5th Problem Set is up. Comments appreciated.

Should those be distinct integers in number 6?

No, that is not necessary.

haha, nice signature.

What do you guys think about the 4th PSet?

That's a good idea...If I do this kind of thing, I could probably actually make a successful math club!

The problems are good.  For the regular AoPSer, they're quite easy, but for one who is new to this, they'd be very fun to do!

Problem is, Mandelbrot starts soon, so I don't have much time left...looks like I have just get it started, and can really get it going next year....perhaps just AMC, AIME, USAMO etc this year...

Yeah all you need in order to have a successful math club is some one with initiative and the right resources.   :wink:

Yeah most of them are pretty easy. Did you like number 14 though?

Any other comments on the 4th set? I'm going to post the 5th one soon.


all integers in the sequence are $0.3recurring(10^n-1)$ (multipy out the brackets dont add first). But I think I can leave the sum now for you to do because I dont know what to do next :(  :?   :oops: ? Can anyone help me? :(  :?  :spider:

cool solution beats my attempt :(  :blush:  :?


[quote="8isgreat"]Let point E be the midpoint of side BC of triangle ABC; BD the altitude of the triangle. Find R[size=59]BDE[/size] if AB=30 cm, BC=26 cm, and AC=28 cm[/quote]

$ P_{ABC}\equal{}\frac{30\plus{}26\plus{}28}{2}\equal{}42$
$ S_{ABC}\equal{}\sqrt{42(42\minus{}30)(42\minus{}26)(42\minus{}28)}\equal{}336$   ---> $ BD\equal{}\frac{2S_{ABC}}{AC}\equal{}\frac{2*336}{28}\equal{}24$   ---> $ \angle BDC\equal{}90^{0}  \minus{}\minus{}\minus{}> BE\equal{}EC\equal{}DE\equal{}\frac{BC}{2}\equal{}13$ --->
$ P_{BDE}\equal{}\frac{DE\plus{}BE\plus{}BD}{2}\equal{}\frac{13\plus{}13\plus{}24}{2}\equal{}25$
  $ S_{BDC}\equal{}\sqrt{25(25\minus{}13)(25\minus{}13)(25\minus{}24)}\equal{}60$ ----->
$ R_{BDE}\equal{}\frac{abc}{4S}\equal{}\frac{13*13*24}{4*60}\equal{}\frac{507}{30}$


Find integers $x,y,z$ that satisfy
$\frac{13}{x^2}+\frac{1996}{y^2} = \frac{z}{1997}$


Suppose that the triangle ABC has:
$ \{\begin{array}{l}tanA\cdot tanB=3\\tanB\cdot tanC=6\end{array}$
Prove that:  $ \widehat{A}=\frac{\pi}{4}$

[hide]Dividing both relations, we get $ \tan C \equal{} 2 \tan A$. From the first equation we have $ \tan B \equal{} \frac{3}{\tan A}$. Therefore, making $ \tan A \equal{} x$,

$ x \plus{} \frac3x \plus{} 2x \equal{} 6x$

from which we get x = 1.[/hide]

[quote="Carcul"]$ x \plus{} \frac3x \plus{} 2x \equal{} 6x$[/quote]
Exactly [i]how[/i] did you get this equation?

It's known that in $ \triangle ABC$ holds:

$ tgA\cdot tgB\cdot tgC\equal{}tgA\plus{}tgB\plus{}tgC$
,so that's how he got the equation .


[quote="bane"]solve: $\pi\cos x=|x-\frac{\pi}{2}|$

is there any algebraic method to find the second value ??
(apart from the obvious one which is $x=\frac{\pi}{2}$)[/quote]

I don't know, but we can consider case work.

$\pi\cos x=x-\frac{\pi}{2}$ or $\pi\cos x=\frac{\pi}{2}-x$

If anyone finds an algebraic solution please post it.


linear algebra
matrix
vector
inequalities
quadratics
function
triangle inequality

Suppose $ A$ and $ B$ are two $ 2\times2$ symmetric matrices over $ \mathbb{R}$ which are both positive definite. And let $ \det \left(A\right) \equal{} \det \left(B\right)$. Prove that $ \det \left(A \minus{} B\right) \le0$.

Since both matrices are symmetric, and at least one of the matrices is positive definite, there is $ U \in O(n)$ such that
\[ U^*AU \equal{} \Lambda_A \quad \text{ and } \quad U^*BU \equal{} \Lambda_B,
\]
where $ \Lambda_A$ (resp., $ \Lambda_B$) is a diagonal matrix with eigenvalues of $ A$ (resp., of $ B$) on the diagonal.

We have
\[ \det(A \minus{} B) \equal{} \det(U^*(\Lambda_A \minus{} \Lambda_B)U) \equal{} \det(\Lambda_A \minus{} \Lambda_B).
\]
So your problem is reduced to the case where both $ A$ and $ B$ are diagonal. In this particular case, the statement is obviously true. So we are done.

Since the matrices $ A$ and $ B$ are symmetric, we have $ A\equal{}\left(\begin{array}{cc}a&c\\ c&b\end{array}\right)$ and $ B\equal{}\left(\begin{array}{cc}x&z\\ z&y\end{array}\right)$ for some reals $ a$, $ b$, $ c$, $ x$, $ y$, $ z$. The reals $ a$, $ b$, $ x$, $ y$ are positive (in fact, the matrix $ A$ is positive definite, so that $ v^TAv>0$ for all vectors $ v\in\mathbb{R}^2$ with $ v\neq 0$, so particularly $ e_1^TAe_1>0$, but $ e_1^TAe_1\equal{}e_1^T\left(\begin{array}{cc}a&c\\ c&b\end{array}\right)e_1\equal{}a$, and thus $ a>0$, and similarly $ b>0$, $ x>0$ and $ y>0$).

Let $ t\equal{}\det A\equal{}\det B$. Then, $ t\equal{}\det A\equal{}\det\left(\begin{array}{cc}a&c\\ c&b\end{array}\right)\equal{}ab\minus{}c^2$, so that $ c^2\equal{}ab\minus{}t$ and thus $ \left|c\right|\equal{}\sqrt{ab\minus{}t}$. Similarly, $ \left|z\right|\equal{}\sqrt{xy\minus{}t}$. Note that $ t>0$, because $ t\equal{}\det A$ and $ \det A>0$ (in fact, $ A$ is positive definite and thus has positive determinant).

Now, by the triangle inequality,

$ \left|c\minus{}z\right|\geq\left|\left|c\right|\minus{}\left|z\right|\right|\equal{}\left|\sqrt{ab\minus{}t}\minus{}\sqrt{xy\minus{}t}\right|\equal{}\left|\frac{\left(ab\minus{}t\right)\minus{}\left(xy\minus{}t\right)}{\sqrt{ab\minus{}t}\plus{}\sqrt{xy\minus{}t}}\right|$
$ \equal{}\left|\frac{ab\minus{}xy}{\sqrt{ab\minus{}t}\plus{}\sqrt{xy\minus{}t}}\right|\equal{}\frac{\left|ab\minus{}xy\right|}{\sqrt{ab\minus{}t}\plus{}\sqrt{xy\minus{}t}}$
$ \geq\frac{\left|ab\minus{}xy\right|}{\sqrt{ab}\plus{}\sqrt{xy}}$ (since $ \left|ab\minus{}xy\right|\geq 0$ and $ \sqrt{ab\minus{}t}\plus{}\sqrt{xy\minus{}t}\leq\sqrt{ab}\plus{}\sqrt{xy}$, the latter because of $ t\geq 0$)
$ \equal{}\left|\frac{ab\minus{}xy}{\sqrt{ab}\plus{}\sqrt{xy}}\right|\equal{}\left|\sqrt{ab}\minus{}\sqrt{xy}\right|$.

Since both sides of this inequality are nonnegative, we can square it:

$ \left|c\minus{}z\right|^2\geq\left|\sqrt{ab}\minus{}\sqrt{xy}\right|^2$.

In other words,

$ \left(c\minus{}z\right)^2\geq\left(\sqrt{ab}\minus{}\sqrt{xy}\right)^2$.

But $ \left(\sqrt{ab}\minus{}\sqrt{xy}\right)^2\equal{}ab\plus{}xy\minus{}2\sqrt{ay\cdot bx}$. Since the AM-GM inequality yields $ 2\sqrt{ay\cdot bx}\leq ay\plus{}bx$, we thus have $ \left(\sqrt{ab}\minus{}\sqrt{xy}\right)^2\geq ab\plus{}xy\minus{}\left(ay\plus{}bx\right)$. Thus,

$ \left(c\minus{}z\right)^2\geq\left(\sqrt{ab}\minus{}\sqrt{xy}\right)^2\geq ab\plus{}xy\minus{}\left(ay\plus{}bx\right)\equal{}\left(a\minus{}x\right)\left(b\minus{}y\right)$.

Thus, $ \left(a\minus{}x\right)\left(b\minus{}y\right)\minus{}\left(c\minus{}z\right)^2\leq 0$.

Hence,

$ \det\left(A\minus{}B\right)\equal{}\det\left(\left(\begin{array}{cc}a&c\\ c&b\end{array}\right)\minus{}\left(\begin{array}{cc}x&z\\ z&y\end{array}\right)\right)\equal{}\det\left(\begin{array}{cc}a\minus{}x&c\minus{}z\\ c\minus{}z&b\minus{}y\end{array}\right)$
$ \equal{}\left(a\minus{}x\right)\left(b\minus{}y\right)\minus{}\left(c\minus{}z\right)^2\leq 0$,

qed.

  Darij

Hi Darij; your proof is too complicate.
There exists an invertible real matrix $ P$ s.t. $ P^TAP\equal{}I$ and $ P^TBP\equal{}D$ where $ D\equal{}diag(\lambda_1,\lambda_2),\lambda_1>0,\lambda_1\lambda_2\equal{}1$.(simultaneous diagonalization).
$ signum(det(A\minus{}B))\equal{}signum(det(I\minus{}D))$. Then it's easy.

Thanks, now I understand your (and Volodja's) proof! What you use is the Sylvester theorem (every symmetric matrix over $ \mathbb{R}$ can be diagonalized by an orthogonal matrix) in disguise ("orthogonal" replaced by "orthogonal with respect to $ A$").
This should also answer Moubinool's question.

  darij

Hi Darij,volodja;
1) the Volodja's proof is false (except if $ AB\equal{}BA$); in particular, in the simultaneous diagonalization, $ P$ isn't an unitary matrix and the eigenvalues are modified.
2) the result is also valid if $ A,B$ are non negative symmetric matrices.(use a continuity argument ).

[quote="darij grinberg"][quote="loup blanc"]1) the Volodja's proof is false (except if $ AB \equal{} BA$);[/quote]

Ah, I see, but it becomes correct upon replacing $ U \in O\left(n\right)$ by $ U \in \text{GL}\left(n\right)$ (and $ U$ by $ U^{ \minus{} 1}$ in the formula). (And removing the part about eigenvalues, which is unnecessary.)

  darij[/quote]

Oops, excuse my having been inaccurate. I wanted to use well-known theorem saying that given two symmetric matrices $ A,B \in M_n(\mathbf R)$, one of which is positive-definite, there exists $ T \in \mathrm{GL}(n,\mathbf R)$ such that both $ T^{ \minus{} 1}AT$ and $ T^{ \minus{} 1}BT$ are diagonal with eigenvalues on the diagonal. So one should replace $ U$ and $ U^*$ in my previous post with this $ T$ and $ T^{\minus{}1}$ (and thus $ O(n)$ with $ \mathrm{GL}(n,\mathbf R)$).

Hi Volodja,
your result is false. The simultaneous diagonalization doesn't concern linear applications but bilinear forms:
Let $ A$ be  real SPD and $ B$ be real symmetric; then there exists an invertible real matrix $ P$ s.t. $ P^TAP\equal{}I$ and $ P^TBP\equal{}D$ where $ D$ is a diagonal matrix and $ signature(B)\equal{}signature(D)$.
It's a basis change of bilinear forms! You haven't any matrix inverse.

Ok, now I see my mistake. :) I completely forgot the underlying theorem. Thanks.

The proof should be much simpler (and correct  :wink: ): Since $ A\minus{}B$ is symmetric, we can determine an orthogonal matrix $ Q$ and a diagonal matrix $ D$ such that $ A\minus{}B\equal{}QDQ^{\minus{}1}$ holds. The case $ \det(A\minus{}B)>0$ would imply $ D$ to be strictly positive definite or strictly negative definite, which would yield $ A>B$ or $ B>A$, when seen as quadratic functions over $ \mathbb{R}$ But then $ \det A>\det B$ or $ \det B> \det A$, in both cases a contradiction.


[b]Parse error:[/b] parse error in [b]/var/www/Forum/latexrender/phpbb_hook_2.php[/b] on line [b]489[/b]

and no images get rendered.

Someone should take a look at this i guess... and yes I mean an admin :P

what topics ? testing $ tt $ $ ttt $ and \[ tttt \]

It happened to me too about half an hour ago, but it went away just as fast :).

yup seems nothing's wrong now :-D

[i]Posted from a Sony Ericsson K700i [/i]


calculus
integration
function
complex numbers
calculus computations

I had difficulties in solving the 2 questions below. Can anyone help?

1) Let $ f: [a,b] \mapsto \mathbb{R}$ be a bounded function. Assume there is a partition Q of $ [a,b]$ such that $ L(Q,f) \equal{} U(Q,f)$. Show that f is riemann integrable on $ [a,b]$. [Hint: Use the fact that $ L(P,f) \leq L(f) \leq U(f) \leq U(P,f)$ for any partition P of $ [a,b]$.

2) Let $ g: [1,3] \mapsto \mathbb{R}$ be a continuous function. If the upper Riemann integral of $ g$ on $ [a,b], U(g)$, is $ \minus{} 6$, compute the lower riemann integral $ L(g)$. Justify your answer.

Edit: Thank You JRav. I forgot how to use \mathbb{R}. Glad you point this out.

Note: Don't use $ \Re$.  That is the symbol for the real part of a complex number.  If you want to map to the real numbers, use $ \mathbb{R}$, which is \mathbb{R}.

For 2, what is the relation between [a,b] and [1,3]? What is the relationship between upper and lower Riemann integrals for Riemann integrable functions :?:

Hi SouthStar, do you think the question is wrong? There is no additional details for the Q2. That is the exact question I extract from a book in the library.

We have $ L(Q,f) \leq L(f) \leq U(f) \leq U(Q,f)$ and $ L(Q,f) \equal{} U(Q,f) \implies L(f) \equal{} U(f)$.

I seem to remember seing something similar in Spivak.


At a dance party, a group of boys and girls exchange dances the following way: the first boy dances with 5 girls, the second with 6, and so on. Let $ \ b$ represent boys and $ \ g$ represent girls. Find $ \ b$ in terms of $ \ g$
Aops Volume 1 page 24.

Let there be $ b$ boys, then the $ bth$ boy dansed with $ b\plus{}4$ girls. So, at most, there are $ 5 \plus{} 6 \plus{} ... \plus{} (b\plus{}3) \plus{}  (b\plus{}4) \equal{} \frac{b(b\plus{}9)}{2}$ girls.

Now we just express $ b$ in terms of $ g$.

$ g \equal{} \frac{b(b\plus{}9)}{2}$

$ 2g \plus{} \frac{81}{4} \equal{} b^2 \plus{} 9b \plus{} \frac{81}{4}$

$ 2g \plus{} \frac{81}{4} \equal{} (b\plus{}\frac{9}{2})^2$

$ b \equal{} \sqrt{2g \plus{} \frac{81}{4}} \minus{} \frac{9}{2}$

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1526066#1526066[/url]

In the future, please check the Contests/Resources section for the problem before posting it again. This way, we can keep the discussion in one place.


function
trigonometry
algebra
polynomial
national olympiad

Vietnam olympiad mathematics 2003.
                              Day1
1.let fuction $f: R-\to R$ such that
             $f(cotgx)=sin2x+cos2x$ for all $x$ be long to $(0;\pi)$
     Find maximal and minimal of $g(x)=f(sin^{2}x)f(cos^{2}x)$
                              Day2.
1.does there exist  $x,y,u,v,t$ is iteger such that
                    $x^2+ y^2 =(x+1)^2+u^2=(x+2)^2+v^2=(x+3)^2+t^2$
3.let $a$ is real distinct 0 and a sequence ${x_n}$,$n=1,2,...$ is definded by $x_1 =0 and x_{n+1}(x_n+a)=a+1$ for all$n$
a)find expression of $x_n$.
b)prove that ${x_n}$ has finite limit when n goes to infinite.

3. a) Solve the recurrence directly, we have

For $a=-1,\ x_n=0\ (n\geq1)$ or $x_1=0$ and $x_n=1\ (n\geq 2).$

For $a\neq -1$ and $a\neq -2,\ x_n=1+\frac{a+2}{(-a-1)^n-1}\ (n\geq 1).$

For $a=-2,\ x_n=1-\frac{1}{n}\ (n\geq 1).$

\[\text{Day 1}\]
1. http://www.mathlinks.ro/Forum/viewtopic.php?p=779769#779769
Let  $f: \mathbb{R}\to\mathbb{R}$ is a function such that $f( \cot x ) = \cos 2x+\sin 2x$ for all $0 < x < \pi$. Define $g(x) = f(x) f(1-x)$ for $-1 \leq x \leq 1$. Find the maximum and minimum values of $g$ on the closed interval $[-1, 1].$

2. http://www.mathlinks.ro/Forum/viewtopic.php?p=779773#779773
The circles $C_{1}$ and $C_{2}$ touch externally at $M$ and the radius of $C_{2}$ is larger than that of $C_{1}$. $A$ is any point on $C_{2}$ which does not lie on the line joining the centers of the circles. $B$ and $C$ are points on $C_{1}$ such that $AB$ and $AC$ are tangent to $C_{1}$. The lines $BM, CM$ intersect $C_{2}$ again at $E, F$ respectively. $D$ is the intersection of the tangent at $A$ and the line $EF$. Show that the locus of $D$ as $A$ varies is a straight line.

3. http://www.mathlinks.ro/Forum/viewtopic.php?p=779774#779774
Let $S_{n}$ be the number of permutations $(a_{1}, a_{2}, ... , a_{n})$ of $(1, 2, ... , n)$ such that $1 \leq |a_{k}-k | \leq 2$ for all  $k$. Show that $\frac{7}{4}S_{n-1}< S_{n}< 2 S_{n-1}$ for $n > 6.$
\[\text{Day 2}\]
4.  http://www.mathlinks.ro/Forum/viewtopic.php?p=779779#779779
Find the largest positive integer $n$ such that the following equations have integer solutions in $x, y_{1}, y_{2}, ... , y_{n}$ :
$(x+1)^{2}+y_{1}^{2}= (x+2)^{2}+y_{2}^{2}= ... = (x+n)^{2}+y_{n}^{2}.$

5. http://www.mathlinks.ro/Forum/viewtopic.php?p=779781#779781
Define $p(x) = 4x^{3}-2x^{2}-15x+9, q(x) = 12x^{3}+6x^{2}-7x+1$. Show that each polynomial has just three distinct real roots. Let $A$ be the largest root of $p(x)$ and $B$ the largest root of $q(x)$. Show that $A^{2}+3 B^{2}= 4$.

6. http://www.mathlinks.ro/Forum/viewtopic.php?p=779784#779784
Let $\mathcal{F}$ be the set of all functions $f : (0,\infty)\to (0,\infty)$ such that $f(3x) \geq f( f(2x) )+x$ for all $x$. Find the largest $A$ such that $f(x) \geq A x$ for all $f\in\mathcal{F}$ and all $x$.

From http://www.kalva.demon.co.uk/vietnam/viet03.html


Let the sequence of random variables $ \{ X_m, \; m \geq 0\ \}, \; X_0=0$, be an infinite random walk on the set of nonnegative integers with transition probabilities \[ p_i=P(X_{m+1}=i+1 \mid X_m=i) >0, \; i \geq 0 \,\] \[ q_i=P(X_{m+1}=i-1 \mid X_m=i ) >0, \; i>0.\] Prove that for arbitrary $ k >0$ there is an $ \alpha_k > 1$ such that \[ P_n(k)=P \left ( \max_{0 \leq j \leq n} X_j =k \right)\] satisfies the limit relation \[ \lim_{L \rightarrow \infty} \frac 1L  \sum_{n=1}^L P_n(k) \alpha_k ^n < \infty.\]

[i]J. Tomko[/i]

can anybody solve this problem? or at least give a hint  :s


I'm guessing a lot of you is in California but since a lot of people don't fill out their location (or rather, "in a sensible manner" :lol:) and it's fairly difficult to scope out active members through Memberlist, I started a poll about it. Please post details if you want (or the country if you're outside U.S.)

I am from Texas as serveral here are from Texas I see./*

you need to add the choice "somewhere else in the US"

confuted, can you change Florida to "Other states within the U.S.A."? Thanks.

Oh yeah and if you're from Massachusetts please say so, like me (I'm in metropolitan Boston, near two ends of Green Line I believe :))

just wondering (i don't really care about footbal) are the new england patriots in massachusetts or are they in a different state.

Why change Florida to "other states?"  It is superior to all others.

Um, no.

The Patriots play near Boston.

I'm living in Mass., but I'm from (and identify with) New York, so that was my vote.

I live in southeastern lower Michigan in the township of Bedford. Pretty small. Only one high shool and one junior high, but several Elementary Schools. If anyone else on AoPS lives near here I would be interested in knowing!

Max, why is so bad that when people think Texas they think cowboy. I like that reference. It defines us sort of like Georgia and peaches. Its cool being in Texas and Fort Worth is called cowtown thats why Im proud to be a suburb of Ft. Worth. I like animals and I like land. I want to live on a ranch someday.

central pennsylvania, close to state college (penn state main campus)

I go to school in Ithaca, New York but I am originally from Great Neck, NY (located in Long Island)

Umm.. I've moved a lot. I presently live in Ramsey, NJ, go to school in Hackensack, NJ.. but a couple years ago I lived in Rockland County, NY.. and before that in Ottawa, Canada. And before that, it was China.

what happened to the northwest??? Northwest PRIDE!!! :D

Practically no AoPSer live there...meep my five words (which will probably kill me later on but oh well :))

Florida had no vote at the time...sorry.

Our English teacher actually told us of an ironic society...I'll post it as a riddle in Fun and Games if you're interested.

I live in Kabul, Afghanistan.  Just to clear that up, I don't really live in Tokyo.

Ohio.  Kinda boring, but nice nonetheless.

what if we live in the US part of the time and outside of it the other part of the time. and I demand that North Carolina has to be added to the list. how could anyone have a serious poll without north carolina. sheesh. you know north carolina got 5th in the nation at MC last year. so ha.

Who cares about North Carolina...then we'll have to put South Carolina, and there are more popular states then them.

I don't know [b]how[/b] I got Tokyo mixed up with Afghanistan, maybe I am crazy :lol:

San Jose, CA

Messed up weather here a lot.

I like the carolinas. Actually, I lived in NC for about 6 months when I was 1. Meow.

Nevada.  I won't even bother asking you to put it on the poll because nobody else will vote for it...

[quote]Practically no AoPSer live there...meep my five words (which will probably kill me later on but oh well ) [/quote]

Actually, I think Eugevir and Kamior both live in the pacific northwest....


I'm writing a research paper about Maxwell's equations, so I need to have a construct which will have an integral with "closed surface" beneath it in a normal font. I worked on figuring it out for a while, and currently I have
\def\cint#1{\ensuremath{\stackrel{\displaystyle \int{#1}}{\textnormal{\tiny{closed surface}}}}}
which is exactly what I want, except that LaTeX is aligning other things in the equation with the bottom of that construct, when it would be much better to align them to the middle or top of it. Here's an example of what I mean:
[tex]\phi_B=\stackrel{\displaystyle \int{\vec{B}\cdot d\vec{A}}}{\textnormal{\tiny{closed surface}}}[/tex]
Obviously, that looks nice, except that the Phi B = would look much better in the middle there. Anyone with any ideas?

This isn't bad. The construction is dreadful, but it looks fairly decent.

[tex]\stackrel{\big{\phi_B=}}{\big{\;}}[/tex][tex]\stackrel{\displaystyle \int{\vec{B}\cdot d\vec{A}}}{\textnormal{\tiny{closed surface}}}[/tex]

[quote="zscool"]what about [tex] \oint [/tex][/quote]
I'm saving that for closed loop.

[quote="stevem"]The problem is that \stackrel is meant for small text on top of a symbol such as [tex]A \stackrel{\alpha}{\longrightarrow}B[/tex] so LaTeX lines up A and B with the arrow which is the main part.

You need \underset which is the counterpart, providing small text below symbols (you need to use amsmath package but you probably do anyway).
[tex]\phi_B=\underset{\text{\tiny{closed surface}}}{\displaystyle \int{\vec{B}\cdot d\vec{A}}}[/tex][/quote]
Beautiful. Much thanks.

This is the alternate form of Ampere's Law I was telling you about the other night...

[tex]\displaystyle{\vec{\nabla}}\times \vec{\mathbf{B}} = \mu_0 \vec{\mathbf{J}}}[/tex]

Now explaining that to your class would be quite some fun!

note- Maxwell's fix/extension to this equation was left off... I'm feeling lazy right now!

[edit] - yes I mean't \mathbf, thanks!

[quote="Spoon"][tex]\displaystyle{\vec{\nabla}}\times \vec{\mathBB{B}} = \mu_0 \vec{\mathBB{J}}}[/tex][/quote]

Do you mean? [tex]\vec{\nabla}}\times \vec{\mathbf{B}} = \mu_0 \vec{\mathbf{J}}[/tex]

ie use \mathbf rather than \mathbb which is for blackboard bold as in [tex]\mathbb{R}[/tex]. \displaystyle isn't necessary for this formula which will save typing :)

Being a bit picky now... I don't suppose anyone will be able to fix this, but it's worth a try.
[tex]\underset{\text{\tiny{closed surface} }}{\displaystyle \int{}}[/tex]
looks beautiful. "Closed surface" is centered underneath the integral sign, just like it would be written.

But when things are put "into" the integral, "closed surface" shifts over to the right:
[tex]\underset{\text{\tiny{closed surface} }}{\displaystyle \int{\vec{B}\cdot\emph{d}\vec{A}}}=0[/tex]

If I was to use an even longer integrand (I don't need to, but in theory), the closed surface would be centered beneath the integral-integrand expression. It'd be better if it'd stay centered under the integral. But then perhaps equations like
[tex]\phi_B=\underset{\text{\tiny{closed surface} }}{\displaystyle \int{\vec{B}\cdot\emph{d}\vec{A}}}[/tex]
would present a problem to the LaTeX renderer, since part of "closed surface" would be under the equals sign.

Anyway, does anyone have any ideas on how to fix this (relatively trivial) problem?

[tex]\displaystyle\underset{\text{\tiny{closed surface}}}{\int} \mspace{-25 mu} \vec{B}\cdot \emph{d}\vec{A}=0[/tex]

Is that what you want?

[tex]
\displaystyle
\underset{\substack{ \text{\tiny{closed}} \\ \text{\tiny{surface}} } }{\int}\vec{B}\cdot\emph{d}\vec{A}=0
[/tex]

It may save some space to have "closed surface" on top of each other instead of all on one line.

LaTeX and y'all's mastery of it has not yet ceased to amaze me. A combination of the two things you posted works exquisitely with everything I've thrown at it. Yes, exquisitely. Thanks.


Let $ a,b,c$ positive reals and $ ab\plus{}bc\plus{}ca\equal{}1$. Prove that

$ \frac{a^3\plus{}1}{a\plus{}b}\plus{}\frac{b^3\plus{}1}{b\plus{}c}\plus{}\frac{c^3\plus{}1}{c\plus{}a}\geq\frac{11}{4}\plus{}\frac{1}{2(a\plus{}b\plus{}c)}$

[quote="Ligouras"]Let $ a,b,c$ positive reals and $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that

$ \frac {a^3 \plus{} 1}{a \plus{} b} \plus{} \frac {b^3 \plus{} 1}{b \plus{} c} \plus{} \frac {c^3 \plus{} 1}{c \plus{} a}\geq\frac {11}{4} \plus{} \frac {1}{2(a \plus{} b \plus{} c)}$[/quote]

[size=200]prove [/size]
by Cauchy-Schwartz

$ \frac {a^3 }{a \plus{} b} \plus{} \frac {b^3 }{b \plus{} c} \plus{} \frac {c^3 }{c \plus{} a}\geq \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{a(a \plus{} b) \plus{} b(b \plus{} c) \plus{} c(c \plus{} a)} \equal{} \frac {(a^2 \plus{} b^2 \plus{} c^2)}{a^2 \plus{} b^2 \plus{} c^2 \plus{} 1}$

$ \frac { 1}{a \plus{} b} \plus{} \frac { 1}{b \plus{} c} \plus{} \frac { 1}{c \plus{} a}\geq \frac {9}{2(a \plus{} b \plus{} c)}$

Therefore, we just need to prove

$ \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{a^2 \plus{} b^2 \plus{} c^2 \plus{} 1} \plus{} \frac {9}{2(a \plus{} b \plus{} c)}\geq\frac {11}{4} \plus{} \frac {1}{2(a \plus{} b \plus{} c)}$

or
$ \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{a^2 \plus{} b^2 \plus{} c^2 \plus{} 1} \plus{} \frac {4}{(a \plus{} b \plus{} c)}\geq\frac {11}{4}$

and
$ \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{a^2 \plus{} b^2 \plus{} c^2 \plus{} 1} \plus{} \frac {4}{(a \plus{} b \plus{} c)}\geq \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{a^2 \plus{} b^2 \plus{} c^2 \plus{} 1} \plus{} \frac {4}{\sqrt {3}\sqrt {a^2 \plus{} b^2 \plus{} c^2}}$

we just need to prove

$ \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{a^2 \plus{} b^2 \plus{} c^2 \plus{} 1} \plus{} \frac {4}{\sqrt {3}\sqrt {a^2 \plus{} b^2 \plus{} c^2}}\geq\frac {11}{4}$

by AM-GM,$ a^2 \plus{} b^2 \plus{} c^2\geq ab \plus{} bc \plus{} ca \equal{} 1$

 assume $ t^2 \equal{} a^2 \plus{} b^2 \plus{} c^2$         we only to prove while $ t\geq 1$
$ \frac {t^4}{t^2 \plus{} 1} \plus{} \frac {4}{\sqrt {3}t}\geq\frac {11}{4}$
or
$ \minus{} 33t^3 \minus{} 33t \plus{} 12t^5 \plus{} 163\sqrt {3}t^2 \plus{} 16\sqrt {3}\geq 0$
assume $ u \equal{} t \minus{} 1$,then only prove
$ 87u^3 \plus{} 21u^2 \minus{} 72u \minus{} 54 \plus{} 12u^5 \plus{} 60u^4 \plus{} 16\sqrt {3}u^2 \plus{} 32\sqrt {3}u \plus{} 32\sqrt {3}\geq 0$

$ 87u^3 \plus{} 21u^2 \minus{} 72u \minus{} 54 \plus{} 12u^5 \plus{} 60u^4 \plus{} 16\sqrt {3}u^2 \plus{} 32\sqrt {3}u \plus{} 32\sqrt {3}\geq (21 \plus{} 16\sqrt {3})u^2 \minus{} (72 \minus{} 32\sqrt {3})u \plus{} 32\sqrt {3} \minus{} 54$

it easy to prove
$ (21 \plus{} 16\sqrt {3})u^2 \minus{} (72 \minus{} 32\sqrt {3})u \plus{} 32\sqrt {3} \minus{} 54\geq 0$


find real $ m$ such as the system has unique real solution
$ x \ge y^{2}+m$
$ y \ge x^{2}+m$

we take the only $ a\in\mathbb{R}$ such that $ a-a^{2}=\max_{x\in \mathbb{R}}(x-x^{2})$

if $ (x,y)$ is solution with $ y-x^{2}-m>0$
so $ \exists \epsilon >0: \ (x+\epsilon,y)\ is\ solution$
(we take $ \epsilon$ such that $ 2\epsilon x+\epsilon^{2}\le (y-x^{2}-m)$)
so we must have $ y=x^{2}+m\ and\ x=y^{2}+m$
then $ x^{2}-y^{2}+x-y=0\to x=y \ ou\ x=1-y$
if $ m<\max_{x\in \mathbb{R}}(x-x^{2})$
$ (x,x)$ is solution $ \forall x\in\{x\in\mathbb{R}|\ x-x^{2}\ge m\}$
then we must have $ m\ge max_{x\in \mathbb{R}}(x-x^{2})$
if $ m=max_{x\in \mathbb{R}}(x-x^{2})$ then $ x\neq y$ $ (a,a)$ and $ (a,1-a)$ are solution
if $ m>max_{x\in \mathbb{R}}(x-x^{2})$ then $ x\neq y$
so $ x=1-y$ and $ m=x-(x-1)^{2}=x^{2}-x+1$
then $ m=1+\max_{x\in \mathbb{R}}(x-x^{2})$ where the only solution is $ (a,1-a)$


Ring Theory
abstract algebra
calculus
integration
algebra
function
domain

Suppose that $ R$ is a noetherian integral domain, and $ x \in R \setminus 0$ with $ (x) \in \text{Spec}(R)$. Show that $ R$ is a UFD if and only if $ R_x$ is a UFD, where $ R_x$ is $ R$ localized at $ \{1, x, x^2, \ldots\}$.

Is there a way to prove this without getting down and dirty with factorizations.

Use the fact that a noetherian domain is a UFD iff every height-1 prime ideal is principal. owk


geometry
3D geometry
number theory theorems
number theory

When $ a_1,a_2,\cdots,a_n,k$ are positive integers, we can say
$ \sqrt{a_1}\plus{}\sqrt{a_2}\plus{}\cdots\plus{}\sqrt{a_n}\equal{}k\Rightarrow a_i\equal{}b_i^2(\forall 1\le i \le n,b_i \in N)$?
How about when $ a_i,b_i,k$ are positive rational numbers?
I think these are true(and maybe it is a (well-known) theorem), but I can't prove it. Can someone help me?

[quote="ZetaX"]Multiply with a common denominator to reduce it to the integer case.[/quote]
OK, I know. but how can we say in case which $ a_i,b_i,k$are integer? and can we expand it(about square root->cube root)?

For the square roots: by a good amount of tricky algebra.
For the cube roots: by a very good amount of very tricky algebra.
In general: By nonelementary means (i.e. Kummer theory).

See http://www.mathlinks.ro/viewtopic.php?t=27089 .


calculus
integration
function
limit
real analysis
real analysis unsolved

Suppose that $ f: [0,1] \rightarrow \mathbb{R}$ is continuous. For each $ n \in \mathbb{N}$, define $ g_n : [0,1] \rightarrow \mathbb{R}$ by $ g_n(x)\equal{}f(x^n)$ for all $ x \in [0,1]$. Prove that the sequence $ (\int^1_0 g_n(x))$ converges to $ f(0)$.
Attempt
This question really confuses me. I don't see how this sequence converges to $ f(0)$. A few helpful hints on this one would be appreciated. Thanks in advance.

Note that $ g_n$ converges pointwise to the function $ g$ defined by $ g(x) \equal{} \begin{cases} f(0) & x < 1 \\ f(1) & x \equal{} 1 \end{cases}$.

Although pointwise limit and integration is generally not interchangable, this problem requires you to show that this is possible under the given condition, since if this is true, we have $ \textstyle \lim \int_{0}^{1} g_n \equal{} \int_{0}^{1} \lim g_n \equal{} \int_{0}^{1} g \equal{} f(0)$.

Lebesgue dominated covergence theorem.
note that $ g_n$ bounded by some $ Const > 0$ since $ f$ is continuous, also we integrate on the compact interval, thus we can use
Lebesgue dominated covergence theorem

What happens for the case $ x\equal{}1$? Can we overlook the case $ x\equal{}1$ or does this case also satisfy the requirement that $ (\int^1_0 g_n(x))$ converges to $ f(0)?$

$ \int_{0}^{1} g_n(x) \, dx$ isn't a function of $ x$.


The difference between the larger root and the smaller root of

\[ x^2 - px + \frac{p^2-1}{4} = 0 \]
is

(A) 0
(B) 1
(C) 2
(D) $p$
(E) $p+1$

I believe the answer is B.  I went through with the quadratic equation and it ended being +- 1/2 so the difference in p would be 1.

Is there a nice easy way to see that this is indeed the answer without working through the algebra of the quadratic formula?

B is correct.

The best way to do this problem is to think about how the roots of a quadratic are related.

In fact, if we let x^2 + ax + b = (x-p)(x-q) = 0, where p,q are the roots, it's not hard to see that:

a = -(p+q)
b = pq

Now, lets see if we can find (p-q) in terms of a and b.

It's probably best to multiply that factored expression by a leading coefficient since not all quadratics have a lead coefficient of 1.  That way you can see the complete relationship between the roots and coefficients.  This was however exactly the idea I was trying to evoke.

Well, if you multiply the original equation by 4, then you get the equation $4x^2-4px+p^2-1=0$, which can be factored to $(p-2x)^2-1=0$, so $p-2x=\pm1$, leading to $x=\frac{\pm1-p}{-2}$. The difference of these two roots is 1, or (B).

Nicely done.  We could also examine the sum and product of the roots and make a similar argument.

[b]
In my opinion, the best way is to find the roots of the equaion.
Since we have $c=\frac{p^2-1}{4}$ and $ a=1 $
Then we get $4ac=p^2-1$
And then $b^2-4ac=1$
Because it has a $\frac{something}{4}$ as a term, i immediatly come up with $b^2-4ac$![/b]

By using the sums and products of roots, we can get
x1+x2=p
x1x2=(p 2 -1)/4
(x1-x2) 2 =p 2 -4(p 2 -1)/4
(x1-x2) 2 =1
So the answer is B.


To improve:

1) Take out, say, a geometry book. Go through and derive all of the theorems presented no matter how long it takes. After you derive each theorem, look back and see what did I do that helped me? What could I have done to make my proof better?

2) Take one problem and solve it in many different ways.

3) Do lots of math problems and not just do them, but look back and see what you did and what problems it can be applied to.

4) Once you are advanced, get a higher level AOPS book and begin to work on it.

do problems, lots and lots of problems.
The more ways you figure out how not to do a problem, the better. If you figure out how to do the problem, go bakck and try it a different way. See how many ways there are not to do the problem... and why they don't work/
do this a lot. don't get discouraged. You'll get it ;p

Play chess  :D ..oh, and do lots and lots of problems :)

[quote="Totally Zealous"][quote="mindboggling"]Hi everyone, this is my first time on this forum and so far it has been a very pleasant and intriguing experience reading all the threads.

There is a question in my mind that troubles me. Firstly, I love math, it's almost like taking a sneak peek into god's thoughts. However, Math is a topic i'm terrible and slow at. Hence, I decided to work harder. Through this experienced i learnt that what i lacked wasn't exaclty effort, but the ability to SEE the method to a problem. My question is, what are some of the ways to improve this skill? How can you develop the necessary perpsectives to enhance the possibility of solving a problem ( logic, math, lateral or any problems in general).

For example, I tried to become less stubborn in life and as a result, i was more open to different ideas while solving problems.

Thanks for all the help![/quote]

Do lots of problems.
Then do them again.
and again...
and again...
...
until you have memorized them.[/quote]
what's the point in doing the same problem over and over again and memorizing it?

Are the books "How to solve it" by Polya and " The Art and Craft of Problem Solving" by Zeitz any helpful?

they are good starters if you are not familiar with some things

Neither of those books are really "starters."
Here's a long list of all kinds of math books, sorted into categories and levels. http://www.artofproblemsolving.com/Wiki/index.php/Math_books :D

The Art and Craft of PS is [u]hard[/u] :D

I would start with the Art of Problem Solving vol. 1, which can be bought on this website. Then, when you have learned all of that, buy the Art of Problem Solving vol. 2.


floor function
trigonometry
linear algebra
matrix
search
calculus
calculus computations

I'm looking for a way to solve the following problem without using

$ \prod_{k\equal{}1}^{\lfloor \frac{n\minus{}1}{2}\rfloor} \tan\left(\frac{k\pi}{n}\right)\equal{}\left\{ \begin{matrix} \sqrt{n} & \text{for} & n \;\; \text{odd} \\
\\
1 & \text{for} & n\;\; \text{even} \end{matrix} \right.$

(where $ n \equal{} 9$).

Problem: Calculate the value of the expression

\[ \tan{20^{\circ}}\tan{40^{\circ}}\tan{60^{\circ}}\tan{80^{\circ}}.\]

Search makes wonders sometimes:

[url]http://www.mathlinks.ro/viewtopic.php?t=210333[/url]
[url]http://www.mathlinks.ro/viewtopic.php?t=172461[/url]


A diagonal of a 5-gon is "good" if it is parallel to one of the sides of the 5-gon. If four are "good", show the last has to be, too.

Denote the vertices of the pentagon by $A,B,C,D,E$ counterclockwise and assume $BC\parallel AD, CD\parallel BE, DE\parallel CA, EA\parallel DB.$ Since $\overrightarrow{AD}$ is parallel to $\overrightarrow{BC}$, there exists a scalar $\lambda$ such that $\overrightarrow{AD}=(\lambda+1)\overrightarrow{BC}$. Substituting $\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}$, we get

$\overrightarrow{AB}+\overrightarrow{CD}=\lambda\overrightarrow{BC}\quad(1)$

Similarly,

$\overrightarrow{BC}+\overrightarrow{DE}=\mu\overrightarrow{CD}\quad(2)$
$\overrightarrow{CD}+\overrightarrow{EA}=\nu\overrightarrow{DE}\quad(3)$
$\overrightarrow{DE}+\overrightarrow{AB}=\xi\overrightarrow{EA}\quad(4)$

and we have to prove $EC\parallel AB$ or, put differently, that there exists a scalar $\rho$ such that

$\overrightarrow{EA}+\overrightarrow{BC}=\rho\overrightarrow{AB}\quad(\dag)$

Also,

$(\lambda+1)\overrightarrow{BC}=\overrightarrow{AD}=-(\overrightarrow{DE}+\overrightarrow{EA})$, which gives

$\overrightarrow{DE}+\overrightarrow{EA}=-(\lambda+1)\overrightarrow{BC}\quad(5)$

Expressing $\overrightarrow{CD}$ from $(1)$ and substituting into $(2)$, then expressing $\overrightarrow{DE}$ from that and substituting into $(3)$, and finally expressing $\overrightarrow{EA}$ from that, we get

$\overrightarrow{CD}=-\overrightarrow{AB}+\lambda\overrightarrow{BC}$
$\overrightarrow{DE}=-\mu\overrightarrow{AB}+(\lambda\mu-1)\overrightarrow{BC}\quad(*)$
$\overrightarrow{EA}=(-\mu\nu+1)\overrightarrow{AB}+(\lambda\mu\nu-\lambda-\nu)\overrightarrow{BC}\quad(**)$

Substituting $(*)$ and $(**)$ into $(4)$ we get

$(-\mu+1+\mu\nu\xi-\xi)\overrightarrow{AB}+(\lambda\mu-1-\lambda\mu\nu\xi+\lambda\xi+\nu\xi)\overrightarrow{BC}=\overrightarrow{0}$

Since $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are linearly independent (that is, they aren't collinear), this is equivalent to

$-\mu+1+\mu\nu\xi-\xi=0\quad(6)$
$\lambda\mu-1-\lambda\mu\nu\xi+\lambda\xi+\nu\xi=0\quad(7)$

Substituting $(*)$ and $(**)$ into $(5)$ we get

$(-\mu-\mu\nu+1)\overrightarrow{AB}+(\lambda\mu+\lambda\mu\nu-\nu)\overrightarrow{BC}=\overrightarrow{0}$

Again, by the reason of linear independency, we get

$-\mu(1+\nu)+1=0\quad(8)$
$\lambda\mu(1+\nu)-\nu=0\quad(9)$

From $(8)$ we get $\mu=\frac{1}{1+\nu}$. Substituting that into $(9)$ we get $\lambda-\nu=0$ or $\lambda=\nu$

From $(6)$ we get $\xi=\frac{1-\mu}{1-\mu\nu}=\frac{1-\frac{1}{1+\nu}}{1-\frac{\nu}{1+\nu}}=\nu$

Substituting $\lambda=\xi=\nu$ into $(7)$, we get

$\mu\nu-1-\mu\nu^3+2\nu^2=0\iff \mu\nu(1-\nu^2)+2\nu^2-1=0$

Since $\mu(1-\nu^2)=1-\nu$, we get

$\nu^2+\nu-1=0\quad (10)$

Having in mind that $\nu+1>0$ because of $\overrightarrow{CA}=(\nu+1)\overrightarrow{DE}$ where those two vectors have the same direction, we get

$\nu=\frac{\sqrt{5}-1}{2}$

Now the relation $\mu=\frac{1}{1+\nu}$ gives $\mu=\frac{\sqrt{5}-1}{2}=\nu$

Therefore $\lambda=\mu=\nu=\xi$. Now from $(**)$ we obtain

$\overrightarrow{EA}+\overrightarrow{BC}=(-\nu^2+1)\overrightarrow{AB}+(\nu^3-2\nu+1)\overrightarrow{BC}$

Since $(10)\implies -\nu^2+1=\nu$ and $(10)\implies \nu^3-2\nu+1=(\nu-1)(\nu^2+\nu-1)=0$, we get

$\overrightarrow{EA}+\overrightarrow{BC}=\nu\overrightarrow{AB}$, which means that $(\dag)$ is proven: $\rho=\nu$ and $EC\parallel AB.$

$\boxed{Note}$: Does $\lambda=\mu=\nu=\xi=\rho=\frac{\sqrt{5}-1}{2}$ imply that $ABCDE$ is a regular pentagon?


trigonometry
limit
function
calculus
derivative
calculus computations

IF $ f$ is one-one differentiable, and $ f^{\prime}\neq0$ ; then $ f^{\prime}$ is continuous

thank you.

ps1:

f may be just  one times differentiable

ps 2:

$ f(x) = \{\begin{array}{rl} 0, & x = 0 \\
x^2\sin \frac {1}{x}, & x\neq 0 \\
\end{array}$

This $ f$ is not one-one function.

you might wanna wait a tiny little bit longer. :D

[quote="ICQ-2008"]IF $ f$ is one-one differentiable
[/quote]

If this expression means: $ f$ two times differentiable

then use this impilcation(well known):

$ \forall g$: $ g$ differentiable $ \implies$ $ g$ continous.

apply it to f' in your exercice

hmmm...
what if:

$ f(x)=\{
\begin{array}{rl}
0,&x=0\\
x^2\sin \frac{1}{x},&x\neq 0\\
\end{array}$

then obviously $ f'$ isn't continuous. :maybe:

[quote="BaBaK Ghalebi"]hmmm...
what if:

$ f(x) = \{ \begin{array}{rl} 0, & x = 0 \\
x^2\sin \frac {1}{x}, & x\neq 0 \\
\end{array}$

then obviously $ f'$ isn't continuous. :maybe:[/quote]

why f is not continous?

because:

$ f^\prime (x)=\{
\begin{array}{rl}
0,&x=0\\
2x\sin \frac{1}{x}-\cos \frac{1}{x},&x\neq 0\\
\end{array}$

which isn't continuous at $ 0$.

how did you compute $ \lim_{x \to 0} f'(x)$ I couldn't compute it, But the statement I gave is always true, I am sure and I can give a proof to it.

[quote="ICQ-2008"]Is this statement wrong? Why nobody can solve?  :rotfl:

ps :

$ f(x) = \{\begin{array}{rl} 0, & x = 0 \\
x^2\sin \frac {1}{x}, & x\neq 0 \\
\end{array}$

This $ f$ is not one-one function.[/quote]
oh,sorry didn't notice the 1-1 condition.

@FOURRIER:your statement isn't useful here,note that one-one means that for every $ a,b\in D_f$ we have $ f(a)\neq f(b)$.

[quote="BaBaK Ghalebi"]
@FOURRIER:your statement isn't useful here,note that one-one means that for every $ a,b\in D_f$ we have $ f(a)\neq f(b)$.[/quote]

That's injective function.

$ f(x) \equal{} 10x \plus{} x^2 \sin \frac {1}{x}$ for $ x \ne 0$, with $ f(0) \equal{} 0$ is not one-one function please. check once!

What about $ f(x) \equal{} 10x \plus{} x^2 \sin \frac{1}{x}$ for $ x \ne 0$, with $ f(0) \equal{} 0$?  That should be a counterexample.

Is this statement wrong? Why nobody can solve?  :rotfl: :D

[quote="Ravi B"]Hmm, I believe my function is one-to-one.  According to my calculations, its derivative is always positive, which means that the function is increasing.  But I don't have time to give the argument here.[/quote]

I like your joke. In fact , I believe i can fly.

Okay, here is a proof that my function is strictly increasing.  By the product rule, its derivative is $ f'(x) \equal{} 10 \plus{} 2x \sin \frac{1}{x} \minus{} \cos \frac{1}{x}$ for $ x \ne 0$, with $ f'(0) \equal{} 10$.  Because $ \left| \sin t \right| \le \left| t \right|$ for all $ t$, the term $ 2x\sin \frac{1}{x}$ is at least $ \minus{}2$.  Of course $ \cos \frac{1}{x}$ is at most 1.  So the derivative is everywhere at least $ 10 \minus{} 2 \minus{} 1 \equal{} 7$, which is positive.  So $ f$ itself is increasing.

ICQ-2008, you will have to teach me how to fly. :)


geometric inequality
combinatorics
combinatorial inequality
point set
Harm Derksen
IMO
IMO 1989

Let $ n$ and $ k$ be positive integers and let $ S$ be a set of $ n$ points in the plane such that

[b]i.)[/b] no three points of $ S$ are collinear, and

[b]ii.)[/b] for every point $ P$ of $ S$ there are at least $ k$ points of $ S$ equidistant from $ P.$

Prove that:
\[ k < \frac {1}{2} \plus{} \sqrt {2 \cdot n}
\]

I still remember how unbelievably happy I was when I managed to solve this one a very long time ago :).

Let $\{A_i\}_{i=1}^n$ be our set of points. We count the pairs $(A_i,\{A_j,A_\ell\}),\ i\ne j\ne \ell\ne i$ such that $A_iA_j=A_iA_\ell$. There are $n$ choices for $A_i$, and for each $A_i$ there are $\ge\frac{k(k-1)}2$ choices for $\{A_j,A_\ell\}$, so the number of such pairs is $\ge\frac{nk(k-1)}2$. On the other hand, there are $\frac{n(n-1)}2$ choices for $\{A_j,A_\ell\}$, and for each $\{A_j,A_\ell\}$ there are at most two choices for $A_i$ (because we can't have $\ge 3$ points on the perpendicular bisector of $A_jA_\ell$). This means that the number of pairs is $\le n(n-1)$. We have thus found $n(n-1)\ge\frac{nk(k-1)}2\Rightarrow 2(n-1)\ge k(k-1)\ (*)$. If $k\ge \sqrt{2n}+\frac 12$, then $k(k-1)\ge 2n-\frac 14>2n-2$, contradicting the last inequality in $(*)$, so we're done.

I solved it yesterday when I was trying to sleep; two possible approaches came to my mind. First of all, I tried to get an upper bound for $k$ by counting $nk$ points (at least $k$ equidistant points for any of the $n$ points) and then subtracting $2\binom{n}{2}$, since two points may share exactly two other points, equidistant from both of them. But this yields a horrible bound for $k$, namely $k \leqslant n$, which is even worse than the obvious $k \leqslant n-1$. But this was actually not really surprising, since I haven't used the fact that no three points are collinear.

So I tried to find another way of counting the $k$ equidistant points and then subtracting the possible "common" points, incorporating the fact that no three points are collinear. Due to my last observation (namely that two circles intersect in at most two points; imagine the $k$ equidistant points as a circle) I started to think of pairs of points, since this idea includes the fact that no three points are collinear (a pair of points belongs to at most two of these "circles"). The rest is the same as grobber did already more than 4 years ago. :)

One could also visualise this by setting up an incidence matrix as follows. Let $A=(a_{i\mbox{,}j})$ be an $n\times n$ matrix, where every row and column represtents one of the $n$ points. An entry $a_{i\mbox{,}j}$ is $1$, if point $P_j$ is one of the $k$ (or more) equidistant points from $P_i$. \[\begin{pmatrix} 0 & 0 & 1 & 1 & \cdots & 1 \\ 1 & 0 & 1 & 0 & \cdots & 0 \\ 1 & 0 & 0 & 0 & \cdots & 1 \\ & \vdots & & & \ddots & \vdots \\ 1 & 0 & 0 & 1 & \cdots & 1 \\ 0 & 1 & 1 & 1& \cdots & 0  \end{pmatrix}\]
To solve the problems, it's enough to count the pairs on $1$'s in each row in two different ways. Therefore, let $\mathcal{T}$ be the set of all unordered pairs of $1$'s that lie in the same row. Counting by rows: since there are at least $k$ equidistant points in each row, we have at least $n\binom{k}{2}$ pairs of ones in total. Hence $|\mathcal{T}| \geqslant n\binom{k}{2}$. Counting by columns: for any two columns $i$ and $j$, there are at most two pairs of $1$'s (since no three points are collinear). Thus, $|\mathcal{T}| \leqslant 2\binom{n}{2}$. The result follows.

The solution I found is just like grobber's except using expected values (since probability is more "natural" to think of in problems like these).

However, the solution hinted at in the pdf attached [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=121&t=23507]here[/url] seems to use an approach that's only slightly different but avoids the noncollinearity condition altogether.

So I'm wondering if the noncollinearity condition is used subtly in some way I'm not seeing, or if it's actually not necessary in this solution:

[hide="Non-noncollinear solution"]So obvious simplification and the fact that $n$ is an integer gives that what we want is $n\ge\dbinom{k}{2}+1$
Now for each point, draw a circle around it that contains at least $k$ points in $S$.
For any point $P$ in $S$, let $d_P$ be the number of circles it's on.
For any point $O$ in $S$, let $f_O$ be the number of points on the defined circle with center $O$, so $f_O\ge k$ for all $O$.
Now $\sum_{P\in S}d_P = \sum_{O\in S}f_O \ge nk$, since each point-circle pair counted in the right side is also counted in the left side. So the expected value of $d_P$, $E(d)$, is $\ge nk/n=k$.
But notice that since any two circles intersect in at most two points, $\sum_{P\in S}\dbinom{d_P}{2}\le 2\dbinom{n}{2}$, since the left side counts the number of circle-pair intersections (with possible point multiplicity) in $P$ while the right side upper-bounds the number of circle-pair intersections (with possible point multiplicity) in general. So now $E(\dbinom{d}{2})\le (2/n)((n-1)(n)/2)=n-1$. The LHS is at least $\dbinom{E(d)}{2}=\dbinom{k}{2}$ since the function $\dbinom{d}{2}$ is convex (midpoint "sums" underestimate convex integrals; Jensen's if we have finite cases), which is what we want.[/hide]

For comparison, here's my solution, which does use the noncollinear condition. I don't see how the first solution manages to do without it, since every inequality seems to be weaker here than there...
[hide="Noncollinear solution"]
So obvious simplification and the fact that $n$ is an integer gives that what we want is $n\ge\dbinom{k}{2}+1$
Now for each point, draw a circle around it that contains at least $k$ points in $S$.
For any point $P$ in $S$, let $d_P$ be the number of circles it's on.
For any point $O$ in $S$, let $f_O$ be the number of points on the defined circle with center $O$, so $f_O\ge k$ for all $O$. Since the function $\dbinom{x}{2}$ is increasing for $x\ge 1$ an integer, $\dbinom{f_O}{2}\ge \dbinom{k}{2}$ for all $O$. So $E(\dbinom{f}{2})\ge \dbinom{k}{2}$.
But notice that since any pair of two points shares at most 2 circles ([u]otherwise we have 3 circles with centers on their perpendicular bisector, which is not allowed[/u]), $\sum_{O\in S}\dbinom{f_O}{2}\le 2\dbinom{n}{2}$, since the left side counts the number of point-pairs that share a circle (with possible circle multiplicity) in $P$ while the right side upper-bounds the number of point-pairs that can share a circle (with possible circle multiplicity). So now $\dbinom{k}{2}\le E(\dbinom{f}{2})\le (2/n)((n-1)(n)/2)=n-1$, which is what we want.[/hide]

Maybe it's the fact that the first solution uses a (rather trivial) convexity argument (which can be explained geometrically), that the noncollinearity is given for the ease of a completely (safely) elementary solution?

(Also... as a sidenote, notice how in the two solutions we basically switch the words "point" and "circle" (along with some minor changes to the basic inequalities)... combinatorial duality between circles and points... almost eerie, more strange than projective point-line duality or inversive line-circle duality.)

A very good point. To better understand what is happening here, let us first completely remove the problem from its intuitive geometrical setting, and re-state it in only terms of element/set incidence.

[color=#0000FF]Let $ n$ and $ k$ be positive integers, and let $ S$ be a set of $ n$ elements, such that to every element $ p$ of $ S$ there is associated a set $S_p \subseteq S\setminus \{p\}$ of at least $k$ elements.[/color]

Of course this does not induce any restriction on the largest possible size of $k$, other than $k\leq n-1$. Let us make the notation $C_p = \{q \in S \mid p \in S_q\}$. The original problem introduced an extra requirement (non-collinearity $3$ by $3$), which in the course of the proof was used in the form

[color=#4000FF]Constraint 1. For any pair $p\neq q \in S$ we have $|C_p\cap C_q| \leq 2$.[/color] (that being because non-collinearity of any three points). However, the pure geometrical setting induces the following

[color=#FF0000]Constraint 2. For any pair $p\neq q \in S$ we have $|S_p\cap S_q| \leq 2$.[/color] (that being because non-identical circles meet at two points at most). This alone is also sufficient for a (similar) proof, due to the [b]duality[/b] of the incidence relations, as shown just above.

Now everything is clear. Any of the two constraints is equivalent in finding the required uper bound for $k$. The geometrical setting however fooled (?) the proposer into introducing Constraint 1., maybe because his mind was set on that proof, when Constraint 2. was in fact automatically fulfilled.

This is just an easier version of China TST 2011 #2

First consider $X$, the number of ordered triples $(P,Q,R) \in S$ where $PQ=PR$.
For every pair $(Q,R)$ at most $2$ $P$'s exist by perpendicular bisector theorem. Thus, $X\le 2n(n-1)$.
Now from (b), for every $P\in S$ there are at least $k(k-1)$ $(Q,R)$'s and thus $nk(k-1)\le X\le 2n(n-1) \Rightarrow k(k-1) \le 2(n-1)$.
Now for the sake of contradiction suppose that $k \ge 1/2+\sqrt{2n}$.
Then $(1/2+\sqrt{2n})(-1/2+\sqrt{2n})=2n-1/4 \le 2n-2$, a contradiction.

[b]Solution.[/b]
We count the number of triples of the form $(center,\text{ a point on the circle}, \text{a point on the circle})$. For every center, there are at least $k$ points, so there are at least $n\binom{k}{2}$ triples. Note that for every two points on a circle there are at most $2$ points which can be the center since no three points are collinear (the two points lie on the perpendicular bisector). So there are at most $2 \binom{n}{2}$ triples.
$$n\binom{k}{2}\leq 2\binom{n}{2}\implies k \leq \frac{1+\sqrt{8n-7}}{2} < \frac{1}{2}+\sqrt{2n}$$

Consider the segments joining every two points in the set $S$. Let the total number of these segments be $T$. Now clearly $T=\binom{n}{2}$. \\
Now by the second condition we can say that for each point $P$ there at least $k$ points lying on the circumference of the circle with center $P$. Thus, we get at least $\binom{k}{2}$ segments on each circle. Now since, there are $n$ points $P$, we get that there are $n \cdot \binom{k}{2}$ segments in total, but observe that we have over counted the segments that lie on two circles.\\
Now since, each circle can have only one chord in common and there are $n$ circles, we have over counted $\binom{n}{2}$ segments. This implies :
$$ T \geq n \cdot \binom{k}{2}- \binom{n}{2} \implies \binom{n}{2} \geq n \cdot \binom{k}{2}- \binom{n}{2} \implies n \cdot \binom{k}{2} \leq 2 \cdot \binom{n}{2}$$. \\
From this we get that :
$$n(\frac{k!}{(k-2)! \cdot 2!})-2(\frac{n!}{(n-2)! \cdot 2!}) = \frac{nk(k-1)}{2}-n(n-1) \leq 0$$
$$\implies nk(k-1)-2n(n-1)=n(k^2 - k - 2(n-1)) \leq 0$$
$$\implies k^2 - k - 2(n-1) \leq 0$$
Thus, \\
$$k \leq \frac{1+\sqrt{1+8(n-1)}}{2}=\frac{1}{2}+\frac{\sqrt{8n-7}}{2}<\frac{1}{2}+\frac{\sqrt{8n}}{2}=\frac{1}{2}+\sqrt{2n}$$
And we are done!


Well, I'm definitely seeing more activity on this (AoPS) than on the exetermath forum...who is having trouble getting on here?

how do you get to the exeter math forum? i dont seem to be able to find it.

haha sara, i just got on, i don't think i was going to the right place
probably should use this forum
perhaps mass email is needed to remind people...?

[quote="saracampbellsoup"]sara sheehan was.. i think she emailed you?[/quote]

Oh yeah...I added her but forgot to specifically tell her that I did...I guess she's here now.

[quote="saracampbellsoup"]what forum are we using for vacation? i am way too easily confused to keep going between the two websites. aops is great if everyone can get on (at least one person is having troubles) but maybe the other one is easier. i don't care... somebody make a decision! (club heads??)[/quote]Who can't get on AoPS? Let us know and we'll try to fix his/her problems. Other than that, have a nice stay here ;)

[quote="Valentin Vornicu"]Who can't get on AoPS? Let us know and we'll try to fix his/her problems. [/quote]

Never mind; it's fixed.

AoPS all the way. I actually haven't even heard of the other website before, so I suggest to stick with this one :).


If you're short on time in a MATHCOUNTS state competition and you have a few problems left that you know you can't do, is it safe to guess? I mean, I know there's no penalty for guessing, but is it a good idea? What should you do if you're low on time and you have a few problems left to go?

If you have less than a minute left, I'd guess. If you have about 5 minutes, maybe less, I'd check over the ones you're not sure about the most (I usually mark ones I'm not sure on with stars), or, if you've already checked or just have like one or two problems, work on those, they could make a huge difference.

My advice giving skills need to get better.

LOL what's the point of guessing if you really don't know the answer? the chance you'd get it right wouuld be like almost 1 over 1 million or something low like that... :maybe:

What's the point of leaving them blank? I only said guess if there's like a dew seconds left, otherwise do the problems.

well you know that many of the answers can only be in a certain range of say 20-43. So you have a much better chance than 1 in a million.

Btw, what I do, is if I have 1 minute left is just try to quickly finish another problem.

If I have 5, then check the ones that you already did that you aren't entirely sure about.

Once I guessed a problem right (a #30) on our school's qualifying test last year, since I knew it had to be pretty small, and 2 just didn't seem to fit in.

Obviously, don't randomly guess a number $ x$, such that $ \minus{}\infty < x < \infty$. Read the problem, identify a good range where the answer might be, and take a guess. Keep in mind the limits (express your answer to the nearest tenth, as a reduced fraction, etc.) and cross your fingers.

Guessing probably wont be the best idea, since you could use that time to try and solve other problems you dont know, but I did guess a target problem i was completely clueless about correctly in Nationals last year  :)

Yeah

Guessing isn't the best option

But, I did guess a problem correctly at Mathcounts Nats last yr.  It was an educated guess based on the numbers in the problem (multiples of seven).

uhh you could guess and check your answers.

guessing only takeslike 5 seconds...

it takes like a second to write down an answer...

Guessing pays off sometimes. Last year at state MC on the target question #8:

A teenager is someone who is 13,14,15,16,17,18, or 19 years old. The product of the ages of a particular group of teenagers is 705,600. What is the mean of their ages? Express your answer as a decimal to the nearest tenth.

I got WAY  too caught up on #7. I had enough time to figure out that 3 ages had to be 15, 15, and 14, so I randomly guessed 14.8 and it turned out to be right! So not randomly guessing, but educated guessing works.

[quote="7h3.D3m0n.117"]Guess and check doesn't work for MATHCOUNTS, especially since the problems require a written answer. This tactic might work for the AMC though.[/quote]

you misinterpreted what i said.

guess and check not at the same time with the same problem.

I mean guess all the ones you dont know and then check the ones you do know (the ones youve done already but may have made an arithmetic error on).

not guess and check the ones you dont know..that would be pointless.

sorry, the way i wrote it might have seemed  ambigious, but if you read the posts before me youll know they were arguing about wheter to guess or to check so i just said guess and check

I guessed #8 on last year's Chapter Target...and got it right :lol: . I went on to win the chapter. If there's a chance...go for it if you really have to.

If you're aiming for a perfect score... check the last few problems. Otherwise, guess everything you can.  Of course, read all the problems before time is up, and you should probably guess answers even before the end so that if you haven't gotten that far, you'll still have an answer.  Remember to circle the problems so you know which ones you guessed on.  Then do them after getting all of the ones you're sure about.

Yeah, I circle the ones I think I got wrong.
So basically, if I finish, I start checking over from question 1.
As soon as the proctor says 5 minutes left, I stop checking the ones I didn't circle and check the ones I did.

[quote="illusiat"]guessing helped me get in top 10 on this year's state! :lol:[/quote]

thanks...everybody...mroe pressure on me for making state cd.

[quote="gauss1181"]Well, you didn't really have to guess for the teenagers age problem.

You just have to use your math computation skills fast to get the answer...and know about means and stuff.

Well, it sometimes is good to guess for the team round even though it may not work...My team guessed on the last problem in the chapter team round but we got it wrong :mad:[/quote]

Well, you're right, normally you wouldn't guess, but I just spent basically until they called time on the one right before it (which I got wrong, somehow. :wallbash_red: ). I didn't even look at that question until  about 30 seconds before they called time (I know it was stupid, but I was only in 6th grade.). I ended up getting 20th or so.  :blush:

20th in states?

thats rather ironic when u just guess

Well, yeah it was like 21st or something in the low twenties.

I have a pretty easy state, which helps.

GUESSING WORKS!!!
On #30 on a Chapter sprint round, time was running out, and i just guessed a random number...
I GOT IT RIGHT!!!

Anyways, people who say there's a 1 in a million chance for getting a problem right, that is WRONG...
It's easy to find some insight, and get a number range... That greatly increased my chances.
As for the person above me who won chapter CD in 7th grade,

I AM IN SEVENTH GRADE, TOO!!!!
I WON COUNTDOWN ROUND, TOO!!!!

Quote thecheezinator:

As for the person above me who won chapter CD in 7th grade,

I AM IN SEVENTH GRADE, TOO!!!!
I WON COUNTDOWN ROUND, TOO!!!!

Woot. Go 7th graders who win CD!

Quote: 7h3.D3m0n.117

Just because it happened once doesn't mean it works everytime. You got lucky. Congratulations.

In general, guess only if you're running out of time. Never do it if you can at least attempt the problem.

Couldn't have said it better myself. It worked for me after the stupid mistake of not even looking at the other Target problem.

[quote="alanchou"][quote="thacheezinator"]
I AM IN SEVENTH GRADE, TOO!!!!
I WON COUNTDOWN ROUND, TOO!!!![/quote]

rather unnecessary, huh?

as for guessing, go ahead if you absolutely need to, but if you can do the easy problems really fast and get them right, you might find the solution in the middle of the test[/quote]
why unnecessary?

how is it possible that u guess a hard problem and get it right?

I kinda meant that part right over there.  It takes attention away from the topic

If you want to know how it's possible to guess a hard question's answer correctly, then it's called luck.  Although you can limit the cases.

I never make a wild guess :) only make a guess if the chances are worth it for using up 5 seconds. The only time I remember guessing on MC ...well it was a 50-50 chance...but I got it wrong :(

AIME is a different story.I managed to answer all the questions! :lol: I got a --

I don't think you can say whether or not you answered all the questions Jason.

Of course either I was able to do all of them or I guessed all the ones I couldn't do. anyone who hasn't taken the AIME still has no clue how hard they are :)


trigonometry
Pythagorean Theorem
geometry
similar triangles

Line segments drawn from the vertex opposite the hypotenuse of a right triangle to the points trisecting the hypotenuse have lengths $\sin x$ and $\cos x$, where $x$ is a real number such that $0<x<\frac{\pi}2$. The length of the hypotenuse is
$\text{(A)} \ \frac 43 \qquad \text{(B)} \ \frac 32 \qquad \text{(C)} \ \frac{3\sqrt{5}}{5} \qquad \text{(D)} \ \frac{2\sqrt{5}}{3} \qquad \text{(E)} \ \text{not uniquely determined}$

[quote="4everwise"]Line segments drawn from the vertex opposite the hypotenuse of a right triangle to the points trisecting the hypotenuse have lengths $\sin x$ and $\cos x$, where $x$ is a real number such that $0<x<\frac{\pi}2$. The length of the hypotenuse is
$\text{(A)} \ \frac 43 \qquad \text{(B)} \ \frac 32 \qquad \text{(C)} \ \frac{2\sqrt{5}}{5} \qquad \text{(D)} \ \frac{2\sqrt{5}}{3} \qquad \text{(E)} \ \text{not uniquely determined}$[/quote]Eh? I didn't get one of the answers. I can't find what I'm doing wrong though and dinner calls...[hide]Suppose there exists $\bigtriangleup ABC$ such that $m\angle ABC=90^\circ$. Let points $M$ and $N$ trisect segment $AC$ such that $M$ is closer to vertex $A$ than point $N$. Let $MB$ have length $\sin x$ and $NB$ have length $\cos x$ for some $0<x<\frac{\pi}2$. Let $AM=MN=NC$ have length $y$. Thus, $AC$ has length $3y$. Let $AB$ have length $c$ and $CB$ have length $a$. Now we can finally begin the actual problem...

From Stewart's Theorem on $\bigtriangleup ANB$ with cevian $MB$, \[ (c)(y)(c)+(\cos x)(\cos x)(y)=(\sin x)(\sin x)(2y)+(y)(y)(2y),  \] \[ (c^2+\cos^2x)(y)=(2y)(\sin^2 x+y^2).  \]
From Stewart's Theorem on $\bigtriangleup BMC$ with cevian $NB$, \[ (a)(y)(a)+(\sin x)(\sin x)(y)=(\cos x)(\cos x)(2y)+(y)(y)(2y),  \] \[ (a^2+\sin^2x)(y)=(2y)(\cos^2 x+y^2).  \]
Adding these two equations, \[ (a^2+c^2+\cos^2x+\sin^2x)(y)=(2y)(\sin^2 x+\cos^2 x+y^2+y^2).  \] But from the Pythagorean Theorem, $a^2+c^2=(3y)^2=9y^2$ and from trigonometry identities $\sin^2x+\cos^2x$ for all $x$. Thus, \[ (9y^2+1)(y)=(2y)(1+2y^2).  \] Since $y$ is clearly positive, divide by $y$ and simplify: \[ 5y^2=1,  \] \[ 9y^2=\frac{9}{5}, \] \[ 3y=\frac{3\sqrt{5}}{5}.  \] Hence, the length of the hypotenuse is $\frac{3\sqrt{5}}{5}$.[/hide]

[hide="Solution"]Answer: $\text{(C)}$

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=536[/img]

Applying the Pythagorean Theorem on triangles $CDF$ and $CEG$, we get

$4a^2+b^2=\sin^2 x$
$a^2+4b^2=\cos^2 x$

Since $\sin^2 x+ \cos^2 x=1$, why not add these?  :D

Adding and simplifying gives $a^2+b^2=\frac 15$.

Now, the hypotenuse is $3\sqrt{a^2+b^2}=3\frac{\sqrt{5}}{5}$.

[/hide]

[hide=Median Formula]
Let $BC$ be trisected by $D$ and $E$ and let $BD = DE = EC = a$. Also, let $AB = b$ and $AC = c$. Assume that $AD = \sin x$ and $AE = \cos x$.

By the Pythagorean Theorem, we have that $b^2 + c^2 = 9a^2$. Since $AD$ is the median of triangle $ABE$ and $AE$ is the median of triangle $ACD$, by the median formula we get that \[\sin^2 x = \frac{2b^2 + 2\cos^2 x - 4a^2}{4}\] \[\cos^2 x = \frac{2c^2 + 2\sin^2 x - 4a^2}{4}.\] Adding the latter two equations, we get that \[1 = \frac{2b^2 + 2c^2 + 2 - 8a^2}{4}.\] Using the fact that $b^2 + c^2 = 9a^2$, we get the equation \[1 = \frac{5a^2 + 1}{2},\] which gives us that $a = \frac{1}{\sqrt{5}}$. The answer is $3a = \boxed{\frac{3\sqrt{5}}{5}}$.
[/hide]

[quote=minsoens][hide="another solution?"]

Assume you have triangle ABC, with M and N as trisecting points in the hypotenuse AB.
Draw line segments parallel to BC from M and N so they meet AC;
draw line segments parallel to AC from M and N so they meet BC
Then, by similar triangles, AC and BC are trisected.

Let $\frac{AC}{3}=y$, $\frac{BC}{3}=x$.
Draw line segments CM and CN. Then you have two right triangles:
By Pythagoras,
$x^2+(2y)^2=\sin^2x$
$(2x)^2+y^2=\cos^2x$
Add the equations together:
$5x^2+5y^2=1$
But $\sqrt{(3x)^2+(3y)^2}$ is the hypotenuse, so
$\sqrt{(3x)^2+(3y)^2}=\sqrt{9\cdot \frac{1}{5}}=\boxed{\frac{3\sqrt{5}}{5}}$

[/hide][/quote]

Thank you for sharing as your solution is much cleaner than mine. I will keep in mind to drop perpendicular lines to create similar triangles.

@below I see no problem with bumping posts with new content on a math-related site.

@above casually bumping a 17 year old thread


Έστω $ a_{0},a_{1},...,a_{n},...$ μια ακολουθία αριθμών που ικανοποιούν $ \left(3\minus{}a_{n\plus{}1}\right)\cdot\left(6\plus{}a_{n}\right)\equal{}18$ και $ a_{0}\equal{}3$. Υπολογίστε το $ \sum^{n}_{i\equal{}0}\frac{1}{a_{i}}$.

[quote="r_boris"]Μετά τις πράξεις
$ \displaystyle \frac {1}{a_{n \plus{} 1}} \equal{} \frac {1}{3} \plus{} \frac {2}{a_n} \Rightarrow \frac {1}{2^{n \plus{} 1}a_{n \plus{} 1}} \equal{} \frac {1}{3.2^{n \plus{} 1}} \plus{} \frac {1}{2^na_n}\Rightarrow \frac {1}{2^na_n} \minus{} \frac {1}{3} \equal{} \frac {1}{3} \minus{} \frac {1}{3.2^n}\Rightarrow \frac {1}{a_n} \equal{} \frac {2^{n \plus{} 1}}{3} \minus{} \frac {1}{3}\Rightarrow S_n \equal{} \frac {2(2^{n \plus{} 1} \minus{} 1)}{3} \minus{} \frac {n \plus{} 1}{3}, n \equal{} 0,1,2,..$[/quote]

Ένας άλλος τρόπος να το αποδείξουμε είναι να θέσουμε στην παραπάνω $ b_{n} \equal{} \frac {1}{a_{n}}$. Για δοκιμάστε και αυτό λοιπόν...


Sorry for the rather vague title; I didn't want it to be too long.

Also I apologize if this is the wrong section of the forums to post this question, but I found it the most appropriate so here goes:

What is the greatest value of "n" such that [img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/b/5/3/b535a45d48945c0c8f9be357d7c3a4eee25ea5d3.gif[/img] will evenly divide [img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/9/e/6/9e6e6024bd2867c47972d9f2fbff42ed4955512f.gif[/img]?

I need to be able to do this problem in under 1 minute without the aid of a calculator. Any shortcuts?

[hide]
to calculate the maximum power of n in k!:

divide k by n and floor.  repeat until result is 0.  then add up all the results.

[/hide]

[quote="JBL"][quote="CircleSquared"]to calculate the maximum power of n in k!:

divide k by n and floor.  repeat until result is 0.  then add up all the results.[/quote]

So, by your method, what's the largest power of 4 that divides $ 15!$?  (Or the largest power of 6, or 8, or 10, or 12?)[/quote]

Any specific reason you're asking me for evens?

[quote="Altheman"]His point is that your method only works for prime numbers.[/quote]

... dang.
well i've never really been asked about n being a composite number, so i never had to worry about this.

[quote="pythag011"]First of all, we ccan ignore the 125 because it is not a multi;ple of 2. There are 8 even numbers from 1-16. However, some of these are also multiples of 4, and there are 4 of these. Similiarly, there are 2 divisible by 8 and 1 divisible by 16.

It happens there are 8+4+2+1 twos. Therefore, the answer is 15. Generally, n! has (# of multiples of 2)+(# of multiples of 4)+....+(# of multiples  of the greatest power of 2 that is smaller or equal to n.[/quote]

Thanks. I should've thought of that. It makes sense now.

[quote="CircleSquared"]well i've never really been asked about n being a composite number, so i never had to worry about this.[/quote]

That's not true -- I just asked you above! ;) And, in fact, you almost certainly have been asked about composite numbers: "How many $ 0$s are in the final tail of $ 15!$?", for example, is a pretty standard question.  (As one of many, many instances on the forum, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128935]here,[/url] where I asked a question similar to the ones above.)  The questions I've asked are just modest variations: "How many 0s does $ 15!$ end in in base $ b$, $ b \in\{4, 6, 8, 10, 12\}$?"  The original question asked for base 2.

A few other places this has been discussed:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24684
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=173261
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106297


geometry
geometric transformation
reflection
conics
parabola
function
absolute value

$ \frac{1}{1+|x|-2}$

I tried taking cases but it turns out to complicated  :blush:

Cases are exactly the way to go.  And why have you separated the two constant terms $ 1,-2$?  Should one of them be in the absolute value?

sry forgot about this qn

the qn is supposed to be

$ \frac{1}{x^{2}+|x|-2}$

instead of taking cases can you do it like this?

$ \frac{1}{x^{2}+|x|-2}$ = $ \frac{1}{x^{2}+\sqrt{x^{2}}-2}$

does $ |x|$ = $ \sqrt{x^{2}}$ = x?!?!?!

well in genreal if u can sketch the graph of $ f(x)$ then the graph of $ f(|x|)$ is just the reflection of graph of $ f(x),x>0$ along the $ Y$ axis along with the graph of $ f(x),x>0$
here the expression can be written as
$ \frac{1}{|x|^{2}+|x|-2}$
i think this should help u

OK, if sketching the curve is all that you want to do,
i don't see why it has to be difficult... break it down
into manageable steps... :maybe:

(i) first draw $ y_{0}=|x|$, which is "V"shaped with it's vertex at (0,0)
(ii) then move it down by 2 units to get $ y_{1}=|x|-2$
(iii) now draw the parabola $ y_{2}=x^{2}$ which is "U" shaped with it's vertex at (0,0)
(iv) then add the 2 curves $ y_{1}$ and $ y_{2}$, to draw $ y_{3}=x^{2}+|x|-2$
(v) finally, take the reciprocal of $ y_{3}$, and draw $ y_{4}=\frac{1}{x^{2}+|x|-2}$

[quote="bos1234"]does $ |x|$ = $ \sqrt{x^{2}}$ = x?!?!?![/quote]

Yes, for real numbers.  I doubt this will help you graph it, however.

It does make it clear that this is an even function; you can sketch the graph for positive $ x$, and then reflect that sketch.


[hide] A quarter and a nickel. The nickel isn't a quarter, so one of your coins isn't a quater. [/hide]

We are obviously living in 1875 or 1876, and there are twenty-cent coins going around (minted in Carson City, Nevada). The other coin is a dime.

http://en.wikipedia.org/wiki/Twenty-cent_piece_%28U.S._coin%29

:D

Or, the old, common, uncreative (IMHO) answer: [hide]One coin is not a quarter; the other is. The two coins are a nickel and a quarter.[/hide]

[quote="math92"]If i have two coins that total 30 cents and one is not a quarter, then what are the two coins?[/quote]
[hide="answer"]a quater and a nickel :) [/hide]

one quater and uno nickel
another riddle:

a plane carrying people from only America is on a flight to Brazil
it crashes in the Bahamas
where are the survivors buried?

[hide]The survivors are alive so they aren't buried.[/hide]

We are obviously living in barbaric times, when people are buried alive. Therefore...

The point: There can be alternate answers! Like the 20-cent piece from 1875...

[quote="perfect628"][hide]In the ground?[/hide][/quote]
 :rotfl:
No.

[quote="lotrgreengrapes7926"]A plane crashes on the border of Ghana and Benin, where are the dead buried? :D[/quote]

Hey!!! That's [b]my[/b] joke!!!!! :mad:
[hide]Benin and Ghana have Togo between them.[/hide]
Except I use Nepal and Bangladesh.

Technically Benin and Ghana do not border each other... :roll:

[quote="lotrgreengrapes7926"]A plane crashes on the border of Ghana and Benin, where are the dead buried? :D[/quote]

[hide]What a bad joke.  Heard this one before.

It is Togo, I beleve, since Togo is between them.

What a bad riddle.

i heard of this with Nepal and Bangladesh.  Except the answer is India for this one.[/hide]

It's not that bad... I've seen the one with Nepal and Bangladesh too. I think it's more of a riddle to see if you know your geography than a joke. You could also use states, like Kansas and Arkansas if the person's really dumb.

Okay it is not bad, but it isn't funny.  As you said, it just tested your geography skills.


geometry
incenter
circumcircle
inradius
vector
trigonometry
geometric transformation

Let I,O,H denote respectively the incenter,the circumcenter and the orthocenter of a triangle with sides a,b,c and the inradius r.
Prove that the area A of the triangle IOH is

                  A=|(a-b)(b-c)(c-a)|/8r

I'll use vectors to solve this problem.
Let R be the radius of the circumcircle.
In the following, MN denotes the vector with its origin in M and endpoint in N, not the segment joining M and N.
It is known that OH=OA+OB+OC and OI=(aOA+bOB+cOC)/(a+b+c)=(aOA+bOB+cOC)/(2p).
I'll denote by MN*PQ the vector product (not the scalar product) of vectors MN and PQ.
It is known that is we have a triangle ABC then 2*area[ABC]=|AB*AC|, where |MN| is the length of the segment supporting MN.

So 2*area[IOH]=|OH*OI|=(1/2p)*|(OA+OB+OC)*(aOA+bOB+cOC)|=(1/(2p))|OA*OB*(b-a)+OB*OC*(c-b)+OC*OA*(a-c)| (1).

OA, OB, OC are coplanar so OA*OB, OB*OC, OC*OA are all vectors perpendicular to the plane of OABC, so they are colinear. Because I make all the vector products in the same sense (clockwise or counterclockwise - I prefer to draw the triangle counterclockwise), all the vector products have the same sense (they are on the same side of the plane ABC). So there is a versor (unit vector- |u|=1) u such that OA*OB=|OA*OB|*u, OB*OC=|OB*OC|*u, OC*OA=|OC*OA|*u, so (1) comes to:
2*area=(1/(2p))*|R 2sin(2C)*u+...|=(R 2/(2p))|(b-a)sin(2C)+(c-b)sin(2A)+(a-c)sin(2B)|=(R 2/(2p)) *|(2(b-a)sinCcosC+...|=(R/(2p))*|(b-a)ccosC+(c-b)acosA+(a-c)bcosB|. (I used the Sine Law for the last)

Let T=(b-a)cosC+(c-b)cosA+(a-c)cosB
We have T=(b-a)c(a 2 +b 2 -c 2)/(2ab)+...=(1/(2abc))*[(b-a)cc(aa+bb-cc)+(c-b)aa(bb+cc-aa)+(a-c)bb(aa+cc-bb)].
So 2*area=(R/(4abcp))|(b-a)cc(aa+bb-cc)+(c-b)aa(bb+cc-aa)+(a-c)bb(aa+cc-bb)|[b](2)[/b].

Let S=(b-a)cc(aa+bb-cc)+(c-b)aa(bb+cc-aa)+(a-c)bb(aa+cc-bb).
We have a-c=-(b-a)-(c-b), so
S=(b-a)(aacc-cccc-aabb+bbbb)+(c-b)(aacc-bbcc-aaaa+bbbb)=(b-a)(bb-cc)(bb+cc-aa)+(c-b)(bb-aa)(bb+aa-cc)=(b-a)(b-c)[(b+c)(bb+cc-aa)-(b+a)(bb+aa-cc)]=(b-a)(b-c)[(aa+bb+cc)(c-a)-2aa(b+c)+2cc(a+b)|=(b-a)(b-c)[(aa+bb+cc)(c-a)-2ac(a-c)-2b(a-c)(a+c)]=(b-a)(b-c)(c-a)(aa+bb+cc+2ab+2ac+2bc)=(b-a)(b-c)(c-a)(a+b+c) 2=4p 2(b-a)(b-c)(c-a).

So back in (2) we have 2*area=(Rp/abc)|(a-b)(b-c)(c-a)|. we have to prove 2*area=2*|(a-b)(b-c)(c-a)|/(8r).
I suffices to show (I say suffices and not equivalent because we have problems if the triangle is isosceles) that Rp/(abc)=1/(4r) <=> pr=abc/(4R). The LSH and thr RHS are equal because they both equal area[ABC].

So area[IOH]=|(a-b)(b-c)(c-a)|/(8r).

Your proof is correct and very nice.

I would like to know if it is possible to prove this result using geometric transformations.

Let w be the midpoint of OH. w is the center of the nine points circle. By Feurbach, we get Iw=R/2-r.
Formulas for OH and OI are well known.
From these a formula for IH can de found and so we know the lenghts of all the sides of OIH and we can found the area from Peron's formula.

The proof for Feurbach requires inversion. The fact that w is the center of the 9 points center can be proved by similarity (I don't know if homothety is a word). The formula for OI can be found by inversion, so there are lots of geometric transformations in the computations of the sidelengths for OIH, but I think that the computations for the area are uglier than mine.

OH 2 =9R 2 -aa-bb-cc.
OI 2 =R 2 -2rR.
Iw=R/2-r and is the median in OIH. From this we will get
HI 2 =4RR+2rr-(1/2)(aa+bb+cc).

Now lets look at Perron's formula for the area of a triangle whose sidelengths are a,b,c.
It is 1/4 \sqrt ((a+b+c)(-a+b+c)(a-b+c)(a+b-c))=1/4 \sqrt (((b+c) 2-a 2 )(a 2 -(b-c) 2) =...=1/4 \sqrt (2aabb+2aacc+2bbcc-aaaa-bbbb-cccc), so to compute the area it sufices to know the squares of the sidelengths.

I don't want to finish the computations because my previous solution had plenty of it.

Let us use areal coordinates (normalized barycentric coordinates) with respect to $ \triangle ABC.$ Coordinates of the centroid, orthocenter and incenter are given by:

$ G \left(\frac {1}{3}: \frac {1}{3}: \frac {1}{3} \right ) \ , \ H(\cot B \cot C: \cot A \cot C: \cot A \cot B) \ , \ I \left ( \frac {a}{2p}: \frac {b}{2p}: \frac {c}{2p} \right )$

${\left(\begin{array}{ccc}\frac {1}{3} & \frac {1}{3} & \frac {1}{3} \\
 \\
\cot B \cot C & \cot A \cot C & \cot A \cot B \\
 \\
\frac {a}{2p} & \frac {b}{2p} & \frac {c}{2p} \end{array}\right) = \frac {[\triangle IGH]}{[\triangle ABC]} \ \Longrightarrow}$

$ 6p \cdot \frac {[\triangle IGH]}{[\triangle ABC]} = \sum_{\text{cyclic}} a(\cot A\cot C - \cot A\cot B)$

Make the substitution:

$ \cot A = \frac {b^2 + c^2 - a^2}{4[ABC]} \ , \ \cot B = \frac {a^2 + c^2 - b^2}{4[ABC]} \ , \ \cot C = \frac {a^2 + b^2 - c^2}{4[ABC]}$

$ \frac {48[\triangle IHG] \cdot [ \triangle ABC]^2}{r} = \sum_{\text{cyclic}} a^4(c - b) +  a^3(c^2 - b^2)$

$ 48[\triangle IHG] \cdot [\triangle ABC]^2 = 2rp[a^3(c - b) + b^3(a - c) + c^3(b - a)]$

$ \Longrightarrow \ [\triangle IHG]$ $ = \frac {a^3(c - b) + b^3(a - c) + c^3(b - a)}{24[\triangle ABC]}$

Since $ [\triangle IGO] = \frac {1}{2}[\triangle IHG]  \Longrightarrow \ [\triangle IGO] = \frac {a^3(c - b) + b^3(a - c) + c^3(b - a)}{48[\triangle ABC]}$

Using difference of cubes and substituting $ [\triangle ABC] = rp$ gives

$ [\triangle IGO] = \frac {|(a - b)(b - c)(c - a)|}{24r}  \Longrightarrow \  [\triangle IOH] = \frac {|(a - b)(b - c)(c - a)|}{8r}$


probability
Alcumus
videos
LaTeX
geometry
3D geometry
sphere

can people write some forumlas that would be useful for the chapter competition. like what the addition of arithmetic sequences or what sum of the first blank odd numers would be

also can someone write an overview of what we should know for the chapter competion.

There's so much that it's not really possible to list everything. What subjects do you have trouble with?

i don't think there are that many counting and probability formulas to know. in any case, the alcumus videos cover the vast majority of it.

can someone give like 1 or 2 forumulas that would e useful in chapter competiton or one they used  a lot last year or in the school round. Like whats the formula for the sum of the first blank odd numbers.

Responding to your example formulas-

Sum of terms in arithmetic sequence
number of terms(first term+ last term)/2

Sum of first blank odd positive integers
(number of terms)^2

For the sum of the first blank odd integers. I thought that was for even...

$ 1 \plus{} 3 \plus{} 5 \plus{} 7... \plus{} (2n \minus{} 1) \equal{} \boxed{n^2}$

$ 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} ...n \equal{} \frac {n(n \plus{} 1)}{2}$

uhh ill post formulas for others later.

[quote="nikeballa96"]$ 1 \plus{} 3 \plus{} 5 \plus{} 7... \plus{} (2n \minus{} 1) \equal{} \boxed{n^2}$

$ 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} ...n \equal{} \frac {n(n \plus{} 1)}{2}$

uhh ill post formulas for $ \binom{n}{p}$ later.[/quote]

Fixed your Latex :)

haha sorry. i was in a rush and just hit enter without previewing...anyways, formulas.

$ \binom{n}{p}\equal{}\boxed{\frac{n!}{(n\minus{}p)!p!}}$

$ n$ permutations of $ p$ (i don't know the proper way to express in latex): $ \boxed{\frac{n!}{p!}}$

$ 1^2\plus{}2^2\plus{}3^2\plus{}...n^2\equal{}\boxed{\frac{n(n\plus{}1)(2n\plus{}1)}{6}}$

$ 1\plus{}3\plus{}5\plus{}7\plus{}...\plus{}(2n\minus{}1)\equal{}\boxed{n^2}$

$ 1\plus{}2\plus{}3\plus{}4\plus{}...\plus{}n\equal{}\boxed{\frac{n(n\plus{}1)}{2}}$

those are the ones that are used most commonly, i cant remember any others.

Volume of cylinder : (area of base) * height or pi * r^2 * h
Volume of sphere : 4/3 * pi * r^3

Sum of first n numbers:
$ \frac{n(n \plus{} 1)}{2}$

hmm. so whats sum of first blank even numbers?

just outta curiosity, is there a way to answer questions that are like "in how many ways can this sum of money be created with dimes, quarters, pennies, nickels?" without brute forcing?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1396323#1396323]Is searching that hard?[/url] (I searched "pennies nickels dimes quarters".) In general, I don't believe that there's a simple formula.


When folded, O, C, A, and B all border X, so the only one left, which is the opposite of X, is R.

The side opposite will be $ 3$ "sides" away from the original.


If you like intelligence tests you should also try:
http://www.worldofiq.co.cc

You need to complete the demo test to be allowed to register and start the actual test.

Things you might like:
- questions with different answers for each user
- a top with the most intelligent - so answer carefully because you lose points each time you answer incorrectly

BOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO!!!!!!!!!!!!!!!!!!! This is  :spam:  :spam:  :spam:

I just got a perfect. Not too hard.

But when I tried to register it seems that my email died.


use these numbers:  15, 4, 10, 19, and 22 one time with any combination of the operations *,+,-, and / to make a true equation.  u can use parenthesis.

example    19-4-15+22-10=12

can u make the numbers 0 through 7?

I could only get 2, 4, and 6

HELP!!!! PLEASE!!!!!!

Whoa, I think you posted this twice, because I put all of the answers to your question except # 5 in another thread.


x :^3:  + y :^3: = 22z :^3:  for x,y,z coprime positive integers.

I know for a fact that there is only one solution, with x,y,z each having about 5 digits. I'm just wondering if there is a way to find and prove uniqueness without using analytic number theory. I have no clue myself.

Try a google search. This is a very famous problem. I just read something about it recently on the web, but I can't remember where.

Okay I remembered where I saw it. It was mentioned as an aside in this long rambling paper by Louis De Branges about the Riemann Hypothesis. He claimed that it was given to him as a youth and he spent the better part of a year solving it.

Actually, note that there are multiple solutions; but the other ones are just cube multiples of the one you are referring to.

Here's a site with some information on this general type of equation:
http://www.mathsoft.com/mathresources/problems/article/0,,2186,00.html

In general these types of diophanitine questions are very hard. I doubt that there is any "elementary" solution to it, or even an easy way to show that it has only one solution. I really don't know enough to say for sure though.

From a general (and quite spectacular) result that Gerd Faltings found in 1983, it's known that all such cubic equations have at most a finite number of (independent) solutions in both integers and rational numbers. This is a consequence of very deep results in Algebraic Geometry.


x,y,z are non-negative real numbers.Prove that

$3\geq\sqrt{x^2+1}-x+\sqrt{y^2+1}-y +\sqrt{z^2+1}-z$

I guess it's pretty obvious that $\sqrt{x^2 + 1} \leq x + 1$ (square both sides...)


There was a teacher who held up a piece of paper and asked how many squares were on it. They all said 6 and were right. Then the teacher held up the same piece of paper again and asked the same question and all the students said 9 which was right again. How?

[hide]49 is $7^{2}$, and 4 and 9 can be separated to be $2^{2}$ and $3^{2}$. Same with 169, 144, etc.[/hide]

[hide]
There are 6 squares on one side of the piece of paper, and 9 squares on the other side.
[/hide]

[quote="box of rocks"]*** Spoiler Ahead ***

to mustafa's "hidden  comment content"

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I see the 6 squares being:  49, 7^2, 4, 9, 2^2, and 3^2.

I don't see where you described what constitutes the 9 squares.[/quote]

That was just an example of what the squares could be. For an actual answer, try

[hide]1, 4, 9, 16, 49, 144[/hide]

mustafa and rocks I didn't expect such a complicated answer.

e pi is right.

Ihatepie,

yes, I read the intended answer.

However, you invite a gamut of candidate solutions when the parameters in your puzzle are as relatively broad as in yours.


$ a,b,c,d$ positive number prove:
$ (a^{a}b^{b}c^{c}d^{d})^{2}\geq a^{b\plus{}d}b^{a\plus{}c}c^{d\plus{}b}d^{a\plus{}c}$

ln(a^ab^bc^cd^d)²>ln(a^(b+d)*b^(a+c)*c^(d+b)*d^(a+c))
<===>2alna+2blnb+2clnc+2dlnd>(b+d)ln(a)+(a+c)lnb+(b+d)lnc+(a+c)lnd
ona (alna+blnb+clnc+dlnd)>=b*lna+c*ln(b)+b*ln(c)+a*lnd (you can prove that with the inequalite of reordonnemnt)
and (alna+blnb+clnc+dlnd)>dln(a)+a*ln(b)+d*ln(c)+c*ln(d) with reodonnemnt
puis on deduit la resultat4
i'm sorry of i use the french becauce i don't bruch well the english

[quote="zaya_yc"]$ a,b,c,d$ positive number prove:
$ (a^{a}b^{b}c^{c}d^{d})^{2}\geq a^{b \plus{} d}b^{a \plus{} c}c^{d \plus{} b}d^{a \plus{} c}$[/quote]

By Rearrangement, $ a^ab^bc^cd^d\ge a^bb^cc^dd^a$ and $ a^ab^bc^cd^d\ge a^db^ac^bd^c$. Multiplying these two ineqaulties we get the desired one.


Evaluate $ \displaystyle{\sum_{i\equal{}1}^8 {\frac{2}{\sqrt{i}\plus{}\sqrt{i\plus{}1}}}}$

[quote="JRav"]$ \frac {2(\sqrt {i} \minus{} \sqrt {i \plus{} 1})}{i \minus{} (i \plus{} 1)} \equal{} 2(\sqrt {i} \minus{} \sqrt {i \plus{} 1})$[/quote]

[hide]Actually $ 2(\sqrt{i\plus{}1}\minus{}\sqrt{i})$.

$ 2(\sqrt{2}\minus{}\sqrt{1}\plus{}\sqrt{3}\minus{}\sqrt{2}\plus{}\hdots\plus{}\sqrt{9}\minus{}\sqrt{8})\equal{}2(\sqrt{9}\minus{}\sqrt{1})\equal{}4$.[/hide]


trigonometry
geometry
integration
limit
function
calculus
calculus computations

Let
$V = \int\limits_2^{\scriptstyle \hfill \atop \scriptstyle \infty \hfill} {\pi \frac{1}{{(x - 1)^2 }}dx} = \pi \mathop {\lim }\limits_{t \to \infty } \int\limits_2^t {\frac{1}{{(x - 1)^2 }}dx = } \pi \mathop {\lim [}\limits_{t \to \infty } ( - \frac{1}{{t - 1}}) - ( - \frac{1}{{2 - 1}})] \\ \\ = \pi (0 + 1) = \pi \\$
Why  the volume generated by rotating the infinite area $\int\limits_2^{\scriptstyle \hfill \atop \scriptstyle \infty \hfill} {\frac{1}{{x - 1}}dx} =\infty$
about x-axis is finite ?

Have you heard of Gabriel's Horn? http://mathworld.wolfram.com/GabrielsHorn.html  Consider the solid generated by rotating the function $f(x) = \frac{1}{x}$ about the x-axis from $x=1$ to $\infty$.  The volume is $\pi$, but the surface area is infinite.  Meaning that it couldn't hold enough paint to paint itself.  :)


ratio
algorithm
floor function
probability
geometric sequence
geometric series
irrational number

1)show that 1/10  belong to the Cantor set.
2) How can we check whether any given real no. belongs to the Cantor set.

I personally dont feel that ternery expansion will help as for example [ :sqrt: e].here ternery expansion becomes difficult.

You're completely wrong. First, $\frac1{10}=.002200220022\dots_3$ is in the Cantor set. Second, finding the ternary expansion is basically the only way to determine whether a number is in the set. Since $\sqrt{e}-1=.122111220\dots_3$ has ones in its ternary expansion, it is not in the Cantor set.

Finding expansions in whatever base is ugly by hand, but routine calculation for a computer.

o is there no other way. But I was almost certain that 1/10 donot belong to cantor set because this problem came in 1 of the entrance exams of 1 of our institutes(i mean the 1st part)

[quote="the game"]But I was almost certain that 1/10 donot belong to cantor set because this problem came in ...[/quote]
What are you going to believe? Direct, convincing evidence, or the vague authority of what it said (or might have said) on some exam? The thing about arguing mathematics is that no one has the authority to overrule convincing evidence.

$x$ is in the Cantor set if and only if it has a ternary expansion that consists only of 0's and 2's. That expansion is correct: $\frac1{10}=.002200220022\dots_3.$

Put another way,

$\frac2{3^3}+\frac2{3^4}+\frac2{3^7}+\frac2{3^8} +\cdots$ is a geometric sequence with ratio $\frac1{3^4}$.

Its sum $=\frac{\frac2{3^3}+\frac2{3^4}}{1-\frac1{3^4}}=\frac{6+2}{81-1}=\frac8{80} =\frac1{10}.$

No as a student of mathematics I only believe in hard,concrete proofs.Sorry about that.
By the way Kent can you please explain your proof using geometric series.

Things to remember about geometric series:

1. If you identify a series as a geometric series, you [b]must[/b] be able to say what the ratio is. The ratio is what you multiply any term by to get the next term.

2. For a finite geometric series, if the ratio is any number other than 1,

$\text{sum} =\frac{\text{(first term)}-\text{(next term after the last)}}{1-\text{ratio}}$

3. If $|\text{ratio}|<1$, we can take the limit in that to get that the sum of the infinite geometric series is

$\text{sum} =\frac{\text{(first term)}}{1-\text{ratio}}$

For the series in this problem, let

$\frac2{2^3}+\frac2{2^4}+\frac2{2^7}+\frac2{2^8}+\frac2{2^{11}}+ \frac2{2^{12}}+\cdots$

$=\left(\frac2{2^3}+\frac2{2^4}\right)+ \frac1{2^4}\left(\frac2{2^3}+\frac2{2^4}\right) + \frac1{2^8}\left(\frac2{2^3}+\frac2{2^4}\right)+\cdots$

We identify this as a geometric series with ratio $\frac1{2^4}$ and first term $\left(\frac2{2^3}+\frac2{2^4}\right).$ the rest follows from the formula in #3.

Ok i have edited my post I should have done it a long time ago. I know what the sum of a geometric series is.I have no problem about the fact that its sum is 1/10.then how does it help me to show that 1/10 belong to Cantor set.Moreover jmerry's expansion though undoubtedly correct will not help me o tackle the problem in the exam-hall where I wont have any access to computer or a calculator.So can you tell me of any method that will help me to do so without the use of calculator/computer.

You can just use the 'greedy' algorithm,

$1/10<1/3$
$1/10<1/3^2$

$1/10>2/3^3$, etc...
$1/10-2/3^3=7/270>2/3^4$

so $1/10=(.0022.........)_3$

of course you need more terms than that, but if you think you have a pattern, just try to expand it in geometric series.

Frankly speaking I didnt quite undersand the algorithm :(  but one thing interests me is how the geometric progression helps to show that 1/10 belongs to the cantor set.

I just use the greedy algorithm as a general way to classify procedures in which you take the largest thing possible (in this case, multiples of powers of 3) at every possible step.  That is how you calculate expansions in bases other than 10 (and in base 10 too; it's just that we do it instinctively)

But I am not very satisfied with this method.Is there no other better & elegant method.I still havnt understood how that geometric series helps in showing the existence of 1/10 in the cantor set

?

Let $\mathcal{C}$ be the cantor set.

We guess that $1/10$ takes the form $(.00220022\ldots)_3$.  We want to show that a ternary expansion of this form lies in $\mathcal{C}$.

Clearly $(.00200020002\ldots)_3\in\mathcal{C}$.  This is a geometric series with sum $3/40$
Clearly $(.000200020002\ldots)_3\in\mathcal{C}$.  This is a geometric series with sum $1/40$

Summing, we get $1/10=(.00220022\ldots)_3\in\mathcal{C}$, done.

It's somewhat significant to note the thought process.  We guess that $1/10$ can be expressed as a sum of geometric series.  We use some numerical legwork to guess at the form that it would take.  Now we calculuate the individual geometric sums making up this guess.  If they sum to the original number, we're in the clear.

[quote="blahblahblah"]?

We guess that $1/10$ takes the form $(.00220022\ldots)_3$.
.[/quote]
How do you make that guess.Thats why this idea seems so far-fetched to me?I said in an exam-hall you dont get access to computer/calculator.
I have perfectly understood what you are trying to say.

You can do it with pencil and paper. Just commit yourself to writing everything in base 3, but use the division algorithm that you know. Since we're writing everything in base 3, 10 becomes 101. The division would look something like this:

[code]       . 0 0 2 2 0 0 2
      __________________
101  |1. 0 0 0 0 0 0 0
         2 0 2
           2 1 0
           2 0 2
               1 0 0 0
                 2 0 2
                   2 1[/code]
You can see that we've already started to repeat ourselves.

I showed you how to get the first 4 decimals in post 8

You just take the largest power of $3$ out at every step, subtract, repeat the process on the new number.  You can look this stuff up on the internet as well - there are much better tutorials on change of base than on some explanation I'll give you.

Ok its very clear but can you find a elegant proof because how do you use this method to find whther a irrational no belongs to the set or not.

Realistically, the only way to show whether a number belongs to the Cantor set is to determine its ternary expansion. Think of this as an infinite sequence of questions:

1. Is $\frac13<x<\frac23$ ? If yes, then $x\not\in K.$ If no, continue.

Let $y_1=3x-\lfloor 3x \rfloor.$

2. Is $\frac13<y_1<\frac23$ ? If yes, then $x\not\in K.$ If no, continue.

Let $y_2=3y_1-\lfloor 3y_1 \rfloor.$

and so on.

If $x$ is a rational number, then you can employ the division algorithm to determine the full repeating pattern of the ternary expansion and thus determine membership in $K$ in a finite amount of time, one way or the other.

What happens if $x$ is an irrational number? With probablilty 1, the process given above will terminate in a finite amount of time and tell us that $x\not\in K.$ But it is possible (albeit an event of probability zero) the the process does not terminate. Hence, while it is relatively easy to prove that a particular rational number is not in the Cantor set, it may be difficult to prove that it is a member.

As an obvious example, it's easy to create irrational numbers which lie in the Cantor set and which do not lie in the Cantor set.

For example, if we consider

$(.202002000200002\ldots)_3$ it is obviously irrational but in the cantor set; similarly
$(.101001000100001\ldots)_3$ is obviously irrational but not in the cantor set.

Hmm its all very clear.Thanks Kent & blahblahblah

Now I think I found something.I am going through few nos & i found that the nos which donot belong to cantor set,have their 1st 1 in the ternery representation  at a the kth place(say) after the pt. then the no of partion of the interval [0,1] to check that the no donot belong is also k.Is my finding correct.If so how do I prove it.If the no of steps is not exactly k then can the condition be replaced by the condition "that the no of steps reqd is atmost/atleast k.


linear algebra
matrix
algebra unsolved
algebra

A binary matrix or a $(0,1)$-matrix is a matrix whose entries are all either $0$ or $1$.
1) How many $n$×$n$ $(0,1)$-matrices $A$ are there such that $A^{2}$ is also a $(0,1)$-matrix?
2) How many $n$×$n$ $(0,1)$-matrices $A$ are there such that for all positive integers $k$, $A^{k}$ is also a $(0,1)$-matrix?

If $A^{k}$ is (0,1) matrice for any $k=1,2,...,n$, then A is permutation. There are n! permutations.

Let Un the set of (n,n) and {0,1} matrices A s.t. for all positive integers k: A^k is a (0,1) matrix.
   We show that: if A is an invertible matrix of Un, then A is a permutation:
The set of (n,n) and {0,1} matrices is finite; thus there exists k,l s.t. A^k=A^l therefore A^(k-l)=Id or A^(-1)=A^(k-l-1).
LEMMA (easy): if B and B^(-1) have non negative elements, then in each row and in each column of B there exists one and only one element non zero.
Conclusion: A is a permutation.
  Unfortunately there exist very much elements in Un which are not invertible. (9 if n=2 for example).
   Remark: if n=2 the results for questions 1) and 2) are 11.