let n is a positive interger. n>1, n doesnot divide by 5.
         Prove : (3^n +2^n) doesnot divide by n.

It is very easy .
A hint
Consider the least prime divisor of n.


Euler
limit
function
calculus
logarithms
derivative
real analysis

find the limit

${\lim_{n\to \infty }\, \left(\frac{n!}{\sqrt{2 \pi n}\left(\frac{n}{e}\right)^{n}}+\frac{11}{12 n}\right)^{n}}$

the result it's not too obvioulsy but simple

First note that Stirling's approximation may be written as:

$n! = \sqrt{2 \pi n}\left(\frac{n}{e}\right)^{n}(1+\frac{1}{12 n}+\frac{1}{288 n^{2}}+\cdots)$

Then your limit may be rewritten as:

$\lim_{n \to \infty}\left( 1+\frac{1}{12 n}+\frac{1}{288 n^{2}}+\cdots+\frac{11}{12 n}\right)^{n}$

$=\lim_{n \to \infty}\left( 1+\frac{1}{n}+\frac{1}{288 n^{2}}+\cdots \right)^{n}$

$= e$.

ou bien?

what a little obivious that i used stirling approximation  :blush:  hehehe

but you solved it rigth

I was wondering if it every true that

If ${\lim_{n \to \infty }\, f(n)=1}$

So ${\lim_{n\to \infty }\, \left(f(n)+\frac{1}{n}\right)^{n}=e^{c_{1}}}$,
or in another words, this limit exist for all function $f$ and it's exponential

$\lim_{n\to\infty}f(n)=1$ isn't good enough to claim what you're trying to claim. Instead, you would need $(f(n)-1)=o(n^{-1})$ as $n\to\infty.$ In other words, $\lim_{n\to\infty}n(f(n)-1)=0.$

Just let $f(n)=1+\frac1{\sqrt{n}}$ to see why you need that.

in another words, "f-1 need to reach zero before 1/n "?
hm...

${f(n)=1+\frac{1}{n}}$
$\lim_{n\to\infty}n(f(n)-1)=1$

i think we just need to assume that the limit $\lim_{n\to\infty}n(f(n)-1)$ exist...

Kent. look at this

$f$ is such as:
${\lim_{n\to \infty }\, (f(n))=1}$
${\lim_{n\to \infty }\, (n (f(n)-1))=L}$  - Exist

So
${\lim_{n\to \infty }\, \left(f(n)+\frac{1}{n}\right)^{n}= \text{Exp}\left(1+\lim_{n\to \infty }\, (n (f(n)-1))\right)}$

i solved this using induction. but for a lot of complicate examples it hold. i dont know how to proove or disproove.

but is interesting, why some (or all f) have this form?

That $+\frac1n$ that you've been including all along is a confusing waste of space, and should simply be folded into your $f(n).$ The confusion surrounding that has prevented you from stating the theorem efficiently, and is partly to blame for my less-than-comprehensive initial response to you.

Let's state this correctly:

If $\lim_{n\to\infty}n(f(n)-1)=L,$ then $\lim_{n\to\infty}[f(n)]^{n}=e^{L}.$

This can be expressed better in little-o notation:

$\lim_{n\to\infty}\left(1+\frac{L}{n}+o\left(\frac1n\right)\right)^{n}=e^{L}.$

The second statement means [i]exactly[/i] the same thing as the first statement.

The standard advice given in a calculus class for dealing with with a $1^{\infty}$ indeterminate case (or $0^{0},$ or $\infty^{0}$) is to take the logarithm. Since $f(n)=1+\frac{L}{n}+o\left(\frac1n\right)$ tends to $1,$ it will eventually be positive. We will only bother taking logarithms after that eventuality. Note also that $\lim_{x\to0}\frac{\ln(1+x)}{x}=x+o(x)$ as $x\to 0.$ This is simply a restatement of the fact that $g(x)=\ln(1+x)$ is differentiable at zero with derivative 1.

$\ln\left[\left(1+\frac{L}{n}+o\left(\frac1n\right)\right)^{n}\right]= n\ln\left(1+\frac{L}{n}+o\left(\frac1n\right)\right)$

$=n\left(\frac{L}{n}+o\left(\frac1n\right)+o\left(\frac1n\right)\right) =n\left(\frac{L}{n}+o\left(\frac1n\right)\right)$

$=L+o(1),$ meaning that it tends to $L$ as $n\to\infty$

Since the logarithm is a continuous function with a continuous inverse, that means that the original limit is $e^{L}.$

[quote="Kent Merryfield"]That $+\frac1n$ that you've been including all along is a confusing waste of space, and should simply be folded into your $f(n).$
[/quote]

i dont agree. you wasted 2 lines saying nothing just because of that. (paradox)

I just put the 1/n to be obviously that it is inspirated in Euler (esthetical purpose)

${\lim_{n\to \infty }\, \left(1+\frac{1}{n}\right)^{n}}= e$

i think my way is more elegant  :D

[quote]Let's state this correctly: [/quote]

If my way is wrong your way is too. because is the same thing.
It's the same thing that " a+a= 2a but b+b is not 2 b, because you used b insted of a"

but to solve more easily we can hide it, to not waste paper or bits

if you get confused just because a little (1/n) is not my falt. i think my way is more elegant, if you dont agree with me dont blame.

i'm not good with little o and big O notation. so...

but thaks for your inexpressive solution, it's nice  :lol:  i liked it


function
algebra
polynomial
LaTeX
partial fractions
geometric series

Wondering what people think of this problem:

The fraction $ \dfrac{2}{x^2\minus{}3x\plus{}2}$ can be written as an infinite series.  Find the sum of the first four terms of the series expansion for $ x\equal{}\minus{}1$ and $ x\equal{}\minus{}2$

Probably a nicer way...

[hide="Solution"]
$ \frac{a}{1\minus{}r}$ is the formula for infinite geometric series...$ a\equal{}2$ and $ r\equal{}\minus{}x^2\plus{}3x\minus{}1$, which is the common ratio.

If $ x\equal{}\minus{}1$, then this is $ \minus{}1\minus{}3\minus{}1\equal{}\minus{}5$, so $ 2\plus{}2(\minus{}5)\plus{}2(\minus{}5)^2\plus{}2(\minus{}5)^3\equal{}\minus{}208$.

If $ x\equal{}\minus{}2$, then this is $ \minus{}4\minus{}6\minus{}1\equal{}\minus{}11$, so $ 2\plus{}2(\minus{}11)\plus{}2(\minus{}11)^2\plus{}2(\minus{}11)^3\equal{}\minus{}2440$.
[/hide]

Right, but that formula only works for $ \minus{}1<r<1$ right?  At both of those cases $ r$ isn't within those values...

Does anyone think the wording is sort of ambiguous?

Erm...if r isn't within those values, the sum would be either positive or negative infinity (or zero, if a is zero).

Hi. Using partial fractions, or by inspection,

$ \frac{2}{x^{2}\minus{}3x\plus{}2} \equal{} \frac{2}{(x\minus{}1)(x\minus{}2)} \equal{} \frac{2}{x\minus{}2} \minus{}\frac{2}{x\minus{}1}$

$ \equal{} 2(1\minus{}x)^{\minus{}1} \minus{} (1\minus{}\frac{x}{2})^{\minus{}1}$, valid for $ \minus{}1 < x < 1$,

$ \equal{} 2\plus{}2x\plus{}2x^{2} \plus{}2x^{3} \plus{} ... \minus{} (1\plus{} \frac{x}{2} \plus{} \frac{x^{2}}{4} \plus{} \frac{x^{3}}{8} \plus{} ...)$

$ \equal{} 1\plus{}\frac{3x}{2} \plus{} \frac{7x^{2}}{4} \plus{} \frac{15x^{3}}{8} \plus{} ...$.

But this is not valid for $ x\equal{}\minus{}1$ or $ x\equal{}\minus{}2$

???

@ confused audience: read the question carefully. It's not the whole series, hence convergence has nothing to do with this problem. ;)

[quote="randomguy64"]Convergence has EVERYTHING to do with this problem.

The only way the question makes sense is if 2/(x^2-3x+2)=a+ar+ar^r+... and 0 < |r|< 1. If not, 2/(x^2-3x+2) must equal zero, and obviously x=-1 and x=-2 aren't roots, so that can't be it. Also, the fraction obviously doesn't form an infinite series, since there is absolutely no way to get the next term while the "x=-1 and x=-2" part makes sense.

@fishy: How does that work? More accurately, how did you get that formula out of the fraction given?[/quote]
[quote="Quickster94"]Find the sum of the first four terms of the series expansion for $ x \equal{} \minus{} 1$ and $ x \equal{} \minus{} 2$[/quote]
You absolutely do not need the concept of convergence to answer this question. Read the question, it's asking for the sum of the FIRST FOUR TERMS. Did I make myself clear?

[quote="10000th User"][quote="randomguy64"]quote[/quote]
You absolutely do not need the concept of convergence to answer this question. Read the question, it's asking for the sum of the FIRST FOUR TERMS. Did I make myself clear?[/quote]

How are you planning to find the first four terms without realizing that the infinite series is a converging one, and therefore use the formula for it?

@fishythefish: ignore my previous question; I figured it out.

@azjps: you're right...

[quote="azjps"]That point is moot because the question is vague as written, as an "infinite series" implies that the desired fraction is the convergence of the partial sums of a certain sequence, of which there are certainly many sequences that could apply (eg, one can easily make the first four terms sum to whatever one likes - what's wrong with, say, the infinite sequence $ \pi,e,\pi^e,5,f(x) \minus{} (\pi \plus{} e \plus{} \pi^e \plus{} 5),0,0,0,\ldots$?). I would presume that something like "power [Maclaurin] series" was intended, in which case see AndrewTom's post (though that series indeed does not converge).[/quote]You do have a valuable point and this is perhaps what randomguy64 may be trying to say in a not-so-clear way.

Using power series, we could define a new function $ f(x)$, where $ f$ is the first four terms of the power series $ p(x)$. Finding $ f(\minus{}1)$ and $ f(\minus{}2)$ is now a feasible thing.

Quickster94: if the question is from an old book, it is possibly implied that the infinite series 'converges wherever its base point might be at'. Power series is usually the choice for representing a real function and thus, you can follow AndrewTom's series expansions.

[quote="Quickster94"]Wondering what people think of this problem:

The fraction $ \dfrac{2}{x^2 \minus{} 3x \plus{} 2}$ can be written as an infinite series.  Find the sum of the first four terms of the series expansion for $ x \equal{} \minus{} 1$ and $ x \equal{} \minus{} 2$[/quote]

There is only one ambiguity in this question, and that is the nature of the "infinite series."  A function could have a variety of infinite series representations.
That said, however, it may be assumed that the problem is asking for the appropriate power series representation; i.e.,
$ f(x) \equal{} \sum_{k\equal{}0}^\infty a_k x^k$
for suitable real constants $ a_k$.
If we make this assumption, then the problem is completely unambiguous.  Since
$ x^2 \minus{} 3x \plus{} 2 \equal{} (x\minus{}1)(x\minus{}2)$
it follows that
$ \frac{2}{x^2 \minus{} 3x \plus{} 2} \equal{} 2\left(\frac{1}{x\minus{}2} \minus{} \frac{1}{x\minus{}1}\right) \equal{} \frac{2}{1\minus{}x} \minus{} \frac{1}{1\minus{}(x/2)}$.
Since
$ \frac{1}{1\minus{}x} \equal{} 1\plus{}x\plus{}x^2\plus{}\cdots \equal{} \sum_{k\equal{}0}^\infty x^k$,
we find
$ \frac{2}{x^2 \minus{} 3x \plus{} 2} \equal{} 2 \sum_{k\equal{}0}^\infty x^k \minus{} \sum_{k\equal{}0}^\infty (x/2)^k \equal{} \sum_{k\equal{}0}^\infty (2\minus{}2^{\minus{}k}) x^k$.
Thus as a formal power series the given function has coefficients of the form $ a_k \equal{} 2\minus{}2^{\minus{}k}$.  We are not concerned with convergence because we are not interested in evaluating the power series in its entirety.  All we are being asked to do is compute a partial sum of the formal power series:
$ g(x) \equal{} \sum_{k\equal{}0}^3 a_k x^k \equal{} \sum_{k\equal{}0}^3 (2\minus{}2^{\minus{}k}) x^k$.
It is a straightforward exercise to compute $ g(\minus{}1)$ and $ g(\minus{}2)$.  Again, it doesn't matter that the series is divergent for these values, because we are not trying to equate it to the value of $ 2/(x^2 \minus{} 3x \plus{} 2)$ at $ x \equal{} \minus{}1$ or $ x \equal{} \minus{}2$.

Debate over convergence aside, I'm still wondering what you all think of my solution. The question did not imply that we had to find the general expansion for the fraction, so I plugged in the values of $ x$ first, and then used infinite geometric series that I know to equal the results. After that, finding the first four terms is trivial.

BTW: Using this method, you can [b]make[/b] the series converge.  :P

[quote="Quickster94"]This is interesting

So it seems there's a lot of ambiguity about the question, which is what I thought.  Anyway, "official solution" is problem one of below link:

[url]http://www.vtmathcoalition.org/talent-search/TS0910Solutions1.pdf[/url]

Can anyone prove that this is the only solution or that more than one solution exists?
I feel like there must be more than one infinite series that sums to that in closed form...[/quote]

Well, as I already mentioned, there are any number of "infinite series" representations of the given function--Fourier series, Laurent series, power series...the particular series expansion that was employed in the solution is a type of power series.  One could see it as an expansion in powers of the reciprocal of x.

Their expansion is no more "correct" than the one I provided in my previous post.  If the question had originally asked for x=-2 only, then some justification could be made in that the series provided in their solution is convergent for such a value.  But at x = -1, neither power series is convergent.

It's a badly worded question, no doubt.

Can somebody explain how the infinite series in the "real solution" is equal to the fraction?

They arrive at it using polynomial long division.
I don't really have the LaTeX skills to show out the work for that, so if more explanation is needed, some1 else will have to step in :D


One-inch cubic blocks are used to build a solid $ 4^{\prime\prime} \times 4^{\prime\prime} \times 4^{\prime\prime}$ cube. The outside of the cube is painted blue. When the one-inch cubic blocks are counted, what is the ratio of the number of one-inch cubic blocks painted on three faces to the number of one-inch cubic blocks painted on two faces? Express your answer as a common fraction.

There are $ 12$ edges, each of which will be painted twice. There are $ 8$ corners, each of which will be painted three times. The ratio is $ \frac8{12}\equal{}\boxed{\frac23}$.

[hide=Solution]To find the number of painted cubes in the cube, we can do
\[4^3-(4-2)^3=56\]
[hide=Why?]There are $4^3$ cubes in total, and the innermost cube will be unpainted. There will be a 1 cube border on each side, so the side of that cube is 4-2=2. The volume is therefore $(4-2)^3$[/hide]
We know that there are $(4-2)^2*6=24$ 1 by 1 cubes. (1 by 1 cubes make up the center square of each face, and there are 6 faces.)
This means there are 32 cubes that have paint on either 2 or 3 sides. The corners will be painted three times, and there are 8 corners, hence 24 cubes with paint on 2 sides.

Our desired ratio is $\frac{8}{24}=\frac{1}{3}$.[/hide]


analytic geometry
geometry
trapezoid
vector
linear algebra

let A B C D be 4 points in R3 which are not coplanar. Let M be the midpoint of AC, N be the midpoint of BD.
if BM.MD=0 (dot product) and AN.NC=0
show that |AC|=|BD|

I was planning to show that |ABxCD|=0 (cross product),
which means AB || CD. and hence |AC| = |BD|

but then the math is quite messy....I want to make sure I'm in the right track, or should I try other approach?

First, it's impossible for the segments AB and CD to be parallel, since that implies the four points are coplanar (it's given that they're not).
Second, proving that those two segments are parallel does not immediately imply that AC = BD; consider every non-isosceles trapezoid (maybe you manipulated the other conditions to arrive at this conclusion, but I can't see it right away).

This is a geometry problem. What do the zero dot products tell you about those pairs of segments? Then look at triangles ANC and BMD.

This is easy for its appearance, Vector is very useful for such a poof.

Let $ \overrightarrow{AB} \equal{} \overrightarrow{b}, \overrightarrow{AC} \equal{} \overrightarrow{c}, \overrightarrow{AD} \equal{} \overrightarrow{d}$, we have

$ \overrightarrow{BM}\cdot \overrightarrow{MD} \equal{} (\overrightarrow{AM} \minus{} \overrightarrow{AB})\cdot (\overrightarrow{AD} \minus{} \overrightarrow{AM})\equal{} \left(\frac 12\overrightarrow{c} \minus{} \overrightarrow{b}\right)\cdot \left(\overrightarrow{d} \minus{} \frac 12\overrightarrow{c}\right) \equal{} 0$

$ \Longleftrightarrow 2\overrightarrow{c}\cdot \overrightarrow{d} \minus{} |\overrightarrow{c}|^2 \minus{} 4\overrightarrow{b}\cdot \overrightarrow{d} \plus{} 2\overrightarrow{b}\cdot \overrightarrow{c} \equal{} 0\ \cdots [1]$

Similarly, $ \overrightarrow{AN}\cdot \overrightarrow{NC} \equal{} 0\Longleftrightarrow 2\overrightarrow{b}\cdot \overrightarrow{c} \minus{} |\overrightarrow{b}|^2 \minus{} 2\overrightarrow{b}\cdot \overrightarrow{d} \plus{} 2\overrightarrow{c}\cdot \overrightarrow{d} \minus{} |\overrightarrow{d}|^2 \equal{} 0\ \cdots [2]$

$ \therefore |\overrightarrow{AC}|^2 \minus{} |\overrightarrow{BD}|^2 \equal{} |\overrightarrow{AC}|^2 \minus{} |\overrightarrow{AD} \minus{} \overrightarrow{AB}|^2 \equal{} |\overrightarrow{c}|^2 \minus{} |\overrightarrow{b}|^2 \minus{} |\overrightarrow{d}|^2 \plus{} 2\overrightarrow{b}\cdot \overrightarrow{d} \equal{} 0$, which is obtained by $ [2] \minus{} [1]$. Q.E.D.

I was just thinking that in right triangles ANC and BMD, the length of the median to the hypotenuses are half the lengths of those hypotenuses. So |MN| = |AC|/2 = |BD|/2.


Solve the equation: $2\cos^{4}x-2\cos 2x=\left(\sin x-\cos^{2}x\right)^{2}$

Let $t = \sin x$. Then \[\begin{aligned}\left(t^{2}+t-1\right)^{2}= \left(\sin x-\cos^{2}x\right)^{2}= 2\cos^{4}x-2\cos2x = 2\left(\cos^{4}-2\cos^{2}x+1\right) = 2\left(1-\cos^{2}x\right)^{2}= 2t^{4}\end{aligned}\]Therefore, $t^{2}+t-1 = \pm\sqrt{2}t^{2}$. But we know that $t \leq 1$, so $t^{2}+t-1 \leq t^{2}$. Therefore, $t^{2}+t-1 = \sqrt{2}t^{2}$ is impossible, so we know that $(1+\sqrt{2})t^{2}+t-1 = 0$. And yeah, just solve it as a quadratic. Doesn't seem to be a nice answer...


Who give me usaco traning ''contact'' problem all tests?(Please send zip file).

1. Wrong forum.
2. What are you talking about? (I do USACO, btw, so I'm not completely oblivious, I just don't understand your question)

usaco section 2.1 ''contact'' problem test data's

those are so big to paste and no time to make zip!
 :(


ive decided that i suck at posting in my blog so im allowing my contributors to take over for a bit.i am opening 3 spots in my contributing list and if you would like to join in go to my blog and read the instructions

how do u contribute like add stuff? i guess i will


inequalities
calculus
integration
inequalities unsolved

Let $x,y$ be positive real numbers, $n,m$ be posistive integers. Prove that
 $(n-1)(m-1)(x^{n+m}+y^{n+m}) +(n+m-1)(x^ny^m+x^my^n) \geq mn(x^{n+m-1}y+xy^{n+m-1}) $.
 it was posted but i haven't seen a solution. :P
who can find out   :roll:

What is
[quote="harazi"]Chebyshev integral inequality.[/quote]
?

[quote]I have already posted solution using Chebyshev integral inequality.[/quote]
 Woo, please harazi ;) . I'll try to find a nice solution for this one.


geometry
incenter
vector
trigonometry
circumcircle
complex numbers
romania

Another one which can be solved with polars and poles, huh? :)

$N$ is on $BC$ and $B'C'$, which are the polars of $A'$ and $A$ respectively (wrt the incircle). This means that $AA'$ is the polar of $N$. In the same way we find that $BB'$ is the polar of $M$, so $P=AA'\cap BB'$ must be the pole of $MN$, so $PI\perp MN$ (it's well-known that the polar of a point wrt a circle is perpendicular to the line joining the point to the center of the circle).

This time the solution is suitable, because the problem was proposed to the seniors. :)

I knew that you would solve it in a jippy with polars :D ;) andrei negut did it in the same way.

I think we can do it by vectors. Choose any point as the origin, the vector IP
will be proporional to u*OA+v*OB+w*OC, MN to (v-w)OA+(w-u)OB+(u-v)OC
u-a-b,v=b-c,w=c-a.
By multiplying them you must get 0.
But still, does there exist a pure geometrical solution?

[quote="Valentin Vornicu"]The problem is taken from Kvant, where a pole-polar solution was given (I think). However, almost all the contestants who solved it, used complex numbers or vectors and a lot of computations :D[/quote]

Almost is well said. I didn't solve it during the contest. I'm such an idiot not computing it with complex numbers. It's really not that hard. It's NOTHING SMART in computing but you get 7 points for it. I was afraid (in the contest) that I will run out of time if I start computing. But when I saw the results of Rom TST I was very very very ... sad (and Valentin can confirm this) because I didn't have the courage to compute. Well that's that. There is allways next year  :(   :(   :( (No there isn't  :)   :(  :( )
P.S. A piece of advice :
When you see a problem which can be computed with vectors, complex numbers etc. DO IT otherwise you MIGHT regret that you didn't LATER.

yes, that is a well-known fact, at least for some of us ;) :D

Please iura, would you explain why the vector $\vec{IP}$ is proportional to what you say?

[quote="Valentin Vornicu"]Let $I$ be the incenter of the non-isosceles triangle $ABC$ and let $A',B',C'$ be the tangency points of the incircle with the sides $BC,CA,AB$ respectively. The lines $AA'$ and $BB'$ intersect in $P$, the lines $AC$ and $A'C'$ in $M$ and the lines $B'C'$ and $BC$ intersect in $N$. Prove that the lines $IP$ and $MN$ are perpendicular.[/quote]

If you do not like polar map. You can also invert with the inccircle. More interesting fact: the two Soddy points are on IP (so this is called the Soddy line) and is perpendicular to the Georgon (spelling?) line. Use nice inversions solve all this.  (I have a short written piece on this.) The computational proofs of this also appeared on American Math Monthly.

[quote="sbos"]I have that $\vec{MN}$ is collinear with $(b-c)OA+(c-a)OB+(a-b)OC$ and
$\vec{IP}$ is collinear with $\sum[(b+c)(a+c-b)(a+b-c)-2a^2(b+c-a)]OA$, but I can't see why do we have $MN\cdot IP=0$. Can someone help me, please, I'm not very common with vectors?[/quote]

sbos, how did you find those formulas?

I used barycentrics, got the first one right, but the second one is very messy... :(

I think that a straightforward check of $\overrightarrow{MN} \cdot \overrightarrow{IP}=0$ works, i.e. take $O=C$, expand everything, use the cosine theorem etc.; now it's just a potentially monstrous identity to check (if you have some computer algebra software, you can do it; otherwise i think it would a waste of time). Perhaps others can find a nicer approach to proving this...

Related question: Say $\overrightarrow{m}=x\overrightarrow{OA}+y\overrightarrow{OB}+z\overrightarrow{OC}$ and $\overrightarrow{n}=t\overrightarrow{OA}+u\overrightarrow{OB}+v\overrightarrow{OC}$. Are there any necessary/sufficient symmetric conditions on $x,y,z,t,u,v$ in order to have $\overrightarrow{m} \cdot \overrightarrow{n}=0$?  [b] [edit nr. 2] [/b] I have found in Paul Yiu's "Introduction to Triangle Geometry" a way to do this: consider the barycentrics of the infinite points of $IP$ and $MN$; then check that $\sum S_A f f' = 0$, where $S (f:g:h)$ and $T(f':g':h')$ are the infinite points I was talking about, and $S_A = \frac12 \left( b^2+c^2-a^2 \right)$ etc; thus, the problem reduces to proving $\sum f(a,b,c) =0$ (I can't find the paper where I wrote $f$ right now).

[b][edit nr. 1][/b] just wondering: is there really a solution using only vectors or complex numbers? It seems impossible to get the perpendicularity from what I obtain :(

[b]I note the points [/b]$X\in AA'\cap IN,\ Y\in BB'\cap IM$. [b]From the wellknown relations [/b]
$IN\perp AA',\ IM\perp BB'$ [b]results [/b]$IX\cdot IN=IY\cdot IM=r^2$, [b]i.e. the line [/b]$MN$
[b]is the image of the circumcircle with the diameter [/b]$[IP]$ [b]through the inversion
with the pole I and the constant $r^2$. Therefore, [/b]$IP\perp MN.$

The lines A'A passes through the intersection A of the tangents CA, AB to the circumcircle (I) of the contact triangle $\triangle A'B'C'$ at the vertices B', C', hence, it is its A'-symmedian. Likewise, B'B is its B'-symmedian, C'C its C'-symmedian and the 3 symmedians concur at its symmedian point P. The tangent CA to the circumcircle (I) of the $\triangle A'B'C'$ at the vertex B' meets the opposite side C'A' at the center M of its B'-Apollonius circle (M), the tangent BC at the vertex A' meets the opposite side B'C' at the center N of its A'-Apollonius circle (N) and the tangent AB at the vertex C' meets the opposite side A'B' at the center K of its C'-Apollonius circle (K). The B'- and A'-Apollonius circles (M), (N) meet at 2 points $Q_1, Q_2$ and by transitivity of equivalence, the C'-Apollonius circle (K) also passes through these 2 points. Hence, their centers M, N, K are collinear and $Q_1Q_2 \perp MN$ is their common radical axis. The circumcenter $I \in Q_1Q_2$ of the contact  triangle $\triangle A'B'C'$ lies on this radical axis, because the circumcircle (I) is perpendicular to all 3 Apollonius circles (M), (N), (K). Let the A'-symmedian meet the circle (I) again at A''. Then $A''A \equiv A'A$ is the A''-symmedian of the triangle $\triangle A''B'C'$ and the triangles $\triangle A'B'C', \triangle A''B'C$ also have the common circumcircle (I) and the A'-, A''-Apollonius circle (N). Hence, the common symmedian $A'A \equiv A'A''$ is the radical axis of the circles (I), (N). Likewise, the symmedians B'B and C'C are radical axes of the circle pairs (I), (M) and (I), (K). By transitivity of equivalence, the power of the symmedian point P to all 3 Apollonius circles is the same as the power of this point to the circle (I), hence, $P \in Q_1Q_2$ lies on their radical axis. As a result, $IP \equiv Q_1Q_2 \perp MN$. The line IP is the Brocard axis of the triangle $\triangle A'B'C'$.

Dear Armpist and Mathlinkers,
I discovered the problem now...
ABC and A'B'C' are perspective and bilogic. Accordling to Sondat's theorem, we are done...
See for example: httrp://perso.orange.fr/jl.ayme    vol. 1  Le theorem de Sondat
Thank you again, dear Armpist...
Sincerely
Jean-Louis

Let AA’ and BB’ intersect 2nd time the circle I at Q and R respectively; since the tangents at B’ and C’ concur on AQ, A’B’QC’ is harmonic, therefore the tangents at A’ and Q intersect on B’C’, hence at N. similarly, AC, A’C’ and the tangent at R concur at M.
Now, I has equal power to the circles (M, MB’) and (N,NA’) – IB’ and IA’ are tangents to these circles, but they are inradii of DABC, and the same for P, since PA’.PQ = PB’.PR (from the power of P to (A’B’C’), hence PI is the radical axis of the 2 circles centered at M and N, consequently PI and MN are perpendicular.

Best regards,
sunken rock

Here my solution:
Let $ AA' \cap (I)=A'' $ and $ BB' \cap (I)=B'' $.We know that $ A''A' $ and $ B''B' $ are polars of $ M $ and $ N $ , respectively.If $ B''A'' \cap A'B'=T $ and $ A'B'' \cap A''B'=S $ , then $ {I,P,T,S} $ is orthocentric system.Now, apply Pascal's theorem to non-convex hexagon $ A''B''A'A'B'A'' $ , we get $ A'A' \cap A''A''=N $,$ B''A' \cap A''B'=S $ and $ B'A' \cap B''A''=T $ are collinear.Similarly, we have $ M \in ST $.So $ IP \perp ST=MN $.DONE !

[quote=Valentin Vornicu]The problem is taken from Kvant, where a pole-polar solution was given (I think). However, almost all the contestants who solved it, used complex numbers or vectors and a lot of computations :D[/quote]

Can you tell me exactly which Kvant have you taken this question? I mean  which year's Kvant magazine?

[hide=Complex bash]WLOG incircle is unit circle. Let $AA'\cap \text{incircle}=K$ and $L=BB'\cap \text{incircle}$.Let $T=B'C'\cap BC=T$. Using the intersection of chords formula we get $t=\frac{a'^2(b'+c')-2a'b'c'}{a'^2-b'c'}=\frac{k^2(b'+c')-2kb'c'}{k^2-b'c'}$ so we get $k=\frac{2b'c'-a'(b'+c')}{b'+c'}$. Doing the same for $L$ we get $P=\frac{2(a'b'c'-a'b'(a'+b')-c'(a'^2+b'^2)}{a'b'(a'+b')-c'a'(c'+a')-c'b'(c'+b')}$.It's easy to get $m=\frac{b'^2(a'+c')-2a'b'c}{b'^2-a'c'}$ and $n=\frac{a'^2(b'+c')-2a'b'c}{a'^2-b'c'}$(since $M=B'B'\cap A'C'$).And $i=0$. So it's easy to check that $IP\perp MN$

Dear Mathlinkers,
also at

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=383421

Sincerely
Jean-Louis

Poles and Polars are back!
[quote=Valentin Vornicu]Let $I$ be the incenter of the non-isosceles triangle $ABC$ and let $A',B',C'$ be the tangency points of the incircle with the sides $BC,CA,AB$ respectively. The lines $AA'$ and $BB'$ intersect in $P$, the lines $AC$ and $A'C'$ in $M$ and the lines $B'C'$ and $BC$ intersect in $N$. Prove that the lines $IP$ and $MN$ are perpendicular.

[i]Alternative formulation.[/i] The incircle of a non-isosceles triangle $ABC$ has center $I$ and touches the sides $BC$, $CA$ and $AB$ in $A^{\prime}$, $B^{\prime}$ and $C^{\prime}$, respectively. The lines $AA^{\prime}$ and $BB^{\prime}$ intersect in $P$, the lines $AC$ and $A^{\prime}C^{\prime}$ intersect in $M$, and the lines $BC$ and $B^{\prime}C^{\prime}$ intersect in $N$. Prove that the lines $IP$ and $MN$ are perpendicular.[/quote]

Say $AA' \cap \odot(I)=A'',BB' \cap \odot(I)=B''$
Clearly $A''B'A'C',B''A'B'C'$ are harmonic.
So the tangents to the incircle through $A',A''$ meet at $N$,So $AA'$ is the polar of $N$ and similarly $BB'$ is the polar of $M$,by La Hire $MN$ is the polar of $AA' \cap BB'=P$ w.r.t $\odot(I)$ so $IP \perp MN$ $\blacksquare$

We use poles and polars wrt to the incircle
Note that: $a \equiv B'C' \implies a \equiv N$ by La Hire’s Theorem $n \equiv A \implies P \in n \implies p \equiv N$
Similarly $P \in m \implies p \equiv M$
Then: $MN \equiv p \implies IP \perp MN$.$\blacksquare$


Find all positive integers $a$ and $b$ such that
\[ {a^2+b\over b^2-a}\quad\mbox{and}\quad{b^2+a\over a^2-b} \]
are both integers.

[quote="barcelona"]From $ \frac{a^{2}+b}{b^{2}-a} $,we get:$a^2+b\geb^2-a => b^2-b-a^2-a\le0$
$D=1+4a^2+4a=(2a+1)^2$ which means that :$b\in[1;a+1]$
From $\frac{b^{2}+a}{a^{2}-b} $,we get that $b^2+a\gea^2-b => b^2+b+a-a^2\ge0$
$D=1-4a+4a^2=(2a-1)^2$ which means that $b\in[a-1;\infty ]$.
So finaly we get $b\in[a-1;a+1]$
[b][u]1case:[/u][/b]
$b=a-1$.Here we get the solutions $(a,b)=\{(2,1),(3,2)\}$
[b][u]2case:[/u][/b]
$a=b$..Here we get the solutions $(a,b)=\{(2,2),(3,3)\}$
[b][u]3case:[/u][/b]
$b=a+1$..Here we get the solutions $(a,b)=\{(3,2),(3,4)\}$[/quote]

[quote="ophiophagous"]

WLOG $a \ge b$

$b^2 + a \ge a^2 - b$

$a^2 - b^2 = (a+b)(a-b) \le a+b$

$a-b \le 1$

and the rest follows[/quote]

[quote="barcelona"]
$b=a+1$..Here we get the solutions $(a,b)=\{(3,2),(3,4)\}$[/quote]
$(3, 4)$ is [b]not[/b] a solution!

What's wrong with this solution?

[hide="huh"]

Notice that if $\frac{x}{y}$ is a positive integer for expressions $x, y$ then we have $y \le x$. By this logic, we can write

$b^2-a \le a^2+b$

$b^2-a^2 = (b+a)(b-a) \le b+a$

It's obvious that $b+a$ is nonzero since the problem tells us that $a, b$ are positive integers. Then we can divide both sides by $b+a$ and we get $b \le a$.

Similarly, we have

$a^2-b \le b^2+a$

$a^2-b^2 = (a+b)(a-b) \le a+b$

$a \le b$

Since we got $b \le a$ and $a \le b$ to be simultaneously true, the only way in which this can occur is if $a=b$. We then must have

$\frac{a^2+a}{a^2-a}$

is an integer. Since $a$ is nonzero, this is equivalent to saying that

$\frac{a+1}{a-1} =1 + \frac{2}{a-1}$ is an integer. Hence, $a=2$ or $a=3$, and it follows that the only solutions are $\boxed{(2,2),(3,3)}$.

Where did I go wrong?

[/hide]

EDIT: Oops I forgot to account for the fact that the given expressions didn't necessarily have to be positive. I get it now :)

Here's another solution:

Let first WLOG be $a \ge b$. Because of $a^{2} - b | b^{2} + a$
we have that $a^{2} - b \le b^{2} + a$ so $a^{2} - a \le b^{2} + b \le b^{2} + a$,
or equivalently $a^{2} - 2a \le b^{2}$ so either $a^{2} - 2a = b^{2}$ or $(a-1)^{2} \le b^{2} \le a^{2} $
So we have three cases:
$1) a = b$ which implies $a^{2} - a | a^{2} + a$ and also $a^{2} - a | 2a$ so $a \le 3$ which gives the solutions $(2,2)$ and $(3,3)$.
$2) b = a - 1$ which implies $a^{2} - 3a + 1 | a^{2} + a - 1$ so $a^{2} - 3a + 1 | 4a - 2$ so $1 \le a \le 6$, and checking these values
gives the solutions $(2,1)$, $(3,2)$ and also $(1,2)$ and $(2,3)$ by symmetry.
$3) b^{2} = a^{2} - 2a$ or $(b - a +1)*(b + a - 1) = -1$, but this equation has no solutions in natural numbers so we are done.

Here's a proof sketch that's similar to steppewolf's:

Use division bounds to get three cases: a=b+1, b=a+1 and a=b. First two cases are symmetric, so all solutions can be derived from the second and third cases alone.

Case 3: a=b.

Both expressions simplify to $\frac{a+1}{a-1}$, whose gcd must be either 1 or 2 by the euclidean alg. The former yields (2,2) and the latter yields (3,3).

Case 2: b=a+1. One of the expressions simplifies to 1, so ignore that. The other simplifies to $\frac{a^2+3a+1}{a^2-a-1} = 1+\frac{4a+2}{a^2-a-1}$. Use division bounds again to get that a=1,2,3,4. From this, just plug in until you get solutions of (2,1), (1,2), (2,3), (3,2). And you're done.

Since $a^2+b$ and $b^2+a$ are more than zero, we have $a^2+b\geq b^2-a$, and $b^2+a \geq a^2-b$. These cases reduce to $a \geq b-1$ and $b+1 \geq a$ respectively, so we see that $b+1 \geq a \geq b-1$. We now have three cases, $a=b-1$, $a=b$ and $a=b+1$ of which the latter case is the mirror of the first, so we essentially have two cases.
[b]Case One: $a=b-1$[/b] We see that the fractions reduce to $1$ and $\frac{b^2+b-1}{b^2-3b+1}$. Since $b^2+b-1 \geq b^2-3b+1$, we see that $4b-2 \geq b^2-3b+1$, yielding that $b \leq 6$. Now, we must simply check $b \in (1,2,3,4,5,6)$. It is easy to see through testing that the only two solution here are $(1,2)$ and $(2,3)$.
[b]Case Two: $a=b$[/b] We see that $\frac{b^2+b}{b^2-b}$ must be an integer, so $2b \geq b^2-b$, yielding $b \in (1,2,3)$, from which we see that $(2,2)$ and $(3,3)$ are solutions.

Hence, the only solution are $(1,2)$, $(2,3)$, $(2,1)$, $(3,2)$, $(2,2)$ and $(3,3)$ are the only solutions. $\square$

Because $ b^2 - a | a^2 + b $
so we have $ b^2 - a \le a^2 + b $
and then $ b \le a + 1 $
which means $ b = a + 1 $ or $ b = a $ so we can easily work it out

[quote=AopsUser101]We know that:
$$a^2+b \ge b^2-a \Longleftrightarrow a^2+a \ge b^2 -b \Longleftrightarrow \left(a + \frac{1}{2} \right)^2 \ge \left(b - \frac{1}{2} \right)^2 \Longleftrightarrow a + \frac{1}{2} \ge b - \frac{1}{2} \Longleftrightarrow a -b \ge -1$$$$b^2 + a \ge a^2 -b \Longleftrightarrow b^2 + b \ge a^2 -a \Longleftrightarrow \left(b + \frac{1}{2} \right)^2 \ge \left(a - \frac{1}{2} \right)^2 \Longleftrightarrow b + \frac{1}{2} \ge a- \frac{1}{2} \Longleftrightarrow a-b \le 1$$Case 1: $a = b + 1$
$$\frac{a^2+b}{b^2 -a} = \frac{b^2 + 3b + 1}{b^2 - b - 1} = 1 + \frac{4b + 2}{b^2-b-1}$$Note that when $b \ge 7$, $4b + 2 < b^2 -b -1$, so all we have to do is manually check all $b$ from $0$ to $6$, which gives solutions: $(a, b) = \boxed{(1, 2), (2, 3)}.$
Case 2: $a = b $
$$\frac{a^2+b}{b^2-a}=\frac{a^2+a}{a^2-a} = \frac{a+1}{a-1} = 1 + \frac{2}{a-1}$$This gives solutions $(a , b ) = \boxed{(2,2),(3,3)}$
Case 3: $b = a + 1$
$$\frac{b^2+a}{a^2-b} = \frac{a^2+3a+1}{a^2-a-1} = 1 + \frac{4a+2}{a^2-a-1}$$which by the same reasoning as Case 1, gives $(a , b) = \boxed{(2,1),(3,2)}$.[/quote]

Hey! I believe you forgot to account for $a=b$ since the inequalities are satisfied there as well. In my opinion the problem is basically trivialised once we consider inequalites since first one gives $b-a \leq 1$ and second one gives $a-b \leq 1$ so it is clear that we can only have $b=a+1$ or $b=a$ ( I am writing for $b$ only since it doesn't matter as the equations are symmetric). And then the equality case is clearly super easy while the other one gives an upper bound of $a \leq 5$ which is simple case analysis. Simple problem :).

Sincerely,
Aayam

[quote=Maths_1729]Observe equation is symmetric so $(a, b) =(b, a)$
$a^2+b\geq b^2-a$ gives $a-b\geq -1$ and
$b^2+a\geq a^2-b$ gives $a-b\leq 1$
Hence $-1\leq a-b \leq 1$

Case 1-: when $a-b=0$ we get $\frac{a+1}{a-1}\in N$ $\implies a=2, a=3$

Case 2-: when $a=b+1$ we get $\frac{b^2+3b+1}{b^2-b-1} \in N$ $\implies b=2, a=1$

Case 3 -: when $a=b-1$ we get $\frac{b^2+b-1}{b^2-3b-1} \in N$ $\implies b=3, a=2$[/quote]

nice one

[hide=not my solution, for storage][img]https://i.ibb.co/WPVh77V/APMO-2002-P2-SOLUTION.png[/img][/hide]


The values of $ a$, $ b$, $ c$ and $ d$ are 1, 2, 3 and 4, but not necessarily in that order. What is the largest possible value of the sum of the four products $ ab$, $ bc$, $ cd$ and $ da$?

Let $ a\equal{}1$.  Then, we need to maximize $ b\plus{}bc\plus{}cd\plus{}d\equal{}(c\plus{}1)(b\plus{}d)$.

If we want to make this product as large as possible let $ c\equal{}4$ so the product becomes $ (4\plus{}1)(2\plus{}3)\equal{}\boxed{25}$


AMC
AIME
USA(J)MO
USAMO
AMC 12
Duke
college

Sweet, first vote, first reply.
This is my second year doing the AMC competitions, last year (8th grade) I got to the AIME and promptly failed miserably with  a score of 1. But I did get a resonably good score on the AMC 10.
This year, I'm probably going to get to the AIME and die again. Go me.

I'm an 8th grader and I haven't gotten to the AIME yet.
I hope I can do the latter next week.

I took the USAMO last year as an 8th grader.

-Ravi

Made AIME for the second time this year, as a junior.  If I can make USAMO once in my life, I'll be ecstatic...

Btw, isn't it supposed to be MOSP rather than MOPS?

Oops. typo.  :blush:
Is there a way to edit polls?

I think so, but I've never done one, so I couldn't tell you how...

I cant find a way.  :(

anyway. I was kind of looking for a bell curve like figure that is centered around the AIME/AMC10/12 area.

You might get that, although it's hard to know for sure, because AoPS is a rather skewed sample...

There is no "S" in MOP.   :whistling:

(I went to MOP two years ago under the Freshman rule. It was awesome)

really? I always thought it was MOSP.  :maybe:

I assure you, and everyone that went to MOP will too, that there is no "S" in MOP.

(ok, it's officially known as "MOSP" but everyone that goes there calls it "MOP")

[quote="buzzer11"][quote="bpms"][quote="buzzer11"]Wondering?[/quote]
Isn't this poll favoring arnav?
It's funny when multiple people vote for the bottom option.[/quote]

arnav?[/quote]
Dude, do you not know who arnav is?

He accomplished the bottom option.

[quote="buzzer11"][quote="MysticTerminator"]I'm from Arnav's school. I'm not sure what he got as he ran away immediately after the contest ended :/. H...[/quote]

 :huh:

I thought he WAS Arnav.[/quote]

That was in a quote in Arnav's post.  It's in the AMC 12 discussion thread...

[quote="buzzer11"][quote="MysticTerminator"]I'm from Arnav's school. I'm not sure what he got as he ran away immediately after the contest ended :/. H...[/quote]

 :huh:

I thought he WAS Arnav.[/quote]
Actually, n^4+4 said that.

[quote="jasonwingofcarmelin"]what did arnav get on the amc12>[/quote]
i'm guessing perfect score

huh apparently I was never a n00b at any point in my life that is pretty darn tricky

everyone couldn't talk at one point in their life, wherever it was, including in their mother's stomach, right when they're born, etc.

so yeah everyones been a noob at one point, whatever that means.

arnav is very nubby sometimes, i mean come on

he's a unc fan and got banned in the NC forum

but he's no noob at math (other than the fact that he was 2nd 2 years in a row in the state math contest and never got a perfect Duke math meet score)

he pwns all amc-series tests though

....

I am usually very noobish though too

I'm guessing the highest score possible.

[quote="Karth"][quote="MysticTerminator"]eh I could take the NC forum any time I want

seriously now what

I'm pretty sure I did get a perfect at duke at some point

since I know I got a perfect at one of these ARML type things

and it certainly wasn't at ARML

and after beating anders I stopped caring about state and just sort of went to eat donuts[/quote]

eh, you shouldn't eat too many donuts... either that, or you should exercise (so that you could've actually attempted to tackle me at duke, heheh)[/quote]

...

soooo, you're going to HMMT, right?

Unfortunately, I can't, because I live in Georgia, and I have to drive like 800+ miles to HMMT. Not to mention that I think I'd have to miss school for it. And I have like 1238 AP's that I'm taking, so probably not.

You can still (try to, but probably not succeed) tackle me at ARML or something.


geometry
parameterization
calculus
calculus computations

I have 3 questions dealing with areas that I need help with.

1. Clara the cow has been tied to a circular silo with radius r by a rope just long enough to reach a point diametrically opposite to the point where she is tied. If she goes to the left side of the silo, she can stand far away from the silo, while at the right side, she can only graze right next to the silo.  We wish to compute the total area of the region upon which she can graze.  How far from the silo can the cow stand when she is to its left?

Assume that a very long rope is wound around the circular silo, and then unwound while being held taut.  The curve traced by the end of the rope is called the involute of the circle.  if the silo has radius r and center O, and if the parameter theta is chosen, show that the parametric equations of the involute are:
x = r(cos(theta) + theta sin(theta)
y = r(sin(theta) - theta cos(theta)

What is the area of the grazing region available to the cow?


Let a triangle ABC. $ D\in BC, E\in CA, F\in AB$. M,N,P be the midpoints of segments AD,BE,CF. Prove that S(DEF)=4S(MNP).

It is easy to brute force with coordinates. Let the vertices of ABC be (0,0) (1,0) and (a,b), and F,E, and D are (ma,mb), (n,0), and (p, $ \frac {b(p \minus{} a)}{a \minus{} 1}$). Then, you can caclulate M,N, and P and use the shoelace formula on both triangles. It is quite messy, but only takes about 5 minutes.

[quote="Wickedestjr"]What is S?[/quote]
I'm sure that S denotes the area.

S(DEF)=4S(MNP)

But wait, does it really matter what S is, because if you divide by S on both sides you get;

(DEF) = 4(MNP)


superior algebra
superior algebra unsolved

Let $ A$ be a commutative ring with an odd number of elements. We denote: $ M \equal{} U(A)$ and $ N \equal{} \left\{x\in A|x^2 \equal{} x\right\}$. Prove that $ |N|/|M|$.

[quote]
Since you write $ U(A)$, are we assuming that this ring has a $ 1$?[/quote]
Yes.


Andrew was born in the year $ 19xy$. In $ 2010$, $ y$ and $ x$ will be the tens and units digits of his age. What is $ x\plus{}y$?

$ 2010\minus{}19xy\equal{}yx$

the last digit of 2010 is 0.So 0 will be changed to 10 when subtracting(carry over)

So 10-y=x

So x+y=10

Andrew was born in the year 19xy . In 2010 ,  and  will be the tens and units digits of his age. What is ?
Let me try this:
 2010-(1900 + 10x +y)=10y+x
110-10x-y=10y + x
110=11(x+y)
x+y=10


inequalities
induction
inequalities unsolved

I don't see why you need induction... it obviously must hold for p,q in the rationals so substitute for the a_i a_i^1/q...
then subtract rhs to get a zero and difference, then factor or rearrangement should do...

[quote="sjgaudih"]I tried to solve this inequality by induction method, but I have hard time prooving the first step of that....can you prove this one without use muirhead inequality??

let $n>2, a_{1}>a_{2}>.....>a_{n}>0$ and $p>q>0$
show that
${a_{1}}^{p}{a_{2}}^{q}+{a_{2}}^{p}{a_{3}}^{q}+....+{a_{n}}^{p}{a_{1}}^{q}\geq{a_{1}}^{q}{a_{2}}^{p}+{a_{2}}^{q}{a_{3}}^{p}.....{a_{n}}^{q}{a_{1}}^{p}$[/quote][hide]Indeed, \[\{(a_{1}a_{2})^{q}, (a_{2}a_{3})^{q}, \ldots, (a_{n-1}a_{n})^{q},(a_{n}a_{1})^{q}\},\] is sorted the same way as \[\{a_{1}^{p-q}, a_{2}^{p-q}, \ldots, a_{n-1}^{p-q},a_{n}^{p-q}\}.\] Hence, the sum of the term-wise product is greater than or equal to the sum of the term-wise product of any permutation, by the Rearrangement Inequality. Note the RHS is one result of such a permutation, since \[a_{1}^{q}a_{2}^{p}=(a_{1}a_{2})^{q}\cdot a_{2}^{p-q},\] etc.[/hide]


Solve for $a$ in terms of $b$ if two more than twice $a$ is three less than the square of the number which is one less than $b$.

[quote="pascal_1623"][hide]
$2a+3=(b-1)^2$
$2a+3=b^2-2b+1$
$a=\dfrac{(((b^2)-2b)-2)}{2}=\boxed{\dfrac{b^2}{2}-b-1}$ I don't think that's right :? [/hide][/quote]

It was $2$ more than twice of $a$, not just twice $a$


Oh, that's handy. I've always just used \mathrm{mod}~m .

~ in math mode artificially inserts a space.

[quote="nr1337"]I know there are two ways to do mods in LaTeX.  I know one is [tex]\pmod m[/tex] (\pmod), but what is the other?[/quote]
There are at least 4 mod functions:

[tex]\begin{array}{ll}
\text{bmod} & z \equiv x+y \bmod{n^2} \\
\text{pmod} & z \equiv x+y \pmod{n^2} \\
\text{mod} & z \equiv x+y \mod{n^2} \\
\text{pod} & z \equiv x+y \pod{n^2} \\
\end{array}[/tex]

The 2 ways you are thinking of in LaTeX are \bmod & \pmod whereas \mod & \pod are AMS-LaTeX

Thank you!  Is there any place I can find a list of all the AMS-LaTeX macros? The only one I know is \begin{proof}, which I'm not even sure if you need AMS-LaTeX to use.

I'm not sure there's a list as such, but all the macros are listed in the various documents near the bottom of http://www.ams.org/tex/amslatex.html , with an explanation of how they are used.

Have a look particularly at
amsmath user's guide (amsldoc.pdf)
Short Math Guide for LaTeX
AMS theorem package user's guide (amsthdoc.pdf) (this is the one with proof)

Thank you.  Combined, those files are over 45 pages.  I'll have to start reading...
[tex]
2^2 \equiv 1 \bmod 3

4^4 \equiv 1 \pmod 5

6^6 \equiv 1 \mod 7

10^{10} \equiv 1 \pod {11}
[/tex]


Show that the closure of a set $ A$ is the smallest closed set containing A.

what is your definition of closure? where does $ A$ live?
usually (in general topology) the definition of closure is exactly what you're asking for  :)

$ A$ lie in the real line.Closure is defined as the intersection of all closed sets containing A.How can I say that it will be the smallest closed set?

ok, this is a general thing.

in all (topological) settings you define the closure of a subset $ A$ of a (topological) space $ X$ to be the intersection of all closed subsets containing $ A$. now, call $ F$ the smallest closed set containing $ A$ and $ \overline{A}$ the closure of $ A$.

since $ \overline{A}\equal{}\bigcap C$ over all $ A \subset C$ closed, in particular $ F$ is one of those set, then $ \overline{A}\equal{}\overline{A}\cap F$, and so $ \overline{A}\subset F$.
now, suppose there's something in $ F\setminus \overline{A}$: this means that there's a set $ C$ which is closed and contains $ A$ such that $ F\not\subset C$. but then taking the intersection $ C\cap F$ yields a closed subset of $ X$ containing $ A$, and strictly smaller than $ F$, contradiction.
so we've shown both inclusions, and $ F\equal{}\overline{A}$.


Given $ a,b \in \math{N}$ that $ 3a+12b=2mn$
Find $ a,b$ that $ ab$ is min.

[quote="bibobeo"]Given $ a,b \in \math{N}$ that $ 3a + 12b = 2mn$
Find $ a,b$ that $ ab$ is min.[/quote]
[hide="Alternate Solution"]
By the AM-GM, we have that
\[ \frac {3a + 12b}{2}\ge \sqrt {36ab}
\]

\[ \frac {2ab}{2}\ge 6\sqrt {ab}
\]

\[ ab\ge 6\sqrt {ab}
\]

\[ a^2b^2\ge 36ab
\]

\[ ab\ge 36
\]
So, we call ab=36 since that is the smallest value of ab.  Thus,
\[ 3a + 12b = 2ab = 2(36) = 72
\]

\[ a + 4b = 24
\]

\[ b = 6 - \frac {a}{4}
\]
Okay, so $ ab = 36$.  Thus,
\[ (a)(6 - \frac {a}{4}) = 36
\]

\[ 6a - \frac {a^2}{4} = 36
\]

\[ a^2 - 24a + 144 = 0
\]

\[ a = 12
\]

\[ b = 6 - \frac {12}{3} = 3
\]

\[ \boxed{(a,b) = (12,3)}
\]
Yeah, so it doesn't matter whether a and b are integers or not.  As long as they are positive, the values of a and b are still 12 and 3 respectively.  [/hide]

I came across this problem while doing some physics exercises.
Although this problem is easy, I want to find the best way to solve it since I think if the equation has 100 solutions then we have to try every $ a,b$ to find out the min $ ab$?


Solve the solution to this compund ineuqality
2x+5y≥1
3x-4y≤2

need a little help

2x ≥ 1 - 5y
x ≥ (1 - 5y) / 2

Substitute that into the second one, and you can find the intersection point of these two inequalities. Then test a point in each one and find the solution set.


inequalities
vector
inequalities proposed

Probably your inequality can be extended to a much more general result, which is still a conjecture for me:

For any k positive reals $s_1$, $s_2$, ..., $s_k$ and for every real r, let the number $\sqrt[r]{\frac{s_1^r+s_2^r+...+s_k^r}{k}}$ be called the [b]r-power mean[/b] of the numbers $s_1$, $s_2$, ..., $s_k$.

Then, the general conjecture would be:

[i]If $a_1$, $a_2$, ..., $a_n$ are n positive reals, and u and v are two nonnegative reals such that $u\leq v$ (I am not sure whether the non-negativity of u and v is really imporant, but let it be), then the u-power mean of all pairwise v-power means of the numbers $a_1$, $a_2$, ..., $a_n$ is $\geq$ to the v-power mean of the pairwise u-power means of the numbers $a_1$, $a_2$, ..., $a_n$.[/i]

Your inequality is the particular case of this for n = 3, u = 1 and v = 2.

The particular case for u = 0 and v = 1 was discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=18574 .

I tried to prove the conjecture for u = 1, v = 2 and generic n, but ended up in an absolute mess of binomial coefficients. Maybe I will try again...

Some days ago I had a short discussion on the above general conjecture with Christian Reiher. Stupidly enough, we didn't really elaborate and so I still don't know whether the conjecture is true or not. Anyway, I'm 90% sure that it's true.

  Darij

This comes from the final round of Polish Olympiad 2003 I guess

I haven't thought about Darij's conjecture, but this problem is quite straightforward.

First note that we may assume that $a, b, c \geq 0$, as all the instances of the variables on the LHS are squared.

By squaring, we see that the ineq is equivalent to
\[
2\sum_\mathrm{cyc} \sqrt{(a^2+b^2)(a^2+c^2)} \geq \sum_{cyc} a^2 + 3\sum_{cyc} ab
\]

Using cauchy, we find that
\[
2\sum_\mathrm{cyc} \sqrt{(a^2+b^2)(a^2+c^2)} \geq 2\sum_\mathrm{cyc} (a^2+bc)
\]
and the rest follows by Muirhead (or simple manipulation).

Oh,I think It has a more trivial solution!
Let
$S_a=\sqrt{2b^2+2c^2}+b+c$
$ S_b=\sqrt{2c^2+2a^2}+c+a$
$ S_c=\sqrt{2a^2+2b^2}+a+b$
$ S=\sqrt{3(a+b)^2+3(b+c)^2+3(c+a)^2}+2(a+b+c)$

Then,the ineq

$\leftrightarrow LHS-2(a+b+c) \ge RHS-2(a+b+c)$

$\leftrightarrow \sum \frac{(a-b)^2}{S_c} \ge \frac{\sum {(a-b)^2}}{S}$

But $S_a,S_b,S_c \le S$,then we are done!

I think that if you can prove this :

For $x=(x_1,x_2,x_3),y=(y_1,y_2,y_3),z=(z_1,z_2,z_3)$ vectors in three dimensional euclidean space we have

$\sqrt{2}(mod(x)+\mod(y)+\mod(z) ) \geq \sqrt{3}\mod(x+y+z)$

where $\mod(x) = \sqrt{x_1^2+x_2^2+x_3^2}$

you have a more natural proof.

Is that true?
Can you prove it?

Thank you very much.

[quote="darij grinberg"]Probably your inequality can be extended to a much more general result, which is still a conjecture for me:

For any k positive reals $s_1$, $s_2$, ..., $s_k$ and for every real r, let the number $\sqrt[r]{\frac{s_1^r+s_2^r+...+s_k^r}{k}}$ be called the [b]r-power mean[/b] of the numbers $s_1$, $s_2$, ..., $s_k$.

Then, the general conjecture would be:

[i]If $a_1$, $a_2$, ..., $a_n$ are n positive reals, and u and v are two nonnegative reals such that $u\leq v$ (I am not sure whether the non-negativity of u and v is really imporant, but let it be), then the u-power mean of all pairwise v-power means of the numbers $a_1$, $a_2$, ..., $a_n$ is $\geq$ to the v-power mean of the pairwise u-power means of the numbers $a_1$, $a_2$, ..., $a_n$.[/i][/quote]

As I already pointed out in http://www.mathlinks.ro/Forum/viewtopic.php?t=18574 , this conjecture is wrong for every $n\geq 4$. Maybe it is true for n = 3. Maybe even the following corrected version of this conjecture holds:

[i]If $a_1$, $a_2$, ..., $a_n$ are n positive reals, and u and v are two nonnegative reals such that $u\leq v$ (I am not sure whether the non-negativity of u and v is really imporant, but let it be), then the u-power mean of all pairwise v-power means of the numbers $a_1$, $a_2$, ..., $a_n$ is $\geq$ to the v-power mean of the (n-1)-tuple-wise u-power means of the numbers $a_1$, $a_2$, ..., $a_n$.[/i]

  darij


Care este conditia necesara si suficienta pentru matricile $A,B^$ patratice $nxn$ cu elemente complexe,ca sa existe doua matrici complexe $X,Y$ astfel incat $AX+BY=I_n$?

Pacat, nimeni nu e interesat. Voi face un experiment: fie $A$ o matrice $3x3$ cu elemente rationale si astfel incat $det(A-\sqrt[3]{2}I_3)=0$. Calculati $detA$. Pot sa pariez ca pana maine gasesc cel putin 7 oameni care sa o rezolve... :D

Cred ca e putin vag enuntul (de la prima problema, nu de la a doua :)). Ce fel de conditie? aca noyam cu $Im(M)$ spatiul vectorial $\{x\in\mathbb C^n|\exists y\in\mathbb C^n,\ x=My\}$, atunci conditia este $Im(A)+Im(B)=\mathbb C^n$, dar probabil ca e vreo conditie care se exprima mai elegant? Poate numai in functie de matricile $A$ si $B$, de rangurile lor, sau ceva de genul asta? :?

Pai am spus conditie necesara si suficienta. Daca dai una si arati ca e necesara si suficienta, ai rezolvat problema.

Pai aia pe care am scris-o ar cam fi necesara si suficienta, nu? :)


Let $p$ be a prime number. Is the polynomial $P(x) = x^{p-1}+x^{p-2}+...+x+1$ irreducible in $Q[x]$? In $R[x]$?

I know it is  irreducible in $\mathbb{Q}[x]$, by apply Eisenstein's criterion for $P(x+1)$. For $\mathbb{R}$, that is strange question for me  :maybe:

By Gaus theorem  irreducible in $\mathbb{Q}[x] \Longleftrightarrow$   irreducible in $\mathbb{Z}[x]$. That polynomial is cyclotimc polynomial and it is ireducible in $\mathbb{Z}[x]$.

Oh  :oops: , You are right!
[quote="Svejk"]If $p\geq 4$ it can't be irreducible in $\mathbb R[X]$,because any polynomial with real coefficients and degree greater than 2 can be written as the product of polynomials of degree I and II with real coefficients[/quote]


inequalities
prime numbers
number theory

Let $ p_1,p_2,\ldots,p_n$ be pairwise distinct odd primes such that $ \displaystyle p_i \mid \left(\frac {\displaystyle \prod_{j \equal{} 1}^n{p_j}}{p_i}\right) \minus{} 1$ for all $ 1 \le i \le n$.

I guess that $ n \equal{} 1.$ Try to (dis-)prove this conjecture.

[quote="bboypa"]Let $ p_1,p_2,\ldots,p_n$ be pairwise distinct odd primes such that $ \displaystyle p_i \mid \left(\frac {\displaystyle \prod_{j \equal{} 1}^n{p_j}}{p_i}\right) \minus{} 1$ for all $ 1 \le i \le n$.[/quote]
This condition can be restated as
\[ p_1\cdot p_2\cdots p_n \mid \minus{} 1 \plus{} p_1\cdot p_2\cdots p_n\cdot \sum_{i \equal{} 1}^n \frac {1}{p_i}.\]
Therefore, for $ n>1$, there exists integer $ k\geq 1$ such that
\[ k\cdot p_1\cdot p_2\cdots p_n \equal{} \minus{} 1 \plus{} p_1\cdot p_2\cdots p_n\cdot \sum_{i \equal{} 1}^n \frac {1}{p_i}\]
or
\[ \sum_{i \equal{} 1}^n \frac {1}{p_i} \equal{} \frac {k\cdot p_1\cdot p_2\cdots p_n \plus{} 1}{p_1\cdot p_2\cdots p_n}.\]
and, in particular,
\[ \sum_{i \equal{} 1}^n \frac {1}{p_i} > k \geq 1.\]
Minimal set of odd primes satisfying this inequality has $ n \equal{} 9$ (for primes $ 3,5,\dots,29$). Therefore, if solution for $ n > 1$ exists then $ n\geq 9$.

It turns out that [b]bboypa[/b]'s question is equivalent to [url=http://mathworld.wolfram.com/GiugasConjecture.html]Giuga's conjecture[/url]. Namely, if such primes exist then their product would represent a counterexample to Giuga's conjecture.

A couple of relevant papers:
[url=http://www.math.uwo.ca/~dborwein/cv/giuga.pdf]Giuga's conjecture on primality[/url]
[url=http://arxiv.org/abs/0712.3954]Cyclic systems of simultaneous congruences[/url]


logarithms
real analysis
real analysis unsolved

$x_0>0$
$x_{n+1}=x_n-e^{\frac{-1}{x_n^2}}$

Find an equivalent
Find asymptotic expansion with two terms or three terms

:idea: . How can this help? I believe that some equivalent should be something like $\frac{1}{\sqrt{lnn}}$ but not only I cannot prove it, I'm not even sure.

[quote="harazi"]:idea: I believe that some equivalent should be something like $\frac{1}{\sqrt{\ln n}}$ but not only I cannot prove it, I'm not even sure.[/quote]
You're right, harazi! :) Indeed, the equivalent is $\frac 1{\sqrt{\ln n}}$. Since you "don't like my hints", let me, for diversity, post a complete proof.

Fix $a>0$ and consider $y_n=\exp\left\{\frac a{x_n^2}\right\}$. Clearly, $x_n\to 0$, so $y_n\to+\infty$. The question is how fast. Note that, as $x\to 0$,
\[
\frac a{(x-e^{-1/x^2})^2}= \frac a {x^2} + \frac {(2+o(1)) a e^{-1/x^2}}{x^3}\,,
\]
and the second term tends to $0$, so
\[y_{n+1}=e^{a/x_n^2}\left(1+ (2+o(1))a e^{-1/x_n^2}/x_n^3\right)\\
=y_n+(2+o(1))a e^{(a-1)/x_n^2}/x_n^3=y_n+z_n.\]
Note that, for $a>1$, we have $z_n\to+\infty$ and, therefore, $y_n>n$ for all large $n$. This allows to conclude that, for all $a>1$, one has $a/x_n^2>\ln n$ for large $n$, i.e., that $\limsup_{n\to\infty} x_n\sqrt{\ln n}\leqslant 1$. Also, if $a<1$, we have $z_n\to 0$, whence $y_n<n$ for all large $n$ and, finally,  $\liminf_{n\to\infty} x_n\sqrt{\ln n}\geqslant 1$.

Now, can anyone guess the next term?  ;)


Prove that for every positive integer $n$ there exists a unique ordered pair $(a,b)$ of positive integers such that
\[ n = \frac{1}{2}(a + b - 1)(a + b - 2) + a .  \]

Let $T(m)$ be $m$-th triangular number, i.e. $T(m)=\frac{m(m+1)}{2}$. It's obvious that for any given $n$ there's exactly one $m_0$ such that $T(m_0)<n\leq T(m_0+1)$. (The strict algebraic proof isn't difficult and will be supplied if needed.)

Now we'll let $a+b-2=m_0$ and $a=n-T(m_0)$. All that remains to proove is that such chosen $a$ and $b$ are always positive:

From $T(m_0)<n$ it follows $a=n-T(m_0)>0.$ On the other hand,
\begin{eqnarray*}b=m_0-a+2=m_0-n+T(m_0)+2=(T(m_0)+m_0+1)-n+1=T(m_0+1)-n+1\end{eqnarray*}. From $n\leq T(m_0+1)$ it follows $T(m_0+1)-n+1\geq 1$. Hence both $a$ and $b$ are positive, and their uniqueness is guaranteed by the uniqueness of $m_0$. QED

[quote="Sathej"]Hello,
    This can also be approached in a different way.
   Consider the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ........
It can be easily shown that if the nth term of this sequence is a/b, then n equals the required expression in a and b(given in the problem). Uniqueness  can also be shown as a/b occurs only once in the sequence. This would be clear if one can see how the sequence is actually built.
Sanjith[/quote]

very nice idea presented sathej..i would like to know how it crossed your mind.. :?:

It's a well known bijection between the rationals and the naturals - some sort of a diagonalisation. So that's probably it.

[quote=fire123]Check out PSS Chapter 6, Problem 70. It is almost exact same problem. (Just re-factor and divide by two)[/quote]

PSS? Problem Solving Strategies(Arthur Engel)??


An "n-pointed star" is formed as follows: the sides of a convex polygon are numbered consecutively $1,2,\cdots,k,\cdots,n$, $n\geq 5$; for all $n$ values of $k$, sides $k$ and $k+2$ are non-parallel, sides $n+1$ and $n+2$ being respectively identical with sides $1$ and $2$; prolong the $n$ pairs of sides numbered $k$ and $k+2$ until they meet. (A figure is shown for the case $n=5$)

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=704&sid=8da93909c5939e037aa99c429b2d157a[/img]

Let $S$ be the degree-sum of the interior angles at the $n$ points of the star; then $S$ equals:

$\text{(A)} \ 180 \qquad \text{(B)} \ 360 \qquad \text{(C)} \ 180(n+2) \qquad \text{(D)} \ 180(n-2) \qquad \text{(E)} \ 180(n-4)$

[hide]Noticing that each vertex point of the original polygon forms opposite angles, one is forced to conclude that each angle formed by a star's point is the same.  There's n triangles surrounding the n-sided polygon that forms the star.  Each triangle has 180 degrees so that's 180n degrees.  Next, label the angles of the polygon $a_1,a_2,a_3,...a_n$.  The unknown, unfixed angles in the triangles are $180-a_1,180-a_2,180-a_3,...180-a_n$, however, each angle appears twice.  Then, there's the n fixed point angles.  That's nk+360n-360(n-2)=180n.  Rearranged, that's n(180-k)=720.  Setting n=5 shows that k=36 so 5(36)=180.  n=6 causes k=60 so 6(60)=360.  n=8 causes k=90.  In other words k=180(n-4)/n.  Therefore, nk=180(n-4)[/hide]


let $ B$ be a point on a circle $ S_1$ and let $ A$ be a point distinct from $ B$ on the tangent at $ B$ to $ S_1$.let $ C$ be a point not on $ S_1$ such that the line segment $ AC$ meets $ S_1$ at two distinct points .let $ S_2$ be the circle touching $ AC$ at $ C$ and touching $ S_1$ at a point $ D$ on the opposite side of $ AC$ from $ B$ .prove that the circumcenter of triangle $ BCD$ lies on the circumcircle of triangle $ ABC$

You can find a number of proofs here: http://www.mathlinks.ro/viewtopic.php?p=118667#118667
Enjoy!

See my solution (with diagram) here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1365985#1365985


find all primes $ x,y,z$ $ \frac{1}{x} \plus{} \frac{1}{y} \plus{} \frac{1}{z} \equal{} 1$

[quote="aadil"][quote]Through pigeonhole principle if x,y, and z are distinct and we consider the expression
x =< y =< z just so that the principle yields some solution, we get (2,3,6)
6 is not prime so any permutation of this solution is wrong and the only possible triplet is when
x= y= z
(3,3,3)[/quote]
please explain how?[/quote]

[hide="solution"]
Suppose cases
1.$ z \ge y \ge x$
2.$ y \ge x \ge z$
3.$ x \ge z \ge y$

[b]Case 1.[/b]We have  $ 1\equal{}\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\leq \frac{3}{x}$ therefore $ x\leq 3$ so we consider values of $ x$

$ x\equal{}1$ it's obvious that $ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\equal{}1$ doesn't have solutions in $ N$

$ x\equal{}2$ we have $ \frac{1}{y}\plus{}\frac{1}{z} \equal{} \frac{1}{2}$ and consider the cases:

$ y\equal{}2$ there's no solution.
$ y\equal{}3>x \implies \frac{1}{z}\equal{}\frac{1}{6} \implies z\equal{}6>x$
$ y\equal{}4>x \implies z\equal{}4>x$

$ y \ge 5 \implies \frac{1}{y}\plus{}\frac{1}{z}\leq \frac{2}{y} \leq \frac{1}{2} \leq \frac{2}{3}$ so there's no solution.

[b]Case 2[/b] and [b]Case 3[/b] are similarly founded so:

$ x,y,z\in N$$ x,y,z\in N$ only solutions are:

$ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\implies (x,y,z)\equal{}(2,3,6);(2,6,3);(3,6,2);(3,2,6);(6,3,2)(6,2,3);(2,4,4)(4,2,4);(4,4,2)$ and $ (3,3,3)$

therefore $ (3,3,3)$ is the only prime triplets. :D[/hide]

you don't have to treat all those cases, $ x,y$ and $ z$ are symetric then we can assume that, $ x\geq y\geq z$
from that we get $ z\geq 3 \Rightarrow z\equal{}2 or z\equal{}3$
$ z\equal{}2 \Leftrightarrow \frac{1}{x}\plus{}\frac{1}{y}\equal{}2 \Leftrightarrow (x\minus{}2)(y\minus{}2)\equal{}4 \Rightarrow (x,y)\equal{}{(4,4);(6,3)}$
$ z\equal{}3 \Leftrightarrow \frac{1}{x}\plus{}\frac{1}{y}\equal{}\frac{2}{3} \Leftrightarrow (2x\minus{}3)(2y\minus{}3)\equal{}9 \Rightarrow (x,y)\equal{}{(3,3)}$
then $ (x,y,z)\equal{}{(6,3,2);(4,4,2);(3,3,3)}$ and all there symetric permutation, from where we deduct that $ (x,y,z)\equal{}(3,3,3)$ is the only prime solution,

[quote="peine"]you don't have to treat all those cases[/quote]

Yes,of course but I thought it would be easier to understand  :D :blush:


Registration is over.  This is the thread for you to post your totals from your games.  [b]Do not post anything else here.[/b]  I might set up a place to do it elsewhere, but for now, post your wins and losses here.

The qualifications round, since it consists of only 2 teams, [b]must be completed by Friday[/b].  I will PM everyone soon.  The qualification round will be as follows:
It will consist of 2-3 games:  the first 2 games will be ranking based Normal games with 7 questions each, and 45 seconds for each problem.  They must be viewable.
If no team wins both games, a countdown will be held between any one player from each team.
Make sure totals from [b]each[/b] game are posted here.

In the qualification round are only: [b]Indiana[/b] (8)[b] vs. SagginWolley [/b](9).  Everyone else gets to wait until next round.

All right.  Finally back.

Everyone in the 2 teams recieved (and opened) their PMs.  You have the rest of the week to complete the first round.  Things have been going very slow lately, much due to my laziness, but now that that's over let's get back to this.


Working together, Jo and Ralph can do the garden chores in 6 hours. It takes Jo twice as long as Ralph to do the work alone. How many hours does it take Jo working alone?

Let the amount of work be $ 1$ to ease up things a bit. Then, let $ j,r$ be Jo and Ralphs' times, respectively. Then we have $ \frac2{\frac1j\plus{}\frac1r}\equal{}6$, and $ j\equal{}2r$. Solving for $ j$, we find that $ j\equal{}\boxed{18}$.

[hide="A Faster Solution"]
We can visualize it as Jo and Ralph taking $6$ hours to do parts of the work, and the addition of the parts of the work make the whole job. Since Jo takes twice as long has Ralph to do the work and they finish in the same amount of time, Ralph is doing twice the amount of work she is.
If she does $j$ of the work, Ralph is doing $2j$ of it. The sum is all of the work, or $1$
Since $j +2j = 1$, Jo does $\frac{1}{3}$ of the work in $6$ hours. She does all of it in $6\cdot3 = \boxed{18}$ hours
[/hide]


geometry
3D geometry
number theory open
number theory

Prove or disprove that there are infinitely many numbers of the form $ x^{3}+y^{3}-1$ inexpressible as the sum of three positive cubes.

[quote="s.tringali"]It's immediate if we can prove there're infinitely many $ x \in \mathbb{N}= \{1, 2, \ldots\}$ such that $ x^{3}$ is not expressible as the sum of the cubes of three positive integers - btw, this would answer as well one last of Nbach's unsolved problems ([url=http://www.mathlinks.ro/Forum/viewtopic.php?t=162361]here[/url], check for the case $ i = 2$). In fact, I think no power of $ 2$ can be expressed as the sum of the cubes of three positive integers, but till now I was able neither to prove it nor to find any counterexample. Who can? :)[/quote]
Um, Nbach's problem is from GMO 2006.


http://www.artofproblemsolving.com/NewClassroom/ClientDoc.php

very bottom of the page:

"Monitor
Monitor with priviledges such as gag/ungag, kick, open/close room, etc."

should be "privileges"

We fixed it, thanks for lettings us know :)


I found this problem in the book called problems, problems, problems.
In the diagram, $\Delta ABC$ (right-angled) has $AC=b$, $BC=a$, and $AB=c$. Initially the triangle is located in square $WXYZ$ of side c, as shown. The triangle moves counterclockwise inside the square so that the vertices A and B are always on the sides of the square. This motion continues until the triangle returns to its initial position. Determine the total distance in terms of a and b travelled by the point C.

Could a mod move this if this if it's too hard. thanks :)

Claim:
As the diagram stands, C always lies on CZ when A is on WZ

consider when A is on WZ and B is on BY

$\angle ACB=90^\circ=\angle WZY$

then $\angle ACB+\angle WZY=180^\circ$

so $ACBZ$ is cyclic

then by inscribed angles, $\angle ABC=\angle WZC$

That means that $\angle WZC$ is constant

since C is inside the square, it must lie on the line $\mathcal{L}$ such that for any point, $P$, on $\mathcal{L}$ (inside the square), we have that $\angle ABC$

Similar logic follows for when the triangle moves onto other sides of the square

Now the problem is much easier, just draw the triangle on all 3 sides of the square and compute the perimeter of the inner square which is just

$4|a-b|$

I posted it here, but since no one replied I posted it again in pre-olympiad. Altheman, I think you missed something in your solution.
[quote="srulikbd"]nice problem  :)  do we need to use analytic geometry? cause it seems that C goes in curved line...[/quote]
The problem can be done without. :)
You can try though, you might get an interesting solution.


Three hollow and concentric spheres have radii $R_{1}= 1m$, $R_{2}= 2m$, and $R_{3}= 3m$. The smaller sphere was charged with $q_{1}= 1 \mu C$ and the bigger one with $q_{3}=-2 \mu C$, while the middle sphere was connected to the Earth.

(a) Compute the potential in each sphere.
(b) Next, the middle sphere was unconnected from the Earth, and the two smaller spheres were connected through a conducting wire. Compute the new charge and potential on each sphere.

It can be solved easily using linear superposition.


Let $ a;b;c;k > 0$ and $ abc \equal{} 1$ . Prove that:

$ x^2 \plus{} y^2 \plus{} z^2 \ge\ \frac {(k \plus{} 1)x}{yz \plus{} kxz} \plus{} \frac {(k \plus{} 1)y}{zx \plus{} kyx} \plus{} \frac {(k \plus{} 1)z}{xy \plus{} kzy}$

[quote="nguoivn"]But it's easy. After using Am-Gm, it's became:
$ x^2 \plus{} y^2 \plus{} z^2 \geq\ xy \plus{} yz \plus{} zx$[/quote]
Can U detail it ? I don't understand your ideal. It 's very easy??? :maybe: I don't think so

Let $ x\equal{}\frac{a}{b},.....$
the ineq is
$ \sum \frac{a^2}{b^2}\ge (k\plus{}1)\sum \frac{a^2}{b^2\plus{}kac}$
which is true by SS :lol:

[quote="Allnames"]Let $ x \equal{} \frac {a}{b},.....$
the ineq is
$ \sum \frac {a^2}{b^2}\ge (k \plus{} 1)\sum \frac {a^2}{b^2 \plus{} kac}$
which is true by SS :lol:[/quote]
Why do U say method ? I don't need method  :(  I want your full proof

[quote="onlylove_math"][quote="nguoivn"]$ \frac {(k \plus{} 1)x}{z(y \plus{} kx)} \equal{} \frac {(k \plus{} 1)x^2y}{kx \plus{} y}$
I think you understood  :)[/quote]
Continous please ! It is only a small part of problem  :) Thanks![/quote]
I don't think so. It's a half of solution :D
By Am-Gm, we have:
$ \frac{(k\plus{}1)x^2y}{kx\plus{}y} \leq\ \frac{1}{k\plus{}1}x^2y(\frac{k}{x}\plus{}\frac{1}{y})\equal{}\frac{kxy\plus{}x^2}{k\plus{}1}$
So, it's becomes: $ x^2\plus{}y^2\plus{}z^2 \geq\ xy\plus{}yz\plus{}zx$
I think you understood :D

[quote="nguoivn"][quote="onlylove_math"][quote="nguoivn"]$ \frac {(k \plus{} 1)x}{z(y \plus{} kx)} \equal{} \frac {(k \plus{} 1)x^2y}{kx \plus{} y}$
I think you understood  :)[/quote]
Continous please ! It is only a small part of problem  :) Thanks![/quote]
I don't think so. It's a half of solution :D
By Am-Gm, we have:
$ \frac {(k \plus{} 1)x^2y}{kx \plus{} y} \leq\ \frac {1}{k \plus{} 1}x^2y(\frac {k}{x} \plus{} \frac {1}{y}) \equal{} \frac {kxy \plus{} x^2}{k \plus{} 1}$
So, it's becomes: $ x^2 \plus{} y^2 \plus{} z^2 \geq\ xy \plus{} yz \plus{} zx$
I think you understood :D[/quote]
Very nice; Thank U ! Sorry if my ideal make U  :mad: because my proof use SS and it don't simply  :)

[quote="nguoivn"]No, I like your emotion. You share to ask the things which you don't understand.
Certainly, you can improve fastly in ineq and the other fields.
I can't do as you. So, I' m only an amateur people :D[/quote]
Thank nguoivn for everything ! Follow me, I see U is professional people  :lol:

your problem is not very nice.I think you should make your problem with more new ideal base on method!
like that we can use SS if N+2M >=0 but I'm sure that you are pro at ineq! :wink:

[quote="Math pro"]your problem is not very nice.I think you should make your problem with more new ideal base on method!
like that we can use SS if $ N + 2M \ge\ 0$ but I'm sure that you are pro at ineq! :wink:[/quote]
Thank for your ideal . But my problem isn't simply  :wink:
I 'll try follow your ideal but .Can you prove that  Let $ a;b;c \ge\ 0$and $ c = max$ or $ min$ {$ a;b;c$}
 $ N(a - b)^2 + 2M(a - c)(b - c) \ge\0$  with $ N + 2M \ge\ 0$  :wink:


number theory
least common multiple
group theory
prime numbers

I've no idea of the difficulty of this or if its solvable.

Given a positive integer n, maximise the lowest common multiple of a set of positive integers (allowing repeats) which add to n.

Or for those of you who know some group theory: Find the maximal order of an element of the symmetric group S_n.

[edit - fixed. thanks.]

The only thing that I've got is that this partition of $n$ must be made out only from prime numbers. And this took me to the hypothesis that the first k prime numbers should be added up to the biggest smallest than $n$.
Uhh.. and also 2 is the first one. I couldn't prove that 3 must come after 2.

n=12 has the best partition of 3,4,5. :)

ok, sorry, i made a horrible mistake. I was maximising the number of prime numbers in the LCM instead of maximising the LCM itself.

Now that I think about this, its easy to prove that there exists a maximal arrangement where each term is a prime power. I doubt its solvable from there.


Find all functions $f: \mathbb{N}\rightarrow\mathbb{R}$ satisfying:
$f(0)=0 \quad f(1)=1$
$f(0)\leq f(1) \leq f(2) \leq \cdots$
$f(x^2+y^2)=f(x)^2+f(y)^2 \quad \forall x, y \in \mathbb{N}$

I understand you assume $0 \in \mathbb{N}$ ? Then $f ( m )^2 = f ( m^2 )$ for all positive integer $m.$

Suppose there is an $m \in \mathbb{N}$ such that $f ( m ) = f ( m + 1 )$.
Then $f ( m )^2 = f ( m + 1 )^2$, hence $f \left( m^2 \right) = f \left( m^2 + 2 m + 1 \right)$.
Since $f$ is increasing $f ( m^2 ) = f ( m^2 + 2 m + 1 )$ implies $f ( m^2 ) = f ( m^2 + 1 ) = \ldots = f ( m^2 + 2 m + 1 )$
Remark that $f ( m^2 + 1 ) = f ( m )^2 + f ( 1 )^2 = f ( m^2 ) + 1 > f ( m^2 )$. A contradiction.
Therefore $f ( 0 ) < f ( 1 ) < f ( 2 ) < \ldots$

Now remark that $f ( 5 ) = f ( 2^2 + 1 ) = f ( 2 )^2 + 1 = 2^2 + 1 = 5$.
Moreoever $f ( 5^{2^n} ) = f ( 5 )^{2^n} = 5^{2^n}$ for all positive integer $n \in \mathbb{N}$.
Hence $0 = f ( 0 ) < f ( 1 ) < f ( 2 ) < \ldots < f ( 5^{2^n} ) = 5^{2^n}$.
This implies that $f ( m ) = m$ for all positive integer $m$.

How do you know, that $f(5^n)=5^n$ :?:. It is surely true, when $n$ is a power of two. But what with other $n$

You are absolutely correct. It should be $f(5^{2n})=5^{2n}$ rather than $f(5^n)=5^n$. I went too fast and forgot about it :(.

$5^n$ may be change to $x_n=2^{2^n}$, because $f(m^2)=f(m)^2 \ \Longrightarrow f(x_n)=x_n$.

I don't understand the last your step. Could you explain.
And it should be $f(5^{2^n})=5^{2^n}$


Dear Mathlinkers,
what are the different proofs of Urquhart's theorem?
Sincerely
Jean-Louis

I think that this (esay to prove) result is coming from the past but, I didn't know that it is called as [b][size=100]Urquhart's theorem[/size][/b].

( Finally, all the simple results which are used for the solution of other problems, already they have been listed as personal theorems ).

Kostas Vittas.

Dear Mathlinkers,
I come again with the same question: how many are there different proofs of Urquhart's theorem?Sincerely
Jean-Louis

I think it is easy to prove with pure geometry theorems,
but it is hard to prove by calculating.

So I uploaded this proof.

Dear Frankvista and Mathlinkers,
I know that there is a trigonometric proof that I haven't in my archives.
I am waiting for your geometrical proof.
Sincerely
Jean-Louis

Also, I will post here later the elementary proof I have in mind of the [b][size=100]Urguhart's theorem[/size][/b], if it is different from the one of [b][size=100]frankvista[/size][/b].

Kostas Vittas.


[quote="wiosna"]$ 9^ 8^^^7^^^6^^^5^^^4^^^3^^^2 \mod 9!$[/quote]
First of all I don't think that this is a "stupid question".

Here is the outline solution:
Let $ x={9^ 8^^^7^^^6^^^5^^^4^^^3^^^2}$
At first notice that, $ 9!=2^7\times 3^4 \times 5 \times 7$
Obviously, $ x \equiv 0\mod 3^4$
And $ x\equiv 1 \mod 2^7$ (It can be seen from many thing, you can also notice that $ \phi(2^7)=2^6$ and $ (9,2)=1$)
And $ x \equiv 1 \mod 5$
Last of all, $ {9^ 8^^^7^^^6^^^5^^^4^^^3^^^2}$ $ \equiv {2^ 8^^^7^^^6^^^5^^^4^^^3^^^2}$ $ \equiv 4 \mod 7$ (from $ 2^3 \equiv 1 \mod 7$ and $ 8^\text{odd} \equiv 2 \mod 3$)
We just need to solve the following system using Chinese remainder theorem. :)
\begin{align*}
x & \equiv 1 \mod 2^7 \\
x & \equiv 0 \mod 3^4 \\
x & \equiv 1 \mod 5 \\
x & \equiv 4 \mod 7
\end{align*}

Sorry, some people like it to set a "!" after their posts.
And I thought, that you only mean mod9 and so I wondered^^

[quote="wiosna"]Question is......How much it is $ 9^^8^^7^^...^^2$$ mod9!$[/quote]
You just need to find it using Chinese remainder theorem. :)


algebra
polynomial
greatest common divisor
superior algebra
superior algebra unsolved

Let \[ f(x)\equal{}a_0\plus{}a_1x\plus{}a_2x^2\plus{}a_{10}x^{10}\plus{}a_{11}x^{11}\plus{}a_{12}x^{12}\plus{}a_{13}x^{13} \; (a_{13} \not\equal{}0) \] and \[ g(x)\equal{}b_0\plus{}b_1x\plus{}b_2x^2\plus{}b_{3}x^{3}\plus{}b_{11}x^{11}\plus{}b_{12}x^{12}\plus{}b_{13}x^{13} \; (b_{3} \not\equal{}0) \]
be polynomials over the same field. Prove that the degree of their greatest common divisor is at least $ 6$.

[i]L. Redei[/i]

Euclidean algorithm, basically the notation $(f_1;g_1)$ keeps track of which coefficients may not be zero when we subtract a multiple of one of the polynomials from the other. We replace $f,g$ by their differences alternatively first by replacing $f$ with $b_{13}f-a_{13}g$ etc. The coefficients should look like
$$(f;g) = (0\sim 2, 10\sim 13;0\sim 3,11\sim 13) \to (0\sim 3, 10\sim 12;0\sim 3, 11\sim 13) \to (0\sim 3, 10\sim 12;0\sim 4, 11\sim 12)\to (0\sim 4, 10\sim 11; 0\sim 4, 11\sim 12) \to (0\sim 4, 10\sim 11; 0\sim 5, 11) \to (0\sim 5, 10; 0\sim 5, 11)\to (0\sim 5, 10;0\sim 6)$$
so this terminates with the replacement of $g$ having degree $\le 6$.

Also I think the problem statement should read "at most $6$" in the end.

I don't really understand Stroller's reasoning but he's right: $degree(gcd(f,g))\leq 6$.
If we randomly choose $f,g$ , then they are coprime.
It was not asked to show that there exists an example with $degree(gcd)=6$.

The maximal degree is obtained -for example- when $K=\mathbb{Q}$ for
$f=x^{13}-x^{12}-x^{11}+x^{10},g= 4x^{13}-5x^{11}+x^3$; then  $gcd(f,g)=(x^3-x^2-x+1)x^3$.
,
To find the above example I canceled the terms of degree $<3$ and I randomly tested the remaining coefficients.

However, I think that the author of the question had a very specific idea in mind and that he did not find
this result by chance. Does somebody have an idea ?


i'm just wondering if there is a formula/pattern or something to calculate all the Pythagorean  triplets so i dont have to memorize them

Of course there is...

let the lengths of the two legs be $2mn$ and $m^2-n^2$
The hypotenuse is $m^2+n^2$

Then start plugging in integral values of m and n, where m is greater than n. :)

[hide]Go me!! I remembered this!!  :) [/hide]

Oh, one more thing. If you want to find triples, it's easy to start working with the 2mn term.

[i]All [/i]primitive  triplets are gven by

$a=2mn$
$b=m^2-n^2$
$c=m^2+n^2$

Choose any  integers $m$ and $n$  and you get the triplet, eg m=1,n=2 gives (3,4,5)

To make sure the triangle is primitive, jus have m,n prime to each other (one even and one odd)

primitive here means - the sides do not have a common factor, eg (3,4,5) is primitive (6,8,10) is not (as it is just the double of the first triplet)

(So general formala - once you get (a,b,c) from above, any muliple of that triplet (eg (ka,kb,kc) ) is also a triplet.

It's one thing to remember it, it's a whole other thing to actually prove this  :) , if, tianle, you can't remember formulas as well as 4everwise.

$a^2+b^2=c^2$
$(m^2-n^2)+(2mn)^2=(m^2+n^2)^2$
$m^4-2m^2n^2+n^4+4m^2n^2=m^4+2m^2n^2+n^4$
$m^4+2m^2n^2+n^4=m^4+2m^2n^2+n^4$

In all cases, this formula works. :)

Let us denote a set of pythagorean triplets by $(x,y,z)$.  I took a look a few triangles and say a pattern.  The first pythagorean triplet has side length that is odd and the two others are consecutive numbers.  The smallest, odd side can be denoted by $y=2(n+1)-1=2n+1$, which is the opposite side. The adjacent side is$x$, where it equals $x=(2n+1)n+n=2n^2+n+n=2n^2+2n=2n(n+1)$.  And finally, the hypotenuse is 1 larger than the adjacent side, which is $z=x+1=2n(n+1)+1$, as a result, the triple could be denoted by $(2n(n+1),2n+1,2n(n+1)+1)$.

Example, the 3rd set of pythagorean triples is:
$(24,7,25)$.

This is what I have noticed.

Masoud Zargar


Does the intermediate counting/probibility class cover or review all the material in the introductory. If i understand the counting in both the aops books will that teach me enough to directly skip to intermediate?

on the classes website you can take the post test of that class, and if you're able to do a high percentage of those questions without struggling too badly then you probably can skip it.

[quote="Calculus"]Does the intermediate counting/probibility class cover or review all the material in the introductory. [/quote]No -- the intermediate class picks up right where the introductory class leaves off.  In other words, throughout the intermediate class, it is assumed that you know the material covered in the introductory class.

You should attempt to solve the problems on the [url=http://www.artofproblemsolving.com/Classes/IntermCounting/PreTest.pdf]Pre-Test[/url] for the Intermediate Counting class.  If you can solve these problems without much difficulty, then you're ready for the Intermediate Class.


Let $ G$ be a group, $ N\triangleleft G$ a $ p'\minus{}$ group, $ P\leq G$ a $ p\minus{}$ group. Prove that $ C_{\overline{G}}(\overline{P})\equal{}\overline{C_G(P)}$. The one subset is easy. The hint for the other way is to let $ \overline{C}\equal{}C_{\overline{G}}(\overline{P})$ and then to show that $ [P,C]\equal{}1$.

Does this make sense to anyone? We defined $ \overline{C}$ to be a certain group that is not necessarily a quotient group mod $ N$ like we would normally notate. How do we then find $ C$ for that commutator? In other words how do we remove the bar? Normally to do that you would just multiply by $ N$.

I think by correspondence theorem, your $ C$ exists. Multiplying N should do.

So $ C\equal{}C_G(PN)$? Because then it is trivial, since $ C$ would then commute with everything in $ P$.

I guess $ C \equal{} \{xn : xN \in \bar{C}, n \in N\}$


[quote="jmccray"]At the time when mining asteroids have become feasible, astronauts have connected a line between their 3500kg space tug and 6200kg asteroid.  Using their ships engine, they pull on the asteroid with a force of 490N.  Initially the tug and the asteroid are at rest, 450m apart.  How much time does it take for the ship and asteroid to meet?[/quote]

M=astroid mass
m=tug mass
X=450m
F=490n
$M*A=F$        $m*a=F$
we have M,m,F so we're able to calculate A&a
$A_{tot}$ is the relative A for the system  $A_{tot}$=A-a
be careful that you should consider the direction of each vector :lol:
we have this formula: :arrow:
 X=$/frac{A_{tot}*t^2}{2}$+$v_0*t$
$v_0$=0
so we can find t=time


Alright, I am making a new hurt and heal, because the top soda one ended. I did get permission from the mods to do this one. Basically, it is all about things on AoPS. Heal = +1, and Hurt = -2. Highlight what you are hurting in red, and highlight what you are healing in green.

Let the hurt and heal begin;

AoPS Blogs
Contests (The section that includes lots of math problems.)
Mafia
Bookstore
Online Classes
Math Jams
AoPSWiki
LaTeX
Middle School Section
High School Section
Contests & Programs Section
Olympiad Section
College Playground Section
Arts & Sciences Section
Everything Else in the Forum Index (Including G & FF, Online School, AoPS FTW, etc.)

[color=green]Middle School Section - 16[/color]
AoPS Blogs
Contests  - 15
Mafia  - 15
Bookstore  - 15
Online Classes - 15
Math Jams  - 15
AoPSWiki  - 15
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College Playground Section  - 15
Arts & Sciences Section - 15
Everything Else in the Forum Index - 15
[color=red]LaTeX  - 13[/color]

Middle School Section - 16
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LaTeX - 13
Bookstore - 13

Middle School Section - 16
[color=green]Everything Else in the Forum Index - 16 [/color]
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Please highlight them.

Middle School Section - 16
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[b][color=green]AoPS Blogs - 15[/color]
[color=red]Everything Else in the Forum Index - 14[/color][/b]
LaTeX - 13
Bookstore - 13

Middle School Section - 16
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[color=green]Everything Else in the Forum Index - 15[/color]
[color=red]Online Classes - 13[/color]
LaTeX - 13
Bookstore - 13

NOOO! NOT ANOTHER HURT N HEAL!!! NOOOOO!!!
bah, w/e

[color=green]Mafia - 16[/color]
Middle School Section - 16
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[color=red]AoPSWiki - 13 [/color]

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eggy, please highlight

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Online Classes - 13
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College Playground Section - 11

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Hey, there's no "Everything Else in the Forum Index" in Wickedstjr's post.

[color=green]Everything Else In the Forum Index-17[/color]
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Everything Else In the Forum Index - 17
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Isn't there a rule that says no more hurt'n heals? just use a poll. Jeez.

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dragon96, please read the thread, I clearly stated that I got permission from the mods.
eggy - Please highlight your changes.
evilhamster - ernie already added everything else that was in the forum index that I didn't have before.
                   it is not necessary

[quote name="RedFireTruck" url="/community/p17790118"]
^THIS POST WAS MADE ONE DAY AFTER MY BIRTHDAY. :rotfl:
[/quote]

wut how did u find this thread XD

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geometry
incenter
perpendicular bisector
algebra
system of equations
geometry unsolved

Given is ABCD, a parallelogram. Escribed circle of ABD touches the extensions of AB and AD in M and N. MN intersects BC and CD in P and Q. Prove that P and Q lie on incircle of triangle BCD.

[color=darkblue]Let $ I$ be incenter of $ BCD$, $ O$ excenter of $ ABD$. Denote $ \{E\} \equal{} DI \cap BO$ and $ \{F\} \equal{} BI \cap OD$. Let $ P'$ and $ Q'$ be foot of the perpendiculars from $ I$ to $ BC$ and $ CD$ respectively. Then

$ \angle{BED} \equal{} 180^\circ \minus{} \dfrac{\angle{CDB}}{2} \minus{} \angle{DBO} \equal{} 180^\circ \minus{} \dfrac{\angle{ABD}}{2} \minus{} \bigg(90^\circ \minus{} \dfrac{\angle{ABD}}{2}\bigg) \equal{} 90^\circ$

Similiarly $ \angle{BFD} \equal{} 90^\circ$. Let's prove that points $ E,P',Q',F$ are colliniear. Quadrilateral $ IP'Q'C$ is cyclic, so

$ \angle{IP'Q'} \equal{} \dfrac{\angle{BCD}}{2}$ $ (1)$

Quadrilateral $ BIP'E$ is also cyclic, so

$ \angle{EP'I} \equal{} 180^\circ \minus{} \angle{IBE} \equal{} 180^\circ \minus{} \bigg(90^\circ \minus{} \frac {\angle{CBD} \minus{} \angle{DBC}}{2}\bigg) \equal{}$
$ \equal{} 90^\circ \plus{} \frac {\angle{CBD} \plus{} \angle{DBC}}{2}$ $ (2)$

By summing $ (1)$ and $ (2)$ we get

$ \angle{EP'Q'} \equal{} 180^\circ$

Similiarly points $ P',Q',F$ are collinear.
Thus, we need to prove that points $ M,E,P'$ and $ Q',F,N$ are collinear (then points $ M,E,P',Q',F,N$ will be collinear and $ P \equal{} P'$, $ Q \equal{} Q'$).

Because $ DE \perp BO$ and $ BF\perp OD$, $ I$ is orthocenter of $ OBD$, so $ OI \perp BD$. Let $ \{H\} \equal{} OI\cap BD$. Also note that $ E$ lies on the perpendicular bisector of $ MH$ because $ OM \equal{} OH$ and $ BM \equal{} BH$. Pentagon $ BEP'IH$ is cyclic and $ \angle{HEI} \equal{} \angle{IEP'}$ (because $ P'I \equal{} IH$). We want to prove that $ \angle{MEB} \equal{} \angle{P'EO}$. But

$ \angle{MEB} \equal{} \angle{BEH} \equal{} 90^\circ \minus{} \angle{HEI} \equal{} 90^\circ \minus{} \angle{IEP'} \equal{} \angle{P'EO}$.[/color]

[img]http://pic.ipicture.ru/uploads/090123/wup3F4x5m4.png[/img]

Let R be the point where the excircle touches BD.

Lemma 1: $ QC \equal{} CP$
Proof: Since AM and AN are tangents to the excircle, $ AM \equal{} AN$, and triangle MAN is isosceles. Thus $ \angle M \equal{} \angle N$. Since AD is parallel to BC $ \angle N \equal{} \angle BPM \equal{} \angle QPC$. $ \angle A \equal{} \angle C$, so by AA similarity $ \triangle MAN \sim \triangle QCP$, and triangle QCP is isosceles with $ QC \equal{} CP$.
--end lemma--

Lemma 2: $ DR \equal{} DQ$
Proof: $ \angle QPC \equal{} \angle PQC \equal{} \angle DQN$, so $ \angle M \equal{} \angle N \equal{} \angle QPC \equal{} \angle DQN$. Triangles AMN and DQN share $ \angle N$, so by AA similarity, they are similar. Triangle DQN is isosceles with $ DN \equal{} DQ$. Since DR and DN are tangent to the excircle $ DR \equal{} DN \equal{} DQ$.
--end lemma--

Lemma 3: $ BP \equal{} BR$
Since AD and BC are parallel $ \angle A \equal{} \angle MBP$. Triangles MAN and MBP share an angle, so by AA similarity they are similar. Triangle MBP is isosceles with $ MB \equal{} BP$. Since MB and BR are tangents to the excircle $ MB \equal{} BR$, so $ BP \equal{} MB \equal{} BR$.
--end lemma--

By the three lemmas triangles BPR, CPQ, and QDR are all isosceles, so the perpendicular bisectors of RP, PQ, and QR bisect B, C, and D respectively. Thus they are concurrent at I, the incenter of triangle BCD.

By SAS $ \triangle CPI \cong \triangle CQI$, $ \triangle DQI \cong \triangle DRI$, and $ \triangle BRI \cong \triangle BPI$.

Let $ \angle CPI \equal{} \angle CQI \equal{} \alpha$, $ \angle DQI \equal{} \angle DRI \equal{} \beta$, and $ \angle BRI \equal{} \angle BPI \equal{} \gamma$. $ \angle CQI \plus{} \angle DQI \equal{} 180^\circ$, $ \angle DRI \plus{} \angle BRI \equal{} 180^\circ$, and $ \angle BPI \plus{} \angle CPI \equal{} 180^\circ$, so $ \alpha \plus{} \beta \equal{} 180^\circ$, $ \beta \plus{} \gamma \equal{} 180^\circ$, and $ \gamma \plus{} \alpha \equal{} 180^\circ$. Solving this system of equations we get $ \alpha \equal{} \beta \equal{} \gamma \equal{} 90^\circ$.
IP, IQ, and IR are perpendicular to BC, CD, and DB, respectively, so the incircle of BCD touches BC, CD, and DB at P, Q, and R, respectively. Thus P and Q are on the incircle of BCD.


Let $ a,b,c$ be non-negative real numbers with sum $ 3$, prove that
\[ (1 \plus{} a^2b)(1 \plus{} b^2c)(1 \plus{} c^2a) \le 5 \plus{} 3abc
\]
and
\[ (1 \plus{} a\sqrt {b})(1 \plus{} b\sqrt {c})(1 \plus{} c\sqrt {a}) \le 3 \plus{} 5\sqrt {abc}.
\]
My best wish for every Mathlinkers, luck, health, happiness and success!

Your old friend  :)

merry Christmas and happy new year to you too, hungkhtn!
for the moment I have a proof for the firs one:
$ (1 \plus{} a^2b)(1 \plus{} b^2c)(1 \plus{} c^2a)\le 5 \plus{} 3abc$ is equivalent to
$ 1 \plus{} (a^2b \plus{} b^2c \plus{} c^2a) \plus{} abc(ab^2 \plus{} bc^2 \plus{} ca^2) \plus{} (abc)^3\le 5 \plus{} 3abc$ so using $ x^2y \plus{} y^2z \plus{} z^2x \plus{} xyz\le 4$ for $ (a,b,c)$ and $ (c,b,a)$
we need to prove $ abc \equal{} t$ and
$ 1 \plus{} (4 \minus{} t) \plus{} t(4 \minus{} t) \plus{} t^3\le 5 \plus{} 3t$
but the last one is equivalent to
$ t^2(1 \minus{} t)\geq 0$ which is true as $ 1 \equal{} \frac {a \plus{} b \plus{} c}{3}\geq \sqrt [3]{t}$  :)

That is correct! Albanian Eagle!

How could you prove the stronger one?

\[ abc(ab^2\plus{}bc^2\plus{}ca^2) \plus{}a^2b\plus{}b^2c\plus{}c^2a\le 4\plus{}2abc.\]

I think I proved that:
$ 2\geq (abc\plus{}1)(a^2b\plus{}b^2c\plus{}c^2a\minus{}2)$
this one seems to be really strong.

[quote="Albanian Eagle"]I think I proved that:
$ 2\geq (abc \plus{} 1)(a^2b \plus{} b^2c \plus{} c^2a \minus{} 2)$
this one seems to be really strong.[/quote]

This is not true, unfortunately,

@Sorry all. Did anyone know how to send solution and problem to Crux? I tried to ask and send to crux@cms.math.ca but I got no sign of answer.

:roll:
Ooops!
Ok then, give us a hint, I'm out of ideas.

OK. The hint is to consider two case: $ ab^2\plus{}bc^2\plus{}ca^2\le 3$ or $ \ge 3$. Then using a clever estimation.

yes exactly my approach!
Of course the case $ a^2b\plus{}b^2c\plus{}c^2a\le 3$ is obvious and we only need the other case, but I can't come with "the clever estimate"

Hi, Eagle. Here it is

\[ (ab^2\plus{}bc^2\plus{}ca^2\minus{}3)(abc\minus{}1) \le 0.\]