ratio
geometry
circumcircle
geometric transformation
rotation
parallelogram
homothety

Point $ D$ lies inside triangle $ ABC$ such that $ \angle DAC\equal{}\angle DCA\equal{}30^{\circ}$ and $ \angle DBA\equal{}60^{\circ}$. Point $ E$ is the midpoint of segment $ BC$. Point $ F$ lies on segment $ AC$ with $ AF\equal{}2FC$. Prove that $ DE$ is perpendicular to $ EF$

$ H$ is a point on $ [AC]$ such that $ BD$ is the internal bisector of angle $ BAD$ and $ G$ is the midpoint of $ [AC]$. $ ABDH$ is cyclic so $ m(ADH)\equal{}30$ and $ m(DHC)\equal{}60$. $ 2.AH\equal{}2.DH\equal{}CH$ hence $ CF\equal{}FH\equal{}DF$. $ EF$ is paralel to $ BH$ and $ EG$ is paralel to $ AB$. so $ m(FEG)\equal{}m(FDG)\equal{}30$ so $ EFDG$ is cyclic. result follows.

Draw the equilateral triangle DF'F", with F' and F" on (AC), see F'=F and DF=FC.
Denote O the circumcenter of ABD; a rotation of 120 deg around D matches C to A and F to O; we get DCFO parallellogram, so the middle of DF is the middle of CO too (denote it by P). Then PE is half BO which is the radius of the circle (ABD) and equal to DO and DF; so the median EP is half of DF and we are done.

Best regards,
sunken rock

[quote="easternlatincup"]Point $ D$ lies inside triangle $ ABC$ such that $ \angle DAC \equal{} \angle DCA \equal{} 30^{\circ}$ and $ \angle DBA \equal{} 60^{\circ}$. Point $ E$ is the midpoint of segment $ BC$. Point $ F$ lies on segment $ AC$ with $ AF \equal{} 2FC$. Prove that $ DE$ is perpendicular to $ EF$[/quote]
[hide="Solution"]
Let $ F'$ be the point on $ AC$ such that $ F'D\perp AD$, extend $ CD$ to meet the circumcircle of $ \triangle BDA$ at $ B'$, and let the midpoint of $ DC$ be $ P$.  Now, since $ \angle DAF' \equal{} 30$ and $ \angle F'DA \equal{} 90$, we have that $ AF' \equal{} \frac {2AD}{\sqrt {3}}$.  Also, since $ \angle DCA \equal{} \angle DAC \equal{} 30$, we have that $ AC \equal{} DA\sqrt {3}$, so $ 3CF \equal{} CF \plus{} AF \equal{} AC \equal{} DA\sqrt {3}\implies AF \equal{} 2CF \equal{} \frac {2DA}{\sqrt {3}} \equal{} AF'$, so $ F \equal{} F'$.  Thus, $ \angle FDA \equal{} 90\implies \angle FDC \equal{} \angle ADC \minus{} \angle FDA \equal{} 120 \minus{} 90 \equal{} 30 \equal{} \angle DCA$, meaning that $ \triangle DCA$ is isoceles, so $ FP\perp DC$.  Furthermore, $ \angle PFD \equal{} 90 \minus{} \angle CDF \equal{} 60$.  Now, $ \angle B'DA \equal{} \angle DCA \plus{} \angle DAC \equal{} 30 \plus{} 30 \equal{} 60 \equal{} \angle DBA \equal{} \angle DB'A$, so $ \triangle DB'A$ is equilateral, and thus, $ \angle CAB' \equal{} \angle DAC \plus{} \angle DAB' \equal{} 30 \plus{} 60 \equal{} 90$, which means that $ \triangle B'AC$ is right.  Now, $ DA \equal{} DC \equal{} DB'$, so $ D$ is the midpoint of $ B'C$, $ P$ is the midpoint of $ DC$, and $ E$ is the midpoint of $ BC$.  Hence, $ DE$ is a midline of $ \triangle B'CB$ and $ PE$ is a midline of $ \triangle BDC$.  It follows that $ DE\parallel BB'$ and $ PE\parallel BD$, so $ \angle DEP \equal{} \angle B'BD \equal{} \angle B'AD \equal{} 60 \equal{} \angle DFP$, so $ DEFP$ is cyclic, meaning that $ \angle DEF \equal{} 180 \minus{} \angle DPF \equal{} 180 \minus{} 90 \equal{} 90\implies DE\perp EF$. [/hide]

Let $ \omega$ be the circumcircle of $ \triangle ABD$. Let $ \omega$ intersect $ AC$ at $ F'$. $ \angle DF'A\equal{}120$, $ \angle F'DA\equal{}30$, $ \angle F'DC\equal{}90$, it follows that $ AF'\equal{}F'D\equal{}\frac{1}{2}\cdot F'C$, so $ 2AF'\equal{}F'C$. So $ \frac{1}{2}\cdot F'C\equal{}FC$. (1)

Let $ DC$ intersect $ \omega$ again at $ D'$. Indeed, $ \angle ADD'\equal{}30\plus{}30\equal{}60$, $ \angle DD'A\equal{}60$, so $ DD'A$ is equilateral, and $ DD'\equal{}DA\equal{}DC$, so $ \frac{1}{2}\cdot D'C\equal{}DC$. (2)

By the midpoint property, $ \frac{1}{2}\cdot BC\equal{}BE$ (3)

Note that $ \angle D'AC\equal{}60\plus{}30\equal{}90$ so $ D'F'$ is a diameter in $ \omega$. Thus $ \angle D'BF'\equal{}90$. (4)

Consider a homothety, $ \mathcal{H}$, with ratio $ \frac{1}{2}$ center $ C$. Indeed by (1), (2), and (3), $ \mathcal{H}$ maps $ D'BF'$ to $ DEF$ so $ \angle D'BF'\equal{}\angle DEF$ which is equal to $ 90$ by (4). The conclusion follows.


I've make a mistake into the old topic!
That's the problem:

solve the differential equation:
f:R->R,
f(x)=xf(f(x))

f(x)=x is valid, but i can't prove that  :(

This is not a differential equtions, at least from my point of view. So I'll move this to the Algebra section.

Some ideas:
If [tex]f[/tex] is continuous and we have a root [tex]>0[/tex], then [tex]f(x)\equiv 0[/tex] is the only solution.

It can be easily seen that f(0)=0

If [tex]f(f(x)),f(f(y))\neq 0[/tex]  then the following holds:
[tex]f(x)=f(y)\;\Rightarrow\;f(f(x))=f(f(y))[/tex]
[tex]\Rightarrow\; x=\dfrac{f(x)}{f(f(x))}=\dfrac{f(y)}{f(f(y))}=y[/tex]

Thus if we have a root [tex]x> 0[/tex] f has to vanish identically in [tex][0;x][/tex] by the intermediate value theorem.

Now let [tex][0;x_0][/tex] be the largest neighbourhood of 0 in which f vanishes identically. Now take [tex]\xi[/tex] slightly larger than [tex]x_0[/tex] and assume [tex]f(\xi)>0[/tex]. By continuity We can choose [tex]\xi[/tex] so, that [tex]f(\xi)\leq x_0[/tex], thus [tex]f(\xi)=\xi f(f(\xi))=0[/tex]. Contradiction; The case [tex]f(\xi)<0[/tex] can be done similarly ( perhaps it's a little bit more complicated).

Now it's up to you to make this rigorous  :D


function
trigonometry
algebra unsolved
algebra

Let $ f(x)\equal{}\frac{1}{3}(x\plus{}\frac{1}{x})$. Solve the following system of equation:

$ x_2\equal{}f(x_1)$
$ x_3\equal{}f(x_2)$
...
$ x_n\equal{}f(x_{n\minus{}1})$
$ x_1\equal{}f(x_n)$

If $ \vec{x}\equal{}(x_1,...x_n) \in \mathbb{R}^n, n \in \mathbb{N}$ is a solution then also $ \minus{}\vec{x}$ is.
Naturally if $ sgn$ is the sign-function then $ sgn(x_1)\equal{}sgn(x_2)\equal{}...\equal{}sgn(x_n)$, so wlog $ x_i >0, 1\le i \le n$.
Now, if $ x_1> \sin{\frac{\pi}{4}}$ then $ x_1<x_n<x_{n\minus{}1}<...<x_2<x_1$, contradiction.
If $ 0<x_1<\sin{\frac{\pi}{4}}$ then $ x_1>x_n>x_{n\minus{}1}>...>x_2>x_1$, contradiction.
So the unique solutions are $ \vec{x}\equal{}k \vec{I}$, where $ |k|\equal{}\sin{\frac{\pi}{4}}$


trigonometry
inequalities unsolved
inequalities

prove that $ |\sin 3n\minus{}2|\plus{}|\sin 3n\minus{}1|\plus{}|\sin 3n| > \frac {8}{5}$

$ |sin3n\minus{}2|\geq 1;|1\minus{}sin3n|\plus{}|sin3n|\geq 1\Rightarrow LHS\geq 2$

$ \text{The above argument may not be right!!}$

$ sin3n\minus{}1  \text{ should be } sin (3n\minus{}1)$

 $ sin3n\minus{}2  \text{ should be } sin (3n\minus{}2)$

For all real $ x$ we obtain: $ |\sin x| \plus{} |\sin(x \plus{} 1)| \plus{} |\sin(x \plus{} 2)| > \frac {8}{5}.$  :ninja:

[quote="vinskman"]$ \text{For all real x,} |\sin x| \plus{} |\sin(x \plus{} 1)| \plus{} |\sin(x \plus{} 2)| > \frac {8}{5}.$

$ \text {How to prove it ?}$[/quote]
It's very ugly. :(
Let $ f(x)\equal{}|\sin x| \plus{} |\sin(x \plus{} 1)| \plus{} |\sin(x \plus{} 2)|.$
Since, $ f(x\plus{}\pi)\equal{}f(x)$ we obtain three cases.
Case 1. $ 0\leq x\leq\pi\minus{}2.$
Hence, $ f(x)\equal{}\sin x \plus{} \sin(x \plus{} 1) \plus{} \sin(x \plus{} 2)$
and $ \min_{[0,\pi\minus{}2]}f\equal{}\min\{f(0),f(\pi\minus{}2)\}>\frac{8}{5}.$
Case 2. $ \pi\minus{}2\leq x\leq\pi\minus{}1.$
Hence, $ f(x)\equal{}\sin x \plus{} \sin(x \plus{} 1) \minus{} \sin(x \plus{} 2)$ and $ \min_{[\pi\minus{}2,\pi\minus{}1]}f>\frac{8}{5}.$
Case 3. $ x\in[\pi\minus{}1,\pi].$
Hence, $ f(x)\equal{}\sin x \minus{} \sin(x \plus{} 1) \minus{} \sin(x \plus{} 2)$ and $ \min_{[\pi\minus{}1,\pi]}f>\frac{8}{5}.$


[img]http://i227.photobucket.com/albums/dd72/TPFC/chienthan.png[/img]
Denote $ \Delta_2\cap \Delta_3 \equal{} G,\Delta_3\cap AC \equal{} H,\Delta_3\cap AE \equal{} M$, $ N,L$ be midpoint of $ DM,AB$, $ O$ is circumcentre of triangle $ ABC$. We have:
$ M\in (ABD)$ (because $ \angle{MDE} \equal{} \angle{MAB}$), then $ \angle{ING} \equal{} 90^0 \equal{} \angle{IEG}$.
Hence $ \angle{GIE} \equal{} \angle{GNE} \equal{} 2\angle{MDE} \equal{} \angle{BAC} \equal{} \angle{COE}$
It mean two triangle $ IEG,OEC$ are similarly, hence two triangle $ ECG,EOI$ are similarly.
But $ \angle{IEG} \equal{} \angle{OEC} \equal{} 90^0$ then $ IO,CG$ are perpendicular.
It mean $ CG,AB$ are parallel! Q.E.D

[quote="chien than"]Let $ ABC$ be a triangle with $ AB \equal{} AC$.$ E$ is midpoint of $ BC$.$ D$ is a point on $ BC$.$ I$ is center of circumcircle of  triangle $ ABD$.A line $ \Delta_1// AB$  and throughs $ C$ .A line $ \Delta_2 \perp AC$  and throughs $ D$.A line $ \Delta_3 \perp IE$ throughs $ E$.Prove that $ \Delta_1;\Delta_2;\Delta_3$ are concurrent.[/quote]
Let $ H,K \in BC$ so that  $ IH,LK\perp BC$
We have
$ A\equal{}CI^2\minus{}CL^2\plus{}DL^2\minus{}DE^2\plus{}EE^2\minus{}EI^2\equal{}CH^2\minus{}EH^2\plus{}DK^2\minus{}DC^2\minus{}DE^2\equal{}\overline{CE}.\overline{EH}\plus{}2\overline{DC}.\overline{EK}$
And $ \overline{DC}\equal{}\overline{DB}\plus{}\overline{BC}\equal{}2(\overline{HB}\plus{}\overline{BE})\equal{}\minus{}2\overline{EH}$
$ 2\overline{EK}\equal{}\overline{CE}$
$ \Rightarrow   A\equal{}0$
So $ \Delta_1,\Delta_2,\Delta_1$ are concurrent
 :D


calculus
integration
function
trigonometry
real analysis
real analysis unsolved

How can I solve this integral:

$\int_{-\theta}^{\theta}e^{c\cdot\cos(t)}\,dt$

where $c$ is real number.

I know when $\theta=\pi$, this integral is actually
the modified bessel function (of $c$). But how about the
general case of $\theta$ ? Can we still have closed form expressions ?

The antiderivative of $e^{c\cos t}$ is not an elementary function, so there is no nice closed form expression for this integral. You can try to get a power series expansion in $c$, but I'm not sure it will help much. On the other hand, if you need the asymptotics for $c\to+\infty$, say, you can easily do it with the Laplace asymptotic formula. And, of course, you can always compute the integral numerically for each particular pair $c,\theta$.


calculus
integration
logarithms
symmetry
calculus computations

A variant on the symmetry theme? What happens if you "flip it around" meaning make the substitution $u=12-x?$

I don't know about the symmetry theme (is it a thread?).

What do I get if I set $u=12-x$?

Thanks Dr. Kent.

I want you to try that. Note that $u=12-x$ means that $x=12-u.$

What about the following integral? Can it be calculated exactly?

$\int_{4}^{8}\frac{\ln(9-x)\ln(x-3)}{\ln(9-x)+\ln(x-3)}dx=?$

Did you mean to have a $+$ in the numerator of that, Didi?


You certainly are!!!

 :rotfl: :rotfl: :rotfl:

YES I AM! lol I LOVE READING INTERESTING TOPICS!  :lol:

What is the point of this thread? It's about as random as one of my posts.


superior algebra
superior algebra unsolved

Let L/K be an algebraic extension. Prove that extension L(X)/K(X) is also algebraic and [L(X):K(X)]=[L:K].

An element in $ L$ is algebraic over $ K(X)$. $ X$ is automatically algebraic over $ K(X)$, so the minimal subfield containing them, this time being $ L(X)$, is also algebraic over $ K(X)$.

For basis, let $ a_1,\cdots,a_n$ be a basis of $ L/K$. We show that it is linearly independent over $ K(X)$. WLOG, we may assume the coefficient to be in $ K[X]$ by clearing the denominator. Then we can compare the coefficients of each power of $ X$, and see that the coefficients have to be all 0.

Then we need to show that $ a_1,\cdots,a_n$ generates $ L(X)/K(X)$. Let $ f/g \in L(X)$ with $ f,g \in L[X]$. For $ f,g \in L[X]$, clearly we can express it as a linear combination of $ a_i$ over $ K(X)$. For $ 1/g$, since it is algebraic over $ K(X)$, we can write it as a $ K(X)$-linear combination of powers of $ g$. Each power of $ g$ is in $ L[X]$, so is a linear combination of $ a_i$ over $ K(X)$. So clearly $ f/g$ can be written as a linear combination of $ a_i$ over $ K(X)$.


First problem
[quote]For  $a,b,c$ positive reals prove that

$8(x^3+y^3+z^3)^2 \geq 9(x^2+yz)(y^2+zx)(z^2+xy)$[/quote]

Second problem
[quote]Determine all real $x,y,z$  such that

$xa^2+yb^2+zc^2 \leq 0 $ is true whenever $a,b,c$ are the sides of a triangle [/quote]

So I think it is correct.

I think second problem is not true
Here's counterexample: Suppose $c=max(a,b,c)$ - consider $x,y=1$ and $z=-1$ we have $a^2 + b^2 \leq c^2$ which is obviously true

Maybe it should be: "at least one of $x,y,z \leq 0$" but anyway it's obvious

Manlio, I bet you know that $m_a\geq l_a$!

I have formulated erroneously the second problem, sorry very much.

I will correct it.

The first problem is so much idiot just like me when I posted it. :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:
Sorry very much, Harazi. I will replace it with another inequality I found on my notes, but not created by me. I don't remember if it was already posted, but probably yes. Really I don't remember. However it is simple and nice as it can be solved in many different ways.

I have corrected the second inequality and changed the first so now you can enjoy to solve these two very easy problem (I say they are very easy as I have been able to solve them).

Maybe I misunderstood something but second problem is now obvious ! There's nothing to prove - are you sure about this $x,y,z \leq 0$ ?

Sorry for my bad english. Now I hope it is correct.


A friendship candle burns wax at a rate of 10 mm^3/min. What is the number of cubic centimeters in the volume of a candle that will burn for exactly one week?

I don't have time to post my solution, but my answer is...[hide]100.8cm^3[/hide]

-jorian


Show that $ ax^m \plus{} by^n \equiv c \mod p$ has the same number of solutions as $ ax^s \plus{} by^t \equiv c \mod p$, where $ s \equal{} \gcd(m, p \minus{} 1)$ and $ t \equal{} \gcd(n, p \minus{} 1)$.

What's the right-hand side of the second equation?

I have to find a bijection between the sets of solutions from one congruence equation to the next but this is where I get stuck.

Hint: You need to find a bijection $ \mathbb{F}_p\to\mathbb{F}_p$ which maps $ x^m$ to $ x^s$ for every $ x\in\mathbb{F}_p$. Do you see how to do this?

  darij

I don't see how to do it? Can someone finish this problem?


Ten persons went to a bookstore. It is known that :
1) Every person has bought 3 kinds of books.
2) For every 2 persons, there is at least one kind of books which they both have bought.
Let $m_i$ be the number of the persons who bought  the i th kind of books and $M = max \{m_i\}$.
Find the smallest possible value of $M$ .

it's hard to explain the solution of a combinatoric problem (expecially for my POOR english...)...
now i'm treating the people as element of $A=\{a,b,c,d,e,f,g,h,i,l\}$ and the owning a particular book is the belonging to a particular subset.

every element belongs to EXACTLY three subsets.
in every subset, so, he must meet the other 9 element, so the number we are searching $(min\{max\{m_i\}\})$ is geather than or equal four.
now let's try to find some subsets of four elements that satisfy the text conditions.
for what we said first, every element must meet each element EXACTLY ONE time (if not, we will have a 5 or more elements subset).
WLOG, let's suppose there are these subsets
$\{abcd\}$
$\{aefg\}$
$\{ahil\}$
in the other his two apparition, $b$ must meet $e,f,g,h,i,l$, but for every partition of this set in two three-elements subset, there exist at least a couple of elements that meets for the second time.so, it's impossible that the number we are searching $(min\{max\{m_i\}\})$ is four.
The number we are searching is five because it's possible to create five-elements subsets that satisfy the text condition:here are a possible situation:
$\{abcef\}$
$\{acdgh\}$
$\{abdil\}$
$\{befgh\}$
$\{cghil\}$
$\{defil\}$

i'm sorry for my BAD english
bye bye

ITALIANS LOVE COMBINATORICS........

Let $L$ be the desired minimum.

-------------------------------
[b]Lemma 1:[/b] We must have $L\geq 4$.

[b]Proof:[/b] Suppose for the sake of contradiction that $L<4$. Consider one of the people and their three books. They can have in common each of their books with at most $2$ others, hence there can only be $6$ others, contradiction. $\square$

-------------------------
[b]Lemma 2:[/b] If $L=4$, each book is owned by exactly $4$ people.

[b]Proof:[/b] Suppose for the sake of contradiction that there is a book called "Book" which at most $3$ people own. Consider one of these people, call him Buk. Buk shares ([i]Note:[/i] by "shares" I mean has in common) "Book" with only two people, but he has seven people left to share books with. Hence one of his books must be shared with at least four others, meaning $L\geq 5$, contradiction. $\square$

-----------------------
[b]Lemma 3:[/b] We cannot have $L=4$.

[b]Proof:[/b] Suppose for the sake of contradiction that $L=4$. We count the number of owned books in two ways. In the first way, we are explicitly given that there are $30$ books. In the second way, we use the second lemma. Since each book must be owned $4$ times, the number of books is a multiple of $4$. However, $30$ is not a multiple of $4$, contradiction. $\square$

---------------------------
Now since $L\neq 4$ we can strengthen our bound on $L$ to $L\geq 5$. It remains to find an example of when $L=5$, which is doable. Number the books $1$ to $6$. The people can have books $165,165,213,236,236,241,245,416,435,435$, so in fact $L=5$. $\blacksquare$

-----------------------
Does anyone know of a good method to find an example for $L=5$? I just kept trying things until I found something that worked but it really shouldn't have taken as long as it did.

For constructing $L = 5$, first drop the idea of having two students get the same books (multisets are no fun). Then notice that you have 20 ways of assigning 3 out of 6 books, and the intersection condition means you can't pick one of these ways and its complement. That means for each of the 20 combinations of 3 books, you need a student to have that combination or its complement. So one thing you can do is start with a simple configuration with no duplicate combinations and the intersection condition satisfied, and tweak it by replacing some book combinations with their complements. This method of tweaking guarantees the intersection condition will remain satisfied.

Specifically: start with the configuration where all students have book 1 and their remaining two books are one of the ten ways to have two of books 2-6. Then have the students with 123, 134, 145, 156, 126 exchange their books for the other three. (Notice that 23, 34, 45, 56, 26 form a 5-cycle when treated as graph edges.) This gives you the working configuration
\[ 124, 125, 135, 136, 146, 234, 345, 456, 256, 236. \]

Ah, this is what I tried, but I struggled to find a $5$ sets to take the complement of. I see why taking the 5-cycles will work, but was there a motivating factor to consider them other than nice symmetry? Specifically, is there a graph theoretic interpretation that you considered?

Out of the five books 2-6, you need five complements, each book must be represented twice, and each complement contains two books out of 2-6. If you have a $5 \times 5$ grid with two marked squares in each row/column and all distinct rows/columns, the only arrangement that works is the equivalent of a 5-cycle.


Given real number x and 0<x<pi/2.Prove that
                         (sinx)^(sinx)<(cosx)^(cosx)


I am a 10th grader right now I signed up for AP Physics B next year.
However, I was wondering what the differences between Physics C and B are (from what I hear, C is still an introductory coruse and does not require knowledge learned in physics B).
What are some prequisites required to take Physics C?
Thanks

Calculus is a pretty big part of Physics C.


Define f(N)=N!. Let a=f(1)f(2)f(3)...In other words, to optain the decimal representation of a write the decimal representation of f(1), f(2), f(3)... in a row.Is a rational?

[quote="Arne"]No. If it is rational, it must have a period. Look at the zeros at the end of the factorial numbers. It is easy to see that there can't be a period (since the numbers of zeros increases and can be made arbitrarily large)
Now very rigorous but correct - I am sure you can rewrite it "mathematically"[/quote]
This is rather idea than a solution, but I will try to apply it

[quote="Arne"]I can elaborate on it if you want me to do so. Just ask, if necessary.[/quote]
please try

Any rational number must have a periodic set of digits which (possibly after some finite number of other ones) are repeated indefinitely.  So suppose that this number is rational, and has a repeating section.  Suppose that section is of length $n$.  $f(k)$ will end in at least $n$ $0's$ for all $k>10^n$ (this is an extremely loose bound), so there are infinitely many blocks of $n$ zeroes, and hence the entire repeated section must be $0$ - but this cannot happen, since $f(k)$ starts with some non-zero digit for each $k$, and hence there are infinitely many of these too - a contradiction.

(Sorry if I'm treading on your toes here Arne, but I presume that's much the same as what you would have said.)

[quote="Arne"][quote="Diarmuid"]Sorry if I'm treading on your toes here Arne, but I presume that's much the same as what you would have said.[/quote]

Hey, that's no problem at all! :)

And yes, I had pretty much the same argument :)[/quote]
And ever since I read your idea I put it in exactly thesame `package`.

[quote="Arne"][quote="Makaveli"]And ever since I read your idea I put it in exactly thesame `package`.[/quote]
What do you mean?[/quote]
I was trying to say that the idea with the zeros may be put forward in a different way(at last I believe so)but this is the most obvious one.


I feel things might look a bit better if you add a sidebar and some other image on the top.....  :)

I protest.
Whenever we kids make a website there is always an irritating bit between the name and the dot com.(e.g. vf11)
Not fair.

But how does that matter in any case ??  :?  And, well, if you're ready to pay, I dont think there's a problem between the website name and the $.com$.....Anyway, as long as you and I continue using free webhosting services, we dont have a choice but to get that part between the name and the $.com$... so I guess there's no use cribbing about it........Besides, it's pretty harmless text....I dont get why anyone should find that irratating... :(

I don't mean it's ugly or horrid, but won't it be nice to have something like [url]http://www.karthik.com[/url] or [url]http://www.math92.com[/url]?

[quote="math92"]i need help. My site is missing something...tell me what you think of it and let me know. thanks
heres the site:
http://math92.vf11.com[/quote]

I think that it is missing posters.  ;)

$.com$ is the awesomeness. Guess what I spent part of the $\$$50 I recently won at MAML on.

I think what's missing is some original idea or content that will actually get posters to join.

[quote="math92"]don't you have to pay for the .com ending?[/quote]

You do.....that's why you and I end up only with sites with something like vf11 while we stick to free hosting...:(

hehee cool, i like how theres only one topic in each section and 10 topics in games and its all weird and written by you =/

What's that i see below - [b]Get your own free forum???[/b]
I sense something fishy!

The site's in peices.

Try to merge things.
Fill the white space.
Change the skin/theme to something like AOPS.

i updated a couple things.  Could you guys join it and add some suggestions?

Put in an avatar if you can. I have already joined it. Once you have a hundred or so members it will get going, don't worry.

i got an avatar.  I only have 10 members!!!!!!!!
2 of them are me
2 of them are my sister
6 of them from aops

where do i get new members!

sorry!
you just have to wait...and wait.....and wait.........and a day will come when you will have twenty new memebers each day.....just like AoPS!!

ok, i submitted my website to google, so it might come up on google...

that i am afraid won't help much ... people don't generally go about looking for these sites - there are plenty of them I guess ... but they join the ones that catch their eyes ...

Query your friends about what they search for on google like 'math problem solver' or 'free games' or 'homework help'

Change your site name from hang out central to something including those keywords.

Then the site will be like 1 or 2 in such google searches.

hang out central?? I thought it was fuzzy bunnies or something!

it was fuzzzy bunnies but then i changed it to attract more people. ARGH. i dont know what to call it!!

follow Vihang's idea.
there's smething in it.

you have to submit your site to google (i think), so i did and put my site under those keywords

Keywords and site name are two different things

Try a google search

example crash
You will get sites with 'Crash' in the sitename before sites with 'crash' just as a keyword.

So a related name is a better choice.

[quote="bubka"]and who will want to search for 'crash' (even 'cash' makes sense!)?[/quote]
I think that there is a movie called crash.

there could indeed be many things called crash, seeing that it is an English Word, but one look at math92's site will tell you that it is not calculated to please somebody looking for information on the movie crash, so let it pass.


A little help with the following TC problem:
http://www.topcoder.com/stat?c=problem_statement&pm=7332

Hm, doesn't look too much like a DP problem when you look at the constraints... if you know which SRM / Tournament this was, you could try finding it's editorial on TopCoder webpage.

Well, I found the editorial, but the explicit reccurent relationship isn't given in the solution, it only says it's solvable using DP-on-subsets.

I looked at the discussion and now I understand :). OK, the solution goes like this:

Since k <= 6, the i-th participant has only at most 13 possibilities of placement. Suppose that you take first participant and try all the possibilities - he can end up first, second... k-th. After you choose placement of the first participant, the second participant has some possibilities based on k and on what position took the first participant. (Two people can't share the same place.) You can do the DP on how many people are left to choose and what positions are already taken (at most 100 times 2^13 possibilities):
[code]int dp(int peopleLeft, int positionsTakenBitMap) {
	for (int i = -k; i <= k; i ++) {
		if (not positionTaken[i]) {
			take_position(i);
			result += dp(peopleLeft - 1, newBitMap);
		}
	}
}
[/code]The new bitmap can be easily calculated from the current one and from what position the i-th player took.

To add on, the motivation here is that the evolution from state to state is constrained by a fairly small portion of the history, which means we should just be able to keep track of all of it in some way. If we continue along these lines of thought, we arrive at the given DP, which is to consider which of the locations within 6 of you are open. We need not worry about skipping over locations, as we will eventually reach a problem later on since permutations are one-to-one and onto, and then that branch of the memoization will return 0.


What is the slope of a line perpendicular to the line whose equation is $ \frac{x}4\minus{}\frac{y}5\equal{}1$? Express your answer as a common fraction.

$ \frac{x}{4}\minus{}\frac{y}{5}\equal{}1 \rightarrow 5x\minus{}4y\equal{}20 \rightarrow 4y\equal{}5x\minus{}20 \rightarrow y\equal{}\frac{5}{4}x\minus{}5$.

$ \frac{5}{4}x\equal{}\minus{}1 \rightarrow x\equal{}\minus{}\frac{4}{5}$.

Answer: $ \minus{}\frac{4}{5}$


\[ a,b,c,d\ are\ positive\ real\ numbers\ such\ that\ \ \sum{\frac {1}{a^4 \plus{} 1}} \equal{} 3\ \ \ \ \ prove\ that\ \sum{a^2b^2\le 2}\]

It's Iran's second round problem with a little change................
why there isn't any reply? :blush:

Let $ a^4 \equal{} \frac {x}{y \plus{} z \plus{} t}$ and $ b^4 \equal{} \frac {y}{z \plus{} t \plus{} x}$ and etc..

The inequality may be written as :

$ \sum_{sym}\sqrt {\frac {xy}{(y \plus{} z \plus{} t)(z \plus{} t \plus{} x)}} \le 2$

Using AM-GM inequality

$ \sum_{sym}\sqrt {\frac {xy}{(y \plus{} z \plus{} t)(z \plus{} t \plus{} x)}} \le \sum_{sym}\frac {1}{2}\left(\frac {y}{y \plus{} z \plus{} t} \plus{} \frac {x}{x \plus{} t \plus{} z}\right) \equal{} \sum_{cyc}\frac {1}{2}\left(\frac {y}{y \plus{} z \plus{} t} \plus{} \frac {z}{y \plus{} z \plus{} t} \plus{} \frac {t}{y \plus{} z \plus{} t}\right) \equal{} 2$


A fitness center charges a membership fee plus a fixed amount per day for each day the center is used. If Claire paid $ \$68$ for 22 days of use, and Adrian paid $ \$71$ for 24 days of use, how many dollars is the membership fee?

This is how I solve these kinds of problems.

We take $ \$71$ and subtract $ \$68$ from it, getting $ 3$. There was a difference of $ 2$ days, so we divide: $ \frac{3}{2} \equal{} 1.5$

So each day costs $ \$1.50$. We multiply $ 1.5$ by $ 22$, and get $ 33$.

This means that not including the membership fee, $ 22$ days would cost $ \$33$. We just subtract this from $ 68$ to find how much the membership fee is.

$ 68 \minus{} 33 \equal{} 35$

The membership fee is $ \$35$.

Systems of equations method:

Let $a$ be the fixed amount and $b$ be the amount per day. We have the following equations:
\begin{align*}
a+22b=68 \\
a+24b=71 \\
\end{align*}

Subtracting the two equations, we have $-2b=-3$, or $b=\dfrac32$. Plugging in $b=\dfrac32$ into the first equation, we get $a+33=68$ and $\boxed{a=35}$.


Given that each of the numbers p and q lies between 0 and 1, prove that the numbers p(1-q) and q(1-p) is less than or equal to 1/4.


conics parabola calculus derivative trigonometry algebra polynomial

[quote="isabella2296"]The only ways I can think of are

- practice

- graphing calculator[/quote]

Thank you very much for that. I guess I really have to get myself a graphic calculator.  :oops:

just as it is with all geometric figures, it is impossible to draw a parabola with 100% accuracy. (even with a calculator)
besides, in all classwork and competitions, you do not need a perfect parabola to solve a problem

Or you can use asymptote.  It's a lot easier, and you can print your work.

What I often do is use vertex and two points to its left and two points to its right. By doing this, you at least get the bottom part of the parabola right and that is usually helpful.

But as you'll discover, graphing an accurate parabola doesn't get you too far in problem solving anyway (i.e. there isn't that much of advantage).

[quote="I Hate Watermelons"]I personally find it hard to graph parabolas accurately using only the x/y intercepts and the vertex. It might be because I am really bad at drawing curves, or my hands are too shaky. Any "tricks" to graph with 100% accuracy (using pen and paper) without having to plot too much points?

Thanks.[/quote]

Take arts

[quote="moogra"][quote="Poincare"]If you know calculus then derivatives are a LOT of help (my opinion).

If you don't do what silverfalcon says and don't just plot random points what i would do if i didn't know calculus would be to plot two or three points close together and one or two farther away so you get the main idea of the curve if you keep practicing this then you'll be drawing like  :play_ball:  i know that cause i just couldn't draw a sine or cosine curve then i just did like about 75 problems on them and now i like drawing sine and cosine curves  :D[/quote]

Um no. Calculus for parabolas? That's useless. My suggestion is to use the latus rectum.[/quote]

Calculus can help to find the vertex (for those who know calculus, this is done by setting the first derivative to zero) but not really much beyond that. Calculus can help you which sides going up and down but since this is parabola, it's trivial and doesn't help much.

The only reason using calculus could be considered is to get the general idea of which direction the ends of the parabola point towards. For higher level polynomial functions, it can be used to find the minima and maxima.

[quote="not_trig"][quote="Silverfalcon"]

Calculus can help to find the vertex (for those who know calculus, this is done by setting the first derivative to zero) but not really much beyond that. Calculus can help you which sides going up and down but since this is parabola, it's trivial and doesn't help much.[/quote]

Finding the vertex requires no calculus - it's just halfway in between the roots. Also, I think calculus could be slightly useful if you want to use Euler's method (but that would be very tedious.)[/quote]

Look what I wrote carefully. I said that calculus [b]can[/b] help. I never said that it is required. Yes, I know that it is halfway between the roots and there are many algebraic ways for it. I was just giving one example --- taking first derivative. There is really no "required" in math. You can usually go one way or another and arrive at one answer.

If you really want to, you can probably use the directrix definition of a parabola and bits of string or something along those lines.  Of course, given a parabola in standard form you'd then have to [i]compute[/i] its directrix...

In other words, it's not worth it.  I usually graph $ y \equal{} x^2$ on graph paper (by plotting the points $ (\pm 1, 1), (\pm 2, 4)$, ...) and then relabel the axes.

my only tip: if you are plotting consecutive integer x values, then the y-values change by the leading coefficient*next odd number - for $ y \equal{} ax^2 \plus{} bx \plus{} c$, if the change between $ x \equal{} 1$ and $ x \equal{} 2$ is $ a(2n \plus{} 1)$, and we are moving away from the vertex, then the change between $ x \equal{} 2$ and $ x \equal{} 3$ is $ a(2n \plus{} 3)$, the change between $ x \equal{} 3$ and $ x \equal{} 4$ is $ a(2n\plus{}5)$, etc (this is assuming that $ \minus{} b/2a$ is an integer.  This technique works on $ x \equal{} \minus{} b/2a \pm g$ where g is an integer, so it is easy to adapt to plot non-integer x's if you want to)


What is the method behind the probability problems that have an infinite sample space (one that cannot be determined)?

For example, what is the probability that when successively flipping a coin, one gets a chain of three consecutive heads before 2 consecutive tails?

For these problems, I don't know where to start, since basically the problem can start anywhere. Yet, these problems come up all the time in difficult contests. Could someone shed some light? Thanks.

Because each toss is indepandant of the other, and the chance of getting either heads or tails is 1/2, then you would say for the 3 consecutive heads, itwould be 1/2*1/2*1/2=1/8.  Then for the 2 consecutive tails would be 1/2*1/2=1/4.  then, you miltiply them together to find the probability, whichy is 1/4*1/8=1/32

I think you misunderstood the problem.  He is not asking for the probability that if you flip 5 coins, what is the probability that they come out HHHTT.  He is asking, if you keep flipping coins, what is the probability that you see a HHH before you ever see a TT.

A good way to solve such problems is by drawing a state diagram, and assigning one variable per state.  See the following links to previous topics:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13501]coins are cool[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10180]Question[/url]


logarithms
inequalities unsolved
inequalities

prove that

$ \bf{(\forall n \in \mathbb{N}^*)n^{\frac{1}{n}}<3^{\frac{1}{3}}}$

[quote="mathmen"]prove that

$ \bf{(\forall n \in \mathbb{N}^*)n^{\frac {1}{n}} < 3^{\frac {1}{3}}}$[/quote]

$ f(x) = x^{\frac {1}{x}},x > 0$, we have $ f'(x) = x^{\frac {1}{x} - 2}\cdot(1 - \ln{x})$, thus when $ x < e$ get $ f'(x) > 0$, when $ x > e$ get $ f'(x) < 0$, but $ 2 < e < 3$, so when $ n\in \mathbb{N}^ +$, we have $ f(n) < \max\{f(2),f(3)\}$, just we need to check that ${ f(2)\le 3^{\frac {1}{3}}}$ and ${ f(3) \le 3^{\frac {1}{3}}}$, which is easy.


Should there be a topic just denoted to events, like meteor showers, eclipses, etc.?
Maybe like a sticky topic?
Instead of making a bunch of seperate topics?

I think that's a great idea. That way, people can easily see what astronomical events are coming up or are already occurring. But, maybe the sticky could be used for announcements only, and discussions about the events could be reserved for a new forum topic.

Okay. :)
The only thing about the sticky topic is that an administrator would need to make it.
What can of annoucements do you mean?

Anyone else have opinions?

By announcements I meant things such as "Leonids peak tonight" and other notifications, but that topics such as "here are my pictures of the Leonids" could be put in their own threads, just to make it easier to find information about upcoming events without having to sort through lots of replies.


[b]ABCD[/b] are concyclic, [b]E[/b] is an arbitrary point on the plane,[b] P,Q,M,N,R,S[/b] are the projections of [b]E [/b]toward the lines and diagonals
show that the midpoints of [b]MP,NQ,RS[/b] are collinear

Let (O) be the circumcenter of ABCD.
There are three cases.
[b]Case 1:[/b] If $ E\equiv O$ then $ M,N,P,Q,R,S$ are the midpoints of $ AB,BC,CD,DA,AC,BD$. The midpoints of $ MP,NQ,RS$ is the centroid G of quadrilateral ABCD.
[b]Case 2:[/b] M lies on (O).
Applying Simson's line, we have $ (M,S,Q), (M,R,N), (Q,P,R),(S,P,N)$ are collinear.
Applying Gauss's line for quadrilateral $ MSPR$, we get the midpoints of $ MP,NQ,RS$ are collinear.
[b]Case 3:[/b] M doesn't lie on (O) and $ M \not\equiv O$
[b]Lemma:[/b] Given two lines $ d_1, d_2. A_1,A_2,A_3$ lie on $ d_1, B_1,B_2,B_3$ lie on $ d_2. C_1,C_2,C_3$ lie on $ A_1B_1, A_2B_2, A_3B_3$, respectively. If $ \frac {\overline{A_1A_2}}{\overline{A_1A_3}} \equal{} \frac {\overline{B_1B_2}}{\overline{B_1B_3}} \equal{} k$ and $ \frac {\overline{C_1A_1}}{\overline{C_1B_1}} \equal{} \frac {\overline{C_2A_2}}{\overline{C_2B_2}} \equal{} \frac {\overline{C_3A_3}}{\overline{C_3B_3}}$ then $ C_1,C_2,C_3$ are collinear and $ \frac {\overline{C_1C_2}}{\overline{C_1C_3}} \equal{} k$
This lemma can be solved by using Thales's theorem. I give it to the readers :D
Return to case 3.
Let $ X,Y,Z$ are the midpoints of $ MP, NQ,RS$. Let $ EO\cap (O) \equal{} \{E'\}$
$ M',N',P',Q',R',S'$ are the projections of E' on $ AB,BC,CD,DA,AC,BD. C',Y',Z'$ are the midpoints of $ M'P', N'Q', R'S'.G$ is the centroid of $ ABCD$.
Applying that [b]lemma[/b] above, we get $ (G,X,X'), (G,Y,Y'), (G,Z,Z')$ are collinear and $ \frac {\overline{GX}}{\overline{GX'}} \equal{} \frac {\overline{GY}}{\overline{GY'}} \equal{} \frac {\overline{GZ}}{\overline{GZ'}} \equal{} \frac {\overline{OE}}{\overline{OE'}}$
But $ X',Y',Z'$ are collinear then applying Thales's theorem, we obtain $ X,Y,Z$ are collinear.
QED

[quote="livetolove212"]But $ X',Y',Z'$ are collinear then applying Thalet's theorem, we obtain $ X,Y,Z$ are collinear.
QED[/quote]
Could you please explain what Thalet's theorem is? and how do you prove your lemma?  :|

Oh sorry, It's a mistake in typo.It must be "[b]Thales[/b]'s theorem" :D
This is the solution for my lemma:
Construct two lines which pass through $ C_1$ and parallel to $ A_1A_3$ and $ B_1B_3$.
As you can see: $ A_2X_2//A_3X_3//A_1C_1, B_2Y_2//B_3Y_3//B_1C_1$
$ \Rightarrow A_2X_2//B_2Y_2$
But $ \frac{\overline{A_2X_2}}{\overline{B_2Y_2}}\equal{}\frac{\overline{A_1C_1}}{\overline{B_1C_1}}\equal{}\frac{\overline{A_2C_2}}{\overline{B_2C_2}}$ then $ X_2,C_2,Y_2$ are collinear.
Similarly $ X_3,C_3,Y_3$ are collinear.
On the other side, $ \frac{\overline{C_1X_2}}{\overline{C_1X_3}}\equal{}\frac{\overline{A_1A_2}}{\overline{A_1A_3}}\equal{}\frac{\overline{B_1B_2}}{\overline{B_1B_3}}\equal{}\frac{\overline{C_1Y_2}}{\overline{C_1Y_3}}$
And $ \frac{\overline{C_2X_2}}{\overline{C_2Y_2}}\equal{}\frac{\overline{C_3X_3}}{\overline{C_3Y_3}}$
$ \rightarrow C_1,C_2,C_3$ are collinear and $ \frac{\overline{C_1C_2}}{\overline{C_1C_3}}\equal{}\frac{\overline{C_1X_2}}{\overline{C_1X_3}}\equal{}\frac{\overline{A_1A_2}}{\overline{A_1A_3}}\equal{}k$

[quote="vittasko"]This problem has been posted before in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=227863]Collinear[/url], of [b][size=100]Butterfly[/size][/b], who gave us this nice solution based on the [b][size=100]Lemma[/size][/b] you mentioned dear  [b][size=100]livetolove212[/size][/b].

Tell me please if you have a different reference about this [b][size=100]Lemma[/size][/b], which is what I have called as [b][size=100]ERIQ Theorem[/size][/b].

Thank you in advance, Kostas Vittas.[/quote]
Dear Kostas Vittas,
This [b]lemma[/b] is very useful. You can see it in my friend, Pham Minh Khoa's solution  for [b]Viet Nam TST 2009, Problem 5[/b] :
http://www.mathlinks.ro/viewtopic.php?p=1474889#1474889

Thank you dear [b][size=100]livetolove212[/size][/b] for the reference.

The solution of this powerfull theorem you presented above, is the same as the one I posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=97493]Here[/url] ( [b][size=100]proof 2[/size][/b], post #11 ).

Best regards, Kostas Vittas.


geometry
perimeter
number theory
prime numbers

A scalene triangle has side lengths which are prime numbers and the
length of its perimeter is also prime. What is its smallest possible
perimeter?

the lengths are 5,7,and 11, which add up to 23.
hence, the answer is $ \boxed {23}$

Note that you have to use a little guess and check to arrive at the answer.

That is a degenerate triangle because 3 + 5 is not greater than 11.


according to my previous post. when I seeking for the answer (but I can't find it in my books  :( ). I found this statement:
$f$ is differentiable, if $\lim_{x \rightarrow \infty} f(x) = 0$ $\Rightarrow$ $\lim_{x \rightarrow \infty} f'(x) = 0$

I've prove that if  $\lim_{x \rightarrow \infty} f(x) = L$ then $\lim_{x \rightarrow \infty} f'(x) = 0$ , but I use mean value theorem.

the book put that problem (statement) in earlier section than the mean value section  :(
is there any other way to prove this?

[quote]$f$ is differentiable, if $\lim_{x \rightarrow \infty} f(x) = 0 \Rightarrow \lim_{x \rightarrow \infty} f'(x) = 0$ [/quote]
This is false. What you can say is this: if $\lim_{x\to\infty}f(x)$ exists (whether it's zero or some other number makes no difference) and if $\lim_{x\to\infty}f'(x)$ exists, then $\lim_{x\to\infty}f'(x)=0.$

But the limit of $f'(x)$ may not exist. Here's an example:

For $x>1,\,f(x)=\frac1x\sin(x^3).$ Then $\lim_{x\to\infty}f(x)=0$ but $f'(x)=3x\cos(x^3)-\frac{\sin(x^3)}{x^2}$ does not have a limit as $x\to\infty.$

Ok, now I add the condition $\lim_{x \rightarrow \infty} f'(x)$ exist, then can you prove (just give a hint) that the limit is $0$ without using mean value theorem. just for the case $\lim_{x \rightarrow \infty}f(x) = 0$


ABCD be a quadrilateral. E and F are the midpoints of AD and BC. join A,F; F,D; B,E; E,C.
AF and BE intersects each other at X and DF and CE intersects each other at Y.
Prove that,  triangle(ABX) + triangle(CYD) = quadrilateral(EXFY)

Note $ A(ABC)$ area of triangle $ ABC$.
We have:
$ A(ABX)\equal{}A(ABE)\minus{}A(AXE),\ A(CDY)\equal{}A(CDE)\minus{}A(EYD)$
Then $ A(ABX)\plus{}A(CDY)\equal{}A(ABE)\plus{}A(CDE)\minus{}(A(AXE)\plus{}A(EYD))$  (1)
$ A(EXFY)\equal{}A(AFD)\minus{}(A(AXE)\plus{}A(EYD))$  (2)
So, from (1) and (2) we have to prove that $ A(AFD)\equal{}A(ABE)\plus{}A(ECD)$.
Let $ BB_{1}\perp AD,FF_{1}\perp AD,CC_{1}\perp AD$. Then $  FF_{1}\equal{}\frac{BB_{1}\plus{}CC_{1}}{2}$
$  A(ABE)\plus{}A(CED)\equal{}\frac{AE\cdot BB_{1}}{2}\plus{}\frac{DE\cdot CC_{1}}{2}\equal{}$
$  \equal{}\frac{AD}{2}\left(\frac{BB_{1}\plus{}CC_{1}}{2}\right)\equal{}\frac{AD\cdot FF_{1}}{2}\equal{}A(AFD)$

Since the area of AFCD =  area of BFDE = 1/2 area of ABCD (split them into triangles), we apply the carpets theorem to the two quadrilaterals and we are done.

[quote="gotztahbeazn"]can you explain that more not trig

and also for red dog, how did you know $ FF_{1}\equal{}\frac{BB_{1}\plus{}CC_{1}}{2}$? is this just common knowledge that i dont know or do you have to prove it[/quote]
$ BB_{1}C_{1}C$ is a trapezoid and $ FF_{1}$ is the median line.

[quote="gotztahbeazn"]can you explain that more not trig

AFCD doesn't even look like 1/2 of ABCD, in my diagram it looks like much more than 1/2

and also for red dog, how did you know $ FF_{1}\equal{}\frac{BB_{1}\plus{}CC_{1}}{2}$? is this just common knowledge that i dont know or do you have to prove it[/quote]

oh! sorry! should be AFED :blush:

[hide]Let [EXFY]=I, [ABX]=II, [CYD]=III

We can express $ II\plus{}III \equal{} I\plus{}[ABE]\plus{}[BAF]\plus{}[DCE]\plus{}[CDF]\minus{}[ABCD]$.(%)

$ [ABE]\equal{}[ABD]/2,[BAF]\equal{}[BAC]/2,[DCE]\equal{}[DCA]/2,[CDF]\equal{}[CDB]/2]$.

$ [ABCD]\equal{}[ABD]\plus{}[CDB]\equal{}[BAC]\plus{}[CDB]\equal{}\frac{[ABD]\plus{}[CDB]\plus{}[BAC]\plus{}[CDB]}{/}2$.

Substituting in to (%), we get II+III=I.[/hide]


What is the probability that a randomly selected factor of  $66^{11}$ is a multiple of  $66^{6}$ ? Express your answer as a common fraction.

EDIT: I accidentally posted in the wrong forum. :blush: Can a mod please move it? Thaks

find the number of factors of $6^{11-6}=66^{5}$there are$6\times6=36$

[quote="usaha"]is it 343/1728, you know all multiples of 66^6 are multiples of 66^11, there are 1728 multiples of 66^11, and out of them 343 are multiples of 66^6, so it is 343/1728[/quote]

do you mean that a multiple of 6^11 is a multiple of 6^6


A beautiful series for Cotangent but I have no idea to prove it : \[\sum_{k=-\infty}^\infty\frac1{\omega+2\pi i.k}=\frac{i}{2}\cot\frac{i\omega}2\]

Using the Fourier series for cos ax and we get \[\pi \frac{\text{cos }ax}{\text{sin }a\pi}=\frac{1}{a}-2a\sum_{k=1}^\infty \frac{(-1)^{k}}{k^{2}-a^{2}}\text{ cos }kx\] for every $a\not \in \mathbb{Z}$. Taking $x=\pi$ to get \[\pi \text{ cot }\pi a=\frac{1}{a}-2a\sum_{k=1}^\infty \frac{1}{k^{2}-a^{2}}.\] Since \[\frac{1}{a-k}+\frac{1}{a+k}=\frac{2a}{a^{2}-k^{2}},\] we have \[\pi\text{ cot }\pi a=\sum_{k=-\infty}^\infty \frac{1}{a-k}.\] Choose $a=i \omega/(2\pi)$ to get \[\pi \text{ cot }\frac{i \omega}{2}=\sum_{k=-\infty}^\infty \frac{2 \pi}{i \omega-2\pi k},\] so \[-\frac{\pi}{i}\text{ cot }\frac{i \omega}{2}=\sum_{k=-\infty}^\infty \frac{2 \pi}{ \omega+2\pi i k},\] or \[\frac{i}{2}\text{ cot }\frac{i \omega}{2}=\sum_{k=-\infty}^\infty \frac{1}{ \omega+2\pi i k}.\] (Note that by writing 2\pi i \cdot k you can write the dot in the middle of the line height)


Find all solutions to the functional equation $f(1-x)=f(x)+1-2x$.

Hmm, I would think that substituting $x$ for $1-x$ and solving for $f(x)$ would get it, but you end up with identity...

but I did notice $f(x)=x$ is a solution.

I noticed that $x^2$ works since the difference of consecutive squares is $2x-1$. And I also tried substituting $1-x$ but yeah it didn't work.

Anyway, [hide="hint"]separate the terms in a useful way[/hide]

yes this one is tricky!

[hide="huge hint"]define a new function g(x)=f(x)-x[/hide]

[hide="Cool"]
Notice the rather awesome rearrangement

$f(1 - x) - (1-x) = f(x) - x$

Define $g(x) = f(x) - x$.  Then $g(x) = g(1-x)$.  So $g(x)$ can be any function even about $x = \frac{1}{2}$.
[/hide]

Nice! Or an even better continuation:

[hide]
Rewrite it as $g(1/2-1/2+x)=g(1/2+1/2-x)$ and define $h(x)=g(1/2+x)$. Then we get $h(x-1/2)=h(-(x-1/2))$. So $h$ is any even function. Thus we can take an even function $h(x)$ and construct a solution $f(x)=g(x)+x=h(x-1/2)+x$.
[/hide]

Or something like that, right? :P

nice, i feel so stupid, completely missed those solutions

I got this b4 looking at any hints.
Rewrite as $f(1-x)-(1-x)=f(x)-x$. Define $g(x)=f(x)-x$. We know that $g(1-x)=g(x)$ for all x. Thus g is a constant function, i.e., $g(x)=f(x)-x=c$. Thus $f(x)=x+c$ is the general solution. You can plug back in to the original to verify.

[quote="drunner2007"]We know that $g(1-x)=g(x)$ for all x. Thus g is a constant function,[/quote]

 :huh:

Look at the other posts again.

Sorry about that. I guess that was wrong to say it would equal a constant. so its an even function about x=1/2. interesting...


geometry
circumcircle
Euler
geometry unsolved

Let triangle ABC has T inside triangle ABC such that :
      angle BTC = angle CTA = angle ATB = 120 degree.
Let A' ; B' ; C' are the circumcenter of (TBC) ; (TCA) ; (TAB) .
Prove that two triangles ABC and A'B'C' have the same centroid

We will prove A'G is median of triangle $A'B'C'$
Draw euqaliteral triangle $BCA''$ outside then $A,T,A''$ are collinear, according property of centroid easily see $A'G//TA\perp B'C'\Rightarrow A'G\perp B'C'$ according Napoleon's problem $\triangle A'B'C'$ is equaliteral $\Rightarrow A'G$ is median of $A'B'C'$

Another nice result:Three Euler line of three triangle $TBC,TCA,TAB$ are concurrent! :)

Napoleon problem (maybe):Given $\triangle ABC$ draw three equaliteral base on three side and outside $\triangle ABC$ then centers of them are three vertex of a equaliteral triangle! :D


A thermodynamics professor had written a take home exam for his graduate students. It had one question:
"Is hell exothermic or endothermic? Support your answer with a proof."

Most of the students wrote proofs of their beliefs using Boyle's Law or some variant. One student, however wrote the following:

First, we postulate that if souls exist, then they must have some mass. If they do, then a mole of souls can also have a mass. So, at what rate are souls moving into hell and at what rate are souls leaving? I think that we can safely assume that once a soul gets to hell, it will not leave. Therefore, no souls are leaving.

As for souls entering hell, lets look at the different religions that exist in the world today. Some of these religions state that if you are not a member of their religion, you will go to hell. Since there are more than one of these religions and people do not belong to more than one religion, we can project that all people and all souls go to hell.

With birth and death rates as they are, we can expect the number of souls in hell to increase exponentially.

Now, we look at the rate of change in volume in hell. Boyle's Law states that in order for the temperature and pressure in hell to stay the same, the ratio of the mass of souls and volume needs to stay constant.

So, if hell is expanding at a slower rate than the rate at which souls enter hell, then the temperature and pressure in hell will increase until all hell breaks loose.

Of course, if hell is expanding at a rate faster than the increase of souls in hell, than the temperature and pressure will drop until hell freezes over.

It was not revealed what grade the student got.

I will not reveal my thermodynamics grade either.
[b][i][u]Edit:  "Hell" means: (Not heaven, but the other place)[/u][/i][/b]

My math/ physics teacher has a very similar story posted in her room, except on this one the student got the only A.

And the student proved that - hadn't frozen over because a girl told him that she'd sleep with him when - freezes over, and she hadn't.


limit
inequalities
number theory proposed
number theory

Prove that $\lim_{n \to \infty} \frac{\tau(n)}{\sigma(n)}=0$ where $\tau(n)$ & $\sigma(n)$ denote the number of and the sum of all divisors of $n$, respectively.

It is easy to show that $\tau(n) \leq 2\sqrt{n}$ and $\sigma(n) > n$.

$\sigma (n) >n$ is trivial for $n>1$.
To prove $\tau (n) \leq 2 \sqrt n$, take any divisor $a$ of $n$. Then either $a$ or $\frac{n}{a}$ is $\leq \sqrt n$, thus you can pair up the divisors of $n$ such that in every pair at least one of them is $\leq \sqrt n$, giving the result.

[quote="mathematica"]isn't $\tau(n)$ the [b]number[/b] of divisors??[/quote]

As ZetaX explained, you can always pair up the divisors so that the smaller is not greater than $\sqrt{n}$ and the larger is not less than $\sqrt{n}.$ Hence, you can have at most $\sqrt{n}$ smaller divisors (since neither is greater than $\sqrt{n}$), which means at most $2\sqrt{n}$ divisors in total.

This is how is solved it myself:

$\sigma(n)=\sum\limits_{d|n}d=\sum\limits_{d|n}\frac{n}{d}=n\sum\limits_{d|n}\frac{1}{d}$

therefore we have

$\frac{[\sigma(n)]^2}{n}=(\sum\limits_{d|n}d)(\sum\limits_{d|n}\frac{1}{d})$

now with applying the cauchy-schwarz inequality we get:

$\frac{[\sigma(n)]^2}{n}\geq[\tau(n)]^2$

therefore:

$0\leq\frac{\tau(n)}{\sigma(n)}\leq\frac{1}{\sqrt{n}}$

now according to the pressure principle we get:

$\lim_{n \to \infty} \frac{\tau(n)}{\sigma(n)}=0$


I am very stuck on this problem! Someone please help:

THE BUCKLED RAILROAD TRACK

Railroad designers are always looking for ways to improve the safety and comfort of train travel.

Suppose technology made it possible to make and transport straight rails one mile long. If one winter two pairs of these rails were installed along a two mile portion of track in the desert, then each rail would expand approximately one foot in length in the summer heat. Suppose that instead of buckling, these rails keep their (straight) shape. In this case, the ends where they meet could jut upward to form a triangle. How high above the ground would the ends of the rails be?

Let H be the height (in feet) of the tracks above the ground where the two rails come together.

a) How large do you think H is? Is it big enough for you to stick your arm between the ground and the tracks? Is it big enough walk through? Could you drive a car under the buckled tracks? Make a guess.

b) Draw a picture and calculate H. (Note: a mile is 5280 feet long.)

c) How does your calculated value for H compare with your guess in part (a)?

Thanks a lot!

Draw a picture, and use the Pythagorean Theorem.

Already posted here: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=120414[/url]


integration
trigonometry
floor function
calculus
calculus computations

Let $p>0$ be a real number and let $n\ge{0}$ be an integer. Evaluate

$ u_{n}(p)=\int_{0}^{\infty}e^{-px}\sin^{n}(x)dx$.

[hide="what i have so far"]
$ u_{n}(p)=\left(\frac{n^{2}-n}{p^{2}+n^{2}}\right)u_{n-2}(p)$

what am i supposed to do next since I have no idea how to get a general formula for $u_{n}(p)$
[/hide]

The numerator is not so bad: it's the product of $k(k-1)$ over $k\le n$, which is simply $n!$. The denominator is tricky: I set $p=1$, searched in Sloane's Encyclopedia and got [url=http://www.research.att.com/~njas/sequences/A006228]this[/url].

Remember that the antiderivative of $\sin^{n}x$ is usually represented by a recursion, too.

any new thoughts? i am still unable to solve this. :mad:


Show: For all real numbers $ a,b,c$ with $ 0<a,b,c<1$ is:
\[ \sqrt{a^2bc\plus{}ab^2c\plus{}abc^2}\plus{}\sqrt{(1\minus{}a)^2(1\minus{}b)(1\minus{}c)\plus{}(1\minus{}a)(1\minus{}b)^2(1\minus{}c)\plus{}(1\minus{}a)(1\minus{}b)(1\minus{}c)^2}<\sqrt{3}.\]

[quote="valerie"]Show: For all real numbers $ a,b,c$ with $ 0 < a,b,c < 1$ is:
$ \sqrt {a^2bc \plus{} ab^2c \plus{} abc^2} \plus{} \sqrt {(1 \minus{} a)^2(1 \minus{} b)(1 \minus{} c) \plus{} (1 \minus{} a)(1 \minus{} b)^2(1 \minus{} c) \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)^2} < \sqrt {3}$[/quote]
We need to prove that:
$ \frac {\sqrt {a^2bc \plus{} ab^2c \plus{} abc^2}}{\sqrt {3}} \plus{} \frac {\sqrt {(1 \minus{} a)^2(1 \minus{} b)(1 \minus{} c) \plus{} (1 \minus{} a)(1 \minus{} b)^2(1 \minus{} c) \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)^2}}{\sqrt {3}} < 1$
but:
$ LHS < \frac {\sqrt [4]{a^2bc \plus{} ab^2c \plus{} abc^2}}{\sqrt [4]{3}} \plus{} \frac {\sqrt [4]{(1 \minus{} a)^2(1 \minus{} b)(1 \minus{} c) \plus{} (1 \minus{} a)(1 \minus{} b)^2(1 \minus{} c) \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)^2}}{\sqrt [4]{3}} \leq \frac {a \plus{} b \plus{} c \plus{} \frac {a \plus{} b \plus{} c}{3}}{4} \plus{} \frac {1 \minus{} a \plus{} 1 \minus{} b \plus{} 1 \minus{} c \plus{} \frac {1 \minus{} a \plus{} 1 \minus{} b \plus{} 1 \minus{} c}{3}}{4} \equal{} 1$
Q.E.D

$ \sqrt{a\plus{}b\plus{}c}$ and $ \sqrt{3\minus{}a\minus{}b\minus{}c}$ are smaller than $ \sqrt{3}$  so all we need to prove is
$ \sqrt{abc}\plus{}\sqrt{(1\minus{}a)(1\minus{}b)(1\minus{}c)}$ plugging $ a\equal{}{sin}^2x$ b=sin^2y c=sin^2z with $ 0<x,y,z<90$
so it becomes $ sinxsinysinz\plus{}cosxcosycosz<3$
$ sinxsinysinz\plus{}cosxcosycosz<sinxsiny\plus{}cosxcosy\equal{}cos(x\minus{}y)<1$ we are done

$ \sqrt{abc}\sqrt{a\plus{}b\plus{}c}\plus{}\sqrt{ (1\minus{}a)(1\minus{}b)(1\minus{}c)}.\sqrt{3\minus{}a\minus{}b\minus{}c} \le \sqrt{(1\minus{}a)(1\minus{}b)(1\minus{}c)\plus{}abc}.\sqrt{3} \le \sqrt{3}$

Letting $ \frac {a + b + c}{3} = d$, we'll prove a stronger inequality
$ \sqrt {abcd} + \sqrt {(1 - a)(1 - b)(1 - c)(1 - d)} < 1$
which holds trivially by

$ \sqrt {abcd} + \sqrt {(1 - a)(1 - b)(1 - c)(1 - d)}$
$ < \sqrt [4]{abcd} + \sqrt [4]{(1 - a)(1 - b)(1 - c)(1 - d)$ (trivial)
$ \leq \sqrt [4]{(a + (1 - a))(b + (1 - b))(c + (1 - c))(d + (1 - d))$ (Cauchy)
$ = 1$

You forgot closing "$ \}$". $ \Longrightarrow$ "potentially dangerous"  :lol: [quote="qwerty414 more or less"]Letting $ \frac {a \plus{} b \plus{} c}{3} \equal{} d$, we'll prove a stronger inequality
$ \sqrt {abcd} \plus{} \sqrt {(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d)} < 1$
which holds trivially by

$ \sqrt {abcd} \plus{} \sqrt {(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d)}$

$ < \sqrt [4]{abcd} \plus{} \sqrt [4]{(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d)}$ (trivial)

$ \leq \sqrt [4]{(a \plus{} (1 \minus{} a))(b \plus{} (1 \minus{} b))(c \plus{} (1 \minus{} c))(d \plus{} (1 \minus{} d))}$ (Cauchy)

$ \equal{} 1$[/quote]

Nice! But note that it's not stronger. Just everything divided by $ \sqrt {3}$ and written in a nicer form :)

[quote="valerie"]Show: For all real numbers $ a,b,c$ with $ 0 < a,b,c < 1$ is:
$ \sqrt {a^2bc \plus{} ab^2c \plus{} abc^2} \plus{} \sqrt {(1 \minus{} a)^2(1 \minus{} b)(1 \minus{} c) \plus{} (1 \minus{} a)(1 \minus{} b)^2(1 \minus{} c) \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)^2} < \sqrt {3}$[/quote]

My solution :

Rewrite given inequality as follows :

\[ 2\sqrt{\frac{abc(a\plus{}b\plus{}c)}{3}} \plus{}2\sqrt{\frac{(1\minus{}a)(1\minus{}b)(1\minus{}c)(3\minus{}a\minus{}b\minus{}c)}{3}} < 2\]

Then we apply the AM-GM inequality to obtain :

\[ LHS <abc \plus{}\frac{a\plus{}b\plus{}c}{3} \plus{}{(1\minus{}a)(1\minus{}b)(1\minus{}c)}  \plus{}1 \minus{}\frac{a\plus{}b\plus{}c}{3} < 2\]
\[ \iff ab \plus{}bc \plus{}ca \minus{}a \minus{}b \minus{}c \plus{}1<1\]

Which is true because $ a\plus{}b\plus{}c >ab \plus{}bc \plus{}ca$ ,since $ a,b,c  <1$

[i] How many rounds have Austrian Mathematical Olympiad , for which grade this problem was proposed ?  [/i]

[quote="enndb0x"][i] How many rounds have Austrian Mathematical Olympiad , for which grade this problem was proposed ?  [/i][/quote] Sorry for answering that late^^ We have three rounds and there is no grade distinction. :) (this was the first one and thus very easy)


you mean the 'blog options' in profile?

yes.

Do you have more than 1 tag? If not, there ain't going to be a cloud.


MATHCOUNTS
geometry
perimeter
videos
summer program
MathPath
AMC

OK, nats is over, and we can make comparisons for next year.
Who do you think will win?

Kevin Chen most likely, but anyone can win it.

I think it would be more exciting if we had a "Who will get 2nd place?" poll next year... but then again, who thought Daesun was going to win this year?  It's a tossup.

I think I will fail 8th grade and go back to Nats and win it all.

[quote="mathgeniuse^ln(x)"]Actually, you will just fail 8th grade and won't even make it to Mathcounts Nationals.[/quote]

I concur.

[quote="biffanddoc"]Its gonna be an unecpected again, someone who got in 58-100 range this year.[/quote]

Hopefully not, especially not 90th, ugh god, if 90th this year wins I will kill myself. I am dead serious here.

Oh, I will totally win  ;)

Well it depends on what you mean by who will win. Kevin will win the written round, no question. (Although there is a question, since everyone thought Nathan woiuld win written this year)

Well we can't really predict now, becvause we don't even know who's going

Maybe in some really strong state there's a 7th grader this year that just missed nationals.

I think it was Ajay Radakrishnan from Connecticut, but I don't know.

Matthew Vengalil was 92nd.

I hope someone from Michigan does win though, particularly from our school.
Cough Cough

Would that be Allen?

Elizabeth Synge came in 24th, she was in 7th grade.

Wait I just checked the yearbook. Neil, don't you mean Ajay [i]Ravindran[/i]?

I guess so.

Allen got 70th Place.

Well,yeah.

watch out for jeff feng :P  he got 16th or something at state this year.

I got a 35 at states, and a 33 at nats. For some reason that is really odd. Both were complete chokes. I am also pretty good at countdown, but speed was my downfall at nats.
Also why do people think preet patel will make countdown? I know him personally and PREET PATEL is a complete idiot. I think he has less of a chance of making countdown than Neil does to not qualify for the AIME.

[quote="robinhe"][quote="Treething"][quote="SplashD"]
Do you go to [b]Chinese school[/b] or did some guy teach it to you?[/quote]

I take chinese in school[/quote]

Do you know korean better than you know chinese? :P[/quote]

Ha ha.  u are real funny. ;)

Treething did tell me he was Korean at MathPath though, if you were actually being serious.

[quote="pgpatel"]I got a 35 at states, and a 33 at nats. For some reason that is really odd. Both were complete chokes. I am also pretty good at countdown, but speed was my downfall at nats.[/quote]

But, the thing is Alan improved by 6 points from Chapters to States.  Usually people from our state deprove by that much, but he improved.  That is just scary.

Does anyone know the story of what happened during the Masters Round of '98, when Ryan Timmons was in it?  My math teacher said that something happened that inflicted Ryan in a negative way during that round.

I just took the AMC10 and am  :mad: because I got a 125 and would have qualified for the AIME. I wish my school did  the AMC10 :(.

Well, I wish I took AMC 10A, and I would've gotten a 150.0, opposed to 146.5 on AMC 10B.  Also, if I took AMC 12B, I would've gotten a 139.5, opposed to 116.0 on AMC 12A, though even though I did these questions again and I got a 150.0. :)

[quote="biffanddoc"]im really sad that I didnt do MATHCOUNTS in 6th grade because i got 46/46/46/46 and Im also sad that I didnt do chapters this or last year, because I got 46 on both. Im also sad that I didnt do MATHCOUNTS in any year from 1984 to 2003 because on those tests I got 46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46
Im also sad that I wasnt in countdown any of those years (except in 1984 when it didnt exist) becuase I could have answered all of those questions in <0 seconds. Same with masters round for all those years. And Im mad that I wasnt on the DE team before 2005 because I got 10 on all of them, with my eyes closed, while being mutilated with electric drills and having my teeth and fingernails ripped out. Wow I am so mad[/quote]

I think your big long 46/46/46/... made this page really really wide.

[quote="Treething"][/quote][quote="mathgeniuse^ln(x)"][quote="Ignite168"][quote="robinhe"]Mei also means beautiful so yeah...[/quote]

But you would never said Mei Mei as Mei meaning beautiful, or as a negation. I have never seen it used for anything else.

Guy's I'm sorry, I got 1 guy's place on my team wrong! He got 186th, not 188th.[/quote]

Darn, I wish I understood Chinese. :)[/quote][quote="Treething"]

Mei as beautiful and Meimei as in little sister are pronounced very differently, though...

wow this i sad, I'm korean and I know this[/quote]

Yeah but you can't tell that when you're writing it online unless you write 1 2 3 4 or nothing after "Mei"

[quote="biffanddoc"]
[hide="text"]
im really sad that I didnt do MATHCOUNTS in 6th grade because i got 46/46/46/46 and Im also sad that I didnt do chapters this or last year, because I got 46 on both. Im also sad that I didnt do MATHCOUNTS in any year from 1984 to 2003 because on those tests I got 46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46/46
Im also sad that I wasnt in countdown any of those years (except in 1984 when it didnt exist) becuase I could have answered all of those questions in <0 seconds. Same with masters round for all those years. And Im mad that I wasnt on the DE team before 2005 because I got 10 on all of them, with my eyes closed, while being mutilated with electric drills and having my teeth and fingernails ripped out. Wow I am so mad[/hide][/quote]

Man you're cool, is there any way I can learn to be like you?

[quote="mathgeniuse^ln(x)"]Me too.

Treething, I know that I am stupid, but I don't quite get your signature.[/quote]

It's meant to be that way, I'm sure about 99% of the AoPS population doesn't understand my signature.

Actually, that was the life of a famous person in the world...

I'm pretty sure that it is also the life of many other not so famous people in the world...

Oh, yeah, there [i]were[/i] only about 10000000 in this world who satisfy the condition given in Treething's signature.


function
real analysis
real analysis unsolved

Let $f: [a,b]\to\mathbb{R}$ such that:

    i) $f$ is differentiable on $(a,b)$;
   ii) $f$ is continous and non-constant on $[a,b]$;
   iii) $f(a)=f(b)$.

  Prove that there is $c\in (a,b)$ such that $f'(c)=f(c)$.

EDIT:

[quote="Cezar Lupu"]Let $f: [a,b]\to\mathbb{R}$ such that:

    i) $f$ is differentiable on $(a,b)$;
   ii) $f$ is continous on $[a,b]$;
   iii) $f(a)=f(b)$.

  Prove that there is $c\in (a,b)$ such that $f'(c)=f(c)$.[/quote]

Consider $f(x)\equiv1, x\in[a,b]$

Even with the edit, I still don't think the problem is correct.  Consider $f(x) = 1+x^{2}$, where $a =-\frac{1}{2}$ and $b = \frac{1}{2}$.

Should be either $f(a)=f(b)=0$, or, more generally, $f(b)=e^{b-a}f(a)$. This is one of many standard problems where you just have to apply the classical Rolle's theorem to some fancy function (in this case, to $e^{-x}f(x)$).


If $ P \equal{} \frac{s}{(1 \plus{} k)^n}$ then $ n$ equals:

$ \textbf{(A)}\ \frac{\log{\left(\frac{s}{P}\right)}}{\log{(1 \plus{} k)}}\qquad
\textbf{(B)}\ \log{\left(\frac{s}{P(1 \plus{} k)}\right)}\qquad
\textbf{(C)}\ \log{\left(\frac{s \minus{} P}{1 \plus{} k}\right)}\qquad \\
\textbf{(D)}\ \log{\left(\frac{s}{P}\right)} \plus{} \log{(1 \plus{} k)}\qquad
\textbf{(E)}\ \frac{\log{(s)}}{\log{(P(1 \plus{} k))}}$

hello, we have $ (1\plus{}k)^n\equal{}\frac{s}{P}$, taking the logarithm of both sides we get $ n\log(1\plus{}k)\equal{}\log(\frac{s}{P})$ or $ n\equal{}\frac{\log(s)\minus{}\log(P)}{\log(1\plus{}k)}$.
Sonnhard.

[hide="Slightly different"]

$ (1\plus{}k)^n\equal{}\frac{s}{P}$.

Taking the logarithm of this in base 1+k, $ n\equal{}\log_{1\plus{}k}\frac{s}{P}\equal{}\frac{\log \frac{s}{P}}{\log (1\plus{}k)}$, so $ \boxed{\textbf{A}}$

[/hide]


calculus
derivative
quadratics
conics
parabola
function
absolute value

1.Find the derivative of y=ax^2+bx+c
2.Find the second derivative of this quadratic
3.explain how your answer to#2 tells you about the shape of a parabola and the leading coefficient a

4.Find the derivative of y=ax^3+bx^2+cx+d
5.Find the second derivative of this quadratic
6.Explain what the second derivative tells you about the cubic function

1. $ (ax^2+bx+c)'=2ax+b$
2. $ (ax^2+bx+c)''=2a$
3. If the second derivative is positive, the parabola opens up, otherwise it opens down. The magnitude of the second derivative (absolute value) tells how wide/narrow the parabola is; the greater the value of the absolute value, the narrower the parabola will be. If the second derivative is $ 0$, the graph of the function is a line.
4. $ (ax^3+bx^2+cx+d)'=3ax^2+2bx+c$
5. $ (ax^3+bx^2+cx+d)''=6ax+2b$
6. The value of $ x$ for which the second derivative is $ 0$ is the value of $ x$ at which the point of inflection occurs. Also,  the function defined by the second derivative has the same limit ($ \+\infty$ or $ -\infty$) when $ x$ approaches $ \+\infty$ or $ -\infty$ as the original cubic function.


algebra
polynomial
number theory unsolved
number theory

15. p is a prime. Determine the maximal degree of a polynomal T(x) whose coefficients belong to {0,1,2,...,p-1} whose degree is less that p, and which satisfies T(n)=T(m) mod p   if and only if   n=m mod p

We can find such polynomiuals of degree $p-2$: just take $T(x)=x^{p-2}$.

Assume now there is such a polynomial $T(x)=x^{p-1}+P(x)$, where $P$ is a polynomial of degree at most $p-2$. We may as well assume that $P(0)=0$. We thus have $T(0)=0,T(x)=1+P(x),\ \forall x\ne 0$ (we're working in $\mathbb Z_p$ here). This means that $P$ induces a permutation on $1,2,\ldots,p-1$, so there is an $x\in\{1,2,\ldots,p-1\}$ s.t. $P(x)=-1$. For this $x$ we get $T(x)=T(0)=0$, so we have a contradiction.

This shows that $p-1$ isn't good, so we're left with $p-2$.


Prove, that there are infinitely many triples $(n,n+1,n+2)$ for which each number belonging to the triple is a sum of two positive squares.

[hide="solution"]Choose $n=2k^{2}(k+1)^{2}$ :D (of my friend not of mine) [/hide]


AMC
USA(J)MO
USAMO
inequalities proposed
inequalities

If $ a, b, c\geq0$ and $ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \equal{} 4$, show that $ ab \plus{} ac \plus{} bc \minus{} abc\leq2$.

Easy but it can be a bit chalenging!

That is 2001 USAMO problem.

[url=http://www.mathlinks.ro/viewtopic.php?p=96705#96705]See here[/url]


Find all the real numbers $ N$ that satisfy these requirements:

1. Only two of the digits of $ N$ are distinct from $ 0$, and one of them is $ 3$.

2. $ N$ is a perfect square.

Since this is number theory I believe you meant natural numbers N.

If $ N$ is divisible by 10, it is divisible by 100, so keep dividing through by 100 until it is not divisible by 10, so that we finally end up with $ N \equal{} 3000..0a$ or $ a000..003$, with $ a$ an integer, $ 1\le a\le 9$.

If there are no 0's, then only $ 36$ works. Otherwise, assume there is at least 1 0 inbetween $ a$ and $ 3$.

The only square residues mod 9 are $ 0,1,4,7$, so the only possible values of $ a$ are $ 6,7,1,4$ since $ N$ mod 9 is $ 3 \plus{} a$.

mod 4, $ a000..003$ is congruent to 3, so is not a square.

Therefore, the only possible candidates are $ 3000..01$, $ 3000..04$, $ 3000..06$, $ 3000..07$. However, the one ending in 6 and the one ending in 7 don't work mod 4, so only the first 2 are possible candidates.

$ 3*10^{n} \equal{} x^2 \minus{} 1 \equal{} (x \minus{} 1)(x \plus{} 1)$, boring casework should rule this out, since $ gcd(x \minus{} 1,x \plus{} 1) \equal{} 1,2$, similarly for the other one, since $ gcd(x \minus{} 2,x \plus{} 2) \equal{} 1,2,4$, someone who is not as tired, or who enjoys doing such casework can complete the solution :P.

Therefore $ 36*100^{n}$ are the only integers that satisfy it, for $ n$ being any integer $ \ge 0$.

[quote=Rofler]
$ 3*10^{n} \equal{} x^2 \minus{} 1 \equal{} (x \minus{} 1)(x \plus{} 1)$, boring casework should rule this out, since $ gcd(x \minus{} 1,x \plus{} 1) \equal{} 1,2$, similarly for the other one, since $ gcd(x \minus{} 2,x \plus{} 2) \equal{} 1,2,4$, someone who is not as tired, or who enjoys doing such casework can complete the solution :P.
[/quote]

I solved this recently. Started the same way you did. I didn't use GCD but still completed the solution, don't know if this is correct, but here's what I did:

$3 \cdot 10^n$ = (x-1)(x+1) or (x-2)(x+2), so I thought of splitting $2^n \cdot 3 \cdot 5^n$ into two sets of factors which are either 2 apart or 4 apart.

Since they are 2 or 4 apart they both need to be even (or odd but that's not possible because the product would be odd) so I put a two into each of them.

Now if we put a 5 into each of them then both numbers will be divisible by 10 and cannot be 2 or 4 apart, so that case does not work, therefore we need to put every single 5 into one of the factors, and that would leave it way larger than the other factor and therefore more than 2 or 4 apart (easy to prove).

It feels too simple but it might work.


function
inequalities
real analysis
real analysis unsolved

Supposed $ f$ is a continuous and twice differentiable function in the interval $ [4,6]$ such that $ |f(x)| \leq 1$ and $ |f''(x)| \leq 1$, then prove that $ |f'(x)| \leq 2$. Thank you for the solution given before.

We have (by Taylor's theorem), for every $ a\in [4,6]$:
\[ f(6) \equal{} f(a) \plus{} f'(a) (6\minus{}a) \plus{} f''(\xi_1) \frac{(6\minus{}a)^2}{2} \geq f(a) \plus{} f'(a) (6\minus{}a) \minus{}\frac{(6\minus{}a)^2}{2},\]
and
\[ f(4) \equal{} f(a) \plus{} f'(a) (4\minus{}a) \plus{} f''(\xi_2) \frac{(4\minus{}a)^2}{2} \leq f(a) \plus{} f'(a) (4\minus{}a) \plus{} \frac{(4\minus{}a)^2}{2},\]
for some $ \xi_1, \xi_2\in (4,6)$. Subtracting the second inequality from the first, we get
\[ 2 \geq f(6) \minus{} f(4) \geq 2 f'(a) \minus{} ((6\minus{}a)^2/2 \plus{} (4\minus{}a)^2/2) \geq 2 f'(a) \minus{} 2.\]
We conclude that $ f'(a) \leq 2$. Similarly, $ f'(a) \geq \minus{}2$.


geometry
3D geometry
number theory proposed
number theory

Prove that if $ a>1$ then there exist natural $ n$ such that $ 4(a^{n}\plus{}1)$ is not cube of positive integer

Suppose for  each n then $ a^n + 1 = 2x_n^3$
Lemma 1 : [u]$ a+1=2^m$[/u]
Suppose there exist a odd prime p such that $ p|a + 1$
Chose n is odd then .
$ v_{p}(a^n + 1) = v_p(a + 1) + v_p(n)$
From the condition then we must have
$ v_p(a^n + 1)\equiv 0 (\mod 3)$
But $ v_{p}(n)$ take all value on N so this gives contradiction.
So $ a + 1 = 2^m$
Chose n ssuch that $ n\equiv 1 (\mod 2)$ then we have
$ \gcd(a + 1,\frac {a^n + 1}} = \gcd(a + 1,n) = 1$
So $ a + 1 = 2b^3,\frac {a^n + 1}{a + 1} = b_n^3$
So for all n is odd then
$ \frac {a^n + 1}{a + 1} = b_n^3$
Take $ n = 7$ it gives contradiction .
So there is a integer number n such that $ 4(a^n + 1)$ is not a cube.

Other way:
Take $ n\equal{}2m$ where $ m\equiv 1 (\mod 2)$
Then $ a^n\plus{}1\equal{}(a^2)^m\plus{}1$
Prove as above we have $ a^2\plus{}1\equal{}2^{m'}$
So $ a\plus{}1\equal{}2^m,a^2\plus{}1\equal{}2^{m'}$
It follow that $ a\equal{}1$

Hi friends!

I have an easy solution!

We know that  $ 4(a^3\plus{}1)$ and $ 4(a^9\plus{}1)$ are perfect cubes, so their quotient  $ a^6\minus{}a^3\plus{}1$ is a
perfect cube. But if $ a>1$,  then $ (a^2\minus{}1)^3<a^6\minus{}a^3\plus{}1<a^6$, so it is between two consecutive
cube, which is a contradiction!

Mohsen J

General :
$ a,m$ is constant natural number such that $ a>1$
Prove that there exist infinite $ n\in N$ such that $ m(a^n\plus{}1)$ is not power of any natural number with exponent great than 1 .

No one ?
I will post my solution if no one reply it  :)


Hey people! So I started this thread with a question...What is your most favorite online game???? If you would like leave a link to the game and tell us what its about!

Heres mine...
2 Minute Football 3D. Its really fun!!!@@@
[url]http://www.shockwave.com/gamelanding/football3d.jsp[/url]

NO D'UH! Starcraft!

That question was redundant.

$ \text{bestest online game ever}\minus{}>\text{Starcraft}$

Learn your math.  :D

Well, I'm pretty sure everybody knows starcraft. Classic RTS game.

seconded :D though my parents think i've _completely_ stopped playing it.. i started playing a little a few months ago. on and off..

I enjoyed playing [url=http://armorgames.com/play/841/grid16]grid16[/url]. It's made up of 16 minigames, each done for two seconds at a time.

[quote="kimmystar94"]uh... no duh. BnB and Kartrider from Nexon pwn all.... :D[/quote]

Mario Kart is way more hardcore.

sodaplay.com!!!
It's cool. You can build stuff. What's not to like?

lol yeah sodaplay.. i built a really good amoeba once, though. Just didn't know how to use it in races (?)

AoPS FTW.

 :) I don't know why but I like to try and solve questions.

Twelve Sky. A martial arts MMORPG. It is heavily PvP based, with constant warfare going on between the three factions you can play.

http://12-sky.aeriagames.com

Oh. My favorite game is plants versus zombies. I don't know how to post links.

wow. nice reviving a 2.5 year old thread (sarcasm)


It's practically automatic: each term $ \frac{1}{j^{n}k^{n}}$ with $ j < k$ appears once on the left and gets doubled and appears twice on the right when we square $ \zeta(n)$; each term $ \frac{1}{k^{2n}}$ appears once on the left and gets doubled and appears twice on the right, once when we square $ \zeta(n)$ and once from $ \zeta(2n)$.  More generally, we can replace $ k^{n}$ with any positive sequence having convergent sum, replace $ \zeta(n)$ with the sum of the sequence, replace $ H_{k, n}$ with the partial sums and replace $ \zeta(2n)$ with the sum of the squares of the terms of the sequence.


What is the maximum number of three element subsets of {1, 2, 3, 4, 5, 6, 7, 8}, which have the property that any two of the subsets have at most one element in common.

We discuss the general situation with {1,2,...,n}. Each number can appear at most [(n-1)/2] times, where [] denotes the floor function. Hence the subsets can use, in total, at most n*[(n-1)/2] numbers. So the max number of subsets, X, satisfies: X<=n*[(n-1)/2]/3.

Setting n=8, we have X<=8. And an example of the subsets demonstrates sufficiency:

{1,2,3}
{1,4,5}
{1,6,7}
{2,4,7}
{2,5,8}
{3,5,6}
{3,7,8}
{4,6,8}

However, I think the bound I gave above is too weak for large n.

your formula:

X \leq n*[(n-1)/2]/3

can be rewritten as:
for n even: X \leq n/6 - n/3
for n odd: X \leq n/6 - n/6

this isn't always integer, so we may write: [n/6 - n/3] and [n/6 - n/6]

now I think we should simply apply any method to reach this, with the great advantage that we need no proof that it's optimal. If we find a way to reach it, then it is optimal by the preceding proof :)


BDEF is a rhombus
FC=4cm
BC=10cm
Find AB.

Refer to picture below.

[hide]Hey look! It's more similar triangles  :D ![/hide]

Because FC=4 and BC=10, BF=6. And because BDEF is a rhombus so DE=6. $ADE$ and $CFE$ are similar triangles so $\frac{FC}{DE}=\frac{FE}{DA}=\frac{2}{3}$. Hence, AD = 9. Therefore, AB=9+6=15. ;)


geometry
parallelogram
trigonometry
trig identities
Law of Cosines

Given that a parallelogram has a side length of $4d$ and diagonals of $6d$ and $4d$ find the sum of the other two sides in terms of $d$.

[hide="answer"]
Is it $2d\sqrt{10}$?
[/hide]

[hide="Solution"] Let $\alpha$ be the corner angle opposite to the $6d$-diagonal and so $\pi-\alpha$ is that opposite to the $4d$-diagonal. Let $x$ denote what we are searching for. Use Law of Cosines twice:
\[ \begin{aligned} 36d^2 & = 16d^2 + x^2 - 8xd\cos\alpha \\ 16d^2 & = 16d^2 + x^2 - 8xd\cos(\pi-\alpha)\end{aligned}  \]
Simplifying that last one to $0 = x^2 + 8xd\cos\alpha$ and adding that to the first equality yields $36d^2 = 16d^2 + 2x^2$, so $x^2 = 10d^2$, and $\boxed{2x=2\sqrt{10d}}$, as desired.
(Edit: Thanks, Hellstar, for remarking my little mistake at the end  :D)
[/hide]

[hide]Let $x$ be one side of the parallelogram.

Applying the Cosine Law in the isoseles triangle, finding the cos of the opposite angle to the the side $4d$, say $cosA$

$cosA=\frac{x^2+16d^2-16d^2}{2x*4d}=\frac{x^2}{8xd}=\frac{x}{8d}$

Since two diagonals of the parrallelogram cut each other at the midpoint of each, applying cosine law again to the small triangle created by half of each diagonal.

$cosA=\frac{x^2+4d^2-9d^2}{2x*2d}=\frac{x^2-5d^2}{4xd}$

Then we will have:

$\frac{x}{8d}=\frac{x^2-5d^2}{4xd}$

Solving that gives us: $x=d\sqrt{10}$. Then $2x=2d\sqrt{10}$

Q.E.D
[/hide]

Kurt, I think you just forgot to multiply your answer by 2, since the question was asking for the sum of the sides.  :D

In a parallelogram, the sum of the squares of the diagonals equals the sum of the squares of the sides. (This is basically just a generalization of what Kurt and YooSam have done. It can be easily proven by the cosine law.) So if we let the missing side be $x$, we have:

$16d^2+36d^2=32d^2+2x^2$

From this, we find that $x=d\sqrt{10}$ so $2x = 2d\sqrt{10}$

[hide="my sol"]
Solve this,
$3\sqrt{16d^2-h^2}=\sqrt{36d^2-h^2}$
Find $h^2$ which is $\frac{27d^2}{2}$
Substitute in,
$\frac{x}{2}=\sqrt{16d^2-h^2}$
to get $x$ as the side,
then times by $2$ to get $\boxed{2d\sqrt{10}}$
[/hide]

By the way use this formula next time its pretty helpful for parallelograms: $2(a^2+b^2)={d_1}^2+{d_2}^2$ where $a$ and $b$ are sides, and $d_1$ and $d_2$ are diagonals.


can anyone show me how to use hash tables in a problem? for example, in the IOI, at same problems it says that we can improve complexity by using them

thank you

I can't really think of a problem that would require use of hash tables. Perhaps if you posted the problem you read about?

Take this task from USACO:

[quote]Problem 3: Arithmetic Progressions [Brian Dean, 2004]

FJ has implanted a new machine-readable RFID chip inside each of his cows
so he can monitor them to ensure they all come home after a hard day
feeding in the pastures.

He was intrigued to notice one night that the cows often lined up with
many of their serial numbers in an arithmetic progression. One after
he saw five cows go by with these serial numbers:
		1 4 3 5 7
A quick inspection shows that cows numbered 1, 3, 5, and 7 were in
the order of an arithmetic progression (each member of the progression
differs from the previous progression's member by the same integer).

Help FJ learn if this is coincidence.  Given a list of N (1 <= N <=
2,000) serial numbers (nonnegative integers less than one billion),
find the longest subsequence of this list that forms an arithmetic
progression (progressions can run from large to small or small to
large) and report its length.

PROBLEM NAME: arithprg

INPUT FORMAT:

* Line 1: A single line with the integer N

* Lines 2..N+1: N lines, each with a single integer. The first line
        represents the first cow's serial number, etc.

SAMPLE INPUT (file arithprg.in):

5
1
4
3
5
7

OUTPUT FORMAT:

* Line 1: A single integer that is the length of the longest
        progression.

SAMPLE OUTPUT (file arithprg.out):

4[/quote]

Let A[i][j] be the length of the longest progression that has it's last two elements i and j (i < j <= N). Obviously, if the cows' serial numbers are in array V[], the element before V[i] and V[j] should be V[i] - (V[j]-V[i]) to form a progression. So, if you hash the entire array V at the beginning you cand find the position k where V[i] - (V[j]-V[i]) is (if it exists) in O(1), and update A[i][j] as A[k][i] + 1. This gives us an O(N^2) algorithm. I believe very similiar ideea can be used for task FROG from IOI 2002. Since you are Romanian (perfect_radio) i suggest you should also read the article http://info.devnet.ro/articole.php?page=art&art=27 (is in Romanian).


8. My three-digit code is 023. Reckha can’t choose a code that is the same as mine in two or more of the three digit-positions, nor that is the same as mine except for switching the positions of two digits (so 320 and 203, for example, are forbidden, but 302 is fine). Reckha can otherwise choose any three-digit code where each digit is in the set {0, 1, 2, ..., 9}. How many codes are available for Reckha?

[hide]Well, first there are 10*10*10, or 1000, ways for him to make a three digit code. There would be 10 ways for him to choose one that has two of the same digits for a certain pair, so 30 ways in all. But 023 is counted 3 times so subtract 2 for 28. There are 3 ways in which the numbers of her code are switched. In total so, it's 1000-28-3 which is 969.[/hide]

[quote="andrewmath"]There would be 10 ways for him to choose one that has two of the same digits for a certain pair, so 30 ways in all. But 023 is counted 3 times so subtract 2 for 28. There are 3 ways in which the numbers of her code are switched. In total so, it's 1000-28-3 which is 969[/quote]

I sort of lost you. Did you do $\binom52$ for the $10$ ways for him to choose one with 2 of the same digits? But that doesn't really make sense.
Same thing with how 023 is counted 3 times >_<


I'm pretty sure that it means you're given an arithmetic sequence. The first term is 2 and the difference between each term is 3.

We need to find out if one of the terms is a perfect square. The terms can be from 2 to $\infty$


linear algebra
matrix
linear algebra unsolved

[color=red]
Let $A$ be a symmetric matrix of order $3$ such that $det(A^{2}+I_{3})=1.$ Prove that $A^{3}=O_{3}.$ ;)
[/color]

If $A$ is symmetric then we can write $A=C^{-1}DC$, where D is a diagonal matrice such that the elements in the diagonal are the eingenvalues of $A$. Let $z_1$, $z_2$ and $z_3$ this eingenvalues. So we´ll have that $({z_1}^{2}+1)({z_2}^{2}+1)({z_3}^{2}+1)=1$. But this eigenvalues are reals, because A is symmetric. Hence we must have $z_1=z_2=z_3=0$. So we have that $A=0_3$.

It can be easily prove if the order of $A$ is $n$, where n is any natural number.

[quote="cezar lupu"]
Let $A$ be a symmetric matrix of order $3$ such that $det(A^{2}+I_{3})=1.$ Prove that $A^{3}=O_3$[/quote]

It does not say that A is real matrix, it could be a complex symetric matrix.

Then it's pretty obvious that the conclusion does not hold.

yes the result does not hold if A is complex symetric for instance

$A=diag(\frac{i}{\sqrt{2}},1,0)=^{t}A$
$det(A^2+I_3)=1$
but $A^3 \neq 0$


function
integration
calculus
calculus computations

[quote="mdk"]let f:[0,1] imply R be an integrable function, prove that  $ \int_{0}^{1}\;x^{n}\;f(f(x)\;dx=0$[/quote]

That is false, what about $ f(x)=x$ then $ \int_{0}^{1}x^{1}\cdot x \ dx \not = 0$.
I think you forgot a condition.


geometry
inequalities
geometric transformation
number theory
algebra
functional equation
IMO

Hi MathLinkers,

Just like you did last year, I think it would be fun to try to guess how this year's IMO paper will be.
My prediction (but I'm so bad at guessing  :D ):

1) Number Theory
2) Combinatorics
3) Geometry
4) Algebra (a functional equation maybe)
5) Number Theory
6) Combinatorics

Give your opinion please!

I think there is at least one inequality every year for the past $ n$ years that I can remember

My guess:
1. Inequality
2. Combinatorics
3. Geometry
4. Combinatorics
5. Number Theory
6. Algebra

my guess:

1)number theory
2)algebra
3)geometry
4)combinatorics
5)geometry
6)number theory

Here is my:
1)Combinatorics
2)Algebra
3)Geometry
4)Algebra
5)Geometry
6)Number Theory

When first tour starts? At what time? :)

One of the contestants has just told me that today it has been:
1) Geometry
2) Inequality
3) Number Theory

[color=green]Sorry, I don't know how to translate it.

1) Sea ABC un triángulo acutángulo y H el ortocentro. Consideramos las 3 circunferencias con centro el los puntos medios de cada lado y que pasan por H. La correspondiente a cada lado corta a ese lado en 2 puntos. Demostrar que los 6 puntos están todos sobre una misma circunferencia.

2)a) Demostrar que $ \frac {x^2}{(x\minus{}1)^2} \plus{} \frac {y^2}{(y\minus{}1)^2} \plus{} \frac {z^2}{(z\minus{}1)^2} \geq 1$
b) Demostrar que hay infinitas ternas x,y,z de racionales que dan la igualdad en a).

3) Demostrar que para infinitos n el numero n²+ 1 tiene un divisor primo mayor que 2n + sqrt(2n).[/color]

Good translation for the first one, darij.  For the second one, the missing condition is $ xyz\equal{}1$...

[quote="darij grinberg"]1) Let me guess, without translating it completely: Is this one of the Droz-Farny theorems at
http://www.pandd.demon.nl/drozf.htm ?
It's even in Honsberger's book (19th and 20th century euclidean geometry)...

2) Some condition is obviously missing... Let me bet that once it is inserted, the problem turns into something like http://www.mathlinks.ro/viewtopic.php?t=37317 , and you have to show that some equation of degree $ \approx 3$ has infinitely many rational solutions.

Thanks for the problems! Now I'm wondering whether imo-2008.es are too ashamed or why else they still don't have them published...

  darij[/quote]

OK so Albanian Eagle and TomciO are leading with 1 correct guess each, the rest of us scores 0 points   :( :P

haha! yeah... well a little disappointing after last year where I guessed all 6 of them. :D

[quote]Thanks for the problems! Now I'm wondering whether imo-2008.es are too ashamed or why else they still don't have them published...

darij[/quote]
I was thinking the same thing  :roll:

4. A set $ X$ of positive integers is called {\it nice} if for each pair
$ a$, $ b\in X$ exactly one of the numbers $ a\plus{}b$ and $ |a\minus{}b|$ belongs to
$ X$ (the numbers $ a$ and $ b$ may be equal). Determine the number
of nice sets containing the number 2008.

5. A group of persons is called {\it good} if its members can be distributed
to several rooms so that nobody is acquainted with any person in the same room
but it is possible to choose a person from each room so that all
the chosen persons are acquainted with each other.

   A group is called {\it perfect} if it is good and every set of its members
is also good.

   A perfect group planned a party. However one of its members, Alice,
brought here acquaintance Bob, who was not originally expected,
and introduced him to all her other acquaintances. Prove that
the new group is also perfect.

6. A convex hexagon is given. Let $ s$ be the sum of the lengths of
the three segments connecting the midpoints of its opposite sides.
Prove that there is a point in the hexagon such that the sum of its distances
to the lines containing the sides of the hexagon does not exceed $ s$.

:D Two combinatorics. Looking on the problems, I think only a very intelligent person can get 42 points.

Thanks to [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=4906]Kęstutis Česnavičius, Lithuania[/url] for proposing a problem! Yeah!  :)  And for the proposer countries:

1. Russia
2. Austria
3. Kęstutis Česnavičius, Lithuania
4. South Korea
5. France
6. Russia

[quote="Yuriy Solovyov"]So, I think today will be

4) Combinatorics
5) Functional equation
6) ... Maybe Hard geometry[/quote]

You are right :blush:

Ok so these are the final results of our little game

uglysolutions: 1 point (P4)
hendrata01: 0 points
Albanian Eagle: 1 point (P2)
TomciO: 2 points (P2, P4)

Congrats TomciO!

Isn't it interesting that all of our points are on Algebra problems?

[quote="Fedor Petrov"]4. A set $ X$ of positive integers is called {\it nice} if for each pair
$ a$, $ b\in X$ exactly one of the numbers $ a \plus{} b$ and $ |a \minus{} b|$ belongs to
$ X$ (the numbers $ a$ and $ b$ may be equal). Determine the number
of nice sets containing the number 2008.

5. A group of persons is called {\it good} if its members can be distributed
to several rooms so that nobody is acquainted with any person in the same room
but it is possible to choose a person from each room so that all
the chosen persons are acquainted with each other.

   A group is called {\it perfect} if it is good and every set of its members
is also good.

   A perfect group planned a party. However one of its members, Alice,
brought here acquaintance Bob, who was not originally expected,
and introduced him to all her other acquaintances. Prove that
the new group is also perfect.

6. A convex hexagon is given. Let $ s$ be the sum of the lengths of
the three segments connecting the midpoints of its opposite sides.
Prove that there is a point in the hexagon such that the sum of its distances
to the lines containing the sides of the hexagon does not exceed $ s$.[/quote]

What is this?

[quote="Fedor Petrov"]4. A set $ X$ of positive integers is called {\it nice} if for each pair
$ a$, $ b\in X$ exactly one of the numbers $ a \plus{} b$ and $ |a \minus{} b|$ belongs to
$ X$ (the numbers $ a$ and $ b$ may be equal). Determine the number
of nice sets containing the number 2008.

5. A group of persons is called {\it good} if its members can be distributed
to several rooms so that nobody is acquainted with any person in the same room
but it is possible to choose a person from each room so that all
the chosen persons are acquainted with each other.

   A group is called {\it perfect} if it is good and every set of its members
is also good.

   A perfect group planned a party. However one of its members, Alice,
brought here acquaintance Bob, who was not originally expected,
and introduced him to all her other acquaintances. Prove that
the new group is also perfect.

6. A convex hexagon is given. Let $ s$ be the sum of the lengths of
the three segments connecting the midpoints of its opposite sides.
Prove that there is a point in the hexagon such that the sum of its distances
to the lines containing the sides of the hexagon does not exceed $ s$.[/quote]

 :wow:  :read:  :wow:  :spam:  :spam:  :spam:

Whaw  :rotfl:

:spider: :D  Strange to post those problems ;)

I have to apologize for the joke. These three are from Tuymaada 2008.

[color=green]I found a picture where it says which problems from the shortlist were chosen for the test.
http://www.imo-2008.es/album12/eng/slides/P1010211.html
1st day: G1 - A2 - N6
2nd day: A1 - C4 - G7

I also found one where it says "Beauty Contest" ( http://www.imo-2008.es/album12/eng/slides/P1010176.html )
do you know anything about that?
[/color]

Fedor Petrov wrote:
4. A set of positive integers is called {\it nice} if for each pair
, exactly one of the numbers and belongs to
(the numbers and may be equal). Determine the number
of nice sets containing the number 2008.

5. A group of persons is called {\it good} if its members can be distributed
to several rooms so that nobody is acquainted with any person in the same room
but it is possible to choose a person from each room so that all
the chosen persons are acquainted with each other.

A group is called {\it perfect} if it is good and every set of its members
is also good.

A perfect group planned a party. However one of its members, Alice,
brought here acquaintance Bob, who was not originally expected,
and introduced him to all her other acquaintances. Prove that
the new group is also perfect.

6. A convex hexagon is given. Let be the sum of the lengths of
the three segments connecting the midpoints of its opposite sides.
Prove that there is a point in the hexagon such that the sum of its distances
to the lines containing the sides of the hexagon does not exceed .

Can somebody solve these problems from Tuymaada 2008.

Thecool, create specific topics and discuss the problems there. I know the solutions of all three, though I did not succeed in solving all of them by myself...


vector
induction
analytic geometry
linear algebra
linear algebra unsolved

This is a generalization of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105991]this problem[/url]:
Suppose $V\subset\mathbb R^{n}$ is an affine space of dimension $k$ on $\mathbb R$; Also assume that $S_{1},S_{2},\dots,S_{n}$ are subsets of $\mathbb R$ that $|S_{1}|\geq|S_{2}|\geq\dots\geq|S_{n}|$. Prove that \[|V\cap S_{1}\times S_{2}\times\dots\times S_{n}|\leq \prod_{i=1}^{k}|S_{i}|\]

I'm going to assume that $t=n$, because otherwise I don't really know what $S_{1}\times\ldots\times S_{t}$ means :?. Also, denote $S_{1}\times\ldots\times S_{n}$ by $\mathcal S$.

It is more convenient to prove it for affine spaces instead of vector spaces, and we do this by induction on $n+k$. The initial cases, when $n+k$ is small ($\le 2$, for example), are easy to take care of, so asume the assertion is proven for values smaller than our current $n+k$.

Let $S_{1}=\{a_{1},\ldots,a_{u}\}, u=|S_{1}|$, and consider the affine hyperplanes $\mathcal H_{i},\ i=\overline{1,u}$ defined by the equations $x_{1}=a_{i}$ respectively ($x_{1},\ldots,x_{n}$ are the coordinates in $\mathbb R^{n}$).

If the affine space $V$ is contained in one of the $\mathcal H_{i}$, which we may assume WLOG is $\mathcal H_{1}$, then we can assume that $n$ is one unit smaller and we are working only with the sets $S_{2},\ldots,S_{n}$. It follows from the induction hypothesis that $|V\cap\mathcal S|\le\prod_{i=1}^{k}|S_{i+1}|\le\prod_{i=1}^{k}|S_{i}|$.

If $V$ is not contained in any of the $\mathcal H_{i}$, then its intersection with each $\mathcal H_{i}$ is a $k-1$-dimensional affine subspace of $\mathcal H_{i}$. From the induction hypothesis applied in each $\mathcal H_{i}$ to the sets $S_{2},\ldots,S_{n}$, it follows that each $V\cap\mathcal H_{i}\cap\mathcal S$ has at most $\prod_{i=2}^{k}|S_{i}|$ elements. Since there are $|S_{1}|$ such sets and together they cover $V\cap\mathcal S$, we get the desired result.

You're right I corrected it and replaced [i]vector space[/i] by [i]affine space[/i].
 P.S. My solution is the same as yours.


Given $ a,b,c >0$ and $ a^5\plus{}b^5\plus{}c^5\equal{}3$.Prove that:
$ \frac{a^3\plus{}abc}{(b\plus{}c)^2}\plus{}\frac{b^3\plus{}abc}{(c\plus{}a)^2}\plus{}\frac{c^3\plus{}abc}{(a\plus{}b)^2} \ge\ \frac{3}{2}$
 :)

[quote="nguoivn"]My brother, I think you shouldn't creat an ineq which based on a unsolve problem  :wink:
After use Voni-Schur, we obtain a "unsolve problem" . Can you solve it?[/quote]
but arqady solved it :)

Arqady, can you give me the link of your proof for:
$ \frac {a^2}{b \plus{} c} \plus{} \frac {b^2}{c \plus{} a} \plus{} \frac {c^2}{a \plus{} b} \geq\ \frac {3}{2}. \sqrt [5]{\frac {a^5 \plus{} b^5 \plus{} c^5}{3}}$
Indeed, I haven't had any proof for it yet :(
Thank you very much, my friend.


Two trains leave from New York and Seattle at the same time. If the distance between New York and Seattle is 3720 miles and one train is traveling 30 miles per hour and the other train is traveling 90 miles per hour, how long, in hours, will it take for the trains to cross each other?

How would you do this problem?

[hide]Together, they travel 120 miles per hour. $3720\div120=31\: \text{hours.}$[/hide]


Where can I find the BAMM tests of ingenuity for 2003-2006?
Thanks,
bpms

1994-2002: [url]http://bamm.usfca.edu/toi.html[/url]

2005: [url]http://usfca.edu/math/bamm/home2005b.html[/url]

Um, I already had seen those, that's why I was asking for 03-06...