factorial function calculus integration search

How does one compute gamma(x) (the extension of the factorial function) for real x without using integration? (In short, is there a simple series that can give an approximation to an arbitrary degree of accuracy?)

[quote="Fermatprime"]is there a simple series that can give an approximation to an arbitrary degree of accuracy?[/quote]

Stirling's Approximation is a well-known approximation for the factorial function.  I suggest you search the web for "Stirling's Approximation" or for "Stirling's Series".

!!!

That looks pretty cool, but there's just two things.

1) The formula for the coefficients of z^(-n) is pretty complicated and

2) It's asymptotic, which isn't good enough if you're writing a program to test for zeros of the zeta function and you need to calculate gamma(3/4 + 15000000000i), really fast.

I should point out that it's not too hard to prove something that comes close to Stirling's formula:

Consider log(n!)= :Sigma: (log(k)) (k=1 to n). We can bound this by two integralsto get something like e(n/e)^n<n!<en(n/e)^n. Taking the geometric mean of the two bounds gives something very close to Stirling's formula since :sqrt:(2:pi:) is close to e.


simplify in mod 6 (x^5 + 3x^4 - 2x^2 + 5x - 8)

[hide]The product of any three consecutive integers is always divisible by 6, since at least one of them is even and exactly one is divisible by three. Hence

\begin{eqnarray*}(x-1)x(x+1)\equiv 0\pmod{6} &\Longrightarrow& x^3-x\equiv 0\pmod{6}\\ &\Longrightarrow& x^3\equiv x\pmod{6}\\ &\Longrightarrow& x^4\equiv x^2\pmod{6}\\ &\Longrightarrow& x^5\equiv x^3\equiv x\pmod{6}\end{eqnarray*}.

Now we have

\begin{eqnarray*}x^5+3x^4-2x^2+5x-8 &\equiv& x+3x^2-2x^2+5x-2\pmod{6}\\ &\equiv& x^2+6x-2\pmod{6}\\ &\equiv& x^2-2\pmod{6}\end{eqnarray*}[/hide]


Let $S$ be infinite set.
Is it true that $S\times S$ has the same cardinality as $S$?

yes it is true,u can construct bijection between them :assume that $C$ is collection of pairs $(B,\phi)$ with $B\subset   S$and $\phi_B:B\rightarrow B*B$ a bijection.it turns out that $C$ has partial ordering.{$(B_1,\phi_1)<(B_2,\phi_2)<--->B_1\subset B_2$ and $\phi_1$ is restriction of$\phi_2$so by Zorn lemma this $C$ has maximal ,name it $(C,\phi)$ now prove that $C=S$


inequalities
MIT
college
logarithms
function
calculus
real analysis

Prove that \[ \frac{a+b}{2} > \frac{b-a}{\ln b - \ln a} > \sqrt{ab}  \] holds for all $a,b \in \mathbb R_+^\ast$, $a \neq b$.

I searched for it, but I failed to find it.

We can assume that $b>a$, and then let $\ln a =x,\; \ln b =y$. Then $y-x=t>0.$ Now our inequality has the view:
\[ \frac{e^x+e^y}{2}\ge\frac{e^y-e^x}{y-x}\ge\sqrt{e^{x+y}}. \] Which is equivalent to \[ \frac{1+e^t}{2}\ge \frac{e^t-1}{t}\ge e^{\frac{t}{2}} \]
The right sing is obvious: \[ f(t)=e^t-1-te^{\frac{t}{2}}\ge 0 \], because $f'(t)=e^{\frac{t}{2}}(e^{\frac{t}{2}}-1-\frac{t}{2})\ge 0$ and $f(0)=0$.

About the left sign:
$f(t)=(t-2)e^t+2+t$. Thus $f'(t)=e^t+(t-2)e^t+1$,$f'(0)=0$ and $f''(t)=te^t>0$. It follows that $f'(t)>0$. Then $f(t)>0$ because $f(0)=0$

How do you get the equivalence of
\[ \sqrt{e^{x + y}}  \]
and of
\[ e^{\frac{t}{2}}  \]
?

Suppose that $b>a$ and consider the function $f(x)= \ln\,x -2 \frac{x-1}{x+1} , \, x \in [1, \infty)$.

For $x>1$ it is $f'(x)=\frac{(x-1)^2}{x(x+1)^2} >0$ so for $x>1$ it is $f(x)>f(1)$.

But since $f(1)=0$ and $\frac{b}{a} >1$ we take $f(\frac{b}{a})>0$.

This gives $\frac{a+b}{2} > \frac{b-a}{\ln b - \ln a}$.

Now consider the function $f(x)= \sqrt{x} \ln\,x -x +1 , \, x \in [1, \infty)$.

It is $f'(x)=\frac{\ln\,x -2 \sqrt{x}+2}{2 \sqrt{x}}$ and $f''(x)=- \frac{\ln\,x}{4 x^{3/2}}<0$ for $x>1$.

Then $f'(x)<f'(1)=0$ therefore $f(x)<f(1)=0$ for $x>1$.

If we take $x = \frac{b}{a} >1$ we get $f(\frac{b}{a})<0$ and this gives $\frac{b-a}{\ln b - \ln a} > \sqrt{ab}$.

 :)

By the way can you prove this ? :  :lol:

${\displaystyle \frac{a+b}{2}>\frac{1}{e}\left(\frac{a^a}{b^{b}}\right)^{\frac{1}{a-b}}>\frac{b-a}{\ln b - \ln a}}$

We have $(\sinh{x} - x \cosh{x})' = -x \sinh{x} < 0$ and $(x - \sinh{x})' = 1 - \cosh{x} < 0$ for all positive $x$, so $\tanh{x} < x < \sinh{x}, \ x > 0$.

Let $x = \ln{\sqrt{\frac{b}{a}}}$.

This gives us $\frac{\sqrt{\frac{b}{a}}-\sqrt{\frac{a}{b}}}{\sqrt{\frac{b}{a}}+\sqrt{\frac{a}{b}}} < \ln{\sqrt{\frac{b}{a}}} < \frac{\sqrt{\frac{b}{a}}-\sqrt{\frac{a}{b}}}{2}$ which is equivalent to the main inequality.

[quote]By the way can you prove this ? :lol:

${\displaystyle \frac{a+b}{2}>\frac{1}{e}\left(\frac{a^a}{b^{b}}\right)^{\frac{1}{a-b}}>\frac{b-a}{\ln b - \ln a}}$[/quote]

It follows from the inequality $\frac{x+1}{2} > \frac{1}{e} x^{ \frac{x}{x-1}} > \frac{x-1}{ \ln{x} }$ for $x >1$ .

Prove this using calculus and then set $x= \frac{b}{a} ,\, b >a$. :)


geometry
3D geometry
modular arithmetic
number theory solved
number theory

Show that if p>3 is prime then p^n cannot be the sum of two positive cubes for any n>= 1.

What about p=2 or p=3?

Well, quickly for p=2 and 3, 2^1=1+1 and 3^2=1+8.

yeah! but u forgot the first question , so i get the point.
Assume that $p^n=a^3+b^3$ where $p$ is prime and $p>3$ and $n\ge 1$ , $a,b$ are positive integer.We can consider that $gcd(a,b)=1$ This means $a+b$ is divisible by $p$ and cause $a,b$ are not equal similiuosly to 1 so $a^2-ab+b^2>1$ so $a^2-ab+b^2>1$ is also divisible by $p$ . From the fact that $a\equiv b \pmod{p}$ so $a^2-ab+b^2\equiv 3b^2\pmod{p}$ this implies $p=3$ or $p|b$ . Both are contract!
This done!

Here is a generzied version :

$p , q$ are distinct odd primes and $p^n = a^q+b^q$ where $n,a,b$ are positive integers , has no solution.


FROM THE COMPLETE WORKS OF SWAMI VIVKANANDA

THE ART AND SCIENCE OF MUSIC, EASTERN AND WESTERN

[Shri Priya Nath Sinha]

It was an evening in July 1898, at the Math, in Nilambar Mukerjee's garden-house, Belur. Swamiji with all his disciples had been meditating, and at the close of the meditation came out and sat in one of the rooms. As it was raining hard and a cold wind was blowing, he shut the door and began to sing to the accompaniment of Tânpurâ. The singing being over, a long conversation on music followed. Swami Shivananda asked him, "What is Western music like?"

Swamiji: Oh, it is very good; there is in it a perfection of harmony, which we have not attained. Only, to our untrained ears, it does not sound well, hence we do not like it, and think that the singers howl like jackals. I also had the same sort of impression, but when I began to listen to the music with attention and study it minutely, I came more and more to understand it, and I was lost in admiration. Such is the case with every art. In glancing at a highly finished painting we cannot understand where its beauty lies. Moreover, unless the eye is, to a certain extent, trained, one cannot appreciate the subtle touches and blendings, the inner genius of a work of art. What real music we have lies in Kirtana and Dhrupada; the rest has been spoiled by being modulated according to the Islamic methods. Do you think that singing the short and light airs of Tappâ songs in a nasal voice and flitting like lightning from one note to another by fits and starts are the best things in the world of music? Not so. Unless each note is given full play in every scale, all the science of music is marred. In painting, by keeping in touch with nature, you can make it as artistic as you like; there is no harm in doing that, and the result will be nothing but good. Similarly, in music, you can display any amount of skill by keeping to science, and it will be pleasing to the ear. The Mohammedans took up the different Râgas and Râginis after coming into India. But they put such a stamp of their own colouring on the art of Tappa songs that all the science in music was destroyed.

Q. Why, Mahârâj (sir)? Who has not a liking for music in Tappa?

Swamiji: The chirping of crickets sounds very good to some. The Santâls think their music also to be the best of all. You do not seem to understand that when one note comes upon another in such quick succession, it not only robs music of all grace, but, on the other hand, creates discordance rather. Do not the permutation and combination of the seven keynotes form one or other of the different melodies of music, known as Ragas and Raginis? Now, in Tappa, if one slurs over a whole melody (Raga) and creates a new tune, and over and above that, if the voice is raised to the highest pitch by tremulous modulation, say, how can the Raga be kept intact? Again, the poetry of music is completely destroyed if there be in it such profuse use of light and short strains just for effect. To sing by keeping to the idea, meant to be conveyed by a song, totally disappeared from our country when Tappas came into vogue. Nowadays, it seems, the true art is reviving a little with the improvement in theatres; but, on. the other hand, all regard for Ragas and Raginis is being more and more flung to the winds.

Accordingly, to those who are past masters in the art of singing Dhrupada, it is painful to hear Tappas. But in our music the cadence, or a duly regulated rise and fall of voice or sound, is very good. The French detected and appreciated this trait first, and tried to adapt and introduce it in their music. After their doing this, the whole of Europe has now thoroughly mastered it.

Q. Maharaj, their music seems to be pre-eminently martial, whereas that element appears to be altogether absent in ours.

Swamiji: Oh, no, we have it also. In martial music, harmony is greatly needed. We sadly lack harmony, hence it does not show itself so much. Our music had been improving steadily. But when the Mohammedans came, they took possession of it in such a way that the tree of music could grow no further. The music of the Westerners is much advanced. They have the sentiment of pathos as well as of heroism in their music, which is as it should be. But our antique musical instrument made from the gourd has been improved no further.

Q. Which of the Ragas and Raginis are martial in tune?

Swamiji: Every Raga may be made martial if it is set in harmony and the instruments are tuned accordingly. Some of the Raginis can also become martial.

The conversation was then closed, as it was time for supper. After supper, Swamiji enquired as to the sleeping arrangements for the guests who had come from Calcutta to the Math to pass the night, and he then retired to his bedroom.

Hello,
I think Swamiji primarily refers to Hindusthani music as being influenced by Mohammadeans.Here down south,the art form has largely remained unchanged and has only two main sources 1)The Krithis,(not Kirthanas) of the Trinity and few others,2)The Pann concept borrowed from Ancient Tamil Arts.As for Western Music,well,the Manodharma aspect is not so pronounced.That is why here in Chennai,in the December season,many westerners marvel at Neravals and Swarams. The violinist,Mridhangist and other accompanists are absolutely unaware of what the main artist would sing.No rehearsals.Many a time,it happens that they have never even met before.Yet,they respond then and there,on the stage,with perfect poise,accuracy and speed.It comes only with long,hard assidous practice and of course some innate skills are required.
In Hindusthani music,more stress is on Alapana.Fewer compositions are rendered.Memory is another vital aspect of Carnatic music.No artist would use paper to remember lyrics.There are thousands of Krithis of the Trinity that have to be memorised.The job is all the more difficult for an accompanist as he/she has to be ready for any Krithi to be sung at any time at any Kutcheri(concert).
Sathej

PANEM ET CIRCENSES i agree with you.However dhrupad and veena vadan have not changed much here too.

Hello,
    By the way, what do those Latin phrases in this post as well as in the other post mean?I know Cogito Ergo sum means I think,therefore I am.But the other phrase Panem et Circenses seems out of context here.
Sathej

May I ask a favor here? (Because I am the moderator of the forum, I hope you say yes.  :wink:  ) For the many readers of this forum who don't live in India, could you refer to sources (and especially music recordings) that are available in the wider world? There is a forum on AoPS specifically for discussion of issues pertaining to India, and I want to make sure that this Music and Art Forum generally hosts threads that have a broad interest to anyone from anywhere interested in music or art. So far I am finding this thread interesting, but I want to make sure it isn't just a private conversation on a public forum.


function
inequalities
algebra unsolved
algebra

Do there exist a bounded function $f: \mathbb{R}\to\mathbb{R}$ such that $f(1)>0$ and $f(x)$ satisfies an inequality $f^2(x+y)\ge f^2(x)+2f(xy)+f^2(y)$?

More general, suppose  there is $a > 0$ such that  $f(a) > 0 $. Let $S$ be such that  $\ \displaystyle S= \sup_{x\neq 0}(f^2(x))$ . We can write :
 $f^2(x) + 2f(a) + f^2(\frac{a}{x})\leq  f^2(x+\frac{a}{x})\leq S \Longrightarrow  S + 2f(a) \leq S   \Longrightarrow f(a) \leq 0 $ (false  !!)
:cool:  Manifestly a such function does'nt existe ! :cool:

You say

$\displaystyle f^2(x) + 2f(a) + f^2(\frac{a}{x})\leq f^2(x+\frac{a}{x})\leq S \Longrightarrow S + 2f(a) \leq S$

I don't understand... could someone elaborate?
Also, am I right to assume that
$\displaystyle f^2(x)$ means $(f(x))^2$?

[quote="dystopia"]You say

$\displaystyle f^2(x) + 2f(a) + f^2(\frac{a}{x})\leq f^2(x+\frac{a}{x})\leq S \Longrightarrow S + 2f(a) \leq S$

I don't understand... could someone elaborate?
[/quote]
Do you agree that : $\displaystyle f^2(x) + 2f(a) + f^2(\frac{a}{x})\leq f^2(x+\frac{a}{x})\leq S, \forall_{x\neq 0}$  ?
Do you agree that  :$\displaystyle f^2(x) + 2f(a) \leq S, \forall_{x\neq 0}$ ?
Do you agree  that the real sub-set  $\{ f^2(x) | x\neq 0\}$ is bounded ?
Then, from the définition of $Sup$ (superior bound of a bounded real sub-set), you have to agree $S + 2f(a) \leq S$ ..
 :cool: Do you follow me ?:cool:

thanks
I must've been tired or something...
I saw why $2S+f(a)\le S$, but not why $S+f(a)\le S$
 :blush:

[quote="dystopia"]
Also, am I right to assume that
$\displaystyle f^2(x)$ means $(f(x))^2$?[/quote]
$f^2(x)=(f(x))^2$  :)

Really liked Diogene's idea of looking at the supremum of $f^2$... Anyways... Problem can also be solved, through looking at infimum, and supremum, separately, as demonstrated below:

Let, $M \geq \{f(x):x\in\mathbb{R}\}$. Since, $f(\mathbb{R})$ is bounded, it has a supremum (also, an infimum), and let, $M=\sup_{x\in\mathbb{R}}f(x)$; and let, $-m=\inf_{x\in\mathbb{R}}f(x)$ (here, the $-$ sign is auxiliary in the beginning, but, we will show that, the infimum is indeed negative). Clearly, $f(1)>0\implies M>0$. Let, $P(x,y)$, be the assertion of the problem.

Now, let, $f(1)=\epsilon>0$. Fix a $\delta\in(0,\epsilon)$; and, using the definition of supremum, take an $x\in\mathbb{R}$, such that, $f(x)>M-\delta$ (in particular, $f(x)>0$). $P(x,1)$, gives us,
\begin{align*}
f(x+1)^2 &\geq f(x)^2+2f(x)+f(1)^2\\
&\geq (M-\delta)^2 + 2(M-\delta)+\epsilon^2\\
&>(M-\delta)^2 + 2(M-\delta)+\delta^2\\
&=M^2\\
&\implies f(x+1)>M \text{ or } f(x+1)<-M.
\end{align*}
Clearly, the first possibility is contradictory, hence, $f(x+1)<-M$. Now, since, $-m=\inf_x f(x) \leq f(x+1)<-M$; we have, $m>M$.

Next, we fix a $\gamma$, and select a corresponding $x$, such that, $f(x)<-m+\gamma$. The existence is justified, through the definition of infimum.

Now, we will look at, $P\left(x,\frac{1}{x}\right)$ (select, $x$ such that, $x\neq 0$):
\begin{align*}
f\left(x+\frac{1}{x}\right)^2 &\geq f(x)^2 + 2f(1)+f\left(\frac{1}{x}\right)^2\\
&\geq f(x)^2 + 2f(1)\\
&> (m-\gamma)^2 + 2\epsilon.
\end{align*}
Now, we will show that, $\gamma>0$ can be selected in such a way that, $(m-\gamma)^2 +2\epsilon>m^2$.
$$
(m-\gamma)^2+2\epsilon>m^2 \iff \gamma^2 -2m\gamma+ 2\epsilon>0.
$$
The discriminant of this quadratic is, $\Delta=4m^2-8\epsilon$. Hence, its roots are,
$$
\gamma{1,2}=\frac{2m\pm \sqrt{4m^2-8\epsilon}}{2}=m\pm\sqrt{m^2-2\epsilon}.
$$
If the discriminant is less than $0$, then, life is good; otherwise, it suffices to select, $\gamma<m-\sqrt{m^2-2\epsilon}$. Noting that, this is a positive quantity, therefore, says that, it suffices to pick $\gamma$ arbitrarily small.

Finally, we get,
$$
f\left(x+\frac{1}{x}\right)^2 >m^2.
$$
If, $f\left(x+\frac{1}{x}\right)>m$, we get a contradiction, since, $m>M$; conversely, if, $f\left(x+\frac{1}{x}\right)<-m$, then, we again, arrive at a contradiction, at the infimum side. Thus, no such function exists.

[quote]Do there exist a [b][u]bounded[/u][/b] function $f: \mathbb{R}\to\mathbb{R}$ such that $f(1)>0$ and $f(x)$ satisfies an inequality $f^2(x+y)\ge f^2(x)+2f(xy)+f^2(y)$?[/quote]


function
calculus
derivative
calculus computations

Hi, I know what the answer is, I am just looking for the theory behind the solution.

4.  Let $ k$ be the number of real solutions to the equation $ e^x \plus{} x \minus{} 2 \equal{} 0$ in $ [0,1]$, and let $ n$ be the number of real solutions that are [u]not[/u] in $ [0,1]$.  What are the values of $ k$ and $ n$?

What I did was apply the intermediate value theorem since it is obviously continuous.  I got a negative value at $ x\equal{}0$ and a positive value at $ x\equal{}1$.  Hence, obviously there is one real solution in $ [0,1]$.  What I am struggling with is knowing that there is ONLY 1 solution in that interval and that there are no solutions anywhere else.  Also, obviously no calculator allowed :p.

General principle (for roots of well-behaved real functions): if you want to prove that there are at least so-and-so many roots, use the intermediate value theorem. (You've done that already). If you want to prove that there are no more than so-and-so many roots, use an argument from the mean value theorem/Rolle's Theorem family (including simply analyzing intervals of increase and decrease, which is also in that family.)

$ f(x)\equal{}e^x\plus{}x\minus{}2.$

$ f'(x)\equal{}e^x\plus{}1.$

Since $ f'(x)>0$ always, the function is strictly increasing and can have at most one root. Or, for an alternative way to phrase the same argument: if $ f$ had two distinct roots, then by Rolle's Theorem, its derivative would have a zero somewhere in between. But since the derivative has no zeros, this can't happen.

Great response!  I knew there was some solid theory somewhere behind this, but I had long forgotten it.  Thank you so much!

[quote="Kent Merryfield"]General principle (for roots of well-behaved real functions): if you want to prove that there are at least so-and-so many roots, use the intermediate value theorem. (You've done that already). If you want to prove that there are no more than so-and-so many roots, use an argument from the mean value theorem/Rolle's Theorem family (including simply analyzing intervals of increase and decrease, which is also in that family.)

$ f(x) \equal{} e^x \plus{} x \minus{} 2.$

$ f'(x) \equal{} e^x \plus{} 1.$

Since $ f'(x) > 0$ always, the function is strictly increasing and can have at most one root. Or, for an alternative way to phrase the same argument: if $ f$ had two distinct roots, then by Rolle's Theorem, its derivative would have a zero somewhere in between. But since the derivative has no zeros, this can't happen.[/quote]

Could you have used the second derivative to see if there are any inflection points? If you visualize it, the only way that a function can have two roots in an interval is if it goes upwards/downwards and then downwards/upwards, intersecting the x-axis both times. Thus, there must be an inflection point somewhere in between there.

Concavity and inflection points add nothing to our knowledge here. That would be a waste of time.

By the way, the GRE has more than 60 questions and only about 3 minutes per question; it is really quite important not to waste time. They're structured as multiple choice. The original poster didn't give the answer choices, although it's not hard to guess what they were likely to have been.


calculus
integration
trigonometry
trig identities
calculus computations

How do I integrate sin^3(2x), that's sin of 2x, cubed.
I've tried a lot, I've converted sin^2 to (1-cos^2) and multiplied through.  I've used double angle and half angle identities, but I still cannot integrate it by substitution.

Calculus shouldnt be posted in this forum...

but while we're here...

Lets let $u = 2x$.
$du = 2dx$
$1/2 du = dx$

We are now at integral of $1/2\sin^3(u)$

Now use pythagorean identity to integrate and then substituting in v. Replace v with u and then replace u with x.

Mod can move this thread.

Thanks, but I don't know how to integrate the expression sin(u)-cos^2(u)sin(u)du.  I've tried a lot of things, but I just don't see it.  I have not been integrating with trig identities very long and i might be a little rusty :mad: .

$\int\sin^3(2x)\,dx$

Start with a "little" substitution, just to clear away a little clutter: let $u=2x,$ so $dx=\frac12du.$

$\frac12\int\sin^3u\,du=\frac12\int\sin u\,(1-\cos^2u)\,du$

Now make the substitution $v=\cos u$. We have that $dv=-\sin u\,du,$ which we have available.

$-\frac12\int(1-v^2)\,dv=-\frac12\left(v-\frac{v^3}3\right)+C= \frac{v^3}6-\frac v2+C$

$=\frac{\cos^3u}6-\frac{\cos u}2+C$

$=\frac{\cos^3(2x)}6-\frac{\cos(2x)}2+C.$

You can use the following identity.

$\sin 3\theta =3\sin \theta -4\sin ^ 3 \theta\Longleftrightarrow \sin ^ 3 \theta =\frac{1}{4}(3\sin \theta -\sin 3\theta).$


geometry
power of a point
radical axis
geometry solved

Sol: Let a,b,c be the interior angles at A,B,C.Let F denote the intersection of the cirle Y and the segment AC.The angle between the chord FB and the tangent line AB is equal to
the circumferential angle over this chord. It follows the angle FBA=c, and AFB=b. So the quadrilateral ABDF is cyclic and the angle DFC=b. Using the defination of the circle X we have ED is perpendicular to AD and <EDC=pi/2 - b= <ECD. So the circle X passes through the point C.and <DEC=2b.(since <DFC=b and E is center) The circles X and Y
intersect in the points F and C, but F lies on AC by construction.

[quote="kim77"]:The point D on the side BC of the acute triangle ABC is such that AB=AD.
Let E be a point on the altitude from C of triangle ABC such that the circle X with center E is tangent to the line AD in the point D. Denote with Y the circle through C that is tangent to the line AB in the point B.
Prove that the point A lies on the line through the intersections of the circles X and Y.[/quote]

Have I drunk too much, have I slept too little or is this "hard" geometry question trivial? Since the circle X touches the line AD at the point D, the power of the point A with respect to this circle X equals $AD^2$. Since the circle Y touches the line AB at the point B, the power of the point A with respect to this circle Y equals $AB^2$. But AB = AD and thus $AB^2=AD^2$; hence, the powers of the point A with respect to the circles X and Y are equal. And therefore, the point A lies on the radical axis of the circles X and Y, i. e. on the line joining the intersections of these circles X and Y.

  darij


Concider a set of 1985 positive interger, not necessarily distinct, and none with prime factors bigger than 23. Prove taht there must exist 4 integers in the set whoes product is equal to the fourth power of an integer.

Solution is here:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366595#p366595[/url]


geometry
geometric transformation
reflection
parallelogram
trigonometry

haha really cheap, nonrigorous solution:

[hide]
Assume its a 30-60-90 triangle, as the angles work out if A is the right angle.  Then its clear that the answer is 1/2.[/hide]

the diagram is not uniquely defined...here is a quick solution...light trig...

reflect $A$ over $E$ to $A'$, it follows $ABA'C$ is a parallelogram, and $\angle BA'A=a$ [we see here that it is easy to construct many triangles for $ABA'$ that satisfy the conditions...]

WLOG $AB=1$, note that $A'B=2\cos a$ by law of sines...then $DE=\frac{1}{2}AA'-AB=\frac{1}{2}*(\cos 2a+2\cos^{2}a)-\cos 2a=\frac{1}{2}$, as desired

You are rihgt.
nateharman1234 if you don't like you can solve that i asked some at olimpic chapter. :lol: And i don't like giving value.
[hide][img]http://img243.imageshack.us/img243/3249/sdfdfyy9.png[/img][/hide]
[hide][img]http://img412.imageshack.us/img412/9503/dffgnc6.png[/img][/hide]


As the figure shown,

The first three terms are in continuing process which can develop  many small and big triangles each time ,

$ { a_n  }$ represents the total no.of generating big and small triangles individually.

$ a_1=1, a_2=5 ,$ and $ a_3=13 ,\cdots ,a_n =\cdots$

Find $ a_{20}$ , and $ a_n$

This is sequence [url=http://www.research.att.com/~njas/sequences/A002717]A002717[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].


Can anyone solve this?

You are downstairs in a lobby of a house. There are 3  switches, all in the "off" position. Upstairs, there is a room with a lightbulb that is turned off. One and only one of the three switches controls the bulb. You want to discover which switch controls the bulb, but you are only allowed to go upstairs once. How do you do it? (No fancy strings, telescopes, etc. allowed. You cannot see the upstairs room from dawnstairs. The lightbulb is a standard 100 - watt bulb.)

[hide="this is a classic"]

turn on the first switch for a long time, and turn it off. turn on the second switch, and go upstairs. if the bulb is hot but not lit, the first switch controls the light. if the bulb is on, the second switch controls the bulb. if the bulb is off and not hot, the third switch controls the bulb.

[/hide]


linear algebra
matrix
linear algebra unsolved

Slove equation matrices in $ M_2{\mathbb{(Z)}}$ :
1) $ A(A \plus{} E)(A \plus{} 7E)(A \plus{} 8E) \equal{} B^2$
2) $ AB \equal{} 2A \plus{} 3B$
 Here $ A,B \in M_2{\mathbb{(Z)}}$, $ \det(A) \equal{} \det(B) \equal{} 1$,$ E$ is $ 2$x$ 2$ identity matrix

1) has no solutions.
Let $ spectrum(A)\equal{}\{\lambda,\mu\}$. $ \lambda\mu\equal{}1$ and if $ k\equal{}1,7$ or $ 8$ then
$ (\lambda\plus{}k)(\mu\plus{}k)\equal{}1\plus{}k.trace(A)\plus{}k^2\equal{}\pm{1}$. It is impossible.
2) Let $ spectrum(A)\equal{}\{\lambda,\mu\}$. $ (A\minus{}3I)B\equal{}2A$ thus $ A\minus{}3I$ is invertible and $ B\equal{}(A\minus{}3I)^{\minus{}1}2A$. Moreover $ (\lambda\minus{}3)(\mu\minus{}3)\equal{}4$ and $ trace(A)\equal{}2$. Therefore $ (A\minus{}I)^2\equal{}0$ and $ A\equal{}I\plus{}N$ where $ N^2\equal{}0$. We deduce $ (A\minus{}3I)^{\minus{}1}\equal{}\minus{}1/4(A\plus{}I)$. Finally $ B\equal{}1/2(I\minus{}3A)$. The converse is easy.
Conclusion: $ A\equal{}I\plus{}N,B\equal{}\minus{}I\minus{}(3/2)N$ where $ N\equal{}\begin{pmatrix}u&v\\w&\minus{}u\end{pmatrix}$ where $ u,v,w$ are even integers s.t. $ u^2\plus{}vw\equal{}0$.


The majority of plants with leaves have their stomata on the underside to prevent evaporation and water loss. However, some plants will have this reversed (for example, the pitcher plant), seeing as the water is enclosed on the inside so gas exchange requires for the stomata to be on the outer side. This leaves a question though, how does light pass through the spongy mesophyll to the palisade mesophyll (which have a main purpose of photosynthesis)?

I don't really get how this is related to the stomata.  The light would get through just as it would in any other plant.

[quote="JRav"]The light would get through just as it would in any other plant.[/quote]

Right, this is because the epidermis is transparent and the light is able to hit the layer underneath; the palisade mesophyll. However, the spongy mesophyll does not ordinarily perform mesophyll and it would be a baracade for the palisade, since it's not transparent.

Is it maybe that the pitcher plant also receives energy from the insects it traps and digests? It wouldn't need to have the most efficient photosynthesis if it got glucose from another source as well. If the design of the leaves helps to trap food, then perhaps the evolutionary advantage of that is greater than that of photosynthetic efficiency.

@7h3--2 reasons
1.leaf is quite thin and cells or cuticle don't act as barriers
2.light can also b reflected from ground

@krazy--there r other examples where we have epistomatous leaves

then perhaps the evolutionary advantage of that is greater than that of photosynthetic efficiency.
[url=http://www.math123xyz.com]Interactive Math Resources[/url]


calculus
integration
trigonometry
calculus computations

use this :
$ \int_{0}^{\plus{}oo}\frac {\sin ax}{e^{x} \plus{} 1}dx \equal{} \sum_{k \equal{} 1}^{ \plus{} oo}\int_{0}^{\plus{}oo}( \minus{} 1)^{k \minus{} 1}e^{ \minus{} kx}\sin ax dx$

I have used residue theorem and I get
$ \frac{1}{2a}\minus{}\frac{\pi}{2\sinh\pi a}$.


geometry
circumcircle
combinatorics solved
combinatorics

Each point in the plane is coloured with one of two distinct colours. Prove that there is an equilateral triangle whose vertices have the same color.

Consider a regular hexagon (with circumradius $1$, say) and its center. You'll find that all points distance $2$ apart are the same color, and from here the conclusion will follow.


USAMTS AMC USA(J)MO USAMO geometry trigonometry AIME

So who is doing the 2008-09 USAMTS??? I am pretty sure i am, see here for details [url]http://www.usamts.org[/url]

i'm sorry i knew you would be disappointed, here is the thing that says that. [url]http://www.usamts.org/Rules/U_RulesEligibility.php[/url]

dude

undefined is like japanese citizenship?

wow

or wait

it should be like korean

since he was born there

and he went to japan

like a few months back

oh wait i'm doing usamts
i'm predicting gettin' a 0

you forgot to type an additional one and zero.

001??? just kidding, they don't grade as hard as USAMO, on this one geometry problem i didn't know how to do it so this was my solution, "I drew the picture the scale and measured $ 50^\circ$, the answer was 47 and i got a 1/5 for it. it was pretty amazing. trust me #H34N1, u'll do fine.....

dude cognos

i remember that problem

that was beast

i did a bunch of inversion

and like

actualy got that problem with like trig bash

yeah that one looked really hard for me..... i could probly do it now though. yeah that was the lowest score i got on any problem in round 1. i got a 5/5/3/1/[b]5[/b] for a 19. what sux is after i sent in my answers i'm like "okay i got a 11 or 12 maybe lower so there is not point in doing the other rounds" but after the second round came out, i got my score and i was like "19?!?!?!?!??!, i don't deserve this" so ya. i spent like a week on round 1 and didn't do the other rounds.

Dude #3 was like easy

was that the arctan one? just set like bounds and whatever; there was an aime problem like that this year I think

ya i know number 8 or something. anyone who did USAMTS should not have missed that problem, luckily i didn't.

i thought the order of difficulty from ezest to hardest on USAMTS rd. 1 last year was:

2,1,5,3,4 or 1,2,5,3,4.

First of all, #2 was way easier than #1. #2 took me like 3 minutes and #1 took like 15 minutes since I kept messing up on like multiplication. #3 was easier than #5 and #5 was harder than #4.

maybe ill do it...ive never done one before but they always send me the first set of questions...i just put it to the side and dont bother...

but seeing that these things are important and actually help maybe ill attempt to do one but i suck at proofs and writing out answers...

It'll be especially easy for me to do it because of the long time limit.

I can just look at my AoPS books or online to review strategies for solving certain problems if I'm stuck on one.  :lol:

Because if you try everything possible, for example, on a geometry problem (adding parallel lines, building right triangles/triangles, trigonometry, inversion, homothethy, projection, similar and congruent triangles, etc.), eventually the answer should be in reach.

Like, for example, I forgot the fact that $ gcd(m,n)\cdot lcm[m,n] = m\cdot n$, where $ gcd(a,b)$ and $ lcm[a,b]$ denote the greatest common divisor and least common multiple of two integers, respectively, and $ m$ and $ n$ are integers.
Of course, I could derive this fact any time I want to in a math competition, but in something with strict time limits such as AMC/AIME, I wouldn't be able to remember it.
However, if I'm faced with a number theory problem involving divisibility on the USAMTS, I can search through all the divisibility tools that exist and use the appropriate one (and then I'd remember facts such as the one stated in the above paragraph).

So, yeah, of course I'm doing it (hopefully I'll get Mathematica for free so I can make it calculate one billion factorial again!).  Last year, more 9th graders got perfects than anyone else :huh: ...  So hopefully I'll get lucky since I'll be a 9th grader...  I love having lots of time for math problems.

I spent like 3 hours last year on all but round 1, which I spent like 5 hours on..which accounts astronomically for my score :huh:

didn't do round 4

this year i think i will do all of the rounds, and attempt to get 40+  :D . right would it be weird if USAMO was multiple????

Theres a problem loading page for the usamts link... anyways i dont think so as i have no idea what it is.

I'm doing RCML though :) its sorta based (or whatever the correct word is) in my city so i do it.

logic did you hit disable smilies in this post

O MY GOSH GUYS AND GALS, ONE MORE DAY, wait, we can submit our solutions any time tomorrow, right? i'm gonna pull an all-nighter and finish the problems. i'll be like a basket-case tomorrow. YAY....

[quote="the future"][quote="unimpossible"]lol. im not even doing it[/quote]

i am i just did all the problems today lol...

btw who are all these freshman noobs at the high school now that think they are good at math (besides lyndon+youkow)[/quote]

HEY! WOAH! ya jk i suck @ math. the other "noob freshmen" (logic, mathand, mustit) are from my school, no offense logic and mustit just kinda suck at math, mathandknowledge is at least as good as i am (probly better), but then again that doesn't say much.. :P

so how do you guys (who did it) think you did? for me i think around 20 or 21. maybe lower maybe higher. idk

OMG results will be out THIS WEEK. i am so excited. and plus,

$ \frac{42}{1337} \approx \frac{\pi}{100}$

cognos, gonna win your grade at rose-hulman?

lol, i hope so. i am aiming for a 95 or 100, but i'll probly end up getting in the 60's or 70's  :( . plus i haven't done much contest prep. just been doing USAMO problems so i might fail.... r u gonna win your grade @ rose-hulman?

NOPE
atleast you are doing math, all ive been doing is sat reading comp

ever since school has started, i have been doing very little math outside of WOOT. so i might suck, or i might do really well. i am hoping for the latter.  :) (althought the former is more likely  :wink: )


calculus
algebra
polynomial
derivative
calculus computations

[quote="bos1234"]Prove that for a polynomial of degree n, the (n+1)th root vanish but the nth root does not[/quote]

Do you mean that a polynomial of degree $n$ cannot have $n+1$ roots?


algebra
polynomial
search
LaTeX
function
number theory
number theory proposed

Let $f(x)$ be a polynomial with integer coefficeints with degree $n$.
For any prime number like $p$ , there exist a prime number $q$ and the natural number $m$ such that \[f(p)=q^{m}\] Prove that $f(x)=x^{n}$

[quote="lomos_lupin"]Erh ,I searched for it 10 times ,Its always saying

No topics or posts met your search criteria

I even searched for $f(x)=x^{n}$, whats wrong with this search ?

Zetax what did you search for ?[/quote]

az Valentin said 2 month ago,You cannot search for $\LaTeX$ yet , this bug will correct soon.
 :(

Yes, searching here is hard without searching for $\LaTeX$-code.

I searched for "prime power" and me as author (I tried prime power only first, but that didn't work, even when restricting to number theory only).

In Russian forum proved better result:
If P(x) polinomial with integer coefficients, and exist N, suth that for all distinct primes p>N,q>N $(P(p),P(q))=1$, then $P(x)=\pm x^{n}$.

Not harder to prove:

Let $p$ be any prime. Assume there is a prime $r|P(p)$, but $r \neq p$.
Then we take a prime $q \equiv p \mod r$ but $\neq p$, so that we get $P(q) \equiv P(p) \equiv 0 \mod r$. Thus $r| \gcd(P(p),P(q))$, contradicting the condition.

Thus we have that $f(p)=\pm p^{g(p)}$ for all primes $p$ and $g(p)$ being some function of $p$. Since all nonconstant (Rust: you forgot $P(x) = \pm 1$) polynomials get strictly monotonic from some point, we w.l.o.g. assume $P(p)=p^{g(m)}$ for all big primes $p$.

Now for all big numbers $m$ we have $m^{n-1}< P(m) < m^{n+1}$ with $n$ being the polynomial's degree.
This gives that we have $P(p)=p^{n}$ for all big primes $p$. So this polynomial and the polynomial $x^{n}$ have infinitely many points together, thus are equal.

[quote="ZetaX"]Rust: you forgot $P(x) = \pm 1$[/quote]
I don't forget, it mean $P(x)=\pm x^{n}$, were n=0.


geometry
power of a point
radical axis
geometry unsolved

Construct the radical axis $ d$ of two given circles $ (O_1)$, $ (O_2)$ (probably you know how to do it).
Let $ M$ be the midpoint of $ O_1O_2$. The circle of diameter $ MP$ meets the line $ d$ at $ K$. The line $ KP$ is just the wanted line which cuts equal length chords on the two circles. See at the picture.
Tell me if you need explanation for my constructions in some topics.  :wink:


I have a round bar with a radius of 6 cm.How do I find the area of the round cross section of this bar ?

Please don't post the same question multiple times.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=116816[/url]


Si $a,b,c >0$ son numeros reales.

demostrar

$\frac{1}{b(a+b)} + \frac{1}{c(b+c)} + \frac{1}{a(a+c)} \geq \frac{27}{2(a+b+c)^2}$

Fabiola        :clap:

Bertolt Br._Peru

mi solucion:

[hide]Por $MA -MG$ se tiene que:
$\frac{1}{b(a+b)} + \frac{1}{c(b+c)} + \frac{1}{a(a+c)} \geq 3 \sqrt [3] {\frac{1}{a(b+c)b(a+c)c(a+b)}}$
Entonces se reduce a probar que:
$\frac{2(a+b+c)^2}{9} \geq \sqrt [3]{a(b+c)b(a+c)c(a+b)}$

Y por $MA - MG$ se tiene:
$\frac{2(ab+ac+bc)}{3}\geq \sqrt [3] {a(b+c)b(a+c)c(a+b)}$

Entonces es suficiente probar que:
$\frac{(a+b+c)^2} {3}\geq ab+ac+bc$

Que es lo mismo que probar la concida desigualdad:
$a^2 + b^2 + c^2 \geq ab + bc +ac$[/hide][/code]

Mi solucion:

Sea $s=a+b+c$, por reordenamiento con $1/(s-a), 1/(s-b), 1/(s-c)$ y $1/a,1/b,1/c$ se tiene que la suma inicial es mayor que:

$\sum_{sym} 1/a(s-a)$ que por Jensen con $f(x)=1/x(s-x)$ con $x$ entre $0$ y $s$ se tiene que esta suma es mayor que:

$\frac{3}{(s/3)(s-s/3)}$ que es justamente $27/2s^2$ lo que prueba la desigualdad.

Nota: Cuando las desigualdades las podemos hacer simetricas son mucho mas faciles de atacar que cuando son ciclicas!


Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius 20 feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point 10 vertical feet above the bottom?

[hide]When the rider is 10 vertical feet from the bottom, he is also 10 vertical feet from the center of the ferris wheel. The rider is 20 feet from the center (because the radius is 20); by Pythagorean Theorem the rider is $10\sqrt{3}$ horizontal feet from the center. Recognizing that these lengths form a 30-60-90 triangle, the rider has traveled 60 degrees, or 1/6 of a minute = $10$ seconds.

Ugg, it sounds so much clearer with a diagram...[/hide]

You mixed up the angles for the 30-60-90 triangle.

Sorry if my explanation is unclear, what I mean is that the central angle cut off by the rider and the bottom is 60 degrees.


[hide]
Let $a=x$ and $b=2x-3$.
Note that $12x(x-1)=4(x+2x-3)=4(a+b)$.

We can thus rewrite the equation as $\sqrt[3]{a}+\sqrt[3]{b} =\sqrt[3]{4(a+b)}$.
Cubing both sides to get $a+b+ 3\sqrt[3]{ab}({\sqrt[3]{a}+\sqrt[3]{b}) = 4(a+b)}$
Simplifying and rearranging terms, we get $\sqrt[3]{a}+\sqrt[3]{b} =\frac{a+b}{\sqrt[3]{ab}}$.

Since the RHS of this equation and the RHS of the original equation are both equal to $\sqrt[3]{a}+\sqrt[3]{b}$, we obtain $\sqrt[3]{4(a+b)} = \frac{a+b}{\sqrt[3]{ab}} \Longleftrightarrow \sqrt[3]{4ab}= \sqrt[3]{(a+b)^2}$
Cubing both sides, $4ab=a^2+2ab+b^2\Longleftrightarrow (a-b)^2=0$, so $a=b$.

Hence $x=2x-3\Longleftrightarrow \boxed{x=1}$
(If we substitute $x=1$ into the original equation, we get $0=0$, so this works.)
[/hide]

Edit- I made a mistake in the 2nd line...
Never mind, I don't think my approach works very nicely.

There's another solution as well.  Check your second line frt, that should be $4a\left(a+b\right)$.

Oops :oops:

I tried to fix my solution, but it became kinda ugly.
There must be a better way to solve this (unless I'm missing something)...

[quote="shadysaysurspammed"]Anyone......[/quote]

Well, when no one can solve it, then you are due to post a hint

No hint yet, this problem is not as hard as it looks. Try harder, just because you can't solve it at first glance it doesn't mean you need an immediate hint.

Do the square brackets represent floor function or parentheses?


I've applied and don't know if I've been accepted yet, but in the mean time, I had some questions I'd like answered by students that have attended the program.

*What are the dorms like?
*what furniture is provided?
*Do we need to pack anything for our dorms, i.e. sheets, pillowcases, etc.?
*What's the bathroom situation like?
*What about showers?
*What is student life like? i.e. during breaks, lunch, hanging out
*What do you recommend packing?
*What's up with the washer/dryer situation?
*How much money should I bring? I have family in the area, could they deliver any extra money to me if I run out?
*What clothing is recommended? This may seem like a stupid question, but I look like a standard rocker (emo, Goth, metalhead, whatever) to the naked eye, and I'm guessing my spiked gauntlets are a little too extreme. Will I need formal/business attire, etc.?
*What's the weather like?

My email is available on request, which is the best way to contact me. If there's anything else you think I'd like to know about the HCSSiM program, it would be greatly appreciated!

Toodles!
Pockets

I doubt they'll require
any formal attire;
if they do I will be in a mire:
for suits, $ \neg \exists$ is my quantifier.

But I don't really know, I'm not an alumn. I can also tell you, however, that I would be happy to see your spiked gauntlets make an appearance at HCSSiM 2008. And good luck on the application, by the way.

Unfortunately, I was rejected this year.

:o

I was really stoked for your spiked gauntlets. This is bad news.


Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.
[asy]unitsize(.3cm);
defaultpen(linewidth(.8pt));
path c=Circle((0,2),1);
filldraw(Circle((0,0),3),grey,black);
filldraw(Circle((0,0),1),white,black);
filldraw(c,white,black);
filldraw(rotate(60)*c,white,black);
filldraw(rotate(120)*c,white,black);
filldraw(rotate(180)*c,white,black);
filldraw(rotate(240)*c,white,black);
filldraw(rotate(300)*c,white,black);[/asy]$ \textbf{(A)}\ \pi \qquad \textbf{(B)}\ 1.5\pi \qquad \textbf{(C)}\ 2\pi \qquad \textbf{(D)}\ 3\pi \qquad \textbf{(E)}\ 3.5\pi$

[hide]So the big circle has radius 6/2=3. THe area is then $ 9\pi\minus{}7(\pi)\equal{}2\pi$. C.[/hide]

Nice preperation for tomorrow.

The big circle has area of $9\pi$ and the inner circles have area $\pi$. So the answer is $9\pi - 7\pi = 2\pi \Rightarrow \boxed {C}$.


MATHCOUNTS
geometry
analytic geometry
Ring Theory
probability
Euler
ratio

Hi. Mathcounts is based on mainly 3 curriculums in math: Algebra 1, Geometry, and Algebra 2. Sometimes it is based on logic, but it varies a lot.

If you want to study for Mathcounts, I suggest you study the following things:

Coordinate Planes (Like y=mx+b)
Order of Operations
Property of Circles
Volumes and Surface Areas
System of Linear Equations

And what else? Help me here...

I think the people that went to Nationals could help you.

Er, he was asking for previous test problems, which can be found [url=https://mathcounts.org/Page.aspx?pid=295]here[/url] and [url=http://www.mnspe.org/mathcounts/coach.htm]here[/url]..

Anyway, MATHCOUNTS is [b]not[/b] based on Algebra 1 & 2 and Geometry.  There are four main topics: Algebra, Geometry, Combinatorics, and Number Theory (though NT is not stressed much in MATHCOUNTS).

If I had to say MATHCOUNTS was based on some certain "curricula", then I would say Algebra 1, Geometry, and Discrete Math (which is normally a high school and college course)...

He was asking for problems??? But he said he wanted to KNOW about some.

[quote="rick8huang"]I just want to check out what type of questions there are. These can be from the 2008 Chapter, Sprint, or Countdown rounds. Please I really want to know about some.[/quote]
He wants to check out what type of questions there are. If you want sample tests, see AIME15's post. But for topics, I think probability and statistics would be included as well.

Those given in the Handbook are:

Algebraic Expressions & Equations
Sequences & Series
Measurement
Number Theory
Solid Geometry
Plane Geometry
Coordinate Geometry
Logic
Probability, Counting & Combinatorics
Percents/Fraction
Pattern Recognition
Proportional Reasoning
Problem Solving (Misc.)
General Math
Statistics

To get a copy of the Handbook Vol. I [url=http://mathcounts.org/Document.Doc?id=57]click here[/url]

[quote="AIME15"]Er, he was asking for previous test problems, which can be found [url=https://mathcounts.org/Page.aspx?pid=295]here[/url] and [url=http://www.mnspe.org/mathcounts/coach.htm]here[/url]..
[/quote]

Why Mathcounts only publishes Chapter and State problems, but no National problems?

MathCounts also puts some School Competitions on their websites.

[quote="danb80"]Those given in the Handbook are:

Algebraic Expressions & Equations
........
Proportional Reasoning
............
[/quote]

Can anyone give an example for a problem in "Proportional Reasoning"?

I think it would be like similar triangles. I got that from "proportion".

Proportional stuff:

When dribbling a basketball up a court, Gloria dribbles at a rate of two dribbles for every three steps she takes. At this rate, how many dribbles will she complete during her 51 steps to the other end of the court?

34! 34!
i think that algebra questions on mc are always trivial, while the geo can get difficult (mc difficult-requiring thought).

Following are two examples of questions on proportional reasoning (from the handbook)

Starting at the town of Euler and traveling 40 miles to the town of
Pythagoras, Rashid travels at the rate of 2 miles every 15 minutes.
Returning from Pythagoras to Euler, he travels 2 miles every 3
minutes. What was Rashid’s average speed, in miles per hour, for the
entire trip? Express your answer as a mixed number.

The ratio of the length of the sides of square ABCD to the length of the sides
of square PQRS is 3:1. If the area of square ABCD is 9 square units, what is the
length of side PS?

[quote="PowerOfPi"] But for topics, I think probability and statistics would be included as well.[/quote]

This is part of combinatorics :wink:

Oh, I didn't know that.

Also, the Pythagorean Theorem comes up A LOT. You should memorize the common ones.

Another list of content is the [url=http://www.unidata.ucar.edu/staff/russ/mathcounts/diaz.html]MathCounts Bible[/url]

[quote="tiger21"]Why Mathcounts only publishes Chapter and State problems, but no National problems?[/quote]

They [url=http://secure.sportsawardsonline.com/applications/default/store/index.asp?product_category_id=1000009]do publish the Nationals problems[/url] in the yearly sets. It's just not [i]free[/i].

Maybe that is why people are asking for them. :D


Εστω S ενα συνολο και * μια διμελης πραξη του S που ικανοποιει τους παρακατω κανονες $ \forall$$ (x,y)\in S^2$
$ x*(x*y) \equal{} y$
$ (y*x)*x \equal{} y$
ΝΔΟ η $ *$ ειναι αντιμετα8ετικη αλλα οχι απαραιτητα προσαιτεριστικη

[quote="DIMITRIS T"]Εστω S ενα συνολο και * μια διμελης πραξη του S που ικανοποιει τους παρακατω κανονες $ \forall$$ (x,y)\in S^2$
$ x*(x*y) \equal{} y$
$ (y*x)*x \equal{} y$
ΝΔΟ η $ *$ ειναι αντιμετα8ετικη αλλα οχι απαραιτητα επιμεριστικη[/quote]

Επιμεριστική ως προς ποιά πράξη στο $ S$?

Cheerio,

Durandal 1707

re mages deite ligo afth thn askhsh giati dn mu bgainei :surrender:

Γράφω τη λύση μου με κάποια επιφύλαξη γιατί έχω να δω τέτοια πράγματα από το 1990 που ήμουν Γ λυκείου.

Λοιπόν από εκφώνηση ισχύουν οι σχέσεις:

x * (x * y) = y     (1)
(y * x) * x = y     (2)

Θέλουμε να δείξουμε ότι x * y = y * x

Ξεκινάω από την (1)
x * (x * y) = y
Κάνω την πράξη * με (x * y) από δεξιά.
(x * (x * y)) * (x * y)   = y  * (x * y)
Εφαρμόζω στο πρώτο μέλος της προηγούμενης σχέσης τη (2) με (x * y) στη θέση του x και x στη θέση του y. Οπότε προκύπτει:
x = y  * (x * y)
Τώρα κάνω την πράξη * με y από αριστερά. Έτσι έχω
y * x = y * (y  * (x * y))
Εφαρμόζω στο δεύτερο μέλος την (1) με y στη θέση του x και (x * y) στη θέση του y. Οπότε προκύπτει το ζητούμενο
y * x = x * y

Ελπίζω να μην ξέφυγε κάτι.

Αλλη μια λυση
Η σχεση $ x*(x*y)\equal{}y$ για $ x\minus{}>x*y$ του $ S$ δινει $ (x*y)*[(x*y)*y]\equal{}y$ οποτε $ (x*y)*x\equal{}y$ και για $ y\minus{}>x*y$ εχουμε
$ [x*(x*y)]*x\equal{}x*y$ η $ y*x\equal{}x*y$
Θα δειξω οτι δεν ισχυει απαραιτητα η προσεταιριστικη
Εστω οτι ισχυει $ (a*b)*c\equal{}a*(b*c)$  Για $ c\minus{}>b*a$ εχουμε $ (a*b)*(b*a)\equal{}a*(b*(b*a))$ η $ (a*b)*(b*a)\equal{}a*a$ η $ (b*a)*(a*b)\equal{}a*a$  η $ [(b*a)*a]*b\equal{}a*a$ η $ b*b\equal{}a*a\equal{}e$ που δεν οφειλει να ισχυει για καθε $ a,b$ στο $ S$ γιατι ..με αντιπαραδειγμα συνολου με τις παραπανω ιδιοτητες...ειναι αργα για να το βρω..

κ στις δυο λυσεις υπάρχει ενα προβλημα.
Δεν μπορουμε να κανουμε κπ αντικαταστασεις πχ δεν μπορουμε να βαλουμε οπου χ το χ*y
δοιτι δεν ξερουμε αν  η πραξη *  ειναι κλειστη.(δηλ. μπορει τα χ,y να ανηκουν στο S αλλα δν ξερουμε αν ανηκει κ το x*y sto S) :wacko:

[quote="cretanman"]Δημήτρη, η διμελής πράξη $ \star$ στο S (binary operation on S) ορίζεται να είναι μία απεικόνιση $ \star : S\times S \to S$ άρα αναγκαστικά για δύο στοιχεία $ s_1,s_2 \in S$ ισχύει $ s_1\star s_2 \in S$.

Οπότε τέτοιο πρόβλημα δεν υφίσταται. Αλοίμονο εάν κάθε φορά στον ορισμό μίας ομάδας $ (G,\star)$ λέγαμε ότι η πράξη $ \star$ είναι εσωτερική (διμελής σχέση). Περνάμε αμέσως και λέμε τις ιδιότητες που έχει μία ομάδα με βάση την πράξη $ \star$ θεωρώντας δεδομένο ότι για δύο στοιχεία $ g_1,g_2 \in G$, ισχύει $ g_1\star g_2 \in G$. Παρατήρησέ το στα διάφορα βιβλία.

Για ένα αντιπαράδειγμα στην προσεταιριστική ιδιότητα πάρε το $ S$ να είναι το σύνολο των ακεραίων και όρισε $ x\star y \equal{} \minus{} x \minus{} y$. Τότε

$ x\star (y\star z) \equal{} \minus{} x \plus{} y \plus{} z$, και $ (x\star y)\star z \equal{} x \plus{} y \minus{} z$. Όπως φαίνεται από τις 2 τελευταίες σχέσεις, οι αρχικές συνθήκες για την πράξη $ \star$ ικανοποιούνται, ενώ αν απλά διαλέξουμε $ x\neq z$, τότε δεν ισχύει ο προσεταιρισμός.

Αλέξανδρος[/quote]
Ευχαριστω πολυ Αλεξανδρε
καλα που μαου το ειπες διοτι χωρις αυτη την παρατηρηση δν μπορουσα να λυσω την ασκηση.(Δεν εχω διαβασει σχεδον κα8ολου βασικη αλγεβρα,ακομη πρωτο ετος ειμαι :P )

Αγαπητε Δημητρη αυτο που αναφερεις το προσεξα πρην ασχοληθω με την ασκηση δηλ.
αν δεν ηταν κλειστο το$ S$ ως προς την * δηλαδη το $ x*y$ να μην ανηκει στο $ S$ τοτε πως οριστηκε το $ x*(x*y)$ που δινει η εκφωνηση .Με λιγα λογια θελω να πω οτι απ τα δεδομενα της εφωνησης καλυπτομαστε ως προς την κλειστοτητα....
Αλαξανδρε καλο το αντιπαραδειγμα σου


group theory
abstract algebra
vector
superior algebra
superior algebra unsolved

YES! I was wrong. I've made a calculation error.

Now it's solved.

But I still have another question:

How to prove that GL ( 3, F2) is simple???????

[quote="math beginner"]G = SL ( 2, F3)

HOw many Sylow 3-subgroups does G have??

By Sylow theorem, n3 = 1, 4, or 16. I'm pretty sure it's not 1 (As I found 2 of them at least). And somehow I think it's not 16 neither. But I don't know how to prove that it's not 16. Can someone help???????

Or am I wrong??[/quote]

If SL ( 2, F3) denotes the special linear group of a two-dimensional vector space over $ F_{3}$ (transformations with determinant 1), I think its order is $ 24$.
Thus the number of Sylow 3-subgroups is $ 1$ or $ 4$.
There are at least two : one with generator [1, 1; 0, 1] and one with generator [1, 0; 1, 1] (elementary transvections).
Thus the answer is $ 4$.

For the second question : GL ( 2, F3)   is of order $ 48$ and the least simple group with composite order is of order 60, thus GL ( 2, F3) is not simple.

Friendly,
P.

[quote="math beginner"]Thank you . But for the second quesiton,
it's GL (3, F2) instead of GL (2, F3)[/quote]

Sorry.
For a vector space $ V$ over $ F_{2}$, $ GL(V) \equal{} SL(V)$ and $ SZ(3, F_{2})$ has order $ 1$ (Rotman, theor. 8. 10, p. 222), thus $ PSL(3, F_{2})$, which is the factor group  $ SL(3, F_{2})/SZ(3, F_{2})$, is isomorphic to $ GL(3, F_{2})$.
But $ PSL(3, F_{2})$ is simple (Rotman, theor. 9.46, p. 279) so $ GL(3, F_{2})$ is simple.

Friendly,
P.

"Special center":
The matrixes of type $ M \equal{} a \cdot E_n$ and $ \mathrm{det}(M) \equal{} 1$.


Together my sister and I have $ \$1000$. If $ 5\%$ of my money is $ \$6$ more than $ 6\%$ of my sister's money, how many dollars does my sister have?

suppose we let s = the amount of money the sister has, and m = the money the "main" person has. we set up two equations.

$ s \plus{} m \equal{} \$1,000$
$ \frac{5}{100}m \equal{} \frac{6}{100}s \plus{} \$6$

in the second equation, multiply by $ 100$ on both sides to clear denominators.

$ 5m \equal{} 6s \plus{} \$600$

then isolate $ m$ by dividing by $ 5$ on both sides.

$ m \equal{} \frac{6s \plus{} \$600}{5}$

substitute it into the first equation to get

$ s \plus{} \frac{6s \plus{} \$600}{5} \equal{} \$1,000$

simplifying and rearranging gives

$ 11s \plus{} \$600 \equal{} \$5,000$

solving this equation gives

$ 11s \equal{} \$4,400$ from which we get $ s \equal{} \$400$


For a given positive integer $n$, find all pairs of integers $(a,b)$ such that

${a-n\over b-n}={a^n\over b^n}$

1. a=b give solution for all n. If n=1 have not other solutions.
2. $a\not=b,n>1.$
Let $x=\frac ab$ - rational number. Then $\frac{bx-n}{b-n}=x^n$. Therefore
(1) $b=n+\frac{n(x-1)}{x^n-x}, a=nx+\frac{n(x-1)}{x^{n-1}-1}.$
Let $x=\frac pq , (p,q)=1$. Because b is integer, we have $\frac{p(p^{n-1}-q^{n-1}}{p-q} =p(p^{n-2}+\dots+q^{n-2})|n$.
Because a is integer $q(p^{n-2}+\dots+q^{n-2})|n$ Therefore a and b are integer only if
$pq(p^{n-2}+\dots+q^{n-2})|n$.
2.1 n=2 all oher ($a\not=b$) solutions give by (1), when x=2, or x=1/2.
2.2. n>2 $pq(p^{n-2}+\dots+q^{n-2})|n \Longrightarrow p=q=1.$ Therefore are not other solutions.

What if $p<0$ or $q<0$? The problem asks for integer solutions, not positive integer solutions.


cool prob

[hide]
take natural log of both sides

$\frac{x}{2}\ln \big((x^{2}-8x+9)-(x^{2}-8x+7)\big) = \Big(1+\frac{x}{4}\Big) \ln 2$

$\frac{x}{2}= 1+\frac{x}{4}$

$x = 4$

[/hide]

[quote="Kurt Gödel"][quote="gotztahbeazn"][hide]
take natural log of both sides
$\frac{x}{2}\ln \big((x^{2}-8x+9)-(x^{2}-8x+7)\big) = \Big(1+\frac{x}{4}\Big) \ln 2$
[/hide][/quote]

Huh? How did you get thát?  :|[/quote]

Don't get excited, his solution got on a wrong foot immediately, because he took $\ln(ab)$ instead of $\ln(a+b)$.

Ah, okay. I didn't want to sound excited though  :lol:

Do you have a nice solution, Farenhajt?

I do, but shall we give others some more time? Here's a hint though.

[hide="A bit of a spoiler"]Substitute to get a two-equations-with-three-variables system[/hide]

[hide="Oh right, you were the one that posted the original problem... so its not taboo to post solutions..."]
\begin{eqnarray*}(\sqrt{x^{2}-8x+9}+\sqrt{x^{2}-8x+7})^{x/2}+(\sqrt{x^{2}-8x+9}-\sqrt{x^{2}-8x+7})^{x/2}= 2^{1+(x/4)}\\ (\sqrt{x^{2}-8x+9}+\sqrt{x^{2}-8x+7})^{x}+(\sqrt{x^{2}-8x+9}-\sqrt{x^{2}-8x+7})^{x}+2(2^{x/2}) = 2^{2(1+(x/4))}\\ (\sqrt{x^{2}-8x+9}+\sqrt{x^{2}-8x+7})^{x}+(\sqrt{x^{2}-8x+9}-\sqrt{x^{2}-8x+7})^{x}=2^{1+(x/2)}\end{eqnarray*}
Hmm... $A=\sqrt{x^{2}-8x+9}+\sqrt{x^{2}-8x+7}, B=\sqrt{x^{2}-8x+9}-\sqrt{x^{2}-8x+7}$. $AB=2$
$A^{x}+B^{x}=2(AB)^{x/2}\implies^{\text{div.}{(AB)^{x}}}\implies \frac{1}{A^{x}}+\frac{1}{B^{x}}=2(AB)^{-x/2}\implies \left(\frac{A}{B}\right)^{x}+\left(\frac{B}{A}\right)^{x}+2 = 4 \implies Y+\frac{1}{Y}= 2 \implies \boxed{Y=1}\ \ \ \text{By AM-GM, or simply mult. thru by Y}$ ... now we say oh wow
Now before the mega step we should weed out bad cases like:
[b]1) x=0[/b] This is in fact a solution after checking!
[b]2) A=0[/b] This can never happen because \begin{eqnarray*}\sqrt{x^{2}-8x+9}=\sqrt{(x-9)(x+1)}=0 \\ \implies \sqrt{x^{2}-8x+7}=\sqrt{(x-1)(x-7)}\not =0 \end{eqnarray*}
[b]3) B=0[/b] Or in other words, $x^{2}-8x+9=x^{2}-8x+7$ which is never true.
After removing these cases, we are fine to say:
$Y=1 \implies A=B \implies \sqrt{x^{2}-8x+7}=0 \implies \boxed{x\in \{1, 7\}}$
So if I did this right, $x\in \{0, 1 , 7\}$ is the solution set.
[/hide]

[hide="Yup - that is the idea"]Substitute $A$ and $B$ and then find $A^{x}, B^{x}$ from $A^{x}+B^{x}=2^{1+x/2}, A^{x}B^{x}=2^{x}$[/hide]

wow thats so much more complicated than my... seemingly wrong solution

but wait... i thought

$\ln a+\ln b = \ln ab$?

in which case i had:

$\ln \big(\sqrt{x^{2}-8x+9}+\sqrt{x^{2}-8x+7}\big)+\ln \big( \sqrt{x^{2}-8x+9}-\sqrt{x^{2}-8x+7}\big)$

so this is then...

$\ln \Big[\big(\sqrt{x^{2}-8x+9}+\sqrt{x^{2}-8x+7}\big) \cdot \big( \sqrt{x^{2}-8x+9}-\sqrt{x^{2}-8x+7}\big)\Big]$

which is...
$\ln \big( (x^{2}-8x+9)-(x^{2}-8x+7) \big) = \ln 2$

can someone tell me where i went wrong?

oh oops one of my friends told me what i did wrong... hahah wow  :blush:

yea thanks for the help

[quote="srulikbd"][quote="Altheman"]blah...

$y=x^{2}-8x+8$, and $4x\rightarrow x$

$(\sqrt{y+1}+\sqrt{y-1})^{2x}+(\sqrt{y+1}-\sqrt{y-1})^{2x}=2\cdot 2^{x}$ and $\frac{1}{ (\sqrt{y+1}+\sqrt{y-1})^{2x}}+\frac{1}{ (\sqrt{y+1}-\sqrt{y-1})^{2x}}=\frac{2}{2^{x}}$

$\frac{ (\sqrt{y+1}+\sqrt{y-1})^{2x}}{ (\sqrt{y+1}-\sqrt{y-1})^{2x}}+\frac{ (\sqrt{y+1}-\sqrt{y-1})^{2x}}{ (\sqrt{y+1}+\sqrt{y-1})^{2x}}=2$

blah blah z+1/z=2 implies z=1

$\frac{ (\sqrt{y+1}+\sqrt{y-1})^{2x}}{ (\sqrt{y+1}-\sqrt{y-1})^{2x}}=1$

$(\sqrt{y+1}+\sqrt{y-1})=\pm (\sqrt{y+1}-\sqrt{y-1})$

$y=\pm 1$

then solve and check to make sure everything works out...[/quote]
nice solution, but how did you reach from
 $\frac{1}{ (\sqrt{y+1}+\sqrt{y-1})^{2x}}+\frac{1}{ (\sqrt{y+1}-\sqrt{y-1})^{2x}}=\frac{2}{2^{x}}$
to
$\frac{ (\sqrt{y+1}+\sqrt{y-1})^{2x}}{ (\sqrt{y+1}-\sqrt{y-1})^{2x}}+\frac{ (\sqrt{y+1}-\sqrt{y-1})^{2x}}{ (\sqrt{y+1}+\sqrt{y-1})^{2x}}=2$
thanks[/quote]

Multiply the first two equations. Then you get $(a+b)({1\over a}+{1\over b})=4\iff{a\over b}+{b\over a}+2=4\iff{a\over b}+{b\over a}=2$.

But in the end he lost a possibility: $x=0$ (since $t^{2x}=1\iff t=\pm 1\lor x=0$)


First find $g'(x)$ with the product rule of the differentiation, then you can find the measure of the tangent:$g'(2).$
Secondly confirm the tangent point $(2,\ g(2)).$ Now you can find the tangent line.

[quote="Kle"]Let f be a differentiable function with f(2)=3 and f'(2)=-5, and let g be the function defined by g(x)=xf(x). what is the equation of the tangent line to the graph of g at the point where x=2.[/quote]First, \[ g(2)=2\cdot f(2)=2\cdot(3)=6 \]Also, \[ g'(x)=xf'(x)+f(x), \] from the Product Rule. So then \[ g'(2)=2\cdot-5+3=-7. \] Hence, the tangent line is \[ y=-7x+C. \] But it must pass through $(2, 6)$. Thus, the final equation is \[ y=-7x+20. \]

Your answer is correct, towersfreak2006.

i would use the same method shown above except i would have left the answer in point-slpoe form. (which is acceptable for AP exams)

$y-6=-7(x-2)$ this way is basically fool proof since you wont chance making a basic arithmatic mistake.


probability
geometry
rectangle
AMC
AIME
function
number theory

since i have been getting on aops lately dynamo seems to be not on most of teh times, and hurdy,
O__o  which is very weird and i know what weird is because i am weird888

I try to not waste time.

Sigh, AL forums is all spam now......ok attempt to make more doable problems and despamify:
1.what is probability of rolling a multiple of 3 on a on a roll of four 7 sided dice
2.factor:
a.$ x^3 \plus{} 1$.
b. 1001
c. 999999
d. 10101
e. $ x^6 \minus{} 1$
f. $ x^4 \plus{} x^2 \plus{} 1$
3. compute with a calculator (or, if you are like me, with paper): $ \frac{1}{89}$. What do you notice? Can you prove the pattern?
4.factor:89999
5. Let $ \phi \equal{} \frac{1 \plus{} \sqrt 5}{2}$. compute the following powers by hand(unless you can somehow do this with a calculator, which may be very possible since i know nothing about calculators):$ \phi^2, \phi^3, \phi^4, \phi^5, \phi^6, \phi^7$ [b]in terms of $ \phi$.[/b]
[hide="1/3 of the answer"] $ \phi^2 \equal{} \phi \plus{} 1$. $ \phi^3 \equal{} 2\phi \plus{} 1.$ [/hide]

What do you notice?

now compute: $ \frac{\phi^n \minus{} (\frac{\minus{}1}{\phi})}{\sqrt 5}$. eh... ok use the calculator.

darn, those problems got somewhat on the hard side...I can't seem to pick easier problems that I still find interesting....

I challenge the ppl in AL to try to prove some of the patterns.

Sorry... I've been having a life lately, and we had district and academic team tourn.s
IM ACTUALLY GOING TO DO MATH!

2.
a. $ (x\plus{}1)$ $ (x^{2}\minus{}x\plus{}1)$
b-d How do you factor just a number? Prime factorization?
e. $ (x\plus{}1)$ $ (x\minus{}1)$ $ (x^{4}\plus{}x^{2}\plus{}1)$
f. prime? unless you do something with i that im forgetting

3. .011235 (paper)
...
FIBBINOCCI!!!!
I dont know why this is

Is this the golden rectangle or whatever?
lemme guesss...

to the fourth is 3 times it plus one
to the fifth is 4 times it plus one
to the sixth is five times it plus one.....

f. get owned dynamo!!!!!
i can do it!!!!
ok here i go
(x^2+1)^2-x^2
(x^2-x+1)(x^2+x+1)

[quote="dynamo729"]Sorry... I've been having a life lately, and we had district and academic team tourn.s
IM ACTUALLY GOING TO DO MATH!

2.
a. $ (x \plus{} 1)$ $ (x^{2} \minus{} x \plus{} 1)$
b-d How do you factor just a number? Prime factorization?
e. $ (x \plus{} 1)$ $ (x \minus{} 1)$ $ (x^{4} \plus{} x^{2} \plus{} 1)$
f. prime? unless you do something with i that im forgetting

3. .011235 (paper)
...
FIBBINOCCI!!!!
I dont know why this is

Is this the golden rectangle or whatever?
lemme guesss...

to the fourth is 3 times it plus one
to the fifth is 4 times it plus one
to the sixth is five times it plus one.....[/quote]

comments: ok woops i did not see this response til now, for 2 b-d, yea i meant prime factorization. For 2f, its not "prime" whatever that means in algebra. The expression can be factored as weird did it, and I was aiming for you to note that $ x^4 \plus{} x^2 \plus{} 1$ is a geometric sequence and can thus be expressed as $ \frac{x^6 \minus{} 1}{ x^2 \minus{} 1}$ , and hoped you would see that HEY I JUST FACTORED THE NUMERATOR and it can be rewritten as $ \frac{(x\plus{}1)(x\minus{}1)(x^2 \plus{} x \plus{} 1)(x^2 \minus{} x \minus{} 1)}{(x\plus{}1)(x\minus{}1)} \equal{} (x^2 \plus{} x \plus{} 1)(x^2 \minus{} x \plus{} 1)$. Of course, it is probably much easier to see weird's equally good, if not better, way of doing it

ok so i got got bored today:

for 3, your right its fibonacci and it doesnt stop at 3. Add the following (easiest done as decimals...) = $ \frac{1}{100} \plus{} \frac{1}{1000} \plus{} \frac{2}{10^4} \plus{} \frac{3}{10^5} \plus{} \frac{5}{10^6} \plus{} \frac{8}{10^7} \plus{} \frac{13}{10^8} \plus{} \frac{21}{10^9} \plus{} \cdots$. Then [hide="dont look til youve done that for like 10 decimals places"] compute $ \frac{1}{89}$ to 10 decimal places. Then be amazed :). Then [hide="why it works"] try to find a closed form expression for the infinite series that i wrote in fraction form above [hide="hint for doing that"] use a problem solving strategy identical to the way you derive a formula for an infinte geo series...on a side note, the fibonacci sequence, like every other linear recurrence (i.e. each $ a_n$ in the recursive sequence can be expressed as some linear combination of previous terms ... ie $ 3a_{n\minus{}1} \plus{} 4a_{n \minus{} 2} \minus{} 99a_{n \minus{} 5}$) gets closer and closer to becoming a geometric sequence.

In fact, it is interesting to compute 1/1, 2/1, 3/2, 5/3, 8/5 13/8 21/8 etc as decimals (and remember that $ \phi$ is about 1.618

ok i got a little off topic[/hide][/hide][/hide]

so far no answer for number 4...

and dynamo...unfortunately, the last 3 lines for number 5 you wrote were all wrong.

weird, is that your final finished answer of the complete prime factorization?

I encourage anyone who doesn't already know what's going on in the unanswered problems above to do some exploration, or at least answer the questions

btw GL on aime to you too weird. I am going to fail because I am sick :( and we have to take in Spring Break.

[quote="weird888"]#4)  89999

answer is (900000-1)=
              (300+1)(300-1)=
              (301)(299)
^^ thank you very much[/quote]

okay, through trial and error, i get (23)(13)(7)(43)
NOW i have the right answer!!!!

good job, :)

as a side note (to dynamo), guessing the powers of $ \phi$ to be $ \phi \plus{} 1, 2\phi \plus{}1, 3\phi \plus{} 1, 4\phi \plus{} 1$ is not very wise, since I think that an exponential function cannot be linear unless the base is like 1 or 0.


the Barclays English Premier League starts 2morrow.

Will Manchester United retain the crown to get 4 consecutive title without Ronaldo or Tevez to beat Liverpool's 18th title?

Will Chelsea win? They are the favorites 4 most ppl.

Will Liverpool win to get the 19th title? Can Gerrard and Torres stay fit?

Can Arsenal win the league eventhough they lost 2 important players?

Can Manchester City and the money win the league with the likes of Barry, Adebayor, Tevez, Santa Cruz?

P.S I hope Manchester United win

City is a failure, getting good players doesn't mean much. They need to play as a team, and it's very difficult (in any team) to find a strategy that will take the best out of players, and M.C. isn't going to do much.

Arsenal was always a underdog in the 4 teams clash, and they haven't got much to change, so no Arsenal.

Liverpool, Chelsea play good football and they are in the close tie with M.U (I cheer for Liverpool!)

M.U. isn't going to lose much with C. Ronaldo gone, he wasn't playing his best last season, and yet M.U. was on the top.

Hope: Liverpool
Prediction: M.U. or Chelsea or Liverpool

Liverpool lost Xabi Alonso, their best passer. and they replaced him with a injury prone midfielder not proven in PL.

Chelsea didn't get weakend, so thats a plus. they're rlly the favorites now.

Arsenal, no offense but i think they will fight for the top four spot. they lack depth, and lost a striker who can score a lot and a good defender.

Manchester United will play as a TEAM now, not a 1 man show. i expect rooney to score 25+ goals, and berbatov to step up his game. luis antonio valencia will CREATE goals, not score. nani and anderson needs to step up. im happy with the squad we have, but i wouldnt a signing or two, espically the Center Midfield. I hope we sign Van Der Vaart (heard he will cost 5 million or less), sneijder, or silva. they rest of the money we have can go to the debt we have.

City, well they have no chemistry and TOO many strikers.

GLORY GLORY MAN UTD

EDIT: wouldn't mind*

btw, here's the last year's standing:

1. Man Utd WOOT
2. Liverpool
3. Chelsea.
4. Arsenal

i added man city to the poll because the have lots of money and bought some nice players. though theyre not going to do much imo

Although I h-a-t-e Man Utd after the 1999 CL finals, I'll have to be objective and say that the odds are on their side. Chelsea and Liverpool do have the potential to challenge Man Utd on their way to the title, but Man Utd seems more compact and they seem to easily maintain good form in the Premiership, which one can't say for Chelsea, for instance. Arsenal and City are Europa League material this year, but nowhere near the top 3 spots, IMHO. City could make something out of their investments in a year or two, but I doubt even that - this 'Galacticos' tactic with buying the cremé players and making such a heterogeneous team can easily backfire, just like it backfired on Real Madrid in a number of occasions.

That's my 2c.

Everton 1 - 6 Arsenal
[/sarcasm] Yeah, Arsenal are definitely Europa league material. [sarcasm]

Why does everyone write off Arsenal even before the season has well and truly begun ?

yeah, but everton were terrible. and its the first real competitive match since months, anything can happen. their lack of depth will show in January.

omg omg omg omg omg omg omg omg omg omg we lost 1-0 to burnley omg omg omg omg omg omg omg omg omg omg

we need a cleanup.
get out!:
Carrick=cannot run. his passes were terrible 2day. he was invisible

Anderson=cannot shoot, cannot cross, does nothing. i know he's young, but he had enough time to prove himself.

Ji-Sung Park=cannot control a pass, cannot pass, cannot shoot. ship him off or something

Wes Brown=below average at best.

Michael Owen=he was for free but missed too many chances. he better score 5-10 goals in premier league or ship him off or someting

plz get:
van der vaart/silva/sneijder=creativity in midfield needed. goals needed too.

gojko kacar=an unnoticed talent. he is a defensive midfielder. would cost about 10 million. hargreaves is too injury prone

a striker. prefer sergio aguero (even though he is cup-tied, we have no chance of winning champions league, so why not) would be amazing. aguero=goals or great saves from the keeper.

we need to practice on our penalty kicks (pk) too. anderson should take the pk, so he can get confident boost and has scored every pk so far. i have a feeling this is going to be a rebuilding year.

Owen was a risky investment. It is true that Michael was world-class 8-10 years ago, but have in mind that his speed was a great advantage on his side. Today, he's not fast as he's used to be. But I gues Sir Alex and co. know what they're doing.

It's not a surprise that you're criticizing Carrick and Brown, I always wondered how did they end up in the first squad at all (but keep in mind that I always watched Man Utd on the critical side, the last player in Man Utd I liked was Keane  :lol:  ).

As a Bayern fan, I know Hargreaves' glass legs  :rotfl: I'm happy to see someone's been watching Bundesliga (you've noticed Gojko Kačar, well done  :wink:  )

Man Utd will soon have to face the change of generations: Giggs, Scholes, van der Sar will leave. Giggs and Scholes are already spending their time on the substitute bench, but who is going to replace Edwin? I'm afraid of the Bosnich and co. scenario after the Schmeichel's leave.

i hope we buy gojko kacar before his price tag skyrockets. he would be awesome for us.

a creative midfielder is a must. rooney has to tracke back wayyyy to much. nobody is feeding our strikers. our midfield is simply terrible. except for evra and rooney, our players are scared of taking on players 1 on 1. if we cant beat a team that will most likely to get relegated (meaning they will be in the bottom 3 out of 20), how will we be first???????????? ship some players that are not good enough to play for us and buy some players that are or will be.

hsiljak, bundesliga was awesome last year. i usually dont watch bundesliga cause bayern usually gets the title. bayern also gets best players from their rivals for some reason (olic, mario gomez to name a couple).

Arsenal 2:Celtic 0
it seems that Arsenal has a Great starting with its young players like 2 season ago.
if its good players doesn't get Injured it will be Champion. :lol:

_ Arsenal fan

[quote="hjoon0125"]
hsiljak, bundesliga was awesome last year. i usually dont watch bundesliga cause bayern usually gets the title. bayern also gets best players from their rivals for some reason (olic, mario gomez to name a couple).[/quote]

This Bayern tactic on the transfer market really annoys me. It is fairly obvious they're making their opponents weaker that way, but they're not making their team much stronger - none of the best players in recent FCB history didn't come from a title-contending Bundesliga team, but they did manage to ruin a few perfectly good players by taking them out of a team in which they played in a superb form, and then leaving them to rot on the sub bench.

Wolfsburg keeps their victory run in this season, they sure look serious - and they have almost the same team as the last year.

BTW, if people from Man Utd management ever read this  :lol: or get the idea of their own to buy Kačar, it would be an interesting Serbian triple in United: Vidić, Kačar and Ljajić  :lol:

Oh, I managed to forgot Zoran (he was in the package with Ljajić, right?). That's probably because I'm fan of the bigger and better team from Belgrade  :P

Back to the topic: Seems like Man Utd is serious about Miranda capture... and the van der Vaart transfer should go through - but you never know before they put the pen on paper.

OMG, what a match. United's victory was fully deserved  :wink:

[Bayern is on a great run, I hope they'll keep it up]

That was a really good match... But United always (I saw many matches) wins in the last minutes. Concentration, luck or what? But United is strong at the end

Chelsea - MU - MC - Liverpool are in front

WHAT A MATCH. ALL CAPS NEEDED.

anybody who says there was 2 much extra time is a hater. bellamy celebrated for 1.5 minutes, and the substitution took a minute, so it was fully deserved.

bugi, chelsea won 3 matches this season thanks to injury time goals.

True... for Chelsea I mean. But MU won soooo many such matches last year, this year... Their concentration and motivation in the last minutes when they are in a lose or draw situation is for respect...

as long as manutd win im happy.

im rlly disappointed with foster though. he's making too many silly mistakes. the first goal is his fault.

[quote=hsiljak]Man Utd will soon have to face the change of generations: Giggs, Scholes, van der Sar will leave. Giggs and Scholes are already spending their time on the substitute bench, but who is going to replace Edwin? I'm afraid of the Bosnich and co. scenario after the Schmeichel's leave.[/quote]

Foster looks just like Mark Bosnich used to be.

(Situation with GKs in Bayern isn't bright either: Rensing lost his place in the first team to Butt, and it doesn't look like he'll get back on goal soon - even if he accepts Serbian NT offer, van Gaal still won't let him play  :lol: : but Akinfeev is on the wishlist...)

chelsea 3-0 tottenham, thats game over. looks like my prediction was rite

[quote="Bugi"]Rensing will play for Serbia?[/quote]

Well, there's a chance: allegedly, Antić showed interest for Rensing, and his mother is Serbian.

ah, michael owen saves the day, whew! im rlly glad manutd didnt sign tevez, he runs like a headless chicken and scores only 5 goals a season. thats it. even i could do that.

manutd 2-2 sunderland. gah rlly a disappointing performance.

Positives:
-what a goal by berbatov!!!!!
-we got a point.
-we showed our character. we play till the final whistle, and got ourselves a goal

Negatives:
-i thought scholes was playing 4 sunderland, atleast 50% of his passes were given to a sunderland player.
-except 4 the last 20 minutes, nani was way off
-welbeck was terrible.
-another poor performance by foster. i had it with him. i would be surprised if he doesnt make a mistake which costs us a game.
-rooney was off.

but i think they had a bad day. all teams have a bad day sometimes

manuntd FTW!!!!!!!!!

arsenal second place FTW!!!!!!

arsenal will not be 2nd. they might look good now, but wait till u get 2 winter when injuries mount and theres a hectic schedules.

chelsea 2-0 liverpool

liverpool already lost more games this season so far (8 games played) than last season (38 games played)


Given a quadrilateral ABCD, we trisect AB,CD,BC, DA by the points P_1, P_2, Q_1, Q_2, R_1 , R_2 , S_1 , S_2 . Let N,M,L,K be the intersection points of Q_1 P_1 and S_2 R_2, Q_2 P_2 and S_2 R_2, Q_2 P_2 and S_1 R_1, Q_1 P_1 and S_1 R_1. Prove that KLMN/ABCD = 1/9.

[quote="jin"](1)$\frac{P_1P_2Q_2Q_1}{ABCD}=\frac{1}{3}$;[/quote]

How to prove it?
Thanks.


See Mewto55555's blog.  I personally doubt it.  Let's see what y'all have to say!

I am. See the link in my sig by the way.

And how long does really leaving last for?

come on mew mew. we need you. the aops community needs every single member. each and every one is important.

NOOOOOOOOOOOOOOOOOOOOOOOOoo

MEW CANT LEAVE

hmm doing this for attention or are you really leaving?

[quote="james4l"]All "legitimate" AoPSers represent important constituencies to the AoPS community. However if someone wants to quit, just be it. Anyways, he has also stated explicitly that he will stay in the Games Forum.[/quote]
I agree.
Bye Mew. We will miss you :).

[quote="james4l"]All "legitimate" AoPSers represent important constituencies to the AoPS community. However if someone wants to quit, just be it. Anyways, he has also stated explicitly that he will stay in the Games Forum.[/quote]

Hmmm... I remember once that I quit in May.  I came back in June, however...

NOOO MEWTO your one of our best posters on the forums. WE NEED YOU. and if you are leaving you could give away all your points to other kids who need a boost to be more active on FTW. Showing that you are willing to go down to 1200 might convince me more that you will leave. Anyway don't! its going to be miserable to post here without someone who seems to hate me.. lol anyway we need to keep as many of our old posters untill the new generation springs up.

No more free rating on FTW anymore mahtblitz/

[quote="pythag011"]If it is true, bye mew.  :(  :(  :([/quote]

that rhymes. anyways, i dont think it'll last. aops and ftw and alcumus have become such large parts of all of our lives, that i don't think anyone can simply quit. i can see him reducing the number of hours he spends on aops but i dont think hes actually gonna [i]quit[/i]

I think that if he says he's going to quit, then he will quit. He might come back though.

I think he's gonna come back as another user and be all good, etc.

I just saw him on FTW, I knew he wouldn't leave.

[quote="Mewto55555"]Clearly, 23 people were wrong.[/quote]

AHA MEW!! YOU ARE NOT LEAVING AFTER ALL!!!

[/img]

Actually, Mewto55555 [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=252005]is leaving[/url]...

wheres the kinda option?

anyway...

I voted YES because he is KINDA leaving. It says in his blog that he will leave AoPS except the games forum.

He is gone now. Since he can't post or pm anymore.

you cannot judge it yet you must wait and see who knows what will happen

This topic was a long time ago. Please do not revive old threads.

Okay. One last try. He posts a lot on his blog...

I would doubt he would have performed the way he did at MC if he had quit. After 6 months, he still hasn't quit.

[quote="EggyLv.999"]he is playing/modding mafia... he is still here[/quote]

He is also playing Epic Space Wars (remade)

Lock this, Mewto55555 is clearly not leaving.

since when has a Fun Factory thread been locked...?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=258691

There have been older posts such as the Box Game which I'm too lazy to dig up.

[quote="ytrewq"]Please don't click the submit button twice either.[/quote]

And clicking the sumbit button twice is when your computer lags a lot, and you think the message hasn't submitted yet. So you click again. It wasn't jjx's fault (hopefully)


Let be $ a,b,c\in \mathbb{R}$ such that $ a^2\plus{}b^2\plus{}c^2\equal{}3$ . Prove that :

$ |a|\plus{}|b|\plus{}|c|\minus{}abc\le 4$

How can we sort out these problems?  :P
Firstly, by observing that
$ |a| \plus{} |b|\plus{}|c| \le 3$
Furthermore, we have
$ |abc| \le 1$ which implies to the following critical point
$ abc \plus{} 1 \ge 0$
We end our proof!

[quote="great math"]How can we sort out these problems?  :P
Firstly, by observing that
$ |a| \plus{} |b| \plus{} |c| \le 3$
Furthermore, we have
$ |abc| \le 1$ which implies to the following critical point
$ abc \plus{} 1 \ge 0$
We end our proof![/quote]
What about the following?  Find max of $ P \equal{} a\plus{}b\plus{}c\minus{}abc$ ,great math :wink:

If I am not wrong, the maximum value is 2 when $ a\equal{}1, b\equal{}1, c\equal{}\minus{}1$

[quote="great math"]If I am not wrong, the maximum value is 2 when $ a \equal{} 1, b \equal{} 1, c \equal{} \minus{} 1$[/quote]
Please post solution

[quote="arqady"][quote="great math"]Math is a holy subject[/quote]
I think, also Math is a holy object.  :wink:[/quote]
But the AGM inequality stated in the signature........
"HOLY OBJECT??"
Please let's talk ON-TOPIC....... we can exchange words through pm. :oops:


limit
trigonometry
algebra unsolved
algebra


[quote="mysteryfigure"]:) :)

Waiting for your reply.[/quote]
You can find it in APMO 1999 or 2000.!Good luck! ;)
APMO in [url=http://camel.math.ca/CMS/Olympiads/Competitions/apmo.html]camel.math.ca/CMS/Olympiads/Competitions/apmo.html[/url]

thank you.
but i can't find the server so can't catch a glance at it. :(

take a look here:
http://www.kalva.demon.co.uk/usa/usoln/usol815.html


Could someone urgently help me with the following  two questions please? Thanks

1) Find the maximum area of an isosceles triangle whose equal sides are 10cm in length. Use half the
third side of the triangle as the variable in terms of which to express the area of the triangle.

2) A drug is injected into a patient at a rate given by r(t) = a t (e^−bt) ml/sec, where t is in seconds since the injection
started, 0 <= t <= 5 and a and b are constants. The maximum rate of 0.3 ml/sec occurs half a second after the
injection starts. Find the values of a and b.

the area can be expressed as $\frac{1}{2}absinC$, so if a and b are each 10, the area is

$\frac{1}{2}(10)(10)sinC=50sinC$

this is maximized when sinC is maximized.  so, C=90 degrees, since it has a sine of 1.  therefore, the area is 50.  (i'm not exactly sure what you mean by using half the other side to express the area...)

 :)


group theory
abstract algebra
inequalities
Galois Theory
number theory
relatively prime
superior algebra

How to prove the following:

If $ G$ is a solvable group of order $ mn$, where $ m, n$ are relatively prime natural numbers, then $ G$ has a subgroup of order $ m$.

Here is a proof not using [url=http://en.wikipedia.org/wiki/Schur%E2%80%93Zassenhaus_theorem]Schur-Zassenhaus[/url].

Let $ G$ be a minimal counterexample.
Let $ N\unlhd G$ be maximal with $ |N|$ coprime to $ m$, so $ |N|$ divides $ n$.
Then clearly $ G\not\equal{}N$ and $ G/N$ has some minimal normal subgroup $ L/N\not\equal{}1$.
But $ G/N$ is solvable, so $ L/N$ is an elementary abelian $ p$-group for some prime $ p$.
Then $ L\equal{}NP$ for some Sylow $ p$-subgroup $ P$ of $ L$, and by maximal choice of $ N$
we must have $ p|m$, and then $ |P|$ also divides $ m$.
Then using the [url=http://en.wikipedia.org/wiki/Frattini_argument]Frattini argument[/url] we get $ G\equal{}LN_G(P)\equal{}NPN_G(P)\equal{}NN_G(P)$
and therefore $ |G: N_G(P)|\equal{}|N: N\cap N_G(P)|$ divides $ n$.
But this gives $ |G|>|N_G(P)/P|\equal{}{|G|\over |P|\cdot |G: N_G(P)|}\equal{}{m\over |P|}\cdot{n\over |G: N_G(P)|}$
and by minimal choice of $ G$ we find some subgroup $ H/P$ of $ N_G(P)/P$ of order $ {m\over |P|}$.
This gives $ H\le G$ of order $ m$.
Contradiction!

[quote="olorin"]Here is a proof not using so $ L/N$ is an elementary abelian $ p$-group for some prime $ p$.[/quote]
could you explain this?

For any minimal normal solvable subgroup $ N\not\equal{}1$ of a finite group $ G$
we have the commutator $ N'$ also normal in $ G$ with $ N'<N$. So $ N'\equal{}1$ and $ N$ is abelian.
Then for any prime $ p| |N|$ we have $ 1\not\equal{}\Omega_p(N): \equal{}\{x\in N: x^p\equal{}1\}$ also a normal subgroup of $ G$ with $ \Omega_p(N)\le N$.
So $ N\equal{}\Omega_p(N)$ is an elementary abelian $ p$-group.


inequalities
vector
calculus
function
real analysis
real analysis unsolved

This was a homework problem that I did not complete in time to turn it in, but I still would like to know how to get the right inequality, as it has frusterated me to the point of giving up. The idea I tried to work with was that S was simply that portion of the plane below the line y(x)=x. The distance from an arbitrary point p= (x,y) of S to the line y(x)=x is  (x-y)/sqrt(2).

Thus I considered the open ball of center p and radius (x-y)/sqrt(2), and tried to show that for any point q= (z,w) in this open ball, we have z>w.
Drawing this on paper makes its truth obvious, but I was unable to get from the inequality d(p,q) < (x-y)/sqrt(2) to the inequality I want, z>w.
Whats going on here? Thanks.

of consider $ f(x,y)\equal{}x\minus{}y$ and notice that $ S\equal{}f^{\minus{}1}( (0;\plus{}\infty) )$

Thanks for the swift replies. I'm still thinking about what alekk said, but as for using different metrics, in the problem it is assumed that it is the normal euclidean metric. I should have said so before.

Yes, i know, but it really doesn't matter, because the three metric are equivalent, so they induce the same topology. Which means that even though the balls aren't the same, the open sets are.

Or a different way to get at the same thing rmfgcampos is saying:

You don't have to have a "perfectly tight" result. In other words, you don't have to find the [i]largest[/i] ball centered at $ (x,y)$ that fits into $ S,$ you only have to find [i]some[/i] such ball.

So let your radius $ r$ be $ \frac {x \minus{} y}2.$ Or $ \frac {x \minus{} y}{10^{10}}.$ (I had some teachers who were fond of overkill.) Say we went with $ r \equal{} \frac {x \minus{} y}2.$ Then if $ d((z,w),(x,y)) < r,$ it is clear that $ w \minus{} y < \frac {x \minus{} y}2$ so that $ w < \frac {x \plus{} y}2$ and $ z \minus{} x > \minus{} \frac {x \minus{} y}2$ so that $ z > \frac {x\plus{}y}2.$ From that, you clearly have $ z > w,$ which is all you need.

Now here's the unnecessarily tight argument. We don't need it, but it's a nice bit of linear algebra. The hint: when thinking of Euclidean norms, also think about dot products. And the Cauchy-Schwarz inequality.

Claim: if $ x > y$ and $ d((z,w),(x,y)) < \frac {x \minus{} y}{\sqrt {2}},$ then $ z > w.$

Proof:

Start by writing $ z \minus{} w \equal{} (z \minus{} x) \minus{} (w \minus{} y) \plus{} (x \minus{} y).$

Now think of $ (z \minus{} x) \minus{} (w \minus{} y)$ as the inner product $ \langle z \minus{} x,w \minus{} y\rangle\cdot\langle 1, \minus{} 1\rangle.$ By Cauchy-Schwarz:

$ |(z \minus{} x) \minus{} (w \minus{} y)|\le \sqrt {(z \minus{} x)^2 \plus{} (w \minus{} y)^2}\cdot\sqrt {1 \plus{} 1}$

$ < \frac {x \minus{} y}{\sqrt {2}}\cdot\sqrt {2} \equal{} x \minus{} y.$

Hence $ (z \minus{} x) \minus{} (w \minus{} y) > \minus{} (x \minus{} y)$ and thus $ (z \minus{} x) \minus{} (w \minus{} y) \plus{} (x \minus{} y) > 0.$

And we have proved that $ z \minus{} w > 0$ or $ z > w.$

Thanks kent, that was the continuation of my original idea that I was hoping for. The other responses are perfectly fine but I would have been uneasy if no one could show why the inequality must follow from the radius I proposed.

If you had been working with, say, $ x>2y$ instead of $ x>y,$ you'd just have to use the vector $ \langle 1,\minus{}2\rangle$ where I used $ \langle 1,\minus{}1\rangle$ in the above argument. This argument involves orthogonality, projections, and so forth.

[quote="alyosha"]I'm still thinking about what alekk said.[/quote]He uses the fact that if $ f$ is continuous, then $ f^{\minus{}1}(U)$ is open for every open $ U$. Here $ f(x,y)\equal{}x\minus{}y$ is obviously continuous, and the open interval $ ]0,\infty[$ is certainly open. Now, since $ x\minus{}y>0$ is equivalent with $ x>y$, we have $ S\equal{}f^{\minus{}1}(]0,\infty[)$.

Alekk's idea is probably the fastest, but at this point in the course (advanced calculus) we had not covered the preservation of openness by the inverse image of continuous functions---continuous functions aren't even mentioned until the next chapter; we are using Rosenlicht's "Introduction to Analysis".

Well, my advice to you would be "learn some topology". Calculus starts making more sense after that.

[quote="alyosha"]at this point in the course (advanced calculus) we had not covered the preservation of openness by the inverse image of continuous functions---continuous functions aren't even mentioned until the next chapter; we are using Rosenlicht's "Introduction to Analysis".[/quote]Ah, you will see it soon, at page 70, in the chapter where continuity is defined. Once you know that inverse images of continuous functions preserve openness, this is almost always the fastest way to go. Of course the estimate also works, and it is always instrutive to see multiple solutions.


let $ {x1,x2 \in R}$ be the two roots of equation $ {x^2\plus{}ax\plus{}2b\minus{}2\equal{} 0}$; and $ {x1<x2; x1\in(0,1);x2\in(1,2)}$; then find the range of $ {\frac{b\minus{}4}{a\minus{}1}}$.

[quote="CCMath1"]let $ {x1,x2 \in R}$ be the two roots of equation $ {x^2 \plus{} ax \plus{} 2b \minus{} 2 \equal{} 0}$; and $ {x1 < x2; x1\in(0,1);x2\in(1,2)}$; then find the range of $ {\frac {b \minus{} 4}{a \minus{} 1}}$.[/quote]

$ a\equal{}\minus{}x_1\minus{}x_2$ and $ b\equal{}1\plus{}\frac{x_1x_2}2$ and so $ {\frac {b \minus{} 4}{a \minus{} 1}}\equal{}\frac 12\frac{6\minus{}x_1x_2}{x_1\plus{}x_2\plus{}1}$ which is clearly a decreasing function of both $ x_1$ and $ x_2$

So minimum is achieved for $ x_1\to 1$ and $ x_2\to 2$ and is $ \frac 12$ and maximum is achieved $ x_1\to 0$ and $ x_2\to 1$ and is $ \frac 32$

So, range of $ {\frac {b \minus{} 4}{a \minus{} 1}}$ is $ (\frac 12,\frac 32)$


Εστω τριγωνο $ ABC$ και $ P$ τυχαιο σημειο του υψους $ AD$. Εστω οτι η $ BP$ τεμνει την $ AC$ στο $ K$ και η $ CP$ την $ AB$ στο $ L$. Φερω την $ LK$ που τεμνει την $ BC$ στο $ S$. Η μεσοκαθετος της $ BC$ τεμνει τις $ AS,AB,AC$ στα $ X,Y,Z$. ΝΔΟ $ X$ μεσο του $ YZ$.

Βάζω ένα "μικρό" hint :wink:
[hide]$ A(S,B,D,C) \equal{} 1$ (ή -1, όπως νομίζετε :P )[/hide]

Αν δεν κάνω λαθος, αρκεί αντί για μεσοκάθετο να πούμε "κάθετος στη BC"

Επίσης, αντί για το ύψος $ AD,$ μπορεί να θεωρηθεί οποιαδήποτε $ A$-σεβιανή, αρκεί η ευθεία των $ X,\ Y,\ Z,$ να είναι παράλληλη προς την $ AD.$

Κώστας Βήττας.


There is an object in the water.The depth of water is H.(Object is in in the depth H).The man is seeing the object at the depth h=1m.The distance between the eyes of the man and the object is s=5m.The man is in the height x=1.5m of water level.The refractive index of water is n=1.33 and of air n=1.Find the depth of water H.

P.S.please everyone,help me,we had this problem in National Olimpiad.I want to check my answer with yours.Please answer both numerical and equation.

The solution is rather long, and I don't have time now to write it all down... The expression you need is
\[h = H\frac{n^{2}\cos^{3}\theta}{(n^{2}-\sin^{2}\theta)^{3/2}}, \]
where $\theta$ is the angle between the line connecting the eye and the image and the vertical. This is for $n_{0}= 1$, I don't have a more general form...

i reached a 4 grade ecuation in H and with a program i found H=2 m(aprox)

what did u reach?
what grade is the problem?

The idea for solving this problem is fairly standard; you take two close rays coming from the object, refract them on the surface of the water, find the intersection of the new rays and there's the image. I used some unusual Taylor expansions to simplify things when I solved this one, and the calculation gets somewhat lengthy, but it's straightforward...

@[b]Namiqfizik:[/b] Could you please post the other problems from your national olympiad?

My answer to this problem was H=1.86m  .I think we can take H=2m approximately.


Prove that for each prime p the equation 2 p  + 3p = an has no integer solutions (a,n) with a, n > 1.


geometry
geometric transformation
homothety

Are you asking for the area of the square? or what...

If the square touches the side of length $ c$, and the altitude in the triangle to that side has length $ h$, then the side length of the square is $ ch/(c\plus{}h)$. You can derive this from the homothety centered at C taking one of the square's sides to the side of the triangle opposite C.

Similar triangles is another way, if I'm thinking about the right question.


calculus
integration
trigonometry
function
derivative
logarithms
analytic geometry

First of all, i will study the behavior of the function with the a.
$\int_{0}^{\pi/2}cos(x)|\frac{2sen(x)}{1+sen^{2}(x)}-a|$
if a<0 this is an easy calculation which gives:
$ln(2)+a$
If a>=1 easy case again:
$a-ln(2)$
The problem is with $0\leq a\leq 1$. Since $\frac{2sen(x)}{1+sen^{2}(x)}$ has a positive derivative in [0,pi/2]we will have:
${\int_{0}^{t}cos(x)(a-\frac{2sen(x)}{1+sen^{2}(x)})+\int_{t}^{\pi/2}cos(x)(\frac{2sen(x)}{1+sen^{2}(x)}-a})$ where t is $t=arcsen( \frac{1-\sqrt{1-a^{2}}}{a})$
I hope i didn´t messed up anything (again eheheh)

[quote="kunny"]For real number $a,$ find the minimum value of $\int_{0}^{\frac{\pi}{2}}\left|\frac{\sin 2x}{1+\sin^{2}x}-a\cos x\right| dx.$[/quote]
[u][b]My solution:[/b][/u]
$I=\int_{0}^{\frac{\pi}{2}}\left|\frac{\sin 2x}{1+\sin^{2}x}-a\cos x\right| dx$
$=\int_{0}^{\frac{\pi}{2}}\left ( \frac{\cos x}{1+\sin^2 x  \right )} \left|a\sin^2 x-2\sin x + a\right| dx \overset{t=\sin x}{\rightarrow}$
$=\int_{0}^{1}\frac{\left | at^2-2t+a \right |}{t^2+1}dt;$
1)  If $a=0$ it becomes $I=\int_{0}^{1}\frac{d(t^2+1)}{t^2+1} =\boxed{\ln 2}.$
2) If $a \neq 0, $we have: $\Delta' =1-a^2$
 if  $\Delta' \le 0\Rightarrow |a| \ge 1$.
2.1) If $ a \ge1 \Rightarrow at^2-2t+a \ge 0 \Rightarrow  $
$I=\int_{0}^{1} \frac{at^2-2t+a}{t^2+1}dt$
$=\int_{0}^{1} \left (a-\frac{2t}{t^2+1}  \right ) dt=\boxed{a-\ln 2}$ .
2.2) if $a \le -1 \Rightarrow at^2-2t+a<0\Rightarrow  $
$I=-\int_{0}^{1} \frac{at^2-2t+a}{t^2+1}dt$
$=-\int_{0}^{1} \left (a-\frac{2t}{t^2+1}  \right ) dt =\boxed{-(a-\ln 2)}$.
3) $\Delta '>0\Rightarrow t_{1,2}= \frac{1\pm \sqrt{1-a^2}
}{a} \overset{Viete}{\rightarrow}$
$t_1t_2=1\Rightarrow t_1;t_2<0$ or $t_1;t_2>0 $;
3.1) If $-1<a<0$ then $t_1,t_2 \notin [0;1]\Rightarrow $
$I=\left |\int_{0}^{1} \frac{at^2-2t+a}{t^2+1}dt  \right |
=\left |a-\ln2  \right |=\boxed{\ln2-a}.$
3.2) If $0<a<1$ then $t_1=\frac{1 + \sqrt{1-a^2}}{a} >1$ and
$t_2=\frac{1 - \sqrt{1-a^2}}{a} \in (0;1)$
$I=\left |\int_{0}^{\frac{1 - \sqrt{1-a^2}}{a}}
\frac{at^2-2t+a}{t^2+1}dt  \right |+\left |\int_{\frac{1 - \sqrt{1-a^2}}{a}}^{1}
\frac{at^2-2t+a}{t^2+1}dt  \right |$
$=\left |(at-\ln (t^2+1))|_{0}^{\frac{1 - \sqrt{1-a^2}}{a}}  \right |+\left |(at-\ln (t^2+1))|_{\frac{1 - \sqrt{1-a^2}}{a}}^{1}  \right |$
$=(\ln (t^2+1)-at)|_{0}^{\frac{1 - \sqrt{1-a^2}}{a}}
+(at-\ln (t^2+1))|_{\frac{1 - \sqrt{1-a^2}}{a}}^{1} $
$=\boxed{a+ \ln 2-4\ln a -2(1-\sqrt{1-a^2})+2\ln (1+a^2-\sqrt{1-a^2})}$.
[b]P/s:[/b]
Now, the problem of finding min of integration becomes simple. This problem must consider many cases, Especially, we must compare with $a$ value of $0, 1$, we will solve the problem.

That's　a Long work.

My solution is same as Blambira's.

We can make more shorter than the solution.

$I(a)=\int_0^{\frac{\pi}{2}} \left|\frac{\sin 2x}{1+\sin ^ 2 x}-a\cos x\right|dx$

$=\int_0^1 \left|\frac{2t}{1+t^2}-a\right|dt\ (t=\sin x)$.

When $a<0$, $I(a)$ is a linear function with negative slope, when $a>1$, $I(a)$ is a linear function with positive slope, thus

by the continuity of $I(a)$, We should consider the case $0\leq a\leq1$.


1.) The real numbers $u$ and $v$ satisfy:

$u^3-3u^2+5u-17=0$
$v^3-3v^2+5v+11=0$

Determine $u+v$

2.) The two shortest sides of a right angled triangle, $a$, and $b$, satisfy the following inequality:

$\sqrt{a^2-6a\sqrt{2}+19} + \sqrt{b^2-4b\sqrt{3}+16} \leq 3$

Determine $a^2+b^2+c^2$

Spoiler for problem 1:
[hide]
Problem 1 is problem 322:
[url]http://www.cms.math.ca/Competitions/MOCP/2004/prob_jul.html[/url]
[url]http://www.cms.math.ca/Competitions/MOCP/2004/sol_jul.html[/url]
[/hide]

Problem 2 is problem 353 (essentially):
[url]http://www.cms.math.ca/Competitions/MOCP/2005/prob_jan.pdf[/url]
I should add that it is still before the due date of this problem set (February 15).

Hmm.. okay then...I didnt know about that..

uhh... isn't the first equation missing a +5u... no wonder I was having so much trouble :)

[quote]In that case, please do not solve number two...(as I didnt know about that)[/quote]
Too bad . . . I just came up with a rather clever solution.


just wanted to know ... still need to do my mathpath thing i think XD

dont you need to get 20+ in amc 8 to participate, if that is it i got invited but i am not going.


vector
real analysis
real analysis unsolved

if ll ll  be a norm on E.E is a vector space.show that we have: $ %Error. "display" is a bad command.
\forall$ u,v
$ %Error. "display" is a bad command.
\in$ E   $ %Error. "display" is a bad command.
\mid$ llull_llvll$ %Error. "display" is a bad command.
\mid  \le$llu_vll.

use the definition of the absolute value and the triangular identity.


inequalities
logarithms
inequalities unsolved

[quote="arqady"][quote="quanghoa"]For a,b,c >= 2 prove that
$log_{a+b}^{c}+log_{c+a}^{b}+log_{c+b}^{a}\geq 1.5$  :!:[/quote]
Since, $a\geq2$ and $b\geq2$ the following inequality holds:
$(a-1)(b-1)\geq1\Leftrightarrow ab\geq a+b\Leftrightarrow\ln(a+b)\leq\ln a+\ln b.$
Hence, $\sum_{cyc}\log_{a+b}c=\sum_{cyc}\frac{\ln c}{\ln(a+b)}\geq\sum_{cyc}\frac{\ln c}{\ln a+\ln b}\geq\frac{3}{2}.$[/quote]
Thanks a lot


inequalities
function
search
inequalities theorems

People, can somebody post some links to solved problrms which ilustrate the application of this inequality in a good way?

Use function search, example

 http://www.mathlinks.ro/Forum/viewtopic.php?highlight=popoviciu&t=21786

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=popoviciu&t=19097

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=popoviciu&t=22364

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=popoviciu&t=19812

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=popoviciu&t=19962

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=popoviciu&t=16084

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=popoviciu&t=5633

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=popoviciu&t=2105


The 2007 AMC 8 begins today, Tuesday, November 13, so please, no discussion of the problems or solutions until Wednesday, November 21, 2007.  (See the Teachers' Manual, Section IX, page 6 for the dates of official adminstration, Nov 13-20, 2007.)

Thank you, your cooperation makes it posible to have an interesting contest for all participants.

Steve Dunbar
MAA Director, American Mathematics Competitions

yeah, im taking it tomorrow.
please don't spoil the fun.  :lol:

What the hell, it's AMC season already??
Time sure flies...

Wait, so that means we can discuss tomorrow rite?

or do we have to wait until 11:59 PM on the 21st then wait a minute then discuss? :P