trigonometry
inequalities
inequalities proposed

Let $a,b,c$ be positive reals s.t. $a^2+b^2+c^2+2abc=1$. Prove that \[a+b+c\leq\frac{3}{2}\]

somebody know where can I find topic with proof of $\cos ^2A+\cos ^2B+\cos ^2C+2\cos A \cos B \cos C =1$ ? I'm sure I've seen it before ...  :?

But it is very simple... I am sure you can prove it yourself.

Actually this inequality comes from this one:
Let $x,y,z$ be positive reals greater than 1 s.t. $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$. Prove that \[\sqrt{x+y+z}\geq\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\]This one is easy from Cauchy.

[quote="mecrazywong"]Actually this inequality comes from this one:
Let $x,y,z$ be positive reals greater than 1 s.t. $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$. Prove that \[\sqrt{x+y+z}\geq\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\][/quote]

Actually, the equivalence of this to the initial inequality $a+b+c\leq\frac32$ for positive reals a, b, c satisfying $a^2+b^2+c^2+2abc=1$ is not too obvious: One has to square the inequality $\sqrt{x+y+z}\geq\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$, to cancel terms and then to apply the substitution

$a=\sqrt{\left(y-1\right)\left(z-1\right)}$; $b=\sqrt{\left(z-1\right)\left(x-1\right)}$; $c=\sqrt{\left(x-1\right)\left(y-1\right)}$;

for the other direction, one has to substitute

$x=\frac{bc+a}{a}$; $y=\frac{ca+b}{b}$; $z=\frac{ab+c}{c}$.

Showing that the conditions $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$ and $a^2+b^2+c^2+2abc=1$ are equivalent is an easy calculation.

Note that the inequality $\sqrt{x+y+z}\geq\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$ was also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=25650 and the inequality $a^2+b^2+c^2+2abc=1$ at http://www.mathlinks.ro/Forum/viewtopic.php?t=23888 and (in a slightly modified form) at http://www.mathlinks.ro/Forum/viewtopic.php?t=25977 .

Now, the actual reason why I am writing this message is to show that the initial inequality $a+b+c\leq\frac32$ for positive reals a, b, c satisfying $a^2+b^2+c^2+2abc=1$ can also be proved directly (i. e. without any substitution):

Assume, for the sake of contradiction, that $a+b+c>\frac32$. Then, if we succeed to prove that $a^2+b^2+c^2+2abc>1$, then we will get a contradiction with the condition $a^2+b^2+c^2+2abc=1$, so that our assumption must be wrong, and the inequality is proven. So all we have to do is to show that $a^2+b^2+c^2+2abc>1$.

At first,

$2\left(a+b+c\right)\left(a^2+b^2+c^2\right)+9abc-\left(a+b+c\right)^3$
$=a\left(a-b\right)\left(a-c\right)+b\left(b-c\right)\left(b-a\right)+c\left(c-a\right)\left(c-b\right)\geq 0$

by Schur's inequality. Thus,

$2\left(a+b+c\right)\left(a^2+b^2+c^2\right)+9abc\geq\left(a+b+c\right)^3$.

Thus,

$2\left(a+b+c\right)\left(a^2+b^2+c^2\right)+6abc\left(a+b+c\right)$
$>2\left(a+b+c\right)\left(a^2+b^2+c^2\right)+6abc\cdot\frac32$       (since $a+b+c>\frac32$)
$=2\left(a+b+c\right)\left(a^2+b^2+c^2\right)+9abc\geq\left(a+b+c\right)^3$.

Dividing this inequality by a + b + c, we get

$2\left(a^2+b^2+c^2\right)+6abc>\left(a+b+c\right)^2$.

Applying $a+b+c>\frac32$ again, we see that

$2\left(a^2+b^2+c^2\right)+6abc>\left(\frac32\right)^2=\frac94$.

On the other hand, by Cauchy-Schwarz, $\left(1^2+1^2+1^2\right)\cdot\left(a^2+b^2+c^2\right)\geq\left(1\cdot a+1\cdot b+1\cdot c\right)^2$, so that $3\cdot\left(a^2+b^2+c^2\right)\geq\left(a+b+c\right)^2$, and thus $a^2+b^2+c^2\geq\frac{\left(a+b+c\right)^2}{3}>\frac{\left(\frac32\right)^2}{3}=\frac34$. Adding this to the inequality $2\left(a^2+b^2+c^2\right)+6abc>\frac94$ obtained above, we get

$3\left(a^2+b^2+c^2\right)+6abc>3$.

Division by 3 yields $a^2+b^2+c^2+2abc>1$. This yields the required contradiction, and thus our inequality is proven.

  darij


This is my first time coaching a Mathcounts team and naturually, I hope my team makes it to the State level   :)

How hard it is to make it to the State level must vary depending on what Chapter you are in.  I live in Nassau County, NY where the top 3 teams go to the State competition.
I'm giving last year's competition to my students for practice.  How many correct on the 2006 Chapter competition would be "average?"  What would be the minimum score of a student that made it to the State competition?

I doubt I can get a definitive answer, but I'm just looking for a ballpark figure.  For instance, the Sprint round has 30 problems.  Does 20 correct put a student in line to possibly make it to States or is 20 average?

It's definitely above average, but probably too low to place in the top 10 or so individuals (that depends on your chapter though).

On last year's chapter competition, here is some data from the chapter containing Albany, NY.  They had 143 students compete.  Their average sprint score was 15.7; their average target score was 9.5.

Regarding making it to states, the team score matters more than the individual score, at least in New York.  (In each NY chapter, the top three teams automatically qualify for states; in addition, the top two other students qualify.)  In the chapter containing Albany, the third-best team scored a 52.75 last year.  In Manhattan, the third-best team (the team I coached) scored a 56.50.  Because those two counties are probably the most competitive in New York, I'm guessing that a lower score would qualify in Nassau County.

It depends. In my county a 20/7 or better is enough to be almost guaranteed for countdown, and this year I predict it'll only take a 30 to make top 10. Last year the lowest score was $33 \pm 2$ and the highest was a 45. I know that third written had a 37. Also, the top two teams get to go two state, and usually a score of like 47 or higher is good enough to make it. In my county doing well isn't hard even though last year two people on the nationals team were from it (different schools) and 4 people over all were in countdown at states (1, 3, 5, 6).
However in the sugarland county in Texas the competition is fierce. Teams have gotten scores of up to 65 before. Usually there are 3 to four teams in the sixties. Last year a 44 was the minimum to make countdown.

[quote="jhredsox"]anyways, 34 people from my chapters make it to nationals.
-jorian[/quote]

I think you mean states.  :wink:

oh yah, this is like the 3rd time i've done this :P
-jorian

I agree that the chapter you live in matters (imagine living in a district where there are only 3 people competing :D) but also the particular test also matters. Some years are definitely harder than others.

In my chapter, to go to state in individuals, you needed 38.  HOwever 6 teams made States, so all you need to do is make top 6 teams.

[quote="jhredsox"]I think you made a typo on the 5th line, 4th word.

anyways, 34 people from my chapters make it to nationals.
-jorian[/quote]
No, that isn't a typo, read the sentence carefully.

[quote="indianamath"]HOwever 6 teams made States, so all you need to do is make top 6 teams.[/quote]

Do you mean in your chapter or in general? Because like it varies by chapter, in Massachusetts one chapter was like top 3 teams/individuals, mine was top 7. In Delaware, chapter doesn't even exist.

Last year I believe the State Individual cutoff was 39 in my chapter (7th place), and the team cutoff was around 51.

But it really depends on your state and chapter, some are much stronger than others.

In my chapter (somewhere in the bay area) the cutoff for individuals making it to states was 44 (The top 4 teams and next highest 5 individuals made it). I don't know what the team cutoff was. The top team in our chapter won both chapter and state, they got 64.5 at chapters and a 58.25 at states.
I think our chapter is harder than a normal chapter so I have no idea what the cutoff for a normal chapter would be.

in our chapter, north florida (big bend), 10th was a 38.  (24/7). and 4th was a 42.  (28/7).

Our chapter is very small (60 or so competitors) and there was a kid who had 32 total and still got 10th, but in NY I doubt it would be that easy to get in the top 10. At our chapter the lowest person that actually made it to state as an individual had 34 or so, so 20 on the sprint round probably wouldn't cut it :wink:

[quote="13375P34K43V312"][quote="Ubemaya"]The top team in our chapter won both chapter and state, they got 64.5 at chapters and a 58.25 at states.
[/quote]
Wow 64.5? That's pretty good. That's like 46 45 44 43 + 20 on the team round. That was your team, and don't make sure random comment saying you got a 17 or something at chapter  :)[/quote]

No the teams score was 46 46 something, they had two out of the three perfects in our chapter.
I got a 26 at chapter, it was above average :) .

I just realized... I probably have to get like a 43 or 44 to reach my goals.


[quote="ch1n353ch3s54a1l"][quote="Arvind_sn"]Joe has a fair 6-sided die. He rolls it 4 times. What is the probability of the numbers being 1 ,2, 3, and 4, in that order? What are the odds of the numbers being 1 ,2, 3, and 4, in that order? Explain. :D[/quote]

Isn't the first and second question the same thing?
[hide]$\displaystyle\left(1/6\right)^4=\frac{1}{1296}$[/hide][/quote]

Odds are expresesd in a different way, I think odds would be $\boxed {1296: 1}$ Not too sure though as I sort of forgot how to express them.

[hide]Probability is $\frac{1}{6}^4=\frac{1}{1296}$  Odds are $1: 1295$[/hide]

[quote][hide]Probability is the amount of chances a certain outcome wanted over the total number of possiblities.
Odds is the amount of chanes wanted over the amount of chances not wanted.
Probability is 1/1296
Odds are 1/1295[/hide][/quote]

Couldn't have put it better. :D

Actually, for a more descriptive way to describe odds, you all should visit http://en.wikipedia.org/wiki/Odds...

[quote="ragnarok23"][quote="Iron Fist"][quote="ch1n353ch3s54a1l"]Well...that's BASICALLY the same thing.
Wouldn't it be $1295: 1$?[/quote]

>_> he's the only correct guy here.

btw the AoPS book has a good explanation on odds :p .[/quote]
It actually depends, whether it is the odds in favor of that happening or against that happening.[/quote]

Odds against is what's usually assumed.


limit
integration
calculus
calculus computations

For $ n$ an integer, evaluate
\[ \lim_{n \rightarrow \infty} \left( \frac{1}{\sqrt{n^2 \minus{} 0^2}} \plus{} \frac{1}{\sqrt{n^2 \minus{} 1^2}} \plus{} \cdots \plus{} \frac{1}{\sqrt{n^2 \minus{} (n \minus{} 1)^2}} \right).
\]

[quote="aidan"]For $ n$ an integer, evaluate
\[ \lim_{n \rightarrow \infty} \left( \frac {1}{\sqrt {n^2 \minus{} 0^2}} \plus{} \frac {1}{\sqrt {n^2 \minus{} 1^2}} \plus{} \cdots \plus{} \frac {1}{\sqrt {n^2 \minus{} (n \minus{} 1)^2}} \right).
\]
[/quote]
$ L\equal{}\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k\equal{}0}^{n\minus{}1}\frac{1}{\sqrt{1\minus{}\left(\frac{k}{n}\right)^{2}}}\equal{}\int_{0}^{1}\frac{dx}{\sqrt{1\minus{}x^{2}}}\equal{}\frac{\pi}{2}$ :)

Could you further elaborate? I'm lost as to how that's converted to an integral. I thought the factor of $ \frac{1}{n}$ would become zero as $ n$ approached infinity.

It's straightforward from these 2 links :
http://en.wikipedia.org/wiki/Riemann_sum
http://en.wikipedia.org/wiki/Riemann_integral


calculus
integration
trigonometry
real analysis
real analysis unsolved

Suppose
$\int_0^1 |f(x)| dx < \infty$ and $\int_1^\infty \frac{|f(x)|}{x^2} dx < \infty$
Show that
$\int_0^\infty \sin ax dx \int_a^\infty f(y)e^{-xy}dy = a \int_a^\infty \frac{f(y)}{a^2+y^2}dy.$

There's a notational problem involving the variable $x$. If it is a dummy variable inside an integral, then it should not appear elsewhere. If we interpret the left hand side as

$\int_0^{\infty}\sin ax\left(\int_a^{\infty}f(y)e^{-xy}dy\right)dx$

then the whole thing is just a matter of justifying the use of Fubini's Theorem.

yes, all we need to do is to clarify the validity of the exchange of integral signs...
i just met a problem on that...


inequalities function logarithms trigonometry geometry circumcircle inradius

[size=150]prove that

 If sin (A/2) sin (B/2) sin (C/2) = 1/8
then  triangle ABC is equilateral[/size]

Well-known. By Jensen on, for example, the function $\ln{\sin{x}}$ we get that \[\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}} \leq \frac{1}{8}\] with equality iff $ABC$ is equilateral.

http://www.mathlinks.ro/Forum/viewtopic.php?t=45005

Can you see the relations between my problem and yours?  :)

[quote="Arne"]\[\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}} \leq \frac{1}{8}\][/quote]

There are dozens of ways to prove this inequality. For instance, it follows from http://www.mathlinks.ro/Forum/viewtopic.php?t=30785 , applied to $a=\frac{A}{2}$, $b=\frac{B}{2}$, $c=\frac{C}{2}$. Or note that $r=4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$, where R is the circumradius and r is the inradius of triangle ABC, so $\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\leq\frac18$ is equivalent to $r\leq 4R\cdot\frac18$, i. e. to $r\leq\frac{R}{2}$, what is Euler's inequality. Another proof was given in http://www.mathlinks.ro/Forum/viewtopic.php?t=42758 post #4 or http://www.mathlinks.ro/Forum/viewtopic.php?t=21188 .

  darij


modular arithmetic
number theory
prime numbers

$ p \plus{} 1 \equal{} x^2 \\
p \equal{} (x \minus{} 1)(x \plus{} 1)$
given p is prime it has only two divisors so
$ x \minus{} 1 \equal{} 1 \\
x \plus{} 1 \equal{} p \\
p \equal{} 3$

[quote="AndrewTom"]Which prime numbers are 1 greater than a square? Which are 1 less than a square? Which are 1 more than a number to the powe of n and which are 1 less than a number to the power of n?[/quote]

$ \boxed{\text{Which prime numbers are 1 greater than a square?}}$

There are too many primes like that   ,e.g.

$ 2 \equal{} 1^2 \plus{} 1 ,5\equal{}2^2\plus{}1,17 \equal{} 4^2 \plus{} 1,37\equal{}6^2\plus{}1,101 \equal{} 10^2 \plus{} 1,197 \equal{} 14^2 \plus{} 1,257 \equal{} 16^2 \plus{} 1,\cdots$

:starwars: :D

[hide="1 less than $ x^n$"]
$ p\equal{}x^n\minus{}1$
$ p\equal{}(x\minus{}1)(x^{n\minus{}1}\plus{}x^{n\minus{}2}\plus{}...\plus{}x\plus{}1)$
Prime only has two divisors, so $ (x\minus{}1)\equal{}1$, $ p\equal{}2^{n\minus{}1}\plus{}2^{n\minus{}2}\plus{}...\plus{}2\plus{}1$

[url]http://en.wikipedia.org/wiki/Mersenne_prime[/url]
[/hide]

[hide="hint on 1 more than $ x^2$"]If $ p\equal{}x^2\plus{}1$, then, for any $ k<x$, $ x\not \equiv \sqrt{\minus{}1} \pmod{k}$[/hide]


integration
function
limit
real analysis
real analysis unsolved

Let $f: [0,1]\to R, f_{n}(x)=\int_{0}^{x}f(t)e^{nt}dt,n\geq 1$. If the sequence $(f_{n}(x))_{n\geq 1}$ is bounded $\forall x\in [0,1]$ then $f=0$.

The function $f$ has to be  continous (else the statement is false) . We have $\forall n, |f_{n}(x)| \leq B (x)< \infty$
Let it suppose that $f$ is not identically zero, so there is $0 < a<b<1$ such that $\forall x\in [a,b] , |f(x)|\neq 0$ , so the minimum of $|f|$ on $[a,b]$ is not zero.
Let $m>0$ be this minimum : $m=\min_{t\in [a,b]}|f(t)|$ .Because of $f$ is continous, there is $c_{n}\in [a,b]$ such that  $|\int_{a}^{b}f(t) e^{nt}dt|=|f(c_{n})\int_{a}^{b}e^{nt}dt| \geq m\int_{a}^{b}e^{nt}dt$
We can write $m\int_{a}^{b}e^{nt}dt \leq |\int_{a}^{b}f(t)e^{nt}dt| =|\int_{0}^{b}f(t) e^{nt}dt-\int_{0}^{a}f(t) e^{nt}dt|$$\leq |\int_{0}^{b}f(t) e^{nt}dt|+|\int_{0}^{a}f(t) e^{nt}dt| \leq B(b)+B(a)$
But $\lim_{n\rightarrow \infty}\int_{a}^{b}e^{nt}dt = \infty$ , so $m=0$ (contradiction !) . Finally $f$ has to be identically zero .. q.e.d..
 :cool:

Here's slightly different proof.

Suppose the contrary, then since $f \in C^{0}([0,1])$, WLOG one may assume that $f(t) > 0$ in some $(a,b) \subset [0,1]$. Since $n \mapsto f(t)e^{nt}$ is $\ge0$ and is increasing to $+\infty$ for $t\in(a,b)$, by Monotone Convergence Theorem we get that
\[\lim_{n\to\infty}|f_{n}(b)-f_{n}(a)| = \lim_{n\to\infty}\int_{a}^{b}f(t)e^{nt}dt =+\infty, \]
contradicting the fact that $\sup |f_{n}(b)-f_{n}(a)| <+\infty$.


$ ABC$ is an right triangle such that $ AB$ is chord. the length of altitude $ CD$ is $ h$.$ M$ and $ N$ are midpoints of $ AD$ and $ BD$  respectively.find the distance of vertex $ C$ to the orthocenter of $ CMN$ traingle.


inequalities
trigonometry
geometry solved
geometry

It is pretty similar to the previous inequality you posted.
 The point is showing $(\sum cosA)^2 \geq \sum sinBsinC$

[color=blue][b]Problem.[/b] If ABC is an acute-angled triangle, prove that $\sum\frac{\cos B\cos C}{\sin\left(2A\right)}\geq\frac{\sqrt3}{2}$, where the $\sum$ sign stands for cyclic sums.[/color]

Another [i]solution:[/i] Since $\sin\left(2A\right)=2\sin A\cos A$, the inequality in question, $\sum\frac{\cos B\cos C}{\sin\left(2A\right)}\geq\frac{\sqrt3}{2}$, rewrites as $\sum\frac{\cos B\cos C}{2\sin A\cos A}\geq\frac{\sqrt3}{2}$. Upon multiplication by 2, this becomes $\sum\frac{\cos B\cos C}{\sin A\cos A}\geq\sqrt3$. Since both sides of this inequality are positive (because ABC is an acute-angled triangle and thus cos A, cos B, cos C are positive), squaring it is an equivalence transformation; thus, it remains to prove that $\left(\sum\frac{\cos B\cos C}{\sin A\cos A}\right)^{2}\geq 3$. Now, for any three reals x, y, z, the inequality $\left(x+y+z\right)^{2}\geq 3\left(yz+zx+xy\right)$ holds, because

$\left(x+y+z\right)^{2}-3\left(yz+zx+xy\right)=\frac12\left(\left(y-z\right)^{2}+\left(z-x\right)^{2}+\left(x-y\right)^{2}\right)\geq 0$.

Using the $\sum$ sign, this inequality rewrites as $\left(\sum x\right)^{2}\geq 3\cdot\sum yz$.

Applying this to $x=\frac{\cos B\cos C}{\sin A\cos A}$, $y=\frac{\cos C\cos A}{\sin B\cos B}$ and $z=\frac{\cos A\cos B}{\sin C\cos C}$, we obtain $\left(\sum\frac{\cos B\cos C}{\sin A\cos A}\right)^{2}\geq 3\cdot\sum\frac{\cos C\cos A}{\sin B\cos B}\cdot\frac{\cos A\cos B}{\sin C\cos C}$, so that $\left(\sum\frac{\cos B\cos C}{\sin A\cos A}\right)^{2}\geq 3\cdot\sum\frac{\cos^{2}A}{\sin B\sin C}$. But in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=110333#110333]http://www.mathlinks.ro/Forum/viewtopic.php?t=15558 post #4[/url] I showed that $\sum\frac{\cos^{2}A}{\sin B\sin C}\geq 1$, so that

$\left(\sum\frac{\cos B\cos C}{\sin A\cos A}\right)^{2}\geq 3\cdot\sum\frac{\cos^{2}A}{\sin B\sin C}\geq 3\cdot 1=3$,

qed.

  Darij

[quote="James Potter"]Prove that for every acute triangle $\Delta ABC$, we have:$\sum_{cyclic}\frac{\cos B\cos C}{\sin 2A}\ge \frac{\sqrt{3}}{2}$[/quote]
[b] Lemma.[/b] Let be given a triangle $ABC$, we have: \[\frac{MA}{a}+\frac{MB}{b}+\frac{MC}{c}\ge\sqrt 3\quad \forall\; M\]
And then apply this [b]Lemma.[/b] for triangle $A'B'C'$ and the point $H$
where: $A'\in BC: AA'\perp BC,\,B'\in CA: BB'\perp CA,\,C'\in AB: CC'\perp AB$ and $H=AA'\cap BB'\cap CC'$


az doostane aziz mikham ke hamanande maratone hendese dar in maratone tarkibiat ke faghat ekhtesas be soalate dar hade marhale 2 dare hozoore fa'ali dashte bashan.

SOALE 1;  2n noghte dar safhe darim.va yek noghgteye P darim.sabet konid  noghteye P daroone tedade zoji az mosallashaye tashkil shode az in noghat gharar darad.( lotfan rahe hal kamel neveshte she :wink: )

soal: $8$adade mosbat darim ke kuchektar mosaviye 2004 hastand, sabet konid $4$az anha mesle $a,b,c,d$ra mitavan tori entekhab kard ke: $4+d=<a+b+c=<4d$

rasti kasi damorde soale oveis hich ideyi nadare? :(

yek jadvale binahayt dar binahayat darim ke ba adade $1,2,\dots,n^{2}$ pro shode va dar har satr az har adad haddeaghal yekbar oomade va dar har jadvale $n*n$ hich 2 adadi baham barabar nistan sabet konid sotooi vojood darad ke dar an daghighan $n$ adade motemaiez amade...

[b]2n [/b]noghta ro be ham vasl mikonim agar noghteye dar nazar gerefte shode kharej az shekle mohadabe be voojood amade bashe ke too hich mosalasi nist.
[b]2n[/b] zeli mohadabe be voojood oomade be tedadi teke tavasote ghotraye [b]2n[/b] zeli taghsim shode kafie sabet konim har tike joze [b]zojta[/b] mosalase:
1 tika ro dar nazar migirim farz mikonam ke joze i ta mosalas bashe bad be tikeye kenarie oon tike mirim ke ba tikeye avali dar tikeyi az ghotre [b]b[/b] moshtarake,oon tike  joze mosalasaye ghabli ke ye zeleshoon ghotr[b] b[/b] boode nist dar avaz joze mosalashaye jadidi mishe ke baz ham ghotre[b] b[/b] ast.hala oon mosalasayi ke kam vaziad shodan ekhtelafeshoon [b]zoje[/b] chon :
ye ghotr az [b]2n[/b] zeli ro dar nazar begirid mikahim ba in zel mosalas besazim[u] 2n-2-a[/u] ta ye taraf va [u]a[/u] ta ye tarafe digeye ghotre morede nazar gharar migiran.ke ekhtelafeshoon zoje.hala kafie sabet konim hadeaghal ye tike joze zojta mosalase.
hala ye zele t az [b]2n[/b] zeli ro dar nazar migirim mosalasi ro ke kam tarin masahat ro dare va ye zelesh  t hast dagigan joze [b]2n-2[/b] mosalase. pas  hameye tike ha joze zojta mosalasan.
kheili rize kari dasht omidvaram ideye halesho gerefte bashid :wink: [/b]


I don't know if that's correct :

$ n^{2}+n=2k^{2}\Leftrightarrow n(n+1)=2k^{2}$

Indeed, $ k^{2}\mid n$

Because $ n^{2}+n=2k^{2}$ we have $ n < 2k^{2}\Leftrightarrow \frac{n}{k^{2}}<2$

But, $ \frac{n}{k^{2}}$ must be intenger. Indeed, $ n = k = 1$ and $ n = 0$

Then, the solutions are $ (n,k)=(0,0), (1,1)$

$ k^{2}|n$ is wrong in general (and your "proof" is a bit strange).
Take $ k=6,n=8$ for another solution (and there are infinitely many).

[quote="cancer"]Find all the solutions to $ n^{2}+n=2k^{2}$ ,where $ n,k$ are positive integers.[/quote]

$ n(n+1) = 2k^{2}$

If $ n$ is even we have,
$ \frac{n}{2}\cdot (n+1) = k^{2}$
So since $ \gcd(n,n+1)=1$ we must have,
$ \frac{n}{2}= a^{2}$
And,
$ n+1=b^{2}$
So,
$ 2a^{2}+1=b^{2}$
This is Pellian and use the post the pco did..

If $ n$ is odd we have,
$ n\cdot \frac{n+1}{2}= k^{2}$
So,
$ n=a^{2}$ and $ n+1=2b^{2}$ and again we have Pellian.

One last, if I may:

$ 4n^{2}+4n+1 = 2(2k)^{2}+1 \implies$
$ (2n+1)^{2}-2(2k)^{2}= 1$

Which is a direct way of getting a Pell equation.


[quote="BFG"]Are you going to do it if nebula42 makes a mock Mathcount (Sprint and Target, State/National Level)?[/quote]

of course, that is, if he makes one

[quote="BFG"]Are you going to do it if nebula42 makes a mock Mathcount (Sprint and Target, State/National Level)?[/quote]
Why not? :)

No need to make two (nearly) identical topics. I think his point is, he wants those two guys two write problems because they performed well at Mathcounts (you are sorely mistaken about Treething though  :P  :wink: ).

[quote="Ignite168"][quote="anirudh"]Yes. I would of course do it. I think Ignite168 is a naysayer, no offense.[/quote]

What the HECK is a naysayer?[/quote]
Probably something derogatory...

A naysayer is someone who is habitually contrary and grumpy.


function
algebra
functional equation
algebra open

If I solve some functional equation (looking for function $f(x)$), and I gussed one solution. Name the guessed solution as function $h(x)$. Sometimes is advisable to set $f(x)=g(x)+h(x)$ and then see what happen in the given equation.

And my question is about this: what if after rewriting changes the given equation (in f) into the same equation in $h(x)$?

E.G. $f(x+y)=f(x)+f(y)$ we can guess solution $f(x)=tx$ and then set $f(x)=tx+h(x)$ and than we obtain the same equation in $h(x)$ like the previous one in $f(x)$. What to do with this?

Call $S$ the set of all functions $f(x)$ such that $f(x+y) = f(x) + f(y)$.  Since it is easily shown that $f(x) = cx$, for a constant c, fits the equation, then setting $f(x) = cx + h(x)$ gets you $f(x+y) = c(x+y) + h(x+y) = cx + h(x) + cy + h(y) \implies h(x+y) = h(x) + h(y)$.  All this tells you is that $f(x), g(x) \in S \implies f(x) + g(x) \in S$.

You get nothing new, but by choosing appreciate constant c, you can get h(1) = 0. Sometimes it help. Of course, in our case, it means nothing.


In the plane, $\Gamma$ is a circle with centre $O$ and radius $r$, $P$ and $Q$ are distinct points on $\Gamma$, $A$ is a point outside $\Gamma$, $M$ and $N$ are the midpoints of $PQ$ and $AO$ respectively. Suppose $OA=2a$ and $\angle{PAQ}$ is a right angle. Find the length of $MN$ in terms of $r$ and $a$

$\sqrt{\frac{r^2}{2}-a^2}$.
Anyway, it is unnecessary for $A$ to be outside the circle.

Here is an analytic solution (luckily, there is no complicated calculation).
Let $O$ be origin and $OM$ be $x$-axis. Suppose $P(x_0,y_0),Q(x_0,-y_0)$ and $N(x,y)$. Thus $M(x_0,0)$ and $A(2x,2y)$. Since $AP$ is perpendicular to $AQ$, $\overrightarrow{AP}\cdot\overrightarrow{AQ}=0$, which is
$(2x-x_0)(2x-x_0) + (2y-y_0)(2y+y_0)=0$
Expand it and apply $x^2+y^2=a^2$
we get $2x_0x = \frac{4a^2+x_0^2-y_0^2}{2}$.

$MN^2 = (x-x_0)^2+y^2 = x^2-2x_0x+x_0^2+y^2 = a^2 - 2x_0x+x_0^2 = a^2 - \frac{4a^2+x_0^2-y_0^2}{2} + x_0^2 = \frac{x_0^2+y_0^2}{2}-a^2=\frac{r^2}{2}-a^2$.


[size=200][color=darkred][b]PREFACE[/b][/color][/size]
Last time, we left off with the epic start to the series of the Thirteenth Moon. We finished off with our brave heros fighting off the eipians with valor. Eventually, through a series of tough physics and guts, our moderators were saved, and order was restored in ForTheWinia...

dragon96 was strolling down the road, hand in hand with isabella2296. The skipped and danced and frolicked around under the starry sky. Soon, they had tired themselves out and layed themselves out on the fresh cut grass. "What a perfect night" izzy breathed, and recieved a quick nod from dragon96. "It's getting quite late out" dragon said, "let's go back to the hero's return party". They started to walk of into the distance as yellow eyes stared them down.

They walked through the woods, once again chatting rapidly. They noted 13 species of owl, and 7 species of trees. "Look there's another octagonal redwood!" dragon exclaimed, "and there's a trignometris hooter!". The pair walked down the path and as soon as they saw the welcoming lights of the party, they started sprinting off, only to be stopped by two darts, straight into the side of the neck.

[size=200][color=green][b]CHAPTER I[/b][/color][/size]
The next morning, a concerned party group found the two in the middle of the road, smiles still etched onto their faces in enjoyment, tranquilized with a fast acting paralyzer. With a quick dial to 911, the two were on stretchers and carried to the local hospital. After a quick checkup, the doctor looked had a short chat with reknowned military captain, AIME15.

"You may want to look into this captain. Take a quick look at the darts we removed from their necks!"

"These look calcgorian in nature, am I correct doctor?"

"Absolutely! These came from the 12th moon, a land savaged by constant volcano eruptions and noted for a lack of oxygen while containing massive amounts of carbon."

"Very interesting doctor, I'll look into creating a mission to go to calcgoria", AIME15 lied through gritted teeth. He had no time to create a mission, for he was busy working on the paperwork that he had created from the last mission he had completed.

He walked out into the lobby. On the TV was the news channel. The headline of the day: ADMINS CAPTURED!!! CALCGORIANS HELD RESPONSIBLE!!!

[size=59](sorry for the spliting between posts. My comp is being stupid and for some reason, I can't type past one box.)[/size]

AIME scrambled to the office and orginized a group of 13 men, (and himself) as follows:

1. EggyLv.999- 1229 MRU = 229
2. isabella2296- 1703 MRU = 703
3. FlyAgaric- 1200 MRU = 200
4. AIME15- 2053 MRU = 999
5. lotsofmath- 1658 MRU = 658
6. r15s11z55y89w21- 1637 MRU = 637
7. spaceguy524- 1188 MRU = 188
8. louiethechewy123- 1037 MRU = 37
9. Chompy- 1200 MRU = 200
10. vahalla- 1218 MRU = 218
11. Bugi- 1520 MRU = 520
12. tennis123- 1175 MRU = 175
13. dragon96 - 1518 MRU = 518
14. algebra2000- 1300 MRU = 300

More details later.

I thought I joined!  :mad: :wallbash_red:

Anyhow, dojo should be corrected as: 13 people. :P

EDIT:make that 14 since I went in.

Are there going to be positions? I saw some when I was reading LTM (the first).

[hide="Ranks In Order Of MRU"]
\[ \begin{tabular}{|c|c|c|}\hline Rank & Username & MRU
\\ \hline 1 & AIME15 & 999
\\ \hline 2 & isabella2296 & 703
\\ \hline 3 & lotsofmath & 658
\\ \hline 4 & r15s11z55y89w21 & 637
\\ \hline 5 & Bugi & 520
\\ \hline 6 & dragon96 & 518
\\ \hline 7 & algebra2000 & 300
\\ \hline 8 & EggyLv.999 & 229
\\ \hline 9 & vahalla & 218
\\ \hline 10 & FlyAgaric & 200
\\ \hline 10 & Chompy & 200
\\ \hline 12 & spaceguy524 & 188
\\ \hline 13 & tennis123 & 175
\\ \hline 14 & louiethechewy123 & 37
\\ \hline \end{tabular}\]
[/hide]

Dojo, is there a leader of the group?

hopefully not AIME...remember all those "bad leader" comments in one?
Otherwise, AIME15 is okay...

So whens chap. 2 come out?

Ok... :P you put my lower bound for FTW rating (never mind). Wait, what's MRU? And is FTW important for this?

MRU is health, at least in the original LTM.

MRU is determined by dojo with your FTW rating, 999 is highest possible.

its determined by how much rating points you have over 1000.

so if your rating is 900, you can't join.

as for the preface: dragon wishes :P

when are we actually going to start playing?

When SOMEONE figures out I signed up but they forgot to add me to da list. Either that or LTM is dead and people don't like it anymore so they don't want to play it.

So what are we supposed to do right now?

[quote="algebra2000"]as for the preface: dragon wishes :P [/quote]

Har har. You are quite hilarious.

[quote="dragon96"]
Har har. You are quite hilarious.[/quote]

yes i am :)

so... you need a replacement?  :(

Use a different comp, then.

ok dojo, relax, let the waves of energy flow through you, revitalizing your soul. oh...wait...your computers broken? oh well...

so, since dojo's computer died, LTM II will be postponned? :(

[quote="Bugi"]Why does everyone has no connection, dying computer or some other problem when modding a game?  :wink:  :wink:  :wink:[/quote]

quite true, but you forgot to include vacation.

Uh oh :(.

I only have 187 MRU...

At least I don't have the least amount of MRU

During our mission, can you gain MRU????? SAY YES

louiethechewy is going to die very quickly...wow now my rating has grown to 1430 or so...since the game hasnt rly started can you change my MRU dojo?

Is this going to be put on hold until Dojo has computer access again? If so, we have another 2 weeks to wait...

I thought I joined?!??!   :(  My FTW rating is 1188.  (If you don't believe I tried to join, check the joins thread.  I tried....)

Perhaps you should have read the rest of that thread.

Dojo is at MathZoom, and has limited access for now.

Dojo, when you come back, make a new thread.....

[quote="Bugi"]Why does everyone has no connection, dying computer or some other problem when modding a game?  :wink:  :wink:  :wink:
[/quote]

My theory is that everybody has these problems, but they are not noticed when you aren't modding a game.

[quote="AIME15"]Two weeks, I believe.

Posted: Fri Jul 24, 2009 6:17 pm  [/quote]

*cough* hasnt been two weeks yet *cough*

[quote="Dojo"]Sorry, just got back from the math camp.[/quote]

To Dojo: make a new thread since the entire 2nd page is spam.


Yes, the inequality is inverse. I have edited. Thanks for appointing my mistake.

[quote="Inequalities Master"]Let $ a_1,a_2,...,a_n > 0$. Prove that:
$ a_1^{a_1}a_2^{a_2}...a_n^{a_n}\geq (a_1a_2...a_n)^{\frac {a_1 \plus{} a_2 \plus{} ...a_n}{n}}$[/quote]
is it yours
[hide="hint"]first,rewrite it in the form $ ln (RHS)\le ln(LHS)$Assume $ a_1\le ..a_n$ and use Chebysev ineq
It is quite old,Inequalities Master[/hide]


HMMT MIT college email geometry calculus Princeton

OK,

I don't think SC will be sending a team to HMMT (we's a po' state).  However, I would like to go, and I know we have a few other strong students in the state who would also like to go.  If worst comes to worst, we can do some fundraising (aka begging).  So what do teams look like?  Who is in QUAGGA?  I assume that's Florida?  Who are the NC people going with?  Also, if anyone on a decent team has an open spot for me, PM me and we can discuss it (I can pay my own way).

Yeah, I actually looked into that a few days ago and according to the website the deadline's passed already.  I'm not all that involved in the contest so you'll have to e-mail hmmt-request at mit dot edu and ask them.

Where did you see that?  I looked at the website, and you can pay up until the day of the contest...but I can't find any deadlines.  I will email them and find out, I guess.

But anyway, the website says you can still change your roster, so if a team has an open spot...

Yeah, you can still change your roster.  I guess I was replying to the "I don't think SC will be sending a team".  If there are still open spots then you could probably still sign up, but I don't know if the cap was reached or not.

Yeah, I emailed him, and December 7th was the deadline.  In fact, some teams that registered on time were waitlisted.  Oh well, I'll just have to find a team that will let me have a spot.  Even if I'm not on a killer team, I can still place individually in Algebra or perhaps Combinatorics.

[quote="mysmartmouth"][quote="Cicatriz"]Vestavia is going as a school, like it should be[/quote]

Well the rules say you don't need to be a school...and think about it: Vestavia vs. a normal high school isn't fair.  We can't take classes called "Math Team."

But I think I may have found a team, we'll have to work out the details though.[/quote]

thatd suck.  at our middle school, math competitions is an elective.  its the best class ever.

[quote="mysmartmouth"][quote="Cicatriz"]Vestavia is going as a school, like it should be[/quote]

Well the rules say you don't need to be a school...and think about it: Vestavia vs. a normal high school isn't fair.  We can't take classes called "Math Team."

But I think I may have found a team, we'll have to work out the details though.[/quote]

and Vestavia vs. both states of Carolina or the combined power of Florida and choice picks from Illinois/DC is any more fair?

a) Why not?

There are advantages to having a strong enough school to field a decent HMMT team.  Since you're all living around the same area and go to the same school it is much easier to get together for practice.

b) SC/NC isn't forming a team for HMMT.  Even if they were that wouldn't be any more "unfair" than NC by itself.

Honestly,

SC + NC < NC

because teams were picked subjectively.  We didn't take the 10 best overall, we took the 7 best SCers, and the 3 best NCers.  However, NCer #10 is probably better than SC #1.

Yeah, note how Jeremy Hahn was on the central MD team.

that was actually something different

they passed up noah blach though wtf

I just think it's more interesting to compare either schools or states. Obviously one school can have strengths over other school but that kind of variety makes it interesting. If you want something completely fair, we can just have one team per division and they will compete against their own score 10 times, so it will always be a 10-way tie for first. That'd make everyone happy, right?

I was on a conglomerate team, so I can't say it's completely unforgivable, but I couldn't really feel that much pride about our success as a team.

Who's on "Woot minus Quagga"?

[quote="Cicatriz"]Who's on "Woot minus Quagga"?[/quote]

$\bullet$ Calculus
$\bullet$ chess64
$\bullet$ E^(pi*i)=-1
$\bullet$ K81o7
$\bullet$ n^4+4
$\bullet$ NeverOddOrEven
$\bullet$ pkerichang
$\bullet$ SplashD

It would be so much better if Dnas agreed to come, and also if we had the Quagga WOOTlings...

Hmm. I think our team is pretty good.

We're only recruiting 2-3 people for our first team. And well, one of them is a time honored tradition. Yay, Arnav! It is only a competition. Its giving an opportunity to people who otherwise wouldn't have it; to go with a team that has a good chance of placing in the top 5. I think you all underestimate the actual strength of the Florida team, when we have our best people. Notice that at Princeton, several of our 2nd team members placed in the top 5 or 10.

There are a couple people who have asked to be on a Florida team, and I'm totally cool with that. I'm delegating them to our 2nd and 3rd team. People just like us, I guess. As far as the NC+SC thing goes, more power to them. The WOOT team idea is just a bit ridiculous though. There isn't even like a state behind that. But we'll see how it goes.

EDIT: Giving it more thought: I've decided that for all you stragglers without a team, we have 0 spots on our third team. I suggest you all come up with a team name that I approve of.

EDIT 2: Wow that went fast.

[quote="Pakman2012"]The WOOT team idea is just a bit ridiculous though. There isn't even like a state behind that.[/quote]

What... it's clearly MD+NC+NY+NJ+CA.

I don't really see what the big deal is with teams consisting of people from all over.  Instead of making things less fair I can argue it does the opposite.  What if I were from a school that was unable to field an HMMT team and thus would not go unless I joined another team?  On the other hand, it gives people or schools that never win a chance to do so.  Yeah, we all know TJ, Exeter, AAST, Vestavia, etc are good schools.  You've won/placed over and over again.  At the least, we don't want you to grow complacent at the top ;)

I'd hardly compare Vestavia's ability at HMMT with TJ or Exeter...

[quote="mysmartmouth"][quote="chess64"][quote="Pakman2012"]The WOOT team idea is just a bit ridiculous though. There isn't even like a state behind that.[/quote]

What... it's clearly MD+NC+NY+NJ+SC.[/quote]

Whoa, who's the WOOT person from SC?  If you don't have one, and want one... :wink:[/quote]

Woops, that was a typo. No one from SC... actually wait we do have someone from CA though... so yeah, replace the SC with CA.

[quote="Phelpedo"]Vestavia is mostly notable for its performance on Mu Alpha Theta, isn't it?[/quote]

Yea but our ability to perform drops quickly when you focus on tournaments like arml/hmmt.

dude we should send a team called quagga IMO team


$\sin x + \sin y = m, \cos x +\cos y = n$
Find $\tan{(x+y)}$

Based on identity $\frac {\sin{x}+\sin{y}}{\cos{x}+\cos{y}}=\tan{(\frac{x+y}{2})}$ , we get $\tan({\frac{1}{2} (x+y)})=\frac{m}{n}$

Let $(x+y)=A$ therefore based on identity $\tan{(2.(\frac{1}{2}A))}=\frac{2 \tan{\frac{1}{2}A}}{1-\tan^2{\frac{1}{2}A}}$

We get $\tan(x+y)=\frac{2mn}{n^2-m^2}$


Given $\{ a_{n}\}$, $a_{0}= \frac{1}{3}$, $a_{n}= \sqrt{\frac{1+a_{n-1}}{2}}$. Show that $\{ a_{n}\}$ is monotonically increasing.

you can show that 0<a_n<1

then make a trig substitution $a_{n}=\cos\theta_{n}$ where $0<\theta<\frac{\pi}{2}$

the condition is equivalently $a_{n+1}=\cos\theta_{n+1}=\cos\frac{\theta_{n}}{2}$

so $\theta_{n+1}=\frac{1}{2}\cdot\theta_{n}$

since $\theta_{n}$ approaches $0$, (where $\theta_{n}$ is strictly decreasing)

then cosine is a decreasing function so since the domain decreases, the sequence is strictly increasing

Here are two more solutions.

Solution one:
Solar Plexsus has shown that $a_{n}< 1$, so $a_{n+1}-a_{n}=  \sqrt{\frac{1+a_{n}}{2}}-a_{n}$. This expression is greater than zero if and only if:

$ a_{n}^{2}\leq \frac{1+a_{n}}{2}\Longleftrightarrow 2a_{n}^{2}-a_{n}-1 \leq 0 \Longleftrightarrow (2a_{n}+1)(a_{n}-1) \leq 0$, which is obviously true.

Solution two:
$ a_{n+1}-a_{n}= \sqrt{\frac{1+a_{n}}{2}}-\sqrt{\frac{1+a_{n-1}}{2}}= \frac{1}{\sqrt{2}}\cdot \frac{a_{n}-a_{n-1}}{\sqrt{1+a_{n}}+\sqrt{1+a_{n-1}}}$, which means that $a_{n+1}-a_{n}$ and $a_{n}-a_{n-1}$ has the same sign. Similarly we get that all $a_{i}-a_{i-1}$ have the same sign. Since $a_{0}= \frac{1}{3}$ and $a_{1}= \sqrt{\frac{2}{3}}> \frac{1}{3}= a_{0}$, so we always have $a_{n+1}> a_{n}$, which means $\{ a_{n}\}$ is monotonically increasing.


function
calculus
derivative
algebra unsolved
algebra

[quote="jagdish"][b](1) find  Min value of (x^2+y^2)^1/2 +(x^2+(y-1))^1/2 =
(x^2+y^2)^1/2  square root of  (x square +y square) [/b][/quote]

We are looking for the minimum value of $ \sqrt{x^2\plus{}y^2}\plus{}\sqrt{x^2\plus{}y\minus{}1}$

Wlog say $ x\geq 0$

1) if $ x\in[0,1)$
We need then $ y\geq 1\minus{}x^2\geq 0$ and then ($ y\geq 0$) $ \sqrt{x^2\plus{}y^2}\plus{}\sqrt{x^2\plus{}y\minus{}1}$ is an increasing function of $ y$ and minimum is reached for $ y\equal{}1\minus{}x^2$

This minimum is $ \sqrt{x^2\plus{}(1\minus{}x^2)^2}$ $ \equal{}\sqrt{(x^2\minus{}\frac 12)^2 \plus{}\frac 34}$ and this minimum is reached for $ x^2\equal{}\frac 12$
So, if $ x\in[0,1)$, minimum is reached when $ x^2\equal{}y\equal{}\frac 12$ and the value is $ \sqrt{\frac 12 \plus{}\frac 14}\equal{}\frac{\sqrt 3}2$

2) if $ x\geq 1$ and $ y\geq 0$
Then $ \sqrt{x^2\plus{}y^2}\plus{}\sqrt{x^2\plus{}y\minus{}1}$ is an increasing function of $ x$ and minimum is reached for $ x\equal{}1$ and the value is $ \sqrt{y^2\plus{}1}\plus{}\sqrt{y}$ whose minimum is reached for $ y\equal{}0$ and value is $ 1$

3) if $ x\geq 1$ and $ y\leq 0$
Then $ \sqrt{x^2\plus{}y^2}\plus{}\sqrt{x^2\plus{}y\minus{}1}$ is an increasing function of $ x$ and minimum is reached for $ x\equal{}\sqrt{1\minus{}y}$ and the value is $ \sqrt{y^2\plus{}1\minus{}y}$ whose minimum is reached for $ y\equal{}0$ and is $ 1$

So the minimum value is obtained when $ x^2\equal{}y\equal{}\frac 12$ and the value is $ \boxed{\frac{\sqrt 3}2}$

[b](1) Thanks PCO for giving a NICE solution.
but actually my question is (1)
find Min value of (x^2+y^2)^1/2 +(x^2+(y-1)^1/2)^1/2 =
(x^2+y^2)^1/2 square root of (x square +y square)[/b]

So new problem (please try to give the good questions on the first try) :(

[quote="jagdish"][b](1) but actually my question is (1)
find Min value of (x^2+y^2)^1/2 +(x^2+(y-1)^1/2)^1/2 [/quote]

We are looking for the minimum value of $ \sqrt {x^2 \plus{} y^2} \plus{} \sqrt {x^2 \plus{} \sqrt{y \minus{} 1}}$

Wlog say $ x\geq 0$
$ y\geq 1$

For $ x\geq 0$ and $ y\geq 1$, $ \sqrt {x^2 \plus{} y^2} \plus{} \sqrt {x^2 \plus{} \sqrt{y \minus{} 1}}$ is an increasing function of both variables $ x$ and $ y$ and so minimum is reached when $ x\equal{}0$ and $ y\equal{}1$ and is $ 1$

If you are allowed to use Calculus, you could use derivitives to find critical values and apply sign chart. Minimum occurs when left side is negative and right side is positive. But taking derivitive is a booger.


calculus
integration
ratio
floor function
quadratics
geometric sequence
algebra

For a positive number such as 3.27, 3 is referred to as the integral part of the number and .27 as the decimal part. Find a positive number such that its decimal part, its integral part, and the number itself form a geometric progression.

[hide]It's $ \phi\equal{}\frac{1\plus{}\sqrt5}{2}$.

If the integral part is greater than 1, then the common ratio will be too large.  So the integral part is one.
If $ x$ is the decimal portion, then $ \frac{1}{x}\equal{}\frac{x\plus{}1}{1}$.  Solve for $ x\plus{}1$ to get $ \phi$.[/hide]

Nice. I actually got that as well except I somehow did algebra wrong and stared at it until I saw your solution.  :blush:

Here's the same problem with a different wording.

Let $ x$ be a real number and $ \lfloor x \rfloor$ be the floor function.  For which values of $ x$ do $ x\minus{}\lfloor x \rfloor, \lfloor x \rfloor, x$ form an increasing geometric sequence?

Nice, it seems your wording makes the problem appears harder though.  :D

[hide]Let $ d$ be the decimal part (which is not zero), and $ r$ be the geometric progression common ratio. The integral part is $ dr$ and the number itself is $ dr^2$ (the next term in the geometric sequence). But also $ dr^2\equal{}dr\plus{}d$. Solving the quadratic equation $ r^2\equal{}r\plus{}1$ gives us the same positive answer.[/hide]


calculus
integration
calculus computations

Hi there i am trying to make this equation look exact.

$(Cos(2y)-Sin (x)) dx-2 Tan (x) Sin (2y) dy = 0$

What I've done so far is take the partial with respect to x and y.

So, my

$M_{y}$ is equal to $-2 Sin (2y)-0$  and,

my $N_{x}$ is equal to $-2(Sec^{2}(x)) Sin (2y)$

Which makes it not exact. So, then I tried using
$\frac{M_{y}-N_{x}}{-N}$ and,

here is where I have tried so many times to find out a way to find an Integration Factor (I.F.)

Any suggestions will help, thanks for your time.

hello, i have got the following solution
$ y(x)\equal{}\frac{1}{2}\arccos\left(\frac{\sin(x)^2\plus{}C}{2\sin(x)}\right)$
Sonnhard.


logarithms
floor function
irrational number
number theory unsolved
number theory

Hey all,
I have some difficulties solving a problem.

Prove that there exist a power of 2 such that its decimal representation starts with 2009.

Looks very interesting and I believe that it is a popular problem, but I couldn't google it.

More generally

[color=blue]Prove that given any positive integer $ N$, there are infinitely many powers of $ 2$ whose decimal representation start with $ N$.[/color]

We look for integers $ n,k$ such that $ N10^n < 2^k < (N\plus{}1)10^n$; then the decimal representation of $ 2^k$ will start with $ N$. Take decimal logarithms to get $ n\plus{} \log N < k\log 2 < n \plus{} \log(N\plus{}1)$. But $ \log 2$ is irrational (easy to prove), and some celebrated theorems (Dirichlet, Kronecker, Weyl) state that the fractional parts of the integer multiples of any irrational number are dense in $ (0,1)$. Therefore there exist infinitely many positive integers $ \ell$ such that $ \ell\log2 \minus{} \lfloor \ell\log2 \rfloor \in (0, \log (N\plus{}1) \minus{} \log N)$. Denoting $ m \equal{} \lfloor \ell\log 2 \rfloor$, this is equivalent to $ 0 < \ell\log 2 \minus{} m < \log (N\plus{}1) \minus{} \log N$. Take now some integer $ M$ large enough so that $ M \plus{} \log N > 1$, and take $ \displaystyle t \equal{} \left \lfloor \frac {M \plus{} \log N} {\ell\log 2 \minus{} m} \right \rfloor$. Then $ M \plus{} \log N < (t\plus{}1)(\ell\log 2 \minus{} m) < M \plus{} \log (N\plus{}1)$. All that is left is take $ n \equal{} M \plus{} (t\plus{}1)m$ and $ k \equal{} (t\plus{}1)\ell$.

Moreover, this proof works for powers of any positive integer $ d$ other than $ 2$, with the exception of $ 10$ and its powers (which obviously start with $ 100\ldots$). This ensures that $ \log d$ is irrational, and the proof goes the same.

Thanks a lot. I also found a solution for 123454321 instead of 2009, but I wasn't quite convinced by that solution.

I followed mavropnevma's steps and it produced n=129932880.  The first digits of 2^n are 200913254645...


During which years were calculators allowed on the AMC/AHSME?

why did they stop allowing the use of it?

It removed the unfair advantage students with high tech calculators had over those with low tech ones. Additionally, it is not possible to just bash numbers for answers anymore.


Find $ x$ if $ \displaystyle \frac{2}{x} \minus{} \frac{3}{5} \plus{} \frac{1}{x} \equal{} \frac{1}{5}$.

First simplify this equation.

$ \frac{2}{x} \plus{} \frac{1}{x} \equal{} \frac{4}{5}$

We want to eliminate the denominators, so multiply both sides by $ 5x$.

$ 10 \plus{} 5 \equal{} 4x$

$ 15 \equal{} 4x$

$ \boxed{\frac{15}{4}} \equal{} x$


conics
parabola
analytic geometry
algebra unsolved
algebra

Let p and q be positive numbers such that y=x^2 -2px + q has no common point with the x-axis. Prove that there exists points A and B on the parabola such that the segment AB is parallel to to the x axis and <AOB (O is the coordinate origin) if and only if p^2<q<=1/4. Find the values of p and q for which the points A and B are defined in an unique way.

What's the condition about  "<AOB"  (angle ?)


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

$ \sum\sqrt {\frac{a(b+c)}{a^2+bc}} \leq \sqrt{(\sum \sqrt{a})(\sum {\frac{1}{\sqrt{a}})}}$

[quote="manlio"]For $ a,b,c$ positive reals prove that

$ \sum\sqrt {\frac {a(b + c)}{a^2 + bc}} \leq \sqrt {(\sum \sqrt {a})(\sum {\frac {1}{\sqrt {a}})}}$[/quote]
[url=See here.]http://www.mathlinks.ro/viewtopic.php?t=119080[/url]


Given x,y,z>0.Prove that
$ \frac{x(y\plus{}z)^2}{2x\plus{}y\plus{}z}\plus{}\frac{y(x\plus{}z)^2}{x\plus{}2y\plus{}z}\plus{}\frac{z(x\plus{}y)^2}{x\plus{}y\plus{}2z}$≥$ \sqrt(3xyz(x\plus{}y\plus{}z))$

[quote="SUPERMAN2"]Given x,y,z>0.Prove that
$ \frac {x(y \plus{} z)^2}{2x \plus{} y \plus{} z} \plus{} \frac {y(x \plus{} z)^2}{x \plus{} 2y \plus{} z} \plus{} \frac {z(x \plus{} y)^2}{x \plus{} y \plus{} 2z}\geq\sqrt{3xyz(x \plus{} y \plus{} z)}$[/quote]
But it's obvious by SOS. :wink:

$ \sum_{cyc}\frac {x(y \plus{} z)^2}{2x \plus{} y \plus{} z} \minus{} \sqrt {3xyz(x \plus{} y \plus{} z)} \equal{}$
$ \equal{} \sum_{cyc}\left(\frac {x(y \plus{} z)^2}{2x \plus{} y \plus{} z} \minus{} yz\right) \plus{} xy \plus{} xz \plus{} yz \minus{} \sqrt {3xyz(x \plus{} y \plus{} z)} \equal{}$
$ \equal{} \sum_{cyc}\frac {z^2(x \minus{} y) \minus{} y^2(z \minus{} x)}{2x \plus{} y \plus{} z} \plus{} \sum_{cyc}\frac {z^2(x \minus{} y)^2}{2\left(xy \plus{} xz \plus{} yz \plus{} \sqrt {3xyz(x \plus{} y \plus{} z)}\right)} \equal{}$
$ \equal{} \sum_{cyc}(x \minus{} y)\left(\frac {z^2}{2x \plus{} y \plus{} z} \minus{} \frac {z^2}{2y \plus{} x \plus{} z}\right) \plus{}$
$ \plus{} \sum_{cyc}\frac {z^2(x \minus{} y)^2}{2\left(xy \plus{} xz \plus{} yz \plus{} \sqrt {3xyz(x \plus{} y \plus{} z)}\right)} \equal{}$
$ \equal{} \sum_{cyc}(x \minus{} y)^2\left(\frac {z^2}{2\left(xy \plus{} xz \plus{} yz \plus{} \sqrt {3xyz(x \plus{} y \plus{} z)}\right)} \minus{} \frac {z^2}{(2x \plus{} y \plus{} z)(2y \plus{} x \plus{} z)}\right).$
Thus, it remains to prove that
$ (2x \plus{} y \plus{} z)(2y \plus{} x \plus{} z)\geq2\left(xy \plus{} xz \plus{} yz \plus{} \sqrt {3xyz(x \plus{} y \plus{} z)}\right).$
But $ (2x \plus{} y \plus{} z)(2y \plus{} x \plus{} z)\geq2\left(xy \plus{} xz \plus{} yz \plus{} \sqrt {3xyz(x \plus{} y \plus{} z)}\right)\Leftrightarrow$
$ \Leftrightarrow2x^2 \plus{} 2y^2 \plus{} z^2 \plus{} 3xy \plus{} xz \plus{} yz\geq2\sqrt {3xyz(x \plus{} y \plus{} z)},$
which is true because $ x^2 \plus{} y^2 \plus{} z^2\geq xy \plus{} xz \plus{} yz\geq\sqrt {3xyz(x \plus{} y \plus{} z)}.$

It seems that the following inequality is true too.
Let $ x,$ $ y$ and $ z$ are positive numbers. Prove that:
\[ \frac {x(y \plus{} z)^2}{3x \plus{} 2y \plus{} 2z} \plus{} \frac {y(x \plus{} z)^2}{2x \plus{} 3y \plus{} 2z} \plus{} \frac {z(x \plus{} y)^2}{2x \plus{} 2y \plus{} 3z}\geq\frac{4}{7}\sqrt{3xyz(x \plus{} y \plus{} z)}\]


HI, I am a high school student. I did really well on the AP CALC BC exam. But my school doesnt offer any math class to me b/c i finished calc bc. And I studied multi-calc on my own. SO , im just wondering if u are really good with calc, are there any OFFICIAL calculus contests, exams, or competitions?

ALSO, are there any official and outstanding math research competitions for high school juniors?
INTEL is for seniors only. And i dont wanna form a team...

The Siemens Westinghouse competition, and you can do Intel on your own if you want.


At Jefferson Summer Camp, $ 60\%$ of the children play soccer, $ 30\%$ of the children swim, and $ 40\%$ of the soccer players swim. To the nearest whole percent, what percent of the non-swimmers play soccer?

$ \textbf{(A)}\ 30\% \qquad
\textbf{(B)}\ 40\% \qquad
\textbf{(C)}\ 49\% \qquad
\textbf{(D)}\ 51\% \qquad
\textbf{(E)}\ 70\%$

[hide="Click for solution"]
Let us assume there are $ 50$ children at the camp.  We know that:

$ 0.6 \times 50\equal{}30$ children play soccer.

$ 0.3 \times 50\equal{}15$ children swim.

$ 0.4 \times 30\equal{}12$ children play soccer and swim.

There are $ 50\minus{}15\equal{}35$ children who do not swim and $ 35\minus{}17\equal{}18$ of them play soccer.  The desired percentage is thus $ 100 \times \frac{18}{35}$, which is just over $ 50 \%$, so the answer is $ \boxed{\textbf{(D)}}$.
[/hide]

[quote="AIME15"][hide="Click for solution"]
Let us assume there are $ 50$ children at the camp.  We know that:

$ 0.6 \times 50 \equal{} 30$ children play soccer.

$ 0.3 \times 50 \equal{} 15$ children swim.

$ 0.4 \times 30 \equal{} 12$ children play soccer and swim.

There are $ 50 \minus{} 15 \equal{} 35$ children who do not swim and $ 35 \minus{} 17 \equal{} 18$ of them play soccer.  The desired percentage is thus $ 100 \times \frac {18}{35}$, which is just over $ 50 \%$, so the answer is $ \boxed{\textbf{(D)}}$.
[/hide][/quote]
Where did u get the 17 to subtract from 35?

Assuming $ 50$ kids, you know that there are $ 30$ kids that play soccer, therefore if $ 12$ play both sports, you know $ 18$ swims only. Similarly, you know that there are only $ 3$ kids who only swim. That sums to $ 12 \plus{} 18 \plus{} 3 \equal{} 33 \rightarrow50 \minus{} 33 \equal{} 17$ kids who don't do either sport, which you need to subtract from the total kids after subtracting swimmers.

Or you can just stop at saying that there are $ 30$ kids that play soccer and $ 12$ swim, so only $ 30\minus{}12\equal{}18$ play soccer and don't swim.

I got confused with the wording of "non-swimmers". I thought it was anyone who didn't swim, so I got my fraction to be 36/46 because the 36 kids who play soccer and don't swim plus the 10 kids who didn't do anything is 46. I'm basing these numbers off of a camp with 100 children.

You are counting wrong.  Soccer players can also be swimmers...

I still don't get this, can someone please explain this thoroughly?
I don't get where you are getting all the numbers from.

Assume that there are 100 children in the camp.
$60\% \times 100=60$ of them play soccer.
$30\% \times 100=30$ of the children swim.
Out of the 60 soccer players, $40\% \times 60=24$ of them swim.
So $60-24=36$ play soccer but do not swim.
There are 30 children who swim, so there are $100-30=70$ total non-swimmers.
We are looking for the percent of non-swimmers playing soccer, which is $\frac{36}{70}$, which is slightly over $\frac{35}{70}=\frac{1}{2}=50\%$. So the answer is (D) 51%.

It is really helpful if you draw a Venn Diagram.

[hide=Solution]Assume that there are $1000$ children in the camp.
$600$ play soccer, $300$ swim, and $40\%$ of $600=240$ play soccer and swim. So there are $700$ non-swimmers in the camp.
$600-240=360$ non-swimmers play soccer, so $\frac{360}{700}$ of the non-swimmers play soccer. This can be simplified to $\frac{18}{35}$, which is about $\boxed{\textbf{(D)}\ 51\% \qquad}$ [/hide]

[quote=HrishiP][quote=ethanlanc]7.5 year bump :rotfl:[/quote]

megarnie is probably going to bump every single AMC problem thread ever so expect more of that :P. Although seriously I know I can't make you not post solutions (you're allowed to) but this is getting a little annoying to see what I mean just scroll down C&P and it's all AMCs (mostly).[/quote]

I won't bump all amc problems only some


Find the Roots of $ 8x^{4}\minus{}8x^{2}\plus{}1$

i will just write the answer down, someone else please show how you work them out.

$ x\equal{}\frac{\pm\sqrt{\sqrt{2}\plus{}2}}{2}$ or $ \frac{\pm{\sqrt{2\minus{}\sqrt{2}}}}{2}$

[hide="Hint"]
Let $ x^{2}\equal{}a$ and then just use the quadratic formula
[/hide]


Its idea is Intermediate but since it has a lot of square roots, I decide to post on Pre-Olympiad. It's very easy for Pre-Olympiad level so have fun on "easy solving."

Number 1

For what integers $b$ and $c$ is $x = \sqrt{19} + \sqrt{98}$ a root of the equation $x^4+bx^2+c = 0$?

[hide="View"]
I got $b = -234, c = 6241$ if anyone wanted the answer.[/hide]

The equation is a quadratic in $x^2$ with roots $x^2 = \frac{-b \pm \sqrt{b^2 - 4c}}{2}$.  We have that $117 + 14\sqrt{38}$ is the square of a root.  Thus, for some integers b and c, we have $234 + 28\sqrt{38} = -b \pm \sqrt{b^2 - 4c}$.

Now it is clear $b = -234$ etc.

Very nice method, Singular.

From his answer, it is pretty easy to find $c$ because once you found what $b$ equals:

You have $x^4-234x^2+c = 0$ and putting $\sqrt{19} + \sqrt{98}$ for $x$ gives $-6241+c = 0$ which tells $c = 6241$.

I got the answer by first setting $x-(\sqrt{19} + \sqrt{98})$ and multiplying it by $x+(\sqrt{19}+\sqrt{98})$ and then I'll have quadratic equation. Using some playing with positive and negative sign, you can come up with quartic equation, $x^4-234x^2+6241 = 0$, telling $b = -234$ and $c=6241$.

Again, nice easy method, Singular. :D


For the last couple days my Aops hasn't been updating properly.  Basically, even through there have been new posts made in certain sub-forums, I can only view the old posts from 1 -2 days ago.  This is happening in getting started, mathcounts and some other topics.. What can I do?

EDIT- nvm, i see what the problem is- I have to open the topic and the refresh, I use to be able to refresh the homepage and everything else would be updated as well....

Deeply werid.  Try deleting all your old internet files, and perhaps your cookies.


Does there exist a function $ f\colon \mathbb{R}\to\mathbb{R}$ such that for all x the following equality holds
$ f(sinx)\plus{}f(sin^2x)\plus{}f(sin^3x)\equal{}f(cosx)\plus{}f(cos^2x)\plus{}f(cos^3x)$

_______________________________________
Azerbaijan Land of the Fire   :ninja:  :ninja:

[quote="Pirkuliyev Rovsen"]Does there exist a function $ f\colon \mathbb{R}\to\mathbb{R}$ such that for all x the following equality holds
$ f(sinx) \plus{} f(sin^2x) \plus{} f(sin^3x) \equal{} f(cosx) \plus{} f(cos^2x) \plus{} f(cos^3x)$   [/quote]

Yes, it does : $ f(x)\equal{}a$


algebra
polynomial
calculus
derivative
search
algebra proposed

Let $ n>2$ be an integer. Show, without calculus, that the polynomial $ P(x)\equal{}x^n\minus{}nx^{n\minus{}1}\plus{}1$ has $ n$ distinct zeros.

[hide]I think it's also true that $ P(x)\equal{}0$ has $ n$ distinct real roots, but I could not find a way of proving it. Any help on this will be greatly appreciated. :) [/hide]

That depends on what you mean by "calculus."  I can define the [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=225324]formal derivative[/url] without any knowledge of calculus and show that it has the power to detect repeated roots (and in fact, the only reason we want to define the derivative formally is to detect repeated roots in fields where we do not have a natural notion of limit).

One can in fact say something much stronger:  $ P(x)$ is [i]irreducible[/i] for $ n > 2$ by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1088214531&t=150]Perron's criterion[/url] and, since $ \mathbb{Q}$ is a [url=http://en.wikipedia.org/wiki/Separable_extension]perfect field[/url], cannot have repeated roots.  (I do not claim to have satisfied your constraint; Perron's criterion requires a fair bit of analysis to prove, and the proof that fields of characteristic zero are perfect requires the use of the formal derivative.)

Edit:  Your conjecture fails for $ n \equal{} 4$ but suggests an interesting exercise.  Let $ Q(x)$ be a polynomial of degree $ n$ with $ n$ distinct real roots.  Prove or disprove that $ P(x) \equal{} Q(x) \minus{} Q'(x)$ also has $ n$ distinct real roots.

Thanks for your comments. By "without calculus" here I meant that, "use only plain and simple algebra, no advanced things i.e. do it with no knowledge of calculus" (kinda like the rule at IMO as [url=http://en.wikipedia.org/wiki/International_Mathematical_Olympiad]Wikipedia[/url] says:)

[quote="Wikipedia"]They require no knowledge of higher mathematics such as calculus and analysis[/quote]

The $ nx^{n\minus{}1}$ surely makes one believe that the solution to this problem must have something to do with derivatives. But I just put that expression for fun, it's nothing special really! I can give you examples of hundreds of polynomials which have nothing to do with derivatives. :)

That's not what I mean.

[b]Theorem:[/b]  A polynomial $ P(x)$ has no repeated roots if and only if $ \gcd(P(x), P'(x)) \equal{} 1$, and this is true over [i]any[/i] field.

This has nothing to do with the form of the problem; it's simply both the most basic and the most useful tool we have to deal with repeated roots.  The calculation becomes trivial:  $ P'(x) \equal{} nx^{n\minus{}1} \minus{} n(n\minus{}1) x^{n\minus{}2}$ which has roots $ 0$ and $ n\minus{}1$, neither of which are roots of the original equation - and voila!

I know it isn't very hard if we use derivatives, I also found the same solution as yours. But I have another solution where I don't use any calculus at all. I was looking for such a solution. That's why I said, "no calculus!" Hope you understand what I mean. :)

I believe that [b] Nayel [/b] is correct. I 'm still working on it.

@[b] t0rajir0u [/b] Did you mean that we could define the derivative of a polynomial as:

Definition: Let $ f(x) \equal{} \sum_{i\equal{}0}^{n} {a_i x^i} \in F[x]$. The derivative of $ f$ is defined as:
\[ f'(x)  \equal{} \sum_{i\equal{}0}^{n} {i a_i x^i}\].

From this definition, we could prove the following properties:
i) $ (f\plus{}g)' \equal{} f' \plus{} g'$
ii) $ [f(g)]' \equal{} f'(g) g'$
etc.

And then applying these properties to prove that:

$ (x\minus{}\lambda)^k \| f(x)$ if and only if $ f^{i} (\lambda) \equal{} 0$ for all $ 0 \leq i \leq k\minus{}1$ and $ f^{k} (\lambda) \not\equal{} 0$.

@[b] Nayel [/b] I don't think that the terms $ n x^{n\minus{}1}$ could be replaced arbitrarily.

Yes.  Usually I believe the defining properties are something along these lines:

- The derivative operator $ \frac{d}{dx}$ takes polynomials to polynomials and is linear,
- $ \frac {d}{dx} x \equal{} 1$, and
- $ \frac {d}{dx} fg \equal{} f \frac {dg}{dx} \plus{} g \frac {df}{dx}$.

From these three properties everything else follows.  As for your second comment, $ nx^{n \minus{} 1}$ can be replaced by $ kx^{n \minus{} 1}$ for $ |k| > 2$; see my comment.


ARML geometry calculus integration trigonometry rotation

1.
In right triangle $ABC$, $\angle C = 90^\circ$, $c=3$, and $a+b=\sqrt{17}$.  Compute the area of the triangle.

2.
If $(a+b):(b+c):(c+a) = 6:7:8$, and $a+b+c=14$, compute the value of $c$.

3.
Let $z$ be a root of $z^5-1=0$, with $z \not= 1$.  Compute the value of

\[z^{15} + z^{16} + z^{17} + \cdots + z^{50}. \]

4.
A regular polygon of $2n$ sides is inscribed in a circle of radius 5.  The sum of the squares of the distances from any point on the circle to the vertices of the polygon is 600.  Compute the value of the integer $n$.

5.
The circles, each of radius 5, lie in perpendicular planes.  They intersect in two points, $X$ and $Y$, thus determining a common chord $\overline{XY}$.  If $XY = 8$, compute the distance between the centers of the circles.

6.
How many positive integers less than 54 are equal to the product of their proper divisors?

7.
Find all positive integers $n$, less than 17, for which $n! + (n+1)! + (n+2)!$ is an integral multiple of 49.

8.
In triangle $ABC$, $a \ge b \ge c$.  If $\displaystyle \frac{a^3+b^3+c^3}{\sin^3{A} + \sin^3{B} + \sin^3{C}} = 7$, compute the maximum possible value of $a$.

[hide="Solution to #3"]We can write the solutions to $z^5=1$ as $e^{.4k\pi i}$ for $k\in \{1,2,3,4\}$. WLOG assume k=1.
\[z^{15}+z^{16}+z^{17}+\cdots+z^{50}\]
\[=\frac{z^{15}(z^{36}-1)}{z-1}\]
\[=\frac{e^{14.4\pi i}-1}{e^{.4\pi i}-1}\]
Note that $e^{14.4\pi i}$ is $e^{.4\pi i}$ rotated around the origin in the Argand diagram in the Complex Plane 7 times. Therefore we have $\frac{e^{.4\pi i}-1}{e^{.4\pi i}-1}=\boxed{1}$[/hide]

That was one of my current ARML homework problems. Luckily, I had solved it yesterday.


I am one of those uber-uncreative people who can't solve.  :(

Isn't there another poll like this already?

Well, not on the official polls, so I thought it would be ok.

There might be one somewhere else though...

This is funny $ \rightarrow$ :stretcher:

EDIT: Some people have discussed this before, but it wasn't put on a poll, according to my search.

There might already be one in Math Strategy Games.  Anyway I  can solve it in about 6-7 minutes [u][i]with[/u][/i] a Seven-Step guide that came with it.  I'd be lost without it. :D

2-5 is unrealistic
it should be like sub 20
20-40
40-1 min
1min-2 min
2min+
can't solve
anyone can do it in 2 min with practice

I could never solve the Rubik's Cube, no matter how often I tried, since I was five years old. :(  :rotfl:

My single record is maybe about 8 seconds...(ie a very lucky one)

My best average of 12 is about 19 sec.

Need to have more poll options for fast people (sub 20 people). My record is 26.5 (actually 26.49) seconds.

3:44 is my fastest so far. I used the seven-step solution guide, but I was a little more creative and made it a 6-step solution guide with shortcuts.

Edit: 3:31.689

Edit 2: 2:35.99

My record is 1:21. I got lucky and it skipped a few steps for me.

However my average falls between 1:40 and 2 min.

[quote="7h3.D3m0n.117"][quote="1=2"]Who here CAN use the solutions guide and do it under three minutes?[/quote]

When I first solved the Cube, I did it with the solutions (F2L Beginners method) and averaged a little under 3 minutes. Then I started memorizing easy algorithms and my time halved.[/quote]

How do you average under three minutes first time around? I have a rubix cube that doesn't turn easily, is that the issue?

P.S. 1:41.85

[quote="1=2"]
How do you average under three minutes first time around? I have a rubix cube that doesn't turn easily, is that the issue?[/quote]

You should see my cube...it can't turn, the stickers are dying, and it's amazing how I could solve the cube in my lifetime. I'm trying to get another cube but here's my method:

Before learning to solve the actual cube I got bored and started this fun competition between me and my sister. The goal was just to solve one side (the edges didn't matter). Eventually we added multiple sides (independant of each other) and we averaged about less than 20 seconds per side (it's not really solved because of the edges). We then made it challenging and solved the side for real (with edges). Our average didn't increase that much because we knew how the algorithms. When it was time to solve the whole cube, the first step (cross + corners) takes no time at all.

Also, my moves with the cube have no hesitation to them (other than converting steps and analyzing my states). So to solve for the second layer I hardly do any pausing at all.

49 secs. Haven't done it in a while though.

I memorized the solution guide and solved in 1:52.
Way better than before.
:D

I tried solving my friends' at CTY, but i just couldn't....
and I thought people who SOLVED rubix cubes were amazing..
i didn't know there were people who could cube in like 45 seconds..

i know how to make a checkerboard pattern though...
about ALL i know..

my friend used a 5x5 cube and made a heart out of it...
she is skilled..

i solved mine in less than one second or as i like to call "instantly"

all you have to do is take it out of the box.  :P  Just kidding!

yay i got lucky and skipped a few steps!
still got 1min30 tho...

[quote="moogra"][quote="1=2"]Who here CAN use the solutions guide and do it under three minutes?[/quote]

I got 36 seconds just now using the beginners method. :)[/quote]

Wow that's impressive with the beginners  :) . If I'm unfortunate, I'll average about 36 seconds just to get the first side and end up with a little under two minutes as my time.

My best was 56.something secs.  Haven't done it in a long time though...


The relation $ x^2(x^2 \minus{} 1)\ge 0$ is true only for:

$ \textbf{(A)}\ x \ge 1\qquad \textbf{(B)}\ \minus{} 1 \le x \le 1\qquad \textbf{(C)}\ x \equal{} 0,\, x \equal{} 1,\, x \equal{} \minus{} 1\qquad \\\textbf{(D)}\ x \equal{} 0,\, x \le \minus{} 1,\, x \ge 1\qquad \textbf{(E)}\ x \ge 0$

[quote="sunehra"]The relation $ x^2(x^2 \minus{} 1)\ge 0$ is true only for:

$ \textbf{(A)}\ x \ge 1\qquad \textbf{(B)}\ \minus{} 1 \le x \le 1\qquad \textbf{(C)}\ x \equal{} 0,\, x \equal{} 1,\, x \equal{} \minus{} 1\qquad \\
\textbf{(D)}\ x \equal{} 0,\, x \le \minus{} 1,\, x \ge 1\qquad \textbf{(E)}\ x \ge 0$[/quote]

$ x^2(x^2 \minus{} 1)\ge 0 \Leftrightarrow x\equal{}0$ or $ x^2\minus{}1 \ge 0$ , so the answer is $ D$

[hide="Algebraic Solution"]
We have $ x^2 \ge 0$ for all $ x$, so we must have $ x^2 \minus{} 1 \ge 0\implies x \le \minus{} 1$ and $ x \ge 1$ and $ x \equal{} 0$ (for equality). Therefore, $ \fbox{(D)}$ is the answer.
[/hide]
[hide="Graphing Solution"]
If you can get a general idea of how it looks, then this problem is very straight forward. We use the fact that an even power graph does up at the left and up at the right. The graph will intersect for the first time at $ x \equal{} \minus{} 1$, then it will be tangent to the origin as it works its way up, and finally it will work its way down and then up again intersecting the $ x$-axis for a final time at $ x \equal{} 1$. The graph is above the $ x \minus{} axis$ or touching it at $ x \equal{} 0$, $ x \ge 1$, and $ x \le \minus{} 1$ making the answer $ \fbox{(D)}$.
[/hide]


inequalities
induction
function
inequalities theorems

If $n \geq 2$ and $a_1, \ldots, a_n > -1$, and that all the a_i's have the same sign, then \[\prod_{i=1}^n (1+a_i) \geq 1+ \sum_{i=1}^n a_i\]

[quote]what happens if some of ai are 0?[/quote]
Alright, just to be safe, I've changed it to include equality.

extra 0's doesn't matter.

the product gets x1 so doesn't change, the sum gets +0 so doesn't change :)

Yes, that works. There are a lot of working methods though :)

[quote="perfect_radio"]Let $x > -1 \neq 0$, $a \in \mathbb{R}$.
Then:

$(1+x)^a>1+ax$, if $a>1$ or $a<0$

$(1+x)^a<1+ax$, if $0<a<1$

How do you prove this?
for natural numbers induction works, but about reals?[/quote]

calculus. Analyse $f(x)=(1+x)^a -1-ax$. $f'(x)=a(1+x)^{a-1} - a$. Extremum occurs when x=0. $f''(x)=a(a-1)(1+x)^{a-2}$. When $a>1$ or $a<0$, $f''(x)>0$, so $x=0$ is the minimum of the function which is 0. Otherwise, $f''(x)<0$, so $x=0$ is the maximum of the function.

Is there any non-analysis appraoch. Surely!
Though I like Langrange multipliers...

Bomb

Isnt this solvable by AMGM  :D  :D

Bomb

[quote="bomb"]To Billzhao, Isn't your proposed inequality Wierstrass' inequality...

Bomb[/quote]No... It's bernouilli's inequality. As the title says. ;)

It's not Bernoulli, I'm pretty sure it is Wierstrass when all a's >= 0 but when 0>=a's>=-1, it becomes Boniferri's inequalty....

Bomb


Okay, here's how it works.  I'm going to say a phrase, and you guys have to ask yes or no questions to try and solve it.  If you think you've solved it, I'll ask you 3 yes or no questions to see if you really have solved.  Once someone has solved the puzzle, I or the problem solver will explain the puzzle.  I know it sounds vague, but I can't think of a better way to word it.

The phrase is:
[size=150][b]"Jeanette & Giuseppe"[/b][/size]

Oh, I'm pretty sure this belongs here since it was in a book called [i]The Art of Problem Solving[/i], but not exactly sure.

[quote="JS1527"]Does it have something to do with the similarities in their names?  Like, "jea" and "gui" sound the same, then they both have "ette" and "eppe"?[/quote]

no

[quote="jelyman"]Could this be any more vague?[/quote]

yes

[quote="jelyman"]Could this be any more vague?[/quote]

Yes

[quote="TripleM"]Is "Guiseppe" spelt right? Its not Giuseppe?[/quote]

To answer your questions realisticly, I must have typed it wrong, I really don't know.

And to answer the questions as it if was in the puzzle, No.

Do the answers to questions have something to do with the initial phrase (I mean, if the phrase had changed would you answer some questions in a differentmanner)?


inequalities
algebra
polynomial
inequalities proposed

If $ a,b,c$ are nonnegative real numbers, then

$ \sum (a\minus{}b)(a\minus{}c)(a\minus{}2b)(a\minus{}2c)(a\minus{}3b)(a\minus{}3c) \ge 0$,

with equality for $ a\equal{}b\equal{}c$, for $ \frac {a}{2}\equal{}b\equal{}c$ or any cyclic permutation, and again for  $ \frac a{3}\equal{}b\equal{}c$ or any cyclic permutation.

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ \sum (a \minus{} b)(a \minus{} c)(a \minus{} 2b)(a \minus{} 2c)(a \minus{} 3b)(a \minus{} 3c) \ge 0$,

with equality for $ a \equal{} b \equal{} c$, for $ \frac {a}{2} \equal{} b \equal{} c$ or any cyclic permutation, and again for  $ \frac a{3} \equal{} b \equal{} c$ or any cyclic permutation.[/quote]

Actually, the following stronger and extended inequality holds.

If $ a,b,c$ are real numbers, then

$ \sum (a \minus{} b)(a \minus{} c)(a \minus{} 2b)(a \minus{} 2c)(a \minus{} 3b)(a \minus{} 3c) \ge 3(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2$,

with equality for $ a \equal{} b \equal{} c$, for $ \frac {a}{2} \equal{} b \equal{} c$ or any cyclic permutation, and again for  $ \frac a{3} \equal{} b \equal{} c$ or any cyclic permutation

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ \sum (a \minus{} b)(a \minus{} c)(a \minus{} 2b)(a \minus{} 2c)(a \minus{} 3b)(a \minus{} 3c) \ge 0$,

with equality for $ a \equal{} b \equal{} c$, for $ \frac {a}{2} \equal{} b \equal{} c$ or any cyclic permutation, and again for  $ \frac a{3} \equal{} b \equal{} c$ or any cyclic permutation.[/quote]

it is equivalent to :

$ 1 \minus{} 12q \plus{} 26r \plus{} 44q^2 \minus{} 286qr \plus{} 819r^2 \minus{} 12q^3 \geq 0$

hence if : $ \delta \equal{} 13(28q \minus{} 5)(27q^2 \minus{} 38q \plus{} 10) \leq 0$  ==> $ q\leq \frac {5}{28}$ we are done otherwise if $ q \geq \frac {5}{28}$ then:

$ \displaystyle ( \sum{a^2} \minus{} \sum{ab} ) (LHS) \equal{} \frac {13(a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2 ( 18\sum(ab) \minus{} 5\sum(a^2))}{4} \plus{} ( \sum{a^4} \minus{} \frac {7\sum{ab(a \plus{} b)}}{2} \plus{} 11\sum{ (ab)^2} \minus{} 5abc\sum{a})^2$

this solution is perhaps wrong, so  someone check it please  :)

I think it is right (for $ p\equal{}a\plus{}b\plus{}c\equal{}1$). Nice, Anas. :lol:

[quote="anas"]Thank you VasC, I think  it works for the stronger one  :wink:

P.S: what is your solution please ? :)[/quote]
I have 4 solutions to this ineguality, based on 4 new theorems concerning the sixth degree symmetric homogeneous polynomial inequality $ f_6(x,y,z)\ge 0$: for either real or nonnegative $ x,y,z$; with either necessary and sufficient conditions, or only sufficient conditions. Notice that the theorems with sufficient conditions are much more practical. The theorem below is for nonnegative real numbers, with necessary and sufficient conditions.

[b]Theorem. [/b]Let $ f_6(x,y,z)\ge 0$ be a sixth degree symmetric homogeneous polynomial inequality, written in the form
\[ Ar^2 \plus{} g_1(p,q)r \plus{} g_2(p,q)\ge 0,\]
where $ A > 0$, $ p \equal{} x \plus{} y \plus{} z$, $ q \equal{} xy \plus{} yz \plus{} zx$, $ r \equal{} xyz$. Let
\[ d(p,q) \equal{} g_1^2(p,q) \minus{} 4Ag_2(p,q),\]

\[ \delta_0 \equal{} d(t,1),\ \ \ \delta(t) \equal{} d(t \plus{} 2,2t \plus{} 1),\ \ \ h(t) \equal{} 2At \plus{} g_1(t \plus{} 2,2t \plus{} 1).\]
Suppose that $ f_6(x,1,1)\ge 0$ and $ f_6(0,y,z) \ge 0$ for any nonnegative real numbers $ x,y,z$. The inequality $ f_6(x,y,z) \ge 0$ holds for all nonnegative real numbers $ x,y,z$ if and only if one of the following two conditions is fulfilled:

(a)  if $ t\ge 1$ and $ \delta(t) > 0$, then $ h(t)\le 0$;

(b) if $ 0 < t\le 1$ and $ \delta(t) > 0$, then $ h(t)\ge 0$, and if $ t\ge 2$ and $ \delta_0(t) > 0$, then $ g_1(t,1)\ge 0$.


analytic geometry
vector
number theory unsolved
number theory

Solve the equation $ x! \plus{} y! \plus{} z! \equal{} 2^{u!}$

[b]Claim:[/b] up to permutation of the first $ 3$ coordinates the solution vectors $ (x,y,z,u)$ are:
$ (0,0,2,2),(0,1,2,2),(1,1,2,2)$

[i]Proof:[/i] Since $ x!\plus{}y!\plus{}z!\geq 3$ holds, we have $ u>1$. Up to permuation we can assume $ x\leq y\leq z$, therefore $ x!|2^{u!}$, i.e. $ x<3$. The case $ x\equal{}0,1$ requires $ y\equal{}0,1$ and yields $ z!\equal{}2^{u!}\minus{}2$. Since $ 2^{u!}\minus{}2\equiv 2\bmod{4}$ we must have $ z<4$, which gives $ z\equal{}u\equal{}2$.

The case $ x\equal{}2$ requires $ y!|2^{u!}\minus{}2$, i.e. $ y<4$. The sub-case $ y\equal{}2$ yields $ z!\equal{}2^{u!}\minus{}4$, i.e. $ z\geq 4,u\geq 3$, hence $ z<6$, since $ 2^{u!}\minus{}4\equiv 4\bmod{8}$. But $ z\equal{}4,5$ do not permit $ u$.

The subcase $ y\equal{}3$ yields $ z!\equal{}2^{u!}\minus{}10$, i.e. $ z<4$. But the only candidate $ z\equal{}3$ does not permit a valid $ u$.


algebra
polynomial
function
real analysis
real analysis unsolved

1.Does anybody know nice asymptotics for $ F(x) \equal{} \sum \limits_{l \equal{} 0}^{k} (J_l^{(0,1)}(x))^2$ where $ J_l$ are normalized  Jacobi polynomials orthogonal with respect to $ (1 \plus{} x)$?
2. Prove that $ max_{[ \minus{} 1,1]} (1 \plus{} x)\sum \limits_{l \equal{} 0}^{k} (J_l^{(0,1)}(x))^2 \equal{} 2F(1)$
Thanks
Edit. For the second one I've just found nice general Sego theorem. It says that if we have the sequence of orthogonal polynomials on the segment $ [a,b]$ with respect to non-decreasing weight $ \omega (x)$ then the function $ |p_n (x)\parallel{}\omega (x)|^{\frac{1}{2}}$ attains maximum at the point $ b$

I wonder if you have taken a look at the book "Orthogonal polynomials" by Gabor Szego. There is a lot of such stuff there :) (not that I remember much of it nowdays :().

Yes, I've actually found Szego theorem there. But for the first question I still did not get what I need.

OK, I'll ask a few friends of mine who should know such things. Is the normalization condition that the $ L^2((1\plus{}x)\,dx)$-norm of each polynomial is $ 1$, or something else?

Yes, you are absolutely right (their norm in $ L_2((1\plus{}x))dx$ is $ 1$).
Thanks a lot.


When you cut a certain rectangle in half, you obtain two rectangles that are both similar to the original rectangle. Find the ratio of the longer side length to the shorter side length.

[hide="Solution"]
Assume that the rectangle's dimensions are x and y, where $ x > y$.
Then $ \frac{\frac{x}{2}}{y} \equal{} \frac{y}{x}$.
Then $ x^2 \equal{} 2 y^2$
so $ x \equal{} \sqrt{2} y$

$ \sqrt{2} : 1$
[/hide]
which happens to be more or less the ratio of the A4 and related sheets. They designed it that way, so you could photocopy two A4's onto one A4 with the right ratio. Um, I got distracted.


Asertain the digits $ \{x,y,z\}\subset\overline {0,9}$ so that $ \overline {yz}^2 \equal{} \overline {xyz}$ , i.e. $ (10y \plus{} z)^2 \equal{} 100x \plus{} 10y \plus{} z$ .

We have $ \overline{yz}(\overline{yz}\minus{}1)\equal{}100x$ then we have two case:
-) $ \overline{yz}\vdots 25,\overline{yz}\minus{}1\vdots 4$: easy to see $ \overline{yz}\equal{}25$
-) $ \overline{yz}\vdots 4,\overline{yz}\minus{}1\vdots 25$: easy to see no result!


function
real analysis
real analysis unsolved

Why must a periodic function $f: \mathbb{R}\rightarrow\mathbb{R}$ be uniformly continuous in order to be expressed as $f\left(x\right)=\sum_{n\in\mathbb{Z}}a_{n}e^{2\pi inx}$?

Depends on what you mean by "expressed". An infinite series of functions can converge in a number of ways. If it converges to $f$ uniformly, then the limit must be a uniformly continuous function by a standard theorem of real analysis. Other types of convergence (pointwise, a.e., $L^{2}$, ...) do not require continuity of $f$.


vector
induction
linear algebra
matrix
inequalities
linear algebra unsolved

Hello. Please try to prove the following:

Let $ T : V \rightarrow V$ be a linear operator in a unitary space $ V$.
Prove that if
(1) For all $ \lambda$ eigenvalue of $ T$ $ |\lambda|=1$.
(2) For each vector $ a$ in $ V$ $ \parallel T(a)\parallel \leq \parallel a\parallel $.
then $ T$ is an isometry.

(Here $ \parallel x\parallel $ is the length of x.)
Please give a hint for a proof, or a proof.
I know one proof using induction on V's dimention, but I would like to see another proof.
Thanks in advance for any ideas/hints.

I assume that you mean "complex inner product space" when you say "unitary space". This also means that all eigenvalues exist, and we don't get anything funny from that.

For all $ x$, $ \langle x, T^{*}Tx\rangle =\langle Tx,Tx\rangle=\|Tx\|^{2}\le \|x\|^{2}$. Since $ T^{*}T$ is self-adjoint and positive semidefinite, it is similar to a real diagonal operator; there is an orthonormal basis $ B$ in which the matrix of $ T$ is diagonal with entries in $ [0,\infty)$. By the norm condition, all diagonal entries in this basis must be $ \le 1$.

On the other hand, the determinant of $ T$ is a unit since all of its eigenvalues have magnitude 1, and this means that the determinant of $ T^{*}T$ is 1. The only way for this to happen with our other inequalities is for $ T^{*}T$ to be the identity, and that makes $ T$ an isometry.

:thumbup:  Thank you, it's a very nice proof!

(and yes, you figured out correctly what I meant.)


[url=http://bildr.no/view/509704][img]http://bildr.no/thumb/509704.jpeg[/img][/url]

I can not understand the sentence denoted by red line.

I feel a little shy to ask so many questions.
I am studying by myself. I need friends and teachers.
So I think I will ofter ask questions at the beginning of self-studying.

Don't laugh at me.

Thanks very much.

The misunderstanding stems from the vicious way (that I noticed on other occasions in your posts)  the authors of those articles use the fraction symbol. So, when they wrote, just before the red-underlined passage,
[quote]... $ y \geq x \plus{} 1/f(f(x)) \minus{} 1$ ...[/quote]
what they actually meant was $ y \geq (x \plus{} 1)/(f(f(x)) \minus{} 1)$.
Now the passage you underlined becomes intelligible, since $ f(x\plus{}y) \geq f(x) \plus{} yf(f(x))$ is just the functional relation, to which we apply $ f(x) \geq 0$, and $ yf(f(x)) \geq x \plus{} 1 \plus{}y$, which is just a rewrite of the relation I edited.

By the way, I sent you an e-mail on your personal address, with a pdf document attached, in which there is a lot about this very problem (I am the author of a much more difficult version of this - which was asked at some Italian contest)  :)


Prove or disprove that:for any $ n \geq 2$ there exist 2n distinct positive integer $ a_1,...,a_{2n}$ such that:$ a_1^2\plus{}...\plus{}a_n^2\equal{}a_{n\plus{}1}^2\plus{}...\plus{}a_{2n}^2$.

How about this?
Prove or disprove that:for any $ n \geq 2$ there exist $ 2n\plus{}1$ distinct positive integer $ a_1,a_2,...,a_{2n\plus{}1}$ such that:
$ a_{1}^{2}\plus{}...\plus{}a_{n}^{2}\equal{}a_{n\plus{}1}^{2}\plus{}...\plus{}a_{2n}^{2}\equal{}a_{2n\plus{}1}^2$

[quote="k.l.l4ever"]How about this? [...][/quote]

You're welcome
I'm glad to have helped you.

And, btw, for you second problem :
Let $ u\equal{}\sum_{k\equal{}1}^{n\minus{}1}(15k\plus{}5)^2$
Let $ v\equal{}\sum_{k\equal{}1}^{n\minus{}1}(15k\plus{}3)^2$

$ \sum_{k\equal{}1}^{n\minus{}1}(2(v\plus{}1)(15k\plus{}5))^2\plus{}((v\plus{}1)(u\minus{}1))^2$ $ \equal{}\sum_{k\equal{}1}^{n\minus{}1}(2(u\plus{}1)(15k\plus{}3))^2\plus{}((u\plus{}1)(v\minus{}1))^2$ $ \equal{}((v\plus{}1)(u\plus{}1))^2$

[quote="pco"][quote="k.l.l4ever"]Prove or disprove that:for any $ n \geq 2$ there exist 2n distinct positive integer $ a_1,...,a_{2n}$ such that:$ a_1^2 \plus{} ... \plus{} a_n^2 \equal{} a_{n \plus{} 1}^2 \plus{} ... \plus{} a_{2n}^2$.[/quote]

Since $ (a\plus{}2)^2\plus{}(2a\minus{}1)^2\equal{}(a\minus{}2)^2\plus{}(2a\plus{}1)^2$ $ \forall a$, it's obvious that if the property is true for $ n$, it's true for $ n\plus{}2$. So we just have to show for $ n\equal{}2$ and $ n\equal{}3$

$ n\equal{}2$ : $ 6^2\plus{}7^2\equal{}2^2\plus{}9^2$
$ n\equal{}3$ : $ 4^2\plus{}6^2\plus{}19^2\equal{}2^2\plus{}3^2\plus{}20^2$

Hence the result[/quote]
Hello, what do you mean with showing the cases $n=2$ and $n=3?$ i understood it so that if $n$ is even then we can use the identity which you showed for the pairs, and if $n$ is odd then we can replace six of them so: $4; 6; 19$ -->RHS and $2; 3; 20$-->LHS and then we have remained even amount of numbers on both sides which we have to replace and we do it with the useful identity $ (a\plus{}2)^2\plus{}(2a\minus{}1)^2\equal{}(a\minus{}2)^2\plus{}(2a\plus{}1)^2$ is it so?


inequalities
number theory unsolved
number theory

Let n be a positive integer, and let d(n) denote the sum of the positive divisors of n, including 1 and n itself. Prove that:
      d(1)/1+d(2)/2+d(3)/3+......+d(n)/n<=2n

let k be a numebr smaller than $n$, then let f(k) be the numebr of multiples of $k$ in the set $(1,2,3....,n)$. since $d(m)/m=\sum_{d|n}1/d$ we have that k contributes $f(k)/k$ to this sum, but $f(k)$ is at most $n/k$ so we get our sum is at most $n(1+1/2^2+1/3^2....)<2n$

[quote="Pascual2005"]since $d(m)/m=\sum_{d|n}1/d$ [/quote]
what is m? :read:

...and you may use $\frac{\pi^2}{6}n$ instead of $2n$...


A merchant buys a number of oranges at $ 3$ for $ 10$ cents and an equal number at $ 5$ for $ 20$ cents.  To "break even" he must sell all at:

$ \textbf{(A)}\ \text{8 for 30 cents} \qquad
\textbf{(B)}\ \text{3 for 11 cents} \qquad
\textbf{(C)}\ \text{5 for 18 cents} \\
\textbf{(D)}\ \text{11 for 40 cents} \qquad
\textbf{(E)}\ \text{13 for 50 cents}$

[hide="Solution"]He buys $ n$ oranges.
He spends $ 10n/3\plus{}4n\equal{}22n/3$ cents.
If he sells them for $ c$ cents per orange, he makes $ 2cn$ cents.
$ 2cn\equal{}22n/3\implies c\equal{}11/3$.
So the answer is $ \boxed{\textbf{(B)}}$.
[/hide]


When I add a second entry to a category, it doesn't show up when I click on the category. How can I add more entries to categories?

EDIT: Ah, never mind. It seems to be working now.

Did you remember to click "Submit" after selecting the category you wanted?  :)


geometry
perimeter
circumcircle
geometry proposed

Given an acute angled triangle with perimeter $L$, if its pedal triangle (the inscribed triangle of smallest perimeter) has perimeter $P$, then $\frac{P}{L} \leq \frac{1}{2} $. The equality holds iff the triangle is equilateral.

I cannot find a purely geometric solution. :?

Let ABC be our triangle and A', B', C' are the projections.
O the center of circumcircle and H the ortocenter.
Using AO is perpendicular to B'C' then
Area(ABC) = 1/2 AO*B'C' + 1/2 BO*A'C' + 1/2 CO*A'B'.
hense L*r=P*R or P/L=r/R ≤ 1/2

Regard $A''$, $B''$, $C''$ as the midpoints of sides $BC$, $CA$, $AB$ respectively, then we get the triangle $A''B''C''$ has perimeter $M$.
Obviously, $M = \frac{1}{2} L$, and for $P \leq M$, we have $\frac{P}{L} \leq \frac12$, the equality holds iff $A'=A''$, $B'=B''$, $C'=C''$, mean the triangle is equilateral.


conics
parabola
algebra proposed
algebra

Your answer is incorrect. We don't need to find all of $a_{n},\ b_{n},\ c_{n}.$


limit
logarithms
calculus
calculus computations

$ a_1,a_2,...,a_n>0$,Find:
$ \lim_{x\longrightarrow 0}(\dfrac{\sum a_n^x}{n})^\dfrac{1}{x}$

For big $ x$ we have:

$ \sum a_n^x \sim \left(\max a_n\right)^x$

This means that:

$ \left(\frac{1}{n} \sum a_n^x\right)^{\frac{1}{x}} \sim \left(\frac{1}{n} \left(\max a_n\right)^x \right)^{\frac{1}{x}} \equal{} \max a_n \sqrt[x]{n}$

Finally:

$ \lim _{x\to\infty}\left(\frac{1}{n} \sum a_n^x\right)^{\frac{1}{x}} \equal{} \max a_n  \lim _{x\to\infty}  \sqrt[x]{n} \equal{} \max a_n$

[quote="milin"]For big $ x$ we have:

$ \sum a_n^x \sim \left(\max a_n\right)^x$
[/quote]

How about when $ a_1\equal{}a_2\equal{}...\equal{}a_n\equal{}1$,then the equality means $ n\sim 1$ which i can't understand。
and i think the answer to the problem is $ \sqrt[n]{a_1a_2..a_n}$by the power ineq:
$ f(2)\ge f(1)\ge \sqrt[n]{a_1a_2..a_n}\ge f(\minus{}1)$

i am sorry if i made any mistake

When $ \forall n \,a_n \equal{} 1$ then you have:

$ \left(\frac {1}{n}\sum_n a_n^x\right)^{\frac {1}{x}} \equal{} \left(\frac {1}{n}\sum_n 1\right)^{\frac {1}{x}} \equal{} \left(1\right)^{\frac {1}{x}} \equal{} 1 \equal{} \max a_n$

But you are right.. It should have been only

$ \left(\frac {1}{n}\sum_n a_n^x\right)^{\frac {1}{x}} \sim \left(\frac {1}{n} \left( \max a_n\right)^x \right)^{\frac {1}{x}}$

without the remark of the sum being similar to the maximal value.

Also, as for the $ \frac{1}{n}$ factor - you can remove it immediately because $ \lim _{ x\to \infty} \frac{1}{\sqrt[x]{n}} \equal{} 1$.

[quote="milin"]For big $ x$ [/quote]  We have $ x \to 0$, not $ x \to \infty$ ;)

[hide="answer (no solution)"]$ \sqrt[n]{a_1 \cdot a_2 \cdots a_n}$[/hide]

so can anyone give the solution?
thanks so much~~

One way to do this, is to take the logarithm of the expression, and then
you have to deal with a limit of the form "$ \frac 0 0$", which you probably know how do,
e.g. using differentiation.


I'm taking those three SAT II's this saturday, almost without notice. I'm completing AP Calc BC, AP Chem, and AP Physics C. If anyone here has taken them before, can you give me advice as to what I should review/skim; especially physics and chem. thanks

Math should be easy as pi for you.

If you're ready for AP Chem, you'll be ready for SAT chem.  I took SAT II chem half way through the year and got a 770.  The AP exam covers so much more material than the AP exam.

I haven't taken physics so I don't know.

Just look at sparknotes.com or check some books out at the library and look them over a little bit.  But if you're ready for the AP chem and AP physics you shouldn't have much of a problem.

Yeah, the SAT II's are way easier than the AP. I think the questions are more general too (at least for the American History one).

Just make sure you look at the structure of the nasty true/false section.  It gets so misleading, because overall it will be a wrong statement, but part i and part ii by themselves might be completely fine.

[quote="Bictor717"]Chem should be a breeze compared to the AP.
Note that AP Physics C does not cover everything on the SATII, like vibrations, I think. Grab a review book and check.[/quote]

Ditto the AP Physics C vs. SAT II stuff. The SAT II mostly covers AP Physics B material and Physics C is MUCH more limited in scope, so don't walk into the test expecting to see all Mechanics and E&M questions. There will be questions on literally EVERYTHING. So do some review.

yeah, thanks for the advice. i finished today.

the math was a lot easier than i expected. i thought that they would turn it up a notch for the sat ii, level 2. its kind of sad when you compare it to the mathematics entrance exams at japanese universities (like the ones frt has posted).

the chemistry wasn't as bad as i thought. u guys were right that the ap test covers so much more. the test was basic chemistry that i could confidently work through.

unfortunately, you guys were also right about the way the physics compares to physics c. i was aware that there would be some optics and waves problems, but there were a lot more than i expected. i only had like 2 days for any kind of review, since i only knew i was going to take the test last thursday. i went over optics in physics b last year, but it is the area of physics i have most neglected. since i want atleast a 750 on the physics, i'll probably need to take this again.

once again, thanks for the advice anyways.

Do you guys think that these SAT II's would be too hard for someone who hasn't taken physics yet (not offered except to seniors  :mad: ) and whose chemistry class was utterly worthless?  (we've learned nothing, at least in terms of chemistry.  I have gotten better at sudoku and cribbage...)

for physics it would be too hard if you havn't had the class. if you really want to, you could learn it on your own, but you would have to put ample time into reading and understanding concepts from a preparation book or text book.

for chemistry its the same kind of thing

overall the tests aren't that hard if you take time to study for them directly. you can do it if you work at it.


analytic geometry
graphing lines
slope
calculus
calculus computations

Find the necessary and sufficient condition such that 2 curves $ ax^2 \minus{} by^2 \equal{} 1, cx^2 \minus{} dy^2 \equal{} 1\ (abcd\neq 0)$ have a intersecion of point and their tangent line at $ P$ are perpendicular to each other.

Slope of the first curve$ \equal{}m_{1}\equal{}\frac{ax}{by}$

Slope of the second curve$ \equal{}m_{2}\equal{}\frac{cx}{dy}$

Since the two tangents are perpendicular,we have
$ m_{1} . m_{2}\equal{}\minus{}1$

which gives
$ \frac{acx^2}{bdy^2}\equal{}\minus{}1$..................(1)

Also,$ ax^2 \minus{} by^2 \equal{} cx^2 \minus{} dy^2$\\since (x,y) is the point of intersection

$ \frac{x^2}{y^2}\equal{}\frac{b\minus{}d}{a\minus{}c}$...................(2)

Eliminating x and y from (1) and (2),we get

$ \frac{1}{a}\plus{}\frac{1}{b}\equal{}\frac{1}{c}\plus{}\frac{1}{d}$


Let $ a,$ $ b,$ $ c$ and $ d$ are positive numbers. Prove that
$ \left(\frac{3a}{a\plus{}2b\plus{}2c\plus{}2d}\right)^{\frac{3}{4}}\plus{}\left(\frac{4b}{b\plus{}3c\plus{}3d\plus{}3a}\right)^{\frac{2}{3}}\plus{}$
$ \plus{}\left(\frac{6c}{c\plus{}5d\plus{}5a\plus{}5b}\right)^{\frac{3}{5}}\plus{}\left(\frac{d}{a\plus{}b\plus{}c}\right)^{\frac{1}{2}}\geq2$

[quote="arqady"]Let $ a,$ $ b,$ $ c$ and $ d$ are positive numbers. Prove that
$ \left(\frac {3a}{a \plus{} 2b \plus{} 2c \plus{} 2d}\right)^{\frac {3}{4}} \plus{} \left(\frac {4b}{b \plus{} 3c \plus{} 3d \plus{} 3a}\right)^{\frac {2}{3}} \plus{}$
$ \plus{} \left(\frac {6c}{c \plus{} 5d \plus{} 5a \plus{} 5b}\right)^{\frac {3}{5}} \plus{} \left(\frac {d}{a \plus{} b \plus{} c}\right)^{\frac {1}{2}}\geq2$[/quote]
Just by AM-GM:
$ \left(\frac {3a}{a \plus{} 2b \plus{} 2c \plus{} 2d}\right)^{\frac {3}{4}} \ge \frac{2a}{a\plus{}b\plus{}c\plus{}d}$
$ \left(\frac {4b}{b \plus{} 3c \plus{} 3d \plus{} 3a}\right)^{\frac {2}{3}} \ge \frac{2b}{a\plus{}b\plus{}c\plus{}d}$
$ \left(\frac {6c}{c \plus{} 5d \plus{} 5a \plus{} 5b}\right)^{\frac {3}{5}} \ge \frac{2c}{a\plus{}b\plus{}c\plus{}d}$
$ \left(\frac {d}{a \plus{} b \plus{} c}\right)^{\frac {1}{2}} \ge \frac{2d}{a\plus{}b\plus{}c\plus{}d}$

[quote="Yuriy Solovyov"][quote="ehku"][quote="arqady"]Let $ a,$ $ b,$ $ c$ and $ d$ are positive numbers. Prove that
$ \left(\frac {3a}{a \plus{} 2b \plus{} 2c \plus{} 2d}\right)^{\frac {3}{4}} \plus{} \left(\frac {4b}{b \plus{} 3c \plus{} 3d \plus{} 3a}\right)^{\frac {2}{3}} \plus{}$
$ \plus{} \left(\frac {6c}{c \plus{} 5d \plus{} 5a \plus{} 5b}\right)^{\frac {3}{5}} \plus{} \left(\frac {d}{a \plus{} b \plus{} c}\right)^{\frac {1}{2}}\geq2$[/quote]
Just by AM-GM:
$ \left(\frac {3a}{a \plus{} 2b \plus{} 2c \plus{} 2d}\right)^{\frac {3}{4}} \ge \frac {2a}{a \plus{} b \plus{} c \plus{} d}$
$ \left(\frac {4b}{b \plus{} 3c \plus{} 3d \plus{} 3a}\right)^{\frac {2}{3}} \ge \frac {2b}{a \plus{} b \plus{} c \plus{} d}$
$ \left(\frac {6c}{c \plus{} 5d \plus{} 5a \plus{} 5b}\right)^{\frac {3}{5}} \ge \frac {2c}{a \plus{} b \plus{} c \plus{} d}$
$ \left(\frac {d}{a \plus{} b \plus{} c}\right)^{\frac {1}{2}} \ge \frac {2d}{a \plus{} b \plus{} c \plus{} d}$[/quote]

Can you explain, please? First, for example :)[/quote]
By AM-GM inequality we have
$ \sqrt[4]{\left(\frac{a\plus{}2b\plus{}2c\plus{}2d}{3a}\right)^3} \\
\equal{}\sqrt[4]{1.\left(\frac{a\plus{}2b\plus{}2c\plus{}2d}{3a}\right)\left(\frac{a\plus{}2b\plus{}2c\plus{}2d}{3a}\right)\left(\frac{a\plus{}2b\plus{}2c\plus{}2d}{3a}\right)} \\ \le \frac{1}{4}\left(1\plus{}3.\left(\frac{a\plus{}2b\plus{}2c\plus{}2d}{3a}\right)\right) \\ \equal{} \frac{a\plus{}b\plus{}c\plus{}d}{2a}$
Hence,
$ \left(\frac {3a}{a \plus{} 2b \plus{} 2c \plus{} 2d}\right)^{\frac {3}{4}} \ge \frac {2a}{a \plus{} b \plus{} c \plus{} d}$


Hello all first time poster here.

I'm having a bit of a problem with the question in the pic.
[url=http://img373.imageshack.us/my.php?image=img6716xn0.jpg][img]http://img373.imageshack.us/img373/2046/img6716xn0.th.jpg[/img][/url]
Now I'm 96% sure that Car1 is ahead of car2 I'm just unsure how to work this kind of problem out, any tips? -thanks a bunch.

Welcome!

The trick you can use is the fact that area under the graph represents the displacement of the car.


1. Sa se determine aria marginita de graficele functiilor $\ f_n $ si $\ f_{n+1} $, unde $\ f_n $:[0, $\infty $) $ \to $ $\mathbb{R}$ (n $\in \mathbb{N}^*$), $\ f_n(x) $ = $\ x^n lnx$ (x>0) si $\ f_n $ (0)=0 .

2. Fie F(x)=$\int _{x}^{2x}{\frac {t^2}{t^2+\sin^2{t}}dt}$, $x \in \mathbb{R}^* $. Atunci F este impara.

1. Se arata usor ca $f_{n}$ sunt continue in $0$, $f_{n}$ si $f_{n+1}$ au puncte comune $\left(  0,0\right)  $ si $\left(  1,0\right)  $. Deducem ca aria cautata este
\[
\underset{a\searrow0}{\lim}\int_{a}^{1}\left|  f_{n+1}\left(  t\right)
-f_{n}\left(  t\right)  \right|  dt=\frac{1}{\left(  n+1\right)  ^{2}}%
-\frac{1}{\left(  n+2\right)  ^{2}}.
\]
(vezi poza atasata)
2. Daca $f$ e o functie para, avem
\[
\int_{-x}^{0}f\left(  t\right)  dt=\int_{0}^{x}f\left(  t\right)  dt,
\]
pentru orice $x.$ Atunci
\[
F\left(  -x\right)  =\int_{-x}^{-2x}f=\int_{-x}^{0}f+\int_{0}^{-2x}f=\int
_{0}^{x}f+\int_{2x}^{0}f=-\int_{x}^{2x}f=-F\left(  x\right)  .
\]


LaTeX
superior algebra
superior algebra unsolved

Please write your question here in LaTeX. Not every one has (or wants to use) Word...

I think [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=159144]this[/url] is an answer. This is made less clear by the lack of a subscript on your $ \aleph$.

Incidentally, we can read plaintext pseudo-markup stuff just fine. It's having to download and open something outside the browser that really drives people away.

Most $ \text{\LaTeX}$ codes are pretty intuitive; \aleph is $ \aleph$, \subset is $ \subset$ (or \subseteq is $ \subseteq$), and so on. The forum interprets the codes, so it's the standard around here. If you feel like learning it, there's a lot of good stuff in [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]this subforum[/url].


Let a1 = 1, and an = (n - 1)an-1 + 1, for n > 1.

For what values of n is an divisible by n?

So far I've found primes 2, 5, 13, 37, 463, 83507 306759, 412021, and 466123 computationally, plus 4.
Any product of the above also works.
However, I can't ascertain which primes will work, even though I can see that the general solution is an = SUMk=0n-1 (n-1)!/k!

Any ideas?

Something is wrong. 83507, 306759, 412021, and 466123 are not primes.
2, 5, 13, 37, 463 are the only currently known primes that satisfy the required property.
See http://oeis.org/A064384 for further details and references.
Not necessary prime $ n$'s are given in http://oeis.org/A064383


superior algebra
superior algebra unsolved

For what positive integers $ n$ is it true that "every $ g$ in a finitely generated group $ G$ satisfies $ g^n \equal{} 1$" implies $ G$ is finite?

This is obvious when $ n \equal{} 2$ and has been assigned to me when $ n \equal{} 3$ (although I couldn't prove it), but I don't know the general answer.  A proof for the $ n \equal{} 3$ case would be nice.

There are counterexamples for bigger $ n$, see http://en.wikipedia.org/wiki/Burnside%27s_problem .


function
number theory proposed
number theory

In the following $ {\mathbb N} $ denotes the set of positive integers and   $ a,b $ are  given integers in  ${\mathbb N}. $
By $ \mbox{\rm Card}(M) $   is denoted  the numbers of elements of a certain set $ M $.
Let us consider two functions  $ f,g : {\mathbb  N}\to {\mathbb N}  $ such that  $ f(xy)=g(ax+by) $ for all $ x,y \in {\mathbb N} .$

[b]1).   Find an upper-bound for   $ \mbox{\rm Card}\left(f({\mathbb N})\right)\; . $
2) .  In case $ a=b=1 $, it's true that   $ \mbox{\rm Card}\left(f({\mathbb N} )\right)\le 3 \;\;   $ ? [/b]


Buna ziua,

Nu am avut prea mult timp sa invat cum pot edita text matematic in acest forum. Ma puteti indruma va rog? (Eu folosesc Word si Equation Editor, se poate utiliza aici?)

Multumesc

Nu, aici folosim Latex. Uita-te pe forum, vezi ca este "Latex help". ;)