michaewong

----------------
dont wry bogtro i think ur tite

Ok... It's been a while, and so far I have 2 people (Me and Verizonwires). minicon has been more successful, so I'm aborting this. Sorry.


A town in India has a population of 20,000 people. 5 per cent of them are one-legged, and half the others go barefoot. How many sandals are worn in the town ?

 :lol:

So we know that everyone that has one legs wears one sandal. And half the others wear 2 sandals. Thus we have $ 20000\cdot.05\plus{}20000\cdot\frac{19}{20}\cdot\frac{1}{2}\cdot2\equal{}\boxed{20000}$.

BTW, town of India?

But people who have 2 legs wear 2 sandals so 20,000 x 0.05 (the sandals on the people that have 1 foot)  +  20,000 x 0.95 (the sandals on the people that have 2 feet) = 20,000 sandals!

5% of the people in the town of 20,000 people are one-legged, and 5% is 1/20, so 1/20*20,000=1,000 people are one-legged. That means there are 19,000 other people. Since half of these go barefoot, that means half of them wear 2 sandals. There are 19,000 sandals worn by normal people. 19,000+1,000 is 20,000 sandals in all.


How do we determine whether one complex number is greater than or less than another?   :huh:

Maybe we are to find the loci of the point $(x,y)$ such that $z=x+yi\ (x,y\in\mathbb{R}).$

Must the following conditions of sense :
$z^2+z\in R$ for $(1)\ ,$ $z+\frac 4z\in R$ for $(2)$ and $z+\frac{a^2}{z}\in R$ for $(3)\ .$

Virgil, you have run into a contradiction.  You have proven that you were wrong by referring to this problem as "easy" from the follow-up posts.  Why refer at all to it then?

[quote="Virgil Nicula"]Solve the following inequations with a complex variable $z\in C\ :$
$1.\blacktriangleright\ z^2+z\le 0\ .$
$2.\blacktriangleright\ 4\le z+\frac 4z\le 5\ .$
$3.\blacktriangleright\ -2a\le z+\frac{a^2}{z}\le 2a\ ,$ where $a\in (0,\infty )\ .$[/quote][color=blue][b][u]Proof.[/u][/b]

$1.\blacktriangleright$ The inequation has a sense if and only if $z^2+z\in R$ $\Longleftrightarrow$
$z^2+z=\overline z^2+\overline z$ $\Longleftrightarrow$ $(z-\overline z)(z+\overline z+1)=0\ .$
[u]The first case.[/u] $z=\overline z$ $\Longrightarrow$ $z\in R$ and $z^2+z\le 0$ $\Longleftrightarrow$ $\boxed {\ z\in [-1,0]\ }\ .$
[u]The second case.[/u] $z+\overline z=-1$ $\Longrightarrow$ $Re\ z=-\frac 12$ and $z(z+1)\le 0$ $\Longleftrightarrow$
$z=-\frac 12+\lambda i\ ,\ \lambda\in R$ and $-z\overline z\le 0$ $\Longleftrightarrow$ $\boxed {z\in \left\{-\frac 12+\lambda i\ |\ \lambda \in R\right\}}\ .$
Therefore, $\boxed {z\in [-1,0]\cup \left\{-\frac 12+\lambda i\ |\ \lambda \in R\ \right\}}\equiv [AO]\cup d\ ,$ i.e. the reunion between the segment $[AO]$ of the axis $Ox\ ,$ where $A(-1,0)$ and the parallel line $d$ to the axis $Oy$ with the equation $x=-\frac 12\ .$

$2.\blacktriangleright$ The inequation has a sense if and only if $z\ne 0$ and $z+\frac 4z\in R$ $\Longleftrightarrow$
$z\ne 0$ and $z+\frac 4z=\overline z+\frac{4}{\overline z}$ $\Longleftrightarrow$ $z\ne 0$ and $(z-\overline z)(z\overline z-4)=0\ .$
[u]The first case.[/u] $z\ne 0$ and $z=\overline z$ $\Longrightarrow$ $z\in R^*$ and $z+\frac 4z\in [4,5]$ $\Longleftrightarrow$
$z\in R^*$ and $\left(z+\frac 4z-4\right)\left(z+\frac 4z-5\right)\le 0$ $\Longleftrightarrow$ $\boxed {z\in [1,4]}\ .$
[u]The second case.[/u] $z\ne 0$ and $z\overline z=4$ $\Longleftrightarrow$ $z\ne 0$ and $\frac 4z=\overline z$ $\Longrightarrow$
$|z|=2$ and $Re\ z\in \left[2,\frac 52\right]$ $\Longleftrightarrow$ $\boxed {\ z=2\ }\ .$
Therefore, $\boxed {\ z\in [1,4]\ }\ .$

$3.\blacktriangleright$ The inequation has a sense if and only if
$z\ne 0$ and $z+\frac{a^2}{z}=\overline z +\frac{a^2}{\overline z}$ $\Longleftrightarrow$ $z\ne 0$ and $(z-\overline z)(z\overline z-a^2)=0\ .$
[u]The first case.[/u] $z\ne 0$ and $z=\overline z$ $\Longleftrightarrow$ $z\in R^*$ $\Longrightarrow$
$z\in R^*$ and $\left(z+\frac{a^2}{z}+2a\right)\left(z+\frac{a^2}{z}-2a\right)\le 0$ $\Longrightarrow$
$z\in R^*$ and $(z+a)^2(z-a)^2\le 0$ $\Longrightarrow$ $\boxed {\ z\in \left\{ -a,a\right\}\ }\ .$
[u]The second case.[/u] $z\ne 0$ and $z\overline z=a^2$ $\Longleftrightarrow$ $|z|=a$ $\Longrightarrow$ $|z|=a$ and $-2a\le z+\overline z\le 2a$ $\Longrightarrow$ $|z|=a$ and $-a\le Re\ z\le a$ $\Longrightarrow$ $|z|=a\ .$ Therefore, $z\in \left\{z\in C\ |\ |z|=a\ \right\}\ ,$ i.e. the circle with the equation $x^2+y^2=a^2\ .$[/color]


AMC
AIME
geometry
USA(J)MO
USAMO
AMC 10
AMC 10 B

[color=green]Here are the answers. Original post follows below. -mod[/color]

1. 840
2. 697
3. 011
4. 177
5. 072
6. 412
7. 041
8. 398
9. 420
10. 346
11. 600
12. 011
13. 090
14. 905
15. 150

While proctoring, I tried working them. I wasn't having a good day thinking about geometry problems, and I haven't re-looked at anything afterwards. Anyway, this is what my notes say:

1. 807 [color=red][Wrong][/color]
2. 697
3. 011
4. 177
5. -
6. 786 [color=red][Wrong][/color]
7. 041
8. 393 [color=red][Wrong][/color]
9. 420
10. 346
11. 552
12. -
13. 090
14. 875 [color=red][Need to check this again][/color]
15. -

1. 840
2. 205
3. 011
4. 032 (guess)
5. 120 (guess)
6. 412
7. 041
8. 398
9. 456
10. 346
11. 600
12. 077 (guess)
13. 090
14. 743 (guess)
15. 126 (guess)

I am fairly certain that numbers two and nine are wrong, but quite a few people agreed with me on where we [i]dis[/i]agree.

Hopeful score: 8.

P.S. This AIME 12 was very similar to 1999 AIME 12.

[quote="1=2"]WATR got at LEAST a 13. I am very sure.[/quote]

Not MY answers, those were supposed to be the answers.

I think I got a 9 or 10.

The answers should be

1. 840
2. 697
3. 011
4. 177
5. 072
6. 412
7. 041
8. 398
9. 420
10. 346
11. 600
12. 011
13. 090
14. 905
15. 150

Many people put 807 for #1, forgetting that 9,6,4 is a geometric sequence. Also, number 13 is very easy to get the right answer if you assume that it repeats after the first two numbers (you obtain the condition $ a_1a_2 \equal{} 2009$), but actually proving it is rather amusing. The best I got was to elegantly prove that $ a_1\plus{}a_2 \geq 88$, then bash out the $ 88$ and $ 89$ cases.

Heh, for number 1, I got 931 and 124 also.  When I subtracted, I got 707 :rotfl:

[quote="1=2"]1. 840
2. 205
3. 011
4. 032 (guess)
5. 120 (guess)
6. 412
7. 041
8. 398
9. 456
10. 346
11. 600
12. 077 (guess)
13. 090
14. 743 (guess)
15. 126 (guess)

Looks like you got an 8 nice

I am fairly certain that numbers two and nine are wrong, but quite a few people agreed with me on where we [i]dis[/i]agree.

Hopeful score: 8.

P.S. This AIME 12 was very similar to 1999 AIME 12.[/quote]

[quote="Hamster1800"]The answers should be

1. 840
2. 697
3. 011
4. 177
5. 072
6. 412
7. 041
8. 398
9. 420
10. 346
11. 600
12. 011
13. 090
14. 905
15. 150

Many people put 807 for #1, forgetting that 9,6,4 is a geometric sequence. Also, number 13 is very easy to get the right answer if you assume that it repeats after the first two numbers (you obtain the condition $ a_1a_2 \equal{} 2009$), but actually proving it is rather amusing. The best I got was to elegantly prove that $ a_1 \plus{} a_2 \geq 88$, then bash out the $ 88$ and $ 89$ cases.[/quote]

You can prove elegantly that $ a_1a_2 \ge 2009$ and $ a_1a_2 \le 2009,$ so $ a_1a_2 \equal{} 2009.$

I didn't record answers, but some of tehm that I remember:

1. 840
5. 072 (somehow, this was the only problem I didn't do)
6. 412
10. 346
11. 600
12. 011
13. 090
15. 150

[quote="Walk Around The River"]Eh...I think it's something like:

1. 840
2. 697
3. 011
4. 177
5. 072
6. 412
7. 041
8. 398
9. 420
10. 346
11. 600
12. 11
13. 90
14. 905
15. ???

I missed number 6 because 4^4 isn't 64, but since I got #14, I knew what it was then...  :|[/quote]

I swear everyone who missed that problem(#6) did that. i did too. :P

Eh, I added wrong on that problem.  I got 411, and my friend got 410.  :huh:

I haven't done problem #15, but I agree with Hamster's answers on problems 1 through 14.

Is there a compendium of all the 2009 [i]problems[/i] together somewhere on the internet (i.e. like a pdf I can print out)?

In the [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2009]resources section[/url].

[quote="cognos599"]i just took it for practice since i am taking AIME II. These were my answers.

1. 840
2. 697
3. 011
4. 177
5. 072 <-- I totally bsed this one
6. 412
7. 041
8. 398
9. 138
11. 024

so that puts me at an 8/15. hmm. i guess that would get me into USAMO with an index of 221. grah i wish i had taken this test. i bet AIME II will be really hard and watch me get a 6 and get an index of 201 and the cutoff will be like 201.5 or something.[/quote]
how did you pass with a 108 on the amc10B???

i've gotten like 10000000 questions about this, its supposed to be a joke, i actually got a 141 on AMC 10B.  :)

you know what, im gonna change my sig.

If I didnt bubble anything in wrong, I got a 10 (1-9, 11).  I missed number 10 and almost guessed 12 right (guessed 111, wouldve guessed 011 had it not been an answer already)

NO!!!!!!!!!!

Stupid mistakes......I hate them...

I would've qualified for USAMO had I written down the right answer...I got 177 in my notes but wrote down 119 for some reason....

I missed 200 by 5 points......

...sigh.... I could've made it if I didn't make stupid mistakes... I just found out that I read #11 incorrectly (noninteger coordinates  :rotfl: ) I wasted sooooo much time trying to figure out why there weren't an infinite number of x coordinates whose difference was a multiple of 2.....

I think I got a 9.

I got 2,3,6,7,8,9,10,11,13

#1 I was stupid
#4 I got 176 when it was 177, no idea how
#5 I put 252 because I thought it wanted the whole length (180+72)
#14 Arithmetic error while solving, easy problem though

I got all these about 15 minutes after time was up

I guessed on 12 and 15

Could've easily gotten 13, oh well my index is 229.5, that should be alright

[quote="abacadaea"][quote="Walk Around The River"]...

I missed number 6 because 4^4 isn't 64, but since I got #14, I knew what it was then...  :|[/quote]

I swear everyone who missed that problem(#6) did that. i did too. :P[/quote]

QFT

So for #4, couldn't you make angle A = 0, so the parallelogram would be squished into a line? It'd make it a lot simpler.

Yes, that's what alot of people did. That seems to be a bit dangerous to assume, but it works.

Another "safe" thing you could do is assume it is a rectangle.


how do you find the vertex of y=-5x^2 + 4x - 6?

i know you complete the square but how do i do that?

pls and thx :lol:

There's a better way to find the vertex. $\frac{-b}{2a}$. You know the line of symetry is between the two roots and that it intersects the vertex, so what's between $\frac{-b+\sqrt{b^{2}-4ac}}{2a}$ and $\frac{-b-\sqrt{b^{2}-4ac}}{2a}$?

Hmm. We are doing conic sections in math right now. Does $-\frac{b}{2a}$ still work for hyperbolas, ellipses, etc.?

uh, dude,
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=140657

you don't need to post questions in multiple forums...


Will there be any upcoming calculus classes in the near future? If yes, when?

Not in the near future.  Maybe someday, but not soon.


linear algebra
matrix
algebra
polynomial
invariant
linear algebra theorems

Two $\displaystyle n \times n$ matrices $\displaystyle A$ and $\displaystyle B$ over the field $\displaystyle K$ are called similar if there exists an invertible $\displaystyle n \times n$ matrix $\displaystyle P$ over $\displaystyle K$ such that \[ P^{-1} A P = B .  \]

Similar matrices share many properties: they have the same rank, the same determinant, the same trace, the same eigenvalues (but not necessarily the same eigenvectors), the same characteristic polynomial and the same minimal polynomial.

My question is: how do you prove that they have the same minimal polynomial?

Edit: I thought a little more on this and I realised that it is very easy after all.  :blush:

Just looking at the largest invariant factors solves the problem.


Αν $ a,b,c$ θετικοί ακέραιοι, να λυθεί η εξίσωση: $ (\frac{a}{b})^2\plus{}4\plus{}(\frac{1}{b})^2\equal{}c^2$

LEMMA:Aν $ p \equal{} 3mod4$ και $ p/x^2 \plus{} y^2$ με $ x,y$ ακεραιοι τοτε $ p/x$ και $ p/y$ αποδυκνειεται ευκολα με πολλουσ τροπους οπως με το συμβολο legendre που ισχυει οτι $ ( \minus{} 1/p) \equal{} ( \minus{} 1)^{(p \minus{} 1)/2}$
H ασκηση μας μετατρεπετε $ b^2(c^2 \minus{} 4) \equal{} a^2 \plus{} 1$
Συμφωνα με το λημμα (αποφευγοντας τις τετριμενες λυσεις) θα πρεπει $ b \equal{} 1mod4,c \equal{} 3mod4,a \equal{} 0mod2$
Κανοντας λιγες πραξεις εχουμε οτι $ (a \plus{} 1)^2 \equal{} b^2(c^2 \minus{} 4) \plus{} 4a$
Eρευνω την εξισωση στο $ mod8$ οποτε εχουμε $ b^2 \equal{} 1mod8,c^2 \minus{} 4 \equal{} 5mod8,4a \equal{} 0mod8 \equal{} > b^2(c^2 \minus{} 4) \plus{} 4a \equal{} 5mod8 \equal{} > (a \plus{} 1)^2 \equal{} 5mod8$ ΑΤΟΠΟ γιατι τα τετραγωνικα υπολοιπα του $ 8$ ειναι $ 0,1,4,$.(Μενει να ερευνησω οπωσ ειπα και παραπανω τις τετριμενες λυσεις που ειναι $ (a,b,c)\equal{}(2,1,3)$)

Εχετε κανει τυπογραφικο λαθος, εκει που εχετε γραψει 4a επρεπε να ειχατε 2a, οποτε δεν προκειπτει ατοπο δουλευοντας modulo καποιον αριθμο (εξ αλλου αν υπηρχε τροπος να καταληξουμε σε ατοπο δουλευοντας modulo καποιον αριθμο τοτε δεν θα ειχαμε καθολου λυσεις)

Η εξισωση λυνεται διαφορετικα και εχει απειρες λυσεις οι οποιες προκειπτουν απο μια εξισωση Pell:

http://www.mathlinks.ro/viewtopic.php?t=297746


In line 3 of the soln. where has the r dissappeared to when calculating Q

since Q=pA  where p is charge density and A is area (2$ \pi$rh)

[img]http://i33.tinypic.com/2drrigo.jpg[/img]
[img]http://i36.tinypic.com/2i1godk.jpg[/img]

As the charge density is variable you can't say that the charge is the product of the volume versus the density;[b] which value of the variable density[/b]???
Unless you use the average density, but to get it you would need to pass by the integral, anyway.


Find all$ : f : R\R\rightarrow R$ such that$ : xf(y)-yf(x)=f(\frac{y}{x})$

[quote="DCTPKTCSPHN"]Find all$ : f : R\R\rightarrow R$ such that$ : xf(y) - yf(x) = f(\frac {y}{x})$[/quote]

[hide="My solution"]
Let $ P(x,y)$ be the assertion $ xf(y)-yf(x)=f(\frac yx)$ that we supposed true $ \forall x\neq 0$, $ \forall y$

$ P(2,0)$ $ \implies$ $ f(0)=0$

$ P(x,1)$ $ \implies$ $ f(\frac 1x)=-f(x)$

$ P(\frac 1x,y)$ $ \implies$ $ \frac{f(y)}x+yf(x)=f(xy)$

$ P(\frac 1y,x)$ $ \implies$ $ \frac{f(x)}y+xf(y)=f(xy)$

And, from these last two equalities, considering $ x,y\notin\{0,1\}$, we get $ \frac{f(y)}x+yf(x)=\frac{f(x)}y+xf(y)$ and so $ \frac{f(x)}{x-\frac 1x}=\frac{f(y)}{y-\frac 1y}$

So $ f(x)=a(x-\frac 1x)$ $ \forall x\notin\{0,1\}$
$ P(1,1)$ $ \implies$ $ f(1)=0$ and the above equality is true for $ x=1$
Plugging back in the original equation, we see that this indeed is a solution.

Hence the set of solutions :
$ f(0)=0$
$ f(x)=a(x-\frac 1x)$ $ \forall x\neq 0$ for any real $ a$
[/hide]

[quote="Edward_Tur"]http://www.mathlinks.ro/viewtopic.php?t=183261[/quote]

 :lol:  and I already answered this first post (and forgot it).

Fortunately, the results are the same (and the demo too)


geometry
circumcircle
inequalities
inequalities proposed

[quote="Markelov"]For the triangle $ ABC$ let $ {R}$ be the circumcircle radius, $ r$ = incircle radius,  $ s$ = semiperimeter (half of the perimeter).

[b]1[/b]. Prove, that for any triangle
\[ \frac {3^5}{s} \leq \frac {5^3}{2R} \plus{} \frac {2^4}{r}
\]
[b]2[/b]. Prove that the above inequality becomes an equality if $ ABC$ is isosceles triangle with the angle $ \arccos{\frac {4}{5}}$ at the base (e.g. for $ AB \equal{} AC \equal{} 5$, and $ BC \equal{} 8$). For all other triangles (including the equilateral one) $ \leq$ becomes $ <$.

I've worked a lot to find this inequality, and expect it to be hard. If you have found the solution or have any comment, I would like to hear it. Please, post it here or e-mail me.

Thank you!

Sergey Markelov
markelov@mccme.ru

P.S. Note that this effect (inequality becomes equality for non-equilateral triangle) cannot happen with the inequaltities of the type $ xR \plus{} y\,r \plus{} z\,s \geq 0$, see [u][url=http://www.mathlinks.ro/viewtopic.php?t=207461]here[/url][/u] for details.

[img]http://i13.photobucket.com/albums/a287/Gaz_v_pol/inequality.jpg[/img][/quote]
My solution... :)


A certian building is in the shape of a hemisphere of radius 20 ft., with the circular base of radius 20 ft. being the floor. A 24 ft. board is leaning against the building so that the base of the board is on the ground. Assume the center of the base of the board is located 5 ft. from the edge of the building. Find the distance from the center of the base of the board to the point where the board is touching.

What does it mean when its 5 feet away from the edge of the building? I didn't know a hemisphere had an edge.

[quote="archimedes1"]I think it means 5 ft. from the circle that lays on the floor.

[hide]Let's take a cross-section:  We have a semicircle and a line tangent to it.  If a line is tangent to a circle, then the radius meets the line at a right angle.  So, we can make a right triangle.  The hypotenuse, on the floor and streching from the center of the circle to our line, is $20+5=25$.  One leg is the radius, $20$.  Now use the Pythagorean theorem to solve for the last side, which is $15$.  This length is what the problem is asking for.[/hide][/quote]
Thats right!. I just didn't understand when it said 5 feet away from the edge.


conics
ellipse
geometry unsolved
geometry

Turn D 90 degrees around C clockwise (F).Draw a circle diameter AF(center O).The line CO cut he circle at M and N.Line paralel to AM trough C define the major axis direction and the line parale to AN trough C define minor axis direction.CM=b and CN=a.The foci is achivied


analytic geometry
graphing lines
slope
floor function
greatest common divisor

A line passes through $P(2,2)$ and $Q(50,998)$. How many other points with integer coordinates lie on line PQ that are strictly between P and Q?

[hide]
The slope of the line is $\frac{996}{48}$, or $\frac{62}{3}$.  So every increase of 3 in the x coordinate gives a 62 increase in the y coordinate.
So the answer is $\lfloor\frac{50}{3}\rfloor-\lfloor\frac{2}{3}\rfloor$, or 16.[/hide]

I hope I did that right.

[quote="math92"][quote="undefined117"]shouldn't it be 15, not 16?[/quote]
50/3=16.6666
floor 16.6666 and you get 16
the floor of 2/3 is 0
16-0=16
 :wink:[/quote]
ohh oops, i didn't see that :blush:

[quote="undefined117"][quote="asdff"]Again, shoudnt it be 11?[/quote]
could you show how you came to that conclusion?[/quote]

gcf-1

[quote="CircleSquared"]
The slope of the line is $\frac{996}{48}$, or $\frac{62}{3}$.  So every increase of 3 in the x coordinate gives a 62 increase in the y coordinate.
So the answer is $\lfloor\frac{50}{3}\rfloor-\lfloor\frac{2}{3}\rfloor$, or 16.

I hope I did that right.[/quote]

Sorry I dont understand that.  You have to note the y-intercept IS NOT an integer if I understand what you are saying.  I too get 11.

What's this about the y-intercept?  My solution doesn't mention the y-intercept; and no, it isn't an integer.
Umm, could someone please post a [b]solution[/b] that gives 11?
Then maybe I could see where I went wrong. :read:

move the entire thing 2 left and 2 down, so the line passes from $(0,0)$ to $(48,996)$.

the slope is $\frac{83}{4}$ so for every multiple of 4 as the x value, there is an integer y value.

48/4 = 12

so answer is $12-1 = \boxed{11}$.

EDIT: haha i guess that first sentence wasn't really necessary.

THAAAAAT explains it. :oops_sign:
Okay, it is 11, I agree.


I am having a hard time to answer these problems, if I have a turing machine M and symbol a does M ever write the symbol a when started on the empty tape? Is this a decideable or undecideable problem?

I think it's undecidable. Otherwise you could solve the halting problem by modyfing the turing machine to print a certain letter just before it halts (some new letter it never uses anywhere else).

that is kind of true, what is actually the language represented by this ?


Prove that if $ 0<q<1$ then
\[ (1\plus{}q)(1\plus{}q^2)(1\plus{}q^3)\dots\equal{}\frac{1}{(1\minus{}q)(1\minus{}q^3)(1\minus{}q^5)\dots}\]

The LHS counts the number of partitions of $ n$ into distinct parts and the RHS counts the number of partitions of $ n$ into odd parts.  Given a partition $ \lambda \vdash n$ with distinct, write each part as $ 2^p q$ where $ q$ is odd and construct the partition $ \lambda'$ where $ q$ appears $ \sum_{2^p q \in \lambda} 2^p$ times.  Each value of $ p$ appears at most once by assumption, so this defines the unique base-$ 2$ representation of the number of times $ q$ appears in an arbitrary partition $ \lambda'$ of $ n$ into odd parts.  The result follows (formally).  Convergence can be established from the following

[b]Lemma:[/b]  If a sequence $ (a_n)$ satisfies $ \sum |a_n|^2 < \infty$, then $ \sum a_n$ converges iff $ \prod (1 \plus{} a_n)$ converges.


Two perfect logicians, S and P, are told that integers x and y have been chosen such that 1 < x < y and x+y < 100.  S is given the value x+y and P is given the value xy.  They then have the following conversation.

P:  I cannot determine the two numbers.
S:  I knew that.
P:  Now I can determine them.
S:  So can I.

Given that the above statements are true, what are the two numbers?


I typed a paper on Microsoft word, and then I changed it to Wingdings 3 font on Microsoft Word 2000. After I saved it, and exited out of the program, and then re-entered, I could not convert the wingdings 3 font back to Times new Roman. It just converted to tons of boxes with no letters.

What should I do? I'm frustrated.

Now I can't even read what I wrote.

Save your document in the html format ("Save as", "Web Page"). Then try to recover the text from the html file.

Learn how to read Wingdings fluently and then convert it back manually.

Huh. If Yury's suggestion doesn't work, try to redownload Times New Roman.

[url]http://office.microsoft.com/en-us/help/HA010347301033.aspx[/url] has directions.

Open the file in notepad. When I tried it, i saw, among a lot of other random-looking codes, the message with a weird character that looks like this ð in between each letter. Paste it into the word document, hit CTRL + F to get the to the find menu; go to replace and replace each occurence of the ð with nothing to delete them all.

I tried the previous suggestion, and the find and replace didn't work.

I saved it as webpage from Yuri's suggestion, but I don't understand how to make it HTML like Yuri said. The save as webpage part just made the screen of microsoft word Wider than before. I don't get HTML format or file; I'm not what you call a computer person.  :blush:

[quote="Casa De Mathematica"]I tried the previous suggestion, and the find and replace didn't work.
[/quote]

It didn't? just copy the strange symbol (the symbol might be different on your computer; in fact, it might look like a space), then hit ctrl + f, go to the replace tab, copy the strange symbol into the "Find what: " box, leave the "replace" box blank hit replace, and it should work...

Additionally, you could try doing the replacing in Notepad itself; go to Edit and then click replace.

[quote="Yury"][quote="Casa De Mathematica"] I don't get HTML format or file;[/quote]
You should get an html file. Try this once again. Then you need to open this file in a web browser. If this doesn't help, open it in Notebook.[/quote]
I could be completely wrong about this, but I think that, by "get", he means understand.

[quote="roadnottaken"]I could be completely wrong about this, but I think that, by "get", he means understand.[/quote]
I see.  :maybe:  I tried what I suggested on my computer and it worked.

to open it in your web browser, put it in notepad and save as a yourtexthere.txt file. Then, rename it as yourtexthere.html

you know, that's just a glitch in word?
 :rotfl:  :lol:


number is in the form:   ABCDEF   ,where a,b,c,d,e,f   are digits of the number.let the no.   be   N
N*2=CDEFAB
N*3=BCDEFA
N*4=EFABCD
N*5=FABCDE
N*6=DEFABC
  THE NUMBER IS VERY COMMONAND YOU'LL QUITE FAMILIAR WITH IT.
EASY ONE. :)

:huh: solve it . dont make guesses   and its not 123456

yes, it is. but solve it.  and if you can delete your post rofler so that others solve it without knowing ans,

[hide="Note"]Consider the decimal expansion of $ 1/7$[/hide]

It's pretty straightforward, actually -- you have six linear systems in six variables.

:huh: but thats now that you know the solution that you can relate with other things as 1/7
 but solve for when you are finding the answer not knowing the answer :!:  :!:

you have quite complicated it.    its a very simple problem
2*6=12   . so we'll get a 7 digit no. if a=2,, so a=1  and the rest can found in a min.

@analytic Can you please explain the answer a bit more?

@analytic knows that $ A$ must be $ 1$ because otherwise $ 6N$ would be a 7-digit number rather than the 6-digit number $ DEFABC$. The rest hinges on:[hide]If you look at the first digit in each number you realize that neither one of the digits can be zero. Also, since the numbers $ N$, $ 2N$, $ 3N$, etc increase and differ by more than 100000, then $ A<C<B<E<F<D$.[/hide]and[hide]If you look at the last digits of each number, you realize that $ B$ and $ C$ are even $ ($last digits of $ 2N$, and $ 6N$, which are obviously multiples of $ 2)$.
You also figure out that $ CD$ is a multiple of $ 4$, because those are the last digits of $ 4N$ $ ($multiple of $ 4)$.
Similarly, you realize that $ E$ must be $ 5$, because it is the last digit in a multiple of $ 5$ $ (5N\equal{}FABCDE)$.
Finally, since the digits in a multiple of $ 3$, $ ($like $ 3N)$ add up to a multiple of $ 3$, you have a clue about $ F$.[/hide]

I got it:

When multipled by the highest number, 6, the no.of digits is still six i.e. ABCDEF * 6 = DEFABC. So, DEFABC can at most be 987654. Divide this by 6 to get ABCDEF and this would mean that A has to be 1.

Rest is easy. ABCDEF * 3 = BCDEFA. Since A is 1, F must be 7 so that 7*3 = 21, leaving A as the last digit. With F known, do I need to add anything? ABCDEF * 4= EFABCD. So D must be 8 since F=7, and so on. Given below are the 5 numbers which we get when we multiple 142857 by 2,3,4,5,6.


Given a natural number n such that $ n>4$. Prove that:
$ 1.3<\frac{1}{n\plus{}1}\plus{}\frac{1}{n\plus{}2}\plus{}...\plus{}\frac{1}{4n}<1.4$

may be not the simplest way but works, and also it allows you to solve a lot of problems:

With $ e\equal{}\lim_{n\rightarrow\infty}{(1\plus{}\frac{1}{n})^n}$
[hide="first prove that"]
[hide="a"]
For all integers $ n\ge1$
$ (1\plus{}\frac{1}{n})^n\le 3$
[/hide]
[hide="b"]
For all integers $ n\ge1$
$ (1\plus{}\frac{1}{n})^n<(1\plus{}\frac{1}{n\plus{}1})^{n\plus{}1}$
[/hide]
[hide="c"]
For all integers $ n\ge1$
$ (1\plus{}\frac{1}{n\plus{}1})^{n\plus{}2}<(1\plus{}\frac{1}{n})^{n\plus{}1}$
[/hide]
[hide="d"]
$ \lim_{n\rightarrow\infty}{(1\plus{}\frac{1}{n})^n}\equal{}\lim_{n\rightarrow\infty}{(1\plus{}\frac{1}{n})^{n\plus{}1}}$
[/hide]
[/hide]
Now we are ready:
[hide="solution"]
$ (1\plus{}\frac{1}{n})^n<e<(1\plus{}\frac{1}{n\minus{}1})^n$ for all $ n>1$
$ 1\plus{}\frac{1}{n}<e^{\frac{1}{n}}<1\plus{}\frac{1}{n\minus{}1}$
$ \ln{(n\plus{}1)}\minus{}\ln{n}<\frac{1}{n\plus{}1}<\ln{n}\minus{}\ln{(n\minus{}1)}$

So:
$ \sum_{k\equal{}n\plus{}1}^{4n}{\ln{(k\plus{}1)}\minus{}\ln{k}}<\sum_{n\plus{}1}^{4n}{\frac{1}{k}}<\sum_{k\equal{}n\plus{}1}^{4n}{\ln{k}\minus{}\ln{(k\minus{}1)}}$
$ \ln{(4n\plus{}1)}\minus{}\ln{(n\plus{}1)}<\sum_{n\plus{}1}^{4n}{\frac{1}{k}}<\ln{4n}\minus{}\ln{n}$
$ \ln{(4\minus{}\frac{3}{n\plus{}1})}<\sum_{n\plus{}1}^{4n}{\frac{1}{k}}<\ln4$

Now you will only need at least a logarithm table to conclude
(note that the left hand increases as $ n$ increase so it must be changed to $ \ln(4\minus{}\frac{3}{5})$)

Wich should work if I haven't made an arithmetic mistake. Note that you can sharpen the bounds all you want by taking some terms out of the sum.
[/hide]

 :)


John wants to cut a 6cm x 7cm x 8cm block of cheese into 1cm x 1 cm x 1cm cubes, WHat is the mimimum number of cuts she has to make?

[quote="PowerOfPi"][hide="Solution"]
He has to make 5 cuts on the side 6cm long, 6 on the side 7cm long, and 7 on the side 8cm long, for a total of $ 5 \plus{} 6 \plus{} 7 \equal{} \boxed{18}$.
[/hide][/quote]

The thing is he can stack the blocks after each cut to minimize the number of cuts he has to make.


I have created a website using the CMS xoops and I want to install LaTeX (Latexrender of course) in my server in order to use it on my site.
Is there a detailed solution for this problem? Because I don't have any idea how to proceed.

Thanks.

Assuming that you have installed LaTeX, ImageMagick and Ghostscript on your server and have downloaded [url=http://www.mayer.dial.pipex.com/tex.htm]LatexRender[/url] look at the readme.txt file in the otherPHP/latexrender directory in the download. The only thing that it won't say is exacty where to enter the lines [code]include_once('/full_path_here_to/latexrender/latex.php');
$latextext=latex_content($latextext);[/code] as that depends on the program that uses latexrender. It's best to seek help for that part on a xoops forum.

[quote="stevem"]Assuming that you have installed LaTeX, ImageMagick and Ghostscript on your server [/quote]
This is exactly what I m looking for. how can I install software on a server?
Thanks.

You need root access to the server, so effectively you need a dedicated server (or a virtual private server). It depends on what operating system it uses and whether there's a root control panel (like [url=http://www.cpanel.net/products/cPanelandWHM/linux/whmov.htm]WHM[/url]) which helps to install software. Different flavours of Linux have different ways of installing packages. RedHat and SuSE use rpms and Ubuntu uses a package manager for example; these can be installed from the command line if there's no suitable control panel facility.

If you don't have root access you won't be able to do this and have to persuade your host to do it for you. If not then you'll need to use [url=http://www.forkosh.com/mimetex.html]mimeTeX[/url] instead.


Let $ a,b,c$ be positive real numbers with $ a^3\plus{}b^3\plus{}c^3\equal{}3.$ Determine the largest value of
$ a^4\plus{}b^4\plus{}c^4\plus{}a^2b^2\plus{}b^2c^2\plus{}c^2a^2.$

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers with $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3.$ Determine the largest value of
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2.$[/quote]

$ (a^4 \plus{} b^4 \plus{} c^4 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)_{min} \equal{} 3\sqrt [3]{3},$

$ (a^4 \plus{} b^4 \plus{} c^4 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)_{max} \equal{} 6$.

[quote="fjwxcsl"][quote="can_hang2007"]Let $ a,b,c$ be positive real numbers with $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3.$ Determine the largest value of
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2.$[/quote]

$ (a^4 \plus{} b^4 \plus{} c^4 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)_{max} \equal{} 3\sqrt {3}$.[/quote]
what will happen when $ a\equal{}b\equal{}c\equal{}1,$ dear fjwxcsl?

[quote="can_hang2007"][quote="fjwxcsl"][quote="can_hang2007"]Let $ a,b,c$ be positive real numbers with $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3.$ Determine the largest value of
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2.$[/quote]

$ (a^4 \plus{} b^4 \plus{} c^4 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)_{max} \equal{} 3\sqrt [3] {3}$.[/quote]
what will happen when $ a \equal{} b \equal{} c \equal{} 1,$ dear fjwxcsl?[/quote]

sorry! $ 3\sqrt [3]{3}\le{a^4 \plus{} b^4 \plus{} c^4 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}\le{6}.$

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers with $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3.$ Determine the largest value of
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2.$[/quote]

Hi Can, my proof is based on AM-GM inequality.  :)

Using AM-GM, we have the followings:

$ a^6\plus{}a^3\plus{}a^3 \geq 3a^4$
$ b^6\plus{}b^3\plus{}b^3 \geq 3b^4$
$ c^6\plus{}c^3\plus{}c^3 \geq 3c^4$
$ a^3b^3\plus{}a^3b^3\plus{}1 \geq 3a^2b^2$
$ b^3c^3\plus{}b^3c^3\plus{}1\geq 3b^2c^2$
$ c^3a^3\plus{}c^3a^3\plus{}1 \geq 3c^2a^2$

adding all 6 inequalities will yield:

$ a^6\plus{}b^6\plus{}c^6\plus{}2(a^3b^3\plus{}b^3c^3\plus{}c^3a^3)\plus{}2(a^3\plus{}b^3\plus{}c^3)\plus{}3\geq 3(a^4\plus{}b^4\plus{}c^4\plus{}a^2b^2\plus{}b^2c^2\plus{}c^2a^2)$

and that is equivalent to:

$ (a^3\plus{}b^3\plus{}c^3)^2\plus{}2(a^3\plus{}b^3\plus{}c^3)\plus{}3\geq 3(a^4\plus{}b^4\plus{}c^4\plus{}a^2b^2\plus{}b^2c^2\plus{}c^2a^2)$

$ \Leftrightarrow 18 \geq 3(a^4\plus{}b^4\plus{}c^4\plus{}a^2b^2\plus{}b^2c^2\plus{}c^2a^2)$

$ \Leftrightarrow 6 \geq a^4\plus{}b^4\plus{}c^4\plus{}a^2b^2\plus{}b^2c^2\plus{}c^2a^2$  :D

Let $ a,b,c$ be positive real numbers with $ a^3+b^2c +bc^2= 3.$ Prove that
$$3\sqrt [3]{3}\leq a^4 + b^4 + c^4 + a^2b^2 + b^2c^2 + c^2a^2\leq 6$$


Because few Ohio people know about AoPS..., thus resulting in virtually no posts.

I posted a lot of problems to attract attention.

which also increases your post count by a TON and is considered to be spam by many, including me.

and also, Ohio is definitely not the least active.
Try Kansas,Arizona,Vermont,Penn,Minesota,Missouri,Nevada,New Mexico,Utah,and Virginia forums.

Sorry about that. :blush:

Its still less active than SC...

Well, South Carolina has all the people from joml88's school and stuff.

And I personally don't really know anyone from Ohio who's into math, except for nr1337.  Or anyone from any of the schools that do well in contests.

Another reason is that even though I am from Ohio, this is the first time I have seen this forum.

[quote=".:Logic:."]I will be active.. if there is something to respond to.. :P  :)[/quote]

There are quite a bit of unsolved problems in this forum. We could work on those.

And we just had the OCTM this morning.  We could always talk about that...

I'll start posting stuff and spamming like Ani.


Given the list of six positive integers 9, 13, a, 12, b, and 8, how many ordered pairs (a,b) will result in a mean of 10 for the entire list?

If the mean of the 6 positive integers is 10, this implies that the sum of all the numbers is 60. Adding the numbers up gives $ 9\plus{}13\plus{}a\plus{}12\plus{}b\plus{}8\equal{}60\rightarrow42\plus{}a\plus{}b\equal{}60\implies a\plus{}b\equal{}18$. Thus there are $ \boxed{17}$ pairs of $ (a,b)$.

normally for n ( an integer) there will be n+1 ways for it to be represented as a sum of two numbers from (0,1,2,3,4......n) so in this case for 18, there are 19 representations, but you must take (0,18) an (18,0) out because 0 is not a positive (nor negative) integer. so once you get to a+b=18 then you do 19-2 so then you get 17. I hope this helps


Find all real-valued functions $f$ on the reals such that $f(-x) = -f(x)$, $f(x+1) = f(x) + 1$ for all $x$, and $f\left(\dfrac{1}{x}\right) = \dfrac{f(x)}{x^2}$ for $x \not = 0$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=54871
Aren't they the same?  :)

Yes, they are exactly the same  :D I just haven't noticed this thread posted by Rushil. Topic closed - discussion about this problem is taken to topic made by Rushil.


Today, during one of my qualifying exams, I realized that I have a very difficult time working through complex reasoning problems under time pressure-- it feels like can't think clearly (or at all), or my mind is seizing up.  Since many of you are highly successful problem solvers under time pressure, could you offer me some advice for how I can get myself to think more clearly?

Hi,

I realize that this is a fairly difficult thing to do. I also struggled a lot (still kind of do).

First, I just get a good look at a difficult problem for some time, namely 45 secs or so, and subsequently, just think about it. And the just go on from there. However, I think that this matures with practice. The more you practice, the more clever ways you can develop quickly under time constraints. The second one, the one about practice worked a charm for me. There is this book [i]Mathematical Circles: The Russian Experience[/i], which consists of a lot of very creative and good problems. They could also be great practice if done in appropriate timed conditions.

Again, these are just somethings that have worked for me.

Regards,
limac

[quote="PKLove"]If you can't work under pressure, you don't know the material. Learn the material.[/quote]
That's quite a bold assumption to make. There have definitely been tests that I've taken where "knowing the material" was not enough to do well, and more complex reasoning techniques are necessary. That being said, good knowledge of the material will also give you confidence and help you focus and stay calm despite the time pressure.

As far as working on questions under pressure, I find that the easiest way to do this is to keep calm and not worry excessively about the time limit. It's important to be aware of time constraints, but do not force yourself to solve problems in a specific amount of time--often, if problems are hard for you, they are probably hard for everyone else as well, and tests may not be designed to be completed. When I work on problems during a test, I tend to go in sequential order, but move on if I cannot proceed in a reasonable amount of time so at least I get a chance to think about every single question and decide which ones to do first.

Out of curiousity: are these written or oral qualifying exams?  (I was about to make some comments about oral exams, but I realized they might be completely unhelpful to you :).)

[quote]Out of curiousity: are these written or oral qualifying exams? (I was about to make some comments about oral exams, but I realized they might be completely unhelpful to you Smile.)[/quote]

Written, unfortunately.  Four hours, enough questions so that you have to start writing on each one almost immediately in order to finish in time.  B- minimum passing grade, but you can carry a C in one of the four subjects.  One opportunity for retakes.  I'm in linguistics, not math, but I would imagine the basic idea is largely the same-- you spend your time in class learning sorts of constructions and tools to solve them, and on the exam you're given variations on the constructions and asked to analyze them.  Or, demonstrate some of the shortcomings of the tools you have, or how some tools work better in some circumstances than others.

It sounds like oral exam skills could be helpful for Q&A sessions after conference talks, or proposal/dissertation defenses, and if they are, I'd be happy to hear any advice you have  :)

Universal advice: Find some old versions of the test, and get familiar with the format and the style of questions, timing yourself.

This isn't a perfect simulation. On the real thing, you'll be full of adrenaline, unlike in practice. From your first post, it seems that adrenaline hurts you. Minimize it; practice to make the test feel as routine as possible, and keep yourself calm.

There is always the tried and true music performance trick. Eat a banana before (and possibly during?). They have natural beta blockers to help keep you from panicking.

I dont see the point of timed tests. In reality....aren't real research problems solved over a matter of years? Most people who are used to doing things fast fizzle out when they are faced with a research problem they cant solve.

The point of timed tests is not necessarily to simulate "real research."  And the "most" in your third sentence is an assertion in serious need of supporting data.  (Of course, very few people used to doing things at [i]any[/i] speed end up persuing research of any sort.)

@ Osud:  Unfortunately, the (small amount of) advice I have about written quals is somewhat specific to the medium of an oral exam on prepared knowledge, which is quite a different experience than giving a research talk.  (And I should say that, unlike many of my European colleagues, my qualifying exams were the first serious oral exam I have ever taken.  And with any luck, the last :).)  I guess two comments I can make with general applicability are that "silence in response to a question is your enemy" and "honesty is your friend" -- it seems to me substantially less awkward and embarrassing to simply cop to not knowing the answer to something than to stand silently or to flail around without any hope of touching on the right answer.

I hope the rest of your quals have gone/are going well!


Find 2007-valued integer n such, that number n^n+(n+1)^{n+1} is compound

I assume that 'compound' means composite :wink:

Let $ p$ be any odd prime.
Then $ p$ is coprime with both $ 2(p\minus{}1)$ and $ 2p\minus{}1$.

Let $ A_p \equal{}(2(p\minus{}1))^{2(p\minus{}1)} \plus{} (2(p\minus{}1)\plus{}1)^{2(p\minus{}1)\plus{}1}$.

Now, modulo $ p$, using Fermat's little theorem, we have
$ A_p \equal{} 1\plus{}(2(p\minus{}1)\plus{}1) \equal{} 0$.
Since $ p<A_p$ it follows that $ A_p$ is composite, and we have found an infinite family of the desired kind.

Pierre.


geometry
combinatorics unsolved
combinatorics

Two persons take turns to choose point on the plane.After each step,the area of the triangle formed by any
three choosed point is not less than 1 nor larger than 1000,the people who can't go on lost.Who has the
winning strategy?

The second player wins.

Hint:
[hide]
Use symmetry.
[/hide]

Let us call the first player A, the second player B.
B picks an arbitrary point $O$ different from the first point chosen by A. For each point $P$ denote the point symmetric to $P$ with respect to $O$ by $P'$.
Each time A chooses a point $X$, B chooses the point $X'$. Why is his move valid? Consider a triangle $ABX'$ (where points $A$ and $B$ are some already chosen points). First suppose that neither $A$ nor $B$ equals to $X$. Then the area of the triangle $ABX'$ equals to the area of the triangle $A'B'X$. Hence the area of $ABX'$ lies in the valid range. Now suppose that, say, $B = X$. Then the area of the triangle $AXX'$ is twice the area of $AOX$, so it equals to the area of $AA'X$, which lies in the valid range.


geometry
trapezoid
geometry proposed
pentagon

Let $ ABCDE$ be a convex pentagon such that $ AB\plus{}CD\equal{}BC\plus{}DE$ and $ k$ a circle with center on side $ AE$ that touches the sides $ AB$, $ BC$, $ CD$ and $ DE$ at points $ P$,  $ Q$, $ R$ and $ S$ (different from vertices of the pentagon) respectively. Prove that lines $ PS$ and $ AE$ are parallel.

The condition $ AB\plus{}CD\equal{}BC\plus{}DE$ implies that $ AP\equal{}SE$.
Denote by $ U$ -the intersection of the segment $ AO$ with the circle $ k$.
Denote by $ V$ -the intersection of the segment $ CO$ with the circle $ k$.
Suppose that $ R$ is the radius of the circle $ k$.
From the power of point $ A$ we get that $ AP^2\equal{}AU\cdot (AU\plus{}2R)$.
From the power of point $ E$ we get that $ SE^2\equal{}EV\cdot (EV\plus{}2R)$.
Since $ AP\equal{}SE$ we get that $ AU\cdot (AU\plus{}2R)\equal{}EV\cdot (EV\plus{}2R)\iff EV\equal{}AU\iff AO\equal{}CO$.
So $ \triangle OPA\equiv \triangle OSE\iff \angle AES \equiv \angle EAP$.
It's pretty obvious now that $ APSE$ is an issoscelles trapezoid, so $ PS\parallel AE$.

[quote="Marius Mainea"]It is obvious that  AP=ES

 If O is center of circle then $ \triangle AOP\equiv\triangle EOS$ and $ \angle OAP\equiv\angle OES$

 Because $ \angle APS\equiv\angle PSE$ yields the conclusion.[/quote] My solution is the same. In my opinion this is far too easy problem for such a competition like JBMO.

[quote="ridgers"]can somebody please post the figure?[/quote]

Here is the figure

[quote="Marius Mainea"]It is obvious that  AP=ES

 If O is center of circle then $ \triangle AOP\equiv\triangle EOS$ and $ \angle OAP\equiv\angle OES$

 Because $ \angle APS\equiv\angle PSE$ yields the conclusion.[/quote]

Can somebody give further explaination about this, especially how did you get AP=ES ?

[quote="Dr N0"]Can somebody give further explaination about this, especially how did you get AP=ES ?[/quote]
First Step:
Since $AB$, $BC$, $CD$ and $DE$ are tangents and since [i]tangents from the same point are equal[/i], we get:
$BP=BQ$
$CQ=CR$
$DR=DS$
Now we add $CR=CQ$ and $DR=DS \Rightarrow CR+DR=CQ+DS \Rightarrow CD=CQ+DS$.
Also $AB+CD=BC+DE \Rightarrow CD=BC+DE-AB$.
We subtitute the second in the first and get $BC+DE-AB=CQ+DS \Rightarrow AB=BC-CQ+DE-DS \Rightarrow AB=BQ+SE$.
Since $BP=BQ \Rightarrow AB=BP+SE \Rightarrow AB-BP=SE \Rightarrow AP=ES$

Second Step:
Now we have to prove that $\triangle{AOP}=\triangle{EOS}$
i. $AP=ES$ as we found above
ii. $OP=OS$ since they are both radii
iii. $\angle{APO}=\angle{ESO}=90$ since the angle between a tangent and a radius is a right angle
Now we know that $\triangle{AOP}=\triangle{EOS}$.
Let $\angle{AOP}=\angle{EOS}=x \Rightarrow \angle{POS}=180-2x$.
$\triangle{POS}$ is isosceles since both $OP$ and $OS$ are radii.
$\triangle{POS}:$ $\angle{OPS}=\angle{OSP}=\dfrac{180-\angle{POS}}{2}=\dfrac{180-180+2x}{2}=\dfrac{2x}{2}=x$
Now we have found that both angles $AOP$ and $OPS$ are equal to $x$ and since they are alternate we can now conclude that $ {PS}\parallel{AE}$

The key is to use the condition $$AB+CD = BC+DE \iff AP+BP+CP+RD = BQ+QC+DS+SE \iff AP = S,$$ so $\triangle OPA \cong \triangle OSE$, and $\angle A = \angle E$. Thus $PSEA$ is an isosceles trapezoid, and $\overline{PS} \parallel \overline{AE}$.

Use equal tangents to line segments to find that AP=ES. With a well drawn diagram, we try to prove that <EPO=<POA. OP=OS, PA=SE, and angles APO and ESO are both right angles, so we have SAS congruency. So, <POS=180-2<POA (CPCTC), or <SPO=<POA, as desired. $\blacksquare$


conics
ellipse
geometry proposed
geometry

Think ellipse that two focus are $ F$, $ F'$ respectively.
$ D$ is arbitrary point on ellipse, and $ C$ is a circle that its radius is $ \bar{F'D} \plus{} \bar{FD}$.
$ \vec{F'D} \cap C$ is $ E$.$ C'$is circle that radius $ \bar{DF}$ and center is $ D$
Then, $ C'$ and $ C$ are tangent at E.

Prove of disprove please


How do I use continued fractions to approximate a given value?

I'll assume you know how to compute the continued fraction. After you have computed it, you can cut it off at any point (i.e. ignore the 1+1/etc stuff since it won't be very big). So if you have a continued fraction 1+1/(2+1/(3+1/(4+1/...))), you can take 1+1/(2+1/3) as an approximation of your number.

What if I am given a value in decimal like 0.702154626108? How do I compute it to a continued fraction?

Remember that continued fraction means an expansion like a0 + 1/(a1 + 1/(a2 + ...)).  Here each ai is an integer (and positive for i :ge: 1).

Let x be the number you want to expand as a continued fraction.  Clearly a0 = floor(x).  By rearranging the continued fraction, we get 1/(x - a0) = a1 + 1/(a2 + ...).  In other words, we get a continued fraction for 1/(x - a0).

By repeating the process above, we get the following algorithm.  Keep repeating the following two steps:
[code]output floor(x)
x := 1 / (x - floor(x))
[/code]
Then we will output each of the ai in order.  You can stop if x becomes an integer or you get tired.  Try it on the number you asked about.

See  [url]http://mathworld.wolfram.com/ContinuedFraction.html[/url] for more details.


$ n>1$  natural number. Prove:

$ \sqrt{2\sqrt{3\sqrt{4...\sqrt{n}}}}<3$

I believe that this is a sum from Engel where he uses the concept of reverse induction :)

Wouldn't reverse induction just prove it for $ n$ less than a specific fixed bound?


algebra
polynomial
number theory unsolved
number theory

Prove that for any natural number n there exist a polynomianl like f such that f belongs to Z[x] and deg f=n and f(0),f(1),....,f(n) are distinct power of 2 .

I have a little idea, I hope it will be helpful... :oops:

Let $ \pi: \{1,2,...,n\}\mapsto \{1,2,...,n\}$ be a bijection maps $ k\mapsto \pi(k)$ and $ f(k)\equal{}2^{\pi(k)},\forall k\equal{}\overline{0,n}$. By the Largrange's interpolation formula we deduce that:

$ f(x)\equal{}\sum_{k\equal{}0}^n \left(\prod_{i\equal{}0,i\neq k}^n 2^{\pi(k)}.\frac{x\minus{}i}{k\minus{}i} \right) \equal{} \sum_{k\equal{}0}^n \left(\frac{(\minus{}1)^{n\minus{}k}.2^{\pi(k)}}{k!(n\minus{}k)!}.\prod_{i\equal{}0,i\neq k}^n (x\minus{}i) \right)$

Then by some way of thinking (may be the way of calculating any coefficients of $ f(x)$ in the formula above :rotfl: ) we will state that there exists a $ \pi$ such that every coefficients of $ f(x)$ are integers and the problem will be solved.

http://www.mathlinks.ro/viewtopic.php?t=1366
http://www.mathlinks.ro/viewtopic.php?p=191725#191725 (see post #4)

Similar problems:
http://www.mathlinks.ro/viewtopic.php?t=50899
http://www.mathlinks.ro/viewtopic.php?t=254980


geometry probability calculus AMC AIME USA(J)MO USAMO

Question: What extra knowledge does one have to know for AMC 10? How about AMC 12?

I am comparing this to a typical high school level textbook on Algebra 2 and Geometry (and Precalculus- if you're talking about AMC 12). This also implies no extra knowledge that Aops gives you :D .

I heard that you can use typical high school material to get up to question number 20 in the AMC 10, but I'm not sure if that's true...

Concepts are more important than level. Some things, like counting, probability, and number theory aren't covered thoroughly in schools. You should focus on those, then increase your knowledge on all subjects.

[url=http://www.unl.edu/amc/f-miscellaneous/faq.shtml#amc1012]Here[/url] (#2) is a table of problem content for past AMC 12s.

[quote]Question: What extra knowledge does one have to know for AMC 10? How about AMC 12? [/quote]

Frequently Asked Questions AMC 1012/ #2, #5 and #10
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions

Out of curiosity, what would be a sample problem likely to appear on the amc 12 which could be grouped under "Calculus & Analysis?"

[quote]what would be a sample problem likely to appear on the amc 12 which could be grouped under "Calculus & Analysis?[/quote]

Teachers' Manual, Page 30, (2009 AMC 12 B, #24)
See the MathWorld.com classification at the bottom of the page.

Classification as Calculus & Analysis is an artifact of the highest level classification in the heirarchical tree of classifications of mathematical topics used by mathworld.com
From an advanced mathematical point of view it makes sense, from the narrower  point of view of the AMC 12 alone it does not necessarily make sense.

Steve Dunbar
MAA Director, American Mathematics Competitions

Thanks for all the quick responses :)

Just wondering
[quote]Q, AIME #2. If you take both A and B tests, but one score is higher than the other, do you get to choose which score to use with your AIME score?
A. We use the higher numerical score, wherever it came from, whatever test it was achieved on.
Q, AIME #3. Can a student take both the AIME and the Alternate AIME (AIME II)?
A. No.  Students are only allowed to take one AIME.  There are no circumstances in which this may be waived. [/quote]

Clearly, question #2 refers to the AMC test. What happens if you take both an AMC 10 and an AMC 12, your AMC 10 score is higher, and both AMC 12 and AMC 10 scores allow you to qualify for AIME? (i.e. what score determines your USAMO index? How about the USAJMO or USAMO? If both qualify, you do get to take both?)

AMC 10 score determines USAJMO (unless your AIME score is >=11)
AMC 12 score determines USAMO

If you make USAMO, you cannot take USAJMO.


For each positive integer $ n$, define teh polynomial

$ P_n(z) : \equal{} 1^3 z \plus{} 2^3 z^2 \plus{} \dots \plus{} n^3 z^n$

Do the zeroes of $ P_n$ lie inside, outside, or on the unit circle $ |z|\equal{}1$?

Well, unless if I'm much mistaken, one of the zeroes is 0 itself, which happens to be in the unit circle.  Assuming that the other zeroes follow suit, I would say the answer is "insde the unit circle." :P

Thanks for the insight, Math Geek.

Posted before: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=527921899&t=12915[/url]

I assume you got this from ACoPS #H34N1. I made a similar post a few days ago. :P :D

[quote="CatalystOfNostalgia"]Thanks for the insight, Math Geek.[/quote]

Hmm, is this veiled sarcasm? :P  Cuz if it is, I'm sorry, but my insight can only go as far as this. :|


In the wiki, "http://en.wikipedia.org/wiki/Binary_relation"

set-like: for every x in X, the class of all y such that yRx is a set. (This makes sense only if we allow relations on proper classes.) The usual ordering < on the class of ordinal numbers is set-like, while its inverse > is not.

I can't understand "This makes sense only if we allow relations on proper classes."...
And what the "ordinar numbers" above paragraph?

Don't worry - just subtleties of fundamenta of set theory! And it is "ordinal", not "ordinar" (it refers to Cantor's ordinals, for example $ \omega$ is the first transfinite one).
But again, don't worry. If interested, read an introduction to set theory book.


agar A,B,C se rase mosalasi bashand sbet konid:
sinA+sinB+sinC>=3/2(sqrt3)

$0<A,B,C<\pi\implies \frac{d^{2}}{dx^{2}}\sin x=-\sin x<0\implies$ concave.  By Jensen's Inequality,
\[\sin A+\sin B+\sin C\leq 3\sin\left(\frac{A+B+C}{3}\right)=\frac{3\sqrt{3}}{2}\]


During the course of this morning (9AM - approx 12 EST), AoPS was moving excruciatingly slow for me. Just around 30 minutes ago (12:15 PM EST), the site didn't respond at all.

XP and Firefox.

Same thing happened for me at that time today.  It also happened yesterday just before 6 PM EST for over ten minutes or so.
Vista IE.

EDIT Then, again, from 1:30 PM to 2:10PM EST.


function
calculus
derivative
real analysis
real analysis unsolved

I know this is really old, but is this the exact statement of the problem?
We don't assume the continuity of f''?
Also, is it possible that there's a typo and f' got switched with f''?

Uh... did you read the question [i]carefully[/i]? It says f is twice differentiable...

as I understand it, that means that the first and second derivatives exist.
So f' must be continuous, but f'' may not be.

That's why I said if you read and thought carefully because differentiability DOES imply continuity. Do think you know the difference between derivative and differentiability?

Every definition I can find says that differentiable means the derivative EXISTS, not that it must be continuous. So,
f' exists, and f'' exists.
f' is continuous, since it's derivative exists.
Why must f'' be continuous?
Unless you're working under a different definition of "differentiable".

And yes, I know that if a function is differentiable, it's continuous.
And of course I know the difference between the derivative of a function and differentiability.

10000th User, there's no need to be unnecessarily pejorative; let's just focus on the mathematics.  I think the question is that we don't know that the second derivative is differentiable, so how do we know that it is continuous?  The function and its derivative are differentiable and therefore continuous, but does that imply that the function's second derivative is continuous?

My questions are legitimate and certainly in no way being derogatory, and I do not understand your use of the word 'pejorative' but I'll let it slide.

The way I see this (this is hard problem BTW), you do not need the continuity of $f''$. Yes I agree what you are saying, that the continuity of $f''$ is ambiguous.

[EDIT] I just talked to my cousin and he said that $f''$ is continuous up to [i]discontinuities of second kind[/i].... I'm sorry I don't know what discontinuities of second kind means. Perhaps I should learn more from him...

actually, now I think I have a counterexample.
Forgive my current ignorance of Latex.

try

f(x) = 1/2*x^2 - 2x + 2     for 0 < x < 1/2

5/2*x^2 - 4x + 5/2            for 1/2 < x < 1

darn, I'm wrong.

f''(1/2) would not exist.

[quote="Peter VDD"]The function $f: \mathbb{R}\rightarrow\mathbb{R}$ is twice differentiable and satisfies $f(0)=2,f'(0)=-2,f(1)=1$.
Prove that there is a $\xi \in ]0,1[$ for which we have $f(\xi)\cdot f'(\xi)+f''(\xi)=0$.[/quote]

Did you gave enough information? $f'(1) = -1/2$ or not ?

The problem is correct as it stands.

The midvalue theorem works for any derivative, even if not continuous, if that's what you're asking. Not knowing this costed me an alike problem last year too, so it does come back from time to time. :)

The problem is correct.
Take $g(x)=\frac{1}{2}f^2(x)+f'(x)$...clearly $g(0)=0$.
Now take the function $h(x)=\frac{x}{2}-\frac{1}{f(x)}$ notice that $h(1)=h(0)=-\frac{1}{2}$ so by Rolle`s theorem there exists $c\in (0,1)$ such that $h'(c)=0\longrightarrow f^2(c)+2f'(c)=0\longrightarrow g(c)=0$ so now apply Rolle on $(0,c)$ for $g$.

[quote="Peter VDD"]The function $f: \mathbb{R}\rightarrow\mathbb{R}$ is twice differentiable and satisfies $f(0)=2,f'(0)=-2,f(1)=1$.
Prove that there is a $\xi \in ]0,1[$ for which we have $f(\xi)\cdot f'(\xi)+f''(\xi)=0$.[/quote]
Let $g(x) = f'(x) + \frac{1}{2}f^2(x)$. Then $g(0) = 0$ by assumption, and the challenge is to show that $g'(\xi) = f''(\xi) + f(\xi)f'(\xi) = 0$ for some $\xi \in ]0,1[$.  Clearly this will follow if we can show that $g(c) = 0$ for some $c \in ]0,1[$, by Rolle's theorem. Note that $g$ is defined on $[0,1]$ and is differentiable there.

Let's also set $h(x) = \frac{2}{x+1}$. Then $h(0) = 2, \, h(1) = 1, \, h'(x) + \frac{1}{2}h^2(x) = 0$.

Suppose now $g(x) \ne 0$ for all $x \in ]0,1[$. This means that either $g(x)> 0$ or $g(x) < 0$ for all $x\in ]0,1[$.

Suppose $g(x) > 0$ for all $x \in ]0,1[$, i.e. $f'(x) + \frac{1}{2}f^2(x) > 0$ on $]0,1[$.  Then $f(x)>h(x)$ on some maximal interval $]0,d[$, since $f(1) = h(1)$, and $f(d) = h(d), \, f'(d)\le h'(d)$. But then $-\frac{1}{2}f^2(d) < f'(d) \le h'(d) = -\frac{1}{2}h^2(d) = -\frac{1}{2}f^2(d)$, which is impossible.

Hence $g(x) > 0$ on $]0,1[$ is impossible. For the same reason, $g(x)<0$ on $]0,1[$ is also impossible.

Hence there exists $c \in ]0,1[$ such that $g(c) = 0$ and $\xi \in ]0,c[$ such that $g'(\xi) = f(\xi) f'(\xi) + f''(\xi) = 0$. ;)

[quote="10000th User"][EDIT] I just talked to my cousin and he said that $f''$ is continuous up to [i]discontinuities of second kind[/i].... I'm sorry I don't know what discontinuities of second kind means. Perhaps I should learn more from him...[/quote]

Since $f''$ is the derivative of a function it has Darboux property so it`s discontinuities are of the second kind. Discontinuities of the second kind are those that are not of the first kind aka jump discontinuities.

[quote="spix"]The problem is correct ...[/quote]Nice and elegant  :)

LOL...but I shall post a solution. I LOVED THE PROBLEM....
Take the function $g(x) = \frac{1}{f(x)}-x/2$ If $f$ has no zero in $[0,1]$ then $g$ is cont and diff in the domain and so $\exists c\in(0,1)$so that $g(c) = 0$ as $g(0) = g(1)$ Now take $h(x) = f'(x) +f(x)^2/2$ Note that $h(c) = 0$ and also $h(0) = 0 $  so apply rolles's theorem to get the desired result. Now say $S$ be the set of zeroes of $f$ in the domain $(0,1)$ as $S$ is bounded so it has an infimum $a$ and a supremum $b$ and as$f$ is cont so $f(a) = f(b) = 0 $ Now note that as $f(0)  = 2$ so $f'(a)$ needs to be $\le 0$ and similarly $f'(b)\ge 0 $ so as $h$ is continuous by IVP $ h(e) = 0$ for some $e\in[a,b]$ now again apply Rolle's theorem on $(0,e)$


Astronomers have estimated that there are 200 billion stars in our galaxy. They believe that only 1 star in 25 might have planets suitable for life. Using this ratio, calculate how many billion stars in our galaxy might have planets suitable for life.

Since $ \frac{1}{25}$ of the stars "work", the answer is $ 200\text{billion}\times\frac{1}{25}\equal{}\boxed{8}\text{billion}$. Remember, it asks how many [b]billion[/b], not the actual number.


How many factors are both in $5^2 + 4^2 + 3^2 + 2^2 + 10$ and $10(5*4*3*2*1)^2$?
Hey guys, this is my first latex text... hehehe...

i think you made a mistake. it should be like this:
[hide]
you're right that the first one is $2^6$. however $(2^6,10\cdot5!^2)\neq2$

it should be $2^6$ So there are 7 common factors, however only one of them is prime. So if you were looking for prime factors, then it would only be 1.

Remember that $a\mid b$ and $a\mid c$ does not imply that $(b,c)=a$
[/hide]

[hide]Oops, I messed up on the second one, we combine the two numbers, and we have to find $\GCF{(64,144000)}=2^6$  7 factors[/hide]


$ x,y,z > 0,$
$ 0\leq {\frac {\left( y \plus{} z \right) ^{2}}{2\, \left( x \plus{} y \right) ^{2} \plus{} 2 \, \left( y \plus{} z \right) ^{2} \minus{} \left( z \plus{} x \right) ^{2}}} \plus{} {\frac {\left( z \plus{} x \right) ^{2}}{2\, \left( z \plus{} x \right) ^{2} \plus{} 2\, \left( x \plus{} y \right) ^ {2} \minus{} \left( y \plus{} z \right) ^{2}}} \plus{} {\frac {\left( x \plus{} y \right) ^{2}}{2\, \left( y \plus{} z \right) ^{2} \plus{} 2\, \left( z \plus{} x \right) ^{2} \minus{} \left( x \plus{} y \right) ^{2}}} \minus{} {\frac {9}{16}}$
BQ

[quote="xzlbq"]$ x,y,z > 0,$
$ 0\leq {\frac {\left( y \plus{} z \right) ^{2}}{2\, \left( x \plus{} y \right) ^{2} \plus{} 2 \, \left( y \plus{} z \right) ^{2} \minus{} \left( z \plus{} x \right) ^{2}}} \plus{} {\frac {\left( z \plus{} x \right) ^{2}}{2\, \left( z \plus{} x \right) ^{2} \plus{} 2\, \left( x \plus{} y \right) ^ {2} \minus{} \left( y \plus{} z \right) ^{2}}}$ $ \plus{} {\frac {\left( x \plus{} y \right) ^{2}}{2\, \left( y \plus{} z \right) ^{2} \plus{} 2\, \left( z \plus{} x \right) ^{2} \minus{} \left( x \plus{} y \right) ^{2}}} \minus{} {\frac {9}{16}}$
BQ[/quote]

$ \sum {\frac {(x \plus{} y)^2}{2(y \plus{} z)^2 \plus{} 2(z \plus{} x)^2 \minus{} (x \plus{} y)^2}}\geq \frac {(2x \plus{} 2y \plus{} 2z)^2}{3(x \plus{} y)^2 \plus{} 3(y \plus{} z)^2 \plus{} 3(z \plus{} x)^2}$

only prove

$ \frac {(2x \plus{} 2y \plus{} 2z)^2}{3(x \plus{} y)^2 \plus{} 3(y \plus{} z)^2 \plus{} 3(z \plus{} x)^2}\geq \frac {9}{16}$

 $ \Longleftrightarrow 5x^2 \plus{} 5y^2 \plus{} 5z^2 \plus{} 37xy \plus{} 37yz \plus{} 37zx\geq 0$

should be:
$ 0\leq {\frac { \left( y\plus{}z \right) ^{2}}{2\, \left( x\plus{}y \right) ^{2}\plus{}2
\, \left( y\plus{}z \right) ^{2}\minus{} \left( z\plus{}x \right) ^{2}}}\plus{}{\frac { \left(
z\plus{}x \right) ^{2}}{2\, \left( y\plus{}z \right) ^{2}\plus{}2\, \left( z\plus{}x \right) ^
{2}\minus{} \left( x\plus{}y \right) ^{2}}}\plus{}{\frac { \left( x\plus{}y \right) ^{2}}{2\,
 \left( z\plus{}x \right) ^{2}\plus{}2\, \left( x\plus{}y \right) ^{2}\minus{} \left( y\plus{}z
 \right) ^{2}}}\minus{}1$
BQ


Question: "The solutions of the equation x ^ 2 + y ^ 3 = z ^ 2 is finite, infinite or empty?"

I did it this way:

If x = z=a
we have, -y^3=0, so y=0, and the solutions are (a,0,a), a is a real number, so knowing that there are infinite real numbers, we have infinite (x, y, z)

It had 10 points!

Or do it like this:

One solution for the equation is $ 1^2\plus{}2^3\equal{}3^2$, so any ordered triple like $ (k^3,2k^2,3k^3)$ does the trick. This method is one of general methods for proving that the given equation has infinitely many solutions (first find one solution, then you get a family of solutions - note that this $ k$ I brought in will cancel because we will get $ k^6$ everywhere)

General solution:

Let's assume we found a solution to the equation $ x^a\plus{}y^b\equal{}z^c$ (this can be applied for more equations, but I took this as an example), let it be $ p,q,r$

Then a family of solutions is given by $ (pk^{\frac ma},qk^{\frac mb},rk^{\frac mc})$, where $ m$ is any number which is divisible by $ a,b,c$ (for example their least common multiple).


Let $ a,b,$ and $ c$ be digits with $ a\ne0.$ The three-digit integer $ abc$ lies one third of the way from the square of a positive integer to the square of the next larger integer. The integer $ acb$ lies two thirds of the way between the same two squares. What is $ a \plus{} b \plus{} c$?

$ \textbf{(A)}\ 10 \qquad \textbf{(B)}\ 13 \qquad \textbf{(C)}\ 16 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 21$

[hide="Solution"]
Let the perfect square be $s$.
Their difference is $(s+1)^{2}-s^{2}$, or $2s+1$
The first number, $100a+10b+c$, is 1/3 the difference, or
$100a+10b+c = \frac{2s+1}{3}$
The second number, $100a+10c+b$, is 2/3 the difference, or
$100a+10c+b = \frac{2(2s+1)}{3}$

Subtract eqn. 1 from eqn. 2

$9(c-b) = \frac{2s+1}{3}$
$27(c-b) = 2s+1$

We have to make sure that s remains an integer, as it originally was.

So we know that c - b cannot be even, and it cannot be zero.
We know then, that they must differ by one (we're in base ten, and they cannot differ by more than two, or else we'd have to carry and have a four digit number)

Thus $27 = 2s+1$, and s = 13.
Thus, our perfect squares are 169 and 196.
The number 1/3 between the two is 178,
giving $a+b+c = 1+7+8 = \boxed{16}$, which is Choice C
[/hide]

Lucky you. Some of us do our algebra the old fashioned way.
(Not a personal insult)
 :P

[hide="Words"] $abc-acb$ is a multiple of $9$ that is one-third the distance between two consecutive squares, so that distance must be a multiple of $27$.  Letting that distance be $27$ the squares are $13, 14$ (easy lemma: $(n+1)^{2}-n^{2}= 2n+1$). [/hide]

[quote="1=2"]I just picked a few squares...

[hide="Does this work?"]"HEY! It does!"

13^2=169
14^2=196

178, 187, so 16.[/hide][/quote]That's what I did too.  I started doing the algebra but then I realized that it was a waste of time because there would only be a couple possibilities, so I did guess-and-check and my first guess (13 and 14) was right! :P


A B C D are distinct coplanar points, no three of which are collinear. If E is a point not in the plane of A, B, C, and D, how many distinct planes are determined by the five points?

[hide]Three non-collinear points determine a plane. Through E, there are $\binom{4}{2}=6$ planes possible by choosing two of the other points. There is one more plane determined by ABCD, so there are a total of $7$ planes.[/hide]

Or

[hide]Three non-collinear points determine a plane. There are $\binom{5}{3}=10$ combinations of three points among the five points. However, ABCD are coplanar so the $\binom{4}{3}=4$ planes defined by the different combinations of these points are the same; we overcount by 3. $10-3=7$.[/hide]


algebra
polynomial
calculus
integration
algebra proposed

Prove that the polynomial $ p(x)$ will be divisible by $ D$ for every integral $ x$ if it is divisible for any $ deg(p)\plus{}1$ consecutive numbers.

Clarified question: "Let $ P(x)$ be a polynomial of degree $ k$, and suppose that $ D \mid P(x)$ for $ k \plus{} 1$ consecutive integers $ x$. Prove that $ D \mid P(x)$ for all integral $ x$."

This is pretty much destroyed by [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=227941]finite difference[/url] tables. $ \Delta^m P(0)$ is divisible by $ d$ whenever $ 0 \leq m \leq k$, and every $ P(x)$ can be written as an integral combination of $ P(0), \Delta P(0) \ldots \Delta^k P(0)$.


Prove that any polygon with an odd number of sides can be constructed from its midpoints (i.e. given the midpoints of a polygon with an odd number of sides, prove that you can construct the polygon with a ruler and compass.)

EDIT: Oh, I did not realize that this was in Engel...

Algebraically, it's a bit of linear algebra: the $ n\times n$ circulant matrix $ \frac12\begin{bmatrix}1&1&0&\cdots&0&0\\0&1&1&\cdots&0&0\\ 0&0&1&\cdots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&1&1\\ 1&0&0&\cdots&0&1\end{bmatrix}$ is invertible if $ n$ is odd. The inverse has rational entries, and rational arithmetic operations are constructible.

What is that inverse?
It's $ \begin{bmatrix}1&\minus{}1&1&\cdots&\minus{}1&1\\1&1&\minus{}1&\cdots&1&\minus{}1\\ \minus{}1&1&1&\cdots&\minus{}1&1\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ \minus{}1&1&\minus{}1&\cdots&1&\minus{}1\\ 1&\minus{}1&1&\cdots&1&1\end{bmatrix}$. The coordinates of the vertices are alternating sums of the coordinates of the midpoints.

It is an easy system to solve without linear algebra...


the inequality is equivalent to
$ (a^{2}\plus{}b^{2}\plus{}c^{2})(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})\geq9\frac{(a^{2}\plus{}b^{2}\plus{}c^{2})}{a\plus{}b\plus{}c}$ which is true since
$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\geq\frac{9}{a\plus{}b\plus{}c}$ when $ a, b, c$ are positive


calculus
integration
analytic geometry
graphing lines
slope
geometry
derivative

Let the curve $ C: y \equal{} |x^2 \plus{} 2x \minus{} 3|$ and the line $ l$ passing through the point $ ( \minus{} 3,\ 0)$ with a slope $ m$ in $ x\minus{}y$ plane. Suppose that $ C$ intersects to $ l$ at distinct two points other than the point $ ( \minus{} 3,\ 0)$, find the value of $ m$ for which the area of the figure bounded by $ C$ and $ l$ is minimized.

I can't think of any other way.
Beat the problem with a stick.

The line $ l$ is given by $ y\equal{}m(x\plus{}3)$
$ C$ intersects with $ l$ when $ |x^2 \plus{} 2x \minus{} 3| \equal{} m(x\plus{}3)$
That has solutions at
$ x\equal{}\minus{}3$, $ x\equal{}1\minus{}m$, $ x\equal{}1\plus{}m$

letting $ y_1 \equal{} x^2\plus{}2x\minus{}3$ and $ y_2 \equal{} m(x\plus{}3)$

$ A\equal{}\int_{\minus{}3}^{1\minus{}m} \minus{}y_1(x) \minus{} y_2(x)dx \plus{} \int_{1\minus{}m}^{1}y_2(x) \plus{} y_1(x)dx \plus{} \int_{1}^{1\plus{}m}y_2(x) \minus{} y_1(x)dx$

$ \frac{dA}{dm} \equal{} y_1(1\minus{}m) \plus{} y_2(1\minus{}m) \plus{} \int_{\minus{}3}^{1\minus{}m}\minus{}(x\plus{}3) dx \plus{} y_2(1\minus{}m) \plus{} y_1(1\minus{}m) \plus{} \int_{1\minus{}m}^{1}x\plus{}3 dx \plus{} y_2(1\plus{}m) \minus{} y_1(1\plus{}m) \plus{} \int_{1}^{1\plus{}m}x\plus{}3 dx$

$ \frac{dA}{dm} \equal{} 2y_1(1\minus{}m) \minus{} y_1(1\plus{}m) \plus{} 2y_2(1\minus{}m) \plus{} y_2(1\plus{}m) \minus{} \int_{\minus{}3}^{1\minus{}m}x\plus{}3 dx \plus{} \int_{1\minus{}m}^{1\plus{}m}x\plus{}3 dx$

Noting that $ \minus{}y_1(1\minus{}m) \equal{} y_2(1\minus{}m)$ and $ y_1(1\plus{}m) \equal{} y_2(1\plus{}m)$

$ \frac{dA}{dm} \equal{} 0 \minus{} \left[\frac{x^2}{2} \plus{} 3x\right]^{1\minus{}m}_{\minus{}3} \plus{} \left[\frac{x^2}{2} \plus{} 3x\right]^{1\plus{}m}_{1\minus{}m}$

$ \frac{dA}{dm} \equal{} \minus{}\frac{m^2}{2}\plus{}12m \minus{} 8$

Setting equal to 0 to find the extrema

$ m^2 \minus{} 24m \plus{} 16 \equal{} 0$

$ m \equal{} 12\pm 8\sqrt{2}$
The second derivative test shows that the plus yields a relative max while the minus yields the min

thus
$ \boxed{m \equal{} 12 \minus{} 8\sqrt{2}}$

What's the range of $ m$ for which the given condition is satified?

Ah yes, that is important.

$ 0<m<4$

Taking the derivative of $ C$ and looking at left and right hand limits around $ x=-3$, we see:
For $ m<-4$, there are two intersections
For $ -4\leqq<0$, there is exactly one intersection
For $ m=0$, there are exactly 2
For $ m\geq 0$ there are only two intersections

The positive root I had did not hold the condition of three distinct intersections.

That's right.  or from [quote="chickendude"]I can't think of any other way.
Beat the problem with a stick.

The line $ l$ is given by $ y \equal{} m(x \plus{} 3)$
$ C$ intersects with $ l$ when $ |x^2 \plus{} 2x \minus{} 3| \equal{} m(x \plus{} 3)$
That has solutions at
$ x \equal{} \minus{} 3$, $ x \equal{} 1 \minus{} m$, $ x \equal{} 1 \plus{} m$[/quote]

$ \minus{} 3 < 1 \minus{} m < 1 \plus{} m$ gives the range of $ m.$

Now our problem is completed. :)


My question is this: do you think time travel will ever be possible for mankind, and if so, what would it be like?

Here's a thought experiment to go along with the question: (taken from wikipedia.org) Suppose a man travelled back in time and killed his biological grandfather before the latter met the traveler's grandmother. As a result, one of the traveler's parents (and by extension the traveller himself) would never have been conceived. This would imply that he could not have travelled back in time after all, which in turn implies the grandfather would still be alive, and the traveller would have been conceived allowing him to travel back in time and kill his grandfather. Thus it becomes a paradox.

Personally, I believe time travel will never really be possible for humans. First of all, there's the problem raised by the thought experiment: if you travelled back in time, you'd end up changing it, but if you change enough of it, it implies you never travelled back in the first place, thus leading into the paradox.

But that's not all; recently, I read "Darkness Creeping: Twenty Twisted Tales" by Neal Shusterman, and another theory came forth: Suppose you travelled forward (or backwards, it doesn't matter) in time a certain amount. For consistency's sake, let's say you travel forward in time exactly one year. After you've travelled that one year, you'll end it in the exact same place you started in. However, every moment, the galaxies are moving away from each other, the planets are shifting, and the universe is expanding. So if you arrived in the exact same spot in the universe you were in one year ago, the earth would've long ago moved on, and you'd arrive in empty space, therefore suffocating and dying. My, what a cheerful thought.  :)

What do you guys think?

The theme of this thread reminds me of the great cult movie:[b]Donnie Darko[/b]

As for the thought experiment, I think the universe would be sort of self-healing. What I mean is, if a man goes back and kills his grandfather, there would be two completely parallel universes formed at that moment.  The first is the universe the man originally came from, where his grandfather lives and the man is able to time-travel back.  The second universe is one where the man's grandfather is killed by the man and the man continues to exist in.  Paradox averted in both universes.  Please tell me if I missed some logical problem here though  :)

As for the possibility of time travel, I think travel to the past may someday be possible.  As the past is fixed, it already happened, it may be possible to travel to one of those fixed points in time.  However, the only way I can see future time travel to be possible is if some sort of Fate exists, that the future is already predetermined. Otherwise, the future is dependent upon countless variables, and so there is the problem of which of the futures you are travelling to.

As for the time travelling into some random part of space, that's very interesting.  It would of course, imply some definite definition of "location" in the universe.  If the popular theory is right that the universe has a center and is expanding, this is possible by a 3-D coordinate method.  However, there still is the problem of the expanding universe:  what happens if you time travel back so far that the universe hadn't expanded to your distance outward yet?  Where do you appear? Is there even a where to talk about?

Anyway, those are my thoughts.  :)  Love the topic!

TT into the future: Could be pretty easy if you can hit high speeds (einstein's theory of relativity)

I think that it is THEORETICALLY possible to travel forwards because of the theory of relativit. (like Mewto55555 said.)  But if you went backwards, then you couldn't change anything.  Everything has already happened in the furutre, so the past cannot change.  The past has already happened where you came from.

If this makes no sense, sorry.  It's really hard to grasp.   It's making my brain go in a circle!  :(

[quote="shentang"]There is also about if time traveling was possible then we would see a time traveler right now or some time before, but we haven't. Or possibly they are very secretive  :ninja:[/quote]

Of course, if I were some time traveler from the future, I wouldn't pick this time period as the one to come back to.  More ancient history would be more interesting by far.  Although I suppose ancient is relative...

[quote="treeswitheyes"]I don't think travel to the past will happen because, it umm already happened[/quote]

Why would this eliminate the possibility of past travel? I would think the fact that its already happened would make time travel all the more feasible.

Still, though, it seems possible that time travelers would pick very distant time periods to travel to.  And if someone did come back from the future to today, I would think one of the following would happen: 1) They have to avoid causing a disturbance and so avoid people or 2) They wouldn't exactly traipse around telling people "Hey, I'm from the future." Really, how many people would believe them?

[quote="westiepaw"]But if you went backwards, then you couldn't change anything.[/quote]
This seems like the most likely reason that travel backwards is ruled out.  If you were to travel to a past, a point in time that is fixed, then you would already break that fixation by your mere presence.  At that point, you simply didn't exist, but then you appear and change the air around you, or step on an ant, or something else that may have a ridiculous chain of events.

Now that I say it, though, it seems possible that the above paragraph I said isn't really a conflict if when time travelers go to a point in the past, their presence there was already a property of that point in time, and so nothing disastrous occurs. And this also would be fine if my first paragraph were correct.

[quote="treeswitheyes"]I don't think travel to the past will happen because, it umm already happened, and travel to future would be kind of useless for us if you couldn't travel back, also the idea that there is already a future we can travel to suggested that we don't have choices, a little like the three fates in Greek mythology[/quote]

Yes, but in a lot of movies and books the "future" is only the absolute future if the decisions people are making continue to be decisions and enacted until the time of the "future"

It seems like a lot of people here say that only time travel forwards would be possible..... I believe the exact opposite. Ignoring the fact that the universe and galaxies are constantly shifting, I believe it'd only be possible to travel BACK in earth's history. Then, if you change something substantial in the past, like someone else in the thread, I believe a parallel universe of sorts will be created. In this sense, every single time a decision is made (it doesn't have to be related to TT), a whole other universe is created. Here's a very simple example: If I were given the choice of vanilla or chocolate ice cream, if I chose vanilla, that'd lead off to one universe. However, there'll be another universe created as if I chose chocolate, and then things would be different in bother universes. Or something like that. Meh.

But I just don't think time travel to the future will ever be possible because, well, the future isn't set in stone. We create out own futures. Sometimes I say I believe in "fate", but not "destiny". For everything good that happens to you in life, something bad is sure to follow. So time isn't random, it has some order in it, like "if x happens, then later y will happen". Or "If y happens, then either w or z will happen". So it's not random, but it's not like certain things are set to happen. =)

@dreambold: As you typed that post you were travelling into the future.

[quote="cyberspace"]We're all born with our own DNA and genes that dictate how we work and react to things.[/quote]  When someone throws a ball at you and you duck, I assure you that the decision to do so was not made at the level of DNA ;)

[quote="cyberspace"]When you think about it, the future is pretty much fixed.  We're all born with our own DNA and genes that dictate how we work and react to things.  So, in a way, we don't really have free will over own actions, as our human nature that we were born with dictates our actions no matter what.  The same goes for any living thing, and nonliving actions are the result of some causing force.  In this manner, the future is already predetermined.[/quote]
What does the philosophical question of free will have to do with time traveling forward? Even if the future is fixed (which you seem to be quite certain of), what if it was predetermined that you would time travel forward, thus altering the future?

The "grandfather paradox" arguments against time travel has a few possible resolutions: 1) a theoretical time traveling device might need a receiving end; thus, any time traveling could not start until the first time machine was constructed (after which the chaos begins?), 2) perhaps traveling forward only is possible, in which case this paradox has no meaning, and 3) the [url=http://en.wikipedia.org/wiki/Novikov_self-consistency_principle]Novikov self-consistency principle[/url], which basically argues that the universe will "conspire" to stay logically consistent, e.g., if the traveler tried to shoot his grandfather, the gun could jam.

@JBL: Well, sure :oops: what I meant though is that its part of how we're wired as humans. Like if you threw a ball at someone in the exact same conditions at two different times, they would react the same. (Or something like that I'm not sure how to explain what I mean)

@Brut3Forc3: Free will has alot to do with it! If all our actions are predetermined, then there is exactly one possible future. This would eliminate the problem of [i]which[/i] future you are traveling to and so would make future travel seem more plausible.

[quote="Brut3Forc3"]Even if the future is fixed (which you seem to be quite certain of), what if it was predetermined that you would time travel forward, thus altering the future?[/quote]

I'm not sure what you mean here.  :huh: If it was predetermined, then the future that is inevitable would be the one where you travel forward into it.

well aren't we assuming that the universe runs on human reasoning and logic?

since the overall speed of our space-time worms are probably constant, the rate at which we travel in time decreases as the rate at which we move in the first three dimensions increases. we are most likely traveling forward in time as quickly as we can when we are still in the first three dimensions (which we are far from being, of course). Similarly, the maximum speed that we can go backwards in time is the same as the speed we can go forward in time, again achieved by being still.
so i believe that instantaneous time travel is impossible, but time travel in general still is. to go back in time, one simply turns their space time worm around in time and continue normal activity in the first three dimensions. however, to achieve a "travel into the future" effect, one must ditch earth and attempt to stay much more still in space than anyone on earth possibly could, so that one goes forward in time more quickly than those on earth.
[quote="cyberspace"]When you think about it, the future is pretty much fixed.  We're all born with our own DNA and genes that dictate how we work and react to things.  So, in a way, we don't really have free will over own actions, as our human nature that we were born with dictates our actions no matter what.  The same goes for any living thing, and nonliving actions are the result of some causing force.  In this manner, the future is already predetermined.[/quote]
this philosophy goes hand-in-hand with the theory that our consciousness is simply moving around in a fixed four dimensional space, and motion in the fourth dimension achieves the "time" effect, as we seem to have no control over it.
but if this were the case, what is the point of heaven and hell? think about it.

[quote="JBL"]Because of this:
[quote="shentang"]There is also about if time traveling was possible then we would see a time traveler right now or some time before, but we haven't. Or possibly they are very secretive  :ninja:[/quote]  (The grammar of the first sentence is only barely comprehensible, but I think the overall meaning is clear.)[/quote]

I've thought about this at some point, and the theory I heard was that at some point a machine (or physical phenomenon) needs to be invented or appear which has the capacity for receiving time travelers. Then perhaps from that point on time travel will be possible. An interesting idea that brings up is that, at the moment such a machine comes into existence, what will happen to the world? It seems like that would such a significant point in time that future time travelers would want to visit it, so who knows what could occur...

Also, travelling in time, going to the same point in a different time wold be very limited.  The earth moves !  You would end up in empty space.