calculus
integration
derivative
calculus computations

i dont get the substitution process.. can somebody xplain it to me,

for example this problem n then i can try to figure out te others...

antiderivative (6x^2)dx / ((2x^3+7)^(3/2))

ust run through how to let u=?.... and how to come up with an answer and waht i am looking for

$ \int\frac{6x^2}{(2x^3\plus{}7)^{3/2}}\,dx$

Since substitution in an integral amounts to the chain rule read backwards, look for composite functions. The "natural" substitution is for "the thing inside." Two caveats:

1. There's no guarantee that it will work.

2. There are plenty of "unnatural" substitutions that may be worth trying.

But in this case, let's try the natural substitution. That is, let $ u\equal{}2x^3\plus{}3.$ As an instant reflex action, every time you write down a proposed substitution, you take its derivative before you even think about what to do next. In this case, you write down $ du \equal{} 6x^2\, dx$

Now, in this case (clearly a very "rigged" problem), we happen to have $ 6x^2\,dx$ just sitting right there. So that's $ du.$

The integral becomes

$ \int\frac{du}{u^{3/2}}\equal{}\int u^{\minus{}3/2}\,du\equal{}\minus{}2u^{\minus{}1/2}\equal{}\minus{}\frac{2}{\sqrt{3x^3\plus{}3}}\plus{}C.$

Try differentiating that to check it. Notice how you use the chain rule in that, and what happens when you use the chain rule. Remember that - substitution is the chain rule read backwards.


inequalities
geometry
incenter
inradius
geometry proposed

Given $\triangle ABC$ incenter $I$, inradius $r$and $A',B',C'$ is pedal triangle of $I$, circumcentre $O$ prove that
$OA'+OB'+OC' \ge 3(r-OI)$

Since C'OI is a triangle we have $OC'+OI\geq IC'$ so $OC'+OI\geq r$.We add and we get the result. :)

oh yes it too easy  :lol: ,really it is
$OA'+OB'+OC'\ge 3(r-\frac{OI}{2})$
thank you!

A challenge for you $OA'+OB'+OC'\geq R+r$ :D

yes thank you  :D you are right because follow Carnot theorem
$OA'+OB'+OC'\ge OA''+OB''+OC'' =R+r$ with  $A'',B'',C''$ is midpoint of $BC,CA,AB$  :D but I think I has a big mistake thank you again!


find the exact value of:
$cos(20)cos(40)cos(80)$
(of course, do this without a calculator)

Please do not respond to a post saying "nice solution" or something equivalent.  If you want to express your awe or gratitude, PM the person.  Of course, if you have something else CONSTRUCTIVE to say along with it, you can express admiration for someone else's solution.


Let positive integers $ n,x,y,z$ satisfying $ x^3\plus{}y^3\plus{}z^3\equal{}nx^2y^2z^2$. Prove that $ n\equal{}1$ or $ n\equal{}3$.

[hide="Solution"]WLOG $ x\ge y\ge z$

$ 3x^3\ge x^3\plus{}y^3\plus{}z^3\equal{}nx^2y^2z^2$
$ 3x\ge ny^2z^2\cdots (i)$

From the main eqn. $ x^2|y^3\plus{}z^3$
$ x^2\le y^3\plus{}z^3$
$ (3x)^2\le 9(y^3\plus{}z^3)\cdots (ii)$

From $ (i)$ and $ (ii)$
$ 9(y^3\plus{}z^3)\ge n^2y^4z^4$
But $ 2y^3\ge y^3\plus{}z^3$
Thus $ 18y^3\ge n^2y^4z^4$
or $ 18\ge n^2yz^4\cdots (iii)$

If $ n\ge 5$ this clearly can't hold.
If $ n\equal{}4$ then $ y\equal{}z\equal{}1$ from $ (iii)$
putting in the main eqn $ x^3\plus{}2\equal{}4x^2$ which clearly has no integer solutions.
if $ n\equal{}2$ then $ (iii)$ gives $ z\equal{}1$ and $ y\equal{}1\text{ or }2\text{ or }3\text{ or }4$
putting these cases in the main equation there is no solutions.

So $ n$ can't be $ 2,4,\ge 5$
hence it must be $ 1$ or $ 3$.[/hide]


geometry
3D geometry
summer program
Mathcamp
MathPath
videos
AwesomeMath

There's already a topic about this somewhere else.

I solve rubix cube in about 3 min. I learned how to solve it by reading the thing that tells you how to solve it that comes with the rubix cube when you buy it. Its the 7-step solution thing. My fastest was 2 min. 34 secs. I've solved it about 7 times since day before yesterday (that was the first time that i solved a rubix cube)

1. somewhere around 2-  2.5 minutes.
2. i found a guide on the links page of speedcubing.com
3. too many to count. maybe 100?

[quote="GoBraves"][quote="chesspro"]I'm speaking for a person I met here, at MathCamp (although I can solve a Rubik's cube).

This person holds a few world records and I saw him get it twice yesterday - 13 and 19 seconds. He's gotten the cube in 25 moves (23 is the minimal amount in any situation). He's done it blindfolded and single-handedly, as well as when juggling two other cubes.[/quote]

Actually the world record for fastest time is around 12 seconds (by Macky) ..I think you are talking about Macky too.[/quote]

Yes, I am talking about Macky (I didn't want to put his name down). Of course, it would be very hard to get 12 seconds every time; he must go slower at other times.

[quote="chesspro"]Of course, it would be very hard to get 12 seconds every time; he must go slower at other times.[/quote]
He also goes faster at other times..

I learned how to solve a Rubik's cube from my dad. he taught me the way he figured out 25 years ago. And... I don't know how many times I've solved it.. And I posted on the other thread already.

i have gotten about 45 minut. (i'm slow)
There is a topic in Math and Strategy games

I learned by looking up a solution online.  My best non-lucky (not skipping steps) time is currently 1:06.95.

Edit: Now it's 56.14s.  I can generally do it in a little over 1:10.

I can do it in 5 to 10 minutes  :blush: with the guidebook. I don't have a cube myself though... I used a friend's.

[quote="math92"][quote="mathclass"] I don't have a cube myself though... I used a friend's.[/quote]
you should get a cube of your own.  there are lots of websites you can go tho and they have them at Target or Wal-Mart i think.[/quote]

It was addicting at first... but then I kept forgetting the combo's, and now I think it's pretty much a waste of time (well a FUN waste of time :P)

if you come to think about it, it would be easier to peel off the stickers and rearrange them......

i tried that and the sticker ripped... :lol:

Oh, I think it's quite tricky. I've seen such a thing as a micky mouse-head. Nobody could solve it. But the results sometimes looked quite funny.

An another question for the real 2-D cube: I can't imagine how to move the different pieces of the square in the plane without cutting the 2-D cube. I think it's not possible, or am I wrong?

I just started learning how to solve a rubiks cube 4 weeks ago. My record is 45 seconds but my average is closer to 1 min 05 secs

A "2D" cube would prolly be something like those number slider things.. where you hafta put it in order from 1 to 15. Like this one: http://www.cee.hw.ac.uk/~ceehsa1/languages/javascript/puzzle2/

[quote="Drunken_Math"]Please only post here if you can solve a Rubix Cube!  :)

How fast can you solve a Rubix Cube?
How'd you learn how to solve it?
How many times have you solved a cube (estimate if you did it too many times already)?

I can solve a Rubix Cube in 5 minutes
I learned it from Dr. Zeitz's breakout group sessions at Mathpath.
I solved it a lot of times so far, maybe around 20 times...[/quote]

1 minute
I learned from the rubiks hint thingy
I did it like... 200 times? more?

Last avergae I took was like 19...and move average like 40 for the 3x3x3
I havesolved it a lot...but I don't anymore...

Maybe I should have gone to mathcamp instead of promys...nah...

And chesspro, was the 25 moves lucky, or a consistent average?

That was for fewest moves, not fastest time. So Macky had that planned out.  :)

at awesomemath:

"let's see who can solve a rubik's cube underwater"

vincent: i'll do it
20 seconds later he comes up with a solved rubik's cube

andy: let me try
30 seconds later he comes up

daesun: wtf, let me try
a minute and 40 seconds later he comes up with a solved rubik's cube

i used to be pro at mathpath, now i suck (i tried to learn the petrus method but i gave up)

I hate rubik's cubes mines stickers are totally crap!

well replace the stickers

Oh, but if you have it planned out, you could make it as few moves as you wanted, right?

oh yes!  :D  I could rip the stickers off and then play rubik's cube

[quote="math92"]oh, well i have seen a 2-D cube before.  THey are so easy to solve :D
it is just like a rubix cube exept it has 4 squares on one side.[/quote]

really? its the same...

I can do the cube in one min. I learned how and amsp.(HI AMSPERS!!) Ive done it at least 2383 times. I have been doing it since I learned how.

There is a topic about this in brainteasers...

i learned about 9 days ago and 1 min is my best, 1:30 is my average.

nice. my range is now like 48 seconds-1:30


Let S be a set which has finitely prime.
Show that there are finitely pair of positive integers (a, b,c)
such that
$a>b>c$ and factors of $(ab+1)(ac+1)$ are elements of S

sorry for my VBE (very bad english.)

Hmmm, I don't know if the following result was proved or not, but in 1996 it was still a conjecture: if $P(x)$ is the greatest prime divisor of $x$ and $a,b,c$ are distinct positive integers with $max(a,b,c)$ tends to $\infty$, then $P((ab+1)(bc+1)(ca+1))$ also goes to infinity. So, good luck in solving this!


trigonometry
AMC
AIME
trig identities
Law of Cosines

A regular 12-gon is inscribed in a circle of radius 12.  The sum of the lengths of all sides and diagonals of the 12-gon can be written in the form
\[ a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}, \]
where $a$, $b$, $c$, and $d$ are positive integers.  Find $a + b + c + d$.

[hide]If we label the vertices $V_{1}, V_{2}, \cdots, V_{12},$ call a segment, $V_{i}, V_{i+k},$ as $L_{k}.$ Using the Law of Cosines, we find all $L_{i}.$ Adding up all values of $L_{i}$ and multiplying by $12$ yields:

$144(2+\sqrt{2}+\sqrt{3}+\sqrt{6})$[/hide]

Im confused can you explain taht a little more, liek why you multiply by 12 and stuff

Well, he's using Law of Cosines to determine the length of every type of diagonal. And then there are twelve of each, excluding the diameters. So what you get is $6(24)+288(\sin 15+\sin 30+\sin 45+\sin 60+\sin 75)$, I think.

He's not excluding them.  He's just bringing them outside the sum because there are only $ 6$ of them, as opposed to $ 12$ of every other kind of diagonal.

When you apply the law of cosine on the 150 degrees angle you get x^2=288-288cos150, solve for x we get the square root of (2+square root of 3), how did you guys simplify this into the square root of 6?

Multiply by 4 to get $ 8 \plus{} 4\sqrt3 \equal{} 6 \plus{} 2 \plus{} 2\sqrt2\sqrt6 \equal{} (\sqrt6 \plus{} \sqrt2)^2$
(BTW what's the way in general to simplify expressions like these? it took me a while to realize that trick of multiplying by 4... on the actual aime with the stinky time pressure i might not even have time to think...)

Dang this problem is sooo easy for #12 but i still got it wrong (somehow i got stupid and got the coefficient of $ \sqrt6$ as 288 when it should have been 144 and missed the ans by 144... yeah the computations sucked)  :wallbash:  :dry:  :wallbash_red:


On a clock with hour, minute, and second hands, the second hand is 14 cm long. In 71 minutes, the tip of the second hand travels a distance of $k\pi$ cm. What is the value of $k$?

Only one I missed on this round. :mad:

[hide]Isn't it just $k = 71 \cdot 28 = 1988$?[/hide]

Yes :blush:

I had some function $f(m)=\frac{m}{60} \cdot 28\pi$ :?


combinatorial geometry
geometry
3D geometry
counting
Intersection
IMO
IMO 1964

Supppose five points in a plane are situated so that no two of the straight lines joining them are parallel, perpendicular, or coincident. From each point perpendiculars are drawn to all the lines joining the other four points. Determine the maxium number of intersections that these perpendiculars can have.

[hide="Solution"]We have 10 lines between these 5 points, and between any 4 points, there are 6 lines. Thus, from each point, we draw 6 perpendiculars, giving us 30 perpendiculars total. This gives us $ \binom{30}{2}\equal{}435$ intersection points total.
But clearly, we have overcounted. From every point, there are 6 lines emerging from it; we have counted $ \binom{6}{2}\equal{}15$ intersections between these lines, when there is only 1. Thus we have an over count of 14 for each 5 points, giving us $ 435\minus{}70\equal{}365$ remaining intersections.
We have also failed to account for parallel lines; each of the 10 lines has 3 perpendiculars which are parallel, giving an overcount of 30. We now have $ 365\minus{}30\equal{}335$ remaining points of intersection.
Finally, we have failed to count when 3 perpendiculars (coming from distinct points) coincide. We know that this happens at each of the $ \binom{5}{3}\equal{}10$ triangles formed from our original triangles, at their orthocenter. We count the intersection 3 times when there is only 1, giving an overcount of $ 20$, and reducing our maximum to $ 315$ points.[/hide]

[quote=comb2019]Could you help me with finding the example which fits the value of 315, please, or maybe how to build the example?[/quote]

To build an example take very ugly irrational numbers for which the slopes can't be equal or a rational multiple of the others. This will ensure that none of them are parallel or perpendicular.
One such example can be $(1,1),(e,\pi),(e^2,\pi^2),(e^3,\pi^3),(e^4,\pi^4)$. This should work. Perhaps, Other powers of $e,\pi$ will also work. The fact is that you have to take care of the given conditions.


trigonometry
function
calculus
calculus computations

Evaluate: $ \sum_{n \equal{} 1}^{k}\tan^{\minus{}1}(\frac{1}{n^2 \plus{} n \plus{} 1})$

Here's a hint:

$ \tan^{\minus{}1} a \plus{} \tan^{\minus{}1} b \equal{} \tan^{\minus{}1}\left(\frac{a\plus{}b}{1\minus{}ab}\right)$.

[hide]If the above was not sufficient, then try to find a function $ f(n)$ such that $ \tan^{\minus{}1} f(n\plus{}1) \minus{} \tan^{\minus{}1} f(n) \equal{} \tan^{\minus{}1} \frac{1}{n^2\plus{}n\plus{}1}$.

[hide]Observe that the above result gives a telescoping series.[/hide]
[/hide]

$ \sum_{n\equal{}1}^k \arctan\left(\frac{1}{n^2\plus{}n\plus{}1}\right) \equal{} \sum_{n\equal{}1}^k \left[\arctan\left(\frac{1}{n}\right) \minus{} \arctan\left(\frac{1}{n\plus{}1}\right)\right]$

$ \equal{} \arctan(1) \minus{} \arctan\left(\frac{1}{k\plus{}1}\right) \equal{} \frac{\pi}{4} \minus{} \arctan\left(\frac{1}{k\plus{}1}\right)$.

As an additional exercise, try to find

$ S \equal{} \sum_{n\equal{}1}^{\plus{} \infty}\arctan\left(\frac{2}{n^2}\right)$

This one is a modification of the original question, with the choice

$ \tan^{\minus{}1} (n\plus{}1) \minus{} \tan^{\minus{}1} (n\minus{}1) \equal{} \tan^{\minus{}1} \left(\frac{2}{n^2}\right)$.


Prove that

$ 6(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)\leq 27abc\plus{}10(a^2\plus{}b^2\plus{}c^2)^ {\frac{3}{2}}$

with $ a,b,c \in R$

[quote="Math pro"]let a^2+b^2+c^2=1
then use p,q,r really nice! :lol:[/quote]
Sorry, can you detail it ,detail it please
I cant proved it with  your way,i think it is very hard
My proof is quite short

[quote="Allnames"]Prove that

$ 6(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)\leq 27abc \plus{} 10(a^2 \plus{} b^2 \plus{} c^2)^ {\frac {3}{2}}$

with $ a,b,c \in R$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=52
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122166


From the University of Maryland Mathematics Competition, Part 1, 2001:

Tom Sawyer and Huck Finn want to paint a fence. Tom can paint the fence by himself in 3 hours, and Huck can paint the fence by himself in 4 hours. At 12:00 noon they start painting the fence together. However, at some point they get into a fight. They fight for 10 minutes, during which time no painting gets done. After the fight, Huck leaves and Tom finishes painting alone. If Tom finishes painting at 2:25 pm, at what time did the fight begin?
a. 12:30   b. 1:00   c. 1:10   d. 1:15   e. 1:30

[hide]Let $t$ be the number of hours after noon until the fight. The total time painting was (2 hours 25 minutes)-(10 minutes)=2 hours 15 minutes=9/4 hours. Tom can paint 1/3 of the fence in 1 hour and Huck can paint 1/4 of the fence in 1 hour, so
$t\left( \frac{1}{3}+\frac{1}{4} \right) +\frac{1}{3}\left( \frac{9}{4}-t\right)=\frac{t}{4}+\frac{3}{4}=1$. This gives $t=1$, so fight began at 1:00. [b](B)[/b][/hide]


[hide="Solution"]There are $ 6$ possible ways she can purchase $ 2$ bars:

Milk, Milk
Milk, Dark
Milk, Bittersweet
Dark, Dark
Dark, Bittersweet
Bittersweet, Bittersweet

Of those $ 6$, the first $ 2$ don't work, but the other $ 4$ do.

$ \frac {4}{6} \equal{} \frac {2}{3}\quad\to\quad\mathrm{E}$[/hide]

[quote="8parks11"]Type of chocolate bar Percent cocoa by weight:
Milk 35
Dark 50
Bittersweet 70

A website sells 3 types of chocolate bars of equal weight. If Serena orders 2 chocolate bars at random form the website, the melts them together, what is the probability that the resulting mix contains at least 50% cocoa by weight?

$ A. 1/9$
$ B. 1/3$
$ C. 4/9$
$ D. 1/2$
$ E. 2/3$

I'm tutoring several students in SAT Math and I am getting E as the answer but the book is saying D without any explanation. Who is wrong?[/quote]
I think the book is wrong and you're right.
The only ways of getting less than 50% is when its either milk-dark, milk-milk, or dark-milk.
that's 3 of $ 3^2$ ways. $ 1\minus{}\frac{3}{9}\equal{}\frac{2}{3}$


be nazare man bad nist yek kami dar morede masaele mohasebati bahs va tabadole nazare bokonim.
lets start. :D

[quote="ehsan2004"]manzoret az masaleye mohasebati chiye , hendeseye mohasebati ,ya .....[/quote]
albateh bishtar manzooram hendeseye mohasebatie vali dar kol manzooram kolle masaele.

be nazare khode man masaele jalebi hastand.kheili jaleb.

in ham yek soale mohasebatie rahat:
if $a>b>c$ then prove that $a+h_a>b+h_b>c+h_c$

[quote="amir2"]in ham yek soale mohasebatie khafan:

agar dar yek 6 zelei parekhati ke vasate 2 zelei rooberoo ra be ham vasl mikonand$\frac{\sqrt 3}{2}$ majmooe an 2 zel bashad angah sabet konid Zaveiye haye 6 zelei hame baham barabarand.

           $\mbox{IMO}\text{2003}$[/quote]
in masale ro man ghablan,  ba estefade as bordar (vector), hal karde boodam, albate fekr konam ye rah e hal e asoon tar ham dareh.

[quote="BaBaK Ghalebi"]jensen jan in soal ba yezare mosalasat bazi hal mishe manzooram sin va cos ast
sadast ziadam kharkari nadare[/quote]
mer30 az rahnamieet vali khodam midoonam sadast vali az oonja ke amir joon ye soale mohasebati khast va man moteasefane dar astinam yek soale sakht nadashtam leza ino ferestadam. :D

ye masaleye mohasebati bahal.
char zelyiye kamele $abcdef$dade shode do ghate delkhahe $ep$va $fm$
yekdigar ra bar roye ac ghat mikonand neshan dahid $nc,td,ap$hamresand.
(emtedade $ab,cd$ yekdigar ra dar E va $ac,bd$ yekdigar ra dar f ghat mikonand.
$ep,fm$be tartib $bc,cd$ ra dar P va M ghat mikonand)

inam yek masaleye dige az rosiye:
masahate 4 mosalase sabz ba ham barabar ast sabet konid masahate 3 charzelyie abi niz ba ham barabar ast


I'm writing my CV in LaTeX Here is a part of my CV
\textbf{Programming languages}

\begin{tabular}{lll}
& C & intermediate knowledge\\
& Java & intermediate knowledge\\
& Visual Basic & intermediate knowledge\\
\end{tabular}

$ \,$

\textbf{Markup languages}

\begin{tabular}{lll}
& CSS & basics\\
& HTML & intermediate knowledge\\
& XML & basics\\
\end{tabular}

But I would like to see those tables such that the knowleges-colums are aligned. Now those markup languages starts too left. And if I combine those tables, I don't see how I can put the title "Markup languages" in-between the tables.

You can fix the width of any colmn using p{[i]width[/i]} instead of l in \begin{tabular}{lll}. However, it would probably be easier to merge the cells for the heading as in:[code]\begin{tabular}{lll}
\multicolumn{3}{l}{\textbf{Programming languages}} \\
& C & intermediate knowledge\\
& Java & intermediate knowledge\\
& Visual Basic & intermediate knowledge\\
\multicolumn{3}{l}{\textbf{Markup languages}}\\
& CSS & basics\\
& HTML & intermediate knowledge\\
& XML & basics\\
\end{tabular}\end{tabular}[/code]
\begin{tabular}{lll}
\multicolumn{3}{l}{\textbf{Programming languages}} \\
& C & intermediate knowledge\\
& Java & intermediate knowledge\\
& Visual Basic & intermediate knowledge\\
\multicolumn{3}{l}{\textbf{Markup languages}}\\
& CSS & basics\\
& HTML & intermediate knowledge\\
& XML & basics\\
\end{tabular}

You can change the l to c in \multicolumn{3}{l} to centre the heading if you wish.

Better than previous version. I have written my CV by using flushleft-tags. Now the table should be aligned a little bit more left, because now Programming languages are a little bit too right.
[code]

\begin{flushleft}
\textbf{IT-knowledge}

\textbf{Operating systems} Mandrake Linux 10.0, Microsoft Windows 95, Microsoft Windows XP Professional, Ubuntu Linux 6.06--9.04
\end{flushleft}

$\,$

\begin{flushleft}
\textbf{Software} Microsoft Office Excel, \LaTeX, OpenOffice, Microsoft Word, Microsoft Visio
\end{flushleft}

\begin{flushleft}
\begin{tabular}{lll}
The code Stevem wrote.
\end{tabular}
\end{flushleft}
[/code]

@{} will suppress the intercolumn spacing that would normally be there. \noindent is simpler than flushleft. Here are some examples and you can experiment further.
Click on each example to see the code

$ \noindent \text{Example 1} \\
\begin{tabular}{@{}lll}
\multicolumn{3}{@{}l}{\textbf{Programming languages}} \\
& C & intermediate knowledge\\
& Java & intermediate knowledge\\
& Visual Basic & intermediate knowledge\\
\multicolumn{3}{@{}l}{\textbf{Markup languages}}\\
& CSS & basics\\
& HTML & intermediate knowledge\\
& XML & basics\\
\end{tabular}$

$ \noindent \text{Example 2} \\
\begin{tabular}{@{}l@{}ll}
\multicolumn{3}{l}{\textbf{Programming languages}} \\
& C & intermediate knowledge\\
& Java & intermediate knowledge\\
& Visual Basic & intermediate knowledge\\
\multicolumn{3}{l}{\textbf{Markup languages}}\\
& CSS & basics\\
& HTML & intermediate knowledge\\
& XML & basics\\
\end{tabular}$

$ \noindent \text{Example 3}\\
\begin{tabular}{@{}l@{}ll}
\multicolumn{3}{@{}l}{\textbf{Programming languages}} \\
& C & intermediate knowledge\\
& Java & intermediate knowledge\\
& Visual Basic & intermediate knowledge\\
\multicolumn{3}{@{}l}{\textbf{Markup languages}}\\
& CSS & basics\\
& HTML & intermediate knowledge\\
& XML & basics\\
\end{tabular}$

Note that \text{} is not required in a document - but is needed for this forum


Since it has been solved, are you taking it away as a post count 'level?'

I hope it changes, because they already have a conjecture. hodge

Maybe a neater one will come out.  :lol:

No, the seven levels (not including new member) are the 7 millenium problems, which were all open in the beginning of 2001.

P=NP
Hodge Conjecture
Poincare Conjecture
Birch and Swinnerton Dyer Conjecture
Riemann Hypothesis
Navier-Stokes Equations
Yang-Mills Theory

Gregory Perelman proved the Poincare Conjecture just last year in an apprently badly written paper. I believe Yau came close, but didn't actually do it.

[quote="13375P34K43V312"]Gregory Perelman proved the Poincare Conjecture just last year in an apprently badly written paper. I believe Yau came close, but didn't actually do it.[/quote]
This has been discussed many times on the forum. But what you said is kinda inaccurate so I wanted to point this out.

1) Perelman's paper was not badly written. It was just really concise so people had to work pretty hard to figure out what he did.

2) Yau didn't do anything. After Perelman published his papers, a lot of mathematicians got cracking on checking it and trying to understand what he did. Yau and two of his students (?) were among these, and they published a book about it. There was some controversy about how much they contributed to the proof. There was a New Yorker (or something) article that bashed Yau quite a bit. I don't think Yau or his students or anyone else has contributed anything to the conjecture since Perelman's papers besides to understanding Perelman's proof. Perelman proved it.

If you mean Yau tried, that may be true, I don't know. But if so, his proof wasn't "almost" as in it was the same as Perelman's until some point where Yau got stuck. It just didn't work.

I don't mean to bash Yau here. I just feel like people don't give Perelman enough credit, simply because he's trying to be as inconspicuous as possible and not be made out into one of the greatest mathematicians in the world. Perelman's not saying "oh I didn't actually do anything." He's saying "it doesn't matter that it was me." There's a big difference.

3) Perelman's work was following ideas of Hamilton. Hamilton had ideas about a way to go about proving it, and Perelman read Hamilton's work and actually carried these out. If you want someone to give partial credit to, it's Hamilton.

hmmm... I made an unroved formula for the area of an n sided polygon with side length s, but someone could prove it... Now I don't have any good ones. :(

wikipedia is your best friend forever.
[url]http://en.wikipedia.org/wiki/Millenium_problems[/url]


Nobody interested? Maybe I should give a hint?

If I am not wrong  a stronger result is
[quote="Rzeszut"]Prove that if $a,b,c,d>0$ and $a^{2}+b^{2}+c^{2}+d^{2}=1$, then \[\sum\left( \sqrt[4]{a(1+a^{2})}\cdot\sqrt{b+c+d}\right) \leq \frac{7(a+b+c+d)+5}{4}\] and it is proven by AM-GM  :|

[quote="silouan"]If I am not wrong  a stronger result is
[quote="Rzeszut"]Prove that if $a,b,c,d>0$ and $a^{2}+b^{2}+c^{2}+d^{2}=1$, then \[\sum\left( \sqrt[4]{a(1+a^{2})}\cdot\sqrt{b+c+d}\right) \leq \frac{7(a+b+c+d)+5}{4}\] [/quote]
and it is proven by AM-GM  :|[/quote]
Yes, you're right.  :blush:  :blush:  :blush:  I knew that I had a very weak estimation (besised of $\mathrm{AM-GM}$ I used $\mathrm{QM-AM}$), but I didn't suppose that so weak. It was my first inequality problem, I'm not experienced in creating inequalities.


What is the Simson Line and how do you prove its existence?

It's the degenerate pedal triangle formed when the pedal point (is that what you call it?) lies on the circumcircle. Or... go to wikipedia? or mathworld?


linear algebra
matrix
geometry
3D geometry
algebra
polynomial
Galois Theory

I've started investigating matrix roots of unity; that is, the $ n \times n$ matrices $ \mathbf{A}$ such that $ \mathbf{A}^k \equal{} \mathbf{I}$ for some $ k$.  I haven't done much, but so far I've got explicit forms for the $ 2 \times 2$ square and cube roots of unity (although I may have missed a case regarding the latter).

I have not, however, been able to find information regarding this subject online.  Is this a well-known aspect of matrix rings, and if so, where can I find more information on the subject?  Specifically, I'm interested in a classification of such matrices.

I'm aware that for a given $ k, n$ the corresponding matrix roots of unity form a multiplicative group, and that any matrix similar to such a matrix is also a matrix root of unity.  What other results can be deduced?

The minimal polynomial of such a matrix divides $ X^k \minus{} 1$, hence it has only simple roots.

That's equivalent to $ A$ being diagonalizable...

[hide]Look [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=189490&search_id=984059581#189490]here[/url].[/hide]

[i]Note.[/i] Over finite fields, complications appear :) I was assuming that the field was $ \mathbb C$.

My thoughts: This is pretty closely tied to Galois theory- it's not very interesting over an algebraically closed field of characteristic zero, because there every such matrix is diagonalizable. If either of those two fails, we get other matrices.

Example 1: $ 2\times 2$ matrices over $ \mathbb{R}$ of order 3 (not algebraically closed).
We get $ I$ and all matrices with characteristic and minimal polynomial $ x^2\plus{}x\plus{}1$; this is one conjugacy class for $ I$ and one for the rest.

Example 2: $ 2\times 2$ matrices over $ F$ of order $ p$, where $ p$ is the characteristic of $ F$ (not characteristic zero).
Here, we get $ \pm I\plus{}N$, where $ N$ is nilpotent. That's a total of four conjugacy classes, depending on whether $ N$ is zero and which sign we take in $ \pm$.

In the alg. closed and char. zero case, there are $ \binom{n\plus{}k\minus{}1}{n}$ conjugacy classes- this is generally larger than in the other cases.


Let $ a,b,c\in\mathbb{R}and a,b,c>0$such that $ a^{2}+b^{2}+c^{2}=3$.
     Prove that $ \frac{5}{a+b+c}+\frac{3}{abc}\geq 18$.

Are you sure?????Shouldn't it be $ 4\frac{2}{3}$???????

[quote="duy_nguyen"]Let $ a,b,c\in\mathbb{R}and a,b,c > 0$such that $ a^{2}+b^{2}+c^{2}= 3$.
     Prove that $ \frac{5}{a+b+c}+\frac{3}{abc}\geq 18$.[/quote]
Maybe $ 5(a+b+c)+\frac{3}{abc}\geq 18$ $ ?$
If so see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86904


About how much time will need to be spent on the courses Algebra 3 and Precalc?

I hope to know because I need to balance my time.

Thanks.

I can't say anything about the Precalc (my guess is that it will change from previous years since there is an accompanying text now), but I can say something about Algebra 3. Assuming you have never done algebra at the AIME/Olympiad level, then I'd say around 5 - 10 hours a week including book work and message board problems (I strongly recommend these for practice when they aren't "exercises", i.e. problems you can do in your head). But there's a lot of variation based on how much effort/enthusiasm you put towards the class. It is not necessary to do the message board problems, but it is your only opportunity to get (possibly) critiqued by fellow classmates and/or the teacher or assistant, other than in the open classroom times. But then again, it all depends on you. What I recommend is going through the book way ahead of the class. From my experience of AoPS classes, it's just easier if you are a bunch of chapters ahead of the class so that you don't fall behind when you are burned out from math or get suddenly busy with extracurriculars and/or a lot of school work.

Also, get the challenge sets done a week or two before they're actually due. This way, you can post on the message board regarding queries you have about the challenging problems and ask for hints. You can learn more if you collaborate. The challenge sets are yet another reason to go ahead in the book. If you are ahead, then you'll know the concepts that can be used to solve the problems.

Take Algebra 3 before Precalc.

Expect 5-8 hours/week per course to get the most out of the classes.

[quote="sunehra"]
Also, get the challenge sets done a week or two before they're actually due. This way, you can post on the message board regarding queries you have about the challenging problems and ask for hints. You can learn more if you collaborate. The challenge sets are yet another reason to go ahead in the book. If you are ahead, then you'll know the concepts that can be used to solve the problems.[/quote]

lol sunehra obviously a lesson learned from all of us frantically working right before the deadline.

[quote="Blazer123"]Thanks!

Should I take Algebra 3 first or Precalc first?  (I will be taking Algebra 2 in the upcoming school year and I hope I can skip precalc in school)[/quote]

Algebra 3 will help aLOT more in intermediate algebra, so do that first.


function
algebra
polynomial
geometric series

I have been trying to learn about generating functions, and I just read about characteristic equations.  The thing is, it seems to me like the method is backwards.  Can anyone point me to a proof?

For instance, I'm trying to solve $ a_n\equal{}2a_{n\minus{}2}\minus{}a_{n\minus{}1}$.  So I found $ G(x) \equal{} 2x^2G(x) \minus{} xG(x) \plus{} a_0 \plus{} x(a_1\plus{}a_0)$ ($ 2x^2G(x)$ from $ a_{n}$, $ \minus{}xG(x)$ from $ a_{n\minus{}1}$ and $ a_0\plus{}x(a_1\plus{}a_0)$ to adjust it to start with the correct values).

This implies $ G(x) \equal{} \frac{(a_1\plus{}a_0)x\plus{}a_0}{1\plus{}x\minus{}2x^2}$.  The only thing I can thing of is that $ G(1/x)$ gives the supposedly correct characteristic equation, but this is not a valid move.  Did I do something wrong?

You are correct about the relationship between the denominator of the generating function and the characteristic equation, but it's a relationship that needs to be proven.  See a special case [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=178530&search_id=2069231291]here[/url].

I won't give the full proof, so here's a sketch:  it will turn out that the denominator of the generating function has roots that are the reciprocal of the roots of the characteristic polynomial (which is what you want to prove).  Call these roots $ r_1, r_2, ... r_n$.  Then the denominator factors as $ a(1 \minus{} r_1 x)(1 \minus{} r_2 x) ...(1 \minus{} r_n x)$.  Now use partial fraction decomposition and expand the geometric series as in my demonstration above.

Interesting... I think I understand it now.  So the characteristic equation is NOT the denominator of the generating function?  I think that is where I was confused.  Anyway, I think that, from the method of characteristic equations, it's pretty obvious that the characteristic equation is equivalent (or at least proportional to) the denominator of the $ G(1/x)$, since to find the generating function $ G$ we multiply by $ x$ to shift the terms of the sequence to greater indices but with characteristic equations, we replace the greater indices with the higher powers of x, the exact reverse.  But, as you pointed out, the power series of the partial fraction decomposition ends up taking the reciprocals of the roots to powers, which we know are the roots of the characteristic equation.

Am I explaining this correctly?


algebra
polynomial
integration
algebra solved

A polynomial in x has integer coefficients and takes the value 1 for four distinct integer values of x. Prove that there is no integer value of x for which the polynomial takes the value 24.

[quote="Arne"]Let p(x) be the given polynomial.

Then p(x) - 1 has fours distinct integers, say a1, a2, a3, a4 among its roots.

Hence there exists a polynomial q such that

p(x) - 1 = q(x)(x - a1)(x - a2)(x - a3)(x - a4).

Suppose that, for a certain integer t, p(t) = 24.

Then q(t)(t - a1)(t - a2)(t - a3)(t - a4) = 23.

Since 23 is prime, this is obviously impossible since the LHS has more than three distinct factors (because a1, a2, a3, a4 are distinct).

The result follows.[/quote]
Same hypothesis
There is no integer t such that p(t)=2004 if so then
q(t)(t-a1)(t-a2)(t-a3)(t-a4)=2003, 2003 is prime and so on...
in fact two distincts integers a1,a2 is enough to get a contradiction
 \cap  \cap  \emptyset  \cup \beta  \int


How many pythagorean triples contain the number 12? (ex. 5-12-13)

nice problem. :)
let 12,a,b (a<b) the sides of the right triangle, with a,b E N*(natural)
Then if 12>a,b      a^2+b^2=144     (1)
and   if 12<a  or a<12<b       a^2+144=b^2     (2)
a,b are integers
I think the only solution is 5-12-13 from (2).
We have (b+a)(b-a)=144 and both a,b should be even....

[hide="Answer"]In a base Pythagorean triple, one leg is odd, one leg is even, and the hypotenuse is odd. Therefore, if the hypotenuse is 12, the base hypotenuse must be 3, which is impossible, so we must look at the cases where one leg is 12. We can find the two legs of a base triangle by using the formulae $m^2-n^2$ and $2mn$, for any distinct integers $m$ and $n$; since the difference between squares is always odd, we must use the $2mn$ formula to find an even side, so we can get $12$ with $(1,6)$ and $(2,3)$, which are the Pythagorean triples $(5,12,13)$ and $(12,35,37)$. However, these are just the base triangles, so we must find the multiples. We obviously have the $(12,16,20)$ and $(9,12,15)$ triangles, but are there any others? The only other factor of 12 that we have not used is $6$, and we can get a side of six only with the pair $(1,3)$, which is the triangle $(6,8,10)$, and that will result in the $(12,16,20)$ triangle that we already discovered.

So, the answer is four. :)[/hide]

OH yeah multiples!!! i totally forgot and that's why I only found 2!!!  :oops:

Err... true, I was thinking consecutive squares. However, when the difference between two squares is $12$, you have the $(12,16,20)$ triangle that I mentioned anyway, so it doesn't change the answer.

[quote="chess64"][quote="JesusFreak197"]...the difference between squares is always odd...[/quote]

:huuh: Hmmm... the difference between any two even squares is obviously even....[/quote]
i think he means consecutive squares, but i'm not sure.

[hide]i didnt know any complicated ways to do it...

but i know most of the pythagorean triples....and i only got 4 (5-12-13), (12-16-20), (9-12-15), (12-35-37)[/hide]

Another way to  look at  [hide="it"] Assuming you know that 12  is one of the leg then
 $12^2= c^2-a^2=(c+a)(c-a)$ to get interger values all you have to do is to:  find the number of even  (or both odd) factors of 144. ( 2*72,4*36,6*24 and 8*18)  or 4 possible values.[/hide]

Yes that works unless you want to find the actual triples.

[quote]Yes that works unless you want to find the actual triples[/quote]

Don't know if you are refering to my message, but if you are, you do find actual triples ( One is side is 12,  amd other (two)  sides are, as mentioned above.   [hide="something like "] (p+q)/2 and (p-q)/2 where  p and q  are the (even) factors of 144).[/hide]


Circle O has radius 12 cm and angle 1 has a measure of 60 degrees. Find the number of square centimeters in the area of the shaded region. Express your answer to the nearest tenth of a square centimeter.

[asy]import olympiad;
size(75);
draw(Circle((0,0), 5));
point((0,0));
label("O", (0,0), SW);

draw((0,0)--(5,0)--(2.5, 5*sqrt(3)/2)--cycle);

fill((5,0)--(2.5, 5*sqrt(3)/2)--(5, 5*sqrt(3)/2)--cycle);
clip(Circle((0,0), 5));

draw(anglemark((5,0), (0,0), (2.5, 5*sqrt(3)/2), 25));
label("1", (.75, sqrt(3)/4), ENE);[/asy]

Area of sector-area of equilateral triangle= 144pi/6-144sqrt(3)/4 = 24pi-36sqrt(3) which is approximately 13.0

[quote="AIME_is_hard"]area of shaded = Area of sector - area of equilateral triangle = $ \frac{144\pi}{6}\minus{}\frac{144\sqrt{3}}{4} \equal{} 24\pi \minus{}36\sqrt{3}$ which is approximately $ 13.0$[/quote]

please write your solutions so they are a little easier to read


geometry
circumcircle
incenter
power of a point
radical axis
geometry proposed

$ K \in BC, L \in CA, M \in AB$ are feet of the internal angle bisectors of a $ \triangle ABC.$ $ F_a, F_b, F_c$ are tangency points of its 9-point circle with the excircles against $ A, B, C.$ Show that the triangles $ \triangle KLM \sim \triangle F_aF_bF_c$ are similar.

This is perhaps redundant, but this is also proven in the reference mentioned by jayme in this post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=276344[/url]

It is also shown that these two triangles are perspective.

[b]nsato[/b]: Yes, the proof of $ \triangle KLM \sim \triangle F_aF_bF_c.$ is there, I've read the reference by now, and it is quite a calculation.
_______________________

Let $ (O)$ be circumcircle, $ (N)$ 9-point circle, $ H$ orthocenter, $ I$ incenter, and $ I_a, I_b, I_c$ excenters of the triangle $ \triangle ABC.$ Let $ (O_a)$ be circumcircle of the triangle $ \triangle II_bI_c.$ $ (O)$ is its 9-point circle, $ I_a$ its orthocenter, and $ LM \perp O_aOI_a$ its orthic axis; the radical axis of $ (O_a), (O).$ Similarly, $ MK \perp OI_b, KL \perp OI_c$ (well known). Thus the altitudes of the triangle $ \triangle KLM$ form the same angles as the 3 ex-diacentral lines $ OI_a, OI_b, OI_c.$

By 2nd Fontene theorem (see [url]http://www.mathlinks.ro/viewtopic.php?t=173756[/url], etc.), pedal circle of any point $ P \in OI_a$ WRT $ \triangle ABC$ goes through the orthopole of $ OI_a$ WRT $ \triangle ABC,$ which lies on its 9-point circle $ (N).$ When $ P \equiv I_a,$ this pedal circle is the A-excircle $ (I_a),$ having only one tangency point  $ F_a$ in common with $ (N),$ which is therefore the orthopole of $ OI_a.$ Similarly, $ F_b, F_c$ are orthopoles of $ OI_b, OI_c.$

Let the rays $ (HF_a, (HF_b, (HF_c$ cut $ (O)$ at $ P_a, P_b, P_c.$ The triangles $ \triangle F_aF_bF_c \sim \triangle P_aP_bP_c$ are centrally similar with center $ H$ and coefficient $ \frac {_1}{^2}$. But $ P_a, P_b, P_c$ are poles of Simson lines $ p_a, p_b, p_c$ WRT $ \triangle ABC,$ perpendicular to $ OI_a, OI_b, OI_c,$ i.e., parallel to $ LM, MK, ML,$ and pairwise intersecting at $ Q_a \equiv p_b \cap p_c,$ $ Q_b \equiv p_c \cap p_a,$ $ Q_c \equiv p_a \cap p_b.$ Consequently, the triangles $ \triangle Q_aQ_bQ_c \sim \triangle KLM$ are centrally similar. Since $ \angle Q_cQ_aQ_b \equal{} \angle (p_c, p_b) \equal{} \frac {_1}{^2} \angle P_cOP_b \equal{} \angle P_cP_aP_b,$ etc., the triangles $ \triangle P_aP_bP_c \sim \triangle Q_aQ_bQ_c$ are also similar. As a result, $ \triangle F_aF_bF_c \sim \triangle P_aP_bP_c \sim \triangle Q_aQ_bQ_c \sim \triangle KLM.$


m = log^-1 (4.03)

solve for m
the answer is NOT 1.65

hello, did you meant
$ m\equal{}\frac{1}{\log(4.03)}$?
Sonnhard.

[quote="eddd83"]no. it is not 1/log(4.03)[/quote]
Thank you very much  :(

Domi

img256.imageshack.us/my.php?image=logproblemtt5.jpg

here's a picture of my problem.


Let $n$ be natural number and $1=d_1<d_2<\ldots <d_k=n$ be the positive divisors of $n$.

Find all $n$ such that $2n = d_5^2+ d_6^2 -1$.

divide by $n$ to acquire equivalent form:

$2=\frac{d_5}{d_{k-4}}+\frac{d_6}{d_{k-5}}-\frac{1}{n}$ and so we conclude that $k\leq 10$. As we have $k \geq 7$ (actually $k \geq 6$ but $k=6$ clearly yields no solution).  So we are to check four cases - using $d(\prod p_i ^{\alpha _i}=\prod(\alpha _i +1)$ it is quite easy


Find all numbers of the form $\overline{ab}$ in base $10$ and natural numbers $n$, which satisfy the equality \[ (n-1) \overline{ab2} + n = 2006 .  \]

[i]Ion Safta[/i]

Is it possible that there are no solutions?

[hide]We have $(n-1)\underline{ab2}+n-1+1=2006$, or
$(n-1)\underline{ab3}=2005$.
This means that all factors of $2005$ that have a units digit of $3$ are valid choices for $\underline{ab3}$.
We find that the prime factorization of $2005$ is $2005=5\times401$, and from this we can easily see that there are no factors of $2005$ that have a units digit of $3$, so there are [b]no solutions[/b].[/hide]


Consider a standard $ 8 \times 8$ chessboard consisting of $ 64$ small squares coloured in the usual pattern, so $ 32$ are black and $ 32$ are white. Azig-zag path across the board is a collection of eight white squares, one in each row, which meet at their corners. How many zig-zag paths are there?

[hide]
Write out the number of paths at each square starting from one at each white square on the top, and then summing downwards ... which takes no time and gives $ 296$.

We could try recursion:

Consider the number of paths on a $ 8\times n$ board. Let $ a_{k,n}$, $ k \equal{} 1,2,3,4$ be the number of paths reaching the $ k$th white square from the left on the $ n$th row from the top. WLOG suppose the top left corner is white.

Then $ a_{k,1} \equal{} 1$, and we easily find the recursions $ a_{1,2m \plus{} 1} \equal{} a_{1,2m}$ and $ a_{k,2m \plus{} 1} \equal{} a_{k \minus{} 1,2m} \plus{} a_{k,2m}$ for $ k > 1$, and $ a_{4,2m} \equal{} a_{4,2m \minus{} 1}$ and $ a_{k,2m} \equal{} a_{k,2m \minus{} 1} \plus{} a_{k \plus{} 1,2m \minus{} 1}$ for $ k < 4$.

Then let $ S_{n} \equal{} a_{1,n} \plus{} a_{2,n} \plus{} a_{3,n} \plus{} a_{4,n}$; so $ S_{2m \plus{} 1} \equal{} 2S_{2m} \minus{} a_{4,2m}$ and $ S_{2m} \equal{} 2S_{2m \minus{} 1} \minus{} a_{1,2m \minus{} 1}$, which doesn't really help, but solving gives $ S_8 \equal{} 296$.

[/hide]


For $ a,b,c$ be sidelengths of a triangle, show that
\[ \frac a{a^2\plus{}5bc}\plus{}\frac b{b^2\plus{}5ca}\plus{}\frac c{c^2\plus{}5ab}\ge\frac{a\plus{}b\plus{}c}{2(ab\plus{}bc\plus{}ca)}\]

[quote="dduclam"]For $ a,b,c$ be sidelengths of a triangle, show that
\[ \frac a{a^2 \plus{} 5bc} \plus{} \frac b{b^2 \plus{} 5ca} \plus{} \frac c{c^2 \plus{} 5ab}\ge\frac {a \plus{} b \plus{} c}{2(ab \plus{} bc \plus{} ca)}\]
[/quote]
After using the Cauchy-Schwarz Inequality, we see that it suffices to show that the following inequality holds
$ 2(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca) \ge a^3\plus{}b^3\plus{}c^3\plus{}15abc,$
which is true and it is not hard to prove.

[quote="mestav"]Can_hang2007 please prove your inequality. I replaced a => x+y, b=>y+z, c=>x+z and multiplied and simplied. I have (3,0,0)+(1,1,1) >= 2 (2,1,0) at your inequality. I think it is false[/quote]
Dear friend,
The inequality
$ \sum_{sym} x^3 \plus{} \sum_{sym} xyz \ge 2\sum_{sym} x^2y$
is true and it is known as Schur's Inequality.

We can prove it by writing it as
$ x(x\minus{}y)(x\minus{}z)\plus{}y(y\minus{}z)(y\minus{}x)\plus{}z(z\minus{}x)(z\minus{}y) \ge 0,$
or
$ (x\minus{}y)^2(x\plus{}y\minus{}z)\plus{}z(x\minus{}z)(y\minus{}z) \ge 0.$
Now, by assuming $ z\equal{}\min\{x,y,z\},$ we deduce the result.

[quote="can_hang2007"][quote="dduclam"]For $ a,b,c$ be sidelengths of a triangle, show that
\[ \frac a{a^2 \plus{} 5bc} \plus{} \frac b{b^2 \plus{} 5ca} \plus{} \frac c{c^2 \plus{} 5ab}\ge\frac {a \plus{} b \plus{} c}{2(ab \plus{} bc \plus{} ca)}\]
[/quote]
After using the Cauchy-Schwarz Inequality, we see that it suffices to show that the following inequality holds
$ 2(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \ge a^3 \plus{} b^3 \plus{} c^3 \plus{} 15abc,$
which is true and it is not hard to prove.[/quote]

You are right, Can. The final inequality equivalent to $ (3c\minus{}a\minus{}b)(a\minus{}b)^2\plus{}(a\plus{}b\minus{}c)(a\minus{}c)(b\minus{}c)\ge0$, which obviously if we assume that $ c\equal{}\max\{a,b,c\}$.

Now, what do you think about the following inequality?
If $ a,b,c$ be sidelenghths of an acute triangle, then
\[ \frac a{a^2 \plus{} 3bc} \plus{} \frac b{b^2 \plus{} 3ca} \plus{} \frac c{c^2 \plus{} 3ab}\ge\frac 1{2}\left(\frac1{a\plus{}b}\plus{}\frac1{b\plus{}c}\plus{}\frac1{c\plus{}a}\right)\]


Don't shy away from this, beginners

Prove that 7^n, where n is any natural number, is equivalent to 1 mod 6.  In other words, 7^n divided by 6 leaves remainder 1.  Hint: [hide]Try to show that it's true for 7^1, and then using that knowledge show that it's true for 7^2, and then 7^3, etc.  Other methods are available but this is the most beginner-friendly (I think)[/hide]

Another hint: [hide]Express numbers in a way to include 6, so when you do things, you can get rid of the multiples of 6, since they would leave remainder 0 when divided by 6.[/hide]

BHorseMath wrote:yay! a simple problem...I think.

[hide] 7^1=7/6=1 with a remainder of 1.  7^2=49/6=8 with a remiander of 1.  7^3= 343/6=57 with a remainder of 1.  and so on.  There for it is apparent that 7^n would have a remainder of one when divided by 6.[/hide]

Did I do this right?

How can you show that, say, 7^314159265358979 has a remainder of 1 when divided by 6?

use a calculator j/k j/k.

[hide] is is because anytime 7 is divided by 6 theres a remainder of 1?[/hide]

But then if we have 6*7/6, we have remainder 0.

Hint: [hide]7=6+1.  7^n=6k+1.  If this is true (and k is some natural number), then show that 7^(n+1) has remainder 1.  Hint within a hint: What do you do to 7^n to get 7^(n+1)?[/hide]

BHorseMath wrote:okay...

[hide] To get 7^n to equal 7^(n+1) you just add a 1 right?  So if 7^n=6k+1 then 7^(2+1)=343/6=57 with a remainder of 1.  So 7^3=6k3+1?  Okay I'm stuck.[/hide]

First, check your first sentence.  You "add" a 1, but what does that 1 do?  It's like saying 5*(5+1).  It's not 5*5 + 1; it's 5*5 + 5.  Also, if you want to work with specific examles, think this way:  [hide]7^3 = 343/6 = 57 remainder 1.  Then that means it's 57*6 + 1.  Don't calculate it yet.  Instead, try keeping that as it is.  Then do something to 7^3 to get 7^4.  Then do the same thing to 57*6 + 1 since that's equal to 7^3.  But instead of calculating it, try keeping it in the same form (i.e., blah + blah).  Don't try to calculate the result and then separate again.  It's already separated, so keep it that way.[/hide]

BHorseMath wrote:okay[hide]  7^3*(6+1)=7^4. [/hide]

OK you're really, really close.

Quote:[hide] 57*(6+1)*(6+1).[/hide]

Fix this up a slight bit:  [hide]57*6+1 does not equal 57*(6+1).  Now just multiply it out after you fix that little bit.  You know how to distribute or "FOIL" right?  Well do that here and don't calculate everything yet.  Take a look at what you have.[/hide]

[hide] 57*6+1*(6+1)=343*7=2401[/hide]

I think I messed something up again....

Yes, try doing what Osiris said.  And then multiply it out like this:

[hide](a+b)(c+d) = ac + ad + bc + bd by distributiveness (that was a made up word btw)[/hide].

After you multiply like I said, don't calculate [i]anything[/i] (i.e., don't say 5*5 = 25, just leave it as 5*5, if that were the case).  Just take a look at what you have, and what would happen if you divided by 6.

BHorseMath wrote:[hide] 57*6+1*(6+1)=343*7=2401[/hide]

[hide]Ok so you know that (57*6+1)(6+1) = 7^4.  So now multiply out the left hand side.  Use the fact that (a+b)(c+d) = ac + ad + bc + bd (I assume you know this already).  So use the numbers in your multiplication instead of a, b, c, and d.  So multiply first instead of adding first.  That way you get to keep the sixes separate.  And then you end up with four multiplications (ac, ad, bc, bd) that you have to add up.  Look closely at each multiplications, and don't add them up yet.  Instead, look for divisibility by 6 and you notice something.[/hide]

342 + 57 + 42 + 7 doesn't equal (57 * 6 + 1)(6 + 1).

Don't calculate those numbers:

[hide]Expand (57 * 6 + 1)(6 + 1) as 57 * 6 * 6 + 57 * 6 + 1 * 6 + 1 * 1.[/hide]

What do you see now?

BHorseMath wrote:[hide] 342+57+42+7.  342 and 42 are dicisible by 6 and 57 and 7 aren't.  But what does that say?  They 342 and 42 have 2 of the same numbers and 57 and 7 have one of the same numbers but I don't think thats right.[/hide]

You mis-multiplied.  This is what you should be getting:

[hide]((57*6)+1)(6+1) = (57*6)*6 + (57*6)*1 + 1*6 + 1*1

Now if we divided the whole thing by 6, then what's the remainder going to be?  Don't use your calculator or paper to actually calculate everything.  The problem is a lot simpler just by looking at it in this form.  Try dividing certain parts by six and think like your subtracting sixes until you can subtract no more.  Well subtract sixes in bunches from different parts of this expression, and find what the remainder is, but DO NOT actually calculate the values!![/hide]

[quote="BHorseMath"]The only numbers divisible by six are six?[/quote]

What??  There are plenty of numbers divisible by 6!!  6*1, 6*2, 6*3, ... to name a few.  Look at the terms closely.  Look for ways to subtract a bunch of sixes.  6*x, for example, is x sixes (obviously!)  After subtracting all the sixes you can, what's left?

First way to do this:

[spoiler]7^n=(6+1)^n
=6^n+(n)6^(n-1)+(nC2)6^n+...n(6)+1
6(6^(n-1)+n(6^(n-2)+...n)+1
=6k+1, where k=6^(n-1)+n(6^(n-2)+...n[/spoiler]

Second way to do this:
[spoiler]
7(mod 6) :equiv: 1(mod 6)
By the property of modulus
If m(mod 6) :equiv: n(mod 6)
then m^k(mod 6) :equiv:n^k(mod 6)
Thus 7^n(mod 6) :equiv: 1^n(mod 6) :equiv: 1[/spoiler]

Third way to do this:

[spoiler]Note that a^n-1=(a-1)(1+a+a^2+...a^(n-1))
Thus 7^n-1=(6)(1+7+7^2+..7^(n-1)
Thus 7^n(mod 6) :equiv: 1 (mod 6)[/spoiler]

Your second way is identical to the first.  But anyways the beginners should do it by induction because many don't know/understand the binomial theorem.

I did it by induction. Maybe BHorseMath doesn't know what she should do because she isn't aware that you're trying to make her do the question through induction, probability1.01?

[quote="BHorseMath"]57's and 1's.[/quote]

I have no clue what you mean.  Please explain everything.  What do you have left when you take away all the sixes?  You're just subtracting sixes from the number (57*6)*6 + (57*6)*1 + 1*6 + 1*1.  Subtract them bunches at a time.

Every natural number can be expressed as 6*m + n, with 0 :le: n :le: 5.  You're taking away the 6*m part and finding the n.  So what you do is you try to get rid of sixes.  You can get rid of multiples of 6 within numbers because they won't change the remainder.  Like if you have 6*m + n and 6*(m+5) + n, the remainder is the same, even though you added a multiple of six (namely, 5*6 = 30).

Bhorsemath,
Now assume when [tex]7^n[/tex] divided by 6 has remainder of 1, try to show
when [tex](7^n)*7[/tex] divided by 6 also  has remainder of 1.

Don't just do specific number!

[quote="BHorseMath"][spoiler] If 7^n has a remainder of 1 then we know that (7^n)*7 has a remainder of 1 when divided by 6 because it would be the same as (7*7)^n.[/spoiler][/quote]

This is not true and not sufficient.  Try what I said in my last post.  After getting rid of the multiples of 6, what are we left with?  It helps to be familiar with the fact that a(b+c) = ab + ac to solve this problem.

[spoiler] (57*6)*6+(57*6)*1+1*6+1*1
                  -6    -6        -6         -6
leaves (57)+(57)*1=1+1*1
            [/spoiler]

I really don't know what I am supposed to do or even if that was right.

No you were trying to divide by 6.  Basically the number 57*6*6 + 57*6 + 6 + 1 has three terms sharing a factor of 6.  That means it's just 6*(a bunch of stuff) + 1.  Now I'm pretty sure you know what the remainder is if you divide that by 6.  Now generalize for things other than 7^3.

wouldn't it be...

[spoiler]7^1=7
7 mod 6 :equiv: 1

so... 7^n=6k+1
7^(n+1)=7(7^n)=7(6k+1)
7^(n+1)=42k+7
(42k+7)/6=7k+1 with a remainder or 1.

There, proof by induction.

[/spoiler]

[spoiler]
Using the fact that a^c  :equiv: b^c mod m,

7  :equiv: 1 mod 6
7^n  :equiv: 1^n  :equiv: 1 mod 6.

QED
[/spoiler]

[quote="BobofBobs"]very nice. I, however, don't have much practice with mods, so i didn't think to do that.[/quote]

Remember, facts come from somewhere.  He provided a fact.  You provided a good proof of the fact.  (Another good proof is by using the binomial expansion.)

Find the coefficients when expanding (x+y)^n.  I'll explain how, but maybe you should try to figure it out a little (this might be hard for a beginner, if you are one).

You can try to find the pattern of the coefficients yourself.

HINT:
[tex](x+a_1)(x+a_2)...(x+a_n)[/tex]
[tex]=x^n+(a_1+a_2+...a_n)x^{n-1}+(a_1a_2+a_1a_3..)x^{n-2}+...a_1(a_2)...(a_n)[/tex]
When [tex]a_1=a_2=...=a_n[/tex]
We would have [tex](x+a_1)^n[/tex]

No, you need to find the coefficients of [tex](x+a_1)^n[/tex]

Dum de dum...Can you just say that 7/6 gives a remainder of 1?  So even if it was 7 :^2:  it would still give the same remainder?

You would need to demonstrate that that was true first.  Just because it might "seem" true doesn't mean anything, mathematically speaking.

Let n equal 1.  7*1=7, and when 7 is divided by 6, the remainder is indeed 1.  Now let n equal 2.  7*7=49, and when 49 is divided by 6, the remainder is still 1.  Thus, we can conclude that since 7 to the first power and 7 :^2: , and 7 to the nth power are all multiples of 7, they will all leave a remainder of one when divided by 6, and that as long as a number is 7 raised to some power it will leave a remainder of 1 when divided by 6.  Um...what makes 343 leave a remainder of one when divided by 6 you ask?  Well, 6's prime factors are 2 and 3, and since 7 is one more than 6...actually I have no clue where I'm getting at here =/

(This was my first proof by the way.)

Explain to me what part of that shows that 343 is one more than a multiple of six?


function
calculus
integration
floor function
derivative
absolute value
number theory unsolved

Given $\epsilon>0$ and $p>1$, prove that for all sufficiently large $K$ there exists a laticial point whose distance to the curve $ x^p+y^p=K^p$ is smaller than $\epsilon$.

Let’s work only with $x,y\in[0,K]$. This allows us to define a decreasing function $f_K:[0,K]\to[0,K]$ by $f_K(x)=(K^p-x^p)^{\frac 1p}$. It’s also continuously differentiable on $[0,K)$, with $f'(0)=0$.

Take $\delta>0$ so small that $|f_1'|<\varepsilon$ on $[0,\delta]$, and $K$ so large that $f_K(0)-f_K(K\delta)>1\ (*)$. Fix such a large non-integral $K$ and define the following points: $y_0=\lfloor K\rfloor,x_0=f_K^{-1}(y_0),x_1=\lfloor x_0\rfloor,y_1=f(x_1)$. The distance between the point $(x_1,y_0)$ and the curve is at most $y_1-y_0$. On the other hand, due to $(*)$, we have $x_1,x_0<K\delta$, and since $\frac{y_1-y_0}{x_1-x_0}$ is the derivative of a point in $(x_1,x_0)$, it must have absolute value $<\varepsilon$. $|x_1-x_0|\le 1$, so $y_1-y_0<\varepsilon$, and we’re done.

Is it correct?


1. When you need to find say the potential of Cr4+/Cr and you know Cr4+/Cr3+ and Cr3+/Cr, why can you not simply add standard potentials?

2. In a mercury barometer, if you introduce water, you find vapor pressure from water. But on a phase diagram, it seems to me that vapor pressure of water only comes when you hit a certain atmospheric pressure?

3. As you go from liquid to gas and you hit boiling point, there's a dynamic equilibrium. As you add more heat, what is happening on the molecular level as the equilibrium shifts towards the gas phase completely?

Thanks!

1. Wait, you can't? I always thought that you could just add the reduction potentials. If you can't, then we can just exploit the difference in energy input/output and repeat the redox reaction in both directions and get free energy out of it.

2. Everything has vapor pressure, even solids.

Vapor pressure exists because at a certain temperature, not everything has the same kinetic energy. Some particles will have very low kinetic energy while some will have energies so high that they are effectively gaseous.

Think of vapor pressure as the partial pressure exerted by the percentage of particles in a sample that have KE higher than the critical value (boiling point). KE of particles follow the Maxwell-Boltzmann distribution, so the vapor pressure from a sample can be visualized as thus:

Say x is the average temperature of a sample. Find the distribution corresponding to that temperature and call the "function" f(k) as a function of temperature.

The amount of particles in the vapor phase can be visualized as the amount of particles corresponding to the improper integral of f(k) with respect to k from the boiling point of the substance to infinity.

http://en.wikipedia.org/wiki/Maxwell-Boltzmann_distribution

3. On the molecular level, the addition of more heat causes the overall energy in the system to increase. For the substance to boil, the "hydrogen bonds", or the electrostatic attraction between the hydrogen (with a local positive charge) and Oxygen (with a local negative charge) of neighboring water molecules must be broken. Once it is broken, the individual H2O molecules will have enough freedom of movement to become gaseous, or move nearly independently of other gaseous water particles.

Think of it as:

$ H_2O_{(l)} \plus{} 40.65 \mbox{kJ} \minus{} > H_2O_{(g)}$

1. lol no. you have to add the free energies and then convert back to E to find the total voltage. I just don't understand why you can't add potentials anyways :/

2. mm. im more confused about how vapor pressure of solids/liquids fits in with phase diagrams.

Like in a closed system, you always have to account for vapor pressure of water. But in a phase diagram, it seems to suggest that you can only have liquid/vapor coexisting at certain atmospheric pressures... textbooks always say like "in the liquid region, only one phase exists: liquid."

Somehow that just doesnt click with me. Doesn't that mean that in most closed systems, water doesn't have vapor pressure since atm pressure is not exactly right?

3. thanks =)

[quote="adchia"]1. lol no. you have to add the free energies and then convert back to E to find the total voltage. I just don't understand why you can't add potentials anyways :/

2. mm. im more confused about how vapor pressure of solids/liquids fits in with phase diagrams.

Like in a closed system, you always have to account for vapor pressure of water. But in a phase diagram, it seems to suggest that you can only have liquid/vapor coexisting at certain atmospheric pressures... textbooks always say like "in the liquid region, only one phase exists: liquid."

Somehow that just doesnt click with me. Doesn't that mean that in most closed systems, water doesn't have vapor pressure since atm pressure is not exactly right?[/quote]

Huh... You would have to walk me through on 1. I always just added the reduction potentials on tests and got the right answer. (Like when they give me Fe3+/Fe2+ and Fe2+/Fe)

2. Phase diagrams show the GENERAL state of the sample. So essentially, if a point on a phase diagram falls within the liquid region, it says that: "Macroscopically, the sample will appear liquid."

See THIS not-to-scale picture that I made haphzardly with horrible aliasing artifacts >.>

http://i40.tinypic.com/2lw24uw.png

Anyways, I picked a point in the liquid region. Ignore the pressure axis because I don't want to make a 3-D plot of everything, so we regard the pressure within the sample as constant.

Now, we have the familiar Maxwell-Boltzmann distribution of particles there. Here, we see that although it is a liquid, a lot of the particles have KE that is either very low as to fall in the solid state, or KE so high that it falls within the gaseous phase.

To understand why we don't have chunks of ice floating around in our cups of water, we must recognize the fact that the KE of all particles in a sample are constantly CHANGING. That means a particle with high KE at this moment might transfer it to another particle in the next. Since there is a substantial fraction of particles falling in the liquid phase, on average the particles will have KE in the liquid region (which is what we defined in the first place), so nothing ever remains solid long enough and have enough other solid particles accumulating on it for the phenomenon to be observed macroscopically.

Now, notice the red part. That corresponds to the fraction of particles in the gas phase. Unlike liquid/solid state particles, gaseous particles will LEAVE the sample occasionally, generating the familiar vapor pressure.

Phase diagrams are confusing like that; you have to remember that nothing just spontaneously changes phases with a 0.001% change in temperature/pressure. It's a gradual process requiring a lot of heat input/output.

Boiling occurs when the vapor pressure of the sample reaches atmospheric pressure; that's what the borderline over there between liquid and gas represents.

On the border, nothing is too special; it just means that the equilibrium constant of the reaction $ A_{(l)} \rightarrow A_{(g)}$ is 1. (I think... I'm not 100% sure. But in any case, it just means that both liquid and gas coexist in equal proportions.)

thanks. your explanation was very helpful

As for the reduction potentials thing:

http://www.siliconfareast.com/ox_potential.htm

Note that the potential of Fe3+/Fe is -0.04, Fe2+/Fe is -0.44

http://books.google.com/books?id=nt-crAJEtVYC&pg=PA192&lpg=PA192&dq=Fe3%2B/Fe2%2B+reduction+potential&source=bl&ots=ucCu_TEfBZ&sig=BiDCAyr0UxXs-IacNpe8pi3zLhQ&hl=en&ei=qgr1SfDONcaFtgfj7d2oDw&sa=X&oi=book_result&ct=result&resnum=9#PPA192,M1

shows that Fe3+/Fe2+ is 0.77

To relate everything, Fe3+/Fe2+ has free energy of -1*96500*0.77, Fe2+/Fe has free energy -2*96500*-.44

Adding the free energies, 10 518 should equal -3*96500*E_Fe3+/Fe. Dividing through indeed confirms that Fe3+/Fe should be -0.04.

(This is also in atkins. The SRP are also confirmed by atkins.)

Hope that helps!

[quote="adchia"]1. When you need to find say the potential of Cr4+/Cr and you know Cr4+/Cr3+ and Cr3+/Cr, why can you not simply add standard potentials?

[/quote]

In my knowledge, and I can be wrong on this one, you can only directly add/substract if your resultant process has no electrons in it.

Your particular question gives a resultant half-equation: $ \ce{Cr}^{3\plus{}} \plus{}3e^{\minus{}} \rightleftharpoons \ce{Cr}$ hence you cannot directly use the reduction potentials.

ye that makes sense. im fairly sure that's right banished.

And so my original question was asking why can you not add them. Conceptually, it seems that adding them would be fine. You reduce say Fe3+ to Fe2+, generating some voltage, and then take the Fe2+ to Fe, generating some more voltage, which SEEMS to suggest that you can just add voltages...but you cant  :huh:

As I indicated earlier, I don't understand why the voltages would be different. If they are, it seems that you can keep just doing/reversing the reaction and exploit the voltage difference, therefore accumulate the energy difference, and thus effectively generating free energy (thereby powering an infinite motion machine and violating the first law of thermodynamics.)

I am not sure I understand what you are asking yanglunj. :(

As for adchia:

Recall the equation $ \Delta G^{\circ} \equal{} \minus{}nFE^{\circ}$.

Say we have the processes $ \ce{Fe}^{3\plus{}} \plus{} 3e^{\minus{}} \rightleftharpoons \ce{Fe}$ and $ \ce{Ag\plus{}} \plus{}e^{\minus{}} \rightleftharpoons \ce{Ag}$ with reduction potentials $ E^{\circ}_1$ and $ E^{\circ}_2$ respectively.

Converting these values to standard Gibbs free energy gives: $ \Delta G^{\circ}_1 \equal{} \minus{}3FE^{\circ}_1$ and $ \Delta G^{\circ}_2 \equal{} \minus{}FE^{\circ}_2$.

Now energy we can add/subtract and the overall free energy change (standard) is $ \Delta G^{\circ} \equal{} \Delta G^{\circ}_1 \minus{} 3\Delta G^{\circ}_2 \equal{}\minus{}3FE^{\circ} \equal{} \minus{}3FE^{\circ}_1 \plus{}3FE^{\circ}_2$ (for example, it can be the other way around too).

Hence $ E^{\circ} \equal{} E^{\circ}_1 \minus{} E^{\circ}_2$ and that hopefully explains the fallacy of "adding/subtracting" voltage.


quadratics
number theory theorems
number theory

I can only tell you how to solve linear recurrence relationships, of the form $au_{n+2} = bu_{n+1} + cu_{n}$, you start of by assuming the formula is of the form $u_n = \lambda^n$. Substitute in and you get $a\lambda^2 - b\lambda - c = 0$. Solve the quadratic to find two roots of $\lambda$ which are $\alpha$ and $\beta$ say.

Then your formula is $u_n = A\alpha^n + B\beta^n$. To work out $A$ and $B$ you need to use the given starting values of $u_0$ and $u_1$ and solve the simultaneous equations.

I suspect your formula may involve a $A\alpha^{3n}$ term but I'm not sure.

[quote="conceive"][quote="Lovasz"]My question may be old.
I found some problems in books, where the authors gave some methods to find general formula of a sequence, but I don't understand how did they do in some ones, for example:
$u_{n+1}=u_n^3+u_{n-1}$
Thanks for help. :)[/quote]
And what is given?[/quote]
_conceive, that is an example for my requirement.
_Centy, thanks. But I know this type. I asked you, i.e. $u_{n+1}=u_n^3+u_{n-1}$.
 :)


calculus
derivative
function
induction
algebra
polynomial
calculus computations

Define function f(x)=$ e^{\minus{}\frac{1}{x^2}}$ when x is not zero and 0 when x is zero. Prove that nth derivative of the function at x=0 is always 0 for positive integers n.

You could go about showing that the nth derivative at any point $ \neq 0$ is of the form $ \sum a_i \frac{e^{\minus{}\frac{1}{x^2}}}{x^i}$ and then use induction to get what you want.
(the basic idea is that that exponential function will always 'win the race' against any polynomial)


Ah.Which numbers can x and y ?
x!+y!=(x+y)!
Um :oops:  :roll:

$ x\equal{}y\equal{}1$
Only this is valid.
it's easy to check, if $ x \geq y$
that $ x$ should equel $ y$ or $ y\plus{}1$
and then by checking this $ 2$ cases, we have only one one pair of solution.

[quote="boyvietnam"]But how do you know that[/quote]
[hide="Index"]
Assume that there is some solution such $ x\minus{}y \geq 2$, then we have:

$ \frac{x!}{y!}\plus{}1\equal{}\frac {(x\plus{}y)!}{y!}$
We have that LHS - is odd and RHS - is even, so we have contradiction.

PS: i forgot one more pair of solution $ x\equal{}y\equal{}0$[/hide]

[quote="SaYaT"]
 i forgot one more pair of solution $ x \equal{} y \equal{} 0$[/quote]
Hello
$ x\equal{}y\equal{}0$ can't be the solution.
LHS=$ 2$ where as RHS=$ 1$
Thank u.

[quote="kabi"][quote="SaYaT"]
 i forgot one more pair of solution $ x \equal{} y \equal{} 0$[/quote]
Hello
$ x \equal{} y \equal{} 0$ can't be the solution.
LHS=$ 2$ where as RHS=$ 1$
Thank u.[/quote]

Hi=)
Thanks , so....then..i didn't forgot anything :lol:

Well, combinatorially speaking, $ n!$ is the number of ways to permute $ n$ objects in a row, so since there is $ 1$ way to permute $ 0$ objects in a row, $ 0!\equal{}1$.

[url]http://mathforum.org/dr.math/faq/faq.0factorial.html[/url]
that should help you.

[quote="eli140"]Why does factorial 0 equal one?[/quote]

I heard that it was mostly just to make combinatorics work out better. There might be another "real" reason why 0!=1, but I think that's mainly it.

I think it's because there's only one way to arrange 0 objects - and that's to not arrange them. I'm not totally sure if this is right, but I think it is...

Hello
this topic has been discussed earlier.
Check this::
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=217127
Thank u


:mad: What is it with people double posting? http://www.artofproblemsolving.com/Forum/viewtopic.php?t=316047


geometry
power of a point
radical axis
geometry proposed

$ M,N$ lie on $ AB,AC$ such that $ MN$ is parallel to $ BC.$ $ K,L$ lie $ MN$ such that $ BL,CK$ intersect at $ P.$ Circles $ MPK$ and $ NPL$ intersect at $ P,Q.$ Prove that points $ P,Q,A$ lie on a line.

If $M' \equiv MQ \cap BC$ and $N '\equiv NQ \cap BC,$ then $ QPCM',QPBN'$ are both cyclic since $ \angle PCM' \equal{} \angle LKP \equal{} \angle PQM$ and $ \angle PBN' \equal{} \angle KLP \equal{}\angle PQN.$ Thus $ \odot(MQN)$ cuts $ AC,AB$ at $ D,E$ lying respectively on $ \odot(QPC)$ and $ \odot(QPB),$ due to $\angle NDQ \equal{}\angle NMQ \equal{} \angle QPC$ and $ \angle MEQ \equal{} \angle MNQ \equal{} \angle BPQ.$ From power of $ A$ WRT $ \odot(MQN)$ and $ \triangle ABC \sim \triangle AMN,$ we have

$ AD \cdot AN \equal{} AE \cdot AM  \Longrightarrow  AD \cdot \frac {AC \cdot MN}{BC}\equal{} AE \cdot \frac {AB \cdot MN}{BC}$

$\Longrightarrow  AD \cdot AC \equal{} AE \cdot AB$ $\Longrightarrow$ $ A$ is on the radical axis $ PQ$ of  $ \odot(QPC),\odot(QPB).$

here is simple solution :P

$ AP \cap MN,BC \equal{} I,Q'$. then $ \frac{{IL}}{{IK}} \equal{} \frac{{BQ'}}{{BC}} \equal{} \frac{{IM}}{{IN}} \to \overline {IL} .\overline {IN}  \equal{} \overline {IM} .\overline {IK}  \Rightarrow {P_{I/(NPL)}} \equal{} {P_{I/(PKM)}}$
then $ AI$ is radical axis , follow $ P,Q,A$ lie on a line.


we are given n>3 points in the plane no three of which lie on a line.
Does there exist a circle that passes through at least three of the given points
and contains none of the other given points in its interior?

yes it exists and u don't need the condition" no three of them lie on a line " u should just have this condition:" all of them doesn't lie on a line" for a proof see Problem-Solving Strategies,Problem 22, Extremal Principle.

use extremal principal!
[hide="more"] Consider two points $ A$ and $ B$ that have the minimum distance,another idea is to consider $ \Delta ABC$ with minimum circumradius[/hide]


Princeton
college
geometry
geometric transformation
Science Olympiad

This is the topic for discussing the upcoming Physics Olympiad, to be held in Isfahan, Iran, in July 2007. Here are a few questions for the participants, to get the discussion started:
- How are you preparing for the IPhO?
- What is the selection process like in your country?
- Do you hope to win a medal? :roll:
- Given the current situation in the region, do you think that you will be secure in Iran?

I've already responded to the first two questions in the topic discussing IPhO '06. I still don't want to answer the third one :) But I would really like to hear your opinion on the fourth question...

I have a feeling that this years IPHO will not take place. Iran will surely go to war with the US. and GBr. till then. the war will last probably for 5 or 6 years. So there wont be any IPHO for the next 7 years :( . Lets hope this doesnt happen though :lol:

That's a bit pessimistic, isn't it? I cannot comment on the war, but I like the logic: a full-scale war $\Rightarrow$ no more IPhO's for some time :)

If there will be a war against USA than because of the obviously more advanced war technology there will not be a war at a big scale,only in the countries that USA will consider enemies so i don't think Ipho in the next years will  be intrerrupted because of a war against America.I'm not very informed about what's going on in middle Asia but it's pity and revolting that there still exists such conflicts today.

As I said, I don't feel competent to discuss the possible war. I'm just asking (among other things) how do you feel about going to Iran and do you think you'd be safe there?

Btw, I've studied the list of participating countries on the official IPhO site, [url]http://www.ipho2007.ir[/url]. Reportedly, 75 countries have applied so far, but there's no sign of USA or Canada. Great Britain is there, though, as are most of the other European countries. Interesting...

Djole,is your full name Radicevic Djordje ?

Yes, it is :) I did say this somewhere on the forum, I believe, but how did you find it?

I was looking at silver medallists in the web site ipho2006.org and I saw Djordje  from Serbia. :) But I can't remember you from the olympiad.Do you know the results of my team in Singapore(not only mine)?How many points did you get from theoretical part and experimental part?

I remember a guy named Azar who won a gold medal, but I remember him only because he was active at this forum prior to the Olympiad. I don't know about the others. Looking at the preliminary list of medalists, I see that you have won a silver medal, right? But I don't remember you, either...

Concerning my points... Well, I feel that I could've done better. As far as I remember, I've had around 18 on the theory and 11 on the experiment.

What do you think about this year?Can you get gold? :)

I can only hope so :) But I'll be happy with any medal... What about you?

I also hope so.But I want gold.Hey do you know about the new olympiad?We heard that next year Mr.Waldemar begins European Physics Olympiad.

Really? I haven't heard of that. Well, of course it will start next year, when I'm in college :( Just like the junior science olympiad which started as soon as I started high school and became too old...   :mad: Who will be the host? Poland, probably.

I also think that firstly it must be Poland.*Motherland of Mr.Waldemar :lol: ).Hey in which university do you want to study?Abroad or Serbia?Don't you want to go Singaporian NTU?I'll go there.

No, not Singapore :) I don't like sea food and the weather is awful, so as tempting as it might be I won't go there. I'm going to Princeton :)

good luck all , good luck Geo Team

does someone know not official results or cuttofs or it is too early?

Who knows the results of IPhO 2007? Thank you.

well although the 'result' word is there in the official web page but yet there is no hyperlink to it

[b]China[/b] won with all gold I think and the host ([b]Iran[/b]) placed 7 with
5 medals of which two were gold, 2 silver, 1 bronze. [b]South Korea[/b] was 2nd and 3rd came
[b]Russia[/b]. The [b]US[/b] had some gold medals (all of them got a medal) but I
don't remember their ranking.
There will most surely be a complete list out on the official web page soon, be
patient :lol:

[url=http://www.ipho2007.ir/Results/results.asp]Congratulations ![/url]

 :clap2:

great djole has got a gold. :thumbup:  :coolspeak:

Well done Djole, congratulations :w00t:  :first:  :coolspeak:

Edit: I can't find Australia, did we even participate or did we just get badly owned?

Thanks for the congratulations ;) As for the other mathlinkers, [b]inteluser[/b] got a gold, too, [b]namiqfizik[/b] got bronze and [b]iostream.h[/b] got silver (I think). Australia didn't participate. The problems were fairly easy -- what's your opinion?

i found 1,2 easy.three will try later :)(was having a overview the day i posted the problems)

Congratulations Djole! It was clear to me that you would win gold!  :lol:

Hehe, thanks, it wasn't that clear to me but everything turned out all right... By the way, congratulations to everybody who participated, and especially to those who have won a medal!

Does anyone have more pictures to share? Swedish team, Iranian team,
and others? Please post them.

I see the theory was quite easy. Shame that I didn't go :( Maybe next year...
Congratulations to all medalists :)  :clap2:

Hey, does anyone of you go to IPhO 2008 ?
I am a member of the Moldovan team for IPhO 2008 and would like to chat to some people who have already been there or would be. My yahoo id is : victorbuza


In the pictures below, it consists of 3 pieces of paper. When we swap 2 pieces in the top of the first picture, we obtain the second picture. The thing is there are 13 men in the first, but 12 men in the other.

Nice!

Yeah I know this one. On les-mathematiques.net those are dwarves instead of men ...

Uh...did you remove it?  Although I see empty file description slots, I don't see any file attached...or is that a problem on my browser?


calculus
integration
trigonometry
calculus computations

Calculate the following indefinite integrals.

[1] $\int \frac{dx}{1+\cos x}$

[2] $\int x\sqrt{x^2-1}dx$

[3] $\int a^{-\frac{x}{2}}dx\ \ (a>0,a\neq 1)$

[4] $\int \frac{\sin ^ 3 x}{1+\cos x}dx$

[5] $\int e^{4x}\sin 2x dx$

[quote="kunny"]Calculate the following indefinite integrals.

[1] $\int \frac{dx}{1+\cos x}$
[/quote]
[hide]Use [tex]u=\tan\left(\frac{x}{2}\right)[/tex] to get [tex]\tan\left(\frac{x}{2}\right)+C[/tex][/hide]
[quote]
[2] $\int x\sqrt{x^2-1}dx$[/quote]
[hide][tex]u=x^2-1[/tex] to get [tex]\frac{1}{3}(x^2-1)^{3/2}[/tex][/hide]
[quote]
[3] $\int a^{-\frac{x}{2}}dx\ \ (a>0,a\neq 1)$[/quote]
[hide][tex]a^{-x/2}=\exp\left(-\frac{1}{2}x\ln(x)\right)[/tex] which gives [tex]-\frac{2a^{-x/2}}{\ln(a)}[/tex][/hide]
[quote]
No time for the others now...:)

Thank you for your reply, dheardie.

For #1, :(

You don't like the substiution? I thought it was kinda neat

Or the fact that you have to use $\cos(2\arctan(u))=\frac{2}{u^2+1}-1$

I admit it may not be the neatest substitution, but it was the first one that came to mind...

[b]Edit[/b]: It is correct yes...

[quote="liyi"]Maybe he likes to do it in this way
$\int \frac{dx}{1+\cos x} = \int \frac{1-\cos x}{\sin^2 x}dx = \int \frac{dx}{\sin^2 x} - \int \frac{d(\sin x)}{\sin^2 x} = -\cot x + \csc x + c$[/quote]

Yes! :)

[quote="litoxe2718281828"]My hundredth post! =)

[4]$\int \frac{\sin^3 x}{1+\cos x}dx=\int\frac{\sin x(1-\cos^2 x)}{1+\cos x}dx=\int(-\sin x\cos x+\sin x)dx$
$=\frac{\cos^2 x}{2}-\cos x+C$[/quote]

O.K. :)

[quote="kunny"][quote="amirhtlusa"]#5 is by parts, twice :)
answer:
$\frac{2}{3}(\frac{-e^{4x}\cos{(2x)}}{2}+\frac{e^{4x}\sin{(2x)}}{4})$[/quote]

You are wrong. :([/quote]

Method is correct
$-\frac{1}{10}e^{4x}(2\sin(2x)-cos(2x))$

[1]agian:
$\int \frac{dx}{1+\cos x}=\int \frac{dx}{2\cos^2 \frac{ x}{2}}=\int \frac{d \frac{x}{2}}{\cos^2 \frac{ x}{2}}=\tan \frac{x}{2} +c$

[quote="kunny"]Calculate the following indefinite integrals.

[4] $\int \frac{\sin ^ 3 x}{1+\cos x}dx$[/quote]

Generalization
Calculate

 $\int \frac{\sin ^ {2n+1} x}{(1+\cos x)^n}dx$

[quote="kunny"]Calculate the following indefinite integrals.

[1] $\int \frac{dx}{1+\cos x}$[/quote]
Generalization
Calculate

$\int \frac{a(1-\cos x)^{n}+b(1+\cos x)^{n}}{(1+\cos x)^{n+1}}dx$


does anyone here familiar about markov's theorem for polynomials????

http://www.mathlinks.ro/Forum/viewtopic.php?t=2716

And try this : http://www.mathlinks.ro/Forum/viewtopic.php?t=17198

can you show me a proof of this theorem please???? i couldnt find a book which tackles this matter...


Υπάρχει συνάρτηση $ f: \mathbb N \rightarrow \mathbb N$ έτσι ώτσε $ f(f(n\minus{}1))\equal{}f(n\plus{}1)\minus{}f(n)$ για όλα τα $ n\ge 2?$

Θα δείξουμε ότι δεν υπάρχει συνάρτηση $ f: \mathbb{N}\to\mathbb{N}$ τέτοια ώστε
\[ f(f(n \minus{} 1) \equal{} f(n \plus{} 1) \minus{} f(n)\]
Καταρχάς η παραπάνω σχέση συνεπάγεται ότι $ f(n \plus{} 1) \minus{} f(n) > 0$. Μ' άλλα λόγια η $ f$ is γνησίως άυξουσα. Αυτό συνεπάγεται ότι:
\[ f(n)\geq n \quad (1)\]
για κάθε $ n\in\mathbb{N}$.

Απ' την άλλη, η σχέση $ f(n \plus{} 1) \equal{} f(n) \plus{} f(f(n \minus{} 1))$ συνεπάγεται $ f(n \plus{} 1) > f(f(n \minus{} 1))$ η οποία, λόγω της μονοτονίας, δίνει:
$ n \plus{} 1 > f(n \minus{} 1)$ ή
\[ f(n \minus{} 1)\leq n\quad (2)\]
Οι $ (1)$ και $ (2)$ δίνουν ότι $ f(n) \equal{} n$ ή $ f(n) \equal{} n \plus{} 1$ για κάθε $ n$. Αυτό όμως είναι άτοπο γιατί τότε το δεξιό σκέλος της
ισότητας
\[ f(f(n \minus{} 1)) \equal{} f(n \plus{} 1) \minus{} f(n)\]
είναι φραγμένο ενώ το αριστερό δεν είναι.


The numbers from 1 to 150, inclusive, are placed in a bag and a number is randomly selected from the bag. What is the probability it is not a perfect power (integers that can be expressed as $ x^{y}$ where $ x$ is an integer and $ y$ is an integer greater than 1. for example, $ 2^{4}\equal{}16$ is a perfect power, while $ 2\times3\equal{}6$ is not a perfect power)? Express your answer as a common fraction.

Perfect powers under 150: (* means repeated)

Squares:
1
4
9
16
25
36
49
64
81
100
121
144

Cubes:
1*
8
27
64*
125

Quads?:
1*
16*
81*

5th-7th power:

1*
32
64*
128

EDIT: EDIT: I counted 17 non starred numbers, so the probability is 133/150. Thanks fishy and ernie!

The problem asks for the probability that the number is NOT a perfect power, so the answer is $ \frac{133}{150}$.

epic revive (sorry), but I just wanted to say that the No_sources should be corrected. http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=524477636&t=210374

If this is spam sorry, but I don't think it is.


inequalities proposed
inequalities
algebra
China

[quote="nguoivn"]Given $ A, B, C$ are three sides of a triangle. Find Minimum:
\[ P \equal{} 2(sinA \plus{} sinB \plus{} sinC) \plus{} cotg\frac {A}{2} \plus{} cotg\frac {B}{2} \plus{} cotg\frac {C}{2}\]
:)[/quote]
Let $ tan\frac{A/2} \equal{} a,... \equal{} b,... \equal{} c \equal{} > ab \plus{} bc \plus{} ca \equal{} 1(a,b,c > 0)$
We have $ P \equal{} \frac {1}{abc} \plus{} 4( \sum{\frac {a}{1 \plus{} a^2}})$
$ P \equal{} \frac {1}{abc} \plus{} \frac {8}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$ (because $ ab \plus{} bc \plus{} ca \equal{} 1$)
By AM-GM,we have
$ P \geq \frac {4 \sqrt {2}}{\sqrt {abc(a \plus{} b)(b \plus{} c)(c \plus{} a)}}$
and $ abc(a \plus{} b)(b \plus{} c)(c \plus{} a) \leq \frac {8(ab \plus{} bc \plus{} ca)^3}{27}$
$ \equal{} > P \geq 6\sqrt {3}$
$ ' \equal{} ' < \equal{} > A \equal{} B \equal{} C \equal{} 60o$

[quote=nguoivn]Given $ A, B, C$ are three sides of a triangle. Find Minimum:
\[ P \equal{} 2(sinA\plus{}sinB\plus{}sinC)\plus{}cotg\frac{A}{2}\plus{}cotg\frac{B}{2}\plus{}cotg\frac{C}{2}\]
 :)[/quote]

$$sinA\plus{}sinB\plus{}sinC \plus{}\frac{1}{2}\left(cot\frac{A}{2}\plus{}cot\frac{B}{2}\plus{}cot\frac{C}{2}\right)\geq 2\sqrt {3}$$
$$\implies$$
$$sinA+sinB+sinC+cotA+cotB+cotC\ge \frac{5\sqrt{3}}{2}.$$
[hide=here]$$ cotA+cotB+cotC\ge tan\frac{A}{2}+tan\frac{B}{2}+tan\frac{C}{2}\geq \sqrt {3}.$$
https://artofproblemsolving.com/community/c6h457155p2568905
https://artofproblemsolving.com/community/c6h591663p3506210[/hide]

[quote=sqing]$$ cotA+cotB+cotC\ge tan\frac{A}{2}+tan\frac{B}{2}+tan\frac{C}{2}\geq \sqrt {3}.$$[/quote]
$$ cotA+cotB+cotC\ge \frac{\sqrt {2}}{2}\left(\sqrt {tan^2\frac{A}{2}+tan^2\frac{B}{2}}+\sqrt {tan^2\frac{B}{2}+tan^2\frac{C}{2}}+\sqrt {tan^2\frac{C}{2}+tan^2\frac{A}{2}}\right).$$
(xzlbq)


algebra
number theory
polynomial
composite numbers
IMO Shortlist

If you want only a solution, please download file Shortlist2005 on this site. Do you want an another solution?  :maybe:

I'm actually posting this for the sake of completeness, since I have noticed that it is not included in the ressources-page yet; and I also could not find it with the search function which made me assuming that it has not been posted yet (correct me, if I am mistaken).

It was posted before, but I'm unable to find it, too. I remember it was posted by some iranian member, possible with first letter "s" (if that helps anyone  :roll: ).
But lets just discuss here, I think there was no answer given.

http://www.mathlinks.ro/Forum/viewtopic.php?t=87836

but the formulation of the problem is too different to merge, in my opinion.

  darij

There has to be some mistake in this, but I am probably to tired to figure it out.

Can't you just use big $m$ that is greater then $a_{0}$. Then $a_{0}|P(m!)$ as it divides all it's factors, and you can always pick such $m$ that $P(m!)>a_{0}$ (there are at most $n$ $m$'s such that $P(m!)=a_{0}$), so it's composite...

Thanx in advance.

Let $ n$ be the degree of $ P$.

We begin with the observation that by Wilson's theorem, $ (q \minus{} 1)! \equiv \minus{} 1 \mod q$ for primes $ q$, so $ (q \minus{} k)! \equiv ( \minus{} 1)^{k} \cdot \frac {1}{(k \minus{} 1)!} \mod q$ for $ 1 \le k < q$. (We use the convention that fractions mod $ q$ are interpreted as division in the field.)

It is then useful to consider the polynomial $ Q \equal{} x^n P(1/x)$, i.e. the polynomial with the coefficients of $ P$ "reversed." Set $ c_k \equal{} ( \minus{} 1)^k (k \minus{} 1)!$. We have

$ P((q \minus{} k)!) \equiv P(1/c_k) \equiv Q(c_k) \cdot c_k^{ \minus{} n} \mod q$.

This in particular means that $ Q(c_k) \equiv 0 \mod q \implies P((q \minus{} k)!) \equiv 0 \mod q$ as long as $ q > k\minus{}1$. For sufficiently large $ k$, $ Q(c_k)$ must have a prime divisor $ p > k \minus{} 1$, since $ \gcd(Q(c_k), c_k)$ divides the constant term of $ Q$. (Recall that $ c_k$ is just $ (k \minus{} 1)!$ up to a sign, so that $ Q(c_k)$ can not have only factors less than $ k$ when $ k$ is large.)

We now define $ L(t)$ to be the least common multiple of $ 1, 2, \ldots, t$. If we set $ x \minus{} 1 \equal{} L(t) \plus{} 1$ with $ t$ sufficiently large, we know that $ x, x \plus{} 1, \ldots, x \plus{} t \minus{} 1$ are composite, so we have that $ Q(c_x)$ has a prime divisor $ p > x \plus{} t \minus{} 1$. Then,

$ Q(c_x) \equiv 0 \mod p \implies P((p \minus{} x)!) \equiv 0 \mod p$,

and

$ (p \minus{} x)! \ge t! \plus{} 2p \minus{} 2t \minus{} 2x$
$ > 3L(t) \plus{} 2p \minus{} 2t \minus{} 2x$
$ > 2p$,

where the inequalities may only hold for sufficiently large $ t$. Nonetheless, such a $ p$ exists for arbitrarily large $ x$, so that we may take $ p$ sufficiently large so as to make $ (p \minus{} x)!$ large and guarantee $ |P((p \minus{} x)!)| > p$. Combining with $ P((p \minus{} x)!) \equiv 0 \mod p$, this means that $ P((p \minus{} x)!)$ is composite, as desired.

[quote="bolpoin"][quote]For sufficiently large $ k$,  $ Q(c_k)$ must have a prime divisor $ p>k\minus{}1$ , since $ \text{gcd}(Q(c_k),c_k)$ divides the constant term of $ Q$ . (Recall that $ c_k$ is just  $ (k\minus{}1)!$ up to a sign, so that $ Q(c_k)$  can not have only factors less than  $ k$ when $ k$ is large.) [/quote]

can someone explain more about this step? i can't understand why  $ \text{gcd}(Q(c_k),c_k)$ divides the constant term of $ Q$
and  $ Q(c_k)$  can not have only factors less than  $ k$ when $ k$ is large.
Thanks.[/quote]

Me,too. But, I think that the proposition "For sufficiently large $k$,  $Q(c_k)$ must have a prime divisor $ p>k-1$" is true. So I think that probability1.01's answer is correct.
I will post this proposition's proof.

Proof:Since $k$ is sufficiently large, specially $k>2a_{n}$, We have $c_k = (-1)^{k} \cdot (k-1)! \equiv 0 \pmod{a_{n}^{2}}$.
Therefore $Q(c_k) \equiv a_{n} \pmod{a_{n}^{2}}$, or there exists a positive integer $s$ such that
$Q(c_k) = s\cdot a_{n}^{2}+a_{n} = a_{n}(s\cdot a_{n}+1) $.
Since $|Q(c_k)|$ is sufficiently larger than $a_{n}$, $|s\cdot a_{n}+1|$ is larger than 1.
Therefore, $s\cdot a_{n}+1$ must have a prime divisor.
Let $p$ be a prime divisor of $s\cdot a_{n}+1$, $p$ must be strictly more than $k-1$.
Otherwise, $p \le k-1$ or $c_k = (-1)^{k} \cdot (k-1)! \equiv 0 \pmod{p}$.
Therefore $Q(c_k) \equiv a_{n} \equiv 0 \pmod{p}$, or $0 \equiv s\cdot a_{n}+1 \equiv 1  \pmod{p}$. This is a contradiction.

[quote=Yimin Ge]Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots+a_{0}$, where $a_{0},\ldots,a_{n}$ are integers, $a_{n}>0$, $n\geq 2$. Prove that there exists a positive integer $m$ such that $P(m!)$ is a composite number.[/quote]

The main idea is to note $(p-k)!(k-1)! \equiv (-1)^k \pmod{p}$. Indeed, $(p-1)! \equiv -1 \pmod{p}$ by Wilson's Theorem and $$(p-(k-1)) \cdot \dots \cdot (p-1) \equiv (-1)^{k+1}(k-1)! \pmod{p}.$$ Now define $Q(x)=x^nP(1/x)$. Define $x_k=(-1)^k(k-1)!$ and for $k$ large enough, we'll pick a prime $p_k \mid Q(x_k)$. Clearly, we see $p_k \mid P((p_k-k)!)$. Now we claim that for some large $k$, we have $(p_k-k)!-p_k>M$ for any fixed $M>0$.

Let $m$ be a large number with $m \equiv 1 \pmod{6}$. Observe that $m!+j$ has no prime factor exceeding $m!+1$ for any $2 \le j \le m+3$. Hence any prime $p \mid m!+j$ will satisfy $p \mid (m!+1)!$. For $k=m!+2$ if $p_k \mid m!+j$ for any $2 \le j \le m+3$ then $p_k \mid x_k$ hence $p_k \mid Q(0)$ (also leading coefficient of $P(x)$), false for $k$ large since $p>(k-1)$. Hence $p_k>k+(m+1)$. Hence $(p_k-k)!-p_k>M$ for any fixed $M>0$ by taking $k$ large.

Finally, $P((p_k-k)!)=p_k$ but $P((p_k-k)!)>(p_k-k)!+P(0)$ so we get the desired contradiction! $\blacksquare$

[b]Note.[/b] The mod $6$ restriction is only to ensure $m+1, m+2, m+3$ do not have a prime among them.

EDIT. So I forgot to add $p_k>(k-1)$, indeed $\gcd(Q(x_k), x_k)=Q(0)$ for $k$ large.

EDIT. So to make clear why $(p_k-k)!-p_k>M$, note that $$(p_k-k)!-p_k=(p_k-k)((p_k-k-1)!-1)-k>(m+1)(m!-1)-(m!+2)$$ and we are good to go.

Suppose $\deg P = d$. For now, we will assume $P$ has positive leading coefficient (if its negative just do $P\to -P$). Let $Q(x)=x^dQ(1/x)$. Note that essentially by Wilson's theorem, we have $(p-k-1)!\equiv 1/k!\pmod{p}$ for odd $k$ (for even $k$ we have the opposite sign relation, but we won't be using that) and any prime $p$. Thus, for $k$ even, we have
\[Q(k!)\equiv P((p-k-1)!)\pmod{p}\]
for any prime $p$.

Let $f(M):=\mathrm{lcm}(1,2,\ldots,M+1)+1$. Pick $M$ sufficiently large, and choose it so that $M+1$ is not prime. Then we have (for large enough $M$) that $Q(0)^2\mid f(M)!$, so $Q(f(M)!)/Q(0)$ has all its prime factors larger than $f(M)$. But since $f(M)+1,\ldots,f(M)+M$ are all composite, we must have that any prime factor $p$ of $Q(f(M)!)/Q(0)$ is at least $f(M)+M+1$. Suppose $p=f(M)+M+r$ for $r\ge 1$. By construction, $Q(f(M)!)\equiv 0\pmod{p}$, and since $f(M)$ is even, we have $p\mid P((p-f(M)-1)!)$.

It now suffices to show $P((p-f(M)-1)!)>p$. We see that for large enough $x$, we have $P(x)>x+P(0)$. Note that $(p-f(M)-1)!=(M+r-1)!$. We claim $(M+r-1)!>f(M)+M+r$. We have that $f(M)=\mathrm{lcm}(1,2,\ldots,M)+1$ since $M+1$ not prime, so $f(M)<M!/2$ for large enough $M$. Thus,
\[p=f(M)+M+r<M!/2+M+r<(M+r-1)!,\]
so $(p-f(M)-1)!>1.1p$ for large enough $M$. Also eventually $P$ is increasing, so we have then that $P((p-f(M)-1)!)>P(p)>1.1p+P(0)>p$ for large enough $M$. But we also $p\mid P((p-f(M)-1)!)$, so $P((p-f(M)-1)!)$ is not prime, as desired.

I will show that in fact there are infinitely many integers.

Suppose there exists an $N$ such that $f(n!)$ is not composite for $n \geq N$.

[b][color=blue]Key Step:[/b][/color] Note that $$f((p-k)!) \equiv 0 \pmod p \iff f((k-1)!) \equiv 0 \pmod p.$$ Set $$x_k = a_0(k-1)^d + a_1(k-1)^{d-1} (-1)^k + \cdots + a_d(-1)^{kd}.$$ Pick a large $k$ such that $a_d^2 \mid (k-1)!$, and set $p_k \mid x_k / a_d$. Note that $p_k^2 \nmid x_k / a_d$, and also $p_k \geq k$.

Now fix $k = a(N+1)! + 2$ for some $a \geq 1$. Relaying $p_k$ from earlier, note that as $p_k \mid f((p_k - k)!)$ and $p_k - k \geq N$, we have $f((p_k-k)!) = p_k$.

At this point we're basically done: there exists a sequence $\{x_a\} = \{p_k - k\}$ with $f(x_a!) = x_a + a(N+1)! + 2$. This creates size issues as $f((x_a+1)!) - f(x_a!) \leq (N+1)! + 1$, but $f$ is increasing for sufficiently large $N$.

By Wilson's theorem and induction on $k$, we find that $(p-1-i)! \equiv \frac{(-1)^{i+1}}{i!} \pmod{p}$ for $1 \leq i \leq p$.

Pick $q$ to be a massive prime and the difference between $q$ and the gap between $q$ and the prime greater than it is large as well. Now consider the "reverse polynomial" $Q(x)=x^nP(1/x)$. If $q$ is taken to be very large, we can force $\nu_p(a_n)<\nu_p(q!)$ for all primes $p$, since the LHS is fixed while the RHS is unbounded. Then consider the quantity $Q(q!)/a_n$, which must be an integer and is coprime to $q!$. If $q$ is made large, this quantity must be large as well, hence it should be divisible by some prime $p>q$ (so $p$ obviously divides $Q(q!)$ too). Then
$$P\left(\frac{(-1)^{q+1}}{q!}\right)\equiv P\left(\frac{1}{q!}\right)\equiv \frac{Q(q!)}{(q!)^n} \equiv 0 \pmod{p} \implies P((p-q-1)!) \equiv 0 \pmod{p},$$
but since we selected $q$ such that any primes after it are much larger, $p-q-1$ can be made to be unbounded, hence at some point $P((p-q-1)!)$ must obviously exceed $p$, hence we find that $P((p-q-1)!)$ is composite. $\blacksquare$

EDIT: Actually we need $p-q-1$ to grow quickly enough, not just for it to be unbounded. Since the $n$-th prime is asymptotic to $n \log n$, we can pick $q$ such that no primes lie in $[q,q+2\log q]$, hence $p-q-1 \geq \log p$. Then $(\log p)!=\Omega((\log p)^{\log p})$ by Stirling's which dominates by $p$ by taking logs.

[hide=Solution] Let WLOG $P(x)$ has a positive leading coefficient. We will find an appropriate even $k$, such that there exists a prime $p>k$, depending on $k$, such that $P((p-k)!)$ is divisible by $p$ and greater than $p$ (then $n=p-k$ will work).

Let's firstly focus on the divisibility part. Using the congruence $(p-k)!(k-1)! \equiv 1 \pmod p$ (which holds for even $k$) and the "reverse" polynomial $Q(x)=x^nP(\frac{1}{x})$, we have $p\mid P((p-k)!) \iff p\mid Q((k-1)!)$, so we want to choose $k$ so that $Q((k-1)!)$ has a prime factor greater than $k$. Let $Q(0)=c$; then $\frac{Q((k-1)!)}{c} \equiv 1 \pmod{\frac{(k-1)!}{c}}$, so if $k$ is sufficiently large, then all primes less than $k$ will divide $\frac{(k-1)!}{c}$ and this will force $Q((k-1)!)$ to have a prime divisor $p>k$. To conclude this part, any large even $k$ ensures the existence of a prime $p$, satisfying the divisibility $p \mid P((p-k)!)$.

Now, we should choose $k$, so that the prime $p$ we obtain forces $P((p-k)!)>p$. Since $P$ has degree at least $2$, then $P((p-k)!)>(p-k)!$ if  we make sure that $p-k$ can be arbitrarily large. Notice that we can indeed force $p-k \geq t$ to be large by ensuring that $k+1, k+2, \ldots, k+t-1$ are composite and $k=(t-1)!$ does exactly that. Hence, we only need $(p-k)! \geq p$, which indeed holds for large $t$: $(p-k)! \geq t!>t+(t-1)!$. This $k$ obviously satisfies the property required in the previous paragraph, so it indeed works.

Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$ and $Q(x)=x^nP\left(\frac1x\right)=a_0x^n+a_1x^{n-1}+\dots+a_{n-1}x+a_n$. Let $r(x)=\prod_{p\mid x}{p}$ and let $N$ be a positive integer such that
$a_n\cdot r(N!)\mid N!$. Since $\nu_p\left(\tfrac{N!}{r(n!)}\right)$ is non-decreasing, for all $k\ge N$, $a_n\cdot r(k!)\mid k!$. We have
\[\frac{Q(k!)}{a_n}=a_0\left(\frac{k!^n}{a_n}\right)+a_1\left(\frac{(k)!^{n-1}}{a_n}\right)+\dots + a_{n-1}\left(\frac{k!}{a_n}\right)+1\]
Note that everything except the $1$ must be divisible by $r(k!)$, so $k!$ is coprime with $Q(k!)$. Now, let $p_k\mid Q(k!)$ for some $p_k>k$. Then, $p_k\mid P\left(\tfrac{1}{k!}\right)$. By Wilson's, we have
\[\frac{(-1)^{k-1}}{k!}\equiv (p_k-k-1)!\pmod p_k\] so for odd $k$, $p_k\mid P((p_k-k-1)!)$, so $P((p_k-k-1)!)=p_k$. Now, by Chinese Remainder Theorem, there exists unboundedly large sequences of consecutive composite numbers, so $p_k-k-1$ can be made to be infinitely large, and thus $P((p_k-k-1)!)$ will far exceed $O(x)$, so it cannot be equal to $p_k$.


Round Two

7. There are 11 men and 24 women at a dance. Lisa and Claire each insist upon having a male partner(any one will do) for every song. If there is no partner changing during songs, what is the minimum number of songs that must be played so that every woman can dance with every man?

8. The sides of $\triangle ABC$ are extended to $A'$, $B'$, and $C'$ so that \[ \frac{AA'}{AB} = \frac{BB'}{BC} = \frac{CC'}{AC} = 4 \]
If the area of $\triangle ABC$ is 1, then find the area of $A'B'C'$.

[quote="K81o7"][hide="8"]$125$?  The area of $\triangle{A'BC}$ must be $5$, having a base 5 times as large as $\triangle{ABC}$.  By similarly logic the area of $\triangle{A'B'C}$ is $25$.  Then, because $\frac{AC'}{AC}=5$, $\frac{\triangle{A'B'C'}}{\triangle{A'B'C}}=5 \implies \triangle{A'B'C'}=125$.[/hide]

It seems too easy...[/quote]

I think you're wrong. $125$ seems WAY too big..

#7
[hide]
So $2$ fixed people are in every song. first pick $9$ from $22$. The first series of dances need $11$ songs for all $11$ people, including $2$ fixed to dance with all males. Since we have $13$ left that haven't danced, get $9$ more and play song $11$ more times. We have $4$ left and the same procedure follows.
$11+11+11=\boxed{33}$.
[/hide]

#8
[hide]
Basically we have sides $a,b$ and $c$ corresponding to $BC,AC,AB$ respectively.
The area of triangle $ABC$ is $\frac{ab\sin C}{2}=\frac{bc\sin A}{2}=\frac{ac\sin B}{2}$.
to find area of $A'B'C'$, we need to sum of areas of $BB'C',CC'A', AA'B', ABC$.
Which is $6bc\sin A +6ac\sin B +6ab\sin C=36[ABC]$.
So $36[ABC]+[ABC]=37[ABC]=\boxed{37}$.[/hide]

[quote="agolsme"][hide="on 7"]
I counted 33 too, but the solutions booklet  says 27 for some reason -- with a really odd explanation attached:

[code]
Suppose there are $s$ songs played, so that a total of $11s$ dances betweeen couples are possible. Demonstrate that the stipulations of the problem require at least 11(22) + $2s$ dances to occur
[/code]
[/hide]

on 8-
k81o7: you are overcounting
jimhu: I am not quite sure how you you got those equations but $37$ is not the correct answer.
[/code][/quote]

Sorry I didnt explan it that well. If a triangle is there and given sides $a$ and $b$, and angle $C$ between them, the area of triangle is $A=\frac{ab\sin C}{2}$.
So basically $[ABC]=\frac{ab\sin C}{2}=1$. and $[BB'C']=[AA'B']=[CC'A]=\frac{3a*4b\sin C}{2}=12[ABC]=12$. So the answer is $37$ without doubt... unless you worded the problem wrong..

Oh yeah i see what I did...

Then, [hide] using area=absinC, we have
$\triangle{AA'B'}=\triangle{BB'C'}=\triangle{CC'A}=20$, so we get $\boxed{61}$, because $\sin{\theta}=\sin{180^{\circ}-\theta}$.[/hide]

I get 37 for 8, too.  I did the same thing as jimhu.  What are you doing there, k81o7?

The answer for #8 depends on the locations of the points.

Take the locations of $A$, $A'$ and $B$ for example.
If $A'$, $A$ and $B$ are collinear in this order, then $61$ is the correct answer.
However, if $B$ is between $A$ and $A'$, then $37$ is the correct answer.

By the way, you can just use area ratios to compute the answer.
(It's simply $1+3(4\cdot 5)=61$ or $1+3(3\cdot 4)=37$ ;) )

I put C between A and C', B between A and A' and between C and B'. So, the result is (A'B'C')=12.

I think the result may change, depending in the position on the points.