I am a 4th year High school student and am thinking about participating in the Informatics Olympiad(the discussion regarding learning languages ec. is in the "Reality Check?" post in this forum). I Believe the most important thing(besides knowing th elanguages of course) is the Algorithms. There are a number of books out there and though most people think"Introuction to Algorithms" by Cormen is the authoritative book, I don't think I have the time to read 1200 pages. I have considered the "Art of Computer Programming" series(the BIBLE) and have got the first part. Which other bokks on Algorithms would you recommend? Please suggest books that are not tooo long ( Probably < 600- 700 pages) and are introductory while good for the IOI. Are the books recommended at the official site ( [url=http://olympiads.win.tue.nl]olympiads.win.tue.nl[/url] Good enough? Which are the other books that might be used for the Olympiad???

I think there are more forums of the same character as this one, and I have already posted my opinions in those.

Don't start reading books, start coding, do lots of PRACTICE.

And the way of thinking that "I don't have time to read a 1200 page book" seems a little odd to me. If you don't have time to read a book, you don't have time to practice. And then, you probably don't have enough enthusiasm into programming. You either like it and devote your time to it, or you don't.

There is no recipe to learn to get gold at IOI overnight. Sorry.

Although you are write that practice is THE KEY, I am a total beginner and as you might know, I cant writing codes starightaway!! That is why I am asking for books on algorithms. Also, which programming language would you most recommnd? Where can I get it and its learning documentation? Is Java good? What's so special in Java? ALso, can I get Visual C++???

In that case, do not start reading books on algorithms. Try handling the basics (I mean - really basics) of the language first from any kind of a book about a language or web documentation (google: c++ tutorial), and then start coding, from the simplest algos. USACO is good for this.

I started with Pascal, this year switched to C++. Maybe Pascal is a little easier to learn for a beginner but their essence is almost the same.

Is Java good? Does it have wide applicabiliy??? I have a book on C++ and a web tutorial . Should I continue with it? Pascal is a bit less well known. Is it because it had some shortcoming. Also, which is the best compiler for C++. When should I start studying algorithms? I was thinking if reading "knuth" alongside learning languages!!!

Java is a really useful language, if I was starting to learn a language I'd definitely pick Java. They say it's easier to learn, easier to use, it is supposed to be secure and reliable and platform independent and multimedial and all kinds of things that you could use if you were writing everyday real-world applications... however, it's main disadvantage at algorithm competitions is that it runs much slower. Fortunately, the people that run these competitions are aware of it and they usually set time limits for Java submits higher (5x the normal time limit or so...). Not all the competitions support Java submissions though.
As for Pascal, it is not good for anything except for the fact that there is Delphi, a (kind of) free development enviroment with which you can create windows GUI applications easily.
C/C++ is considered [i]the[/i] language for algorithm competitions. (At least by me :).) This would be my choice of language to learn if you're planning to compete at IOI-like competitions. As I've already written at a different forum, a very good compiler + development enviroment is Dev-C++ which is completely free. That is for Windows. Under Linux, C/C++ compilers are usually integral part of the operating system.

Does Knuths "Art of Computer Programming: Volume 1(Fundamental Algorithms) " cover all algorithm related issues that might appear in competitions or besides olympiads, does the book cover all you need to know[hide]( by 'all' I obviously don't mean what it says-- just 'all' means covering matter useful for atleast 80% of the population)[/hide] about algorithms ,etc??? Many people recommended the "Algorithm Design Manual" by Skiena. Which obe would you prefer. ALso, does "Practice of Programming" by Kernighan include algorithms only?? What is its use in IOI? I got "Programming Pearls" -- is it good???


induction
ceiling function
algebra proposed
algebra

The real sequence is defined as follows: $ x_0 \equal{} 1, x_1 \equal{} 1 \plus{} k$, where $ k$ is a positive real, $ x_{2n\plus{}1} \minus{} x_{2n} \equal{} x_{2n} \minus{} x_{2n\minus{}1}$, and $ \frac{x_{2n}}{x_{2n\minus{}1}} \equal{}\frac{x_{2n\minus{}1}}{x_{2n\minus{}2}}$. Show that $ x_n > 1994$ for all sufficiently large $ n$.

But can a sequence be a gp as well as an ap???? only a,a,a,.... is possible as far as i know.....

doesn't such a definition generate them
P.S hell_ever i have sent u a Private Message please check it at the top right hand side of this page

It is simple enough to prove by induction the formula
\[ a_{2n}\equal{}(1\plus{}nk)^2,a_{2n\plus{}1}\equal{}(1\plus{}nk)(1\plus{}n(k\plus{}1))\]
This gives $ a_M>1994$ for $ M>2 \left\lceil \frac{\sqrt{1994}\minus{}1}{k} \right\rceil$.


calculus
integration
trigonometry
limit
derivative
calculus computations

compute $\int_0^{+\infty} \frac{sin(tx) dx}{x(x^2+1)}$

We assume that $t>0$ here.
Let $F(t) = \int_0^{+\infty} \frac{\sin(t x) dx}{x(x^2+1)}$.
It is clear that $F(t)$ exists for all $t>0$.
and
$\int_0^{+\infty} \frac{\partial}{\partial t}\left(\frac{\sin(t x)}{x(x^2+1)}\right) dx = \int_0^{+\infty} \frac{\cos tx}{x^2+1} dx$
converges uniformly for all $t>0$. (from Weierstrass's Test).
Hence
$F'(t) = \int_0^{+\infty} \frac{\cos tx}{x^2+1} dx$.
Now let us consider $F''(t)$.
$\int_0^{+\infty} \frac{\partial}{\partial t}\frac{\cos tx}{x^2+1} dx = -\int_0^{+\infty} \frac{x\sin tx}{x^2+1} dx$ which uniformly converges for all $t\in [a,A]$ (from Dirichlet's Test).
Thus
$F''(t) = -\int_0^{+\infty} \frac{x\sin tx}{x^2+1} dx$.
Now we see that
$F''(t)-F(t) = -\int_0^\infty \frac{\sin{tx}}{x}dx = -\frac{\pi}{2}$.
Solve the differential equation,
$F(t) = \frac{\pi}{2} + C_1 e^t + C_2 e^{-t}$.
From $F(0) = 0$, we have $C_1 + C_2 = -\frac{\pi}{2}$.
From $\lim_{t\to\infty} F(t) = 0$ (using Riemann's Theorem) , we know that $C_1 = 0$, and then $C_2 = -\frac{\pi}{2}$.
And $F(t) = \frac{\pi}{2}(1-e^{-t})$.

The Residue Theorem!

Consider an indented (around the origin) semicircular contour in the upper half plane.  We know that the contribution along the upper part goes to zero by the Jordan Lemma.

As the indented portion shrinks to the origin, the residue at $0$ is $1$ (independent of $t$), so

$\int^{\infty}_{-\infty} \frac {e^{iz}dz}{z(z^2+1)}=\pi i+2\pi i\text{Res}_{z=i}f(z)=\pi i-\pi ie^{-t}$

Use the evenness of the integral to get $\frac {\pi}{2}(1-e^{-t})$

Just as a note, I used

$\text{Res}_{z=z_0} \frac {e^{itz}}{z(z^2+1)}=\frac {e^{itz}}{3z^2+1}$ to calculate the residues.

I was trying not to use complex analysis  :oops:

of course that is the idea Liyi. Nevertheless, though it is not hard to prove, I can't understand how you justify the computation of $F\"(t)$. (what is this dirichlet test ?).

Exchanging differentiation and integration signs need some conditions. In the frame of mathematical analysis (Lebesgue's integral is not introduced), the condition is strong. We have the following theorem.

[u]Theorem[/u]. Suppose $f(x,y)$ and $f'_y(x,y)$ are continuous on $[a,\infty)\times[c,d]$ and $\int_a^\infty f(x,y)dx$ converges for all $y\in[c,d]$; $\int_a^\infty f'_y(x,y)dx$ uniformly converges for all $y\in[c,d]$. Then we have that $I(y) = \int_a^\infty f(x,y)dx$ is differentiable on $[c,d]$ and \[I'(y) = \int_a^\infty f'_y(x,y)dx\].

[u]Dirichlet's Test[/u]. If $f(x,y)$ and $g(x,y)$ satisfies
(1) $\int_a^A f(x,y) dx$ is uniformly bounded;
(2) $g(x,y)$ is monotone about $x$ (for a fixed $y$);
(3) $g(x,y)$ uniformly tends to 0 about $y$ as $x$ tends to $\infty$
then  $\int_a^\infty f(x,y)g(x,y)dx$ uniformly converges for all $y\in [c,d]$.

[quote="liyi"]...  $\int_a^\infty f'_y(x,y)dx$ uniformly converges for all $y\in[c,d]$. ...[/quote]

I don't understand what you mean here.

Okay.
$\int_a^\infty f(x,y)dx$ uniformly converges for all $y\in S$ means that

for $\epsilon > 0$, there exists $A>a$ such that $\left|\int_A^\infty f(x,y)dx\right| < \epsilon$ for all $y\in S$.


function
quadratics
number theory solved
number theory

Let $g(n)$ be define as follows: $g(1)=0,g(2)=1$ and
$g(n+2)=g(n)+g(n+1)+1$, ($n\geq 1$).
Prove that if $n>5$ is a prime, then $n$ divides $g(n)(g(n)+1)$.

Some problems like this one have been posted beforer on the forum. As usual, I'm stuck half-way :). In my notation $g(n)=a_n$ (I'm more comfortable with that :)).

By computing the generating function of the sequence (pretty boring stuff; I don't want to do it here), we can find $a_n=\frac {(-1)^n}{\sqrt 5}(r_+^{n+1}-r_-^{n+1})$, where $r_+=\frac {-1+\sqrt 5} 2$ and $r_-=\frac {-1-\sqrt 5} 2$.

We have two cases: if (a) $(\frac 5p)=1$ then we can regard $r_{\pm}$ as numbers in $\mathbb Z_p$, and we easily reach the conclusion that $p|a_p$.

On the other hand, if  (b) $(\frac 5p)=-1$ then we have to look at the $r$'s in the field (it's now a field) $\mathbb Z_p[\sqrt 5]$. We should show that $p|a_p+1$, but in order to do this we must show that in $\mathbb Z_p[\sqrt 5]$ we have $r^{2(p+1)}=1$, where $r$ is any of $r_{\pm}$. I don't know how to do this because, as I've mentioned, $\mathbb Z_p[\sqrt 5]$ is a field so it has $p^2-1$ invertible elements, and $2(p+1)$ is much smaller than $p^2-1$.

Another approach might work.

There I go again missing obvious things :):

In the case (b) we just lift $r_+$ to the power $p+1$ if we get $\pm 1$ we're Ok. We also use the fact that $\binom {p+1}k=1 (\mod p)$ for $k=0,1,p,p+1$ and $0 (\mod p)$ otherwise. We get, after all the computations, $\frac{1-\sqrt 5-5^{\frac {p-1}2}\sqrt 5+5^{\frac {p+1}2}}4$. Because $5$ isn't a quadratic residue of $p$, we have $5^{\frac {p-1}2}=-1 (\mod p)$, so that thing is equal to $-1 (\mod p)$, so $r^{p+1}=-1\Rightarrow r^{2(p+1)}=1$, Q.E.D.

We don't need any freaky fields or rings or anything like that :).

I think you may find smth interesting here:
[url]http://www.mathlinks.ro/viewtopic.php?t=1428[/url]

For grobber: I think you are not doing well working with $a_n$ as linear combination of (n+1)th powers of $r^+$ and $r^-$. nth powers are easier to work with because you know that $(a+b)^p=a^p+b^p$ when working mod p. Of course you can adapt, but nth powers are just nicer.


algebra
function
domain
algorithm
polynomial
Ring Theory
superior algebra

Let $K$ a comutative field. Find all prime ideals from $\mathbb{K} [X].$. ;)

Yeah- $K[X]$ is a principal ideal domain by the divison algorithm, so the prime ideals are precisely the ideals generated by irreducible polynomials.

$\mathbb{K}[x]$ is an euclidean domain (by the degree of the polynomials), so also a principal ideal one. Now if $\mathbf{o}$ is an ideal of $\mathbb{K}[x]$, there is an $a \in \mathbb{K}[x]$ with $\mathbf{o} = a \mathbb{K}[x]$.
If $\mathbf{o}$ is prime, we get that $a$ is prime $=$ irreducible.
Now if $xy=az \in a\mathbb{K}[x]$, you get that $a$ divides $x$ or $y$, so the inverse also holds.


Is there any way to make text smaller than \tiny?

It depends on the font you are using. If you are using a standard  font then the [url=http://tug.ctan.org/pkg/fix-cm]fix-cm package[/url] will scale them to any size.
[code]\usepackage{fix-cm}
...
{\tiny This is tiny}\\
\fontsize{3}{5pt}\selectfont This is size 3[/code]


$S$ is a subset of $\left\{1,2,3,...,2n\right\}$ with $n+1$ elements. Show that we can always choose $a,b \in S$ so that $a | b$.

This has been posted before.  We write each number in the form $2^{a}b$ where $b$ is odd and $a$ is a nonnegative integer.  Then we have every even number in the set is a power of 2 times some odd number in the set.  Then, we have only n odd numbers, so we have $n$ psosible values for $b$.  Then two numbers written this way have the same $b$, as desired.

Indeed. I heared this question from one of the Dutch team members on IMC.

(Did I give you this problem in fact, Jan? I only remember giving it to Arne and Christophe?)

An absolute Erdös classic. I've known this problem since... 4 years or something.

We dealed with this problem on the training weekend in Beersel  :)

And there will always be two such that $gcd(a,b)=1$

That holds because there will always be two consecutive integers.


geometry
circumcircle
Euler
incenter
geometry unsolved

$\triangle ABC$ is scalene with circumcircle $(O)$ and incircle $(I).$ Let $X,Y,Z$ be the midpoints of the arcs $BC,CA,AB$ of $(O).$ Show that the inverse lines of the circles with diameters $IX,$ $IY,$ $IZ,$ under the inversion with respect to $(I),$ cut the corresponding sidelines of the medial triangle of $\triangle ABC$ at three collinear points. Furthermore, show that this line also passes through the pole of the Euler line of $\triangle ABC$ WRT $(I).$

After massive calculations, I found that all three points satisfy the equation
\[ XI^2 \minus{} XG^2 \equal{} \frac{16Rr \plus{} r^2 \minus{} s^2}{9},\]
where $ I$ is the incenter and $ G$ is the centroid.  Thus, all three points lie on a line that is perpendicular to $ GI$.  I don't think the pole of the Euler line lies on this line.  It would be nice to see a synthetic solution.

Thanks for your interest dear nsato, the problem is indeed correct.

[url=http://www.xtec.cat/~qcastell/ttw/ttweng/resultats/r276.html]Lemma 1[/url]. If $ M_c, M_b$ are  midpoints of  $ AB$ and $ AC,$ then the pole of $ M_bM_c$ WRT $(I)$ is the orthocenter $ H_1$ of $\triangle IBC.$

[url=http://www.xtec.cat/~qcastell/ttw/ttweng/definicions/d_Schiffler_p.html]Lemma 2[/url]. Euler lines of $\triangle ABC,\triangle IBC,\triangle ICA,\triangle IAB$ concur at Shiffler's point of triangle $\triangle  ABC.$

The inverse line $p_a$ of the circle with diameter $ IX,$ under inversion WRT $(I),$ is the polar of $X$ WRT $(I).$ The polar of $ H_1$ WRT $(I)$ is the A-midline $M_bM_c$ of $\triangle ABC.$ Therefore, $J_A \equiv M_bM_c \cap p_a$ is the pole of $XH_1,$  the Euler line of $\triangle IBC.$ Similarly, $J_B,J_C$ are the poles of the Euler lines of $\triangle ICA,$ $\triangle IAB$ $\Longrightarrow$ $J_A,J_B,J_C$ are collinear on the polar of the Schiffler's point of $\triangle ABC$ WRT $(I)$ $\Longrightarrow$ $J_AJ_BJ_C$ passes through the pole of the Euler line WRT $(I),$ as desired.

First, I misread the question.  But I still got three collinear points.  :|  I'll be posting it as another problem.

Second, thanks for your very nice problem and solution.


سلام به همه من سانازم . من 4 روزه دارم رو این سوال فکر می کنم کسی است کمکم کنه ؟
R ----->R+

F(x^2+y^2)=F(x^2-y^2)+F(2xy)

 So F(x) = ? .....

لطفا کمکم کنید ...

age gharar bedim $ x\equal{}0$ natije mishe $ f(y^2)\equal{}f(\minus{}y^2)$ yeni dar vaghe baraye har $ x\in\mathbb{R}$ darim $ f(x)\equal{}f(\minus{}x)$ khob pas kafie $ f$e aadade mosbat ro hesab konim:

gharar bedid $ x^2\minus{}y^2\equal{}a,2xy\equal{}b$ khob hala deghat konid ke darim:

$ a^2\plus{}b^2\equal{}x^4\plus{}y^4\minus{}2x^2y^2\plus{}4x^2y^2\equal{}x^4\plus{}y^4\plus{}2x^2y^2\equal{}(x^2\plus{}y^2)^2$

pas dar vaghe:

$ x^2\plus{}y^2\equal{}\sqrt{a^2\plus{}b^2}$

pas ba tavajoh be farze masale khahim dasht:

$ f(\sqrt{a^2\plus{}b^2})\equal{}f(a)\plus{}f(b)$

ya be ebarati baraye har $ a,b\in\mathbb{R}$ darim:

$ f(\sqrt{a\plus{}b})\equal{}f(\sqrt{a})\plus{}f(\sqrt{b})$

khob hala gharar bedid $ g(x)\equal{}f(\sqrt{x})$ khob pas avvalan $ g: \mathbb{R}^\plus{}\to\mathbb{R}^\plus{}$ hamchenin $ g(a\plus{}b)\equal{}g(a)\plus{}g(b)$ khob hala inam moadele kooshye,ke ye moadele maroofe,dar har soorat gharar bedid $ b\equal{}\epsilon\to 0^\plus{}$ pas darim:

$ g(a\plus{}\epsilon)\equal{}g(a)\plus{}g(\epsilon)>g(a)$

pas dar vaghe tabamoon soudie,khob pas javabamoon hast $ g(a)\equal{}ca$ yeni $ f(x)\equal{}cx^2$... :wink:

آیا کسی  راه حل دیگری دارد ؟


calculus
integration
logarithms
trigonometry
special factorizations
calculus computations

Find the following integrals:

1) Find the indefinite integral   $ \int \frac{3^{\tanh x}}{\cosh ^{2}x}dx$.

2) Find the indefinite integral     $ \int \frac{dx}{x\sqrt{1\plus{}(\ln x)^{2}}}$.

3) Find the indefinite integral    $ \int e^{2x}\sin ^{2}xdx$.

4) Find the indefinite integral     $ \int \frac{x\plus{}1}{x^{2}\plus{}x\plus{}1}dx$.

Hi.

For 3 I would write $ \sin^{2} x$ in terms of $ \cos 2x$ and use integration by parts.

For 4. I would split the integrand into $ \frac{1}{2}\frac{2x\plus{}1}{x^{2}\plus{}x\plus{}1} \plus{}\frac{1}{2}\frac{1}{x^{2} \plus{} x \plus{} 1}$. The first one is a log and the second, after completing the square is an $ \tan^{\minus{}1}$, I think.

Sorry, gotta go.

For 2 we first substitute lnx=u so we have
$ \begin{array}{l}
 I \equal{} \int {\frac{{du}}{{\sqrt {1 \plus{} {u^2}} }}} \mathop  \equal{} \limits_{du \equal{} \frac{1}{{{{\cos }^2}t}}dt}^{u \equal{} \tan t} \int {\frac{{\cos t}}{{{{\cos }^2}t}}dt \equal{} } \int {\frac{{{{\left( {\sin t} \right)}^\prime }}}{{\left( {1 \minus{} \sin t} \right)\left( {1 \plus{} \sin t} \right)}}dt \equal{} }  \\
  \equal{} \frac{1}{2}\left[ {\ln \left( {1 \plus{} \sin t} \right) \minus{} \ln \left( {1 \minus{} \sin t} \right)} \right] \plus{} c \equal{}  \\
  \equal{} \frac{1}{2}\ln \left( {\frac{{1 \plus{} \sin t}}{{1 \minus{} \sin t}}} \right) \plus{} c \equal{}  \\
  \equal{} \frac{1}{2}\ln \left( {\frac{{1 \plus{} \sin (arc\tan u)}}{{1 \minus{} \sin (arc\tan u)}}} \right) \plus{} c \equal{}  \\
  \equal{} \frac{1}{2}\ln \left( {\frac{{1 \plus{} \sin (arc\tan (\ln x))}}{{1 \minus{} \sin (arc\tan (\ln x))}}} \right) \plus{} c \\
 \end{array}$

$ \frac{d}{dx}(\tanh x)\equal{}\frac{1}{\cosh ^2x}$.


[i]Source[/i]:[b][u]Spanish Math Olympiad[/u][/b]

Show that $\sqrt{x} + \sqrt{y} + \sqrt{xy} = \sqrt{x} + \sqrt{(y + xy + 2y\sqrt{x} )}$.

When you square and multiple $ \sqrt{y} and\ \sqrt {xy} $ , I belive, the signs of x and y come into picture. Rest is ok. Isn't it?

I think probably it has got more to it but it is form an olympiad, that is weird!

I think it is very obvious that it is equivalent  :roll:

[hide] $\sqrt{(\sqrt{y} + \sqrt{xy})^2} = \sqrt{y+xy+2y\sqrt{x}}$ [/hide]

Hi,
Is this really an exercise from Olympiad tests??? :o
It's very easy from the start we have [tex]\displaystyle \sqrt{x} + \sqrt{y} + \sqrt{xy} = \sqrt{x} + \sqrt{(y + xy + 2y\sqrt{x} )}[/tex] we can simplify by [tex] \sqrt{x} [/tex] the rest is more difficult :D
So how can this be an Olympiad test :P

A catch is that when we have $ 2 \times \sqrt {y} \times \sqrt {xy} $ , we can write it as $2  \sqrt{ x y^2 } $ if and only if at least one of $ y $ & $ xy $ is positive. Maybe thats what they are asking for

No,
When [tex] \sqrt{x}[/tex] and [tex] \sqrt{y} [/tex] are defined so [tex] x \ge 0 [/tex] and [tex] y \ge 0 [/tex]

If you can write $x$ in such way that $\sqrt{x}$ is real , then definitely we can say $x$ is positive . :P

Ah yes, Maybe the goal is to show the other equality It's more difficult. Did you show it?

of course , my point is valid in $ \mathbb{C} $ nos. Also, the 2nd equality is by no means difficult.


What is the probability that a randomly-selected two-digit
number is a multiple of 14? Express your answer as a common
fraction.

There are 90 two digit numbers. Out of these, 14*1 thru 14*7, or 7 numbers, are multiples of 14. This is $ \boxed{7/90}$

[hide]1.there are 90 2 digit #'s from 1-100 '

2.there are 14*1-14*7 multiple of 14 from 1-99 and thats 7

3.so we get our answer as 7/90[/hide]

 :spam:  :spam:  :spam:  :spam:


The vertices of triangle BAC are lattice points (they have integer coordinates). There are no other lattice points on the boundary of the triangle, but there is exactly one lattice point inside the triangle. Prove that it must be the centroid of ABC.


induction
number theory unsolved
number theory

Prove that for any integers $ n \ge 2$ there is a set $ A_n$ of $ n$ distinct positive integers such that for any two distinct elements $ i,j \in A_n, |i\minus{}j|$ divides $ i^2\plus{}j^2.$

[quote="moldovan"]Prove that for any integers $ n \ge 2$ there is a set $ A_n$ of $ n$ distinct positive integers such that for any two distinct elements $ i,j \in A_n, |i \minus{} j|$ divides $ i^2 \plus{} j^2.$[/quote]
This problem has been posted before. You should notice some ideas :
 Let A be a set of numbers , we defind  $ A \plus{} k \equal{} \{x \plus{} k|x\in A\}$ .
1) If $ A$ be a good set then $ A \plus{} d$ also is a good set where $ d \equal{} \prod_{i < j}(j \minus{} i)$ .
2)If A be a good set ,then $ A \plus{} \{0\}$ is  a good set .
Induction on N will do the rest .


geometry
perimeter
inradius
circumcircle
HCSSiM
incenter
area of a triangle

[hide]The area of the circle will be 4pi. The area of the triangle will be 17 because using the formula rs. r=inradius,s=semi perimter[/hide]

[quote="mecrazywong"][quote="Altheman"]for the other question...

note that $p=a+b+c$, $A=\frac{1}{2}rp$, $R=\frac{abc}{4A}=\frac{abc}{2rp}$,

by am-gm, $\frac{a+b+c}{3}\ge\sqrt[3]{abc}$

$\frac{p^{3}}{27}\ge 2rpR$

$R\le \frac{p^{2}}{54r}$ with equality when $a=b=c$

p.s. this is nice![/quote]
Sorry to spoil the fun, you have only found a upper bound for the circumradius of $\triangle ABC$.[/quote]

And... that's what I had asked for in my follow-up question, if you hadn't read it in the post above.

[quote="I"][i]Another question:[/i] With the same perimeter and inradius, what's the maximum radius of the circumcircle?[/quote]

[quote="erik-the-red"][quote="Shivar"]The circle inscribed in a triangle with perimeter 17 cm. has a radius of 2 cm.  What is the area of the circle?  What is the area of the triangle?[/quote]

[hide]

The area of the circle is trivial. The radius is given, so $\pi \cdot r^{2}= 4\pi$.

I'm still thinking about the triangle, though.

[/hide][/quote]

Use the fact that the area of a triangle is $A=rs$, where $A$ is the area, $r$ is the inscribed circle's radius, and $s$ is the semiperimeter.

Haha the alternate solution to the problem is to realize that this is a HCSSiM problem that has a numerical answer, so clearly the answer must be 17 heh.

erik-the-red, this formula (A=rs) comes from drawing the line from the incenter of the triangle to the three verticies. This creates 3 triangles, all of whose altitude is r, the radius of the incircle. The bases of these three triangles add up to the perimeter. multiplying r*p/2 to get the area of all the triangles (you can try assigning other variables to the other sides and adding them up individually but it will eventually come down to this) gives the area of the entire triangle.

It doesn't make sense to me to write it as A=rs just because s=p/2. what does the semiperimeter have anything to do with the area of this triangle from the way this formula is derived?

[quote="cincodemayo5590"]Wow, I'm sorry about my above post ^. Somehow I ended up editing my original post, instead of making a new one to respond to [b]mecrazywong[/b]; but I can't edit the post anymore because of the forum restrictions.  :ninja:[/quote]

 :ninja:

[quote="Altheman"][quote="erik-the-red"][hide="Question"]
I take it semiperimeter is half the perimeter?

If so, then $A = rs = (2) \cdot \frac{17}{2}= 17$?
[/hide][/quote]

yes...[/quote]

Just making sure.


calculus
integration
search
algorithm
linear algebra
matrix
function

I would appreciate help with this problem and also an explanation of what a line integral is:

Evaluate explicitly each of the line integrals
\[ \int (x \textup{ d}x \plus{} y \textup{ d}y \plus{} z \textup{ d}z),
\int (y \textup{ d}x \plus{} x \textup{ d}y \plus{} \textup{ d}z),
\int (y \textup{ d}x \minus{} x \textup{ d}y \plus{} e^{x\plus{}y} \textup{ d}z),\] along (i) the straight line path joining the origin to $ x \equal{} y \equal{} z \equal{} 1$, and (ii) the parabolic path given parametrically by $ x \equal{} t, y \equal{} t, z \equal{} t^2$ with $ 0 \le t \le 1$.

For which of these integrals do the two paths give the same results, and why?

[quote="aidan"]I would appreciate help with this problem and also an explanation of what a line integral is[/quote]  It seems to me that you have these written down in the wrong order.  The appropriate thing to do here is to read the section on line integrals in any standard multivariable calculus text and/or a few of the websites (like wikipedia) that pop up when you do a search in your favorite search engine for the words "line integral."  A few minutes of reading will be enough to uncover an algorithm that will allow you to evaluate the integrals in question; a somewhat more thorough effort will introduce you to the idea of a conservative field, which is what the "why" part is asking about.

Wikipedia tells me the line integral along a piecewise smooth curve $ C$ is defined as
\[ \int_C f \textup{ d}s\]
where $ \textup{ d}s$ is the length of an infinitestimal segment of the curve and $ f$ is a scalar field. But in the integrals given, e.g. $ \int (x\textup{ d}x \plus{} y\textup{ d}y \plus{} z\textup{ d}z)$ how do I deal with the differentials dx, dy, dz or convert this to one where there is only one differential, say dt?

Edit: Wait, do you write them as
\[ \int \begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} \textup{ d}x \\ \textup{ d}y \\ \textup{ d}z \end{pmatrix}, \, \int \begin{pmatrix} y \\ x \\ 1 \end{pmatrix} \cdot \begin{pmatrix} \textup{ d}x \\ \textup{ d}y \\ \textup{ d}z \end{pmatrix}, \, \int \begin{pmatrix} y \\ \minus{}x \\ e^{x\plus{}y} \end{pmatrix} \cdot \begin{pmatrix} \textup{ d}x \\ \textup{ d}y \\ \textup{ d}z \end{pmatrix}?\]
Is it correct to say that $ \textup{ d}s \equal{} \left|\begin{pmatrix} \textup{ d}x \\ \textup{ d}y \\ \textup{ d}z \end{pmatrix}\right|$?

Keep reading :).  That's the line integral of a scalar function -- you have here the line integral of a vector field, which is the next step down the page.  What you wrote in your edit is correct, but we don't need to have $ ds$ in this context.  The next thing you need to do is to translate into the parametrization of the curves over which you're integrating.

Ok, since
\[ \int_{C}\mathbf{F}(\mathbf{r}) \cdot \textup{ d}\mathbf{r} \equal{} \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r'} \textup{ d}t\]we can consider things in the light of $ x \equal{} y \equal{} z \equal{} t$ for the first integral. Then $ \textup{ d}x \equal{} \textup{ d}y \equal{} \textup{ d}z \equal{} \textup{ d}t$. So the first integral simply becomes $ \int_0^1 3t \textup{ d}t$?

I have another question at this point: say you have a vector field $ \mathbf{r}$ of a curve dependent on a parameter $ t$. Later on in most courses they will want to parametrize the field in terms of the arc length $ s$ and consider the function as $ \mathbf{r}(s(t))$. What is the purpose of this?

[quote="aidan"]So the first integral simply becomes $ \int_0^1 3t \textup{ d}t$?[/quote]  Yes.  (And similarly with the others.)

[quote="aidan"]I have another question at this point: say you have a vector field $ \mathbf{r}$ of a curve dependent on a parameter $ t$. Later on in most courses they will want to parametrize the field in terms of the arc length $ s$ and consider the function as $ \mathbf{r}(s(t))$. What is the purpose of this?[/quote]  In the notation of the first part of your post, the vector field is $ \textbf{F}(\textbf{r})$ and the curve is parametrized by $ \textbf{r} \equal{} \textbf{r}(t)$.  Arc-length (or constant-speed) parametrization has a number of nice features.  Notably, the tangent vector $ \textbf{r}'(t)$ is always a unit vector and so the second derivative $ \textbf{r}''(t)$ is a normal vector to the curve.  This allows a canonical coordinate frame at every point of the curve.

I see.. that's nice indeed. And earlier there was something about conservative forces - that means the integral depends only on the endpoints and not on the path taken?

Yes -- some vector fields have this property and others do not, and there is a nice characterization of those fields that do have this property.


Find all naturals $ x,y,z$, which satisfy the equation $ 1\plus{}{2^{x}}{3^{y}}\equal{}z^{2}$

[quote="aleksandar"]Find all naturals $ x,y,z$, which satisfy the equation $ 1 \plus{} {2^{x}}{3^{y}} \equal{} z^{2}$[/quote]
$ 2^x3^y\equal{}(z\minus{}1)(z\plus{}1)$

1) $ x\equal{}0$ $ \implies$ $ 3^y\equal{}(z\minus{}1)(z\plus{}1)$
Both $ z\minus{}1$ and $ z\plus{}1$ cant be multiple of $ 3$, so one is $ 1$ and the other is $ 3^y$ $ \implies$ $ 1\plus{}2^03^1\equal{}2^2$

2) $ x>0$ $ \implies$ $ z$ even $ \implies$ $ x\geq 2$
Both $ z\minus{}1$ and $ z\plus{}1$ cant be multiple of $ 3$
Both $ z\minus{}1$ and $ z\plus{}1$ cant be multiple of $ 4$
So 4 cases :

2.1) $ z\minus{}1\equal{}3^y2^{x\minus{}1}$ and $ z\plus{}1\equal{}2$ impossible

2.2) $ z\minus{}1\equal{}3^y2$ and $ z\plus{}1\equal{}2^{x\minus{}1}$, so $ ^2\cdot 3^y\plus{}2\equal{}2^{x\minus{}1}$ so $ 3^y\plus{}1\equal{}2^{x\minus{}2}$
If $ y\equal{}0$, we have the solution $ 1\plus{}2^33^0\equal{}3^2$
If $ y>0$, the equation $ 3^y\plus{}1\equal{}2^{x\minus{}2}$ mod 3 implies $ x\minus{}2\equal{}2v$ even and then $ 3^y\equal{}(2^v\minus{}1)(2^v\plus{}1)$
Only one of the two factors may be divisible by 3 and so $ 2^v\minus{}1\equal{}1$ and the solution $ 1\plus{}2^43^1\equal{}7^2$

2.3) $ z\minus{}1\equal{}2^{x\minus{}1}$ and $ z\plus{}1\equal{}2\cdot 3^y$, so $ ^2\cdot 3^y\minus{}2\equal{}2^{x\minus{}1}$ so $ 3^y\minus{}1\equal{}2^{x\minus{}2}$
If $ x\equal{}2$, this gives no solution
If $ x\equal{}3$, this gives the solution $ 1\plus{}2^33^1\equal{}5^2$
If $ x\geq 4$, the equation $ 3^y\minus{}1\equal{}2^{x\minus{}2}$ mod 4 implies $ y\equal{}2v$ odd and so $ (3^v\minus{}1)(3^v\plus{}1)\equal{}2^{x\minus{}2}$
Only one of the two factors may be divisible by $ 4$ and so the other is $ 2$, so $ 3^v\minus{}1\equal{}2$ and the solution $ 1\plus{}2^53^2\equal{}17^2$

2.4) $ z\minus{}1\equal{}2$ and $ z\plus{}1\equal{}2^{x\minus{}1}3^y$ and the solution $ 1\plus{}2^33^0$ already found.

So the five solutions :
$ 1\plus{}2^03^1\equal{}2^2$
$ 1\plus{}2^33^0\equal{}3^2$
$ 1\plus{}2^43^1\equal{}7^2$
$ 1\plus{}2^33^1\equal{}5^2$
$ 1\plus{}2^53^2\equal{}17^2$


Let $ a,b,c$ be positive real numbers and $ k\ge 0$. Prove that
\[ \frac{a}{\sqrt{a+kb}}+\frac{b}{\sqrt{b+kc}}+\frac{c}{\sqrt{c+ka}}\le \sqrt{\frac{3(a+b+c)}{k+1}}\]

In our topic, VASC proposed the general problem for my inequality

[quote="Vasc"][quote="nayel"]I think I just proved that the following inequality holds for $ k,a,b,c>0$ and $ a+b+c=1$
\[ \frac{a}{\sqrt{a+kb}}+\frac{b}{\sqrt{b+kc}}+\frac{c}{\sqrt{c+ka}}\le \sqrt{\frac{3}{k+1}}\]
[/quote]
Actually, the following inequality holds for $ k,a,b,c>0$ :
\[ \frac{a}{\sqrt{a+kb}}+\frac{b}{\sqrt{b+kc}}+\frac{c}{\sqrt{c+ka}}<\sqrt{\frac{(k+1)(a+b+c)}{k}}\]
[/quote]

And Nayel, also in this topic, you show a solution to it (actually, the solution is totally the same as the particular case). But at the moment, I think your problem is wrong, It is wrong just for $ k=2$. I think you  can check it easily.


1. I can paint the floor alone in 2 hours. My dad can paint it alone in 5 hours. How long will it take us to paint the floor together?

2. In four hours, Adam, Ben, and Chris can paint the whole house. If Adam works alone, he can paint the house in 28 hours. If Ben works alone, he can paint it in 7 hours. How long does it take Chris to paint the house alone?

3. Amber and Brittney can make a bucket of paint in 3 hours. Amber and Chelsea can make a bucket of paint in 4 hours. Brittney and Chelsea can make a bucket in 6 hours. How long will it take all 3 girls to make a bucket of paint if they work together?

4. Pump A can fill a pool in 6 hours. Pump B can fill the pool in 4 hours. Pump C can drain the pool in 5 hours. Pump A is opened and begins filling an empty pool. 2 hours later, pump C is opened and starts draining. An hour after that, pump B is opened. How many more hours will it take to fill up the pool?

5. 4 people can make 10 pounds of flour in 3 days. How many days will it take 3 people to make 4 pounds of flour?

That's it for now. These are the types of questions (work/rate/time) that you should never miss on a Mathcounts test.

[hide="hint for 1-4"]
How much of the house/bucket/pool is made/filled every hour?
[/hide]

[hide="hint for 5"]
The people and flour are directly proportional
The people and days are in inverse proportion
The flour and days are directly proportional.
[/hide]

I may be wrong on some...

[hide="Answer"]1.$ \frac{10}7$ hours

2.$ 14$ hours

3.$ 40$ minutes ??

4.$ \frac37$hours

5.$ \frac85$days??[/hide]

Fixed..I think..

[hide]3.$ \frac{8}{3}$hours

4.$ \frac{18}{37}$hours[/hide]

[quote="Chinaboy"]I don't think I got those as well. Someone correct me if I'm wrong. I just made these questions out of the blue.[/quote]
You can't be wrong....

I got a new answer for 4..(miscomprehension of a problem..and I still might be wrong)

[hide]$ \frac{13}{42}$hour[/hide]

[quote="Chinaboy"]
3. Amber and Brittney can make a bucket of paint in 3 hours. Amber and Chelsea can make a bucket of paint in 4 hours. Brittney and Chelsea can make a bucket in 6 hours. How long will it take all 3 girls to make a bucket of paint if they work together?

[/quote]

[hide="What I did on #3"]let $ A$=Amber, $ B$=Brittney, $ C$=Chelsea.

$ A\plus{}B\equal{}\frac13$

$ A\plus{}C\equal{}\frac14$

$ B\plus{}C\equal{}\frac16$

Adding those up, we get $ 2A\plus{}2B\plus{}2C\equal{}\frac{4\plus{}3\plus{}2}{12}\equal{}\frac9{12}\equal{}\frac34$

Divide by 2 to get $ A\plus{}B\plus{}C\equal{}\frac{3}{8}$

We want $ (A\plus{}B\plus{}C)x\equal{}1$, so $ x\equal{}\boxed{\frac83}$[/hide]

[quote="Chinaboy"]4. Pump A can fill a pool in 6 hours. Pump B can fill the pool in 4 hours. Pump C can drain the pool in 5 hours. Pump A is opened and begins filling an empty pool. 2 hours later, pump C is opened and starts draining. An hour after that, pump B is opened. How many more hours will it take to fill up the pool? [/quote]

hmm... would this be it?

[hide]$ \frac16(3)\minus{}\frac15\plus{}(\frac16\plus{}\frac14\minus{}\frac15)x\equal{}1$

$ \frac12\minus{}\frac15\plus{}(\frac{10\plus{}15\minus{}12}{60})x\equal{}1$

$ \frac3{10}\plus{}(\frac{13}{60})x\equal{}1$

$ (\frac{13}{60})x\equal{}\frac7{10}$

$ x\equal{}\frac7{10}\times \frac{60}{13}\equal{} \boxed{\frac{42}{13}}$[/hide]


conics
parabola
geometry
perpendicular bisector
geometry proposed

Hi, everyone.

Prove that triangles subtended at the focus of a parabola by any two tangents are similar.

[b]with synthetic proofs (no analytic, no trigonometric)[/b]

Thanks in advance. :blush:

Let $ F$ be the parabola focus and $ d$ its directrix. Parabola main axis $ m \perp d$ through $ F$ cuts the directrix at $ D$ and the perpendicular bisector $ v$ of $ FD$ is the parabola vertex tangent. Let arbitrary parabola tangents $ s, t$ at $ S, T$ intersect at $ X.$ Let $ K, L \in d$ be feet of perpendiculars from $ S, T$; $ SK \equal{} SF, TL \equal{} TF.$ Parabola tangents $ s, t$ bisect the angles $ \angle FSK, \angle FTL$ $ \Longrightarrow$ they cut the segments $ FK, FL$ at their midpoints $ M, N \in v.$ Since $ FM \perp SMX$ and $ FN \perp TNX,$ quadrilateral $ FMXN$ is cyclic. Then $ \angle FXT \equal{} \angle FXN \equal{} \angle FMN \equal{} \angle MSK \equal{} \angle FSM \equal{} \angle FSX$ and similarly, $ \angle FXS \equal{} \angle FTX$ $ \Longrightarrow$ $ \triangle FSX \sim \triangle FXT$ are similar.


Solve this equation:
$x\in \mathbb{R}$
$3^x+4^x+5^x=6^x$ :)

Let $f(x)=(3/6)^x+(4/6)^x+(5/6)^x$. We have $f^{'}(x)<0$. Therefore equation had unique solution x=3.

why the solution is $3$ when we have $f^{'}(x)<0$? :)
can you explain more?

That means that the function is strictly decreasing, thus is injective.


For all positive integer $ n\geq 2$, prove that product of all prime numbers less or equal than $ n$ is smaller than $ 4^{n}$.

It equavalent to $ \sum_{p\le n} ln(p)<nln4$. It is known, that $ \sum_{p\le x}ln(p)\equal{}x(1\plus{}o(1)).$

Please, and understandable and elementary solution.

Elementary solution is Chebechev's method: $ \prod_{k<p\le 2k}p\le C_{2k}^k\equal{}\theta \frac{2^{2k}}{\sqrt{\pi k}}.$

First note that if $ n > p > \frac{n}{2}$ then $ \prod_{\frac{n}{2} < p < n} p < {n \choose \frac{n}{2}}$
By applying Stirling's formula you can see $ {n \choose \frac{n}{2}} < 4^{\frac{n}{2}}$
So then we have $ \prod_{p} p \equal{} (\prod_{\frac{n}{2} < p < n} p)(\prod_{\frac{n}{4} < p < \frac{n}{2}} p) (\ldots) < 4^{\frac{n}{2}}4^{\frac{n}{4}} \ldots \equal{} 4^{n}$


I just had a dream that I didn't qualify for the AIME last year and was really sad and decided to go on AoPS and talk about how I didn't make it but then AoPS was acing really weird and I didn't know how to go to any of the forums besides Fun and Games.  It was really weird.

There is a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159549]thread[/url] for this you know.

Okay, so tell me again why you have to tell us about it???

OMG, i didnt get all As :ewpu: Call the police!


abstract algebra
group theory
superior algebra
superior algebra unsolved

-the group has order 143.

-is Z24 isomorphic to S4?

not sure of the source.

1) Note 143 = 11*13. There is a theorem about this.

2) Note Z24 is abelian whereas S4 is not.

sorry, but for 1), could you point me @ the theorem?

new to group theory, but i'm looking at these sylow theorems and trying to figure it out...feeling stupid. anymore tips?

well apply sylow theorems. in the case of a group G of order 143, we have two normal sylow subgroups order 11 and 13, which must be cyclic (prime order!), and G is the direct product of these, hence cyclic (11 and 13 are coprime).


Find the value(s) of $ x$ which make the series converge.

$ \displaystyle\sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^n(2x\minus{}3)^n}{4^n}$

$ \sum_{n \equal{} 0}^{\infty} r^{n}$ converges $ \Longleftrightarrow |r| < 1$.


[quote]a parabola only has 1 variable squared) [/quote]

Not necessary : for example  x^2+y^2+2xy=16 is a  parabola.
(You look at   b^2-4*a*c   value  assuming the given eq is ( a*x^2 +b*x*y + c*y^2 + .other_.terms  =0) amd see if it is  positive ( Hyparaobla), zero (parabola) or negative (ellipse).. .. circle is special case when  a=c and b=0 etc..

prob.  the first thing one learns when one does  conic section  .. :maybe:
Edited later to remove typos

[quote="marucho8d7"]My bad. sorry[/quote]

Never mind. I won't give the answer to this problem, though.

[quote="Yrael"]It's an ellipse. Here's why:

There are both X^2 and Y^2 terms, thus it is not a parabola (a parabola only has 1 variable squared)
Then look at the signs of the X^2 and Y^2 terms - they are both positive, thus you know it's not a hyperbola (in a hyperbola, they are of different signs)
Lastly look at the coefficients of the X^2 and Y^2 terms - they are not the same, thus you know it's not a circle.

That leaves an ellipse.[/quote]

[quote="Zratonx"]Conic in Alg 1... Either it is something else or TX educations is declining we didn't do this stuff until midterm Alg 2...[/quote]

The way Yreal said it, I could prove with what I learned in midterm Geometry.

[quote="Gyan"][quote]a parabola only has 1 variable squared) [/quote]

Not necessary : for example  x^2+y^2+2xy=16 is a  parabola.
(You look at   b^2-4*a*c   value  assuming the given eq is ( a*x^2 -b*x*y + c*x^2 + .other_.terms  =0) amd see if it is  positive ( Hyparaobla), zero (parabola) or negative (ellipse).. .. circle is special case when  a=c and b=0 etc..

prob.  the first thing one learns when one does  conic section  .. :maybe:[/quote]

your method seems to be good
25X^2 + Y^2 - 100X - 2Y + 76 = 0
you said [i]the given eq is ( a*x^2 -b*x*y + c*x^2 + .other_.terms  =0)[/i]
I think it should be a*x^2 -b*x*y + c*[b]y[/b]^2 + .other_.terms  =0
and a=25, b=0, c=2, b^2-4*a*c=-4*25*2=-200, an ellipse.

P.  - Thanks, typo is corrected in the original  post.

[quote="marucho8d7"]What conic does 25X^2 + Y^2 - 100X - 2Y + 76 = 0 represent?

A: PARABOLA
B: CIRCLE
C: ELLIPSE
D: HYPERBOLA
E: NOT ENOUGH INFORMATION[/quote]

25x2 + y2 - 100x - 2y + 76 = 0
25[x2 - 4x] + y2 - 2y + 76 = 0
25[(x - 2)2 - 4] + (y - 1)2 - 1 + 76 = 0
25(x - 2)2 - 100 + (y - 1)2 - 1 + 76 = 0
25(x - 2)2 + (y - 1)2 = 52

Because the coefficient of the x brackets is different to the y bracket the shape will be an ellipse, if they are equal it would be a circle.


Prove that the equation $ x^4\minus{}x^3\minus{}1\equal{}0$ has exactly two roots. If $ a$ and $ b$ denote their sum and product respectively, prove that $ b<\minus{}\frac{11}{10}$ and $ a> \frac{6}{11}$.

Can someone please post a solution for this problem?
I can't seem to link the bounds for $ a,b,c$ with the condition for the occurence of a pair of complex conjugates as roots.

Let $ x^4\minus{}x^3\minus{}1\equal{}(x^2\minus{}ax\plus{}b)(x^2\minus{}(1\minus{}a)x\minus{}\frac 1b)$.
Then $ a(1\minus{}a)\plus{}b\minus{}\frac 1b \equal{}0$ and $ \frac ab\equal{}b(1\minus{}a)$ or
$ b^2\equal{}\frac{a}{1\minus{}a}, \frac 1b \minus{}b\equal{}a(1\minus{}a)$. Therefore $ 0<a<1$ and $ 0<\frac 1b \minus{}b\equal{}c\equal{}a(1\minus{}a)\le\frac 14$
If $ b>0$, then $ \frac{\sqrt{67}\minus{}1}{8}\le b<1$. It give $ 4b>a^2$ and $ x^2\minus{}ax\plus{}b$ had not solution. Therefore $ b<0$ and $ \frac{\minus{}1\minus{}\sqrt{67}}{8}\le b<\minus{}1$. It give $ (1\minus{}a)^2\plus{}\frac{4}{b}<0$ - only 2 real solution.
Therefore $ 1<\frac{a}{1\minus{}a}\le \frac{34\plus{}\sqrt{67}}{68\plus{}\sqrt{67}}<\frac 59$. Therefore $ \frac{20}{81}<\frac 1b \minus{}b\le \frac 14$.
It give $ b<\minus{}\frac{10\plus{}\sqrt{6661}}{81}\equal{}\minus{}1.131048761...<\minus{}\frac 98<\minus{}\frac{11}{10}$ and $ a\equal{}\frac{b^2}{1\plus{}b^2}>\frac{1}{2}\plus{}\frac{5}{\sqrt{6661}}$ or $ a>0.561263286..>\frac 59 >\frac{6}{11}$

[quote="Agr_94_Math"]Prove that the equation $ x^4 \minus{} x^3 \minus{} 1 \equal{} 0$ has exactly two roots. If $ a$ and $ b$ denote their sum and product respectively, prove that $ b < \minus{} \frac {11}{10}$ and $ a > \frac {6}{11}$.[/quote]

Let $ P(x)\equal{}x^4\minus{}x^3\minus{}1$

1) two real roots
$ P'(x)\equal{}4x^3\minus{}3x^2\equal{}4x^2(x\minus{}\frac 34)$ and so $ P(x)$ is decreasing on $ (\minus{}\infty,\frac 34)$ and increasing on $ (\frac 34,\plus{}\infty)$
Since $ \lim_{x\to\minus{}\infty}P(x)\equal{}\plus{}\infty$ and $ \lim_{x\to\plus{}\infty}P(x)\equal{}\plus{}\infty$ and $ P(\frac 34)\equal{}\minus{}\frac{283}{256}<0$, we have exacty two real roots $ x_1<\frac 34<x_2$

2) bounds
$ P(\frac {91}{66})\equal{}\minus{}\frac{135461}{66^4}<0$ $ \implies$ $ x_2>\frac{91}{66}$

$ P(\minus{}\frac 45)\equal{}\minus{}\frac{49}{5^4}<0$ $ \implies$ $ x_1<\minus{}\frac 45$

$ P(\minus{}\frac {55}{66})\equal{}\frac{9559}{2376^2}>0$ $ \implies$ $ x_1>\minus{}\frac {55}{66}$

So $ a\equal{}x_1\plus{}x_2>\frac{91}{66}\minus{}\frac {55}{66}\equal{}\frac 6{11}$

And $ b\equal{}x_1x_2<\minus{}\frac{91}{66}\frac 45<\minus{}\frac {11}{10}$

Q.E.D.

[quote="pco"]
2) bounds
$ P(\frac {91}{66}) \equal{} \minus{} \frac {135461}{66^4} < 0$ $ \implies$ $ x_2 > \frac {91}{66}$

$ P( \minus{} \frac 45) \equal{} \minus{} \frac {49}{5^4} < 0$ $ \implies$ $ x_1 < \minus{} \frac 45$

$ P( \minus{} \frac {55}{66}) \equal{} \frac {9559}{2376^2} > 0$ $ \implies$ $ x_1 > \minus{} \frac {55}{66}$

[/quote]
How these numbers came to your mind dear [b]pco[/b]???

Dimitris

[quote="Dimitris X"]How these numbers came to your mind dear [b]pco[/b]???[/quote]

 :blush: I computed the exact roots thru calculus and then looked for approx values not too complicated to use


Find the focus for  y=3x^2+6x+4

how do u find the focus...? is there an equation to use..

In geometry, the foci (singular focus) are a pair of special points used in describing conic sections. The four types of conic sections are the circle, parabola, ellipse, and hyperbola.
$Source: Wikipedia$

that equation is equivalent to
$y=3(x+3)^{2}-5$
so the vertex is at (-3, -5)
remember the equation
$y=4px$ where $p$ is the focus?
translating the vertex back to the origin, we find the focus
$y=3x^{2}$ is the base parabola, so $3x^{2}=4px$
so $p=\frac{3x}{4}$
...and I can't remember what to do from here. sorry - I didn't take my first term of precal when they covered this stuff.

thanks... but can anyone finish showing me..?

[hide="probably wrong..."]$y=3x^{2}+6x+4=3(x+1)^{2}+1$. The general form for a parabola is $y-k=\frac{1}{4a}(x-h)^{2}$(?), where (h,k) is the vertex, and a is the distance from vertex to focus.
$(-1,-1)$ is the vertex.
$a=1/12$. So the focus is at $(-1,-11/12)$(?).[/hide]

The vertex is actually $(-1,1)$, so the focus is at [hide]$(-1,\frac{13}{12})$.[/hide]

yea the answer is (-1,13/12) but how did you get that could you do it from squaring it.. i keep trying it but cant get that 13/12 focus is (h,p+k) right..?

The focus is $(h,k+p)$, and $(h,k)$ is the vertex, which is $(-1,1)$. Since $\frac{1}{4p}=3$ you get $p=\frac{1}{12}$.

Thus, the focus is at $(-1,1+\frac{1}{12})$ or $\boxed{(-1,\frac{13}{12})}$.

wait how u get 1/4p=3 ? doesnt that equal to 12 and not 1/12??

$\frac{1}{4p}=3$ comes from the the factored equation $y-1=3(x+1)^{2}$. Compare this with $y-k=\frac{1}{4p}(x-h)^{2}$.


geometry
3D geometry
geometric transformation
rotation

Each of the faces of a cube is colored by a different color. How many of the coloring are distinct?

Here's another Burnside's Lemma approach.
[hide]
Let $ G$ be a group of all rotations on a cube. Now $ |G|=24$ because there are 6 ways to place the face down and another 4 different positions after that.

Let $ X$ be the set of all possible positions of the cube, not counting rotations. So $ |X|=6!=720$. We say two arrangements in $ x_{1},x_{2}\in X$ are equivalent if there is some rotations that brings $ x_{1}$ to $ x_{2}$, i.e. $ x_{1},x_{2}$ are in the same orbit under $ G$ acting by rotations on $ X$.

Now if $ g=e$, i.e. the identity element then $ X_{g}= X$ and so $ |X_{g}|=720$. If $ g\not = e$ then $ |X_{g}|=0$ because all other positions are changed.

As explained above we are counting the total number of [b]orbits[/b] and by Burnside's Lemma we have:
$ \frac{1}{|G|}\cdot \sum_{g\in G}|X_{g}| = \frac{1}{24}\cdot \left(720+0+...+0 \right) = 30$.
[/hide]

While the above solution is certainly more educative and easier to generalize with, below is the most basic solution to this particular problem:

[hide]Without loss of generality, let the face with color 1 be the top face.

There are 5 remaining possibilities for the bottom face's color.
There are 4! ways to color the four sides.
With the top face locked, there are 4 rotational positions.

$ N=\frac{5 \times 4!}{4}=30$[/hide]

yep, choose a specific color for the upper face of the cube, say A. Then we have five choices for colouring the lower face of the cube, say with color B. Rotate the cube so that some color C is facing us. Now the remaining sides are fixed with respect to these three. We can distribute the three colors in 3 ×2 ×1 = 6 ways, giving 5 ×6 = 30 possibilities.


algebra polynomial induction geometry calculus

[Possibly posted before.]

Solve

\begin{eqnarray*}1-{x\over 1}+\frac{x(x-1)}{1\cdot 2}-\frac{x(x-1)(x-2)}{1\cdot 2\cdot 3}+\dots+(-1)^{n}\frac{x(x-1)(x-2)\dots(x-n+1)}{n!}=0\end{eqnarray*}

I believe you mean \begin{eqnarray*}1-{x\over 1}+\frac{x(x-1)}{1\cdot 2}-\frac{x(x-1)(x-2)}{1\cdot 2\cdot 3}+\dots+(-1)^{n}\frac{x(x-1)(x-2)\dots(x-n+1)}{n!}=0\end{eqnarray*}

[i]Solution:[/i]

We will prove by induction that \begin{eqnarray*}1-{x\over 1}+\frac{x(x-1)}{1\cdot 2}-\frac{x(x-1)(x-2)}{1\cdot 2\cdot 3}+\dots+(-1)^{n}\frac{x(x-1)(x-2)\dots(x-n+1)}{n!}=(-1)^{n}\frac{(x-1)(x-2)\dots(x-n)}{n!}\end{eqnarray*}
For $n=1$, we have $1-x=-(x-1)$ which is true.  Assume that it is true for all $n$.  By adding $(-1)^{n+1}\frac{x(x-1)(x-2)\dots(x-n)}{(n+1)!}$ to both sides of the equation, we see that after simplification our conjecture is proven by the principle of mathematical induction.  As a result, we are to solve the for $x$ in \[(-1)^{n}\frac{(x-1)(x-2)\dots(x-n)}{n!}=0\] Clearly, we have that $x=1,2,\dots,n$.

Masoud Zargar

Yup, there was a typo in the statement which is now edited.

im dumber  :lol:

I am only taking geometry, and am still working hard for tournaments lol.

You already are probably taking pre cal or calculus or even higher

[quote="neelnal"]What is induction. [/quote]

We have a statement that we want to prove is true for all positive integers $n$.  This usually takes the form of showing that some sequence has some form; i.e. $s_{n}= f(n)$.

First, we verify that it is true for $n = 1$, that is, $s_{1}= f(1)$.

Then we show that

[b]If[/b] the statement is true for $n = k$,

[b]Then[/b] the statement must also be true for $n = k+1$.

Because it is true for $n = 1$, it is true for $n = 2$, which means it is true for $n = 3$, ... which means it is true for all positive integers.

[hide="A more formal response"] Call $S \subseteq \mathbb{N}$ an [b]inductive set[/b] if $a \in S \implies a+1 \in S$.

The Principle of Induction states that any inductive set that contains $1$ is $\mathbb{N}$. [/hide]

Edit:  A more direct solution to the problem is not hard, either.  We suppose $x = 1, 2, 3, ... n$.  We see that after the $n^{th}$ term the numerator is zero, so for each $x$ value we are really dealing with the summation

${x \choose 0}-{x \choose 1}+{x \choose 2}-+...$

Which is, of course, $P(1)$, where

$P(t) = (t-1)^{x}$

Of course, this is identically equal to $0$.

Moreover, the original polynomial has degree $n$, so we have found all the roots.


1) f is an injective function N -> N such that:

f(f(n)) $\leq$ (n+f(n))/2.

Determine all possibilities for f.

2) Find all integers n such that  :sqrt: ((4n-2)/(n+5)) is rational.

I get the answers to be [color=white] (1) f(n) = n and (2) n = 13 only [/color].

What do people think?

I got the same answer for 2.

$\frac{4n-2}{n+5}=\frac{a^2}{b^2}$ with (a,b)=1.
$n=\frac{2b^2+5a^2}{4b^2-a^2}$
We see from $2*(2b^2+5a^2)-(4b^2-a^2)=11a^2$ that $d=(2b^2+5a^2, 4b^2-a^2)|(11a^2)$.  However, because $(a,b)=1$, $2b^2+5a^2$ cannot be divisible by $a^2$.  So $d|11$, which means that d=1 or 11.  We then set $4b^2-a^2=1$, to which a simple mod 4ing reveals that there there are no solutions, and $4b^2-a^2=11$, whose only positive solution is $a=5, b=3$.  This implies that $n=13$.

[quote]1) f is an injective function N -> N such that:
f(f(n)) $\leq$ (n+f(n))/2. [/quote]

. if f(n) < n then f(f(n)) < n and by induction 0 <= f^k(n) < n which is impossible since the f^k(n) are all different. So f(n) >= n for all n.

. if f(n) > f(n) then f(n) < n, we have just seen it's not possible. So f(n) = f(n) for all n and f(n)=n since f is injective.

So f(n)=n is the only solution.

Hope it's correct.


function
algebra
domain
calculus
derivative
analytic geometry
graphing lines

It is true that Q(x)=x^5+x^3+x is a one to one function whose domain and range are all numbers.

Suppose that R is the function inverse to Q, There is no simpe algebraic way to compute values of R. compute R(3), R'(3), and R''(3).

Well the first derivative is
5x^4+3x^2+1
and you can see that setting x^2 to.. X, the new equation 5X^2 + 3X + 1 has no real roots when equated to 0.
That is, its determinate (is that the right word? ) b^2 - 4ac = 9 - 20 < 0.
Therefore the function has no critical points so though it could increase or decrease (figure out where if possible)
anytime f(x) = f(y), x=y. That is, the graph never slopes "back down" to a 'y' already visited.

So yes it is 1:1.

Also, f(-inf) = -inf
f(+inf) = +inf
Therefore domain and range are both the Real set.
By observation F^1(3) = 1.  Because F(1) = 3.
I've given F`up top, so try it's inverse question if possible. Though you might need a restriction on the domain/range.


Assume that penguins live with a density of 1,000 penguins per square mile and can run at an average speed of 7 miles per hour on land and swim at 20 miles per hour. Also assume that a polar bear has a territory of 10 square miles, can run at 25 miles per hour and swim at 10 miles per hour.

How many penguins will an average polar bear eat in any given month, remembering that a polar bear could, as a maximum, only eat one penguin per hour and 7% of the land is next to the sea.

[quote="chess64"][quote="jli"]you only need to post the topic once and please be more discriptive with your title[/quote]

You're a mod; just change the title. (title changed from ??? to Penguins and Polar Bears by me)[/quote]

Sorry about that...


combinatorial geometry
point set
combinatorics
Triangle
IMO Shortlist

Determine whether or nor there exist two disjoint infinite sets $ A$ and $ B$ of points in the plane satisfying the following conditions:

a.) No three points in $ A \cup B$ are collinear, and the distance between any two points in $ A \cup B$ is at least 1.

b.) There is a point of $ A$ in any triangle whose vertices are in $ B,$ and there is a point of $ B$ in any triangle whose vertices are in $ A.$

[hide]This is not possible.

First lemma:  There exist four points in $ A$ that are the vertices of a quadrilateral, convex or concave, which contains a fifth point in $ A$ inside it.  That is, there exists some quadrilateral (whose vertices are in $ A$) that has another point in $ A$ inside it.
Proof:  Choose any five points in $ A$.  Either these five satisfy the lemma, or the five points are the vertices of a convex pentagon.  If the five points form a convex pentagon $ VWXYZ$, then each of the three disjoint triangles $ VWX$, $ VXY$, and $ VYZ$ must contain one point in $ B$.  These three points in $ B$ form a triangle in which there must be a point in $ A$.  But this new point in $ A$ is inside $ VWXYZ$!  So we will be able to find four vertices of the pentagon which form a quadrilateral that contain this new point.  Thus the lemma is proved.

Second lemma:  If a quadrilateral $ VWXY$ contains a point $ P$ in its interior as in the first lemma (All five points are in set $ A$), then there exists another point in set $ A$ that lies in the convex hull of the interior of the quadrilateral.
Proof:   If $ VWXY$ is concave with $ \angle WXY > \pi$, then the three disjoint triangles $ VWX$, $ VXY$, $ WXY$ each contain a point in $ B$.  These three points in turn determine a triangle in which a new point in $ A$ lies.  (If this point happens to be $ P$, then we just consider triangle $ PWX$ instead of $ VWX$, or triangle $ PXY$ instead of $ VXY$, depending on where $ P$ lies.)
If $ VWXY$ is convex, then the three triangles $ VWP$, $ WXP$, $ XYP$ determine three points in $ B$ which in turn determine a new point in $ A$ in the interior.

After applying the second lemma to some five points, we will be left with a quadrilateral with two points in its interior.  We can now choose one of the points such that the remaining points satisfy lemma 1.  The new quadrilateral will then be no larger than the first.   wee can repeatedly aply lemma 2 and keep getting smaller quadrilaterals.  But all of this must stop eventually, because there is a minimum distance between any two points!  So this is impossible. [/hide]

[hide]
Lemma 1:  Given any convex set of $5$ points in $A$, there is another point inside the points that is in $A$.
Proof: Triangulate the set, we have enough points to form a triangle with points in $B$, thus there must be a another point in $A$. $\boxed{}$.

Take $25$ points that belong to $A$. If we take the convex hull, by Lemma 1 there are at least five points inside this convex hull.  Assume that are $a$ points in this convex hull that belong to $A$ (including the points on the convex hull) and let $b$ be the number of points in the convex hull that are in $B$.  (Both of these numbers are finite since there is at most one point in any disc of diameter one.)

Lemma 2: Given a set of $d$ points with $c$ points in the convex hull (not on the convex hull), there exists a triangulation of the $d$ points with $d-2+c$ triangles.
Proof:  If $c=0$, we do a usual triangulation.
If $c=1$, draw a line from each point in the convex hull to the point.
We can prove it for $c\ge 2$ by adding another point $P$ inside this triangulation and drawing lines from that $P$ to the verticies of the triangle that it is in. $\boxed{}$

Since there are at least $5$ points in $A$ in the convex hull (not counting the points on the convex hull), there are at least $(a-2)+5$ triangles in the triangulation of these $a$ points in $A$.  Thus, $b\ge a+3$.  If we triangulate the $b$ points in $B$, there are at least $b-2$ triangles.  Thus, $a\ge b-2$.  In conclusion, $b\ge a+3\ge b+1$, contradiction.

edit: So no such sets of points exists.[/hide]

@archimedes1,
I am sorry if this is dumb question, but is it possible for the quadrilateral to decrease every time but decrease "asymptotically," so it decreases but stays above side length 1? Thanks!

MathPanda1, I see what you are saying. If you look at each of these "concentric" quadrilaterals individually, then it is not obvious that any of them have side length less than 1. Instead, note that archimedes1's lemma 2 proves that any given quadrilateral $WXYZ$ with finite area must have an infinite amount of distinct points inside of it. From here it is easy to see that there must be two points with distance less than 1. :)

Ah, I see, that is really nice, thank you so much for all your help!


geometry
calculus
integration
real analysis
real analysis unsolved

This problem requires the knowledge of multivariable calculus. If it is too advanced for this part of the forum feel free to move it.

Prove that the average area of the two dimensional orthogonal projections (i.e. shadows) of a convex three dimensional body is $1/4$ its surface area.

It looks like my poor little topic has gotten lost in this forum. :( Maybe someone can move it to Unanswered and Proposed Problems.

Read some books on Integral Geometry. This is the 3D case of the general theorem.


Here are two bets that are offered in craps:
Big Six: A six will be rolled before a seven.
Big Eight: An eight will be rolled before a seven.
Both bets pay out at 1:1.  This means that you will either win 1 for every 1 that you bet (plus you keep the original bet), or you will lose that bet.

One dollar is bet on Big Six and one dollar is bet on Big Eight.
Compute the probability that:
1. Both bets will lose.
2. Only one of the two bets will win.
3. Both bets will win.

Are you rolling two dice and taking the sum, or what?

*punches self in face*
Yes.  You are rolling two fair six-sided dice and taking their sum.  Sorry, I forgot to include that.


Let $ a,b,c>0$. Prove that

$ (a\plus{}b\plus{}c)^5\geq81(a\plus{}b\minus{}c)(a\minus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)$

------------------
 :blush:  Panagiote LIGOURAS

$ (a\plus{}b\plus{}c)^2 \ge 3(ab\plus{}bc\plus{}ac)$, so we must prove that $ (a\plus{}b\plus{}c)^3\ge 27(a\plus{}b\minus{}c)(a\minus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c)$,
let $ a\equal{}x\plus{}y,b\equal{}y\plus{}z,c\equal{}x\plus{}z$, so our inequality becomes $ (2(x\plus{}y\plus{}z))^3\ge 27.8xyz$, and this is true by AM-GM

[quote="Ligouras"]Let $ a,b,c > 0$. Prove that

$ (a \plus{} b \plus{} c)^5\geq81(a \plus{} b \minus{} c)(a \minus{} b \plus{} c)( \minus{} a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$

------------------
 :blush:  Panagiote LIGOURAS[/quote]

your inequality is equivalent as: if $ a,b,c$ are the sides of triangle

$ ((a\plus{}b\minus{}c)\plus{}(a\minus{}b\plus{}c)\plus{}(\minus{}a\plus{}b\plus{}c))^3(a\plus{}b\plus{}c)^2\geq3^3(a \plus{} b \minus{} c)(a \minus{} b \plus{} c)( \minus{} a \plus{} b \plus{} c)3(ab \plus{} bc \plus{} ca)$

which is obviously true !!  :oops:


symmetry
geometry
3D geometry
tetrahedron
octahedron
icosahedron
dodecahedron

Prove that

[b]The group of rotational symmetries of a tetrahedron is
A(4).

The full group of symmetries of a tetrahedron is
S(4).

The group of rotational symmetries of a cube or of an octahedron
is S(4).

The group of rotational symmetries of an icosahedron or of a dodecahedron is A(5).[/b]

based on:

A regular solid is a 3-
dimensional polyhedron in which each face is a regular polygon. Any two
faces as well as any two vertices can be matched by an isometry (a rigid
motion) of the 3-dimensional space. It is convenient to use a symbol
{p, q}

for a regular solid whose faces are regular
p-gons with q of them situated
around each vertex. We have proved that there are FIVE regular solids:

•
TETRAHEDRON {3, 3},

•
CUBE {4, 3},

•
OCTAHEDRON {3, 4},

•
DODECAHEDRON {5, 3},

•
ICOSAHEDRON {3, 5}.

Any regular solid may be inscribed in a sphere, and then any symmetry of
any regular solid will leave the center of the sphere fixed and will transform
the surface of the sphere onto itself. We call a rotational symmetry of a
regular solid any rotation of the sphere (with respect to an axis passing
through its center) mapping the regular solid into inself. A symmetry of a
regular solid is either a rotational symmetry or a reflection with respect to
a plane (also passing through the center of the sphere), mapping the regular
solid onto itself.

Group theory goes [url=http://www.artofproblemsolving.com/Forum/index.php?f=8]here[/url].


Maybe you mean  $ 1\minus{}a$ instead of $ a\minus{}1$. Then the answer is $ 2$. :ninja:

Actually, it is correct as is.
[hide="hint"]If $ X \equal{} p\plus{}q$, then:
$ X^3 \equal{} p^3\plus{}3p^2q\plus{}3pq^2\plus{}q^3$
$ X^3 \equal{} p^3\plus{}q^3\plus{}3pq(p\plus{}q)$
$ X^3 \equal{} p^3\plus{}q^3\plus{}3pqX$
[/hide]

@panos lo: i think it's the same, or maybe i'm mistaken

i've found it,
$ \sqrt[3]{a\plus{}\frac{a\plus{}8}{3}\sqrt{\frac{a\minus{}1}{3}}}\plus{}\sqrt[3]{a\minus{}\frac{a\plus{}8}{3}\sqrt{\frac{a\minus{}1}{3}}}\equal{}x$

We know that if $ a\plus{}b\plus{}c\equal{}0$ we have $ a^3\plus{}b^3\plus{}c^3\equal{}3abc$
Let, $ \sqrt[3]{a\plus{}\frac{a\plus{}8}{3}\sqrt{\frac{a\minus{}1}{3}}}\equal{}a,\sqrt[3]{a\minus{}\frac{a\plus{}8}{3}\sqrt{\frac{a\minus{}1}{3}}}\equal{}b,x\equal{}c$
Then $ c\minus{}a\minus{}b\equal{}0$, so $ c^3\minus{}a^3\minus{}b^3\equal{}3abc$
$ x^3\minus{}2a\equal{}3x.\sqrt[3]{a\plus{}\frac{a\plus{}8}{3}\sqrt{\frac{a\minus{}1}{3}}}.\sqrt[3]{a\minus{}\frac{a\plus{}8}{3}\sqrt{\frac{a\minus{}1}{3}}}$
Simplifying this, we'll get $ x^3\minus{}2a\equal{}4x\minus{}ax$
$ x^3\minus{}4x\minus{}a(2\minus{}x)\equal{}0$
$ x(x\plus{}2)(x\minus{}2)\plus{}a(x\minus{}2)\equal{}0$
$ (x\minus{}2)(x^2\plus{}2x\plus{}a)\equal{}0$
 We get $ x\equal{}2$ or $ x^2\plus{}2x\plus{}a\equal{}0$, but this doesn't satisfy the equation since $ a\le 1$ which is contradict with $ a\ge 1$, so $ \sqrt[3]{a\plus{}\frac{a\plus{}8}{3}\sqrt{\frac{a\minus{}1}{3}}}\plus{}\sqrt[3]{a\minus{}\frac{a\plus{}8}{3}\sqrt{\frac{a\minus{}1}{3}}}\equal{}2$


Does anyone happen to know whether the following problem was solved or not (and if yes, could you please indicate the solution):
   For a fixed n>1 find all functions f defined on the positive integers with values in the same set such that f(a^n+b^n)=f^n(a)+f^n(b). For n=2 we find a very nice French problem (I think it was given in 1994). I'm really interested in this problem, but I'm not smart enough to solve it even when n=3. Any help?

Maybe it is a time to move it to Open Questions.


I came upon this message when I tried to reply to TripleM's post under the now non-existant (which I didn't cause) thread "nominations":
Sorry, but only can reply to posts in this forum.

I'm pretty positive this was because just when I tried to reply to the post someone deleted the entire thread, is that the case?

That's what happens when someone deletes a post while you're writing a reply -- it happened to me when I clicked on one of the posts that was deleted between the time I entered the forum and the time I clicked on the message.


abstract algebra
group theory
superior algebra
superior algebra solved

Show that a finite abelian group is not cyclic iff it contains a subgroup
  isomorphic to $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ for some prime $p$.

Let the group be $G$.

If $G$ is cyclic, so are it's subgroups. So if such a subgroup of type $(\mathbb{Z}/p \mathbb{Z})^{2}$ exists, it has to be cyclic. But every element has at most order $p$, never giving all the $p^{2}$ elements.
Thus if $G$ is cyclic, such subgroups will not occure.

Now write $G$ as product of finite cyclic groups by the structure theorem. By the chinese remainder theorem, we know $\mathbb{Z}/a \mathbb{Z}\times \mathbb{Z}/b \mathbb{Z}\cong \mathbb{Z}/(ab) \mathbb{Z}$ for coprime $a,b$. Thus we merge factors $\mathbb{Z}/a \mathbb{Z}$ and $\mathbb{Z}/b \mathbb{Z}$ untill it's not possible anymore, meaning that all factors $\mathbb{Z}/a \mathbb{Z}$ and $\mathbb{Z}/b \mathbb{Z}$ are such that $a,b$ have a common prime divisor $p$.
If there is only one factor left, $G$ is cyclic, so assume there is more than one factor. Then choose any two factors, let them be $\mathbb{Z}/a \mathbb{Z}$ and $\mathbb{Z}/b \mathbb{Z}$, and take prime $p$ such that $p|a,b$. Then $\mathbb{Z}/a \mathbb{Z}\times\mathbb{Z}/b \mathbb{Z}$ is a subgroup of $G$ and since $\mathbb{Z}/p \mathbb{Z}$ is a subgroup of $\mathbb{Z}/a \mathbb{Z}$ and $\mathbb{Z}/b \mathbb{Z}$, we get that $(\mathbb{Z}/p \mathbb{Z})^{2}$ is a subgroup of $G$.


[quote="enndb0x"]If $ a,b$are posivite Real numbers then prove the inequality :
 $ \frac {a^2}{b^2} \plus{} 1 > 2ab \minus{} \frac {b^2}{b^2 \plus{} 1}$[/quote]

your inequality is symply absurd.suppose it is true.then make $ a\equal{}b\equal{}1000$ .then we´ll have that $ 2>2*{1000}^2\minus{}\frac{1000^2}{1000^2\plus{}1}$.absurd.

[quote="nikeballa96"][quote="v235711"][quote="enndb0x"]If $ a,b$are posivite Real numbers then prove the inequality :
 $ \frac {a^2}{b^2} \plus{} 1 > 2ab \minus{} \frac {b^2}{b^2 \plus{} 1}$[/quote]

your inequality is symply absurd.suppose it is true.then make $ a \equal{} b \equal{} 1000$ .then we´ll have that $ 2 > 2*{1000}^2 \minus{} \frac {1000^2}{1000^2 \plus{} 1}$.absurd.[/quote]

you've shown that it's true for one value of $ a$ and $ b$. now how do we prove that it's true for ALL values of $ a$ and $ b$?[/quote]

i did not show it is true. i simply shown a counterexample.in other words, this inequality does not work for all posiible $ a$ and $ b$. this inequality is false.


[hide="i think"]We have $\frac{B}{2}=2A\Longrightarrow \cos{\frac{B}{2}}=\cos{2A}\Longrightarrow \cos^{2}{\frac{B}{2}}=\cos^{2}{2A}\Longrightarrow \frac{1+\cos{B}}{2}=(2\cos^{2}{A}-1)^{2}$. Now use cosin law.[/hide]

N.T.TUAN could you complete your solution ? I tried to follow your hint but...
I also found that the relation is equivalent to to prove that

$a^{4}-2a^{2}b^{2}+b^{4}+a^{3}c-ab^{2}c-a^{2}c^{2}-ac^{3}=0$

Could anyone help in solving this ?

If your problem is true then my hint will works, I am sure. But it can long. Please try again!  :wink:

It is known that when in $\triangle ABC \angle B=2\angle A$ $b^{2}-a^{2}=ac$.
Let now $\angle B=3\angle A$. Let $M \in AC$ is such that $\angle CBM=\angle A$. As $\angle ABM=2\angle A$ $AM^{2}-BM^{2}=BM.c$. As $\triangle ACB \sim \triangle BCM$ we obtain $BM=\frac{ac}{b}$, $CM=\frac{a^{2}}{b}$ and $AM=b-CM=\frac{b^{2}-a^{2}}{b}$. Now sustitute in $AM^{2}-BM^{2}=BM.c$ to get

$(b^{2}-a^{2})(b-a)=ac^{2}$ - true when $\angle B= 3\angle A$.

When $\angle B= 4\angle A$ take $K$ on $AC$ such that $\angle CBK=\angle A$. Now $(AK^{2}-BK^{2})(AK-BK)=BK.c^{2}$. Replace $BK=\frac{ac}{b}$ and $AK=\frac{b^{2}-a^{2}}{b}$ to obtain the desired equality.
It is evident that this process can be continued the same way to obtain more and more compicated formulas when $\angle B=5\angle A$ etc.


can anyone show me the relation between poisson process and semi


isnt the post test to intro geometry almost identical to the pretest to olympiad geometry?

There is definitely a level in between Intro Geometry and Olympiad and unfortunately we do not yet have a course for that level.  Students interested in this level of geometry might consider taking the AIME Problem Series which covers a lot of hard AIME geometry which is around the level we might call Intermediate Geometry.  It's not as good as a whole course, but it's part of a good comprehensive class.

There are 3 days spent on geometry and one spent on trigonometry and related algebraic geometry topics during the problem series.

[quote="Pakman2012"]I was wondering if this course was being thought about.  I was looking at the Winter schedule, and aside from Olympiad classes and Problem Series, nothing else was being offered.  I don't think I'm that able at geometry to the point where I could do Olympiad Geometry, but I do feel I can do Introduction to Geometry post test with ease.  I would be left behind in Olympiad Geometry but bored in Introduction to Geometry.

I think more than a few people feel the same way.

Just curious.[/quote]

Unfortunately, it will be a little while before we have such a course - I'll be developing it concurrently with the Intermediate Geometry textbook I'll be writing, which I hope to begin this winter.

Just a suggestion as far as scheduling is concerned.  The Introduction to Geometry Course should be offered this spring/summer, then Intermediate Geometry in the summer/fall (i.e. Now), and then Olympiad Geometry in the winter as planned.

I just wish I could use it to prepare for this year's AIME rather than next lol.
Self studying just isn't as fun


geometry
Columbia
probability
HCSSiM
Ross Mathematics Program
number theory

Does anyone know of any math (math- not science) programs in the NYC metro area? I am very interested in Math, am a junior, and am a girl. I applied for the columbia summer program "A Survey for Modern Mathematics" but that is only a month (july), and I would like to do more- during the schoolyear and August, perhaps.

Any ideas?
thanks for any help  :)

I just googled it, and it seems like a math competition in NYC that is mainly run through high schools...

I think that I'm looking more for a program than a math competition, mostly because this is my situation:
I had always been somewhat proficient in math compared to others in my grade- I was a year ahead. Then, last year I became really interested in math, so I skipped to extended honors, which I am taking this year, and which is my best subject. I'll be taking BC calc next year.
So, I am way below par compared to those who participate in math competitions. I'm not sure that would be the best option.

I don't live in NYC (and don't go to a NYC school either,) but I just figured that the programs I was looking for would most likely exist there.
Any other suggestions?

The City College of New York offers a free five-week summer program in math, but I believe it is open only to students who live in New York City.  Here is a link:
[url]http://www.collegenow.cuny.edu/students/summer/ccny.html[/url].

Unfortunately, that program runs through July, which is when I hope to go to the columbia program. I'll keep it in mind if I can't go, though.

Any other suggestions? ;)

Where do you live (and go to school)?  Many programs are run by public institutions (cities, states, or their universities) and thus only accept students who live in the particular region (examples: the CCNY program and the New Jersey Governor's School for Science).  Also, many programs are afterschool, which you might not be able to reach if you lived to far away.  (I also gave you an answer on the NYC forum, although it doesn't contain much that people haven't said here already.)

Ask your guidance counselor about college classes after school. I will be taking a Probability and Statistics course at Queens College starting in about 3 weeks. Last term, I did Number Theory at Columbia. CUNY also offers College Now, college courses that can be taken for credit. Most of the College Now classes aren't too difficult.
Most summer programs are only a month long. :(  I did CCNY three years ago and had a great time. I went to HCSSiM two years ago, which was 6 weeks long. I wanted to go to the Ross summer program at U.Chicago last year, but it started too soon, and I needed to take Regents and didn't want to take the August ones.
Do some volunteer work or travel or independent study in August.


Find the shape and position of pentomino, that according to the same rule, should fit into the empty square. Please, send the solution and rule (logic behind the solution).

[img]http://img86.imageshack.us/img86/7025/pentominopuzzle4rb.jpg[/img]

I know what the piece is
I just don't know how it's supposed to go in.

[quote="sapphyre571"]
It's a w-shaped piece in C3, C4, D4, D5, and E5
[/quote]
How did you figure it out? :?

my mom is a pre-k teacher so she has all these cool brain-developing things for kids
I play with them when I'm bored
the w is missing
i just looked for a pattern as to is placement
[hide="placement solution"]
each vertical row is set up that if the three 5-by-5 squares (with pentominos included) is overlapped,
then none of the pentominos overlap.
Using this as a basis,
the only possible placement for a w that satisfies this condition
is the place I specified earlier.[/hide]

[quote="JBL"]Why the W and not the cross?  No one has argued for the uniqueness of this solution which seems like a fundamental part of solving this problem.[/quote]

I have the official solution, and it is unique. In other words, there is a simple logic that eliminates all other possibilities and leaves only sapphyre571's solution which is correct. As you say, although sapphyre571's logic seems pretty, unfortunately it leaves multiple solutions. Maybe someone can find rule that will lead to only one solution?  :)

I guess I kinda cheated - the bottom-right squre looks more darkened so I knew that it would have to cover that square. The cross cannot do that. Working on complete solution...


Where can I find solutions to the IMO Team Selection Tests?


geometry
3D geometry
tetrahedron
perimeter
inequalities
function
geometry unsolved

Given a tetrahedron $ABCD$, $E$, $F$, $G$, are on the respectively on the segments $AB$, $AC$ and $AD$. Prove that:

i) area $EFG \leq$ max{area $ABC$,area $ABD$,area $ACD$,area $BCD$}.
ii) The same as above replacing "area" for "perimeter".

Fix $E,F$, and move $G$ on $AD$. If we regard the area/perimeter of $EFG$ as a function of $x=AG$, then it's easy to see that this function is convex, so it reaches its maximum in one of the endpoints of the segment $AD$. We can thus move $G$ to a vertex of the tetrahedron and increase the area/perimeter. We repeat the procedure until all three vertices $E,F,G$ are placed in some vertices of the tetrahedron, and the conclusion follows.


geometry
circumcircle
incenter
trigonometry
geometric transformation
reflection
parallelogram

The exinscribed circle of a triangle $ABC$ corresponding to its vertex $A$ touches the sidelines $AB$ and $AC$ in the points $M$ and $P$, respectively, and touches its side $BC$ in the point $N$. Show that if the midpoint of the segment $MP$ lies on the circumcircle of triangle $ABC$, then the points $O$, $N$, $I$ are collinear, where $I$ is the incenter and $O$ is the circumcenter of triangle $ABC$.

Maybe, but while trying to solve it I also tried not to spoil it with trilinears or trigonometry :).

Cuzzez! I have been thinking upon this problem for two hours, and the only solution I found is a kind of trig bash. However, since nobody has posted any synthetic solution, here is mine:

We know that if $I_{a}$, $I_{b}$, $I_{c}$ are the excenters of triangle ABC, then

- the lines $AI_{a}$, $BI_{b}$, $CI_{c}$ are the three altitudes of the triangle $I_{a}I_{b}I_{c}$;
- the triangle ABC is the orthic triangle of the triangle $I_{a}I_{b}I_{c}$;
- the incenter I of the triangle ABC is the orthocenter of the triangle $I_{a}I_{b}I_{c}$;
- the circumcircle of the triangle ABC is the nine-point circle of triangle $I_{a}I_{b}I_{c}$.

We also know that

- the circumcenter V of the triangle $I_{a}I_{b}I_{c}$ lies on the line $I_{a}N$ (and the two similar lines through $I_{b}$ and $I_{c}$);
- this circumcenter V is the reflection of the incenter I of triangle ABC in the circumcenter O of triangle ABC;
- the circumradius of triangle $I_{a}I_{b}I_{c}$ equals $I_{a}V = 2R$, where R is the circumradius of triangle ABC.

(Note that the circumcenter V of triangle $I_{a}I_{b}I_{c}$ is called the [i]Bevan point[/i] of triangle ABC.)

Since the points M and P are the points of tangency of the A-excircle of triangle ABC with the sidelines AB and AC, these points M and P must be symmetric to each other with respect to the angle bisector of the angle CAB. Hence, the midpoint of the segment MP must lie on the angle bisector of the angle CAB. But this angle bisector is the line $AI_{a}$, and this line is the $I_{a}$-altitude of triangle $I_{a}I_{b}I_{c}$. So we see that the midpoint of the segment MP lies on the $I_{a}$-altitude of triangle $I_{a}I_{b}I_{c}$. On the other hand, by our hypothesis, the midpoint of the segment MP lies on the circumcircle of triangle ABC, i. e. on the nine-point circle of triangle $I_{a}I_{b}I_{c}$. Hence, the midpoint of the segment MP is the point of intersection of the $I_{a}$-altitude of triangle $I_{a}I_{b}I_{c}$ with the nine-point circle of this triangle. But, by the definition of a nine-point circle, the $I_{a}$-altitude of triangle $I_{a}I_{b}I_{c}$ intersects the nine-point circle of this triangle at two points: at the foot A of the $I_{a}$-altitude, and at the midpoint of the segment joining the vertex $I_{a}$ of the triangle $I_{a}I_{b}I_{c}$ with the orthocenter I of this triangle. Since the midpoint of the segment MP is distinct from A, we can therefore conclude that the midpoint of the segment MP is the midpoint of the segment joining the vertex $I_{a}$ of the triangle $I_{a}I_{b}I_{c}$ with the orthocenter I of this triangle. In other words, the midpoint of the segment MP is the midpoint of the segment $I_{a}I$.

Therefore, the quadrilateral $M I_{a}P I$ is a parallelogram, and $MI\parallel I_{a}P$. But $I_{a}P \perp CA$; hence, $MI \perp CA$, and thus the triangle AYM is right-angled, where Y is the point where the incircle of triangle ABC touches the side CA. Hence, $AY=AM\cdot \cos \measuredangle YAM=AM\cdot \cos A$. Now, since the points M and P are symmetric to each other with respect to the angle bisector of the angle CAB, we have AM = AP, so that $AY=AP\cdot \cos A$, and

$\cos A=\frac{AY}{AP}$.

But Thales shows

$\frac{AY}{AP}=\frac{r}{r_{a}}$,

where r, $r_{a}$, $r_{b}$, $r_{c}$ are the inradius and the exradii of triangle ABC. Thus,

$\cos A=\frac{r}{r_{a}}$.

Now, what do we want to prove? We want to show that the points O, N, I are collinear. This will obviously hold if we show that N = V (since we know that the points O, V, I are collinear, for V is the reflection of I in O). The point V lies on the line $I_{a}N$; hence, in order to obtain N = V, it is enough to show that the segments $I_{a}N$ and $I_{a}V$ are equal. But $I_{a}N = r_{a}$ and $I_{a}V = 2R$. Hence, we must show that $r_{a}= 2R$.

Here the battle begins. Denoting x = cos A, y = cos B, z = cos C, we know the formulas

$r=R\left( x+y+z-1\right)$;
$r_{a}=R\left(-x+y+z+1\right)$.

Also, we have given

$x=\cos A=\frac{r}{r_{a}}$,

so that $r=xr_{a}$. Thus, we can substitute the above formulas to get

R (x + y + z - 1) = x R (- x + y + z + 1).

After some algebra, this simplifies to

(x - 1) (y + z - x - 1) = 0.

Now, since $x-1\neq 0$ (the cosine of a triangle's angle is never 1), this becomes y + z - x - 1 = 0, so that y + z = x + 1, and

$r_{a}=R\left(-x+y+z+1\right) =R\left(-x+x+1+1\right) =2R$,

qed..

The converse of the problem also holds and can be proved just by inverting the observations above.

[b]EDIT:[/b] See also http://www.mathlinks.ro/Forum/viewtopic.php?t=105101 for a discussion of this problem.

  Darij

[hide="Only for the fans and not for the professionals."][color=darkred]The $A$- exincircle $C_{a}=C(I_{a},r_{a})$ of a triangle $ABC$ touches the sidelines $AB$, $BC$ and $AC$ in the points $R$, $N$ and $P$ respectively.
The its incircle $C(I,r)$ touches the lines $BC$, $CA$, $AB$ in the points $D$, $E$, $F$ respectively. Denote : the circumcircle $w=C(O,R)$
of the triangle $ABC$ ; the middlepoint $M$ of the side $[BC]$ ; the orthocenter $H$ ; the middlepoint $L$ of the segment $PR$.
Prove that $H\in IM$ $\Longleftrightarrow$ $N\in OI$ $\Longleftrightarrow$ $\cos A=\frac{r}{2R}$ $\Longleftrightarrow$ $r_{a}=2R$ $\Longleftrightarrow$ $L\in w$ $\Longleftrightarrow$ $H\in EF\ .$[/color]

[color=darkblue][b]Proof.[/b] Suppose w.l.o.g. that $b>c$. Then,

$1.\blacktriangleright\ DB=NC=p-b\ ;\ MD=MN=\frac{b-c}{2}\ ;\ ID=r\ ;\ OM=R\cos A$.
Therefore, $\boxed{\ N\in OI\ }$ $\Longleftrightarrow$ $ID=2\cdot OM$ $\Longleftrightarrow$ $\boxed{\ \cos A=\frac{r}{2R}\ }\ .$

$2.\blacktriangleright$ If $l_{a}$ is the length of the bisector of the angle $\widehat{BAC}$, then $\boxed{\ L\in w\ }\Longleftrightarrow AL\cdot l_{a}=bc\Longleftrightarrow$
$p\cos\frac{A}{2}\cdot \frac{2bc\cos\frac{A}{2}}{b+c}=bc$ $\Longleftrightarrow$ $p(1+\cos A)=b+c$ $\Longleftrightarrow$ $\cos A=\frac{p-a}{p}$ $\Longleftrightarrow$ $\boxed{\ \cos A=\frac{r}{r_{a}}\ }\ .$

$3.\blacktriangleright$ Denote the intersections $S\in AH\cap IM$, $V\in AI\cap BC$ and the projection $U$ of the point $A$ on the line $BC$.
Apply the Menelaus' theorem to the transversal $\overline{MIS}$ for the triangle $AUV$ : $\frac{MV}{MU}\cdot \frac{SU}{SA}\cdot \frac{IA}{IV}=1$ $\Longleftrightarrow$
$\frac{\frac{a(b-c)}{2(b+c)}}{\frac{b^{2}-c^{2}}{2a}}\cdot \frac{SU}{SA}\cdot\frac{b+c}{a}=1$ $\Longleftrightarrow$ $\frac{SA}{a}=\frac{SU}{b+c}=\frac{h_{a}}{2p}$ $\Longleftrightarrow$ $SA=\frac{ah_{a}}{2p}=\frac{2pr}{2p}$ $\Longleftrightarrow$ $SA=r\ .$
Therefore, $\boxed{\ H\in IM\ }$ $\Longleftrightarrow$ $AH=AS$ $\Longleftrightarrow$ $2R\cos A=r$ $\Longleftrightarrow$ $\boxed{\ \cos A=\frac{r}{2R}\ }\ .$

$4.\blacktriangleright\ \boxed{\ r_{a}=2R\ }$ $\Longleftrightarrow$ $(p-b)(p-c)=2Rr$ $\Longleftrightarrow$ $p_{w}(D)=p_{w}(I)$ $\Longleftrightarrow$ $OI=OD$ $\Longleftrightarrow$
$2\cdot OM=ID$ $\Longleftrightarrow$ $\boxed{\ \cos A=\frac{r}{2R}\ }$, where $p_{w}(X)$ is the power of the point $X$ w.r.t. the circle $w\ .$

$5.\blacktriangleright\ H\in EF$ $\Longleftrightarrow$ $\frac{2R\cos A}{\cos\frac{A}{2}}=\frac{p-a}{\sin \left (B+\frac{A}{2}\right)}$ $\Longleftrightarrow$ $2R\cos A\sin\left(B+\frac{A}{2}\right)=(p-a)\cos\frac{A}{2}$ $\Longleftrightarrow$ $2R\cos A\sin\left(B+\frac{A}{2}\right)\cdot 2\cos \frac{A}{2}=(p-a)\cos\frac{A}{2}\cdot 2\cos\frac{A}{2}$ $\Longleftrightarrow$ $2R\cos A(\sin C+\sin B)=(p-a)(1+\cos A)$
$\Longleftrightarrow$ $(b+c)\cos A=(p-a)(1+\cos A)$ $\Longleftrightarrow$ $\cos A=\frac{p-a}{p}$ $\Longleftrightarrow$ $\cos A=\frac{r}{r_{a}}\ .$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=5397[/color][/hide]

Let $X$ be midpoint of $MP$ ,WE know $I_{a}X.I_{a}A=OI_{a}^{2}-R^{2}=2Rr_{a}=r_{a}^{2}\rightarrow r_{a}=2R$
let $D,D'$ be reflection of  $I,O$ on the $BC$ we have $OX\perp BC$ and  $D'D=D'N$ so cause $OX=R\ ,\ I_{a}N=2R$ and $X$ is midpoint of $II_{a}$ hence $O,I,N$ are collinear.

Inversion in the excircle $(I_{a})$ carries the vertices A, B, C into the midpoints X, Y, Z of the excircle chords MP, MN, PN and the circumcircle (O) of $\triangle ABC$ into the 9-point circle of $\triangle MNP.$ If $X \in (O),$ the circumcircle is carried into itself, hence it is the 9-point circle of $\triangle MNP$ (and $r_{a}= 2R)$. But (O) is also the 9-point circle of the triangle $\triangle I_{b}II_{c},$ always centrally similar to the triangle $\triangle MNP$ (with parallel sides). Thus in case $X \in (O),$ these 2 triangles are congruent.  The external bisector $I_{b}I_{c}\perp AI_{a}\equiv AX$ of $\angle A$ meets (O) again at the midpoint Q of $I_{b}I_{c}$ and XQ is a diameter of (O), hence the midpoint O of XQ is the similarity center of $\triangle MNP \cong \triangle I_{b}II_{c},$ so that I, N, O are collinear and O is the midpoint of IN.

[quote="Virgil Nicula"]

$2.\blacktriangleright$ If $l_{a}$ is the length of the bisector of the angle $\widehat{BAC}$, then $\boxed{\ L\in w\ }\Longleftrightarrow AL\cdot l_{a}=bc\Longleftrightarrow$
$p\cos\frac{A}{2}\cdot \frac{2bc\cos\frac{A}{2}}{b+c}=bc$ [/quote]

Why is this?

$2.\blacktriangleright$ If $l_{a}$ is the length of the bisector of the angle $\widehat{BAC}$, then is well-known that $l_{a}=\frac{2bc}{b+c}\cdot\cos \frac{A}{2}$. Denote $S\in AI\cap BC$. Then $\boxed{\ L\in w\ }$  $\Longleftrightarrow$ $\left\{\begin{array}{c}\triangle ABL\sim \triangle ASC\\\ LA=RA\cdot \cos \frac{A}{2}\\\ RA=p\end{array}\right\|$ $\Longleftrightarrow$ $\left\{\begin{array}{c}AL\cdot l_{a}=bc\\\ LA=p\cdot \cos \frac{A}{2}\end{array}\right\|$ $\Longleftrightarrow$
$p\cos\frac{A}{2}\cdot \frac{2bc\cos\frac{A}{2}}{b+c}=bc$.


calculus
integration
inequalities
function
derivative
probability
expected value

Let $f : \left[ 0, 1 \right] \to \mathbb R$ be a continuous function and $g : \left[ 0, 1 \right] \to \left( 0, \infty \right)$.
Prove that if $f$ is increasing, then
\[\int_{0}^{t}f(x) g(x) \, dx \cdot \int_{0}^{1}g(x) \, dx \leq \int_{0}^{t}g(x) \, dx \cdot \int_{0}^{1}f(x) g(x) \, dx .\]

Let $f: [0,1]\to \mathbb{R}$ and $g: [0,1]\to \mathbb{R}^{+}$ be continuous functions. Suppose $f$ is an increasing function.

Now define $h$ on $[0,1]$ as

$h(t) = \int_{0}^{t}g(x) \, dx \cdot \int_{0}^{1}f(x)g(x) \, dx-\int_{0}^{1}g(x) \, dx \cdot \int_{0}^{t}f(x)g(x) \, dx$.

Then $h$ is differentiable and the derivative is given by

$h'(t) = g(t) \int_{0}^{1}f(x)g(x) \, dx-f(t)g(t) \int_{0}^{1}g(x) \, dx = g(t) \int_{0}^{1}(f(x)-f(t))g(x) \, dx$.

Since $\frac{h'(t)}{g(t)}= \int_{0}^{1}(f(x)-f(t))g(x) \, dx$ is a (continuous) decreasing function with $\frac{h'(0)}{g(0)}> 0$ and $\frac{h'(0)}{g(0)}< 0$, there is only one point $c \in (0, 1)$ satisfying $h'(c) = 0$. Moreover,

(1) $h$ is increasing on $[0, c]$, so $0 = h(0) \leq h(t)$ for $t \in [0, c]$
(2) $h$ is decreasing on $[c, 1]$, so $h(t) \geq h(1) = 0$ for $t \in [c, 1]$

In any case, $h(t) \geq 0$.

I was kind of blind with this problem :) I started in the contest with the idea of sos440 but I could not finish...My solution is a little different. I took H the antiderivate of fg, G the antiderivate of g and the inequality was equivalent with:
$\frac{H(t)-H(0)}{G(t)-G(0)}\leq \frac{H(1)-H(0)}{G(1)-G(0)}$ ( I assumed t different of 0 and 1)
Even not I was not able to find a solution using the dervative so applying the trick from the convex function I got that the inequality is equivalent with:
$\frac{H(t)-H(0)}{G(t)-G(0)}\leq \frac{H(1)-H(t)}{G(1)-G(t)}$
Now just apply cauchy for the 2 expresions...the first is f(a) with a in (0,t) and the second f(b) with b in (t,1). The inequality is thus proved :)

[quote="perfect_radio"]Let $f : \left[ 0, 1 \right] \to \mathbb R$ be a continuous function and $g : \left[ 0, 1 \right] \to \left( 0, \infty \right)$.
Prove that if $f$ is increasing, then
\[\int_{0}^{t}f(x) g(x) \, dx \cdot \int_{0}^{1}g(x) \, dx \leq \int_{0}^{t}g(x) \, dx \cdot \int_{0}^{1}f(x) g(x) \, dx . \]
[/quote]

I assume g is integrable, though you didn't say so.. :p

Use the additivity of interval theorem to reduce the given inequality to:
\[\int_{0}^{t}f(x) g(x) \, dx \cdot \int_{t}^{1}g(x) \, dx \leq \int_{0}^{t}g(x) \, dx \cdot \int_{t}^{1}f(x) g(x) \, dx . \]
from here convert
\[\int_{0}^{t}f(x) g(x) \, dx \]
to a Riemann Stieltjes integral with f as the integrand. Note that the integrator is of bounded variation since g is everywhere positive. Do the same for
\[\int_{t}^{1}f(x) g(x) \, dx \]
to have f as the integrator. (to be definite, the integrator in both cases is the integral of g over (0, x))

Then the mean value theorem  for Riemann Stieltjes integrals tells you that you can write:
\[f(c) \int_{0}^{t}g(x) \, dx \cdot \int_{t}^{1}g(x) \, dx \leq \int_{0}^{t}g(x) \, dx \cdot f(d) \int_{t}^{1}g(x) \, dx . \]
(where I have converted the Rieman Stieltjes integral back to a Riemann integral for clarity).

But (if you know the proof of the mean value theorem for Riemann Stieltjes integrals) you will notice that 0 <= c <= t and t  <= d <= 1. Since f is increasing, f(c) <= f(d) and the inequality is proven.

Of course if you really wanted to prove it you had better work backwards from what I've done. Heh. Hurrah for the first problem on this forum that I even remotely understsand! This is a great problem for anyone learning the Riemann integral. It certainly took me a while to figure it out. The main trick was to split up the intervals of integration on both sides. That seems to be the only way to justify that c < d.

P.S. I can't believe I just solved my first ever Olympiad question. There must be some error here. In the meanwhile, I shall sleep well. :)

$g$ is continuous (but can be chosen to be only Riemann integrable, I think) and $t$ is in $[0,1]$.
For some reason, I forgot to write these conditions :blush:

Anyway, your solution is fine. Actually, on a close inspection, you'll discover that it's equivalent to ciprian's solution.

BTW, here's my version:
Let $h(t) = \left( \int_{0}^{t}f(x) g(x) \, dx \right) \left\slash \left( \int_{0}^{t}g(x) \, dx \right) \right.$. The conclusion is equivalent to proving that $h(t) \leq h(1)$.
We can prove more, $h$ is actually increasing. Now compute $h^\prime(t)$:
\[\left( f(t) g(t) \int_{0}^{t}g(x) \, dx-g(t) \int_{0}^{t}f(x) g(x) \, dx \right) \left\slash \left( \int_{0}^{t}g(x) \, dx \right)^{2}\right. . \]
$g$ is positive (and integrable), so the Mean Value Theorem works here.
Now just use the fact that $f$ is increasing to get that $h^\prime \geq 0$.

[quote="perfect_radio"]BTW, here's my version:
Let $h(t) = \left( \int_{0}^{t}f(x) g(x) \, dx \right) \left\slash \left( \int_{0}^{t}g(x) \, dx \right) \right.$[/quote]

This seems to be by far the most natural way to approach the problem to me.

Here's another approach.  Let $h : [0, 1] \to \mathbb{R}$ be the function defined by
\[h(x) = \begin{cases}1 & \textrm{if }x \le t; \\ 0 & \textrm{otherwise}. \end{cases}\]
Note that $h$ is a decreasing, piecewise-continuous function.  Then the inequality can be rewritten as
\[\int_{0}^{1}g(x) f(x)h(x) \, dx \cdot \int_{0}^{1}g(x) \, dx \le \int_{0}^{1}g(x) f(x) \, dx \cdot \int_{0}^{1}g(x) h(x) \, dx. \]
You might already recognize this inequality as Chebyshev's inequality.  If not, note that because $f$ is increasing and $h$ is decreasing, we have
\[g(x) g(y) (f(x)-f(y)) (h(x)-h(y)) \le 0 \]
for all $x$ and $y$ in $[0, 1]$.  Now integrate both sides over all $x$ and $y$ in $[0, 1]$, and you will get the desired inequality.

Or you might like to view the inequality via probability.  By scaling, we may assume that
\[\int_{0}^{1}g(x) \, dx = 1. \]
In other words, $g$ is a probability density function.  Let $X$ be a random variable with density function $g$.  Then our inequality becomes
\[E(f(X)h(X)) \le E(f(X)) E(h(X)) \, . \]
Again, this is Chebyshev's inequality.  For a proof from scratch, start with the inequality
\[(f(X)-f(Y))(h(X)-h(Y)) \le 0, \]
where $Y$ is an independent copy of $X$.  Now take the expected value of both sides.


function
summer program
Mathcamp
set theory

Prove, using only mathematics (no altering of the equation), that 1=1. Basically,

Prove: 1=1.

Read my signature if you find this hard.:rotfl:

many axioms are based on logic and therefore proven in teh human mind.  Think about logical things in mathematics such as set theory (simple ones).

[quote="boxedexe"]many axioms are based on logic and therefore proven in teh human mind.[/quote]What are you talking about?

Couldn't there be an axiom like $(\forall{a})(a=a)$?

How do you know 2=2?
And the successor of 2 isn't 1...

equality is transitive, so as long as $1$ is well defined, we have $1=1$.

[quote="boxedexe"][quote="eryaman"][quote="Singular"]http://mathworld.wolfram.com/PeanosAxioms.html[/quote]Then you could say... since the successor of 1 is 2, and the successor of 1 is 2, then $1=1$[/quote]

You meant to say, since the successor of 1 is 2 and the successor of 2 is 1, $1=1$, right?

Masoud Zargar[/quote]

Nah, he meant exactly what he said.  Remember, if 2 is the successor of 1, then 1 is not the successor, but the predecessor of 2.
And in any case, this is a postulate upon which everything is built, so you'd need some other form of math to prove it.  I remember an interesting conversation with my friends about a universe in which the reflexive property does not hold...But it's like Godel said: in any system of mathematics, there will be something which cannot be proven by something else in the system.

This is a beautiful solution, which is very, yet logical and correct.  There are other proofs using set theory.

Okay well set theoretically (if I still remember from 8th grade):

1 is defined as set of the null set.

Now 1=1 becuase we can define an isomorphism that maps every element to another element in 1, namely the identity function!

[quote="AntonioMainenti"][quote="zeus_three"]Proof by Contradiction

Assume this is not true, that $1\ne1$ or 1 is "NOT 1"

Then every time i have 1 cup of coffee
im having NOT 1 cup of coffee which means
im having NOT NOT 1 cup of coffee...
which is a beatiful paradox, because NOT NOT, is a double negative and so 1=1, and thus proven. (but i guess the double negative is itself an axiom? - or at least derived from one?)

I wonder if this logic works?[/quote]This should be submitted to a site about comically fallacious proofs.[/quote]

 :rotfl:  :rotfl: , Zeus_three, I think my teacher would love to have you in his philosophy class  :lol: .

By the way, about the proof $1=1$, I don't remember actually :lol: , but I once heard John Conway talk about that [b]beautiful proof[/b] in mathcamp this year. I wish I had remembered  :( .

I thought i could chip in my dumb thoughts...
0=0
0+1=0+1
1=1 :|

I know my sign says think twice before you speak :D


analytic geometry
graphing lines
slope
calculus
function
derivative

This is from some AHSME....

If $x^2+y^2=14x+6y+6$,what is the largest possible value of $3x+4y$?

[hide="Answer"]$x^2+y^2=14x+6y+6\Rightarrow x^2-14x+y^2-6y=6 \Rightarrow (x-7)^2+(y-3)^2=8^2$
Therefore, we have a circle with center at $(7,3)$ and $r=8$. Okay... I know there's another way to solve it, but it's not coming to mind right now, so I'm going to take the derivative. $f'(x)=2x-14+2y\frac{dy}{dx}-6\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{14-2x}{2y-6}$. Therefore, our maximum is at $x=7$ and our answer is $3(7)+4(11)=\boxed{65}$.[/hide]

Hope I did that right...

[hide]i dont know if this is the exact approach, but consider the line $3x+4y=n$, we want to maximize $n$ clearly, the line will be tangent to the circle (and be above),
this line has a slope of $\frac{-3}{4}$, and if this line is tangent to the circle, then the line through the center to the point of tangency will be perpendicular to that line, and the slope will $\frac{4}{3}$, and since it passes through the center, the equation will be $y-3=\frac{4}{3}(x-7)$
so if we take the intersection of that line, and the circle, we will have 2 point of intersection, and if we take our maximized point, the answer follows
(note, you do not actually have to go through the algebra to solve those 2 equations, you can just draw the x and y components of that line) but anyway, we get our point to be $\left(\frac{59}{5},\frac{47}{5}\right)$ and
$3x+4y=\boxed{73}$[/hide]

AHSME problems never require calculus.  Your solution is wrong anyway.  You totally disregarded that we were trying to maximize 3x+4y...it's nowhere in your solution.  Altheman's solution is the intended solution (but please use the Hide tags next time...I put them in for you).

sorry about that, i will remember to hide my solutions now

[quote="JesusFreak197"]I know that AHSME problems never require Calculus, but there's nothing wrong about using it when applicable.

As for my answer... I know we're trying to maximize $3x+4y$, and I did use that at the end with the values of $x$ and $y$ that I got, but my mistake was that the maximum of the circle equation is not the maximum of $3x+4y$...[/quote]

The main problem was that you used calculus blindly.  It should have been obvious that the two values you got would be the 'top' and 'bottom' of the circle, which you could have found by inspection anyways.

[quote="blahblahblah"]Using Cauchy Schwarz always makes me happy

[hide]$8^2\cdot 5^2=((x-7)^2+(y-3)^2)(3^2+4^2)\geq (3(x-7)+4(y-3))^2$

This is the same as $40\geq 3x-21+4y-12\implies 73\geq 3x+4y$[/hide][/quote]

That's pretty brilliant!

[quote="JesusFreak197"]I know that AHSME problems never require Calculus, but there's nothing wrong about using it when applicable.

As for my answer... I know we're trying to maximize $3x+4y$, and I did use that at the end with the values of $x$ and $y$ that I got, but my mistake was that the maximum of the circle equation is not the maximum of $3x+4y$...[/quote]
Yes, you want to maximize $3x+4y$ not the other one eg $(x-7)^2+(y-3)^2-8^2$ (which BTW you know is  always zero!).

Ok mods close your eyes and don't look at the hidden part. :) (Did not want to dabble in calculus, just commenting on what was wrong in the method)

[hide]So what you have to do is  find y in terms of x  from the second equation, substitute it so you $3x+4y$ becomes a function of $x$ and now do your thing with calculus to maximize it.

This does become a  little messy, an easy method, if you have done Lagrange Multiplier Method would be to write:
$z=3x+4y - \lambda ((x-7)^2+(y-3)^2-8^2)$  Now take partial derivative wrt $x,y, \lambda$ equate each to zero and  find $x,y,\lambda$ substitute and you are done. [/hide]

Above part is calculus, I would delete it in a day or so after  JesusFreak197 had a look at it.. hope that is useful.

[quote="blahblahblah"]Using Cauchy Schwarz always makes me happy

[hide]$8^2\cdot 5^2=((x-7)^2+(y-3)^2)(3^2+4^2)\geq (3(x-7)+4(y-3))^2$

This is the same as $40\geq 3x-21+4y-12\implies 73\geq 3x+4y$[/hide][/quote]

I dind't understand...Why start with $8^2*5^2$?...and what about the equation of the question,where is that coming in use in your solution blahblahblah?

The 'equation of the question' is just $(x-7)^2+(y-3)^2=8^2$.  I know that if I use Cauchy-Schwarz appropriately, I can extract a $3x+4y$ somehow.  That is where the $3^2+4^2=5^2$ comes in.

[quote="JesusFreak197"]Hmm... I've learned Cauchy-Schwarz, but apparently I don't really have much practice with it, so I'm not sure what you did. The basic formula is $\displaystyle\left(\sum_{k=1}^na_kb_k\right)^2\le \left(\sum_{k=1}^na_k^2\right)\left(\sum_{k=1}^nb_k^2\right)$; what numbers did you put in where?

Gyan, thanks for the clarification on what I did wrong. I had put $y$ in terms of $x$ and $m$ at one point, but I saw that I had one variable and one unknown, which didn't quite seem congenial to a derivative, so I didn't use that. :oops:[/quote]

He used $a_1 = x-7, a_2 = y-3$ and $b_1 = 3, b_2 = 4$ to get $((x-7)^2+(y-3)^2)(3^2+4^2) \ge (3(x-7)+4(y-3))^2$.