AMC
AIME
USAMTS
MATHCOUNTS
AMC 10
geometry
trapezoid

I got all 5s for the first time! I am very surprised :D

What about you guys?

24 :D I'm very proud because I actually beat a rival (from MA, guess who :roll: )

And also because I improved an entire 11 points from the first round!
That's right, [size=200][color=green]11[/color][/size]! :lol:

55554 -> I fail at anything involving computation, such as saying how the sum evaluates to $ 1506\cot\frac {\pi}{2008}$.

25.

Though I realized at the last second that I did some unnescessary copmlex number stuff at the 5th problem.

Uh, is there a reason why my solution to number 5 was completely discounted by the judge.  I gave the source code for my java program and proved that it indeed gave the right solution, and all the judge said was:

"You have the right answer, but didn't prove very much.  See the solution."

He gave me two points for having the right answer and didn't even bother to consider my justification for it.  wtf

Is this supposed to happen whenever a person uses a computer program to solve a question?  Cuz that's not what the rules say...

(I know I should be protesting this elsewhere, but can someone clarify before I make a fool out of myself?)

The absolute worst that could happen is your appeal gets denied. Seriously, don't hesitate, unless it truly is frivolous.

Yes, but the problem is that my chance of winning the protest is far too slim.  The criteria never even mentions if we can use a computer program to brute-force the problem, so what do I do?

And the protest page warns that if I protest, I should acknowledge the risk that my score may go down as well. :(

cough 16.... well, i don't feel too bad, because i spent less than a week on this...

5-5-5-5-5
My first perfect score!
Kinda surprising, as i didn't expect to recieve full points on number 3(but looking at the rubric, i fulfilled all the criteria)...
This round will count towards AIME qualification, right?

[quote="mathemonster"]5/5/5/5/0 ( :blush: )
So 20, and combined I have a 39.
Owned Yongyi lol jk[/quote]

Technically, you did own me because my total score stands 37 as of now.

This is what it says
[quote]In the algebra, deduct 1 point for a key step performed via Mathematica (or other
computer/calculator program) [b]without veriﬁcation[/b]. [/quote]

So if you can prove that your program gives the right answer, you would still get full points, I think.

You see, something like a Java program will most likely not spit out something in the form $ 1506\cot{\pi/2008}$, but rather a decimal. So there is really no way to truly verify a Java program as compared to your proposed answer.

Just saying that "it is only within 0.001 off" doesn't cut it rigorously.

@Yongyi, yes, but that does not justify taking off more than half of the question's alloted points for that isn't fair in my mind.  And anyway, that wasn't the reason the grader took off so many points.

[quote="jvenezuela716"][quote="Math Geek"]Yes, but the problem is that my chance of winning the protest is far too slim.  The criteria never even mentions if we can use a computer program to brute-force the problem, so what do I do?[/quote]

Umm... did you even read the grading criteria? It clearly says that solutions using a computer algebra system, etc., would get points off...[/quote]

Um, did [b]you[/b] even read the grading criteria?  Cuz if you did, then let me tell you that it "clearly" says that using a computer or other program in the [b]algebra [/b] would get a point off, like Anderson said.  So yeah, good call, jvenezuela716...

I'm pretty sure the person who graded my paper wasn't familiar with the rules.  So yeah...  Do the fundamental rules of USAMTS that every judge should consider count as "grading criteria?" :huh:

Judges grade very differently. Two people I know had the same (incomplete) solution. One got a 1 (or was it a 2?) and the other person got a 4. The solutions were the same, and the person who got a 1 wrote his solution better than the person who got a 4. So it really depends on whether you get a very strict judge.

Yay! I actually didn't fail! I did *TONS* better than I did in round 1! :D

I got a 20 (4/5/4/5/2)

Quite a surprise, though the grader didn't seem to like my solutions for #1 and #5 since they weren't well written, rigorous, or easy to follow. :|

Lol well, I hope I don't get on the fail boat again for Round 3 even though I kinda think I did...geez mathcounts crap got in the way.

5/5/0/0/2=12
i am protesting #4, because i want 1 point
blargh...i suck at math

[quote="mathemonster"]No I suck at math- I just FAILED the AMC10 A, despite having a 40 for USAMTS after 2 rounds.[/quote]
Don't worry, you're not alone. I failed too, even though I made the AIME easily. I still think I suck at AMC 10.

and to make this non-spam, how is it decided what the cutoff score for an USAMTS round is for a score group to be on the Leaderboard?

how is making AIME easily fail? nw I feel worse

i lost.
SCRATCH THAT: 33 points after getting my 1 point for #4 that i protested for

i got a 9 when i was expecting a 13  :( altho i think taht with the grading criteria i shouldve gotten an 11 but before i make a fool of myself i would like ur opinions to makes sure i shouldve gotten a 11 so here is my round 2 and 3 but i only want my round 2 checked

I think that when they told me i should prove my shape was a trapezoid it was actually quit obvious so thats one point back
I know it says all diagram are at the bottom of the page but i printed it off and hand drew them so there not available on my pdf ><

When they say prove 2^a=3^b-1 is only true for 8,9 (which i didnt) they took off two points when i think they shouldve taken off one but im not sure because i didnt directly find it in the grading criteria

I got an 18, boo.

total = 22 + 18 = 40 so far.

[quote="peregrinefalcon88"]
When they say prove 2^a=3^b-1 is only true for 8,9 (which i didnt) they took off two points when i think they shouldve taken off one but im not sure because i didnt directly find it in the grading criteria[/quote]

In my opinion, proving that is basically half the problem.

I didn't resubmit, but I probably should have.

Ha I feel pretty embarrassed: I tried to appeal my 3 on number 3, and I chose to appeal 2 instead!  :blush: I got 55342 for a 19, along with 22 for rd. 1 to give 41.

[quote="Yongyi781"]At least my 24 on Round 2 brought my total score to an ok range (37-41) as opposed to a mediocre range (0-36).[/quote]

YESS!!!! i made it to your idea of a good nonmediocre score.

I fail more than anyone else here  :lol:
5/5/0/0/0=10
But to be fair I did them all last minute and barely attempted anything other than #1 and #2.

I screwed up the LCM's on #4 and got 13 instead of 4, but I proved my answer (with the screwed up LCM) nevertheless, so shouldn't I have gotten... something?

Oh I meant "4 instead of 13"
Heh, not thinking clearly today...

5/4/0/5/0

Wow, everyone time I do a set of problems at this level, I fail at the ones that involve algebra (if you group competition math into algebra/inequalities, NT, combinatorics, and geo, so trig falls under algebra), but find the geo and NT and counting ones really easy (this means I don't learn anything at all in school? note: the only reason I do ok on geo is because I spent a lot of this summer studying basic geo since geo in school had nothing to do with proofs....or anything....sorta). I missed 3 and 5 this round completely (just like on round 1, where I missed 5 after trying to work on it for at least 12 hours and almost missed 3 had I not half-solved it at the last second) with 0 points on both.

Same thing on round 3, where I was totally clueless on number 5, but found the others to be easy.

I have yet to start on round 4 :(

On number 2 of this round, I did not prove that the quadrilateral was an isosceles trapezoid since I thought it was kind of obvious, but I guess that is my fault...

I guess I need to study some algebra....

[quote="grn_trtle"]Actually I just noticed that all I said was "so and so are parallel, so this is isosceles," and yet I got a 5.
Honestly, I put no effort into round 2 at all (and I totally skipped rounds 1 and 3).
But I'm almost done with round 4, so that's good  :lol:[/quote]

yea thats way early, but a waste if you messed up r1 + r3

I'm now seriously considering taking AIME through USAMTS, because I totally bombed AMC12 yesterday. :(

[quote="ProtestanT"]I got a 25 on round 2 as well! However I'm not as optimistic about round 3...since I only submitted solutions for problems 1 and 2 of round 3...[/quote] what??? why?

[quote="peregrinefalcon88"]i got a 9 when i was expecting a 13  :( [/quote]

Don't expect a 13. it's bad luck, which is probably why you got a 9.  :P

[quote="mathemonster"] what??? why?[/quote]
I didn't have time to do the others...

23. I got the right triple on #4 (Yay Mihailescu's Theorem!), but somehow ended up with 10. Also used a rectangle in #2 which I didn't prove was a rectangle...btw I lost.

Total of 47 so far, which I suppose isn't too bad...


function
limit
integration
real analysis
real analysis unsolved

Give an example of a sequence of continuous functions $f_{n}$ on $[0,1]$ that converge pointwise to zero but s.t. $\lim_{n\to \infty}\int_{0}^{1}f_{n}(x)dx$ is not zero.


function
inequalities
geometry
geometric transformation
reflection
inequalities proposed

Let $a,b,c,d$ be non-negative numbers.
(a) If $a+b+c+d=2$, then
            $\sum\frac 1 {1+a^2} \geq 3$;
(b) If $a+b+c+d=4$, then
            $\sum\frac 1 {1+a^2} \geq 2$;
(c) What is the minimum value $E_4(s)$ of
         $E_4=\sum\frac 1 {1+a^2}$
for $a+b+c+d=s$, where s>0 is given;
(d) Idem (c) for $n$ numbers.

AC method???

$\frac{1}{1+x^2}\geq\frac{2-x}{2}$.

  Darij

[quote="Vasc"](b) If $a+b+c+d=4$, then
            $\sum\frac 1 {1+a^2} \geq 2$.[/quote]
Yes, Darij: $\sum_{cyc}\frac{1}{1+a^2}\geq2\Leftrightarrow\sum_{cyc}(\frac{1}{1+a^2}-\frac{1}{2}+\frac{1}{2}(a-1))\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{a(a-1)^2}{1+a^2}\geq0.$ :)

In order to answer problem c, replace $s$ with $4s$ (it's less messy)
note that $\frac{1}{x^2+1} \geq \frac{3s^2+1-2sx}{s^2+1}^2$
is equivalent to $(x-s)^2(2sx+s^2-1) \geq 0$ which is true if $s \geq 1$
So minimum happens when $a=b=c=d=s$.

But if $s<1$ I don't really know.

The general problem is

SUppose that $a_1+a_2+...+a_n=n$. Find the best constant $k$ st
\[ \sum_{k=1}^n\frac{1}{k+a^2} \ge \frac{n}{1+k} \]

It's not difficult and same as a problem of my post here along time ago (unsolved)

Find the best constant $k$ st
\[ \sum_{k=1}^n\frac{a_i}{k+a_i^2} \ge \frac{n}{1+k} \]

[quote="hungkhtn"]The general problem is
Suppose that $a_1+a_2+...+a_n=n$. Find the best constant $k$ st
\[ \sum_{k=1}^n\frac{1}{k+a^2} \ge \frac{n}{1+k}  \][/quote]
Actually, a more general problem is to find the minimum value of
$\sum_{k=1}^n\frac{1}{k+a^2}$ for any given $k>0$.
This is equivalent to my question (d):
If $a_1+a_2+...+a_n=s$, then
$\sum_{k=1}^n\frac{1}{1+a_k^2} \geq n-s^2\max_{1\leq k\leq n} \frac{k}{k^2+s^2}$ .

[quote="Gibbenergy"][quote="Vasc"](b) If $a+b+c+d=4$, then
            $\sum\frac 1 {1+a^2} \geq 2$.[/quote]
I am not sure about this but I am inspired from the sols founded in http://www.mathlinks.ro/Forum/topic-82021.html
I guess half concave and convex can solve it .Right?[/quote]
This problem too easy, but in relation to this
[quote="Vasc"]If $a+b+c+d=\sqrt{12},$ then
$\sum \frac 1 {1+a^2} \geq \frac{16}{7}$[/quote]
 you are absolutely right.
Zhaobin's theorem kills it. ;)

Ac method? first time I hear :) could you tell more about it?

I called the AC-Theorem the following statement:
[b]Arithmetic Compensation Theorem (AC-Theorem).[/b]Let $s>0$, and let $F(x_1,x_2,...,x_n)$ be
a symmetrical continuous function on the compact set in $R^n$,
$S=((x_1,x_2,...,x_n): x_1+x_2+...+x_n=s, x_1 \geq 0,..., x_n \geq 0)$.
If $F(x_1,x_2,...,x_n) \leq \max{(F(\frac{x_1+x_2}{2},\frac{x_1+x_2}{2}, x_3,...,x_n),F(0,x_1+x_2,x_3,...,x_n))}$
for all $(x_1,x_2,...,x_n)\in S$ with   $x_1>x_2>0$, then
$F(x_1,x_2,...,x_n) \leq \max_{1\leq k\leq n}{F(\frac{s}{k},..., \frac{s}{k},0,...,0)}$  for all $(x_1,x_2,...,x_n)\in S$.

I don't know  :)  why but this's exacly the theorem my friend sent to me in this time I complete this book. His name is Boby_steven (Dinh Ngoc An) in Viet Nam and of course It's in my book. Sprising, Vasc invents it too, but Boby even invents it so longtime ago. He calls it UMV (Undefind Mixing Variable).

I don't negate that Boby_steven used this way longtime ago.
I used it two years ago in Crux Mathematicorom to prove and generalize a very difficult inequality. Moreover,
in my book Algebraic Inequalities there is a chapter destined to this very interesting method, including the theorem proof. ;)

Yes, My book is too. Even that SMV theorem and UMV theorem (as I say) can be proved by very elementary ways, not anything advance.

thank everybody to tell more about AC method. Could anyone give a online book or reference for me? I'm interested in it.

[quote="Vasc"]I rounded the problem with (c) and (d) questions.
A particular and more difficult case of (c) - compared with (a) and (b), is the following:
If $a+b+c+d=\sqrt{12}$, then
$\sum \frac 1 {1+a^2} \geq \frac{16}{7}$.[/quote]
Application of zhaobin's theorem.
Let $a=\frac{x\sqrt3}{2},$ $b=\frac{y\sqrt3}{2},$ $c=\frac{z\sqrt3}{2},$ $d=\frac{t\sqrt3}{2}.$
Hence, $x+y+z+t=4$ and we have to prove that $\sum_{cyc}\frac{1}{3x^2+4}\geq\frac{4}{7}.$
Let $f(x)= \frac{1}{3x^2+4}.$ Then $f''(x)=\frac{6(9x^2-4)}{(3x^2+4)^3}.$
Hence, $f$ is convex on $[\frac{2}{3},4]$ and $f$ is concave on $[0,\frac{2}{3}].$
Case 1. $\frac{2}{3}\leq x\leq y\leq z\leq t.$
Here $f(x)+f(y)+f(z)+f(t)\geq4f\left(\frac{x+y+z+t}{4}\right)=4f(1)=\frac{4}{7}.$
Case 2. $x\leq\frac{2}{3}\leq y\leq z\leq t.$
Hence, $\sum_{cyc}f(x)\geq f(x)+3f\left(\frac{y+z+t}{3}\right)=f(x)+3f\left(\frac{4-x}{3}\right).$
Remain to prove here that $f(x)+3f\left(\frac{4-x}{3}\right)\geq\frac{4}{7}.$
But $f(x)+3f\left(\frac{4-x}{3}\right)\geq\frac{4}{7}\Leftrightarrow\frac{1}{3x^2+4}+\frac{9}{x^2-8x+28}\geq\frac{4}{7}\Leftrightarrow$
$\Leftrightarrow x(x-1)^2(x-6)\leq0,$ which true.
Case 3.1 $x\leq y\leq\frac{2}{3}\leq z\leq t$ and $x+y\leq\frac{2}{3}.$
Then $f(x)+f(y)\geq f(0)+f(x+y)$ and $f(z)+f(t)\geq2f\left(\frac{z+t}{2}\right).$
Let $x+y=u,$ where $u\in[0,\frac{2}{3}].$
Hence, remain to prove here that $f(0)+f(u)+2f\left(\frac{4-u}{2}\right)\geq\frac{4}{7}.$
But $f(0)+f(u)+2f\left(\frac{4-u}{2}\right)\geq\frac{4}{7}\Leftrightarrow(3u-4)^2(3u^2-16u-8)\leq0.$
The last inequality is true for $u\in[0,\frac{2}{3}].$
Case 3.2 $x\leq y\leq\frac{2}{3}\leq z\leq t$ and $x+y\geq\frac{2}{3}.$
Here $f(x)+f(y)\geq f\left(x+y-\frac{2}{3}\right)+f\left(\frac{2}{3}\right)=f\left(u-\frac{2}{3}\right)+f\left(\frac{2}{3}\right)$
and $f(z)+f(t)\geq2f\left(\frac{z+t}{2}\right)=2f\left(\frac{4-u}{2}\right).$
Hence, remain to prove here that $f\left(u-\frac{2}{3}\right)+f\left(\frac{2}{3}\right)+2f\left(\frac{4-u}{2}\right)\geq\frac{4}{7}.$
But $f\left(u-\frac{2}{3}\right)+f\left(\frac{2}{3}\right)+2f\left(\frac{4-u}{2}\right)\geq\frac{4}{7}\Leftrightarrow f\left(u-\frac{2}{3}\right)+2f\left(\frac{4-u}{2}\right)\geq\frac{1}{7}.$
This is obviously true, since $u\in[\frac{2}{3},\frac{4}{3}]$ and $f\left(u-\frac{2}{3}\right)\geq f\left(\frac{4}{3}-\frac{2}{3}\right)=\frac{3}{16}>\frac{1}{7}.$
Case 4.1 $x\leq y\leq z\leq\frac{2}{3}\leq t$ and $x+y+z\leq\frac{2}{3}.$
Hence, $f(x)+f(y)+f(z)\geq2f(0)+f(z).$
Remain to prove here that
$2f(0)+f(z)+f(4-z)\geq\frac{4}{7}\Leftrightarrow\frac{1}{3z^2+4}+\frac{1}{3(4-z)^2+4}\geq\frac{1}{14},$  where $z\leq\frac{2}{3}.$
But $\frac{1}{3z^2+4}\geq f\left(\frac{2}{3}\right)=\frac{3}{16}>\frac{1}{14}.$
Case 4.2 $x\leq y\leq z\leq\frac{2}{3}\leq t$ and $x+y+z\geq\frac{2}{3}.$
Hence, $\sum_{cyc}f(x)\geq f(x)+f(y)+f(z)\geq f(0)+f(y)+f\left(\frac{2}{3}\right)\geq\frac{1}{4}+\frac{3}{16}+\frac{3}{16}>\frac{4}{7}.$ :)
Welcome back zhaobin!

[quote="conceive"]Note that

[quote="darij grinberg"]$\frac{1}{1+x^2}\geq\frac{2-x}{2}$[/quote]

is equivalent with $x(x-1)^2\geq0$ what is obvious for all $x\geq0$.

conceive[/quote]
And $\sum_{cyc}\frac{1}{1+x^2}\geq\sum_{cyc}\frac{2-x}{2}=4-\sqrt3$ $?$
But  $\frac{16}{7}>4-\sqrt3.$ ;)

[quote="Vasc"]I called the AC-Theorem the following statement:
[b]Arithmetic Compensation Theorem (AC-Theorem).[/b]Let $s>0$, and let $F(x_{1},x_{2},...,x_{n})$ be
a symmetrical continuous function on the compact set in $R^{n}$,
$S=((x_{1},x_{2},...,x_{n}): x_{1}+x_{2}+...+x_{n}=s, x_{1}\geq 0,..., x_{n}\geq 0)$.
If $F(x_{1},x_{2},...,x_{n}) \leq \max{(F(\frac{x_{1}+x_{2}}{2},\frac{x_{1}+x_{2}}{2}, x_{3},...,x_{n}),F(0,x_{1}+x_{2},x_{3},...,x_{n}))}$
for all $(x_{1},x_{2},...,x_{n})\in S$ with   $x_{1}>x_{2}>0$, then
$F(x_{1},x_{2},...,x_{n}) \leq \max_{1\leq k\leq n}{F(\frac{s}{k},..., \frac{s}{k},0,...,0)}$  for all $(x_{1},x_{2},...,x_{n})\in S$.[/quote]

I've read your article about AC method in mathematical reflection. You say in that article that if $f(a_{1},a_{2},\cdots,a_{n})$ is the maximum of the function then $f(a_{1},a_{2},\cdots,a_{n})=f(\frac{a_{1}+a_{2}}{2},\frac{a_{1}+a_{2}}{2},a_{3},\cdots,a_{n})$ or $f(a_{1},a_{2},\cdots,a_{n})=f(0,a_{1}+a_{2},a_{3},\cdots,a_{n})$ and you say that using this process repeatedly then eventualy there is $(x_{1},\cdots,x_{n})$ where $x_{i}\in\{0,s/k\}$ for which $f(x_{1},\cdots,x_{n})=f(a_{1},\cdots,a_{n})$. Is this that obvious? For me it is not obvious at all and I think we need some kind of limiting argument as in the proof of SMV theorem by hungkhtn.

[quote="arqady"][quote="Vasc"]I rounded the problem with (c) and (d) questions.
A particular and more difficult case of (c) - compared with (a) and (b), is the following:
If $a+b+c+d=\sqrt{12}$, then
$\sum \frac 1{1+a^{2}}\geq \frac{16}{7}$.[/quote]
Application of zhaobin's theorem.
Let $a=\frac{x\sqrt3}{2},$ $b=\frac{y\sqrt3}{2},$ $c=\frac{z\sqrt3}{2},$ $d=\frac{t\sqrt3}{2}.$
Hence, $x+y+z+t=4$ and we have to prove that $\sum_{cyc}\frac{1}{3x^{2}+4}\geq\frac{4}{7}.$
Let $f(x)= \frac{1}{3x^{2}+4}.$ Then $f''(x)=\frac{6(9x^{2}-4)}{(3x^{2}+4)^{3}}.$
Hence, $f$ is convex on $[\frac{2}{3},4]$ and $f$ is concave on $[0,\frac{2}{3}].$
Case 1. $\frac{2}{3}\leq x\leq y\leq z\leq t.$
Here $f(x)+f(y)+f(z)+f(t)\geq4f\left(\frac{x+y+z+t}{4}\right)=4f(1)=\frac{4}{7}.$
Case 2. $x\leq\frac{2}{3}\leq y\leq z\leq t.$
Hence, $\sum_{cyc}f(x)\geq f(x)+3f\left(\frac{y+z+t}{3}\right)=f(x)+3f\left(\frac{4-x}{3}\right).$
Remain to prove here that $f(x)+3f\left(\frac{4-x}{3}\right)\geq\frac{4}{7}.$
But $f(x)+3f\left(\frac{4-x}{3}\right)\geq\frac{4}{7}\Leftrightarrow\frac{1}{3x^{2}+4}+\frac{9}{x^{2}-8x+28}\geq\frac{4}{7}\Leftrightarrow$
$\Leftrightarrow x(x-1)^{2}(x-6)\leq0,$ which true.
Case 3.1 $x\leq y\leq\frac{2}{3}\leq z\leq t$ and $x+y\leq\frac{2}{3}.$
Then $f(x)+f(y)\geq f(0)+f(x+y)$ and $f(z)+f(t)\geq2f\left(\frac{z+t}{2}\right).$
Let $x+y=u,$ where $u\in[0,\frac{2}{3}].$
Hence, remain to prove here that $f(0)+f(u)+2f\left(\frac{4-u}{2}\right)\geq\frac{4}{7}.$
But $f(0)+f(u)+2f\left(\frac{4-u}{2}\right)\geq\frac{4}{7}\Leftrightarrow(3u-4)^{2}(3u^{2}-16u-8)\leq0.$
The last inequality is true for $u\in[0,\frac{2}{3}].$
Case 3.2 $x\leq y\leq\frac{2}{3}\leq z\leq t$ and $x+y\geq\frac{2}{3}.$
Here $f(x)+f(y)\geq f\left(x+y-\frac{2}{3}\right)+f\left(\frac{2}{3}\right)=f\left(u-\frac{2}{3}\right)+f\left(\frac{2}{3}\right)$
and $f(z)+f(t)\geq2f\left(\frac{z+t}{2}\right)=2f\left(\frac{4-u}{2}\right).$
Hence, remain to prove here that $f\left(u-\frac{2}{3}\right)+f\left(\frac{2}{3}\right)+2f\left(\frac{4-u}{2}\right)\geq\frac{4}{7}.$
But $f\left(u-\frac{2}{3}\right)+f\left(\frac{2}{3}\right)+2f\left(\frac{4-u}{2}\right)\geq\frac{4}{7}\Leftrightarrow f\left(u-\frac{2}{3}\right)+2f\left(\frac{4-u}{2}\right)\geq\frac{1}{7}.$
This is obviously true, since $u\in[\frac{2}{3},\frac{4}{3}]$ and $f\left(u-\frac{2}{3}\right)\geq f\left(\frac{4}{3}-\frac{2}{3}\right)=\frac{3}{16}>\frac{1}{7}.$
Case 4.1 $x\leq y\leq z\leq\frac{2}{3}\leq t$ and $x+y+z\leq\frac{2}{3}.$
Hence, $f(x)+f(y)+f(z)\geq2f(0)+f(z).$
Remain to prove here that
$2f(0)+f(z)+f(4-z)\geq\frac{4}{7}\Leftrightarrow\frac{1}{3z^{2}+4}+\frac{1}{3(4-z)^{2}+4}\geq\frac{1}{14},$  where $z\leq\frac{2}{3}.$
But $\frac{1}{3z^{2}+4}\geq f\left(\frac{2}{3}\right)=\frac{3}{16}>\frac{1}{14}.$
Case 4.2 $x\leq y\leq z\leq\frac{2}{3}\leq t$ and $x+y+z\geq\frac{2}{3}.$
Hence, $\sum_{cyc}f(x)\geq f(x)+f(y)+f(z)\geq f(0)+f(y)+f\left(\frac{2}{3}\right)\geq\frac{1}{4}+\frac{3}{16}+\frac{3}{16}>\frac{4}{7}.$ :)
Welcome back zhaobin![/quote]
What is Zhaobin's theorem  :?:  :?:

[quote="barra"][quote="Vasc"]I called the AC-Theorem the following statement:
[b]Arithmetic Compensation Theorem (AC-Theorem).[/b]Let $s>0$, and let $F(x_{1},x_{2},...,x_{n})$ be
a symmetrical continuous function on the compact set in $R^{n}$,
$S=((x_{1},x_{2},...,x_{n}): x_{1}+x_{2}+...+x_{n}=s, x_{1}\geq 0,..., x_{n}\geq 0)$.
If $F(x_{1},x_{2},...,x_{n}) \leq \max{(F(\frac{x_{1}+x_{2}}{2},\frac{x_{1}+x_{2}}{2}, x_{3},...,x_{n}),F(0,x_{1}+x_{2},x_{3},...,x_{n}))}$
for all $(x_{1},x_{2},...,x_{n})\in S$ with   $x_{1}>x_{2}>0$, then
$F(x_{1},x_{2},...,x_{n}) \leq \max_{1\leq k\leq n}{F(\frac{s}{k},..., \frac{s}{k},0,...,0)}$  for all $(x_{1},x_{2},...,x_{n})\in S$.[/quote]
Is this that obvious? For me it is not obvious at all and I think we need some kind of limiting argument as in the proof of SMV theorem by hungkhtn.[/quote]
I think you are right, but only in the extended case of the theorem. This is in accordance with my assertion "using this process repeatedly then eventualy (finally) there is $(x_{1},\cdots,x_{n})$ where $x_{i}\in\{0,s/k\}$ for which $f(x_{1},\cdots,x_{n})=f(a_{1},\cdots,a_{n})$".

[quote="Vasc"]
I think you are right, but only in the extended case of the theorem. This is in accordance with my assertion "using this process repeatedly then eventualy (finally) there is $(x_{1},\cdots,x_{n})$ where $x_{i}\in\{0,s/k\}$ for which $f(x_{1},\cdots,x_{n})=f(a_{1},\cdots,a_{n})$".[/quote]

My question is: is this assertion abvious and doesn't deserve any proof or you just didn't include the proof because of any other special consideration (to make the paper short, or maybe any other reason)

[quote="Vasc"][quote="barra"] My question is: is this assertion abvious and doesn't deserve any proof or you just didn't include the proof because of any other special consideration (to make the paper short, or maybe any other reason)[/quote]
In my opinion, this assertion is about obvious and many readers know it. I saw one or two rigorous proofs for it in Crux Mathematicorum.
On the other hand, Mathematical Reflections is not a journal of theoretical mathematics. Thus, because of the space restriction or of the difficulty restriction, even the introductory inequality remainded unsolved in my article. The result unpublished is the following.
Let $a_{1},a_{2},...,a_{n}$ be non-negative numbers such that $a_{1}+a_{2}+...+a_{n}=n$, and let
$F=\frac 1{p-a_{2}a_{3}...a_{n}}+...+\frac 1{p-a_{1}a_{2}...a_{n-1}}$.
If $e_{n-1}=(1+\frac 1{n-1})^{n-1}$ and $q=\frac{(n-1)e_{n-1}}{n-e_{n-1}}$, then
(a) $F \leq \frac n{p-1}$ for $p\geq q$;
(b) $F \leq \frac{n-1}{p}+\frac 1{p-e_{n-1}}$ for $e_{n-1}<p\leq q$.[/quote]
Can you post your solution?

[quote="eagerkill"]
What is Zhaobin's theorem  [/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64933

Here is my problem. Do we can use AC theorem to prove it?
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=495675

See a complete proof of the AC Theorem in the paper
ON THE ARITHMETIC COMPENSATION METHOD
published in 2012 in
http://www.ijpam.eu/contents/2012-80-3/index.html
or
http://www.ijpam.eu/contents/2012-80-3/6/6.pdf


For how many ordered pairs of positive integers $ (x,y)$ is $ x\plus{}2y \equal{}
100$?

This is a diophantine equation...
Making a parametric equation,
$ x_1 \equal{} 20 \plus{} 2 t$
$ y_1 \equal{} 40 \minus{} t$
t can range from 1 to 39, or else $ y_1$ will not be positive.
Therefore, there are $ \boxed{39}$ pairs.

You fail. What is 49*2? Well 50 doesn't work because then the other variable ends up being zero. So 49 ways.

chengp, please don't make rude comments to other users, like "You fail".  You should say, "I disagree with you because...".

We have $x + 2y = 100$.  There are a few possibilities:  $x = 2$ and $y = 49$, $x = 4$ and $y = 48$, and so on up to $x = 98$ and $y = 1$.  So the answer is indeed $\boxed{49}$.

2004 AMC 12A Problem 3.

I initially misread the "positive integers" and put down 51. (I included (100, 0) and (0, 50)).


search
Ramsey Theory
combinatorics unsolved
combinatorics

For $n$ people, if it is known that

(a) there exist two people knowing each other among any three people, and

(b) there exist two people not knowing each other among any four people.

Find the maximum of $n$.

Here, we assume that if $A$ knows $B$, then $B$ knows $A$.

This is asking for the value of the ramsey number R(3,4) it is 9 so the maximum wanted is 8. You can search the web for proof of this.

we will prove that $ 8\ge n$

(*)if someone A knows at least 6 people, by Ramsey's theorem, there exist 3 people among them who dont know each other in every three students, or there exist 3 people they know each other. A and the three people form a group of four people such that every two of them know each other. contradiction
(**)if some one A knows at most n-5 people, then in the remaining there are at least 4 people, none of knows A. Thus every two of the four people know each other, contradiction.
when $ n\ge10$ one of (*) and (**) must occur
if $ n \equal{} 9$ in order to avoid (*) and (**), each person knows exactly 5 pther people. Thus the number of pairs knowing each other make a contradiction.This contradiction implies $ 8\ge n$.Hence, the maximum value is 8.

Here is the picture for $n=8$ (my sol is the same as aboves):

[img] https://lh3.googleusercontent.com/ieAlRuBqTkMAdxyQqakQI5PHGtRrAlZl0efUNx0hvR1PDNZzu3NJ06DamgO-jRJKxru2vg=s100 [/img]

Here, where $A_1 , A_2 , \dots, A_8$ represent the $8$ students. The line segment between $A_i$ and $A_j$ means $A_i$ and $A_j$ know each other.


geometry
perpendicular bisector
geometry open

ABC is an acute-angle triangle.M,N is on AB and BC sides ,respectively(not A,B,C,D).K is midpoint of MN.B,D be intersection of CAN and BCM of circumcirle.CD line through cimcumcenter of ABC.
Prove that the perpendicular bisector of AB throug K lie.

It might be easy, provided the proper wording/notation was given!

Best regards,
sunken rock


quadratics
group theory
abstract algebra
vector
arithmetic sequence
number theory
relatively prime

Can you show that the set $ M: \equal{}\{\sqrt{p} \,\, | \,\, p \,\; \text{is prime} \}$ is $ \Bbb{Q}\minus{}$ linear independent?

Yes.  Suppose to the contrary that there exists a $ \mathbb{Z}$-linear relation

$ \sum_{k \equal{} 1}^{n} c_k \sqrt {p_k} \equal{} 0$

(which is equivalent to the existence of a $ \mathbb{Q}$-linear relation).  By quadratic reciprocity, for every $ p_i$ there exists an arithmetic progression with common difference $ 2p_i$ such that the primes $ q$ such that $ \left( \frac {p_i}{q} \right) \equal{} 1$ are precisely the primes in this arithmetic progression, and similarly there exists an arithmetic progression with common difference $ 2p_i$ such that the primes $ q$ such that $ \left( \frac {p_n}{q} \right) \equal{} \minus{} 1$ are precisely the primes in this arithmetic progression.

By CRT, the union of all of these arithmetic progressions, each of which has the property that the common difference is relatively prime to the first term, is an arithmetic progression with the same property, and by Dirichlet's theorem this progression contains a prime $ q$ that does not divide $ c_n$.

Now consider the homomorphism $ \phi : \mathbb{Z}[ \sqrt {p_1}, ... ] \mapsto \mathbb{F}_q$ defined by identifying $ \sqrt {p_k}, k < n$ with the corresponding square root $ \bmod q$ and identifying $ c_n \sqrt {p_n}$ with the value implied by the relation.  Then $ \sqrt {p_n}$ exists $ \bmod q$; contradiction.

Edit:  Here is a second proof.

The splitting field of $ P(x) \equal{} \prod_{k \equal{} 1}^{n} (x^2 \minus{} p_k)$ is the field $ \mathbb{Q}[ \sqrt {p_1}, ... \sqrt {p_n} ]$ and we want to show that it has degree $ 2^n$ over $ \mathbb{Q}$.  By a basic property of the degree of field extensions, this is true if and only if for every $ i$ the field $ \mathbb{Q}[ \sqrt {p_1}, ... \sqrt {p_{i \minus{} 1}} ]$ does not contain $ \sqrt {p_i}$.

The Galois group of the above extension is $ (\mathbb{Z}/2\mathbb{Z})^{i \minus{} 1}$, so its subgroups are vector spaces over $ \mathbb{F}_2$ of dimension at most $ i \minus{} 1$.  By the fundamental theorem of Galois theory the only intermediate fields of degree $ 2$ over $ \mathbb{Q}$ are the fixed fields of subgroups corresponding to vector spaces of codimension $ 1$, and these are precisely the fields

$ \mathbb{Q}[ \prod_{j \in S} \sqrt {p_j} ]$

where $ S$ is a non-empty subset of $ \{ 1, 2, ... i \minus{} 1 \}$.  Clearly none of these fields are the field $ \mathbb{Q}[ \sqrt {p_i} ]$.

The second method is more general (and proves a stronger result!), but the first method is, I think, better suited for small cases, since in those cases it's a finite calculation to exhibit a prime $ q$ that works.

[quote="t0rajir0u"]By quadratic reciprocity, for every $ p_i$ there exists an arithmetic progression with common difference $ 2p_i$ such that the primes $ q$ such that $ \left( \frac {p_i}{q} \right) \equal{} 1$ are precisely the primes in this arithmetic progression, and similarly there exists an arithmetic progression with common difference $ 2p_i$ such that the primes $ q$ such that $ \left( \frac {p_n}{q} \right) \equal{} \minus{} 1$ are precisely the primes in this arithmetic progression.[/quote]

Please could you explain how the existence of the arithmetic progression follows from quadratic reciprocity. I've been looking at it for a while, but I can't quite spot it.

Thank you

Quadratic reciprocity gives $ \left( \frac {p_i}{q} \right) \equal{} \left( \frac {q}{p_i} \right) ( \minus{} 1)^{\frac {(q \minus{} 1)(p_i \minus{} 1)}{4} }$.  The factor $ \left( \frac {q}{p_i} \right)$ is periodic of period $ p_i$ in $ q$ and the other factor is periodic of period $ 4$ in $ q$, so the RHS is periodic of period $ 4p_i$ (not $ 2p_i$ as I wrote).  (The proof is slightly different when $ p_i \equal{} 2$, but it's essentially the same.)  This is just a generalization of results like "the primes such that $ \minus{} 1$ is a square are precisely the ones congruent to $ 1 \bmod 4$" that quadratic reciprocity is supposed to explain.  (Also, it's actually a union of disjoint arithmetic progressions, but that doesn't matter for this problem; pick one.)

Also, my last step is not rigorous as written:  the homomorphism should actually be into $ \mathbb{F}_{q^2} \equal{} \mathbb{F}_q[ \sqrt{p_n} ]$ and the contradiction is that $ \sqrt{p_n}$ is sent to the prime field $ \mathbb{F}_q$.

[quote="t0rajir0u"](Also, it's actually a union of disjoint arithmetic progressions, but that doesn't matter for this problem; pick one.)[/quote]

This was exactly the problem that I couldn't quite work out. Thanks you very much for your time, it is much appreciated.


function
inequalities
algebra unsolved
algebra

Let $f(x)=ax^{2}+bx+c$ that $|f(x)| \leq 1$ for all x satisfying $|x| \leq 1$.
Prove that:
$|f(x)| \leq 17 \Leftrightarrow x \leq 3$

The correct conclusion is |$f(x)| \leq 17$ when $|x| \leq 3$
Is this right? Linkin_Park? I think you should check (phone to some of your friend, Eg: Green Tea ,,  :D )


rotation
geometry
geometric transformation
number theory
relatively prime

Prove that $m | (m-1)!+1 \Rightarrow m \in \mathbb{P}^{+}.$

[hide="Solution"]Assume the contrary, that $m = ab$ for $a, b > 1$ and $gcd(a, b) = 1.$ Since $a, b$ are relatively prime, we may prove that $(ab-1)!+1 \equiv 0 \bmod{a}$ and $(ab-1)!+1 \equiv 0 \bmod{b}$ in order to prove that $m | (m-1)!+1.$ Now, note that $ab-1 > a-1 \ge a \Leftrightarrow a | (ab-1)!,$ which contradicts the equation $(ab-1)!+1 \equiv 0 \bmod{a}.$ Thus, our assumption that $a, b > 1$ is false. Since $a,b > 0$ irrespective of the value of $m$ being prime or not, we know that either $a$ or $b$ is $1,$ a contradiction.[/hide]

A couple of things:

1. This isn't Wilson's Theorem; it's the converse.

2. Your proof fails for $m = 4$, such $a$ and $b$ don't exist. However, it is only necessary that $(m-1)!$ contain any divisor $> 1$ of $m$, so it's fine.

3. Lastly, it's false for $m = 1$. But that's just being picky.

Well, if you want to prove this one, it suffices to show $(m, [m-1]!) \neq 1$...and if m is not prime, this is clearly the case...though this isn't Wilson's Theorem, as Paladin pointed out.  The proof of Wilson's Theorem involves pairing with multiplicative inverses...(ie every element 1, 2, 3, ..., p-1 has a unique multiplicative inverse in that set, so you can create a pairing scheme, except 1 and p-1 are their own multiplicative inverses, so you put them in separately).

The full statement of Wilson's theorem is in fact

$m > 1, m | (m-1)!+1 \Leftrightarrow m \in \mathbb{P}$

I don't suppose anyone has come up with a combinatorial proof?

[quote="Ravi B"]There is a combinatorial proof that if $p$ is a prime, then $(p-1)!+1$ is divisible by $p$.  Are you interested in seeing it?[/quote]

Sure  :)

Here's a combinatorial proof of Wilson's Theorem which uses stellated polgons.

If $p=2$ then the theorem clearly holds. So we consider on the odd primes. Consider $p$ points evenly distributed on a circle. Now we consider how many stellated $p$-gons we can form. First we know that the number of permutations is $P^{p}_{p}$ but since each of the $p$-gons can be described in $2p$ different ways so we really only have $\frac{p!}{2p}$. But then we have some $p$-gons that are unchanged when we rotate them by $\frac{2 \pi}{p}$ and we have $\frac{p-1}{2}$ of these. So the rest of the $p$-gons, $\frac{p!}{2p}-\frac{p-1}{2}$ fall naturally into sets of $p$-elements and each member of the set can be obtained by succesive rotations by $\frac{2 \pi}{p}$. Thus the total number of sets are $\frac{(p-1)!-(p-1)}{2p}$ and hence $2p | [(p-1)!-p+1]$.

Therefore: $p | (p-1)!+1$.

Note: There's probably some typos in here.


We have the inequality

$\sqrt{x-1}+\sqrt{y-1}\leq\sqrt{xy}$  for  $x,y\geq 1$ .

Now use this for the numbers $a,b$  and  then  to  $ab+1,c$

and you will have what  you want


Hi. Reading another topic i think this one would be interesting to share: What does your country's name means? In which language is it? Which is it's oficial name?

In the case of Mexico, the official name is "Estados Unidos Mexicanos" (United States of Mexico). Mexico is a nahuatl word and means: "Place of the center of the moon". It was originally the name of Mexico-Tenochtitlan (Wich is now the Mexico City) The whole country took the name of its capital (funny thing, isn't it?) after the independence (Before Mexico was part of the New Spain). The official name was influenciated by the U.S.

China, or People's Republic of China.

Named for its fine china. :P

Argentina comes from "Argentinum", which means "plata"

"Argentinum"="silver" in latin I thought¿
(or does "plata" mean this; at least it sounds more like "flat" or "square")

Yes, "silver", but the colour... (because of the "Río de la Plata" a nice-coloured river)

Río de la Plata: Isn't it possible that the river and the country are named for the precious metal (chemical symbol Ag)? During colonial times, the river, along with the ports near its mouth, were a significant export route for silver mined in what is now Bolivia.

Bosnia and Herzegovina. Official name, in Bosnian/Croatian/Serbian: Bosna i Hercegovina. Origin of the name Bosna (Bosnia) isn't known. some historians think that Bosnia was named that name because of a Slavic tribe which inhabited Bosnia in the 7th century-the name was first mentioned in the X century (Constantin Porphirogenetis: De administrando imperio). Herzegovina got its name much later (late 1440s). It was named after the title of its ruler, Herceg (Herzog) Stjepan Kosaca. Until 1878, this area was often called just Bosnia, and it was its official name (Banovina Bosna, Kingdom of Bosnia, Sanjak of Bosnia, Eyalet of Bosnia). After the Austro-Hungarian occupation in 1878, Bosnia became The Province of Bosnia and Herzegovina. Later, in the SHS kingdom, Kingdom of Yugoslavia and socialist Yugoslavia, it was called the province of Bosnia and Herzegovina (Later: (Socialist) Republic of Bosnia and Herzegovina).

[quote="mathnerd314"]United States of America :sleeping:

Although there's another topic discussing the usage of "America"...[/quote]

You may have noticed that this topic explicitly developed from that one.

The United States of America arises, of course, because the nation was formed by uniting a number of states, and these states were in America, thus United States of America.  "America" is derived from Amerigo Vespucci, an Italian cartographer.

I was thinking that, since we have a large number of Americans who now will have very little to add to this conversation, they might consider doing the names etc. of their individual states.

never knew that

anyways, i live in the US

I live in the US. The Americas were named after Amerigo Vespucci, an Italian explorer, who made voyages in the late 1400s and early 1500s.

[quote="JBL"]
I was thinking that, since we have a large number of Americans who now will have very little to add to this conversation, they might consider doing the names etc. of their individual states.[/quote]
I think that's a great Idea. So, anyone who lives in the U.S. please post your state name

Where I live: Texas (Don't know); New Jersey (named for Jersey; I don't know where that name comes from though)

Where I was born: England (I don't know :( ); Great Britain, however, is from the Roman name for the island, Britannia.

Where my family is from: Ireland (from the Gaelic name, I guess, but beyond that I don't know) and Israel (Biblical of course).

Vermont literally means green mountains in french

[quote="kimmystar94"]anyone know for Korea? i'm pretty sure it's this:

In korean it's written, 대한민국. Which means large korean country of the people.
which is more like, The Republic of Korea

대한민국 comes from chinese characters..
大韓民國

anyone nice want to translate for me?[/quote]
Big Korean Republic.

China in Chinese means "Middle Kingdom" as American textbooks put it. I like to say "Middle Country" because I personally think the textbooks overcomplicate Chinese things to try and sound better.
Yes we Chinese are very egocentric, so we thought that our country was in the middle of the world :)

[quote="ლევანი"] ... (qartvel - Georgian)... [/quote]
I think qartvel[b]i[/b]-georgian.
also I note that it's [b]not[/b] state of georgia.
And in Georgia is: Georgian language.
    I think, it will be good if there, in National forum will be  also Georgian forum, because much people from Georgia can't speak in English. Yes , I agree , they can try , but...   :maybe:

I live in Canada, and I think I am one of the few on this site. The majority seem to be americans

[quote="kimmystar94"]yeah.. cuz isn't it like
中國??
I remember now!! :)

didn't the chinese think they were the middle of the world?[/quote]

yes. we're egotists :D

guess what nationality

South Georgia and the South Sandwich Islands..

Your flags under your post count and stuff say that.
Of course I'm not going to believe that...

Bangladesh; comes from an old region named Banga -> Bengal -> Bangla, and Desh which means country. In the period 1947-1971, following the end of the British rule India was partitioned into three parts: one of which was East Bengal (later named as East Pakistan). Bangladesh was named after the Liberation war in 1971.

Before the invasion of the west, Chinese Empire was the political center of the East Asia (maybe except Japan), and Chinese Emperor was the only legal emperor in that region. The title of the leaders of other countries was only KING. And since long long time ago, the center of China was called Middle Earth (interesting). That was the origin of MIddle Country.

Colombia, a.k.a  "República de Colombia".

I believe that the name comes from Cristopher Columbus.

In South Carolina, Carolina comes from Carolus, which is the Latin name for Charles after the King Charles who ruled England at the time...

Go figure :P

[quote="pacman2812"]I believe that the name comes from Cristopher Columbus.[/quote]
Well, yes, who else would it be besides Christopher Columbus (Cristofero Colombo, Cristóbal Colón)? The man's name also appears in:

The city of Columbia, South Carolina.
The city of Columbia, Missouri.
The Columbia River in the Pacific Northwest of North America (although technically, the river seems to have been named for a ship.)
The city of Colón, Panama.

And probably a hundred other places.

... all dedicated to a man who was so badly wrong about estimating the circumference of the Earth that it's no surprise the Portuguese told him to go away.

District of Columbia, British Columbia, I suspect also...

sLOVEnia(probably never heard of this country) :rotfl:

whoa actually I have I bet the capital is ljubljana


prove for  all positive integers p $\geq 3$  exist  a positive integers $n_1 , n_2 ,..., n_p$ such that $\frac{1}{n_1} + \frac{1}{n_2} + ...+ \frac{1}{n_p} = 1$

See : http://www.mathlinks.ro/Forum/viewtopic.php?t=65445 . :)


induction
combinatorics unsolved
combinatorics

graph $G_{n}$ has $n!$ points which represents all permutation of set ${1,2,...,n}$, and there is a line between two points if one permutation can be gathered by changing the mutual position of only two numbers. is there a hamiltonian circuit in this graph $G_{n}$?

It is just induction: assume it is true for $n-1$ and then just move $n$ from the far right to far left, one at a time, through the set of all permutations, and the proof will hold.


function
calculus
derivative
probability
topology
calculus computations

Dear mathlinkers i am looking for functions continus but no derivable in any number.
coold u help me.

There are at least two senses in which "most" continuous functions are nowhere differentiable. The first is Baire category: the functions that have a derivative at any point are of category 1 (countable union of nowhere dense subsets) in the complete metric space $ C[a,b].$ The second is measure, in the form of probability. If we use the notion of Brownian motion to give us a probability measure on the set of continuous functions on a certain interval with a certain base point, then the subset of those functions with a derivative at any point is a set of probability zero.


We need to put n cents of stamps on an envelop, but we have only (an unlimited supply of) 5 cents and 12 cents stamps. Prove that we can perform the task if n>=44.

[hide] It is possible to make all values greater than $ mn-m-n$ using only stamps of value $ m$ or $ n$.  Therefore you can have all values greater than $ 5\times 12-5-12=43$, or all values greater than or equal to 44.[/hide]


Given a string of x numbers in the range 1 to y (e.g. if x is 5 and y is 4, a string might be 4, 3, 4, 2, 2) perform any operation on any two adjacent numbers following the order of operations to achieve a given final, z. With the string 4, 3, 4, 2, 2, you could perform the operations 4+3*4/2-2 which, using the order of operations, gives 10.

For all strings of numbers x long within range 1 to y, what is the number of combinations which yield a given number z when any operations are performed?

For example, if x is 3, y is 4, the combinations you get are
1 1 1
1 2 2
...
4 4 4

Say you let z=5.

How many of those strings of numbers from 1, 1, 1 to 4, 4, 4 can you perform operations between any two adjacent terms such that the number that comes out at the end is 5? What is the percentage of strings with this characteristic?

This gets really big computational-wise. Anyone have any ideas about how to go about making a formula for such an idea?

$4+\frac{3 \cdot 4}{2} - 2 = 8$

Do the operations need to be in a set order?
If so, then you might be able to formulate something relatively easy, depending on whether you run into difficulties and how ordered you allow the operations to be.

If you are allowed to make things complicated (for example $x=6$ and $y = 7$, then $\frac{7^3}{\frac{3+4}{5 \cdot 2}}$ would be an example of something that is very nasty looking.

How did you get that? The operations can be in any order, and the string satisfies the condition if at least one order of operations gives the desired number in the end. Could you put the steps to how you got it and the number of strings which satisfy the condition in variable form?

Thanks!

I was giving you an example of a string, of course I made it appear nasty in sheer geometry, but the evaluation of the fraction is easy.

I still don't understand exactly what type of operations are allowed, in general your question is not well defined.  Mabye if you gave a very simple case, I would understand the question.

Let x= the length of the string
y= the upper bound for the terms in the string (no term may be above y, we'll assume the lowest number is 1)
z= some number we want to get to

In the case x=2 y=2 z=1
String     Operations
1 1 ---> 1+1=2    1-1=0      1*1=[b]1[/b]      1/1=[b]1[/b]
1 2 ---> 1+2=3    1-2=-1      1*2=2      1/2=.5
2 1 ---> 2+1=3    2-1=[b]1[/b]      2*1=2      2/1=2
2 2 ---> 2+2=4    2-2=0      2*2=4      2/2=[b]1[/b]

The number of strings satisfying z=1 for x=2 and y=2 is 3 out of 4. Notice that order is important as an operation is inserted between two numbers acting upon the second, and note that the order of operations should be followed in higher cases (2, 3, 2 could have one possible order of operations 2+3*2 which would give 8, not 10).

The number of strings is not the number of different order of operations performed on the strings. All that is important for the different order of operations is that one such order for a given string yields the predetermined z.

Another question I thought of, determine the z that is most possible for given x,y.

For x=3 y=2 all of the strings are
1 1 1
1 1 2
1 2 1
1 2 2
2 1 1
2 1 2
2 2 1
2 2 2
And then perform operations on each of those strings to get a given z.

The only allowed operations are +-/* and you may not add parentheses.


function
advanced fields
advanced fields unsolved

It is known that every linear function defined on a finite dimensional set is continuous. I'm looking for a counterexample to refute it in infinite dimensional case. Does any one have an idea?

A linear function $f: \E \to \RR$ is continuous if, and only if, it is bounded on the unit ball of the normed vectorial space $E$. From here it is very easy to find non-continuous functional if $E$ is infinite dimensional.
(e.g. : $e_{n}\to n$ where $e_{n}$ is an orthonormal basis of $E$)

The functional defined by alekk works fine if $E$ is not assumed to be complete, of if we do not require the functional to be defined on the entire space. To "find" an unbounded linear functional defined at every point of a complete normed space, one must use Zorn's lemma.


real analysis
probability
real analysis unsolved

$\mathbb T = \mathbb{R}/ \mathbb{Z}$ the usual torus with the induced Lebesgue measure. Suppose that $A_{k}$ are measurable subsets of $\mathbb T$ such that $\sum_{k>0}m(A_{k}) =+\infty$. Prove that there exists a sequence of real numbers $x_{k}$ such that for almost every $x \in \mathbb{T}$, $x$ belongs to infinitely many $x_{k}+A_{k}$. (here, $x_{k}+A_{k}$ is the translated set, modulo $\mathbb{Z}$ of course)

PS: looks like [url=http://planetmath.org/encyclopedia/BorelCantelliLemma.html]Borel Cantelli lemma[/url] :)

Yep. Choose $x_{k}\in\mathbb T$ uniformly and independently: by Borel-Cantelli you'll get the sequence you want with probability $1$.


combinatorial geometry
combinatorics unsolved
combinatorics

Determine whether or not there exist two disjoint infinite sets $ A$ and $ B$ of points in the plane satisfying the following: when all points in $ A$ are coloured blue, and points in $ B$ coloured red:
conditions:
1- No three points in $ A \cup B$ are collinear, and the minimum distance between any two points in $ A \cup B$ is 1.
2- There is a red point of A in every triangle whose vertices are blue, and vica versa.

Such sets do not exist. Suppose we have A and B with given properties. Since A is discrete we can  find 5 points of A such that in convex hull of this 5 points there is no other point from A.

CASE 1 Convex hull is pentagon $ A_{1}..A_{5}$. But then in all triangles $ A_{1}A_{2}A_{3}$,  $ A_{1}A_{3}A_{4}$ and $ A_{1}A_{4}A_{5}$ there is a point from B. The triangle formed from these 3 points from B is in the pentagon $ A_{1}..A_{5}$, so it does not contain point from A. Contradiction.

CASE 2 Convex hull is quadrilateral ABCD, and fifth point X is inside it. Then consider the triangles XAB, XAD, XBC and XCD. In all of them there is a point from B, Simmillary to the CASE 1  we get a contradiction.

CASE 3  Convex hull is triangle. The same as above.


2006 #30:
[img]http://img18.imageshack.us/img18/8597/88464176.png[/img]

The correct answer is B. But when I try assuming fast equilibrium, I get $ Rate \equal{} k[H_2][NO]^2/[H_2O]$

why does the H2O part of it just disappear?

2006#49

[img]http://img22.imageshack.us/img22/8084/88847356.png[/img]

Correct answer is A. I don't understand that at all so somebody please explain :)

Thanks!

For 49, I was thinking that repulsion would be higher for smaller atoms, but that gives the opposite answer... so I went and googled it:

"It would be important for you to
understand hybridization theory. The bond angle of H-O-H is 105, and that
of H-
S-H is 92. This would suggest that S is hardly hybridized at all, and in fact
using p-orbitals (with little s-characteristics) to form the S-H bonds,
whereas O is sp3 hybridized.

To understand why S is not hybridized, we need to understand that the large
size of S allows the electron pairs to be far from each other so that the
energy incurred from repulsion is not very large and need not be minimized.
Since O is a lot smaller, a more stable, less energetic configuration, would
have to be made - hence the hybridization."

http://www.newton.dep.anl.gov/askasci/chem03/chem03447.htm

So apparently, as you go down in periods, the larger size of the atoms results in less hybridization, thus the angle approaches 90 degrees.


conics
hyperbola
geometry
perpendicular bisector
geometry proposed

Let $ABCD$ be a convex quadrilateral ($AB>AD$) inscribed in the circle with the diameter $[AC]$. I consider two mobile points $X\in [BC]$, $Y\in [DC]$ so that $\widehat {BAX}\equiv \widehat {DAY}$. Ascertain the geometrical locus of the middlepoint $L$ of the segment $[XY]$.

May be, I've missed something but, as the triangles ABX and ADY are indirectly similar, the midpoint of XY lies on the perpendicular bisector of BD

Here is my solution(?):

XY envelopes a parabola. So midpoint is on a tangent line through

center of circle.

 Thank you.

 M.T.

[quote="armpist"]Here is my solution(?):

XY envelopes a parabola. So midpoint is on a tangent line through

center of circle.

 Thank you.

  :spider:  :gathering:  :spider:

Let’s prove  my generalization of this problem. :)
[color=green]Problem:Let $ABCD$ be  a convex quadrilateral $(AB>AD)$that
$\angle ABC=\angle ADC$ .I consider two mobile points$X\in{BC},Y\in{DC}$so that$\angle{BAX}=\angle{DAY}$. Ascertain the geometrical locus of the middlepoint$L$of the segment$[XY]$.[/color]
[color=red]Solution:[/color]Let $M$,$N$ be a middlepoint of segments $AC$,$MD$.
We have that $\angle ABC=\angle ADC$and$\angle{BAX}=\angle{DAY}$from my problem(we solved it in this topic[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=38041]www.mathlinks.ro/Forum/viewtopic.php?t=38041[/url])
we can say that $M,N$and$L$ lie in same line $\Rightarrow$ That locus is the line pase through $M$and$N$  ;)

If the condition <ABC = <ADC  is relaxed,  then the locus

   is the 2 branches of a hyperbola.

   With the equality of angles, hyperbola branches degenerate into straight lines,

   with one of them joining  the line at infinity, which is hard to see

   on the computer screen.

   Thank you.

   M.T.


:blush: what does D.I.V.E. mean??? :blush: i never heard of it.

i don't know what DIVE is either :blush:

i aint noe nottin bout D.I.V.E too :spider:


let we have a circle and 3 points inside of it.draw a triangle that it's vertix are on the circle and sides of triangle pass through these points.

[color=darkblue]Also if we have n points inside the circle with can cunstruct a n-gon,that those points lie on sides of n-gon.[/color]

I just recently came across this; it's known as Castillon's problem.  Here are some references:

[url]http://mathworld.wolfram.com/CastillonsProblem.html[/url]

[url]http://www.math.uoc.gr/~pamfilos/eGallery/problems/Castillon_Circle_Triangle.html[/url]


Evaluate $ \sum\limits_{m = 1}^\infty {\sum\limits_{n = 1}^\infty {\frac{1}
{{m^2 n + mn^2 + 2mn}}} }$


Hey everybody. I am wondering where can I get ACoPS in the GTA. Any help will be greatly appreciated. :)

I just used amazon.com... Otherwise you might be able to find a copy at the toronto reference library, near younge and bloor.


If $ a\plus{}b\plus{}c \not\equal{} 0$ and $ \frac{a^2}{b\plus{}c}\plus{}\frac{b^2}{a\plus{}c}\plus{}\frac{c^2}{a\plus{}b}\equal{}0$ then find the value of:

\[ \frac{a}{b\plus{}c}\plus{}\frac{b}{a\plus{}c}\plus{}\frac{c}{a\plus{}b}\]

[quote="moldovan"]If $ a + b + c \not = 0$ and $ \frac {a^2}{b + c} + \frac {b^2}{a + c} + \frac {c^2}{a + b} = 0$ then find the value of:
\[ \frac {a}{b + c} + \frac {b}{a + c} + \frac {c}{a + b}\]
[/quote]

\[ \boxed{(a+b+c)\cdot %Error. "display" is a bad command.
\sum_{cyc}\frac{a}{b+c}} = %Error. "display" is a bad command.
\sum_{cyc}\frac {a^2}{b + c} + %Error. "display" is a bad command.
\sum_{cyc}a(\frac{b}{a+c}+\frac{c}{a+b})= %Error. "display" is a bad command.
\sum_{cyc}(\frac{ab+ac}{b+c})=a+b+c\]
Therefore
\[ \frac {a}{b + c} + \frac {b}{a + c} + \frac {c}{a + b}=1\]


Given $p,q\in\mathbb{R}$ and the equation $x^3-px+q=0$ has three real roots.
Show that:
(1) $p\ge 0$.
(2) all the roots $\le 2\sqrt{\frac{p}{3}}$.

[quote="gotztahbeazn"]wait isn't $p \le 0$ because it's negative in the trinomial?[/quote]

Nope... $p\ge 0$

Can't anyone  do something with the senond one?

[quote="Bob"]1. Suppose the roots are $r_1$, $r_2$, and $r_3$. Then $r_1r_2+r_1r_3+r_2r_3=-p$ and $r_1+r_2+r_3=0$. From the second equation, we have $r_3=-r_1-r_2$. Substituting in the first equation, we have $-r_1^2-r_2^2-r_1r_2=-p$. By Descarte's Rule of signs, the equation has two positive roots and one negative root. Without loss of generality, suppose $r_1$ is the negative root. Applying AM-GM to the above eqution, we have $-p\le0$. So $p\ge0$.

I hope this is right and I'll have to think some more about the second one.  :)[/quote]

but why Descartes' Rule of Signs? we don't know what the sign value of $q$ is. $q$ can be positive, negative, and zero. maybe you need to check more cases...

$(x+1)(x+2)(x-3)=x^3-7x-6=0$
$(x-1)(x-2)(x+3)=x^3-7x+6=0$
$x(x-1)(x+1)=x^3-x+0=0$

Palytoxin was right.  I treated q like it was a number and not a variable.  :blush: I may be mistaken again, but maybe this is correct:

Suppose $p\leq0$. If $q$ is positive, then the equation becomes $x^3+|p|x+|q|=0$. By Descarte's Rule of signs, it has no positive solutions, since $f(x)$ has no sign changes, and at most 1 negative solution, since $f(-x)$ has one sign change. So unless one of the solutions is 0, the equation has no more than 1 real solution, which is a contradiction. Suppose $q$ is negative. Then the equation becomes $x^3+|p|x-|q|=0$. f(x) has one sign change and f(-x) has no sign changes. Contradiction again! Since $q$ can't be zero, $p\geq0$

hmm... interesting contradiction Bob

this is what i did: using the same notation as your previous solution, $r_1,r_2,r_3$ are real roots, then $r_1+r_2+r_3=0$ and $r_1r_2+r_2r_3+r_3r_1=-p$. so $r_3=-(r_1+r_2)$ and put it into the second expression and simplify we get $p=r_1^2+r_2^2+r_1r_2$. but we can simplify it by using AM-GM with absolute values of the roots $|r_1|,|r_2|$ and get
\[p=r_1^2+r_2^2+r_1r_2\geq 2|r_1r_2|+r_1r_2\geq |r_1r_2|\geq 0\]
and this always works too and we dont need to know the nature of the real roots

the second part baffles me. i only got it down to $2p=r_1^2+r_2^2+r_3^2$ after some algebraic manipulation. maybe you can finish it off for me.. :)


geometry
incenter
geometric transformation
homothety
perimeter
circumcircle
reflection

If nothing else works (I can't see anything nice right now :(), you can use a tiny application of Menelaus and show that $DP$ passes through the incenter of $ABC$ (analogously, the other two lines will also pass through the incenter, so this is the point we're looking for).

The $\triangle DEF$ is the medial triangle of the $\triangle ABC$, i.e., centrally similar with the homothety center at the centroid $G$ and homothety coefficient $-1/2$. Let $s$ be the semiperimeter of the $\triangle ABC$. Then

$\frac{FQ}{DQ} \cdot \frac{DR}{RE} \cdot \frac{EP}{FP} = \frac{s - BC}{s - AB} \cdot \frac{s - AC}{s - BC} \cdot \frac{s - AB}{s - AC} = 1$

By Ceva's theorem, the cevians $DP, EQ, FR$ are concurrent. Let $s(\triangle DEF)$ be the semiperimeter of the $\triangle DEF$.

$DE + DQ = BF + [s(\triangle DEF) - BF] = s(\triangle DEF)$

and similarly,

$EF + ER = FD + FP = s(\triangle DEF)$

Hence, the cevians $DP, EQ, FR$ are perimeter splitters of the medial $\triangle DEF$ and as such, they meet at the Nagel point of this triangle. The Nagel point of medial triangle is identical with the incenter of the original $\triangle ABC$.

Because of similarity of the triangles $\triangle AEF, \triangle BFD, \triangle CDE$ with the triangle $\triangle ABC$ (homothety coefficients $1/2$), this concurency point $K$ is the midpoint of the incenter $I$ and the orthocenter $H$ of the $\triangle ABC$.

The distance of the tangency point $P$ of the incircle of the $\triangle AEF$ from the vertex $E$ is

$EP = s(\triangle AEF) - AF = \frac{s(\triangle ABC) - AB}{2}$

and similarly

$FQ = \frac{s(\triangle ABC) - BC}{2}$, $DR = \frac{s(\triangle ABC) - CA}{2}$

Let $P', Q', R'$ be the tangency points of the excircles of the $\triangle DEF$ opposite to the vertices $D, E, F$ with the sides $EF, FD, DE$.  The distance of the tangency point $P'$ from the vertex $E$ is

$EP' = s(\triangle DEF) - DE = \frac{s(\triangle ABC) - AB}{2}$

and similarly,

$FQ' =  \frac{s(\triangle ABC) - BC}{2}$, $DR' = \frac{s(\triangle ABC) - CA}{2}$

The points $P' \equiv P, Q' \equiv Q, R' \equiv R$ are identical. The normals through the incenters of the $\triangle AEF, \triangle BFD, \triangle CDE$ to the sides $EF, FD, DE$ pass through the excenters $X, Y, Z$ of the $\triangle DEF$. Let $M$ be the foot of a normal from the point $K$ to the side $XY$ of the $\triangle XYZ$. The right angle $\triangle XKM, \triangle YKM$ are congruent, because the side $KM$ is common and $\measuredangle XKM = \measuredangle YKM$. This follows from the fact that $KM$  is parallel to the bisector of the $\measuredangle EFD$ and $XK \perp EF, YK \perp FD$, i.e., both angles are equal to $90^o - \frac{\measuredangle EFD}{2}$. Consequently, $KX = KY$ and similarly, $KY = KZ$, i.e., the point $K$ is the circumcenter of the $\triangle XYZ$ - the Bevan point of the $\triangle DEF$.

The Spiker and 9-point circles of the $\triangle ABC$ are the incircle and circumcircle, respectively, of the medial $\triangle DEF$ and the Bevan point of a triangle is the reflection of its incenter in its circumcenter.

[quote="mathmaniack"]D,E and F are the respective midpoints of the sides BC,CA and AB  of triangle ABC. The incircles of triangles AEF,BFD and CDE are tangent to EF,FD and DE at P,Q and R,respectively. Prove that PD,QE and RF are concurrent, and identify their point of concurrency as one of the centers of ABC.[/quote]

Just to prove this problem equivalent to another problem on the board:

As Yetti pointed out, the point of concurrency of the lines PD, QE and RF is the incenter of triangle ABC, so the solution of the problem comes down to showing that the lines PD, QE and RF all pass through the incenter of triangle ABC. Of course, it is enough to show that the line PD passes through the incenter of triangle ABC; the same property for the other two lines will then follow by analogy.

If the incircle of triangle ABC touches the side BC at $X_1$, then the point P is the midpoint of the segment $AX_1$. Why? Well, this is really easy to see: Since the points E and F are the midpoints of the segments CA and AB, the homothety with center A and factor 2 maps these points E and F to the points C and B, respectively; thus, this homothety maps the triangle AEF to the triangle ACB, and consequently, this homothety maps the point where the incircle of triangle AEF touches its side EF to the point where the incircle of triangle ACB touches its side CB. In other words, our homothety with center A and factor 2 maps the point P to the point $X_1$. Hence, the point P is the midpoint of the segment $AX_1$. Remembering that the point D is the midpoint of the segment BC, the fact that the line PD passes through the incenter of triangle ABC now becomes identic with the problem posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=2888 .

By the way, the problem is quite well-known: if my memory doesn't cheat me, it was proposed in Crux Mathematicorum by Antreas P. Hatzipolakis and Paul Yiu.

  Darij


I'm sorry but most of us will only help you if you post your work and specific questions about steps you need help on.

[quote="Donovan"]Divide using synthetic division...
1.) (3x^3-2x^2+2x-1) / (x-1)

Simplify. Assume all varibles are postive...
1.) 4√54
2.) 1/ 3+√5
3.) (5+i)(1+i)

Give the equation in both standard form and slope intercept form...
1.) It passes through (6, -7) and (8,13)

Solve for exact values of x...
1.) 3^x=3^(3x+1)
2.) 36^x=6^(x^2-3)
3.) log(x^2+36)=log 100

Factor completely...
1.) (x^3) +64
2.) (8x^3) -27

Solve for exact values...
1.) x^2 -x-12=0

Solve the inequality...
1.) x^2 +4x-21 <=0
  :( thanks if you can help...[/quote]
Simplify:
1. 12sqrt(6)
Solve for exact values of x:
1. x=-0.5
3. x=[u]+[/u]8
Solve for exact values:
1. x=-3, 4


Let $\{a_n\}$ be a sequence defined recursively as follows:

$\ a_1=1,a_{n+1}=\frac{1}{2}a_n+\frac{n^{2}-2n-1}{n^{2}(n+1)^{2}} $ (n=1,2,3,....)   Find the maximum value of n such that  $\sum_{k=1}^na_k$ is minimal.

Note: negative exponents are not used to make the proof a bit esaier to understand.

[hide]Let .  We then have a sequence defined to be

.

Knowing that ,  we can deduce that the sum of n elements [/hide]

Aargh!!  I just realized that I messed up and I don't want to take the time to correct myself...I'll just post what I have because I know that this is generally the correct way to do it...

At first it may seem hard for a pre-olympian. A careful manipulation of the algebraic expressions reveals that it is not that bad at all.

[b]towbomb[/b]: Can you solve this problem when you come back to the forum? I'm just curious if you can do it since you [b]almost[/b] solved it  :D  :D  :D

It's really not such an ugly problem as thought at first. The recursion formula yields:

$a_{n+1}=\frac{a_n}{2}-\frac{1}{n^2}+\frac{2}{(n+1)^2}$.

By induction it is easy to prove:

$a_{n+k}=\frac{a_n}{2^k}-\frac{1}{2^{k-1}n^2}+\frac{2}{(n+k)^2}$.

Now we set $n=1$ and $k=i-1$:

$a_i=\frac{2^i-i^2}{i^2\cdot2^{i-1}}$.

From these formula we have: $a_1=1$, $a_2=0$, $a_3<0$, $a_4=0$ and $a_k>0$ for $k>4$. So we have the minimal patial sum $\sum_{i=1}^ka_i$ for $k=4$. And its value is $\frac{35}{36}$.

I had a lot of fun with this problem :D  :D  :D . Thanks, kunny.

Misha

Thank you ,Misha! :P

Your solution is exatly right.

Let me answer only one comment,please.

From the formula which you yielded first, we have $a_{n+1}-\frac{2}{(n+1)^2}=\frac{1}{2}\left(a_n-\frac{2}{n^2}\right)$.

Solving this,we obtain $a_n=\frac{2}{n^2}-\left(\frac{1}{2}\right)^{n-1}(n\geqq1)$

kunny

Okay, I came back to solve it my way:

Let $m=\frac{n^2-2n-1}{n^2(n+1)^2}$
$a_1=1,\ a_{n+1}=\frac{1}{2}a_n+m$
$a_1=1$
$a_2=\frac{1}{2}+m$
$a_3=\frac{1}{2}\left(\frac{1}{2}+m\right)+m=\frac{1}{4}+\frac{3m}{2}$
$a_n=\frac{1}{2^{n-1}}+2m-\frac{1}{2^{n-2}}m$ for all $n\ge 2$
Using the formula for the sum of the negative powers of 2, we get:
$S_n=2-\frac{1}{2^{n-1}}+2nm-\left(2-\frac{1}{2^{n-2}}\right)m$
$S_n=2-\frac{1}{2^{n-1}}+\left(2n-2+\frac{1}{2^{n-2}}\right)m$
$S_n=\frac{2^n-1}{2^{n-1}}+\left(\frac{(n-1)2^{n-1}+1}{2^{n-2}}\right)m$
And then substitute for m:
$S_n=\frac{2^n-1}{2^{n-1}}+\left(\frac{(n-1)2^{n-1}+1}{2^{n-2}}\right)\frac{n^2-2n-1}{n^2(n+1)^2}$

Now, the problem becomes finding the minima of this equation (which, if you couldn't tell, is not a nice derivative).

Thank you , towbomb!

Good luck for performance marching band!

I'm looking forward to your solution.

kunny

Thank You!  I am (still) in the process of solving it, but I furthered my solution.

I came home from school today and once again worked on my solution.  It is not nearly as elegant as I had first thought, but it looks like it will still work if I can figure out that messy derivative.

Not to be discouraging, but Mathematica has decided that it can't solve that equation for you (the one that results from taking a derivative, that is).  The fact that this is a problem of positive integers suggests that you're better off comparing $f(x)$ to $f(x+1)$ than to $f(x+\epsilon)$

It's quite typical of students who have recently learned calculus to try to apply calculus to everything possible.  It really isn't feasibly possible to do this, as this problem is concerned more with the discrete than the continuous, and calculus at a basic level only works well when we apply it to continuous functions.

towbomb's way does not make sense. $n$ is an index, you can't just deal with a function of $n$ as if it were some constant.


We can prove the stronger:
Given $ a, b, c > 0$ and $ a\plus{}b\plus{}c\equal{}3$. Prove that: $ (a^2b^2\plus{}b^2c^2)(ab\plus{}bc\plus{}ca)^2 \leq\ 27$
<=> $ q^2(q^2\minus{}6r) \leq\ 27$
Using $ r \geq\ \frac{4q\minus{}9}{3}$, we have: $ RHS \leq\ (q\plus{}1)(q\minus{}3)^3 \leq\ 0$ (obviously trues)  :)


let F be a function from Z+toZ+ where Z+=(1,2,3......) such that F(F(m)+F(n))=m+n for every m,n in Z+. Find all possible values for F(2004)

Let $f$ be such a function.
Clearly, $f$ is injective since if $f(a)=f(b)$ then $a+b=f(f(a)+f(b))=f(f(a)+f(a)) = a+a$.
Now, for each $n \geq 2$, we have :
$f(f(n)+fn)) = 2n = (n-1)+(n+1) = f(f(n-1)+f(n+1))$.
Since $f$ is injective, we deduce that $2f(n) = f(n-1)+f(n+1)$. It easily follows that $f(x) =ax+b$ where $a,b$ are constant.
Conversely, the unique solution with that form is $f(x)=x$.

Thus $f(2004)=2004$.

Pierre.

yes! i did the same thing brothers !
i ll try to solve it with somthing else ! :)
bye


Let $p$ be a prime number and $a,b$ be two positive integer numbers s.t. $p|a^2+ab-b^2$ doesn't hold. Define the sequence $u_n$ as:
$u_1=a;u_2=b;u_n=u_{n-1}+u_{n-2}$. Prove that this sequence is periodic $\mod p$ with period isn't depend on $a,b$.

[i]moderator edit[/i]

I see no movement here.Too bad,can someone find a solution?


function
real analysis
real analysis unsolved

Let $f$ be a real function defined on $S \subset R$.  Under what condition of $f$ do we have

$\bigcap_{n=1}^{\infty}f(E_{n})= f(\bigcap_{n=1}^{\infty}E_{n})$ for any $S \supset E_{1}\supset E_{2}....$ ?

I hope the condition of $f$ is as weak as possible.

The only condition that I can see is hardly a weak one: that $f$ be injective.

It should always be true that $f(\bigcap_{n=1}^{\infty}E_{n}) \subseteq \bigcap_{n=1}^{\infty}f(E_{n});$ it's the opposite inclusion that is more problematic.

No. Consider $E_{n}=[n,\infty)$ and let $f$ be any continuous periodic function.

What about monotone continuous function? this is actually the function I am dealing with for this problem right now.

A strictly monotone function is injective. Continuity is irrelevant.

The necessary and sufficient condition is that $f^{-1}(y)$ be finite for any $y$.

Necessity: if $f^{-1}(y)$ contains distinct points $\{x_{n}\}_{n\ge 1}$, then let $E_{n}=\{x_{k}\colon k\ge n\}$.

Sufficiency: if $y\in\bigcap_{n\ge 1}f(E_{n})$ and $f^{-1}(y)=\{x_{1},\dots,x_{N}\}$ is finite, then at least one $x_{i}$ belongs to infinitely many $E_{n}$s (and therefore to all $E_{n}$s).


How many three-digit numbers are there such that the hundreds digit is less than the tens digit, the tens digit is less than the units digit, and the sum of the digits is a perfect square?

The easiest thing to do is to[hide] add up the digits of the greatest possible number that fills these conditions, 789. This adds up to 24. So the perfect squares we must find are 16,9,4,&1. The only 3-digit number that adds to 1 is 100, which does not fit the problem. The same is true for 121, 112, and 211, which add to 4. This leaves 9 and 16. From here I considered each potential hundreds digit:

9-7=2, and 72something doesn't work

9-6=3, and 63something doesn't work

.

.

.

9-3=6, and 4 could be the tens digit, but there is no fitting ones digit (same trouble with 4)

9-2=7, and 234 works!

9-1=8.    126 works

               135 works

               144 would work, but it is not allowed.

SO we have 3 that add to 9. Now for 16.

16-7=9

16-6=10

16-5=11

16-4=12   457 works

                 466 would, but....

16-3=13    349 works

                  358 works

                  367 works

                  376 would, but.....

16-2=14     21something would need a units of 13

                   .

                   .

                  259 works

                  268 works

                  277 would but yada yada.

16-1=15

                  169 works

                  178 works

So all together we have 3 that add up to 3, and 8 that add up to 16

3+8=11

11 is the answer! (I think) :mrgreen: [/hide] I just put the green guy for no reason.

im 2 years late but it's a good problem worth discussing.

is the posted solution actually right?  I don't think it is.  I got a bit more: 17 possibilities.

we basically need the number of solutions in positive integers to :

x+y+z = 16

and

a+b+c = 9

such that x,y,z and a,b,c are not equal to eachother.  then we divide by 6 to remove the ordering because order does not matter.  and given any 3 unequal digits that satisfy one of those 2 equations, there is a way to order them so that they can serve as the hundreds, tens, and units digits.

I did this and got 17 possibilities.  anyone want to try this?

wait nvm I'm dumb.  i forgot that in getting the solutions to thsoe equations, you have to limit x,y,z <= 9 and then that method of itneger solutions to equations just dont work as well.

sigh systematic enumeration always beats logic  :roll:

I got that too. Basically when you've got digits that equal 16, just increase the tens digit by one, and decrease the unit digit by 1 to keep the same number. But remember that 10's is less than units digits.
ex. from 259, you can obtain 268, but 277 is not avaliable to us, since 7 = 7.

Yeah, thats exactly what I did, as you can see from the ordering of my numbers

Topper's a rather smart boy. He was at Nationals this year, in both Mathcounts and Word Power.

Barnacle...use the edit button.

Great solutions.

yes.  this problem was easy, I only brought it up because I misunderstood something .. the best way is just infererences + enumeration as the other posters have shown.

thanks for teaching me about this edit button too :)


integration
probability
function
real analysis
real analysis theorems

Can you please tell me what is the result (closed form solution) of the series

1) Sum{i = 1->00} (i^k*b^i)
where k positive integer, k>1 and 0<b<1

Also assume a_0 = 1, a_i = (N-1)*N(-2)*....*(N-i) for  1=<i<=N

Find

2) Sum{i=0->N} a_i where N is a fixed positive integer

Solution to either problem will be of much help!
Thanks
Nick

PS If anyone knows a good website summarizing series known results please let me know!

$1.\ \sum_{i=0}^\infty i^kb^i$ with $k$ an positive integer and $0 < b < 1$.

This would not have a closed form in terms of $k$. If you pick a particular $k$, you can get a closed form in terms $b$. But if you wanted a general solution, it would probably involve the Bernoulli numbers or something like that. I can tell you it would be convergent for any particular $b$ and $k$ though! :D

$2.\ \sum_{i=0}^N \frac{(N-1)!}{(N-i-1)!}$. I don't really know what to do off the top of my head and I have to go but I'll come back later and see if it's still an open question.

[quote="Xevarion"] But if you wanted a general solution, it would probably involve the Bernoulli numbers or something like that. I can tell you it would be convergent for any particular $b$ and $k$ though! :D
[/quote]

I don't think Bernoulli numbers get involved here, it has to be something else possibly simpler, and as you said convergence is trivially guaranteed for 0<b<1
Thanks for your time!
Nick

Well, convergence doesn't guarantee anything at all about a simple expression.

$\int_0^1 x^xdx$ is easily shown to be finite, but good luck finding a closed form  :P

The problem is that the way you evaluate that kind of summation is by considering it to be a power series (replacing $b$ by $x$) and then figure out what it's the power series for and put $b$ back in. But the power of $i$ on the front doesn't result in a simple pattern there, and it seems to me to be the sort of thing that spawns the Bernoulli numbers. That's a vague impression though, no guarantee that it would actually happen. But if you doubt, try to find the value of the sum in terms of $b$ for $k = 0, 1, 2, 3$ (and maybe a few more) and see if you can deduce a pattern from there, and then try to prove it backwards.

By the way, if I translated your second question correctly, Maple tells me that there is no closed form for that in terms of $N$ either -- it spits the dreaded "hypergeom" back at me.

edit: speaking of which, Maple and Mathematica are good programs for all kinds of math, which includes summing series. Maple won't give me a closed form for the first problem in general, either.

An important remark: I forgot to say that a_n = n^(k-1)*b^n is a probability mass function and I am just interest in the mean value!
Nick


سلام اینم سوال هفتگی سوم
ثابت کنید از بین هر دویست عدد صحیح می توان صد عدد طوری انتخاب کرد طوری که مجموع آن ها بر صد بخش پذیر باشد
ب)تعمیم دهید
In English:
a)Prove between each 200 integer we can find hundred nums taht the sum of them is divisible by the hundred
b)generalize it.

tnx
ممنون

Salam ! khaste nabashid.
mikhastam ye rahnamaee baraye ghesmate (B) karde basham va begam ke dashtan 199 adad ham kafist:
[hide]1)ebteda baraye adade avale $ p$ sabet konid az bein $ 2p \minus{} 1$ adade sahihe mesle $ a_1 , a_2 , .... ,a_{2p \minus{} 1}$ mishe $ p$ adad entekhab kard ke majmooeshan bar $ p$ bakhshpazir bashe . 2)baraye har $ n$ hokmo sabet konid.
age ghesmate avallo hal nakardid ziad narahat nashid sakhte vali 2 ro mitoonid rahat esbat konid.[/hide]

سلام
بله درسته
من نویسنده وبلاگ
imo.nlogfa.com
خستم و این هم یکی از سوالای هفتگیه
بله فرموده شما درسته
ببخشید میشه بگید لاتک چیه و از کجا می تونم گیرش بیارم
بای

توی پست بالای فروم ایران یک فایلی هست و توضیحات مختصری داده. در مورد خود لاتک هم می تونید [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]این جا[/url] اطلاعات کسب کنید

سلام
خیلی متشکرم
حتما توضیحات رو مطالعه می کنم
ممنون از جواب سریعتون
یا حق


geometry
circumcircle
parallelogram
geometry proposed

The line passing through the ortho center of trinangle $ ABC$ and the midpoint of its side $ BC$ intersects the circumcircle of $ ABC$ in $ A_{1}$ and $ A_{2}$. Prove that the ortho centers of triangles $ ABC$, $ A_{1}BC$, $ A_{2}BC$ are the vertices of a right angled triangle. :)

[b]lemma1:in $ \triangle ABC$ we have $ AH=2OM$ were $ H,O,M$ are the orthocenter,circumcenter and the midpoint of $ BC$ respectively[/b]
---------------------
[b]lemma2:in $ \triangle ABC$ let line $ HM$ intersect the circumcircle at point $ A'$ then $ AA'$ is a diameter of circumcircle[/b]
---------------------
now let $ H_{1},H_{2}$ be the orthocenters of $ \triangle A_{1}BC,\triangle A_{2}BC$ respectively,so we have $ \boxed{AH=A_{1}H_{1}=A_{2}H_{2}=2OM (I)}$
and also we have $ AH \bot BC,A_{1}H_{1}\bot BC,A_{2}H_{2}\bot BC$ thus we have $ \boxed{AH \parallel A_{1}H_{1}\parallel A_{2}H_{2}(II)}$now from $ (I),(II)$ we get that the quadrilateras $ AHH_{2}A_{2},AHH_{1}A_{1}$ are parallelogram so we get that:
$ HH_{2}\parallel AA_{2},HH_{1}\parallel AA_{1},AA_{2}=HH_{2},AA_{1}=HH_{1}$
$ \Rightarrow\triangle A_{1}AA_{2}=\triangle H_{1}HH_{2}$
but according to lemma 2,$ \triangle A_{1}AA_{2}$ is right angled,so is $ \triangle H_{1}HH_{2}$...


inequalities
ratio
linear algebra
linear algebra unsolved

Let $A$ be a real symmetric positive definite matrix.Let $\lambda_{min}$ and $\lambda_{max}$ be its smallest and biggest eigenvalues.Let $X\in R^n$.Show that:
$(^tXX)^2\leq (^tXAX)(^tXA^{-1}X)\leq \frac{1}{4}(\sqrt \frac{\lambda_{max}}{\lambda_{min}}+\sqrt \frac{\lambda_{min}}{\lambda_{max}})^2 (^tXX)^2$.

For the left-side inequality: We can write our $A$ in the form $A=UDU^*$, where $D=diag(\lambda_1,...,\lambda_n)$ and $U$-initary matrix. Now consider the ratio:
$\frac{x^*Ax}{x^*x}=\frac{(U^*x)^*D(U^*x)}{(U^*x)^*(U^*x)}=\frac{y^*Dy}{y^*y}$ (*), where $u=U^*x$. Also it can be easily seen that we can restcrict ourselves to the case when $|x|=1$, because otherwise just write this for $\frac{x}{|x|}$ and all the sides multiply by $|x|^4$ .So $|y|=1$.  The same as (*) we can write for $A^{-1}$:
$\frac{x^*A^{-1}x}{x^*x}=\frac{(U^*x)^*D^{-1}(U^*x)}{(U^*x)^*(U^*x)}=\frac{y^*D^{-1}y}{y^*y}$ (***). Thus according to the said above about the resctriction $|y|=1$ multiplying (*) and (***) we have to prove $(y^*Dy)(y^*D^{-1}y) \geq 1$-but this is equivalent to $(\sum_{i=1}^n \lambda_i |y_i|^2) \left( \sum_{i=1}^n \frac{|y_i|^2}{\lambda_i} \right) \geq 1$, which is Cauchy-Byniakovski inequlaity for $a=(|y_i|\sqrt{\lambda_i}), b=(\frac{|y_i|}{\sqrt{\lambda_i}})$

Now, the right-side inequality:using that was said above we have to prove the inequality:
$(\sum_{i=1}^n \lambda_i |y_i|^2) \left( \sum_{i=1}^n \frac{|y_i|^2}{\lambda_i} \right) \leq \frac{1}{4}(\sqrt{\frac{\lambda_{max}}{\lambda_{min}}}+\sqrt{\frac{\lambda_{min}}{\lambda_{max}}})$-but as far as i know it is known inequality if i not mistaken it is due to Kantorovich


geometry
topology
calculus
function
vector
linear algebra
advanced fields

I'm taking a class in differential geometry next semester and we are using the book Introduction to Differentiable Manifolds and Riemannian Geometry by Boothby. The book looks kind of intimidating (I don't have it yet just saw the table of contents online). If anyone has used this book what would you say I should review before going into the class? Thanks.

I took a graduate class on differential geometry in [url=http://www.math.wustl.edu/]Washington University[/url] where Boothby is now a Professor Emeritus. Strangely enough, the textbook was different, and I never read Boothby's book. Generic advice for those taking differential geometry: know your multivariable calculus (chain rule, implicit function theorem, Green/Stokes) and linear algebra.

I agree with mlok.  And my personal opinion for beginners of differentiable manifold is to keep in mind that everything you do in the beginning is just to paste charts together using partition of unity, that's all.

And I will suggest reading the part of tangent and cotangent vectors if you want to preview the book before the class begins.

To math_music:  What Real I, II refers to...?  But I believe generally no, you don't really need much prerequisities other than multivariables and linear algebra

It's kind of advanced calculus (I and II) i think. Maybe it can be described as theoretical calculus with rigorous proofs...

That would be necessary, if they contains calculus of multivariables.


Given a triangle ABC. We build two angle bisectors in the corners A and
B. Than we build two lines parallel to those ones through the point C.
D and E are intersections of those lines with the bisectors. It happens,
that (DE) line is parallel to (AB).
Prove that the triangle is isosceles.

It takes longer to name the necessary points than to solve :).

Let $I$ be the incenter of $ABC$ and $A',B'$ the feet of the bisectors from $A,B$ respectively. Furthermore, let $M,N$ be the intersections of $CD,CE$ with $AB$ (respectively).

The condition is equivalent to $\frac{CD}{DM}=\frac{CE}{EN}\iff \frac{B'I}{IB}=\frac{A'I}{IA}\iff A'B'\|AB$, and it's now clear that this can only happen when $CA=CB$ (because $I$ lies on a median).


Triangle $ PQR$ is isosceles and the measure of angle $ R$ is $ 40^{\circ}$. The possible measures of angle $ P$ are $ x,y,z$. What is the value of the sum $ x + y + z$?

Case 1:

$ \angle P  \cong \angle Q$

Then we have $ 2 \angle P\plus{}40\equal{}180 \implies \angle P\equal{}70^\circ$

Case 2:

$ \angle P \cong \angle R$

Then we have $ \angle P\equal{}40^\circ$

Case 3:

$ \angle Q \cong \angle R$

Then we have $ \angle P\equal{}180\minus{}40\minus{}40\equal{}100^\circ$

Summing these yields $ \boxed{210^\circ}$.


search
combinatorics unsolved
combinatorics

Consider the set  $ U\equal{}\{1,2,3,...,3n \}$. Prove that for each partition of $ U$ in $ 3$ subsets of $ n$ elements each, we can choose $ 1$ element from each subset, such that one of them is equal to the sum of the other $ 2$ elements chosen.

Posted before.
Look here for a more general result
http://www.mathlinks.ro/viewtopic.php?search_id=838242110&t=3781

Pierre.


Prove that in a triangle $ ABC$

$ \sin A \plus{} \sin B \plus{}\sin C \leq \frac{3\sqrt{3}}{2}$.

$ f(x) \equal{} \sin x$ is concave over $ [0,\pi]$, so $ \sin A \plus{} \sin B \plus{} \sin C \le 3 \sin \frac{A\plus{}B\plus{}C}{3} \equal{} \frac{3\sqrt{3}}{2}$, as desired.

(Actually, I don't know whether concavity is over closed or open intervals, but the Jensen's argument should still work here.)


Prove that $ n^7\plus{}7$ is not square for very natural numbers $ n$.

[hide="Proof"]We have $ n^7 + 7 = t^2$ where $ t$ is positive integer. Obviosly if $ n$ is odd, then $ t$ is even and vice versa.The equality is equivalent to:
  $ t^2 + 11^2 = 2^7 + n^7$
 [b]1case:[/b]$ n$-even, $ t$-odd. The right hand side is $ \equiv 0 (mod8)$. Then we have $ t^2\equiv - 1(mod 8)$-impossible.
  [b]2case:[/b]$ n$-odd, $ t$-even.The right side always has a divisor in the form $ 4k + 3$ ,so there exist prime in the same form in its canonical decomposition. We know that if $ p = 4k + 3$ and $ p|x^2 + y^2\Rightarrow p|x,p|y$. We have $ p = 4k + 3$ and $ p|t^2 + 11^2$ , so $ p|t$ and $ p|11$. So $ t = 11s$. The left side is $ =0(mod11)$, so we have $ n^7\equiv - 7(mod11)$-impossible.[/hide]
 PP Thank you for this nice problem :clap2:  :10:


geometry
geometric transformation
rotation
ratio
AMC
AIME

In triangle $ABC,$ $AB=13,$ $BC=14,$ $AC=15,$ and point $G$ is the intersection of the medians. Points $A',$ $B',$ and $C',$ are the images of $A,$ $B,$ and $C,$ respectively, after a $180^\circ$ rotation about $G.$ What is the area if the union of the two regions enclosed by the triangles $ABC$ and $A'B'C'?$

[hide="solution"]Since $BG = GB'$ and $AG = GA'$, $AB'A'B$ is a parallelogram. The same for $A'C'AC$ and $BCB'C'$. Hence $\Delta A'B'C' \equiv \Delta ABC$. Let $P = B'C' \cap AC$ and $Q = A'B' \cap AC$. Then $\Delta B'PQ$ is the homotetic of the triangle of midpoints with center $G$ and ratio $r$. Let $K$ be the midpoint of $AC$. Then $B'K$ is a median and $B'K = KG$, so $r = \frac{2}{3}$. Hence $\left[B'PQ\right] = \frac{\left[ABC\right]}{9}$. Finally the desidered area is $\frac{2}{3}\left[ABC\right]$ and using Heron we get $56$.[/hide]

You found the intersection.

We are actually looking for the union.

Yeah.  It is twice the intersection apparently.

can someone write a COMPLETE solution to this prob?

thx

Yes, I worked on this problem a few days ago. I believe it is the AIME II 2003, but I could be wrong.

[quote=paladin8]OK, this is going to be hard to describe, but here goes. I would recommend drawing a picture before trying to follow this.

We know $A'B'$ is parallel to $AB$ and also that the distance from $A'B'$ to $G$ is equal to the distance from $AB$ to $G$ (since its a rotation around $G$) and is also equal to $\frac{1}{3}h_{A}$ (the altitude from $A$). This follows from the property that the centroid divides the medians in the ratio $2: 1$.

So... if we let $D$ and $E$ be the intersection of $A'B'$ with $AB$ and $AC$, the triangle $ADE$ is similar to $ABC$ with ratio $\frac{1}{3}$. So it has $\frac{1}{9}$ the area of $ABC$. The same thing can be said for the small triangles at vertices $B$ and $C$, so they sum to an area of $\frac{1}{3}$ the area of $ABC$.

Then the union is just the whole triangle and the three smaller triangles, which is $1+\frac{1}{3}= \frac{4}{3}$ the area of $ABC$, or $\frac{4}{3}(84) = 112$ in this case.[/quote]

How can you tell $A'B'$ is parallel to $AB$?

[quote=whatRthose]@above its a 180 degree rotation, thats prob why.[/quote]

Thanks

[quote=Blackhawk]Hi, can someone explain why in [url="https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_6"]This[/url] diagram, you can conclude that all the triangles are equal?

Thanks![/quote]

If you draw a super accurate picture you could see why.


A craftsmen goes to lunch not long after noon. As he leaves he observes the exact placement of a clock's hands. On his return he notices that the minute and hour have exchanged places. When does he return ?

[size=75]
from book "The Moscow puzzles" by Boris A. Kordemsky

Please hide Your solution . Write inside Hide ... /Hide
[/size]

What is up with the watch problems?

[hide="Solution"]During the lunch hour, the clock hands move a total of 360 degrees. The minute hand moves 12 times as fast as the hour hand, covers $ \frac{12}{13}$, and the hour hand covers $ \frac{1}{13}$. The craftsman was absent $ \frac{12}{13}$ of a hour, or $ 55\frac{5}{13}$ minutes.From noon till x minutes later, when the craftsman goes to lunch, the minute hand moves x minutes past 12 and the hour hand $ \frac{1}{12}x$ minutes past 12. As he leaves the hands are $ \frac{11}{12}x$ apart. But this distance has been shown to be $ \frac{1}{13}x\cdot60$ minutes. Then $ \frac{11}{12}x\equal{}\frac{1}{13}\cdot60$, and $ x\equal{}5\frac{5}{143}$ minutes. The craftsman left for lunch at 12:05 $ 5\frac{5}{143}$ minutes, was absent for $ 55\frac{5}{13}$ minutes, and came back at 1:00 $ \frac{60}{143}$.[/hide]


in triangle ABC,the points D,E,Z,H,Q are the midpoints of the segments BC,AD,BD,ED,EZ,respectivly.

I,is the intersection HQ,AC.then prove that:

1)AK=3CK

2)HK=3HQ

3)BE=3EI

4)S(ABC):S(EQH)=32

yes i am agree with you ashegh this problem is not so not(but the other problems that you sent are nice :D )


They were brought up immersed in all these shameless mediums, the only thing they know and do is lying and brainwashing.
[url]http://www.timesonline.co.uk/article/0,,2089-2271971,00.html[/url]

So, I suppose that means you have evidence to refure the claims of this article ... (waiting)

name the sources that the information from where you would like to accept, then I can see whether there is some joint.
It's much easier to make and spread a lie than refuting one, especially when the lie is conducted by a country, not only a government. And even if you have completed the quest, those lie makers will say the same but in some other time and place. What can people do? Follow Sisyphus?
It's correct maybe, that "[b]You can fool all the people some of the time, and some of the people all the time, but you cannot fool all the people all the time.[/b] ". But the fact is, they don't need to fool all the people all the time, fool some relevant people within some relevant period, that's enough. And this is what those mediums doing.
Will people bother to refute those lies? I don't think so. At least for the Americans. I see this from those tv programs, like the history channel. there was once a program called "Declassified, Chairman Mao", they said "[b]Mao contracts a venereal disease, but women are proud to be infected as proof of their relationship with him[/b]". The only thing I could say is, [b]like audiences, like mediums, and vise versa[/b].
Of course, some mediums are a little bit cleverer and behave like nobles now. They will talk about some [b]evils[/b] on the front page as if they're true. after people accepted these, and have no more interest on them, there will maybe be news on some almost invisible corner saying something like "we're sorry that what we said on *** before is proved to be imprecise". See? they do not lie at all. Cheers for the justice!

So, just to clarify:  you have no evidence whatsoever that any single claim made in that article is false.  Right?

Actually, the article provides an interesting look at why the economic rise of China could end up as a disaster for everyone.


abstract algebra
group theory
induction
superior algebra
superior algebra unsolved

Prove that a solvable group all of whose abelian subgroups are finite must be finite.

Here are two solutions, but in romanian... maybe you can get the idea...
http://www.mateforum.ro/viewtopic.php?t=1490

I haven't read it.
But here is my solution.

Assume $ G'$ is finite by induction on the derived length.
Then $ G/C_G(G')$ is isomorphic to a subgroup of $ Aut(G')$, which is also finite.
Set $ H: \equal{}C_G(G')$ with $ H'\le G'\cap H\le Z(H)$ finite abelian.
There is no infinitely increasing sequence of abelian subgroups $ A_1\lneq A_2\lneq \cdots$,
because otherwise $ \bigcup_{k\ge 1}A_k$ would be an infinite abelian subgroup.
So we find a maximal abelian subgroup $ A\le H$ with $ H'\le A$.
Then $ H/H'$ abelian gives $ A/H'\unlhd H/H'$ and $ A\unlhd H$.
But for $ x\in C_H(A)\setminus A$ we have $ A\lneq \langle A,x\rangle$ also abelian, which contradicts the maximal choice of $ A$.
So $ C_H(A)\equal{}A$ is finite and $ H/C_H(A)$ is isomorphic to a subgroup of $ Aut(A)$, which is also finite.
This gives $ H\equal{}C_G(G')$ finite, and then also $ G$ finite.


Hello. Does AOPS give textbooks ever for discount for Black Friday or anything?


In a triangle $ ABC$, the bisectors of the angles at $ B$ and $ C$ meet the opposite sides $ B_1$ and $ C_1$, respectively. Let $ T$ be the midpoint $ AB_1$. Lines $ BT$ and $ B_1C_1$ meet at $ E$ and lines $ AB$ and $ CE$ meet at $ L$. Prove that the lines $ TL$ and $ B_1C_1$ have a point in common.

We have to prove $AL$≠$LC_1$
We know $(B_1T/AT).(AC_1/BC_1).(C_1M/MB_1)$=1
thus $C_1M/MB_1=a/a+b$
also we have: $AL/C_1L=(AC/B_1C).(B_1M/C_1M)$
thus $AL/C_1L=a^2+ab+ac+bc/a^2$
so $AL$≠$LC_1$
QED


parameterization
geometry
3D geometry
quadratics
Euler
144

Is there a common method that can solve all type of general problem:
$ax^{2}+bx+c=\sqrt{dx+e}$ where $a,b,c,d,e$ are arbitrary numbers and $dx+e\geq 0$
Because I has solved the equation $\sqrt{3x+1}=-4x^{2}+13x-5$
But now i can't solve the equation: $\sqrt{2x+1}=50x^{2}+43x+9$

[quote="hien"]
But now i can't solve the equation: $\sqrt{2x+1}=50x^{2}+43x+9$[/quote]
$x=-\frac{1}{2}$ and one more. :D

But the following equation has a nice roots:
$\sqrt{2x+1}=50x^{2}+59x+17.$ :wink:

[quote]But now i can't solve the equation: $\sqrt{2x+1}=50x^{2}+43x+9$[/quote]
[quote]$x=-\frac{1}{2}$ and one more. :[/quote]

[b]Dear Arqady[/b] What' about $x=-\frac{1}{2}$? It is not a root of the equation?

Just square both sides and rearrange the terms to get a quartic equation.  There is a general formula for quartic equations, although it is a bit messy.

[quote="renatzu"]There is a general formula for quartic equations, although it is a bit messy.[/quote]

A [b]bit[/b] messy?

 :|

The cubic equation is bad enough.  The quartic equation is so cumbersome as to be useless (in my opinion) even as a computer algorithm.

Now, there was a poster who used a very clever method involving constant parameters to solve specific equations of this type, but the general problem is equivalent to solving a quartic, as far as I know.

If we reduce the equation to another quaratic equation, it is a large messy?
In specific one, we can solve easyly to tranfreing miltyequations buy subordinate variables.

I only know method solving a specify type of problem as following:
$\sqrt{ax+b}=c(dx+e)^{2}+\alpha x+\beta$ where $d=ac+\alpha$ and $e=bc+\beta$
Condition $ax+b\geq 0$
[b]The method[/b]
Put $dy+e=\sqrt{ax+b}$
Then the equation transfer to:
$\{\begin{array}{l}c(dy+e)^{2}=acx+bc\\c(dx+e)^{2}=(ac-d)x+dy+bc\end{array}$
For example:
$\sqrt{x+1}=x^{2}+4x+5$
$\sqrt{3x+1}=-4x^{2}+13x-5$

[quote="t0rajir0u"] Now, there was a poster who used ... [/quote]
I guess I know him.  :lol:
[quote]... but the general problem is equivalent to solving a quartic, as far as I know[/quote]
Perfectly yes. In general any solution of 4th degree equation must depend on cubic irrational and this can not be ignored by any method.
The only thing which is possible is to simplify the reducing to cube equation.
Firstly I've shown a trick only but after analizing it I got the another one which makes this reduction to quartic very easy. Any root of this quartic produces immediately 2 quadratic equations. And if in addition this quartic has good root the reduction becomes more simple.
I'm not ready to show this new trick immediately as I need a time to thinking about and also to compare with known ones. Perhaps the discovered matter is known too.
On the 1st thinking it win when compared with methods of reducing due to Ferrary, Hudde and its Euler modifications.

Oh, I was nearly to forget mention that the first easy step of this trick is reducing general equation namely to the form:
$(ax^{2}+bx+c)^{2}= dx+e$

[quote="hien"]Yes, welcome Digger. I wait for your method that solving my open problem. In general, I can't solve because the cubic equation is large messy.[/quote]

You are waiting my cubic would occur much better?

Well, look.

Consider equation

(1) $(ax^{2}+bx+c)^{2}= dx+e$

Compose cubic equation:

(2) $2a\lambda^{3}+(4ac-b^{2})\lambda^{2}+(2ae-bd)\lambda-(\frac{d}{2})^{2}=0$

and choose any its root $\lambda$.
Althought it is of no matter which root to choose we may assume it being real and positive because we can consider $a>0$ in (1), and also for Hien equality taking -RHS if necessary - it produces change restriction on LHS only.

Then the equation (1) is decomposable in two ones:

(3) $ax^{2}+bx+c+\lambda = \pm \sqrt{2a\lambda}(x+\frac{b}{2a}+\frac{d}{4a})$

For example for the equation of [b]Arqady[/b] the cubic equation is as follows: $100\lambda^{3}-81\lambda^{2}-18\lambda-1=0$ and has nice root $\lambda=1$, two others are bad and are not real. So his equality is decomposed into 2 equation: $50x^{2}+49x+12=0$ and $50x^{2}+69x+24=0$
The 2nd one doesn't have real roots. Both roots $x=-\frac{1}{2}$ and $x=-\frac{12}{25}$ of the first one satisfy the restriction $50x^{2}+59x+17\ge 0$

For the equation  $(50x^{2}+43x+9)^{2}= 2x+1$ the cubic equations turns to be

$100\lambda^{3}-601\lambda^{2}+14\lambda-1=0$

and has unique real root involving cubic irrational $\sqrt[3]{146953914}$ in it's expression. Very nice is it?  :lol:
The evaluation by maple gives $\lambda \approx 5.986894585$
This produces bad decomposition of this equation but both parts of decomposition can be easily evaluate using simple calculator only.

As I've mentioned before any general method of solving quartic(*) must include the solution of some cubic equation just because the roots involve some fixed cubic irrational. There can be some differences in that cubic equation but roots of them should  contain this fixed cubic irrational hence they should be "almost the same".
I have no time yet to check whether or not my cubic is the same whith those occuring in Ferrary's method or it is "almost the same" only.

(*) In the above posting I've used the word "quartic" in the sence "cubic equation". There is a great lie that the international language of communication is English ...
It is broken English.

[quote="digger"]... and has unique real root involving cubic irrational $\sqrt[3]{146953914}$ in it's expression. [/quote]
Mad fantasy.  :cursing:
I was intendent to say that the expression of $\lambda$ has $\sqrt{146953914}$ under $\sqrt[3]{\hspace{0.1in}}$

But your method still can't sovle my equation:
$\sqrt{2x+1}=50x^{2}+43x+9$

What you mean by solving?
My trick is brutal force and decide any 4th degree equation. But of course it can't make roots better then they are.
Your equation has 2 negative roots: $-\frac{1}{2}$ and irrational root of cube equation.
It is the very time to remember what wrote [b]t0rajir0u[/b]:

[quote]The cubic equation is bad enough.  The quartic equation is so cumbersome as to be useless (in my opinion) even as a computer algorithm.[/quote]

Because of ignoring rational root $-\frac{1}{2}$ my trick works even worse here. The bad root of my cubic equation leads to bad squared equations and give no  possibility to see even rational root $-\frac{1}{2}$. So the way is more long when compared with reducing to cube equation if we simply devide by $2x+1$.

The only difference may be between the trick I've found and methods I knew before is easy reducing to cube equation.


Here's an  example from vol.2(9.6). I'm got confused on the second step, any solutions would be appreciated.

Find all cube roots of 4+4(3i)^.5

In the book they did something a little different. First they changed 4+4(3i)^.5 into trig form (here's where I'm confused):
8(cos60+sin60). It would be nice if you could explain how they got that.

[quote="Inspired By Nature"]by the way, the cube roots aren't complex  :blush:[/quote]

All numbers are complex. You mean imaginary.  ;)

I think you mean $4+4i(3)^{.5}$. There's a slight difference in notation which turns into a big difference in the problem.

I think there may have been a typo in the book. It should be converted into $8(\cos 60^\circ +i\sin 60^\circ)$.

I thought it was a typo too. Thanks guys