geometry
incenter
analytic geometry
geometry proposed

$\triangle ABC$ a line $l$ through centroid $G$ $l\cap$ lines $BC,CA,AB$ at $A',B',C'$ prove that
$\frac{a^{2}}{\overline{A'B}\cdot\overline{A'C}}+\frac{b^{2}}{\overline{B'C}\cdot\overline{B'A}}+\frac{c^{2}}{\overline{C'A}\cdot\overline{C'B}}=-9$
with any $\triangle ABC$ and any $l$! :P
(here $\overline{XY}$ show directed length of segment!)

[color=darkred][b]Lemma[/b] (well-known). Given are the lines $\overline{ABC}$, $\overline{A'B'C'}$ and a point $P$ so that $P\in AA'\cap BB'\cap CC'\ .$
Then there is the following remarkable relation : $\boxed{\ \frac{\overline{PA}}{\overline{PA'}}\cdot\overline{BC}+\frac{\overline{PB}}{\overline{PB'}}\cdot\overline{CA}+\frac{\overline{PC}}{\overline{PC'}}\cdot\overline{AB}=0\ }\ .$[/color]

[color=darkblue][b]Proof of the proposed problem.[/b]I'll work with orientate segments, although I'll use conveniently $XY$ as well as $\overline{XY}$. Therefore,
$\frac{AB}{C'A}+\frac{AC}{B'A}=-3$ a.s.o. $\Longrightarrow$ $\sum\left(\frac{AB}{C'A}+\frac{AC}{B'A}\right)=-9$ $\Longrightarrow$ $\sum \left(\frac{AB}{C'A}+\frac{BA}{C'B}\right)=-9$ $\Longrightarrow$
$\sum \frac{AB\cdot (C'B-C'A)}{C'A\cdot C'B}=-9$ $\Longrightarrow$ $\sum \frac{c^{2}}{C'A\cdot C'B}=-9\ .$[/color]

ok that is a good solution Virgil I like your Lemma!
There is a similar problem if keep all assumtion but change centroid $G$ by $I$ incenter prove that with any $l$
$\frac{a}{\overline{A'B}\cdot\overline{A'C}}+\frac{b}{\overline{B'C}\cdot\overline{B'A}}+\frac{c}{\overline{C'A}\cdot\overline{C'B}}=-\frac{p}{Rr}=const$
 :P

Are you sure about this?  I did the calculations, and it does not work out to be a constant.

[quote="gemath"]ok that is a good solution Virgil I like your Lemma!
There is a similar problem if keep all assumtion but change centroid $G$ by $I$ incenter prove that with any $l$
$\frac{a}{\overline{A'B}\cdot\overline{A'C}}+\frac{b}{\overline{B'C}\cdot\overline{B'A}}+\frac{c}{\overline{C'A}\cdot\overline{C'B}}=-\frac{2(a+b+c)}{Rr}=const$
 :P[/quote]

[quote="nsato"]Are you sure about this?  I did the calculations, and it does not work out to be a constant.[/quote][color=darkblue]Yes, this is and the my opinion. I found generally for a point $P(x,y,z)$ (with barycentrical coordinates) the following relation : $\boxed{\ \sum_{\mathrm{cyc}}[\ \frac{\overline{AB}}{\overline{C'A}\cdot \overline{C'B}}\cdot(x\cdot \overline{C'A}-y\cdot\overline{C'B})\ ]=3\ }\ .$
[b]Proof.[/b] $y\cdot \frac{\overline{AB}}{\overline{AC'}}+z\cdot\frac{\overline{AC}}{\overline{AB'}}=1$ a.s.o. $\Longrightarrow$ $\sum_{\mathrm{cyc}}(y\cdot\frac{\overline{AB}}{\overline{AC'}}+z\cdot\frac{\overline{AC}}{\overline{AB'}})=3$ $\Longrightarrow$
$\sum_{\mathrm{cyc}}(y\cdot\frac{\overline{AB}}{\overline{AC'}}+x\cdot\frac{\overline{BA}}{\overline{BC'}})=3$ $\Longrightarrow$ $\sum_{\mathrm{cyc}}[\frac{\overline{AB}}{\overline{C'A}\cdot\overline{C'B}}\cdot(x\cdot\overline{C'A}}-y\cdot{\overline{C'B})]=3$
In the particular case $P: =G(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ obtain the Gemath's relation : $\sum_{\mathrm{cyc}}[\frac{c^{2}}{\overline{C'A}\cdot\overline{C'B}}]=-9\ .$
If $P: =I$, then $\sum_{\mathrm{cyc}}[\frac{\overline{AB}}{\overline{C'A}\cdot\overline{C'B}}\cdot(a\cdot \overline{C'A}-b\cdot\overline{C'B})]=6p$ and hence ... work you because I could  obtain nothing.
[b]Remark.[/b] For any $X\ ,\ Y\ ,\ Z\ \ : \ \ \overline{XY}=-\overline{YX}\mathrm{\ \ ;\ \ }$ $\overline{XY}=\overline{ZY}-\overline{ZX}=\overline{XZ}-\overline{YZ}\ \ ;\ \ Z\in XY\Longrightarrow$ $(\overrightarrow{ZX}\ ,\ \overrightarrow{ZY})=\overline{ZX}\cdot\overline{ZY}$,
the left hand is a scalar product and the right  hand is multiply of two real numbers,
i.e. if the points $X$, $Y$ belong to the same line and $X(x)$, $Y(y)$, then $\overline{XY}=y-x\ .$

[b]I think Gemath used wrongly and/or imattentively these commonplace relations ![/b][/color]

Oh I still see it is true, I only have mistake it equal $-\frac{a+b+c}{2Rr}$ but it is still equal const when $l$ around $I$ here is my fig by Cabri !
Here $const=1.36$ and in my calculations I consider $\overline{A'B}\cdot\overline{A'C}=\vec{A'B}\cdot\vec{A'C}$
(scalar product)  :)

[quote="gemath"]Oh I still see it is true, I only have mistake it equal $-\frac{a+b+c}{2Rr}$ but it is still equal const when $l$ around $I$ here is my fig $\boxed{\mathrm{by\ Cabri\ }\ }$ !
Here $const=1.36$ and in my calculations I consider $\overline{A'B}\cdot\overline{A'C}=\vec{A'B}\cdot\vec{A'C}$
(scalar product)  :)[/quote]What mean you ...[b][u]"by Cabri"[/u][/b] ... ?

Ah Virgil Cabri is software of geometry
[url=http://www.cabri.com]here[/url] is its web

You're right, I thought the nuemrators were still $a^{2}$, etc.  Nice problem!


I can't send any programs to my new TI-89 Titanium! :( I just got it today, so unlike my old TI-83+SE, I don't know anything about it. All the programs I try to send end in .89z, but when I send to the calculator with TI-Connect, it says, "Incompatible Type." Am I doing something wrong??? :?

Use TILP.  It almost always works now, and is generally faster than TI-Connect.

Links:
[url=http://www.ticalc.org/archives/files/fileinfo/374/37481.html]Windows version[/url]
[url=http://www.ticalc.org/archives/files/fileinfo/374/37480.html]*nix version[/url]

I tried using TiLP, but I don't know how to use it with a blacklink. When you say "custom" or "null" for port (in setup->communication) it doesn't work :( And I can't figure out TiLP II either...

i have the same problem with my 84 plus silver. Sometimes it just doesn't register, but most of the programs work. Did you choose the right type cuz for 83 the file is .83 but for 83z plus its .8xz

hope you dont have to return it, the 89 is pretty nice ;)

Yeah, I'm pretty sure it's .89z but then again I don't know much about TI-89 Titaniums, so ... :(

edit: I got it to work with my Silverlink! Thanks for the help! :)

(I love TI-Chess :D)

Most programs come with instructions you could just read those


A convex quadrilateral PQRS has properties such that PS = RS = 19, $ \angle PQS \equal{} u, \angle RQS \equal{} v, \angle PSQ \equal{} 2v,$ and $ \angle RSQ \equal{} 2u$. If $ \angle PQS \plus{} \angle RQS \plus{} \angle PSQ \equal{} \angle RSQ \equal{} 180^\circ,$ find the remainder when $ (PQ)(RQ)$ is divided by 1000.

Do you mean:

[quote="Silverfalcon"]$ \angle PQS \plus{} \angle RQS \plus{} \angle PSQ \plus{} \angle RSQ \equal{} 180^\circ,$[/quote]

If so,

[hide]

$ u \plus{} v \plus{} 2u \plus{} 2v \equal{} 3u \plus{} 3v \equal{} 180^\circ$, so $ u \plus{} v \equal{} \angle{PQR} \equal{} 60^\circ$. Also, $ 2u \plus{} 2v \equal{} \angle{PSR} \equal{} 120^\circ$. As the opposite angles sum up to $ 180^\circ$, $ PQRS$ is a cyclic quadrilateral. Hence, $ \angle{PQS} \equal{} \angle{PRS} \equal{} u$, $ \angle{RQS} \equal{} \angle{RPS} \equal{} v$, $ \angle{RSQ} \equal{} \angle{QPR} \equal{} 2u$, and $ \angle{PSQ} \equal{} \angle{PRQ} \equal{} 2v$. Now, since $ PSR$ is isosceles, $ \angle{RPS} \equal{} \angle{PRS} \equal{} u \equal{} v \equal{} 30^\circ$ Then, $ \angle{QPS} \equal{} \angle{QRS} \equal{} 90^\circ$. So, $ \triangle{QSP}\cong\triangle{QSR}$ and are 30-60-90 triangles. Thus, $ QP \equal{} QR \equal{} 19\sqrt3$. $ (QP)(QR) \equal{} 361\cdot 3 \equal{} 1083\equiv 83\pmod {1000}$

[/hide]

Sorry if that's not what you meant.

Edit: thanks.

What you have is correct (including answer) but just two errors in your solution.

It should be $ u \plus{} v \equal{} \angle PQR$ and $ \angle QPR \equal{} 2u$.

There are a lot of ways to do this problem. Anyone wants to do it other ways?


integration
calculus
function
calculus computations

Help me!! Find $ \int \frac{{e}^{x}dx}{x\minus{}1}$

First, welcome to this forum.

Second, try to learn and use Latex.

Third, your indefinite integral, $ \int \frac{e^x}{x\minus{}1}\,dx$, cannot be expressed with elementary functions. In other words, the function $ \frac{e^x}{x\minus{}1}$ does not have an elementary antiderivative.

May be $ \int {\frac{{e^x dx}}
{{x \minus{} 1}}}$??

So there is no solution for this problem. Thanks Carcul and Sara90.

hello, we have $ \int\frac{e^x}{x\minus{}1}\,dx\equal{}\minus{}e\text{Ei}(1,1\minus{}x)$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, we have $ \int\frac {e^x}{x \minus{} 1}\,dx \equal{} \minus{} e\text{Ei}(1,1 \minus{} x)$.
Sonnhard.[/quote]

Hello Sonnhard, would you please tell me how to get that result ? Thanks.

That's a result that doesn't really say anything; it's just a name for this function, extending our usual collection of names. The definition is a very similar integral.


what is the sum of the digits in the decimal representation of $(789,000,003,450,000)^{2}$

[hide="hint"]There are a LOT of 0s in the expansion.[/hide]

[hide]54 by brute force :D [/hide]

See if you can use this answer to figure out a formula

How long did you take to brute force? If its over 2 mins, its probably too long for mathcounts

he meant brute force as in entering it in calculator and adding the digits.

No, I dont think he meant that... because the calculator wouldnt fit all that..

How would you do it if your calculator only fit 8 digits?

[quote="usaha"][quote]My computer's calc fits 27 digits [/quote]

What? What type of calculator- I gotta get one of those[/quote]

His [u][i][b]computer's[/b][/i][/u] calculator

Well, I did (789000000000000+3450000)^2,

So essentialy, you are trying to find the sum of the digits in a^2, b^2 and 2ab, so 789^2=622521, 345^2=119025, and 2ab=544410, you totally disregard the zeros.. ok add them up to get 54.
EDIT: I dont know if this is correct, as I see someone has allready typed it into their calculator to get 52, not 54

Art of Owna, your answer agrees with the calculator answer!

But, you can't totally disregard the zeros, they are important.

Would the answer be the same if you were calculating

(789 + 345)^2 ?

[quote]Would the answer be the same if you were calculating

(789 + 345)^2 ?[/quote]

No, because now the zeros matter.  The only reason the zeros  dont matter because it was in this  type of form(not exact but a basic outline)-when you expand (a+b)^2 for a=789000000000000 and b=3450000

    abcd0000000000000000000   this is a^2
    0000000abcd000000000000   2ab
 + 0000000000000abcd000000   b^2

where a, b,c, d are digits(there can be more, just no overlapping)

so you can just striaght up add all the digits as there is no overlapping

but (789+345)^2 is in the form

    abcd000000   this is 789^2
    00abcd0000   this is 2*789*345
+  0000abcd00   this is 345^2

You can see the digits overlap, so you cant just add all of them up


modular arithmetic
number theory unsolved
number theory

show that the equation $ y^2 \plus{} 2^2 \equal{} x^3 \plus{} 3^3$ has no integer solutions

Is there an easy way to do this, that is an elliptic curve [url]http://en.wikipedia.org/wiki/Elliptic_curve[/url].

Maybe this lemma will help:
If $ p | a^2\plus{}b^2, p \in P, p \equal{} 4k \plus{} 3$, we will have: $ p | a, b.$

Now, if x is even, y must be odd.
$ LHS \equiv 1 (mod 4)$
$ RHS \equiv 3 (mod 4)$
This can't be right.

We have: $ y^2 \plus{} 4 \equal{} (x\plus{}3)(x^2 \plus{} 3x \plus{} 9)$
If x = 4k + 3:
Because $ x^2 \plus{} 3x \plus{} 9 \equiv 3 (mod 4)$,  #p| y^2 + 2^2 (p=4q + 3)[/tex]. Then 2^2 = 4 will be divisible by p => Wrong.

If x= 4k + 1: I'm working on it... :D

[quote="iTAQ"]We have: $ y^2 \plus{} 4 \equal{} (x \plus{} 3)(x^2 \plus{} 3x \plus{} 9)$
[/quote]

$ y^2 \plus{} 4 \equal{} (x \plus{} 3)(x^2 \minus{} 3x \plus{} 9)$  :wink:

Then, $ x\equal{}4k\plus{}1$ follows as you said for $ x\equal{}4k\plus{}3$.  If $ x\equal{}4k\plus{}3$, then $ (x\plus{}3)(x^2\minus{}3x\plus{}9)\equiv 2\pmod{4}$ which is impossible since $ y^2\plus{}4\equiv0,1\pmod{4}$.


if a,bis natural and (a,b)=1 prove that:
[b](a+b,(a^2)-ab+(b^2))=1or3[/b][/b]

If $d|a+b$ and $d|a^{2}-ab+b^{2}= (a+b)^{2}-3ab$ then $d|3ab$. If some prime divisor $p$ of $d$ is such that $p|a$ then $p|b$ (because $p|a+b$) - contradiction with (a,b)=1. The same would happen if $p|b$. So $(d,ab)=1$. Hence, $d|3ab \Rightarrow d|3 \Rightarrow d=1$ or $d=3$.


What is the 7th term in $(4a-3b)^{11}$

Please don't evaluate and then count for the 7th term...

[quote="B4k4ka"][hide]the answer is... $256a^{7}$?[/hide][/quote]
Answer is wrong, I let you do it one more time

[quote="Drunken_Math"][quote="math92"][quote="236factorial"][hide="lol answer"]$86220288a^5b^6$?[/hide][/quote]
how did you do that?[/quote]

There's a formula:

The rth term in the binomial $(a+b)^n$ is

$\frac{n!}{(n-r+1)!(r-1)!}*a^{n-r+1}*b^{r-1}$[/quote]
do you do the same thing even if it's minus B?

[quote="B4k4ka"][/quote][quote="Drunken_Math"][quote="math92"][quote="236factorial"][hide="lol answer"]$86220288a^5b^6$?[/hide][/quote]
how did you do that?[/quote]

There's a formula:

The rth term in the binomial $(a+b)^n$ is

$\frac{n!}{(n-r+1)!(r-1)!}*a^{n-r+1}*b^{r-1}$[/quote][quote="B4k4ka"]
do you do the same thing even if it's minus B?[/quote]

If it's $-b$, then you would substitute $-b$ for $b$  ;)

[quote="Drunken_Math"][/quote][quote="B4k4ka"][/quote][quote="Drunken_Math"][quote="math92"][quote="236factorial"][hide="lol answer"]$86220288a^5b^6$?[/hide][/quote]
how did you do that?[/quote]

There's a formula:

The rth term in the binomial $(a+b)^n$ is

$\frac{n!}{(n-r+1)!(r-1)!}*a^{n-r+1}*b^{r-1}$[/quote][quote="B4k4ka"]
do you do the same thing even if it's minus B?[/quote][quote="Drunken_Math"]

If it's $-b$, then you would substitute $-b$ for $b$  ;)[/quote]
oh... *osaka face*


AMC
AIME
USA(J)MO
USAMO
MATHCOUNTS
AMC 8
geometry

My biggest plan for AMC this year is to JUST QUALIFY for the AIME. I know there are these people in this site that has intelligency that I didn't even think the humans could have and compare to that, my goal seems small. I don't have any (well, I wish I can but that's not really possible) hope on making after AIME but I'm really nervous about making AMC-10. It's my first year and well, I don't want to have something that happened in my MC year (which I blew up my first year -.-;;;) so hopefully, I can qualify to AIME this year and maybe next year, prepare to take AIME more seriously and qualify for the USAMO (I really wish). I know I did all the stuffs I can, like bought all the past AMC-10 exams (there are only few but it worthed it) and do many many LOTS of problems but I just can't take off my nervousness. Perhaps, someone can tell the cure to it. :P My school really, well, we didn't do ANYTHING yet. So, it's really up to myself ONLY if I WANT to pass AMC.

I want to know what you guys' plan are and what you're thinking about it. I hope I can be in the discussion after a month or so about passing and considering for AIME but who knows? ;) You can't be sure for anything.

I'm a senior.  This is my first, last, and only year to take the AMC 12.  I hope to pass the AMC 12, and be the first person at my school in the last three years to take the AIME. (My friend has been the sole AMC-taker the past two years, and although he passed both times, he doesn't have a green card, so he can't do AIME--at least that's what I think it was). If I make it to AIME I want to get a couple questions right--I just wish I was able to do this last year (I supposed to, but there was miscommunication between my friend, the school testing coordinator, and AMC--resulting in him taking the test, and me using the proctor's booklet and an AMC 10 answer sheet--the problem relates to this--and taking the test as a practice.  I wouldn't have passed--answered exactly eleven, and there was no way they were all correct.)

So yeah, pass AMC 12, take AIME, and actually solve two or three of them---that's my AMC goal.

my goal is just to pass the amc-10 to get into the AIME. i'm in 9th grade now and this is my second year taking the amc-10. 8th grade year was a good year in mathcounts for me, but i bombed the amc 10. silverfalcon, you seem to have prepared a whole lot for the amc-10. i think you're fine and just don't stress out. The main factor in lower scores are careless mistakes (in my opinion). so just try and watch out for them. Good luck!

I'd really like to be able to qualify for AIME with the AMC 10, but I don't know how well that will work out.  And if I qualify, for the AIME, I'd like to be able to solve at least 2 or 3 problems on it.  I know the Mock AIME's are supossed to be really hard, but whenever I see them I feel like I can't even solve the first one, which is also why I never turn my answers.  And if I'm really lucky, making it to the USAMO would be nice, but I don't think that will happen.

My goal is to qualify for USAMO as well, my average on 3 old AIMEs I took for practice is 8.33, so I have a very very very small chance of making it this year, but I'll still practice and hopefully improve!

oops I accidentally voted for pass USAMO :blush:

[quote="tetrahedr0n"][quote]doesn't have a green card, so he can't do AIME[/quote]

A green card is not required to take the AIME. It may (or may not be) required to take the USAMO.[/quote]

I don't think it's even required for USAMO...

well I think you need to permanent resident at least (of USA or one of those other places, like Puerto Rico or Canadia or w/e)

my goal is to not break any elevators this year

You do need a green card (permenant residency) for the USAMO.  Not for AIME, though.  (I had some non-green-carded acquaintances who took AIME, anyhow.  I suppose you could always lie about it, too.  I wouldn't recommend that, though.)

I'm a ninth grader and most likely, if i make usamo, i'll make mop(they took ninth graders last year) so that's what i voted for. My amc 12 score is usually around 125-130(after rigorous no-life practicing) and my aime score is right on the border(7-8). I was so happy that i got a nine once, but that was on one of the old ones. Most of the time I don't bother to look at the problems at the end even though i could probably do mbe one of them and i mess up on one of the easy ones.

As I said above, my average on 3 old AIMEs I did is a 8.3333. However I want to make it a 9 or 10 so that I can qualify for USAMO comfortably.  How should I practice in the next 1 Month, 3 Weeks, 2 Days? I am weak at counting and I am a little better at geometry and  algebra...

Feiqi, I think you already know the best ways to practice...read AoPS and other books, do old practice tests, and so on.  Just stop worrying - you're still a freshman; you have plenty of time.  Don't put so much pressure on yourself.

My goal as a coach is to have at least one of my team members qualify for AIME this year, and to score better than the guy who qualified last year.

For next year and the years following, my goals are a bit more ambitious. Somewhere in there, I don't know what year, comes qualifying for USAMO, and I hope sometime one of the team members qualifies for MOP.

Good luck to all of you guys and gals this year.

i took amc and aime before (but obviously not allowed to take usamo...) and aime is pretty hard. i especially didn't like geometry. good luck whoever is taking!

i'll be happy with just making the AIME.
i can't make the USAMO because 1. my math isn't so great and 2. i'm not a US citizen or permanent resident.

[quote="shinwoo"]My goal is not even on the list...

My goal is to pass AMC8 and get an invitation for AMC10..[/quote]

You don't need an invite to take AMC10

Yeah, but if you take the AMC 8, and get a high enough score, then you get an invitation to take the AMC 10.  His school might use that to decide who should take the AMC 10.

Good luck to everyone taking the AIME.  In my case I'm hoping that I can get around a 10 on the AIME and just qualify for the USAMO.  That would be perfect  :D .  I've taken the AIME the past two years and got 3's but I've worked my butt off and can get 7's on the more recent AIMEs consistently.  Hopefully, I can improve those three points before the test in March.

BTW: Does anyone have other suggestions of what to do to prepare other than keep practicing the old AIME's?

[quote="MathWhiz1286"]BTW: Does anyone have other suggestions of what to do to prepare other than keep practicing the old AIME's?[/quote]

AoPS is offering a special [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php#class4]AIME Problem Seminar[/url] on the 26th and 27th of February 2005.

My goal is qualifying for the USAMO. i'm hoping a score of 7-8 will get me in with my AMC score.

This is my first year taking the AMC tests. As a junior, I am sure of qualifying for the AIME with 130+ on 12, but on past AIMEs I have averaged about 6. I know this is not good enough, but I set my goals upon USAMO, because maybe a month's studying will improve that score by a couple of points.

Now that I've qualified for AIME, I'm changing my goal to qualifying for USAMO.

My goal was to make it to AIME, so im happy
Next year, im hoping to make USAMO

Well, this is my first year taking the AMCs.  I only got 122 on the 12A (on practice tests I was getting more like 135 every time)  :( .  My goal is to make it to the AMO, but that will be very tough to do with such a low AMC score.  But I'll attempt to get 9 or 10 on the AIME, and MAYBE that will get me to the AMO, but I'm not sure what the cutoff index usually is.  I'm a junior, so I hope that I can make it to the AMO this year so I can put it on my college applications during the winter and fall next year.

[quote="juggle7"]I'm the only one from my school that has ever taken AMC tests.  Everybody has been drawn to the "average-kid, only-school-curriculum" UIL tests that Max has complained about.  Anyway, most all I do outside of school (and inside really) is math.  So, most all I am missing is knowing the nuances of the tests and maybe a bit of topology.  My goal is to get to USAMO and possibly MOP.[/quote]

Wow to that; I understand the situation.  Only I live in a hick town enough that I see a plain no-one-cares attitude for all of it--and other things.

My goal is, of course, the common one:  To make USAMO.  My prediction is that I will come close, not more than ten points on either side.  My thought is that I should care--

But then, there are English papers and the anti-synchronies of normal life.

Has anyone seeing this actually [i]qualified[/i] for the USAMO?  All I see are a bunch of dreamers wishing, hope against hope, that they will qualify.  And does anyone know what the USAMO index (10*AIME + AMC) to qualify was in 2004?  It seems to me that everyone is shooting for a magical barrier ~200.  I cannot find this posted anywhere.

Last year, the index was a 210.0 - it said it on the AIME results sheet, and on the AMC website.

A lot of the ninth graders here on AoPS may be shooting for "red" MOP, which of course requires qualifying for USAMO, but other than that doesn't require an enormous USAMO score, except to possibly get past other ninth graders. And last year some ninth graders qualified for MOP "the hard way," by being in the top group after USAMO scores were taken into account.

Qualifying for USAMO would be a big deal for the two AIME-qualifiers in our local group.


[url]http://www.cs.rochester.edu/users/faculty/nelson/cesium/cesium_faq.html[/url]

Discuss.

The joke was too stupid, I had to laugh.

To the uninitiated, the flame appears purplish rather than blue; however, after sufficient meditation, study of the holy writs, and blissful hours spent in reverence before Cesium burning in her naked glory, the true blue nature of her flame is revealed.

Bernic Lake is also the site of the annual Cesium festival on February 24 where tons of Cesium are burned in the snow, and other Cesium madness prevails.

 :rotfl:  :rotfl:

eh, i've seen that joke before... =D
and 1=2, you are taking it far too literally... i think it is cesium though.

I'm just joking. I checked the periodic table. I don't leave home without it!!!  :P

And it's Caesium.

[quote="1=2"]I found a few errors that made the joke:

[hide="spoilers:"]It's spelled Caesium.
It melts at 28.
It boils at 671.
It has a specific gravity of 1.8785.
It has an atomic weight of 132.90543
That bad word is Fermium.
The answer to Q7 is DEFINITELY NOT!!!!!!
The first one in Q11 can replace the grenade.[/hide][/quote]

1: That's an alternate spelling, as sulfur and sulphur are two different spellings of the same thing.
6: How do you know?
7: Why not?  Consider it a rite of purification.

Sweet...so I'm actually not the only one who thinks cesium is awesome!

I thought the bad word is "francium."

All alkali metals are cool.

"The longest lasting isotope, Fr233 has a half life of only 22 minutes".


Hi!
my problem is that the minus sign $ \minus{}$ sometimes isn't shown when i write a LaTeX equation.
In its place, there appears a symbol looking like an 'a' with a cap.
I'm sure i write the - clearly in the equation.

Why does this happen/ any remedy?

Thank You!

We can't diagnose the problem unless you give a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url]. It might be a font problem, who knows without more detail?

I think it's an mdash versus ndash problem -- LaTeX only converts one of them into a minus sign.  Other people have had a similar problem in the past, too -- in http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1171066938&t=227617 a ... got turned into $ \hat a$, and in this thread http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1171066938&t=111326 it got fixed by a moderator.  Use the short dash and you'll be fine.

$ - x$
$ -x$

For the first line I used the usual - on my keyboard. For the second, I used (on my Mac keyboard) option -, or –.

The has been a persistent problem with me, and I'm familiar with it. I only fairly recently started using $ \text{\TeX},$ and I have s considerable number of old documents that I wrote in Word. I've updated a number of these old documents, usually by pasting a paragraph of so at a time from Word into $ \text{\LaTeX},$ and then going through and fixing all the problems and replacing all the formulas. But I almost always used –, the long dash, for the minus sign in Word, and if I don't fix that, the problem seen above happens consistently. (Yes, I'm aware that I should more consistently use a search and replace addressed to this specific issue.)

There's no doubt in my mind that this is what is happening to the O.P. - which is exactly what JBL said.


For $ m,n,a_{ij}$ positive real numbers, where $ i,j\equal{}1,2,...,k$ and $ k\in N, k\geq2$, prove that

$ \left.\frac{a^{m\plus{}n}_{11}\plus{}a^{m\plus{}n}_{12}\plus{}\cdots\plus{}a^{m\plus{}n}_{1k}}{a^m_{11}a^n_{21}\plus{}a^m_{12}a^n_{22}\plus{}\cdots\plus{}a^m_{1k}a^n_{2k}}\plus{}\frac{a^{m\plus{}n}_{21}\plus{}a^{m\plus{}n}_{22}\plus{}\cdots\plus{}a^{m\plus{}n}_{2k}}{a^m_{21}a^n_{31}\plus{}a^m_{22}a^n_{32}\plus{}\cdots\plus{}a^m_{2k}a^n_{3k}}\plus{}\cdots\plus{}\frac{a^{m\plus{}n}_{k1}\plus{}a^{m\plus{}n}_{k2}\plus{}\cdots\plus{}a^{m\plus{}n}_{kk}}{a^m_{k1}a^n_{11}\plus{}a^m_{k2}a^n_{12}\plus{}\cdots\plus{}a^m_{kk}a^n_{1k}}\geq k\right.$

[b]REMARK:[/b]
This problem generalizes some of my inequalities and the previous inequalities proposed by [b]fjwxcsl[/b].

We can even prove the inequality a bit more general for $ k\cdot l$ numbers $ a_{ij}$ where $ 1\leqslant i\leqslant k$ and $ 1\leqslant j\leqslant l$, i.e.
\[ \sum_{i=1}^k\frac{\sum_{j=1}^la_{ij}^{m+n}}{\sum_{j=1}^la_{ij}^ma_{i+1j}^n}\geqslant k\mbox{,}\]
where all indices $ i,j$ are taken modulo $ k,l$, respectively.

By AM-GM, it suffices to prove
\[ \prod_{i=1}^k\sum_{j=1}^la_{ij}^{m+n}\geqslant\prod_{i=1}^k\sum_{j=1}^la_{ij}^ma_{i+1j}^n\mbox{.}\]
This inequality is eqivalent to
\[{ \prod_{i=1}^k[(\sum_{j=1}^la_{ij}^{m+n})^{\frac{m}{m+n}}\cdot(\sum_{j=1}^la_{i+1j}^{m+n})^{\frac{n}{m+n}}}]\geqslant\prod_{i=1}^k[\sum_{j=1}^la_{ij}^ma_{i+1j}^n]\mbox{,}\]
which is true by Hölder's inequality.  :)

As you already know, I do love generalizations. Each generalization contains effort and work, passions and hopes.
This inequality is the final generalization for the following problems
http://www.mathlinks.ro/viewtopic.php?t=321108
http://www.mathlinks.ro/viewtopic.php?t=321143
http://www.mathlinks.ro/viewtopic.php?t=321179
http://www.mathlinks.ro/viewtopic.php?t=321337
http://www.mathlinks.ro/viewtopic.php?t=321510
http://www.mathlinks.ro/viewtopic.php?t=321515
And not only for them. I'm sure about that.
It's really interesting to notice that there are nice proofs for all those particular cases. I have to admit that I didn't know this generalization before. This result comes from your remarks.
[b]We can consider it as a teamwork[/b]. Any solved problem on MathLinks  could be considered as a teamwork. Thank you very much for your solutions and remarks.
I wish you all the best, health, happiness and...
A HAPPY NEW YEAR !

P.S. [b]MathLinks is great![/b]

This inequlity can be still further generalised by using positive real numbers $ \alpha,\beta,\dots,\lambda$ instead of $ m,n$, but it's a bit wyrd to type up.


probability
geometry
3D geometry
octahedron
pyramid
AMC 10
AMC

Everyone likes AMC 10, right?
Nothin' like a neverending marathon of it.

Make sure to [b]HIDE YOUR SOLUTIONS.[/b]

Lets start with an easy one:

Three tiles are marked X and two other tiles are marked O. The five tiles are
randomly arranged in a row. What is the probability that the arrangement
reads XOXOX?

(A) 1/12 (B) 1/10 (C) 1/6 (D) 1/4 (E) 1/3

[hide="Easy way but not very rigorous..."]
There are $ \dfrac{5!}{2!3!}\equal{}10$ ways to arrange these five tiles, so the answer must be $ \boxed{(B) \dfrac{1}{10}}$ [/hide]
[hide="Answer this and u get 20 dollars"][hide="Made you click"]Centers of adjacent faces of a unit cube are joined to form a regular octahedron. What is the volume of this octahedron?
(A)1/8 (B) 1/6 (C) 1/4 (D) 1/3 (E) 1/2
[/hide][/hide]

[hide="Solution (:"]Because the octahedron is formed by two square pyramids, we only have to calculate the volume of the pyramids $ \cdot$ 2.  Because the cube is a unit cube, the length of the square base of the square pyramid is $ 2/ \sqrt 2$.  Therefore, the area of the pyramid base is 1/2.  The height is clearly 1/2, so the volume of one of the pyramids is $ 1/4 \cdot 1/3$ = 1/12.  1/12 times 2 is 1/6, so the volume of the octahedron is (B) 1/6.[/hide]

[hide="New Problem"]
For each positive integer n, let S(n) denote the sum of the digits of n. For how many values of n is n+S(n)+S(S(n)) = 2007?

$ \text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$[/hide]

[quote="mathwizarddude"][hide]$ S(n)\equiv n\mod 9\implies n\equiv0\mod3$
$ s(n)\le1 \plus{} 9 \plus{} 9 \plus{} 9 \equal{} 28$
$ s_2(n)\le10\implies n\ge1969$.
$ 1971\implies s \equal{} 18\implies s_2 \equal{} 9$
$ 1974\implies s \equal{} 21\implies s_2 \equal{} 3$
$ 1977\implies s \equal{} 24\implies s_2 \equal{} 6$ WORKS
$ 1980\implies s \equal{} 18\implies s_2 \equal{} 9$ WORKS
$ 1983\implies s \equal{} 21\implies s_2 \equal{} 3$ WORKS
(The rest of 198x won't work since all n, s(n), s(s(n)) are strictly increasing;
the 199x won't work since $ s > 19$
$ 2001\implies s \equal{} 3\implies s_2 \equal{} 3$ WORKS

So the answer is $ 4$[/hide]

Define a block move to be taking any set of numbers that appear consecutively and put them back in any spot as a block. For example, $ 6 5 3 4 2 1 \mapsto 4 2 6 5 3 1$ is a valid block move for 6 numbers.
Find the total number of permutations of $ 1,2,3,4,5$ that can be formed from it using only block moves.[/quote]

Why don't you just say find the ways you can use 1,2,3,4 and 5 to make a 5 digit number? :P

[hide="Solution"]
5 choices on where to put the 5, 4 choices for 4 and so on, so:
$ 5!\equal{}120$. 120 is the answer.
[/hide]

Oops, I meant only [b]one[/b] block move... don't know what I was thinking when typing...

[hide]I'm pretty sure it's 20. [/hide]

[hide="New Problem"]Mr. Jones has eight children of different ages. On a family trip his olderst child, who is 9, spots a license plate with a 4-digit number in which each of two digitrs appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?

$ \textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$[/hide]


induction
combinatorics unsolved
combinatorics

$200n$ diagonals are drawn in a convex $n$-gon. Prove that one of them intersects at least 10000 others.

[b]proposed by D. Karpov, S. Berlov[/b]

Obviously $200n\leq \frac {n(n-3)}2$ so $n\geq 403$.
More generally, we prove that if $n>150$ and if no diagonal intersects more than $10000$ others, we have drawn at most $200(n-51)$ diagonals.
Proof
If $n<350$ we have $\frac {n(n-3)}2\leq 200(n-51)$ and it's obvious.
Now if $n>350$ and we have drawn more than $200(n-51)$ diagonals then  we have a vertex from which exit at least $350$ diagonals. Take the $200$th diagonal.
Now it breaks the polygon into two smaller ones.
 Let the polygons have sides $k,l$($k+l=n+2)$. We note $k,l>150$
Then by induction hypothesis the number of diagonals in first polygon is at most $200(k-51)$ and in second at most $200(l-51)$.
Summing since we have at least $200(n-51)$ diagonals we get at least $10000$ diagonals intorsecting the $200th$ diagonal contradiction
QED

[quote="iura"]
If $n<350$ we have $\frac {n(n-3)}2\leq 200(n-51)$ and it's obvious.
[/quote]
For $n=349$ it is wrong. But I think that this solution may be corrected, though we did not know such solution.

Well, we can replace that $n=348$ and do a little bit corrections.
But I think this doesn't matter much..

I think, it is quite incorrect question on the forum, which is devoted to problem solving...  ;)  :D


Solve the following system

sinx+siny+sinz=1
cosx+cosy+cosz=1
sin3x+sin3y+sin3z=2
x,y,z are real numbers

I tried with complex numbers but didnt obtain any big deal.

Well, if you try harder, it works with complex numbers!

how can i solve this by using complex numbers??  :blush:


size big 1 squares = 4*5=20
size big 1 squares = 16+center 1 =17
size 4 squares = 9
size 9 squares = 4
size 16 squares = 1

20+17+9+4+1=51

answer ; 51


${f=\sqrt{z+\sqrt{y+\sqrt{x}}}+\sqrt{y+\sqrt{z+\sqrt{x}}}+\sqrt{z+\sqrt{x+\sqrt{y}}}+}$  ${\sqrt{x+\sqrt{z+\sqrt{y}}}+\sqrt{y+\sqrt{x+\sqrt{z}}}+\sqrt{x+\sqrt{y+\sqrt{z}}}}$

${(x+y+z) (x y+z y+x z)=\theta^{3}9}$
${(x,y,z,\theta)>0}$

${f\geq 6 \sqrt{\theta+\sqrt{\theta+\sqrt{\theta }}}}$

Proove this since $\theta$ is a positive constant

At this time, I can slove this problem, but it is long  :(  and with same method as problem of you before. But, I haven't short solution!  :oops:


[hide="trivial"]97 (it divides 100! but not 50!)[/hide]

[quote="Iversonfan2005"]What is the greatest two digit prime factor of

$\binom{100}{50}$?[/quote]

What does the question mean ?

ok, here's anothe/slightly different this time:

$\binom{200}{100}$, what's the greatest two digit prime factor?

[hide]
This number is $\frac{200!}{(100!)^2}$. If $p$ is a prime less than 100, then $p$ divides 100! so there are at least two factors of $p$ in the denominator. Then there must be at least three factors of $p$ in the numerator, so the prime must be less than 200/3. The largest prime that satisfies this is 61. Since there are three factors in the numerator but only two in the denominator (61>50 so there is only one factor of 61 in 100!), the largest such prime is $\boxed{61}$.
[/hide]


A,B and C are different integers less than or equal to 10. The arithmetic mean of A and B is 9. The geometric mean of A and C is 6*sqrt2. The harmonic mean of B and C is what?

[hide="Solution"]From what we are given, we know $ A \plus{} B \equal{} 18$ and $ AC \equal{} 72$. For the first equation, the only possible $ (A,B)$ are $ (8,10)$ and $ (10,8)$ because the numbers are distinct. For the second equation, the only possible $ (A,C)$ are $ (8,9)$ and $ (9,8)$. Because the $ A$ values are the same in both, we get $ A \equal{} 8$ and therefore $ B \equal{} 10$ and $ C \equal{} 9$. The [url=http://en.wikipedia.org/wiki/Harmonic_mean]harmonic mean[/url] is $ \frac {2}{\frac {1}{10} \plus{} \frac {1}{9}} \equal{} \frac {180}{19}$.[/hide]


It was fun, I didn't do too well due to not being in a school in it, other than officially, and I had no team, and wasn't in a class. :P
I recommend it to anyone, it's fun even if your the worst one there.

Nice post but u can post this on a well-read thread and not waste a spot telling what u can say better elsewhere.
 :P Glad u like mathcounts though

I'm really glad you liked MathCounts!!!  :D

I loved it... Even though I failed xP.

The only reason I did well, is that there were like 70 people there, and I got like somewhere around 50th place, and there were less than 20 people my age.  :rotfl:


calculus
integration
inequalities
function
limit
trigonometry
inequalities unsolved

[quote="primenumber"]let $ f: [0,\infty)\to(0,\infty)$ be integrable function such that $ \lim_{t\to\infty}f(t) \equal{} 0$.
Prove the following inequalities
\[ (1)\qquad \sum_{k \equal{} 1}^{\infty}( \minus{} 1)^kf(k\pi) < \int\limits_{0}^{\infty}\!\! f(t)\cos t\: dt < \sum_{k \equal{} 0}^{\infty}( \minus{} 1)^kf(k\pi)\]

\[ (2)\qquad \sum_{k \equal{} 1}^{\infty}( \minus{} 1)^k f\left(k\pi \plus{} \frac {\pi}{2}\right) < \int\limits_{0}^{\infty} f(t)\sin t dt < f(0) \plus{} \sum_{k \equal{} 0}^{\infty}( \minus{} 1)^k f\left(k\pi \plus{} \frac {\pi}{2}\right)\]
[/quote]
I think there must be some assumption about monotonicity. Otherwise you can take something like

$ f\equiv0$ (almost) except for a peak around $ x \equal{} 2\pi$, e.g. $ f(x)\equal{}e^{\minus{}k(x\minus{}2\pi)^2} , k$ big enough

to see that it doesn't hold.

But if $ f$ is monotonously decreasing, the ineqs are easy to prove by looking at the positive and negative parts of the integrands and partial integration.

this counterexample isn't good because $ f$ should have only positive values  :wink:

please, show one proof (with f  decreasing)
if we take the positive and negative parts of improper integral we first need to know if $ \int_0^{\infty} f(t)|\cos t|dt$ converges, is it trivial ? I don't see  :(

sorry I meant "almost" not in the sense "almost everywhere" but vaguely "almost 0" = very small, like the example I have given.

And by positive and negative parts, I mean cutting the integral (let's take the one with $ \sin$ for convenience) like $ \lim_{N \to\infty}\sum_{k \equal{} 1}^{N}\int^{k\pi}_{(k \minus{} 1)\pi}$. By monotonicity, we have $ \left|\int^{k\pi}_{(k \minus{} 1)\pi}f(t)\sin t dt\right|\ge\left |\int^{(k \plus{} 1)\pi}_{k\pi}f(t)\sin t dt\right|$ (perform $ t\mapsto t \minus{} \pi$), so the limit of this alternating sum exists.

Note that $ \int_{0}^{\infty}f(t)|\sin t|dt$ need not converge.


inequalities
induction
inequalities unsolved

[quote="prowler"]This is a famous Hardy inequality:
1/a1+2/(a1+a2)+...+n/(a1++an)< 2(1/a1 + 1/a2 + 1/an)[/quote]

this problem is in KVANT M999,I think :P

In this problem we can't replace 2 by c<2.Indeed,we only choose $a_i=i$.And I think this problem is not strange. There are a lot of problems which this problem is as similiar as these one.
1) Prove that$\forall a_1,a_2,...,a_n \in R $we have:
                    $\sum_{k=1}^n (\frac{a_1+a_2+....+a_k}{k})^2 \leq4 \sum_{i=1}^n a_i^2$.We can't replace 4 by c<4
The consequence of this problem is problem 4/USAMO1994
2)Prove that$\forall a_1,a_2,...,a_n>0 $we have:
                   $\sum_{k=1}^n( a_1a_2....a_k)^{\frac{1}{k}}\leq e\sum_{k=1}^n a_i$. We can't replace e by c<e
Source:France,1982
3)Prove that$\forall a_1,a_2,...,a_n \in R $we have:
    $\sum_{k=1}^n(a_1+a_2+....+a_n)^2\leq \frac{1}{4(sin\frac{\pi}{4n+2})^2}\sum_{i=1}^n a_i^2$.We can't replace$ t=\frac{1}{4(sin\frac{\pi}{4n+2})^2}$ by $c<t$
Source:Mathematical and Youth T10\309

[quote="prowler"]
 -[(n^2)/2]/x + (n+1)(n+3)/(x+y) < 4/y (x,y > 0)

or after opening parenthesis : (ny-2x)^2 + xy > 0 c.t.d[/quote]

I think that you have some mistakes because these two inequality are not equivalent.

Here are my calculations darkmaster:

-[n2/2]/x + (n+1)(n+3)/(x+y) < 4/y (x,y > 0)

 is equalent to
n2(x+y)y + 4x(x+y)- (n2+4n+3)xy >0

n2y2 +4x2 (4n-1)xy > 0
or
(ny  2x)2+xy>0

I twice saw this inequality on froum. Inequality with constant 4 was proposed on All-Soviet Union Olympiad.

[quote="prowler"]Here are my calculations darkmaster:

-[n2/2]/x + (n+1)(n+3)/(x+y) < 4/y (x,y > 0)

 is equalent to
n2(x+y)y + 4x(x+y)- (n2+4n+3)xy >0 [/quote]
You forgot the factor (1/2) , the first inequality is equivalent to:
    $ 4x(x+y)+\frac12n^2y(x+y)-(n^2+4n+3)xy > 0$
 and it doesn't lead to the answer :?  :?

Similar questions are being discussed in http://www.mathlinks.ro/viewtopic.php?p=124961


geometry
trigonometry
trig identities
Law of Cosines

is there a way to proove heron's formula? im having a little trouble.

Of course there's a way, then how would we know it was true?

[url]http://jwilson.coe.uga.edu/EMT668/EMAT6680.2000/Umberger/MATH7200/HeronFormulaProject/GeometricProof/geoproof.html[/url]

[url]http://en.wikipedia.org/wiki/Heron's_formula#Proof[/url]

There is also a proof in which we can drop an altitude and use Pythagoras, which can be found in Introduction to Geometry, chapter 6.

It's instructive to derive it yourself. Use the formula $ [ABC] \equal{}\frac{1}{2}ab\sin{C}$ and the Law of Cosines (with angle $ C$ of course).


can some one tell me if dna is water or fat soluable?? please explain... cause i need to knwo what will happen when dna is mixed with water,..

..and also there is one more question if u can answer ......"What happened when the isopropanol was added to the liquid that passed through the filter? What does this tell you about the behaviour of DNA?"------ i knwo when we mix it the dna begins to appear and floats to the surface... but what is its behavious?

thank you-------
- titan

[quote="saturnlife"]I'm not sure, but I'm guessing that DNA would be soluble in water since DNA has a good deal of H-bonds (so I'm thinking that it's polar)[/quote]Yes, DNA is soluble in water.

I noticed that titan posted this in both biology and chemistry forums...  :?

The second part of the question that remains unanswered (I don't know), "Is DNA is also soluble in fat?"

I suspect that it acts as a soap and is soluable in both, forming mycells, with the polar end out in water and polar ends in in fat.

[quote="wizard2378"]I suspect that it acts as a soap and is soluable in both, forming mycells, with [b]the polar end out in water and polar ends in in fat[/b].[/quote]

Could you explain that a bit more?  I understand the usual mechanism for fatty carboxylate soaps to form mycells, but where is DNA sufficiently nonpolar to do that?  (I'm assuming that "[b]polar [/b]ends in [] fat" is a typo.)

how can DNA be  is water soluable? in our cells there is a lot of water! i think that outside (P,des.) isnt and inside (A,T,G,C) it is

The outside is much more water "soluble" than the inside. The sugar (deoxyribose) groups are highly polar, like all sugars, and the phosphate groups that link together the sugar are partly ionized. (Remember that the A in DNA stands for "acid".) The entire DNA molecule, in water, carries a considerable negative charge. It must be accompanied by positively charged compounds (polyamines and basic proteins) in order to be reasonably stable.

On the other hand, for a molecule the size of DNA, "solubility" doesn't mean that much by iteslf. The massively long-chain polymer does not disassociate in water - it remains a molecule of incredible size. But its stability and shape is partly maintained by the aqueous environment around it.

And no, DNA is nothing at all like a soap or detergent.


ratio
algebra
polynomial
function
invariant
geometric sequence
binomial coefficients

If $x+y = xy = 9$, compute the remainder when $x^{2007}+y^{2007}$ is divided by $1000$.

[hide="can someone help me continue?"]
i solved for x and y and I get

$\left(\frac{18}{9\pm3\sqrt{5}}\right)^{2007}+\left(\frac{9\pm3\sqrt{5}}{2}\right)^{2007}$

and im not sure how to continue

thanks
[/hide]

[hide="Big hint for everyone"]

Let f(n)=x^n+y^n
note that

(x+y)(x^(n-1)+y^(n-1))=x^n+y^n+xy(x^(n-2)+y^(n-2))

rearanging, we get:

9(f(n-1)-f(n-2))=f(n)

So starting with f(1)=9, f(0)=x^0+y^0=2, we get a squence. Now just find the last 3 digits of f(2007).
[/hide]

@QuattoMaster

Whatever happened to binomial coefficients?
Those appeared to have disappeared in your expansions

I had the same thing you did, but my summation involved binomial coefficients, making it much harder to compute

I had:
[hide]
$\left(\frac32\right)^{2007}2\left(\sum_{n=0}^{1003}\binom{2007}{2n+1}3^{2n+1}5^{1003-n}\right)$

which is the same thing, with a 3 factored out of the $9\pm3\sqrt5$ and with binomial coefficients
[/hide]

[quote="chickendude"]@QuattoMaster

Whatever happened to binomial coefficients?
Those appeared to have disappeared in your expansions

I had the same thing you did, but my summation involved binomial coefficients, making it much harder to compute

I had:
[hide]
$\left(\frac32\right)^{2}007 2\left(\sum_{n=0}^{1003}\binom{2007}{2n+1}3^{2n+1}5^{1003-n}\right)$

which is the same thing, with a 3 factored out of the $9\pm3\sqrt5$ and with binomial coefficients
[/hide][/quote]
Oh!  I forgot about those!   :blush:

Hmm since everybody reposted the things I had in my post....

Can miyomiyo please post? I think we have just been citing all the common tricks for this kind of problem but the exponent is just too large. I have a feeling that there is no nice none computer heavy solution (well I have a couple ideas that work for calculating it by hand but its still pretty long and messy) and that miyomiyo just made it up for random numbers... (well not random but you know what I mean)

Note the recursions :wink: it is rational ... and an integer ... just another beauty of mathematics.

[b]me@home:[/b] It is true that I don't know the answer to the question, but I saw that it was possible to get $x^{n}+y^{n}$ for any integer $n$, so I chose $2007$.  What do you think I should do to make the problem simpler?

[quote="mathgeniuse^ln(x)"]so 561[/quote]
No. 766.

[quote="rgmds"]Is $\left(\frac{9\pm3\sqrt5}{2}\right)^{2007}+\left(\frac{18}{9\pm3\sqrt5}\right)^{2007}$ not irrational? How do you find the remainder of an irrational divided by $1000$?[/quote]

It's a lot easier to see why this expansion is rational once you realize that

$\frac{9 \pm 3 \sqrt{5}}{2}= \frac{18}{9 \mp 3 \sqrt{5}}$

In other words, once you expand both binomials the irrational parts cancel exactly.  It's a little harder to see that the expansion is integer, but that's what the recursions are for.

so what is an answer for this? expand into series and solve for the sums and then use mods to compute the remainder?

The actual answer is $766$.  It's just a matter of getting it without a calculator now.

[quote="GlenthemanN"]no calculator?!  :ninja:[/quote]
That would trivialize the problem, just plug and chug into a graphing

Let $S_{n+1}=p, S_{n}=q$. We want to find a (polynomial) function such that $f(72p-81q,9p-9q)=f(p,q)$. It's easy to see that there is no degree 1 polynomial that can possibly satisfy this. Furthermore, we can assume wlog that the constant term is zero since it cancels anyway. All second degree polynomials can then be written as $ap^{2}+bq^{2}+cpq+dp+eq$. Thus \[a(72p-81q)^{2}+b(9p-9q)^{2}+c(72p-81q)(9p-9q)+d(72p-81q)+e(9p-9q)=ap^{2}+bq^{2}+cpq+dp+eq\]
Solve that system and congrats you have found out $f(S_{2}007,S_{2}006)$... and with a little more trickery you can find $S_{2}007$ I think.

[quote="me@home"]Let $S_{n+1}=p, S_{n}=q$. We want to find a (polynomial) function such that $f(72p-81q,9p-9q)=f(p,q)$.[/quote]
I don't see where this comes or the recursion you gave earlier. Newton sums give a recursion formula that will solve the problem but to solve it you have to solve a system of 2007 equations.

Just looking for an invariant to help compute the final term.


calculus
derivative
trigonometry
calculus computations

Hello everyone:

I am having trouble with the finding the second derivative of this question as I am self-studying calculus. Any help or hints are appreciated.

Thank you!

---

1. Given the parametric equations $ x \equal{} \cos \theta$ and $ y \equal{} 2 \theta$, determine $ \frac {dy}{dx}$ and $ \frac {d^2y}{dx^2}$.

---

I had no trouble finding the first derivative, $ \minus{} 2 \csc \theta$, so I will not show my work for that.

For the second derivative:

$ \frac {d^2y}{dx^2} \equal{} \frac {\frac {d}{d \theta} \frac {dy}{dx}}{\frac {dx}{d \theta}}$

$ \frac {d}{d \theta} \minus{} 2 \csc \theta \equal{} 2 \cot \theta \csc \theta$

Applying the Product Rule:

$ n'(x) \equal{} \cot \theta( \minus{} \csc \theta \cot \theta) \plus{} \minus{} \csc^2 \theta (\csc \theta)$

$ n'(x) \equal{} \minus{} \csc \theta \cot^2 \theta \minus{} \csc^3 x$

Therefore:

$ \frac {d^2y}{dx^2} \equal{} \frac { \minus{} \csc \theta \cot^2 \theta \minus{} \csc^3 x}{ \minus{} \sin \theta}$

However, the answer provided is [hide]$ \minus{} \frac {2 \cos \theta}{\sin^3 \theta} \minus{} 2 \csc^2 \theta \cot \theta$[/hide].

For convenience let's assume you're dealing with some curve C given by $ x\equal{}x(t)$ and $ y\equal{}y(t)$. Now, the first part's easy; no more difficult than $ \frac{dy}{dx}\equal{}\frac{y^{\prime}(t)}{x^{\prime}(t)}$. Heres the kicker though, that formula can be a little bit misleading; you differentiate $ dy/dx$ with respect to $ t$ and divide that by $ dx/dt$ again. That's your second derivative.

So with $ x\equal{}\cos{t}$, $ y\equal{}2t$.... $ dy/dx\equal{}\minus{}\frac{2}{\sin{t}}\equal{}\minus{}2\csc{t}$.
Differentiate with respect to t and divide by $ \minus{}\sin{t}$ once more to get $ \frac{d^2y}{dx^2}\equal{}\minus{}2\csc^2{t}\cot{t}$

Thanks for your reply, igiul.

I have two further questions:

1. The answer is [hide]$ \minus{} \frac {2 \cos \theta}{\sin^3 \theta} \minus{} 2 \csc^2 \theta \cot \theta$[/hide]. Could it be shown as to how to get the first fraction?

2. My second question is how your final answer is negative, and I may have overlooked something. Here's what I've done:

$ \frac {d}{dt}\left(\cfrac{dy}{dx}\right) for \minus{} 2\csc\theta$

$ f(x) \equal{} \minus{} 2$, $ f'(x) \equal{} 0$

$ g(x) \equal{} \csc \theta$, $ g'(x) \equal{} \minus{} \csc \theta \cot \theta$

Applying the chain rule:

$ \frac {d}{dt}\left(\frac {dy}{dx}\right) \equal{} \minus{} 2( \minus{} \csc \theta \cot \theta)$

Therefore:

$ \frac {d^2y}{dx^2} \equal{} 2 \csc^2 \theta \cot \theta$

Yes, the first fraction is found by breaking down the second into sines and cosines.

$ \frac{dy}{dx}\equal{}\minus{}2\csc{\theta}\Rightarrow \frac{d}{d\theta}\bigg(\frac{dy}{dx}\bigg)\equal{}\minus{}2\cdot\minus{}\cot{\theta}\csc{\theta}\equal{}2\csc{\theta}\cot{\theta}$

Now we divide this new expression by $ \frac{dx}{d\theta}\equal{}\minus{}\sin{\theta}$ to yield $ \frac{d^2y}{dx^2}\equal{}\minus{}2\csc^2{\theta}\cot{\theta}$

Thanks for your reply once again, igiul.

Would you mind showing the steps to arriving at the first fraction?


I am having a hard time with these problems and figuring out where to get started.  I realize that there are two steps but getting the first step has me baffled.

Here's an example:  1 * 2 + 2 * 3 + 3 * 4 + . . .  + n( n+ 1) = n(n + 1)(n + 2)  / 3

I think the Basis step is 1*1 = 1 but I doubt I'm right.

Any help would be greatly appreciated.

The "basis step" is what you would have for the smallest $n$ for which this makes sense. In this case, that would be $n=1.$ Note that the last term of the sum is $n(n+1).$ The "basis step" would be $1\cdot 2=\frac{1\cdot2\cdot3}3,$ which is true.

Also, I'm moving this to a different area of the site.

What about posting this in a lower section? This is not what pre-olympiad is for. ;)

Bucs44: I've been advised that future questions of this sort should be posted in "intermdiate topics." But this one is here for now.

[quote="You"]What about the Inductive step?

1*2*3+n(n+1) = n(n + 1)(n + 2) / 3[/quote]
Does that statement mean anything to you? It has no meaning at all to me. It's just symbols stuck together haphazardly.

I think you need a tutor, and you need to talk to the instructor of your course. I have my doubts that we're going to be able to give you the help you need. But to give it a try ...

The structure of the proof goes like this:

Make the assumption that the $n$th statement is true.

By the $n$th statment, I mean precisely

$1\cdot 2+2\cdot3+\cdots+n(n+1)=\frac{n(n+1)(n+2)}3$

Now, do you understand what that means? In particular, do you understand "$\cdots$"? I'm not going to take your word that you understand. Please write this out in full for the particular cases $n=3$ and $n=4.$

(I think I'll wait for your reply before going any further.)


limit
real analysis
real analysis unsolved

$ \Gamma(z) \equal{}\lim_{m\to\infty}\frac{m!\;\; m^{z}}{z (z\plus{}1)\cdots (z\plus{}m)}$ per definition.

$ \Gamma\left(z\plus{}\frac{k}{n}\right) \equal{}\lim_{m\to\infty}\frac{m!\;\; m^{z\plus{}\frac{k}{n}\minus{}1}}{\left(z\plus{}\frac{k}{n}\right)\,\left(z\plus{}\frac{k}{n}\plus{}1\right)\cdots\left(z\plus{}\frac{k}{n}\plus{}m\minus{}1\right)}$

After replacing $ m!$ by $ \sqrt{2\pi m}\left(\frac{m}{e}\right)^{m}$ and extending the fraction by $ n^{m}$

$ \Gamma\left(z\plus{}\frac{k}{n}\right) \equal{}\lim_{m\to\infty}\frac{\sqrt{2\pi}\left(\frac{mn}{e}\right)^{m}m^{z\plus{}\frac{k}{n}\minus{}\frac{1}{2}}}{(nz\plus{}k) (nz\plus{}k\plus{}n)\cdots (nz\plus{}k\plus{}n(m\minus{}1))}$

$ \Rightarrow n^{nz\minus{}\frac{1}{2}}\prod_{k \equal{} 0}^{n\minus{}1}\Gamma\left(z\plus{}\frac{k}{n}\right) \equal{}\lim_{m\to\infty}\frac{\sqrt{2\pi}^{\,n}\left(\frac{mn}{e}\right)^{mn}(mn)^{nz\minus{}\frac{1}{2}}}{nz (nz\plus{}1)\cdots (nz\plus{}mn\minus{}1)}$

$ \equal{}\sqrt{2\pi}^{\, n\minus{}1}\lim_{m\to\infty}\frac{\sqrt{2\pi}\left(\frac{m}{e}\right)^{m}m^{nz\minus{}\frac{1}2}}{nz (nz\plus{}1)\cdots (nz\plus{}m\minus{}1)}\equal{}\sqrt{2\pi}^{\, n\minus{}1}\Gamma(nz)$


calculus
integration
function
trigonometry
logarithms
limit
calculus computations

I try to do homework and the improper integration got me stucked.

I tried my best but I am not sure wheter this is a right solution or not.

#1. integral (upper/infinite,lower/0) (x*arctan(x))/((1+x^2)^2) dx

 I used the integration by parts and set u=arctan x dv= x/((1+x^2)^2),and I concluded it as divergent. Since I have a slight idea, if someone could give me a detailed solution, thank you very much.

#2. integral (2, 0) z^2*ln(z) dz

 I also used the integration by parts and set u=ln(z) and dv=z^2dz and I got the answer which is (8/3)ln2 - (8/9)

 I am a bit confused about "uv" evaluation, which is supposed to be ln(z)*(1/3)*z^3 evaluating from 0 to 2. A detailed solution would be very welcome.

#3. determine convergent/divergent by comparasion theorem

 integral (infinite, 1) 1/(x+e^(2x))dx

 For this problem, I honestly have no idea how to approach.

Sorry for the mess but I don't know how to picture the integral nicely. Again, Thank you for helping me

I've moved this here, where it belongs - Calculus Computations and Tutorials.

1. $\int_0^{\infty}\frac{x\arctan x}{(1+x^2)^2}\,dx$

What does the integrand look like for large $x$? Note that $\arctan x$ tends to a constant (the constant being $\frac{\pi}2.$) Hence, for large $x,$ the whole function looks like $\frac{C}{x^3}.$ Hence it is limit comparable to a convergent integral, and converges.

If you want to calculate it, see how far the substitution $u=\arctan x,\,x=\tan u$ takes you.

2. $\int_0^2z^2\ln x\,dz$

You've got it right. You're worried about this:

$\left.\frac13z^3\ln z\right|_0^2.$

The lower evaluation is really a limit:

$\lim_{z\to0^+}\frac13z^3\ln z$

Think this: yes, the logarithm does go to $-\infty,$ but slower than any power of $z.$ That $z^3$ overwhelms the logarithm, and the limit is zero.

3. $\int_1^{\infty}\frac1{x+e^{2x}}\,dx$

For large $x,$ that $x$ in the denominator makes very little difference. Compare to $\int_?^{\infty}e^{-2x}\,dx.$ It converges.

Thank you for your help
I'm sorry but could you be more specific about the calculation of number 1?

Thank you, again!  :blush:

Let $\theta=\arctan x$ so that $x=\tan\theta.$

We have $d\theta=\frac{dx}{1+x^2}$ and $1+x^2=\sec^2\theta.$

As $x$ goes from $0$ to $\infty,$ $\theta$ goes from $0$ to $\frac{\pi}2.$

So the integral becomes

$\int_0^{\frac{\pi}2}\frac{\tan\theta\cdot\theta}{\sec^2\theta} \,d\theta = \int_0^{\frac{\pi}2}\theta\sin\theta\cos\theta\, d\theta$

$=\frac12\int_0^{\frac{\pi}2}\theta\sin(2\theta)\,d\theta$

Finish it from there.


USAMO scores are up guys.
[url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2007-ua/2007usamoscores.shtml]here[/url]

311 got an 11.

515 got a 12 :). Only 64 people away from MOP. Must. Work. On. Rigor.

282 got a 14.  YES!!! highest score so far on this thread.  unfortunately, the sample size is like 3 which makes it completely invalid.  meaning no blue MOP.

wow - last year red mop cutoff was 9, i made 7.
wow - this year blue mop cutoff was 17, i made 14.

sad.

I got 17, which makes the blue mop cutoff, but that's irrelevant since I'm a senior. At least I was pleasantly surprised at how well I did (compared to how badly I thought I did right at the end of the USAMO).

Depends on if they knock me down to red or not.

an 8th grader in blue, whos not even in nat mc...
wow...

not everyone does mathcounts

I met at least one boy in blue MOP last year who had never heard of it

I am one point below the blue MOP cutoff. :o

[quote="Hamster1800"]340 got 18

Highest freshman, since no freshmen got HM/winner.[/quote]

!!OMG!! delong was right!   :lol:
and i failed the usamo.....

yeah dude seriously? mathcounts sucks

perhaps he is in blue MOP [i]because[/i] he is not in mathcounts

[i]perhaps[/i]

I'm caught between feeling happy for my large increase from last year and ridiculously sad because I missed the cutoff for Blue MOP by 1 point. Honestly though, I never expected I would ever have a chance at doing so well. and now I'm so close...

darn... 187, got a 15

No MOP for me this year =( looks like my solution to number 2 got owned. Darn. I was hoping for another trip to Nebraska ...

Oh well, congratulations to all the winners and HMs! (I don't mean to be rude or anything to the other winners, but I want to shout out a big gratz to Delong and Eric, for I think they were both really good last year and deserve the win this year :D )

23, not so bad for my first USAMO  :D  .  Although I probably could have written enough for another point or two if I had more time for #2, I'm glad I wrote what i did.  But there were 5 22's...I feel bad for those people...Well I'm glad I beat what 416 got last year.

1 point away from mop last year, 4 points this year! I predict 7 points away next year. :/

8.. have to work really hard  :wink:
where do  you find the cut offs for MOPS

Try not giving up in frustration and going outside. ;)

Seriously, just keep hacking at it until you see it. Eventually, you'll get better at solving them. A good idea is to take a lot of #1s, order them in any order you want, and go through them [i]without skipping around[/i]. If you spend little chunks of time on the USAMO on each problem, you'll get nothing done. Try to get one done without skipping to another one.

hahahahaha

School's can provide random interference.  This year a custodian tried to kick me out of a testing room, and then I think someone threw a party in the stairwell right next to the room... it was rather distracting.

Heh, skipping around is exactly how I always do olympiad problems, and it worked amazingly well for me this time. Of course I also just got lucky, but for me poking at a problem a little, wandering off, and coming back later makes a lot of things pop into my mind that would otherwise be blocked by my preconceptions of which ideas look "useful".

[quote="aspiringmathlete"]Hmmm... first time for me (8th grade)

I scored 9.  Not bad, but #1 was a really easy problem.  I guess my proof for 4 wasn't rigorous enough...

Does anyone know a method for getting better at USAMO problems?  Besides giving up in frustration and going outside?[/quote]

Meh, you'll get better naturally. 9 is a strong score for an 8th grader and if you maintain or improve your skill level you'll almost definitely get into Red MOP.

If you find yourself getting frustrated when you try to sit down and straight-up solve problems (I know I do), just have a couple of unsolved USAMO problems in your pocket at any given time and take a look at them when you're bored—doing that for the past year significantly improved my problem solving abilities.

On the second day, I actually sat with #4 for about two hours before moving on.  Then I came back and got it in like 15 minutes...

I spent 3 hours on number 1, and 3.5 hours on number 4.
Go me?

3 hrs on #1.
5 seconds on #2("No. QED")
20 min on #3(i assumed random things were trivial)
40 min on #4.
3 hours on #5(which i only got up to the factoring)
10 min? decided to prove equality case for #6(never got around to finishing)

100 hours spent looking at the ceiling

What were the highest middle school scores?

[quote="K81o7"]But there were 5 22's...I feel bad for those people...[/quote]

Me too, Krishanu.  :(

But I suppose I should actually feel lucky for doing as well as I did this year.  I was definitely not expecting more than 21.

[quote="SuperThread"]Evan O'Dorney got an HM and he's in eighth grade, so clearly the highest middle school score is between 19 and 22...

Doesn't look like there were any other middle schoolers who made it to MOP, though.[/quote]

Hence the "s" after the end of "score" in my original post.

[quote="dhthstn"]so in individual scoring list i see one got 32;one got 28 and others winner got about 23-26
does anyone know who got 32 point?[/quote]

If I'm not mistaken, there are two 28's.  As far as I know, Lawrence got the 32, and the two 28's were Zhai and Gong.

While we're at it, how about freshmen? Hamster got 18, who else was close to that?

OK, as of right now and until further notice:

[b]Do not discuss other people's scores without their explicit permission.[/b]

The AMC does not match up scores to names for a reason: to protect students' privacy.  They only release the exact score for one person, namely the person who scored the highest.  (Well, actually, they give prizes for the top 3, so 2nd and 3rd are deducible too, so I've left that information in; and of course you can deduce people's score range based on if they are a winner or HM.)

I am going to remove all the posts in this thread that refer to other people's scores, plus a few other posts that are way off topic and/or spammy.


superior algebra
superior algebra unsolved

Let $ F$ be the quotien field of $ \mathbb{C}[x_1,x_2,x_3,x_4,x_5]/<x_1^2\plus{}x_2^2\plus{}x_3^2\minus{}1>.$
Exhibit a transcendence base $ z_1,...,z_d$ for $ F$ over $ \mathbb{C}$, and express $ F$ explicitly as an algebraic extension over $ \mathbb{C}(z_1,...,z_d)$.


Let $ o(n)$ denote the greatest odd divisor of $ n$.

Find the value of $ \sum_{n \equal{} 1}^{\infty}\frac{o(n)}{n^{3}}$.

[hide="Answer"]$ \frac{\pi^{2}}{7}$[/hide]

Let $ n = o(n)\cdot 2^{p(n)}$ such that $ o(n)$ is odd integer and $ p(n)$ is integer. Then

\begin{eqnarray*}\sum_{n = 1}^{\infty}\frac{o(n)}{n^{3}}& = &\sum_{n = 1}^{\infty}\frac{1}{o(n)^{2}\cdot 2^{3p(n)}}\\ & = &\sum_{i=0}^{\infty}\sum_{j = 1, 3,\cdots}^{\infty}\frac{1}{j^{2}\cdot 8^{i}}\\ & = &\frac{8}{7}\cdot\frac{3\zeta(2)}{4}\\ & = &\frac{\pi^{2}}{7}\end{eqnarray*}


Let $ A \equal{} 4, B \equal{} 5, C \equal{} 1$

$ C \equal{} B \minus{} A$
$ C(B \minus{} A) \equal{} (B \minus{} A)^2$
$ CB \minus{} CA \equal{} B^2 \minus{} 2AB \plus{} A^2$
$ CB \minus{} CA \minus{} A^2 \equal{} B^2 \minus{} 2AB$
$ AB \plus{} CB \minus{} CA \minus{} A^2 \equal{} B^2 \minus{} AB$
$ AB \minus{} CA \minus{} A^2 \equal{} B^2 \minus{} CB \minus{} AB$
$ A(B \minus{} C \minus{} A) \equal{} B(B \minus{} C \minus{} A)$
$ A \equal{} B$

$ 4 \equal{} 5$

Like most flawed proofs its a divide by zero (can you see it?)

The $ A(B \minus{} C \minus{} A) \equal{} B(B \minus{} C \minus{} A)$ is true because anything multiplied by $ 0$ equals $ 0$. $ 0\cdot K \equal{} 0\cdot C$ doesn't mean $ C \equal{} K$. You would have to cancel out the zeros by division.

[quote="SimonM"][quote="ZhangPeijin"]You divide by 0...
 :o[/quote]

I already said that... Now this is hypocritical, but please stop adding useless things to threads[/quote]
You don't have to be so critical...
And explain, how your post may be less useless than mine.
Seeing as there's something called a PM.

Yes, and I don't wish to say this publicly, but honestly, you're last like 10 posts were not productive in any sense.

Ok, maybe that is hypocritical, but you should maybe stop soon


Find all rational numbers $r$ such that \[r+\frac1r\] is an integer.

[hide="Easy"]Write $r = \frac{a}{b}$, $a,b \in \mathbb{Z^{*}}$. We get $ab | a^{2}+b^{2}$. Let $p$ be a prime factor of $a$ or $b$. Denote by $\alpha, \beta$ the exponents of $p$ in $a,b$ respectively. If $\alpha \le \beta$, then $p^{\alpha+\beta}| p^{2\beta}| b^{2}$, however $p^{\alpha+\beta}| ab | a^{2}+b^{2}$, so $p^{\alpha+\beta}| a^{2}$, and so $p^{\alpha+\beta}| p^{2\alpha}$, so $\beta \le \alpha$. If $\beta \le \alpha$ we get similarly $\alpha \le \beta$, so we have $\alpha = \beta$ for any prime $p$, hence $a = \pm b$. Therefore $r = \pm 1$.[/hide]

[hide="more easy"]
We want to get $r+\frac{1}{r}=n$, where n is some integer.
Multiply by r on both sides to get $r^{2}-nr+1=0$.
The determinant of the quadratic determines whether we can find integer r.  That happens if and only if the determinant is a perfect square.
The determinant is $n^{2}-4$.  The only times this is a square is when n is 2 or -2.
So substitute these values of n back into the original equation to get $r=1,-1$[/hide]

nice, circlesquared! simple, yet satisfying!
It's hard to find solutions like that.  :)

yes, thank you. both of these are my soutions too. but i prefer the first: the pure number theoretic one. but it can be more simple. we may let $r=\frac{a}{b}$ where $\gcd(a,b)=1$. we get $a^{2}+b^{2}=kab$ for some $k\in \mathbb N$. but this implies $a|b$ and $b|a$. thus $a=\pm b$ and $r=\pm 1$.

Not to beat a dead horsequation, but  just to remind that this is a direct consequence of  "rational roots" theorem.

For example see: [url]http://mathworld.wolfram.com/RationalZeroTheorem.html[/url]

In fact  only integer values of $r^{m}\pm \frac{1}{r^{n}}$ is when $r=\pm 1$ (assuming $r$ is rational, amd both (m,n) are positive integers)

Or in fact  [i]only [/i]possible rational  roots  of
 $x^{n}+a x^{n-1}+b x^{n-2}+\dots \pm 1 = 0$
are $\pm 1$

And it is easy to prove too.. for example the original problem can be proved as:

(Essentially the same argument as the one shown in the above post, but perhaps easy to generalize for a  higher degree
polynomial)
[hide]Assume $r=p/q$ where $p$ and $q$ have no common factors.
Then
$(p/q)+(q/p) = n \implies p^{2}-npq =-q^{2}\implies p(p-nq) = q^{2}\implies p = \pm 1$
(Since p and q has no common factors)
Similarly one can show $q=\pm 1$ [/hide]

to clarify a quick jump in nayel's logic, we have $a^{2}+b^{2}= kab \Leftrightarrow a^{2}= b\left(ka-b\right)$, hence $b \mid a^{2}$.  since $\gcd(a,b) = 1$ we have $b \mid a$ and similarly $a \mid b$


inequalities
integration
trigonometry
logarithms
calculus
calculus computations

Let $f: R\to R, f(x)=(sinx)/x$, for $x \not=0$ and $f(0)=1$
Prove that $\int_{0}^{\pi/2}f(x)dx <1+cos1.$

Since $f(x)\le 1$, it follows that $\int_{0}^{\pi/2}f(x)dx \le \pi/2$, which is tantalizingly close to the estimate that you want... Second try: $\int_{0}^{\pi/2}f(x)dx=\int_{0}^{1}\frac{\sin x}{x}dx+\int_{1}^{\pi/2}\frac{\sin x}{x}dx\le 1+\int_{1}^{\pi/2}\frac{1}{x}dx=1+\ln(\pi/2)$. This is actually smaller than $1+\cos 1$.  :)

Or, to get the result exactly as stated, just estimate $\frac{\sin x}{x}$ by $\sin x$ instead on $1/x$ on $[1,\pi/2]$.


Can someone prove that there exists an Eulerian circuit through a graph with n vertices all of degree at least n/2

This is not true, this graph includes a Hamiltonian cycle not Eulerian.

http://www.mathlinks.ro/viewtopic.php?t=239612

  darij


Am alcatuit o lista de inegalitati uzuale si cateva probleme de inegalitati, sper utile, in vederea pregatirii concursurilor.

[quote="cezar lupu"]Misto lista!
 Mai puteau fi bagate multe altele. ;)[/quote]
Sunt de acord cu tine.
O putem completa.

Inegalitatea mediilor generalizata este valabila pentru $p,q\in{\bf R}$, cu mentiunea ca media geometrica corespunde exponentului $0$(se obtine prin trecere la limita). In alta ordine de ider, cred ca era suficienta denumirea de inegalitatea lui Schur. Oricum initiativa este laudabila, chiar daca, asa cum scria si Cezar Lupu, lista se mai poate completa.

Oricum astfel de inegalitati si sub forma integrala pot fi gasite si in cartea mea scoasa de curand. :D


function
geometry
geometric transformation
reflection
algebra
domain

I need help finding the inverse function of y=sqr(25-x^2) because when I tried to solve it algebraically, the result is the function again. But graphically, if you reflect it on the y=x line, it is a different graph. What is happening here? Thanks.

[quote="21385"]But graphically, if you reflect it on the y=x line, it is a different graph. [/quote]

To the extent that it is a different graph, it should be a different cross section of the circle $ x^2 \plus{} y^2 \equal{} 25$.

The problem is that $ y \equal{} \sqrt {25 \minus{} x^2}$ does not have an inverse unless its domain is restricted to $ [0, 5]$ (and in that case the reflected graph will be the same).  For any $ y$, either $ x$ or $ \minus{} x$ will map to $ y$.  That is, $ \sqrt {25 \minus{} x^2}$ fails the [i]horizontal line test[/i], so it is not one-to-one.

[quote="t0rajir0u"]
The problem is that $ y \equal{} \sqrt {25 \minus{} x^2}$ does not have an inverse unless its domain is restricted to $ [0, 5)$ (and in that case the reflected graph will be the same, except for the endpoints). [/quote]

Shouldn't the domain be $ [\minus{}5,5]$?

[quote="CircleSquared"][quote="t0rajir0u"]
The problem is that $ y \equal{} \sqrt {25 \minus{} x^2}$ does not have an inverse unless its domain is restricted to $ [0, 5)$ (and in that case the reflected graph will be the same, except for the endpoints). [/quote]

Shouldn't the domain be $ [ \minus{} 5,5]$?[/quote]

If you want to make it a inverse function then domain is $ [0,5]$. Since for every $ y$ value only one $ x$ value.

But if you take the function in general then domain is $ [5,\minus{}5]$

The function is also injective on $ [\minus{}5,0]$, so you have two choices for inverse functions.


prove that $ (n!)!$ is divisible by $ (n!)^{(n\minus{}1)!}$

I think this has been posted many times  :D
[hide]
Think of a combinatoric proof of arranging things .
[/hide]

[hide="Hint for algebraic proof"]Prove the lemma that $ n!$ divides the product of any $ n$ consecutive positive integers.[/hide]


function
vector
real analysis
real analysis unsolved

Say that $ m(\Omega)<\infty$ and $ q<p$.  If $ (f_n)_n$ is a sequence of functions in $ L^p(\Omega)$ with $ f_n\rightharpoonup f$ in $ L^p(\Omega)$, then $ f_n\rightharpoonup f$ in $ L^q(\Omega)$.

Just as $ L^p$ is contained in $ L^q$, the dual of $ L^q$ is contained in the dual of $ L^p$. Since $ \langle g,f_n\rangle \to \langle g,f\rangle$ for any $ g\in (L^p)^*$, $ \langle g,f_n\rangle \to \langle g,f\rangle$ for any $ g\in (L^q)^*.$

This uses only the embedding, and would apply to any other pair of topological vector spaces, one embedded in the other.


Extremal combinatorics
combinatorics
counting
IMO Shortlist
IMO Longlist

A subset $S$ of the set of integers 0 - 99 is said to have property $A$ if it is impossible to fill a crossword-puzzle with 2 rows and 2 columns with numbers in $S$ (0 is written as 00, 1 as 01, and so on). Determine the maximal number of elements in the set $S$ with the property $A.$

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

The condition is equivalent to:

No digit may appear both as the first digit of some number and as the last digit of some number.

Let $x$ be the number of digits which occur only as the first digit. Let $y$ be the number which occur only as the last digit. The number of possibilities is then $xy$.
Now we have $x+y=10$, so $xy \leq \frac{1}{4}(x+y)^2 = 25$. Equality occurs when $x=y=5$.


Well, then we can set a new theme for discussion. The question I'm asking is - what are the official IMO solutions REALLY like? I mean, are they the most simple and most beautiful, or they just present the most often solution? Well, as I see, it depends of a proposer of the problem who usually sets the official solution too, but what do you think?

[quote="A. S. K."]Well, then we can set a new theme for discussion. The question I'm asking is - what are the official IMO solutions REAL like? I mean, are they the most simple and most beautiful, or they just present the most often solution? Well, as I see, it depends of a proposer of the problem who usually sets the official solution too, but what do you think?[/quote]

Maybe they first solve the problem, and then they try to show their solution nicer and shorter, in order to be more easy for non-experienced people.

Or, maybe, they solve a generalized problem, and then they make a problem with a special case, much easier to solve.


quadratics
linear algebra
matrix
induction
superior algebra
superior algebra solved

Let q be a quadratic form on R^n with the associated matrix Q = (q_{ik})_{i,k=1}^n and Q_m = (q_{ik})_{i,k=1}^m. Show that q is positive definite if and only if det(Q_m) > 0 for m=1,...,n.

Diagonalize Q to obtain a matrix with positive entries on the main diagonal and 0 in the rest of Q.
This can be done by induction.
q11>0, so you can substract the first line from the rest multiplied nicely to obtain qi1=0 for all i>1.
Do the same for the columns to get q1i=0 for i>1.
Now look at the matrix (qij)2 \leq i,j \leq n. It has the same properties as Q because
det(
1 0 0...0
0 a b...c
............
0 d e...f)=det(
a b...c
.........
d e...f), and because the operations we did on Q did'n change the det of any Qi.
So by induction on n you bring Q to a nice diagonal form where all the elements on the diagonal are positive.


group theory
superior algebra
superior algebra solved

Hi!

Thanks for the answers on my questions. I am working with the book "Introduction to the Galois Correspondence"  by M H Fenrick. Its my first course in algebra and therefore I ask for help to solve the exercises I dont manage to solve myself.

Prove that, if N is normal in G, then G/N is a simple group iff, whenever H is normal in G such that N is contained in H, then either H = N or H = G.

Hi,

Try to show H/N is normal in G/N, and if H <> N or G then you'll have shown G/N is not simple.


http://www.mathlinks.ro/Forum/viewtopic.php?t=36610
http://www.mathlinks.ro/Forum/viewtopic.php?t=29109
http://www.mathlinks.ro/Forum/viewtopic.php?t=86256

  darij


if$2 n+1$ ,$3 n+1$ are both perfect squares show that n is
divisible by $40$

[hide]
$2n+1=k^{2}$ (1)
$3n+1=m^{2}$ (2)

From (1), we can see that $k \equiv 1 \mod 2$.
Also, $2n=(k-1)(k+1)$
As $2$ divides the RHS twice, it must divide the LHS twice, so $n \equiv 0 \mod 2$.

From (2):
$(m-1)(m+1)=3n \equiv 0 \mod 2$
$(m-1)(m+1) \equiv 0 \mod 2$
So $m$ is uneven, and this $4$ divides $(m-1)(m+1)$ and from this $4$ divides $n$.

Also, $3n \equiv 0 \mod 4$. So $3n \equiv 0,4 \mod 8$.
$3n+1 \equiv 1,5 \mod 8$.
So $m^{2}\equiv 1,5 \mod 8$.
But $5$ is not a quadratic residue mod $8$, so $m^{2}\equiv 1 \mod 8$, and $3n \equiv 0 \mod 8$.

Finally, add (1) and (2):
$5n+2 = m^{2}+k^{2}$
$m^{2}+k^{2}\equiv 2 \mod 5$
As the only quadratic residues mod $5$ are $0,1,4$, $m^{2}\equiv k^{2}\equiv 1 \mod 5$. So $m,k \equiv 1,4 \mod 5$. So either $m+k$ or $m-k$ is divisible by $5$. Hence $5$ divides $n$.

$5$ and $8$ divide $n$, so $40$ divides $n$.

[/hide]

[quote="rjt"][hide]

Finally, add (1) and (2):
$5n+2 = m^{2}+k^{2}$
$m^{2}+k^{2}\equiv 2 \mod 5$
As the only quadratic residues mod $5$ are $0,1,4$, $m^{2}\equiv k^{2}\equiv 1 \mod 5$. So $m,k \equiv 1,4 \mod 5$. So either $m+k$ or $m-k$ is divisible by $5$. Hence $5$ divides $n$.

$5$ and $8$ divide $n$, so $40$ divides $n$.

[/hide][/quote]
I don't understand how you get $m^{2}+k^{2}\equiv 2 \mod 5$.

Ummm...it's quite clear that $m^{2}+k^{2}=5n+2$.


it should be $\infty$ since its a 2/3 power minus a 1/2 power

it's easy . but you must explain your solution ;)

when you approach infinity, the 1 becomes insignificant so,

$2^{\frac{1}{3}}*x^{\frac{2}{3}}-2^{\frac{1}{2}}*x^{\frac{1}{2}}$

and that approaches positive infinity because the $\frac{2}{3}$ power takes over since it is the greater power


Hi all, this is a question from the first chapter of Paul Zeitz's Art and Craft of Problem Solving, I'm sure many of you know it well. It wasn't addressed in his online hints section, so I was hoping someone here could at least give me a hint (Hint preferred actually).

Solve: cos(cos(cos(cos(x)))) = sin(sin(sin(sin(x))))

Thanks

It's been posted several times.  You'll gain a lot of insight by graphing both functions (make sure to set your calculator to radians).


function
algebra
polynomial
rational function

what is the 'dot' display on the graphing calculator for rational functions for? Rational function are functions that have one polynomial divided by another, for example:

$f(x)=\frac{x^2+2x-4}{3x+4}$

if you go to MODE on a graphing calculator, there are two choices about halfway down the screen: connected and dot. If u select dot and graph a rational function, then the screen shows dots where a continuous line would be in connected mode. What is the dot mode for?

thanks in advance...

The way the calculator usually draws lines is goes thru each x-pixel and finds the appropriate y-pixel it should use by plugging it into the function.  If you select dot, it merely fills in these pixels.  The default option, connected, draws a line from each point to the point with the x-pixel right next to it, making the set of lines look as close to the actual graph as possible.
There is no difference in how the graph would look, just different ways of drawing it to the screen.

so what is the 'use' of the dot mode? how would it help somebody? :|


advanced fields
advanced fields unsolved

Determine the smallest value for $ P$ so that every graph with $ N$ knots and property $ X_P$
is connected.
Property $ X_P$: Every knot is the root of a path of length $ P$.

Regardless of the exact interpretation, the worst disconnected case is the union of 2 complete graphs with $ n$ vertices (if $ N\equal{}2n$ or $ 2n\plus{}1$). The rest should be straightforward.

Sorry, I did not formulate the problem correctly (due to my lack of knowledge in this field).

Determine the smallest value for $ P( \equal{} P_N)$ so that every graph with $ N$  knots and property  $ X_P$
is connected.
Property $ X_P$: For every knot $ A$, there is a knot $ B$ such that distance$ (A,B) \le P$ .

The new version is just meaningless: $ \text{dist\,}(A,A)\equal{}0$, so the new condition holds in every graph (as long as $ P\ge 0$). You definitely meant something else...

Theorem:
If the minimal length of the longest minimal path is $ P$ (in a graph with $ N$ vertices), then the graph is connected?

What is $ P$?

(A minimal path is a path that represents the shortest distance between its endpoints.)


[quote="mathnovice"]Does anyone know how difficult it is to make a webpage?

Is MS FrontPage easy to learn to use?[/quote]

yes it does, frontpage is easy  :D

Also Dreamwaver is easy, but not that easy. You can also use Word to create web-pages, but it's best to learn XML (rather than HTML which is dying) to create exactly what you want. :)

I dunno... I always use php, working with templates (mainly copying the code from phpBB forum software and modifying it to fit my needs)

I recommend that one starts with a WYSIWYG editor that generates [b]clean[/b] code (such as Mozilla's webpage editor), and then read the code generated to figure out how it works. After one has a solid grip on the inner workings of HTML (or XML as one's case may be), one should start coding webpages in Notepad.

Learning server-side (PHP, ASP, Perl with CGI) and client-side (JavaScript, VBScript) scripting languages should come much later.

i think u should learn basic html codes before u go onto using frontpage or dreamweaver.. cuz nobody really uses codes if he wants to make complicated website.. too much to remember..


If $ x\leq1$, then

$ 1\plus{}96x^{14}\plus{}4x^{16}\geq 53x^{13}\plus{}48x^{15},$

with equality if and only if $ x\equal{}1$.

probably

what's a univariable?
i suck at vocab....

It means it's an inequality in only one variable.

Notice that the real question is to prove the inequality for $ x \in (0,1)$, since showing it for $ 0$, $ 1$ and negative $ x$ are all trivial.

[quote="Ji Chen"]If $ x\leq1$, then

$ 1 \plus{} 96x^{14} \plus{} 4x^{16}\geq 53x^{13} \plus{} 48x^{15},$

with equality if and only if $ x \equal{} 1$.[/quote][quote="TZF"]Notice that the real question is to prove the inequality for $ x \in (0,1)$, since showing it for $ 0$, $ 1$ and negative $ x$ are all trivial.[/quote]$ 1 \plus{} 96x^{14} \plus{} 4x^{16}\minus{} 53x^{13} \equal{} 48x^{15}\equiv f(x)\equal{}f\left(\frac{s}{s\plus{}1}\right) \equiv \frac{g(s)}{(s\plus{}1)^{16}},$

where $ g(s)\equal{}s^{15}\plus{}57s^{14}\plus{}507s^{13}\plus{}1820s^{12}\plus{}4368s^{11}\plus{}8008s^{10}\plus{}11440s^9$

$ \plus{}12870s^8\plus{}11440s^7\plus{}8008s^6\plus{}4368s^5\plus{}1820s^4\plus{}560s^3\plus{}120s^2\plus{}16s\plus{}1>0$

is clearly true for $ s\geq0$.


What is the value of the expression $ \minus{}7\plus{}3\times2^3\minus{}6$?

$-7 + 3 \times 8 -6$

$-7+24-6$

$17-6$

$11$

Thus, the answer is $\boxed{11}$


New Gorillaz video!

Very nice, but I still am more partial to the original (Clint Eastwood)

[url=http://www.clipland.com/Summary/701009103/]Feel Good Inc.[/url]

It looks really good. Hey --- this does look better than the Original ...Anyways, that's individual opinion.  ;)

Best. Band. Evar.

Feel Good, Inc. is used in the new iPod commercial.


okay heres the question:

Janelle averages 40 kilometers per hour biking on level ground. She averages 60% of her level-ground speed riding uphill, and she averages 120% of her level-ground speed riding downhill. The course is level for 5 Kilometers, uphill for 6 Kilometers, and then downhill for 6 Kilometers. How many kilometers per hour is her average speed for the entire course?

[quote="Quickster94"]This is a more complicated version of a d=rt problem.  Find the rates for up and down hill, and then the total time for everything.  So, u can find the avg speed.  I get 34 kmph i think...[/quote]

The only thing that could be wrong with that method was that since it took her a different amount of time riding each path (for example, going uphill would take more time than downhill) you can't just add the speeds and divide by the time, you would have to multiply each speed by the fraction of time she went that fast and then add them up. Its similar to how grades are weighted (such as tests are 50%, homework is 5%).

There was a Math Olympiads that I took in 4th grade (3 years ago) that had a similar problem.

EDIT: By Math Olympiads I mean MMO

For this problem, I used the method of:

$ \text{Average Speed} \equal{} \frac{\text{Total Distance}}{\text{Total Time}}$

Therefore, I got $ \frac{17 \text{mi}}{\frac{1}{2} \text{hr}} \equal{} \boxed{34 \text{mph}}$

[hide]For the ground, she travels it in $ \frac{60}{8}$ minutes. For uphill, she travels in $ \frac{60}{4}$ minutes, and for downhill, she travels in $ \frac{60}{8}$ minutes. So, $ \frac{17}{\frac{1}{2}}\equal{}34$.[/hide]


1) Use Principle of Inclusion Exclusion, getting each of the individual cases by grouping the words together

2) 1, the straight line. But, I'm guessing you want along lattice lines. In that case, let the three directions by Up, Left, and Forward. How many ways can you arrange 6 U's, 6 L's, and 6 F's in a line?

The digits 11234567 are separated into 3 groups.  In how many ways can this be done if the 1s must be in the same group?  What if the 1s must be in different groups?

In how many ways can 8778 be factored if the order of the factors is unimportant.

10 letters of alphabet are selected at random.  What's the probability that none of them are adjacent?

[quote="cdymdcool"]How many arrangements of the letters of the word lumberjack donot contain the words rub,lack or me?

How many shortest paths are there in space from (0,0,0) to (6,6,6)[/quote]

Is the first one 10!-8!-7!-9!+5!+6!+7!-3!

Why are these so urgent?

[hide="Lumberjack one"]
#(rub) + #(lack) + #(me) - #(rub and lack) - #(rub and me) - #(lack and me) + #(rub and lack and me) describes all the words that we don't want.

#(rub) = 8!
#(lack) = 7!
#(me) = 9!
#(rub and lack) = 5!
#(rub and me) = 7!
#(lack and me) = 6!
#(rub and lack and me) = 4!

Answer: $ 10!\minus{}(8!\plus{}7!\plus{}9!\minus{}5!\minus{}7!\minus{}6!\plus{}4!)\equal{}\boxed{3226416}$.
[/hide]

[hide="Shortest paths one"]
Assuming we're working in $ \mathbb{Z}^3$, we simply have to rearrange UUUUUULLLLLLFFFFFF, as rofler said. This happens in $ \frac{18!}{6!6!6!}\equal{}\boxed{17153136}$ ways.
[/hide]

[hide="11234567 one"]
We can "glue" the ones together so that we're partitioning [11],2,3,4,5,6,7 into three groups. Then it's just balls and urns (I'm assuming groups can be empty), which gives $ \binom{7\plus{}2}{2}\equal{}\boxed{36}$.

You can probably figure out the other case yourself.
[/hide]

[hide="8778 one (incomplete)"]
$ 8778\equal{}2^13^17^111^119^1$, so $ d(8778)\equal{}2^5\equal{}32$. I don't think that's the answer, but after this we have to count the ways to combine the divisors or something...
[/hide]

I don't feel like doing the last one, so I'll do it later (sorry, I'm tired...)


Let $ p$ be a prime number and $ a,n$ positive integers. Prove that if $ 2^p\plus{}3^p\equal{}a^n$, then $ n\equal{}1$.

[quote="moldovan"]Let $ p$ be a prime number and $ a,n$ positive integers. Prove that if $ 2^p \plus{} 3^p \equal{} a^n$, then $ n \equal{} 1$.[/quote]

When $ p\equal{}2$, $ 2^2\plus{}3^2\equal{}13$, so the statement is true.
When $ p\equal{}5$, $ 2^5\plus{}3^5\equal{}275\equal{}5^2 \times 11$, so the statement is true in this case.

When $ p \neq 5$ is odd
$ 2^p\plus{}3^p\equal{}2^p\plus{}(5\minus{}2)^p\equal{}.....\plus{}p(5)(2)^{p\minus{}1} \equal{}p(5)(2)^{p\minus{}1} (\mod 5^2)$, as $ 5 \nmid p$,

$ 5|2^p\plus{}3^p$ and $ 5^2 \nmid 2^p\plus{}3^p$,
$ \Rightarrow$ the statement is true in this case also, done.

if p>2
$ 2^p \plus{} 3^p$=$ (2 \plus{} 3)(2^{p \minus{} 1} \plus{} 2^{p \minus{} 2}( \minus{} 3) \plus{} .... \plus{} ( \minus{} 3)^{p \minus{} 1}$ so 5 / ($ 2^p \plus{} 3^p$) since to $ 2 \equal{} \minus{} 3[5]$

so $ 2^{p \minus{} 1} \plus{} 2^{p \minus{} 2}( \minus{} 3) \plus{} ... \plus{} ( \minus{} 3)^{p \minus{} 1}$=$ 2^{p \minus{} 1} \plus{} 2^{p \minus{} 1} \plus{} ...2^{p \minus{} 1}[5]$  (p time)

so $ 2^{p \minus{} 1} \plus{} 2^{p \minus{} 2}( \minus{} 3) \plus{} ... \plus{} ( \minus{} 3)^{p \minus{} 1}$=$ p2^{p \minus{} 1}[5]$ if$ p \ne 5$

so  you can conclude that $ 2^p \plus{} 3^p$ can not be devise by 5²

there for n=1

and you can check for p=2 , 5

[quote="moldovan"]Let $ p$ be a prime number and $ a,n$ positive integers. Prove that if $ 2^p \plus{} 3^p \equal{} a^n$, then $ n \equal{} 1$.[/quote]
Just a funny general problem . Let q be a prime number ,a and b are two integers different from 0 such that $ a \plus{} b \equal{} q$ . Suppose exist $ c,n$ such that $ a^p \plus{} b^p \equal{} c^n$ .
i) Prove that $ p \equal{} q$ if $ n>1$
ii)  $ n\leq 2$

[quote="TTsphn"] Let q be a prime number ,a and b are two integers different from 0 such that $ a \plus{} b \equal{} q$ . Suppose exist $ c,n$ such that $ a^p \plus{} b^p \equal{} c^n$ .
i) Prove that $ p \equal{} q$ if $ n > 1$
ii)  $ n\leq 2$[/quote]

Trivial factoring lemma: $ \upsilon_q(a^p \plus{} q^p) \equal{} \upsilon_q(a \plus{} b) \plus{} \upsilon_q(p)$.


algebra polynomial vector geometry perimeter blogs trigonometry

Going beyond triangles to polygons with more than $3$ sides,

$Question 1.$ Are all equiangular polygons equilateral too?
$Question 2.$ Are all equilateral polygons equiangular too?

$Question 3.$ Will your answer be different if the sides of the polygons are restricted to rational numbers only?

Can you think of some easy ways to construct counterexamples, if any?

[quote="t0rajir0u"][hide="1"] No.  Consider a non-square rectangle. [/hide] [hide="2"] No.  Consider a non-square rhombus. [/hide] [hide="3"] This doesn't affect either counterexample.  This restriction is completely irrelevant to Question 2, and I don't think it hampers Question 1 excessively even for $n > 4$.  For example, for $n = 6$ consider the equiangular hexagon with side lengths $a, a, b, a, a, b$ (in that order).  [/hide][/quote]

A number $p$ is a prime number iff every equiangular polygon of $p$ sides of rational lengths is equilateral.

How do you prove it?

[quote="xxxyyyy"]

A number $p$ is a prime number iff every equiangular polygon of $p$ sides of rational lengths is equilateral.

How do you prove it?[/quote]

I don't think the theorem is right. For example, 5 is a prime, but you can make an equiangular pentagon without it being equilateral--imagine moving any 2 adjacent sides towards the opposite side, or a pentagonal house pressed from above.

[quote="xxxyyyy"]
A number $p$ is a prime number iff every equiangular polygon of $p$ sides of rational lengths is equilateral.

How do you prove it?[/quote]

This also ignores the fact that $2$ is prime.  You can't have a 2-sided polygon.

Or if you say that there are no 2-sided polygons so therefore $2$ automatically fulfills the requirements of the theorem, then the theorem counts $1$ as a prime, which is also incorrect.

[quote="xxxyyyy"]A number $p$ is a prime number iff every equiangular polygon of $p$ sides of rational lengths is equilateral.

How do you prove it?[/quote]

I would look to complex numbers.

[hide="Specifically..."] If a rational equiangular $p$-gon exists, then there exist rational coefficients $r_{0}, r_{1}, ... r_{p-1}$ such that

$r_{0}+r_{1}\omega+...+r_{p-1}\omega^{p-1}= 0$

Where $\omega$ is a primitive $p$-th root of unity.  But then it follows that

$(r_{1}-r_{0})+(r_{2}-r_{1}) \omega+...+(r_{p-1}-r_{p-2}) \omega^{p-2}= 0$

Since $1+\omega+...+\omega^{p-1}$ is in fact the minimal polynomial of $\omega$ across the rationals (true iff $p$ is prime), this is a contradiction unless all of the coefficients of this polynomial are zero; in other words, all of the $r_{i}$ are equal, so the polygon is equilateral.  [/hide]

[quote="rgmds"]I don't think the theorem is right. For example, 5 is a prime, but you can make an equiangular pentagon without it being equilateral--imagine moving any 2 adjacent sides towards the opposite side, or a pentagonal house pressed from above.[/quote]

Yes, but the side lengths produced by your constructions can't be simultaneously rational.

[quote="mets501"]This also ignores the fact that $2$ is prime. [/quote]

An easy problem to fix; just specify $p > 2$.

[quote="t0rajir0u"]If a rational equiangular $p$-gon exists, then there exist rational coefficients $r_{0}, r_{1}, ... r_{p-1}$ such that

$r_{0}+r_{1}\omega+...+r_{p-1}\omega^{p-1}= 0$

Where $\omega$ is a primitive $p$-th root of unity.[/quote]

why is that true? (i'm not very good with these complex interpretations)

$r_{0}, ... r_{p-1}$ are the side lengths.  Think of the sides as vectors that trace out the perimeter of the polygon.  As vectors, they must add up to zero (because the polygon must be closed).

These vectors can be represented as complex numbers; specifically, as $re^{i \theta}$ where $r$ is the length and $\theta$ the direction.  These directions $\theta$ range from $0$ to $\frac{2 \pi}{p}$ evenly because the polygon is equiangular; hence we are dealing with the numbers $e^{i \frac{2 \pi k}{p}}$, the $p^{th}$ roots of unity.

Ah I see now. I have just one more question

[quote="t0rajir0u"]But then it follows that

$(r_{1}-r_{0})+(r_{2}-r_{1}) \omega+...+(r_{p-1}-r_{p-2}) \omega^{p-2}= 0$
[/quote]

How did you get that?

Subtract $r_{0}(1+\omega+...+\omega^{p-1}) = 0$, and divide by $\omega \neq 0$.

Edit:  Ooh, right.  That should actually be

$(r_{1}-r_{0})+(r_{2}-r_{0}) \omega+...+(r_{p-1}-r_{0}) \omega^{p-2}= 0$

But we reach the same conclusion.

Okay I get that now. But I'm still confused about this part; What does it mean by "minimal polynomial"? I don't really understand that paragraph at all.
(hopefully this is my last question)

[quote="t0rajir0u"]Since $1+\omega+...+\omega^{p-1}$ is in fact the minimal polynomial of $\omega$ across the rationals (true iff $p$ is prime), this is a contradiction unless all of the coefficients of this polynomial are zero; in other words, all of the $r_{i}$ are equal, so the polygon is equilateral.[/quote]

Sorry to bump this topic, but an interesting remark:  there is an algebraic interpretation to the stipulation that the side lengths be rational.  If we remove the condition that the side lengths (coefficients) be rational, then there are smaller minimal polynomials.

In particular, for the case $p = 5$ we have the factorization

$(1+x+x^{2}+x^{3}+x^{4}) = (x^{2}-2 \cos \frac{2 \pi}{5}+1)(x^{2}-2 \cos \frac{4 \pi}{5}+1)$

Which corresponds to the first transformation that [b]rgmds[/b] gave.  In other words, when we have

$(r_{1}-r_{0})+(r_{2}-r_{0}) \omega+(r_{3}-r_{0}) \omega^{2}+(r_{4}-r_{0}) \omega^{3}= 0$

We can let $r_{1}-r_{0}= 1, r_{2}-r_{0}=-2 \cos \frac{2\pi}{5}, r_{3}-r_{0}= 1, r_{4}-r_{0}= 0$ which produces an equiangular pentagon with side lengths

$(r_{0}, r_{0}+1, r_{0}-2 \cos \frac{2\pi}{5}, r_{0}+1, r_{0})$

[quote="t0rajir0u"]I wrote a blog post that treats this topic.

http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=134466[/quote]

I'm having trouble understanding your proof of the conjugate root theorem. Your lemma is "if $p(a)=0$ then $m(x)|p(x)$" but what is m(x)? what is a?

$a$ is most generally some element in the splitting field of $F$, where we are considering polynomials in $F[x]$.  Usually $a$ will just be a complex number.  And:

[quote="t0rajir0u"]Another very useful concept that could be applied in several problems I've seen.

[b]Definition:[/b]  Given some $a$ and some [url=http://mathworld.wolfram.com/Field.html]field[/url] (such as the rationals) $F$, then the minimal polynomial $m(x)$ of $a$ over $F[x]$ (the ring of polynomials with coefficients in $F$) is the unique monic polynomial in $F[x]$ of smallest degree such that $m(a) = 0$.[/quote]

Okay. I finally understand what's going on now.

My only question is that I can't figure out how to prove that the minimal polynomial of a pth root of unity is the polynomial of degree p-1.

[quote="t0rajir0u"][quote="mets501"]This also ignores the fact that $2$ is prime. [/quote]

An easy problem to fix; just specify $p > 2$.[/quote]

If you use vectors, you don't even have to fix it. The polygon will have 2 sides. Each angle will be $0^{\circ}$.


Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \left( \sum \frac {a}{b \plus{} c} \minus{} \frac{1}{2}\right)^2 \ge \frac {3}{(a \plus{} b \plus{} c)^2} \sum \frac {a^3 \plus{} abc}{b \plus{} c}
\]
:)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \left( \sum \frac {a}{b \plus{} c} \minus{} \frac {1}{2}\right)^2 \ge \frac {3}{(a \plus{} b \plus{} c)^2} \sum \frac {a^3 \plus{} abc}{b \plus{} c}\]
:)[/quote]
\[ \left( {\sum {\frac{a}{{b \plus{} c}}}  \minus{} \frac{1}{2}} \right)^2  \ge \frac{3}{{\left( {\sum a } \right)^2 }}\sum {\frac{{a^3  \plus{} abc}}{{b \plus{} c}}}
\]

Assume $ a\plus{}b\plus{}c\equal{}1$ then the inequality becomes
\[ \left( {\sum {\frac{a}{{b \plus{} c}}}  \minus{} \frac{1}{2}} \right)^2  \ge 3\sum {\frac{{a^3  \plus{} abc}}{{b \plus{} c}}}
\]
\[ \Leftrightarrow \left[ {2\sum {a\left( {a \plus{} b} \right)\left( {a \plus{} c} \right)}  \minus{} \prod {\left( {a \plus{} b} \right)} } \right]^2  \ge 12\prod {\left( {a \plus{} b} \right)} \sum {\left( {a^3  \plus{} abc} \right)\left( {a \plus{} b} \right)\left( {a \plus{} c} \right)}
\]
\[ \Leftrightarrow \left( {2 \minus{} 5q \plus{} 7r} \right)^2  \ge 12\left( {q \minus{} r} \right)\left( {2q^2  \minus{} 4q \plus{} 6r \minus{} rq \plus{} 1} \right)
\]

By Schur inequality we have $ r \ge \max \left\{ {0,\frac{{4q \minus{} 1}}{9}} \right\}$

If $ q \le \frac{1}{4}$ then $ r \ge 0$ then we need to prove the inequality when $ r\equal{}0$

\[ \Leftrightarrow \left( {2 \minus{} 5q} \right)^2  \ge 12q\left( {2q^2  \minus{} 4q \plus{} 1} \right)
\]
\[ \Leftrightarrow  \minus{} 24q^3  \plus{} 73q^2  \minus{} 32q \plus{} 4 \ge 0 \] (true)

If $ q \ge \frac{1}{4}$ then we have $ r \ge \frac{{4q \minus{} 1}}{9}$ then we will prove the inequality with $ r \equal{} \frac{{4q \minus{} 1}}{9}$
\[ \Leftrightarrow \left[ {2 \minus{} 5q \plus{} \frac{{7\left( {4q \minus{} 1} \right)}}{9}} \right]^2  \ge 12\left( {q \minus{} \frac{{4q \minus{} 1}}{9}} \right)\left( {2q^2  \minus{} 4q \plus{} 1 \plus{} \frac{{\left( {6 \minus{} q} \right)\left( {4q \minus{} 1} \right)}}{9}} \right)
\]
\[ \Leftrightarrow \left( {11 \minus{} 17q} \right)^2  \ge 12\left( {5q \plus{} 1} \right)\left( {14q^2  \minus{} 11q \plus{} 3} \right)
\]
\[ \Leftrightarrow \left( {3q \minus{} 1} \right)\left( { \minus{} 280q^2  \plus{} 167q \minus{} 85} \right) \ge 0
\]

Inequality has been proved.

[quote="Honey_S"]

$ (2 \minus{} 5q \plus{} 7r)^2 \ge 12( q \minus{} r)(2q^2 \minus{} 4q \plus{} 6r \minus{} rq \plus{} 1)$

[/quote]

The inequality is (if my computation is correct) equivalent to

$ f(r) \equal{} r^2(72\minus{}12q) \plus{} r(36q^2\minus{}190q\plus{}40) \plus{} (\minus{}24q^3 \plus{}122q^2\minus{}32q \plus{}4) \geq 0$

and

$ f'(r) \equal{} 2r(72\minus{}12q) \plus{} (36q^2\minus{}190q\plus{}40)$

Dear Honey_S,

why do you state $ f'(r) \geq 0$ ?

Thank you very much.

[quote="Honey_S"]

$ (2 \minus{} 5q \plus{} 7r)^2 \ge 12( q \minus{} r)(2q^2 \minus{} 4q \plus{} 6r \minus{} rq \plus{} 1)$

[/quote]

The inequality is equivalent to (sorry for my previous incorrect msg  :oops: )

$ f(r) \equal{} r^2(121 \minus{} 12q) \plus{} r(36q^2 \minus{} 190q \plus{} 40) \plus{} ( \minus{} 24q^3 \plus{} 73q^2 \minus{} 32q \plus{} 4) \geq 0$

and

$ f'(r) \equal{} 2r(121 \minus{} 12q) \plus{} (36q^2 \minus{} 190q \plus{} 40)$

Dear Honey_S,

why do you state $ f'(r) \geq 0$ ? I cannot see it :blush:

Thank you very much.