I have recently made some updates on the forum to increase speed. please tell me if the loading time has decreased recently. thanks.

I have not noticed any difference in loading times.

I have noticed something, but that might be a placebo effect...

how do you find the speed of the site now? :D

The speed is super! :D ... I have dial up and now it takes only 4 secs for a page to load! :D

magnificient work! :D

hope that everyone else enjoys our new broadband mathlinks too :D:D:D

I haven't had the chance to test it on my SUPER-SLOW dial-up connection yet (:D), but the speed seems ok (I tested it at my aunt's home).

Hm...for me it seems a bit slower than before... :(

Or maybe it's just the effect of needing to click so many times before I can finally see a single thread...

c'mon. all the subforums are directly accesible from the main forum page. actually you have to make exactly the same number of clicks :D

Well, I don't know if there have been new posts in a particular subforum when I look at the main page.  Obviously since I don't want to check every subforum I have to know which ones have the new posts, which means going into each forum.  Then when I hit "backspace" later on it shows the forum page again.  So 2 more pages...

Actually I'm not complaining so much about the new structure of the board; it would be okay if it wasn't for the fact that it seems to take much longer to load for me.  In the past it used to be about 5 seconds, now it's around 10...  :?

i also feel that this forum slows down...

[quote="Valentin Vornicu"]strange. this is casual or usual ... this server has a better outside connection (much better) than the other one.

anyone else from outside Romania can confirm an increase or decrease in speed of loading ?

PS Valiowk do you have dial up or broandband?[/quote]

i hv broadband, and i feel that the speed has slightly decreased recently..

Yeah it seems that the speed decreased.
Valentin can we have the speed of light ?

unfortunately no :) fiber optics still loose some data on the way :D:D
anyway, loosing some speed is also due to the fact that there are more visitators now in the same time.

but for this we have  the new features + unbanning some of our old friends (galois for example) and they  make up for it :)

okay, i've made some traceroots and indeed getting to mathlinks from central USA takes aprox. 0.15 seconds, and before it took only 0.06 seconds. But as we get closer to Europe the balance reverses. I do not know about Asia yet ...

Small note: The board is working much faster now for me.  However I'm posting this at noon, while normally I post on the board at night, so I don't know whether the huge difference is due to the number of people online...

that's most likely the case. I will try to upgrade the memory pack. :)

the last change killed my connection. it takes from 30 seconds to 2 minutes only to load a page....

what last change? :) since when are you experiencing losts in speed? ...

it is possible that you are because of the staticstics.ro script that I have on the main portal page.

I will remove it in the near future, so don't worry about it.

If it's not that .... then Houston we have a problem :D

:( Valentin your pages are still FAR from optimized...

Just look at the source code, you can save like 4K/click, just by removing the comments from your overall_header.tpl file (/templates/subSilver/...)

won't get the square root of the filesize but it'll help :)

since you made this change, you restored the portal... I think you made this change today!

because till yesterday everything was super fast! :D

I have not noticed any changes in the loading speed of the pages. Only the portal page loads slower because of that script.

Won't be for long, don't worry.

About page optimizing -> I need those comments in overall_header to easily make modifications in the code.

When the final version will be released I will purge them.

Nobody seems to ever remember this is still beta :D

well I remember that this is the BETA VERSION! but i guess that we should tell u when things were better or when not!

cheers! :D :D

of course, I was just explaining why I still have comments in overall_header :) ... be aware a new upgrade is due to come in a half of hour :D

to get back on old stuff, it took me over 10 minutes to make this reply :(

Now it's not so slow anymore (although not precisely as quick as before the change of servers, but it is okay)... Thanks Valentin!

  Darij

but I did nothing, neither has the admin of the server.
well I took a look at the bandwidth and RAM usage of the server at that time. Another couple of sites are hosted on the same server, and there were a lot of people on the other sites, so no RAM was available (and no bandwidth either).
Basically we need more RAM, and maybe a new procesor for the server (now it's a 500 Mhz PIII with 256 SDRAM which really sux).

okay, so we moved the data base on a different server (pIII at 1Ghz with 512 Mb RAM, but again it's not a dedicated thing).

hopefully another 256 Mb of RAM will be added to the php content, so that we will have 512 Mb of Ram for the php part too.

and maybe in the future I will find the resources for a dedicated server to host this.

untill then, please post/email any problems regarding speed (and other stuff) you may have.

I noticed that the new server is actually slower than the previous one.Is is it just me noticing these speed differences?


$ \frac{s^2}{2r_a}\equal{}p\minus{}a$, then you can cosctruct the incircle and the A-excircle and the tangent line of these is the side BC.


Try to get 8 on weekdays.

Get 10 on weekends without trying.

25 hours a day.

No, It's more like 7-10.

Probably around 8 on weekdays and maybe a bit more, though usually also 8, on weekends.

Heh.

I found that as I got older I required more sleep <_<.

Those that can get more than 8 are pretty lucky since the body needs a little more than 8 hours of sleep. Those that get just 8 or a bit less are the most unlucky since they spent so much time sleeping and are still going to be very tired in the morning since they didn't get more than 8 hours.

Last year, I got 9 on weekdays and 9 on weekends; this year, it's more like 6-7 on weekdays and 11 on weekends.... If it's a day with minimal homework, then I guess 8 to 9 hours, but that isn't often.

It's been around 8 hours on weekdays, but recently it's been going down to around 6 hours, which isn't good, I know.
Weekends: Around 9-10 hours.

9-10 on weekdays, never below 8, and around 12 during weekend, don't go below 10

9-10 hours usually everyday. Unless when I'm at my friend's house for a sleepover then my sleep amount is nonexistant.

[quote="abacadaea"]yeah whats your guys BEDTIME mines is 9:30 and im in 7th grade. rly anoying[/quote]

Yeah, I know its a bad habit, but my sleep time is usually 11:00 ish.


trigonometry
Pythagorean Theorem
geometry

In triangle ABC, AB=26, BC=30, AC=28. Let D be the midpoint of AC. A semicircle with diameter AC is drawn opposite the triangle. The line through D perpendicular to AB intersects the semicircle at E. The line through D perpendicular to BC intersects the semicircle at F. CF intersects DE at G. Find GE.

For the way I did it, hints:
[hide="1"]Vertical angles[/hide]
[hide="2"]Half angle[/hide]
[hide="3"]Law of Sines[/hide]

[hide="Solution Outline"]
The arithmetic looks really bad, but it's alright and you don't have to do that many intermediate simplifications. Basically we start with the Pythagorean Theorem to find the obvious lengths (without drawing any extra lines). Then, we draw parallels and similar triangle bash. The answer is $ \boxed{\frac {196}{27}}$.
[/hide]


geometry
3D geometry
combinatorics unsolved
combinatorics

There is a cube with length of side $n$.
The cube consists of $n^3$ boxes.
A rook in a cube attacks all the boxes placed in horizontal,vertical and height "line".
1)What is the minimal number of rooks that we need such that all the boxes will be under attack.
2) Where do we need to put the rooks to satisfy this condition?

[hide] for n even: $\frac{n^2}{2}$ and for n odd: $\frac{n^2+1}{2}$
[/hide]

It's been posted before. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=cubic&t=6023]this[/url], for example.

[quote="tim1234133"] for n even: $\frac{n^2}{2}$ and for n odd: $\frac{n^2+1}{2}$
[/quote]
How do I need to put the rooks to get only $\frac{n^2}{2}$ or $\frac{n^2+1}{2}$

The example given in Engel's [i]Problem Solving Strategies[/i]:

For n=8 (as an example), taking a view of the square end of the cube

 -              -                -             -                8	 5	 6	 7
 -              -                -             -              	 7	 8	 5	 6
 -              -                -             -              	 6	 7	 8	 5
 -              -                -              -             	 5	 6	 7	 8
 4	 1	 2	 3	 -	-	-               -
 3	 4	 1	 2	 -	-	-	-
 2	 3	 4	 1	 -	-	-	-
 1	 2	 3	 4	 -	-	-	-

where the number 'n' means a rook in the nth place in the shown row. (and - means no rooks in that row) If n is odd, the two squares in the top right and bottom left
will be of side length $\frac{n-1}{2}$ and $\frac{n+1}{2}$


A charity sells 140 benefit tickets for a total of $ \$2001$. Some tickets sell for full price (a whole dollar amount), and the rest sells for half price. How much money is raised by the full-price tickets?

$ \textbf{(A)} \ \$782 \qquad \textbf{(B)} \ \$986 \qquad \textbf{(C)} \ \$1158 \qquad \textbf{(D)} \ \$1219 \qquad \textbf{(E)} \ \$1449$

http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=63013  :D

[hide]Let $p$ be the price of the full amount ticket, and $x$ be the number of people buying these full price tickets.

We have $px+\frac{1}{2}p(140-x)=2001$.
$p(70+\frac{x}{2})=2001$.

Since $p$ is a whole number amount, p is a factor of 2001. If we want $x$ to be also a positive amount, we must have that the factor $70+\frac{x}{2}$ is greater than 70 (if it isn't, $\frac{x}{2}$ would come out to be a negative number). Another restriction is that $x<140$, thus the factor $70+\frac{x}{2}$ is also less than 140.

The only pair of factors that work for this is $p=23$, $70+\frac{x}{2}=87$
And $x=34$. Thus the price from full price tickets is $23\cdot34=782$. [/hide]

Oh well.. It's in MC forum so I can't control it. I rather prefer it to have it on Getting Started so let's just keep this one as Olympiad Resource problem.

Thanks for pointing out this though.

[hide="Answer"]We have the simultaneous equations $nx+\frac{n}{2}y=2001$ and $x+y=140$. Multiplying the first equation by $2$ and the second by $n$, we have $2nx+ny=4002$ and $nx+ny=140n$, so $nx=4002-140n\Rightarrow n(140+x)=4002$. $4002=2\cdot 3\cdot 23\cdot 29$, so $n$ is one of those numbers. $2$ and $3$ are obviously far too small for the price, so we try $23$. This gives us $x=34$, and so $y=106$. Checking, $34\cdot 23+106\cdot \frac{23}{2}=2001$, so we have our answer. $23\cdot 34=782\Rightarrow \boxed{A}$.[/hide]

Copy and paste answer from the MC thread. :D

[hide]
$p$=price of ticket
$x$=full price tickets
Since 2001 factors into 3x23x29, $(x+\frac{140-x}{2})p$ p has to be either 3,6,23,46,29, or 58.  If all of the tickets were full-price, and it cost 6 for a full-price ticket, the amount paid would be 1040.  That eliminates 3 and 6.  If all of the tickets were half-price, and it cost 46 for a full-price ticket, the amount paid would be 3220, too much.  That eliminates 46 and 58.  We are left with 23 and 29.  140 half-price tickets at $\frac{29}{2}$ dollars each would make 2030 dollars, too much.  We are left with 23 dollars for each full-priced ticket.

$(x+\frac{140-x}{2})23=2001$
$x+\frac{140-x}{2}=87$
$x+140=174$
$x=34$ <-full priced tickets

Multiply by price of ticket:
$34\cdot 23=782$

So 782 dollars.
[/hide]

Why cant we not just set the price to be 2p and then 1/2(2p)=p for the half price. I did that and I couldn't get the system of equations to work?


To the guest who insist about knowing Riemann Hypothesis, please go away and stop asking about Riemann Hypothesis with an [b][i][u]ANONYMOUS or GUEST[/u][/i][/b] login. Speak up and identify yourself. Oh well, it's probably a prank like I did once to Valentin. :P

well why doesn't our admin just disables guest posting then? :?

that's what admins are for, right? :)

I don't disable that because I want this to be an open board. of course there are some natural limitations imposed. I only left guest posting possibility within the non-math topics ...

However if some guest become annoying, his IP will be banned, don't worry about it ;)

this is the oldest topic i can actually bump, is this the longest bump or what?


The numbers c(n,k) are defined for nonnegative integers n>=k such that c(n,0) = c(n,n) = 1 for all n>=0 and
c(n+1,k) = (2^k)(c(n,k)) + c(n,k-1) for n>=k>=1.
Prove that c(n,k) = c(n,n-k) for n>=k>=0

As I remember this one is ISL1998,and the solution can be found in John Scholes' website http://www.kalva.demon.co.uk/short/soln/sh98a4.html

Besides, it has been posted by orl on the forum. Try the "Algebra" section.


Define &a& to be the sum of the cubes of the digits of a, e.g. &123& = $1^3+2^3+3^3$ or 36.

1.  Compute the ordered pair (a,b) if a and b are the only three digit numbers for which &(&a&)& = a.  In other words, find the three digit numbers that are equal to the sum of thet cubes of the digits of the sum of the cubes of its digits.

2.  What is &(&(&(&(&(&(&(&(&(&(123)&)&)&)&)&)&)&)&)&)?

3.  What will happen if you continue &ing a number?

[hide]1. $(153, 371)$

2. 153

3. The number will eventually become stationary.[/hide]

Does that satisfy your wonderences?

[quote="surge"]mbabbitt, how did you get those solutions (for number 1)?
[hide]
$x^3+y^3+z^3=100x+10y+z$?
$\text{for} x,y,z \subseteq{Z}, 0 \le x,y,z \le 9$
0, -1, 1 are also solutions, btw.
 [/hide][/quote]
Are $-1$, $0$, and $1$ three digit integers?  ;)

Yeah! I agree with LordoftheMorons! Try them!

For #1, find a soltution where &a& does not = a but &&a&& does.

Explain #3.  Where will it become stationary?

Consider 352. It doesn't "become stationary." It goes 352, 160, 217, 352, 160, 217...

yep, take every number that divided evenly by 3, and it will reach 153!
prove it!

Call a 2& a number where the & of the & of itself is itself (136, 244)
153 is a 1&
It seems that a number whose digits add to some multiple of 3 will stop at 153.  All numbers whose digits add to some multiple of 5 will end up in a loop of (136, 244).
Digit sums that are multiples of 11 go to 371.

Is this true?

Is there a pattern?

mistake, what about numbers that are multiple of 3 and also 5? like 78...it will reach 153.
I don't know what about numbers that has 1 or 2 remainder when divided by 3.
so please, I still waits for the solutions and for the hints  :P
and a similar ns problem: choose four digit number, such that his digits decrese form small to left, and decrese its
 inverse: 4321-1234=3087
8730-378=8352
8532-2358=6174
7641-1467=6174
is will always reach 6174, cool! what is the proof and what about other nth digit number?
tnx in advance!  :)

[quote="Altheman"]i am pretty sure that for any integer, the sequence will repeat eventually...

clearly, for huge numbers, we will significantly decrease the number of digits until the number is below 10000, then it cannot get above 10000 since 4*9^3<10000, so there are only finitely many numbers it can take on, and after it takes on a repeated number, the sequence repeats[/quote]

Altheman is exactly right.  The sequence is guaranteed to be periodic after a certain point, and if it happens to also hit a number $n$ such that &n& $= n$, it will repeat in a sequence of period $1$.

To be more precise, a $d$-digit number $n$ has value $10^{d-1} \le n < 10^{d}$, but &n& $\le d \times 9^{3}$, so for large values of $d$ the sequence decreases until it must become periodic.


Hello everyone,

Could anyone please explain to me how trigonometry charts are used for angles over 45°?

My table is as such:

[url]http://img511.imageshack.us/my.php?image=trigtable1ja8.jpg[/url]

[url]http://img510.imageshack.us/my.php?image=trigtable2mb4.jpg[/url]

Thank you very much in advance.

You look for the angle you want to find the trig function of and then move across the row to that function for the answer.  :maybe:

Thanks for your reply.

That method only works for angles UNDER 45°. If you look at my table, the angles OVER 45° have the same trigonometric values as to the angles UNDER 45°.

Can anyone explain this?

Do you mean it gives values for the 6 trig functions only for angles up to 45?  So then you have to use $\sin x=\cos(90-x)$, $\csc x=\sec(90-x)$, and $\tan x=\cot(90-x)$.

So for example if you want the cosine of 70 degrees, you know it's the same as the sine of 20.

Thanks for your reply.

It seems to give values over 45° but I don't think I am interpreting them correctly.

For example, with 36°, it lists the sin, cos, etc.. values and on the far right hand corner, there is a 54.0°. However, I don't think tan(36) is  equal to tan (54).

Another example is 38.8°, the values are there for tan(38.8), but it also lists 51.2° on the far right corner. Again, ...(38) seems to be listed as ...(51.2).

Could you please clarify what my table is trying to denote?

I'm pretty sure that is just there to represent the $\sin\theta=\cos90-\theta$, etc.  I'd just ignore it because you don't really need it.

Also, for angles >90, etc, you just have to see what quadrant it is in to determine the correct sign of the angle.

The sign of the function is important.
You can remember the sign like this:
Quadrant 1: A       All functions are positive
Quadrant 2: S       Sin and csc are positive
Quadrant 3: T       Tan and cot are positive
Quadrant 4: C       Cos and sec are positive.

I remember this by saying to myself (ANOTHER STUPID TRIG CHART)!

Thank you very much for all your help and your replies.

Actually, I looked over the trig functions in opposite order: [quote]Look at the very bottom of your table... I guarantee the trig functions will be there, in opposite order. Those are the functions you want to look up for angles > 45.[/quote]

(Smacks forehead)

Thanks again!


geometry
perimeter
trigonometry
area of a triangle
Heron\u0027s formula

Since some people in this forum do not know about the Heron's Formula, this is a brief topic explaining it.

The Heron's Formula is used to find the area of a triangle lest it be equilateral, isoceles, or scalene.

Heron's Formula:

$ p\equal{}$ perimeter of triangle
$ S\equal{}\frac{p}{2}$
$ A\equal{}$ one side
$ B\equal{}$ another side
$ C\equal{}$ the last remaining side

\[ \sqrt{(S)(S\minus{}A)(S\minus{}B)(S\minus{}C)}\]

For exercises to practice this formula and get used to it, PM me!

Also see the entry in the AOPS wiki:

http://www.artofproblemsolving.com/Wiki/index.php/Heron%27s_formula

It's lots more fun proving it rather than stating it.

How would you prove it? I tried to expand s as $ \frac{a\plus{}b\plus{}c}{2}$ and get

$ A\equal{}\frac{\sqrt{(a\plus{}b\plus{}c)(b\plus{}c\minus{}a)(a\plus{}c\minus{}b)(a\plus{}b\minus{}c)}}{4}$. But I don't feel like multiplying that all out..

There's a trig way to do it an a normal geometry way to do it. (Guess which one takes less work?)


tamame p(x,y)ha ra biyabid ke;
p(x+1,y+1)=p(x,y)

[hide="fekr haye avaliye"]
hamaye $p(x,y)=c$ ha dar sharayet sedgh mikonan
$p(x,y)=k(x-y)$ dar sharayet sedeghe mikonad

tavabe e tanavobi ke
dore tanavob e $1$ dar $x$ va $y$
ya doreye tanavobiye $2$ dar $x+y$
dashtebashan dar masale sedgh mikonan.
masalan:

$p(x,y)=\cos{(2\pi\cdot x + 2\pi\cdot y)}$
$p(x,y)=\sin{(2\pi\cdot x)}\cdot \cos{(2\pi\cdot y)}$
$p(x,y)=\sin{(\pi\cdot (x+y))}$

etc.[/hide]


Let $a,b,c>0$. Prove that $( \sum a^2 )( \sum \frac{1}{a} ) \geq 3 \sum a $.

(made by Peita V.)

Just apply the Chebyshev inequality for the oppositely sorted sequences $\left(a^2,\ b^2,\ c^2\right)$ and $\left(\frac{1}{a},\ \frac{1}{b},\ \frac{1}{c}\right)$.

  Darij


$a+b+c=0$ $\leftrightarrow$ $\overline{a+b+c}=0$ $\leftrightarrow$
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$ $\leftrightarrow$ $ab+ac+bc=0$ $\rightarrow$ $a(b+c)=-bc$ $\leftrightarrow$ $a^{2}=\frac{1}{a}$ $\leftrightarrow$ $a^{3}=1$ $\leftrightarrow$ $a^{2004}=1$.Analogously we obtain $b^{2004}=c^{2004}=1$.


inequalities
calculus
geometry
trigonometry
3D geometry
pyramid

Let two constant numbers $ R,r$ with $ R>r$ and let $ x$ be a positive real number such that $ x\leq r$ .

Find the maximum (no calculus  :D  since then it is very easy ) of

$ A\equal{}x(\sqrt{R^2\minus{}x^2}\plus{}\sqrt{r^2\minus{}x^2})$

Construct $ \triangle MNP$ such that $ MN \equal{} r, MP \equal{} R$ and $ x$ is the altitude from $ M$. Then $ A$ is twice the area of that triangle, which is obviously maximized if $ \angle NMP \equal{} 90^\circ$, since $ [MNP]\equal{}{Rr\over 2}\sin\angle NMP$. Hence $ A_{\max} \equal{} Rr$ for $ x_0 \equal{} {Rr\over\sqrt {R^2 \plus{} r^2}}$

Very nice  :lol:  . Now try the following problem . If $ R>r$ and y,z>0  and $ y,z\leq r$ then find the max of

$ B\equal{}yz(\sqrt{R^2\minus{}y^2\minus{}z^2}\plus{}\sqrt{r^2\minus{}y^2\minus{}z^2})$

See the attached diagram.

$ A$ is six times the volume of the pyramid pictured, which also can be calculated as the one third of a product of $ [MNP]$ and spacial altitude from $ Q$. Since $ \triangle MQF$ is also right-angled, its altitude can't exceed $ {MF\over 2}$. Therefore, we can independently maximize $ [MNP]$ and the altitude of $ \triangle MQF$, yielding

\begin{eqnarray*}MF(\max)={Rr\over\sqrt{R^2+r^2}}\implies y_0=z_0={Rr\over\sqrt{2(R^2+r^2)}}\implies A_{\max}={R^2r^2\over 2\sqrt{R^2+r^2}}\end{eqnarray*}


trigonometry
complex numbers
real analysis
real analysis unsolved

Consider the sum Sin(0) + Sin(1) +Sin(2) ... it will never converge to one number but the upper bound seems to be around 1.95. My question is what constant is it? It does not seem to be a multiple of pi. Any help is appricated Btw the website looks cool.

That series is the imaginary part of $\sum_{k=0}^{n}e^{ik}=\frac{e^{i(k+1)}-1}{e^{i}-1}$. The imaginary part of $\frac{e^{i\theta}-1}{e^{i\alpha}-1}$ (for fixed $0<\alpha<2\pi$) is maximized when $\theta=\pi+\frac{\alpha}2$, with value $\text{Im}\frac{-(e^{i\alpha/2}+1)}{e^{i\alpha}-1}= \text{Im}\frac{-1}{e^{i\alpha/2}-1}= \text{Im}\frac{-1(e^{-i\alpha/2}-1)}{|e^{i\alpha/2}-1|^{2}}= \frac{\sin\frac{\alpha}2}{4\sin^{2}\frac{\alpha}{4}}=\frac12\cot\frac{\alpha}4.$

When $\alpha$ is an irrational multiple of $\pi$, $\alpha(n+1)$ comes arbitrarily close to this point. Our upper bound is $\frac12\cot\frac14\approx 1.95815868$.


Show that there exists a unique $ f: \mathbb{R}^{\plus{}}\rightarrow\mathbb{R}^{\plus{}}$ such that $ f(f(x))\equal{}6x\minus{}f(x)$.

[hide="Hint"]For all $ x > 0$, fixed it. And see the sequence $ u_n \equal{} f(...f(x)...)$ ($ n$ of $ f's$).[/hide]

Dude: there is a Putnam forum.  Putnam problems belong there!  Also, in general, when posting contest problems it's nice to give the year and problem number in addition to a descriptive title (it makes searching much easier).


[hide]When factoring, it is often times useful to think of the expression as a function and look for the zeros of the funtion.  In this case, we have \[ f(a,b,c)=(a+b+c)^3-(a^3+b^3+c^3). \]

Notice that $a=-b$ is a zero.  Thus $a+b$ is a factor.  Likewise, $b+c$ and $c+a$ are zeros as well.  Then, you need to find the constant.  You can just do that by multiplying out $(a+b)(b+c)(c+a).$[/hide]


I thought it was a bit easier than usual - it seemed like there were a lot of problems, even near the end, that only required one idea and nothing else.

Hmm I found it trickier than normal, particularly the problems near the middle of the test. That's just me.

I thougth the first 20 were basic.

I couldnt do any of the last 5 during the alloted time.  Although looking back, 22 is a pretty famous type of problem and 25 was just the tangent addition formula.

Did anyone notice there were NO counting/probability probelms on this test???  What was up with that?

And rarely any Euclidean Geometry.

I've only seen the last few problems, but it seems that the last few were easier than normal.

[quote="greentreeroad"]I think it is harder. Especially number 23 and 24 which seem no way to tackle at first.[/quote]

Well, for 24 you just have to take a few logs and the problem quickly becomes easier...

For 24, I experimenting using 2 and get 6, so x seems yielding x+4.

I thought it was a bit easier.. I had ~30 minutes to work on just #23 but flipped a sign, and ended up miscounting both of 24,25 (.. counted period of 25; of course just checking parity would've showed it couldn't be that).

easier. I actually knew what to do on every problem (except #23 but that was because I was like "oh I'm so slow and have only 20 minutes left for last 3 ... let's skip this one it looks sucky after 2 minutes of work")

but... the only actually hard problems were like... 22, 23, 24 in my opinion

I thought it was significantly easier than usual. Number 23 was pretty ugly and so I didn't have time for it, but 25 was trivial if you saw it and 24 and 22 were not too bad at all.

18 and 24 were tricky.

I felt the questions were quite easier, but that could be because there weren't that many counting questions or other easy to mistake questions. 1-22 were all approachable for me, though some of the middle questions were a little trickier than usual, I was going through them slower than I wanted, but made up time because some of the latter ones were easier than usual. I didn't try 23-25, so no comment on them.

(There were too many questions using Vieta's sum-of-roots.)

[quote="pianoforte"](There were too many questions using Vieta's sum-of-roots.)[/quote]

I agree. There were maybe three of them? I checked them all other ways to be sure though.

I think 23 and 24 are the only hard problems. I had trouble with #15 and kinda bruteforced that but otherwise, the first 20 or so problems were pretty quick. Also I think for the most part, this amc is less computational.

The first 22 were all about the difficulty I had expected.  I looked at 23, didn't want to do it, guessed and ended up guessing it right  :P .  It was hard not to be off by 1 on #24 as well.  25 was easy, except that at first I thought that $ tan(60)\equal{}\frac{1}{\sqrt{3}}$ rather than $ tan(30)$  :oops: . So that messed me up.  On 18 it was a bit tricky I thought to see the $ k\equal{}4$ case quickly.

So yea, 23 and 24 were the only unusually hard ones.  And hopefully on the B there will be a few combinatorics problems to compensate

The AMC in my opinion was slightly strange this year. It leaned heavy on polynomials while completely omitting combinatorics/counting problems.
As far as difficulty goes, I thought that the first 13 was a lot easier than usual, while the middle section (14-20) was pretty tricky, especially #18, where I suspect a majority of the test-takers put 6 instead of 7. However, the last 5 were a lot more approachable for me. I got 1-22 correct but given enough time, I think I could've gotten at least 24 and 25. (23 was by far the hardest problem on this test IMO)

I think the test this year was a bit easier than usual, but annoying with all the possibilities for stupid mistakes  :|

I answered 1-22 (and got all of them right) but definitely could have gotten #25 before time was called (had I not made an arithmetic error).
I probably couldn't have done this well if there were some counting problems between 1 and 22.

I hope that the AMC 12B will make up for the lack of counting & geometry problems in the AMC 12A...


projective geometry
geometry unsolved
geometry

Hi folks, im having a little difficulty with this (seemingly) simple quadrilateral problem

ABCD is a quadrilateral with the angles DAB and ABC obtuse. AC and BD are
diagonals meeting at O. Point E lies on BD such that line AE is parallel to BC.
Point  F lies on AC such that line BF is parallel to AD. Prove that EF is parallel to DC

Would really appreciate some help on this one.
Cheers,

Very simple problem. Note that $ \triangle BOF \sim  \triangle EOA$ and $ \triangle AEO \sim \triangle COB $ gives

$\frac {OF}{OA} \equal{} \frac {OB}{OD} \ , \ \frac {OA}{OC} \equal{} \frac {OE}{OB} \Longrightarrow \frac {OD}{OE} \equal{} \frac {OC}{OF} \Longrightarrow EF \parallel DC.$

cheers buddy - I knew it was simple but didn't think it was THAT easy. Feel like a bit of a nob now. Anywho, thanks for the quick reply. Have a good rest of the day :D

Dear Lym and Mathlinkers,
here is also an application of the Pappus 's theorem:
consider the hexagone AEFBCDA built on the lines AC and BD.
Qincerely
Jean-Louis


a 2100 kg truck traveling north at 41 km/h turns east and accerlerates to 51 km/h (a) whats the change in the trucks kinetic energy? What are the magnitude and direction of change in its momentum?

[color=blue]The change in the kinetic energy is $\large\Delta K=\frac{1}{2}m(v_f^2-v_i^2)$

To finnd the change in the momentum, draw apicture and you'll have the answer.

Remember that $\large\vec{\Delta p}=m\vec{v_f}-m\vec{v_i}$[/color]

[quote]draw apicture and you'll have the answer[/quote]

Isn't there any other way of doing it ?? I'm asking since representing the vector as a picture and all that stuff might lead to plenty of errors, especially while you're measuring the dimensions of the resultant in the picture you've drawn........{I'm not really sure if the picture method is accurate enough..}

[color=blue]Because the initial and final vector of the momentum are in two different directions, so in order to find the direction of the change in momentum, you have to draw a picture and find out the needed angle.[/color]

Seems that what phucnv87 want to say is to solve the problem geometrically to find the modulus and an angle that define the direction of $ \large\vec{\Delta p} $ [hide="for instance"] a rectangular triangle [/hide].

[quote="ahaanomegas"][hide="My Solution"]
$
(a)
\large\Delta KE=(1/2)m(v^2 - v_0^2) = (1/2)(2100 kg)((51 m/s)^2 - (41 m/s)^2) = 9.7*10^5 J
$
(b) The change in momentum is:
$
m\large\Delta v
= (2100 kg)(10 m/s) = 2.1 * 10^4 N*s (North)
$
[/hide][/quote]

its 51km/h and not m/s (same for the other value)

[quote="ahaanomegas"][hide="My Solution"]
(b) The change in momentum is:
$
m\large\Delta v
= (2100 kg)(10 m/s) = 2.1 * 10^4 N*s (North)
$
[/hide][/quote]

Also, the truck changed direction from north to east, so $v_i = (41 km/h)\hat{j}$ and $v_f = (51 km/h)\hat{i}$.
(Here, $\hat{i}$ represents the $+x$-axis or east, and $\hat{j}$ represents the $+y$-axis or north.)

Thus, $\Delta v = (51 km/h)\hat{i}-(41 km/h)\hat{j}$ NOT $10 m/s$.


IMO Shortlist
4-variable inequality
cauchy schwarz
Holder
Inequality
algebra

More generaly we prove that if $a_1,\cdots,a_n$ are positive real numbers with sum $s$ and such that $a_1a_2+\cdots+a_na_1=k^2$, then for $r=0,1,2,3$ we have:
\[ s_n= {{a_1^r}\over{s-a_1}}+{{a_2^r}\over{s-a_2}}+\cdots+{{a_n^r}\over{s-a_n}}\geq {{(2k)^{r-1}}\over{ (n-1)n^{r-2}}}. \]
Indeed, the function $x\mapsto {{x^r}\over{s-x}}$ is convex for $s>x\geq 0$ and $r=0,1,2,3$. Hence by Jensen' inequality we have
\[ s_n\geq {{n(s/n)^r}\over{s-s/n}}={{s^{r-1}}\over{(n-1)n^{r-2}}}. \]
On the other hand, we know that $(a_1+\cdots+a_n)^2\geq 4(a_1a_2+\cdots+a_na_1)=4k^2$ and thus
\[ s_n\geq {{(2k)^{r-1}}\over{(n-1)n^{r-2}}}. \]
The stated problem corresponds to: $n=4, r=3$ and $k=1$.

or to write: x^3/(y+z+w)=x^4/x(z+y+w) for all, then use cauchy-schwarz! we will arrive to a simple thing...

 :D

[u][b]Problem:[/b][/u]Let $ a;b;c;d$ be positive real numbers satisfying $ ab\plus{}bc\plus{}ca\plus{}da\equal{}1$.Prove that:
$ \sum\frac{a^{3}}{b\plus{}c\plus{}d}\geq\frac{1}{3}$
[i]My solution[/i]
By AM-GM we have:
$ \frac{a^{3}}{b\plus{}c\plus{}d}{18}\plus{}\frac{a}{6}\plus{}\frac{1}{12}\geq\frac{2a}{3}$
Similar,we have $ \sum\frac{a^{3}}{b\plus{}c\plus{}d}\geq\frac{a\plus{}b\plus{}c\plus{}d}{3}\minus{}\frac{1}{3}$
Note that $ ab\plus{}bc\plus{}ca\plus{}da\equal{}(a\plus{}c)(b\plus{}d)$,so $ (a\plus{}b\plus{}c\plus{}d)^{2}\geq 4(ab\plus{}bc\plus{}cd\plus{}da)\equal{}4$=>$ a\plus{}b\plus{}c\plus{}d\geq 2$
Now,we have $ \sum\frac{a^{3}}{b\plus{}c\plus{}d}\geq\frac{1}{3}$
Done!

Other problem:
Let $ a;b;c;d$ be positive real numbers.Prove that:
$ \frac{a^{3}\plus{}b^{3}\plus{}c^{3}}{b\plus{}c\plus{}d}\geq\sum a^{2}$

[i]Let let $ a$, $ b$, $ c$ and $ d$ positive reals such us $ ab \plus{} bc \plus{} cd \plus{} da \equal{} 1$.
Show that $ {\frac {{a}^{3}}{b \plus{} c \plus{} d}} \plus{} {\frac {{b}^{3}}{c \plus{} d \plus{} a}} \plus{} {\frac {{c}^{3}} {a \plus{} b \plus{} c}} \plus{} {\frac {{d}^{3}} {d \plus{} c \plus{} a}}\ge1/3$[/i]

[b]Approach by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=11748]campos[/url]:[/b]

by hölder we have that

$ \left(\sum \frac {a^3}{b \plus{} c \plus{} d}\right) \left(\sum a(b \plus{} c \plus{} d)\right) \left(\sum 1\right)^2\geq \left(\sum a\right)^4 \\
 \\
\Rightarrow \sum \frac {a^3}{b \plus{} c \plus{} d}\geq \frac {(a \plus{} b \plus{} c \plus{} d)^4}{16(\sum a(b \plus{} c \plus{} d))}$

but it's easy to verify from the condition that
$ (a \plus{} b \plus{} c \plus{} d)^2\geq 4(a \plus{} c)(b \plus{} d) \equal{} 4$
and
$ 3(a \plus{} b \plus{} c \plus{} d)^2\geq 4(2 \plus{} 2ac \plus{} 2bd) \equal{} 4((a \plus{} b \plus{} c \plus{} d)^2 \minus{} (a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)) \\
\Leftrightarrow 4(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)\geq (a \plus{} b \plus{} c \plus{} d)^2$
which is true by cauchy-schwarz

[b]Approach by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=445]Sung-yoon Kim[/url]:[/b]

[quote="cronecker"]Let $ a$, $ b$, $ c$ and $ d$ positive reals such us $ ab \plus{} bc \plus{} cd \plus{} da \equal{} 1$.
Show that $ {\frac {{a}^{3}}{b \plus{} c \plus{} d}} \plus{} {\frac {{b}^{3}}{c \plus{} d \plus{} a}} \plus{} {\frac {{c}^{3}} {a \plus{} b \plus{} c}} \plus{} {\frac {{d}^{3}} {d \plus{} c \plus{} a}}\ge1/3$[/quote]

I think the original form is like this : $ \sum {\frac {a^3}{b \plus{} c \plus{} d}} \geq \frac {1}{3}$. By Cauchy, $ (LHS)(a(b \plus{} c \plus{} d) \plus{} b(c \plus{} d \plus{} a) \plus{} c(d \plus{} a \plus{} b) \plus{} d(a \plus{} b \plus{} c)) \geq (a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)^2$. So $ (LHS)\geq {\frac {(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)^2}{2(ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd)}}$. -(*)

By AM-GM, $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \geq 2(ab \plus{} cd), 2(ac \plus{} bd), 2(ad \plus{} bc)$. So $ (a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)\geq ab \plus{} bc \plus{} cd \plus{} da$ and $ 3(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2) \geq ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd$. Multiplying this, we get that the RHS of (*) is not less than 1/3.

[quote="Fermat -Euler"]Let $ w, x, y, z$ are non-negative reals such that $ wx \plus{} xy \plus{} yz \plus{} zw \equal{} 1$.
Show that $ \frac {w^3}{x \plus{} y \plus{} z} \plus{} \frac {x^3}{w \plus{} y \plus{} z} \plus{} \frac {y^3}{w \plus{} x \plus{} z} \plus{} \frac {z^3}{w \plus{} x \plus{} y}\geq \frac {1}{3}$.[/quote]

  Follows Holder Inequality we have

 $ LHS\ge\frac{(x\plus{}y\plus{}z\plus{}w)^3}{4\cdot3(x\plus{}y\plus{}z\plus{}w)}\equal{}\frac{(x\plus{}y\plus{}z\plus{}w)^2}{4\cdot3(xy\plus{}yz\plus{}zw\plus{}wx)}\ge\frac{1}{3}$

 because $ (x\minus{}y\plus{}z\minus{}w)^2\ge 0$

[quote="manlio"][quote="chien than"]Other problem:
Let $ a;b;c;d$ be positive real numbers.Prove that:
$ \frac {a^{3} \plus{} b^{3} \plus{} c^{3}}{b \plus{} c \plus{} d}\geq\sum a^{2}$[/quote]

Sorry I cannot solve it and I don't remember if it was already posted. :maybe:

[quote="chien than"][u][b]Problem:[/b][/u]Let $ a;b;c;d$ be positive real numbers satisfying $ ab \plus{} bc \plus{} ca \plus{} da \equal{} 1$.Prove that:
$ \sum\frac {a^{3}}{b \plus{} c \plus{} d}\geq\frac {1}{3}$
[i]My solution[/i]
By AM-GM we have:
$ \frac {a^{3}}{b \plus{} c \plus{} d}{18} \plus{} \frac {a}{6} \plus{} \frac {1}{12}\geq\frac {2a}{3}$
[/quote]

How to correct AM-GM ?[/quote]
Dear manlio,
It should be
\[ \sum \frac{a^3\plus{}b^3\plus{}c^3}{b\plus{}c\plus{}d} \ge \sum a^2\]
This inequality is easy with AM-GM Inequality, since
\[ \frac{a^3}{b\plus{}c\plus{}d} \plus{}\frac{a(b\plus{}c\plus{}d)}{9} \ge \frac{2a^2}{3}\]
\[ \Rightarrow \sum \frac{a^3}{b\plus{}c\plus{}d} \ge \frac{2}{3}\sum a^2 \minus{}\frac{2}{9}(ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd) \ge \frac{2}{3}\sum a^2\minus{}\frac{1}{3} \sum a^2 \equal{}\frac{1}{3}\sum a^2\]
Similarly, also by AM-GM,
\[ \frac{b^3}{b\plus{}c\plus{}d} \plus{}\frac{b(b\plus{}c\plus{}d)}{9} \ge \frac{2}{3}b^2\]
\[ \Rightarrow \sum \frac{b^3}{b\plus{}c\plus{}d} \ge \frac{2}{3}\sum a^2 \minus{}\frac{1}{9}\sum a^2 \minus{}\frac{1}{9} (a\plus{}c)(b\plus{}d) \minus{}\frac{2}{9}(ac\plus{}bd) \ge \frac{2}{3}\sum a^2 \minus{}\frac{1}{9}\sum a^2  \minus{}\frac{1}{9}\sum a^2 \minus{}\frac{1}{9}\sum a^2 \equal{}\frac{1}{3}\sum a^2\]
Similarly, we also have
\[ \sum \frac{c^3}{b\plus{}c\plus{}d} \ge \frac{1}{3}\sum a^2\]
Adding up these inequalities, we have done.

Exist a proof which is used Chebysev  ineq ,Can_hang 2007.but your solution is very nice

Thank you very can_hang2007 for your nice solution.  :)

The given condition implies $1 = wx+xy+yz+zx \le w^2+x^2+y^2+z^2$.  Then by Cauchy, \[ \sum_{\text{cyc}} \frac{w^4}{wx+wy+wz} \ge \frac{(w^2+x^2+y^2+z^2)}{2(wx+xy+yz+zx+wy+xz)} \ge \frac{w^2+x^2+y^2+z^2}{2(wx+xy+yz+zx+wy+xz)} \ge \frac{1}{3}. \]

$\sum{\frac{x^3}{y+z+w}}=\sum{\frac{x^4}{xy+xz+xw}} \ge \frac{(\sum{x^2})^2}{2\sum{xy}} (\text{Titu's Lemma})
\ge \frac{2\sum{x^2}-1}{2\sum{xy}} ((\sum{x^2}-1)^2 \ge 0 \Leftrightarrow (\sum{x^2})^2 \ge 2\sum{x^2}-1) \ge \frac{2\sum{x^2}-1}{3\sum{x^2}}$

Now $\frac{2\sum{x^2}-1}{3\sum{x^2}} \ge \frac{1}{3}$
$\Leftrightarrow \sum{x^2} \ge 1$

But this follows with power-mean inequality combined with AM-GM:

$\frac{\sum{x^2}}{4} \ge (\frac{\sum{x}}{4})^2 \ge \frac{4(y+w)(x+z)}{16}=\frac{1}{4}$

from which the result follows...It is easy to see that equality holds if and only if $x=y=z=w=\frac{1}{2}$

By cyclic AM-GM's $x^2+y^2+z^2+w^2\ge 1$, and also $\sum_{\text{sym}} x^2\ge \tfrac{1}{3}\left( \sum _{\text{sym}} xy\right)$.  Hence by Cauchy
\[\tfrac{w^4}{1-(x+z-w)y}+\tfrac{x^4}{1-(y+w-x)z}+\tfrac{y^4}{1-(x+z-y)w}+\tfrac{z^4}{1-(y+w-z)x}\ge\]\[\ge\tfrac{(w^2+x^2+y^2+z^2)^2}{2+2yw+2xz}=\tfrac{\left( \sum_{\text{sym}} x^2\right) ^2}{\sum_{\text{sym}} xy}\ge \tfrac{\sum_{\text{sym}} x^2}{3}\ge \tfrac{1}{3}.\]

We use Titu's Lemma, or Fractional Cauchy-Schwarz. Forming fourth powers in the numerator and applying the Lemma,

$$\sum_{\text{cyc}} \frac{w^4}{w(x+y+z)} \geq \frac{(w^2+x^2+y^2 + z^2)^2}{2(wx+wy+wz+xy+yz+zx)} \geq \frac{1}{3}$$

With the given $wx+xy+yz + zw = 1$, the denominator can be rearranged into $2(1+wy+xz)$. Therefore, it remains to show that

$$\frac{(w^2 + x^2 + y^2 + z^2)^2}{2(1+wy+xz)} \geq \frac{1}{3}$$

It is sufficient to show that $3(w^2 + x^2 + y^2 + z^2) \geq 2(1+wy+xz)$. By AM-GM, we have $w^2 + y^2 \geq 2wy$ and $x^2 + z^2 \geq 2xz$, which yield $w^2 + y^2 + x^2 + z^2 \geq 2(xz + wy)$ when added together. Furthermore, by the well known fact that $a_1^2 + a_2^2 + a_3^2 + a_4^2 \geq a_1a_2 + a_2a_3 + a_3a_4 + a_4a_1$, we have that $w^2 + x^2 + y^2 + z^2 \geq wx + xy + yz + zx =1$. The conclusion follows.

(P.S: Sorry if a similar proof was already posted)

I think this is different enough to warrant a post. Oh well.

By T2 Lemma and the given we get to $\text{Expression}\ge \frac{(w^2+x^2+y^2+z^2)^2}{2(1+wy+xz)}$.
If $wy+zx\le \frac{1}{2}$, then we have, by the rearrangement ineq $w^2+x^2+y^2+z^2\ge wx+xy+zy+zx =1$, $\frac{(w^2+x^2+y^2+z^2)^2}{2(1+wy+xz)}\ge \frac{1}{2+2(wy+xz)}\ge \frac13$.

Otherwise, we have $u=wy+xz\ge \frac12$.
By am-gm, $w^2+x^2+y^2+z^2\ge 2wy+2zx=2u$, so our inequality rewrites as
$\frac{(w^2+x^2+y^2+z^2)^2}{2(1+wy+xz)}\ge \frac{(4u^2)}{2(1+u)}\ge\frac13\iff 6u^2-u-1\ge0\iff (3u+1)(2u-1)\ge 0$ which is true.

Combining both cases, we're done.

Solved using the AM-GM inequality, English is not my native language
$(w+x+y+z)^2 \geq 4(w+y)(x+z) = 4(wx+xy+yz+zw) = 4$
$\Rightarrow w+x+y+z \geq 2$
We want to prove $$\sum_{cycle} \frac{w^3}{x+y+z} \geq \frac{1}{3}$$
By AM-GM we have $\frac{w^3}{x+y+z}+\frac{x+y+z}{a}+\frac{1}{b} \geq \frac{3}{\sqrt[3]{ab}}w$
$$\Rightarrow \sum_{cycle} \frac{w^3}{x+y+z} \geq \frac{3(w+x+y+z)}{\sqrt[3]{ab}}-\frac{3(w+x+y+z)}{a}-\frac{4}{b}$$
$$\Rightarrow \sum_{cycle} \frac{w^3}{x+y+z} \geq 3(w+x+y+z)(\frac{1}{\sqrt[3]{ab}}-\frac{1}{a})-\frac{4}{b}$$
We want $\frac{1}{\sqrt[3]{ab}}-\frac{1}{a}>0 \Rightarrow a^2>b$
$$\Rightarrow \sum_{cycle} \frac{w^3}{x+y+z} \geq \frac{6}{\sqrt[3]{ab}}-\frac{6}{a}-\frac{4}{b}=\frac{1}{3}$$
Let $\frac{1}{3}=\omega-\alpha-\beta$
$\Rightarrow \omega=\frac{6}{\sqrt[3]{ab}}, \alpha=\frac{6}{a}, \beta=\frac{4}{b}$
$\Rightarrow ab=\frac{216}{\omega^{3}}, a=\frac{6}{\alpha}, b=\frac{4}{\beta}$
$\Rightarrow 9\alpha\beta=\omega^{3}$
So any number $\omega, \alpha$ and $\beta$ that satisfy$\begin{cases}
\omega-\alpha-\beta=\frac{1}{3}\\
9\alpha\beta=\omega^{3}\\
a^2>b\end{cases}$ will give us the numbers $x, y$ to prove the inequality
For example
$\begin{cases}\omega=1\\\alpha=\frac{1}{3}\\\beta=\frac{1}{3}\\\end{cases}$ wil give us $\begin{cases}a=18\\b=12\end{cases}$ and we can use that to prove the original inequality
If we plug in any $a$ and $b$ that don't follow this "rule" we will get $$\sum_{cycle} \frac{w^3}{x+y+z} \geq \gamma, \gamma < \frac{1}{3}$$
This took me really long to figure this out, but when I did I was really happy

[quote=Fermat -Euler]Let $ w, x, y, z$ are non-negative reals such that $ wx \plus{} xy \plus{} yz \plus{} zw \equal{} 1$.
Show that $ \frac {w^3}{x \plus{} y \plus{} z} \plus{} \frac {x^3}{w \plus{} y \plus{} z} \plus{} \frac {y^3}{w \plus{} x \plus{} z} \plus{} \frac {z^3}{w \plus{} x \plus{} y}\geq \frac {1}{3}$.[/quote]
[url]https://artofproblemsolving.com/community/c6h548439p3183012[/url]
[url]https://artofproblemsolving.com/community/c6h2152301p15883499[/url]
[size=50]SXTX （7）2008[/size]

It is easy to use Titu's Lemma.
The equation, assume S
$S\geq \frac{(w^2+x^2+y^2+z^2)^2}{2(xw+yw+zw+xy+yz+xz)}$
Since $wx+yx+yz+zw=1, w^2+y^2+z^2+x^2\geq 1; 2xz+2yw\geq 1$
$(w^2+x^2+y^2+z^2)^2\geq (wx+xy+yz+zw)^2=1$ by C-S inequality
so $S\geq \frac{1}{2*1+1}=\frac{1}{3}$

Let $a,b,c,d$ are non-negative reals such that $ a b+a c+a d+b c+bd+c d=1 $. Show that $$ \frac{a^3}{b+c+d}+\frac{b^3}{c+d+a}+\frac{c^3}{a+b+d}+\frac{d^3}{a+b+c}\geq \frac {2}{9}$$


Only the Sugar and Orange bowls left.

Who's everyone rooting for to win, USC or Oklahoma?

I'm personally rooting for USC, since I don't really like Bob Stoops a lot and because my friend likes them and he's influenced me.

OU will beat USC, Virginia Tech I'm hoping will beat Auburn.

Nooooo! Michigan lost... by 1 point on the last play :( :( :(

Haha, to tetrahedon.  I know how you feel, I was hoping Michigan would win.  I live in Austin (where UT is located), so there are crazed longhorns everywhere I look.

*slaps h_s_potter2002*

How could you not want this team to win? A Michigan win would mean so much to everyone (around here at least!) A school-record 2nd Rose Bowl for the awesome Coach Carr. A victory in the last game of top national WC Braylon Edwards.

Credit goes to Young. Seriously, I didn't think Texas was playing great, but Young was awesome. As much as I wanted Michigan to win, that was one of the best performances by a quarterback I've every seen.

[quote="h_s_potter2002"]How can anyone want Michigan to win?[/quote]

This is just a hunch, but are you a Buckeyes fan?  ;)

I've gotta say one thing though, the Iowa-LSU game was awesome. No other words to describe it. Awesome.

I only the last seconds of the Iowa-LSU...I turned on the TV and the first thing I saw was like a 60 yard pass to win the game, I had perfect timing  :P

Go Clemson! wait, they didn't have a bowl game :(
out of whose left, go Auburn and Oklahoma
actually, neither game [b]really[/b] matters to me
i also wanted Michigan to win as well as Boise State. Oh well.

[quote="yif man12"][quote="h_s_potter2002"]How can anyone want Michigan to win?[/quote]

This is just a hunch, but are you a Buckeyes fan?  ;)[/quote]

Yup  :lol:

[quote="furious"]Go Clemson! wait, they didn't have a bowl game :( [/quote]

Well, techinally, they did. But then...

And yeah, I wanted Boise State to win too. But at least Utah won, even if they played Pitt.
Hm, I'm becoming to have a fascination for midwest colleges. Weird.

YEAH!!! USC obviously crushes Oklahoma 55-19!!! WOOHOO!!! :D

Why are the smileys still christmas...

Make the best of the smileys while you can!
 :)  :(  :?  :D  :lol:  ;)  :blush:  :P  :o  :mad:  :oops:  :roll:  :ninja:  :wacko:  :alien:
I put on all the christmas hat ones!


If $M$ is inside of tringle $ABC$ such that $\angle MAB = \angle MBC = \angle MCA = x$. Prove that:

$\frac{1}{sin^{2}x}=\frac{1}{sin^{2}a}+\frac{1}{sin^{2}b}+\frac{1}{sin^{2}c}$

x is the Brocard angle of the triangle, so that

$\cot x = \cot a+\cot b+\cot c$

For example, see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=273]Brocard angle satisfies cot w = cot A + cot B + cot C[/url] or [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107309]11/16 < PA² + PB² + PC² < 1[/url] with a proof. Squaring the above yields

$\frac{1}{\sin^{2}x}-1 = \frac{1}{\sin^{2}a}+\frac{1}{\sin^{2}b}+\frac{1}{\sin^{2}c}-3+2 (\cot a \cot b+\cot b \cot c+\cot c \cot a)$

Since $\cot a \cot b+\cot b \cot c+\cot c \cot a = 1$ for any $a+b+c = \pi,$

$\frac{1}{\sin^{2}x}= \frac{1}{\sin^{2}a}+\frac{1}{\sin^{2}b}+\frac{1}{\sin^{2}c}$

[color=darkblue][b]Lemma.[/b] With the following trigonometrical identities in a triangle $ABC$ you can prove easily the proposed problem :

$\left\{\begin{array}{cc}0\blacktriangleright & \cot A\cot B+\cot B\cot C+\cot C\cot A=1\ .\\\\ 1\blacktriangleright & \sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C\ .\\\\ 2\blacktriangleright & \cos2A+\cos 2B+\cos 2C=-1-4\cos A\cos B\cos C\ .\\\\ 3\blacktriangleright & \cos^{2}A+\cos^{2}B+\cos^{2}C=1-2\cos A\cos B\cos C\ .\\\\ 4\blacktriangleright & \sum\sin^{2}B\sin^{2}C=\prod \sin^{2}A+\left(1+\prod\cos A\right)^{2}\ .\end{array}\right\}$

I"ll prove only the identity $(4)\ .$ Indeed, $(4)\Longleftrightarrow$ $\sum \left(1-\cos^{2}B\right)\left(-\cos^{2}C\right)-$ $\prod\left(1-\cos^{2}A\right)=$ $\left(1+\prod\cos A\right)^{2}$ $\Longleftrightarrow$

$1-2\sum\cos^{2}A+$ $\sum\cos^{2}B\cos^{2}C+\sum\cos^{2}A-$ $\sum\cos^{2}B\cos^{2}C+\prod\cos^{2}A=$ $2\prod\cos A+\prod\cos^{2}A$ $\Longleftrightarrow$

$1-\sum\cos^{2}A=2\prod\cos A\ ,$ i.e. the identity $(3)\ .$ The another identities remain as simple exercises.

[b]Proof.[/b] Apply the [u]Ceva's theorem  (trigonometrical form)[/u] : $\prod \sin (A-x)=\sin^{3}x$ $\Longleftrightarrow$ $\sum\sin (B+C-A-x)=\sin 3x+4\sin^{3}x$ $\Longleftrightarrow$

$\sum\sin (2A+x)=3\sin x$ $\Longleftrightarrow$ $\cos x\cdot\sum\sin 2A=\sin x\cdot\left(3-\sum\cos 2A\right)$ and $(1)$ and $(2)$  $\Longleftrightarrow$ $\cos x\cdot\prod\sin A=\sin x\cdot\left(1+\prod\cos A\right)$ $\Longleftrightarrow$

$\tan x=\frac{\prod\sin A}{1+\prod \cos A}$ $\implies$ $\frac{1}{\sin^{2}x}=\frac{1+\tan^{2}x}{\tan^{2}x}=\frac{\prod\sin^{2}A+\left(1+\prod\cos A\right)^{2}}{\prod\sin^{2}A}$ and $(4)$ $\implies$ $\boxed{\ \frac{1}{\sin^{2}x}=\frac{1}{\sin^{2}A}+\frac{1}{\sin^{2}B}+\frac{1}{\sin^{2}C}\ }\ .$

[b]Remark.[/b] $\frac{1}{\sin^{2}x}=\sum\frac{1}{\sin^{2}A}$ $\Longleftrightarrow$ $\frac{1}{\sin^{2}x}-1=2+\sum \left(\frac{1}{\sin^{2}A}-1\right)$ $\Longleftrightarrow$ $\cot^{2}x=2+\sum \cot^{2}A$ and $(0)$ $\Longleftrightarrow$ $\boxed{\ \cot x=\cot A+\cot B+\cot C\ }\ .$[/color]


Let $ \triangle ABC$ be a fixed triangle inscribed in circle $ (O)$. Let $ (O')$ be an animating circle which is always tangent to $ BC$ at $ M$ and also tangent to arc $ \overarc{BAC}$ of circle $ (O)$. Let $ AT$ be the tangent of $ (O')$ which lies in $ \widehat{BAC}$. The line $ d$ which passes through $ B$ and parallel to $ AT$ intersects line $ AC$ at $ N$. Let $ I,J$ be the midpoint of $ BN,AM$.

Prove or disprove whether the line $ IJ$ passes through a fixed point?

I used Geometry Sketchpad and saw that there is no fixed point.

As I saw in GSP 4.06, the trace of line $ IJ$ has a very special point which is somewhat like an unchanged point, but I'm not sure. Could you check the intersection $ S$ of $ IJ$ and perpendicular line drawn to $ AC$ from the center of the mixitilinear incircle of angle $ A$ of $ \triangle ABC$? :huh:


calculus
integration
limit
function
real analysis
real analysis unsolved

$ f(t) , f_j(t)$ are monotone increasing in index j, monotone decreasing in t
and non-negative. $ \lim_{j\to\infty} f_j(t) \equal{} f(t)$ for all t.

Show that $ \lim_{j\to\infty} \int^c_0 f_j(t) dt \equal{} \int^c_0 f(t) dt$

Help me !!!

[quote="BlueVelvet"]I think you should show that $ f_j$ [url=http://mathworld.wolfram.com/UniformConvergence.html]uniformly converges[/url] to $ f$.[/quote]

i think so too ...
but i don't know how to do..

You can't, because they don't. For example, we could take $ f_j(t) \equal{} t^j$ on $ [0,1]$.

Try showing that $ \sum_k a_kf_j(t_k)$ converges to $ \sum_k a_kf(t_k)$ for any finite collection of points $ t_k$ and coefficients $ a_k$. From this and the fact that the $ f_j$ are increasing in $ t$, derive that the lower and upper sums for $ \int f_j$ with respect to any fixed partition converge to the lower and upper sums for $ \int f$ with respect to that partition.

Another path, which is how I convinced myself this is true, is to note that bounded monotone functions on a bounded interval are Riemann integrable and then apply one of the Lebesgue convergence theorems. I'm not recommending that, because I don't think you've seen those theorems.


linear algebra
matrix
integration
function
advanced fields
advanced fields unsolved

I've been working on this for long, however, till now I still cannot verify it except for the confidence of it being true.

Suppose $ A_n$ is n * n sized real matrix, and the (i, j) element $ a_{ij} \equal{} \int_0^1f ^ {(i)}f ^ {(j)}dx$, where i, j = 0, 1, ... , n - 1, f is an arbitrary real function in $ C^{\infty}[0, 1]$, $ f^{(i)} \equal{} \frac{d^if}{dx^i}$.

My problem is [b]how to verify the serie {$ \lambda_n$} tends to 0, where $ \lambda_n$ is the smallest eigenvalue of $ A_n$[/b], and how fast is the serie tending to 0? An estimation is most welcome.

By the way, by some numerical tests, I am almost sure it is true (tending to 0), however, if otherwise is correct, please also give me an example. Thanks a lot.

This is a redundant thread. Don't double post, and post in the right forum (linear algebra, where the other copy has been moved to).


[quote="shinwoo"]I've been trying to post new problems and tried my best to attract people's attention of this new form, but I don't think it's as popular as the section for middle schoolers who are going to contests :([/quote]Give it some time. Get your friends from school up on the forum, and get your friends' friends to come too, and then you'll get more popular section :)

[quote]Give it some time. Get your friends from school up on the forum, and get your friends' friends to come too, and then you'll get more popular section

[/quote]
I'll do that :D


which state do you guys live in?

sorry if i spelled something wrong

i live in Pennsylvannia

I live in Kansas and if u think thats easy think again, theres a lot of guys from my chapter and the topeka area that are just waiting for March 8th to show off their math skills. :(

I live in Oklahoma.
Most of the kids that do really well at state aren't in my chapter...

Connecticut, from Southwest Chapter. I can't wait for the competition on March 8th!

My chapter is awful.
I'm probably going to fail state, because I will just be working on AMC problems until the 12B...I wonder how much I can do in 9 days.

I live in Puerto Rico, It's not technically a state but it's a territory/colony.

VA PRIDE!!!  :D

I'm just bracing myself to get creamed at state ><

I'm tempted to say Indiana as well. I fail. at life.

TEXAS BEBE!
i actually have a chance 2 place @ state this year...
after all, i have tied kevin chen b4

Imma get owned at States.
They say that MA is the third best State?
Second to Texas and Cali?
Or maybe I'm just being delusional like always.

[size=75][color=darkblue]Texas...

Ugh, it's so hard here.  It was hard enough getting on my school team so I doubt I'll make Nationals (But I'm pretty sure we will win 1st place team at State if  I don't screw everyone else over!).   :lol: [/color][/size]

New York

I hope I don't flunk Nationals again... :(

First California to post! But the Northern CA usually does better than us. I doubt I'll get into nats, but if I do, then I hope I'm not the only Southern Californian.

I'm from Texas.

I can never win state because sugarland is there, and they have won for the past 20 or more years.  Even though i lost to him, bigtime

KEVIN CHEN RULES!!!

GO TEXAS PRIDE!!!
GO ASIAN PRIDE!!!

 :w00t:

[quote="_quizas_"]MAINE! w00t!!! Southern Chapter babeyy!!! already had the sompetition!!! team=2nd individual=6th...psyched for march 15th and states! again, w00t![/quote]

Dood, calm down :).  Anyway, I'm in Georgia if anyone's curious.  Pretty easy state...:P

First Nevadan to post! One of the easiest states to get into nationals from.

wow I havent seen a more pointless topic for years...
just look in the ppls location thingy under their usn.

i live in CA btw which is y i never got into nats :(

Michigan. I and the other three national competitors actually did pretty well, Hustla25. 5th out of 56(?) states/territories is pretty good.

This year, we're going to be... okay. I think.

[quote="kyyuanmathcount"][quote="uldivad9"]Michigan. I and the other three national competitors actually did pretty well, Hustla25. 5th out of 56(?) states/territories is pretty good.[/quote]

5th out of 57  :wink: .  Na, you and the other two national competitors actually did pretty well; I, on the other hand...

California is insanely competitive; there are two state competitions and usually there end up being many-way ties for like, 2nd place (something like 7-way last year?).  It ends up being tie-breakers many times.[/quote]

Tell me about it...

And for countdown there are huge (10+) way ties for 15th place (to be in Countdown).


inequalities
function
induction
limit
real analysis
real analysis unsolved

Let $f$ be a real valued continuous function & $f(x_1/2 + x_2/2)<={f(x_1)/2+f(x_2)}/2$
Then show that $f(x_1/n +... +x_2/n)<={f(x_1)/n+...+f(x_n)}/n$

How is this unsolved?

Anyways, by induction prove it for powers of two, if it holds for $n=k$, then it holds for $n=k-1$ (set the last value equal to the arithmetic mean of the other terms)

I have tried but somehow its not coming.Could you plz explicitly give your proof

Say it holds for $n=k$

Then $\textstyle (\sum^{n-1}_{i=1} f(x_i)+f((x_i+\ldots+x_{n-1})/(n-1)))/n\geq ...$

Now cancel out the extra $f((x_i+\ldots+x_{n-1})/(n-1))$ term that you're going to end up with.

Are you sure but my teacher told me that you ll never reach the answer if you go by this method :(

[quote="zhaobin"]I also think it can.
first $n=2^k$ is true then $n=2^{k+1}$
second $n$ is true then $n-1$[/quote]
Thats something I was trying in the begining.

[quote="blahblahblah"]No, I said it doesn't work in the general case of arbitrary weights which sum to $1$ - it does certainly work for $n$ weights of $1/n$[/quote]
sorry,I can't follow you.
can you explain more :blush:

[quote="blahblahblah"]If we want to prove that

$f$ convex implies $\sum_i \lambda_if(x_i)\geq f(\sum_i \lambda_ix_i)$ then we need a more sophisticated argument, of course.[/quote]
yeah,I know if we want to prove $\sum_i \lambda_if(x_i)\geq f(\sum_i \lambda_ix_i)$ ,the $f$ should have the condition with continuce.
and do you mean $the game$'s have  been sloved?

Well, it's been all but solved, but I've told him enough that I think there's some value in him finishing it from there.

If it's still there in a day I guess I can post a solution.

well,just use $\lambda_i=\lim_{n \to \infty}\frac{[n \lambda_i]}{n}$
right?

Not in the proof that I know.  It is just a straightforward induction with one substitution.

well,I post my solution.(just for $\lambda_1f(x_1)+\lambda _2f(x_2) \geq f(\lambda_1 x_1+\lambda_2 x_2),\lambda_1+\lambda_2=1$,let $\frac{[n \lambda_1]}{n}+\frac{[n \lambda_2]}n=b$
$f(\lambda_1 x_1+\lambda_2 x_2) \leq f(\frac{[n \lambda_1]}{n}x_1+\frac{[n \lambda_2]}{n} x_2) +a \leq \frac{[n \lambda_1]}{bn}f(bx_1) +\frac {[n \lambda_2]}{bn}f(bx_2) +a$
when $n \to +\infty$,we have $b \to 1$,a is any positive number
so we will get the conclusion.right?
and then yours :)

[quote="blahblahblah"]Okay, say that Jensen's holds for three numbers

Set $x_3=(x_1+x_2)/2$.  We want to show that $f(x_1)+f(x_2)\geq 2f(x_3)$

By Jensen's for three variables (do the algebra)

$f(x_1)+f(x_2)+f(x_3)\geq 3f(x_3)$

We may cancel an $f(x_3)$ from both sides

This gives $f(x_1)+f(x_2)\geq 2f(x_3)$ as desired.  The general case is just a slight modification of this method.

The only thing about this method is it doesn't work for the general case, that is, for arbitrary weights which sum to 1[/quote]

That's because it's not true-- a function from an interval into $\mathbb{R}$ is convex iff it is continuous and it satisfies $f(\frac{x+y}{2})\leq \frac{f(x)+f(y)}{2}$

Non-continuous non-convex functions satisfying the latter condition can be "constructed" by "choosing" a basis for $\mathbb{R}$ over $\mathbb{Q}$.

Sure, I know that.  But it's easier to refer to 'Jensen's for three variables' than to say 'say that the condition $3f(\sum_3 x_i)\leq \sum_3 f(x_i)$ holds, even if it's only a Jensen's-like inequality.

I understand what Blahblahblah was trying to say that it is not true for all weights.I think I have the proof & i will post it

[quote="blahblahblah"]Actually, your assertion is untrue - there are discontinuous convex functions.  We need to restrict consideration to open intervals to make what you said true.[/quote]

Whoops, you're right


A particular vending machine randomly chooses from the choices it has no matter what button you press. If the vending machine has 20 choices, what is the expected amount of tries you need to get what you want?

What a bad vending machine.

[hide]OK, my attempt...

The probability of getting the item on the first try is $\frac{1}{20}$.

The probability of getting the item on the second try is $\frac{19}{20}\times\frac{1}{19}=\frac{1}{20}$.

The probability of getting the item on the third try is $\frac{19}{20}\times\frac{18}{19}\times\frac{1}{18}=\frac{1}{20}$.

And the same principle extends to the others, because the fractions will always cancel out. (Handwaving element here.)

Therefore the expected number of tries is $10.5$.[/hide]

Could you use like the geometric distribution here?


geometry
Euler
3D geometry
number theory
prime factorization
algebra open
algebra

could i prove  x^3+y^3=z^3  (x,y,z positiv integers) is impossible because sum of two Nicomedes conchoids is not a Nicomedes conchoid ?
i think this is the real and easy Fermat proof

 :(

What do you mean by the "sum" of two curves?  The $ n \equal{} 3$ case does admit elementary proofs, but this isn't one of them.

[img]http://mathworld.wolfram.com/images/equations/ConchoidofNicomedes/NumberedEquation2.gif[/img]

sum of two trisections is not a trisection
 :(

Again, what do you mean by the "sum" of two trisections?

[b]about history:[/b]

Fermat read a page of Diophant book and make there a proof of this theorem. The proof was 2 lines of words.
What Fermat thought about to make that so easy proof of 2 lines? Only grecq geometry. The only thought possible with grecq geometry was: trisection.

[b]about sum:[/b]
if you make a trisection of angle with Nicomedes conchoid you can't make with this same conchoid another trisection you want by homothetic for exemple. Same for sum of all points of 2 conchoids. You can't define a algebric system with Nicomedes conchoids.

note:

the 2005 proof of this theorem have 100 pages of modern algebra (Fermat didn't know this algebra)
and i think Fermat said the truth about his 2 lines of proof, and i think Fermat now (if alive) would say the 100 pages of proof is ridiculous.
Fermat is :rotfl: about Willes proof in the heaven !

 :(

Just caught this in passing (I don't usually browse the olympiad forums... don't quite have the chops). Anyways, there's good historical evidence to show that Fermat in fact never had this "two line" proof that you keep going on about. I think your last post is a little bit insulting to the impact that Wiles' proof had on mathematics and the amount of effort that mathematicians have put in as a whole to solve this problem.

"there's good historical evidence to show that Fermat in fact never had this "two line" proof that you keep going on about"

there's no historical insulting in your words against Fermat ? Do you say Fermat made bluff with the 2 lines proof ?

 :(

Fermat never said that he had a two line proof, he wrote in the margin of his copy of Diophantus' book "I have a truly marvelous proof of this proposition which this margin is too narrow to contain." He never showed this to anyone, and it was only found after his death. It is likely that he thought he had a proof, but that it was not correct, given how many false proofs of the theorem have been created.

[quote="jrrbc"]What Fermat thought about to make that so easy proof of 2 lines? [/quote]
In light of Kummer's approach to the problem, many believe that Fermat made the false assumption that the rings of integers in cyclotomic fields always had unique prime factorization, which does indeed give a very quick proof of the theorem.  It would have taken rather a lot of foresight to realize, in Fermat's time, that this assumption needed to be questioned.

[quote="jrrbc"]if you make a trisection of angle with Nicomedes conchoid you can't make with this same conchoid another trisection you want by homothetic for exemple. Same for sum of all points of 2 conchoids. You can't define a algebric system with Nicomedes conchoids. [/quote]
I don't see the word "integer" in there anywhere.

"Fermat [b]made the false assumption that the rings of integers in cyclotomic fields always had unique prime factorization[/b], which does indeed give a very quick proof of the theorem"

Last Fermat Theorem can be proved only with grecq geometry, this is what we can know about Fermat "proof".

Do you mean that just with grecq geometry Fermat only could thougt about prime factorization ? Pitagoras theorem can be proved with prime factorization ?

I think about trisection of integers not factorizaction of integers.

 :(

Do you know what Fermat thought?  The standard proof for $ n \equal{} 3$ (and, if I recall correctly, the proof found either by Fermat or Euler) is essentially dependent on $ \mathbb{Z}[\omega]$ having unique prime factorization, and the proof for $ n \equal{} 4$ can be phrased in the same way.  It would have been a very natural generalization for Fermat to suspect, although, of course, we don't know for sure.  What makes you think Fermat spent all his time thinking about greek geometry?

(By the way, as long as we're on this subject, the derivation of the formulas for the Pythagorean triples (i.e. the solutions for $ n \equal{} 2$) [i]can[/i] be proven as a consequence of prime factorization in the integers.)

I think Fermat that time read the Diophantus problem and thougt a greek geometry
That time the Algebra semantic of rings didn't exists
The Pythagoras theorem can be proved with primes factorizaction?

How could Fermat made a two lines of words proof only with "his mistake" about unique primes factorizaction ?
Can you write here that proof with that mistake ? and the same proof to Pythagoras theorem

:(

Still, nobody said there is a two lines proof!
By the way, he more than once said he had a "proof" for something without showing one. It's very likely that in most cases he didn't have one.

[quote]Last Fermat Theorem can be proved only with grecq geometry, this is what we can know about Fermat "proof". [/quote]
We don't know this, and relating this to greek geometry is a bit of random. Greek geometry is not really close to modern (elliptic curve) geometry.

"Greek geometry is not really close to modern (elliptic curve) geometry"

Nicomedes conchoid is not a elliptic curve ?

 :(

It's an equation of degree $ 4$, so not a standard (geometric) elliptic curve.
Additionally, greeks didn't even think of addition on curves, or all the other more modern arithmetic developments.

"not a standard (geometric) elliptic curve"

but is the most aproximate in greek geometry with modern elliptic curve
so: [b]if the Fermat "proof" was correct [/b]it would be about something like standard or non standard elliptic curve.

 :(

There is no reason to assume that Fermat's supposed proof used any geometry, or that it was correct.

"There is no reason to assume that Fermat's supposed proof used any geometry, or that it was correct"

there's one question:  could be proved this theorem at 17 century?
note: the theorem was proposed by Diophantus 19 centuries ago, and was a new greek geometry problem (sum of two cubes volumes with integer length isn't a volume cube with integer length)

If you restrict to cubes: yes.
If you want general powers $ >4$, it's very likely to be not possible (altough there are elementary proofs for some higher exponents).

Fermat was very fond of his method of infinite descent which he had used proving the theorem for the case n=4. In a letter adressed to Christian Huygens, Fermat asks Huygens to tell other mathematicians to work on the problem when n=3 and work on the method of infinite descent as well :!: Eulers proof for this case used descent and so it may be a mistake that Fermat had made when trying to generalize the problem. By the way:
1) There is no information about this two line proof
2) In the end of the letter I just mentioned, Fermat writes: "Perhaps posterity will thank me for having shown that the ancients didn't know everything!!!" So there isn't any reason that you insist Fermat wanted to prove the problem with greek geometry.

Fermat also used to tease other mathematicians about the theorems he found, he asked them to find a proof. If he had found a proof for this, then chances are big he would have asked other mathematicians to find the proof.

There are lots of information in Simon Singh's book, "Fermats last theorem". Very entertaining. It pictures Andrew Wiles way to victory.


Let $ r,n$ be nonnegative integers such that $ r\leq n$.

(a) Prove that $ \frac {n \plus{} 1 \minus{} 2r}{n \plus{} 1 \minus{} r}\binom{n}{r}$ is an integer

(b) Prove that $ \sum_{n \equal{} 0}^{a}\frac {n \plus{} 1 \minus{} 2r}{n \plus{} 1 \minus{} r}\binom{n}{r} < 2^{n \minus{} 2}$ for all $ n\geq 9$, where $ a$ is the greatest integer not more than $ \frac {n}{2}$

[hide="a"]$ \frac{n\plus{}1\minus{}2r}{n\plus{}1\minus{}r}\binom{n}{r}\equal{}\binom{n}{r}\minus{}\binom{n}{r\minus{}1}$[/hide]

Im assuming the $ n$ in your summand is supposed to be an $ r$, or else the question makes no sense.

[hide="b"]From the first result this sum telescopes nicely down to $ \binom{n}{\lfloor\frac{n}{2}\rfloor}$

I tried a counting argement here, to no avail, but was able to do it through hefty induction. I wont regurgitate it here as its basically casework for n odd,even.

[/hide]


[quote="hxy09"][quote="hedeng123"][quote="hxy09"]$ x_1\ge x_2\ge...\gex_n\ge 0$
Prove that:$ \sum_{i = 1}^{n} \sqrt {\frac {x_i^2 + x_{i + 1}^2 + ... + x_n^2}{i}}\le 4\sum_{i = 1}^{n} x_i$
 :D  :D  :D[/quote]

Use computer can be verified $ n = 3,4,5,6$  the inequality holds.[/quote]

Oh,my friend,this inequality is surely hold :wink:
What's more $ 4$ can be improved :lol:[/quote]
I dont know what is your fact, but I found that the inequality follows from the fact
\[ 4n \ge \sqrt{\frac{n}{1}}+\sqrt{\frac{n-1}{2}}+\cdots +\sqrt{\frac{1}{n}}\]
for any positive integer $ n.$

[quote="can_hang2007"][quote="can_hang2007"][quote="hxy09"]$ x_1\ge x_2\ge...\gex_n\ge 0$
Prove that:$ \sum_{i = 1}^{n} \sqrt {\frac {x_i^2 + x_{i + 1}^2 + ... + x_n^2}{i}}\le 4\sum_{i = 1}^{n} x_i$
 :D  :D  :D[/quote]
A sharper result:

If $ x_1 \ge x_2 \ge \cdots \ge x_n \ge 0,$ then
$ \sum_{i = 1}^n\sqrt {\frac {x^2_i + x^2_{i + 1} + \cdots + x^2_n}{i}} \le \frac {\pi}{2} \sum^n_{i = 1} x_i.$[/quote]
And here is the proof:[/quote]

Thank you so much for your nice solution :lol: But I am just wondering how can you come up with such strange ideas?Could you please help me to reveal the [b]hidden facts[/b] in your mind? :lol:

Here is the official solution,also incredible and nice :)


induction
number theory proposed
number theory

$ \{x_n\}_{n\ge 1}$ is a sequence with positive integers.
And $ x_{n \plus{} 2} \equal{} gcd(x_{n \plus{} 1},x_n) \plus{} 2009$

Prove or disprove:There exists such a sequence and it contains at least $ 10^{2009}$ different terms.

[quote="hxy09"]$ \{x_n\}_{n\ge 1}$ is a sequence with positive integers.
And $ x_{n \plus{} 2} \equal{} gcd(x_{n \plus{} 1},x_n) \plus{} 2009$

Nice problem

Prove or disprove:There exists such a sequence and it contains at least $ 10^{2009}$ different terms.[/quote]
Let $ a,b$ two positive integers such that $ b(a\minus{}2009)>a>2\times 2009$ and $ \gcd(a,2009)\equal{}\gcd(b,2009)\equal{}1$

Let then sequence $ u_n$ defined as:

$ u_1\equal{}a$
$ u_2\equal{}b(a\minus{}2009)$
$ u_{n\plus{}2}\equal{}(u_n\minus{}2009)(u_{n\plus{}1}\minus{}2009)$

1) The sequence $ u_n$ is increasing and so all $ u_i$ are pairwise different
Using induction, we have :
$ u_2>u_1>2\times 2009$
Suppose now $ u_{n\plus{}1}>u_n>u_{n\minus{}1} ... >u_2>u_1>2\times 2009$. Then :
$ u_{n\plus{}2}\equal{}(u_n\minus{}2009)(u_{n\plus{}1}\minus{}2009)$ $ >2(u_{n\plus{}1}\minus{}2009)\equal{}u_{n\plus{}1}\plus{}(u_{n\plus{}1}\minus{}2\times 2009))$ $ >u_{n\plus{}1}$
Q.E.D.

2) $ \gcd(u_n,2009)\equal{}1$ $ \forall n$
Using induction, we have :
$ \gcd(a,2009)\equal{}\gcd(b,2009)\equal{}1$, so $ \gcd(a\minus{}2009,2009)\equal{}1$ and so $ \gcd(b(a\minus{}2009),2009)\equal{}1$ and so $ \gcd(u_1,2009)\equal{}\gcd(u_2,2009)\equal{}1$
Suppose now $ \gcd(u_n,2009)\equal{}\gcd(u_{n\plus{}1},2009)\equal{}1$. Then
If prime $ p$ divides $ u_{n\plus{}2}$ and $ 2009$, then $ p$ divides $ 2009$ and either $ u_n\minus{}2009$, either $ u_{n\plus{}1}\minus{}2009$, so $ p$ divides either $ \gcd(u_n,2009)$, either $ \gcd(u_{n\plus{}1},2009)\equal{}1$, hence $ p\equal{}1$
Q.E.D.

3) $ u_n\equal{}\gcd(u_{n\plus{}1},u_{n\plus{}2})\plus{}2009$ $ \forall n>0$
If $ n\equal{}1$ :
$ u_3\equal{}(a\minus{}2009)(b(a\minus{}2009)\minus{}2009)$ and so $ \gcd(u_2,u_3)\equal{}(a\minus{}2009)\gcd(b,b(a\minus{}2009)\minus{}2009)$
But $ \gcd(b,b(a\minus{}2009)\minus{}2009)\equal{}1$ since $ \gcd(b,2009)\equal{}1$
So  $ \gcd(u_2,u_3)\equal{}a\minus{}2009$
So $ u_1\equal{}\gcd(u_2,u_3)\plus{}2009$

If $ n>1$ :
$ u_{n\plus{}1}\equal{}(u_{n\minus{}1}\minus{}2009)(u_n\minus{}2009)$
$ u_{n\plus{}2}\equal{}(u_n\minus{}2009)(u_{n\plus{}1}\minus{}2009)$
So $ \gcd(u_{n\plus{}1},u_{n\plus{}2})\equal{}(u_n\minus{}2009)\gcd(u_{n\plus{}1}\minus{}2009,u_{n\minus{}1}\minus{}2009)$
But $ \gcd(u_{n\plus{}1}\minus{}2009,u_{n\minus{}1}\minus{}2009)$ $ \equal{}\gcd((u_{n\minus{}1}\minus{}2009)(u_n\minus{}2009)\minus{}2009,u_{n\minus{}1}\minus{}2009)\equal{}1$ since $ \gcd(u_{n\minus{}1},2009)\equal{}1$
So $ \gcd(u_{n\plus{}1},u_{n\plus{}2})\equal{}(u_n\minus{}2009)$
So $ u_n\equal{}\gcd(u_{n\plus{}1},u_{n\plus{}2})\plus{}2009$
Q.E.D.

4) We can create sequences $ x_n$ with as much different terms as we want.
Let $ k$ be the number of different $ x_i$ we are looking for.
Just take then $ x_1\equal{}u_k$ and $ x_2\equal{}u_{k\minus{}1}$. Then :
$ x_i\equal{}u_{k\plus{}1\minus{}i}$ $ \forall i\in[1,k]$ (using point 3 above)
All $ x_i$ for $ i\in[1,k]$ are pairwise different (using point 1 above)
Q.E.D.

[quote="TTsphn"]Dear pco ,I think you solved an other problem . The sequence is : $ x_{n \plus{} 2} \equal{} gcd(x_{n},x_{n \plus{} 1}) \plus{} 2009$ , and I guess this sequence has very few different terms , we only need to consider two cases : exist two consecutive which are not relative prime to 2009 and case doesn't exist such terms .  :)[/quote]

Dear TTsphn,
pco is right.Please have a look at (4) :)

Daer pco
This is also the solution I knew. :) Did you know this problem before?I didn't have a reference for it.
I am wondering if there exist some other solutions for this nice problem.I privately think that it is not natural to create $ \{u_n\}$ instead of $ \{x_n\}$,right?  :roll:

[quote="hxy09"] This is also the solution I knew. :) Did you know this problem before?I didn't have a reference for it.
I am wondering if there exist some other solutions for this nice problem.I privately think that it is not natural to create $ \{u_n\}$ instead of $ \{x_n\}$,right?  :roll:[/quote]

I did not know this problem before, so I cant give you any reference.

I first thought as TTsphn, but I realized that taking $ x_1$ and $ x_2$ great enough and $ x_i$ as a decreasing sequence could be a solution, hence the idea to create the reverse sequence u_n


integration
function
probability
trigonometry
calculus
real analysis
advanced fields

On wikipedia,   http://en.wikipedia.org/wiki/Heat_equation,

it says that the solution to the following heat equation:

$ u_t\equal{}ku_{xx}$
$ u(x,0)\equal{}0$
$ u(0,t)\equal{}h(t)$

is:

$ u(x,t)\equal{}\int_0^t \frac{x}{\sqrt{4 \pi k (t\minus{}s)^3}} e^{\minus{}\frac{x^2}{4k(t\minus{}s)}} h(s) ds$

Not having a reference on hand that talks about Green's functions, why is this the correct formula?  Can anyone explain the derivation other than saying that it satisfies the original equation?

If you know some probability, it is very intersting to see it as a consequence of Ito formula. If not, you can try to use a Fourier transform.

First, the formula you wrote down is not correct.

To elaborate on the last post, take the heat equation and apply the Fourier transform in the $ x$ variable only to both sides of the equation.  Depending on your FT conventions, you get something like
\[ \frac {d}{dt}\hat{u}(\xi,t) \equal{} \minus{} \xi^2 \hat{u}(\xi,t)
\]
Then observe that for a fixed value of $ \xi$, this is just an ODE in $ t$ and thus $ \hat{u}(\xi,t)$ can be solved for in terms of the initial condition $ \hat{u}(\xi,0)$.  To get back $ u$ from $ \hat{u}$, just apply the inverse Fourier transform.  The computation then depends on knowing how to take the inverse Fourier transform of the Gaussian distribution.

Morally, this procedure is just the good old separation of variation technique.  If you know nothing about the Fourier transform, but you do know how to solve the heat equation on a *bounded* interval via separation of variables, here's another way to derive the formula:

Consider the heat equation on the interval $ [ \minus{} L,L]$ with periodic boundary condition.  Write down the formula for the solution $ u$ in terms of the initial condition.  (It's easier if you use complex exponentials rather than sines and cosines.)  The formula is an infinite sum over eigenfunctions, whose coefficients are themselves integrals over $ [ \minus{} L,L]$, so you have a big sum over an integral.  Now take $ L\to\infty$, and note that your infinite sum is like a Riemann sum, so as you take $ L\to\infty$, the sum becomes an integral.  If you then compute the integral (same computation as the FT of a Gaussian as above), you'll get the desired formula.  Note that you don't really have to justify this procedure, since it's just motivation for the formula, and you can directly verify that the formula works.  Incidentally, this procedure also motivates the whole idea of the Fourier transform.

I confess I don't even know the Ito formula.  What is it, and how can it be used to solve the heat equation?

P.S.  This is my complaint about Wikipedia.  The sourcing is awful.


conics
ellipse
geometry
geometric transformation
reflection
cyclic quadrilateral
geometry unsolved

Consider ellipse with foci $F$ and $G$. Let lines $k$ and $m$ be tangent to this ellipse at points $A$ and $B$ respectively and denote their intersection as $P$. Let $X$ be the intersection point of segments $FB$ and $GA$. Prove that $XP$ is a bisector of angle $AXB$.

Let $F_1,G_1$ be the reflections of $F,G$ in $PA,PB$ respectively. Because the tangent in a point to the ellipse is the external angle nisector of the angle formed by the lines connecting the point to the foci, $F_1\in AG$ and $G_1\in BF$. Also, $AF_1=AF\Rightarrow GF_1=GA+AF_1=GA+AF$, which is equal to the large axis of the ellipse. In the same way we show that $G_1F$ is equal to the large axis, so $GF_1=G_1F$. We have $PF_1=PF,PG=PG_1$, so the two triangles $PF_1G,PFG_1$ are congruent.

From the above we find, in particular, that $PF_1FX$ is a cyclic quadrilateral with side $PF_1$ equal to the diagonal $PF$ (that's what it looks like on my drawing; it doesn't matter which is which, since either way, the point is that the diagonal and one of the sides starting from $P$ are equal), so the angle the two chords $PF_1,PF$ subtend at $X$ must be supplementary, i.e. exactly what we wanted.


Is it possible to control the width of a tabular? For example, if I had the document:

[quote]\documentclass[11pt]{article}
\begin{document}
\begin{tabular}[t]{lr}
hi&hi \\
\end{tabular}
\end{document}[/quote]

It appears left justified and only about a third of the width of the page. Is it possible to change that?

Add [b]\begin{center}[/b] and [b]\end{center}[/b] around your tabular environment to center it. :)

[code]\documentclass[11pt]{article}
\begin{document}
\begin{center}
\begin{tabular}[t]{lr}
hi&hi \\
\end{tabular}
\end{center}
\end{document}[/code]
There is also a [b]\hfil[/b] command that apparently acts like [b]\hfill[/b] but only to half the width of the page. I'm not sure exactly how it works, however, and I would recommend sticking to centering the tabular. Still, in case you're interested, here's the code:

[code]\documentclass[11pt]{article}
\begin{document}
\hfil
\begin{tabular}[t]{lr}
hi&hi \\
\end{tabular}
\end{document}[/code]

Thanks for the help, but the main thing I need is for it to stretch the width of the whole page.

[code]\documentclass[11pt]{article}
\begin{document}
\begin{tabular*}{\textwidth}[t]{l@{\extracolsep{\fill}}r}
hi& hi \\
\end{tabular*}
\end{document}[/code]gives
$ \begin{tabular*}{\textwidth}[t]{l@{\extracolsep{\fill}}r}
hi& hi \\
\end{tabular*}$

Replace \textwidth by a fixed amount like 10cm if you want.


topology
algebra
function
domain
geometry
3D geometry
sphere

Has anyone done any research on this intriguing topic? If so, can someone please quickly explain what lie groups are to me? I am very interested in them, but at the same time, I am very confused... :)

All help appreciated, as always!

en.wikipedia.org/wiki/Lie_groups

feel free to post specific questions in the Advanced Topics forum

It's not my specialty, but I do know the basics. A Lie group is a smooth manifold which also has a group structure that interacts nicely with the topology: the group operations are smooth.
What is a smooth manifold? It takes a lot of space to define it precisely, but here's an outline: It looks like some $ \mathbb{R}^n$ locally, and we have a notion of differentiabilty associated to this. There is a family of maps from subsets of the manifold $ X$ to subsets of $ \mathbb{R}^n$, which are homeomorphisms. The domains of this family cover all of $ X$, and if $ \phi,\psi$ are two such maps, $ \phi\circ\psi^{\minus{}1}$ is $ C^{\infty}$ where it is defined.
We often include the dimension in the labeling and call this an $ n$-manifold. If a manifold is embedded in some larger smooth manifold, the natural embedded topology often induces a smooth structure as well; this will be used in constructing examples.

Some examples of manifolds: $ \mathbb{R}^n$ with the usual topology, the spheres $ \mathbb{S}^n\equal{}\{x\in \mathbb{R}^{n\plus{}1}: |x|\equal{}1\}$ with the embedded structure, the torus $ T\equal{}\mathbb{S}^1\times \mathbb{S}^1$.
Some of these can be given Lie group structures as well. $ \mathbb{R}^n$ is a Lie group under the usual addition, $ \mathbb{S}^1$ is a Lie group under angle addition, and $ T$ is a Lie group as the direct product of two copies of $ \mathbb{S}^1$. Others can't be handled this way- $ \mathbb{S}^2$ has no Lie group structure.

These examples so far are abelian. There are nonabelian Lie groups as well. Some come from introducing a disconnected part, like the dihedral group on the circle $ O_2$. In more dimensions, we have examples like the matrix groups $ GL_n(\mathbb{R})$ and $ SL_n(\mathbb{R})$.

Each Lie group $ G$ has an associated Lie algebra $ L$: a copy of $ \mathbb{R}^n$ with an associated operation [ ] mapping $ L\times L$ to $ L$. This bracket is bilinear and antisymmetric: $ [ax\plus{}by,z]\equal{}a[x,z]\plus{}b[y,z]$ and $ [x,y]\equal{}\minus{}[y,x]$. It's basically a commutator for $ G$, and $ [x,y]\equal{}0$ for all $ x,y$ when $ G$ is abelian. In fact, for the example $ G\equal{}GL_n(\mathbb{R})$, we can identify $ L$ with the space of all $ n\times n$ matrices, and the bracket in this case is the matrix commutator $ [A,B]\equal{}AB\minus{}BA$.
Another example: $ \mathbb{S}_3$ can be given a group structure using quaternion multiplication. The associated Lie algebra is $ \mathbb{R}^3$ with the cross product.

There is a map from the algebra back to the group, called the exponential. This map is always smooth, and $ \exp(x\plus{}y)\equal{}\exp(x)\exp(y)$ when $ [x,y]\equal{}0$. This means that $ \exp$ is a homomorphism for abelian groups and algebras. Some examples: for $ \mathbb{S}^1$, $ \exp(x)\equal{}e^{ix}$ on $ \mathbb{R}$ (or maybe $ e^{ikx}$ for some $ k$; scale isn't fixed). For $ GL_n(\mathbb{R})$, $ \exp$ is the matrix exponential.

Wow, thank you for the in depth reply! So I think I have the general idea of what Lie groups are... but do they have applications to other subjects? I heard that they have roles in quantum physics, string theory, etc... what kind of relationships do Lie groups have with these sciences?

A Google search with the keywords "Lie groups and quantum mechanics" turns up [url=http://math.ucr.edu/home/baez/lie/lie.html]this[/url].  As a motivating idea, special relativity can be thought of as the study of geometric properties invariant under the [url=http://en.wikipedia.org/wiki/Lorentz_group]Lorentz group[/url], just as Euclidean geometry can be regarded as the study of geometric properties invariant under the [url=http://en.wikipedia.org/wiki/Euclidean_group]Euclidean group[/url].

The Lie bracket gives commutation relations between observables in quantum mechanics and most famously gives rise to derivation of the Heisenberg uncertainty principle.

Much of particle physics is in fact the study of lie algebras in that particles are fundamentally representations of the Poincare group (or the Lorentz group, I forget which is which) in that on the most basic level, they're just things that transform with respect to Lorentz transformations. So directly attacking the Lorentz group head on is pretty tough which is why we have unified theories with respect to much smaller groups, and all of these algebraic properties are encapsulated in Lie theory. I haven't studied any string theory so can't give insight there, sorry.


Set aside 75 minutes without interruption or distraction. Perhaps let your family know that you are taking a practice test. When it comes to the AMC 12, I've never scored higher on a practice test than I did on the actual 12B my freshmen year. I find that most of the time, I don't bother giving my full attention to the test, since for some reason I assume that I'll always make the AIME.

What takes you longer? Finding the crux of a problem or chugging through the calculations? If you're having problems with the latter, you might try to familiarize yourself with a good calculator, such as the TI-89. Although it is only a tool, a calculator can zip through the crude arithmetic, a good advantage on a fast-based test such as the AMC.

I can't tell you what to do about careless mistakes. Organizing your work (if you don't already) is a starting point.

[quote="1234567890"]I am practicing for the AMC/AIME series and I'm looking for tips to improve my AMC score. You see, my AIME scores are improving to average of 6 and often scores of 8 or 9 come up. But on the last few AMC 12 I've taken, I haven't qualified for the AIME (which is kind of wierd, since I got 118.5 on the AMC 12), getting scores in the 90s. And this is on 3 straight tests. I usually answer 18 or so ( I am an extremely slow worker), but get 4 or 5 wrong, which pulls my score down to less than 100. I can answer all but maybe 2 or 3 of the questions when given enough time, but does anyone have any tips to work well in the AMC's?[/quote]

Also, make sure you're taking tests that are from similar years, because some of the earlier AIME's are very easy, and some of the pre-1985 AMC's are very hard.

Practice!

Where did you get these AMCs and AIMEs?

Online or did you order them?

I know you can get all the AIME's from Kalva

I actually am doing the recent ones (96- 01,03,04) and I get those online or had them since last year. :(


Hey guys, I've got an pretty tough algebra question, can any of you guys help me. :)

Assume that an integer [i]n[/i] is no less than 3, for a bunch of all real num: A1, A2, ...An(meaning: A sub n),
record the minimium value of |Ai - Aj|(where 1<= [i]i,j[/i] <=n) as M.
Find out the max value of M when A1^2 +A2^2 +...An^2 = 1

I've got two requests for you guys.

1) Please use LaTeX in your posts.
2) Don't just give an answer, but a complete solution/proof.

Tank you for request but I don't now English well,so I don't writing long text  :?  :idea:

At least try it.

By the way, mathematics is a universal language. You don't even need a lot of text.

HI
that's the correct answer, I can slove it by assuming: Ai=A+j * m(where a is the largest non-positive number)
what's ya method?

By the way, I've got another question.(I can solve part a, but I can't prove the second one)

a) Assume four nature numbers:a,b,c &d. a/b+c/d <1;  a+c= 12
Find the max value of a/b +c/d
b) if a+c is not greater than n(an positive int), prove the max value of a/c+b/d will be 1-1/m
where m is [(2n+1)/3]*([(2n+1)/3]*[n/3]+1)


[url]http://www.deviantart.com/view/14080073/[/url]

It's fun!

it looks like my art on the other computer(except it moves) Yay! I think it is kinda like a thing they show tou in media player..


calculus
integration
function
inequalities
logarithms
inequalities proposed

For   $ -\infty<a<b<\infty $ let $ K(a,b) $  be the set of all functions $ (a,b)\to(0,\infty) $ which are convex on  $ (a,b) $ .  When the right-sides exist , let
\[
||f||_r :=\left(\frac{1}{b-a}\int\limits_{a}^{b}|f(x)|^r dx\right )^{1/r}\; \; ,\; \;
M_p(f) := \left( \frac{|f(a)|^p  + |f(b)|^p}{2} \right)^{1/p} .
\]
QUESTIONS: Are the following assertions true ?
[b]A)[/b] If $ r \in (0,1] $ , then  the inequality   $\; \displaystyle      ||f||_r  \le \displaystyle M_p(f)  , \; \;  \forall  f \in K(a,b) \; ,  $  is verified  if and only if  $      p\ge r\frac{\ln{2}}{\ln{(r+1)}} $.
[b]B)[/b] When $ f(a)$ and $ f(b)$  exist and are positive , the inequality   $  \; \;   ||f||_r  \le \sqrt{f(a)f(b)}  $  holds  for all  $ f \in K(a,b)  $ ,  iff   $ r < -2 $.

I have a question, you said that $K(a,b)$ is the set of all functions $(a,b)\to (0,\infty)$ bu then why did you write $|f(x)|$ in the integral since the function $f$ is in $K$ and has only positive values? Or I miss something?

Your remark, it's  O.K. .

 I don'nt remember exactly my solution. However by positive functions , instead to write e.g.   (f(x))^r  I prefer to insert
  |f(x)|^r .

Isn't it the case that for given boundary values, $||f||_r$ is maximized by letting $f$ be the straight line connecting the endpoints? That would reduce this problem to a calculation involving that linear function.

Right ! But try   in this case, to compare  $||f||_r$  with  $M_p(f)$  ( in order to find  $p,q$).


I know some vectors, but I saw posts like http://www.mathlinks.ro/Forum/viewtopic.php?t=46367 I don't understand. In the book vector is connected between two points, but then when does something like $\vec{B}$ mean? Also, why is  $ (\vec{B} + \vec{C}) \cdot (\vec{B} - \vec{C}) = 0$ ?

If $O$ is the origin, then we define $vec{B}=vec{OB}$.
The second property is valid if OB=OC. Prove it!

I believe the second result that $ ( \vec{B} + \vec{C} ) . ( \vec{B} - \vec{C} ) = 0 $ is true only if magnitude of $ \vec{B} $ is equal to that of $ \vec{C} $ . Am I right??


The points $ (1,2), (1,4), (3,4),$ and $ (3,2)$ are the midpoints of the sides of a square. How many square units are in the area of the square?

The square formed from the midpoints of a square has half the area of the original square. Since the smaller square has area 4, the answer is 4*2=8.


In the introduction to algebra book by mr.ruzczyk (rruzczk on Aops) i don't really understand how to do the mixture problems, can anyone help?

i do not have that book... could you maybe post it?

I don't have that book either... Can you give an example?

Snape is making a potion. It currently consists of 20% DeMuggle juice. How much more DeMuggle juice will he need to make it 40% DeMuggle juice

Say the potion currently has $ x$ ounces total (or some random unit, whatever). Then we have $ 0.2x$ that is  DeMuggle juice, and $ 0.8x$ that is not. From here, we can do it two ways (you will see one is slightly better). Say we're going to add $ y$ ounces of DeMuggle juice.
Method 1: we know that we are going to add $ y$ ounces of DeMuggle juice. This means that after adding, we will have $ 0.2x \plus{} y$ ounces of DeMuggle juice. We also have $ 0.8x$ ounces of stuff that's not DeMuggle juice, so our total number of ounces is $ x \plus{} y$. Thus we set up equations, and we get $ 40\% \equal{} 0.4 \equal{} \frac {0.2x \plus{} y}{x \plus{} y}\implies0.4x \plus{} 0.4y \equal{} 0.2x \plus{} y$
$ \implies0.2x \equal{} 0.6y\implies x \equal{} 3y$. Thus the amount of DeMuggle juice added depends on the amount of potion we started with, which you probably know.
Method 2: we know that we will add $ y$ ounces of DeMuggle juice. That means we will have $ x \plus{} y$ ounces of potion after adding the DeMuggle juice. However, the amount of stuff that's not DeMuggle juice is still the same, $ 0.8x$. We also know that the potion we want is $ 100\% \minus{} 40\% \equal{} 60\% \equal{} 0.6$ stuff that's not DeMuggle juice. So we have $ 0.6 \equal{} \frac {0.8x}{x \plus{} y}\implies0.6x \plus{} 0.6y \equal{} 0.8x\implies0.6y \equal{} 0.2x\implies3y \equal{} x$, as we got with Method 1 (notice this is somewhat less messy compared to 1).
I hope this helps. :)

Feel free to ask more questions.

dude, you're kidding me right? [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=200]yay[/url][/url]


hi evryone,

here is a nice and simple probleme:
suppose that we have a finite set S={a,b,..} of p letters and we want to write
every words of n letters in one big words. For example if S={a,b} and n=2 then the big word W=aaabbbba contains every word of 2 letters, but W=aabba too. The question is: what is the minimum length of W ?

bye.

Did you guys think about this one? It doesn't look easy to me.. I'm pretty sure that if we denote that number by $f(p,n)$, then $f(p,2)=p^2+1$. I'd like to show that $f(p,n)=p^n+n-1$, but I don't even know if it's true. It still holds for $n=3,\ p=2,3$, but that doesn't really say much, does it?

I think I have a procedure for building the sequence in the general case, but I have no idea if it's Ok or not, and since it's kind of hard to explain, maybe I shouldn't try to post it :). Here's a sketch:

Assume we have characters $1,2,\ldots,p$ with the lexicographic order $1>2>\ldots>p$. The number after $p$ is $1$ (as if they were arranged on a circle). Assume $n>2$. We write down all the words with $n$ letters which are formed from a single letter: $11\ldots 122\ldots 2\ldots pp\ldots p$. Then we write down the word formed only from $1$ which has $n-1$ letters. Next, we take the last $n-1$-letter word and add that letter which comes next lexicographically (that word is formed only from $1$'s, and before it we have words formed from $n$ $1$'s and from $n-1$ $1$'s and a $2$, so the number we write down this time is $3$. After this we complete a $n-1$-letter word formed only from $3$'s and continue the process (we add a $5$ and complete the $n-1$-letter word formed from $5$'s; when we're done with $n-1$-letter words we continue with $n-2$-letter words formed from identical letters and so on, and when we have no words to complete by adding the identical digit we just add the digit which comes next lexicographically). I claim that in this manner we obtain a word like the one we're looking form, which has $p^n+n-1$ letters, and this is clearly the minimum attainable.

Any idea if it's correct/how to prove it?

Example: $p=3,\ n=3$.

$111222333$+$11$ (that's the rule)+$33$ (I already have $111$ and $112$ in my sequence, so I take $113$, because $1>2>3$)+$22$ (I have $333$ and $331$ so I want $332$)+$1$ (I start over with $1$)+$2$+$3$ (when I don't [b]have[/b] to write a specific order for the letters I choose the lexicographic order)+$1$ (I have $233$ so the next one is $231$-only the last digit counts in the lexicographic enumeration I'm speaking about)+$2$ (I have $311$ so the next one is $312$) etc.
 In the end we get $11122233311332212312131323211$. You can check the desired property and the sequence has $29=3^3+2$ letters. I've also checked ot for $p=4, n=3$ and it's an extension of what I used for $p=2$, so I think it has good chances of working, although it's purely empirical up until now.

Sorry :(. I think it's completely bogus.. I should've checked the case $n=4$. Anyway, maybe it's helpful (in case anybody is interested :D).

Here's another formulation: Given $p,n$, show that we can arrange $p^n$ symbols from the set $S=\{1,2,\ldots p\}$ on a circle s.t. if we go around the circle in a clockwise direction then we will eventually find exactly one copy of each of the $p^n$ sequences of $n$ characters formed with the symbols in the set $S$.

Take a look at the third problem on [url=http://www.princeton.edu/~mathclub/puzzles6.html]this page[/url]. It's enlisted in the "Impossible Problems" section, so I guess it's not as easy as alekk made it sound :). And it's not even general: it only treats the case $p=2$. I've struggled on that case during a very long train ride and came up with (basicly) nothing :).

It would be nice if Alek  would give us a hint.
I think that when building the sequence, we should just take care that all strings are different. so we will have all, with the restriction that grobber gave a little upwards :D

hi all,

here a way:consider the graph G whose vertices are the subwords of length (n-1) and so that there is a path from vertex A to B iff there is a word of n letters that begins by A and ends by B. Then prove that there is a path that pass through each edges only on time, which is fairly easy because there is p edges that arrives and p that leave each vertex. The conclusion is clear ...  :D

hope it's correct,
bye

Can you prove that there is this path?.
I mean, I know the proof in case of non-directed graph, but not directed.
Can you please help me?

i think the proof is almost the same that with non-oriented graph: there is p edges that arrive to a vertex and p that leave..  so you begin on a vertex V1, then you go to another V2. Because there are as many edges that arrive and leave you can go from V2 to another vertex and so on..

This problem is very similar to C5/isl 2002. It uses the same idea, which is aplied in a very similar way. But I don't find it easy.


Hi - just wondering if someone could help me out with the following question. The question is:

a) There are 8!/(2!)^4 = 2520 distinct arrangements of the symbols aabbccdd. How many of these have no double letters (that is, have no two consecutive identical symbols) ?.

Thanks in advance!

Can someone help please? Or have i posted this in the wrong forum/room?

[hide]
We can find the number of arrangements with no double letters by counting its complement: the number of arrangements with at least one pair of double letters.

There are $4$ possible choices for the pair of double letters. Treating the pair of double letters as one "unit," there are $7!$ ways to arrange the pair and the other $6$ letters. Since we are only dealing with distinct arrangements, we must divide by $2^{3}$ since we did not account for order in the pair of doubled letters.

This, however, double counts the arrangements where there are $2$ pairs of double letters. There are $\binom{4}{2}$ ways to choose those $2$ pairs, $6!$ ways to arrange everything, and we divide by $2^{2}$.

Using PIE to correctly count the rest, the number of arrangements with at least one pair of double letters is \[ \frac{\binom{4}{1}7!}{2^{3}}-\frac{\binom{4}{2}6!}{2^{2}}+\frac{\binom{4}{3}5!}{2^{1}}-\frac{\binom{4}{4}4!}{2^{0}}= 1656\] To find the number of desired arrangements, we subtract this number from the total number of distinct arrangements to get $864$.
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