the point of L and M on the AB and the point of N,P on the BC and the point of K,Q on the AC such that MN||AC and LK||BC ,PQ||AB prove that the point of (AB,KN),(BC,MQ),(AC,PL)are collinear

$N,P\in BC,\ Q,K\in CA,\ L,M\in AB,\ LK\parallel BC,\ MN\parallel CA,\ PQ\parallel AB\Longrightarrow $

$\frac {KC}{KA}=\frac {LB}{LA},\ \frac {NB}{NC}=\frac {MB}{MA},\ \frac {QC}{QA}=\frac {PC}{PB}\ .\ \ (1)$

[b]I note [/b]$X\in BC\cap MQ,\ Y\in CA\cap LP,\ Z\in AB\cap NK$. [b]I will use the Menelaus's theorem[/b]
[b]in the triangle [/b]$ABC$ [b]for the transversal lines [/b]$\overline {XMQ},\ \overline {YPL},\ \overline {ZKN}\ :$

$\frac {XC}{XB}\cdot \frac {MB}{MA}\cdot \frac {QA}{QC}=1\ ;\ \ \ (2)$

$\frac {YA}{YC}\cdot \frac {PC}{PB}\cdot \frac {LB}{LA}=1\ ;\ \ \ (3)$

$\frac {ZB}{ZA}\cdot \frac {KA}{KC}\cdot \frac {NC}{NB}=1\ .\ \ \ (4)$

[b]Fom the product of the relations (1), (2), (3) and (4) results[/b] $\frac {XC}{XB}\cdot \frac {ZB}{ZA}\cdot \frac {YA}{YC}=1\Longrightarrow Z\in XY.$

A slightly different solution:

[color=blue][b]Problem.[/b] Let ABC be a triangle, let N and P be points on the line BC, let Q and K be points on the line CA, and let L and M be points on the line AB, such that KL || BC, MN || CA and PQ || AB. Prove that the points $BC\cap MQ$, $CA\cap PL$ and $AB\cap KN$ are collinear.[/color]

[i]Solution.[/i] We work on the projective plane.

The conditions KL || BC, MN || CA and PQ || AB are equivalent to KL || NP, MN || QK and PQ || LM. Hence, the points $KL\cap NP$, $LM\cap PQ$ and $MN\cap QK$ are infinite points, so they are collinear (they all lie on the line at infinity). Thus, the hextuple of points (K, L, M, N, P, Q) is Pascalian. Thus, by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24392]the Pascal permutation theorem[/url], the permutation (N, P, L, M, Q, K) of this hextuple must also be Pascalian. Hence, the points $NP\cap MQ$, $PL\cap QK$ and $LM\cap KN$ are collinear. But $NP\cap MQ=BC\cap MQ$, $PL\cap QK=CA\cap PL$ and $LM\cap KN=AB\cap KN$. Thus, the points $BC\cap MQ$, $CA\cap PL$ and $AB\cap KN$ are collinear, and the problem is solved.

  darij


$\boxed{2}$, by really hard addition.

this was actually on an algebra 1 mu alpha theta test last year.

Proof:

[img]http://upload.wikimedia.org/wikipedia/en/4/48/PM1%2B1%3D2lemma.jpg[/img]

Should this topic be locked?

2
because it is defined that way
take 1 then add 1,
which means to go to the next 1 number after 1!
count 1, then 2!

[quote="mathnerd314"]Should this topic be locked?[/quote]

probably.  or somebody should move this to college playground  :D


[quote="Temperal"]Show that for positive reals $ a,b$

$ \frac {a \plus{} b}{2}\ge \frac {1}{2a \plus{} 2b}$

EDIT: Okay, yet another problem in which I forgot to specify conditions. See below post.[/quote]
[hide="Solution"]
\[ \frac{a\plus{}b}{2}\ge \frac{1}{2a\plus{}2b}\]  \[ a\plus{}b\ge \frac{1}{a\plus{}b}\] \[ (a\plus{}b)^2\ge 1\] This is true when $ a\plus{}b\ge 1$ which is Temperal's condition, I think... [/hide]


[color=darkblue]Let $ABCD$ be a convex quadrilateral and denote the its area $S=[ABCD]$ and the its semiperimeter $p$.

Then exists the inequality $\boxed{\ 3\sqrt S\le p+\sqrt [4]{abcd}\ }\ ?$ If no (you have at least a counter example), then exists this inequality in  inscribed quadrilaterals ?[/color]

I did a post, it was wrong but I don't know how to erase ir I'll find a geometrical proof


pigeonhole principle
search
number theory unsolved
number theory

For any positive integer $n$,show that $n(n+1)(n+2)...(n+9)$ is not a perfect square.

even $n(n+1)$ cannot be a perfect square for $n\in\mathbb{N}$ .

since $gcd(n,n+1)=1$ so if it is a perfect square then it must be

$n=a^{2}$ , $n+1=b^{2}$ $\implies (b-a)(b+a)=1$ for integer $a,b$
so $a=0$ , $b=\pm 1$ , contradict with $n\in\mathbb{N}$

I know that...
But how can u use the lemma to solve the problem ???  :huh:

This problem is a special case of a theorem due to Erdos.

any one know the solution to Erdos problem ?

In Rushian forum I give solution for equation:
$(y+1)(y+2)...(y+k)=z^{2}.$
You have case k=10.

[quote="Rust"]In Rushian forum I give solution for equation:
$ (y \plus{} 1)(y \plus{} 2)...(y \plus{} k) \equal{} z^{2}.$
You have case k=10.[/quote]

I want to see your solution for $ (y \plus{} 1)(y \plus{} 2)...(y \plus{} k) \equal{} z^{2}.$

I will prove that equation $ (n\plus{}1)(n\plus{}2)...(n\plus{}10)\equal{}A^{2}$ dos not have positive integer solution.

Solution:

Among numbers $ n\plus{}1$, $ n\plus{}2$, ..., $ n\plus{}10$ at most 4 numbers are divisible by 5 or 7. Assume on contrary that for some  $ n$ number  $ (n\plus{}1)(n\plus{}2)...(n\plus{}10)$ is perfect square.

Then among numbers $ n\plus{}1$, $ n\plus{}2$,...,$ n\plus{}10$ there are at least 6 numbers of form $ 2^{x}3^{y}.c$, where all prime factors of number $ c$ are greater than 10 so it follows that $ c$ is perfect square so considered 6 numbers have form: $ 2^{x}3^{y}k^{2}$.

We have to discuss following cases (possible forms of these numbers):

1.$ x$ is even, $ y$ is even

2.$ x$ is even, $ y$ is odd

3.$ x$ is odd, $ y$ is even

4.$ x$ is odd, $ y$ is odd

By pigenhole principle at least two of six considered numbers have same form.

1.(In this case we discuss when there exist two numbers having form 1.)

Both numbers are perfect square. But since $ (x\plus{}1)^{2}\minus{}x^{2}\equal{}2x\plus{}1$ it follows that one of these numbers must be $ 2^{2}$, $ 3^{2}$ or $ 4^{2}$ but in these cases

 $ (n\plus{}1)...(n\plus{}10)$ is not perfect square (because of $ \mod 11, 13, 17$

2. In this case among numbers $ n\plus{}1$, $ n\plus{}2$,...,$ n\plus{}10$ there are two which have form $ 3x^{2}$ and $ 3y^{2}$ but since (WLOG $ x>y$)

 $ 3x^{2}\minus{}3y^{2}\geq 3(y\plus{}1)^{2}\minus{}3y^{2}\equal{}3(2y\plus{}1)$, so we must have $ y\equal{}1$ and again $ (n\plus{}1)...(n\plus{}10)$ is not perfect square.

The other two cases can be discussed in similar way and again we get that $ (n\plus{}1)(n\plus{}2)...(n\plus{}10)$ is not perfect square  CONTRADICTION so our proof is completed.

But i want to know the solution of $ (y \plus{} 1)(y \plus{} 2)...(y \plus{} k) \equal{} z^{2}.$[/quote]
 :(

I just answered to initial ([b]suweijie[/b]'s) question :wink:

here is a topic about [b]The Erdos-Selfridge theorem[/b] which states that [b]the product of consecutive numbers cannot become powers[/b]:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1865593861&t=173365[/url]

[quote="Rust"]In Rushian forum I give solution for equation:
$ (y \plus{} 1)(y \plus{} 2)...(y \plus{} k) \equal{} z^{2}.$
You have case k=10.[/quote]
dear Rust , I'm very interested see you'r solution,can you post it here?or you'r idea?
thanks... :)


[url=http://www.iapt.org.in/index.php?option=com_content&task=view&id=28&Itemid=54]Here[/url] are the solutions to the various National Standard Examinations,2007
Post in ur expected marks

hi there
so did u give the NSE's???

when r the results?

hi.
yes .The results should be out by early next year

Thanks! BTW how did find that link. Its not available anywhere. (I presume you are a student). Results were out on Dec 23-24 last year, dunno about this year though.

Hmm...I feel  one or 2 of their solns may be wrong. Anyways, I've done okay in the objective, slightly bad in the subjective. The cutoffs have usually been around 150-170. I might clear if its near the lower bound but difficult if otherwise.

How much are you getting?

P.S Am referring to physics. I dont really rem what I marked in chem.

same here done well in obj. terrible in subjective.
also last time the results came out in early Jan.(i am sure about this one)

You are right. Jan 4th.

http://www.mathlinks.ro/viewtopic.php?t=126908

How much do you think was the cutoff last year in Chem???

I GOT 13 WRONG AND LEFT 13. SO IM EXPECTING 209. IS DAT GUD ENF??? :(

Well this year was seemingly tougher than last year so I think only 180 should be the cut off.
 :(

i think a few solutions are wrong. doesnt diborane have a 3 centre 2 electron bond or is it 2 centre 3 electron bond.

I think diborane has 4 centre 4 electron bond.
Also questions 5 and 80 have wrong answers.

ques 80 has the wrong answer but 5 is rite

N WHAT ABT Q 17 DOESNT D TARTARIC ACID HAVE AN ALTERNATING AXIS OF SYMMETRY

If it has symmetry how can it be opt. active???  :ninja:   :rotfl:

AN ALTERNATING AXIS OF SYMMETRY DOES NOT MEAN DAT IT ISNT OPTICALLY ACTIVE. ONLY A PLANE OF SYMMETRY DOES. NOT EVEN A CENTRE OF SYMMETRY MEANS DAT THE COMPOUD IS OPTICALLY INACTIVE

I haven't checked my score, but I think you people are mistaken about the cut-offs. Last year, I scored ~95 and the cutoff was ~105 (I got merit certificate). This year as the paper was easier, the cutoff will probably rise to ~130-140. I think 180 is too pessimistic. Either case, I don't hope to go beyond 2-digits, hence best of luck to y'all.

EDIT: All this is for NSEP

Ahem  :D  We were talking abt Chem olympiad
Regarding Physics i think it might just be 150  :P

I really hope these solutions are correct :D  :D  :D


Hi!

I should post this while I still can..

I've been having some trouble trying to log in. Because of my network, I can only browse AoPS (and not ML). Foirst, my old password didn't work. I asked for  new one, and when I changed it with my old one I still couldn't log in (I was using IE and/or Opera) Now I'm logged in but it only worked with Firefox. What's up with this?

Oh, and another thing: I tried to create a new account, but no matter how hard I tried I couldn't because I kept getting the message that the code I entered was incorrect (you know, when you register there is a code you have to read and then enter). I'm pretty sure it wasn't incorrect though..

Sorry for the trouble and for all the mail sent to the admins (I'm sure I was pretty annoying :)).

It appears to be solved, indeed. Thank you!


Say I want to show someone how to type this: $\alpha$.

How would I put actual code (the dollar signs and the \alpha) inside one of those boxes labeled code?

I would do it like this:
[code]\alpha[/code]
So, just the latex without the dollar signs.  Otherwise, you see the alpha character.


function
Gauss
algebra
floor function
ceiling function

What's the difference between ceil function and floor function?
I guess floor function denotes the greatest integer which is less than or equal to X,
and ceil function denotes the least integer which is greater than or equal to X,right?
Don't laugh at me since I ask a so foolish question,because in China we only have Gauss function。

Here are some examples : http://aoki2.si.gunma-u.ac.jp/Hanasi/Algo/diff.html
The floor function : http://mathworld.wolfram.com/FloorFunction.html
The ceiling function : http://mathworld.wolfram.com/CeilingFunction.html
 :)


Sometimes after my PC was shut down, I'm logged out.
(and the canonical auto-login is turned on, cookies, too)

(and still FireFox 1.5.0.4, Windoof XP)

Browser problem. Can't deal with that. Re-log in :)

I didn't have that before the update...
And it reoccured four times till now.

Same story here... and yes, even after resetting all cookies. But haven't found fixed pattern how and when.

Might be a bug though then. Although I repeat I didn't have this problem yet :|

Yeah it's not in fixed pattern... I have no idea what causes it. Maybe something every 48 hours? (which is a fair approximation from what I had... will let you know when it happens again, this is far from urgent, it's not every time, just sometimes... but I don't ever shut down my comp so it's not about shutting down.)

The auto-login key expires after 14 days of not using it. So I really do not understand what is the trouble.

For me it seems to expire all 12 hours or so, I got logged out three times yesterday, but only once before (a week or so away) and today again.

Again now, so still fitting in the 48h assumption.

[edit July 7, just happened again now (around 13.00? my time)... so the pattern doesn't hold but the problem isn't solved]


If $ \&x$ is defined as $ \&x \equal{} x \plus{} 5$ and $ \#x$ is defined as $ \#x \equal{} x^2$ , what is the value of $ \#(\&4)$?

[hide]Just go through all the steps, as it says.

&4=4+5=9
#9=$9^2=81$

DONE[/hide]


inequalities
trigonometry
function
geometry
inequalities proposed

Prove that in any triangle $ABC$ with $A, B, C\in\left(0,\frac{\pi}{2}\right)$ the following inequality is valid:

                          \[ \sqrt{\sin A}+\sqrt{\sin B}+\sqrt{\sin C} >2. \]

The inequality is
\[ \sqrt{a}+\sqrt{b}+\sqrt{c}>\sqrt{8R}  \]
Finally it will remain that
\[ a+b+c>\frac{2abc}{S}  \]
...

Using the fact that function $f(x) = \sqrt{sin(x)}$ is concave on $(0, Pi/2)$ we will have

         $f(A) + f(B) > f(A+B-Pi/2) + f(Pi/2)$ if $A+B \ge Pi/2$
         $f(C) + f(A+B-Pi/2) > f(0) + f(A+B+C-Pi/2)$

and we have done

We have:
\[ 2 < \sqrt{\sin A}+\sqrt{\sin B}+\sqrt{\sin C} < \frac{14}{5} \newline \sqrt{\sin A}+\sqrt{\sin B}+\sqrt{\sin C} > \sin^2 A+\sin^2 B+\sin^2 C > 2  \]

[b]Edit:[/b] Oh, I solved it with $\sin^2 A$, not $\sin A$, sorry for my carelessness.

$sin(A)=\frac{2S}{bc}$ and the analogues.Then all we have to do is prove that $\sqrt{2S}(\frac 1{\sqrt{bc}}+ \frac 1{\sqrt{ab}} + \frac 1{\sqrt{ac})} >2$
And $\sum {\frac 1{\sqrt{bc}} \geq \frac 9{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}} \geq \frac 9{a+b+c}}$.Therefore we have to prove that $\sqrt{2S}\cdot \frac9{a+b+c} > 2 \iff 2S\cdot \frac{81}{(a+b+c)^2} > 4 \iff \frac{81S}{(a+b+c)^2} >2$.But $\frac{81S}{(a+b+c)^2}=\frac{81r}{2(a+b+c)}$ therefore we want to prove that $\frac{81r}{2(a+b+c)} > 2 \iff \frac{r}{a+b+c} >\frac4{81}$ which is preety resonable..
I have denoted $S$=the area of the triangle and $r$=the radius of the incircle.I hope there were no mistakes in my solution.

[quote="luke999"]The inequality is
\[ \sqrt{a}+\sqrt{b}+\sqrt{c}>\sqrt{8R}  \]
Finally it will remain that
\[ a+b+c>frac{2abc}{S}  \]
...[/quote]
   How can you prove this?There are lots of methods to prove it, but I want to see how can you prove this one. :D

Another solution.

We know that $\sqrt{sinA}> sin^2A$. Hence $\sum \sqrt{sinA} > \sum sin^2A = 2 + 2cosAcosBcosC > 2$.

Treegoner solution is the best. I dont know why Lovasz consider sinA + sin B + sin C > 2 as trivial.

Yes, That is right. The following inequality holds in triangle $ABC$
$2\leq \sin^2A+\sin^2B+\sin^2C\leq\frac94$.

We have in all triangles $ABC$
$\sqrt{\cos\tfrac12 A}+\sqrt{\cos \tfrac12 B}+\sqrt{\cos\tfrac12 C}\geq2$.


Let be $ a,b,c$ the sidelenghts of a triangle such that :

$ \frac{a}{b\plus{}c\minus{}a}\plus{}\frac{b}{a\plus{}b\minus{}c}\plus{}\frac{c}{c\plus{}a\minus{}b}\equal{}9\cdot \min\{a\minus{}b^2,b\minus{}c^2,c\minus{}a^2\}$

Determine $ a,b,c$ .

hello, do you mean $ \min\{a^2\minus{}b^2,b^2\minus{}c^2,c^2\minus{}a^2\}$ or $ 9\min\{(a\minus{}b)^2,(b\minus{}c)^2,(c\minus{}a)^2\}$?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, do you mean $ \min\{a^2 \minus{} b^2,b^2 \minus{} c^2,c^2 \minus{} a^2\}$ or $ 9\min\{(a \minus{} b)^2,(b \minus{} c)^2,(c \minus{} a)^2\}$?
Sonnhard.[/quote]

I meant $ 9\min\{a \minus{} b^2,b \minus{} c^2,c \minus{} a^2\}$  :)


Why is the number $ 2^65\plus{}1$ not prime?

[quote="the future"]Why is the number $ 2^65 \plus{} 1$ not prime?[/quote]You mean the number $ 2^510 \plus{} 1$, otherwise known as [hide="..."]$ 321$[/hide]?
I can plainly see that it is a multiple of [hide="..."]$ 3$.[/hide]
I guess there must be a more elegant way of proving it that does not involve knowing by heart the powers of 2, and that would work if it was not such an easy, small number.

I think he means $ 2^{65} \plus{} 1$
In which case; $ 2^{65} \plus{} 1 \equiv ( \minus{} 1)^{65} \plus{} 1 \equiv \minus{} 1 \plus{} 1 \equiv 0 \mod 3$
So $ 2^{65} \plus{} 1$ is a multiple of $ 3$ and hence can't be prime


calculus
integration
calculus computations

Determine whether the improper imtegral $ \int_0^\infty\frac{ln(1\plus{}\alpha^2u^2)}{1\plus{}u^2}du$ where $ \alpha>0$.

Can someone please help. Thank you very much. :P

Let:

$ \int_{0}^{\infty}f(u)du \equal{} \int_0^1 f(u)du \plus{} \int_1^{\infty}f(u)du$

The first clearly converges and the second is asymptotic to:

$ \frac {ln(1 \plus{} \alpha^2 u^2)}{1 \plus{} u^2}\sim 2\frac {ln(\alpha) \plus{} ln(u)}{u^2}$ as $ u\to\infty$ and

$ \int_1^{\infty}\frac {ln(u)}{u^2}du$ converges.  Thus the original integral does so (assuming $ \alpha > 0$).

Oh yea, here's the SOS reference I find useful for these types of problems:

http://www.sosmath.com/calculus/improper/testconv/testconv.html


If $ a,b,c\in \mathbb{R}_\plus{}$ such that $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\le 3$ , prove that :

$ \sum_{cyc}\frac{a^2\plus{}1}{\sqrt{a^2\minus{}a\plus{}1}}\ge 6$

[quote="alex2008"]If $ a,b,c\in \mathbb{R}_ \plus{}$ such that $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\le 3$ , prove that :

$ \sum_{cyc}\frac {a^2 \plus{} 1}{\sqrt {a^2 \minus{} a \plus{} 1}}\ge 6$[/quote]
Enough to prove that $ \sum_{cyc}\left(\frac {a^2 \plus{} 1}{\sqrt {a^2 \minus{} a \plus{} 1}}\minus{}2\right)\ge \sum_{cyc}\left(1\minus{}\frac{1}{a}\right),$
which is true since, $ \frac {a^2 \plus{} 1}{\sqrt {a^2 \minus{} a \plus{} 1}}\plus{}\frac{1}{a}\geq3$ is true for all $ a>0.$


solve in R:

$\sqrt{\frac{x}{y+z}}+\sqrt{\frac{y}{x+z}}+\sqrt{\frac{z}{x+y}}=\frac{3}{2}$

[quote="cckek"][quote="pohoatza"]something is fishy here:

by AM-GM : $\sqrt{\frac{x}{y+z}\cdot 1}\geq \frac{2x}{x+y+z}$, and the analogues.

so: $\sum{\sqrt{\frac{x}{y+z}}}\geq 2$[/quote]
nice:)
but that was an olimpiad problem here in our country>Romania:)[/quote]

lol ?


It is known that $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{d}\equal{}1$ and $ (1\plus{}\frac{a}{bc})(1\plus{}\frac{b}{ca})(1\plus{}\frac{c}{ab})\equal{}\frac{1}{d^2}$, $ a,b,c,d \in R$. Find $ a\plus{}b\plus{}c$

a ,b ,c and are non zero .
Substituting 1/d = 1-(1/a+1/b+1/c) from the 1st equation to the 2nd one results in (a+b+c-1)^2 =0 then a+b+c =1

hello, by the same way i have found
$ (1\plus{}\frac{a}{bc})(1\plus{}\frac{b}{ca})(1\plus{}\frac{c}{ab})\minus{}\frac{(\minus{}bc\minus{}ca\minus{}ab\plus{}abc)^2}{a^2b^2c^2}\equal{}\frac{(a\plus{}b\plus{}c\minus{}1)^2}{abc}\equal{}0$
From here we get $ a\plus{}b\plus{}c\equal{}1$
Sonnhard.

$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\minus{}1\equal{}\minus{}\frac{1}{d} \Rightarrow \frac{1}{d^2}\equal{}\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{2}{ab}\plus{}\frac{1}{c^2}\plus{}1\minus{}\frac{2}{c}\plus{}\frac{2a\plus{}2b\minus{}2ac\minus{}2bc}{abc}$
and
combining this with the second equation we got

$ a^2\plus{}b^2\plus{}c^2\plus{}1\equal{}2a\plus{}2b\plus{}2c\minus{}2ab\minus{}2ac\minus{}2bc  \Rightarrow (a\plus{}b\plus{}c\minus{}1)^2\equal{}0$ so $ a\plus{}b\plus{}c\equal{}1$


algebra
polynomial
linear algebra
matrix
superior algebra
superior algebra unsolved

$1\leq d \leq n$ integers

$C_d=\{A\in M_n(C); deg(\pi_{A}) =d \}$

Prove that $C_d$ is connex by arcs

$\pi_{A}$ is minimal polynomial of $A$

I think we can do like this:

let $M$ be an arbitrary matrix.
Let $a$ be eigenvalue and $d_a$ be the maximum
of the dimensions of jordan cells associated with $a$.
It is easily seen that $d$, the degree of the minimal
polynomial of $M$, is $d_{a_1}+d_{a_2}+..+d_{a_r}$ where
$\{a_1,a_2,..,a_r\}$ are the $r$ distinct eigenvalues.
From that the conclusion follows
(remember that $GL(n,C)$ is arc connected)

Moubi, is your solution different?

Could you please continue, alekk? Is it that simple from here? I remember trying to do the same thing, except that I found some difficulties. How exactly do you turn the Jordan cells of one matrix into those of the other one continuously? (that's all we need to establish after the observation that $GL_n(\mathbb C)$ is path connected). Or am I simply blind? :?

if you have a matrix $M$ with 2 jordan blocks $J_p(a)$ and $J_q(b)$,that are neighbour on the diagonal, I think that one can construct a path towards a matrix where there is a jordan cell $J_{p+q}(c)$. It's hard to explain because i don't know how to write matrices in Latex. I will learn soon  :blush:

Could you at least give a hint as to what a path from, say, $diag(0,1,1,1)$ to $diag(0,0,1,1)$ might look like? :?

[quote="grobber"]How exactly do you turn the Jordan cells of one matrix into those of the other one continuously? (that's all we need to establish after the observation that $GL_n(\mathbb C)$ is path connected). [/quote]

$GL_n(\mathbb{C})$ is the complement of a complex (i.e. real-codimension 2) hypersurface in $M_n(\mathbb{C})$.  To path-disconnect the space you need the excised subspace to have codimension 1, or be "wild".


Let ABC. Find MIN
$ S \equal{} \sum \frac {a}{\sqrt{a^2 \plus{} 3bc}}$

[quote="nguyensilva"]Let ABC. Find MIN
$ S \equal{} \sum \frac {a}{\sqrt {a^2 \plus{} 3bc}}$[/quote]

when $ a \equal{} b \equal{} c$ get $ S \equal{} \frac {3}{2}$, then we will prove that is the min.

to prove that $ \sum_{cyc} \frac {a}{\sqrt {a^2 \plus{} 3bc}} \ge \frac {3}{2}$, by Holder Ineq, we have

$ \sum_{cyc} \frac {a}{\sqrt {a^2 \plus{} 3bc}}$

$ \equal{} \sum_{cyc} \frac {a^{\frac {3}{2}}}{(a^3 \plus{} 3abc)^{\frac {1}{2}}}$

$ \ge \frac {(a \plus{} b \plus{} c)^{\frac {3}{2}}}{(a^3 \plus{} b^3 \plus{} c^3 \plus{} 9abc)^{\frac {1}{2}}},$

so we just need to prove that

$ \frac {(a \plus{} b \plus{} c)^{\frac {3}{2}}}{(a^3 \plus{} b^3 \plus{} c^3 \plus{} 9abc)^{\frac {1}{2}}} \ge \frac {3}{2},$

or

$ 4(a \plus{} b \plus{} c)^3 \ge 9(a^3 \plus{} b^3 \plus{} c^3 \plus{} 9abc),$

for this, we subs $ a \equal{} y \plus{} z,b \equal{} z \plus{} x,c \equal{} x \plus{} y,x,y,z > 0$, ineq becomes

$ 14\sum_{cyc}{x^3} \plus{} 30xyz \ge 12\sum_{sym}x^2y,$

by AM-GM, we have $ 2\sum_{cyc}{x^3} \ge 6xyz$, so we just need to prove that

$ 12\sum_{cyc}{x^3} \plus{} 36xyz \ge 12\sum_{sym}x^2y$

$ \iff 12\sum_{cyc}{x(x \minus{} y)(x \minus{} z)} \ge 0,$

which is Schur Ineq.

done.

[quote="kuing"]

to prove that $ \sum_{cyc} \frac {a}{\sqrt {a^2 \plus{} 3bc}} \ge \frac {3}{2}$, by Holder Ineq, we have

$ \sum_{cyc} \frac {a}{\sqrt {a^2 \plus{} 3bc}}$

$ \equal{} \sum_{cyc} \frac {a^{\frac {3}{2}}}{(a^3 \plus{} 3abc)^{\frac {1}{2}}}$

$ \ge \frac {(a \plus{} b \plus{} c)^{\frac {3}{2}}}{(a^3 \plus{} b^3 \plus{} c^3 \plus{} 9abc)^{\frac {1}{2}}},$

[/quote]

Sorry, why?

[quote="nguyensilva"]$ 4(a \plus{} b \plus{} c)^3 \ge 9(a^3 \plus{} b^3 \plus{} c^3 \plus{} 9abc)$[/quote]
This inequality is not correct. Try $ a\to$ $ \infty^\plus{}$. We can prove that $ \frac{(a\plus{}b\plus{}c)^{\frac{3}{2}}}{(a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}9abc)^{\frac{1}{2}}}>1 \Longleftrightarrow \sum_{cyc}ab(a\plus{}b)>abc$ which is true because $ \sum_{cyc}ab(a\plus{}b)\geq 6abc$ (AM-GM). Hence $ \sum_{cyc}\frac{a}{\sqrt{a^{2}\plus{}3bc}}>1$. But if $ a\to$ $ \infty^\plus{}$ we observe that $ \sum_{cyc}\frac{a}{\sqrt{a^{2}\plus{}3bc}}$ $ \to 1$. So 1 is the best lower bound.

[quote="Flakky"][quote="nguyensilva"]$ 4(a \plus{} b \plus{} c)^3 \ge 9(a^3 \plus{} b^3 \plus{} c^3 \plus{} 9abc)$[/quote]
This inequality is not correct. Try $ a\to$ $ \infty^ \plus{}$. We can prove that $ \frac {(a \plus{} b \plus{} c)^{\frac {3}{2}}}{(a^{3} \plus{} b^{3} \plus{} c^{3} \plus{} 9abc)^{\frac {1}{2}}} > 1 \Longleftrightarrow \sum_{cyc}ab(a \plus{} b) > abc$ which is true because $ \sum_{cyc}ab(a \plus{} b)\geq 6abc$ (AM-GM). Hence $ \sum_{cyc}\frac {a}{\sqrt {a^{2} \plus{} 3bc}} > 1$. But if $ a\to$ $ \infty^ \plus{}$ we observe that $ \sum_{cyc}\frac {a}{\sqrt {a^{2} \plus{} 3bc}}$ $ \to 1$. So 1 is the best lower bound.[/quote]

I don't think so, because in this ineq, $ a,b,c$ are the sides of a triangle... :wink:

Sorry I thought that $ a,b,c \in R^\plus{}$ are arbitrary


Sum the reciprocals of the triangular numbers. This might be a little too simple for this forum. I'm new and half my messages have been moved so far.

[hide]
Since the general form of a triangular number is $\frac{n(n+1)}{2}$, its reciprocal is $\frac{2}{n(n+1)}$.

So we have $2\sum_{k=1}^{n} \frac{1}{k(k+1)}$, which leads to telescoping series. [/hide]

[quote="seethelight"]Actually, the series does converge, and quite nicely. We know this because of basic big O work and the relation between triangular numbers and powers. Try to find the limit![/quote]
Well, based on my previous post, the sum can be written as $\sum_{k=1}^{n} \frac{2}{k(k+1)}$, which is equivalent to $2\left(\left(1-\frac{1}{2}\right)+ \left(\frac{1}{2}-\frac{1}{3}\right)+ \left(\frac{1}{3}-\frac{1}{4}\right) +...\right)$ (telescoping series).

So $\lim_{n\to\infty} \sum_{k=1}^{n} \frac{2}{k(k+1)}=2$.


algebra
polynomial
function
real analysis
real analysis unsolved

Let $H_n$ be the set of all homogenous complex polynomials of $n$ variables. Prove that $H_n$ is the direct sum of the space of polynomials that are multiples of $X_1^2+...+X_n^2$ and those that are harmonic.

Little precision: actually $H_n$ should be read $H(n,p)$ and it is the set of homogenous polynomials of degree $p$ and $n$ variables.  :blush:

it is odd that some classical result appear in contest, i remember that it is proved in Harmonic function theory by Axler.

Sam-n, probably it is odd, but don't forget that at ENS they want future scientists with culture. Whether their method is good or not, I don't know, but since 7 Fields medals are in possesion of students of this school, something makes me think their method works.  ;)


$\bigstar$[b] Problem:[/b] [ZaiZai-Hoang]
Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that:
\[\frac{27}{2}\ge 4(ab+bc+ca)+\frac{a^{2}b^{2}}{a+b}+\frac{b^{2}c^{2}}{b+c}+\frac{c^{2}a^{2}}{a+c}\]

$(a+b)^{2}\geq 4ab$  then
$\frac{ab(a+b)}{4}\geq \frac{a^{2}b^{2}}{a+b}$
So we have
$4(ab+bc+ac)+\frac{a^{2}b^{2}}{a+b}+\frac{b^{2}c^{2}}{b+c}+\frac{a^{2}c^{2}}{a+c}\leq 4(ab+bc+ac)+\frac{ab(a+b)}{4}+\frac{bc(b+c)}{4}+\frac{ac(a+c)}{4}(*)$
Suppose $a>b$
$4(ab+bc+ac)+\frac{ab(a+b)}{4}+\frac{bc(b+c)}{4}+\frac{ac(a+c)}{4}=4(ab+c(a+b))+\frac{ab(a+b)}{4}+\frac{c}{4}(a^{2}+b^{2}+c(a+b))= 4(ab+c(3-c))+\frac{ab(3-c)}{4}+\frac{c}{4}(a^{2}+b^{2}+c(3-c))$
By Sturm method making $a$ and $b$ closer saving sum and $a\geq b$ makes their product and sum of squares bigger. By doing this we can get  $a=b=c$ and $(*)$ will only increase.
So maximum of $(*)$ is when $a=b=c=1$
$4(ab+bc+ac)+\frac{ab(a+b)}{4}+\frac{bc(b+c)}{4}+\frac{ac(a+c)}{4}\leq \frac{27}{2}$

Sturm method, what does this mean? I don't know about Sturm method. Can you show me?  :roll:
But my solution is simpler than above proof, maybe  :wink:
Let try another method  :D

I’m so sorry… My “solution” is wrong
About Sturm’s method
Suppose $a>b$ and their sum is fixed ($a+b=s$). We want to see what is going on with their product while making them closer but saving $a\geq b$. I mean: what is bigger $(a-x)(b+x)$ or $ab$, where $x>0$ and $a-x\geq b+x(x\leq \frac{a-b}{2})$ .
So $(a-x)(b+x)=ab+x(a-b-x)>ab$, because $x\leq \frac{a-b}{2}$ and $a>b$
So making $a$ and $b$ closer(saving their sum and $a\geq b$) makes their product bigger.

Simple example of usage
$a,b,c>0$
$a+b+c=3$
Find maximum of $ab+bc+ac (*)$
Solution
Suppose $a>1>b$ then $ab+bc+ac=ab+c(3-c)$
 making $a$ and $b$ closer makes $(*)$ bigger
We will make one of them =1(we should remember that we must save $a\geq b$ and their sum)
Doing such operations we can make $a=b=c=1$ and $(*)$ will only increase
So $ab+bc+ac\leq 3$

And now the same about sum of squares(We also suppose a>b and their sum is fixed $(a+b=s)$)
We will make them closer(also saving sum and $a\geq b$)
$(a-x)^{2}+(b+x)^{2}<a^{2}+b^{2}$
(And in my proof I told the opposite :oops: )

Zai Zai excuse me I didn't see your ineg before ^^"

From Cauchy we have :

$ \frac{a^{2}b^{2}}{a+b}+\frac{a^{2}c^{2}}{a+c}+\frac{b^{2}c^{2}}{b+c}+4(ab+ac+bc)\leq\frac{(ac+ab+bc)^{2}}{6}+4(ab+ac+bc)$

Set [b]q = ab + ac + bc[/b], [b]p = a + b + c = 3[/b] and [b]r = abc[/b]. We rewrite the ineqality :

$ \frac{a^{2}b^{2}}{a+b}+\frac{a^{2}c^{2}}{a+c}+\frac{b^{2}c^{2}}{b+c}+4(ab+ac+bc)\leq\frac{q^{2}+24q}{6}$

From Schur, t = 1 case, we have

$ a^{1}(a-b)(a-c)+b^{1}(b-a)(b-c)+c^{1}(c-a)(c-b)\geq0\Leftrightarrow p^{3}-4pq+9r\geq0$

Thus,

$ q\leq3r$

But, $ r=abc\leq\frac{(a+b+c)^{3}}{27}=1$ by AM-GM, so $ q\leq3$

Indeed,

$ \frac{q^{2}+24q}{6}\leq\frac{81}{6}=\frac{27}{2}$

[quote="zaizai-hoang"]$ \bigstar$[b] Problem:[/b] [ZaiZai-Hoang]
Let $ a,b,c$ be positive real numbers such that $ a+b+c=3$. Prove that:
\[ \frac{27}{2}\ge 4(ab+bc+ca)+\frac{a^{2}b^{2}}{a+b}+\frac{b^{2}c^{2}}{b+c}+\frac{c^{2}a^{2}}{a+c}\]
[/quote]

We'll use $ \frac{ab(a+b)}{4}\geq \frac{a^{2}b^{2}}{a+b}\leftrightarrow (a-b)^{2}\geq 0$.

So it's enough to prove

$ \frac{27}{2}\geq 4\sum ab+\sum \frac{ab(a+b)}{4}$

With notation $ a+b+c=p$, $ ab+bc+ca=q$, $ abc=r$, last inequality is equivalent with

$ \frac{27}{2}\geq 4q+\frac{pq-3r}{4}$

$ 54 \geq 16q+pq-3r$

$ 3r+54 \geq 19q$$ ...(*)$

Form Schur's inequality we have $ p^{3}+9r\geq 4pq$, but $ p=3$ so $ 3r \geq 4q-9$, and using that in $ (*)$, it's enough to prove $ 4q+45 \geq 19q \leftrightarrow 3\geq q$, but that's trivial since $ a+b+c=3$.

very well, behemont.
You have as same as idea with me :D but I used to them by another way :D
Let $ p=a+b+c,q=ab+bc+ca,r=abc$. The inequality is equivalent to:
\[ \frac{27}{2}\ge 4q+\frac{q^{3}-6pr+3r^{2}}{3q-r}\]
By schur the inequality is equivalent to:
\[ 6p(9-3q)(q+2)\ge 0\]
It is easy to prove. Because: $ 9\ge 3q$ (apply AM-GM inequality)
We are done. This technique has attraction for me, we can solve many probems :wink:


geometry
area of a triangle
geometry proposed

Dear friends,

   Let $ ABC$ be a triangle, $ AD$ - altitude, $ BE$ - bisector of the angle $ \angle ABC$ and $ CF$ - median $ \left(D\in BC, E\in CA, F\in AB\right)$.
1.If $ AD \equal{} BE \equal{} CF$ then the triangle $ ABC$ is equilateral.
2.If one of the angles of the triangle $ ABC$ has $ 60^{\circ}$ and the lines $ AD,BE,CF$ are concurrent then the triangle $ ABC$ is equilateral.  :)

1. By the formula of area of a triangle:
$ (ABC)=\frac{BC.AD}{2}=\frac{AB.SENBFC.CF}{2}$
But: $ \frac{AB.CF}{2}\geq\frac{AB.SENBFC.CF}{2}=\frac{BC.AD}{2}$
Then like $ CF=AD$:
$ AB%Error. "geqBC" is a bad command.
$

Let $ senEBC=senx$.
Like $ EB$ is the bisector:
$ EB=\frac{2.cosx.AB.BC}{AB+BC}$
But like $ AD$ is an altitude:
$ AD=AB.senABC=AB.2.senx.cosx$
And like $ AD=EB$:
$ \frac{2.cosx.AB.BC}{AB+BC}=AB.2.senx.cosx$
$ \implies senx=\frac{BC}{AB+BC}$
$ \implies\frac{1}{2}\geq senx$
$ \implies30^o \geq x$

Like $ AD=CF$ the triangles $ CFB$ and $ ADB$ have the same circumradious.
Then:
$ \frac{BF}{senBCF}=\frac{AB}{senADB}$
But $ AB=2.BF$:
$ \implies senBCF=\frac{1}{2}$
$ \implies BCF=30^o$

Then too:
$ \frac{BC}{senBFC}=\frac{AB}{senADB}$
$ \implies senBFC=\frac{BC}{AB}$
$ \implies sen2x+30^o=\frac{BC}{AB}$

But $ senx=\frac{BC}{AB+BC}$:
$ \implies\frac{senx}{1-senx}=\frac{BC}{AB}$
Then $ \frac{senx}{1-senx}=sen2x+30^o$
So:
$ \frac{senx}{1-senx}=2.senx.cosx.cos30^o+(1-2sen^2x)sen30^o$
$ \frac{senx}{1-senx}-sen30^o=2.senx(cosx.cos30^o-senxsen30^o)$
$ \frac{2.senx-1+senx}{2.senx(1-senx)}=2(cosx+30^o)$
$ \frac{3.senx-1}{2.senx(1-senx)}=2(cosx+30^o)$

Let prove that $ 1\geq\frac{3.senx-1}{2.senx(1-senx)}$
$ \Leftrightarrow 2.senx(1-senx)\geq 3.senx-1$
$ \Leftrightarrow 1\geq 2sen^2x+senx$
And it is true because $ \frac{1}{2} \geq senx$
Then $ 1\geq2(cosx+30^o)$
And that implies:
$ x+30^o\geq60^o$
$ \implies x \geq 30^o$
So $ 30^o \geq x \geq30^o$
Hence $ x=30^o$
And is easy to prove that the triangle is equilateral.

It was a problem for 8-th grade, proposed on all Russian Math Olympiad 1972-73 maybe.

[code]Petry wrote:
If one of the angles of the triangle $ABC$ has $60^o$ and the lines $AD$, $BE$, $CF$ are concurrent then the triangle  is equilateral.
[/code]
It is easy to prove that $ ED\parallel{}AB$.

Then $ \frac {CD}{DB}\equal{}\frac{CE}{EA}\equal{}\frac{BC}{AB}$

Hence $ \frac{cosC}{cosB}\equal{}\frac{CD.AB}{DB.AC}\equal{}\frac{BC}{AC}\equal{}\frac{senA}{senB}$

Then $ cosC.senB\equal{}senA.cosB$

First case:($ C\equal{}60^o$)
$ \frac{1}{2}senB\equal{}senA.cosB$
$ \frac{1}{2}senB\equal{}sen(B\plus{}60^o).cosB$

$ senB\equal{}\frac{\sqrt3.cos^2B}{1\minus{}cosB}$

If $ B\geq60^o$

 $ \implies senB\geq\frac{\sqrt3}{2}$

 Then $ \frac{\sqrt3.cos^2B}{1\minus{}cosB} \geq \frac{\sqrt3}{2}$
 $ \implies(2.cosB\minus{}1)(cosB\plus{}1)\geq0$
 Then $ cosB \geq \frac{1}{2}$
 $ \Leftrightarrow 60^o \geq B$
 $ \Leftrightarrow B\equal{}60^o$

If $ 60^o \geq B$
 $ \implies cosB\geq\frac{1}{2}$

 Then $ \frac{\sqrt3.cos^2B}{1\minus{}cosB} \geq \frac{\sqrt3}{2}$
 $ \implies senB \geq \frac{\sqrt3}{2}$
 $ \Leftrightarrow B \geq 60^o$
 $ \Leftrightarrow B\equal{}60^o$

Second case:($ B\equal{}60^o$)

$ \frac{\sqrt3}{2}cosC\equal{}senA\frac{1}{2}$
$ \sqrt3cosC\equal{}sen(C\plus{}60^o)$
$ \implies\sqrt3cosC\equal{}senC$
$ \Leftrightarrow C\equal{}60^o$

Third case: ($ A\equal{}60^o$)

$ \frac{\sqrt3}{2}cosB\equal{}cosC.senB$

But $ cosC\equal{}\minus{}cos(B\plus{}60^o)$

$ \frac{\sqrt3}{2}cosB\equal{}\minus{}cos(B\plus{}60^o).senB$

$ \implies \frac{\sqrt3.sen^2B}{\sqrt3\plus{}senB}\equal{}cosB$

If $ B\geq60^o$
 $ \implies \frac{1}{2}\geq cosB$

 $ \implies \frac{1}{2} \geq \frac{\sqrt3.sen^2B}{\sqrt3\minus{}senB}$

 $ 0\geq (senB\minus{}\frac{\sqrt3}{2})(2.\sqrt3.senB\plus{}2)$
 $ \Leftrightarrow \frac{\sqrt3}{2}\geq senB$
 $ \Leftrightarrow 60^o \geq B$
 $ \Leftrightarrow 60^o\equal{}B$
If $ 60^o \geq B$
 $ \implies cosB\geq \frac{1}{2}$

 $ \implies \frac{\sqrt3.sen^2B}{\sqrt3\minus{}senB} \geq \frac{1}{2}$

 $ (senB\minus{}\frac{\sqrt3}{2})(2.\sqrt3.senB\plus{}2)\geq0$
 $ \Leftrightarrow senB \geq \frac{\sqrt3}{2}$
 $ \Leftrightarrow B \geq 60^o$
 $ \Leftrightarrow 60^o\equal{}B$
So in all the cases we have that the angles $ A$, $ B$, $ C$ are equal to $ 60^o$:
Then the triangle is equilateral as claimed.


limit
integration
logarithms
LaTeX
calculus
ratio
real analysis

Let $ (x_n)_{n \ge 1}$ be sequences of real numbers such that:
$ \{ \begin{array}{l} x_1=2 \\ x_1+x_2+\cdot\cdot\cdot+x_n=n^2x_n \\ \end{array} $
Evaluate $ lim_{n \to \infty} (n^2x_n)$

[quote="Kouichi Nakagawa"]
$ 0 \equal{} (n \minus{} 1)(n \plus{} 1)x_n \minus{} (n \minus{} 1)^2 x_{n \minus{} 1}$
$ \therefore \ x_n \equal{} \frac {n \minus{} 1}{n \plus{} 1}x_{n \minus{} 1}$[/quote]

$ (n\plus{}1)nx_n\equal{}n(n\minus{}1)x_{n\minus{}1}\equal{}\cdots \equal{}2\cdot 1\cdot x_1\equal{}4$.

Thanks, Kouichi Nakagawa and Kunny.
Problem 2:
Let n is a positive integer and let $ I_n \equal{} \int_{0}^{1} x^n\ln{(1 \plus{} x^2)}dx$
1) Evaluate $ lim_{n\to \infty} I_n$
2) Assume that $ c \in(0,1)$ and let $ A_n \equal{} \int_{0}^{c}x^n\ln{(1 \plus{} x^2)}dx, B_n \equal{} \int_{c}^{1}x^n\ln{(1 \plus{} x^2)}dx$.
Prove that: $ lim_{n \to \infty} \frac {A_n}{B_n}\equal{}0$
Thank, scorpius119!

[quote="NguyenDungTN"]Thanks, Kouichi Nakagawa and Kunny.
Problem 2:
Let n is a positive integer and let $ I_n = \int_{0}^{1} x^n\ln{(1 + x^2)}dx$
1) Evaluate $ lim_{n\to \infty} I_n$
2) Assume that $ c \in(0,1)$ and let $ A_n = \int_{0}^{c}x^n\ln{(1 + x^2)}dx, B_n = \int_{c}^{1}x^n\ln{(1 + x^2)}dx$.
Prove that: $ lim_{n \to \infty} \frac {A_n}{B_n}$[/quote]

The work is a bit too tedious to latex out, but by using integration by parts, I_n reduces to finding the integral ${ F(n) = \int_{0}^{1} (u-1)^n/u} du$ then compute F(n)-F(n-1). It will be very clear from there.

Well, it's just the second part that needs work: $ I_n$ tends to 0 because $ 0\leq I_n\leq \frac{\ln 2}{n\plus{}1}$ for all $ n$.

For (2), integration by parts yields
\[ \int_0^c x^n\ln (1\plus{}x^2)\ dx\equal{}\frac{c^{n\plus{}1}\ln (1\plus{}c^2)}{n\plus{}1}\minus{}\frac{2}{n\plus{}1}\int_0^c \frac{x^{n\plus{}2}}{1\plus{}x^2}\ dx\]
\[ \int_c^1 x^n\ln (1\plus{}x^2)\ dx\equal{}\frac{\ln 2\minus{}c^{n\plus{}1}\ln (1\plus{}c^2)}{n\plus{}1}\minus{}\frac{2}{n\plus{}1}\int_c^1 \frac{x^{n\plus{}2}}{1\plus{}x^2}\ dx\]
So the ratio $ \frac{A_n}{B_n}$ is equal to
\[ \frac{c^{n\plus{}1}\ln (1\plus{}c^2)\minus{}2\int_0^c \frac{x^{n\plus{}2}}{1\plus{}x^2}\ dx}{\ln 2\minus{}c^{n\plus{}1}\ln (1\plus{}c^2)\minus{}2\int_c^1 \frac{x^{n\plus{}2}}{1\plus{}x^2}\ dx}\]
As $ n\rightarrow \infty$, everything except the $ \ln 2$ term drops, since each of the integrals are nonnegative but bounded above by $ \frac{1}{n\plus{}3}$, so the limit is still $ \boxed{0}$.


find all functions f:R into R that:
(i) for every x,y\in R:   f(x)+f(y)+1>=f(x+y)>=f(x)+f(y)
(ii) for every x\in [0,1]: f(x)<=f(0)
(iii) f(-1)=-1 , f(1)=1

replacing $ y\equal{}0: f(x)\geq f(x)\plus{}f(0)$
with $ x\in [0;1]: f(x)\geq f(x)\plus{}f(0)\geq 2f(x)\Rightarrow f(x)\leq 0$
but $ f(1)\equal{}1$ :huh:

u might want to use latex to post problems. That makes them alot easier to read...


AMC
AIME
AMC 12
AMC 12 A
AMC 8
USA(J)MO
USAMO

I guessed on the last two problems because I figured a $\frac{1}{25}$ chance of getting them right wasn't terrible.   :oops:

Depends. If your goal is to make the AIME, read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=130853]this[/url] topic about how many questions to answer (whether guessing or not).

on the AMC 10, because you only get 1.5 for a blank, i just guess unless i have absolutely no idea at all and i can't deduce any of them.

The nature of this question has changed as a result of the new scoring.  In general, I don't usually guess, but it wouldn't be unreasonable.

If you were able to eliminate one answer choice as incorrect, guessing would give you the same expected value as leaving the problem blank.  I think that whether you guess should depend on how well you do on the rest of the test.  For example, if not guessing would leave you just below your goal score (i.e. AIME qualification), then you should guess.  If you are already at or above the score which you are aiming for, you should probably play it safe.  Finally, if guessing can mean the difference between a good score and an outstanding score, and you can afford to lose the points, go ahead.  An example of this could be if you have answered 24 with good certainty (this has been discussed before):  Go for the perfect.

if you don't know how to do a lot of problems, it's best to leave them blank. If it's just 1 or 2,  a good chance at 6 or even 12 points is a lot more alluring than getting a mere 1.5 for each one

ADDED: you should also just leave them blank if you're just a few points away from qualifying for AIME or soemthing like that

I hate leaving questions blank, it makes me feel like a wimp. So if there are <5 questions I'm unsure about, I guess on all of them (bad habit, I know)

I guess when I can eliminate two or more answers, and don't otherwise. When I took the 12A this year, I got to 18 questions before the proctor said I had 30 seconds left, so I just left the other seven questions blank.

I guessed on 4 of the problems randomly filling in bubbles 'A', A, A, A and got 1 right...

Yup, I do it... on the AMC12A there were only a few seconds left after I had solved 24 questions, so for a chance of a perfect score I guessed #22...correctly! However, I had made a calculation error for #25, so it evened out.

My suggestion is that if you are a good problem solver and understand the problem statement, you should guess, because intuition is usually pretty good (at least better than 25%, I think). On the other hand, if there are several questions left and you haven't read them I wouldn't recommend guessing.

I don't guess.  Bad idea—even if you can eliminate one answer choice, you're still more likely to lose points than to gain them by guessing.  I know people say, "guess on the last question so maybe you can get a perfect score", but I suppose I have a slight ethical problem with guessing—if it serves its ends (i.e., you get he question right), then you didn't deserve it.

Edit :  Of course, if you're one question away from qualifying for the AIME, maybe you want to guess, but I think that if you have a nice score, it's kind of immature.

If you are an underclassman and have already qualified for AIME I would suggest guessing on all of the problems unless you feel that you will do really well on the AIME.

I just do whatever is more advantageous to me -- I will do whatever choice maximizes my expected value but
A) All scores below 100 count as 0 because they do not qualify
B) 150 points counts as 160 because it makes my name searchable  :P

I guessed on AMC8 #25 last year... based on the frequency of my filled-in answers :P

and I got it right!  :D

but for 10/12... i'd guess if i got it down to 2 or 3 (since I've qualified for AIME)

if you thilnk 1.5 points per blank will get you to AIME, then don't guess. if you need 6-12 more points, do guess.

okay, I am definitely going to make educated guesses on the test.

I guessed on #25 this year cause I ran out of time

I missed it but it was worth it

Yeah I did the same thing. I ran out of time and I was sure about 24 of the problems, so I guessed on #24 to get the perfect since I didn't have time.

[quote="ark_xm"]If you qualified for AIME already through AMC 12 A, but want to get a higher score on AMC 12 B to improve your chances of getting into USAMO, should you guess? I'm a junior if that matters or anything.[/quote]

Well, it depends on whether you've eliminated choices or not. It's better not to guess if you can't eliminate anything, it doesn't matter whether you guess or not if you've eliminated one, and it's better to guess if you can eliminate two or more answers.

[quote="King Paul"]sometimes, choices can look right.[/quote]
oftentimes they are wrong.

B, C, and D occur MUCH more often than A & E, I've noticed.

Ironically, this year having a bit of extra time screwed me over. On #19 on the AMC10a, I had enough time to plug in answer choice C and see if it would work (I thought it was the most likely correct answer). I made a dumb error in plugging it in and I decided to guess D instead. Turns out the answer was C, so if I didn't have the extra time, I would have guessed C anyways to try to get 150, because I was quite sure of my other 24 answers.

Not complaining just sharing an ironic story.

[quote]
B, C, and D occur MUCH more often than A & E, I've noticed. [/quote]

What are you talking about?

This year's AMC 12A:

A: 5
B: 4
C: 5
D: 7
E: 4

I noticed that I have not filled out an A for the last part of the test.  It made me feel uncomfortable.

It is easier just to not guess because even though pointwise you should break even, if you guess, you will drop that extra little bit behind people around your level.  I have gone for it a few times and each time it cost me.

And Fanatic, I made a dumb error plugging in a calculation in the State Mathcounts Target Round last year.  It was not only once, but many many times in a row, that cost me two points.  So yeah, I am technology deficient too.

AMC 10A
A: 4
B: 5
C: 5
D: 8
E:  3

except consecutive questions often have the same correct answer choice.

Last year's target makes me cry :(.

haha.

But seriously, I'm pretty sure on last year's AMC 10 there was like 8-10 C's and only 2-3 E's. If someone has answer keys from the last few years, we could do a comparison, but from the previous practice tests I've taken and the real deals, I've noticed that E is extremely infrequent and A is less frequent.

And then there will be 10 Es on the Amc B..

2006 AMC 10A
A: 3
B: 5
C: 8
D: 4
E: 5

You were right about the C's.

I heard from two completely different people always to guess C.

Okay, here are the results from the AMC 12A/12B from 2003-2007 (also mixed in AMC 10A from 2007)

A: 42
B: 61
C: 52
D: 58
E: 37
Doubles: 39
Total: 250

The verdict is in -- A and E are indeed less likely responses. However, don't guess C -- guess B or D.

[quote]except consecutive questions often have the same correct answer choice.[/quote]
This has shown to be false above. In a random sample 1/5 of the pairs should be the same answer choice, but the actual proportion of doubles from 2003-2007 is quite a bit less than 1/5.

yay i was right :)!.

I always guess B or D anyways if i don't even get to the question, so doubley that (in my previous post i tried C first because it looked like it was the right answer) :).

i guessed on at least 10 of them.  blind guessing.  because i knew there was no chance id get 100 if i didnt put answers, so i figured itd be to my advantage.  i ended up over 100.  :-)

I am too much of a wimp to guess.

Last year on the AMC 10B, I didn't guess on #23 and decided to be happy with a 146.5.

I would strongly advise not guessing unless you are one question away from the score you want, and you can eliminate 1 choice.

As a freshman, I decided that making the AIME was much more important than getting more points than necessary.   On the 10A, I answered 21 and left the other four blank and figured that I could miss 2 and make it.  It paid off.   :)

i dont guess, unless i am >67.142723% sure i am right

I you don't care about school winner and all that jazz, and you're sure that you qualified for AIME and know that if you didn't guess on the B you'd have a score worse than your A score, then I see no reason not to guess.

i do it on my bio tests when i don't know the material

i guess i should be proud of a 14/31 considering i statistically i shoulda gotten a 6.2

I usually don't guess on the AMCs, but if I guess for any reason, and it is totally random, I guess B

I guessed three questions at Random and Still got two of them right...

I was really lucky... :oops:  :roll:

One of the other students in my school guessed on five questions, got two of them right, and ended up with a 100.5 on the AMC 12b this year.  I think he's the first non-honors student in my school to qualify for AIME :ninja: .

I guessed all 'C''s for the last 5 questions on teh 10a since i ran out of time and I think 3 were right


Canadians can take USAMO.
But can Americans take Canadian Math Olympiad?

Not offically, unless you are a Canadian citizen (I remember there was once a Canadian student living in New Jersey who took the Olympiad.)

Anyways, if the Americans were allowed take the CMO, then you probably wouldn't see too many Canadian names on the rankings. Not that we suck, it's just that we have too few people living on this vast land.

[quote="billzhao"]Jew Jersey[/quote]

That'd be "New Jersey," right?

[quote="mathfanatic"]Also, I'm pretty sure it's called the CMO...[/quote]

Its also often called the CaMO, since the acronym CMO also refers to the China Mathematical Olympiad.

Then what do you call the Cambodian Math Olympiad? Is there one?

And if you want to make that the CamMO, then you have to ask about Cameroon ...

on Kalva, they call Canada CanMO. Lets call Cambodia CaMO, and Camerooun CamMO.

Does it really matter what the acronym is? It all depends on the context. When we mention CMO in Canada, 99% chance that we're talking about our own contest. Besides, the full name isn't that hard to type.

[quote="billzhao"]Does it really matter what the acronym is?[/quote]

Obviously it does.

Cambodia
Cote d'Ivoire
Cuba
Cyprus
Czech Republic
Cameroon
Canada
Cape Verde
Cayman Islands
Central African Republic
Chad
Chile
China
Colombia
Comoros
Congo
Congo, Dem. Rep.
Cook Islands
Costa Rica
Croatia

That's a lot of country.

[quote="ComplexZeta"]I agree. The acronym is extremely important.[/quote]

I think arguably it may in fact be the most important part of the entire competition.

This is such a funny discussion ... hehehe.  http://www.state.gov has somewhere or other a long list of countries with all sorts of fun info (and websites).

Well, come on, when you win the China Mathematical Olympiad and go around bragging about your CMO performance, you don't want people to confuse that with the child's play in the Canada Math Olympiad. ;)

funny, last year I had the impression that USAMO was eaiser for me than the CMO (yes, that is the Canadian one), probably cuz of the excess time alotted on the USAMO.  I hope i feel the same this year also.

[quote=ChebyshevsToast]/pointless bump[/quote]

Thx for bumping it is imperative that we spread this new news

I agree with @3above. USAMO this year was much easier than CMO (I got 9 more points on it).

flex

and apparently now usamo 2003 = usamo 2019

@above

Lol i didn't realize for some reason that he was talking in 2004, OOPS!


number theory
least common multiple
Divisibility Theory
pen

If one $n_i$ is $1$, then all have to be $1$. So lets assume that all are $\neq 1$.

Let $i \in \{1,2,...,k\}$ always and define $n_{k+1} :=n_1$ (indices are seen $\mod k$).
Let $p_i$ be the smallest prime divisor of $n_i>1$ and let $o_i$ be the order of $2 \mod p_i$ and $q_i$ be the smallest prime factor of $o_i$.
We have $o_i|n_{i+1}$ by standard facts on orders, giving $p_{i+1} \leq q_i$. We also have $o_i|p_i-1$, thus $q_i \leq p_i-1$. As a result, $p_{i+1} \leq q_i \leq p_i-1 < p_i$.
But this gives $p_1 < p_2 < ... < p_k <p_1$, contradiction.

Here is an other method.
$ x \equal{} lcm(n_1,..,n_k)$
So $ n_i|2^x \minus{} 1,\forall i \equal{} 1,..,k$
Imply that $ x|2^x \minus{} 1$
So $ x \equal{} 1$ (tradition!)
http://www.mathlinks.ro/Forum/viewtopic.php?p=849217#849217

[quote="TTsphn"]Here is an other method.
$ x \equal{} lcm(n_1,..,n_k)$
So $ n_i|2^x \minus{} 1,\forall i \equal{} 1,..,k$
Imply that $ x|2^x \minus{} 1$
So $ x \equal{} 1$ (tradition!)
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=849217#849217[/quote]
but this only imply $ x|2^{x}-1 $...can you explain how did you obtain $ n_{i}|2^{x}-1 $?

It has been denoted $ x \equal{} \textrm{lcm}(n_1,\ldots,n_k)$.
It follows $ n_i \mid 2^{n_{i-1}} - 1 \mid 2^x \minus{} 1$, for all $i \equal{} 1,\ldots,k$, where $n_0 := n_k$ (indices are taken modulo $k$), since if $a \mid b$, then $2^a - 1 \mid 2^b - 1$.
Therefore $ x \mid 2^x \minus{} 1$.
So $ x \equal{} 1$ (by a well known result, whose proof also uses the least prime dividing $x$, as some method above).

Simply consider the smallest prime $p$ that divides one of the $n_i$. Then we get a smallest prime, leading to a contradiction. Thus there is no prime $p$ that divides any of the $n_i$ giving the desired conclusion.

Let the solution least solution be $(n_{1},n_{2},...,n_{k})$ for some $k$.Let $d_{i}$ be the order of $2$ modulo $n_{i}$.Now $(d_{1},d_{2},...,d_{k})$ will be a smaller solution of unless $n_{2}$
$n_{2}$ is the order of $2$ modulo of $n_{1}$ and like that.Therefore $n_{1}$ divides $\phi(n_{2})$.Therefore $n_{1}$ is less than or equal to $n_{2}$.Similarly $n_{2}$ is less than or equal to $n_{3}$ ,.....,$n_{k}$ is less than or equal to $n_{1}$.Therefore all the numbers must be equal and now $n$ divides $2^{n}-1$ only when $n=1$.

Here I post another slightly different solution to this beautiful problem. :)
[quote=Peter]Suppose that $k \ge 2$ and $n_{1}, n_{2}, \cdots, n_{k}\ge 1$ be natural numbers having the property \[n_{2}\; \vert \; 2^{n_{1}}-1, n_{3}\; \vert \; 2^{n_{2}}-1, \cdots, n_{k}\; \vert \; 2^{n_{k-1}}-1, n_{1}\; \vert \; 2^{n_{k}}-1.\] Show that $n_{1}=n_{2}=\cdots=n_{k}=1$.[/quote]

Consider the following set
$$S=\{(n_1,n_2,\ldots,n_k): n_i\ne 1 \text{for any}~ 1\le i\le k ~\text{and}~ n_{i+1}\mid 2^{n_i}-1~\text{for every i}, \text{where all the indices are taken modulo k}\}$$

Now consider the sum $(n_1+n_2+\ldots+n_k)$. Choose the $n$-tuple from $S$ which has the least sum.
For every $i$, we know that
$$n_i\mid 2^{\phi(n_i)}-1$$
Hence we have that
$$n_i\mid 2^{d_{i-1}}-1$$
where $d_{i-1}=\gcd(n_{i-1},\phi(n_i))$. But we can also say that $d_{i+1}\mid 2^{d_i}-1$.

Now if $d_i<n_{i-1}$ for some $i$, then we can do the following transformation
$$(n_1,n_2,\ldots,n_k)\longrightarrow (n_1,n_2,\ldots,n_{i-1},d_i,n_{i+1},\ldots,n_k)$$
But note that the new tuple has sum smaller than the "mother tuple", a contradiction.

Otherwise if $d_{i}=n_{i-1}$ for all $i$. Then we have $n_i\mid \phi(n_{i+1})\Longrightarrow n_i\le \phi(n_{i+1}) < n_{i+1}$. A contradiction!$~\blacksquare$

if one of $n_i$ s happen to be $1$ then all of them will equal to one. now suppose $p$ is the smallest prime factor of $\prod n_i $

and $p|n_k$ and because $n_k | 2^{n_{k-1}} -1 \implies p|2^{n_{k-1}} -1 \implies 2^{n_{k-1}} \equiv 1 \mod p$

and now as $n_{k-1}$ doesn't have any prime factor less that $p$ so it's co-prime with $p-1$ and because $2^{p-1} \equiv 1 \mod p$

 then $p$ must be $1$ which a contradiction which means we can't chose smallest prime factor of $\prod n_i  \implies \prod n_i =1$


Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree $ n$ with real coefficients is the average of two monic polynomials of degree $ n$ with $ n$ real roots.

There was no solution posted with the link on the contest section-USAMO.

the amc site has solutions: http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2002-ua/02usamosolutions.shtml

[url=http://www.artofproblemsolving.com/Wiki/index.php/2002_USAMO_Problems/Problem_3]A solution is available here[/url].


calculus
derivative
induction
inequalities
geometric series

Evaluate  :sqrt: (1+2 :sqrt: (1+3 :sqrt: (1+4 :sqrt: (1+5 :sqrt: (1+...)...)))). If you have seen it before, please don't post. And don't evaluate it with a calculator because that's just no fun. Enjoy!

Is this really evaluable, like, by hand?

Yuck.  I'll get back to you.  Any not-giving-it-away type hints?

Hm. The only hint I can think of practically gives away the answer. Proving it is still another story, but giving away the answer isn't really much fun.

Okay.  Well, then ...

Is there a general approach for nested radicals that look like continued fractions?  I mean, I can find an expression f(n) so that my fraction becomes :sqrt:(f(1) + :sqrt:(f(2) + :sqrt:(f(3) + ...))) without too much difficulty.  Is there a known way of dealing with the result?

it reminds me of way of finding power series for 1/(1-x)^2, you know 1/(1-x) = 1+x+x^2+... by geometric series, so term-by-term differentiation and get 1+2x+3x^2+4x^3...

Also (1/2)+(2/4)+(3/8)+(4/16)+...=S
S/2=(1/4)+(2/8)+(3/16)+...

S-(S/2)=(1/2)+(1/4)+(1/8)+...=1, hence S=2.

Maybe something to that effect will work? maybe...

there is a formula which i don't know how to prove by rramnujan. It says x+1=sqrt(1+x*sqrt(1+(x+1)*sqrt(1+(x+2)...  Plug in 2 and you get 3.

I was thinking about things like that, but it's so much harder to manipulate things inside the radicals ...

cool method
n(n+2)=n*sqrt(1+(n+1)(n+3))
lets have f(n)=n*(n+2)
so f(n)=n*sqrt(1+(n+1)(n+3))
so f(n)=n*sqrt(1+f(n+1))=keep on doing this
so f(n)=n(n+2) which when plugged in 1 gives a 3

Yeah, that's the solution. Now prove it. That's far from trivial.

You have found what the answer should be.  That's half the battle -- nice work.  But you have not proved that the original nested radical converges to your answer.

Let m and n be positive integers such that m :le: n.  For m = n, define g(m, n) = 1.  For m < n, define g(m, n) = :sqrt:(1 + (m - 1) g(m + 1, n)).  The problem is to evaluate the limit of g(3, n) as n approaches infinity.  We will show that the limit is 3.  I'll put my proof below in spoiler.

[hide]

It's easy to show (by reverse induction on m) that g(m, n) :le: m.  The hard part is to develop a lower bound.  To help, define h(m, n) = g(m, n) / m.  We know that h(m, n) :le: 1.  Thus if m < n, we have h(m, n) = :sqrt:(1/m2 + (1 - 1/m2) h(m + 1, n)) :ge: :sqrt:(h(m + 1, n)).  Applying that inequality n - 3 times, we get

h(3, n) :ge: (1 / n)1/2n - 3.  Thus the limit of h(3, n) is 1.  Hence the limit of g(3, n) is 3.[/hide]

Seen before from a general result from Ramanujan:

x+n+a =  :sqrt: (ax+(n+a) :^2: +x :sqrt: (a(x+n)+(n+a) :^2: +
            (x+n) :sqrt: (a(x+2n)+(n+a) :^2: +(x+2n) :sqrt: (.....

when a=0, n=1 we have:
x+1 =  :sqrt: (1+x :sqrt: (1+(x+1) :sqrt: (1+(x+2) :sqrt: (1+....

put x=2, above

also interesting to some  mkay be this use of  google as a calculator:  [url=http://www.google.com/search?hl=en&ie=UTF-8&oe=UTF-8&q=sqrt%281%2B2sqrt%281%2B3sqrt%281%2B4sqrt%281%2B5sqrt%281%2B6sqrt%281%2B7sqrt%281%2B24%29%29%29%29%29%29%29%29]here[/url]


inequalities
linear algebra
matrix
induction
vector
analytic geometry
linear algebra unsolved

let $a_1 \geq a_2 \geq .. \geq a_n$ and $b_1 \geq b_2 \geq .. \geq b_n$ be some real numbers.
Suppose that:
 $b_1 \geq a_1$,
$b_1 + b_2 \geq a_1+a_2$
...
$b_1+..+b_{n-1} \geq a_1+..+a_{n-1}$ and
$b_1 + ..+ b_n = a_1 + ..+ a_n$
Prove that there is a doubly stochastic matrix $D$ such that $A=DB$ where $A=(a_1,..,a_n)^t$, $B=(b_1,..,b_n)^t$

Just a brief sketch:

Induction on $n$ should work. For $n=2$ it's clear enough. Now assume we've shown the assertion to be true for values smaller than $n$. Let $A_i=\sum_{j=1}^ia_i$, and define $B_i$ similarly.

Now we start decreasing $a_n$ and increasing $a_{n-1}$ (so as to keep their sum constant). Eventually, one of two things will happen: either one of the sums $A_i,\ i<n$ becomes equal to $B_i$, in which case we stop and we're done by the induction hypothesis, which tells us that a ds (doubly stochastic) matrix turns $b$ into the new vector $a$ and the new vector $a$ into the old one, or $a_{n-1}$ becomes equal to $a_{n-2}$. If this happens, then we start decreasing $a_n$ again, and increase $a_{n-1},a_{n-2}$ at the same time, so as to keep their sum constant. We have the same two alternatives as above: either we reach a moment when $A_i=B_i$ for some $i<n$, or $a_{n-1}=a_{n-2}=a_{n-3}$, and so on. If we the first situation is never met, then in the end we'll reach a vector $a$ with all first $n-1$ coordinates equal. Now we continue the process until we do get $A_i=B_i$ for some $i<n$, because proving the theorem for two vectors of this very particular form is easy :).


\[ \sum_{n = 0}^{\infty} {3n \choose n} \left(\frac18\right)^n = \frac{4\sqrt{10}}{5} \sin\frac{\pi + \arccos\frac{3\sqrt{6}}{8}}{3}  \]

Maxal, could you please show me the steps? :)

It follows from the general formula
\[ \sum_{n = 0}^{\infty} {3n \choose n} x^{2n} = \frac{2}{\sqrt{4-27x^2}} \sin\frac{\pi + \arccos\frac{\sqrt{27}}{2}x}{3}  \]
for $x=\sqrt{1/8}$. This formula can be proved using differential equations.

Ok, thanks! :)
By the way, do you know where I can find this proof? (or, if you could prove it, that'd be aweseome!  :D)

[quote="mathmanman"]By the way, do you know where I can find this proof?[/quote]
Take a look at the paper [url=http://www.emis.de/journals/DMTCS/pdfpapers/dmAD0136.pdf]Human and constructive proof ofcombinatorial identities: an example from Romik[/url] by D. Merlini, R. Sprugnoli, and M. C. Verri. Pay special attention there to the Lagrange Inversion Formula in the form of the diagonalization rule at page 384 and the formula for ${3k\choose k}$ at page 387. Then solve the equation $w=t(1+w)^3$ for $w$ and substitute it into $\frac{1+w}{1-2w}$.


Find all values of (x,y,z) such that

xy+x+y=5
xz+x+z=7
yz+y+z=11

[quote="DiscreetFourierTransform"]Now it's time for a little Simon's Favorite Factoring Trick.
Add one to each equation.
$xy+x+y+1 = 6$
$xz+x+z+1 = 8$
$yz+y+z+1 = 12$

Factor.
$(x+1)(y+1) = 6$
$(x+1)(z+1) = 8$
$(y+1)(z+1) = 12$

$(z+1)[(x+1)+(y+1)]=20$
$(z+1) = \frac{20}{(x+y+2)}$

Substitute.
$(x+1)(y+1) = 6$
$(x+1)(\frac{20}{((x+1)+(y+1))}) = 8$
$(y+1)(\frac{20}{(x+1)+(y+1)})=12$
Now we have a system of equations in two variables.
However, I believe that this system has an infinite number of solutions.[/quote]

Hide, please!
[hide="Sol"]DFT, just note that from these equations
$(x+1)(y+1) = 6$
$(x+1)(z+1) = 8$
$(y+1)(z+1)=12$

you get

$(x+1)(y+1)(z+1) = \pm 24$

$x+1 = \pm 2$
$y+1 = \pm 3$
$z+1 = \pm 4$

Thus, $(x,y,z) = (2,3,4) \textrm{ OR }(-3,-4,-5)$[/hide]

Thanks, kohjhsd.

[quote="rebirth"][quote="kohjhsd"][hide]Actually $(x+1)(y+1)(z+1) = \pm 24$
And therefore
$x+1 = \pm 2$
$y+1 = \pm 3$
$z+1 = \pm 4$

So $(x,y,z) = (2,3,4)$ as well as $(-3,-4,-5)$ are solutions.[/hide][/quote]

So when you do that the x+1, y+1, and z+1 they all have to be different?

I mean why cant x+1=1
y+1=1
and z+1=24

I know when you sub back in, it doesnt work, but why?
and why is it + or -24, instead of just +24?

Sorry, Im a little confused[/quote]

[hide]It doesn't work because it doesn't satisfy the restraints:
$(x+1)(y+1) = 6$
$(x+1)(z+1) = 8$
$(y+1)(z+1)=12$

We used the equation $(x+1)(y+1)(z+1) = \pm 24$.
First, we divided this equation by the first equation to get

$z+1 = \pm 4$

Similarly for $x$ and $y$.[/hide]


rotation
conics
ellipse
quadratics
trigonometry
function
analytic geometry

Given the ellipse x^2 -xy + 4y^2 = 36, find the vertices (which may require you to find the angle through which the ellipse is rotated), major axis, and minor axis.  Also, the center of the ellipse is the origin.  The reason I'm posting this problem is some people may not know what to do with the xy term.  Plus, problems like this don't appear too often on tests.

[hide="kind of... ick"]
what you can do is just use the quadratic formula in terms of y and get $y=\frac{x \pm \sqrt{576-15x^2}}{8}$, then the major axis vertices will be at the 2 points where the discriminant is $0$, namely $(\pm \frac{24}{\sqrt{15}}, \pm \frac{\sqrt{15}}{48})$. Then , since the center is the origin, umm... draw a line between major vertices, draw perpendicular at origin, solve.. ick. Then you can just use distance formula, but i dont know a particularly pretty way of doing it. [/hide]

I remember in AoPS V2, it discussed rotating the x and y axes so that the major and minor axes are parallel using trig functions .

But I'm too lazy to either look it up or rederive it.  :P

You can use a linear transformation for rotation.

Indeed, you can rotate the cartesian axes by an angle $\,\,\,\theta\,=\,0.5.arctg\frac{1}{3},\,\,\,$ in order to get the equation in it's easier form ( but the work to do it isn't so easy) and obtain the ellipses axes lengths.

[hide="Click to reveal hidden content"]

the linear transformation is

$\,\,\,x\,=\,x'\,cos\,\theta\,-\,y'\,sin\,\theta\,=\,x'\,\sqrt{\frac{1+\frac{3}{\sqrt10}}{2}}\,-\,y'\,\sqrt{\frac{1-\frac{3}{\sqrt10}}{2}}$

$\,\,\,y\,=\,x'\,sin\,\theta\,+\,y'\,cos\,\theta\,=\,x'\,\sqrt{\frac{1-\frac{3}{\sqrt10}}{2}}\,+\,y'\,\sqrt{\frac{1+\frac{3}{\sqrt10}}{2}}$

This way you will find the vertices (x',y') as $\,\,\,\left(\pm\,6\,\sqrt{\frac{2}{5-\sqrt10}}\,\,\,\,,\,0\,\right)\,\,\,\textrm{and}\,\,\,\,\left(0\,,\,\pm\,6\,\sqrt{\frac{2}{5+\sqrt10}}\,\,\right)$

Now you can put this back in the linear transform system to obtain the required answer.
[/hide]
But this problem is asking for polar coordinates, like $\,\,x\,=\,\rho\,cos\,\theta\,;\,\,\,\,y\,=\,\rho\,sin\,\theta.$

 [hide="Click to reveal hidden content"]

We get $\,\,\rho^2\,(\,5\,-\,3\,cos\,2\theta\,-\,sin\,2\theta\,)\,=\,72.\,\,$

Calling $\,\,\,\, \,\,\,\,\,\,\,\,\,\,sin\,\beta\,=\,\frac{1}{\sqrt10}\,\,\,\,\textrm{and}\,\,\,\,cos\,\beta\,=\,\frac{3}{\sqrt10}\,\,\,,\,\,\,\,\,\,\,\,\,$ the polar equation becomes $\,\,\,\,\,\,\,\rho^2\,=\,\frac{72}{5\,-\,\sqrt10.cos (2\theta\,-\,\beta)}\,\,,\,\,\,$
and we can see that the absolute value of $\,\,\rho\,\,\,$ has his minimum absolute value,

$\,\,\,\,6\,\sqrt{\frac{2}{5+\sqrt10}}\,\,\,\,\textrm{when}\,\,\,\,2\,\theta\,-\,\beta\,=\,\pi\,\,\,\,\rightarrow\,\,\,\,\,\theta\,=\,\frac{\pi}{2}\,+\, \frac{\beta}{2}\,\,\,\,\,\textrm{and maximum}\,\,\,\,\,\,6\,\sqrt{\frac{2}{5-\sqrt10}}\,\,\,\,$
when $\,\,\,\,\, 2\theta\,-\,\beta\,=\,0\,\,\,\,\rightarrow\,\,\,\,\,\theta\,=\,\frac{\beta}{2}$

Transforming back to cartesian you get straightfor the answer.

[/hide]


linear algebra
matrix
linear algebra unsolved

Assume $ S \in M_n(\mathbb{R})$, and $ S^T\equal{}S$,

 $ S_1, S_2 \in M_m(\mathbb{R})$, and $ S_1^T\equal{}S_1,S_2^T\equal{}S_2$

If  $ \begin{pmatrix}
S & \\
 & S_1
\end{pmatrix}$ congruent to $ \begin{pmatrix}
S & \\
 & S_2
\end{pmatrix}$
then $ S_1$ congruent to  $ S_2$

Any real symmetric matrix is congruent to a diagonal matrix in which all of the diagonal entries are $ 1,0,$ or $ \minus{}1.$ Furthermore, this diagonal matrix may be arbitrarily permuted by a further congruence. Hence the only thing we need to know to decide whether two real symmetric matrices are congruent is the "signature" - the number of $ 1$s, $ 0$s, and $ \minus{}1$s in that canonical diagonal matrix. The conditions as stated ensure that $ S_1$ and $ S_2$ have the same signature, so they are congruent.

This is also a particular case of [url=http://en.wikipedia.org/wiki/Witt%27s_theorem#Witt.27s_cancellation_theorem]Witt's cancellation theorem[/url]. But does anyone know a slick proof (I mean no casebash as [url=http://jmilne.org/math/CourseNotes/ala.html]Milne does in ALA[/url]) of the latter?

  darij


What is the equation of the circle in standard form from this matrix?

[x^2+y^2 , x , y, 1; 50,1,7,1;40,6,2,1;52,4,6,1] = 0


Here is an inequality stronger than Popoviciu Inequality. Doesnt it look like Schur inequality?
  For all positive real numbers $ a,b,c$ we have $\sum\frac{1}{3a+1}+\frac{3}{a+b+c+1}\geq\sum\frac{1}{2a+b+1} $, where the second sum has 6 terms.

I don't have the full thing out; I'm posting it despite this because, in any case, at least one of these ideas should be useful.

Let us start by letting $a'=a+\frac{1}{3}, b'=b+\frac{1}{3}, c'=c+\frac{1}{3}$. This gets rid of the ones, and also $a',b',c' \geq \frac{1}{3}$. From here on I'll just use $a,b,c$; I mean $a',b',c'$ throughout. We start with
\[\frac{1}{3a}+\frac{1}{a+b+c}-\frac{1}{2a+b}-\frac{1}{2a+c} =\]
\[\frac{(4a+b+c)((2a+b)(2a+c)-3a(a+b+c))}{3a(a+b+c)(2a+b)(2a+c)}=\]
\[\frac{(a-b)(a-c)}{3a(a+b+c)(2a+b)} + \frac{(a-b)(a-c)}{3a(a+b+c)(2a+c)}\]
We want to prove that when this is cycled through $a,b,c$ its sum is at least zero. Thus we can cancel the threes and $(a+b+c)$, since these are positive and present in each term. Now we regroup the terms as follows:
\[\frac{(a-b)(a-c)}{a(2a+b)}+\frac{(b-a)(b-c)}{b(a+2b)}=\]
\[(a-b)(\frac{a-c}{a(2a+b)}+\frac{(c-b)}{b(a+2b)})=\]
\[\frac{(a-b)^2(2c(a+b)-ab)}{ab(2a+b)(a+2b)}\]
Initially, I thought I had this out, and it's now 10 o'clock at night so I'm too tired to fix it. But it seems almost a certainty from here; trying to make a term negative makes it vanishingly small, perhaps Chebyshev would finish it somehow.

I think this is the correct approach, but how to prove that

$[\frac{(a-b)^2(2c(a+b)-ab)}{ab(2a+b)(a+2b)}]+[\frac{(b-c)^2(2a(b+c)-bc)}{bc(2b+c)(c+2b)}]+[\frac{(c-a)^2(2b(c+a)-ca)}{ca(2c+a)(c+2a)}] \geq 0$

Yes, this is really useful.Now,we just need to do a small step.
Assume that $ a \geq b \geq c $.We need to consider this situation:
$ 2a(b+c)-bc \geq 2b(a+c)-ac \geq 0 \geq 2c(a+b)-ab $.
Note that: $ (a-b)^2 \leq 2(a-c)^2+2(c-b)^2 $,so it is enough to prove that:
$ \frac{4c(a+b)-2ab}{ab(2a+b)(a+2b)}+\frac{2a(b+c)-bc}{bc(2b+c)(c+2b)} \geq 0 $ and the similar one. But $ bc(2b+c)(c+2b) \leq ab(2a+b)(a+2b) $ then
 $ \frac{4c(a+b)-2ab}{ab(2a+b)(a+2b)}+\frac{2a(b+c)-bc}{bc(2b+c)(c+2b)} \geq $
 $ \frac{6ac+3bc}{ab(2a+b)(a+2b)} \geq 0 $ :)

How about this one:
$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{9}{a+b+c} \geq (m+n) \sum \frac{1}{ma+nb} $
where $ m,n $ satisfying:
$ i)m,n>0 $
$ ii)mn \geq (m-n)^2 $. :)


Find all $m,n\in\mathbb{N}$ such that :

$n^{3}=m!+3m-9$

[quote="Rust"]If $m\le 6$ then $m=4,n=3$ or $m=6,n=9$.
If $m>6$, then $m=9k^{3}+3, k\in N, \ n=3kl.$[/quote]

how do you see now that there aren't other solutions, for $m>6?$ :maybe:

[quote="SpongeBob"]Because $n^{3}$ is $0,-1,+1$ modulo $9$, and the left side is $3$ modulo $9$ when $m>6$.
Bye[/quote]

Well, for $m > 6$, we have $RHS \equiv 3m (mod 9)$ which can be $0$ if $3|m$..


I wonder if the following problem is known:

Prove that every [b]circumscribed[/b] quadrilateral can be partitioned into
four [b]circumscribed[/b] quadrilaterals.

The analogous question for [b]cyclic[/b] quadrilaterals is relatively well known
(see Andreescu and Gelca's book) or the link on http://www.cut-the-knot.org.

Your opinions would be greatly appreciated,

Dan

Actually, every circumscribed quadrilateral can be partitioned into four bicentric (both circumscribed and cyclic) quadrilaterals by just taking the radii from the incenter to the tangency points. Along the same lines, every convex quadrilateral can be partitioned into six bicentric quadrilaterals by simply splitting the quad into 2 triangles first and then partitioning each of these into 3 bicentric quads.

The real question is the following:

[b] Can one partition a circumscribed quadrilateral into four circumscribed quads by using
two segments which cross inside the initial quadrilateral? More precisely,
given a circumscribed quadrilateral $ ABCD$, how one can choose points $ M\in (AB)$,
$ N\in (BC)$, $ P\in (CD)$ and $ Q\in (DA)$ , $ MP\cap NQ \equal{}R$ such that
quadrilaterals $ AMRQ$, $ BNRM$, $ CPRN$ and $ DQRP$ are all circumscribed.

I offer $ \$100$ for a solution.

Dan[/b]


tetrahedr0n wrote:What is the largest 2-digit prime factor of the binomial coefficient 200C100?

[hide]For any p, the number of factors that it will have in C(2n,n)is [2n/p]+[2n/p:^2:]+...-2([n/p]+[n/p:^2:]+[n/p:^3:]+...). For any p such that 2n/3<p<n, p will not divide C(2n,n) (there will be one factor in n! and 2 factors in (2n)!). So we know that the p we're looking for is <2n/3=66.666... Also, if n/2<p<2n/3, p|C(2n,n), as there will be one factor in each of the (n!) and 3 factors in the (2n!). So we want the largest p:le:66 which is ... 61?[/hide]

[hide]same, eventually, after much cancelling out all the 2's (i think), you find that there are more 3's on the bottom than the top. Then you know that the answer is a number < 200 divided by 3. The greatest such integer is 61.[/hide]


let ABC be a triangle with <ABC = 60 and <BAC = 40. Let P lie on AB such that <BCP = 70 and let Q lie on AC such that <CBQ = 40. Let BQ intersect CP at R. Prove AR produced, is perpendicular to BC.

[hide="probably my first use of... you can figure it out if you read this"]First get some angles. From sum of angles in triangle, $\angle ACB=80^\circ$. After splitting the angles, $\angle ABQ=20^\circ$ and $\angle ACP=10^\circ$.

Now use angle form of Ceva: let $\theta=\angle BAR$. Then $\sin \theta \sin 40^\circ \sin 10^\circ=\sin (40^\circ -\theta) \sin 20^\circ \sin 70^\circ$.

Use double angle formula on $\sin 40^\circ=2\sin 20^\circ \cos 20^\circ$. Since $\sin 70^\circ=\cos 20^\circ$, these cancel $\Rightarrow 2\sin \theta \sin 10^\circ=\sin (40^\circ -\theta)=\sin 40^\circ \cos \theta-\cos 40^\circ \sin \theta$.

Isolating $\theta$, $\tan \theta=\frac{\sin 40^\circ}{2\sin 10^\circ+\cos 40^\circ}$.

Since $\cos 40^\circ=\sin 50^\circ$, $\sin 10^\circ+\cos 40^\circ=2\sin 30^\circ \cos 20^\circ=\cos 20^\circ$.

Add the other $\sin 10^\circ$; $\sin 10^\circ+\cos 20^\circ=\sin 10^\circ+\sin 70^\circ=2\sin 40^\circ \cos 30^\circ=\sin 40^\circ \sqrt{3}$.

It follows that $\tan \theta=\frac{1}{\sqrt{3}}$, so $\theta=30^\circ$. Then if $AR$ is extended to meet $BC$ at $X$, $\angle AXB=90^\circ \Rightarrow AR\perp BC$.[/hide]


What is the least integer k for which $ 12k \equal{} 5n^3$?

[hide="Solution"]Solve for k to get $ k\equal{}\frac{5n^3}{12}$. Since 5 is obviously not divisible by 12, $ n^3\mid 12$. The smallest such a number is 216. Substituting 216 into the equation for $ n^3$, $ k\equal{}90$. So the smallest k is $ \boxed{90}$.[/hide]


not hard.here is my solution. :)
assume $x_1 \geq x_2 \geq \cdots x_n$
(i)if $x_n \geq 0$,clearly,$x_1 \geq \frac{|x_{1}|+\ldots{|x_{n}|}}{n} \geq \frac{|x_{1}|+\ldots{|x_{n}|}}{2n-1}$
(ii) if   $x_n<0$,we can assume $x_i \geq 0,x_i+1 <0 (i \leq n-1)$
then we have $2x_1+ \cdots +2x_i+(-x_1-x_2- \cdots -x_n)=|x_{1}|+\ldots{|x_{n}|}$
it is also clear.
the equation will be hold,when $x_i=1,(i=1,2, \cdots n-1) x_n=-n$

How  the second  case works ??  I dont  get  it...sorry .

Am I misunderstanding something? L.H.S. = $max {|x_1|,|x_2|,...,|x_n|}$, so the remaining is trivial?


[quote="fjwxcsl"]Let$ {x_1,x_2,...,x_n}\geq{0},x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} n,k\in{N}(k\ge{1}),$prove (or disprove) that

 $ (1 \plus{} x_1^{k \plus{} 1})(1 \plus{} x_2^{k \plus{} 1})...(1 \plus{} x_n^{k \plus{} 1})\ge(1 \plus{} x_1^{k})(1 \plus{} x_2^k)...(1 \plus{} x_n^{k}).$[/quote]

   Using a new  inequality mechanical proving  program,we have proved that the inequality  is holds  for

   $ n \equal{} 3,k \equal{} 1,2,...,30;$
   $ n \equal{} 4,k \equal{} 1,2,...6;$
   $ n \equal{} 5,k \equal{} 1.$


modular arithmetic
algorithm
number theory
prime factorization

how is it found that the following is true: $ 2^{1194622}\equiv 965745\pmod{1194623}$ _without_ using a calculator? Is there maybe a way to use ch. remainder th. here? but then, this would require that one know a factorization for 1194623...

Thanks for your help

If $ 1194623$ is prime, then it can be done very nicely, although your answer is wrong.  By Fermat's Little Theorem,

$ a^{p\minus{}1}\equiv 1\pmod p\Rightarrow 2^{1194622}\equiv 1\pmod {1194623}$

If $ 1194623$ is not prime, we can use Euler's Theorem, $ a^{\phi(m)}\equiv 1\pmod m$ to at least reduce it.

But I don't know how to do it without the prime factorization of $ 1194623$ (that's not to say it can't be done).

hey e, thanks! yes,  i think you would need a way to come up with the prime factorization for $ 1194623$, with pollard-$ \rho$ or something.

The relevant algorithm is called [url=http://en.wikipedia.org/wiki/Exponentiation_by_squaring]exponentiation by squaring[/url].  The whole point of this calculation is to prove that $ 1194623$ is composite (by the Fermat primality test).  Note that we have done so without actually finding any of its factors.


inequalities
Euler
algebra proposed
algebra

Find the minimal $ n$ that $ \frac{1}{1}\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}...\plus{}\frac{1}{n}\ge 1000$.

[quote="zaya_yc"][quote="pelao_malo"]Find the minimal $ n$ that $ \frac {1}{1} \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n}\ge 1000$.[/quote]

$ \frac {1}{n} > ln(1 \plus{} \frac {1}{n})$--------->
$ \frac {1}{1} \plus{} \frac {1}{2} \plus{} ... \plus{} \frac {1}{n} > ln(1 \plus{} \frac {1}{1}) \plus{} ln(1 \plus{} \frac {1}{2}) \plus{} ... \plus{} ln(1 \plus{} \frac {1}{n}) \equal{} ln(\frac {2}{1}\frac {3}{2}\frac {4}{3}...\frac {n \plus{} 1}{n}) \equal{} ln(n \plus{} 1) \equal{} 1000$------>

$ n_{min} \equal{} e^{1000} \minus{} 1$[/quote]
The first you can not  use inequalities to fint out n.
For example we has innequalities: $ \frac {1}{n} \plus{} .. \plus{} \frac {1}{1} > \frac {1}{n}$
So $ \frac {1}{n} > 1000$ which is tradition.
You should use $ \frac {1}{1} \plus{} .. \plus{} \frac {1}{n} \minus{} \ln(n) < C$ where C is Euler constant.
$ 0 < C < 1$
The second : You answer is $ n \equal{} e^{1000} \minus{} 1$ but it is not a integer.

[quote="pelao_malo"]Find the minimal $ n$ that $ \frac {1}{1} \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n}\ge 1000$.[/quote]
$ n\equal{}[e^{1000}]\minus{}(1\ or\ 2)$

[quote="aviateurpilot"][quote="pelao_malo"]Find the minimal $ n$ that $ \frac {1}{1} \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n}\ge 1000$.[/quote]
$ n \equal{} [e^{1000}] \minus{} (1\ or\ 2)$[/quote]
Use \[ \sum_{k\equal{}1}^n\frac 1k \equiv H_n\equal{}ln n\plus{}\gamma \plus{}\frac{1}{2n}\minus{}\sum_{k\equal{}1}^m\frac{B_{2k}}{2kn^{2k}}\minus{}\theta _{m,n}\frac{B_{2m\plus{}2}}{(2m\plus{}2)n^{2m\plus{}2}}.\]
It give, that for minimal n is $ ln(n)\equal{}999,42278433509856713939348790991759756...$


Αφιερωένη στους λάτρες τη γεωμετρίας
Έστω $P$ ένα Ευκελίδιο σημείο στο επίπεδο ενός τριγώνου $ABC$ και έστω $D,E,F$ τα συμμετρικά του $P$ ως προς τις μεσοκαθέτους των τμημάτων $BC,CA,AB$ .Συμβολίζουμε με
$A_{M}$ , $B_{M}$ , $C_{M}$ τα μέσα των τμημάτων $BC,CA,AB$ και με $D_{M},\ E_{M},\ F_{M}$  τα μεσα των τμημάτων $EF,\ FD,\ DE$ .
Έστω $Q$ το ισογώνιο συζηγές του σημείου $P$ ως προς το τρίγωνο $ABC$ και έστω $Q '$ το συμπληρωματικό του $Q$ ως προς το τρίγωνο $ABC$ .Δείξτε οτι οι ευθείες $A_{M}D_{M},\ B_{M}E_{M}, \ C_{M}F_{M}$  περνούν από το σημείο $Q'$

***Η έννοια του συμπληρωματικού .
Έστω S to κέντρο βάρους του τριγώνου $ABC$ και $T$  τυχαίο σημείο στο επίπεδο του τριγώνου .Τότε η εικόνα του σημείου $T$ ύστερα από ομοιοθεσία κέντρου $S$ και λόγου $-\frac{1}{2}$  λέγεται  το συμπληρωματικό του  $T$ ως προς το $ABC$

Καλή διασκέδαση  :)

πρακαλώ τον Σιλουανό να μου κάνει μετάφραση όλων των ασυνηθιστων όρων
τον ευχαριστώ πολύ εκ των προτέρων δείχνει θαυμασια άσκηση

Ποιοι είναι οι ασυνήθιστοι όροι ?   :maybe:
Νομίζω το μόνο που θα μπορούσε να προκαλέσει σύγχηση (το συμπληρωματικό δηλαδή) το εξηγώ .

[b]Ευκλείδειο σημείο[/b]:σημείο (το λέει επιτηδες έτσι για να σε μπερδέψει Αμβρόσιε)
[b]Ισογώνιο συζυγές[/b]:Έστω τρείς σεβιανές που διέρχονται από το P.
Η ημιευθεία Αχ που είναι τ.ω. $\angle BAx=\angle CAP$ καλείται ισογώνια της AP.Κυκλικά ορίζονται και οι άλλες δύο ισογώνιες,Επίσης αποδεικνύεται (με 2 τουλάχιστον τρόπους) ότι οι ισογώνιες σεβιανών είναι και αυτές σεβιανές.Το σημείο από το οποίο διέρχονται οι ισογώνιες των σεβιανών PA,PB,PC καλείται ισογώνιο συζυγές του P.
[b]Συμπληρωματικό σημείo[/b]:ANAΔΟΧΟΣ:SILOUAN!!!!

Ευχαριστώ πάρα πάρα πολύ τον αγαπητό μου Νίκο για την μετάφραση

Αν δεν κάνω λάθος Σιλουανέ, εννοείς το ''complement point'' ενός σημείου σε σχέση με ένα δοσμένο τρίγωνο.

Υπάρχει ένα site με τέτοιους ορισμούς και άλλα ενδιαφέροντα θέματα γεωμετρίας.

http://www.xtec.es/~qcastell/ttw/ttweng/indexs/index_B.html

Κώστας Βήττας.

Γνωρίζω,πλέον τη λύση του προβλήματος αλλά επειδή είναι αρκετά μεγάλη δεν μπορώ να τη βάλω τώρα.Πιστεύω μέχρι την Κυριακή θα την έχω γράψει...

[i]Πρόκειται για πολύ ενδιαφέρον πρόβλημα![/i]

ΜΠΡΑΒΟ ΝΙΚΟ !!!!! :lol:
Περιμένω τη λύση σου .Χρησιμοποιείς διανύσματα καθόλου ?

[quote="silouan"][b]ΑΣΚΗΣΗ ΓΕΩΜΕΤΡΙΑΣ[/b]
Έστω $P$ ένα Ευκελίδιο σημείο στο επίπεδο ενός τριγώνου $ABC$ και έστω $D,E,F$ τα συμμετρικά του $P$ ως προς τις μεσοκαθέτους των τμημάτων $BC,CA,AB$ .Συμβολίζουμε με
$A_{M}$ , $B_{M}$ , $C_{M}$ τα μέσα των τμημάτων $BC,CA,AB$ και με $D_{M},\ E_{M},\ F_{M}$  τα μεσα των τμημάτων $EF,\ FD,\ DE$ .
Έστω $Q$ το ισογώνιο συζηγές του σημείου $P$ ως προς το τρίγωνο $ABC$ και έστω $Q '$ το συμπληρωματικό του $Q$ ως προς το τρίγωνο $ABC$ .Δείξτε οτι οι ευθείες $A_{M}D_{M},\ B_{M}E_{M}, \ C_{M}F_{M}$  περνούν από το σημείο $Q'$

***Η έννοια του συμπληρωματικού .
Έστω S to κέντρο βάρους του τριγώνου $ABC$ και $T$  τυχαίο σημείο στο επίπεδο του τριγώνου .Τότε η εικόνα του σημείου $T$ ύστερα από ομοιοθεσία κέντρου $S$ και λόγου $-\frac{1}{2}$  λέγεται  το συμπληρωματικό του  $T$ ως προς το $ABC$

[/quote]

Η λύση και ένα μικρό σχήμα βρίσκονται στο συννημένο.
Περιμένω τα σχόλιά σας...

[i][b]Νick Rapanos[/b][/i]

Ακριβώς Νίκο μπράβο .Αυτή είναι πάνω κάτω και η λύση μου .


geometry
rectangle
circumcircle
combinatorics proposed
combinatorics

In a square of sidelength $7$, $51$ points are given. Show that there's a disk of radius $1$ covering at least $3$ of these points.

From the upper-left corner of the square divide each of the adjacent sides in 4 segments with length $\sqrt{2}$ and one with length not greater than $\sqrt{2}$ (since $5\sqrt{2}> 7$).
Following horizontal and vertical lines defined by the endpoints of these segments, we divide the square into $25$ rectangles, each of which is a square with side length $\sqrt{2}$ or can be covered with such square.
Note that the circumcircle of a square with side $\sqrt{2}$ has radius $1$. Thus, using these circumcircle, we may cover all the initial square with $25$ circles each with radius $1$ (they are not pairwise disjoint).
The $51$ points belongs to these $25$ circles, so from the pigeon-hole principle, at least $3$ of the points belong to the same circle.

Pierre.

The $ 7 \times 7$ square is covered by a $ 5\sqrt{2} \times 5\sqrt{2}$ square, since $ 5\sqrt{2} > 7$. The larger square is partitioned in 25 $ \sqrt{2} \times \sqrt{2}$ cells. By pigeonhole's principle, at least 3 of the 51 points must be contained in a same cell, at its turn contained in a disk of radius 1.


Θέτουμε για ευκολία $ a_{1} \equal{} a$ και $ a_{2} \equal{} b$. Άρα τώρα πρέπει να αποδείξουμε ότι $ (a \plus{} \frac {1}{a})^{2} \plus{} (b \plus{} \frac {1}{b})^{2}\geq \frac {25}{2}$, με $ a \plus{} b \equal{} 1$

Έχουμε
$ (a \plus{} \frac {1}{a})^{2} \plus{} (b \plus{} \frac {1}{b})^{2}\geq \frac {1}{2}{[(a \plus{} \frac {1}{a}) \plus{} (b \plus{} \frac {1}{b})]}^2 \equal{} \frac {1}{2}[(a \plus{} b) \plus{} (\frac {1}{a} \plus{} \frac {1}{b})]^{2} \equal{} \frac {1}{2}(1 \plus{} \frac {a \plus{} b}{ab})^{2} \equal{} \frac {1}{2}(1 \plus{} \frac {1}{ab})^{2}$    (1)

Αρκεί λοιπόν να αποδείξουμε τη σχέση (1) όπου ισχύει αφού $ (a \plus{} b)^{2}\geq 4ab\Leftrightarrow \frac {1}{ab}\geq 4\Rightarrow 1 \plus{} \frac {1}{ab}\geq 5\Rightarrow (1 \plus{} \frac {1}{ab})^{2}\geq 25\Leftrightarrow \frac {1}{2}(1 \plus{} \frac {1}{ab})^{2}\geq \frac {25}{2}$ :wink:

[quote="mailo"]Αν $ a_1,a_2$ θετικοι πραγματικοι αριθμοι τετοιοι  ωστε $ a_1 a_2\neq 0$ και $ a_1 \plus{} a_2 \equal{} 1$ νδο

$ \left(a_1 \plus{} \frac {1}{a_1}\right)^2 \plus{} \left(a_2 \plus{} \frac {1}{a_2}\right)^2\geq \frac {25}{2}$

Βγαινει με BCS.[/quote]
n apodix8i kati akoma isxirotero...

$ \left(a_1 \plus{} \frac {1}{a_1}\right)^n \plus{} \left(a_2 \plus{} \frac {1}{a_2}\right)^n\geq 2*(\frac{a_1\plus{}a_2}{2}\plus{}\frac{2}{a_1\plus{}a_2})^n$

[quote="silouan"]Just a hint
[hide]Jensen  :D  [/hide][/quote]

Ρε Σιλουανέ, αυτό δεν είναι hint αλλά η λύση της άσκησης. Το μόνο που πρέπει να σκεφτεί κανείς ύστερα είναι να θεσει τη συνάρτηση $ f(x)\equal{}(x\plus{}\frac{1}{x})^{n}$ και η άσκηση βγαίνει. :P


[quote="tranvanluan"]Let $ a;b;c$are the side lengths of a triangle.
and $ a \plus{} b \plus{} c \equal{} 4$
Prove that
$ \frac {1}{a^2 \plus{} 5} \plus{} \frac {1}{b^2 \plus{} 5} \plus{} \frac {1}{c^2 \plus{} 5} < \frac {4}{9}$[/quote]

$ Ineq \leftrightarrow \frac {a^2}{a^2 \plus{} 5} \plus{} \frac {b^2}{b^2 \plus{} 5} \plus{} \frac {c^2}{c^2 \plus{} 5} > \frac {7}{9}$

$ LSH \equal{} \frac {a^2}{a^2 \plus{} 5} \plus{} \frac {b^2}{b^2 \plus{} 5} \plus{} \frac {c^2}{c^2 \plus{} 5}$

$ \equal{} \frac {a^4}{a^4 \plus{} 5a^2} \plus{} \frac {b^4}{b^4 \plus{} 5b^2} \plus{} \frac {c^4}{c^4 \plus{} 5c^2}$

$ LHS \ge\ \frac{(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum a^4 \plus{} 5\sum a^2}$

Need prove  $ \frac{(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum a^4 \plus{} 5\sum a^2} > \frac {7}{9}$

$ \leftrightarrow 2(a^4 \plus{} b^4 \plus{} c^4) \plus{} 18\sum a^2b^2 > 35 \sum a^2$

$ \leftrightarrow 32(a^4 \plus{} b^4 \plus{} c^4) \plus{} 288 \sum a^2b^2 > 35(a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c)^2$

I think we can easy prove it  :)

[quote="Allnames"][quote="onlylove_math"][quote="tranvanluan"]Let $ a;b;c$are the side lengths of a triangle.
and $ a \plus{} b \plus{} c \equal{} 4$
Prove that
$ \frac {1}{a^2 \plus{} 5} \plus{} \frac {1}{b^2 \plus{} 5} \plus{} \frac {1}{c^2 \plus{} 5} < \frac {4}{9}$[/quote]

$ Ineq \leftrightarrow \frac {a^2}{a^2 \plus{} 5} \plus{} \frac {b^2}{b^2 \plus{} 5} \plus{} \frac {c^2}{c^2 \plus{} 5} > \frac {7}{9}$

$ LSH \equal{} \frac {a^2}{a^2 \plus{} 5} \plus{} \frac {b^2}{b^2 \plus{} 5} \plus{} \frac {c^2}{c^2 \plus{} 5}$

$ \equal{} \frac {a^4}{a^4 \plus{} 5a^2} \plus{} \frac {b^4}{b^4 \plus{} 5b^2} \plus{} \frac {c^4}{c^4 \plus{} 5c^2}$

$ LHS \ge\ {(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum a^4 \plus{} 5\sum a^2}$

Need prove  $ {(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum a^4 \plus{} 5\sum a^2} > \frac {7}{9}$

$ \leftrightarrow 2(a^4 \plus{} b^4 \plus{} c^4) \plus{} 18\sum a^2b^2 > 35 \sum a^2$

$ \leftrightarrow 32(a^4 \plus{} b^4 \plus{} c^4) \plus{} 288 \sum a^2b^2 > 35(a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c)^2$

I think we can easy prove it  :)[/quote]
I think The last ineq is not true because $ LHS$ has only  $ 32(a^4 \plus{} b^4 \plus{} c^4)$,but $ RHS$ has$ 35(a^4 \plus{} b^4 \plus{} c^4)$
and i think it is quite hard to prove it.Can you comtinious[/quote]
Yes
$ a\equal{}2;b\equal{}1;c\equal{}1$
$ LHS\equal{}RHS$
But $ a;b;c$are the side lengths of a triangle.
then LHS<RHS
and only_math'proof  is wrong
No one Like it


prove that if ${{a + b} \over a} = {b \over {a + b}}$ is true, then [i]a[/i] and [i]b[/i] cannot both be real.

[hide=" :ninja: considering +ve reals "]$(a+b)^2=ab$

$\Longrightarrow (a+b) = \sqrt{ab}$-----$(1)$

But by [b]A.M-G.M[/b]$\Longrightarrow (a+b)\geq 2\sqrt{ab}$ where equality holds when $a+b=2\sqrt{ab}$ which is true when $a = b$  which gives minimum value of $a+b$

So clearly $(1)$ does not hold for any [b]+ve real a,b[/b]
[/hide]

[hide="considering negative real"]$\text{Let} a>0 ,b<0$

$(a-b)^2= -ab$

$a^2-ab+b^2=0$

$2a^2-2ab+2b^2=0$

$(a-b)^2+a^2+b^2=0$

Which is not possible(for datails,see scorpiuss's post)

Basically,I've just stretched the question unnecesseraly,so go by scorpiuss's or  10000th users methods :D [/hide]
Is this right now?(Yes,what i did earlier was very wrong 10000thuser :blush: ,is this one wrong?)

[quote="shadysaysurspammed"]But by [b]A.M-G.M[/b]$\Longrightarrow (a+b)\geq 2\sqrt{ab}$ where equality holds when $a+b=2\sqrt{ab}$ which is true for all [b]reals.[/b][/quote]You know well this is not true because of the square roots...  :ninja:

[EDIT]You are almost there, just work correctly your arguments

[hide=" ;)  The way I would have done it"] Simplifying you get $a^2+ab+b^2=0$. Clearly both $a,b$ are nonzero. Suppose $b\ne-a$ is real. The discriminant of the expression with respect to $a$ is $b^2-4b^2=-3b^2<0$ for all nonzero reals $b$. This implies that $a$ cannot be real. Since $a$ and $b$ are interchangeable, both cannot be reals simultaneously.[/hide]

[EDIT]Corrected a slight mistake :)

[hide]
Suppose there exists a real solution $(a,b)$. $a\neq 0$ to make the denominator nonzero.

Multiplying by the denominators, $(a+b)^2=ab\Rightarrow a^2+ab+b^2=0$. Multiplying by 2, $a^2+b^2+(a+b)^2=0$.

But by trivial inequality, $a^2>0$ (since $a\neq 0$), $b^2\geq 0$, $(a+b)^2\geq 0\Rightarrow a^2+b^2+(a+b)^2>0$ for all real $(a,b)$ with $a\neq 0$, a contradiction.
[/hide]

[quote="shadysaysurspammed"]Is this right now?(Yes,what i did earlier was very wrong 10000thuser :blush: ,is this one wrong?)[/quote]Not quite correct because in this case AM-GM applies to [b]positive reals[/b]. You have to be careful when you have negative reals. Small example: $a>0$ and $b<0$. Then AM-GM with square roots no longer holds, you can't compare $a+b$ and $2\sqrt{ab}$ because $a+b$ will be a real number but $\sqrt{ab}$ will be an imaginary number. That's why you need change your argument.

[EDIT]scorpius119: since $a+b\ne0$ and $a\ne0$, $b\ne-a$. :)

[quote="10000th User"]Not quite correct because in this case AM-GM applies to [b]positive reals[/b]. [/quote]

Yes ofcourse :wallbash:   looks like I was in a hurry to solve the question.I will take care now.

Thanks a lot 10000th user :lol:


Prove that :for every given positive integer $n$ ,and $\forall r,s\in Z$ we can represent the complex number $r+si$ as the form
$r+si=a_{m}(-n+i)^{m}+a_{m-1}(-n+i)^{m-1}+\dots+a_{1}(-n+i)+a_{0},(0\le a_{i}\le n^{2})$
and the way of the expasion is unique.

ehm.... counterexample:

[hide]$r=n^{2}-1$
$s=-2n$

1) $a_{0}=0$, $a_{1}=0$, $a_{2}=1$, $a_{i}=0$ $\forall i\geq3$
2) $a_{0}=-n^{2}-1$, $a_{1}=-2n$, $a_{i}=0$ $\forall i\geq2$

the first part can be solved so, $a_{0}=r+sn$ and $a_{1}=s$[/hide]

or have i misunderstood the problem?

byebye

Sorry for a typo. :blush:
the coeffients $a_{i}$ should have the property $0\leq a_{i}\leq n^{2},a_{i}\in N$
I have edited my post .
I think This is a famous theorem,But I don't know how to solve it.


function
integration
complex analysis
topology
complex analysis unsolved

Is the following true:

Let $ f$ be a continuous complex function defined on an open set $ O\subset\mathbb{C}$ such that $ \int_{G}f\equal{}0$ for every regular $ n$-gon $ G$ in $ O$. Is $ f$ necessarily holomorphic? Is it true if $ G$ is a non-regular $ n$-gon?

I think both are true, because both of them are path homotopic to a triangle.

A path $ A\to B\to A$ connected by straight segments or any fixed arc is path-homotopic to a triangle, but there are trivially a lot of counterexamples.


geometry
incenter
circumcircle
geometry proposed

Given triangle $ ABC$ with $ AB>AC$. $ l$ is tangent line of the circumcircle of triangle $ ABC$ at $ A$. A circle with center $ A$ and radius $ AC$, intersect $ AB$ at $ D$ and $ l$ at $ E$ and $ F$. Prove that the lines $ DE$ and $ DF$ pass through the incenter and excenter of triangle $ ABC$.

[quote="Raja Oktovin"]Given triangle $ ABC$ with $ AB > AC$. $ l$ is tangent line of the circumcircle of triangle $ ABC$ at $ A$. A circle with center $ A$ and radius $ AC$, intersect $ AB$ at $ D$ and $ l$ at $ E$ and $ F$. Prove that the lines $ DE$ and $ DF$ pass through the incenter and excenter of triangle $ ABC$.[/quote]
Let I be an intersection of ED and the bisector of angle C.
Since $ EA\equal{}AD, \angle EAD\equal{}180^o\minus{}C$ we have $ \angle AED\equal{}\frac{\angle C}{2}\equal{}\angle ACI$
Then $ AECI$ is cyclic we get $ \angle AIC\equal{}\angle AIE\plus{}\angle EIC\equal{}\angle ACE\plus{}\angle EAC$
$ \equal{}\angle ABC\plus{}\frac{180^o\minus{}\angle EAC}{2}\equal{}90^o\plus{}\frac{\angle ABC}{2}$
So I is the incenter of triangle ABC
Let J is an intersection of FD and the external bisector of angle C
We have $ \angle ICJ\equal{}\angle IDJ\equal{}90^o$ hence $ CIDJ$ is cyclic
We obtain $ \angle DJI\equal{}\angle ICD (1)$
On the other hand, $ \angle ADE\equal{}1/2C\equal{} \angle ICB$ then $ CIDB$ is cyclic
So $ \angle ICD\equal{}\angle IBD (2)$
From (1) and (2) we have $ \angle IJD\equal{}\angle IBD$
Then $ IDBJ$ is cyclic we get $ \angle IBJ\equal{}90^o$
Therefore J is the excenter of triangle $ ABC$

[quote="Raja Oktovin"] [color=darkred]Let $ ABC$ be a triangle with $ c>b$ , the incenter $ I$ and the $ A$ - exincenter $ I_a$ . Denote $ D\in (AB)$ for which $ AD\equal{}b$ . Consider $ E$ , $ F$ which

belong to the tangent in $ A$ at the circumcircle of $ \triangle ABC$ so that  $ AE\equal{}AF$ and $ AB$ separates $ I$ , $ F$ . Prove that $ E\in DI$ and $ F\in DI_a$ [/color]. [/quote]

[color=darkblue][b][u]Proof[/u].[/b]

$ \odot\ AD\equal{}AE\equal{}AF\ \implies\ DE\perp DF$ . Observe that $ \left\|\begin{array}{ccc}
CD\equal{}CF & \implies & m(\widehat {ADF})\equal{}90^{\circ}\minus{}\frac C2\\\\
AD\equal{}AC & \implies & m(\widehat{ADI})\equal{}m(\widehat {ACI})\equal{}\frac C2\end{array}\right\|\ \bigoplus\ \implies$

$ \ m(\widehat {FDI})\equal{}$ $ m(\widehat{ADF})\plus{}m(\widehat {ADI})\equal{}90^{\circ}\ \implies\ DI\perp DF$ . Thus, $ DE\perp DF\ \wedge\ DI\perp DF\ \implies\ E\in DI$ .

$ \odot\ SD\equal{}SC\equal{}SB\equal{}SI\equal{}SI_a\ \implies$ the point $ D$ belongs to the circle with diameter $ II_a\ \implies\ DI_a\perp DI\ \stackrel{(DI\perp DF)}{\ \implies\ }\ F\in DI_a$ .

[b][u]Remark[/u].[/b] Prove easily that $ \frac {ID}{IE}\equal{}\frac {a}{b\plus{}c}$ and $ \frac {I_aD}{I_aF}\equal{}\frac ap$ .[/color]


Solve in integers the equation:

    $x^2+y^2+z^2=1993$ ;)


Hey, everyone out there, can you tell me where i can find good books that would help in the Chemistry Olympiad?  Or previous tests or lab practicals with solutions? Thanks

[url]http://www.chemistry.org/portal/a/c/s/1/acsdisplay.html?DOC=education\student\co02.html[/url]

This site has some of the past national exams.


Find all the functions $f: \mathbb N^* \rightarrow \mathbb N^*-\{1\}$ satisfying :
$\forall n \in \mathbb N^*  : f(n)+f(n+1)=f(n+2).f(n+3)-168$

[quote="ali"]Simple indeed.
One can easily see that an -an-2 = an+1 (an+2 - an)
Now since an+1>=2 we get that an -an-2 >= 2(an+2 -an)
this sequence will therefore look like this eventually:
f(0),f(1),f(2),...,f(N),... = a0,a1,....,a,b,a,b,a,b
we should therefore have ab=a+b+168 or that (a-1)(b-1)=13^2
we have (a,b)={(14,14), (2,170)}
going backwards from 14,14 we get the full sequence: 14,14,14,....
that is f(i)=14 for all i.
second case gives 2,170,2,170,.... or 170,2,170,2,...
-Ali[/quote]

I don't think I quite get it.So what do you get eventually?

he gets that $f$ is periodic, with those possible periods

I hope that you posted this exercice after the response time is past :alien:


integration
calculus
calculus computations

If $ I_{n}$ denotes $ \int_{0}^{1} \frac{e^{x}}{(1\plus{}x)^{n}}\ dx$, show that $ nI_{n\plus{}1} \equal{} 1\minus{} (\frac{1}{2})^{n}e\plus{} I_{n}$ and state the values of $ n$ for which this is valid. If $ n$ is a positive integer, what is the disadvantage of trying to use this reduction formula to find $ I_{n}$?

$ u\equal{}(1\plus{}x)^{\minus{}n}, du\equal{}\minus{}n(1\plus{}x)^{\minus{}(n\plus{}1)}, dv\equal{}e^xdx, v\equal{}e^x$

$ \int_0^1 udv \equal{} u(1)v(1)\minus{}u(0)v(0)\minus{}\int_0^1 vdu$

$ \int_0^1 \frac{e^x}{(1\plus{}x)^n}dx\equal{}2^{\minus{}n}e^1\minus{}1^{\minus{}n}e^0 \plus{}n\int_0^1 \frac{e^x}{(1\plus{}x)^{n\plus{}1}}dx$

$ nI_{n\plus{}1}\equal{}I_n\plus{}1\minus{}\frac{e}{2^n}$

so it holds for all n

the disadvantage is that you cant get $ I_1$ from $ I_0$ or else everything would follow nicely from

$ I_n\equal{}\frac{1}{n\minus{}1}\left(1\minus{}\frac{e}{2^{n\minus{}1}}\right)\plus{}\frac{I_{n\minus{}1}}{n\minus{}1}$
$ \equal{}\frac{1}{n\minus{}1}\left(1\minus{}\frac{e}{2^{n\minus{}1}}\right)\plus{}\frac{1}{(n\minus{}1)(n\minus{}2)}\left(1\minus{}\frac{e}{2^{n\minus{}2}}\right)\plus{}\frac{I_{n\minus{}2}}{(n\minus{}1)(n\minus{}2)}$
$ \equal{}\sum_{k\equal{}1}^{n\minus{}1}\frac{1}{\prod_{j\equal{}1}^k (n\minus{}j)}\left(1\minus{}\frac{e}{2^{n\minus{}k}}\right)\plus{}\frac{I_1}{\prod_{j\equal{}1}^k (n\minus{}j)}$


Express $ 3\star(4\star5)$ as a common fraction given $ a\star$$ b\equal{}\frac{ab}{a\plus{}b}$.

Just plug and chug. $ 4 \star 5 \equal{} \frac{20}{9}$, and $ 3 \star \frac{20}{9} \equal{} \frac{3 \cdot \frac{20}{9}}{3 \plus{} \frac{20}{9}} \equal{} \boxed{\frac{60}{47}}$.

[hide="hopefully this is the right answer"]for this problem we just put the #'s in like the #'s inside the equation

(4)(5)/4+5=20/9

(3)(20/9)/3+20/9 =60/47

so the answer is 60/47 or $\boxed{60/47}$[/hide]

[b][i][u]please rate my post thanks!![/u][/i][/b]

please be fair though!!  thanks [=

  :clap2:   :clap2:   :clap2:   :clap2:

may the force be with you


quadratics
function
number theory unsolved
number theory

For any natural number $m=\overline{c_1c_2\dots c_k}$ define the function $L(\overline{c_1c_2\dots c_k})=c_1^2+c_2^2+\dots+c_k^2$ (sum of the squares of the digits, to be perfectly clear). Now define the sequence $\{ a_n \}_{n\in\mathbb{N}}$: $a_1\in\mathbb{N}, a_{n+1}=L(a_n).$

Prove the following: For [i]any[/i] initial $a_1$, the sequence $\{ a_n \}$ will either end up with $1$ (after which it will obviously remain constant), or with $145$, after which it will enter the cycle $145, 42, 20, 4, 16, 37, 58, 89, 145\dots$