trigonometry
geometry
quadratics
function
complex numbers
algebra

I would appreciate it if we could start a trig thread where people post challenge trig questions.  I would appreciate this very much.  I would also appreciate if the questions were kept to AMC/AHSME level.  If someone has an advanced question please post that in the Advanced section.  Could the moderators of this forum try to monitor the level of the problems(there are 5 mods so I hope they will keep this to an intermediate level)?

I'll start with the first question:

How many triangles have area 10 and vertices at (−5, 0), (5, 0), and (5 cos $\theta$, 5 sin $\theta$)
for some angle $\theta$?

Good Idea!!!

Solution to your problem:

[hide]All point  lies on the circle .

Now the base of the triangle is 10. The altitude of the triangle has to be 1/2. Thus we need two lines y=2 OR y=-2 to intersect the circle. y=2 and y=-2 intersect the circle at 4 points. Thus there are 4 such triangles.[/hide]

Wait, I might have found a slight error in beta's response. Doesn't the altitude have to be [hide]2[/hide]?

Here's what I got.

[hide]

10 = 1/2 base*height

10=1/2*10*height

2= height  (which is the y value)

All points (5cos :theta: , 5sin :theta: ) lie on the circle with its center at the origin with radius=5. So, they must lie on x 2 +y 2 =25.

Substituting in :pm:y, we have

x 2 + 4 = 25

x 2 = 21

x =  :pm::sqrt: 21

This still gives us four triangles, but I just wanted to point that out...[/hide]

A tad harder....

Find a closed form for

[tex]\displaystyle{\Sigma_{k=0}^n{\cos{(k\theta)}}}[/tex]

where n is an integer.

[hide]The moral of the story is that complex numbers are often useful with trig.[/hide]

[quote="Magnara"]A tad harder....

Find a closed form for
[tex]\displaystyle{\Sigma_{k=0}^n{\cos{(k\theta)}}}[/tex]
where n is an integer.
[/quote]

[tex]\displaystyle{\Sigma_{k=0}^n{\cos{(k\theta)}}}[/tex] = cos 0 + cos  :theta: + cos 2:theta: + ... + cos k:theta:

cos 0 + cos  :theta: + cos 2:theta: + ... + cos n:theta: = z
2 sin :theta:/2(cos 0 + cos  :theta: + cos 2:theta: + ... + cos n:theta:) = sin (2n+1):theta:/2 - sin (2n-1):theta:/2 + sin (n-1):theta:/2 - ... - sin -:theta:/2 = sin (2n+1):theta:/2 + sin :theta:/2 = 2z sin :theta:/2

sin (2n+1):theta:/2 + sin :theta:/2 = 2z sin :theta:/2
(sin (2n+1):theta:/2 + sin :theta:/2) / 2 sin :theta:/2 = (sin ((2n+1):theta:/2 / sin :theta:/2) + 1)/2 = z.

Our closed form is ((sin (2n+1):theta:/2 / sin :theta:/2) + 1)/2.

By DeMoivre's Theorem,
$(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$
$\displaystyle \sum_{k=0}^{n} \cos k \theta$
= Real part of
$\displaystyle \sum_{k=0}^{n} (\cos \theta+i\sin \theta)^k$
$=\frac{(\cos \theta+i\sin \theta)^{n+1}-1}{(\cos \theta+i\sin \theta)-1}$
Some algebra to get
$\frac{\cos n\theta-\cos\theta-\cos(n+1)\theta+1}{2(1-\cos\theta)}$

prove :
[tex] \cos ( \frac{\pi}{7}) - \cos ( \frac{2 \pi}{7}) + \cos ( \frac{3 \pi}{7}) = \frac{1}{2}[/tex]

[tex] \cos ( \frac{\pi}{9}) \cos ( \frac{4 \pi}{7}) \cos ( \frac{4 \pi}{7}) = \frac{1}{2}[/tex]

[tex] \alpha + \beta + \gamma = \pi [/tex]
find:

[tex] \min{( \tan ( \frac{ \alpha}{2}) + \tan ( \frac{ \beta}{2}) + \tan ( \frac{ \gamma}{2}) )}[/tex]

[tex] \max{( \tan ( \frac{ \alpha}{2}) \tan ( \frac{ \beta}{2}) \tan ( \frac{ \gamma}{2}) )}[/tex]

[quote="Bajoukx"][tex] \cos ( \frac{\pi}{9}) \cos ( \frac{4 \pi}{7}) \cos ( \frac{4 \pi}{7}) = \frac{1}{2}[/tex]
[/quote]
This is false... :D
[tex]\cos^2\frac{4 \pi}{7} = \cos^2\frac{3 \pi}{7}[/tex]
and [tex]\frac{3\pi}{7} > \frac{\pi}{4}[/tex] thus [tex]\cos\frac{3 \pi}{7} < \cos\frac{\pi}{4}[/tex] hence [tex]\cos^2\frac{3 \pi}{7} < \frac{1}{2}[/tex],

[tex]\cos\frac{\pi}{9} \cos \frac{4 \pi}{7} \cos \frac{4 \pi}{7} < \frac{1}{2}[/tex]

[quote="Bajoukx"]
[tex] \alpha + \beta + \gamma = \pi [/tex]
find:

[tex] \min{( \tan ( \frac{ \alpha}{2}) + \tan ( \frac{ \beta}{2}) + \tan ( \frac{ \gamma}{2}) )}[/tex]

[tex] \max{( \tan ( \frac{ \alpha}{2}) \tan ( \frac{ \beta}{2}) \tan ( \frac{ \gamma}{2}) )}[/tex][/quote]

No range constraint of $a,b,c$, like $a,b,c>0$?  :maybe:

sorry,

[tex] \cos ( \frac{ \pi}{9}) \cos ( \frac{2 \pi}{9}) \cos ( \frac{4 \pi}{9}) = \frac{1}{8} [/tex]   :D

and [tex] 0< \alpha , \beta , \gamma < \pi [/tex]

 :)

Here's another:

Evaluate the infinite series [tex]\displaystyle\sum_{n=1}^{\infty}{\arctan{\frac{2}{n^2}}[/tex]

[quote="Bajoukx"][tex] \alpha + \beta + \gamma = \pi [/tex]
find:
(1) [tex] \min{( \tan ( \frac{ \alpha}{2}) + \tan ( \frac{ \beta}{2}) + \tan ( \frac{ \gamma}{2}) )}[/tex]
[/quote]

A general method. Take (1) for example.
Fix $\alpha$, thus $\beta+\gamma$ is a constant.

A Formula (very easy to prove)
\[\tan\theta+\tan\phi = \frac{\tan(\theta+\phi)}{\cos\theta\cos\phi}\]

\[F = \tan \frac{ \alpha}{2} + \tan\frac{ \beta}{2} + \tan\frac{ \gamma}{2} = \tan \frac{ \alpha}{2} + \frac{\frac{\beta+\gamma}{2}}{\cos\frac{\beta}{2}\cos\frac{\gamma}{2}}\]

So the only indefinite item above is [tex]\cos\frac{\beta}{2}\cos\frac{\gamma}{2}[/tex], which is equal to [tex]\frac{1}{2}(\cos\frac{\beta+\gamma}{2}+\cos\frac{\beta-\gamma}{2})[/tex], it is clear to see that when $\beta=\gamma$, [tex]\cos\frac{\beta}{2}\cos\frac{\gamma}{2}[/tex] reaches its maximal value and hence $F$ reaches its minimal value.

Thus, we reduce $F$ to only one variable, since $\beta = \gamma = \frac{\pi-\alpha}{2}$.

We have (denote $\tan(\alpha/4) = y$ with $0<y<1$)
\[F=\tan\frac{\alpha}{2} + 2\tan\frac{\frac{\pi-\alpha}{2}}{2} = \tan2\frac{\alpha}{4}+2\tan(\frac{\pi}{4}-\frac{\alpha}{4}) = \frac{2y}{1-y^2}  + 2\cdot\frac{1-y}{1+y} = 2\cdot\frac{y^2-y+1}{1-y^2}\]

Now consider [tex]G = \frac{y^2-y+1}{1-y^2}[/tex]. We rewrite it as $(1+G)y^2-y+1-G=0$, which is a quadratic equation with regard to $y$, the discriminant $\Delta = 1-4(1-G)(1+G) \geq 0$, we have $G\geq \sqrt3/2$... thus $\min G=\sqrt3/2$ because we find that the root, $y=2-\sqrt3$, does lie in $(0,1)$ when $G=\sqrt3/2$.

In fact, from $y=2-\sqrt3=\tan(\alpha/4)$, we know that $\alpha/4=\pi/12$, thus $\alpha=\pi/3$.

The same idea works for (2). Fix $\alpha$,  and we need to prove that the function reaches its maximal value when $\beta=\gamma$.

It is clear to see that ($x+y$ is a constant)
\[
f(x,y) = \tan x \tan y = \frac{\sin x\sin y}{\cos x\cos y} = \frac{\frac{1}{2}(\cos(x-y)-\cos(x+y))}{\frac{1}{2}(\cos(x-y)+\cos(x+y))} = \frac{\cos(x-y)-\cos(x+y)}{\cos(x-y)+\cos(x+y)}= 1-\frac{2\cos(x+y)}{\cos(x-y)+\cos(x+y)}
\]
reaches its maximal value, when $\frac{2\cos(x+y)}{\cos(x-y)+\cos(x+y)}$ reaches its minimal value, i.e., $\cos(x-y)$ reaches its maximal value (notice that $\cos(x+y)$ is a constant here), when $x=y$.

another way for (2)

we can prove that
\[
\cot\frac{\alpha}{2}\cot\frac{\beta}{2}\cot\frac{\gamma}{2} = \cot\frac{\alpha}{2} + \cot\frac{\beta}{2} + \cot\frac{\gamma}{2}
\]

Note that for [tex]x=\frac{\pi}{9},\sin{x}=\sin{8x}[/tex]
[tex]\sin{8x}=\sin{x}[/tex]
[tex]2\sin{4x}\cos{4x}=\sin{x}[/tex]
[tex]4\sin{2x}\cos{2x}\cos{4x}=\sin{x}[/tex]
[tex]8\sin{x}\cos{x}\cos{2x}\cos{4x}=\sin{x}[/tex]
Obiously for x=pi/9, sin(x) does not equal 0, so we can divide by 8\sin{x} to get:
[tex]\cos{x}\cos{2x}\cos{4x}=1/8[/tex]


function
real analysis
real analysis unsolved

Construct a sequence of continuous functions $ f_n: \mathrm{R}\to \mathrm{R}$ such that $ f_n(x)$ is bounded for each $ x\in \mathrm{Q}$ and is unbounded for each $ x\in \mathrm{R}\symbol{92}\mathrm{Q}$.

I tried to construct such $ f_n$ as (very complicated) piecewise linear functions. But I cannot prove that it will work. Someone please help! Thanx.

The trick is not to try to make it [i]uniformly[/i] bounded for $ x\in\mathbb Q$. That won't work ($ \mathbb Q$ is not $ G_\delta$). Instead, take some enumeration $ q_k$ of rationals and define
\[ f_n(x)\equal{}\min_{1\le k\le n}(k\plus{}n|x\minus{}q_k|)\,,\]
which is probably similar to the sequence of piecewise linear functions you had in mind except I bet you tried to use $ 1$ instead of $ k$ in this formula ;).

Yes! fedja, you read my mind! Thank you so much!


calculus
integration
trigonometry
function
calculus computations

Please help me with this one

$ \int \frac{dx}{\sqrt{(x^2+a^2)^3}}$

where a is some constant.

use
x=tany.
then ull simplify it to forth power of Cosy.
use malty angle formulas to solve it...

Consider
$ I_n \equal{} \int \dfrac{\mathrm dx}{(x^2\plus{}a^2)^n},\qquad n>0$
Integrate by parts to obtain
$ I_n \equal{} \dfrac{x}{(x^2\plus{}a^2)^n} \plus{}2n\int \dfrac{x^2}{(x^2\plus{}a^2)^{n\plus{}1}}\, \mathrm d x$
Transform the second integral as follows:
$ \int \dfrac{x^2\plus{}a^2\minus{}a^2}{(x^2\plus{}a^2)^{n\plus{}1}}\, \mathrm dx \equal{} \int \dfrac{\mathrm dx}{(x^2\plus{}a^2)^{n}} \minus{} a^2\int \dfrac{\mathrm dx}{(x^2\plus{}a^2)^{n\plus{}1}}\equal{}I_n \minus{}a^2I_{n\plus{}1}$

Accordingly, we have the equality
$ I_n \equal{} \dfrac{x}{(x^2\plus{}a^2)^n} \plus{} 2n I_n \minus{} 2na^2 I_{n\plus{}1}$
From where we obtain
$ I_{n\plus{}1} \equal{} \dfrac{x}{2na^2(x^2\plus{}a^2)^n}\plus{} \dfrac{2n\minus{}1}{2na^2}I_n$
And replacing $ n$ with $ n\minus{}1$, we get ultimately a reduction formula:
$ \boxed{I_n \equal{} \dfrac{x}{2a^2(n\minus{}1)(x^2\plus{}a^2)^{n\minus{}1}}\plus{}\dfrac{2n\minus{}3}{2a^2(n\minus{}1)}I_{n\minus{}1}}\qquad n\neq 1$
But for $ n\equal{}1$, we have
$ I_1 \equal{} \int \dfrac{\mathrm dx}{x^2\plus{}a^2} \equal{} \dfrac{1}{a}\tan^{\minus{}1}\dfrac{x}{a}\plus{}C$
Hence, from the reduction formula, we obtain
$ I_2 \equal{} \int \dfrac{\mathrm dx}{(x^2\plus{}a^2)^2} \equal{} \dfrac{x}{2a^2(x^2\plus{}a^2)} \plus{} \dfrac{1}{2a^3}\tan^{\minus{}1}\dfrac{x}{a}\plus{}C$
And finally,
$ I_3 \equal{} \int \dfrac{\mathrm dx}{(x^2\plus{}a^2)^3} \equal{} \dfrac{x}{4a^2(x^2\plus{}a^2)^2} \plus{} \dfrac{3x}{8a^4(x^2\plus{}a^2)} \plus{}\dfrac{3}{8a^5}\tan^{\minus{}1}\dfrac{x}{a}\plus{}C$

$ \int \frac {dx}{ (x^2 \plus{} a^2 )^3 }$

Put $ x \equal{} a \tan\theta$

So the integral becomes $ \int \cos^{4} \theta d\theta$

We know that  $ 2 \cos^{2} \theta \equal{} 1 \plus{} \cos 2\theta.$

Therefore $ 4 \cos^{4} \theta \equal{} 1 \plus{} \cos^{2} 2\theta \plus{} 2 \cos 2\theta$

And  $ \cos^{2} 2\theta \equal{} \frac {1 \plus{} \cos 4\theta}{2}$

Therefore  $ \int \cos^{4} \theta d\theta \equal{} \frac {1}{4}\int 1 \plus{} 2\cos 2\theta \plus{} \frac {1 \plus{}  \cos 4\theta}{2} \equal{} \frac {1}{4} [ \frac {3\theta}{2} \plus{} \sin 2\theta \plus{} \frac {\sin 4\theta }{8} ] \plus{} C$

Now substitute back for $ \theta$ that is $ \theta \equal{} \tan^ { \minus{} 1} (\frac {x}{a} )$

Note that   $ \tan^ { \minus{} 1} (\frac {x}{a} )$ is the inverse function and not $ ( {tan \frac {x}{a} })^{ \minus{} 1}$

[quote="mushkin"]Please help me with this one

$ \int \frac {dx}{\sqrt {(x^2 + a^2)^3}}$

where a is some constant.[/quote]

sorry guys bud i made typo... the integral is

$ \int \frac {dx}{\sqrt {(x^2 + a^2)^3}}$

[quote="mathxl"]Try put x=αtan(u)[/quote]
Why don't?

Try calculate
$ \frac{d}{dx}\frac{x}{a^{2}\sqrt{x^{2}\plus{}a^{2}}}$ :P

The substitution $ x \equal{} a\tan \theta$ reduces the new problem to integrating $ f(\theta) \equal{} \frac{1}{a^2} \cos \theta$.

$ \int\frac {dx}{\sqrt {(x^{2} \plus{} a^{2})^{3}}}$

Put $ x \equal{} a \tan\theta$

So the integral becomes $ \frac{1}{a^2} \int \cos \theta d\theta$

$ \equal{} \frac {1}{ a^2 } \sin \theta .$


How much does winning a gold (silver, bronze) medal at the IPhO increase one's chances of getting admitted to top undergrad schools (such as Princeton, MIT, Harvard)?

Let's say that I'm an international applicant, coming from Croatia, middle class, with great SAT score and a golden medal from the IPhO in my pocket. And let's say that I'm hoping to get into MIT. What are my chances?

If all you have is a high SAT score and a IPhO medal, it won't help that much.  You need the whole package-great extracurricular activities, great letters of recommendation, a high class rank (if your school ranks), AND high SAT scores+IPhO medal.  On its own, the medal doesn't do very much.

I think an International Physics Olympiad gold medal is significantly more important than class rank?

But otherwise, I suppose I agree.

the medal is significant only if compared to those who dont have the medal...
the problem is there is far more than 100 internationals applying to MIT with at least one good metal(gold or high silver) at a sci-oly. and MIT only accepts around 100 internationals.

If you have US green card then your chances are far higher

I would go against the general sentiment on this one and say that those two things give you a very good chance. In my experience (I grew up in Serbia), international gpa is more or less meaningless in most places cause there is no way to check how legitimate it is (you can have straight A's without doing much of anything), and IPhO is a much more unbiased indicator of your abilities. A good SAT score (although not so important for US applicants) demonstrates your English skills, which are also very important for international applicants.

[quote]the problem is there is far more than 100 internationals applying to MIT with at least one good metal(gold or high silver) at a sci-oly. and MIT only accepts around 100 internationals. [/quote]
Then apply to Princeton. The physics department might not be like at MIT, but it's pretty decent here too, especially for the self-motivated student (you can end up doing top-notch graduate level courses and research with great profs by beginning of junior year if your come in with solid classical physics).

:(  I don't have any medals
I think my chances are very low :(


calculus
derivative
integration
real analysis
real analysis unsolved

If u is a sigma-finite meaure and v is a sigma-finite signed meaure s.t. v<<u,
then from radon nikodym thm we know the existence of radon nikodym derivative of v w.r.t u.
Can this be done when u is a sigma-finite signed meaure?
If the answer is yes,can it be proved by using the oringinal radon nikodym thm?

Here's a sketch: Suppose $ \mu$ is a $ \sigma$-finite signed measure, then $ \mu$ can be decomposed $ \mu \equal{} \mu^ \plus{} \minus{} \mu^ \minus{}$ where $ \mu^ \plus{} ,\mu^ \minus{}$ are $ \sigma$-finite positive measures and at least one of them is a finite measure, and moreover there are sets $ P,N$ such that $ P\cup N \equal{} X$, $ P\cap N \equal{} \emptyset$, $ u^ \plus{} (N) \equal{} u^ \minus{} (P) \equal{} 0$ (Hahn-Jordan).

Define the positive $ \sigma$-finite measure $ |\mu| \equal{} \mu^ \plus{} \plus{} \mu^ \minus{}$ (often called the [i]variation of u[/i]), then $ \mu \ll |\mu|$ and hence $ \nu \ll |\mu|$.  By Radon-Nikodym, there is an $ f$ satisfying $ \nu (A) \equal{} \int_A f d|\mu|\equal{} \int_{A} f d\mu ^ \plus{} \plus{} \int_{A} f d\mu ^ \minus{}$.  Let $ g \equal{} \begin{cases} f & x\in P \\
\minus{} f & x\in N \end{cases}$, then $ \nu (A) \equal{} \int_{A}g d\mu^ \plus{} \minus{} \int_{A} gd\mu^ \minus{} \equal{} \int_Agd\mu$.  So we have constructed $ g \equal{} \frac {d\nu}{d\mu}$.

[quote="JoeBlow"]Here's a sketch: Suppose $ \mu$ is a $ \sigma$-finite signed measure, then $ \mu$ can be decomposed $ \mu \equal{} \mu^ \plus{} \minus{} \mu^ \minus{}$ where $ \mu^ \plus{} ,\mu^ \minus{}$ are $ \sigma$-finite positive measures and at least one of them is a finite measure, and moreover there are sets $ P,N$ such that $ P\cup N \equal{} X$, $ P\cap N \equal{} \emptyset$, $ u^ \plus{} (N) \equal{} u^ \minus{} (P) \equal{} 0$ (Hahn-Jordan).

Define the positive $ \sigma$-finite measure $ |\mu| \equal{} \mu^ \plus{} \plus{} \mu^ \minus{}$ (often called the [i]variation of u[/i]), then $ \mu \ll |\mu|$ and hence $ \nu \ll |\mu|$.  By Radon-Nikodym, there is an $ f$ satisfying $ \nu (A) \equal{} \int_A f d|\mu| \equal{} \int_{A} f d\mu ^ \plus{} \plus{} \int_{A} f d\mu ^ \minus{}$.  Let $ g \equal{} \begin{cases} f & x\in P \\
\minus{} f & x\in N \end{cases}$, then $ \nu (A) \equal{} \int_{A}g d\mu^ \plus{} \minus{} \int_{A} gd\mu^ \minus{} \equal{} \int_Agd\mu$.  So we have constructed $ g \equal{} \frac {d\nu}{d\mu}$.[/quote]

Very clear.
Thanks a lot. :lol:

Of course the definition of $ \nu \ll \mu$ is different when $ \mu$ is a signed measure, as noted.


search
induction
algorithm
continued fraction
number theory
Euclidean algorithm
algebra proposed

Nice problem!

Hint:

[hide]The continued fraction of $ a_n$ is encoded in the binary digits of $ n$ with 1 to 1 correspondence.[/hide]

[quote="Lousin Garckz"]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=611926086&t=294846
(Wed, 12, Aug, 2009 while this problem was in Ibero test 23, Sep, 2009, People, just try to do a better work).[/quote]

Unfortunately it seems that has become some kind of trend in this competition -more examples like this can be found in recent papers... In my opinion even if the jury tries to do its duty there might be some countries that are playing foul and do not propose completely original problems... :mad:

However there are many people who are willing to improve the quality of the competition! You can still find original problems, like problem 3 that was proposed by the deputy leader of our country.  :)

Do you have these examples you are talking about? You still don't know what really happened, maybe two people thought of the same thing (which I can guarantee happens sometimes) and you can't blame them as cheaters. Problems are supposed to be proposed way before the competition and maybe there was no time for them to check, so don't judge.

[quote="Jutaro"]Consider the sequence $ \{a_n\}_{n\geq1}$ defined as follows: $ a_1 \equal{} 1$, $ a_{2k} \equal{} 1 \plus{} a_k$ and $ a_{2k \plus{} 1} \equal{} \frac {1}{a_{2k}}$ for every $ k\geq 1$. Prove that every positive rational number appears on the sequence $ \{a_n\}$ exactly once.[/quote]
It is easy to prove by induction that $a_{2^k}=k+1\cdots \cdots (1) $.
In the same we can prove that if $a_m=\frac{a}{b} \iff a_{2^k\cdot m}=k+\frac{a}{b}$.
For any $n \in \mathbb{N}$ we have $a_{2^{n-1}+1}=\frac{1}{n}$. So If we can prove that we can construct $\frac{a+1}{b}$ from any fraction $\frac{a}{b}$ we are done.

We consider 3 cases:
1. $b|a+1$. In this case we can easily construct this from $(1)$.
2. $b<a+1, b\not| a+1$
3. $b>a+1$

In case 2 we can write $a+1=bq+r$. So we have $\frac{a+1}{b}=q+\frac{r}{b}$, which yields that it is enough to prove case 3.

For case 3, if we have $a_m=\frac{r}{b}$, we can say that $m$ is odd and $a_{m-1}=\frac{b}{r}$. Now using the same argument (Euclidean Algorithm) of case 2, and of case 1; we can conclude that we shall get some $a_i=r_j$,

Now we prove the uniqueness of every term. If $a_i=a_j$ with $i>j$. We can can infer from the previous argument that $(a_i,a_j) \to (a_p,a_q)$ for some $p,q$ with $p\not=q$ (because we need the same number of steps for this transformation), and  $a_p=a_q=n \in \mathbb{N}$. But from $(1)$ we get that $p=q$, which is a contradiction.

[quote="Jutaro"]

Unfortunately it seems that has become some kind of trend in this competition -more examples like this can be found in recent papers... In my opinion even if the jury tries to do its duty there might be some countries that are playing foul and do not propose completely original problems... :mad:

[/quote]
Anyways, the problem is folklore. I included it in a textbook I wrote in 1997 and even then I didn't know the source.


consider a triangle $ ABC$ and the line L. such that L intersect the sides of $ ABC$ . BC at $ A_1$  AC at $ B_1$ and AB at $ C_1$
let point P in the triangle ABC and the interections of the lines $ AP,BP,CP$ with the line L  we get $ A_2$,$ B_2$,$ C_2$ respectively. prove that the circumcircles of $ PA_1A_2,PB_1B_2,PC_1C_2$ are concurent in the point except P

See the solution at [url]http://www.mathlinks.ro/viewtopic.php?p=723323#723323[/url].


calculus
integration
derivative
calculus computations

$ \int{9cos(t)e^{9sin(t)} \minus{} e^{cos(t)}sin(t)} dt$

$ \int{ \minus{} t^2sin(t) \plus{} 2tcos(t)} dt$

$ \int{\minus{}8(sin(t))^2 \minus{} 2cos(t)sin(2sin((t)))} dt$

how can I do this?

hello, we have $ \int9\cos(t)e^{9\sin(t)}\,dt\equal{}\int e^u\,du$, when we are setting $ 9\sin(t)\equal{}u$.
Sonnhard.

$ \int{ \minus{} t^2sin(t) \plus{} 2tcos(t)} dt \equal{}\int{ \minus{} t^2sin(t)dt} \plus{} \int{2tcos(t)} dt$

Now you make per partes for each of the integrals.

[quote="EquinoX"]$ \int{9cos(t)e^{9sin(t)} \minus{} e^{cos(t)}sin(t)} dt$

$ \int{ \minus{} t^2sin(t) \plus{} 2tcos(t)} dt$

$ \int{ \minus{} 8(sin(t))^2 \minus{} 2cos(t)sin(2sin((t)))} dt$

how can I do this?[/quote]
i assume you forgot brakets everywhere. :lol:
anyway, for the first , break up into two parts and just substitute 9sin(t) by z_1 and cos(t) by z_2.
second one z_3=$ t^2cos(t)$
3rd :
break up into 2 parts. first one use $ cos2t\equal{}1\minus{}2sin^2t$ and second one is derivative of $ cos(2sin(t))$

hello, by the same way we get $ \minus{} \int e^{\cos(t)}\sin(t)\,dt \equal{} \int e^u\,du$.
Sonnhard.

$ \int{ \minus{} 8(sin(t))^2 \minus{} 2cos(t)sin(2sin((t)))} dt \equal{} \\
\int{ \minus{} 8(sin(t))^2 dt} \minus{} \int{ 2cos(t)sin(2sin((t)))} dt$

$ \int{ \minus{} 8(sin(t))^2 dt} \equal{} \int{ \minus{} 8(1 \minus{} (cos t)^2)} \equal{} \int{ \minus{} 8\left(1 \minus{} \frac {cos (2t) \plus{} 1}{2}\right)dt}$.

$ \int{ 2cos(t)sin(2sin((t)))} dt \equal{} \int{sin {u}du}$, where $ u \equal{} 2sin{t}\Rightarrow du \equal{} 2cos{t}dt$

how do I solve this $ \minus{} 8 \int{ \left(1 \minus{} \frac {cos (2t) \plus{} 1}{2}\right)dt}$.

[quote="EquinoX"]how do I solve this $ \minus{} 8 \int{\left(1 \minus{} \frac {cos (2t) \plus{} 1}{2}\right)dt}$.[/quote]

Read a standard textbook having all theorems of integration. :lol:   :)

$ \int cos mx dx\equal{} (sin mx)/m$.


geometry
trapezoid
symmetry
geometry unsolved

Let $ ABCDE$ be convex pentagon such that $ S(ABC) \equal{} S(BCD) \equal{} S(CDE) \equal{} S(DEA) \equal{} S(EAB)$. Prove that there is a point $ M$ inside pentagon such that $ S(MAB) \equal{} S(MBC) \equal{} S(MCD) \equal{} S(MDE) \equal{} S(MEA)$.

Note the following Lemma:  If in quadrilateral $ XYZT$  $ S(XYT)\equal{}S(XZT)$, than $ YZ\parallel{}XT$.
Proof: Denote by $ a,b$ the distances from points $ Y,Z$ to $ XT$ (Respectively). Than,  $ a\equal{}\frac{2S(XYT)}{XT}\equal{}\frac{2S(XZT)}{XT}\equal{}b$, which means $ a\equal{}b$ and hence, $ YZ\parallel{}XT$. This way the Lemma is proved.
    Apply the Lemma to $ ABCD$, where by the condition, $ S(ABC)\equal{}S(BCD)$. So, $ AD\parallel{}BC$. Analogeously, $ BE\parallel{}CD$, $ AC\parallel{}DE$, $ DB\parallel{}EC$ and $ EC\parallel{}AB$.  Denote $ K\equal{}AC,BE$, $ L\equal{}BD,AC$, $ Q\equal{}CE,BD$, $ N\equal{}AD,CE$, $ P\equal{}AD,BE$.  (Here by $ U\equal{}XY,ZT$ we mean that $ U$ is the intersection point of $ XY$ and $ ZT$).
   Denote $ M\equal{}AQ,KD$. We will prove that $ M$ also lies on lines $ EL,BN,CP$.
   Note that $ ABQE$ and $ ABCN$ are parallelograms. So,$ QE\equal{}CN$ and $ CQ\equal{}EN$. Analogeously, $ EP\equal{}KB$,$ EN\equal{}QC$ and $ AP\equal{}ED$. As $ PN\parallel{}AD\parallel{}BC$, $ \frac{EP}{EB}\equal{}\frac{EN}{EC}$, which means $ \frac{KB}{EB}\equal{}\frac{QC}{EC}$, which imply $ KQ\parallel{}AD$. In trapezoid $ AKQD$ $ L\equal{}AK,DQ$ and $ M\equal{}AQ,KD$. So, by the well-know fact, $ LM$ passes through the midpoints of $ KQ$ and $ AD$. Call this line $ l$. On the other hand, $ KPNQ$ is also trapezoid where $ E\equal{}KP,QN$. So, line joining the midpoints of $ PN$ and $ KQ$, passes through $ E$, but as we've already mentioned, $ AP\equal{}ND$, which imply that the midpoints of $ PN$ and $ AD$ coincide each other and therefore, $ l$ passes through $ E$. So, $ M$ lies on $ EL$. The same way we can prove that $ M$ lies on $ BN$ and $ CP$.
    Now call the midpoint of $ AC$ $ T$. Then, $ S(ABT)\equal{}S(CBT)$ and $ S(AMT)\equal{}S(CMT)$ (As $ BT$ is median in triangles $ ABC$ and $ AMC$). So, $ S(AMB)\equal{}S(ABT)\plus{}S(AMT)\equal{}S(CBT)\plus{}S(CMT)\equal{}S(BMC)$. Analogeously, $ S(BMC)\equal{}S(CMD)$, $ S(CMD)\equal{}S(DME)$,$ S(DME)\equal{}S(MEA)$ and $ S(MEA)\equal{}S(MAB)$.
    Finally, $ S(MBC)\equal{}S(MCD)\equal{}S(MDE)\equal{}S(MEA)\equal{}S(MAB)$ and there exists such an $ M$, as desired.  Lasha Lakirbaia :)

There also exists some (in my opinion) more simple proof, as follows:

Consider $\alpha = S(ABC)=S(BCD)=S(CDE)=S(DEA)=S(EAB)$ and $\beta = S(ABCDE)$, as well as $M = \frac{A+B+C+D+E}{5}$ the center of gravity of $ABCDE$. Since $ABCDE$ is convex, $M$ belongs to its interior.

$S(ABM) = \frac{S(ABA)+S(ABB)+S(ABC)+S(ABD)+S(ABE)}{5}=$ $\frac{2\alpha+S(ABD)}{5}=\frac{2\alpha+(\beta-S(BCD)-S(DEA))}{5}=\frac{2\alpha+\beta-2\alpha}{5}=\frac{\alpha}{5}$.
For reasons of symmetry, we also get $S(BCM)=S(CDM)=S(DEM)=S(EAM)=\frac{\alpha}{5}$ and we are done.


sorry in advance, if this problem should be in another category,

A water skier is pulled over a ramp by a boat going 30mph. If the ramp has a launch angle of 25 deg. and maximum height of 4ft above the water, how far from the end of the ramp does the skier hit the water if he lets go of the rope as he leaves the ramp? The skier's horizontal velocity remains 30mph on the ramp.

Set up a right triangle ... one leg is 30 mph (convert this to  ft/s first, since the answer is in feet .. multiply by 22/15) and the angle next to it 25 deg.

30(22/15)tan(25) = Vy = vertical speed as he is leaving the ramp

0 = (.5)(g)(t^2) + Vy(t) + 4

find t... multiply by 30(22/15) to get the distance

Do we have to count with air drag? If yes How big is the water skier


[quote="Totally Zealous"]ABCD is a square with side of unit length. Points E and F are taken rewpectively on sides AB and AD so that AE=AF and the quadralateral CDFE has maximum area. What is this maximum area?[/quote]
[hide]$\text{Let}$ $x = AE = AF$

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

$\text{Then,}$
$A'(x) =-x+\frac{1}{2}$
$A'(x)$ $\text{changes from positive to negative sign when}$ $x = \frac{1}{2}$. $\text{Thus, the area of CDFE is maximum when}$ $x= \frac{1}{2}$.

$A(\frac{1}{2}) = \boxed{0.625}$[/hide]

[quote="daermon"][hide]I put $e$and $f$ on the midpoints of the sides and got an area of $\frac{5}{8}$[/hide][/quote]
How did you know the maximum would be when they are the midpoints?

I tried it closer to the points in both directions and got smaller answers  :D

[hide]
WE can call AE and AF x

(x^2/2 + (1-x))/2 which is equal to ((1-x) + x^2)/2

This equation is the sum of the areas of the triangles formed by the quadrilateral in the square. To find the area of the quadrilateral we have to subtract this from 1.

(2-(1-x) + x^2)/2 which is equal to (-x^2 + x + 1)/2 which is equal to 1/2 + (x-x^2)/2 = 1/2 + x(1-x)/2

y = x(1-x)/2 is a parabola with maximum value at x=1/2, so the largest possible value for the expression is 1/8

1/8 + 1/2= 5/8[/hide]

[quote="Jhonny Ruiz"]this may sound stupid but can't we say that the greatest possible area of CDFE is 1, which happens when x=0 and F=E=A?[/quote]

That won't be the case because if $AE=AF=0$ then you had be calculating the area of half the square. :)


[quote="mdk"]A city has 10000 different telephone lines numbered by 4-digit numbers. More than
half of the telephone lines are in the downtown. Prove that there are two telephone
numbers in the downtown whose sum is again the number of a downtown telephone
line.[/quote]
It wrong, consider $ 5000,5001,5002,....,10000$ there are $ 5001$ numbers in downtown, which is more than half. But now of these sums are in the downtown.

[quote="sylow_theory"]It wrong, consider $ 5000,5001,5002,....,10000$ there are $ 5001$ numbers in downtown, which is more than half. But now of these sums are in the downtown.[/quote]
sylow_theory, I assume that the valid numbers are $ 0000,0001,\dots,9999$, since $ 10000$ is not a 4-digit number.

[hide="Incomplete Solution"]
I hope somebody can prove this formally.  :maybe:

Instead of attack your case, lets do this with $ \{0,1,2,...,9\}$. So we have a 10 tows with numbers $ 0\leq n\leq 9$. And we want to show that if we have more than half (in this case 6) cities in downtown then there are two whose sum is contained downtown.

Since there have to be at least six, lets do it for exactly 6 towns downtown.

Now concentrate at the [i]smallest[/i] town in numerical value. It is easily seen that if the smallest town is $ 4$ then the only possible towns can be $ 4,5,6,7,8,9$ and then $ 4+5$ is downtown. It cant be bigger than $ 4$, lets go to the next case. Say its $ 3$. So we have $ 3<a_{1}<a_{2}<a_{3}<a_{4}<a_{5}$ where $ 4 \leq a_{i}\leq 9$. Now, it must be that $ 3+a_{i}\leq 9$ because if it isnt then it means $ a_{i}$ is at least 7 which will imply $ a_{5}$ is at least 11, a contradiction. Similarly $ 3+a_{2}\leq 9$. So we have,
$ 4\leq a_{i}\leq 9$ and $ 4\leq 3+a_{1}\leq 9$ and $ 4\leq 3+a_{2}\leq 9$ there are out pigeons the holes are $ \{4,5,...,9\}$ since there are $ 5+1+1 = 7$ pigeons and $ 9-4+1=6$ holes it means two of them end up in the same value.

Now, the next case is when the [i]smallest[/i] town is $ 2$. So we have $ 2<a_{1}<a_{2}<a_{3}<a_{4}<a_{5}$. Now $ 3\leq 2+a_{i}\leq 9$ for $ i=1,2,3$ by similar reasoning above. And $ 3\leq a_{i}\leq 9$ for $ i=1,2,3,4,5$ so we have $ 8$ pigeons and $ 9-3+1=7$ holes it means two of them end up in the same value.

Now, the last case is when the [i]smallest[/i] town is $ 1$. So we have $ 1<a_{1}<a_{2}<a_{3}<a_{4}<a_{5}$. Now $ 2\leq 1+a_{i}\leq 9$ for $ i=1,2,3,4$ and $ 2\leq a_{i}\leq 9$ for $ i=1,2,3,4,5$. We have $ 9$ pigeons and $ 9-2+1=8$ holes so two end up in the same value.

And of course the case with the smallest one being $ 0$ is pointless.

[/hide]

[hide="Formal Solution"]
If $ a_{i}$ are the numbers in downtown then we can monotonize this sequence, i.e. write it in increasing order with $ a_{i}<a_{i+1}$.  With this sequence written out look at the $ 5001$-st element [b]from the right[/b] which has to exist since downtown has at least $ 5001$ numbers. For the sake of being easy to follow let us call this number $ b_{1}$ and the number after that $ b_{2}$ and so on until $ b_{5001}$. The number $ b_{1}$ cant be $ 5000$ for that will imply $ b_{5001}$ must be at least $ 10000$ which is cant be. Now if $ b_{1}=4999$ then $ b_{2}$ [b]must[/b] be $ 5000$ and so on until $ b_{5001}=9999$ and so $ b_{1}+b_{2}=b_{5001}$ and the proof is complete. So lets say $ b_{1}= k$ where $ 1\leq k \leq 4998$. Then it means $ b_{1}\leq b_{1}+b_{j}\leq 9999$ for $ j=2,...,5001-k$ because if not then $ b_{5001-k}$ need to be at least $ 10000-b_{1}$ which will imply (since this sequence is increasing) that $ a_{5001}$ is at least $ 10000$ which is a contraction. Now, $ b_{1}\leq b_{h}\leq 9999$ for $ h=1,2...,5001$. Since the must of pigeons exceeds to holes the proof is complete. And if $ k=0$ its trivial. Q.E.D.

I hope that makes sense. I cant believe I wrote such a monster. :rotfl:

[/hide]


Here is the IMO shortlist problems 1996, I got it from internet.


Determine all positive integers numbers x, y, z that:

$ x.y.z \equal{} 4104$
$ x \plus{} y \plus{} z \equal{} 77$

it is easy.
We have \[ 1404\equal{}2^3.3^3.19\]
So  $ x,y,z$ Is just $ 2^2\;\;;2.3^3\;\;;and \;19$ it is that mean $ 4\;\;; 54\;\;and \;19$
It is only solution.
how do you feel?

hello, i have found
$ x\equal{}3,y\equal{}36z\equal{}38$
$ x\equal{}19,y\equal{}54,z\equal{}4$
$ x\equal{}54,y\equal{}4,z\equal{}19$
$ x\equal{}3,y\equal{}38,z\equal{}36$
$ x\equal{}36,y\equal{}3,z\equal{}39$
$ x\equal{}54,y\equal{}19,z\equal{}4$
$ x\equal{}4,y\equal{}19,z\equal{}54$
$ x\equal{}36,y\equal{}38,z\equal{}3$
$ x\equal{}4,y\equal{}54,z\equal{}19$
$ x\equal{}38,y\equal{}3,z\equal{}36$
$ x\equal{}19,y\equal{}4,z\equal{}54$
$ x\equal{}38,y\equal{}36,z\equal{}3$
Sonnhard.

Are Both of the answers correct ?

Dr Sonnhard Graubner, how do you do this question ?
minhphuc.v, I'm fine, but I think I'm a stupid  :blush:

Thanks.

Determine all positive integers numbers $ x, y, z$ that:
$ x.y.z \equal{} 4104$
$ x \plus{} y \plus{} z \equal{} 77$

Since product is even and sum is odd, two numbers must be even and one odd
Note  $ 4104 \equal{} 2^{3}.3^{3}.19$
Let $ 1\le x\le y\le z$
As $ x \plus{} y \plus{} z \equal{} 77\implies 3z\ge 77\implies 26\le z\le75$

Consider $ z$ odd and $ x,y$ even $ \implies z \equal{} 3^{3}\equal{}27$ or $ 3*19\equal{}57$
$ z \equal{} 27 \equal{} 3^{3} \implies y \equal{} 19*2^{k},k\in\mathbb{N}$  which is not possible as $ y\le z$
and $ z \equal{} 3*19 \equal{} 57\implies x \plus{} y \equal{} 20$ which is also not possible.
$ \implies z$ can't be odd

Now consider $ z$ even $ \implies z \equal{} 2*19 \equal{} 38,2*3^{3} \equal{} 54,2^{2}*3^{2} \equal{} 36$ are only possibilities

[b]case 1[/b], $ z \equal{} 36\implies xy \equal{} 2*3*19\implies y \equal{} 19$ and $ x \equal{} 6$ which does not satisfy $ x \plus{} y \equal{} 41$

[b]case 2[/b], $ z \equal{} 54\implies xy \equal{} 2^{2}*19$ and $ x \plus{} y \equal{} 23\implies y \equal{} 19$ and $ x \equal{} 4$

[b]case 3[/b], $ z \equal{} 38\implies xy \equal{} 2*3^{3}$ and $ x \plus{} y \equal{} 39\implies y \equal{} 36$ and $ x \equal{} 3$

$ x \equal{} 3,y \equal{} 36,z \equal{} 38$ or $ x \equal{} 4,y \equal{} 19,z \equal{} 54$

After considering their permutations we will get $ 2*3! \equal{} 12$ solutions

cant one number be a multiple of 4 and 2 nos. odd?please tell me if i am wrong.

if 2 are odd and one even then sum will become even which is not the case as sum is $ 77$ (an odd number).
Note: odd +odd = even $ \implies$ even + odd + odd  = even :wink:


The Mock AMC E was recently posted but I dont know how to get the problems.  Someone Please help.  Thanks

Nevermind that I found it, I guess it wasnt posted earlier or it wasnt coming up on my computer.


I have just been wondering, how many people usually participate in the USAMTS, and how are the scores divided up?  I got something around a 75 or 80 last year, so I'm wondering how I ranked percentile wise, and also want to know the popularity of the contest (i.e. is it AMC level with millions of people or just USAMO with like 300 people, although the USAMO is popular, just people don't qualify so it's a bad example sorry).

It's usually on the order of 1000 people.

I don't have the stats on past years, but if I remember correctly there were around 300 silver+gold medalists.  This year you'll be able to see score distributions on the USAMTS pages as the contest proceeds.


analytic geometry
function
geometry
integration
3D geometry
calculus
derivative

A closed, cubic container of edge $ l $ is completely filled with a fluid of density $ \rho $. Then, it's put on a car that moves horizontally with constant acceleration $ \vec a $. What's the pressure in the center of the container?

[quote="bus"]Normally, there would be hydrostatic pressure $ \rho g\frac{l}{2} $. Now since the car is accelerating with horizontal acceleration $ a $ (upwards, I assume), the result is $ \rho(g+a)\frac{l}{2} $ because in the noninertial reference frame in which the container with water is stationary, all objects accelerate with extra $ a $ acceleration so it pretty much acts as gravitational acceleration. Which means that the grav. acc. seems to be $ g+a $. I hope this is not too confusing. :)[/quote]

Wait a sec, the gravitational acceleration is not $ g+a $ but $ \vec{g}-\vec{a} $, and since $ \vec{g} \perp \vec{a} $, magnitudes of those two are not equal. (Oh, now I see you assume the "horizontal acceleration" to go upwrads, that's the misunderstanding;-)

I would say that the pressure will be (if $ \vec{a} $ is perpendicular to one of the sides of the box) $ p = \rho g (-\vec{r} \cdot (\vec{g} - \vec{a})) $ where $ \vec{r} $ points from the center of the box towards the upper-front corner, where I assume zero pressure. Therefore $ p = \rho(g+a)\frac{l}{2} $ anyway ... strange ...

Interesting observation, kubus! But can you explain why you invoked a $ \vec{r} $ ?? Can you plz explain?

[quote="andre.l"]A closed, cubic container of edge $l$ is completely filled with a fluid of density $\rho$. Then, it's put on a car that moves horizontally with constant acceleration $\vec a$. What's the pressure in the center of the container?[/quote]The first law of thermodynamics (in differential form) says $\mathrm{d}E = T\mathrm{d}S - P\mathrm{d}V$.

Noting that, $\mathrm{d}E = \dfrac{\partial E}{\partial S}\mathrm{d}S + \dfrac{\partial E}{\partial V}\mathrm{d}V$ we immediately observe $\dfrac{\partial E}{\partial V}= -P$.

The pressure, $P$, is the pressure exerted by the fluid on its surroundings. The pressure the problem asks for is the opposite (opposite due to Newton's third law), the pressure of the surrounding on the fluid, $P_f = -P = \dfrac{\partial E}{\partial V}$.

Now $E$ is the total energy of the system: $E = \dfrac12 m_fv^2 +M_{air}gH +m_fgy_{cm}$.

$\dfrac{\partial E}{\partial V} = \dfrac12v^2\dfrac{\partial m_f}{\partial V} +\dfrac{\partial M_{air}}{\partial V}gH + \dfrac{\partial m_f}{\partial V}gy_{cm}$

$\dfrac{\partial M_{air}}{\partial V} = \rho_{air} \: ; \dfrac{\partial m_f}{\partial V} = \rho_f$

$P_f = \dfrac12 \rho_f v^2 + \rho_{air}gH + \rho_f gy_{cm}$

$\rho_{air}gH = P_{atm}$

$y_{cm} = \dfrac12 l$

$P_f = \dfrac12 \rho_f v^2 +\dfrac12 \rho_f gl + P_{atm}$

Starting from rest with uniform acceleration, $\dfrac12 v^2 =\vec{a}\cdot\Delta\vec{x}$

Therefore,

$\boxed{P_f = \rho_f \vec{a}\cdot\Delta\vec{x} +\dfrac12 \rho_f gl + P_{atm}}$

[quote="kubus"]Oh, now I see you assume the "horizontal acceleration" to go upwrads[/quote]Oops, my bad :) .

[quote="bus"][quote="Dr. No"]$E = \dfrac12 m_fv^2 +M_{air}gH +m_fgy_{cm}$.

$\dfrac{\partial E}{\partial V} = \dfrac12v^2\dfrac{\partial m_f}{\partial V} +\dfrac{\partial M_{air}}{\partial V}gH + \dfrac{\partial m_f}{\partial V}gy_{cm}$

...

$P_f = \dfrac12 \rho_f v^2 +\dfrac12 \rho_f gl + P_{atm}$[/quote]First thing is that I don't like the result - how can the pressure continually increase vith $v$? That's just plainly impossible. Or at least, I think so... :)
Secondly, I don't understand your derivation - but I would like to. Could you kindly explain the symbols you use? More specifically - what are $M_{air}$, $H$ and $y_{cm}$ ? Where are the bounds of the "system" i.e. where does the air you take into account "end"? And finally, what are the $M_{air}gH$ and $m_fgy_{cm}$ parts of $E$? They look like potential energy but there's something odd about them...
Thank you![/quote]First, you're quite right. The above result is only valid if the fluid is moving at constant speed. Since it is not moving with constant speed, the entire kinetic energy expression changes with volume: $\mathrm{d}\left(KE\right) = m_fv\mathrm{d}v = m_f a\mathrm{d}x$.

For the box of fluid travelling down the track $\mathrm{d}V = A\mathrm{d}x = l^2\mathrm{d}x$

Therefore, $\dfrac{\partial \left(KE\right)}{\partial V} = \dfrac{m_f a\mathrm{d}x}{l^2\mathrm{d}x} \Rightarrow \dfrac{\partial \left(KE\right)}{\partial V} = \dfrac{m_f a}{l^2}$.

$m_f = \rho_f l^3 \Rightarrow \dfrac{\partial \left(KE\right)}{\partial V} = \dfrac{\rho_f l^3 a}{l^2}$

$\Rightarrow \dfrac{\partial \left(KE\right)}{\partial V} = \rho_f l a$

Hence the correct expression for the total pressure is:

$\boxed{P_f = \rho_f l\left(a+\dfrac12 g\right) + P_{atm}}$

To answer your other question $y_{cm}$ is the vertical coordinate of the center of mass, $m_fgy_{cm}$ is the potential energy of the fluid above the center of mass; likewise, $M_{air}gH$ is the potential energy of the atmosphere above the fluid.

[quote="bus"]I have some serious problems thinking about $\dfrac{\partial(KE)}{\partial V}$ in situations where $V$ is some heavily constant quantity :P. However, suppose that $g$ and $P_{atm}$ were both zero. Then the pressure is according to you $\rho_f a\l$. But how is this situation different from a nonaccelerating box in a gravitational field with gravitational acceleration $a$ (pointing "left") ? Shouldn't be the result in this case $\dfrac{\rho_f a\l}{2}$ ?[/quote]The factor of $\dfrac12$ that appears in the potential energy term comes because the position of the center of the box is half way down from the top of the box. It doesn't appear in the kinetic energy term because the differential work done is $m_f \,a\; \mathrm{d}x$, but it is done by pushing a fixed cross section $l^2$ through a distance $\mathrm{d}x$ - just like a piston. There is no factor of $\dfrac12$ for the kinetic energy term.

Yes, you're right. But that's not what I've been asking. What I'm saying is that if g=0 and Patm=0 then you can look at the problem from noninertial reference frame in which the water container is stationary and straightforwardly come to answer $\dfrac{\rho_f al}{2}$. This does not match your result.

[quote="bus"]Yes, you're right. But that's not what I've been asking. What I'm saying is that if g=0 and Patm=0 then you can look at the problem from noninertial reference frame in which the water container is stationary [b]and straightforwardly come to answer $\dfrac{\rho_f al}{2}$. This does not match your result.[/b][/quote] How? If you have a cubic block of stuff, and you move it over by a distance $\Delta x$ then the Volume displaced in said movement is$\Delta V = l^2 \Delta x$ if $l$ is the dimension of the box. It's rudimentary mechanics that the change in kinetic energy of a body = the work done on it. With that fact in view, proving that the pressure on the box of fluid is equal to $\rho_f al$ is child's play. Again, the factor of $\dfrac12$ appears because of the position of the center of the box. (Potential energy is a function of position; kinetic energy is not.)

[quote="bus"][quote="Dr. No"]$\boxed{P_f = \rho_f l\left(a+\dfrac12 g\right) + P_{atm}}$[/quote]I say that if you plug in g=0 and P=0 you get $\rho_fla$ while the right answer in this case is $\rho_fla/2$.[/quote]Please prove your result. In the meantime, I'm fairly confident that I've got the geometry right in which case the result (in the absence of atmospheric pressure and gravity) is $\rho_f la$.

[quote="bus"]Consider a noninertial reference frame in which the container is stationary. Since we're in a noninertial system, there is a force $a/\text{d}m$ acting on every piece of mass dm. The container is accelerated by some opposite force so that it is stationary in this frame of reference. However, the particles still feel the $a/\text{d}m$ force and the only thing that prevents them from accelerating is the reaction force of the walls of container - this creates the pressure. It's not hard to see that this situation is completely identical to a situation with gravitational acceleration $\vec a$ in an inertial reference frame with container stationary. And in that case we already know that the solution should be the hydrostatic pressure $\rho_f gl/2$.[/quote]bus,
The gravitational term, $\frac12\rho_fgl$ comes from differentiating the potential energy with respect to volume. $P_{PE} =\dfrac{\partial \left(PE\right)}{\partial V}=\dfrac{\partial}{\partial V}\int_{0}^{l}\int_{0}^{l}\int_{0}^{l} gy \rho_f \mathrm{d}x\mathrm{d}y\mathrm{d}z$

$\Rightarrow \dfrac{\partial \left(PE\right)}{\partial V}= gy \rho_f$

At the vertical center of the cube $y =\dfrac{l}{2}$ (which is provable if I must).

Thus $P_{PE} = \frac12\rho_f gl$.

Now, considering the kinetic energy term $P_{KE} = \dfrac{\partial}{\partial V}\int_{0}^{\Delta x}\vec{F}\cdot\mathrm{d}\vec{x}$.

The differential volume $\mathrm{d}V =\dfrac{\partial V}{\partial x}\mathrm{d}x + \dfrac{\partial V}{\partial y}\mathrm{d}y + \dfrac{\partial V}{\partial z}\mathrm{d}z$

In this case the box of fluid is a cube, $x = y = z = l$, that is moving only along the x-direction.

This implies $\mathrm{d}y = \mathrm{d}z = 0$.

It follows then $\mathrm{d}V =\dfrac{\partial V}{\partial x}\mathrm{d}x = yz\;\mathrm{d}x = l^2\;\mathrm{d}x$.

Evaluating $P_{KE} = \dfrac{\partial}{\partial V}\int_{0}^{\Delta x}\vec{F}\cdot\mathrm{d}\vec{x} = \dfrac{\partial}{l^2\partial x}\int_{0}^{\Delta x}\vec{F}\cdot\mathrm{d}\vec{x} = \dfrac{F}{l^2} = \dfrac{ma}{l^2} =\dfrac{\rho_fl^3a}{l^2} = \rho_fla$.

[quote="bus"]I'm sorry, I didn't want to make you repeat yourself so many times. I see that you came to a different outcome using a different approach and I'm not saying that you're wrong - but I see nothing wrong about my solution either. Can you prove my solution wrong in some more direct way? :?:[/quote]I don't know how much more direct it can get. The definition of pressure is the derivative of energy with respect to volume. Ignoring potential energy terms, that definition reduces to the derivative of work with respect to volume. Work = $\int F\; \mathrm{d}x$ By the chain rule, $\dfrac{\partial W}{\partial V} =\dfrac{\partial W}{\partial x}\dfrac{\partial x}{\partial V} = \dfrac{F}{yz}$.

In this case a is constant and y = z = l.

Therefore, in order for your solution to be correct $\dfrac{ma}{l^2} =\dfrac{\rho_f al}{2} \Rightarrow m = \dfrac{\rho_fl^3}{2}$

My question to you is, is that true?

[quote="bus"]OK, this solution is really simple enough so that I can tell that you have calculated the pressure at the side of the container, not in the middle. The pressure in the center will be half of this (and pressure at the opposite side of the container will be zero).[/quote]That's incorrect. Consider a plane inside the container whose normal is parallel to the direction of travel.
Where you place the plane inside the box is arbitrary, but to satisfy your argument let's say the plane you choose contains the centroid of the box.

As the box itself moves through a distance $\Delta x$, what volume does the plane (containing the centroid) displace? I submit that from the definition of pressure if you compute the work done on the box over the distance $\Delta x$ and divide this by the displaced volume and take the limit of the resulting quotient as $\Delta x$ approaches zero the final result is the pressure on an arbitrary point in the plane (including the centroid).

Exactly! The plane still displaces volume $l^2\Delta x$ but the work changes:

$\dfrac{\partial W}{\partial V}=F=m'a$

where $m'$ is mass of only half of the water. Therefore the pressure is only half of your result.

If you still think I'm wrong, have a look at the far side of the container. You could remove the wall which faces the direction of acceleration and no water will flow out of the container because it's being "pushed inside" - so obviously the pressure at the water surface must be zero.

[quote="bus"]Exactly! The plane still displaces volume $l^2\Delta x$ but the work changes:

$\dfrac{\partial W}{\partial V}=F=m'a$

where $m'$ is mass of only half of the water. Therefore the pressure is only half of your result.[/quote]If you say so, but where'd the other half of the mass go?

Something to think about ...

Last night I had to teach my son's Cub Scout den about Pascal's law (they are working toward a science badge). I only vaguely remembered mention of it from my own freshman physics days. So I looked it up...

[b]Pascal's law:[/b] In a confined fluid, externally applied pressure is transmitted uniformly in all directions. In a static fluid force is transmitted at the speed of sound throughout the fluid and acts at right angles to any surface in or bounding the fluid. -- [i]A Dictionary of Physics [/i][ISBN 0-19-280030-2]

[quote="bus"]However, we're not talking about a pressure externally applied to a confined fluid. You could use the same law to find that the pressure is uniformly distributed throughout a closed container filled with water which is put on a table, this time considering gravity. As you should know, this is wrong because the pressure in fact increases linearly with depth.

I think I'll leave this discussion now, finally aware of the right answer to this problem :P. Strage that it was there the whole time...[/quote]Suit yourself.


Please help with the following problem

Let $ X$ be a set and $ \mathcal D \subset 2^X$, $ \mathcal D$ closed under finite intersection. Denote by $ \mathcal R$ the ring generated by $ \mathcal D$. Let $ \pi$ be the smallest system, $ \mathcal D \subset \pi \subset \mathcal 2^X$ such that $ \pi$ is closed under the following operations:

(i)  finite disjoint unions.

(ii) differences $ A\setminus B$, $ B\subset A$.

Prove that $ \pi \equal{} \mathcal R$.

I have a solution (very complicated), but a solution from book contains the following hint.

We have $ \{A\in \mathcal R | A\cap B \in \pi, B \in \mathcal D\} \equal{} \mathcal R$ since the former is a ring .... I don't really understand why this system is a ring. Help please.

Hint: Show that for every $ B\in\mathcal D$ the set $ M_B\equal{}\{A\in\mathcal R | A\cap B\in\pi\}$ is a ring. For this note that intersection distributes over the symmetric difference. See, e.g., [url]http://en.wikipedia.org/wiki/Ring_of_sets[/url]. Then take the intersection of the family $ M_B$, $ B\in \mathcal D$.

Ok. But my difficulties are still here.

Consider $ A_1,A_2 \in M_B$ so that $ A_1\cap B \in \pi$ and $ A_2 \cap B \in \pi$. We have to show that

(i) $ (A_1 \triangle A_2) \cap B \in \pi$,

(ii) $ (A_1 \cap A_2) \cap B \in \pi$.

Consider (i). You suggest to use $ (A_1 \triangle A_2) \cap B \equal{} (A_1 \cap B) \triangle (A_2 \cap B)$. In spite of  $ C_1 \equal{} A_1\cap B \in \pi$ and $ C_2 \equal{} A_2 \cap B \in \pi$ we can not say that $ C_1 \triangle C_2 \in \pi$ as only $ \mathcal D$ (not $ \pi$) closed under intersection. The same difficulties we have with (ii).

So, I don't know how to prove that $ \pi$ is closed under intersection. May be I'm wrong so explain please.

My own solution uses some general formula for structure of a ring generated by a system of set $ \mathcal D$.

[quote="Oleg1974"]So, I don't know how to prove that $ \pi$ is closed under intersection. May be I'm wrong so explain please.[/quote]
For convenience define a [u]P-system[/u] to be a non-void system of subsets of $ X$ fulfilling the conditions (i) and (ii) above. For any system $ \mathcal A$ of subsets of $ X$ let $ P(\mathcal A)$ the smallest P-system containing $ \mathcal A$. In your notation we have $ \pi\equal{}P(\mathcal D)$.

Claim: Then $ P(\mathcal D)$ is closed under finite intersections if $ \mathcal D$ is.

The proof can be performed by going through the following steps:
(i) Let $ \mathcal P$ be a P-system and $ A$ a subset of $ X$. Then system $ P_A\equal{}\{A\subset X|B\cap B\in \mathcal P\}$ is a P-System.
(ii) For every $ A\in\mathcal D$ it holds $ \mathcal D\subset P_A$, hence $ P(\mathcal D)\subset P_A$. That means that, for $ A\in\mathcal D$, $ B\in P(\mathcal D)$: $ A\cap B\in P(\mathcal D)$.  -- This is already half way of what we want.
(iii) For every $ B\in P(\mathcal D)$ it holds $ \mathcal D\subset P_B$, hence $ P(\mathcal D)\subset P_B$. In other words, for $ A,B\in P(\mathcal D)$: $ A\cap B\in P(\mathcal D)$. ... Done.  :P

Does this help you to complete the proof with my hint in terms of the sets $ M_B$?


An evil king walked into the court of an emperor and challenged him for a duel. The emperor, thinking that he would have to fight against the meek king agreed readily. But the king challenged into a mathematics contest. The king would think of 3 two-digit numbers (x,y,z) and the emperor should tell him any 3 numbers (a,b,c). The king would then tell the emperor ax + by + cz.
From this the emperor should guess the king's 3 numbers. If he guesses right, the king would leave in peace. If he guesses wrong, the emperor would have to forfeit his kingdom.
If you were the emperor what are the numbers you would choose?

Just take a=1,b=100,c=10000.

for eg,
x=12,y=56,z=83.
then the number would be 835612.
Then,he tells x,y,z as the number formed by first two digits,the number formed by next two digits,the number formed by last two digits.
Hope it is right.


geometry trigonometry quadratics algebra

Hi.

Given triangle $ ABC$ with angle $ B$ 90 degrees (the right angle).
Let the hypotenuse equal $ 12$ and the area $ S \equal{} 4$.

Find $ 2\sin{A}$ .

[url=http://img198.imageshack.us/my.php?image=triangle.jpg][img]http://img198.imageshack.us/img198/9529/triangle.jpg[/img][/url]

Thanks.

Ummm....

Let $ BC \equal{} a$ and $ AB \equal{} b$.

We have $ a^2 \plus{} b^2 \equal{} 144$ and $ ab \equal{} 8$. Therefore, $ a^2 \plus{} 2ab \plus{} b^2 \equal{} (a \plus{} b)^2 \equal{} 160$, so $ a \plus{} b \equal{} 4\sqrt {10}$.

Substituting $ b \equal{} 4\sqrt {10} \minus{} a$ in $ ab \equal{} 8$, we have:

$ a(4\sqrt {10} \minus{} a) \equal{} 8 \implies a^2 \minus{} 4\sqrt {10}a \plus{} 8 \equal{} 0 \implies a \equal{} 2\sqrt {10} \pm 4\sqrt {2}$.

Therefore, $ \sin A \equal{} \frac {2\sqrt {10} \pm 4\sqrt {2}}{12} \equal{} \frac {\sqrt {10} \pm 2\sqrt {2}}6$, so $ 2\sin A \equal{} \boxed{\frac {\sqrt {10} \pm 2\sqrt {2}}3}$.

[quote="modularmarc101"]Yeah it's the same as I got. Yours is a little more accurate, since I rounded like 3 times before getting to the last step.[/quote]
When you got to the quadratic equation, you could have factored out 324 from the discriminant, which is a perfect square.

@OP: Are you sure it's not $ \sin 2A$? That answer is a lot nicer looking, and it seems kind of silly to ask for $ 2\sin A$ instead of $ \sin A$.

[quote="Yongyi781"]Anyway,

$ \sin 2A \equal{} 2\sin A \cos A \equal{} \frac {ab}{72} \equal{} \frac 1{72}(2\sqrt {10} \plus{} 4\sqrt {2})(2\sqrt {10} \minus{} 4\sqrt {2}) \equal{} \frac 1{18}(10 \minus{} 8) \equal{} \frac 1{9}$.[/quote]

Which is almost redundant (as a calculation), since $ ab\equal{}2S$ is already given.

[quote="Farenhajt"][quote="Yongyi781"]Anyway,

$ \sin 2A \equal{} 2\sin A \cos A \equal{} \frac {ab}{72} \equal{} \frac 1{72}(2\sqrt {10} \plus{} 4\sqrt {2})(2\sqrt {10} \minus{} 4\sqrt {2}) \equal{} \frac 1{18}(10 \minus{} 8) \equal{} \frac 1{9}$.[/quote]

Which is almost redundant (as a calculation), since $ ab \equal{} 2S$ is already given.[/quote]

How did you do that part? $ 2\sin A \cos A \equal{} \frac {ab}{72}$ ?

$ 2\sin A \cos A \equal{} 2 \frac {bc}{ac} \frac {ab}{ac}$ and $ ac*ac\equal{}144$ . What is $ bc$?

Thanks.

Heh, I let $ a \equal{} AB$ and $ b \equal{} BC$. $ a$ and $ b$ are the side lengths, not the points.


algebra
linear equation
calculus
calculus computations

Find :
$\sum_{n=1}^{\infty}\frac{n}{(2n+1)!!}$ .It is easy to compute ,
$\sum_{n=1}^{\infty}\frac{n}{\frac{(2n+1)!}{n!2^{n}}}$. Then we consider , $xf'(x): =\sum_{n=0}^{\infty}\frac{x^{n}n!}{(2n+1)!}$. Still,I cant think of how to find $f(x)$. Maybe its a product of two series  :huh: , I also tried to write $n!$ as $\prod_{k=1}^{n}k$ and then switching the sum and product ,but I was no luck in finding $f(x)$.Maybe I am missing something obvious :blush: .

Mr Doe,

The value of your series is $\frac{1}{2}$, i.e.,

$\sum_{n=1}^{\infty}\frac{n}{(2n+1)!!}=\frac{1}{2}$.

More generally if $k \in N$ then the value of $\sum_{n=1}^{\infty}\frac{n^{k}}{(2n+1)!!}$ can be calculated exactly by means of recurrence relations.

Let me see if I understood it corectly: You need to calculate

$g(y)=\sum_{n=1}^{\infty}\frac{y^{n}}{(2n+1)!!}$?
I obtained, of course if I made no computational mistakes, that the following diff. eq. holds:

$2yg'(y)=y+(y-1)g(y)$, this is a linear equation which can be solved, but I do not think the solution is nice.

Your $f$ would be given by $xf'(x)=1+g(x/2)$.

Didilica:I dont understand you fully.This is what I did;

$f(x): =\sum_{n=0}^{\infty}\frac{x^{n}}{(2n+1)!!}$.Then is used a supstitution $x: =u^{2}$,Then we have
$uf(u^{2})=\sum_{n=0}^{\infty}\frac{u^{2n+1}}{(2n+1)!!}~/\frac{d}{du}$ ,then we obtain ;
$f(u^{2})+uf'(u^{2})2u=\sum_{n=0}^{\infty}\frac{(2n+1)u^{2n}}{(2n+1)(2n-1)!!}=$
$\sum_{n=0}^{\infty}\frac{u^{2n}}{(2n-1)!!}=\frac{u^{0}}{(-1)!!}+u^{2}\sum_{n=1}^{\infty}\frac{u^{2(n-1)}}{(2(n-1)+1)!!}$.
Then we substitute $t: =n-1$ (but I will still keep on writing $n$ instead of $t$).
$f(u^{2})+2u^{2}f(u^{2})=-1+u^{2}\sum_{n=0}^{\infty}\frac{u^{2n}}{(2n+1)!!}=-1+u^{2}f(u^{2})$.
Then we recall $x=u^{2}$,and we have (something similar to what you got),
$2xf'(x)+(1-x)f(x)+1=0$.

This diff. equation isnt that hard to solve (if we suppose that I didnt have any computational mistakes) :lol: .
I would we grateful Didilica if you could explain you result a little bit further,(concerning $\sum\frac{n}{(2n+1)!!}$ ,and $\sum\frac{n^{k}}{(2n+1)!!}$).

Regards

I used the following idea:

$\frac{n}{(2n+1)!!}=\frac{1}{2}\left(\frac{1}{(2n-1)!!}-\frac{1}{(2n+1)!!}\right)$.

the sum is telescoping.

As for the other sum I used the following trick:

Let $S(k)=\sum_{n=1}^{\infty}\frac{n^{k}}{(2n+1)!!}$.

Now, based on the same decomposition we get that

$S(k+1)=\frac{1}{2}+\frac{a}{2}+\frac{1}{2}\sum_{j=1}^{k}C_{k}^{j}S(j)-\frac{1}{2}S(k)$, where
$a=\sum_{n=1}^{\infty}\frac{1}{(2n+1)!!}$ and $C_{k}^{j}$ is the notation for the binomial coefficient.
I think that, based on this recurrence relation, we may say that these sums are calculated exactly.


How do we get the formula for solving a sequence with higher exponents?

1^n+2^n+3^n...=?

1^n+2^n+...+x^n

if x is very big and you don't want a exact answer you can let (x^(n+1))/(n+1) as the answer

I have tried to find some regulars but failed...

let f(x,n)=1^n+2^n+...+x^n

then f(x,n)=((x(x+1)/2)/(3(n+1)))*(.....)

I make a form of the part(.....) 's coefficient which 1=<n<=9

6
6 3
6 6  0
6 9  1  -1
6 12 3  -3    0
6 15 6  -6    -1       1
6 18 10 -10  -4       4      0
6 21 15 -15  -51/5  51/5  9/5   -9/5
6 24 21 -21  -21     21     9      -9     0

And I can't find any clear regulars :(

hello, i think this will help you
http://mathworld.wolfram.com/PowerSum.html
Sonnhard.


Prove that, the number $3^{105}+4^{105}$ is divisible by $181$


A four-digit number which is a perfect square is created by writing Anne's age in years followed by Tom's age in years. Similarly, in 31 years, their ages in the same order will again form a four-digit perfect square. Determine the present ages of Anne and Tom.

[hide]$\left(m-n\right)\left(m+n\right)=3131\implies m=66, n=35$

From this we have $abcd=1225$ so Anne's age is $\boxed{12}$ and Tom's age is $\boxed{25}$.  This checks out because $66^{2}=4356=\left[12+31\right]\left[25+31\right]$.[/hide]

i don't get how you get $\left(m-n\right)\left(m+n\right)=3131\implies m=66, n=35$ :?

$m^{2}=\left[ab+31\right]\left[cd+31\right]$

$n^{2}=\left[abcd\right]$

The difference gives $m^{2}-n^{2}=\left(m-n\right)\left(m+n\right)=3131$.  Note that the difference is $3131$ because

\[ 1000a+100b+3000+100-\left(1000a+100b\right)=3100  \]

and

\[ 10c+d+30+1-\left(10c+d\right)=31.  \]

Adding these together gives you $3131$ and the only four-digit squares whose difference is $3131$ are $66^{2}$ and $35^{2}$.


Ive always been on the rusty end for changing bases.. can anyone point me in the direction of a comprehensive guide ? thanks

In base 10 the number $7364846$ means:
$7*10^{6}+3*10^{5}+6*10^{4}+4*10^{3}+8*10^{2}+4*10^{1}+6*10^{0}$

In a base 10 number, the right-most digit is the $10^0$ place and as you go left, the exponent is increased by 1.

In a base n number, the right-most digit always is the $n^0$ place and as you go left, the exponent is increased by 1.

The number $7364846$ in base $n$ means:
$7*n^{6}+3*n^{5}+6*n^{4}+4*n^{3}+8*n^{2}+4*n^{1}+6*n^{0}$

When converting bases, unless you are a super genius, always convert to base 10, then to the 2nd base. To convert to the 2nd base, find the largest power of that base that is less than that number. Divide the original number by that power. The quotient is the number in that place and the remainder will be expressed in the remaining places.

Ex:
Convert 843 in base 9 to base 7.
$843$ base 9 = $8*9^{2}+4*9^{1}+3*9^{0}=687$
$687=2*7^{3}+0*7^{2}+0*7^{1}+1*7^{0}$

So our answer is 2001 in base 7.

Bases can sometimes be expressed as a subscript after the number.

Hope this helps.

thanks very much xantos now it's pretty clear:)

Also, you should know base changing tricks.

Change $1001011_2$ to base 4 and base 8
To change from binary to base 4, you can partition the number into pairs like this.

$|01|00|10|11|_2 = 1023_4$

To change from binary to base 8, partition into 3s

$001|001|011_2 = 113_8$

A similar idea works for 3 and 9 etc.

direct number base problems are good, but looking at a seemingly unrelated problem using base 2, etc can lead to a very elegant solution, ex. there are some combinatorics problems that this is helpful

[quote="Agent ';..;'"]thats so cool.. why does that work?[/quote]

Let's say you have a number $N$ and you want to represent it in the base $B$. Then you must have

$N=a_nB^n+a_{n-1}B^{n-1}+\dots+a_1B+a_0$

where $a_k$ are the base-$B$ digits, therefore $0\leqslant a_k\leqslant B-1$

Observe that all the terms on RHS are divisible by $B$ EXCEPT the last one, $a_0$. Since it's smaller than $B$ by definition, it actually represents the remainder in the division $N/B$. When we perform that division and write down the remainder, the quotient will be

$N'=a_nB^{n-1}+\dots+a_2B+a_1$

Now you see that $a_1$ will be the remainder in the new division by $B$, and so on.

[u]Practice Promblems:[/u]
1) What is $121_8$ expressed in base 3?
2) What is $22_3$ expressed in base 6?

[hide]Well, number 1 is definitely 1000, because $2*8^1$ is $16$, not $2$, making it 81, which is 1000 in base 3.

And for the second, I didn't think you could have 55 in base 3?  I thought the highest number you could have was one less than the base i.e. in base 3 you can't have it be higher than 2.  Otherwise it's like counting sixty-ten, etc.[/hide]

[quote="Farenhajt"]And there's also a less brain-consuming procedure:

Keep dividing the base-10 representation by the new base and write down the remainders. Then put the remainders in the opposite order and you get the answer.

Example - represent 245 in base 5

245 / 5 = 49 r. 0
49 / 5 = 9 r. 4
9 / 5 = 1 r. 4
1 / 5 = 0 r. 1

Now you take the remainders and write them bottom-up: 1440. That's the base 5 representation of 245 in base 10.[/quote]

Thanks, I never knew that.

[quote="Apprentice"]A Question:
[hide] Sorry for the mistake... lol I should go back to elementary school...
But i'm still confused why it's 1000:
$1*8^2 + 2*8^1 + 1*8^0 = 64+2+1 = 81$
81/3 = 27 r 0
27/3 = 9 r 0
9/3 = 3 r 0
3/3 = 1 r 0 ??? What now? What would it be?
[/hide][/hide][/quote]
[hide]
then you have 1/3 = 0 R 1. so i believe its 10000

because 81 = 3^4[/hide]

[hide]Yeah, sorry, I left off a zero...

And if number 2 was actually 22, then it's $12$ in base 6.[/hide]


If I put in a hide tag and then a space, the space is ignored. For example,
[code][hide=this]the hide tag getting rid of a following space[/hide] problem is annoying[/code] gives "[hide="this"]the hide tag getting rid of a following space[/hide] problem is annoying" instead of something like "this problem is annoying".

Your post doesn't make any sense to me. The reason it doesn't make any sense is that I [b]am[/b] looking at a following space, so the distinction you're trying to draw attention to doesn't exist.

Perhaps it has something to do with browsers and/or operating systems.

MAC OS 10.5.8, Firefox 3.5.7

And what if I [hide="leave out"](content)[/hide]the following space?

OK, that runs it together - but then I did leave out the space.

My problem is that
[code]
[hide=a]b[/hide]c
[hide=a]b[/hide] c
[/code] give the same thing:
[hide="a"]b[/hide]c
[hide="a"]b[/hide] c
when it should look like:
[color=blue][u]a[/u][/color]c
[color=blue][u]a[/u][/color] c (no hide tag in these ones)

Windows XP, IE8

And I'm telling you that I see two different things in that post. Your pair with the hide tags looks just like your pair without the hide tags.

It must be a problem with IE then, because it works with Firefox. Thanks!  :)


1. $ f_E\minus{}f_A$
$ \equal{}\{1,~x\in E;~~0,~x\notin E\}\minus{}\{1,~x\in A;~~0,~x\notin A\}$
$ \equal{}\{1,~x\in A;~~1,~x\in E\minus{}A;~~0,~x\notin E\}\minus{}\{1,~x\in A;~~0,~x\notin A\}$
$ \equal{}\{0,~x\in A;~~1,~x\in E\minus{}A;~~0,~x\notin E\}$
$ \equal{}\{1,~x\in E\minus{}A;~~0,~x\notin (E\minus{}A)\}$
$ \equal{}f_{E\minus{}A}\equal{}f_{\overline{A}}$.

2a, 2b. [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168304[/url]


AMC
AIME
geometry
area of a triangle
Heron\u0027s formula

how would I prepare for the AIME? please help

1: Do LOTS of past exams, the more recent, the more helpful they are, since that'll probably be what the level is at.
2: Do other problems like the ones that are in the AoPS books.
3: Memorize some commonly used equations, like Hero(n)'s Formula and stuff
4: Visit the board that this topic is on and work on the problems and stuff
5: Say bye-bye to your social life ;)

There have been many other threads on this topic. Check those.

[quote="OKEY"]5: Say bye-bye to your social life ;)[/quote]
So that's why I still can't do well.


calculus
vector
real analysis
real analysis unsolved

i'm reading my calculus course.
i can't understand this rule of calculation with differentials.

it says the following:
if $T\in \mathbb{L(X;Y)}$ and $f(x)=T(x) \forall x\in D$ then
f is differentiable in a and $df(a)=T=f$.

I don't understand it...to me it makes no sens :(
I will probably be posting more questions soon.

If $f: \mathbb{R}^n \to \mathbb{R}^m$, and $a \in \mathbb{R}^n$, and $f$ is differentiable at $a$, then $f'(a)$ is supposed to be a linear transformation $Q$ (for lack of a better letter) with the property that if $h$ is a "small" vector in $\mathbb{R}^n$, then $f(a+h)$ is approximated well by $f(a) + Q(h)$.  (The formal definition of $f'(a)$ expresses this intuitive idea in a way that is precise.)

Suppose that $f$ is actually a linear transformation.  What would be a good approximation to $f(a+h)$?  Well, since $f$ is a linear transformation, $f(a+h)=f(a)+f(h)$ [i]exactly[/i], not just approximately...  So in this case it seems that $f$ itself is a good candidate for being the linear transformation that $f'(a)$ is supposed to be.

This idea could be turned into a rigorous proof that in fact, if $f$ is a linear transformation, then $f'(a)=f$.

I'm not sure if that's exactly what you were asking, but maybe that helps.


advanced fields
advanced fields unsolved

In $ (\mathbb{R}^{2},\plus{})$, let $ A: \equal{}\{(x,0)|x\in\mathbb{R}\}$ and $ B: \equal{}\{(x,y)|x,y\in\mathbb{R},x^{2}\plus{}y^{2}<1\}$. What is $ AB$ and $ BA$?

Presumably as [url=http://mathworld.wolfram.com/RingDirectProduct.html]ring direct products[/url].

That would not be my first guess, because the question was asked about both $ AB$ and $ BA$.  :maybe:

[quote="mlok"]How do you define $ AB$ and $ BA$ in this context?[/quote]
I have no idea what is this problem aiming at. The problem was given in this form and I found no explanation in lecture notes other that standard $ AB \equal{}\{a\plus{}b|a\in A,b\in B\}$. I thought that if $ (x_{1},0)\in A$ and $ (x_{2},y)\in B$ then $ (x_{1},0)\plus{}(x_{2},y) \equal{} (x_{1}\plus{}x_{2},y)$ but I have no idea about the condition $ x^{2}\plus{}y^{2}< 1$ in $ AB$. Does the question make sense if for some topological group $ G$ we have $ A\subset G$ and $ B\subset G$ open? Then at least both $ AB$ and $ BA$ are open subsets of $ G$.

Well, I guess that the answer is $ AB\equal{}BA\equal{}\mathbb{R}\times ]\minus{}1,1[$.

[url=http://en.wikipedia.org/wiki/Minkowski_addition]Minkowski addition[/url] is usually written in the additive form, as $ A\plus{}B$. With this interpretation, your answer is correct. More generally, the Minkowski sum of a set $ A$ with an open ball of radius $ \epsilon$ is the open $ \epsilon$-neighborhood of $ A$.


Apparently, this sentence describes three levels in a cannibal oyster food chain, however, this is just too much for my brain to handle. Someone please explain.

I'm afraid I don't get what you are asking.

1st level: 'Oysters eat.' This is a gramatically correct sentence.
2nd level: 'Oysters oysters eat eat.' This sentence is also gramatically correct. It can be rewritten as 'Oysters that oysters eat eat.' It states that oysters that are eaten by other oysters eat. The subject of the 1st 'eat' is the 2nd 'oysters' and the subject of the 2nd 'eat' is the first 'oysters'.
This can be repeated indefinitely. For example, 'Oysters oysters oysters eat eat eat.' means that oysters tthat are eaten by oysters that, in turn, are eaten by yet other oysters, eat.

This exercise has taught me that it is very difficult to say 'oysters' three times quickly.


By comparing coefficients show that $ P(x \plus{} iy) \equal{} x^{3} \minus{} 3xy^2 \minus{} x \plus{} i(3x^{2}y \minus{} y^{3} \minus{} y)$ is analytic. What does it mean to compare coefficients. I mean we could use the Cauchy Riemann Equations.

Also $ \limsup_{n \to \infty} a_n \equal{} \lim_{n \to \infty} \left(\sup_{k \geq n} a_k \right)$ is a nondecreasing function of $ n$ right?

The limit superior is non-increasing.  As you take away elements of a sequence, it's supremum can either stay the same or get smaller.


It is the...[hide]ONE YEAR ANNIVERSARY OF HIGH SCHOOL MUSICAL[/hide]

WE'RE SOOOOOOOOOOOOARING
FLYIIIIIIIIIIIIIIIIIIIIIIIIING
THERE'S NOT A STAR IN HEAVEEEEEEN THAT WE CAN'T REACHHH

I KNO U GUYZ R ALL GOIN TO CELEBRATE BY WATCHIN UR DVDS 6 TIMES!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!![/hide]

[quote="perfect628"]omg I HATE that movie!!!!!!!!they made us watch it in school, I felt like setting off the fire alarm...[/quote]

we watched it in school too. I thought it was okay.

why did they make you watch it at school?

I'm wondering the same thing.

My sister has the "special edition" DVD.  It's exactly how it originally appeared on TV... with the commercials.

it was a really great movie!  but I think that the 2nd movie will really stink

second movie??

and does nobody remember the name?
"Which team? __________"

No, yesterday was NOT a special day!(except spelling bee, which I flunked.)

Friday is more like it. It was when an award ceremony happened.

Dude, what's so special about that movie? We watched it at school and it wasn't that great... Actually I only remember about half of it. Probably because they kept forwarding it because we didn't have enought time... But seriously, I don't see what's so special about it. It's just a regular movie.

i watched the songs in hindi...not bad dubbing of the lyrics..they are in tune too...

[quote="pianoforte"]why did they make you watch the movie at school?[/quote]

We watched ir for drama class... So pointless... :noo:

my sister won't stop singing those songs. FRUSTRATION RAGES ON. We watched it in band class :huh:

OMFG i LOVE that movie!! lol i know the whole dance to "We're all in this together"! ^^" and the words to all the songs!!

My sister watched it five times in one day. FIVE TIMES. I know all the songs by heart.


inequalities
number theory unsolved
number theory

Solve in positive integers the equation $(a+1)(b+1)(c+1)=2abc$.

Note: the original statement of the problem was $(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})=2$.

Let's assume that $a \le b \le c$.  We have the following inequalities: \begin{eqnarray}1+\frac{1}{a}& < & 2 \\ (1+\frac{1}{a})^{3}& \ge & 2 \\ (1+\frac{1}{a})(1+\frac{1}{b}) & < & 2 \\ (1+\frac{1}{a})(1+\frac{1}{b})^{2}& \ge & 2 \, . \end{eqnarray} From equations (1) and (2), we see that $a$ is either $2$ or $3$.

[b]Case $a = 2$:[/b] From equations (3) and (4), we see that $b$ is either $4$, $5$, or $6$.  The corresponding values of $c$ are $15$, $9$, and $7$, respectively.

[b]Case $a = 3$:[/b] Again from equations (3) and (4), we see that $b$ is either $3$ or $4$.  The corresponding values of $c$ are $8$ and $5$, respectively.

Ah, wonderful. I thought I wasn't getting very far solving it using divisibility. Cheers :D.

Incidentally, if anyone can post a solution exploiting divisibility properties, I would be very interested.


Find all triplets $(x,y,z) \in \mathbb{N}^{3}$ satisfying $x + y + z + 2 = xyz$.

[hide]since you can switch $x$, $y$ and $z$ within the equation, suppose $x \geq y \geq z$
$\\
\Rightarrow 3x-x^3 \geq -2
\\
\Rightarrow x \in \{0,1,2\}$

[u]First case: $x=0$[/u]
Since $x \geq y \geq z$, the only possibility would be $(0,0,0)$ but that just can't be...

[u]Second case: $x=1$[/u]
$\Rightarrow y+z-yz=-3$
Since $y \geq z \Rightarrow 2y-y^2 \geq -3$
$\Rightarrow y \in \{0,1\}$
$y=0$ is impossible and with $y=1$, you have $z+4=z$ which is of course impossible...

[u]Third case: $x=2$[/u]
$\Rightarrow y+z-2yz=-4$
Since $y \geq z \Rightarrow 2y-2y^2 \geq -4$
$\Rightarrow y \in \{0,1,2\}$
$y=0$ is impossible again.
$y=1$ $\Rightarrow$ $z=5$
$y=2$ $\Rightarrow$ $z=2$

So the only option i have left are $(2,1,5)$ and $(2,2,2)$, but i think i made a mistake somewhere, since z turned out to be bigger than both $x$ and $y$ :?  does someone see my mistake? (i probably miss some solutions as well  :( )
[/hide]

What about $(1,3,3)$? And what about permutations of your solutions? Can you show that you found all solutions?  :)

Here's a sketchy solution...
[hide]
First assume WLOG that $x \ge y \ge z$.

$x+y+z+2=xyz$
$x^2+xy+xz+2x=x^2yz$
$x^2+2x+1=(xy-1)(xz-1)$
$(x+1)^2=(xy-1)(xz-1)$

Case 1: $x>2$

So we want to find factors of $(x+1)^2$, but we know we only need to check up to $x+1$ because there will be a complementary factor. Then $(xz-1)|(x+1)^2$ and $xz-1 \le x+1$. Then $z=1$ because if $z \ge 2$ then $xz-1 \ge 2x-1 > x+1$. Thus $(x-1)|(x+1)^2$. It's easy to see that the only common factor $x-1$ and $x+1$ can have is $2$ but since we square $x+1$, $4$ is also possible. So $x-1=2, 4 \Rightarrow x=3,5$. We then get $y=3, 2$. So we have the solutions $(3,3,1)$, $(5,2,1)$, and all permutations of them.

Case 2: $x=1$

If $x=1$, then $1 \ge y \ge z \Rightarrow y=z=1$ which is not a solution.

Case 3: $x=2$

If $x=2$, then $(2+1)^2 = (2y-1)(2z-1)$ and we get the solution $(2,2,2)$ and all of its permutations.[/hide]

[quote="paladin8"][quote="Psycho"]since you can switch $x$, $y$ and $z$ within the equation, suppose $x \geq y \geq z$
$\\
\Rightarrow 3x-x^3 \geq -2
\\
\Rightarrow x \in \{0,1,2\}$
[/quote]

$x^3 \ge xyz$ so you cannot change the $=$ to a $\ge$.[/quote]
woops, yeah  :blush:

and i knew the solution $(1,3,3)$ existed as well, that's why i knew i made a mistake  :?
anyhow, i think Paladin has them all...  :)


Let $ V$ be a complex inner product space and $ L: V\to\mathbb{C}$ a nonzero linear functional defined on $ V.$ Let $ A \equal{} \{x\in V: L(x) \equal{} 1\} \equal{} L^{ \minus{} 1}(\{1\}).$ Suppose that $ A$ has a unique element of least norm, which we call $ z$.

Prove that $ L(x) \equal{} \frac1{\|z\|^2}\langle x,z\rangle$ for all $ x\in V.$

Suppose $ L$ is not continuous.
Then there is a sequence $ x_1,x_2,\cdots$ of elements of $ V$ with $ \parallel{}x_n\parallel{} \equal{} 1$ and $ c_n: \equal{} |L(x_n)| > n$. ($ n\in \mathbb{N}$)
Then $ L(e^{i\phi_n}\frac {x_n}{c_n}) \equal{} 1$ for certain $ \phi_n$, ($ n\in \mathbb{N}$).

The set $ \{e^{i\phi_n}\frac {x_n}{c_n} | n \in \mathbb{N}\} (\subset A)$ has no smallest (in norm) element,
because $ \parallel{}e^{i\phi_n}\frac {x_n}{c_n}\parallel{} < \frac {1}{n}$ for all $ n$.
Conflict. Thus $ L$ has to be continuous.

Then we can write $ L(x) \equal{} < x,a >$ for certain $ a\in V$ (as we all know).
If $ u\in A$, then $ 1 \equal{} L(u) \equal{} < u,a > \le \parallel{}u\parallel{}\cdot \parallel{}a\parallel{}$,
or ($ a\ne 0$!) $ \parallel{}u\parallel{} \ge \frac {1}{\parallel{}a\parallel{}} \equal{} \parallel{}\frac {a}{\parallel{}a\parallel{}^2}\parallel{}$
Note that $ \frac {a}{\parallel{}a\parallel{}^2}$ is also an element of $ A$, and that the previous line shows that it is a smallest, so the! smallest element $ z$ of $ A$ (in norm).
$ z\parallel{}a\parallel{}^2 \equal{} a$, shows that $ \parallel{}z\parallel{}\cdot \parallel{}a\parallel{}^2 \equal{} \parallel{}a\parallel{}$ so that $ \parallel{}a\parallel{} \equal{} \frac {1}{\parallel{}z\parallel{}}$ and $ a \equal{} \frac {z}{\parallel{}z\parallel{}^2}$.
So $ L(x) \equal{} \frac {1}{\parallel{}z\parallel{}^2} < x,z >$ for all $ x \in V$.

It is enough to prove that $ z\in ker(L)^\perp$. Suppose there exists $ \alpha\in \ker(L)$ such that $ Re( < z,\alpha > )\neq 0$ (else take $ i\alpha$).
Now define $ z' \equal{} z \minus{} \alpha$. $ L(z') \equal{} L(z) \equal{} 1$ hence $ z'\in A$.
We want $ \parallel{}z'\parallel{}^2 \equal{} \parallel{}z\parallel{}^2 \minus{} 2Re < z,\alpha > \plus{} \parallel{}\alpha\parallel{}^2 < \parallel{}z\parallel{}^2$ which can be achieved by substituting $ \alpha$ with $ t\alpha$ for $ t$ sufficiently small.
Hence $ z\in \ker(L)^\perp$.
Let $ x\in V$. Then $ L(x \minus{} L(x)z) \equal{} L(x) \minus{} L(x)L(z) \equal{} 0\Rightarrow x \minus{} L(x)z\in\ker(L)\Rightarrow$
$ < x \minus{} L(x)z,z > \equal{} 0\Rightarrow < x,z > \equal{} L(x)\parallel{}z\parallel{}^2$ .

[quote="PeterP"]Then we can write $ L(x) \equal{} < x,a >$ for certain $ a\in V$ (as we all know).[/quote]
My motivation here was to find a way to prove this assertion, so I didn't want to use it - hence, I was looking for Kuba's argument. Thanks.


After homogenizing it sufficies to show that:
$2(a^3+b^3+c^3) \geq a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2$
But:
$2(a^3+b^3+c^3) - \sum a^2b = \sum (a-b)^2(a+b) \geq 0$

Ha ha. It is no hard :D
Put $a=1-b-c$ to the inequality and then expand it. We get:
$V(b,c)= \frac27-b-c+b^2+\frac{16}{7}bc+c^2-\frac97 b^2c-\frac97 bc^2$
Then rewrite it as:
$v(b,c)= (1-\frac97 c)(b+\frac{c}{2} -\frac12)^2+\frac{1}{28}(c+1)(3c-1)^2$
Done :!:

[quote="cezar lupu"]I don't quite like your solution, Gibb.  :)[/quote]
  I don't like it either :D . But its technique is very simple and helps me solve a lot of inequalities. So I use it very often.
  S.O.S is powerful but I don't like it. :P


For what value of $ b$ is the set $ \{2b\minus{}3,b,b\plus{}2\}$ equal to the set
$ \{4,5,6\}$?

If $ b\equal{}4$, then $ 2b\minus{}3\equal{}5$ and $ b\plus{}2\equal{}6$ as desired.


[quote="solafidefarms"][hide]To start with, we make a list of the possible squares:
0
1
4
9
16
25
36
49
64
81
Then we figure out how we can combine them into pairs so that the sum of the pair is less than 100: We will only count the pairs with the smaller number first to avoid overcounting.
We can combine 0 with any of the numbers except itself for 9.
We can combine 1 with any of the numbers for 9.
We can combine 4 with any of the numbers  for 8.
We can combine 9 with any of the numbers for 7.
We can combine 16 with any of the numbers for 6.
We can combine 25 with any of the numbers except 81 for 3.
We can combine 36 with any of the numbers except 81 and 64 for 1.
We can combine 64 with none of the numbers larger for 0.
We can combine 81 with any of the numbers larger for 0.
Thus there are 43 numbers that fit.

So the probability is 43/100.[/hide]

Billy[/quote]

I would say [hide]42/99.  42 instead of 43 because 0+25=9+16.  99 instead of 100, because there are only 99 numbers, not 100.  Simplified, this equals [i][b]14/33[/b][/i].[/hide]
Am I correct?

[quote="MCrawford"]Randomly select a whole number $x$ such that $0 < x < 100$.  What is the probability that $x$ equals the sum of two perfect squares?  Express your answer as a fraction.[/quote]

[hide="solution"]
Something over 99 o_o .

0 is a perfect square, right?
[/hide]

[hide]can't you just do this?
10*9/2
90/2
45
45/99
5/11
>>5/11<<[/hide]
Or is that utterly wrong?

dwx, I can tell you you're wrong :(  I believe the correct answer has been given, and that the answer given on the answer sheet was actually wrong the year of the test, because there are 4 numbers that can be made by the sum of two squares in two different ways(ex. 50 = 25+25 and 49+1)

[quote="dwx314"][quote="mathgeek2006"][hide]I would say 76/99. :?[/hide][/quote]
You know, I'd love to say thats correct and Im wrong and be done with it and get yet another humiliation, but I cant. Could you please show your work?[/quote]

[hide]1+1
1+4
1+9
1+16
1+25
1+36
1+49
1+64
1+81
(9)
4+ = (9)
9+ = (9)
16+ = (9)
25+ = (9)
36+ =  (8)
49 + = (8)
64 + = (5)
81 + = (4)
9+9+9+9+9+8+8+5+4=70

I am hereby changing my answer to 70/99[/hide]

mathgeek,

Wouldn't 4+9 be the same as 9+4???  Seems like you are double counting quite a bit

Yeah Mathgeek, as you go up the line of squares you need to subtract one less from each one like, if you fix a base number as 9 the other numbers should only have 7.

[hide]
A base of zero will have 9 values.
A base of 1 will have 9 values.
A base of 4 will have 8 values.
A base of 9 will have 7 values.
A base of 16 will have 6 values.
A base of 25 will have 4 values.
A base of 36 will have 1 values.
A base of 49 will have 0 values.
and the rest will have 0 values.  So 9+9+8+7+6+4+1=44.  So 44/99=4/9[/hide]

Yeah...Um I knew that all along.... :blush:


calculus
integration
trigonometry
calculus computations

Find the minimum value of $\int_0^{\frac{\pi}{2}} |\cos x -a|\sin x \ dx$

$\left( {\frac{4}{{3\pi }} - 1} \right)^2  $

For every real [b]a[/b] , let $ \, \, \, \, 0 \, \, \le \, \,
f_a \left( x \right) : = \left| {\cos \,x\,\, - a} \right|\,\,sen\,x\,\,\,,\,\,\forall x \in \left[ {0,\,{\pi  \mathord{\left/
 {\vphantom {\pi  2}} \right.
 \kern-\nulldelimiterspace} 2}} \right] $

Then $\, \,
0\,\,\, \le \,\,f_{a_0 } \left( x \right)\,\,\, \le \,\,f_a \left( x \right) \, \, \, \, $  implies \[
\int\limits_0^{{\pi  \mathord{\left/
 {\vphantom {\pi  2}} \right.
 \kern-\nulldelimiterspace} 2}} {f_{a_0 } \left( x \right)} \,\,dx\,\,\, \le \,\,\int\limits_0^{{\pi  \mathord{\left/
 {\vphantom {\pi  2}} \right.
 \kern-\nulldelimiterspace} 2}} {f_a \left( x \right)\,dx} \,\,\,\, \Leftrightarrow \,\,\,\,\int\limits_0^{{\pi  \mathord{\left/
 {\vphantom {\pi  2}} \right.
 \kern-\nulldelimiterspace} 2}} {f_{a_0 }^2 \left( x \right)} \,\,dx\,\,\, \le \,\,\int\limits_0^{{\pi  \mathord{\left/
 {\vphantom {\pi  2}} \right.
 \kern-\nulldelimiterspace} 2}} {f_a^2 \left( x \right)\,dx}
\]  The fact that $\, \,
\int\limits_0^{{\pi  \mathord{\left/
 {\vphantom {\pi  2}} \right.
 \kern-\nulldelimiterspace} 2}} {f_a^2 \left( x \right)\,dx} \,\, = \int\limits_0^{{\pi  \mathord{\left/
 {\vphantom {\pi  2}} \right.
 \kern-\nulldelimiterspace} 2}} {\left( {\cos x - a} \right)^2 sen^2 x\,\,\,dx}  = \frac{\pi }{4}a^2  - \frac{2}{3}a + \frac{\pi }{{16}}  \, \, \, \, \, $

let us conclude that $ \, \, \,
\mathop {\min }\limits_{a\, \in \,{\rm{R}}} \,\,\int\limits_0^{{\pi  \mathord{\left/
 {\vphantom {\pi  2}} \right.
 \kern-\nulldelimiterspace} 2}} {f_a^2 \left( x \right)\, dx} \,  \, \, \, $ occurs when  $\, \, \, \, \, \, a_0  = \frac{4}{{3\pi }} $

Therefore the minimum value asked is
\[
\int\limits_0^\theta  {\left( {\cos x - a_0 } \right)\,\,senx\,\,dx} \,\,\,\, + \int\limits_\theta ^0 {\left( {a_0  - \cos x} \right)\,\,senx\,\,dx}  = 2\,\,\int\limits_0^\theta  {\left( {\cos x - a_0 } \right)\,\,senx\,\,dx} \,\,\,,
\]

where $ \, \, \,
0 \le \,\,\,\theta  = \cos ^{ - 1} \left( {a_0 } \right) \le {{\,\,\pi } \mathord{\left/
 {\vphantom {{\,\,\pi } 2}} \right.
 \kern-\nulldelimiterspace} 2}  \, \, \, \, $. The solution (if correct) follows by usual calculations and substitutions.

totally not sure, but i got 1/4

edit: (when a = 1/2)

[quote="isaacchao"]totally not sure, but i got 1/4[/quote]

Yes, your answer is correct, isaacchao. :)

kunny

Congratulations, isaacchao.

Let me ask you (and kunny) some help... could you spot where my solution is wrong?  Although I am sure it is, I can´t see why!    :(

Thanks in advance.
fskilnik

here's what i did,

 the easy cases are when a is >=1 and <=0.

when a>=1, we find out that integration gives a - 1/2, in which case minimun is 1/2
when a<=0, integration gives 1/2 - a, minimun is 1/2 as well

when 0<a<1, we defind the "pivot point" at a = cosx, x = arccosa

then the integral =

int 0->arccosa  (cosx-a)sinx dx + int arccosa -> pi/2 (a-cosx)sinx dx

which can be evaluated with ease.

we get a^2 - a + 1/2

= (a-1/2)^2 + 1/2 - 1/4

the expression is minimized when a = 1/2, which is in our defined range.

minimun = 1/2-1/4 = 1/4

Thank you both for your responses.  Now I see my mistake and it is (perhaps) nice to put it explicitly:

My "solution" accepts that there exists $\, \, a_{0} \, \, $ such that $ \, \,
f_{a_0 } \left( x \right)\,\,\, \le \,\,f_a \left( x \right) \, \, \, $ for ALL $ \, \,
0 \le x \le {\pi  \mathord{\left/
 {\vphantom {\pi  2}} \right.
 \kern-\nulldelimiterspace} 2} \, \, \, $   but...

$
f_{\frac{1}{2}} \left( 1 \right) < f_{\frac{4}{{3\pi }}} \left( 1 \right)
\, \, \, $ and $ \, \, \, \, \,
f_{\frac{1}{2}} \left( {{3 \mathord{\left/
 {\vphantom {3 2}} \right.
 \kern-\nulldelimiterspace} 2}} \right)\,\, > \,\,\,f_{\frac{4}{{3\pi }}} \left( {{3 \mathord{\left/
 {\vphantom {3 2}} \right.
 \kern-\nulldelimiterspace} 2}} \right) \, \, \, \, $ ....   :oops:


Follows directly from Muirhead, doesn't it?

Edit:  Well, Rearrangement too.  Which is easier to see if you make the substitution $a = u^{2}, b = v^{2}, c = w^{2}$.

[quote="kunny"]Let $a,\ b,\ c$ be positive numbers.Prove that $a^{4}+b^{4}+c^{4}\geq abc(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}).$[/quote]

It is not difficult. Just use $a^{2}+b^{2}+c^{2}\geq ab+bc+ca$, use it three times. :P


algebra polynomial logarithms Vieta search

Let r, s and t be the roots of the cubic polynomial

p(x) = x^3 -2007x + 2002

Determine the value of:
[u]r-1[/u] +   [u]s-1[/u] + [u]t-1[/u]
r+1                   ( s+1 )       (t+1)

You differentiated wrong as well, but your solution is very nice.

Thank you. :lol: I myself have noticed the miscalculation in differentiation.

My solution is same as of boxedexe, and here is another link [hide="Canada 1996"]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1411308186&t=77670[/hide]

[hide]
Slightly more boring, my solution on the day:

$(x-r)(x-s)(x-t)=x^{3}-2007x+2002$
Multiplying out, and equating coefficients:
$rst=-2002$
$r+s+t=0$
$rs+st+rt=-2007$

Note that:
$\frac{r-1}{r+1}+\frac{s-1}{s+1}+\frac{t-1}{t+1}= \frac{3rst+(rs+st+rt)-(r+s+t)-3}{rst+(rt+st+rs)+(r+s+t)+1}=$
$=\frac{-8016}{-4008}=2$
[/hide]


Find all triples $ (a,b,c)$ of real numbers that satisfy:

$ a^{3}\plus{}b^{3}\plus{}c^{3}\equal{}24$ and $ (a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}64$

I think that there are infinitly many reals numbers solutions to the system  in the case :  $ a,b , c$ not necessarly positves.

if $ a, b$ and $ c$ are positives we have $ (a\plus{}b\plus{}c)^3\equal{}a^3\plus{}b^3\plus{}c^3\plus{}3(a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}216$ so $ a\plus{}b\plus{}c\equal{}6.$

now we have the system $ x\plus{}y\plus{}z\equal{}12$, and $ xyz\equal{}64$ where $ x\equal{}a\plus{}b, y\equal{}b\plus{}c$ and $ z\equal{}c\plus{}a$ which have infinitly many

solutions in the first case .

but if $ a , b$ and $ c$ are positive then $ x, y$ and $ z$ are too positives and by AM-GM $ x\equal{}y\equal{}z$ so $ a\equal{}b\equal{}c\equal{}2$.


for positive numbers $a_1,a_2,\cdots,a_n,b_1,b_2,\cdots,b_n$ such that $\sum_{k=1}^na_k=\sum_{k=1}^nb_k$,$n \ge 3$
Prove or disprove:\[ a_1^{n-1}b_1+a_2^{n-1}b_2+\cdots+a_n^{n-1}b_n+a_1a_2\cdots a_n \ge (n+1)b_1b_2\cdots b_n \]

Actually, this inequality is open for $n \geq 4$.
Know somebody to solve it for $n=4$ ?
Its form is so simple! :o

I think equality occurs for $n \geq 4$ only if all $a_i$ and $b_i$ are equal.
A hint for $n \geq 4$ is to use Cauchy-Schwarz Inequality:
$(\frac{1}{b_1}+...+\frac{1}{b_n})(a_1^{n-1}b_1+...+a_n^{n-1}b_n) \geq (a_1^k+...+a_n^k)^2$,
where $k=\frac{n-1}{2}$. Finally, we have to show that
$(a_1^k+...+a_n^k)^2+na_1a_2...a_n \geq n(n+1)$
for $a_1+a_2+...+a_n=1$. To prove this inequality we may use EV-Theorem.
We have to show that
$(x^k+(n-1)y^k)^2 +nxy^{n-1} \geq n(n+1)$
for $0<x\leq y$ and $x+(n-1)y=1$. The proof is enough difficult. :(

thanks very much.this approach is much different from $n=3$. :)


Hello Everyone!

I'm new to the forum, but I plan to stay long.  I am currently in Physics 101 (Mechanics).  Currently, we are on the Rigid Body Equilibrium and Stress/Strain/Moduli.
This is one of the problems I am baffled on.  I'd appreciate it, if someone could show me how to solve it, ASAP.

90. A moonshiner produces pure ethanol late at night and stores in a stainless steel tank in the form of cylinder 0.300m in diameter with a tight-fitting piston at the top.  The total volume of the tank is 250L (0.0250 m^3).  In an attempt to squeeze a little more into the tank, the moonshiner piles 1420kg of lead bricks on top of the piston.  What additional volume of ethanol can the moonshiner squeeze into the tank? (Assume that the wall of the tank is perfectly rigid.)

P.S.  The answer is -0.0541 L.

Thank you!
-Shingo

The pressure inside of the tank will be the atmospheric pressure 100 kPa plus 1420 * 9.81 / 0.3 / 0.3 / pi Pa = 50 kPa, 150 kPa altogether. The temperature should be the same as the outside temperature so you only need to find out how does the density of ethanol change with pressure.... unfortunately, I don't know the answer :(.

I did the same, found the change in volume then couldn't see how the ethanol related to the problem. I have that book too, but havent covered that material yet.

But how did you find the change in volume? Ethanol is under normal conditions liquid right? I wonder if a negative answer means that it is supposed to expand with increasing pressure...


search
college
inequalities proposed
inequalities

If $x, y, z$ are positive real numbers, then
$\frac{1}{x^{2}+y^{2}}+\frac{1}{y^{2}+z^{2}}+\frac{1}{z^{2}+x^{2}}\ge \frac{5}{2(xy+yz+zx)}$

Do you think is it possible to find an elegant solution without full expansion? :maybe:

Many thanks to everybody will think of it.

Two very NICE and ELEGANT solutions. A lot of thanks for Ehku and Shalex. :)

I'm very very sorry for the spelling mistakes which have been corrected.

Another solution of mine; assume $x\geqslant y\geqslant z$, denote
$f(x,y,z)=\frac{1}{x^{2}+y^{2}}+\frac{1}{y^{2}+z^{2}}-\frac{5}{xy+yz+zx}$
It's very easy to show
$f(x,y,z)\geqslant f(0,x+y,z)$
And
$f(0,x+y,z)\geqslant0\\ \Longleftrightarrow\frac{1}{(x+y)^{2}}+\frac{1}{(x+y)^{2}+z^{2}}+\frac{1}{z^{2}}\geqslant\frac{5}{2(x+y)z}\\ \Longleftrightarrow\frac{z}{x+y}+\frac{x+y}{z}+\frac{1}{\frac{z}{x+y}+\frac{x+y}{z}}\geqslant\frac{5}{2}$
With the property of $f(t)=t+\frac{1}{t}$ it's obvious. :wink:


limit
real analysis
real analysis unsolved

hello, the searched limit is $ \frac{e}{2}$.
Sonnhard.

[quote="aktyw19"]$ \lim_{n\rightarrow\infty}n^{2}\left(\left(1 \plus{} \frac {1}{n \plus{} 1}\right)^{n \plus{} 1} \minus{} \left(1 \plus{} \frac {1}{n}\right)^{n}\right)$[/quote]

Hint: $ (1\plus{}x)^{1/x} \equal{} e(1 \minus{} x/2 \plus{} O(x^2))$ for small $ x$.


geometry
rectangle
3D geometry
sphere
Support
topology
analytic geometry

is this posssible:

"Please draw a rectangle with three lines."

Well, if you use the edge of the paper as one side, you can draw the other 3 sides and technically it is a rectangle.

draw a rectangle and 3 lines next to it

like this:
________
|            |   |  |  |
|_______|   |  |  |

doh

I believe the best answer is, indeed, to consider drawing on a sphere: then it's possible. It evidently isn't possible just on the Cartesian plane, but, hey, being as the question is pretty open-ended.

hm, my scetch doesn't work :(
this is a rectangle drawn with 3 lines:
________
|_______| | | |

The response about using Non-Euclidean geometry is pretty interesting.  So I suppose one line would be the width, and then two parallel lines to span all the way around to be the lengths?  I guess the next question would be to draw a triangle with 270 internal degrees...

If you draw it on a klein bottle, I believe you can draw a rectangle with only two lines

Draw a square, and then use three lines off of that square to draw a rectangle. That is what I would do, anyways.

[b]The following is inconclusive, though mildly interesting.[/b]

[hide]$ figure$ [hide="#1"][img]http://img205.imageshack.us/img205/8146/stripvo9.jpg[/img]
In [b]figure #1[/b], $ n=4$ implies $ L=13$. [/hide]
Note that the addition of a rectangle is analogous of the addition of 3 lines, when at least [i]one[/i] rectangle precedes it. The resulting strip of boxes has characteristics obeying
$
\boxed{L=3n+1}
$
It follows immediately that the average number of lines per square is represented by
$
%Error. "display" is a bad command.
\boxed{\bar{L}=\frac{3n+1}{n}}
$
The number of rectangles relates to the average lines per rectangle in such a way that $ \bar{L}\propto n^{-1}$, so as $ n$ increases, $ \bar{L}$ decreases; and ultimately :

$ \bar{L}=\frac{3n+1}{n}=3+\frac{1}{n}$ ; let $ n=\infty$
$ \boxed{\bar{L}_{\infty}=3+\frac{1}{\infty}=3}$

Thus, The average number of lines per square in a rectangle strip consisting of infinite rectangles is 3. My results do not support a 2D construction of a rectangle using only 3 lines.[/hide]

:noo:  You could probably make a real nice topology intro out of it.
(The latter is the only serious statement in this post)


:blush: :blush:
 Find the limit of the following sequence when $n\to+\infty$
$x_{n}=\prod_{k=0}^{n}\frac{2^{2^{k}}+1}{2^{2^{k}}}$

You could have also used induction to prove that identity.

[quote="pleurestique"]notice that $\prod_{k=0}^{n}\frac{x{}^{2{}^{k}}+1}{x{}^{2{}^{k}}}= \prod_{k=0}^{n}(1+\frac{1}{x{}^{2{}^{k}}}) = \sum_{i=0}{}^{2{}^{n+1}-1}\frac{1}{x^{i}}$. this is because in the expansion of $\prod_{k=0}^{n}(1+\frac{1}{x{}^{2{}^{k}}})$, every term is of the form $\frac{1}{x^{i_{0}2^{0}+i_{1}2^{1}+...+i_{n}2^{n}}}$, where each $i_{k}$ equals 0 or 1, and for every such $n+1$-tuple $(i_{0},...,i_{n})$, there is exactly one term of this form; the claim then follows from the fact that every number has a unique binary representation, and so we get all exponents from 0 to $\underbrace{11...1}_{n+1}_{2}= 2^{n+1}-1$.

the limit is therefore $\sum_{i=0}^{\infty}\frac{1}{x^{i}}= \frac{1}{1-\frac{1}{x}}=\frac{x}{x-1}$. in the case of $x=2$ we get $\frac{2}{2-1}=2$ as the limit.[/quote]
thank you for a nice solution,[b]pleurestique[/b]!
:D


Find the minimum strictly positive value of $x^2 + y^2$ given \[ xy(x^2-y^2)=x^2+y^2 \]