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USA(J)MO

First, the people who are allowed to take this round is:

JesusFreak197, dakyru, Andreas, and dk2

Second, idea of some problems were taken from other Olympiad problems. Therefore, these are not PURELY my problems. You have right to look up on the internet for help but please do not try to search the problem itself. Thank you.

From this COP, I think I'm going to use 5/5 grading scale that USAMTS use for the proofs instead of my previous COP proof grading (each problem worthed 1 pt). Here are the problems. You have 1 MONTH to do this. Write the fully-written proof (LaTeX is required; a month is enough for you to learn Latex if you don't know) and pm me.

1. Let $a,b,c$ be real positive numbers. Define $r = \lfloor{a+b+c}\rfloor$. Then prove that:

\[ r + \left(1 + \frac{a}{b}\right)\left(1 + \frac{b}{c}\right)\left(1+\frac{c}{a}\right) \geq 2 \left(1 + \frac{a+b+c}{\sqrt[3] {abc}} + \frac{r}{2} \right)  \]

2. Calculator shows that $\sec 40^\circ + \sec 80^\circ + \sec 160^\circ = 6$. Show how this works.

The deadline WAs 12/04 but I'm extending this to 12/12. Here is why.

For #1, some people simply found the inequaity to be true/false by contradiction and stopped there. This is NOT what I want. I want you to find whether it's true or not and prove it using algebraically (proof of contradicion was weak because most people simply set $a=b=c=\text{some number}$)

Hope this is clear! By the way, I only received two solutions..  :? (one person has to redo it his though, at least for number 1).

Shoot, I kept forgetting about this because I never saw it. I'll try to get the second problem done...

[quote="Silverfalcon"][quote="tµtµ"]For the 1), to disprove something isn't it enough to show a counter-example ?[/quote]

True.. But I just want people to do algebra to support their assertion. Remember, this round is now total of 10 points (each question of 5). So, proof of contradiction will only worth about 2 points.

Of course, some people might say I'm crazy but #1 is a very fun problem.[/quote]

You're being ridiculous.  If something is wrong, a simple counter-example is all you need to rigorously disprove that assertion.  I guarantee you that if there was a question like this on the USAMTS, USAMO, or any other credible Olympiad competition, a simple counter-example would receive full credit.  Perhaps you should try to find a way to restate the problem so that a full, correct answer requires additional work.

OK.

Fixed. It doesn't make much difference anyway (it's pretty easy to see that this is false..)


geometry
function
combinatorial geometry
advanced fields
advanced fields unsolved

Suppose you have a surface in solidworks or in other cad software or a equation of the surface.
I want to make a triangular mesh in this surface.
But I don't want that the mesh has a triangle with area less or greater than a given area, or side greater os less than a given side.

After the mesh is donne, is time to make the net. I want multiple net's that connect together to reform the surface.

My thoughts:
For the mesh (open simple surface): Find the boundary, make equal spaced points with distance $ \frac{MaxLength\minus{}MinLength}{2}$.
Find a point in the surface that is approximately at distance $ \frac{MaxLength\minus{}MinLength}{2}$ of 2 points in the boundary. Find another point in the surface that etc of one point in the boundary and the new point. and do this ...
But I don't know how to make sure if the length of the sides are in a given interval. and how to correct it recursivelly.

For the Net: Find a Triangle with 2 points in the boundary, Get the sides of it and put on a plane. Find a adjacent triangle and put in a plane, and verify if there is self-intersection, repeat it till have, and this is donne.
Find another triangle and do the same until there is no more triangles.
The choice of what point go next to go for the next triangle can be made random.
So you can rum the program a lot of times and select the result with less composed nets.

Appears no one is interested in this, but I will share the solution of the proposition.

For the triangulation you can use [url=http://www-math.mit.edu/~persson/mesh/]distmesh[/url].
Is very simple and with some tweaks it can generate a triangular mesh very good.
You can them extract the points with $ d\equal{}0$ and make the convex hull, so we have a triangular mesh for the surface.

With a list of triangles (DelaunayTri form) you make a function to select a list of adjacent triangles (a small one).

Them I used minimization to find the net of this list.
Suppose we have $ n$ points, we make $ n$ points in 2D space.
Them we make a listo of bars and the correlation of the 2D points and 3D points lengths of this bars.
We sum the square residue of each bar-length.

We do the same with points not joined by a edge.

we them minimize the sum of the 2 (possible putting more weigth in the first, depending on case, always evaluating after the deviation of the net with the mesh).

Later I will put some pictures of a fancy project I made with this algorithm.

that's all, since no-one were willing to help, I doubt anyone will be interested, but maybe after seeing the result. (was intened to be used as art)

anyways


geometry
3D geometry
icosahedron
tetrahedron

We construct a solid figure, through the process of:

$(1)$: Starting with a solid clay cube of side length $8$,

$(2)$: Removing all clay within a smaller cube formed by connecting the centers of each face of the solid cube [i](in this first loop, of the cube from $(1)$)[/i],

$(3)$: Adding back the clay in the shape of an even smaller cube, formed by connecting the centers of the faces of the cube in $(2)$, and

$(4)$: Repeating $(2)$ and $(3)$ ad infinitum.

Whatever clay is removed and left over when reaching step $(4)$ is thrown into a pile, and the figure is compressed into a smaller, solid cube. After creating the $2008^\text{th}$ cube, what volume of clay is in the thrown-out pile?

Wouldn't the "smaller cube" be an icosahedron?  There are 6 center points but there are 8 vertices of a cube...

Oh, right. Lol. I was thinking about how you can inscribe a tetrahedron into a tetrahedron, and didn't give it any thought that it might not work for other shapes.

Sooo... do the same process, i guess, but use tetrahedra instead (starting with length 8).  :P   Sorry about this.


Let AB be a triangle. P is the second point of Brocard. A',B',C' are the proiections of P on the sides BC,CA,AB. Prove that:
A'C 2 +B'A 2 +C'B 2   \geq  (PA'+PB'+PC') 2

[i]unknown source[/i]

This inequality is equivalent to next

(PA^2+PB^2+PC^2)(cos \phi)^2\geq (PA+PB+PC)^2 (sin\phi)^2

it is easy to see,  that  3(PA^2+PB^2+PC^2)\geq (PA+PB+PC)^2 and
it is well known that \phi\leq \pi/6, where \phi --- Brocard angle


(a) Compute the homology groups $ H_n(X, A)$ when X is $ S^2$ or $ S^1 \times S^1$ and A is a finite set of points in X.
(b) Compute the groups $ H_n(X,A)$ and $ H_n(X,B)$ for X a closed orientable surface of genus two with A and B the circles shown. [What are X/A and X/B]? ( please find an attached image file )

I think $ A \rightarrow X$ induces a homological group mapping $ H_0(A) \rightarrow H_0(X)$ and we have an exact sequence

$ H_1(X) \rightarrow H_1(A) \rightarrow H_1(X/A) \rightarrow H_0(X) \rightarrow H_0(A) \rightarrow H_0(X/A)$.

I can't proceed it any further. Any help will be greatly appreciated.

Thanks.

What can you say about $ H_n(A)$, since $ A$ is a finite (and so discrete) set of points? What does this tell us about the relationship between $ H_n(X)$ and $ H_n(X,A)$?

The exact sequence you have is wrong, though; it should be

$ \displaystyle
\dots \to H_1(A) \to H_1(X) \to H_1(X,A) \to H_0(A) \to H_0(X) \to H_0(X,A) \to 0$

[quote="Nukular"]What can you say about $ H_n(A)$, since $ A$ is a finite (and so discrete) set of points?[/quote]
If A has k points, I think $ H_0(A)$ should be $ \mathbb{Z}^k$. For n >=1, $ H_n(A) \equal{} 0$.

[quote] What does this tell us about the relationship between $ H_n(X)$ and $ H_n(X,A)$?[/quote]

For n=0, I guess $ H_0(X/A)$ is 0, but I don't know exactly why. How the points of A and singular 0-simplex of X can be mapped?
For n=1, I am not quite sure but I think the $ H_1(X/A)$ is generated by singular 1-simplices in X, which are paths in X.
For n>=2, $ H_n(X/A) \equal{} H_n(X)$, so  $ H_2(S^2/A) \equal{} \mathbb{Z}$

[quote]The exact sequence you have is wrong, though; it should be

$ \displaystyle \dots \to H_1(A) \to H_1(X) \to H_1(X,A) \to H_0(A) \to H_0(X) \to H_0(X,A) \to 0$[/quote]

right, that was my typing mistake.

I'll appreciate if you give me some details for n=0 and n=1.

Thanks a lot for your help.

[quote="Nukular"]You are correct that $ H_0(X,A) \equal{} 0$. Use the fact that $ A$ contains a point
in each path component of $ X$ (in particular, $ X$ is path connected in this
case).

$ H_1$ is going to be the tricky case, in that you actually will need some information
about $ X$ to compute $ H_1(X,A)$. Try looking at $ X \equal{} S^2$ first. Notice that
$ H_1(X) \equal{} 0$; you then get a short exact sequence

$ 0 \to H_1(X,A) \to H_0(A) \to H_0(X) \to 0$,

and in particular $ H_1(X,A)$ is the kernel of the map $ H_0(A) \to H_0(X)$, which
should be simple for you to compute. Can you do this?

(Then, we can look at $ X \equal{} S^1 \times S^1$).

EDIT: Also, be careful on notation; for "good" $ A$, it is true that $ H_n(X,A) \equal{} \widetilde{H}_n(X/A)$, and $ \widetilde{H}_n(X/A) \equal{} H_n(X/A)$ for $ n > 0$; but
for $ n \equal{} 0$, $ \widetilde{H}_0(X/A)$ has rank one less than $ H_0(X/A)$, which
might be giving you some confusion.[/quote]

Thanks for your reply. It is very helpful for me. Can you check the following answers are correct, please?

Both $ S^2$ and $ S \times S$ are path connected spaces with a single path component.

Thus, $ H_0(S^2)$ and $ H_0(S \times S)$ are isomorphic to an integer group $ \mathbb{Z}$.

For $ X \equal{} S^2$, we have a short exact sequence
$ 0 \to H_1(X,A) \to H_0(A) \to H_0(X) \to 0$. The kernel of the map $ H_0(A) \to H_0(X)$ is isomorphic to $ \mathbb{Z}^{k \minus{} 1}$ where k is the number of points in A.

Thus, $ H_1(X,A) \cong \mathbb{Z}^{k \minus{} 1}$ for $ X \equal{} S^2$.

For $ X \equal{} S \times S$, we have an exact sequence
$ H_1(X) \to H_1(X,A) \to H_0(A) \to H_0(X) \to 0$. Since $ H_1(X) \cong \mathbb{Z} \times \mathbb{Z}$, we have

$ H_1(X,A) \cong \mathbb{Z}^{k \plus{} 1}$ for $ X \equal{} S \times S$.

For the remaining computations,
$ H_2(S^2, A ) \cong \mathbb{Z}$, $ H_2(S \times S, A) \equal{} 0$.
$ H_n(S^2, A ) \equal{} H_n(S \times S, A ) \equal{} 0$ for n >=3.

That looks correct, but you need to prove $ H_1(X,A) \equal{} \mathbb{Z}^{k\plus{}1}$ a bit
more closely -- why does that exact sequence imply that $ H_1(X,A) \equal{} \mathbb{Z}^{k\plus{}1}$ necessarily? Can you justify this?

[quote="Nukular"]That looks correct, but you need to prove $ H_1(X,A) \equal{} \mathbb{Z}^{k \plus{} 1}$ a bit
more closely -- why does that exact sequence imply that $ H_1(X,A) \equal{} \mathbb{Z}^{k \plus{} 1}$ necessarily? Can you justify this?[/quote]

Let $ X\equal{} S \times S$. Since $ H_1(A) \equal{} 0$, we have $ 0 \to H_1(X) \to H_1(X,A) \to H_0(A) \to H_0(X) \to 0$.
Now the short exact sequence form for the above is

$ 0 \to \mathbb{Z}^2 \to H_1(X,A) \to \mathbb{Z}^{k\minus{}1} \to 0$.

It seems like the above is a split exact sequence (is this correct?), so $ H_1(X,A) \cong \mathbb{Z}^2 \oplus \mathbb{Z}^{k \minus{} 1} \cong \mathbb{Z}^{k\plus{}1}$.

Thanks.

It's split exact, since $ \mathbb{Z}^{k\minus{}1}$ is free. Good job.


Okay so who would like to moderate this new forum (and why)?

I'm a graduate student in probability.  Then again, I'm not exactly the most active user these days, but that's hardly unique among some of the existing moderators :)

You're in the College Moderators already so :)

I have question:

If I go [url=http://www.mathlinks.ro/Forum/index.php]here[/url], then I see [i]Probabilities and Statistics[/i] as a separate sub-forum of College Pg.

If I go [url=http://www.mathlinks.ro/Forum/index.php?f=218]here[/url], then I see it under [i]Advanced Fields[/i].

So, which one is it?

It has been previously discussed. It's not a subforum of Advanced Fields, but that's how it shows because it's not a category. Anyway what does it matter?


search
combinatorics unsolved
combinatorics

A set $ P$ consists of $ 2005$ distinct prime numbers.
Let $ A$ be the set of all possible products of $ 1002$ elements of $ P$, and $ B$ be the set of all products of $ 1003$ elements of $ P$.
Find a one-to-one correspondance $ f$ from $ A$ to $ B$ with the property that a divides $ f(a)$ for all $ a\in A$.

[quote="JBL"]A search for "Korea 2005" leads right here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=45929606&t=228870

Please do a search before posting recent contest problems -- many of them are already on the forum.[/quote]

OK. I will post after searching from now on.  :)
I thinked that these problems are not in the forum, since they are not at the resources.

Fair enough -- if you do find a question that is in the forum (either because you find it in a search or you post it yourself) but not in resources, contact one of the resource managers: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1789105149&t=242258


Doing some figuring and have to add the same percent amount to a varying number five or six times.  Like 5000+25%+25%+25%+25%+25%

I've figured it to...
X=5000
Y=.25

A=X+X*Y
B=A+A*Y
C=B+B*Y
D=C+C*Y
E=D+D*Y

E=((((X+X*Y)+(X+X*Y)*Y)+((X+X*Y)+(X+X*Y)*Y)*Y)+(((X+X*Y)+(X+X*Y)*Y)+((X+X*Y)+(X+X*Y)*Y)*Y)*Y)+((((X+X*Y)+(X+X*Y)*Y)+((X+X*Y)+(X+X*Y)*Y)*Y)+(((X+X*Y)+(X+X*Y)*Y)+((X+X*Y)+(X+X*Y)*Y)*Y)*Y)*Y

I've been out of school too long and have lost a few brain cells along the way.  Is there anyway to condense this down to a more managable formula?

$x+25\%x=(1+\frac{25}{100})x=1.25x$

Each time you add $25\%$, you're multiplying by $1.25$.

So you'll have something like $5000\times 1.25\times 1.25\times 1.25\times 1.25\times 1.25=5000\times(1.25)^{5}$.

In general, $X(1+Y)^{n}$.

I thought it would be something easy like that, that's fantastic!

Thank you very much  :lol:


Find a positive integral value $ x$ for which $ \frac{x\minus{}1}{x}\plus{}\frac{x\minus{}2}{x}\plus{}\frac{x\minus{}3}{x}\plus{}\ldots\plus{}\frac{1}{x}\equal{}5$.

[hide]
$ \frac {x \minus{} 1}{x} \plus{} \frac {x \minus{} 2}{x} \plus{} \frac {x \minus{} 3}{x} \plus{} \ldots \plus{} \frac {1}{x} \equal{} 5$.
$ \frac {x*(x \minus{} 1) \minus{} \sum_{k \equal{} 0}^{x \minus{} 1}{k}}{x} \equal{} 5$
$ \frac {2x^2 \minus{} 2x \minus{} x(x \minus{} 1)}{2x} \equal{} \frac {x^2 \minus{} x}{x} \equal{} x \minus{} 1 \equal{} 5$

$ x \equal{} 6$

[/hide]

You dropped the $ 2$ from the denominator too early; $ x \equal{} 6$ gives $ 5/2$. You should get $ x \equal{} 11$. Looks right otherwise. ;)


I'm very confused about this problem, since I'm getting a much stronger result than it states.

An infinite sequence a_1, a_2, a_3,.. of positive integers contains each positive integer exactly once.  For any distinct positive integers m and n, we have

1/1998 < |a_n - a_m|/|n - m| < 1998

Show that |a_n - n| < 2000000 for all n.

My approach:

[hide] Let b_n = k mean that a_k = n.

Then let n = b_b_x and m = b_x. (if b_b_x is not equal to b_x) (*)

We then have that 1/1998 < |b_x - x| / |b_b_x - b_x| < 1998.  Crossmultiplying, we get that b_b_x > x  for all x satisfying (*).

So, suppose that a_a_x = n.  If (*) is satisfied, then, x = b_b_n > n.  For x =1,  we get that n < 1, which is a contradiction.  So , b_b_1 = b_1 .  Thus, b_1 = 1 , so a_1 = 1.  We can now repeat the argument for x = 2 and induct to get that a_x = x for all x, which implies the conclusion.

[/hide]

However, I think that this is clearly ludicrous, since it shows something stronger than any solutions that I have seen, but I can't seem to find my error. Please Help!

[quote="yisun88"]Then let n = b_b_x and m = b_x. (if b_b_x is not equal to b_x) (*)
We then have that 1/1998 < |b_x - x| / |b_b_x - b_x| < 1998.  Crossmultiplying, we get that b_b_x > x  for all x satisfying (*).[/quote]
 :what?:


geometry
incenter
circumcircle
Euler
geometry proposed

Show that the following points are collinear.
(I): the orthocenter of triangle $ ABC$.
(II): the Symmedian point of triangle $ ABC$.
(III): the Symmedian point of orthic triangle of triangle $ ABC$.

[u][b]notation[/b][/u]: P is a point and P' is its Isogonal Conjugate.

Respect to the orthic triangle, the orthocenter is the incenter and the symmedian point the  Mittenpunkt  point.
In the medial triangle of the orthic triangle Mittenpunkt of the orthic triangle is Gergonne, the incenter is Nagel and Lemoine is the isotomic conjugate (in the
medial triangle of the orthic triangle) of the orthocenter of the medial triangle of the orthic triangle. [u]So it's equvalent to prove that Gergonne, Nagel and the isotomic conjugate of the orthocenter are on the same line[/u].

[b][u]Lemma[/u][/b]: in a triangle ABC take a point P and his isotomic conjugate Q. Take a point R on P'Q'. Call T the isotomic cojugate of R', then T stay on PQ.

[b][u]Proof [/u][/b]:
Call N the Nagel point and G the Gergonne point.
As well know G and N are isotomic cojugates and is well know that G' is the internal center of similitude of circumcircle and incircle and N' is the external center of similitude of circumcircle and incircle. So the circumcenter O, the incenter I, G' and N' stay on the same line.
So for the Lemma we can say that the isotomic cojugate of the isogonal cojugate of O stay on GN. But the circumcenter is the isogonal cojugate of the orthocenter H so we can conclude that N, G and the isotomic cojugate of H are align, that is equal to say that Mittenpunkt, Lemione and the incenter are align.
We can also say that the isotomic cojugate of I stay on GN.
Moreover is well know that OI is the euler line of the contact triangle, so we can also say that the isotomic cojugate of the isogonal cojugate of the centroid of the contact triangle and the orthocenter of the contacr triangle stay on GN.

[quote=Zhang Fangyu]But how to prove your lemma?[/quote]

[img]https://scontent.fsgn5-1.fna.fbcdn.net/v/t1.0-9/s851x315/18813410_104343163502380_583651629961605979_n.jpg?oh=f7f94cdc153aa17e8c81f9aac6ddbfbe&oe=59E30C7F[/img]
From here:
https://www.facebook.com/groups/Loicenter/permalink/1526112580795244/

Dear Mathlinkers,
you can see

http://jl.ayme.pagesperso-orange.fr/Docs/La%20droite%20de%20van%20Aubel.pdf  p. 22-23.

Sincerely
Jean-Louis

[quote=¬ ƒ(Gabriel)³²¹º ¼ ]: P is a point and P' is its Isogonal Conjugate.
......
[b][u]Lemma[/u][/b]: in a triangle ABC take a point P and his isotomic conjugate Q. Take a point R on P'Q'. Call T the isotomic cojugate of R', then T stay on PQ.
[/quote]

[quote=Zhang Fangyu]But how to prove your lemma?[/quote]
Lets denote the isotomic of $P$ as $P^t$
we know that the isotomic conjugate  (resp.the isogonal conjucate ) of a line $EF$ is a conic $C(E^t,F^t)$ that pass through $E^t,F^t$
( resp.$C(E',F') $ that pass through $E',F'$)
hence $R \in P'Q'\implies R'\in C(PQ) \implies T=R'^t\in P^tQ^t=PQ$


Let $ n$ be a positive integer. Prove that $ \underbrace{44...4}_{2n}\minus{}\underbrace{88...8}_{n}$ is a perfect square.

I don't know what those curly brackets mean. Could you explain that to me?

It indicates how many digits are in that number.
that would mean 44...444 with 2n 4's and 88...888 with n 8's

[hide]sorry I don't know how to use LaTeX.

if we factor the 4 out it becomes
4 (11...11  - 2( 11...11))              first string of 1's has 2n digits, second string has n
4 (11...11  - (11...11) - (11...11))     first has 2n digits, second and third string both have n digits
4 (11...1100...00 - 11...11)       first string has n 1's and n 0's; second string has n 1's
4 (10^n-1) (11...11)
4 (99...99)(11...11)     string of 9's and string of 1's both have n digits
4*9 (11...11)(11...11)
now we have
36 * (11...11)^2
or (6*11...11)^2
which is a perfect square[/hide]

[hide="With geometric series"]
This is equal to

$ 4(1\plus{}10\plus{}10^2...\plus{}10^{2n})\minus{}8(1\plus{}10\plus{}10^2...\plus{}10^n)\equal{}4\frac{10^{2n}\minus{}1}{9}\minus{}8\frac{10^n\minus{}1}{9}.$

Or

$ \frac{4}{9}(10^{2n}\minus{}1\minus{}2[10^n\minus{}1]\equal{}(\frac{2}{3}(10^n\minus{}1))^2.$

 As we know this is an integer, it must be a perfect square.[/hide]


Let $ S$ be a set with six elements. Let $ \mathcal{P}$ be the set of
all subsets of $ S$. Subsets $ A$ and $ B$ of $ S$, not necessarily
distinct, are chosen independently and at random from $ \mathcal{P}$.
The probability that $ B$ is contained in at least one of $ A$ or
$ S\minus{}A$ is $ \displaystyle\frac{m}{n^r}$, where $ m$, $ n$, and $ r$ are
positive integers, $ n$ is prime, and $ m$ and $ n$ are relatively
prime. Find $ m \plus{}n\plus{}r$. (The set $ S\minus{}A$ is the set of all elements of
$ S$ which are not in $ A$.)

2007 AIME problem.  Here's the solution found on AoPSWiki since I cannot write a decent solution to this problem myself:

[url]http://www.artofproblemsolving.com/Wiki/index.php/2007_AIME_II_Problems/Problem_10[/url]


Come now -- this isn't a homework help center.  Don't you have a textbook?

Giant Tortoise is Galapagos... try doing google searches :)  (I should really know most of these, but geography goes in one ear and out the other... if it ever goes in in the first place...)

What I have so far, but not totally sure.

3. Galapogos Island
5. Kodiac Island
6. Falkland Islands
1. Greenland
2. Midway Island
4. Puerto Rico
7. Grenada
8. Easter Island
9. Pitcairn Island

All I know is that Midway was a turning point (or place, rather) for WWII (June 1942?) and Easter Island is off of Chile quite a bit and has those big faces (Megaliths?).  Sorry

i'm pretty sure Chinaboy's right. all the answers i got were the same as his.


inequalities
geometry
3D geometry
sphere
inequalities proposed

Let x;y;z are real numbers satisfying:$ x^2\plus{}y^2\plus{}z^2\minus{}4x\plus{}2z\leq 0$
Find Max; Min:
  $ F\equal{}2x\plus{}3y\minus{}4z$

hello, we have $ 8\minus{}\sqrt{145}\le 2x\plus{}3y\minus{}4z\le 8\plus{}\sqrt{145}$. The equal sign for the lower bound holds if $ x\equal{}2\minus{}2\sqrt{\frac{5}{29}},y\equal{}\minus{}3\sqrt{\frac{5}{29}},z\equal{}\minus{}1\plus{}4\sqrt{\frac{5}{29}}$ and $ x\equal{}2\plus{}2\sqrt{\frac{5}{29}},y\equal{}3\sqrt{\frac{5}{29}},z\equal{}\minus{}1\minus{}4\sqrt{\frac{5}{29}}$ for the upper bound.
Sonnhard.

Cauchy–Schwarz inequality will help you.

Also you can think about the sphere and the plane in 3-dimension cartesian coordinates. '-'

[quote="karis"]Let x;y;z are real numbers satisfying:$ x^2 \plus{} y^2 \plus{} z^2 \minus{} 4x \plus{} 2z\leq 0$
Find Max; Min:
  $ F \equal{} 2x \plus{} 3y \minus{} 4z$[/quote]
$ x^2 \plus{} y^2 \plus{} z^2 \minus{} 4x \plus{} 2z\leq 0\Leftrightarrow(x \minus{} 2)^2 \plus{} y^2 \plus{} (z \plus{} 1)^2\leq\left(\sqrt5\right)^2,$ which is a ball.
$ 2x \plus{} 3y \minus{} 4z \minus{} F \equal{} 0$ is a plane equation.
Id est, $ \frac {|2\cdot2 \plus{} 3\cdot0 \minus{} 4\cdot(\minus{}1) \minus{} F|}{\sqrt {2^2 \plus{} 3^2 \plus{} ( \minus{} 4)^2}}\leq\sqrt5.$ :wink:


The sequence $ a_{0}, a_{1}, a_{2}, a_{3}, ...,$ is given by

$ a_{0} \equal{} 0, a_{1} \equal{} 1,$

and $ \ a_{n\plus{}1} \plus{} a_{n\minus{}1} \equal{} \frac{4a_{n}}{a_{n}^{2} \plus{}1}$ for $ n \ge 1$.

Prove that $ (a_{n}a_{n\plus{}1}\minus{}1)^{2} \plus{}(a_{n}\minus{}a_{n\plus{}1})^{2} \equal{} 2$,

 that $ a_{n}$ is rational

 and that $ 1\plus{} 4a_{n}^{2} \minus{}a_{n}^{4}$ is the square of a rational number.


Support
geometry
circumcircle
geometric transformation
rotation
similar triangles
geometry proposed

Let $ C$ be the midpoint of the segment $ \overline {AB}$, and the segment $ \overline {MN} \equal{} \overline {AB}$, its support passing through $ C$, and making a $ 60^\circ$ angle with $ AB$, [b]like[/b] in the attached drawing.
If $ P \in AN \cap BM$, $ O$ is the circumcenter of the triangle $ \triangle MNP$ and $ Q$ is the midpoint of $ \overline {MN}$, then $ A, B, O$ and $ Q$ are concyclic.

Best regards,
sunken rock

Hello, here is an idea ( I have to go soon) but I'm sure that it will lead somewhere.

This is a nice instance of rotation. In particular, $ ABC$ maps to $ NMQ$ under a particular rotation (which can be found by taking the intersection of the perpendicular bisectors of $ AN$ and $ BM$. Call this $ S$. By the general theory of spiral similarities, this is nice because we get an additional pair of similar triangles from this construction of $ S$  (that is to say, obviously, $ NMS$ and $ ABS$ are similar. But then $ NAS$ and $ MBS$ are similar.)

Also from the theory of spiral similarity, we see that second pair of similar triangles is similar to $ QCS$ (because $ Q$, $ C$ like in the same proportion relative to $ NM$, $ AB$).

It is also easy enough to see that these triangles are equilateral, that is to say, $ NAS, MBS, QCS$ are three equilateral triangles.

Also I should mention that in the general construction for $ S$, the method is to find point $ P$ as you did (that is to say we look at $ NA$ intersect $ MB$ and draw the circles $ ABP$ and $ NMP$ to find $ S$)

This is a long way of saying that $ ABSP$, $ MSNP$ are cyclic.

Also it follows by spiral similarity that $ OSO'$ is equilateral.

I should say that $ S$ is the second point of intersection of the two circles. Sorry for my scattered throughts but I'll come back to this post when I have time.


A guy walks into a 7-11 store and selects four items to buy.  The clerk at the counter informs the gentleman that the total cost of the four items is $7.11. He was completely surprised that the cost was the same as the name of the store. The clerk informed the man that he simply multiplied the cost of each item and arrived at the total. The customer calmly informed the clerk that the items should be added and not multiplied. The clerk carefully added the items together and informed the customer that the total was still exactly $7.11. (There was no sales tax involved)  Your task is to determine the exact costs of each item?

This is hard for anyone who has not been properly taught in problem-solving skills.  I gave this as a challenge problem to some college math students, and none of them could solve it without writing a computer program to do a systematic search.
Clue: one of the amounts is $3.16.  :!:

[quote="math92"]A guy walks into a 7-11 store and selects four items to buy.  The clerk at the counter informs the gentleman that the total cost of the four items is $\$7.11. He was completely surprised that the cost was the same as the name of the store. The clerk informed the man that he simply multiplied the cost of each item and arrived at the total. The customer calmly informed the clerk that the items should be added and not multiplied. The clerk carefully added the items together and informed the customer that the total was still exactly $\$7.11. (There was no sales tax involved)  Your task is to determine the exact costs of each item?[/quote]

I think you're wrong, xfactor. $\frac{711}{316}=2.25$.

$711=abcd=a+b+c+d$.
$3^2 \cdot 79=abcd = a+b+c+d$

Let's say that $a=79$. Then $3^2=bcd=b+c+d$. That's impossible.
Let's say that $a=79 \cdot 3 =237$. Then $3=bcd=b+c+d$ That's impossible.
Let's say that $a=3$. Then $79 \cdot 3=bcd=b+c+d$. That's impossible.
Let's say that $a=9$. Then $79=bcd=b+c+d$. That's impossible.

No solution? :?

The amounts are, all in dollars, 1.20, 1.25, 1.50 and 3.16.  Both the product and sum are 7.11.  No doubt it is a hard problem.

No, it's not MOEMS level, and that was my point.  To be MOEMS level, it has to be solvable in about 5 minutes.

sorry guys, i didn't think it would be that hard.

chess64, the problem with your solution, is:

You wanted to make the dollar amounts whole numbers by multiplying them by 100. Their sum gets multiplied by 100 to get 711, as you said. But their product gets multiplied by a factor of $100^4$, so it would be 711 000 000. Then you can continue with your method:

$711 000 000=2^6\cdot3^2\cdot5^6\cdot79$

Then you systematically get it close to the answer by dividing the largest number by its lowest factor and multiplying the smallest number by that factor:
$9\cdot64\cdot79\cdot15625\mbox{ Sum: }15777$
$45\cdot64\cdot79\cdot3125\mbox{ Sum: }3313$
$64\cdot79\cdot225\cdot625\mbox{ Sum: }993$
$79\cdot125\cdot225\cdot320\mbox{ Sum: }749$
$125\cdot158\cdot160\cdot225\mbox{ Sum:  668}$
Now as far as I know the only method left is guess and check.
$80\cdot125\cdot225\cdot316\mbox{ Sum: 746}$
$75\cdot125\cdot240\cdot316\mbox{ Sum: 756}$
$120\cdot125\cdot150\cdot316\mbox{ Sum: 711}$


pigeonhole principle
floor function
combinatorics unsolved
combinatorics

thx for any insight.

consider a set of $ n \geq 4$ triples {$ x_i$, $ y_i$, $ z_i$} where $ x_i$, $ y_i$ and $ z_i$ are non-negative integers.
prove you can always choose a subset $ A$ with $ \lfloor (n \plus{} 3)/2 \rfloor$ triples so that

$ \sum_{i\in A} x_i \geq \frac 12 \sum_{i \equal{} 1}^n x_i$, $ \sum_{i\in A} y_i \geq \frac 12 \sum_{i \equal{} 1}^n y_i$, and $ \sum_{i\in A} z_i \geq \frac 12 \sum_{i \equal{} 1}^n z_i$.

here $ \lfloor m \rfloor$ is the floor function.

odd $ n$ can be easily inducted from even $ n$; $ n \equal{} 4$ is easy, got a very messy proof for $ n \equal{} 6$, no idea for $ n \geq 8$.

not sure whether anyone already posted it... heard it is a generalization of a problem from 1992 russian olympiad (the original problem has n=100 with a different way stating the problem), but cannot find it...  once prove the above, it's pretty easy to prove $ \lfloor (n \plus{} 3)/2 \rfloor$ is the lower bound by constructing a counter example.

i am posting the proof for n=4 and 6.  unfortunately i cannot generalize it.

denote $ S_x \equal{} \sum_{i \equal{} 1}^n x_i$, similarly $ S_y$ and $ S_z$.

we will use [i]reductio ad absurdum[/i].

for n=4, assume for each triple {$ x_i$, $ y_i$, $ z_i$}, either (1) $ x_i > S_x/2$, or (2) $ y_i > S_y/2$, or (3) $ z_i > S_z/2$.  there are 3 categories and 4 triples, so two end up in the same category, let's say (1): $ x_i, x_j > S_x/2$.  but this leads to the contradiction: $ S_x > x_i \plus{} x_j > S_x/2 \plus{} S_x/2 \equal{} S_x$.

so we can always find a triple satisfying $ x_i \leq S_x/2$, $ y_i \leq S_y/2$, and $ z_i \leq S_z/2$, and the rest 3 triples form the set we want.

3 cannot be reduced - for example, consider {1,0,0}, {0,1,0}, {0,0,1} and {0,0,1}.

for $ n \equal{} 6$, assume similarly, for each pair of triples: {$ x_i$, $ y_i$, $ z_i$} and {$ x_j$, $ y_j$, $ z_j$}, either (1) $ x_i \plus{} x_j > S_x/2$, or (2) $ y_i \plus{} y_j > S_y/2$, or (3) $ z_i \plus{} z_j > S_z/2$.  there are total of $ C_6^2 \equal{} 15$ pairs of triples, so at least 5 pairs end up in the same category.  now the goal is to prove there has to be two distinctive pairs ($ i$,$ j$) and ($ k$,$ l$) falling in the same category, which leads to the contradiction.

here is the messy part - i use erdos-ko-rado theorem, which says for n=6, there are at most $ C_5^1 \equal{} 5$ intersecting pairs
($ i$,$ j$), and the maximum reaches when the 5 pairs have one common element, for example: (1,2), (1,3), (1,4), (1,5), and (1,6):
http://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Ko%E2%80%93Rado_theorem
the first part means, in order not to have a contradiction, each category has to have exactly 5 pairs.  but the second part says it's a contradiction anyway, because, let's say one category has every pair with $ i$, and another category has every pair with $ j$, but then ($ i$,$ j$) will end up in both categories...

so you can always find a pair of triples with $ x_i \plus{} x_j \leq S_x/2$, $ y_i \plus{} y_j \leq S_y/2$, and $ z_i \plus{} z_j \leq S_z/2$, and the rest 4 triples form the set we want.

easy to see 4 cannot be reduced by considering {1,0,0}, {0,1,0} and four {0,0,1}.

the difficulty is to generalize.  for n=2k, there are $ C_{2k}^{k \minus{} 1}$ distinctive $ (k \minus{} 1)$-size set of triples.  we want to prove they cannot be divided into three categories, with each category containing intersecting subsets of size $ (k \minus{} 1)$.  the simple way of using erdos-ko-rado theorem does not work, because $ 3 C_{2k \minus{} 1}^{k \minus{} 2} > C_{2k}^{k \minus{} 1}$ when k>3...


what place did you get at state?

please vote ONLY if you are from Wisconsin. thanks.

lol ya. i guess you were in wi when this was posted

Whoa, just noticed there are 3 votes for 1-4 place... Cutler/Killian has a account? Or just someone messing around...?

eric and killian both of them. eric=footballrox##### and killian=kkvalvik.

what a creative kid :roll:

You never specified which year...

Mwahaha :harhar:

Although I have nothing to do with WI anymore, who's representing WI at nationals? I know Killian is going, but the rest of last year's team were eighth graders, I believe.

:rotfl:  7 people got the top 4 places in Wisconsin.

[hide]BUNCH A' LIARS [/hide]

Oh wait: [code]please vote ONLY if you are from Wisconsin. thanks.[/code] A rule made to be broken  :D


calculus derivative function trigonometry

Calucate derivative of this function
$ y\equal{}sinx.|cosx|$

I know this is a stupid question, but: Isn't that function not continuous?

It IS continuous, but it's not differentiable everywhere.

This should be in the calculus forum.

$ \sin x |\cos x | \equal{}\sin x\cos x\text{sign}(\cos x) \equal{}\frac{1}{2}\sin (2x)\text{sign}(\cos x)$

The derivative, by the product rule, is $ \frac{1}{2}\sin (2x)\text{sign'}(\cos x)(\minus{}\sin x)\plus{}\cos (2x)\text{sign}(\cos x)$.

For $ \cos x\not \equal{} 0$, $ \text{sign'}\equal{} 0$, since it is constant, giving $ y' \equal{}\cos (2x)\text{sign}(\cos x)$.

At the excluded points, the derivative does not exist.

So, $ \boxed{y' \equal{}\cos (2x)\text{sign}(\cos x)\;\;\text{for}\;\; x\neq\left( k\plus{}\frac{1}{2}\right)\pi}$

[i]Note that $ \text{sign}(x) \equal{}\frac{x}{|x|}$ for nonzero $ x$. It is sometimes defined to equal $ 0$ at $ x \equal{} 0$. Regardless, though, it is not differentiable at $ 0$ and locally constant at all other values.[/i]


Suppose 20% of the eighth-grade students participate in Math Olympics and 95%
of these students get the mathematics teacher of their choice next year. Only
50% of the eighth-grade students who do not participate in Math Olympics get
the mathematics teacher of their choice next year. What percent of eighth-grade
students get the mathematics teacher of their choice next year?
Source: 2009-2010 Warmup 2 problem 9
How would you solve this problem?

[hide="Solution"]
First, we find how many students who do Math Olympics will get the teacher of their choice, and then we find how many students who do not do Math Olympics will get the teacher of their choice.  Call the number of eighth graders $ x$.  We know that $ 0.20x$ is the number of eighth graders that participate in Math Olympics.  Of those students, $ 95\%$ get the mathematics teacher of their choices next year.  That means that $ 0.95\cdot 0.20x \equal{} 0.19x$ students currently participate in Math Olympics and will get the teacher of their choice.

Since $ 0.20x$ is the number of eighth graders that participate in Math Olympics, $ x \minus{} 0.20x \equal{} 0.80x$ is the number of students that do not participate in Math Olympics.  Since $ 50\%$ of those students get the teacher of their choice, $ 0.50\cdot 0.80x \equal{} 0.40x$ is the number of students that currently do not participate in Math Olympics and will get the teacher of their choice.

Therefore, the total number of students that will get the math teacher of their choice next year is $ 0.19x \plus{} 0.40x \equal{} 0.59x$.  This is $ \frac {0.59x}{x} \equal{} \boxed{59\%}$ of all the eighth graders.[/hide]

[hide=" Solution"]

Suppose there are $ 100$ students altogether.

Of the $ 20$ who participate in Maths Olympics, $ 19$ get the teacher of their choice.

Of the $ 80$ who don't participate, $ 40$ get the teacher of their choice.

So out of $ 100$ students, $ 19 \plus{} 40 \equal{} 59$, that is $ 59\%$ get the teacher of their choice.

[/hide]


find all functions $f: \mathbb{R^ {+}} \longrightarrow \mathbb{R^ {+}}$ satisfying the condition
                         $f(x^{y})=f(x)^{f(y)}$ for all $x,y > 0$.

Easy problem though it seems new :), just notice that :
$f(x^{yz} ) = f(x)^{f(yz)} = f(x)^{f(y)f(z)}$ and $f(x^y x^z ) = f(x)^{f(y + z)} = f(x)^{f(y) + f(z)}$


for what  n  is it possible to construct a space polygon of side length :1,...,n(exactly in this order).

such that any three successive sides are pairwise prependicular?


calculus
derivative
real analysis
real analysis solved

Let {x_n} :
 $ x_{n+1} = x_n (2-ax_n) $ , a>0
 Find the value of $ x_0 $ such that : {x_n} converges to $ \frac{1}{a} $

[hide="Short answer"]
If $0<x_0<\frac2a$, then $x_n\to\frac1a.$

If $x_0=0$ or $x_0=\frac2a$, then $x_n\to0.$

If $x_0<0$ or $x_0>\frac2a$, then $x_n\to -\infty.$[/hide]

Can you please tell me how you can find $ \frac{2}{a} $

Hint: The recurrence is equivalent to $ax_{n+1} - 1 = - (ax_n - 1)^2$.

The value $\frac{2}{a}$ is a special value and was find becase for $x_0=\frac{2}{a}$ we are having $x_n=0 \forall n \in N$

In a first-order recurrence $x_{n+1}=f(x_n)$, the fixed points, hence possible limits, are solutions of the equation $x=f(x)$. Fixed points can be stable or unstable, with the deciding factor being the size of the derivative. If $x$ is a fixed point and $|f'(x)|<1$, the fixed point is stable, and all sequences that start in a neighborhood of that point tend to that point in the limit. If $x$ is a fixed point and $|f'(x)|>1$, the fixed point is unstable, and sequences may begin arbitrarily close to that point and yet not remain in the neigborhood. (If $|f'(x)|=1$ or the derivative doesn't exist, then further investigation is needed.)

In this particular example, there are two fixed points: $0$ and $\frac1a.$ The first is unstable; the second is stable.

Where did $\frac2a$ come from? We're looking for initial points that could ever lead to the unstable fixed point at zero. $f^{-1}(\{0\})=\{0,2/a\}$ and $f^{-1}(\{2/a\})$ is empty.


[quote="shobber"]Let $x, y, z$ be real numbers such that $x + y + z \geq xyz$.  Find the smallest possible value of $\frac{x^2+y^2+z^2}{xyz}$.[/quote]
If $x=2,$ $y=-1$ and $z\rightarrow0^{+}$ then $\frac{x^2+y^2+z^2}{xyz}\rightarrow-\infty.$

[quote="silouan"]If $xyz\geq 1$  then $x+y+z\geq xyz$[/quote]why?

Well here's the full quote
[quote="silouan"]I had an idea and I hope I am not wrong .
If $xyz\geq 1$  then $x^2+y^2+z^2\geq x+y+z\geq xyz$
If $xyz\leq 1$  then  $x^2+y^2+z^2\geq 3\sqrt[3]{(xyz)^2}\geq 3xyz$

What do you think?? ;)[/quote]

What he is using is that $xyz\geq 1$ implies $x^2+y^2+z^2\geq x+y+z$. We can see this is true, by looking at all positive $x,y,z$ (if one is negative, it will just decrease the less than side of the inequality).

But... let $x=y=z=2$. Then $8\ge1$ but there stands $6\ge8$, right? :?

right... oh I see his mistake:
he had 1) $x^2+y^2+z^2\geq xyz$
and 2) $x^2+y^2+z^2\geq x+y+z$
then he put them together, which wasn't valid to say that
3) $x^2+y^2+z^2\geq x+y+z\geq xyz$

saying 12>8 and 12>6 doesn't imply 6>8...

I think both you miseread my solution and the problem .
The problem says that $x+y+z\geq xyz$    ;)  . See at shobber's first post .Then I used that if $xyz\geq 1$ then
$x^2+y^2+z^2\geq x+y+z$ .Then by the assuption of the problem that says that $x+y+z\geq xyz$ we have that $x^2+y^2+z^2\geq xyz$ .

Also the counter-example you gave Peter .You put that x=y=z=2 .This contradicts the assuption of the problem $x+y+z\geq xyz$

I hope I am not mistaken

I'm sorry, I indeed misread the statement. But you still assume $xyz>0$ when going from your result to the problem. (without this, the problem is impossible like arqady says, so I'll do this too).

Anyway, I have a sharper bound: $\sqrt{3}$:[list][*]for $xyz\ge3\sqrt{3}$ we have $\frac{x^2+y^2+z^2}{xyz}\ge\frac{(x+y+z)^2}{3xyz}\ge \frac{(xyz)^2}{3xyz} \ge\sqrt{3}$ [*]for $xyz\le3\sqrt{3}$ we have $\frac{x^2+y^2+z^2}{xyz}\ge\frac{3}{\sqrt[3]{xyz}}\ge\sqrt{3}$[/list] giving a $\sqrt{3}$ everywhere. The bound is optimal since attained at $x=y=z=\sqrt{3}$.

Why you say that it is sharper ??The minimum I found in the first case is 1 and $\sqrt 3>1$    :?

You proved that the minimum was at least $1$ somewhere and at least $3$ somewhere else, so a global minimum of at least $1$. This means nothing. Your $1$ is a lower bound, but the value $1$ is never actually reached, the lowest it can go is $\sqrt{3}$. If the art were to find a low lower bound, I can just say $x^2+y^2+z^2\ge0=0xyz$, so I have $0<1$. ;)

Yes you have right this time .  :blush:
I have not noticed that the equality cannot hold.

In fact, this was Balkan MO 2001. There was many topics for it in this area.

[quote="Lovasz"]I think it was posted by mecrazywong before and $x,y,z$ should be positive reals.[/quote]
Indeed, this was posted at [url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=cwmo&t=17142[/url]
Actually even I myself have forgotten posting this problem ;)


lol, there's onlyone of each, how can u make infinity?!

[quote="i_like_pie"][quote="bpms"][quote="espark52"]how many different sums of money can you make using one or more of the following: one penny, one dime, one nickel, one quarter[/quote]
[hide]$\infty$[/hide][/quote]
[quote="mgao"]whole question's glitched, because you said "one or more" which means that we can use, like, 304 pennies and stuff...

There are infinite ways to choose how many pennies and how many nickels so if you put quarters in...[/quote]
You both misunderstood the question. He meant this:

You have 4 coins: a penny, a nickel, a dime and a quarter. How many different amounts of money can you make using one or more coins?

[hide]Penny/nickel/dime/quarter (4), nickel/dime/quarter + penny (3), dime/quarter + nickel (2), dime + quarter (1), penny + nickel + dime/quarter (2), penny/nickel + dime + quarter (2), penny + nickel + dime + quarter (1).

$4+3+2+1+2+2+1=\boxed{15}$[/hide][/quote]
Well he could have worded it better :D

[quote="bpms"][quote="i_like_pie"]You have 4 coins: a penny, a nickel, a dime and a quarter. How many different amounts of money can you make using one or more coins?

[hide]Penny/nickel/dime/quarter (4), nickel/dime/quarter + penny (3), dime/quarter + nickel (2), dime + quarter (1), penny + nickel + dime/quarter (2), penny/nickel + dime + quarter (2), penny + nickel + dime + quarter (1).

$4+3+2+1+2+2+1=\boxed{15}$[/hide][/quote]
Well he could have worded it better :D[/quote]
Well, I understood him. :roll:

[quote="espark52"]how many different sums of money can you make using one or more of the following: one penny, one dime, one nickel, one quarter[/quote]

he never uses the word "four" but does use the term "one or more" :)

:ninja:

-jorian

[quote="jhredsox"]he never uses the word "four" but does use the term "one or more" :)

:ninja:

-jorian[/quote]
All of you are focused on the fact that he said "one or more." Look at the other part:

[quote="espark52"]how many different sums of money can you make using one or more [b]of the following[/b]: [b]one[/b] penny, [b]one[/b] dime, [b]one[/b] nickel, [b]one[/b] quarter[/quote]
He says "of the following" and states the amount of coins you have.

yes, i sware i just saw that right before i saw ur post i like pie (i was gonna post that it made sense now :P )

-jorian

[quote="humblemastermind"][quote="bpms"][quote="jhredsox"]Even though I didn't make the question, I will change it (this is a FUN one for extensions!):

How many different combinations can you make with pennies, nickels, dimes, and quarters if you have to make $\$$ 4.00?

-jorian[/quote]
:vomiting:
This is one case where I would advise plugging it into the formula.
:vomiting:[/quote]

What's this formula?[/quote]
The formula is so ugly that when I saw it I didn't bother to memorize it. To be truthful I don't know it but if you want to you can search it on the forum.

ya, i figured my question was hard :P

-jorian

[quote="i_like_pie"][quote="bpms"][quote="espark52"]how many different sums of money can you make using one or more of the following: one penny, one dime, one nickel, one quarter[/quote]
[hide]$\infty$[/hide][/quote]
[quote="mgao"]whole question's glitched, because you said "one or more" which means that we can use, like, 304 pennies and stuff...

There are infinite ways to choose how many pennies and how many nickels so if you put quarters in...[/quote]
You both misunderstood the question. He meant this:

You have 4 coins: a penny, a nickel, a dime and a quarter. How many different amounts of money can you make using one or more coins?

[hide]Penny/nickel/dime/quarter (4), nickel/dime/quarter + penny (3), dime/quarter + nickel (2), dime + quarter (1), penny + nickel + dime/quarter (2), penny/nickel + dime + quarter (2), penny + nickel + dime + quarter (1).

$4+3+2+1+2+2+1=\boxed{15}$[/hide][/quote]

Yes! Thank you! good job!

Since there's no overlap between amounts of money (i.e. there's no more than one way to get any amount of money), you can get this more simply by realizing it's just:
[hide] $\binom{4}{4}+\binom{4}{3}+\binom{4}{2}+\binom{4}{1}= 1+4+6+4 = \fbox{15}$ [/hide]

Remember, $\sum_{k=0}^{n}{n\choose k}=2^{n}$, so we can just subtract the empty set to get $2^{4}-1=15$.[/hide]


conics
ellipse
analytic geometry
rotation
algebra proposed
algebra

Given an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\ (a>b>0)$, let $F(ae,0)\ (e>0)$ be the focus of this.

If the chords $PQ$ is perpendicular to the chords $RS$ at the point $F$, find the value of $\frac{1}{PF\cdot QF}+  \frac{1}{RF\cdot SF}$

well thats what i would have thought but then it's not necessary to say $e>0$ because eccentricity is $c/a$, $c$ is positive and its given that $a>0$ so... im not sure if any of this matters but i just want to understand the problem before i dive into it.

my idea...
since kunny gave the equation...
well, let's move the coordinate system, to have a system with center on $F$.
(i won't make computations, since they're trivial)
then, let's rotate the coordinate system...
by setting $x = 0$ and $y = 0$ we get one equation in $x$ and one in $y$... if we divide by $x^2$ and $y^2$ the coefficient of degree $0$, we have the product of the two lenghts (with sign) of the axes with the ellypses, so we have $-PF \cdot FQ$ and $-RF \cdot FS$, which is almost what we wanted to have, so it seems like it's done :D

[quote="cats..."]sorry, is $e$ just some constant, or does it stand for something?[/quote]

Yes, $e$ is a constant number. :)

last question, i promise.  $p, q, r, s$ are the points ($\pm a, 0$), $(0, \pm b)$  , or...?

the answer is [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/7afbbd0f4f2060ca5d65aa7adca60154.gif[/img] right? :D


abstract algebra
absolute value
superior algebra
superior algebra unsolved

Let $ C^*$ be the multiplicative group of nonzero complex numbers. What is the kernel of the homomorphism absolute value $ z\to \abs{z}$ of $ R^*$ to itself. What is the image of this homomorphism?

The kernel is $ S^1$ and the image is $ \mathbb R^\plus{}$. Both is immediate from definitions.


MATHCOUNTS geometry AMC USA(J)MO USAMO AIME Duke

To all who read this message,

I would like to invite all middle schools to participate in NEMO (National Enloe Mathematical Olympiad).  It is a mail-in mathematical competiton for students in grades 6 - 8 that enhances problem solving skills; it can be used in preparation for Mathcounts, and other such competitions.  It is run by ex-Mathcounts students (at the nationals level) at Enloe High School (Raleigh, NC), rated #42 in the country.

If you would like to participate in this, or would like more information or a sample of our problems, please contact me at nemosupport@gmail.com.  Thanks for all your time, and I hope to see you participate!

Sincerely,
V. Bhattacharya
Coordinator of NEMO

To all who read this message,

Some of you replied that you need more information.  NEMO tests algebra, geometry, and basic competition-style probability.  It consists of 50 questions in 70 minutes (individual round), 15 questions in 35 minutes (team round, to be done in teams of up to 6), and an "application" round (a basic "proof" round which asks students to explain their reasoning).

Let me go into further detail about the application round.  It poses a mathematical question that must be analyzed and solved by a group of students, but the round has multiple questions to guide students through this analysis, similar to the masters round in Mathcounts.

A very simple example of such a problem:

A subset S of a set T is such that the element in S are elements in T, but the elements of T might not be elements of S.  For example, if T = {1, 2, 3, 4}, S could be {1, 2}, {1}, or { } (empty set).

a) List all the subsets of {1, 2, 3}.
b) How many subsets does {1, 2, 3, 4, 5} have?
c) How many subsets does {1, 2, 3, 4, 5, 6} have?
d) Compare your answers in parts b and c.
e) Try to find an explicit formula for the number of subsets, s, of a set with n distinct elements.

That was a very simple example; the actual round is more original.

PLEASE feel free to e-mail me at nemosupport@gmail.com with questions, comments, or if you want to register.

The entrance fee is $2 per individual round.  We will send you all the round.  Team and application rounds are free.  The test can be administered before/during/after school on any day, preferably before the middle of March.  There is no limit to the number of students from each school who can participate in the individual round, but a max of 3 teams of up to 6 can come frome each school.

I hope this is more information.

Thanks a lot!
V. Bhattacharya

whoa, you're part of a contest. that's pretty cool stuff. congrats, vivek

arnav

To All,

I noticed that 142 people have looked at this site.  It just occurred to me that all these 142 people might not be middle school math teacher; more likely than not, they are middle or high school students.  If any of you are interested in this competition, please tell the middle schools near you.  I would love for them to participate.

Thanks!
V. Bhattacharya

P.S.  Yes, Arnav, I did start this competition; we have 18 schools participating--not bad for the first year!  If you have any problems for the proof round (like Master's Round in Mathcounts), please e-mail me!

To All,

I just wanted to mention that Mikhail Lavrov is helping us write the questions.  This probably does not mean a lot to people outside NC, but North Carolinians probably have heard his name before.  He qualifies for all sorts of competitions.

Thanks!
V. Bhattacharya

Vivek,

Who is Mikhail Lavrov?  If it's somebody obvious, you can tell me and I'll be enlightened by your wisdom... I shouldn't continue that train of thought.

Anyway, I will be participating in NEMO and I will get my opinions and those of my teammates to you in a post tomorrow or Wednesday.  Also, is the "application" round going to require an actual proof? Or will it be something not quite so difficult, like the example? (The formula is 2^n-1, where n is the number of elements in the set.  Mersenne numbers come up yet again!!!)

Daniel Vitek

You mean Enloe as in Enloe high school in raleigh in NC? Cool.  :lol:

Daniel,

Glad to hear you're participating.  The application round will technically be a proof.  However, we will also accept well-written explanations.  It might seem easy now, but wait until you read the problems.   :)

You will have two proofs from which you can choose to do one.

Mikhail Lavrov is a boy who placed in top 15 for National Mathcounts twice, qualified for the USAMO in his freshman year, was invited to the International Mathematical Olympiad, does well on the AIME, etc...  He is really very brilliant.

Thanks for your reply and good luck!
Vivek

Keta,

I am not quite sure who you are, but it is Enloe High School in Raleigh.  Mr. Noland is our coordinator.

Thanks!
Vivek

Vivek,

Yes, I know about Enloe. I live around the region and have attended the Duke math contest.

I'm Interested in the competition too.

Thanks.

Keta,

I attended the Duke Math Meet this year also.  I was just wondering whether I know you or not...

Thanks!
Vivek

Vivek,

I just took it, and it was HARD.  I know that I didn't answer 11 of the problems, and I probably missed at least 5 more.  I'm not going to ask you about any specific questions (because some people may look at this forum who have not yet taken the test), but I may email you about some.  Arjun said that the problems were approximately as difficult as the problems at National Mathcounts.  Is this true?

Also, like I promised, everybody else's reaction was that the problems were very difficult.  Do you know when the sheets will be returned to us?

Daniel Vitek

Daniel,

I had no intention of making it this hard.  I was aiming for 30% chapter round difficulty, 40% state, 20% nationals, and 10% harder than that.  It turned out to be harder than that.  Sorry.

I'll probably grade them by next Monday and send them with my brother.  I have graded yours, but I won't tell you (just to keep you in suspense).  E-mail me at nemosupport@gmail.com for more question.

Good luck on Mathcounts this Saturday!  I'll see you then.

-Vivek

Vivek,

Didn't see you at Mathcounts on Saturday.  If you were there, you know this.  If not, here it is.

Ligon placed 6 people in the top 15.
I got 3rd individual (tied for first, placed 3rd b/c I missed a question earlier than the other two).
Ligon got FIRST TEAM!  (first time ever for Mrs. Heald, I think.
Ligon is sending 5 people to State.  Kevin Li was 13th place, but he wasn't one of the top five individuals.

On to State!

how did phillips do?

vivek - nobody besides anders has been invited to IMO from NC recently, but yes, misha is certainly one of the top students in the country

Phillips got 3rd place team, and they had 3 individuals in the top 15, I think, from what I hear from Ms. McCrea.

Nice job, Daniel!  I don't think Ligon ever got 1st in the chapter; we got it last year at state, and you were there.

Oh, and by the way, Arnav, about Misha: I'm trying to sell a competition here!  :)

Reply soon!
Vivek

Good luck at state, Daniel!  Nice job on Algebra II.

[quote]Oh, and by the way, Arnav, about Misha: I'm trying to sell a competition here! [/quote]Your efforts are commendable, but trying to use my name in doing so is probably not the way to go  ;)  Especially not at this forum where at least half the people are frighteningly smarter than me...

And no, Vivek, I did not make the IMO. And with my USAMO scores this year, I doubt I stand a chance this year either...  :(

Hi  NemoSupport, Misha, Vitek, and MysticTerminator,

Anybody can share with me a copy of Algebra I state finals contest problems (1997 and older)?

Thank you.

cj

[quote="NemoSupport"]You probably will get invited sooner or later, Misha.

By the way, the only person who is smarter than you in math, whom I know, is Arnav, and that's saying something.[/quote]That you don't know many math people? ;)

Oh, and I must have missed something... what do Algebra 1 state finals have to do with NEMO again?

Hey...I thought I'd poke my head in here.  Like I told you earlier Vivek, I have a whole lot of contests sitting around at home.  Unfortunately, I lost my 2004 StateCon, but that's online anyway.  I have a bunch of Wake Tech qualifiers (I suppose the last 5 comprehensive and 1 each for the subject tests), some problems from Duke Math Meet, ARML practice problems, the really old tests from the subject tests.  Whoever was asking for really old Algebra I tests, I think I might have the 1998 Algebra I state final test.  And Daniel, Ligon definitely won chapter my 8th grade year.  I think all 4 of us placed in the top 10 at chapter.  We won states that year too.  But yeah...I have piles of math tests at home and I imagine I won't need many of them after this year so I can leave them with Vivek if he wants them.  For those of you going to ARML, I'll see you there.

Jeremy


let $ f$ a function  from $ R$ to $ R$ such that:
$ (\forall x \in R) (f\circ f)(x) \equal{}\frac{x^9}{(x^2\plus{}1)(x^6\plus{}x^4\plus{}2x^2\plus{}1)}$
prove that $ (\exists !a\in R) f(a)\equal{}a$

[quote="greatestmaths"]let $ f$ a function  from $ R$ to $ R$ such that:
$ (\forall x \in R) (f\circ f)(x) \equal{} \frac {x^9}{(x^2 \plus{} 1)(x^6 \plus{} x^4 \plus{} 2x^2 \plus{} 1)}$
prove that $ (\exists !a\in R) f(a) \equal{} a$[/quote]

Let $ g(x)\equal{}f(f(x))\equal{} \frac {x^9}{(x^2 \plus{} 1)(x^6 \plus{} x^4 \plus{} 2x^2 \plus{} 1)}$
Obviously, $ g(x)\equal{}x$ has a unique solution $ x\equal{}0$ ($ g(x)<x$ $ \forall x>0$ and $ g(x)>x$ $ \forall x<0$).

So $ f(a)\equal{}a\implies f(f(a))\equal{}a\implies a\equal{}0$ and, if such $ a$ exists, it is unique.

Then, let $ b\equal{}f(0)$, we have $ f(b)\equal{}f(f(0))\equal{}g(0)\equal{}0$ and so $ f(f(b))\equal{}f(0)\equal{}b$ and so $ b\equal{}0$

So, it exists a unique $ a$ ($ \equal{}0$) such that $ f(a)\equal{}a$


Fie $\ x,y\geq\ 2$ numere intregi pentru care $\ x^{2x^{2}+y}=y^{x+y}$.Aflati $\ x,y$.

x si y au aceeasi factori primi. Daca un p prim care divide pe x si y are exponentul a in x si b in y, atunci $a(2x^{2}+y)=b(x+y)$ si prin urmare $a<b$, deci $y=kx$. ecuatia se transforma in $x^{2x-1}=k^{k+1}$. Well aici m-am cam blocat. Ma ajuta cineva?

Daca $x>k$ atunci$2x-1<k+1<x+1$ deci $x<2$, contradictie. Daca $x<k$, iteram argumentul de inainte si obtinem ca x divide pe k      (sper ca nu gresesc) si scriind ecuatia se obtine o contradictie. Deci $x=k$ si se obtine $x=2$ si $y=4$. Ma doare ingrozitor capul as ca sunt sanse mari sa fi spus prostii pe aici :(


geometry
3D geometry
sphere
analytic geometry
trigonometry
integration
calculus

Why is the surface area of a sphere $4\pi r^{2}$ and its volume $\frac{4\pi r^{3}}3$?

Using spherical coordinates:
$dxdydz\to r^{2}\sin\phi dr d\phi d\theta$
$S=\int_{0}^{2\pi}\int_{0}^{\pi}R^{2}\sin\phi d\phi d\theta =4\pi R^{2}$
$V=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R}r^{2}\sin\phi dr d\phi d\theta =4\pi R^{3}/3$

Or in other words, if you want a fulfiling answer you need to take calculus.  I haven't ever seen a non-calculus way of proving it.   :maybe: If anyone happens to know one I'd like to see it.

A note to lordWings: obviously, the only people who understand that transformation are people who have already taken multivariable calc, and already know that solution, so it's kind of pointless to post, especially on this forum  :roll:

you can prove the volume formula using a calculus like theorem (the cavaleri principle) (search before asking)

[quote="junggi"]Like altheman said, you can prove the volume with cavalieri's principal and comparing the sphere with some solid(take a cylinder with a gnappe in it. the solid we want is the part of the cylinder not including the gnappe).

As for the surface area, there is a way to prove it without calculus. But we need to do some (simple) limit stuff. Take a sphere and draw a concentric smaller sphere. then we can find the volume(call this V) of the solid outside of the small sphere but inside the big one. As the small sphere's volume approaches the big ones, it will eventually become the surface area. So we need to find the volume of V as the small sphere's volume approaches big one's.[/quote]

Huh?  Wouldn't that limit be 0?

o wait....Gragh I forgot how to prove it....ignore me

eh...you could always try and evaluate the riemann sum for the surface area (i believe that cylinders work)...i don't know if you would call that calculus (it is sort of, but easier ones can be evaluates with convensional means)

an interesting proof for the volume i learned in 7th grade geometry is that you can view the sphere as a bunch of pyramids all meet in the same point. Their height is all r and the sum of their base areas is the surface area, so the total area is SA*r/3 using the pyramid volume (which is proven in Euclid's Elements, so its pretty easy) and the way the geometry book gave to find the SA showed how the way you roll four circles of radius r onto the SA is by considering how a baseball is stiched (that's the idea and i've checked with a college geoemetry professor person and it is certainly not able to be made rigorous at all... there is no way to unroll four circles to the SA of a sphere...)


Let $ \beta_0: H^i(X;Z_p) \rightarrow H^{i\plus{}1}(X;Z)$ be the Bockstein homomorphism associated with the exact sequence $ 0 \rightarrow Z \rightarrow Z \rightarrow Z_p \rightarrow 0$ where $ p$ is prime.

For $ a,b \in H^{*}(X;Z_p)$, show that $ \rho_p \beta_0(ab)\equal{}\beta_0(a)b\plus{}(\minus{}1)^{deg(a)} a\beta_0(b)$ where $ \rho_p$ is reduction mod $ p$.

I will appreciate any help from you nice people.

You can find the proof of this in the book:Representations and Cohomology Par David J. Benson, lemma 4.3.3. you can find it in google:

http://books.google.fr/books?id=5A06yPgrNwMC&pg=PA132&dq=bockstein+homomorphism

good lecture!


The radius of a circle is increased by 100$ \%$. By what percent is
the area of the circle increased?

This is to say that the radius doubles, or the area quadruples = 300% increase.

Suppose $ r\equal{}\text{the initial radius}.$ The original area is $ r^2\pi.$ The new radius is $ 2r$ and the area is $ 2^2r^2\pi\equal{}4r^2\pi.$ The area increased by $ 3r^2\pi$ and is $ \boxed{300\%}$ of the original area.


Let q be a prime number. We define A the set of positive integers X for which : (x,q)=1 and x*q has 3 divisors much than x. Determine if the number :           ______________
                         kpkkpkkpk...kpk with 2004 digits is in the set A.

If (x,q)=1 and x*q has 3 more divisors that x, then x has the form r^2 for some prime r (because the number of divisors is an arithmetical function). So if n=kpkkpkkpk...kpk is in A then n=r^2 for some prime r. But n=kpk*1001001?1=r^2 is impossible. So n is not in A.
I don't know why nobody solved this problem until now.


Let ABC be a triangle, and for each side, draw the intersection of the two angle trisectors closer to that side. (That is, draw the intersection of the trisectors of A and B closer to AB, and so on.) Prove that these three intersections determine an equilateral triangle.

Must we really solve all Coxeter/Greitzer problems here on ML?  :?

What about http://www.cut-the-knot.org/triangle/Morley/index.shtml ?

  Darij

This problem is not an unsolved problem !!!! ;)  ;)
  There are many books mentioning about it .

Dear Mathlinkers,
for an elegant proof,
Robson A., The Mathematical Gazette 11 (1922-23) 310-311
Sincerely
Jean-Louis


What is the greatest possible positive integer value of $ x$ if $ \displaystyle\frac{x^4}{x^2} < 10$?

$ \frac{x^4}{x^2}\equal{}x^2$

So $ x^2<10$

So $ x<\sqrt{10}$

Since $ x$ is an integer, $ x\equal{}3$


function
trigonometry
calculus
limit
real analysis
real analysis theorems

Trigonometric functions $ sin$ and $ cos$ can be defined using the set of functional equations:
1) $ f^2(x) \plus{} g^2(x) \equal{} 1$
2) $ g(x\plus{}y) \equal{} g(x)f(y) \plus{} g(y)f(x)$
3) $ f(x\plus{}y) \equal{} f(x)f(y) \minus{} g(x)g(y)$
How can be proven that that there exist only one pair of functions that satisfy these conditions, that they   are both periodic with the same period, that $ f(x)\equal{}f(\minus{}x)$ and $ g(x)\equal{}\minus{}g(\minus{}x)$ and everything else that is characteristic for $ sin$ and $ cos$?
Can this be done without use of differential equations?
Are there any free online books, texts or articles that deal with this subject?
Thank you in advance.

Note that $ f(x)\equal{}\sin ax,g(x)\equal{}\cos ax$ satisfy these equations for any $ a,$ so these aren't enough conditions to guarantee uniqueness. (There should also be some really wild - that is, non-measurable - solutions as well.)

I'm not sure, but I think that something from calculus needs to be in there - perhaps $ \lim_{x\to 0}\frac{f(x)}{x}\equal{}1.$

As a start, set $ h(x) \equal{} f(x) \plus{} g(x) i$.  Then Statements 2 and 3 lead to $ h(x \plus{} y) \equal{} h(x) h(y)$.  Assuming that $ f$ and $ g$ are real, Statement 1 leads to $ |h(x)| \equal{} 1$.


Find the inverse of f(x) = 1/(x -2).  Graph f(x) = 1/(x - 2) on the same coordinate plane.  Also graph the line y = x.


[quote="tronghieu"]Yes, I have a solution for it by using Cauchy-schwarz . But It isn't a nice solution.  :oops:[/quote]
I hope that it is independent with the Cauchy Schwarz one which I posted on maths.vn! ;)

[quote="tronghieu"]:lol:  I don't know your solution for it.
Anh đưa cho em link dc ko?  :wink:[/quote]
The solution is based on my CYH technique, here is it (sorry, I dont remember the link)
P/s: I'll smile if "your solution" is different from this :), if not...

[quote="tronghieu"]It's similar my solution. I found 1 weeks ago.  :lol:[/quote]
The my solution was posted on maths.vn one year ago ;)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ (ab \plus{} bc \plus{} ca)\left(\frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \plus{} \frac {1}{(a \plus{} b)^2}\right) \ge \frac {9}{4}
\]
This is not a new inequality with someone-Iran TST 1996. And now, proving it is not hard as the past but have you think of find more solutions for it? I think it will be a very interesting work, and luckily, I found 10 proofs for it. In here, I wanna see more solutions from you. Thanks :)
P/s: I expect that the solution dont use SS,SOS, pqr ... :)[/quote]
I've just found the 11th proof. :)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ (ab \plus{} bc \plus{} ca)\left(\frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \plus{} \frac {1}{(a \plus{} b)^2}\right) \ge \frac {9}{4}
\]
This is not a new inequality with someone-Iran TST 1996. And now, proving it is not hard as the past but have you think of find more solutions for it? I think it will be a very interesting work, and luckily, I found 10 proofs for it. In here, I wanna see more solutions from you. Thanks :)
P/s: I expect that the solution dont use SS,SOS, pqr ... :)[/quote]

I know 12 proofs of all, but.. including SOS,pqr...  :D

I succeed to kill this inequality by Schur and Muirhead..

That is :

$ ([5,1,0]\minus{}[4,2,0])\plus{}3([5,1,0]\minus{}[3,3,0])\plus{}([4,1,1]\plus{}[2,2,2]\minus{}2[3,2,1]) \ge 0$

The first two parentheses is Muirhead, and the last is Schur with $ t\equal{}1$ multiplied with $ abc$

can_hang2007, what is :

1. CYH method
2. SMV method
3. EMV method
4. PQR method
5. SS method
6. SOS method

I'm from Indonesia and I want to study this section more and more..

Thanks for your help..  :lol:

I got this proof 3 days ago :) .
$ ( ab \plus{} bc \plus{} ca)(\frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} ) \ge \frac {9}{4}$
<=>$ 4(a \plus{} b)(b \plus{} c)(c \plus{} a)[abc \minus{} (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b) ] \ge (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2$ Which is true by am-gm

I also think it's not nice but it's hard to find an independent proof :)

*$ 2\minus{}\sum \frac{a}{b\plus{}c} \equal{}\frac{ 3abc\plus{}(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b) }{(a\plus{}b)(b\plus{}c)(c\plus{}a) }$
* $ 2 \sum \frac{ab}{(a\plus{}b)^2}\equal{}3\minus{} \sum (\frac{a}{a\plus{}b})^2 \minus{}\sum( \frac{b}{a\plus{}b})^2$
$ \equal{}3\plus{}2\sum  \frac{ a(b\plus{}c)}{(a\plus{}b)(a\plus{}c)}\minus{} (\sum \frac{a}{a\plus{}b})^2 \minus{}(\sum \frac{b}{a\plus{}b})^2$
$ \equal{}5\plus{}\frac{ 8abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\minus{} (\sum \frac{a}{a\plus{}b})^2 \minus{}(\sum \frac{b}{a\plus{}b})^2$
*$ 2(\sum \frac{a}{a\plus{}b})^2 \plus{}2(\sum \frac{b}{a\plus{}b})^2  \equal{}9\plus{}( \sum \frac{a}{a\plus{}b}\minus{}\sum \frac{b}{a\plus{}b})^2$
$ \equal{}1\plus{}\frac{(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2}{(a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2 }$

[quote="red3"]why do you think the last step is am-gm?

$4\prod (a+b)(abc-\prod (a+b-c))\ge \prod (a-b)^2$

but i proved it by Cauchy-Schwarz inequality ,

$\prod (a+b)(abc-\prod (a+b-c))\ge 4\prod (a-b)^2$is also true[/quote]

Woa, your idea is very excellent Zhe.

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ (ab\plus{}bc\plus{}ca)\left(\frac{1}{(b\plus{}c)^2}\plus{}\frac{1}{(c\plus{}a)^2}\plus{}\frac{1}{(a\plus{}b)^2}\right) \ge \frac{9}{4}\]
This is not a new inequality with someone-Iran TST 1996. And now, proving it is not hard as the past but have you think of find more solutions for it? I think it will be a very interesting work, and luckily, I found 10 proofs for it. In here, I wanna see more solutions from you. Thanks :)
P/s: I expect that the solution dont use SS,SOS, pqr ... :)[/quote]
I have a proof but it isn't a nice solution.
http://mathifc.wordpress.com/2010/09/10/inequality-36/

Dear red3,

sorry but I don't understand how to prove this by Cauchy

$\prod (a+b)(abc-\prod (a+b-c))\ge 4\prod (a-b)^2$

Thank you very much

i saw this from Adbek ,this will help,

$(a+b)(b+c)(c+a)(abc-(a+b-c)(b+c-a)(c+a-b)) $
$=\sum ab(a+b)^2(a-b)^2$


In triangle $ ABC$, $ AC \equal{} CD$ and $ \angle CAB \minus{} \angle ABC \equal{} 30^\circ$. Then $ \angle BAD$ is:
[asy]defaultpen(linewidth(.8pt));
unitsize(2.5cm);
pair A = origin;
pair B = (2,0);
pair C = (0.5,0.75);
pair D = midpoint(C--B);

draw(A--B--C--cycle);
draw(A--D);

label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
label("$D$",D,NE);[/asy]$ \textbf{(A)}\ 30^\circ\qquad \textbf{(B)}\ 20^\circ\qquad \textbf{(C)}\ 22\frac {1}{2}^\circ\qquad \textbf{(D)}\ 10^\circ\qquad \textbf{(E)}\ 15^\circ$

[hide="Solution"]Let $ \angle CAD \equal{} \angle CDA \equal{} y$ and $ \angle BAD \equal{} x$.
So $ \angle BDA \equal{} 180 \minus{} y$. $ 180\minus{}y\plus{}x \plus{} \angle ABC \equal{} 180 \Rightarrow \angle ABC \equal{} y\minus{}x$.
Also, $ \angle CAB \equal{} \angle CAD \plus{} \angle BAD \equal{} y\plus{}x$.
$ \angle CAB \equal{} \angle ABC \plus{} 30 \Rightarrow x\plus{}y\equal{}y\minus{}x \plus{} 30$, which gives $ \boxed{x\equal{}15^{\circ}}$, E.[/hide]


Ring Theory
superior algebra
superior algebra unsolved

Let $ A$ be a ring in which $ xA\equal{}x^2A$ for each element $ x$. Show that every primary ideal is maximal.

Sorry for the stupid question but what $ xA$ means?

$ xA$ is the ideal generated by $ x$, also written $ (x)$.

Ok, thanks. But do we suppose that a ring is commutative and it contains $ 1$? If yes I have the solution.

So, suppose that $ I$ is a primary ideal and let $ a$ be an arbitrary element of $ A$ which doesn't belong to $ I$. We want to show that $ (I, a)\equal{}A$ or equivalently that $ (I, a)$ contains a $ 1$. Since $ (a) \equal{} (a^2)$ there is an $ d \in A$ such that $ a \equal{} da^2$ or equivalently $ a(1\minus{}da)\equal{}0$. Since $ I$ is prime ideal, $ a \not \in A$ and $ 0 \in A$ we have that $ 1 \minus{} da \in A$ and now it's clear that $ 1 \in (I, a)$.


Let $(u_n)_{n \in \N}$ be a sequence such that
1)-  $u_0 = u_1 = 1$
2) - $u_{n+1} = \frac{1}{u_n} + \frac{1}{u_{n-1}}$
prove that $(u_n)$ converge (to $\sqrt{2}$)
$\mathrm{Fermat -Euler}$

Just pass to the limit in the recurence and you will get the result. :D

[quote="cezar lupu"]Just pass to the limit in the recurence and you will get the result. [/quote]
Verry fany !:rotfl: ! But how do you prove the convergence ?

 :cool:  Need you some help ? :cool:

[quote="FMako"]How do you solve this thing, I'm getting that it doesn't converge to $\sqrt{2}$.[/quote]
The question is to prove that the sequence is convergent.

Let $M$ be defined by  $M= \max(\max (U_0,U_1),\frac 2{\min (U_0,U_1)})$ . It's easy to show that $U_n$ is bounded : $\frac 2M \leq U_n \leq M$.
So $L$= limit points set of $U_n$ is non empty and bounded , where limit point = limit of a sub-sequence.
Let $I,S$ be the bounds of $L$ :    $I=\inf(L) ,\ S=\sup(L)$ . Manifestly, $\forall \lambda\in L,\ \frac 2S \leq \lambda\leq \frac 2I$
$\lambda = I \Longrightarrow 2 \leq IS ;\ \lambda = S \Longrightarrow IS \leq 2$, so $IS=2$
Let $U_{p_n}$ be a sub-sequence convergent to $S$. Manifestly $U_{p_n-1}, U_{p_n-2}$ have to be convergent to $I$.
In the same way,  $U_{p_n-1}$ is convergent to $I$, so $U_{p_n-2}, U_{p_n-3}$ have to be convergent to $S$
Now it's easy to see that $I=S$, so the sequence $U_n$ is convergent and its limit is $\sqrt{2}$
 :cool:


Eh, 23 fellow-juniors scored below me on the 4th round. I was lazy (read: too busy to devote much time) and only needed 8 points to get Silver, and was impossible to make Gold, so I did 3 problems and got 4/5/4, so yeah. It's interesting being a really low score again; haven't had such a low score since 8th grade I believe....

YAY!! MY FIRST 25

I got 5 5 5 5 5

I got no commends this year =( oh well This is my first year =)

Yea its my first 25 as well! I got one commended and that was for #5 :)


I have some big tab pocket insertable plastic reference dividers that look like the one on this website

http://www.amazon.com/Plastic-Insertable-Organizer-Assorted-AVE11904/dp/B0007M0AOC

but I want to have two big sections for transcripts and weekly problems that I can into smaller sections divide by weeks which I can flip to quickly using the the class outline rather than have a thick, disorganized bunch of transcripts in one part of a binder

What do you recommend I use to divide these sections?

As long as you are organized and everything is easily accessable, anything will work. You don't even necessarily need a binder (you can save trees by not printing everything), I'm hip with all this 21st century stuff so I have everything in a VIRTUAL FOLDER (aka Computer Folder), my transcripts are all copied into word, I also typed class notes (which are essentially the same as the transcripts, but in a more readable manner, example, solution, example, solution, etc.), which also contains the message board problems in an organized place.

you could always have like 2 or 3 binders

Yeah its better not to print them anyway - it takes a good 41 pages for my transcript copied to Word - and even if you printed front and back it'll take tons of paper for the whole thing. Just save as a Word Document or PDF - it's much better that way.

Also, make a cd backup if you decide to go with the computer solution, just in case something happens to them.

[quote="junggi"][quote="archimedes1"]If I copy everything into Word, the LaTeX pics just become black rectangles..[/quote]
eh try clicking on the pic, then scroll up till you cant see it, then scroll back down
it worked for me

rather than copying transcripts into word, i would recommend saving them as html files, which is more pretty, takes less time to save, and you can use [url=http://theccy.com/aops/tp/beta1/]this[/url] (aops user ccy's creation)[/quote]

the transcript player is awesome!!! It makes reviewing and redoing class problems much easier. Thanks for the recommendation! :D


If  a,b,c are  positive numbers, such as $ abc\equal{}1$   then

                                                    $ a^{4}c\plus{}b^{4}a\plus{}c^{4}b\ge ab\plus{}bc\plus{}ca$

the inequality we have to prove is equivalent to
$ \frac{a^{3}}{b}\plus{}\frac{b^{3}}{c}\plus{}\frac{c^{3}}{a}\geq ab\plus{}bc\plus{}ac$
however , $ \frac{a^{3}}{b}\plus{}ab\geq 2a^{2}$
making 2 other similar inequalities and sum them up we get the deserved result

[quote="arqady"][quote="soruz"]If  a,b,c are  positive numbers, such as $ abc \equal{} 1$   then

                                                    $ a^{4}c\plus{}b^{4}a\plus{}c^{4}b\ge ab\plus{}bc\plus{}ca$[/quote]
The following inequalities  hold too:
let $ a,$ $ b,$ and $ c$ are positive numbers such that $ abc \equal{} 1.$ Prove that
1) $ a^{3}c\plus{}b^{3}a\plus{}c^{3}b\geq ab\plus{}ac\plus{}bc,$
2) $ a^{4}c\plus{}b^{4}a\plus{}c^{4}b\geq a^{2}c\plus{}b^{2}a\plus{}c^{2}b,$
3) $ a^{4}c\plus{}b^{4}a\plus{}c^{4}b\geq a^{2}b\plus{}b^{2}c\plus{}c^{2}a.$[/quote]
Nice collection,arqady.After letting $ a\equal{}\frac{x}{y},b\equal{}\frac{y}{z},c\equal{}\frac{z}{x}$,we just use
2.$ 10x^{7}z^{5}\plus{}25y^{7}x^{5}\plus{}4z^{7}y^{5}\geq 39x^{5}y^{5}z^{2}$
3.$ 22x^{7}z^{5}\plus{}16y^{7}x^{5}\plus{}z^{7}y^{5}\geq 39x^{6}y^{3}z^{3}$.
The 1st problem is also solvable by this way. :)


geometry
geometric transformation
rotation

If ABCDE is a regular pentagon, find the smallest rotation about E which maps A to D.

Wouldn't it just be an angle of the pentagon, 108 degrees?

not quite, since we're not rotating about the center, and A and D are not adjacent vertices.

is it 360 - 108 = 252?


function
integration
limit
real analysis
LaTeX
complex analysis
real analysis unsolved

Let $ f$ be a nonnegative measurable function on a measure space $ (X,M,\mu),$ and $ E_1\subset E_2 \subset \cdots$ be measurable subset of $ X.$

Prove that $ \int_{\cup E_n} f d\mu\equal{}\lim_{n\to\infty} \int_{E_n} f d\mu.$

m1(E) = E fd is a measure ,according to a theorem lim m1(En  ) when n ---> = m1 (En) , You can refer to Real and complex analysis ,Walter Rudin ,third edition .I'm sorry ,some symbols maybe don't transfer properly .

The Monotone Convergence Theorem applies to the sequence of functions $ f_n\equal{}f\cdot\chi_{E_n}.$

[quote="christ"]I'm sorry ,some symbols maybe don't transfer properly .[/quote]
They don't, at least for me. The standard for pretty math on this forum is $ \text{\LaTeX}$ [url=http://www.artofproblemsolving.com/Forum/index.php?f=123](here)[/url].


Today maybe the last days of IMO 2007 Vietnam. Some teams may leave even yesterday. I hope that, after this IMO, you - friends all over the world, will have some nice and memorable experiences in Vietnam (not only the very hard questions of IMO). Please share your thought and maybe, some personal experiences.

Final words, have a good flight. Welcome all to return our country one more.

From the organizersø point of view it was a failure but we did have some fun...  :heli: Thanks for everything you have done and sorry for everything iøve done  :D

Each people came to IMO2007 Vietnam has a personal feeling, and keep them as the warmness memory! We are friends, and Mathematics keeps us together!


find all function $f$ which that:
1.$f:R^+\rightarrow R^+$

2.$(x+y)f(f(x)y)=x^2f(f(x)+f(y))$

HI

this may help you solving this schoolmate (but you should have solved it in the exam )

that there dont exist such a function. (try to be creative and dont use the ordinery ways)

I imagined it is the more difficult case. Thanks.

Strange, but nice problem. Let $a=f(1)$. Change x and y and find that $ y^2f(yf(x))=x^2f(xf(y))$. Put $y=1$ and you have $ f(f(x))=x^2f(ax)$. (1). Put $y=1$ in the statement and you have $ x^2f(a+f(x))=f(f(x))(x+1)$ from where $ f(a+f(x))=(x+1)f(ax)$ (2). Now suppose that $f(x)=f(y)$. We find from (1) and (2) that $x^2f(ax)=y^2f(ay)$ and $ (x+1)f(ax)=(y+1)f(ay)$. Make the rapport and you have $(x-y)(xy+x+y)=0$. Thus f is injective. Take now $x>1$ and $y=x^2-x$. From injectivity you have $ (x^2-x)f(x)=f(x)+f(x^2-x)$ and thus $x^2-x>1$ for all $x>1$, false.


Show that all integers $ n$ greater than $ 11$ can be written in the form , $ n\equal{}7a\plus{}3b$, for some non negative integers $ a$ and $ b$.

This is an immediate consequence of a more general fact:
Let $ a, b$ be positive integers such that $ (a,b) \equal{} 1$. Then you can write every positive integer $ \geq$ than $ ab\minus{}a\minus{}b\plus{}1$ as $ ax\plus{}by$ where $ x$ and $ y$ are nonnegative integers.

[quote="alejnaser"]This is an immediate consequence of a more general fact:
Let $ a, b$ be positive integers such that $ (a,b) \equal{} 1$. Then you can write every positive integer $ \geq$ than $ ab \minus{} a \minus{} b \plus{} 1$ as $ ax \plus{} by$ where $ x$ and $ y$ are nonnegative integers.[/quote]

Can you prove that?

Have a look here http://www.mathlinks.ro/Forum/viewtopic.php?p=850055#850055

[quote="alejnaser"]This is an immediate consequence of a more general fact:
Let $ a, b$ be positive integers such that $ (a,b) \equal{} 1$. Then you can write every positive integer $ \geq$ than $ ab \minus{} a \minus{} b \plus{} 1$ as $ ax \plus{} by$ where $ x$ and $ y$ are nonnegative integers.[/quote]

I know about this result, the proof is also known to me, but I have just given a particular case of the result!

If you want a construction of the representation, take $ n \pmod{3}$: if $ n\equiv 0$, then $ a\equal{}0,b\equal{}\frac{n}{3}$. If $ n\equiv 1$, then $ a\equal{}1,b\equal{}\frac{n\minus{}7}{3}$. If $ n\equiv 2$, then $ a\equal{}2,b\equal{}\frac{n\minus{}14}{3}$. Note that $ n>11$ is necessary for this construction.


[quote="mws21"]I plan to take AoPS online classes and I cant seem to select which ones to choose.
Would AMC 12 Problem series and AIME Problem series classes increase both my AMC and AIME scores greater compared to taking intermediate subject classes?

For reference, I have scored 114 on the AMC 12 and 3 :(  on AIME.[/quote]
It depends on what grade you are in.
If you are in 10th-12th grade, then I think you should probably take AIME PS and AMC 12 PS, since your AMC 12 score matters for a lot. I took the AIME PS B, and it's great, you should take it even if you decide not to take the AMC PS, since it helps a ton with doing AIME problems.
As for the subject classes, take one if you are particularly weak in that subject, otherwise the AIME PS and AMC PS should be good for you.

[quote="diophantient"]
As for the subject classes, take one if you are particularly weak in that subject, otherwise the AIME PS and AMC PS should be good for you.[/quote]

If you're wholly focused on contests, this is probably the way to go.  If you work a fair amount during the summer, you'll want to consider WOOT next fall, as well.

We should have our Intermediate Counting & Intermediate Algebra books out in the next 6 months or so; these will help with preparation for the advanced problems in these areas.  AoPS Volume 2 will also help with harder AMC problems and some AIME problems.

Thank you all for suggestions.

[quote="rrusczyk"]
We should have our Intermediate Counting & Intermediate Algebra books out in the next 6 months or so; these will help with preparation for the advanced problems in these areas.  AoPS Volume 2 will also help with harder AMC problems and some AIME problems.[/quote]

Just out of curiousity, would Intermediate books be more in depth than AoPS Volume 2? To which contest do the Intermediate books aimed at?

[quote="rrusczyk"]
We should have our Intermediate Counting & Intermediate Algebra books out in the next 6 months or so;[/quote]

it would be really cool if you could include a page of "certificate of completion" or so in the back... like you did with the online courses. I find it to be very motivating :).

[quote="mws21"]
Just out of curiousity, would Intermediate books be more in depth than AoPS Volume 2? To which contest do the Intermediate books aimed at?[/quote]
Yes, the books will be much more in-depth (together, they'll be many more pages than AoPS 2).  They'll be broader, and more thorough, starting at a more basic level, and ending at a more advanced point.

The books are intended to be a full curriculum for strong students.  They won't just be aimed at contests, but will be effective for students preparing for the AMC, the AIME, and even the beginning steps of national olympiads.


Circles $A$, $B$, and $C$ are externally tangent to each other and internally tangent to circle $D$. Circles $B$ and $C$ are congruent. Circle $A$ has radius 1 and passes through the center of $D$. What is the radius of circle $B$?

[asy]
size(200);
defaultpen(linewidth(0.8));defaultpen(fontsize(8));
draw(Circle(origin, 2));
draw(Circle((-1,0), 1));
draw(Circle((6/9, 8/9), 8/9));
draw(Circle((6/9, -8/9), 8/9));
label("$A$", (-1.2, -0.2), NE);
label("$B$", (6/9, 7/9), N);
label("$C$", (6/9, -7/9), S);
label("$D$", 2*dir(110), dir(110));[/asy]

$ \textbf{(A)}\; \frac23\qquad
\textbf{(B)}\; \frac{\sqrt{3}}2\qquad
\textbf{(C)}\; \frac78\qquad
\textbf{(D)}\; \frac89\qquad
\textbf{(E)}\; \frac{1+\sqrt3}3 $

[quote="#H34N1"][img]http://i96.photobucket.com/albums/l167/adasarathy/P23.gif[/img][/quote]

[hide="solution"]Let the radius of circle $B$ be $r$ and the intersection of $B$ and $C$ be $E$.

The radius of the large circle is 2.

$AB=1+r$ and $BE=r$ so $AE=\sqrt{2r+1}$.

$BD=2-r$ so $DE=\sqrt{4-4r}$ and $AE=1+\sqrt{4-4r}$.

$\sqrt{2r+1}=1+2\sqrt{1-r}$

Now solve for $r$[/hide]

how do you solve for r with the radicals?

Square both sides enough times to eliminate it

$ \sqrt{2r \plus{} 1} \equal{} 1 \plus{} 2\sqrt{1 \minus{} r}$
$ 2r \plus{} 1 \equal{} 1 \plus{} 4\sqrt{1 \minus{} r} \plus{} 4 \minus{} 4r$
$ 6r \minus{} 4 \equal{} 4\sqrt{1 \minus{} r}$
$ 36r^{2} \minus{} 48r \plus{} 16 \equal{} 16 \minus{} 16r$
$ 36r^{2} \minus{} 32r \equal{} 0$
$ 4r(9r \minus{} 8) \equal{} 0$
$ r \equal{} \frac{8}{9}$

If there's a simple solution without trig, why would you want to use it?

There might be a simple solution with trig, or he could be more trig-inclined.

In the solution on this page:

http://www.artofproblemsolving.com/Wiki/index.php/2004_AMC_10A_Problems/Problem_23

How do you know that line O_A E passes through O_D, the center of circle D? I'm having trouble rigorously figuring out why this happens.

[quote="bestwillcui1"]Because $O_b$ and $O_c$ are congruent and circle A passes through $O_d$.[/quote]

So is this just by symmetry? How do you know that the line from the center of circle A through O_d is tangent to both B and C?

Thanks,
ProblemPro

[quote="bobthesmartypants"]Yes, it is just symmetry. Note that if we were to suppose that it wasn't tangent, it must intersect one circle twice and the other circle no times. However, flipping the entire diagram along the line $AO_d$ gives the exact same diagram but with the line passing through the other circle twice. Since the line cannot pass through both circles twice, then our assumption was wrong and it is tangent to both circles.[/quote]

But it isn't passing through both circles twice. If you just flip the diagram, one circle takes the place of another. I don't quite see the contradiction here... Can you clarify?

Thanks,
ProblemPro

you can use ratios with triangles in this problem

[hide=simple solution]Since circle $A$ passes through the center of $D,$ the radius of $D$ is twice that of $A$ or $2$. If circles $B$ and $C$ each have radius $r,$ then Descartes’ Circle Theorem applied to the four circles gives

\begin{align*} \left(\displaystyle\frac{1}{1}+\displaystyle\frac{1}{r}+\displaystyle\frac{1}{r}-\displaystyle\frac{1}{2}\right)^{2}=2\left[\left(\displaystyle\frac{1}{1}\right)^{2}+\left(\displaystyle\frac{1}{r}\right)^{2}+\left(\displaystyle\frac{1}{r}\right)^{2}+\left(-\displaystyle\frac{1}{2}\right)^{2}\right] \\ \left(\displaystyle\frac{1}{2}+\displaystyle\frac{2}{r}\right)^{2}=2\left[1+\displaystyle\frac{2}{r^{2}}+\displaystyle\frac{1}{4}\right] \\ \displaystyle\frac{1}{4}+\displaystyle\frac{2}{r}+\displaystyle\frac{4}{r^{2}}=\displaystyle\frac{5}{2}+\displaystyle\frac{4}{r^{2}} \\ \displaystyle\frac{1}{4}+\displaystyle\frac{2}{r}=\displaystyle\frac{5}{2} \\ \displaystyle\frac{2}{r}=\displaystyle\frac{9}{4} \\ r=\boxed{\textbf{(D)} ~\displaystyle\frac{8}{9}}\end{align*}[/hide]

^^^^Anytime you have 4 circles that are tangent, just use descartes


quadratics
algebra
polynomial
geometry
geometric transformation
algebra proposed

Pete chooses $ 1004$ monic quadratic polynomial $ f_{1},\cdots,f_{1004}$, such that each integer from $ 0$ to $ 2007$ is a root of at least one of them. Vasya considers all equations of the form $ f_{i}\equal{}f_{j}(i\not \equal{}j)$ and computes their roots; for each such root , Pete has to pay to Vasya $ 1$ ruble . Find the least possible value of Vasya's income.

Is it $ 0$?

These polynomials are monic so they are uniquely defined by their two real roots. Basically, you let $ f_{i}$ have roots $ i\minus{}1$ and $ 2008\minus{}i$. Then all the polynomials are merely vertical translations of each other, so they won't intersect.


Can we find the total sum of

$(1\times 2\times\ldots (k+1))~+~(2\times 3\times\ldots (k+2))~+~\ldots ~(n\times (n+1)\times\ldots (n+k+1))$

without referring to the formulas:
$\sum_{k=1}^nk = {1\over 2}n(n+1)$
$\sum_{k=1}^nk^2={1\over 2}n(n+1)(2n+1)$
$\sum_{k=1}^nk^3={1\over 4}n^2(n+1)^2$
ans so on ....?

Write $(i+1)\cdot(i+2)\cdot...\cdot(k+i+1) = (k+1)! \binom{k+i+1}{i}$.
So we are looking for $(k+1)! \sum_{i=0}^{n} \binom{k+i+1}{i} = (k+1)! \binom{k+n+2}{n}$ by the formula $\sum_{i=0}^{n} \binom{k+i}{i}=\binom{n+k+1}{n}$.

Write
\[ m\cdot (m+1)\cdot \ldots \cdot (m+k)=\frac{1}{k+2}\left(m\cdot (m+1)\cdot \ldots \cdot (m+k)\cdot (m+k+1)-(m-1)\cdot m\cdot \ldots \cdot (m+k)\right). \]
The sum telescopes.


How many different four-digit numbers can be obtained by using any four of the digits 2, 3, 4, 4, and 4?

First, break this up into cases:
CASE 1: There are only $ 2$ fours in the number.
That means that the number is $ 2344$ in some order.  We see that we can order this in $ \frac {4!}{1!1!2!} \equal{} \frac {4!}{2!} \equal{} 12$ ways.
CASE 2: All $ 3$ fours are in the number
That means that the number is either $ 2444$ in some order, or $ 3444$ in some order.  We can order $ 2444$ in $ \frac {4!}{3!1!} \equal{} 4$ ways, and same with $ 3444$.  So case 2 has a total of $ 4 \plus{} 4 \equal{} 8$ different ways.

To find the final answer, we add the answer of our cases together to get $ 12 \plus{} 8 \equal{} \boxed{20}$

[hide]There are five choices for the first digit, four for the second, three for the third, and two for the fourth.  This gives us $5!=120$ four-digit numbers containing $2, 3, 4, 4,$ or $4$.  However, the question asks for distinct four-digit number, and because the number $4$ is repeated $3$ times, you simply divide $5!$ by $3!$ to acquire the answer $\boxed{20}$.

$\frac{5!}{3!}=5\times4=\boxed{20}$[/hide]


Let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be a function such that:  For all $a$ and $b$ in $\mathbb{Z} - \{0\}$, $f(ab) \geq f(a) + f(b)$.  Show that for all $a \in \mathbb{Z} - \{0\}$ we have $f(a^n) = nf(a)$ for all $n \in \mathbb{N}$ if and only if $f(a^2) = 2f(a)$

One side of the implication is obvious (put n=2).

Suppose $f(a^2) = 2f(a)$.

Notice that if $f(a^{2^k}) = 2^kf(a)$, it implies $f(a^{2^{k+1}}) = 2f(a^{2^k}) = 2^{k+1}f(a)$, basically the induction step that will show $f(a^{2^k}) = 2^kf(a)$ for naturals k.   (*)

Also, notice $f(a^n) \geq nf(a)$ by $f(a^2 * a^{n-2}) \ge 2f(a) + f(a^{n-2})$ (very obvious).

Less obvious is that $2^k f(a) = f(a^{2^k}) \ge f(a^{2^k - 1}) + f(a)$ so that if $f(a^k) = kf(a)$ for k, then also for all naturals $j \le k$.   But from (*) k can be arbitrarily large.

[quote="Singular"]

Notice that if $f(a^{2^k}) = 2^kf(a)$, it implies $f(a^{2^{k+1}}) = 2f(a^{2^k}) = 2^{k+1}f(a)$[/quote]

Why?

I think it doesn't necessarily imply that..

[quote="Singular"]One side of the implication is obvious (put n=2).
[/quote]

sorry singular i overread this

[quote="peeta"]if $f(a^{2^k})=2^kf(a)$ then $f(a^{2^{k+1}})=f(\left(a^{2^k}\right)^2)=2\cdot f(a^{2^k})=2^{k+1}f(a)$

i dont see the problem, but singular didnt prove both ways only one ...[/quote]

Did I misunderstand the question? I think $f(x^2) = 2f(x)$ works only for $x=a$, not $x = a^{2^k}$

it works for any $a\in\mathbb Z \backslash {0}$

OK.
I thought it mean if we pick $a$ and $f(a^2) = 2f(a)$ then $f(a^n) = nf(a)$


AMC
AMC 12
AMC 12 B
probability
analytic geometry
ratio
geometry

Two lines of length 3 units are mutually perpendicular at their ends (they form an L shape). A point is chosen on each line, at random and independently. These points are then fixed as the centres of circles of unit radius. What is the probability that the circles intersect?

Nice problem!
[hide]Make a graph in the coordinate plane, limited to $ 0 \le y \le 3$ and $ 0 \le x \le 3$.  Let $ y$ represent the height of the center of the circle from the intersection of the perpendicular lines, and let $ x$ represent the distance from the second circle's center to to intersection.  Now, draw a quarter-circle on our graph with radius $ 2$, that is, graph $ x^2\plus{}y^2\equal{}4$.  Do you see why all points $ (x,y)$ on this graph represent a configuration in which the two unit circles are tangent?  (Hint:  Use the Pythagorean Theorem on the right triangle formed by the centers of the circles and the intersection of the lines.)
Hence, for every point $ (x,y)$ on our graph that lies inside the quarter-circle, that is, $ x^2\plus{}y^2 \le 4$, the point represents a configuration in which the circles intersect.  Now we need only find the ratio of the area of the quarter-circle to the area of the whole graph, or $ \frac{\pi 2^2 / 4}{3^2} \equal{} \frac{\pi}{9}$.[/hide]

Nice solution too ;-) Now let me come up with a 3D problem involving continuous probability =P

I thought USAMO consisted of proofs? It will never be this easy =P


So far we have heard someone got 144 in AMC12A and 7 on AIME I, has anyone heard anyone else got AMC12 A >=144 and a 6 on AIME I this year??

I only heard someone got 150 in AMC10 A and also 6 in AIME I.

Thanks!

Please keep all such discussion in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=265708]one thread[/url].


Just seeing if I'm too extreme...(I'm always a perfectionist)
And I know I have weird answer terms...


group theory
abstract algebra
superior algebra
superior algebra unsolved

Whether or not the order of each maximal subgroup of simple group $ Alt_n$ is more than $ n$?

Any subgroup of index $ k$ yields a homomorphism $ A_n\to S_k$ via the operation on the cosets. For $ n\geq 5$ it is injective because $ A_n$ is simple. Because of the orders this is only possible for $ n\leq k$. In fact there is always a subgroup of index $ n$: $ A_{n\minus{}1}$. So $ n$ is the greatest index of a proper subgroup (maximal or not).

For $ n\equal{}4$ the subgroup $ V_4\leq A_4$ has index 3 (and is hence maximal).

For $ n\equal{}3$ we have $ A_3\equal{}\mathbb{Z}/3\mathbb{Z}$. The only proper subgroup is the trivial one and has index 3.

For $ n\equal{}1,2$ we have $ A_n\equal{}1$, so there are no maximal subgroups at all.

I mean that,  for any maximal subgroup $ M$ of $ Alt_n$, whether $ |M|>n$?

Sorry... :blush:

This is also false for small $ n$, e.g. the subgroups of order 3 in $ A_4$ are maximal.

I think for $ n\geq 5$ it is right.

Let $ M\leq A_n$ be maximal with $ |M| \equal{} : k\leq n$. Consider the action of $ A_n$ on the left cosets $ \Omega$. Then $ |\Omega| \equal{} \frac {n!/2}{k}$. Now let $ g: \equal{} gcd(|\Omega\setminus\{M\}|,k)$.

If $ g \equal{} 1$ then $ M$ operates trivially on $ \Omega\setminus\{M\}$, i.e. an element which stabilizes one $ \omega\in\Omega$ stabilizes all. So the operation is regular. This means that the elements of $ \Omega$ are in 1-1-correspondence to the group elements, i.e. $ |\Omega| \equal{} n!/2 \implies k \equal{} 1$.

If $ g > 1$ then $ g \equal{} k$ is in fact a prime number with $ k\leq n\lneq 2k$. Suppose not. Then there would be a prime divisor $ p | k$ less than $ k$ or we would have $ n\geq 2k$. Then $ p$ resp. $ k$ would be divisor of $ \frac {n!/2}{k}$ and hence relative prime to $ \frac {n!/2}{k} \minus{} 1$.
So $ M$ is a group of prime order and hence cyclic. Because of $ k\leq n\lneq 2k$ $ M$ is generated by a $ k$-cycle. If $ k < n$ $ M$ leaves one element fixed and is contained in some $ A_{n \minus{} 1}$ contradicting maximality. So $ n \equal{} k$.

There must be a way to derive a contradiction from this, but I don't see it at the moment...

I try again.

Let $ L$ be a maximal subgroup of $ A_n$ and $ |L|\leq n$. Clearly, the degree $ d(L)$ of $ L$ is less than or equal to $ n$. If  $ d(L) < n$, then $ L\leq A_{d(L)}\times A_{n \minus{} d(L)}$, whose order is more than $ n$ if $ n\geq 6$. If $ d(L) \equal{} n$, then $ L$ is cyclic since $ |L|\leq n$. Obviously, $ n$ must be odd. That is $ A_n$ has an odd order maximal subgroup, which contradicts that $ A_n$ is solvable by well-known Thompson theorem.