Quadrilateral ABCD is cyclic, with perpendicular diagonals AC and BD
intersecting at E. Point M is the midpoint of CD. Prove that the line
through M and E is perpendicular to AB.

Wrong place to ask this type of question in this subforum.

Please see [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=55177[/url]

Mods, please lock this topic, this question can be answered in the above link.

[quote="#H34N1"][hide="Answer"]
You have cyclic quadrilateral $ABCD$. Inscribe it into a circle. Draw both diagonals. $BE=AE$, because those diagonals must be congruent since the quadrilateral is cyclic. You have side $EM'$, which is the point that $\overline{ME}$ will pass through when cutting $AB$ is a common side between $\triangle BM' E$ and $\triangle AM' E$. Then, you have to have $AM'=BM'$, since the quadrilateral is cut by the line that passes through the midpoint of one side and the center of a triangle. By $SSS$ simililatly, $\angle AM' E ~ \triangle BM' E$. Since $AB$ is a line, $\angle AM' E+\angle BM' E$. So, $\angle AM' E=90^\circ$, and $EM\perp AB$.[/hide]

EDIT: I think I'm wrong so please move this. (Its an olympiad and I didn't know)[/quote]It is possible that the diagonals are not congruent and the quadrilateral is still cyclic.

Yeah. That's what I found out. I assumed that it was a rectangle. :oops:


[quote="Inequalities Master"]Let $ x_1,x_2,...,x_n\geq 0$.Prove that:
$ x_1 \plus{} x_2 \plus{} ... \plus{} x_n\geq (n \minus{} 1)\sqrt [n]{x_1x_2...x_n} \plus{} \sqrt {\frac {x_1^2 \plus{} x_2^2 \plus{} ... \plus{} x_n^2}{n}}$[/quote]
It follows from my inequality:
\[ (n\plus{}2)(x_1 \plus{} x_2 \plus{} ... \plus{} x_n)\geq n^2\sqrt [n]{x_1x_2...x_n} \plus{} 2\sqrt {n(x_1^2 \plus{} x_2^2 \plus{} ... \plus{} x_n^2)}\]
For the start prove it:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=90136


the fixed square $ ABCD$ has center  $ O$ and $ OXYZ$ is a variable square $ OX,OZ$ cut $ CD,DB$ in $ K,J$ , Prove that the area of the quadrilateral $ KDJO$ is independent of the position of the square $ OXYZ$

[geogebra]eb8060a0bc401f8f04e50c56d318378201b55544[/geogebra]

Try posting that on the High School Basics forum.

[hide]This question has been posted before.

Rotate $ OXYZ$ about $ O$ by $ 90$ degrees three times and observe that three quadrilaterals identical to $ KDJO$ have been created. The area of each of them is one fourth of that of $ ABCD$[/hide]

tx chrischris , :blush:  i knew that this problem was very well-known . However there is a harder problem related to this one.

With the same notation above now $ ABCD$ is a rectangle , then find the positions of the square $ OXYZ$ such that the area of the quadrilateral $ KDJO$ is maximum and minimum respectively.

I will change the notation a bit because the square in your diagram is $ ABDC$ not $ ABCD$

Let $ AB$ and $ CD$ be the longer sides of the rectangle $ ABCD$.

Construct a smaller square $ EFGH$ with also centre $ O$, each side as long as $ AB$, such that points $ E,F$ are on $ AB$ and $ H,G$ are on $ CD$. As however we rotate $ OXYZ$, we get at least one fourth of the area of $ EFGH$. If we rotate $ OXYZ$ such that $ J$ is on $ CD$ (in fact on $ HG$), and $ K$ is on $ BC$ and extremely close to $ B$. We know the area of $ OJCK$ is maximized because we have maximized the area of the part of rectangle $ ABCD$ outside sqaure $ EFGH$ while keeping it a quadrilateral.
If we rotate $ OXYZ$ such that $ J$ is on $ CD$ (in fact on $ HG$), and $ K$ is on $ BC$ and extremely close to $ C$. Similarly we have minimized the area of the part of $ ABCD$ outside $ EFGH$.


[hide]Let's say $a^x$ has a mod cycle period of y in mod n.  Obviously a to the first power is a.  The last number in this sequence before it repeats must be one since it'll be multiplied next by a to restart the series.  This is the yth number.  Any positive number multiple of y will also work because $a^(ky)=a^y*a^y*a^y*...a^y$ which is 1*1*1*1*...1=1.
Applying this to the original problem, we see that d must be the mod cycle period so it must be equal to 1 in mod n.[/hide]

Solution from Bezout : $d = u*b + v*c$ ($a$ is inversible modulo $n$ so $< 0$ exposents make sense)

[quote="Nea"]Prove if $b>0$, $c>0$, $d=(b,c)$, $a^b \equiv 1(\bmod n)$, and  $a^c \equiv 1(\bmod n)$, then:

$a^d \equiv 1(\bmod n)$[/quote]
Your problem is equivalent to:
$n|a^b-1$ and $n|a^c-1$ $\Rightarrow n|a^{(b,c)-1}$.
you can have a reference [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=358343#p358343]here[/url].


If $ f(n\plus{}3) \equal{} f(n)\minus{} {1\over f(n)}\plus{}1$ and $ f(5)\equal{}10, n\in\mathbb {N}$,
then $ f(2009)\equal{}?$

What makes you think there is a nice answer to this? :huh:

By the way, the answer I get from the program is:
[hide]
$ 673.63328842939274433228898610$[/hide]

Yeah... I agree with Yongyi.  The problem really comes down to doing the same operation 668 times in a row.  It becomes obvious really quickly that there are no patterns.


The positive integers are written consecutively in groups of five so that the first row contains 1, 2, 3, 4, 5; the second row 6, 7, 8, 9, 10; etc. The row which has a sum nearest to the value of 150 is the:

a. 5th row b. 6th row c. 7th row d. 8th row e. 9th row

since it's a consecutive group of 5 integers, you want to set it up such that $ (x\minus{}4)\plus{}(x\minus{}3)\plus{}(x\minus{}2)\plus{}(x\minus{}1)\plus{}x\equal{}150$ where $ \frac{x}{5}$ is the number row. Solving, you get $ x\equal{}32$ so it would be closest to the $ \boxed{6th}$ row.
This can be confirmed by noting that the numbers in the 6th row are $ 25,26,27,28,29,30$ and their sum is $ 140$.


How many positive divisors of 30! are prime?

It's all primes less or equal to 30: 2,3,5,7,11,13,17,19,23,29-10 divisors

You see, we need to find all the primes that are less than 30, so we get 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29 to give us a total of $\boxed{10}$ positive divisors.

[quote=PurpleEnigma]I'm probably being really stupid, but why can primes larger than 30 not divide 30! ?
lol I'm tired right now :)[/quote]

they can try 58

I probably don't know what im doing, but isn't $58$ not a prime?

[quote=HelloToAops][quote=Maxima]Think about it.  $ 30!$ is just $ 30 \cdot 29 \cdot 28 \cdot 27 \cdot \cdots \cdot 1$.  Therefore, $ 31$, could not divide it.  Do you see why?[/quote]

actually yes it can! Do 30! / 31 and see what you get![/quote]

That's not an integer.


Για κάθε $ x_1, x_2, x_3, ... , x_n > 0$ τέτοια ώστε $ \frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} \frac {1}{x_3} \plus{} ... \plus{} \frac {1}{x_n} \le \frac {1}{2}$
να δειχτεί ότι:
1. $ x_1x_2 \plus{} x_2x_3 \plus{} ... \plus{} x_{n \minus{} 1}x_n \plus{} x_nx_1> 4n^2$
2. η παραπάνω ανισότητα δεν βελτιώνεται άλλο

Η εκφώνησή μου έχει typo :blush: . Το άθροισμα των αντιστρόφων μικρότερο ή ίσο του $ \frac {1}{2n}$
και το ζητούμενο είναι $ > 4n^3$

Η λύση που αποδεικνύει και το δεύτερο υποερώτημα στη διαδικασία είναι η εξής:
$ \frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} \frac {1}{x_3} \plus{} ... \plus{} \frac {1}{x_n} \le \frac {1}{2n} \Leftrightarrow$ (AM - HM)
$ (x_1 \plus{} x_2 \plus{} ... \plus{} x_n) \ge (2n^3) (I)$

είναι τώρα $ \frac {a}{x_j} \le \frac {1}{2n} \Leftrightarrow x_j \ge 2 \Rightarrow x_ix_{i \plus{} 1} \ge x_i \plus{} x_{i \plus{} 1}$
με την παράσταση να τείνει στο ίσον όσο πιο κοντά στο 2 βρίσκονται οι αριθμοί
άρα $ x_1x_2 \plus{} x_2x3 \plus{} ... \plus{} x_nx_1 \ge x_1 \plus{} x_2 \plus{} x_2 \plus{} x_3 \plus{} ... \plus{} x_n \plus{} x_1 \equal{} 2(x_1 \plus{} x_2 \plus{} ... \plus{} x_n)$
και, από $ (I)$ είναι $ x_1x_2 \plus{} x_2x3 \plus{} ... \plus{} x_nx_1 \ge 4n^3$
με το πρώτο κι άρα και το τελικό ίσον να μην ισχύουν ποτέ
η παράσταση όμως τείνει στο γενικό ίσον όταν οι αριθμοί τείνουν στο 2,
άρα μπορούμε να δούμε (χρησιμοποιώντας λίγο όρια) ότι αν βρούμε μια βελτίωση $ 4n^3 \plus{} k$
υπάρχουν $ x_1, x_2, ... , x_n$ τέτοια ώστε $ x_1x_2 \plus{} x_2x_3 \plus{} ... \plus{} x_nx_1 \le 4n^3 \plus{} k$
άρα η παράσταση δε βελτιώνεται

Πάντως για την αρχική λανθασμένη που έδωσα πραγματικά έτσι θα έβγαινε.


A groundskeeper must buy bags of fertilizer.  Each bag treats $ 10000$ square feet of ground and costs $ \$27$.  The groundskeeper can buy only whole bags of fertilizer.  How much will it cost to buy the least number of bags necessary to treat $ 6000$ square yards of ground?

6000 square yards is 6000*9=54000 square feet.
The ceiling of 54000/10000 is 6.
6*27=162 dollars

Since there are 3 feet to a yard, there are $3^2=9$ square feet to a square yard. Thus, we need to buy enough fertilizer to treat $6000\cdot9=54000$ square feet of ground. Since we can only buy whole bags of fertilizer, we need to buy 6 of them (able to treat $10000\cdot6=60000$ square feet of ground), costing $27\cdot6=\boxed{162}$ dollars.

Soryy if I am backseat modding, but please do not revive very old threads.


I'd like to know the names of the proposers of the IMO 2006 problems... can we know it right know or it is too early?

Wasn't there a Mathlinker who proposed a problem?

Arne proposed (maybe someone else also? Valentin? Harazi?  :) ) but the question is whether it was used (and which problem it was). Unveil the mystery  :D

My problem hasn't been chosen, which is a pity... My opinion is that it deserved to be chosen instead of a problem of the same category and difficulty which actually made the paper... too bad. :D I'll try again next year.

Will you show us the problem you were going to use?  Or is that the same problem you will use next year?

Well, let me find out first whether it made the shortlist...

[quote="Arne"]Problem 1 - South Korea (I'm pretty sure this one is from Hojoo)
Problem 2 - Serbia
Problem 3 - Ireland
Problem 4 - USA
Problem 5 - Romania
Problem 6 - Serbia[/quote]

And who proposed problem number 5?  :wink:

[quote="Arne"]Problem 1 - South Korea (I'm pretty sure this one is from Hojoo)
Problem 2 - Serbia
Problem 3 - Ireland
Problem 4 - USA
Problem 5 - Romania
Problem 6 - Serbia[/quote]
   Two problems from Serbia?That is very interesting because Serbia is not so big country(they are our first neighbours :)).I can say that I am pretty disapointed with problem of South Korea(even more if it were from Hojoo Lee),that problem could be solved without any knowledge which is needed for IMO and that makes contest somekind unfair for those who know geometry solid.In my opinion problems 3 and 5 were the best for IMO.

How do you propose a problem?  Can anyone do it, or do you have to be asked?

[quote="ideahitme"]Hmm... What I just can say is the fact that the problem 5 is so coooool!
 I also want to know who the proposer of the problem 5 is ![/quote]
And were you the proposer of the 1st problem?  :roll:

I don't know who the proposers are but for problem 5 it might have been Enescu, Val or others. I am not sure about Val otherwise he would have had announced it by now I guess. For problem 4 Zuming Feng might be the proposer.

mysmartmouth, I believe that anyone can do it. TomVerhoeff's report on IMO 2002 (which is available at http://olympiads.win.tue.nl/imo) will be interesting.
shobber, Arne is right. You can find a list of my proposals at my web page at http://my.netian.com/~ideahitme/orange.html

Actually, not anyone can do it. You must talk to your Leader (in case of USA Zuming Feng) and propose to him the problem (usually you also submit the full solution with it). Of course the problem should be [i]IMO difficulty[/i] (although looking at this years problem 1 one might argue what does that phrase mean actually :) ).

PS The Romanian problem is not mine.

two problems from Serbia, that so cool!!!  :rotfl:

Can you tell us who proposed the problem?

I have my shortlist ( :D ) of people who could have done it, but you never know...

Can't reveal it until the next IMO.

And yeah, sorry for the spam. :blush:

Arne: Did your problem reach the shortlist?

And did you made any problem, ideahitme? (Hojoo)

Problems 2 and 6 are from Dušan Đukić, Serbia.

There were better geometry problems in the shortlist. These problems would be more challenging.

2 Mamat.
As I have heard, there are some leaders which are interested more in the sport part of the event than in mathematical. So to reach their aim they followed these rules:

1) Not to let classical geometry be in the task.
2) To make P1 and P4 as easy as possible and other problems as hard as possible.

Then many students score about 14 and a bit more and among them students from those countries appear. As a result:

-Low quality of problems, especially geometry.
-Problem with bronze cut-off Arne stated in one of the topics.


[quote="STARS"]We call a [i]ship[/i] a figure made up of unit squares connected by common edges.
Prove that if there is an odd number of possible different ships consisting of n unit
squares on a $ 10 \times 10$ board, then n is divisible by 4.[/quote]
Could please explain the definition of "different ships"?What kind of ships can be considered as different?I mean,are horizontal and vertical ships consisting of $ 2$ unit squares different?


conics
parabola
trigonometry
geometry proposed
geometry

Let a triangle $ABC$ . At the extension of the sides  $BC$ (to C) ,$CA$ (to A) , $AB$ (to B) we take points  $D,E,F$  such that  $CD=AE=BF$ .
Prove that if the triangle $DEF$  is   equilateral  then $ABC$ is also equilateral

My opinion $ABC$ is equilateral iff \angle DEC=\angle

[quote="abdurashidjon"]My opinion the solution follows from transformation[/quote]

maybe . But you should put your solution

Suppose $\triangle ABC$ is not equilateral. Therefore, he must have one or two longest sides. Let's say $AB$ is one of them and $AC<AB$. So, $CE<AF$ and $\angle CAB\leq \angle ACB$. Hence, $\angle ECD\leq \angle EAF$. Now, considering triangles $ECD$ and $EAF$ it is obvious that $ED<EF$, contradiction.

[quote="abdurashidjon"]My opinion the solution follows from transformation[/quote]

how can u solve it like that?

may be , it cant solve like that. ;)

[color=darkblue][b][u]Lemma.[/u][/b] Given are three parabolas $P_{k}$ with the equations $y=a_{k}x^{2}+b_{k}x+c_{k}\ ,\ k\in \overline{1,3}$. Then

$P_{1}\cap P_{2}\cap P_{3}\ne\emptyset\Longleftrightarrow \left|\begin{array}{ccc}1 & 1 & 1 \\\ a_{1}& a_{2}& a_{3}\\ c_{1}& c_{2}& c_{3}\ \end{array}\right|^{2}=\left|\begin{array}{ccc}1 & 1 & 1 \\\ a_{1}& a_{2}& a_{3}\\ b_{1}& b_{2}& b_{3}\ \end{array}\right|\cdot \left|\begin{array}{ccc}1 & 1 & 1 \\\ b_{1}& b_{2}& b_{3}\\ c_{1}& c_{2}& c_{3}\ \end{array}\right|\ \ \ (1)\ .$

[b]Remark.[/b] Given are the parabolas $P_{k}$ with the equations $y=a_{k}x^{2}+b_{k}x+c_{k}\ ,\ k\in \overline{1,n}\ ,\ n\ge 3$. Then

$P_{1}\cap P_{2}\cap\dots\cap P_{n}\ne\emptyset\Longleftrightarrow rg\left|\begin{array}{cccc}a_{1}& b_{1}& c_{1}& 1\\\ a_{2}& b_{2}& c_{2}& 1\\\ a_{3}& b_{3}& c_{3}& 1\\ \dots & \dots & \dots & \dots\ \end{array}\right|\le 3$ and the condition $(1)\ .$

[b][u]Proof of the proposed problem[/u][/b] (from  http://www.mathlinks.ro/Forum/viewtopic.php?t=88459 ).

Denote $AE=BF=CD=x$ and suppose that the triangle $DEF$ is equilateral,

i.e. $DE=EF=FD=y$. Therefore, the parabolas with the equations

$y=2(1+\cos B)\cdot x^{2}+2a(1+\cos B)\cdot x+a^{2}\ ,$

$y=2(1+\cos C)\cdot x^{2}+2b(1+\cos C)\cdot x+b^{2}\ ,$

$y=2(1+\cos A)\cdot x^{2}+2c(1+\cos A)\cdot x+c^{2}$

have at least a common point. From the previous lemma obtain $a=b=c\ .$[/color]


Find the quotient of two integers whose difference is 10 and whose product is as small as possible.

Smallest value of a(a-10) is -25, so a=5, and a-10=-5. Their quotient is -1.

I don't understand that at all. Did not explain well atall

Let's say we have two numbers, $a$ and $a-10$.

To make their product as small as possible, we can make $-a$ equals to $a-10$.

Solving for $a$, we get $a=5$.

That means we have the two numbers $5$ and $-5$.

Their quotient is $\boxed{-1}$

Or, we could use the vertex of the parabola. Since the quadratic is $a^2-10a,$ and the coefficient of $a^2$ is positive, the minimum value is when $a$ is equal to $-\frac{-10}{2} = 5.$ Thus $a-10 = -5,$ and the quotient is $\boxed{-1}.$


calculus
integration
vector
calculus computations

Does anyone know how to do the following problems? Any help appreciated, thanks.

Let [b]F[/b] = 1x^3 [b]I[/b] + 3y^3 [b]J[/b] and let [b]n[/b] be the outward unit normal vector to the positively oriented circle x^2+y^2=1. Compute the flux integral
[img]http://webwork1.math.utah.edu/math2210fall2005-90/tmp/l2h/set10/8-16201488/16201488img3.gif[/img].


\[ \left\{ \begin{array}{l}
 {x^2}(y \minus{} z) \equal{} \frac{5}{3} \\
 {y^2}(z \minus{} x) \equal{} 3 \\
 {z^2}(x \minus{} y) \equal{} \frac{1}{3} \\
 \end{array} \right.\]

I think there is no real numbers satysfing this system....
I'dont have paper right now( :blush: ) and is really late here so my ''mind computations'' may be mistaken.....

Dimitris

[quote="Dimitris X"]Adding them and making the factorization we get:

$ (x \minus{} y)(y \minus{} z)(z \minus{} x) \equal{} \minus{} 5$.

Multiplying them we get:

$ x^2y^2z^2(x \minus{} y)(y \minus{} z)(z \minus{} x) \equal{} \frac {5}{3} \Longleftrightarrow x^2y^2z^2 \equal{} \minus{} \frac {1}{3}$

which cant be true :wink:[/quote]

i don't understand what you want to prove  :blush:  could you please write clearly  :)

[quote="imedeen"][quote="Dimitris X"]Adding them and making the factorization we get:

$ (x - y)(y - z)(z - x) = - 5$.

Multiplying them we get:

$ x^2y^2z^2(x - y)(y - z)(z - x) = \frac {5}{3} \Longleftrightarrow x^2y^2z^2 = - \frac {1}{3}$

which cant be true :wink:[/quote]

i don't understand what you want to prove  :blush:  could you please write clearly  :)[/quote]

If we add these three equations we take:

$ x^2(y-z)+y^2(z-x)+z^2(x-y)=\frac{5}{3}+3+\frac{1}{3} \Longleftrightarrow x^2(y-z)+y^2z-y^2x+z^2x-z^2y=5 \Longleftrightarrow x^2(y-z)+yz(y-z)-x(y-z)(y+z)=5 \Longleftrightarrow (y-z)(x^2+yz-xy-xz)=5 \Longleftrightarrow (y-z) \left( x (x-y)-z(x-y) \right) =5 \Longleftrightarrow (y-z)(x-y)(x-z)=5 \Longleftrightarrow (x-y)(y-z)[-(z-x)]=5 \Longleftrightarrow (x-y)(y-z)(z-x)=-5$.

BUT if we multiply these three equation we take:

$ x^2y^2z^2(x-y)(y-z)(z-x)=\frac{5}{3} \cdot 3 \cdot \frac{1}{3} \Longleftrightarrow x^2y^2z^2(-5)=\frac{5}{3} \Longleftrightarrow x^2y^2z^2=-\frac{1}{3}.$

Which cant be true....!!!

I hope it is clear enough dear friend now...  :wink:  :)

Can't we just say that from
\[ \left\{\begin{array}{l}{x^{2}}(y\minus{}z) \equal{}\frac{5}{3}\\ {y^{2}}(z\minus{}x) \equal{} 3\\ {z^{2}}(x\minus{}y) \equal{}\frac{1}{3}\\ \end{array}\right.\]
follows that y>z and z > x and x  > y (Since both sides should be positive)
so y > z > x > y
So there are no solutions?

[quote="Brothas"]Can't we just say that from
\[ \left\{\begin{array}{l}{x^{2}}(y \minus{} z) \equal{} \frac {5}{3} \\
{y^{2}}(z \minus{} x) \equal{} 3 \\
{z^{2}}(x \minus{} y) \equal{} \frac {1}{3} \\
\end{array}\right.\]
follows that y>z and z > x and x  > y (Since both sides should be positive)
so y > z > x > y
So there are no solutions?[/quote]

Yeap....
Nice and easy....

I think everyone saw the obvious solution of multiplying the three equations and skipped over the even more obvious proof. :P

[quote="imedeen"]
\[ \left\{\begin{array}{l} {x^2}(y \minus{} z) \equal{} \frac {5}{3} \\
{y^2}(z \minus{} x) \equal{} 3 \\
{z^2}(x \minus{} y) \equal{} \frac {1}{3} \\
\end{array} \right.\]
[/quote]

i rewrite my problem , now it must have solution  :wink:
\[ \left\{\begin{array}{l} {x^2}(y \minus{} z) \equal{} \minus{}\frac {5}{3} \\
{y^2}(z \minus{} x) \equal{} 3 \\
{z^2}(x \minus{} y) \equal{} \frac {1}{3} \\
\end{array} \right.\]

[quote="imedeen"] i rewrite my problem , now it must have solution  :wink:
[/quote]

I dont understand how you could "rewrite" "your" problem and continue to post it in "unsolved" forum :
If this rewritten problem is yours and you know the solution : please post a new topic in the appropriate forum (proposed and own)
Else, how are you sure there is a normal solution (not numeric calculus one) ?

[quote="imedeen"] \[ \left\{\begin{array}{l} {x^2}(y \minus{} z) \equal{} \minus{} \frac {5}{3} \\
{y^2}(z \minus{} x) \equal{} 3 \\
{z^2}(x \minus{} y) \equal{} \frac {1}{3} \\
\end{array} \right.\]
[/quote]

At least two solutions (I dont know if other exist) :

$ (x,y,z)\equal{}\left( \minus{}\sqrt[3]{\frac 23},\minus{}\sqrt[3]{\frac 94},\sqrt[3]{\frac 23}\right)$

$ (x,y,z)\equal{}\left( \minus{}\sqrt[3]{\frac 23},\minus{}\sqrt[3]{18},\minus{}\sqrt[3]{\frac 1{12}}\right)$

you are right mr pco this system has exactly 2 roots


pro.) Prove that if (k, p-1)=d, then the congruence x^k-1 ==0 (mod p)  has d roots.

I'll assume p is a prime :)

Let g be a primitive root modulo p.

Any arbitrary x in the field of integers mod p can be written in g^y for some nonnegative integer y.

The congruence $ x^k \equiv 1 (mod p)$ is equivalent to
$ g^{yk} \equiv 1 (mod p)$. This is equivalent to solving the linear congruence
$ yk \equiv 1 (mod p\minus{}1)$. It is trivial that this congruence yields (k,p-1) solutions.


[quote="pbornsztein"]:D  We used it for our last round of the correspondance training program in France....

Pierre.[/quote]
Isn't this from shortlist 2001

a solution could be found here:

http://www.kalva.demon.co.uk/short/soln/sh01a3.htm


Find all polynomials $ W(x)\in \mathbb R[x]$ such that
\[W(x^2)W(x^3)=W(x)^5 \quad \forall x \in \mathbb R.\]

$ \exists a\in\mathbb{C}: \ W(c) \equal{} 0$
if $ a\not\in \{0,1\}$ we can find $ (a_i)_{i: \in\mathbb{N}}$ such that $ a_0 \equal{} a,a_{i \plus{} 1}^2 \equal{} a_i$ and $ |\{a_i|i\in\mathbb{N}\}| \equal{} \plus{} \infty$
and we have $ W(a_{i \plus{} 1}) \equal{} W(a_{i \plus{} 1}^2)W(a_{i \plus{} 1}^3) \equal{} W(a_i)W(a_{i \plus{} 1}^3)$
so $ W(a_0) \equal{} 0\ and\ W(a_i) \equal{} 0 \Rightarrow W(a_{i \plus{} 1}) \equal{} 0$ gives $ \forall i\in\mathbb{N}: \ W(a_i) \equal{} 0$
then $ \forall x\in\mathbb{R}: \ W(x) \equal{} 0$

if $ W\not\equiv 0$ then $ \forall x\in\mathbb{R}: \ W(x) \equal{} cx^a(x \minus{} 1)^b$ for some $ a,b\in \mathbb{N}$ where $ c^2 \equal{} c^5\neq 0$
and $ W\in \mathbb{R}[X]$ gives $ c \equal{} 1$.
if $ b > 0$ then $ W( \minus{} 1)^5 \equal{} W(( \minus{} 1)^2)W(( \minus{} 1)^3) \equal{} 0$ gives $ X \plus{} 1|X^a(X \minus{} 1)^b$ (imppssible)
then $ b \equal{} 0$ and $ W(X) \equal{} X^a$ for some $ a\in \mathbb{N}$

$ W \equal{} cX^a$ ($ a\in\mathbb{N},c\in\{0,1\}$)

[quote="aviateurpilot"]
if $ a\not\in \{0,1\}$ we can find $ (a_i)_{i: \in\mathbb{N}}$ such that $ a_0 \equal{} a,a_{i \plus{} 1}^2 \equal{} a_i$ and $ |\{a_i|i\in\mathbb{N}\}| \equal{} \plus{} \infty$
and we have $ W(a_{i \plus{} 1}) \equal{} W(a_{i \plus{} 1}^2)W(a_{i \plus{} 1}^3) \equal{} W(a_i)W(a_{i \plus{} 1}^3)$
so $ W(a_0) \equal{} 0\ and\ W(a_i) \equal{} 0 \Rightarrow W(a_{i \plus{} 1}) \equal{} 0$ gives $ \forall i\in\mathbb{N}: \ W(a_i) \equal{} 0$
then $ \forall x\in\mathbb{R}: \ W(x) \equal{} 0$

[/quote]

I think if $ |a|\equal{}1$ but $ a\not\in \{0,1\}$ we still do not have contradiction. The set $ |\{a_i|i\in\mathbb{N}\}|$ might not be infinite.  :)

For $ |a|\equal{}1$ we have $ a\equal{}e^{i\varphi}$ for some $ \varphi\in (0,2\pi]$.
Then $ a_n: \equal{}e^{i\varphi /2^n}$ for $ n\in\mathbb{N}_0$ satisfies
$ a_0\equal{}a$ and $ a_{n\plus{}1}^2\equal{}a_n$ for all $ n\in\mathbb{N}_0$ and $ \#\{a_n: n\in\mathbb{N}_0\}\equal{}\infty$.


Prove that $c^{2}= a^{2}-b^{2}$ on an ellipse when a and b is the major and minor axis respectively, and c is the foci.

I'm not sure if what you wrote is true for your condition...I think I might be wrong though.

You mean $c$ is the distance from the focus to the center.

[hide="Hint"] Remember that one way to define an ellipse is that the ellipse is the locus of points $P$ such that, for foci $f_{1}, f_{2}$, the difference of distances from the foci $| Pf_{1}-Pf_{2}|$ is constant.

What is this quantity at the ends of the major axis?
What is this quantity at the ends of the minor axis? [/hide]

I thought a and b were supposed to be half of the major and minor but I guess not.

Sorry I'm not very good with the conics stuff.


show that if a polynomial P(x) satisfies P(2x^2 - 1)=P^2(x), it must be constant

$ P(2x^{2} \minus{} 1) \equal{} [P(x)]^{2}$

if $ \exists x_{0}\not \equal{} 1,\frac { \minus{} 1}{2},P(x_{0}) \equal{} 0$  $ \implies$  $ P(x )$ has infinitely many roots  $ \implies$  $ P(x ) \equal{} 0$

if $ P(x )$ hasn't infinitely many roots  $ \implies$  $ P(x ) \equal{} c,P(x) \equal{} (x \minus{} 1)c,P(x ) \equal{} (x \plus{} \frac {1}{2}).c,P(x ) \equal{} (x \minus{} 1)(x \plus{} \frac {1}{2})c$

The rest is easy :wink:


geometry
circumcircle
parallelogram
angle bisector
geometry proposed

Given a triangle $ ABC$. Let $ M$ be the midpoint of its side $ BC$. The external angle bisector of the angle $ CAB$ intersects $ BC$ at $ D$. Let the circumcircle of triangle $ ADM$ intersect $ AB$ and $ AC$ again at $ E$ and $ F$, respectively. Let $ N$ be the midpoint of the segment $ EF$. Show that the line $ MN$ is parallel to $ AD$.

A solution of THTT:
Called $K$ is symmetric point of $M$ through $F$
$M$ is midpoint of $BC \Rightarrow FBKC$ is parallelogram $\Rightarrow FC//BK\Rightarrow \angle{BKM=\angle{MFA}}$
$\angle{MFA}=\angle{AEM}\Rightarrow\angle{BKM}=\angle{BEM}\Rightarrow AMKB$ is concyclic quadrilateral $\Rightarrow \angle{BEK}=\angle{FMD}=\angle{FAD}=\angle{DAE}\Rightarrow \angle{BEK}=\angle{DAE}\Leftrightarrow AD//EK(1)$
Because $FN=NE,FM=MK\Rightarrow MN//EK(2)$
From $(1),(2)\Rightarrow MN//AD$  :)

[color=darkblue][b]Lemma (well-known).[/b] Let $ABC$ be a triangle.  Denote the middlepoint $M$ of the side $[BC]\ .$

$1.\blacktriangleright$ Let $E\in AB$, $F\in AC$ be two mobile points so that $BE=CF$ and the line $BC$ [b]doesn't separate[/b] the points $E$, $F\ .$
Then the geometrical locus of the middlepoint of the segment $EF$ is a line $d_{1}$ which is parallelly with the (internal) bisector of the angle $\widehat{BAC}$ and $M\in d_{1}\ .$

$2.\blacktriangleright$ Let $E\in AB$, $F\in AC$ be two mobile points so that $BE=CF$ and the line $BC$ [b]separates[/b] the points $E$, $F\ .$
Then the geometrical locus of the middlepoint of the segment $EF$ is a line $d_{2}$ which is parallelly with the external bisector of the angle $\widehat{BAC}$ and $M\in d_{2}\ .$[/color]

[color=darkred][b]The first problem.[/b] Given triangle ABC with M the  midpoint of BC. The [b](internal) bisector[/b] of the angle A intersects BC in D.
The circumcircle of ADM intersects AB and AC again at E and F. Let N be the midpoint of EF. Show that $MN\parallel AD\ .$[/color]

[color=darkblue][b][u]Proof.[/u][/b] $BE\cdot BA=BD\cdot BM$ $\Longleftrightarrow$ $BE\cdot c=\frac{ac}{b+c}\cdot \frac{a}{2}$ $\Longleftrightarrow$ $\boxed{\ BE=\frac{a^{2}}{2(b+c)}\ }$ and $CF\cdot CA=CD\cdot CM$ $\Longleftrightarrow$
$CF\cdot b=\frac{ab}{b+c}\cdot \frac{a}{2}$ $\Longleftrightarrow$ $\boxed{\ CF=\frac{a^{2}}{2(b+c)}\ }\ .$ Therefore, $BE=CF$ and aplly the first part of the above lemma.[/color]

[color=darkred][b]The second problem.[/b] Given triangle ABC with M the midpoint of BC. The [b]external bisector[/b] of the angle A intersects BCin D.
The circumcircle of ADM intersects AB and AC again at E and F. Let N be the midpoint of EF. Show that $MN\parallel AD\ .$[/color]

[color=darkblue][b][u]Proof.[/u][/b] Suppose w.l.o.g. that $b>c\ .$ $BE\cdot BA=BD\cdot BM$ $\Longleftrightarrow$ $BE\cdot c=\frac{ac}{b-c}\cdot\frac{a}{2}$ $\Longleftrightarrow$ $\boxed{\ BE=\frac{a^{2}}{2(b-c)}\ }$ and $CF\cdot CA=CD\cdot CM$ $\Longleftrightarrow$ $CF\cdot b=\frac{ab}{b-c}\cdot\frac{a}{2}$ $\Longleftrightarrow$ $\boxed{\ CF=\frac{a^{2}}{2(b-c)}\ }\ .$ Therefore, $BE=CF$ and apply the second part of the above lemma.[/color]

[color=darkred][b]Generalization.[/b] Given are the triangle $ABC$ with $M$ the midpoint of $BC$ and two points $X$ and $Y$ which belong to the sideline $BC$ so that $XB=x\cdot XC$, $YB=y\cdot YC$ and $xy=\frac{c}{b}\ .$  The internal bisector of the angle $A$ intersects $BC$ in $D$ and the external bisector of the angle $A$ intersects $BC$ in the point $E$. The circumcircle of the triangle $AXY$ intersects $AB$ and $AC$ again at $P$ and $R$. Let $S$ be the midpoint of $PR$. Show that $MS\parallel AD$ or $MS\parallel AE\ .$[/color]

[color=darkblue][b]Remark 1.[/b] See a larger extension at http://www.mathlinks.ro/Forum/viewtopic.php?t=108013
[b]Remark 2.[/b] [u]A very nice consequence ![/u] What becomes this problem if $x=y=\sqrt{\frac{c}{b}}$, i.e. $X\equiv Y$ ?
The circumcircle of the triangle $AXX$ is tangent to the line $BC$ in the point $X$ (are two same circles !).[/color]

First let's prove that $ \bigtriangleup ABM \sim \bigtriangleup FMN$
From equal arcs we have $ \angle MAE=\angle MFE=\angle MFN$.So it is sufficient to prove
$ \frac{AB}{AM}=\frac{FM}{FN}$
Let circumradiuses of $ \bigtriangleup ABC$ and $ \bigtriangleup ADM$ be $ R_1$ and $ R_2$ respevtively.Then
$ FM=2R_2 \cdot sin\angle CAM , FN=\frac{1}{2}FE=R_2 \cdot sin\angle A , AB=2R_1 \cdot sin\angle C$ and $ AM=\frac{CM}{sin\angle CAM} \cdot sin \angle C=\frac{R_1 \cdot sin \angle A\cdot sin \angle C}{sin\angle CAM}$

$ \Rightarrow \frac{FM}{FN}=\frac{2sin\angle CAM}{sin\angle A }=\frac{AB}{AM}\Rightarrow \bigtriangleup ABM \sim \bigtriangleup FMN \Rightarrow \angle AMB=\angle MNF$

Let $ O$ be circumcentre of $ \bigtriangleup ADM$

$ FD=DE$ and $ FN=NE \Rightarrow FE \perp DN \Rightarrow O \in DN$
$ \Rightarrow \angle NDA=\angle ODA=90-\angle AED=90-\angle AMD=90-\angle AMB=90-\angle MNF=180-(90+\angle MNF)=180-MND$
So $ MN$ is parallel to $ AD$


Starting from the opposite ends of a street, each of two messenger boys skated at his top speed towards the other's starting point. From the time they passed each other, one messenger took 1 more minute, and the other took 2 more minutes, to reach their respective destinations. How many minutes did it take the faster messenger to skate the entire distance?

[hide="too lazy for geogebra"]call the faster dude's speed x and the slower one's y. let X represent the faster guy and Y the slower guy draw a line sorta like this:

X[color=white]---------------------------------------------------[/color]Y
O---------------------------------------------------O

[color=white]...............................................[/color]Y
[color=white]...............................................[/color]<-----------------
O---------------------------------------------------O
---------------------------------->
[color=white]................................................[/color]X

now the remaining distance X has to cover is x and for Y its 2y. therefore, the whole thing is x+2y. Now call the original time elapsed in minutes t. now we have x=ty and 2y=tx. substitution yields $ 2y\equal{}t^2y$ so $ t\equal{}\sqrt{2}$. the whole distance, in terms of x, is $ x\plus{}\frac{\sqrt{2}x}{2}\cdot2\equal{}x(\sqrt{2}\plus{}1)$ so $ \sqrt{2}\plus{}1$ min. that doesnt even look right....[/hide]


geometry
rhombus
3D geometry
tetrahedron
sphere
perimeter
inequalities

Problems of Chinese Mathmetical Olympiad 1998.

Problems of Chinese Mathmetical Olympiad 1999.

Problems of Chinese Mathmetical Olympiad 2000.

Problems of Chinese Mathmetical Olympiad 2001.

Problems of Chinese Mathmetical Olympiad 2002.

Problems of Chinese Mathmetical Olympiad 2003.

謝謝!
還有更多的呢! :)
[url=http://www.sftw.umac.mo/~fstitl/cmo/questions.html]http://www.sftw.umac.mo/~fstitl/cmo/questions.html[/url]
(請選擇繁體中文檢視 Please view in Traditional Chinese)

Okay, I'll do the 1986's to 1993's first. :D

[u][b]1st CMO 1986[/b][/u]
Problem 1
Let $a_1,a_2,\cdots ,a_n$ be real numbers. Prove that the following two statements are equivalent:
(i) $a_i+a_j\geq 0$ for any $i$ and $j$ with $i\neq j$
(ii) $a_1x_1+a_2x_2+\cdots +a_nx_n\geq a_1x_1^2+a_2x_2^2+\cdots +a_nx_n^2$, where $x_1,x_2,\cdots ,x_n$ are any non-negative real numbers satisfying $x_1+x_2+\cdots +x_n=1$.

Problem 2
$D$, $E$ and $F$ are points on $BC$ such that $AD$, $AE$ and $AF$ are respectively an altitude, an angle-bisector and a median of $\triangle ABC$. If $AD=12$, $AE=13$ and $AF=m$, determine the values of $m$ for which $\angle BAC$ is acute, right and obtuse respectively?

Problem 3
Let $z_1,z_2,\cdots ,z_n$ be complex numbers such that $|z_1|+|z_2|+\cdots +|z_n|=1$. Prove that the sum $S$ of a subset of $\{z_1,z_2,\cdots ,z_n\}$ satisfies $|S|\geq\frac{1}{6}$

Problem 4
$PQRS$ is a convex quadrilateral inside $\triangle ABC$. Prove that the area of one of $\triangle PQR$, $\triangle PQS$, $\triangle PRS$ and $\triangle QRS$ is not greater than $\frac{1}{4}$ the area of $\triangle ABC$.

Problem 5
Is there a permutation of the number $1,1,2,2,3,3,\cdots ,1986,1986$ such that for any $k$, there are exactly $k$ other numbers between the two $k$'s?

Problem 6
Each point on a plane is coloured black or white arbitrarily. Prove that there exists an equilateral triangle of side $1$ or $\sqrt{3}$ such that all three vertices have the same colour.

[b][u]2nd CMO 1987[/u][/b]
Problem 1
Let $n$ be a positive integer. Prove that $z^{n+1}-z^n-1=0$ has a root satisfying $|z|=1$ if only if $n+2$ is divisible $6$.

Problem 2
An equilateral triangle $ABC$ of side $n$ is divided into $n^2$ equilateral triangles of side $1$ by lines parallel to the sides of $\triangle ABC$. Each point which is a vertex of st least one of these nit triangles is labelled with a real number. The points $A$, $B$ and $C$ are labelled with $a$,$b$ and $c$ respectively. In each rhombus composed of two unit triangles with a common side, the sums of the labels on the two sets of opposite vertices are equal.
(i) Determine the shortest distance between a point with the largest label and a point with the smallest label.
(ii) Determine the sum of all labels.

Problem 3
In a tournament, every two players have exactly one game between them. There are no ties. A player $A$ is awarded a prize if for every other player $B$, either $A$ defeats $B$ or $A$ defeats some player $C$ who defeats $B$. Prove that if only one player is awarded a prize, then this player defeats every other player.

Problem 4
Given are 5 points inside an equilateral triangle of area $1$. Prove that there exist three equilateral traingles such that they are homothetic to the given triangle, each of the 5 points is inside at least one of these triangles, and their combined area is at most $0.64$.

Problem 5
A tetrahedron is such that a sphere centred at a point $O$ is tangent to all 6 of its sides. Moreover, four spheres with the vertices as their respective centres are externally tangent to one another, and all are tangent to another sphere with centre $O$. Prove that the tetrahedron is regular.

Problem 6
The sum of $m$ distinct positive even integers and $n$ distinct positive odd integers is $1987$. What is the maximum value of $3m+4n$?

[u][b]3rd CMO 1988[/b][/u]
Problem 1
Let $a_1,a_2,\cdots ,a_n$ be given real numbers, not all being $0$. The real numbers $r_1,r_2,\cdots ,r_n$ are such that for all real numbers $x_1,x_2,\cdots ,x_n$, $r_1(x_1-a_1)+r_2(x_2-a_2)+\cdots +r_n(x_n-a_n)$ is less than or equal to $\sqrt{x_1^2+x_2^2+\cdots +x_n^2}-\sqrt{a_1^2+a_2^2+\cdots +a_n^2}$. Determine $r_1,r_2,\cdots ,r_n$.

Problem 2
Given are two concentric circles one having twice the radius of the other. Inscribed in the smaller circle is a convex quadrilateral $ABCD$. The extensions of $AB$, $BC$, $CD$ and $DA$ intersect the larger circle at $C_1$, $D_1$, $A_1$ and $B_1$  respectively. Prove that the perimeter of $A_1B_1C_1D_1$ is not less than twice that of $ABCD$, and determine when equality holds.

Problem 3
Given is a sequence of $n$ real numbers. A dragon is defined to be a block of consecutive terms whose average is greater than $1988$, and the head of this dragon is the first term in the block. A single term greater than $1988$ is also considered to be a dragon as well as its head. Suppose at least one dragon exosts. Prove that the average of all terms which are the heads of some dragons is greater than $1988$.

Problem 4
(i) Suppose $a_1$, $a_2$ and $a_3$ are positive real numbers satisfying
\[ (a_1^2+a_2^2+a_3^2)^2>2(a_1^4+a_2^4+a_3^4) \].
Prove that $a_1$, $a_2$ and $a_3$ are the three sides of some triangle.
(ii) Let $n$ be an integer greater than $3$. Suppose the inequality
\[ (a_1^2+a_2^2+\cdots +a_n^2)^2>(n-1)(a_1^4+a_2^4+\cdots +a_n^4) \]
holds for some positive real numbers $a_1,a_2,\cdots ,a_n$. Prove that $a_i$, $a_j$ are $a_k$ are the three sides of some triangle for all $i$, $j$ and $k$.

Problem 5
Given are three tetrahedrons $A_iB_iC_iD_i$, $1\leq i\leq 3$. Through the points $B_i$, $C_i$ and $D_i$ are drawn three planes $\beta_i$, $\gamma_i$ and $\delta_i$ perpendicular to $A_iB_i$, $A_iC_i$ and $A_iD_i$ respectively. If these nine planes meet at a point and $A_1$, $A_2$ and $A_3$ lie on a line, determine the intersection of the circumspheres of the three tetrahedra.

Problem 6
For any integer $n\geq 3$, let $f(n)$ denote the least positive integer which is not a divisor of $n$. Let $f^{(1)}(n)=f(n)$. If $f^{(k)}(n)\geq 3$, $f(f^{(k)}(n))$ is meaningful and is defined as $f^{(k+1)}(n)$. For each integer $n\geq 3$, determine $k$ where $f^{(k)}(n)=2$.

[u][b]4th CMO 1989[/b][/u]
Problem 1
Each of $A$ and $B$ is a union of pairwise disjoint arcs on the unit circle. Moreover, each arc in $B$ is of length $\frac{\pi}{m}$ where $m$ is s fixed positive integer. Denote by $A^j$ the set obtained from $A$ by rotating it about the origin through $\frac{j\pi}{m}$ radians in the counter-clockwise direction. Prove that there exists a positive integer $k$ such that $l(A^k\cap B)\geq \frac{l(A)l(B)}{2\pi}$, where $l(X)$ denotes the total length of a set $X$ of disjoint pairs.

Problem 2
Suppose $x_1+x_2+\cdots +x_n=1$ where $x_1, x_2, \cdots ,x_n$ are positive real numbers. Prove that
\[ \frac{x_1}{\sqrt{1-x_1}}+\frac{x_2}{\sqrt{1-x_2}}+\cdots +\frac{x_n}{\sqrt{1-x_n}}\geq\frac{\sqrt{x_1}+\sqrt{x_2}+\cdots +\sqrt{x_n}}{\sqrt{n-1}} \].

Problem 3
Let $S$ be the unit circle in the complex plane. Let $f: S\rightarrow S$ be defined by $f(z)=z^m$, where $m$ is a positive integer. Define $f^{(0)}(z)=z$ and $f^{(k+1)}(z)=f(f^{(k)}(z))$ for $k\geq 0$. The smallest positive integer $n$ such that $f^{(n)}(z)=z$ is called the period of $z$. Determine the total number of points in $S$ of period $1989$.

Problem 4
$\triangle ABC$ has inradius $r$. $D$, $E$ and $F$ are points on $BC$, $CA$ and $AB$ respectively. If the inradii of $\triangle AEF$, $\triangle BFD$ and $\triangle CDE$ are all equal, prove that their common value is $r-r'$, where $r'$ is the inradius of $\triangle DEF$.

Problem 5
Given are 1989 points in the plane, no $3$ on a line. How should they be partitioned into $30$ groups of distinct sizes so as to maximize the total number of triangles with vertices in different groups?

Problem 6
Determine all functions $f: S\rightarrow S$, where $S$ is the set of all real numbers greater than $1$, such that $f(x^,y^n)\leq f(x)^{\frac{1}{4m}}f(y)^{\frac{1}{4n}}$ for all real numbers $x,y>1$ and $m,n>0$.

How do I express problem 5 of 1994 in Latex?

[quote="shobber"]I am sorry I only have Jin's file......[/quote]
Ok,can you make the the problem in a picture file,then send me? :)

[b]CMO 1994[/b]
1:$ABCD$ is a trapzoid with $AB || CD$. $E$ is point on segment $AB$, $F$ is point on segment $CD$. Segment $CE$ and $BF$ meets at $H$. Segment $DE$ and $AF$ meets at $G$.
Prove: $S_{EHFG} \leq \frac{1}{4}\cdot S_{ABCD}$

2:There are not less than $4$ candies in $n$($n \geq 4$) plates. Take out a candy from each of the any two of these plates and put it into another plate. We call this One Operation. Can we move all the candies into one plate through finite operations? Prove your answer.

3:Find all functions $f: \left[ 1, + \infty \right) \to \left[1, +\infty \right)$ such that:
(1) $f(x) \leq 2(x+1)$

(2) $f(x+1) = \frac{ \left( f(x) \right)^2 -1 }{x}$

4:Given a $n$-th degree, complex coefficient polynomial $f(z)= C_0z^n + C_1z^{n-1}+ \cdots + C_{n-1}z + C_n$.
Prove that there exists a complex number $z_0$,  $|z_0| \leq 1$, satisfies $|f(z_0)| \geq |C_0|+|C_n|$.

5:Prove that
\[ \sum_{k=0}^{n} \binom{n}{k} 2^k \binom{n-k}{\left[ \frac{n-k}{2} \right]} = \binom{2n+1}{n}  \]
for any natural number $n$.

6:$M: (1994p, 7 \times 1994p)$ is point on the plane where $p$ is a prime number. Find the number of right triangles which satisfy the following condition:
(1) All the three vertex have integer coordinats and $\angle{M}=90^o$.
(2) The incenter of the triangle is the origin.

[quote="shobber"]I think the problems of CMO 2004 are already on this forum. You can check the Natianal Olympiad Section or use the search function.[/quote]

But I can't find it  :(  Can you help me?  :wink:

[quote="shobber"][quote="Jumbler"]Will it be OK if the problems are posted in the Olympiad Section, so that we can discuss?[/quote]
Ask Valentin.[/quote]The answer to that question is yes.

I'd archive them, but I'm not experienced yet to be tackling these problems. Maybe in a couple months.  :)

There are copyright problems. So you'd better ask Valentin.


linear algebra
matrix
LaTeX
vector
superior algebra
superior algebra solved

I feel bad having my first post be a question, but I am having difficulties with this.

A square matrix is called nilpotent if there is a non-negative integer k such that Ak equals a zero matrix.  Prove that a nilpotent matrix has no non-zero eigenvalues.

I have:
Let A be an n x n matrix with eigenvalue t.  Let x be a non-zero vector.  Then,
Ax = tx
and
AAx = ttx = A^2x = t^2x
and so
A^j=t^j x
for any nonzero integer, j.  Therefore, in order for
A^j=0
to be true
t^jx=0
must be true, and this is only true if t=0.

It just sounds clunky.

Your proof looks fine to me.

Welcome to the site.

Your proof is fine. As part of the welcome to the site, I'd like to direct your attention to the resources and tutorials here regarding the LaTeX typesetting system as implemented here. As an exhibit, here is your post with the most minor of changes:

[quote]A square matrix is called nilpotent if there is a non-negative integer $k$ such that $A^k$ equals a zero matrix. Prove that a nilpotent matrix has no non-zero eigenvalues.

I have:
Let $A$ be an $n \times n$ matrix with eigenvalue $t.$ Let $x$ be a non-zero vector. Then,
$Ax = tx$
and
$A\cdot Ax = t\cdot tx = A^2x = t^2x$
and so
$A^j=t^j x$
for any nonzero integer, $j$. Therefore, in order for
$A^j=0$
to be true
$t^jx=0$
must be true, and this is only true if $t=0.$

It just sounds clunky.[/quote]
The only change I'd make is in the first line: Let $x$ be a nonzero [b]eigen[/b]vector of $A.$ That's obviously what you meant anyway. That, and I'd omit the apology at the end.

Well thanks for the help.  I've gone around the board and tried to find some other parts I can help in, but most everythign is either over my head or already answered, but I'll keep trying :p

And i'll look into the latex thing also.


geometry
invariant
analytic geometry
geometry unsolved

How to prove that inversion preserves angles?

I've managed to do it using analytic geometry, incidentally I also proved that inversion tranform circles and lines into circles and lines. It wasn't so hard as it seemed to be, but is there a non-analytic proof?

I found this proof in a book. Without loss of generality we may consider two circles $\mathcal{A}$ and $\mathcal{B}$, which intersect at a point $P$. Let $P'$ be the inverse of $P$ and let $\mathcal{A'}$ and $\mathcal{B'}$ two circles that satisfy:

:arrow: $\mathcal{A}$ and $\mathcal{A'}$ are tangent at $P$.
:arrow: $\mathcal{B}$ and $\mathcal{B'}$ are tangent at $P'$.
:arrow: $\mathcal{A'}$ and $\mathcal{B'}$ pass through $P'$.

Since circles $\mathcal{A'}$ and $\mathcal{B'}$ pass through inverse points $P$ and $P'$, they're orthogonal to the circle of inversion, hence they're invariant under inversion. Therefore the angle between the curves at $P$ is equal to the angle of the inverse curves at $P'$. :)

I am assuming that $B$ and $B'$ are tangent at $P$ not $P'$. Could you explain the proof in a more detailed way :?:

Here you have a figure :)

[url=http://imageshack.us][img]http://img251.imageshack.us/img251/3218/inversionld6.png[/img][/url]

OK, now about the proof. There exists a theorem that says that every circle that passes through two inverse points $P$ and $P'$ is orthogonal to the circle of inversion. But there's another theorem that tell us that every circle orthogonal to the circle of inversion is invariant under the inversion.

That's all, as you can see. You may want to check such theorems, they're not difficult to prove :P

I see now  :oops:   it was really easy. Thanks for your patience.  As an excuse I can only say that I've been learning inversion for only few days.

I would be grateful if you helped with another inversion problem:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=131841[/url].

(If you have the same problem as me please try to solve it in your own way)


modular arithmetic
combinatorics proposed
combinatorics

[color=darkblue]Let $ X$ be a group of people, where any two people are friends or enemies. Each pair of friends from $ X$ doesn't have any common friends, and any two enemies have exactly two common friends. Prove that each person from $ X$ has the same number of friends as others.[/color]

Take two members of $ X$ society, $ \text{A}$ and $ \text{B}$. First has $ a$ friends, and second has $ b$ friends, and let $ \text{A}$ and $ \text{B}$ be friends. Mark with $ A_f$ and $ B_f$ sets of their friends excluding $ B$ and $ A$, respectively.
Members of $ A_f$ are $ \text{B}$'s enemies, since they have a common friend, $ \text{A}$. Each one of the members of $ A_f$ has exactly one friend in $ B_f$ then, and using the same logic ($ B_f$ are $ \text{A}$'s enemies) vice versa holds too.
Since they can't 'share' a friend in the opposite set, this is a bijection, so $ A_f$ and $ B_f$ have the same number of elements, and since $ a \equal{} |A_f| \plus{} 1$, $ b \equal{} |B_f| \plus{} 1$, $ a \equal{} b$.
Using this logic, every friend of $ \text{B}$ or $ \text{A}$ and consecutively, friend of their friends has the same number of friends. Is every member of $ X$ taken into account through this? Yes, because if we had a member who is an enemy to all these members, he'd have to have common friends with them- contradiction.

We can compute this common number of friends for each person.

Denote $ |X| \equal{} n$, and $ m$ the number of mutual pairs of friends. A chevron is a pair $ (u,\{v,w\})$ such that $ u,v$ and $ u,w$ are friends, but (of course) $ v,w$ are not. Then, according to the givens, the total number of chevrons is $ \displaystyle \sum_{v \in X} \frac {1} {2} \deg v (\deg v \minus{}1) \equal{} 2\Big ( \frac {1} {2} n(n\minus{}1) \minus{} m \Big ) \equal{} 2\Big ( \frac {1} {2} n(n\minus{}1) \minus{} \frac {1} {2} \sum_{v \in X} \deg v \Big )$, so $ \displaystyle \sum_{v \in X} \deg v (\deg v \plus{}1) \equal{} 2n(n\minus{}1)$. Knowing that all $ \deg v \equal{} d$, the common number of friends for each person, this yields $ nd(d\plus{}1) \equal{} 2n(n\minus{}1)$, so $ \boxed{n\equal{}\frac {1} {2} d(d\plus{}1)\plus{}1}$. (However, when $ d$ is odd, we need $ n$ even, so this implies $ d \not \equiv 3 \pmod{4}$.)

Hmmm... idk if this is correct but:

Let there be  $ n$ people and call them $ V_{i}$, $ 1\leq i\leq n$. We create a graph with $ n$ vertices (each vertex representing a person). We draw a line between two friends and a dotted line between two non-friends.

We have that the vertices in every dotted line share exactly two more vertices. Therefore, every dotted line is contained in exactly one $ 4$-cycle of straight lines (where both diagonals of the quadrilateral formed are obviously dotted).

We now take a vertex $ V_{i}$ and two dotted lines that are joined to it and $ V_{j},V_{k}$, respectively. Then $ V_{j}V_{k}$ is obviously dotted, meaning it is contained in exactly one $ 4\minus{}cycle$. Then every pair of lines connected to a specific vertex are members of exactly one $ 4$-cycle.

Take a random vertex $ V_{i}$. We will now establish a bijection. Since every pair of lines joined to $ V_{i}$ is member of exactly one $ 4$-cycle, and the diagonals of a $ 4$ cycle are dotted, exactly one dotted line from $ V_{i}$ is in that $ 4$ cycle. Furthermore, since every dotted line belongs to exactly one $ 4$-cycle, every dotted line joined to $ V_{i}$ corresponds to a $ 4$ cycle with a pair of lines from $ V_{i}$. Therefore, the number of pairs of lines joined to $ V_{i}$ equals the number of dotted lines joined to $ V_{i}$.

We can establish the following equation, letting $ |V_{i}|$ be the number of lines connected to $ V_{i}$ and $ n\minus{}1\minus{}|V_{i}|$ the number of dotted lines connected to $ V_{i}$:

$ \frac{(|V_{i}|\minus{}1)(|V_{i}|)}{2}\equal{}n\minus{}1\minus{}|V_{i}|$

Solving this equation gives a constant number of straight lines $ V_{i}$ is connected to.


calculus
integration
trigonometry
calculus computations

I am having problems with this proof.

Prove that, for even powers of sine that:

the integral with a upper limit of pi/2 and a lower limit of 0 of
sin^(2n)xdx = (1*3*5***(2n - 1) / 2*4*6***2n)pi/2.

I hope you understand the problem its kind of hard to type it in.
* = times

Thanks Gary

This is a job for the reduction formulas. Integrate by parts twice- you will express $\int_{0}^{\frac{\pi}{2}}\sin^{2n}x\,dx$ in terms of itself, some boundary terms (which will evaluate to zero), and $\int_{0}^{\frac{\pi}{2}}\sin^{2n-2}x\,dx$. Simplifying this, you get a recursive formula.

Actually, you only need to integrate by parts once in the proof of the reduction formula. Let $u=\sin^{2n-1}x$ and $dv=\sin x\,dx$ in the integration by parts. Once you've done that, express $\cos^{2}x$ as $1-\sin^{2}x,$ split one integral into two, then move one of those integrals back to the left side as part of an algebraic solution.


John has 40% fewer marbles than Alicia.  Miguel has 40% more marbles than Alicia.  If John, Alicia, and Miguel have a total of 180 marbles, how many marbles does Miguel have?

[hide]Let $J$, $A$, and $M$ represent the number of marbles that John, Alicia, and Miguel have, respectively.

From the given information we have the following system of equations:

$J=(1-0.4)A=0.6A$
$M=(1+0.4)A=1.4A$
$J+A+M=180$

Substituting the first and second equations into the third yields:

$0.6A+A+1.4A=180$
$3A=180$
$A=60$

Thus, Alicia has 60 marbles and Miguel has $1.4A=\boxed{84}$ marbles.[/hide]


Hint: there's a finite geometric sum involved.


A sphere is inscribed in a cube. Given that one edge of the cube is 6 inches, how many cubic inches are in the volume of the inscribed sphere? Express your answer in terms of $ \pi$.

The radius of the sphere is $ 3$ inches. Thus its volume is

$ \frac{4}{3}\pi(3)^3\equal{}\boxed{36\pi}$ cubic inches.


linear algebra
matrix
linear algebra unsolved

Find all matrixes  A such that A is commutitive with all invertible matirixes T (means AT=TA for all T).

That means $ A$ commutitive with all matrices of size $ n \times n$
becasue:
$ AB\equal{}BA  \iff A(B\plus{}kI_n)\equal{}(B\plus{}kI_n)A$

That argument is insufficient in a finite field (over the field with two elements, neither $ \begin{bmatrix}1&0\\0&0\end{bmatrix}$ nor $ \begin{bmatrix}1&0\\0&0\end{bmatrix}\plus{}I$ is invertible.

The result is still true, because the invertible matrices span everything. Can you prove this?

Thank you.
An ugly proof.
A matrix which satisfies there is $ x \in \mathbb{F}$ such that $ B \minus{} xI$ is invertible we can the matrix have property $ P$.

If a matrix doesn't satisfy property must have eigenvalues $ 0,1,\cdots,p\minus{}1$
Thus $ E_{ij}$ ($ i \ne j$) have the property $ P$,and also $ E_{11}\plus{}E_{12}$ must have the property $ P$.
The result follows obviously.
Can you check my proof,thank you :)

[Post have been edited]

My solution is to consider $ E_n$ and then $ E_{ij}$ by exchange $ i^{th},j^{th}$ columns. However, idea of Zhaobin is very nice,

Thank you.
And in the solution for jmerry's:
I have something wrong,because $ E_{11}\plus{}E_{12}$ may not has property$ P$.We may should consider $ E_{11}\plus{}E_{12}\plus{}E_{21}$ instead.

Yes, that works.

My version: Assume $ n>1$; it's trivial for $ n\equal{}1$. For each pair $ (i,j)$, find a permutation $ \sigma(i,j)$ such that $ \sigma(i)\neq j$. Let $ T_{\sigma}$ have entries $ t_{kl}\equal{}\begin{cases}1&l\equal{}\sigma(k)\\ 0&\text{otherwise}\end{cases}$.
The matrices $ T_{\sigma(i,j)}$ and $ T_{\sigma(i,j)}\plus{}E_{ij}$ are both invertible, and $ E_{ij}$ is clearly in their span.


inequalities
search
inequalities theorems

A simple [i]search[/i] would have provided you :
http://www.mathlinks.ro/Forum/viewtopic.php?t=29899
http://www.mathlinks.ro/Forum/viewtopic.php?t=42161
http://www.mathlinks.ro/Forum/viewtopic.php?t=67723

 :roll:  :)


There are n point in the plane.It's known that for five point of this,there are exist two line contain all of them.Prove that  there are exist two line contain all of n point.

Sorry,my english is too poor

So there are $n$ points in the plane such that, for any five of them, there are two lines containing them all. This is true iff, among any five points, there are three collinear: indeed, if among some five points no three are collinear, then any line can contain at most $2$, so at most $4$ in total; on the other hand, if there are three collinear, we can take a line containing them and another line containing the other two.

Err... now that I look at this closer... is the problem true? Take six points on the plane $A, B, C, D, E, F$ such that: $ABCD$ is a convex quadrilateral, $E$ is on the segment $AC$, and $F$ is the intersection of $BC$ and $DE$. Then lines $BC, DF$ contain $B, C, D, E, F$; $AC, DF$ contain $A, C, D, E, F$; $AB, DF$ contain $A, B, D, E, F$; $AC, BC$ contain $A, B, C, E, F$; $AD, BC$ contain $A, B, C, D, F$; $AC, BD$ contain $A, B, C, D, E$. But I don't see two lines containing all the points.

I think it’s true for large enough $n$. This is how the problem was proposed in Mathematics Magazine (by Laszlo Lovasz and Paul Erdos; problem MM 1285). In fact, they asked for proof that $n=8$ isn’t large enough, but $n\ge 9$ is. At least that’s how I read it somewhere. However, unless I’m missing something obvious, I think it should be $n\ge 10$.

For a counterexample with $n=8$ points, let $A_1A_2A_3A_4$ be a convex quadrilateral in the most general position (i.e. it has no pairs of parallel sides), take $A_5=A_1A_3\cap A_2A_4,\ A_6=A_1A_2\cap A_3A_4$, and let $A_7,A_8$ be the intersections of $A_5A_6$ with $A_1A_4,A_2A_3$ respectively. Now, I don’t see what’s wrong if we add the point $A_9=A_1A_4\cap A_2A_3$.

I tried to prove it for $n\ge 10$, but it got to boring case consideration, so I let it go :).


trigonometry
geometry
circumcircle
power of a point
radical axis
geometry proposed

Let $ABC$ be acute triangle where $BE(hb)$ and $CF(hc)$ are altitudes. Prove that the radical axis of the circles $(B,hb)$ and $(C,hc)$ passes through midpoint of $EF$.

Let $T$ be the midpoint of $EF$.

The conclusion is equivalent to $CT^{2}-BT^{2}= CF^{2}-BE^{2}$.

We have that $\angle FEC = 180^\circ-\angle CBA$ and that $\angle BFE = 180^\circ-\angle ACB$.

Using the cosine theorem in the triangles $CET,BFT$, we see that $CT^{2}-BT^{2}= BC^{2}\left( \cos^{2}\hat C-\cos^{2}\hat B \right)$.

Also, $CF^{2}-BE^{2}= BC^{2}\left( \sin^{2}\hat B-\sin^{2}\hat C \right)$.

The conclusion is now pretty obvious.

Right or  wrong?

It's OK. Another case:
With your notations. If we consider the circumcircle $(O)$ of $ABC$. Prove that the radical axis of circles $(O),(B,hb)$ passes through $T$.

Let S be the midpoint of DE. Let TS cut BA and BC at X and Y. It's easy to see that X and Y are feet of perpendiculars from D to BA and BC. The power of X wrt circle centered at B is just $DX^{2}= XB \cdot XC$. Similarly, $DY^{2}= YB \cdot YA$. Thus XY is the radical axis, and T lies on it.

[color=darkblue]Denote for a triangle $ABC$ : the projections $E$, $F$ of the orthocenter $H$ to the opposite side-lines $CA$, $AB$ respectively ;
the middlepoint $T$ of the segment $[EF]$ ;  the circumcircle $\alpha=C(O,R)$ ; the circles $\beta=C(B,h_{b})$ and $\gamma=C(C,h_{c})$ ; the circle $\delta$ with the diameter $BC$ ;
the circles $v=C(B,BF)$ and $w=C(C,CE)$ ; [b]the power $\rho_{x}(X)$ of the point $X$ w.r.t. the circle[/b] $x$. Then there are the following relations : \[\boxed{\ \rho_{\alpha}(T)=\rho_{\beta}(T)=\rho_{\gamma}(T)=\frac{1}{2}\cdot (a^{2}-h^{2}_{b}-h^{2}_{c})-\frac{1}{4}\cdot EF^{2}\ \ ;\ \ \rho_{v}(T)=\rho_{w}(T)=\frac{1}{2}\cdot (h^{2}_{b}+h^{2}_{c}-a^{2})-\frac{1}{4}\cdot EF^{2}\ }\ .\] [b]Proof.[/b] The conclusion of this problem expresses that the point $T$ (the middlepoint of the segment $[EF]$) is the radical  center for the circles $\alpha$, $\beta$, $\gamma$.

$1.\blacktriangleright$ Apply the [b]Stewart's relation[/b] in the triangles $AOB$ and $AOC$ : $\boxed{\ OE^{2}=R^{2}+\overline{EA}\cdot\overline{EC}\ }\ ;\ \boxed{\ OF^{2}=R^{2}+\overline{FA}\cdot\overline{FB}\ }\ .$

$2.\blacktriangleright$ The [b]power of the point $A$ w.r.t. the circle $\delta$[/b] is : $\rho_{\delta}(A)=\overline{AE}\cdot \overline{AC}=\overline{AF}\cdot \overline{AB}=m^{2}_{a}-\frac{a^{2}}{4}=\frac{b^{2}+c^{2}-a^{2}}{2}\ .$

Otherwise, from the [b]generalized Pitagora's theorem[/b] results : $\boxed{\ a^{2}=b^{2}+c^{2}-2\cdot\overline{AE}\cdot\overline{AC}=b^{2}+c^{2}-2\cdot \overline{AF}\cdot\overline{AB}\ }\ .$

$3.\blacktriangleright\ \overline{EA}\cdot \overline{EC}=\overline{EA}\cdot \left(\overline{EA}+\overline{AC}\right)=EA^{2}-\overline{AE}\cdot \overline{AC}=$ $c^{2}-h^{2}_{b}-\frac{b^{2}+c^{2}-a^{2}}{2}=\frac{a^{2}+c^{2}-b^{2}}{2}-h^{2}_{b}$ $\Longrightarrow$

$\boxed{\ 2\cdot\overline{EA}\cdot\overline{EC}=(a^{2}+c^{2}-b^{2})-2h^{2}_{b}\ }\ .$ Obtain analogously the relation $\boxed{\ 2\cdot\overline{FA}\cdot \overline{FB}=(a^{2}+b^{2}-c^{2})-2h^{2}_{c}\ }\ .$

$4.\blacktriangleright\ 4\rho_{\beta}(T)=4\left(TB^{2}-BE^{2}\right)=2\left(BF^{2}+BE^{2}\right)-EF^{2}-4BE^{2}=$ $2\left(BF^{2}-BE^{2}\right)-EF^{2}=2\left(a^{2}-h^{2}_{c}-h^{2}_{b}\right)-EF^{2}\ .$

$4\rho_{\gamma}(T)=4(TC^{2}-CF^{2})=$ $2(CF^{2}+CE^{2})-EF^{2}-4CF^{2}=2(CE^{2}-CF^{2})-EF^{2}=2(a^{2}-h^{2}_{b}-h^{2}_{c})-EF^{2}\ .$ Therefore,

$\boxed{\ \rho_{\beta}(T)=\rho_{\gamma}(T)=\frac{1}{2}\cdot (a^{2}-h^{2}_{b}-h^{2}_{c})-\frac{1}{4}\cdot EF^{2}\ }\ .$

$5.\blacktriangleright\ 4\rho_{\alpha}(T)=4(TO^{2}-R^{2})=2(OE^{2}+OF^{2})-EF^{2}-4R^{2}=$ $2(2R^{2}+\overline{EA}\cdot\overline{EC}+\overline{FA}\cdot\overline{FB})-EF^{2}-4R^{2}=$

$2(\overline{EA}\cdot \overline{EC}+\overline{FA}\cdot \overline{FB})-EF^{2}=$ $(a^{2}+c^{2}-b^{2})-2h^{2}_{b}+(a^{2}+b^{2}-c^{2})-2h^{2}_{c}-EF^{2}=2(a^{2}-h^{2}_{b}-h^{2}_{c})-EF^{2}\ .$

Therefore and $\boxed{\ \rho_{\alpha}(T)=\frac{1}{2}\cdot (a^{2}-h^{2}_{b}-h^{2}_{c})-\frac{1}{4}\cdot EF^{2}\ }\ .$

$6.\blacktriangleright\ 4\rho_{v}(T)=4(TB^{2}-BF^{2})=2(BF^{2}+BE^{2})-EF^{2}_{4}BF^{2}=$ $2(BE^{2}-BF^{2})-EF^{2}=2(h^{2}_{b}+h^{2}_{c}-a^{2})-EF^{2}\ .$

$4\rho_{w}(T)=4(TC^{2}-CE^{2})=2(CF^{2}+CE^{2})-EF^{2}-4CE^{2}=$ $2(CF^{2}-CE^{2})-EF^{2}=2(h^{2}_{c}+h^{2}_{b}-a^{2})-EF^{2}\ .$

Therefore, $\boxed{\ \rho_{v}(T)=\rho_{w}(T)=\frac{1}{2}\cdot (h^{2}_{b}+h^{2}_{c}-a^{2})-\frac{1}{4}\cdot EF^{2}\ }\ .$

[b]Remark.[/b] If in the triangle $ABC$ we have $b>c$ and $A=60^{\circ}$, then the point $T$ is the radical center for the circles $v$, $w$ and $C\left(A,\frac{b-c}{2}\right)\ .$ [/color]

[color=darkblue]In the my last message I added a new property : [b][u] the point $T$ belongs to the radical axis of the circles $C(B,BF)$ and $C(C,CE)$.[/u][/b][/color]

[quote="juancarlos"]Let $ ABC$ be acute triangle where $ BE(hb)$ and $ CF(hc)$ are altitudes. Prove that the radical axis of the circles $ (B,hb)$ and $ (C,hc)$ passes through midpoint of $ EF$.[/quote]
Let $ M,N$ are midponts of $ EF,BC$
We have $ \vec{BM} = \frac {\vec{BE} + \vec{BF}}{2}$ so $ 4BM^2 = BE^2 + BF^2 + 2\vec{BE}\vec{BF} = BE^2 + 3BF^2 + 2\vec{BF}\vec{FE}$

similar $ 4CM^2 = 3CE^2 + CF^2 + 2\vec{EF}\vec{CE}$

and $ BF^2 - CE^2 = BE^2 - CF^2$

so
$ 4(MB^2 - MC^2) = 4(BF^2 - CE^2) + 2\vec{FE}.(\vec{BF} + \vec{CE}) = 4(BF^2 - CE^2) + 4\vec{NM}.\vec{FE} = 4(BF^2 - CE^2)$(Because $ MN\perp EF$)

so $ BM^2 - BF^2 = CM^2 - CE^2 \Rightarrow P_{M|(B)} = P_{M|{(C)}}$
 :D


category theory
abstract algebra
superior algebra
superior algebra unsolved

For a category $ C$ we may consider the "group" $ Aut(C)$ of automorphisms of $ C$, i.e. invertible functors $ C \to C$. I just thought, how can we describe it for familiar categories? For example: What does $ Aut(Set)$ look like?
[hide="Hint"]Think about "inner" automorphisms.[/hide]
I already solved this, so I just want to evoke such questions here. In fact, they are studied sparely in literature.

Do you mean contravariant as well as covariant functors?

 For contravariant ones some subtleties may occur. Take for example $ A$ a zero dimensional local ring and $ \omega_{A}$ its dualizing sheaf (which can be defined as a "minimal" injective module containing the residue field); take the duality functor Hom$ _{A}( \frac{\hspace{0.3cm}}{\hspace{0.3cm}}, \omega_{A})$. It is if i am right a contravariant automorphism of the $ A-$mod category (finitely generated modules).

 Probably you thought covariant. It is tempting, as you did for Set, to "conjugate" everything by inner automorphisms, each one chosen for each object (am i right?). But I don't know if every automorphism of $ A-$mod can be achieved this way, since there are some $ A-$linearity properties to be verified.

functor means covariant functor.

I'm not familiar with sheaves, so that I cannot comment this :). I just know that Aut(R-Mod) is very complicated. what is your exact definition of an inner automorphism?

Sorry, i just meant automorphism.

 By the way, canonical modules can be defined algebraically . It just has applications in algebraic geometry.


limit
induction
algebra proposed
algebra

{a_n}: [tex] a_1=1 , a_2=5  ,  a_{n+1}=a_n + \frac{a_{n-1}}{n(n+1)} [/tex] . Prove that : There exists  [tex] \lim_{n->inf}a_n [/tex]

Now this is just silly: isn't $a_n>0$? And doesn't this mean that $(a_n)$ is increasing? And since every increasing sequence has a limit..

He meant finite limit.

It is sufficient to prove that $a_n$ is bounded.
It is easy to check that $a_k\leq 10(1-1/k)$ for $k=2$ and $k=3$.
Apply simpliest induction:
\[a_{n+1}\leq 10(1-1/n)+\frac{10}{n(n+1)}=10-\frac{10}{n+1}=10(1-1/(n+1)).\]
That's all.

Nice solution , Myth . I have another solution :
 It's easy to show $ a_n >0  $ . So $ a_{n+1} > a_n $
 $ a_{k+1} = \frac{a_{k-1}}{k(k+1)} + \frac{a_{k-2}}{k(k-1)} + .. + \frac{a_1}{2.3} + a_2 $
 So by using induction , we easily prove $ a_k < \frac{5}{3} $

[quote="nttu"]So by using induction , we easily prove $ a_k < \frac{5}{3} $[/quote]
?????????????

Oh , sooory Myth . My solution is right (just for $ a_1 = a_2 =1 $) :blush:


How many triangles can be formed if their sidelengths can only be numbers from the Fibonacci sequence?

[hide]None, if repeats are not allowed[/hide]

[hide]an $\infty$ number of degenerate triangles?[/hide]

I'm talking about triangles, non-degenerate (generate?) ones.

[hide]It's $0$ if you can't repeat terms or $\infty$ if you can repeat.
I don't get it, what other possibility is there?[/hide]

What?  :cursing:  That's like saying How many triangles can be formed if their side lengths are real numbers?

[hide]It took you a while SplashD, the answer is obviously infinity.[/hide]

I'm not talking, I am just typing what I am thinking in my head.

Then stop typing, or [hide="even better"]stop thinking! :P [/hide]

I will try to stop thinking about how long it took you to understand what Treething was trying to point out.

I though you meant it as a trick question. You never said repeats were allowed.

He also never said that repeats were not allowed.

Somehow I had a feeling you were gonna say that. :yankchain:

Extension:
Using just the first 10 Fibonacci Numbers:  1,1,2,3,..,55 exactly how many (non-congruent) triangles can be constructed?

Hint:
[hide]Be careful at both ends. ;) [/hide]

Answer:
[hide]I get 52.[/hide]

Same question for first N Fibonacci numbers? (N>2)

[quote="Treething"][quote="robinhe"][quote="Treething"]How many triangles can be formed [b]if their sidelengths can only be numbers from the Fibonacci sequence[/b]?[/quote]

8-8-8 triangle has all side-lengths in the Fibonacci sequence. ;)[/quote]

Shhhhh... don't give it away to other people :roll:[/quote]

*whispering* You know what I think by the conversstion everyone would have already figured it out.

Heh, never misunderestimate ( :P ) the inability of people to read...

[hide]
Infinity, you can just use 2,2,2 or 1,1,1 and you get infinity.[/hide]
Was this supposed to be a trick question, or something?


function
inequalities proposed
inequalities

[quote="Davron"](x+y/x+y+3z) + (y+z/y+z+3x) + (z+x/z+x+3y) =< 6/5[/quote]
If you mean to $\sum\frac{x+y}{x+y+3z}\leq\frac{6}{5},$ where $x,y,z$ are non-negative real numbers, then this is not true.
Try $z=0,x=y=1.$ ;)

[quote="stergiu"]If you think positive numbers , then put the denominators  = a , b , c , solve the system  to express x , y , z as function of a , b , c and try to prove the new Ineq.If the problem is true , you will have an easy solution , following these steps( I hope so !!!)

    Babis[/quote]
I think if x,y,z is positive number, this ineq is still wrong. Just let $z\rightarrow 0,x=y=1$


Let $ R$ be the set of real numbers. Prove that there is no function $ f: R\rightarrow R$ with $ f(0)>0$, and such that:
\[ f(x+y) \ge f(x)+yf(f(x)) \]
for all $ x$,$ y \in R$.

In fact, a stronger claim is true:
The only function $f:\mathbb{R} \to \mathbb{R}$ satisfying $f(x+y) \ge f(x)+yf(f(x))$ for all real $x,y$ and $f(0) \ge 0$ is the constant function $f(x) \equiv 0$.
See [url=http://artofproblemsolving.com/community/c6h307863]here [/url]for a proof.


When I try to moderate the second page of the forum, a general error ("Invalid_Session") occurs. What am I doing wrong?

Not sure how this slipped past the folks at phpbb (the opensource place that built the code).

As a workaround, notice that on the first page of moderating, there's something like

&sid=9385edcff250d071b9b72803e9b96f24

at the end of your url.

After you go to the next page, you get the invalid_session -> note that the &sid=9385edcff250d071b9b72803e9b96f24 isn't on this url.  Add it to the url and hit return.  Then you'll be set.  I'll dig through the code myself and try to fix it, but I think phpbb is upgrading soon, so I may just wait until then.


inequalities
triangle inequality
geometry unsolved
geometry

On the plane are given $ n \ge 3$ points, not all on the same line. For any point $ M$ on the same plane, $ f(M)$ is defined to be the sum of the distances from $ M$ to these $ n$ points. Suppose that there is a point $ M_1$ such that $ f(M_1)\le f(M)$ for any point $ M$ on the plane. Prove that if a point $ M_2$ satisfies $ f(M_1)\equal{}f(M_2),$ then $ M_1 \equiv M_2.$

as a very brutal force you may use the lemma from here http://www.mathlinks.ro/viewtopic.php?p=1306273#1306273

Let $ P_1,P_2,...,P_n$ the points given.

Suppose that $ M_1 \neq M_2$. Let $ M$ be the midpoint of $ M_1M_2$.  For all $ P_i$, consider the point $ X_i$ such that $ P_iM_1X_iM_2$ is a parallelogram. This gives $ MP_i \equal{} \frac {P_iX}2 \leq \frac {M_1P_i \plus{} M_2P_i}2$, by triangle inequality, with equality iff $ P_i$, $ M_1$ and $ M_2$ are collinear.

Then $ f(M)\leq\frac{M_1P_1 \plus{} M_2P_2}2\plus{}\frac{M_1P_2\plus{}M_2P_2}2\plus{} ... \plus{}\frac { M_1P_n \plus{} M_2P_n}2$.

That gives $ f(M) \leq \frac {f(M_1) \plus{} f(M_2)}2 \equal{} f(M_1) \leq f(M)$.

Then all the inequalities are equalities, and this implies that all the points given are collinear, absurd.


The diameter, height and slant length of a cone coincide respectively with the base, height and side length of an isosceles triangle. What is the volume of the cone when the area of the triangle is 16 units squared, and its height is twice the length of the base?

[hide]We see the triangle is a triangle with it's vertex at the vertex of the cone, and it's base as the diameter, and as the problem says, a side length along the slant height. Let the base be $x$, thus the height is $2x$ and we have the equation $\frac{2x(x)}{2}=16\rightarrow x=4$. The base is 4, the height; 8. We have the diamter of the cone the same length as the base, and the height of the cone is the same as the triangle. The area is $\frac{1}{3}\pi 4(8)=\boxed{\frac{32}{3}\pi}$[/hide]


AMC
USA(J)MO
USAMO
number theory
relatively prime

This is a variation on USAMO 2005 #1 that arose from me misreading the problem :D . I'm pretty sure its easier than the actual problem.

Show that it is not possible to arrange all the proper divisors that are greater than 1 of any postive integer in a circle so that all adjacent proper divisors are relatively prime.

Do the proper divisors of $n$ include $n$? Then it is trivial.

No, by definition, proper divisors do not include $n$ itself.

Well first of all, it's possible for primes because they have no proper divisors... and $n = pq$ for $p$ and $q$ primes works too because all you have is $p$ and $q$.

But otherwise take $p$ to be the smallest prime (or any prime) of $n$ and the number $\frac{n}{p}$ must be in the circle. But the only possible other proper divisor that can be relatively prime to this is $p$, so it can't have numbers on both sides (unless $p$ is on both sides, which is the case $n = pq$) and hence cannot be in the circle.


Let ${x_n}$ be array such that  $x_1=603$, $x_2=102$ and $x_{n+2}=x_{n+1}+x_{n}+2 \sqrt{x_{n+1} \cdot x_n-2}$ for $n \geq 1$.

Prove that:

a) $x_n \in N$ for all $n \in N$;
b) there is infinite number of $k$ such that $x_k \equiv 2003 (mod 10^5)$;
c) doesn't exist $l$ such that $x_l \equiv 2004 (mod 10^5)$


algebra
polynomial
ratio
geometric sequence
arithmetic sequence
algebra solved

t_1,t_2,......,t_k is a G.P. , the common ratio is 10.
f(x) is a polynomial , deg f(x)=n , n is a natural number.
If [color=brown]| f(log t_k) |=93 , for k=1,2,......,n+1[/color]
find all possible n and f(x).

\\alpha

For each natural n, there can be at most 2n+1-2 polynomials of degree n satisfying the required conditions.

First, since {tk} is a geometric progression, i.e. tk=t1*rk-1, r being the common ratio, then log tk=log t1+(k-1)log r. Hence we have an arithmetic progression so we can let ak=a1+(k-1)d, where d=log r.

Second, for each ak, |f(ak)|=93 implies that f(ak)= \pm 93, depending on the choice of the sign.

Third, for each point Pk=(ak, f(ak) ), there always exists a polynomial f(x) of degree n such that each Pk is on the curve of f(x), for k=1,2,...,n+1, under certain condition:

(1) By Lagrange's Interpolation, for all values of f(ak), not all the same value, then there always exists such polynomial f of degree n
(2) If f(ak) are all equal, then f(x)= \pm 93, depending on the choice.

Four, note that the linearity of the coefficients are still preserved, i.e. linear independence is invariant.

Five, we can choose the sign of f(ak)= \pm 93 for each k, and so since there are n+1 distinct points, there can be at most 2n+1 such polynomials. But f is of degree n, so we rule out the two situations where f(ak)= \pm 93. Hence there are at most 2n+1-2 polynomials of degree n.

Lastly, note that a1, i.e. t1, is arbitrary. If we can let t1 vary, we can find even more polynomials, so it is better to keep t1 fixed.


Which of these numbers is largest?

$\text{(A)}\ \sqrt{\sqrt[3]{5\cdot 6}} \qquad \text{(B)}\ \sqrt{6\sqrt[3]{5}} \qquad \text{(C)}\ \sqrt{5 \sqrt[3]{ 6}} \qquad \text{(D)}\ \sqrt [3]{5 \sqrt{6}} \qquad \text{(E)}\ \sqrt[3]{6\sqrt{5}}$

[hide="Solution"]Convert all the numbers to the form $\sqrt[6]x$, where $x$ is some number.

The answer is B[/hide]

[hide]The square root will give a larger number than the cube root, so it has to be one of the square root choices... then the cube root should be on the smallest number to make the solution greatest... so we get b[/hide]


Find the coefficient of $x^{49}$ in the product of $(x-1)(x-3)(x-5)....(x-99)$

[hide="Answer"]
We see that the polynomial is degree 50.  So, for $ax^{49}$, we have $a=-1 + -3 + -5 + \cdots + -99 = -50^2 = \fbox{-2500}$.
[/hide]


AMC
AIME
USA(J)MO
USAMO
AMC 12
AMC 8
AMC 10

I just recently found out about the AMC and AIME and USAMO. I'm a bit confused about the order and requirements to take each test. Is it:

AMC-8: middle schoolers, good scorers invited to take AIME?
AMC-10: high schoolers, good scorers invited to take AIME?
AMC-12: ???
AIME: good scorers invited to take USAMO?
USAMO: ???

Almost.  It is:

AMC-8: Up to 8th grade, basically unconnected from others.
AMC-10: Up to 10th grade, good scorers take AIME
AMC-12: Up to 12th grade, good scorers take AIME
AIME: Good scorers take USAMO
USAMO: Good scorers go to MOP (Math Olympiad summer Program), great scorers take TST (Team Selection Test)
TST: Good scorers go to IMO (International Mathematical Olympiad)

So if you're in 9th or 10th grade you can choose between AMC-10 or AMC-12??

Or you can take both, since both are offered on two days.

And you don't have to be in 9th or 10th grade too take them, you can be younger.

If you have a choice as to which ONE to take, however, I would take the AMC 12, because you only have to get a 100+ to get to the AIME. However on the AMC 10, you have to get a 120. But then, the questions on the AMC 12 are a bit harder. But overall, it's easier to make AIME/USAMO with the AMC 12.

Especially since about half the questions on the AMC12 are on the AMC10.

it depends on if you want a high index or if you just want to qualify

if you just want to qualify, definitely take AMC 12 for obvious reasons

[quote="fireemblem13"]but isn't the AMC 12 gonna be harder than the AMC 10?[/quote]

But you don't have to do as well to qualify.  And it's not much harder, really.  (Not til the last five or ten problems.)

the first 15 or so problems are about the same in amc 10 and 12

That's the main reason as to why you should take the AMC 12 rather than the AMC 10... most of the questions are the same... since this is true, and the qualifying index is 100 for the AMC 12 rather than 120 for the AMC 10, it's much easier to qualify for the AIME for the AMC 12.  :)

Somehow I got a higher score on the AMC-12 than AMC-10...

But if your school offers both the A and B tests you should probably take both.

what about my school only offer one test?

aasdfg1234!!!
Screenname stealer!!! :mad:


solving mathematical problems ex 5.1

prove that the area of a unit cube's projection on any plane equals the length of the cube's projection on the perpendicular of this plane.

can some1 please explain the problem to me, beacuse i am not familiar with projective geometry. i understand the first part i believe, but what is the cube's projection on the perpendicular of the plane?

if you are solving the problem please hide it as i want to attempt it myself first.

cmon, some1 please help, it won't take more then 5 min of your time..

the projection of the cube onto a plane is the union of all the point of intersection of a plane with all the lines formed from a point $ o$ to a point $ p$ on the cube, right? now what is the 'perpendicular of this plane' and the length of the cubes projection on it? don't you need a fixed point $ o$ to create a projection, which is not on the plane... im confused..

perhaps the projection is just the intersection of all the lines passing through the pts of the cube that are perpendicular to the chosen plane? and then the definition is similar for the projection on the plane perpendicular to this one, taking the length as the maximum length of this 2nd projection? this sort of makes sense..

 :?:


geometry
circumcircle
inequalities
trigonometry
cyclic quadrilateral
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14580[/img]

 :(

Easy to show that P are on the arc, which doesn't contain C, of the circle with diameter AB.
E,F are on the opposite rays of the ray AC,BC resp such that AE=BF=AB.
Apply the Ptolemy's theorem to the cyclic quadrilateral ACBP,
$ PC \equal{} PB.\frac {AC}{AB} \plus{} PA.\frac {BC}{AB}$
$ \Rightarrow PA \plus{} PB \plus{} PC \equal{} PB(1 \plus{} \frac {AC}{AB}) \plus{} PA(1 \plus{} \frac {BC}{AB}) \equal{} PB.\frac {AC \plus{} AB}{AB} \plus{} PA. \frac{BC \plus{} AB}{AB}$
$ \equal{} PB.\frac {AC \plus{} AE}{AB} \plus{} PA.\frac{BC \plus{} BF}{AB} \equal{} PB.\frac {CE}{AB} \plus{} PA.\frac{CF}{AB}$
$ \leq \sqrt {(\frac {PA^2 \plus{} PB^2}{AB^2})(CE^2 \plus{} CF^2)} \equal{} EF$
The equality holds if $ \frac {PA}{PB} \equal{} \frac {CF}{CE}$, hence $ \triangle PAB\sim\triangle CFE$, which implies $ \angle PAB \equal{} \angle CFE$

[quote="jrrbc"][color=darkred]Let $ ABC$ be a $ A$ - right triangle with circumcircle $ w \equal{} C(O,R)$ . Consider a mobile point $ P\in w$ so that

$ BC$ separates $ A$ , $ P$ . Ascertain the position of $ P$ for which the sum $ PA \plus{} PB \plus{} PC$ is maximum[/color].[/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote $ PA \equal{} x$ , $ PB \equal{} y$ , $ PC \equal{} z$ . Observe that ([b]Ptolemeu's theorem[/b]) $ by \plus{} cz \equal{} ax$ and $ y^2 \plus{} z^2 \equal{} a^2$ . Thus,

$ PA \plus{} PB \plus{} PC \equal{} y \plus{} z \plus{} \frac {by \plus{} cz}{a} \equal{} \frac 1a\cdot [(a \plus{} b)y \plus{} (a \plus{} c)z]$ . Apply the [b]C.B.S.[/b]' inequality :

$ [(a \plus{} b)y \plus{} (a \plus{} c)z]^2\le \left[(a \plus{} b)^2 \plus{} (a \plus{} c)^2\right]\cdot \left(y^2 \plus{} z^2\right)$ , i.e. $ \boxed {\ PA \plus{} PB \plus{} PC\le \sqrt {(a \plus{} b)^2 \plus{} (a \plus{} c)^2}\ }$ .

We'll have equality if and only if  $ \boxed {\ \frac {y}{a \plus{} b} \equal{} \frac {z}{a \plus{} c} \equal{} \frac {x}{a \plus{} b \plus{} c}\ }$ .

[b][u]Remark.[/u][/b] Observe that $ \widehat {BAP}\equiv\widehat {BCP}$ and $ \tan\widehat {BAP} \equal{} \tan\widehat {BCP} \equal{} \frac yz$ , i.e. $ \boxed {\ \tan\widehat {BAP} \equal{} \frac {a \plus{} b}{a \plus{} c}\equal{}\frac {c(p\minus{}c)}{b(p\minus{}b)}\ }$ .

We can construct easily now the maximum position of the point $ P$ (see nice upper [b]Mr. Danh[/b]'s proof).[/color]

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14609[/img]

 :thumbup:

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=209248][color=red][b]here[/b][/color][/url] a generalization.


The number 115 is divisible by the product of its digits, $ 1\cdot1\cdot5$. How many three-digit numbers between 100 and 130 are divisible by the product of their digits?

In addition to $ 111, 112, 115$, there is also $ 128$ so the real answer is $ \boxed{4}$.


geometry
3D geometry
sphere
logarithms
algorithm
real analysis
real analysis unsolved

I quote a measure theory book:

[quote="Mattila"] Since the unit sphere $S^{n-1}$ is compact there is an integer $N(n)$ with the following property: if $y_{1}, \ldots, y_{k}\in S^{n-1}$ with $|y_{i}-y_{j}| \ge 1$ for $i \ne j$, then $k \le N(n)$.[/quote]
This statement is pretty obvious, but I'd like to know a bit about the constant. Would anyone like to try to provide an estimate for $N(n)$? (Mattila doesn't.)

$N(n)$ is the maximal kissing number in $n$ dimension (i.e. the maximal number of spheres touching a sphere of the same radius, $1/2$ in your formulation). [url=http://www.research.att.com/~njas/lattices/kiss.html]These are known lower bounds.[/url]
To quote  N. J. A. Sloane (``Seven Staggering Sequences''):
[quote]The beginning of the sequence of solutions to the N-dimensional kissing number problem is
2, 6, 12, 24, 40?, 72?, 126?, 240, 306?, 500?, ... , 196560, ...
The entries with question marks are merely lower bounds. If an oracle offered to supply 64 terms of any sequence that I chose, I would pick this one.[/quote]

The following volumetric argument shows that $N(n) \leq 3^{n}$. Let $B_{i}= B(y_{i}, 1/2)$ (i.e. $B_{i}$ is the ball centered at $y_{i}$ of radius $1/2$). By the condition, all the balls $B_{i}$ are disjoint. Therefore the volume of their union $U = \bigcup_{i=1}^{k}B_{i}$ equals \[vol(U) = N(n) \cdot \left(\frac{1}{2}\right)^{n}\sigma_{n},\] where $\sigma_{n}$ is the volume of the unit ball. On the other hand, $U \subset B\bigl(0,\frac{3}{2}\bigr)$. So \[vol(U) \leq vol\left(B\bigl(0,\frac{3}{2}\bigr)\right) = \left(\frac{3}{2}\right)^{n}\sigma_{n}.\] Finally, \[N(n) \leq \left. \left(\frac{3}{2}\right)^{n}\sigma_{n}\right/\left(\left(\frac{1}{2}\right)^{n}\sigma_{n}\right) = 3^{n}.\]

You can put $3^{n}-1$ there. Which would be sharp if we used $\parallel x\parallel _{\infty}=\max(|x_{1}|,\dots,|x_{n}|)$ instead of the Euclidean norm.

Right. By the way, this bound is often quite useful. Also it is not terribly bad, since \[C_{1}n \leq \log N(n) < C_{2}n\] (for some positive constants $C_{1}$ and $C_{2}$). The lower bound follows from the following greedy algorithm: Pick any point $y_{1}\in S^{n-1}$. Remove all points whose distance to $y_{1}$ is less then $1$ from the sphere. Pick any survived point $y_{2}$. Remove all points whose distance to $y_{2}$ is less then $1$. Repeat. Each time the normalized Haar measure of points we remove is exponentially small (it can be easily approximated by replacing the Haar measure on the sphere by the Gaussian measure in the entire space).


induction
combinatorics proposed
combinatorics

In the space are given $2006$ distinct points, such that no $4$ of them are coplanar. One draws a segment between each pair of points.
A natural number $m$ is called [i]good[/i] if one can put on each of these segments a positive integer not larger than $m$, so that every triangle whose three vertices are among the given points has the property that two of this triangle's sides have equal numbers put on, while the third has a larger number put on.
Find the minimum value of a [i]good[/i] number $m$.

Let $A(n)$ be the minimum value of a good number if we have $n$ points in space, which means, here, we look for $A(2006)$. Now look at $n=2006$. In the minimal case, there is obviously a segment on which the number $1$ is put, let its Endpoints be $X$ and $Y$. Each of the other $2004$ points is either linked to $X$ by a segment whose number is $1$ or to $Y$, otherways, we could take a point which is linked neither to $X$ nor to $Y$ by a segment with number $1$, as well as $X$ and $Y$, so we would have a triangle which would not satisfy the condition. Let the set of points which are linked to $X$ by a segment with number $1$ be $A$ and the set of points which are linked to $Y$ by a segment with numer $1$ be $B$. No point belongs to $A$ and $B$, or we would have a triangle with three equal numbers, so $|A| + |B| = 2006$. Take any point in $A$ and any in $B$. On the segment which links both points must be put number $1$, or we could take this two points and $X$ (or $Y$, if $X$ is one  of the two points) and we would have three points wich dont satisfy the condition. So, each point in $A$ is linked by a segment with number $1$ with each point in $B$, but two point in $A$ can't be linked by a segment with numer $1$ or we would have a triangle with three equal numbers on its sides. If we write a number greater than $1$ on each other segment, and if we take two points out of $A$ and one out of $B$, they satisfy the condition, because on two of the three segments would be number $1$ and on the third would be a greater number. Similar, if we take two points out of $B$ and one out of $A$. Now, consider we take three points out of $A$. None of the three sides has a $1$ put on it. We need $A(|A|)$ numbers greater than $1$ for these $|A|$ points, similar, we need $A(|B|)$ numbers for the $|B|$ points in the set $B$, but those could be the same as in set $A$. So, we have (since if wlog $|A| \geq |B|$, then $|A| \geq 1003$):

$A(2006) \geq 1+ A(1003) \geq 2+ A(502) \geq ... \geq 10+ A(2) \geq 11$

It is easy to see that we can really do it with just $11$ numbers if we take $|A|=|B|$ and so on, so $11$ is the value we look for.

We can prove that:
$A(n)=A\left(\left[\frac{n+1}{2}\right]\right)+1$

I suppose you are right, this is just what my calculation of the value $11$ is based on...    ;)

[b]My solution[/b]
Replacing $2006$ by $n$ and $m$ by $A(n)$, we will prove that if $2^k<n\leq 2^{k+1}$, then $A(n)=k+1$.
Indeed, we'll prove that by induction on $k$.
The case $k=1$ is easy.
$k\to k+1$. Suppose $2^{k+1}<n\leq 2^{k+2}$ and $m=A(n)\leq k+1$ numbers put are  $\{0,1,\ldots,k\}$. Since by the induction hypothesis we have for $a=0,1,2,\ldots,k$ there are at most $2^{k-a}$ edges beginning at some vertex $M$ that is put the number $a$.
Therefore the numbers of edge beginning at $M$ is $n-1\leq \sum_{b=0}^{k}2^b\to n\leq 2^{k+1}$ (contradiction).
So $m=A(n)\geq k+2$.
Consider the such putting for $m=k+2$.
Suppose $n$ vertices are$M_1,M_2,\ldots,M_n$. We call the segment $M_kM_l (k<l)$ a 'chord' of length $l-k$. So there are $n-1$ style chord $1,2,\ldots,n-1$. For chords $t=2^{u}(2v+1)<n$ then $u\leq k+1$; we put number $u$ on those chords. Therefore we use $k+2$ put numbers $0,1,\ldots,k+1$.
We can easily see that putting is right.


[quote="Interval"]Which statement is logically equivalent to “If a triangle is an isosceles
triangle, then it has two congruent sides”?

(1) If a triangle does not have two congruent sides, then it is an isosceles triangle.
(2) If a triangle does not have two congruent sides, then it is not an
isosceles triangle.
(3) If a triangle is not an isosceles triangle, then it has two
congruent sides.
(4) If a triangle is an isosceles triangle, then it does not have
two congruent[/quote]

Basic Logic: If $A \implies B$, then $\neg B \implies \neg A$. Your statement is not an $\iff$ statement, so we get that statement $\boxed{2}$ is logically equivalent.

Note that $\implies$ means "implies"; and that $\neg$ means "not".


Fie $a, b, c$ numere reale pozitive astfel incat $a^2+b^2+c^2=3$.
Sa se arate ca, oricare ar fi numerele reale pozitive $x, y, z$, are loc inegalitatea:
\[ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}\geq \sqrt{xy}+\sqrt{yz}+\sqrt{zx}  \].

Folosim inegalitatea mediilor:
$\sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}}\geq \frac{a+b+c}{3}$
De unde, din ipoteza avem $a+b+c\leq\ 3$
Folosind $C-B-S$ avem:
$(\frac{x}{a}+\frac{y}{b}+\frac{z}{c})(a+b+c)\geq\ (\sqrt{x}+\sqrt{y}+\sqrt{z})^{2}$
$(\sqrt{x}+\sqrt{y}+\sqrt{z})^{2}=x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{zx}\geq\ 3(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})$
$a+b+c\leq\ 3$ rezulta $3(\frac{x}{a}+\frac{y}{b}+\frac{z}{c})\geq\ (\frac{x}{a}+\frac{y}{b}+\frac{z}{c})(a+b+c)\geq\ x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{zx}\geq\ 3(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})$
De unde avem $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\geq \sqrt{xy}+\sqrt{yz}+\sqrt{zx}$ q.e.d.

I will put  my solution .

By the condition we  have that  $a+b+c\leq 3$

Now by Cauchy we have ,  $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\geq\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{a+b+c}\geq\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{3}\geq \sqrt{xy}+\sqrt{yz}+\sqrt{zx}$

so we are done.

BTW, the person above was a bit faster

Stormraven de ce nu a scris in latex solutia... :?:

Inegalitatea are loc pentru orice $a,b,c\in \mathbb {R_+}$,cu conditia ca $a^2+b^2+c^2 \leq 3$.

[quote="Stormraven"]Din 7 motive...Primul e ca nu stiu latex...Vrei sa le auzi si pe celelalte?  :D[/quote]
CUm sa nu???? :rotfl:

Intr-adevar n-ar strica sa invat Latex...Cand mai am ceva timp...

Sper...acuma timpul nu imi prea permite...


i) Suppose that f'(x) > g'(x) for all x, and that f(a) = g(a). Show that f(x) >g(x) for x >a and that f(x) <g(x) for all
x < a

ii) Show by an example that these conclusions do not follow without the hypothesis that f(a) = g(a)

1) Hint : Consider $h(x) = f(x)-g(x)$. Then $h(a)=0$ and $h'(x)>0$. What does it imply?

2) Examine the proof of (1).

You have lots of options for how to prove things here.  As liyi suggested, put $h=f-g$

Say that $h(b)\leq 0$ for some $b>a$.  Applying the mean value theorem to $h$, we get that $h'(\xi)\leq 0$ for some $\xi\in(a,b)$.  This is a contradiction.  Proceed similarly for the other direction.

The second part is pretty trivial - take for example $f(x)=e^x$ and $g(x)\equiv 0$.

thanks liyi for your cntribution. I dont fully understand how you deduce that h'(c)<=0

from the arguments you showed. Please reiterate

Thanks


geometry
geometric transformation
reflection
geometry unsolved

In $\triangle ABC$, $D$ is the midpoint of $AB$, $E$ is on $BC$ so that $BE/EC = 2$, and $\angle ADC = \angle BAE$.  Find $\angle BAC$.

$\angle BAC=90$.
A parallel through $C$ to $AB$ intersects line $AE$ at $F$. Then triangles $ABE$ and $FCE$ are similar and we see that $\frac {CF}{AB}=\frac12$, thus $CF=AD$. Therefore $ADFC$ is a parallelogram. Since $\angle BAE=\angle ADC$, its diagonals are equal, so it is a rectangle.

[asy] unitsize(4cm);
pointpen=black; pathpen=black;
pair A = Drawing("A", dir(50), dir(90));
pair B = Drawing("B", dir(180));
pair C = Drawing("C", dir(0));
draw(A--B--C--cycle);
pair D = Drawing("D", midpoint(A--B), dir(90));
pair E = Drawing("E", (0.333333333333333333333333,0));
draw(C--D);
draw(A--E);
pair X = Drawing("X", intersectionpoint(A--E, C--D), dir(60));
pair Br = rotate(180,X)*B;
pair F = Drawing("F", intersectionpoint(B--Br, A--C), dir(0));
draw(B--F);
pair J = Drawing("J", rotate(180,A)*D, dir(0));
draw(B--J--C);
[/asy]

[hide="Solution 1"]Let $X \equiv AE \cap CD$. By Menelaus on $\triangle BCD$, transversal $AXE$, we have $\frac{CX}{XD} = 1$, so $CX=XD=XA$, and it follows $\angle BAC = 90^{\circ}$.[/hide]
[hide="Solution 2"]Let $X \equiv AE \cap CD$ and let $F \equiv BX \cap AC$. Ceva's theorem implies $\frac{AF}{FC} = 2$. By van Aubel, $\frac{CX}{XD} = \frac{CF}{FA} + \frac{CE}{EB} = 1$, so $CX=XD=XA$, and it follows $\angle BAC = 90^{\circ}$.[/hide]
[hide="Solution 3"]Reflect $D$ in $A$ to $J$. From homothecies, $JC \parallel AE$, so $\angle CJD = \angle EAD = \angle CDJ$. It follows $\triangle CDJ$ is isosceles, so $\angle BAC = 90^{\circ}$.  [/hide]


An object is launched at a height of $ h$ meters and at a velocity of $ v$ m/s.  Assuming gravity is 10 m/s^2, what angle of launch $ \theta$ will help the object travel the greatest horizontal distance $ d$ and what is $ d$ in terms of $ h$ and $ v$?

[hide="HINT"]
Maximize the equation
$ d = \frac{v \cos \theta \left( v \sin \theta+\sqrt{ \left( v \sin \theta \right)^{2}+2gh}\right)}{g}$
Where $ g=10$.
[/hide]


How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?

Dude, stop adding tons of questions, and especially dont put a ridiculous poll on it. Do that in the Fun Factory. When you ACTUALLY dont understand something, that's when you put the question here. And, when you do, use the Ask on Forum button; you dont need to do it yourself. :)

I think this would belong in the Alcumus Video discussion, because it is the question from the video Counting with Casework Part 2.

If I remember correctly, this was also a problem.
Anyway,
To get the middle number the average, you need the first and last digit to either both be even or both odd. Once we find all these unique cases, then each one will allow one and only one digit to be in the middle.
For evens, the first digit can be 2, 4, 6, 8
and last can be 0, 2, 4, 6, 8
4*5=20
For odds, 1, 3, 5, 7, 9
1, 3, 5, 7, 9
5*5=25

We combine for $ 20 \plus{} 25 \equal{} \boxed{45}$


I want to start up a math circle in my area but I have no idea how to do it. I've gotten about 50 e-mail addresses for a mailing list, but I don't know where to go from there. I emailed someone from one math circle, and she thought it was a good idea and directed me to another guy, and he told me to email another guy, and that guy didn't email me back.
I bet a lot of people here live in math circle areas so hopefully someone can give me some advice.[tab]

hmm, a long island math circle sounds like a great idea, personally I think it is important to get college professors (or graduate students) or very good high school math teachers to teach at the program.  Great Neck is probably not the optimal location, but maybe you can try contacting math professors at Stony Brook or Hofstra to see if any of them are interested.  Keep me updated because I think it is a great idea.


In a standard deck of cards, let there be 8 "green" cards, 12 "red" cards, and 32 "yellow" cards.  If cards are randomly drawn from the deck, what is the probability that a green card is drawn before any red card, regardless of the number of yellow cards drawn?  Two green cards before any red card?  Three? etc.  And what is the probability that a red card is drawn before any green cards, regardless of yellow cards?  Two red cards before any green card? etc.  And what if we wanted to know the probability of at some point there are 6 red cards drawn and 4 green cards drawn, regardless of yellow cards?

This question is not from any math competition, just something I am interested in figuring out but is giving me trouble, so I hope this is the right forum.  I am mostly interested in the method for figuring out the answers to problems along these lines, not just the answers to these specific questions.

[quote="jtrun150"]In a standard deck of cards, let there be 8 "green" cards, 12 "red" cards, and 32 "yellow" cards.  If cards are randomly drawn from the deck, what is the probability that a green card is drawn before any red card, regardless of the number of yellow cards drawn?  . [/quote]
[hide="hint"]
you have to use series, for this kind of problems
for example the probability would be, to get a green card in the first draw+(yellow-green)+(yellow-yellow-green)...etc[/hide]

Thanks.  If I am understanding this correctly, the answer to the part that you quoted should be 40%, right?

$8\left(\frac{32!}{52!}\right)\displaystyle\sum_{x=0}^{31}\left[\frac{(20+x)!}{(1+x)!}\right]$

That is what I did to solve the problem.  Help with the more complicated parts of that problem would still be appreciated.  Thanks for your help.

[quote="jtrun150"]In a standard deck of cards, let there be 8 "green" cards, 12 "red" cards, and 32 "yellow" cards.  If cards are randomly drawn from the deck, what is the probability that a green card is drawn before any red card, regardless of the number of yellow cards drawn?[/quote]
[hide="Comment"]You don't need series to solve these problems.  The yellow cards do not affect the relative order of the green and red cards.  Thus we can ignore the yellow cards.  So the question becomes, given 8 green cards and 12 red cards, what is the probability that the first card is green?  That probability is clearly $\frac{8}{20}$, which simplifies to $\frac{2}{5}$.[/hide]

[quote="jtrun150"]Two green cards before any red card?[/quote]
[hide="Solution"]Same principle.  Again we can ignore the yellow cards.  So the question becomes, given 8 green cards and 12 red cards, what is the probability that the first two cards are green?  That probability is $\frac{8}{20} \cdot \frac{7}{19}$, which simplifies to $\frac{14}{95}$.[/hide]

Thank you for that help.  So those are somewhat easy problems.

But how about this part?

What if we wanted to know the probability of at some point there are 6 red cards drawn and 4 green cards drawn, regardless of yellow cards?

Does it go something like this?

$\frac{\frac{8!}{\left(8-G\right)!}\frac{12!}{\left(12-R\right)!}}{\frac{20!}{\left[20-\left(G+R\right)\right]!}}$

Where $G$ is $4$ in this case and $R$ is $6$.  But what is it that we need to multiply the whole thing by?

Yes, it is more difficult, but not too bad.  Again, ignoring yellow cards, we get the following equivalent problem.  We are given 8 green cards and 12 red cards, shuffled at random.  What is the probability that the first 10 cards split into 4 greens and 6 reds?  (In that case, the second 10 cards will also split into 4 greens and 6 reds.)

We can solve that problem using binomial coefficients.  I'll let you or someone else take it from here.

Yes, that is the correct answer.  There are simpler ways to write the answer (and see it).  One way to write the answer is
\[\frac{\binom{10}{4} \binom{10}{4}}{\binom{20}{8}}.\]
Another way to write it is
\[\frac{\binom{8}{4} \binom{12}{6}}{\binom{20}{10}}.\]


Given triangle $ABC$, let $ABX$ and $ACY$ be two variable triangles constructed outwardly on sides $AB$ and $AC$ of triangle $ABC$, such that the angles $\angle XAB$ and $\angle YAC$ are fixed and $\angle XBA+\angle YCA=180$. Prove that the line $XY$ passes through a fixed point.

also there is another similar but much easier problem as follow:
[hide]for $\triangle ABC$, triangles $\triangle AFB,\triangle AEC$ are constructed outside $\triangle ABC$ such that $AF=AE$ and $\angle FAB+\angle EAC=\angle BAC$ and also $\angle FBA+\angle ECA=180$ prove that $EF$ passes through $H$...
($H$ is the orthocenter of $\triangle ABC$)[/hide]

I think you will not be satisfied with following solution :D . But here it goes. :)

Let $\mathcal{C}$ be circle and $A,B,C$ fixed points on it and $T$ variable point on $\mathcal{C}$.
Let $p$ and $q$ be fixed lines through $A$.
Let line $TB$ cuts $p$ in $P$ and line $TC$ cuts $q$ in $Q$.
Then line $PQ$ goes through fixed point, since map $P\mapsto Q$ from line $p$ to line $q$ is perspective (it is obviously projective and when $P=A$ then so is $T=A$ and there for $Q=A$, which means that $A$ maps to it self.)

Now apply this to your problem.

Ah, is there any easier way, Number1?  :D

How to prove this problem with Euclidean geometry?


Find all ordered pairs of integers $\ (x,y)$ $\ 1 \leq x < y \leq 100$ such that $\ i^x + i^y$ is a real number.

[hide]
since x cannot equal y we know that x and y are limited to even numbers in order to produce and integer.  Thus it is easy enough to list the ordered pairs of such evens.[/hide]

[hide]As demji mentioned, some of the solutions occur when $x\equiv y\equiv 0\pmod{2}$.  However, additional solutions occur when $x\equiv 1\pmod{4}$ and $y\equiv 3\pmod{4}$ or when $x\equiv 3\pmod{4}$ and $y\equiv 1\pmod{4}$.[/hide]

So i'd say in the $x\equiv{y}\equiv0\pmod{2}$ case, there are $\frac{(48)(49)}{2}=1176$ possible ordered pairs ($x=2\Rightarrow48$, $x=4\Rightarrow47$...)
Same thing when $x\equiv1\pmod{4}$ and $y\equiv3\pmod{4}$.  But when $x\equiv3\pmod{4}$ and $y\equiv1\pmod{4}$ there's only $\frac{(47)(48)}{2}=1128$ possible ordered pairs.
The total number would be $2(1176)+1128=\boxed{3480}$.  Someone else can list them  :lol:

I don't believe that is the correct answer.