calculus
derivative
function
limit
algebra
domain
calculus computations

1) A function $ f(x)$ is so defined that for all real x, $ f(x)^n \equal{} f(nx)$. Prove that $ f(x).f'(nx) \equal{} f'(x).f(nx)$.

2) A function $ f: R \minus{}>R$ satisfies the relation $ f((x\plus{}y)/3) \equal{} \frac{ [f(x) \plus{} f(y) \plus{} f(0)] }{3}$ for all x,y of R. If $ f'(0)$ exists, prove that $ f'(x)$ exists for all x of R.

1) Assuming that $ f$ is differentiable, differentiating both sides of the equality $ f(x)^n\equal{}f(nx)$ gives $ nf(x)^{n\minus{}1}f'(x)\equal{}nf'(nx)$, multiplying both sides by $ f(x)$ then gives $ f(nx)f'(x)\equal{}f'(nx)f(x)$ as required.  It remains, however, to show that $ f$ is necessarily differentiable.

2) Letting $ x\equal{}y\equal{}0$ yields $ f(0)\equal{}0$.  Letting $ x\equal{}3t$, $ y\equal{}0$ also gives $ f(t)\equal{}\frac{1}{3}f(3t)$ for all $ t\in \mathbb{R}$.  Now for all $ x,h\in \mathbb{R}$ we have
\[ \frac{f(x\plus{}h)\minus{}f(x)}{h}\equal{}\frac{f(3x)\plus{}f(3h)\minus{}3f(x)}{3h}\equal{}\frac{f(3h)}{3h}\]
Taking limits as $ h\to 0$ gives
\[ f'(x)\equal{}\lim_{h\to 0}\frac{f(3h)}{3h}\equal{}f'(0)\]
for all $ x\in\mathbb{R}$.

Can someone prove the differentiability in the first question.

It depends on the exact statement of the problem.  Should the given equality hold for [i]any[/i] $ n$, or just some specific $ n$?  And in what domain is $ n$ allowed?  $ n\in \mathbb{N}$? $ n\in \mathbb{Z}$?  $ n\in \mathbb{R}$?  Depending on the answer to this question, it may or may not be true that $ f$ is necessarily differentiable.  For example, if $ n\in \mathbb{N}$ then it is not true that $ f$ must be differentiable, as demonstrated by the following counterexample:
\[ f(x)\equal{}\begin{cases} 0 & x\in \mathbb{Q} \\ 1 & x\notin \mathbb{Q} \end{cases}\]

[quote="JBL"][quote="JoeBlow"]Letting $ x \equal{} y \equal{} 0$ yields $ f(0) \equal{} 0$.[/quote]  I don't think so.[/quote]

Good point, that was pretty silly; let's give it another go then.

For any $ x \in \mathbb{R}$ and $ h\in \mathbb{R}\backslash \lbrace 0\rbrace$ we have
\[ \frac {f(x \plus{} h) \minus{} f(x \minus{} h)}{2h} \equal{} \frac {f(3x) \plus{} f(3h) \minus{} f(3x) \minus{} f( \minus{} 3h)}{6h} \equal{} \frac {f(h) \minus{} f( \minus{} h)}{2h}
\]
Taking limits as $ h\to 0$ we conclude $ f'(x) \equal{} f'(0)$ for all $ x\in \mathbb{R}$.  That seems to work. :)


parameterization
vector
trigonometry
calculus
calculus computations

Dylan is riding a roller coaster whose position is of the form $ \vec{r}(t) \equal{} 30\sin(t)\vec{i} \plus{} 30cos(t)\vec{j} \plus{} 15\cos(4t)\vec{k}$.

a) the scariest part of the ride is when it is traveling fastest. Dylan is nervous and wants to know when this will be so that he can close his eyes. For which value of t>0 does this occur first?

b) Dylan is still a little nervous and wants to know if the roller coaster ever points directly downward.

a) maximize the magnitude of the velocity vector (or its square, this is usually easier):

\[ \left| \vec{r}\,^{\prime}(t)\right|^2\equal{} \left| \left< 30\cos t , \minus{}30\sin t , \minus{}60\sin 4t \right> \right| ^2 \]
\[ \qquad\equal{}  900\cos^2 t \plus{} 900\sin^2 t \plus{} 3600\sin^2 4t \]
\[ \equal{} 900\left(1\plus{}4\sin^2 4t\right) \]

Then the maxima occur when $ \sin^2 4t\equal{}1\Rightarrow t\equal{}\textstyle{\frac{(2k\plus{}1)\pi}{8}},\quad k\equal{}0,1,2,\ldots$, the fist of which is $ t\equal{}\textstyle{\frac{\pi}{8}}\approx 0.39 \mbox{ sec}$.

By the way, this is complete BS... remember Newton's Second: $ \vec{F}\equal{}m\vec{a}$. The scariest part of the ride is likely to occur when acceleration is maximized, or at one of the local maxima thereof, when, say combined with a loop or inversion or almost impacting a large solid object like the ground, other such attributes of roller coasters that distinguish them from say, the Disneyland parking lot tram. Furthermore, I am concerned for Dylan: not only does he need to take a physics class in addition to re-taking his vector calculus class, he's is never going to get laid if he insists on being that lame.  :rotfl:

b) Well, does $ \vec{r}^{\, \prime}(t)\equal{} \left<0,0,c\right>$ for $ c\in\mathbb{R}$ have a solution? We have, $ \vec{r}^{\, \prime}(t)\equal{}\left<30\cos t, \minus{}30\sin t, \minus{}60\sin 4t\right>\neq \left<0,0,c\right>$, so no.


[quote="carlebee"]$ a,b,c\in\mathbb{R}^ \plus{} , a \plus{} b \plus{} c \equal{} 1$, Prove
$ \sum_{cyc}\frac {27a^4 \plus{} 1}{a^2 \minus{} 2a \plus{} 1}\geq9$

Don't use Jensen  :)[/quote]
*We have with $ 0<t<1$ : $ \frac{27t^{4}\plus{}1}{t^{2}\minus{}2t\plus{}1} \geq 18t\minus{}3 \Leftrightarrow (t\minus{}\frac{1}{3})^{2}(27t^{2}\plus{}36) \geq 0$
* $ \sum_{cyc}\frac {27a^4 \plus{} 1}{a^2 \minus{} 2a \plus{} 1} \geq \sum_{cyc}(18a\minus{}3)\equal{}9$
:) :) :)


Find all sequences $ a_1,a_2,...$ of non-negative numbers, such that the sum $ S\equal{}\sum_{k\equal{}1}^{\infty}a_k$ is finite and $ \sum_{k\equal{}1}^{\infty}a_{kn}\equal{}\frac{S}{n^{\alpha}}$ for all positive integers $ n$.


function
limit
Support
real analysis
real analysis unsolved

How to show that $C_{0}((0,1))$ is not complete with respect to the norm \[\parallel u\parallel =\text{max }\{|u(x)|: x\in (0,1)\}?\]

Try to find a sequence of functions $f_{n}$ which converge pointwise to a discontinuous function. They can have the same support.

OK. I got it. The function I used was \[f(x)= \begin{cases}0&0<x\leq \frac{a-1}{2a}\\ ax+\frac{1-a}{2}&\frac{a-1}{2a}<x\leq \frac{a+1}{2a}\\ 1&\frac{a+1}{2a}<x<1\\ \end{cases}\] with assumption that $a>1$.

That can't be the example you're looking for; that sequence isn't Cauchy in the uniform norm, and any function with a discontinuous limit won't be.

$C_{0}$ is not $C$; what is the precise definition?

I believe $C_{0}((0,1))=\{f: (0,1)\to \mathbb{R}: f\text{ continuous and }f(0)=f(1)=0 \}$. Therefore, when I just learnt a definition, I saw another reason why my function won't work

With that definition (or even $\lim_{x\to 0}$ and $\lim_{x\to 1}$ instead), the space is complete. The only version I can see that isn't complete is the space of compactly supported continuous functions on the open interval; those functions are zero outside some closed interval $[a,b]$ contained in $(0,1)$.

I agree with jmerry. Let's denote them by $C_{0}[0,1]$ and $C_{c}[0,1]$ respectively, then $C_{0}[0,1]$ is the completion of $C_{c}[0,1]$.
Take a function $f$ in $C_{c}[0,1]$ with support in $[a,b]$, more requirement, $a=\inf\{x, f(x)\neq 0\}$, then $f$ can be defined in $\mathbb{R}$ naturally. Let $0<a_{n}<a_{n+1}<a$ be such that $\lim a_{n}=a$. Let $f_{n}(x)=f(x+a_{n})$, then $f_{n}\in C_{c}[0,1]$ and $\{f_{n}\}$ is a Cauchy sequence since $f$ is uniformly continuous, but the limit is not in $C_{c}[0,1]$.


$\bigstar$ Let $a,b,c$ are positive real numbers such that: $abc(a+b+c)=1$. Find the minimum of the following expression:
\[N=(a+b)(b+c)\]
When does the equal hold? Let kill it  :wink:

I think, the minimum is $2$.
The equality holds  particularly when $a=c=1$ and $b=\sqrt2-1.$

arqady is right.
$abc(a+b+c)=1$
$ba+cb+b^{2}=\frac{1}{ac}$
$N=b^{2}+ba+bc+ac=ac+\frac{1}{ac}\geq 2$

Another one, praticely the same :

Let be a, b and c positive real. Show that

$(a+c)(a+b)\geq2\sqrt{abc(a+b+c)}$

[hide]
$(a+c)(a+b)=a^{2}+ab+ac+bc=a(a+b+c)+bc$

From AM-GM we have,

$\frac{a(a+b+c)+bc}{2}\geq\sqrt{abc(a+b+c)}$
[/hide]


USAMTS
geometry
geometric transformation
reflection
LaTeX
trigonometry

The USAMTS website [url]http://www.usamts.org[/url] has been slightly redesigned, as we get ready for the 2008-09 academic year.  If you find bugs, broken links, etc. on the website, please post them here.

The Round 1 problems for 2008-09 should be available sometime in the next couple of weeks.  The first due date is October 14.

On the registering page, it says "*Your Grade should reflect your grade during the 2007-08 school year. " Should it be for the 2008-09 year instead now?
Also, if you mess up the password and password confirmation (oops  :blush: ), you have to fill in all the info again. It's not exactly a bug, but it would be nice if that were fixed.  :lol:

One last off-topic question: If we handwrite a solution, can we scan it and print it to .pdf, then upload it?

[quote="Brut3Forc3"]On the registering page, it says "*Your Grade should reflect your grade during the 2007-08 school year. " Should it be for the 2008-09 year instead now?[/quote]It should.  This is now fixed.

[quote]Also, if you mess up the password and password confirmation (oops  :blush: ), you have to fill in all the info again. It's not exactly a bug, but it would be nice if that were fixed.  :lol: [/quote]I agree...I'll let our programmer know.

[quote]One last off-topic question: If we handwrite a solution, can we scan it and print it to .pdf, then upload it?[/quote]Yes.  As long as it is legible, we don't care how you create the PDF file: it can be handwritten, printed off a Word file, LaTeX output, or anything else.

I am sad that there won't be commended solutions anymore :(

I thought they made USAMTS a contest where you didn't just try to solve the problem with any way possible, but one where you would actually try to give a good solution. It was kind of like good style at MOP... but oh well. The reasons for removing them are understandable.

[quote="cognos599"]apparently the first round is coming out in early august, so when is the entry form due?[/quote]

Read the last sentence of the first post of this topic.

Yay!  It's finally going to start again (USAMTS) soon!  But, just like not_trig, I wish commended solutions could still exist... it probably was a good motivator for many of the participants.

[quote="cognos599"]i didn't ask when the round was due i was asking when the entry forms are due? and if i did it last year, do i have to re-submit my entry form?[/quote]

Oh, sorry.  I guess I didn't read carefully enough.  But, if you read carefully on the website, it is indicated that entry forms are due every year, regardless of whether you have participated in the past or not.

As for the due date, I don't know (too lazy to find out).  Does anyone else know?

According to the USAMTS website

[quote]You may submit the Entry Form and the Under 13 Permission Form with your first set of solutions, if you are mailing in your solutions. Otherwise, you can send the Entry and Permission forms separately.[/quote]

So the due date is probably October 14, when the first set of problems is due.

[quote]You may submit the Entry Form and the Under 13 Permission Form with your first set of solutions, if you are mailing in your solutions. Otherwise, you can send the Entry and Permission forms separately.[/quote]
so probably it's due before the first round ends.

I agree with not_trig and MathAndKnowledge. Commended solutions really motivated me to try refining my solutions last year (my second round was 10-page long, in comparison to my 5-page long fourth round).

[quote="DPatrick"][quote="Brut3Forc3"]On the registering page, it says "*Your Grade should reflect your grade during the 2007-08 school year. " Should it be for the 2008-09 year instead now?[/quote]It should.  This is now fixed.
[/quote]

I checked the "Register" page today, and the "Your Grade should reflect your grade during the 2007-08 school year" message still appeared.

I've noticed that you stated that you could type up the solutions on a document and mail them.  I use MathType to write equations and math solutions.  I know you have recommended us to submit the USAMTS solution forms, but could we include a header containing the same info provided on the forms, and then paginate the pages on a word document?

Also, I'm curious if there are new prizes for this year, or you haven't gotten to that point yet?

[quote="iironiic"]I've noticed that you stated that you could type up the solutions on a document and mail them.  I use MathType to write equations and math solutions.  I know you have recommended us to submit the USAMTS solution forms, but could we include a header containing the same info provided on the forms, and then paginate the pages on a word document?[/quote]That's perfectly fine.  We recommend the solution forms for handwritten solutions, but if you want to typeset, print, and mail your solutions, that's fine as long as each page contains the required information (name and USAMTS ID#), and the other submission rules are followed (such as each problem starting on a new page, etc.).

We haven't sorted out the prizes year for 2008-09, but it will likely be similar to previous years; I don't know if there will be any "new" books or not.

What is the difference between an entry form and a permission form?  Under "Current Permissions" in My USAMTS there are links to last year's [url=http://www.usamts.org/Forms/USAMTSPermissionForm.pdf]permission form[/url] which is slightly different from this year's entry form.

Okay, so it's quite obvious that I shouldn't submit a form titled 2007-2008 Entry form.  But the bad links at the bottom of [url]http://www.usamts.org/MyUSAMTS/U_MyProfile.php[/url] should probably still be removed.

It should be fixed now.  There is only one combined Entry/Permission form, it should say "2008-09" near the top.

Don't know if this is the right place to put it:

Why isn't there an automatic login on the usamts site?  My password is really long and i typo it about 1 in 3 times....


function
number theory unsolved
number theory

Determine all functions $f:\mathbb{Z_+} \rightarrow \mathbb{Z_+}$, satisfying condition $f(m^2+n^2)=(f(m))^2+(f(n))^2$, for every $m,n \in \mathbb{Z_+}$.

No comments...

http://www.mathlinks.ro/Forum/viewtopic.php?t=6568


advanced fields
advanced fields unsolved

Suppose $ (X_{k},\rho_{k})$ are totally bounded metric spaces. Define $ Z\equal{}\prod_{k\equal{}1}^{\infty}X_{k}$ with the metric $ \tau(x,y)\equal{}\sum_{k\equal{}1}^{\infty}2^{\minus{}k}\frac{\rho_{k}(x_{k},y_{k})}{1\plus{}\rho_{k}(x_{k},y_{k})}$. Show that $ (Z,\tau)$ is totally bounded.

[hide="Hint"]Use $ diam(\prod_{m>n}X_m)\le 2^{\minus{}n}$ to reduce to the case of a finite product - for which the claim is easy.[/hide]


hey everyone.... yea for those of you that are in high school... you may kno that a certain homecoming dance is coming up!!! depending on wherever everyone lives.... (in the US at least)

so i was talking to a friend about how i should ask a girl and he said to use a math problem... him thinking that im some kind of... super math geek  :P  but yea so i did, of course i wont use it tho but yea, just for kicks

soo ill leave this challenge up to all of you who wish to participate, the point is to make a math problem whose solution is actually a "proposal" to the homecoming dance

and i just wanted to see a big array of different problems and to see what you guys could all come up with... so post your unique problem and a brief to full solution

here's mine as a starter (even though its more like... 3 small problems put together)

Simplify this expression and leave it in terms of its original variables:
\[\displaystyle \frac{(\displaystyle\frac{1}{2} \sum_{n=0}^{\infty}\frac{m}{2^n})^e + \sum_{i=0}^{u-1} \displaystyle {i\choose 0}}{2005} = \frac {c-h}{\displaystyle\sum_{i=h}^{c-1} \frac{1}{i(i+1)}}\]

[hide]

infinite geo series the first part on the LHS, then use combinatoric identity $\sum_{i=1}^{u} \displaystyle {i\choose 0} = \binom {u+1}{1}$ for the next part

and RHS, use telescoping to simplify the bottom

the answer should turn out to be:

$m^e + u = 2005hc$?

cute eh?? haha

[/hide]

I dont' get the joke.  :?

Edit: Ok I get it now.  :lol:

[quote="white_horse_king88"]I dont' get the joke.  :?[/quote]Spell out the letters...$m^e+u=2005hc$.  $\text{Me+You=2005 Homecoming}$.

haha yea at... my school... hc is short for homecoming....

just for those of u who dunno XP

Oh, that's awesome!  :D
I'll have to find one like this lol

haha school hasnt even started yet for me.

solving that (as i just did) is proof positive that you probably wont be going to homecoming anyway. besides, i can think of MAYBE one or two girls in my school who can do this...

I can think o no decently hot girls in my school that can do this. Actually I take that back... NO girls in my school that can do this...

haha yea no girls at my school could solve that either

it really isnt hard.... its just the combinatoric part which requires the identity

but yea ne NEW questions???

Err, as far as I know $\sum_{i=1}^{u-1} \binom{i}{0} = u-1$... since $\binom{i}{0} = 1$ for all $i$ and you have $u-1$ of them... the identity probably starts from $i=0$.

hmm i tried to double check w/ my AOPs book but i couldnt find it

but yea looking at ur logic i think ur right... so my bad haha

lol...clever
but in all fairness, nobody will hot will get it

haha yea

sooo im gonna assume no1 is gonna propose their own hc problem?

As for as I'm concerned... no. I have other uses for my times. I have a knack for math but I'm not a math geek.  ;)

..............picking up girls with math problems...............

...........we've reached a whole new level of sadness.......................


Not very hard, but I'm trying to learn how to write problems!
There are exactly 5 ways to write 899 as a sum of a prime and a square: $ 233 \plus{} 26^2,499 \plus{} 20^2,643 \plus{} 16^2,863 \plus{} 6^2$, and $ 883 \plus{} 4^2$; there are exactly 2 ways to write 901 similarly:  $ 577 \plus{} 18^2$ and $ 757 \plus{} 12^2$. How many ways are there to write 900 as a sum of a prime and a square?

Edit: Blarg, probably goes in HSB. Oh well.

y did u chose 900!!!!
its already a square.
hence 1.

[hide="Solution"]$ 900\equal{}30^2\equal{}n^2\plus{}p\implies (30\minus{}n)(30\plus{}n)\equal{}p$, which is a contradiction unless $ 30\minus{}n\equal{}1$ or $ 30\plus{}n\equal{}1$.
If $ 30\minus{}n\equal{}1$, $ n\equal{}29$, and we have $ p\equal{}59$, which is prime.
If $ 30\plus{}n\equal{}1$, $ n<0$, so this doesn't work.

So the only answer is $ 59\plus{}29^2$, so $ \boxed{1}$.[/hide]


geometry
3D geometry
sphere
graph theory

I'm very fund of the combination informatics-math-problemsolving (going to do engineer computer science + mathematics next year) and I'd like to know if people here know a good place to start for me (probably IOI-like stuff, also coding theory and maybe even graph theory)?
Problem is I only know java and PHP, I think we will learn either C or C++ next year. I have no clue where to start, Flanders hasn't even got an informatics olympiad so I  know no one who can help me with this. Anyone here knowing some good sites/documents/problem sets/...? :)

I think that USACO training portal is a good place to start pracitising computational problem-solving. [url=http://ace.delos.com/usacogate]ace.delos.com/usacogate[/url]

Concerning C/C++, don't worry about it too much - the fundamentals are pretty much the same as in Java. The best thing to do is probably to buy a beginner's book in C++ and read it through, like "Learn C++ in 21 days" or something like that.

Also try Sphere Online Judge, or [url=http://spoj.sphere.pl/]SPOJ[/url] for short.  I haven't used it myself, but it looks like it has some past USACO contest problems on it.  Or you could try [url=http://acm.uva.es/problemset/]ACM[/url], which has more than 1500 problems.  There's also the [url=http://acm.timus.ru/]timus[/url] site.

My favorite, though, is [url=http://ace.delos.com/usacogate]USACO[/url].  If you work on the problems there, you can get very good, very fast  :).  Have fun!  :D


abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a group and let $Z$ be the center of $G$ so that $G/Z$ is cyclic. Show that $G$ is abelian.

Posted many times before. Look at [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=cyclic+abelian&t=101603]here[/url].


LaTeX
inequalities unsolved
inequalities

1)$ \frac{a\plus{}b\plus{}c}{\sqrt{3}}\le \sqrt{a^2\plus{}b^2\plus{}c^2}\Leftrightarrow (a\plus{}b\plus{}c)^2\le 3(a^2\plus{}b^2\plus{}c^2)$

which is Cauchy-Buniakowski-Schwartz :

$ (1\cdot a\plus{}1\cdot b\plus{}1\cdot c)^2\le (1^2\plus{}1^2\plus{}1^2)(a^2\plus{}b^2\plus{}c^2)$

2)Use LATEX . I don't understand  :(

hello, $ \frac{a\plus{}b\plus{}c}{\sqrt{3}}\le \sqrt{a^2\plus{}b^2\plus{}c^2}$ is equivalent with $ ab\plus{}ac\plus{}bc\le a^2\plus{}b^2\plus{}c^2$ which is obviously true.
Sonnhard.


An increasing sequence of integers is said to be alternating if it starts with an odd term, the second term is even, the third term is odd, the fourth is even and so on. The empty sequence (with no terms at all) is considered to be alternating. Let $ A(n)$ denote the number of alternating sequences which only involve integers from the set $ \{1, 2, ..., n\}$. Show that $ A(1) \equal{} 2$ and $ A(2) \equal{} 3$. Find the value of $ A(20)$, and prove that your work is correct.

Do you actually mean $ A(2) \equal{} 2$ and $ A(3) \equal{} 3$?

Hi Soarer. It's question 5 here:  http://www.bmoc.maths.org/home/bmo1-1994.pdf

My bad, I misunderstood the question.

For $ A(1)$, we have the empty sequence and 1.
For $ A(2)$, we have the empty sequence, 1, and 12.
[hide]
For $ A(20)$, there are ten odd numbers and ten even numbers in $ \{1,2,..,.20\}$.

Length 0: 1. (Empty sequence)
Length 1: 10 (Choose 1 from ten odd numbers)
Length 2: 10 * 10 (Choose 1 odd, 1 even)
Length 3: 10 * 10 * 9
Length 4: 10 * 10 * 9 * 9
...
Length 20: 10 * 10 * 9 * 9 * ... * 1 * 1

So the total number is 1 + 10 + 10 * 10 + 10 * 10 * 9 + 10 * 10 * 9 * 9 + ... + 10! * 10!.[/hide]

[quote="AndrewTom"]Hi Soarer.

Would you explain how you got $ 10 * 10; 10 * 10 * 9; 10 * 10 * 9 *9$ ; etc?

Thanks.[/quote]

1 from ten odd numbers, 1 from ten even numbers => 10*10

1 from ten odd numbers, 1 from ten even numbers, 1 from remaining 9 odd numbers => 10*10*9

and so on.

geez, i misread the question again! I'll look into it later. :blush:

Thanks very much Soarer. I'm looking forward to seeing your solution.

OK, first consier $ n$ even and consider an alternating sequence.
We always have the empty sequence.
If it ends with $ n$, remove it, then we get an alternating sequence counted in $ A(n\minus{}1)$ (excluding the empty sequence).
If it does not end with $ n$, it is already an alternating sequence counted in $ A(n\minus{}1)$ (excluding the empty sequence)
So $ A(n) \equal{} 1 \plus{} 2(A(n\minus{}1) \minus{} 1) \equal{} 2A(n\minus{}1) \minus{} 1$.

For $ n$ odd, consider an alternating sequence.
We always have the empty sequence.
If it does not end with $ n$, it is already counted in $ A(n\minus{}1)$ excluding the empty sequence.
If it ends with $ n$, remove it, then we either get an empty sequence or an alternating sequence counted in $ A(n\minus{}1)$.
So $ A(n) \equal{} 1 \plus{} A(n\minus{}1) \minus{} 1 \plus{} A(n\minus{}1) \equal{} 2A(n\minus{}1)$

Therefore, for n even, $ A(n) \equal{} 2A(n\minus{}1) \minus{} 1 \equal{} 4A(n\minus{}2) \minus{} 1$. So $ A(n) \minus{} \frac{1}{3} \equal{} 4(A(n\minus{}2) \minus{} \frac{1}{3})$, meaning that $ A(20) \minus{} \frac{1}{3} \equal{} 4^9 ( A(2) \minus{} \frac{1}{3}) \equal{} \frac{2^{21}}{3}$, so $ A(20) \equal{} \frac{2^{21} \plus{} 1}{3}$.

I hope that I didn't make any careless mistakes again. :ninja:

The above solution is still wrong. If the sequence ends in $n$, removing it does not establish a bijection between these sequences and those in $A(n-1)$, since removing $n$ leaves a sequence that ends in a number of opposite parity to $n$, and there are less of these than counted in $A(n-1)$.

It’s fibonnaci sequence, consider sequences ending in odd and even numbers separately. Although I see I am a little late to the party


Any ideas on how I can do this problem using Rouche Theorem and Cauchy Argument Principle?

[img]http://i3.photobucket.com/albums/y57/lilshantypanty/prob2-1.jpg[/img]

Thanks!

Rouche's theorem should be enough - the functions $ f$ and $ z \minus{} f(z)$ have the same number of zeros in the disk, because $ |f(z)| < |z|$ on the boundary.


let $ a,b,c \ge 0$
$ ab\plus{}bc\plus{}ca\equal{}1$
find MIN $ \sum \frac{1}{\sqrt{a\plus{}b}}$

[quote="nguyensilva"]let $ a,b,c \ge 0$
$ ab \plus{} bc \plus{} ca \equal{} 1$
find MIN $ \sum \frac {1}{\sqrt {a \plus{} b}}$[/quote]

$ 2\plus{}\frac{\sqrt{2}}{2}$

[quote="aadil"]did you use power mean?[/quote]

no  I use computer

[url]http://www.irgoc.org/viewtopic.php?f=10&t=21&sid=f3c5060e6b0d3ede0068f235b7930290[/url]


geometry
circumcircle
inequalities
geometry solved

[quote="Fitim"]Let $a$, $b$, $c$ be the sidelengths of a triangle $ABC$, let $m_{a}$, $m_{b}$, $m_{c}$ be the medians of triangle $ABC$, and let $R$ be the radius of the circumcircle of triangle $ABC$.
Prove the inequality
$\frac{a^{2}+b^{2}}{m_{c}}+\frac{b^{2}+c^{2}}{m_{a}}+\frac{a^{2}+c^{2}}{m_{b}}\leq 12R$.[/quote]

This is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16023]a big lemma[/url] $2Rm_{a}\geq b^{2}+c^{2}$.

If you extend the median $AM$ so that it cuts the circumcircle at $A'$, then $AA' =\frac{b^{2}+c^{2}}{m_{a}}$. The chord $AA'$ is obviously less than or equal to $2R$, and the inequality follows.

[color=red][Moderator edit: See also
http://www.mathlinks.ro/Forum/viewtopic.php?t=13032
http://www.mathlinks.ro/Forum/viewtopic.php?t=76684 ][/color]


Let $c\geq1$ be an integer, and define the sequence $a_1,a_2,...$ by $a_1=2$ and\[a_{n+1}=ca_n+\sqrt{(c^2-1)(a_n^2-4)}\]for positive integer $n$.
Prove that $a_n$ is integer for all $n$

After seeing how the sequence behave for small values, i told to myself that to get rid of the squareroot ,a_n must verifie that (a_n)^2-4=(b_n)^2(c^2-1) for n greater or equal to 3,which can be easily seen by induction:
if (a_n)^2-4=(b_n)^2(c^2-1) then (a_n+1)^2-4=(a_n+cb_n)^2(c^2-1).


function
integration
logarithms
real analysis
real analysis unsolved

Define \[F_{n}(x) = \frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}\right) \sqrt{\pi n}}\int_{-\infty}^{x}\left(1+\frac{y^{2}}{n}\right)^{-\frac12(n+1)}dy.\]

Prove that \[F_{n}(x) \to \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}\exp\left(-\frac{y^{2}}{2}\right)dy\] for all $x$.

maybe $n\to+oo$?
ok
some hint :
(I must be more correct..)

first step:

you can show that

$\frac{\Gamma{\frac{n+1}{2}}}{\Gamma{\frac{n}{2}}}\to \frac{\sqrt{n}}{\sqrt{2}}$

only few calculations and use something from here http://en.wikipedia.org/wiki/Gamma_function

seccond step:

by Taylor
$(1+\frac{y^{2}}{n})^{-\frac{1}{2}(n+1)}=\exp{(-\frac{1}{2}(n+1)\ln{(1+\frac{y^{2}}{n})})}\to \exp{(-\frac{1}{2}(n+1)\frac{y^{2}}{n})}\to \exp{(-\frac{y^{2}}{2})}$

For your second step, seems to me like you are using the dominated convergence theorem.

But what dominating function are you using?

... and your first step isn't clear to me :)

in my firts step
let $\frac{n+1}{2}=a$
and
$\frac{n}{2}=b$
then
by Stirling formula we have

$\Gamma{x+1}$~$\sqrt{2\pi{x}}(x/e)^{x}$
therefore
$\Gamma{a}/\Gamma{b}$~$calculaiting...$~$\frac{1}{\sqrt{e}}e^{(\frac{n-1}{2})\ln{(\frac{n-1}{2})}-(\frac{n}{2}-1)\ln{(\frac{n}{2}-1)}}$~$thinking...$~
$\frac{1}{\sqrt{e}}e^{\frac{1}{2}\ln\frac{n}{2}+\frac{1}{2}}$=$\sqrt{\frac{n}{2}}$


Successive discounts of 10% and 20% are equivalent to a single discount of
\[
\textrm{(A)} 30\%\qquad
\textrm{(B)} 15\%\qquad
\textrm{(C)} 72\%\qquad
\textrm{(D)} 28\%\qquad
\textrm{(E)} \textrm{none of these}
\]

[hide="Answer"]90% of $x$ is $\frac{9x}{10}$. 80% of that is $.8\cdot \frac{9x}{10}=\frac{72x}{100}$, so it is equivalent to a 28% discount, and the answer is $D$.[/hide]

[hide]$x\cdot .9=.9x$
$.9x\cdot .8=.72x$
$x-.72x=.28x$
Therefore, 28% discount[/hide]

Or you could just pick a number, let's do $100$, so then, $10$% discount would make it $90$ and then $20$% discount of that would make it $72$, so therefore $28$% is the answer.


If n is a constant exceeding 3, how many positive solutions x does the equation
e^x = x^n.    have?

thanks

the roots satisfy (by taking logarithm)
\[
x = n \ln x
\]
Take $f(x) = x - n\ln x$, we are going to find how many positive zeroes that $f$ has.
Let $f'(x) = 1-\frac nx = 0$, we get $x=n$, and furthermore, we obtain that $f$ is decreasing when $x<n$ and it is increasing when $x>n$.
Notice that $f(\epsilon) > 0$, $f(X) > 0$, and $f(n) = n(1-\ln n) < 0$
so $f$ has only one zero on [tex] [ \epsilon, n ] [/tex] and another unique zero on [tex] [ n, X] [/tex].
Hence, $f$ has two positive zeroes.


consider the number a=1*2*3...*25. show that the number is divisible by 3^10 and determine the largest positive integer n in the form of 10^n that divides a.

[hide]25! is divisible by three, because it has a three from 3, 6, 12, 15, 21, and 24, and two threes from 9 and 18. Thats a total of 10 threes, so 25! is divisible by $3^10$.

The largest multiple of 10 depends on the number of zeros it has when it's multiplied out. 25! has 6 zeros, so $10^6$ is its largest power-of-10 divisor.  [/hide]


1. How do you prove that log2 is irrational?
2. If a and b are irrational then prove that a^b is irrational too.
I was looking through Acops and i got confused.

#1 is kinda vague: log(100) is definitely rational.

#2 is also kinda weird, cuz if $ a=2^{\frac{1}{\pi}}$ and $ b = \pi$, we run into some trouble.

Well, if $ \log2$ is rational, then there will be integers $ a$ and $ b$ such that $ \frac{a}{b}=\log2$, or $ a=b\log2$.

$ 10^{a}= 10^{b\log2}$
$ 10^{a}= 2\cdot 10^{b}$
$ 10^{a-b}=2$

We have a contradiction.

Question 2 is false as is.  [url=http://mathworld.wolfram.com/GelfondsTheorem.html]Gelfond's Theorem[/url] is a partial salvage, but the proof is not easy at all.


A jeweler covers the diagonal of a unit square with small golden squares in the following way:
- the sides of all squares are parallel to the sides of the unit square
- for each neighbour is their sidelength either half or double of that square (squares are neighbour if they share a vertex)
- each midpoint of a square has distance to the vertex of the unit square equal to $\dfrac12, \dfrac14, \dfrac18, ...$ of the diagonal. (so real length: $\times \sqrt2$)
- all midpoints are on the diagonal

(a) What is the side length of the middle square?
(b) What is the total gold-plated area?

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=281[/img]

Suppose the sidelength of the first square is $k$.  Then the gold plated area from the center to the right is $k^2 + (k/2)^2 + (k/4)^2 + \dots = k^2( \sum\limits_{j=0} 2^{-2}^{j}) = k^2(4/3)$, and so the full gold plated area is twice this minus the center which is $(5/3)k^2$.

Also, half the diagonal of the big square + half the diagonal of the second square is 1/4 the diagonal of the unit square.  So $1/2 = k + (k/2)$ implying $k = 1/(3)$.

Er.. I could not follow that. What are your final answers to (a) and (b)? I think it's not correct.

A jeweler covers the diagonal of a unit square with small golden squares in the following way:
- the sides of all squares are parallel to the sides of the unit square
- for each neighbour is their sidelength either half or double of that square (squares are neighbour if they share a vertex)
- each midpoint of a square has distance to the vertex of the unit square equal to $\dfrac12, \dfrac14, \dfrac18, ...$ of the diagonal. (so real length: $\times \sqrt2$)
- all midpoints are on the diagonal

(a) What is the side length of the middle square?
(b) What is the total gold-plated area?

[img]https://1.bp.blogspot.com/-azJkAVACPvQ/XWuytNaj27I/AAAAAAAAKpo/C6C0NOzoQiYbFXfe41nNjNK8PEDvCxJ6wCK4BGAYYCw/s400/1993%2Bflanders%2Bp2.png[/img]


Since this forum is getting cluttered (again) I think we should redivide the forum into the following:

1. Signup Games, for games like BTM, The Mole, LTM, etc.
2. Nonsignup Games, for games like Mental Dictionary Game or Corrupt a Wish
3. Game Discussion, for topics like Smash Discussion Thread
4. Mafia, (leave as is)
5. Polls, (self-explanatory)
6. Fun Factory, frivolous discussion that doesn't fit anywhere else.

Any suggestions?

I think we should divide the forum into THIS following:

[code]
Forum: Games
   Subforum: Signup Games
   Subforum: Nonsignup Games
   Subforum: Game Discussion
   Subforum: Mafia
Forum: Polls
Forum: Fun Factory
[/code]

How about that?

That sounds fine as well.

Also, at the signup games there can be links for classic games that have been completed as a sticky.


A person has a beekeeping, initially with the bees. The annual birth rate of the bee population is constant and equal to P%. Each year bees die X of that creation. The population of bees is equal to the product (Q.A) at the end of N years.
Determine the value of X in terms of Q, N, P and A.

Or in portuguese question:

"Uma pessoa possui uma criação de abelhas, inicialmente com a abelhas. A taxa de natalidade anual dessa população de abelhas é constante e igual a p%. A cada ano morrem x abelhas dessa criação. A população das abelhas é igual ao produto q.a ao final de n anos.
Determine o valor de x em função de q, n, p e a."

Hi Luiseduardo.

In your first sentence, "com a abelhas" should be translated "with A bees", since the "a" in the Portuguese version which means "the" stands for the initial number of bees. This then ties in with the use of A in the rest of the English version. Both versions say X bees die each year. Should that be X%, which makes it a compound interest question or is that too easy?


how would you factor this expression?

im getting that it is unfactorable

$x^{2}-11xy+60y^{2}$

Did you copy it down right?  It looks like it should be $x^{2}-11xy-60y^{2}$.

If it is what t0rajir0u said, it's:
[hide="sushi"]$(x-15y)(x+4y)$[/hide]


Suma_milli and I both made states! GJ!!!!!

 :D  :D  :D

So guys, if we really wanna meet each other so badly at state, how would we recognize each other if it's just two of us (or actually, four, if you count stargirl240 and jgjgjmath) I'm pretty sure they're gonna come with us to state, as they'd possibly pwn chapter on the 21st. Good luck to them too!

EDIT: LOL, who here thinks suma and I are like twins on AoPS? lol it looks like we run after each other on all the forums :D and we have the same avatar, interested in each other (see our profiles), same signature, same location

lol thanks guassy!

Ok here's the master plan:

We all paint our faces completely green, wear all green, and most importantly write AOPS on our name tag.  :rotfl:  :rotfl:

lol?

That would be totally crazy, cuz then the ppl around us would be like  :huh: OMG GREEN MONSTER INVASION!!!!
Well we don't get nametags at the hotel in Matteson, so maybe write AOPS on our calculators? Yeah...And pencils...And perhaps water bottles too...

LOL.  These crazy ideas are fun but I the easiest way would probably be to tell each other what we're going to be wearing to the competition that day. :P I'll probably be wearing a blue shirt that says my schools name and then ALWAYS STRIVING TO...

It also has the thing that rrsuscyk has as his avatar ( I forgot what it was called )

:P

lol that'd be funny. Wouldn't your teammates be like  :huh: ?

hey and also you forgot to check the box that says "attach signature" before submitting a post!

Whoops it's there now.

Not really because actually it's my school's science olympiad shirt (since we don't get mathcounts shirts).  Maybe we can convince our teachers to get us mathcounts shirts since we'll be going to state though! :D

P.S.  Why does it always say you're offline when you're on?

Yay!

Lol my school never went to state as a team :(

it says offline because i opted to have my online status hidden (marked Anonymous) but don't worry i'm gonna change it back right now :D

they still used normal method to figure out my team score

And b/c the rest of my team phailed the written round, we phailed team and ended up 6th. In fact, I was the only person on the team who did all 10 team questions and the only one who did her best on the written round. :(

What's the normal method?

That sucks,  yeah I'm sure you'll be [i]really[/i] glad if you get into IMSA.

[quote="gauss1181"][quote="Stargirl240"][quote="suma_milli"]That sucks.  How do they find your team score?  Do they use a combination of everyone's written round scores and the team score?  I think we could have done a little better on the team round though.  We solved everything but I think I let the rest of my team do a little too much (because I probably could have solved most of them faster than letting them do it)... :(  But anyway we were checking over our answers and I checked one and found out it was wrong right at 10 seconds and we didn't have time to change it.  :(  We could have but our team let a 6th grader be captain since we all didn't care because it was just captain that turned in answers and he was closest to the end.  Darn!  I knew I should have been captain :P

Oh ok.[/quote]
Team score= 2*team round raw score+average of individual scores[/quote]
That's the normal method of calculating the team score. :D
@Aryth: that green smiley is my profile...im under the name Green Antirazz[/quote]

friended you.

[quote="Aryth"]dude does one of you have a facebook? because while stalking josh's facebook i saw a green smiley and was apprehensively shocked.  :wink: friend me.

http://www.facebook.com/profile.php?id=1149751977&ref=profile[/quote]

LOL.   :rotfl: .  I have an fb but I only add people I've met..so yeah after ARML tryouts I will :P  (I don't know I've been using alot of razz smilies lately. :P ... See?)

@stargirl: Oh ok. :D

Shame on you for using so many razz smilies! :wink:  :D
lol XD

sooo has your chapter said anything yet about state advancement?

lol.

Nope we've called them but they haven't picked up yet.

oh.

So you left them a message?

Hopefully they should send top 4 teams and top 5 individuals... :D

No I didn't leave a message yet.  Maybe they'll pick up on the weekday.  Leaving a message is a good idea but I'll talk to my coaches first so that they can leave an official one.

I hope u make states....*nerves shaking* :maybe:

YAY!!! I FINALLY FOUND OUT TODAY!!! SUMA_MILLI IS GOING TO STATES WITH ME!!! :D  :D  :D

lol i just can't be less hyper than this...O.o

Down two cans of Monster or Red Bull.

Me + AMP (the energy drink, not the math camp) @ MOP =  :jump:  :jump:  :jump:


Find all polynomials $ P(x),Q(x),R(x)$ with real coefficients such that
\[\sqrt{P(x)}-\sqrt{Q(x)}=R(x) \quad \forall x \in \mathbb R.\]

[quote="t0rajir0u"]The last time this problem was posted (and solved), it was locked by megus because it was part of an ongoing contest (or magazine, I'm not sure).[/quote]
MYM in Vietnam see here http://www.mathlinks.ro/viewtopic.php?t=168836

[quote="Rust"]$ Q(x) \equal{} q(x)^2, \ P(x) \equal{} (q(x) \plus{} R(x))^2$.[/quote]

So is this the answer with both $ Q$ and $ P$ perfect squares and $ R$ any real coefficients polynomial?

any solution??? I think it is a good problem


AMC USA(J)MO USAMO AoPSwiki trigonometry inequalities

Posted on the [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2007-ua/2007USAMOsoln.pdf]AMC web site[/url].

Also we are hoping to get good solutions on the [url=http://www.artofproblemsolving.com/Wiki/index.php/2007_USAMO]AoPSWiki[/url].  Note that the wiki is not supposed to be a forum: the goal is to get 1 or 2 [b]good[/b] solutions there.  Also, since it is a wiki, it is perfectly acceptable -- indeed, it is [i]encouraged[/i] -- to edit other people's solutions for clarity and to correct errors, or to post improvements.  You can also post notes or comments to a solution.

hmm interesting that i pretty much did the same for problem #6...admittedly i did not see the altitude thing...but that step seemed pretty random to me...and i thought of the construction as a problem though

it it too bad the inversion was like 10x as difficult as just using trig

Hmm I thought my solution for #1 was a bit bogus, but it's very very similar to their first solution, i don't remember what exactly i had though, but I could possibly get 6 on that when I thought for sure I would only get 1. But now I'm probably overestimating.

[quote="davidyko"]Question about the first solution on #1:
How do we know from:
$\frac{s_{k+1}}{k+1}\le\frac{s_{k}}{k}$ that the sequence $\frac{s_{k}}{k}$ is eventually constant?

Sorry if this is kind of stupid, I'm rather new to olympiad problems.[/quote]
$\frac{s_{k}}{k}$ was mentioned before to be a positive integer. Combine with the inequality and you have a nonincreasing sequence of positive integers. If the sequence didn't ever become constant, that means you have an infinite number of positive integers less than a value of $\frac{s_{k}}{k}$, which is obviously false.

[quote="davidyko"]Question about the first solution on #1:
How do we know from:
$\frac{s_{k+1}}{k+1}\le\frac{s_{k}}{k}$ that the sequence $\frac{s_{k}}{k}$ is eventually constant?

Sorry if this is kind of stupid, I'm rather new to olympiad problems.[/quote]
If $\frac{s_{k}}k=\frac{s_{k+1}}{k+1}$, then $a_{k+1}=\frac{s_{k}}k$, so $s_{k+1}=(k+1)\cdot\frac{s_{k}}k$ and the constant term becomes $\frac{s_{k}}k$, so for all $i\ge k$, $\frac{s_{i}}i=\frac{s_{k}}k$.

If $\frac{s_{k+1}}{k+1}<\frac{s_{k}}k$, then it must always be decreasing, but since $\frac{s_{i}}i$ is a positive integer, the sequence must become constant

[quote="davidyko"]Question about the first solution on #1:
How do we know from:
$\frac{s_{k+1}}{k+1}\le\frac{s_{k}}{k}$ that the sequence $\frac{s_{k}}{k}$ is eventually constant?

Sorry if this is kind of stupid, I'm rather new to olympiad problems.[/quote]

I was pretty surprised myself that they brushed over this detail. I spent extra time in my solution trying to explain why this was true, and...they just said it as obvious?!

umm...it kinda follows trivially from the Well-Ordering Principle (i think?)

but yeah I would have appreciated a solution to #6 that did not involve inversion about a circle, as that is a highly advanced concept (cmon people show the trig if you acknowledge that it exists)

and i felt that #5 was really artificial in the factoring...which a lot of people seem to think as well.

Oookay, thanks. I've got it now. Can't believe I didn't see it before.
And yeah, the factorization for #5 looks a bit contrived to me too.


$ y\equal{}3\tan(15\theta)\minus{}2$ has a period=____ degrees

[hide="EDIT Wrong Period"]
$ \frac{\pi}{15}\cdot\frac{180^{\circ}}{\pi}\equal{} 12^{\circ}$.
[/hide]

[quote="AstroPhys"][hide]
$ \frac{2\pi}{15}\cdot\frac{180^{\circ}}{\pi}\equal{} 24^{\circ}$.
[/hide][/quote]

$ \tan$ has a period of $ \pi$ ;)

[quote="AstroPhys"][hide]
$ \frac{2\pi}{15}\cdot\frac{180^{\circ}}{\pi}\equal{} 24^{\circ}$.
[/hide][/quote]
That is not the right answer. I know the answer but not how to get it.


Consider a grid of red and green points with 20 rows and 20 columns. Two adjacent points (on the same line or column) are connected with a segment. If 2 adjacent points have the same colour, the segment connecting them is also of that colour. If 2 adjacent points have different colours the segment connecting them is black. Now, there are 219 red points in the grid, 39 of which are placed on the borders. No red point is placed in one of the corners. Also there are 237 black segments. Determine the number of green segments.

I think I have an answer.

[hide]There are 181 green points(400-219)

There are 4 corners,

33 non-corner edges,

and 144 non-edge points

Now, total the number of lines coming from green points:

4*2+33*3+144*4 = 683

Subtract the number of black lines

683-237 = 446

Because you are counting both ends, divide by 2

446/2 = 223, your answer.[/hide]

Indeed. Now try either 3-D or more colors.

Oh, I didn't think of generalizing ... I'll try :D

I don't think it's a good idea to use more colours.
But it's easy to change the problem to 2 colours and 3 dimensions.

[color=cyan]You could do a multiple-color problem, but I think you'd have to give so much information as to make it unweildy.  For it to be solvable along similar lines, you would have to have seperate colors for connections between each pair of colors.  That is, at very least, you'd need to know how many 1-2 connections and 2-3 connections there were, as well as the number and distribution for all 3 colors.  I'm not even 100% sure that that would determine the answer, but I think it should.  Then with 4 colors, you have so much more information you'd need . . .
The n-dimensional case for 2 colors can be solved exactly the same as the 2-D case, however.[/color]


Find all ordered triples $ (x,y,z)$ of real numbers which satisfy the following system of equations:
\[ \left\{\begin{array}{rcl} xy & \equal{} & z \minus{} x \minus{} y \\
xz & \equal{} & y \minus{} x \minus{} z \\
yz & \equal{} & x \minus{} y \minus{} z \end{array} \right.
\]

[hide="Hint"] Add two of the equation and then you'll be able to consider two cases.[/hide]

[hide]
Add $ x\plus{}y\plus{}1$ to both sides of the first equation to get $ (x\plus{}1)(y\plus{}1)\equal{}z\plus{}1$. The other equations are similar. Now mutliplying, $ [(x\plus{}1)(y\plus{}1)(z\plus{}1)]^2\equal{}(x\plus{}1)(y\plus{}1)(z\plus{}1)$. But if any of $ x,y,z$ equal -1, they all will (if $ x\plus{}1\equal{}0$, then observe $ (x\plus{}1)(y\plus{}1)\equal{}z\plus{}1$ and $ (x\plus{}1)(z\plus{}1)\equal{}y\plus{}1$).

Otherwise, $ (x\plus{}1)(y\plus{}1)(z\plus{}1)\equal{}1$. Substituting into each of the equations gives that $ x,y,z$ are all roots of $ (t\plus{}1)^2\minus{}1\equal{}t(t\plus{}2)$, so each of $ x,y,z$ is 0 or -2. In addition, there are an even number of -2's required to make $ (x\plus{}1)(y\plus{}1)(z\plus{}1)\equal{}1$ and not -1.

This leaves the solutions $ \boxed{(\minus{}1,\minus{}1,\minus{}1),(0,0,0),(\minus{}2,\minus{}2,0)}$ and permutations, and they all work.
[/hide]

[hide="Another Way"]Adding the equations in pairs gives $ x(y\plus{}z)\equal{}\minus{}2x$, $ y(x\plus{}z)\equal{}\minus{}2y$, $ z(x\plus{}y)\equal{}\minus{}2z$.  If none of the variables are 0, then we have a system of linear equations $ y\plus{}z\equal{}\minus{}2$, $ x\plus{}z\equal{}\minus{}2$, $ x\plus{}y\equal{}\minus{}2$, which has solution $ (1,1,1)$.

Now WLOG assume $ x\equal{}0$.  Then the last two equations become $ yz\equal{}\minus{}2y$ and $ yz\equal{}\minus{}2z$.  Therefore, $ y\equal{}z$, and $ y^{2}\equal{}\minus{}2y$.  This gives us the solutions $ (0,0,0)$ and $ (0,\minus{}2,\minus{}2)$.

Therefore, all our solutions are $ (0,0,0)$, $ (1,1,1)$, and permutations of $ (0,\minus{}2,\minus{}2)$.[/hide]

write it as $(x+1)(y+1)=z+1$, and denote $x+1=a,y+1=b, z+1=c, xy=z, xz=y,yz=x$ with solutions $(0,0,0),(1,1,1),(-1,-1,1)$(very easy, sunt suppose they are not 0 and divide any 2 relations and get $a^2=b^2$, so $a=b$ or $a=-b$ and replacing is first equation we obtain the above solutions)

If there's something wrong in my approach, please tell me...
Why am I getting only two solutions....

wlog $x \ge y \ge z$
$\Rightarrow xy \ge xz \ge yz$

Hence
$xy \ge xz$
$\Rightarrow z-x-y \ge y-x-z$
$\Rightarrow z \ge y$
This with our assumptions give
$z=y$
Similarly $xz \ge yz \Rightarrow x=y$

Thus $x=y=z=k$(let)

substituting in one of the equations we get

$k^2+k=0$
$\Rightarrow k=0,-1$

This gives solutions $(0,0,0),(-1,-1,-1)$

Maths is the doctor of science...

Sayantan..

[quote="sayantanchakraborty"]
wlog $x \ge y \ge z$
$\Rightarrow xy \ge xz \ge yz$
[/quote]
You cannot have that inequality.Because $x,y,z$ are not necessarily positive.

My solution:
Let us label the equations $(1),(2)$ and $(3)$ respectively for our reference. Then $(1) - (2)$ yields
$x(y-z) = 2(z-y)$. Thus either $x=-2$ or $y=z$. Hence we can conclude that either:
(i) $x=y=z$. Thus $x^2 = -x \Longleftrightarrow (x,y,z)=(-1,-1,-1),(0,0,0)$ are the solutions.
or (ii) At least one of the three, say $x= -2$. Hence $(1)$ is $-2y = z+2-y \Longleftrightarrow y+z=-2$. $(3)$ is now $yz = -2 -(y+z) = 0$. Hence $y,z$ are the roots of $x^2 + 2x = 0$, namely they are a permutation of $(-2,0)$.
Hence all the solutions are:
$(x,y,z)=(0,0,0),(-1,-1,-1),(-2,-2,0),(-2,0,-2),(0,-2,-2)$.

[hide=almost mental math solution]
Rearrange the equations into $(x+1)(y+1)=(z+1), (z+1)(x+1)=(y+1), (y+1)(z+1)=(x+1)$. Let $x+1=a, y+1=b, z+1=c$.
If any one of $a, b, c$ is $0$ then one can easily check that $a=b=c=0$ is required. Hence $(-1, -1, -1)$ is a solution.
Else, $abc=1$. $ab=c, bc=a, ca=b \implies |a|, |b|, |c|=1$. Casework yields that the solutions for $(a, b, c)$ are $(1, 1, 1)$ and all permutations of $(1, -1, -1)$.
Thus, our solutions for $(x, y, z)$ are $\boxed{(0, 0, 0), (-1, -1, -1), (0, -2, -2), (-2, 0, -2), (-2, -2, 0)}$.
[/hide]

$(0,0,0)$ is a trivial solution
(1) Adding 2 equations we get that $x+y , y+z, z+x = -2$  iff none of them is 0
$ 2( x+ y + z) = -6 $ so $x,y,z = -1 $ is a solution
Suppose wlog $x=0$ then $y,z = -2 $ by (1)
So all the solutions are $(0,0,0) , (-1,-1,-1) , (0,-2,-2)$ and its permutations

[hide=very nice problem!][size=100][color=#f00]Claim:-[/color]all possible solutions for $(x,y,z)$ is $\boxed{(0,0,0),(-2,-2,0),(-1,-1,-1)}$ it's permutation[/size]\\

it's not hard to observe that the system of equations changes to:\\

$(x+1)(y+1)=z+1 \qquad \qquad (\spadesuit)$\\

$(x+1)(z+1)=y+1 \qquad \qquad (\heartsuit)$\\

$(z+1)(y+1)=x+1 \qquad \qquad (\diamondsuit)$ \\

consider $(x+1)(y+1)(z+1) \neq 0$\\

we have $(x+1)(y+1)(z+1)=1$\\

so from $(\spadesuit)$ equation we get $\frac{1}{z+1}=z+1\implies (z+1)^2=1\implies z=0,-2$\\

if $z=0$ then we have $(x+1)(y+1)=1$ and from second equation we have $x+1=y+1\implies x=y$\\

so we get $(x+1)^2=1\implies x=y=0,-2$\\

if so we get solutions from here as $(x,y,z)\in \{(0,0,0),(-2,-2,0)\}$\\

now if $(x+1)(y+1)(z+1)=0$\\

WLOG $x+1=0 \implies x=-1$\\

so from $(\spadesuit)$ and $(\heartsuit)$ equation we get:\\

$y=z=-1$\\

hence combining we get $\boxed{(0,0,0),(-2,-2,0),(-1,-1,-1)}$ it's permutation as solution for this system of equation
 as claim follows $\blacksquare$ [/hide]


geometry
incenter
circumcircle
geometry proposed

Dear everyone

please read the following

for a scalene triangle ABC, let I, H, E, F be incenter, orthocenter, midpoints of CA, AB respectively.
$ EI\cap AB\equal{}L$ , $ FI\cap CA\equal{}M$ , $ LM\cap BC\equal{}N$
P=circumcenter (triangle BHC)

prove that A, I, P=collinear $ \Longleftrightarrow \triangle{BNL}\equal{}\triangle{CNM}$

thanks for reading my shoddy post. :)

Dear 77ant and mathlinkers

It is equivalent with $ \angle BAC \equal{} 60^{\circ}$ $ \longleftrightarrow \Delta BNL \equal{} \Delta CNM$  :)

Sincerely

In fact this is China 2003. Please take a look at http://www.mathlinks.ro/Forum/viewtopic.php?t=660

Dear Kenny O

thanks for the source. I didn't know it before your reply.


1)let p be a prime.
Let a,b,n be postive integers satisfying
q| A(p,n)=(a p n  +b p n  ), where q is an odd prime and q-1 is not divisible by p n  .
Prove that if A(p,n) is divisible by q, but A(p,n-1) is not divisible by q then a,b are divisible by q.
2) If a<>b  and we replace A(p,n) by (a p n  -b p n), then the conclusion in part 1 still holds.

So now we will talk about your problem.At first of cause this is easy to see that there is something wrong in your question cause iff 1) hold then A(p,n-1) is divisible by q which you assume that this does not hold!!!
I think you generize the problem from some properties of Fermat numbers?I know these!
So let me arrange your problem as i understood,
Problem: a ,b are positive integers;p is prime.Let q be a prime number such that A(p,n) is divisible by q but A(p,n-1) is not [this means n is the smallest positive integer such that A(p,n) is divisible by q].Prove that or both a,b is divisible by q or q == 1(mod p^n)[even 2p^n as i do]

 The solution is simple by Fermart theorem and note that if ?(a,p)=1 then we can find an integer a* such that aa*==1 (mod q)
The second i think the same!
Hope this is your idea!


algebra
polynomial
greatest common divisor
ratio
search
system of equations

I'm having hard times factoring polynomials...

such as these

$6x^5-8x^7+10x^2$

$h=32t-16t^2$

could anyone give me a brief idea of how to factor things completely?

I have no clue how to do the first one... but i guess you find the GCF first and takes it out of the equation... but after that i'm clueless... I don't think it's factorable... but that's just what i think

on the second one.. I guess you just find the GCF which makes it $h = 16t(2-t)$ so $ h = {2,0}$

here's some rules that my teacher told us:

1. Factoring "undoes" multiplication, identifying the greatest common factor... sometimes the GCF is a monomial

2. Sometimes the factors are not monomials, but binomials.

3. To factor a trinomial ( or a binomial with subtraction) think about what two integers add to give he middle term and mutiply to give the third term, focusing on the factors of the third term.

4. Before you start factoring, you HAVE to take out the GCF, there can be no common divisor in the parentheses

and if you want I'll post this little thing our teacher taught us for factoring $ax^2$ trinomials using the "Ratio Method"

Hi guys
Factoring , Factoring , One of the dirties things i know :D
you  wana factorise the $6x^3-8x^5+10$? , well there exist a brute (Really brute  :D) Way of factoring anything you want (uasually it works if you have 3 free hours for calculating  :lol:  ) and it is this :
consider your polynomail with name of $P$, let its degree be $n$ then it can be factorise at maximum up to $n$ polynomial with degree $1$. let these polynomials
be $P_1,...,P_n$, and there are in the form of $(a'_i+b'_ix)$ . So we will have this:

$a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0=P=\prod_{i=1}^n Pi=\prod_{i=1}^n (a'+b'x)$ , so we have this :
$a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0=\prod_{i=1}^n (a'_i+b'_ix)$
and we know that the coefficents of both sides should be equal so keep producting the right side and by this you will  have a system of equations for finding the $a'_i ,b'_i$s by $a_i$s.solve it and you will have your polynomail factorised.

P.S: :read:  :ewpu: :basicly if you wana solve that system  by brute force , you may reach a polynomail by degree $n$ again :P  , so at least try to solve it elegantly :D .

[quote="shinwoo"]I think I get it now

Lucy's method was correct for the second one

though I didn't get lomos_lupin 's solution :P

for the first one
it's given that
$6x^5-8x^7+10x^2$
$-8x^7+6x^5+10x^2$
$-8*(x^5*x^2)+6*(x^2*x^3)+10*(x^2)$
$-8x^5*x^2+x^2*6x^3+10x^2$
$x^2*(-8x^5+6x^3+10)$
$2x^2*(-4x^5+3x^3+5)$
$2x^2*(3x^3-4x^5+5)$[/quote]

after that you can find factors by hit and trial method which is really time consuming.

To post 8 please

OK, I am going to be exact this time .
We want to factorise $-8x^7+6x^5+10x^2$, simply we have

$-8x^7+6x^5+10x^2=2x^2(-4x^5+3x^3+1)$ so we should just factorise :

$-4x^5+3x^3+5$

[quote="lomos_lupin"]OK, I am going to be exact this time .
We want to factorise $-8x^7+6x^5+10x^2$, simply we have

$-8x^7+6x^5+10x^2=2x^2(-4x^5+3x^3+1)$ so we should just factorise :

$-4x^5+3x^3+1$

$!!$ One of the good theorems that we have is this :

If the sum of the coefficents of a polynomail is zero then it is divisable by $(x-1)$
$-4x^5+3x^3+1$ is divisable by $x-1$, so we have :

$-4x^5+3x^3+1=(x-1)(-4x^4-4x^3-x^2-x-1)$ so up till now we have

$-8x^7+6x^5+10x^2=2x(x-1)(-4x^4-4x^3-x^2-x-1)$

now we should just factorise the:

$(-4x^4-4x^3-x^2-x-1)$[/quote]

I think you made a mistake,
the thing is not equal to $-4x^5+3x^3+1$

:oops_sign:  ye , so the whole thing is useless,i delete the wrong part  :blush
For fractoring a polynomails you should find its roots .for example
Let $P(x)$ be a polynomail with degree $n$ this form $\displaystyle a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ ,and  let $b_1,...b_m$ be its roots then we have
$P(x)=(x-b_1)(x-b_2)...(x-b_m)G(x)$
where $G(x)$ is not fractisable polynomial with degree $n-m$.
and if $C=\frac{r}{s}.r,s \in \mathbb{Z} (r,s)=1$  be a rational root of $P$ then:
$ r \mid a_0 , s \mid a_n$

For our case we wana fractor $P(x)=-4x^5+3x^3+5$ so we shall find the roots of $P$ . basically we first search for rational roots  and we know that the rational roots must be $\in \pm \{5,\frac{5}{2},\frac{5}{4},\frac{1}{4},\frac{1}{2},1, \}$

check that $P(x),x \in \pm \{5,\frac{5}{2},\frac{5}{4},\frac{1}{4},\frac{1}{2},1, \}$
is zero or not ..

hope that it get zero :lol:

but you are on the right track, but factorising polynomials like this takes alot of time.

exactly , this not a logical way for fractoring . The most intersting way for fractoring is to start playing with the polynomail to make it divisable to the smaller polynomails(which usually is a little hard to do :D )

[quote="LuCy4EvA"]on the second one.. I guess you just find the GCF which makes it $h = 16t(2-t)$ so $ h = {2,0}$[/quote]I think you meant $t=0,2$, but anyway just to get technical the goal here is to factor the polynomials; finding the zeros is an unnecessary extra step.

[quote="shinwoo"]uh...

my step only took few seconds after I learned how to do it[/quote]There's a slight misunderstanding here. You simply factored out $2x^2$, which is the right way to start. Other people were looking for ways to factor further.


[quote="ifai"]Let ABC be an isosceles triangle, D be the mid-point of the base AB. Let E be a point on the side AB, and O be a circum-center of the triangle ACE. Prove that
the line through D and perpendicular to DO,
the line through E and perpendicular to BC, and
the line through E and parallel to AC,
 all meet at the same point.[/quote]

Of course you mean for the 3rd line
"the line through B and parallel to AC"


You could be able to insert malicious code using latex commands to affect the server if there wasn't for that trigger. This code is not parsed as latex code, but as [unparseable ... bla bla bla] :D

[quote="joml88"]^\circ  will make a degree symbol

e.g.   $360^\circ=2\pi$

I don't really know for the second one.[/quote]

Is there any difference between $360^\circ$ and $360^{o}$??

Yes, there is a difference - and for this purpose, \circ looks much better than the letter "o".

Try this:

$\$360^{\text{o}}\$$ which is not really as easy as $\$360^\circ\$$


calculus
integration
analytic geometry
geometry
calculus computations

It was a test question that  I have thought tonight. Solutions won't be up for a while. I just wondering how to approch it.

An agricultural sprinkler distributes water in a circular pattern of radius 100 ft. It supplies water to a depth of e^(-r)feet per hour at a distance of r feet from the sprinkler.

a. What is the total amount of water supplied per hour to the region inside the circle of radius R centered at the sprinkler?

b. Determine an expression for the average amount of water per hour per square foot supplied to the region inside the circle of radius R.

First, we know that r goes from 0 to 100, and theta goes from 0 to 2pi (a complete circle)

So the integral setup is $\displaystyle \int^{100}_0\int^{2\pi}_0 e^{-r} d\theta dr$.

For the second part, recall from 2-D calc that the average value was the integral from a to b / (b-a).  So since b-a was the "length" in 2-D, we want the area in 3-D, so divide the above integral by the area of a circle with a radius of 100.

sorry, the first integral should be from 0 to 100.

[Moderator edit: I fixed that for you. Just remember when you're using numbers with more than one digit as a superscript or subscript, you need to enclose the whole thing in braces: ^{100}. - K.M.]


Euler
modular arithmetic
number theory
relatively prime

If they aren't relatively prime, as far as I know, it's useless.

yes, it only works for reletively prime numbers, i believe.

Its a number theory formula. If 2 numbers, a and b, are relatively prime, the a^(phi of b) = 1 mod b, where the phi of b is the amount of numbers less than b that are relatively prime to b.

P.S. Very useful in contest Maths

[quote="Kab^2"]How do you use the Euler-Fermat theorem if 2 numbers, a and b, aren't relatively prime.[/quote]
The Euler-Fermat Theorem only applies to two numbers $a$ and $b$ if $(a, b)=1$ (i.e. $a$ and $b$ are relatively prime).

If the two numbers aren't relatively prime (i.e. $(a, b) = g$ for some $g > 1$), then it is impossible that
$a^{k}\equiv 1 \pmod{b}$
for any integer $k$ because $a^{k}$ is a multiple of $a$ and therefore is a multiple of $g$, so $a^{k}$ would have to be some multiple of $g$ in modulo $b$, and since $g > 1$, it cannot be equal to $1$.


Prove that: \[\det(A)=\frac{1}{n!}\left| \begin{array}{llllll}\mbox{tr}(A) & 1 & 0 & \ldots & \ldots & 0 \\ \mbox{tr}(A^{2}) & \mbox{tr}(A) & 2 & 0 & \ldots & 0 \\ \mbox{tr}(A^{3}) & \mbox{tr}(A^{2}) & \mbox{tr}(A) & 3 & & \vdots \\ \vdots & & & & & n-1 \\ \mbox{tr}(A^{n}) & \mbox{tr}(A^{n-1}) & \mbox{tr}(A^{n-2}) & \ldots & \ldots & \mbox{tr}(A) \end{array}\right|\]

Just another way to write Newton's formulae for the eigenvalues of $A$. Just interpret the relations of Newton as a linear system and solve it using Cramer's rule.

You're right. I also have a generalization. Let $A$ be a $k\times k$ matrix. Then
\[b=\frac{1}{n!}\left| \begin{array}{llllll}\mbox{tr}(A) & 1 & 0 & \ldots & \ldots & 0 \\ \mbox{tr}(A^{2}) & \mbox{tr}(A) & 2 & 0 & \ldots & 0 \\ \mbox{tr}(A^{3}) & \mbox{tr}(A^{2}) & \mbox{tr}(A) & 3 & & \vdots \\ \vdots & & & & & n-1 \\ \mbox{tr}(A^{n}) & \mbox{tr}(A^{n-1}) & \mbox{tr}(A^{n-2}) & \ldots & \ldots & \mbox{tr}(A) \end{array}\right|\]
where $b$ is the coefficient of $x^{k-n}$ (if $n>k$, then $b$ is zero.) in $p(x)=\det(xI+A)$

[quote="harazi"]Just another way to write Newton's formulae for the eigenvalues of $A$. [/quote]
What is this?

The sum of powers of eigenvalues.
http://en.wikipedia.org/wiki/Newton's_identities

This problem was used at oral examination at ENS École Normale Supérieure in 2019

Can someone provide more insight and hint for this problem?


Let $x$ real then
$\frac{1}{4sin^6x+cos^2x}+\frac{1}{4cos^6x+sin^2x}\le 2$

.


Diophantine equation
algebra
linear equation
counting
IMO Shortlist

Let $ a_1 \geq a_2 \geq a_3 \in \mathbb{Z}^\plus{}$ be given and let N$ (a_1, a_2, a_3)$ be the number of solutions $ (x_1, x_2, x_3)$ of the equation

\[ \sum^3_{k\equal{}1} \frac{a_k}{x_k} \equal{} 1.\]

where $ x_1, x_2,$ and $ x_3$ are positive integers. Prove that \[ N(a_1, a_2, a_3) \leq 6 a_1 a_2 (3 \plus{} ln(2 a_1)).\]

[quote=DreamComeTrue]any idea[/quote]
See [url=https://artofproblemsolving.com/community/c6h1170169]here [/url]for a solution.


invariant
pentagon
algorithm
combinatorics
game strategy
IMO
IMO 1986

To each vertex of a regular pentagon an integer is assigned, so that the sum of all five numbers is positive. If three consecutive vertices are assigned the numbers $x,y,z$ respectively, and $y<0$, then the following operation is allowed: $x,y,z$ are replaced by $x+y,-y,z+y$ respectively. Such an operation is performed repeatedly as long as at least one of the five numbers is negative. Determine whether this procedure necessarily comes to an end after a finite number of steps.

I like this problem a lot also :).

Let \[ S(a_1,a_2,a_3,a_4,a_5) \equal{} \sum_{S \subseteq \{1,2,3,4,5\}} \left|\sum_{i\in S}a_i\right|\].

Observe that under the given transformation, S is strictly decreasing, but no strictly decreasing sequence of positive integers is infinite so we are done.

Proof: Compare S' = S(a+b,-b,b+c,d,e) with S=(a,b,c,d,e).

Most of the terms are the same. However, the different terms we get are S-S' = |a+c+d+e|-|a+2b+c+d+e| > 0 because b is negative, but a+b+c+d+e is positive, so a+c+d+e is positive.

[quote="Ilthigore"]Let
\[ S(a_1,a_2,a_3,a_4,a_5) \equal{} \sum_{S \subseteq \{1,2,3,4,5\}} \left|\sum_{i\in S}a_i\right|\]
.[/quote]

I don't understand your notation. Could you clarify please?

[quote="Agr_94_Math"]Principle of FIniteness of a Decreasing Sequence on positive integers[/quote]

http://en.wikipedia.org/wiki/Well-ordering_principle

Similar problem was proposed at Kosova Mathematical Olympiad

See it [url=http://www.mathlinks.ro/viewtopic.php?t=294347]here[/url]

[quote="Agr_94_Math"]Another way to solve this problem by introducing a different function:

Consider $ x_i$ for $ i \equal{} 1,2,3,4,5$ as  the five numbers in the pentagon. Now, consider $ x_4 \equal{} y < 0$, and by the problem hypothesis $ S \equal{} \sum_{i \equal{} 1}^n x_i > 0$, we have by the function  $ f(x_1, x_2, x_3, x_4, x_5) \equal{} \sum_{i \equal{} 1}^n (x_{i \plus{} 2} \minus{} x_i) ^2$ where $ x_6$ can be thought of as  $ x_1$ and $ x_7$ as $ x_2$.
 Obviously, if we consider $ f_2$ as the new value of $ f(x_i, i \equal{} 1,2,3,..5)$ after the transformation and $ f_1$ as the initial one , we get that  $ f_2 \minus{} f_1 \equal{} 2s x < 0$ by the given condition...... So,  the function is decreasing on the positive integers , by the Principle of FIniteness of a Decreasing Sequence on positive integers, we get that the operation has to stop at some point.[/quote]

WHAT IF THERE ARE N NUMBERS:REPLY QUICKLY

See [url=http://www.math.uchicago.edu/~qendrim/looping-poset.pdf]here[/url] for a paper.

See also [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=294347[/url] for both a generalization to $n$ numbers, as well as a discussion on why the result stays the same if the numbers are only known to be real, not integer.

The algorithm always stops. Consider the function \[f(x_1, x_2, x_3, x_4, x_5)=\displaystyle\sum_{i=1}^5(x_i-x_{i+2})^2, x_6=x_1, x_7=x_2.\] Clearly $f>0$ always and $f$ is integer valued. Suppose, WLOG, that $y=x_4<0$. Then $f_{\text{new}}-f_{\text{old}}=2Sx_4<0$ since $S>0$. Thus if the algorithm does not stop, we can find an infinite decreasing sequence of nonnegative integers $f_0>f_1>f_2>\cdots$. This is impossible, so the algorithm must stop. $\Box$

In PSS,Engel wrote,'All [u][b]but one[/b][/u] of the eleven students who solved the problem found the same function."Can one point out whats the function the remaining participant exhibited??

tc1729 posted the same solution as Agr_94_Math.Well,no point in posting the same solutions.

I have been reading PSS, and I have two questions about this problem:

1. What is the motivation for constructing the function $f(x_1, x_2, x_3, x_4, x_5)=\displaystyle\sum_{i=1}^5(x_i-x_{i+2})^2, x_6=x_1, x_7=x_2.$ How do we guess that it will decrease each step?

2. How do we determine $f_{\text{new}}-f_{\text{old}}=2Sx_4$? Do we have to expand out the function and simplify, or is there an easier way?

For more information on the history, alternative solutions, and generalizations of the problem see http://www.mathe.tu-freiberg.de/~wegert/Papers/Pentagon.pdf. (This is the manuscript of E.Wegert and Ch.Reiher, Relaxation procedures on graphs, Discrete Applied Mathematics 157, 2207-2216 (2009)).
If you would like to learn who the problem arose, click "http://www.mathe.tu-freiberg.de/~wegert/Talks/2012Batumi.pdf" for a talk with some background material.

@JWK750 For your second question, I expanded, but not fully; a lot of things cancel out right away, and the algebra isn't that bad. For your first question, I'm wondering the same thing. Can someone explain the [b]motivation[/b] behind constructing that function? (Like how would someone think of doing that?)

@JWK750 and @azmath333 For your first question: the lacking motivation for this function is the part which made the problem hard (IMO level). When I created the problem (see the second link in my post from Aug 11, 2016), I just "saw" this function. It is very specific, and does not work for all n-gons. To get some deeper understanding you may read the paper addressed in the second link.

$S(a,b,c,d,e)=(a–d) ^2+(b–e) ^2+(b–d) ^2+(b–e) ^2+(c–e) ^2\ge 0$ ,for all $a, b, c, d, e$ are real

And let $a, b, d, e>0,c<0$

And as per transformation $(x, y, z) =(x+y,–y, y+z)$

We get $S(a,b,c,d,e)=S(a,b+c,–c, d+c,e)$
$\implies S(k)=(a+c)^2+(a–d –c) ^2+(c+e)^2+(b–e) ^2+(b–d) ^2=S(a,b,c,d,e)+2c(a+b+c+d+e)<0$
Since $c<0$ ,$a+b+c+d+e>0$  ,so by infinite descent we are done.

Actually the motivation of this done by second moment.
 Actually the $n-gon$ is given the vertices
$v(G)=({a, b, c, d, e})$

[quote=ftheftics]Let, denote the vertices of pentagone is $x_i \forall I =1,2,...,5$
now ,let in first step $(x_1,x_2,x_3,x_4,x_5) \to (x_1,x_2+x_3,-x_3,x_4+x_3,x_5)$
In this case every thing changes but s(sum) of the integer is invariant.
So ,our claim is that we can got a mono variant funtion that decreases.
Let,$f_0(x_1,x_2,x_3,x_4,x_5)=$                 $\sum _{I=1}^5 x_i^2 -x_ix_{i+1}$
So ,$f_0(x_1,x_2,x_3,x_4,x_5)=$
$ \frac{1}{2} \sum_{i=1}^5 (x_{i+1}-x_i)^2$

now let $f_0 \to f_k $ denote operation on $f_0$ for k times and resulting $f_k$.

So,$f_1-f_0\le 0$
I.e $f_1<f_0$
So,f is decreasingby property of integer it is naturally that the operation  will stop after finite steps.    $\square$.       QED[/quote]

Plz can you explain the motivation behind defining the following function...

This might be my favorite problem too! My solution is similar to what was posted above, but seems to be unique (as far as I've seen on the internet for now) and hopefully a little more motivated. Sorry for no latex.

WLOG let x_2 be the negative number.
After a bit of playing around, we see that the sum is constant and that (1) the sum of x_i^2 seems to be a decreasing function. It turns out it isn't, but continuing on we take (2) the sum of (x_i+x_(i+1))^2 before and after the operation.

Neither (1) or (2) is sufficient, but adding them gives us a change of 2*x_2*(sum of x_i) after the operation. This is clearly a negative number. As the sum of 2 squares always decreases after the operation, we see that the operation must end at one point since the sum cannot be less than 0.

[quote=Gorav]If $ y$ < $ 0$, only then we can apply this transformation, but then if we have $ y$ < $ 0$ initially, we have reached the end of procedure and if there is not any $ y$ < $ 0$, we can't apply this transformation. Does it make sense, please clarify?[/quote]
No, because maybe y is now positive, but x and z are smaller - they can be negative. And also two other numbers can be negative.

[quote=KI_HG]Let $a_i$ be the number on vertex $i$, $i=1,2,3,4,5$. $S=a_1+2a_2+3a_3+4a_4+5a_5$. $S$ decreases if and only if we operate on vertex 1.
Assume the process never ends. This implies we operate the move on a vertex infinitely many times. WLOG, this vertex is vertex 1. This contradicts that $S$ has a lower bound.[/quote]

How did you show that $S$ has a lower bound?

It was said that we could change the condition ‘integer’ into ‘real number’ and the conclusion is still true.
But how to prove that?
One argues that we could introduce an number to count disorder pairs of number? But I cannot understand it.
I am grateful if one could tell me how.


What is the remainder when 22222^5555 - 55555^2222 is divided by 7?

Thank you!

[quote="cincodemayo5590"][i]To the forum:[/i] Is there another quick way of doing this? Is my solution correct?[/quote]
Well Fermat's Little Theorem says $a^{p}=a \mod p$ where p is prime.

Euler's Totient theorem (which is a generalization of the last theorem) says that $a^{\phi x}=1 \mod x$ where $\phi x$ is the number of numbers less than x that are relatively prime to x and gcf(a, x)=1.


function
induction
algebra unsolved
algebra

Find all the function f:N*-->N* satisfly:
      $f(x^{2}+f(y))=(f(x))^{2}+y$  for all the positive integers x,y[/hide]

[quote="vietnamesegauss89"]Find all the function $f: \mathbb{N}^{*} \to \mathbb{N}^{*}$ satisfly:
       $f(x^{2}+f(y))=(f(x))^{2}+y$  for all the positive integers $x,y$[/quote]

TeXed.

Find all the function $f: \mathbb{Z}^+ \to \mathbb{Z}^+$ that satisfy $f(x^{2}+f(y))=(f(x))^{2}+y$  for all $x, y \in \mathbb{Z}^+$.

Suppose that $f(a) = f(b)$. Substituting in $a, b$ for $y$ in the given condition, we find that $f(x)^2 + a = f(x^2 + f(a)) = f(x^2 + f(b)) = f(x)^2 + b \Rightarrow a = b$, so $f$ is injective. Thus, suppose that $y > f(2)^2 - f(1)^2$ (***) We can then write $f(1 + f(y)) = f(1)^2 + y$ $=$ $f(2)^2 + (y + f(1)^2 - f(2)^2)$ $=$ $f(4 + f(y + f(1)^2 - f(2)^2))$ $\Rightarrow$ $1 + f(y)$ $=$ $4 + f(y + f(1)^2 - f(2)^2)$. Writing $a = f(2)^2 - f(1)^2$, we find that $y > a$ implies $f(y) = 3 + f(y - a)$. Note that we clearly cannot have $a$ zero, and if $a$ were negative, that would imply that some very large number gets mapped to a negative number under $f$. Thus $a$ is positive and we can simplify the above equation by writing $x = y - a$ to find that if $x \in \mathbb{Z}^+$, $f(x) + 3 = f(x + a)$. Now, in the given equation, we can write $f(x)^2 + (y + a) = f(x^2 + f(y + a)) = f(x^2 + f(y) + 3)$; by repeating this, we can prove by induction that $f(x)^2 + (y + na) = f(x^2 + f(y) + 3n)$; taking $n$ to be $a$ yields $f(x)^2 + y + a^2$ $=$ $f(x^2 + f(y) + 3a)$ $=$ $f(x^2 + f(y) + 2a) + 3$ $=$ $f(x^2 + f(y)) + 9$ $=$ $f(x)^2 + y + 9$, so $a = 3$; thus, we have $f(x) + 3 = f(x + 3)$. Note that when we chose $y$ at the step marked (***), there was nothing special about the numbers $1, 2$ - in particular, we could have chosen $y = f(3)^2 - f(1)^2$ and the same analysis would yield $f(x) + 8 = f(x + 8)$. This gives us that $f(x + 1) = f(x + 4) - 3 = f(x + 16) - 15 = f(x + 8) - 7 = f(x) + 1$ so $f(x) = c + x$ for some fixed $c \in \mathbb{Z}_0^+$. We can now plug this back into the given condition to check that it will always be true only if $c = 0$, so the answer is $f(x) = x$; conversely, it is simple to check that this works.


Is this true?

For $a,b,c$ positive reals prove or disprove that

$a^4+b^4+c^4 +3abc(a+b+c) \geq 4(a^2b^2+b^2c^2+c^2a^2)$

The first counterexample is a = 1, b = 2, c = 2. Sorry.  :blush:

  Darij

Thank you very much, Darij.
You are very quick to find counterexamples.

[quote]I think this is the fastest method to solve inequalities ;)[/quote]


vector
search
function
set theory
linear algebra
linear algebra unsolved

Sorry if it has been posted before. I wish to know the proof of the theorem:

[b] Suppose that B and B' are 2 bases for a vector space V. Then there exists a bijection from B to B'. [/b]

I haven't read any courses in advanced set theory so I want to ask for an elementary solution (at least, as elementary as possible).

[url=http://www.mathlinks.ro/viewtopic.php?t=211543]Proof for finite dimensional spaces[/url]
[url=http://www.mathlinks.ro/viewtopic.php?search_id=60293920&t=173999]Proof for infinite dimensional spaces[/url]

Hi [b] The Captain [/b],

Following the link you gave for the case of infinite-dimensional vector space, I do not really understand what you really means. Could you clarify the following for me ?

i) Why is it the case that $ \bigcup_{b} U_b \equal{} I$ ? [After analyzing, I get this now]

ii) You stated that $ |B| \equal{} |I|$ from the fact that each $ U_b$ is finite. Does it mean that you assume the following "fact": Any infinite collection of finite subsets of an infinite set $ A$ has the same cardinality with $ A$.

iii) What is the symbol $ C$ and the symbol that look likes N near the end of your proof ? If it is the cardinality of the natural numbers then you must have used the fact that |N x X| = |X| for any infinite set X. It requires a proof.

Thank you.

The last line of my proof in that post should be
$ |A|\equal{}|I|\leq\aleph|B|\equal{}|B|$. (A moderator may kindly edit that post.)
Now everything should be clear. $ \aleph$ stands for the cardinality of the set of natural numbers, and I've used the fact that $ \kappa\aleph \equal{}\kappa$, for any infinite cardinality $ \kappa$. (This, easily, follows from the definition of product of two given cardinalities.)

The cardinality of the natural numbers is universally denoted $ \aleph_0$.

[quote="The Captain"]The last line of my proof in that post should be
$ |A| \equal{} |I|\leq\aleph|B| \equal{} |B|$. (A moderator may kindly edit that post.)
Now everything should be clear. $ \aleph$ stands for the cardinality of the set of natural numbers, and I've used the fact that $ \kappa\aleph \equal{} \kappa$, for any infinite cardinality $ \kappa$. (This, easily, follows from the definition of product of two given cardinalities.)[/quote]

Then what is your definition of cardinality?

My definition is: The two set has the same cardinality if and only if there is a bijection between them.

In my definition, it does not obviously follow that $ \kappa\aleph \equal{} \kappa$. Sorry for my lack of knowledge in set theory.

@randomgraph You are right. But I prefer to use $ \aleph_m$ in other contexts.
@mathangel Let $ \kappa$ be an infinite cardinality and let $ \aleph$ denote the cardinality of the set of natural numbers. Then Zorn's lemma yields $ \kappa ^2\equal{}\kappa$. (Please refer to a logic book to see a proof of this result.) And hence, $ \kappa\leq \kappa\aleph\leq \kappa^2\equal{}\kappa$, ie $ \kappa\aleph\equal{}\kappa$.

I got the following solution from my friend. I would like to share it with you.

[hide] Let $ X \equal{} \{ (S,f) :  S \subseteq B, f \text{ injective} : S \rightarrow B', (B\minus{}S) \cup f(S) \text{ forms a basis for V} \}$ . In other words $ (B\minus{}S) \cup f(S)$ is a mixed basis in which a part of $ B$ (namely $ S$) has been replaced by elements of $ B'$ and the remainder $ (B\minus{}S)$ is kept. Each basis is equipotent to $ B$ and $ X \not\equal{} \emptyset$ (take $ S\equal{}\emptyset$). The idea is to use them to migrate to $ B'$.

Define a partial ordering as follows:
\[ (S,f) \leqslant (T,g)\]
iff $ S$ is a subset of $ T$ and $ f$ is identical with $ g$ when restricted to $ S$.

To use Zorn's lemma we need to prove every chain $ C$ has an upper bound. Define the upper bound $ (T,g)$ as follows: $ T \equal{} \bigcup_{(S,f) \in C} {S}$ and $ g(x) \equal{} f(x)$ if $ x \in S$ and $ (S,f) \in C$ (by definition of $ \leqslant$, the function $ g$ is well-defined). It is easy to verify that $ (T,g)$ is an upper bound for $ C$. Applying Zorn's lemma yields a maximal element for $ X$, call it $ (M,h)$.

Suppose $ B\minus{}M \not\equal{} \emptyset \Longrightarrow h(M) \not\equal{} B'$ (otherwise $ h(M)$ would be a basis by itself). Let $ b'\in B'\minus{}h(M)$. It follows that $ b'$ is a linear combination of some elements from $ B\minus{}M$ and $ h(M)$, and at least one of these elements $ b \in B\minus{}M$ (otherwise there would be a linear dependence among elements purely of $ B'$):

\[ b' \equal{} ab \plus{} \sum_{i\equal{}1, b_i \in B\minus{}M}^{n}{\lambda_i b_i} \plus{} \sum_{j\equal{}1, b'_j \in h(M)}^{n}{\mu_j b'_j}\]

Equivalently, algebraic manipulation gives us:

\[ b \equal{} a^{\minus{}1} b' \minus{} \sum_{i\equal{}1, b_i \in B\minus{}M}^{n}{a^{\minus{}1} \lambda_i b_i} \minus{} \sum_{j\equal{}1, b'_j \in h(M)}^{n}{a^{\minus{}1} \mu_j b'_j}\]

Notice that the set $ M \cup \{b\}$ and the function $ h_1: M \cup \{b\} \rightarrow B'$ defined as: $ h_1(x) \equal{} h(x) \forall x \in M$ and $ h_1(b) \equal{} b'$ will gives us a larger element of $ X$. This contradicts the fact that $ (M,h)$ is the maximal element of $ X$. So we conclude that $ M\equal{}B$ and $ h(M)\equal{}B' \Longrightarrow h$ is surjective. Therefore $ h: B \rightarrow B'$ is a bijection between the two bases. [/hide]


Let $a \in \left[ 0, 1 \right)$ be a number written in decimal form. We construct the sequence $\left\{ x_{n}\right\}_{n \geq 1}$ as follows: $x_{1}$ is obtained by arbitrarily permuting the first three decimals of $a$, $x_{2}$ is obtained by arbitrarily permuting the second, the third and the fourth decimals of $x_{1}$ and, in general, $x_{n+1}$ is obtained by arbitrarily permuting the $n+1$th, the $n+2$th and the $n+3$th decimals of $x_{n}$.

a) Prove that any such sequence converges to a limit $b$.
b) Is it possible for $a$ to be rational and for $b$ to be irrational?
c) Find a number $a$ such that any sequence $\left\{ x_{n}\right\}_{n \geq 1}$ constructed from it has an irrational limit.

PS: This was given to 11th graders and it's much more interesting than the usual Cesaro-Stolz problem.

does $x_{i}$ consist of only a permutation of the $i^{th}, (i+1)^{th}, (i+2)^{th}$ digits of $a$, or the rest of $a$ in the original order as well? In the latter case, shouldn't $\{x_{n}\}_{n \geq 1}$ converge to $a$ itself?

[quote="perfect_radio"]$x_{1}$ is obtained by arbitrarily permuting the first three decimals of $a$, $x_{2}$ is obtained by arbitrarily permuting the second, the third and the fourth decimals of $x_{1}$ and, in general, $x_{n+1}$ is obtained by arbitrarily permuting the $n+1$th, the $n+2$th and the $n+3$th decimals of $x_{n}$.[/quote]

That this converges is easy: the difference between $x_{n}$ and $x_{n+1}$ is less than $\frac{1}{10^{n-1}}$.

It is possible for $a$ to be rational and $b$ to be irrational: let $a = 0.01010101\ldots = \frac{1}{99}$.  $x_{n}= x_{n-1}$ unless $n = k!$ for some $k \geq 2$, in which case we are looking at three digits $\ldots101\ldots$ which we change to $\ldots011\ldots$.  Then the difference $a-b$ consists of all zeros, except a 9 in the $k!+1$ decimal place for $k \geq 2$.  This number is irrational (indeed, transcendental).

Haven't thought about the third.

a real $r$ is rational if, and only if, its decimal expansion is periodic. For example try $x=\sum 10^{-k!}$ as an answer for the last question. The idea is that the density of $1$ in its expansion is null, so one will never make the expansion periodic by permuting (in the way described described in the exercice) the decimals of $x$ (because the density of $1$ in a periodic sequence of $0$ and $1$ is strictly positive)


If a; b; c are positive integers such that
0 < a^2 + b^2 - abc =<c
show that a^2 + b^2 - abc is a perfect square
لطفا اين سوال را حل كنيد

حلش كنيد ديگه بابا


Using the definition of $ \binom{n}{r}\equal{}\frac{n!}{r!(n\minus{}r)!}$, we plug in the numbers.
$ \binom{7}{2}\equal{}\frac{7!}{2!(7\minus{}2)!}\equal{}\frac{7!}{2!5!}\equal{}\frac{7\times6}{2}\equal{}\frac{42}{2}\equal{}\boxed{21}$


Please help me with the following problem:
Engine of motor-boat is shut off at t=0.At that moment velocity of the boat was v=2.4 m/s.Assume that drag force is proportional to the speed of the boat.Find the distance that the boat would travel after the engine is shut off if it is known that  at the moment t=3 velocity of the boat was v=1.2m/s.

Here is my question.Assuming that  m is mass of the boat and drag force is F=-bv I found that v(t)=2.4*e^(-bt/m) .
In order to find the answer  I should find such t that v(t)=0.But exponential function v(t)=2.4*e^(-bt/m) is not equal to 0 for any t.
Please correct me if I'm wrong and help to solve this problem.
Thank you.

$\large m\frac{dv}{dt}=-bv \Longrightarrow \frac{dv}{dt}=-\frac{b}{m}dt$

$\large \Longrightarrow v=v_{0}e^{-\frac{b}{m}t}$

At $t=3, v=\frac{v_{0}}{2}\Longrightarrow \frac{b}{m}=\frac{ln2}{3}$

$\large \Longrightarrow v=v_{0}e^{-\frac{ln2}{3}t}$

$\large \lim_{t\to\infty}v=0$

$\large dx=vdt \Longrightarrow x=\int_{0}^{\infty}vdt=\int_{0}^{\infty}v_{0}e^{-\frac{ln2}{3}t}dt=\frac{3}{ln2}v_{0}$

Do I have any mathematical error?


Could anybody post what they got for the first 5 or so?!?!?  I have no confidence in any of my answers.

See my thread for explanations (that are hopefully correct) for all 15 problems.


function
inequalities
inequalities proposed

Prove that in a Acute triangle we have:
[tex](tgA+tgB+tgC)(cosA+cosB+cosC+2(cosAcosB+cosBcosC+cosCcosA))\geq\ 9 \sqrt{3}[/tex].
[tex](tgA+tgB+tgC)(sinA+sinB+sinC)^3\geq\frac {3^6} {8}[/tex]

[quote="spdf"]Prove that in a Acute triangle we have:
[tex](tgA+tgB+tgC)(cosA+cosB+cosC+2(cosAcosB+cosBcosC+cosCcosA))\geq\ 9\sqrt(3)[/tex].
[tex](tgA+tgB+tgC)(sinA+sinB+sinC)^3\geq\ 3^6/8[/tex][/quote]
I thinks this problem very easy
let x = cos(B-C) and we have a function with x..

A short (not really  :D ) solution for the first problem is the combination of Garfulkel and Grindasov 's inequality.
     $\sum cos(B-C) \geq \frac{8}{\sqrt{3}}sinAsinBsinC$
     $tgAtgBtgCsinAsinBsinC \geq \frac{27}{8}$.

[quote="treegoner"]And definitely the second problem is the corollary of the Grindasov 's inequality  ;)[/quote]

what's Grindasov 's inequality? :?

Grindasov 's inequality is the second inequality I posted. :)

Sorry, I forgot this problem for a long time. The solution of Jack Garfulkel 's inequality can be viewed in http://www.mathlinks.ro/Forum/viewtopic.php?t=3849
  About the Grindasov, here is my solution :
     We notice $\frac{cosA}{sinBsinC} + \frac{cosB}{sinCsinA} +\frac{cosC}{sinAsinB} = 2$. The problem is equivalent to proving $\frac{cosA}{sinBsinC}. \frac{cosB}{sinCsinA} .\frac{cosB}{sinAsinC} \leq \frac{8}{27}$. From AM-GM, the result follows.


A person six feet tall walks around the earth. How many feet farther does the top of this person's head travel than the bottoms of his feet? Express your answer in terms of $ \pi$.

Because this person is 6 feet tall, the smaller radius is 6 feet less than the larger. Therefore, the diameter is 12 feet less, and the circumference is [b]12pi[/b] feet less for the smaller circle.


calculus
derivative
induction
trigonometry
calculus computations

By induction or otherwise  prove that if y = $cot^{-1}x$ then the nth derivative of  of y with respect to x is $(-1)^{n}. (n-1)!sinny.(siny)^{n}$

 Hence or otherwise prove that if
 z =$tan^{-1}\frac{xsina}{1+xcosa}$ then the nth derivative of z with respect to x is

$\frac{(-1)^{n-1}.(n-1)!.sin(na-nz).(sin(a-z))^{n}}_{sin^{n}a}$

\sin, \cos, \tan and \cot all look much nicer, as does \cdot in place of ".".


1)We are given 5 lines and two circles in a plane. What is the maximum number of possible intersection points among these seven figures?

2) How many lines are determined by 12 points in a plane, no three which are collinear?

[hide="1"]this may be wrong, but...

the two circles intersect twice.  the first line intersects the circles 4 times, and each line intersects all prior lines.

so $2+4(5)+1+2+3+4=\boxed{32}$[/hide]

[hide="2"]its just $12$ choose $2$, so $\frac{12!}{10!2!}=\frac{(12)(11)}{2}=(11)(6)=\boxed{66}$[/hide]

[quote="13375P34K43V312"][hide="1"] Consider the five lines. Each intersects with everything else in order to maximize points of intersections, so there are ${5\choose2}=10$ intersections of just the lines. Also, each line can intersect both circles twice, so we add another 4 for each line, making a total of $10+4(5)=30$.[/hide]

[hide="2"] A line is determined by 2 points, so it's just ${12\choose2}=66$.[/hide][/quote]

hmm...
i got a different answer for number 1, as you see in my above post.  i think you forgot that the circles can have two intersection points of their own?

13375P34K43V312, you forgot that the 2 circles can intersect each other twice.

Extention, how many [i]planes[/i] are determined by the 12 points where no 3 are collinear?

[hide="basically the same"]

sine 3 non-collinear points determine a plane, its just the same except its ${12\choose3}$.  so $\frac{12!}{9!3!}=\frac{12*11*10}{6}=(2)(110)=\boxed{220}$[/hide]


integration
calculus
logarithms
function
algebra
polynomial
calculus computations

$(1-x^2)y''-2xy'+2y=0$.  Given that $y_1$ and $y_2$ are independent solutions, and $y_1=x$, find $y_2$.  When I try to find $y_2$, I come across $\int \frac{e^{x^2}}{x^2} dx$, which I don't believe has an elementary primitive, but my teacher said we wil learn how to evaluate this integral later in the year. (I still don't think it has a closed form primitive)  I don't want to wait that long so I was wondering if someone can show me how to do this.

I don't get that integral at all. Use a method of reduction of order by setting $y=xu$. We wind up with the equation

$x(1-x^2)u''+2(1-2x^2)u'=0$

After finding the right multiplying factor, we get

$x^2(1-x^2)u''+2x(1-2x^2)u'=\left[x^2(1-x^2)u'\right]'=0.$

Letting the first arbitrary constant of integration be 1, we have

$u'=\frac1{x^2(1-x^2)}=\frac1{x^2}+\frac1{1-x^2}.$

$u=-\frac1x+\frac12\ln\left(\frac{1+x}{1-x}\right)$ so that

$y_2=-1+\frac{x}2\ln\left(\frac{1+x}{1-x}\right).$

The interest in Legendre's equation usually lies with functions that remain bounded at the singular points $x=\pm1.$ That interest would rule out this $y_2$ and bring you back to the polynomial solution.

I made a careless mistake, and now see that the "ugly" integral the teacher was referring to is $\int \frac{1}{x^2(1-x^2)} dx$, which is not difficult at all.

Thanks.


if a, b,  c are non-negative  numbers,then
\[ \sum{(a^{2}-bc)\sqrt{a^{2}+4bc}}\ge 0 \]

[quote="arqady"][quote="Utkirstudios"]if a, b,  c are non-negative  numbers,then
\[ \sum{(a^{2} \minus{} bc)\sqrt {a^{2} \plus{} 4bc}}\ge 0
\]
[/quote]
$ \sum_{cyc}{(a^{2} \minus{} bc)\sqrt {a^{2} \plus{} 4bc}}\ge 0\Leftrightarrow\sum_{cyc}a^{2}\sqrt {a^{2} \plus{} 4bc}\geq bc\sqrt {a^{2} \plus{} 4bc}\Leftrightarrow$

$ \Leftrightarrow\sum_{cyc}\left(a^{6} \plus{} 4a^{4}bc \plus{} 2a^{2}b^{2}\sqrt {(a^{2} \plus{} 4bc)(b^{2} \plus{} 4ac)}\right)\geq$
$ \geq\sum_{cyc}\left(a^{2}b^{2}c^{2} \plus{} 4a^{3}b^{3} \plus{} 2c^{2}ab\sqrt {(a^{2} \plus{} 4bc)(b^{2} \plus{} 4ac)}\right).$
But $ 2a^{2}b^{2}\sqrt {(a^{2} \plus{} 4bc)(b^{2} \plus{} 4ac)}\geq 2a^{3}b^{3} \plus{} 8a^{2}b^{2}c\sqrt {ab}$ and
$ 2c^{2}ab\sqrt {(a^{2} \plus{} 4bc)(b^{2} \plus{} 4ac)}\leq a^{3}c^{2}b \plus{} b^{3}c^{2}a \plus{} 4c^{3}a^{2}b \plus{} 4c^{3}b^{2}a.$
Id est, it remains to prove that
$ \sum_{cyc}\left(a^{6} \minus{} 2a^{3}b^{3} \plus{} 4a^{4}bc \plus{} 8a^{2}b^{2}c\sqrt {ab} \minus{} 5a^{3}b^{2}c \minus{} 5a^{3}c^{2}b \minus{} a^{2}b^{2}c^{2}\right)\geq0.$
We obtain:
$ \sum_{cyc}(a^{6} \minus{} 2a^{3}b^{3} \plus{} a^{4}bc) \equal{} \sum_{cyc}(a^{6} \minus{} a^{5}b \minus{} a^{5}c \plus{} a^{4}bc) \plus{} \sum_{cyc}(a^{5}b \plus{} a^{5}c \minus{} 2a^{3}b^{3})\geq0$
and $ \sum_{cyc}(a^{2}b^{2}c\sqrt {ab} \minus{} a^{2}b^{2}c^{2})\geq0.$
Let's assume $ a \equal{} x^{2},$ $ b \equal{} y^{2}$ and $ c \equal{} z^{2},$ where $ x,$ $ y$ and $ z$ are non-negative numbers.
Hence, it remains to prove that $ \sum_{cyc}(3x^{6} \minus{} 5x^{4}y^{2} \minus{} 5x^{4}z^{2} \plus{} 7x^{3}y^{3})\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(3x^{6} \minus{} 10x^{4}y^{2} \plus{} 14x^{3}y^{3} \minus{} 10x^{2}y^{4} \plus{} 3y^{6})\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x \minus{} y)^{2}(3x^{4} \plus{} 6x^{3}y \minus{} x^{2}y^{2} \plus{} 6xy^{3} \plus{} 3y^{4})\geq0.$  :)[/quote]

But it is ugly. :(

[quote="shfdfzhjj"][quote="arqady"][quote="shfdfzhjj"]

But it is ugly. :([/quote]
I agree with you. Prove better. :wink:[/quote]

Ok, we can use S.O.S to kill it!  :roll:[/quote]
I don't see it. :( What your proof, shfdfzhjj? Thank you!


For the quadratic equation, does the discriminant include the square root symbol?

For example, in the problem x^2+7x+6, is the discriminant sqrt25 (5) or 25?

Hello,

$ b^{2}\minus{}4ac$ would be discriminant.

The square root is not needed because you can determine whether the root exists based only on the sign of the discriminant.

Let $ D$ be the discriminant:

If $ \text{sgn}{D}>0$ then there are two real roots.
If $ \text{sgn}{D}\equal{}0$ then there is one real (double) root
If $ \\text{sgn}{D}<0$ then there are no real roots


inequalities
geometry
circumcircle
inequalities proposed

why does the $ RHS$ in [b]1[/b] seem to be too large? is it really like that?  :(

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

[b]1[/b]. Prove that for all triangles
\[ s^2 \,\leq \,\frac {25}{3}\,r^2 \plus{} \frac {75}{16}\,R^2\]
[b]2[/b]. Prove that $ \leq$ becomes $ \equal{}$ for only one triangle, but [b]NOT[/b] for the equilateral one! Find the angles of this triangle.

Notes:

[b]3[/b]. This effect (inequality becomes equality for non-equilateral triangle) cannot happen with the inequalities of the type $ xR \plus{} yr \plus{} zs \geq 0$, see http://www.mathlinks.ro/viewtopic.php?t=207461

[b]4[/b]. There exists geometric inequalities which becomes equality for [b]two[/b] different triangles, see http://www.mathlinks.ro/viewtopic.php?t=223717

I worked a lot to find this inequality, so I will be happy to see your proof or comment!

Thank you!
Sergey Markelov[/quote]
Let $ a,b,c$ be the sidelengths of the triangle and let $ a\equal{}x\plus{}y,b\equal{}y\plus{}z,c\equal{}z\plus{}x, x,y,z \ge 0$. Then $ s \equal{} x\plus{}y\plus{}z$, $ r \equal{} \frac{T}{s}\equal{} \frac{\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}}{s} \equal{} \sqrt{\frac{xyz}{x\plus{}y\plus{}z}}$, $ R \equal{} \frac{4RT}{4T} \equal{} \frac{(x\plus{}y)(y\plus{}z)(z\plus{}x)}{4\sqrt{(x\plus{}y\plus{}z)xyz}}$.
So we have to prove:
$ (x\plus{}y\plus{}z)^2 \le \frac{25}{3}\frac{xyz}{x\plus{}y\plus{}z} \plus{} \frac{75}{16}\frac{(x\plus{}y)^2(y\plus{}z)^2(z\plus{}x)^2}{16(x\plus{}y\plus{}z)xyz}$
Writing this in terms of $ 3u\equal{}x\plus{}y\plus{}z,3v^2\equal{}xy\plus{}yz\plus{}zx,w^3\equal{}xyz$ we see that it increasing in $ v^2$, and hence we need only to prove it when $ (x\minus{}y)(y\minus{}z)(z\minus{}x)\equal{}0$ and hence: only for isosceles triangles.
Don't know if it helped anyone, but I'm not in the mood for solving it when $ y\equal{}z\equal{}1$ right now :P

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

[b]1[/b]. Prove that for all triangles
\[ s^2 \,\leq \,\frac {25}{3}\,r^2 \plus{} \frac {75}{16}\,R^2\]
[b]2[/b]. Prove that $ \leq$ becomes $ \equal{}$ for only one triangle, but [b]NOT[/b] for the equilateral one! Find the angles of this triangle.

Notes:

[b]3[/b]. This effect (inequality becomes equality for non-equilateral triangle) cannot happen with the inequalities of the type $ xR \plus{} yr \plus{} zs \geq 0$, see http://www.mathlinks.ro/viewtopic.php?t=207461

[b]4[/b]. There exists geometric inequalities which becomes equality for [b]two[/b] different triangles, see http://www.mathlinks.ro/viewtopic.php?t=223717

I worked a lot to find this inequality, so I will be happy to see your proof or comment!

Thank you!
Sergey Markelov[/quote]

$ \boxed{s^2 \le 2R^2\plus{}10Rr\minus{}r^2\plus{}2\sqrt {R} (R\minus{}2r)^{3/2}}$ - Blandon inequality

So we must prove :$ \frac{43}{16}\minus{}10t \plus{}\frac{28}{3}t^2\minus{}2(1\minus{}2t)^{3/2} \ge 0$  , ($ t \equal{} \frac{r}{R} \le  \frac{1}{2}$ )

Equality holdes only for : $ \frac{r}{R}\equal{} \frac{3}{8}$ and  $ \alpha \equal{} 2 *arcsin (\frac{1}{4})$ , $ \beta\equal{}\gamma\equal{}\frac{\pi}{2}\minus{}arcsin(\frac{1}{4})$

More general, the following inequality holds for $ 0<m\le 1$ and $ k\ge 0$:

$ ms^2\le (m \plus{} 1)^2R^2 \minus{} 2(m^2 \minus{} 5m \plus{} 2)Rr \plus{} (4 \minus{} m)r^2 \plus{} k(R \minus{} \frac {2r}{1 \minus{} m^2})^2$.

For $ k \equal{} 0$, equality holds for an equilateral triangle, and for an isosceles triangle with $ b \equal{} c$ and $ \frac a{b} \equal{} 1 \minus{} m$.
For $ k > 0$, equality holds only in the second case.


Evaluate $ y(y\minus{}3x)$ for $ x\equal{}3$ and $ y\equal{}0$.

Well, since you are multiplying $ 2$ numbers, and one of them is $ 0$, then the answer is always $ \boxed{0}$.


geometry
incenter
circumcircle
ratio
geometry unsolved

In triangle $ABC$, let $I$ be the incenter, $O$ be the circumcenter, and $BO$ intersects circumcircle at $M$. Projections from $I$ to $AM$ and $CM$ are $P,Q$ respectively. Find the ratio $\frac{IP}{IQ}$ in terms of sides $a,b,c$

In triangle AIP $cos(I)=cos(\frac{A}{2})=\frac{IP}{AI}.$ So $IP=AIcos(\frac{A}{2}).$
In triangle CIQ $cos(I)=cos(\frac{C}{2})=\frac{IQ}{CI}.$ So $IQ=CIcos(\frac{C}{2}).$
Then $\frac{IP}{IQ}=\frac{AIcos(\frac{A}{2})}{CIcos(\frac{C}{2})}.$
In triangle AIC $\frac{AI}{CI}=\frac{sin(\frac{C}{2})}{sin(\frac{A}{2})}$
So $\frac{IP}{IQ}=\frac{tan(\frac{C}{2})}{tan(\frac{A}{2})}=\frac{p-a}{p-c}=\frac{-a+b+c}{a+b-c}$


group theory
abstract algebra
geometry
geometric transformation
superior algebra
superior algebra unsolved

Let $G$ be a group with $p^km$ elements, where $m<2p$. Prove that it has a normal subgroup whose order is either $p^k$ or $p^{k-1}$.

Hi,

In the original wording of the problem $p$ is prime and $k>1$. I will assume the former, and the latter yields a group of order $pm$ where $m<2p$, meaning either $p$ does not divide $m$ (in which case there is only one $p$-Sylow subgroup, which is normal) or $p=m$, in which case any subgroup of order $p$ will be normal (since a group of order $p^{2}$ is abelian).
Now for $k>1$, we consider the action of $G$ by translations (to the left) on the set of cosets of a $p$-Sylow subgroup $H$; this operation gives an homomorphism $f: G \rightarrow \sigma (G/H)$ whose kernel $K$ is included in $H$, with $p^k*m/|K|=m!$, which gives $|K|=p^{k}$ or $|K|=p^{k-1}$ considering that $p^{2}$ does not divide $m!$.


Let  ABC  be a triangle  and AA' , BB' , CC' three segments ( A' on BC , B' on AC and C' on AB)
so that  each of the segments  AA', BB', CC'  divides the perimeter of the triangle in two equal parts .
Show that  AA' , BB', CC'  are concurrent .

Let $a,b,c$ canonically be the lengths of the sides and $s$ the semiperimeter.
Then $\frac{BA'}{A'C} \cdot \frac{CB'}{B'A} \cdot \frac{AC'}{C'B} = \frac{s-c}{s-b} \cdot \frac{s-a}{s-c} \cdot \frac{s-b}{s-a} = 1$, so they intersect in one point by Ceva's theorem. (the point is called the Nagel-point)


geometry
3D geometry
sphere
analytic geometry
inequalities proposed
inequalities

Consider a box inscribed in the sphere with equation $x^{2}+y^{2}+z^{2}= 1$, such that the coordinates of its vertices are $\left(\pm x, \pm y, \pm z\right)$. The volume of the box is $8xyz$. Of course the volume of the box is strictly smaller than the volume of the sphere, which is $4\pi/3$.

That was also my solution. I was thinking if there is some easy solution not using geometric argument or spherical coordinates.