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This was not solved in the intermediate forum, so I leave it to you all:

Solve [tex]\sqrt{2+\sqrt{4+\sqrt{8+\sqrt{16+\ldots}}}}[/tex]

note, I personally do not know the solution, so I am interested in what is found.

I know this is not the answer you are looking for, but you might be able to do this w/Maple.  I have not actually used it yet but it seems like it would be useful here.

here's my attempt

let...

[tex]
\\
f(n) = \sqrt{2^{n}+\sqrt{2^{n+1}+\cdots}}
\\
[/tex]

the given expression is just f(1), but we have a recursive definition

[tex]
\\
f(n+1) = f(n)^2 - 2^n
[/tex]

lol i have no clue where i'm going w/ this...

[quote="Spoon"]I don't have maple, but I wasn't aware programs could solve problems like this... I am looking for the exact answer, not a decimal, anything calculator can give the decimal.[/quote]

I think Maple's smart enough to give an exact answer. Also, calculators from TI-89 and up can give exact answers. Unfortunately I don't know what question to ask it, since it's not an infinite sum, nor is it a function, in which case a TI-89 could find it's limit.

Define a= :sqrt:(2+:sqrt:(4+:sqrt:(8+...)))
Define b= :sqrt:(2-:sqrt:(4+:sqrt:(8+...)))

when you look at ab and a/b, you get two independent equations in two variables, you can solve. I solved it with the Ti89 and got a solution that is very close to what I was getting by evaluating the expression up to 1024.

Uhh... I don't see how ab or a/b simplifies...

The product or quotient can be written in terms of a and b.

Anyway, I realized the two equations are sneakily not independent and I must have made some arithmetic mistake when I simplified and solved, so ignore what I wrote.

[quote]I think Maple's smart enough to give an exact answer.[/quote]

This is true.  You have to use evalf( to get an approx answer.  Sorry I can't help w/this problem though.  I think w/maple you might just be able to type

sqrt(2+sqrt(4+sqrt(8+...sqrt(2^n), n= :inf:  or something like that and it will evaluate it.  I don't know though.

Getting a decimal approximation on my 89, and playing with it, the answer seems to be very ugly.

I don't think this is an easy thing to get a computer algebra system to do at all.  TI's definitely will not be able to handle it.  I think the kind of idea that schulmannerism is playing with is definitely a good one, but it might be really hard to actually execute, or at least involve lots and lots of trial and error.  For the statend recursion to be helpful, you would need to find at least one value explicitly, which doesn't simplify our job any.  This looks very very hard to me.

It might very well not be solvable... I wish someone could provide some insight proving it not solvable.

If someone can find a way to solve this, here is a formula to evaluate:

[tex]\displaystyle{\lim_{n\to \infty} f(n-1, n)}[/tex]

[tex]\displaystyle{f(n,m)=\sqrt{2^{m-n}+f(n-1,m)}}[/tex]

[tex]\displaystyle{f(0,m)=\sqrt{2^m}}[/tex]

[quote="ComplexZeta"]I thought about this problem for about eight hours yesterday when I was coming home from college. My best attempt was to try to find a function satisfying

[tex]\displaystyle f(x)^2=x+f(2x),[/tex]

but I couldn't think of any finite ones. Clearly it must have a [tex]\sqrt{x}[/tex] term, but then I couldn't find a way to make the smaller terms work out, so I was only able to find

[tex]\displaystyle f(x)=\sqrt{x+\sqrt{2x+\sqrt{4x+\sqrt{8x+\cdots}}}},[/tex]

which isn't very useful.[/quote]

Let's assume that [tex]f(x)[/tex] has a power series expansion [tex]\displaystyle f(x)=\sum_{n=1}^{\infty} c_nx^n[/tex]. The sum starts from [tex]n=1[/tex] since [tex]f(0)=0[/tex]. Then
[tex]\displaystyle \left(\sum_{n=1}^{\infty} c_nx^n \right)^2= x+\sum_{n=1}^{\infty} (2^nc_n)x^n[/tex].

Comparing the coefficient of both sides, we get [tex]c_1=-\frac{1}{2}[/tex] and a recursive relation:
[tex]\displaystyle 2^nc_n = \sum_{i=1}^{n-1}c_ic_{n-i}[/tex] for [tex]n\ge2[/tex].
If we can find [tex]\displaystyle f(1)=\sum_{n=1}^{\infty} c_n[/tex], then we are done since [tex]f(2) = f(1)^2 - 1[/tex].

Warning, this is not rigorous since I don't know whether the power series expansion exists or not. But it at least helps for finding the value. :-)

I thought of this same idea fuzzy, but I don't think it works. First of all, square roots behave very badly as analytic functions around zero - so I'm sure that this function is not analytic at zero either. To support this claim, notice that c1 = -1/2 would mean that f(x) is a decreasing function at zero - making it negative for very small positive values of x. This doesn't seem plausible.

I would actually be happy if someone could just prove that this thing converges (and for which values of x ComplexZeta's generalization of it converges).

[quote="gauss202"]I thought of this same idea fuzzy, ...[/quote]

I am convinced by your reasoning that this is not the right path. Thanks.

Nevermind, Ravi already provided a proof in the other thread that it converges -- and I suspect that his same idea can be used to show that ComplexZeta's generalization converges for all x :ge: 0.

[quote="gauss202"]I thought of this same idea fuzzy, but I don't think it works. First of all, square roots behave very badly as analytic functions around zero ...[/quote]

The function [tex]f(x)[/tex] defined by [b]ComplexZeta[/b] behaves very badly around zero as explained above, but it doesn't behave that bad around 1. So if we define [tex]g(x)=f(x+1)[/tex], then,

[tex]\displaystyle g(x)^2=(x+1)+g(2x+1),[/tex]

Again, let's assume that [tex]g(x)[/tex] has a power series expansion [tex]\displaystyle g(x)=\sum_{n=0}^{\infty} c_nx^n[/tex]. Then
[tex]\displaystyle \left(\sum_{n=0}^{\infty} c_nx^n \right)^2= (x+1) + \sum_{n=0}^{\infty} c_n(2x+1)^n[/tex].

This doesn't look promising. Probably doesn't lead to anywhere...

Consider ComplexZeta's function, [tex]f(x)=\sqrt{x+f(2x)}[/tex].

Ravi showed that f(2) converges to a value less than [tex]\Phi\sqrt{2}\approx 2.288245611[/tex].  In fact, the series converges quickly.  A computation involving just the first 10 terms shows [tex]f(2)\approx 2.179680303[/tex], which I claim is accurate to the precision shown.  (spoon previously provided a slightly higher-precision solution.)

The binomial theorem may provide some insight into the behavior of this function.  The binomial theorem says:  [tex](1+q)^r=1+rq+\frac{1}{2!}r(r-1)q^2+\frac{1}{3!}r(r-1)(r-2)q^3+\ldots[/tex].  We let [tex]r=\frac{1}{2}[/tex], and we factor x out of both terms in the radical, leaving the radical in the form [tex]f(x)=\sqrt{x}(1+\frac{f(2x)}{x})^{1/2}[/tex].

Applying the binomial theorem to get a first-order approximation, we find that as x gets large, [tex]f(x)\approx\sqrt{x}+\sqrt{0.5}[/tex].  I calculate [tex]f(256)\approx 16.71344[/tex]; [tex]f(65536)\approx 256.70751[/tex]; [tex]f(1000000)\approx 1000.70721[/tex].

I have been unable to find a closed form solution for f(x) or f(2).  Is there some reason to think there *should* be a closed form solution?

[quote="fuzzylogic"]Again, let's assume that [tex]g(x)[/tex] has a power series expansion [tex]\displaystyle g(x)=\sum_{n=0}^{\infty} c_nx^n[/tex]. Then
[tex]\displaystyle \left(\sum_{n=0}^{\infty} c_nx^n \right)^2= (x+1) + \sum_{n=0}^{\infty} c_n(2x+1)^n[/tex].
[/quote]
If you assume that series for [tex]g(x)[/tex] converges at [tex]2x+1[/tex] then its radius of convergence should be [tex]\geq2x+1>1[/tex]. But then, it should converge at [tex]x=-1[/tex], which is [tex]f(0)[/tex]. So I don't think this approach is valid.
I thought of another way to deal behaviour of this function at x=0. Maybe we could find some function [tex]h(x)[/tex] (which has closed form expression) that behaves "just like" [tex]f(x)[/tex] (so [tex]f(x)/h(x)[/tex] or [tex]f(x)-h(x)[/tex] has power series expansion at x=0)

Also.  Given [tex]f(x)=\sqrt{x+f(2x)}[/tex].

It is easily shown via induction that [tex]f(x)>1$, $x>0[/tex].

But since [tex]f(0)=\sqrt{0+f(0)}[/tex].  [tex]\big(f(0)\big)^2=f(0)[/tex], which yields [tex]f(0)=0$ or $1[/tex].  We don't even have a function at x=0.

If you are seeking a series expansion, you might have better luck using a technique to shift the origin to a point where the function is continuous and well-behaved.

I was able to program my calculator to find the sum to a finite power of 2. The program prompts for n and then takes the sum sqrt(2+sqrt(4+sqrt(8+...sqrt(2^n)...))). I only have a TI-83, so infinity comes between 2^332 and 2^333. But, for n=10 through n=332, the first 10 digits are identical, and for n=12 through n=332, all 14 digits are equal(my 83 only goes to 14.)If someone who has a better calculator could put in the program and see what happens, that would be great. Here is the program:
:Prompt N
:N+1->C
:0->S
:Repeat C=1
:C-1->C
::sqrt:((2^C)+S)->S
:End
:Disp S

P.S.  -> stands for STO>

I'll try programming it on my TI-89.  I know there's an infinity button on there, but I'm not sure if you can use it for anything besides limits and infinite sums.

Your program returns sqrt(2^n + sqrt( 2^(n-1) ....)), not sqrt(2 + sqrt(4 + sqrt(8...))).  So when I put infinite into the program, I got infinite.

No, my program begins in at the end and works back. Don't use infinity because the very first step will be  :sqrt: (2^ :inf: ) or (1.414...)^ :inf: , so it will be infinity. The program won't work like that. Just put in large N and see if the 89 is able to find an exact answer.

Are you sure?  I plugged in 5 for n and got 4 :rt2:.  The actual answer is 2.something.

Maybe you put the program in wrong. Write out your program and then read mine and see how they differ. I put in 5 and got 2.17922847.

[quote="Spoon"]Umm... we are not interested in a decimal approximation... your efforts are a waste of time, and because this converges so quickly, writing a program is also useless... just type it in to about 1024 and it will give you the exact answer to about 12 decimal places. That however doesn't help us at all. Try not to flood the thread will such posts.

look here:
[quote="rcv"]
Ravi showed that f(2) converges to a value less than [tex]\Phi\sqrt{2}\approx 2.288245611[/tex].  In fact, the series converges quickly.  A computation involving just the first 10 terms shows [tex]f(2)\approx 2.179680303[/tex], which I claim is accurate to the precision shown.  (spoon previously provided a slightly higher-precision solution.)[/quote][/quote]

Writing a program that can compute S is an important step towards finding a recursion, which is better than anything we have as of yet.

Your response is really quite rude and borders on ignorant as well.

[quote="JBL"]Keep it civil, folks.[/quote]

I just don't think it's fair to jump all over someone who has 6 posts like that.  I know I've posted my fair share of errors in and around this forum, and I've always appreciated that I don't catch flak for that - it's just - oh, you did this wrong - not 'what the heck were you thinking?'.

I didn't look at his post count. Blah... I am sorry, I hadn't realized he was new. I am usually quite accepting and helpful to new people.

I am deeply sorry Pork_chop.

Perhaps this link could provide some inspiration:

http://mathworld.wolfram.com/NestedRadical.html

I'm not sure how much that link helps, but, it did help me think of something. We know  :sqrt: (2+ :sqrt: (2+ :sqrt: (2+...  = 2. Can we use this somehow to get to what we want?

I'm "bumping" this thread because it took place before the Mathlinks members joined us. Perhaps one of them has some insight regarding this currently unsolved problem.

[tex]\sqrt{2+\sqrt{4+...}}>\sqrt{1+\sqrt{1+...}}[/tex] And it seems to tend to infinity

This problem has to be sloved by setting up equalities like on the first page:
[tex]f(0)=0  AND  f(x)^2=x+f(2X)[/tex]

We should work this way

What do you think of the following sequence:
********************************************************************
[tex]\forall n\in IN[/tex]:
[tex]  U_(n+1) = \sqrt{1+\sqrt{U_n}}[/tex] with

[tex]U_0 = a[/tex]

********************************************************************

SO [tex]\lim\limits_{n\to\infty} U_n=+\infty[/tex] for [tex]a>1[/tex]

Proof: (i ignore the way you do it in English, but you first prove one case, for n=0, n=1, or n=2, then you prove  (n) true -> (n+1) true and you can say: [tex]\forall n\in IN[/tex] [tex](n)[/tex]true

Let's prove that [tex]U_n>\sqrt(n+1)[/tex] for [tex]a>1[/tex]

For [tex]U_1[/tex] we have [tex]U_1=\sqrt{1+\sqrt{a}}>\sqrt{2}[/tex]

[tex]\exists n \in IN[/tex] :   [tex]  U_n>\sqrt{n+1}[/tex]

Then [tex]U_(n+1)=\sqrt{1+\sqrt{U_n}} > \sqrt{1+\sqrt{n+1}}[/tex]

So, [tex]\forall n\in IN*[/tex] : [tex]   U_n > \sqrt{1+\sqrt{n+1}} > \sqrt{1+\sqrt{n}}[/tex]

[tex]\lim\limits_{n\to\infty} U_n > \lim\limits_{n\to\infty} \sqrt{1+\sqrt{n}}[/tex]

We have: [tex]U_n \to\infty[/tex]

I certainly messed up, if i didn't it would mean [tex]\sqrt{2+\sqrt{4+\sqrt{8+...}}}[/tex] tends to infinity.

HUGO

[quote="icare"]
Then [tex]U_(n+1)=\sqrt{1+\sqrt{U_n}} > \sqrt{1+\sqrt{n+1}}[/tex]
[/quote]

Um, you should have an extra sqrt in there.. U_n is not greater than n+1.
In any case, theres no way you are going to get anywhere trying to prove it diverges, as the above posts seem to show that it converges and extremely quickly.

:blush:  :blush:  :blush:  Right, i've forgotten a sqrt. My proof collapses...

Anyone has an idea ???[/tex]

I found the limit of my sequence (Un):

it's (3+ :sqrt: 5)/2[/tex]

Not so hard if you have the right idea...

We define 3 things
$A = \sqrt{2+\sqrt{4+\sqrt{8+\cdots}}}$
$B = \sqrt{2-\sqrt{4+\sqrt{8+\cdots}}}$
$f(n+1)=f(n)^2-2^n$ (with $f(n) = \sqrt{2^n+\sqrt{2^{n+1}+\sqrt{2^{n+2}+\cdots}}}$)
$A=f(1)$

Comparing $A^2+B^2, A^2-B^2, A^2B^2$ we have 3 linearly independent equations with 3 unknown variables, so simply solving that gives A or f(1), which is requested. :)


For $ a,b,c$ positive reals

\[ \frac {1}{3}\sum\left ( \frac {2a}{b \plus{} c}\right )^{\frac {1}{4}} \geq 1.\]

This inequality does not hold. Simply choose a->0+ ,b=c=1, the LHS will approach 2^(5/4)/3, which is less than 1.


You made some small algebra errors I think. The method you used is correct. Check your work to see if you can find your errors.

[quote]s-s/z=[1/z]/(1-1/z) [/quote]

Wait, how did you get to this? Is there some formula saying that x + x^2 + x^3 + ... = x/(1-x) that I should know about?

That's the sum of an infinite geometric series.

yeah when i was typing it out i saw some, but i thought i fixed them. i'll look later.

In the real world the first thing you should check is whether a series converges absolutely. Otherwise you can perform seemingly sensible manipulations which lead to things that don't make sense. Any series you see in a contest format though will usually converge.

Exactly why do you have the sum of that series memorized? I have to derive it every time.

Right. It shouldn't take 15 minutes to derive it though:

S/x=1+2x+3x^2+...
S=x+2x^2+3x^3+...
S(1/x-1)=1+x+x^2+...=1/(1-x)
S=x/(1-x)^2.

Or another way:

T=1+x+x^2+...=1/(1-x)
dT/dx=1+2x+3x^2+...=1/(1-x)^2
x*dT/dx=x+2x^2+3x^3+...=x/(1-x)^2.

I like the second way better, but they both work.

[quote="Fierytycoon"][quote]Exactly why do you have the sum of that series memorized?[/quote]

Now if darkquantum were here, he'd say that teratomato spent a good 15 minutes deriving that formula, but then posted it on the forum as a mere triviality.  8-)[/quote]

Yup, and here I am:
But i think Hooyoung has all these formulas on hand becasue he's reading up on number theory rite now...

You're a genius! How'd you figure it out?

Right, I have all those power series identities memorized because I'm reading about number theory, and power series have absolutely nothing to do with number theory!

Oh yeah, you're the guy who thinks anything with numeric variables in it is a "number theory" problem.

Forgot about that...

OkOk.
So maybe I don't know what I'm talking about.

o wll...


Hmm...

Just a guess, is it volume, not area?

If it is, you could just push it back twice the depth.

BTW: The question says a square, not an octogon.

It's not, it doesn't give you depth, all it wants you to find is how the carpenter does the person's window. Also, if it were volume, it would have told you the depth of the window.

[asy]draw((0,0)--(1,0)--(1,1)--(0,1)--cycle,blue+linewidth(.3 mm));
draw((1,0)--(1,-1)--(0,0));
draw((0,0)--(1,1)--(2,0)--(1,-1)--cycle,linewidth(.3 mm));[/asy]

Blue square is the original, the black square is the new one

Of course, the carpenter should rotate the window

But how does that diagram "retain the same dimensions"?

[quote="5849206328x"]I assume it means the ratio, otherwise the question doesn't really work at all?[/quote]

Yeah, the problem would be impossible if it meant the same exact dimensions (the square is the most effective quadrilateral).

[quote="joyofpi"]I have another problem... once again... (It appears I'm am feeling a little stupid today  :help: )

A man wants to remodel his house. One wall contains a square window, the dimension of which are one foot high and one foot across. The man tells the carpenter he wants the window to be doubled in area while retaining the same dimensions. The carpenter carries out his wishes- how?

I'm a little confused about the question, since the dimensions makes me wonder if it's about changing the shape, or size.
I also want to know how to go through it, like is there a formula, or whatever.

I was trying to use an octagon, but then I realized that that would probably wouldn't work afterward. I think that the only shape that will work is a quadrilateral, most likely a square or rectangle. That's really all I've figured so far, as well as the new window's area is 2.[/quote]
:huh:
      i dont think this is even possible

[quote="joyofpi"]I have another problem... once again... (It appears I'm am feeling a little stupid today  :help: )

A man wants to remodel his house. One wall contains a square window, the dimension of which are one foot high and one foot across. The man tells the carpenter he wants the window to be doubled in area while retaining the same dimensions. The carpenter carries out his wishes- how?

I'm a little confused about the question, since the dimensions makes me wonder if it's about changing the shape, or size.
I also want to know how to go through it, like is there a formula, or whatever.

I was trying to use an octagon, but then I realized that that would probably wouldn't work afterward. I think that the only shape that will work is a quadrilateral, most likely a square or rectangle. That's really all I've figured so far, as well as the new window's area is 2.[/quote]
Since the title of this thread is 2 windows perhaps he just made another window? :P
But on the more serious side I don't think it is possible unless hee meant it was the same ratio which would mean the side lengths would be $ \sqrt{2}$


[hide="Comprehensive Solution"]It may be helpful to construct a diagram as you read the solution.

To determine the are of one of the small circles, we need the legnth of their radii.

We let the center of the large circle be $O$ and the center of one of the small circles be $P$. We then draw a tangent line from $O$ to circle $P$ at $T$ $\triangle POT$ is a 30-60-90 right triangle because $OP=2OT$.

If you are not convinced, form a regular hexagon by connecting the midpoints of the smaller circles.

If we let $OT=x$, $3x=30$ and $x=10$

Now that we have the radii lengths, the area of the regions is $900 \pi - 600 \pi= 300 \pi$. $\lfloor 300 \pi \rfloor=942$ Our answer is 942.[/hide]

[hide="Answer"]Drawing the tangent radius of the larger circle to any of the smaller circles and the radius that passes through the center of one of the smaller circles, we have a 30-60-90 triangle from the center of the larger circle to the center of the smaller and to the tangent to the smaller circle, because this divides the circle in half, all six circles are congruent, and so $12\cdot m\angle =360$. Therefore, we have $2r=r+n$, where $r$ is the radius of the smaller circle and $n$ is the distance from the center of the larger circle to the edge of the smaller circle at the nearest point. Therefore, $r=n$, and $2r+n=30$, so $r=10$, and we have $\lfloor \pi (30)^2-6\cdot (\pi (10)^2)\rfloor=\lfloor 300\pi\rfloor =\boxed{942}$.[/hide]

First, we must see that we have a ring of 6 congruent circles within a larger circle. We then must go on to realize that in the center of such a ring, there must be one circle congruent to the smaller circles and tangent to each one. Therefore, we see that the diameter of the larger circle is composed of 3 smaller circles. Defining the radius of the larger circle as $R$ and the radius of the smaller circles $r$, we see that $r = \frac{R}{3} = \frac{30}{3} = 10$. Thus, the area of the desired region is $900\pi - 6(100\pi) = 300\pi$. Ergo, our anwer is $\lfloor 300\pi\rfloor \implies \boxed{942}$

[quote="4everwise"]Six circles form a ring with with each circle externally tangent to two circles adjacent to it. All circles are internally tangent to a circle $C$ with radius $30$.[/quote]

how would i draw this? i know that it would look like a ring made of circles but, what does this mean?
circle externally tangent to two circles adjacent to it. All circles are internally tangent to a circle

A ring of six circles, each externally tangent to two circles means that they touch the two adjacent circles at one point. All six circles in the ring are also tangent to a circle inside the middle of that ring.

Or you can just use the $\text{7 Circle Theorem}$ for this problem, in other words, the diameter $60$ is divided by $3$, so each of the little circles have radius $10$, the answer can easily be determined after that.

[quote="Zellex"]What is the 7 circle theorem?[/quote]

If you have 6 circles externally tangent to each other, there can always be a circle you can put in the middle of it with the same diameter.

I don't really think it should be called a theorem though, because it can easily be proven by creating a hexagon w/ the diameters of the circles and stuff..but it's good to know

[quote="4everwise"]Six circles form a ring with with each circle externally tangent to two circles adjacent to it. All circles are internally tangent to a circle $C$ with radius $30$. Let $K$ be the area of the region inside circle $C$ and outside of the six circles in the ring. Find $\lfloor K \rfloor$.[/quote]

Is it stated in the problem that 6 circles are equal or you are assuming it for your convenience?

I constructed an equilateral triangle from c to the center of each smaller circle. This gave
30-x = 2x so 3x=30 x=10 so the answer is [300pi] how do u evaluate that without a calculator?

Learn the digits of pi?  You must know at least $ 3.141$; that's enough.

[hide="Answer"]We can just use the seven circle theorem. So, the radius of each circle is 10. Now, we have answer being $ 900\pi$ for large circle and $ 6\times 100\pi$ for smaller circles. So we have $ \lfloor 900\pi\minus{}600\pi\rfloor \equal{} \lfloor 300\pi \rfloor\equal{}\boxed{942}$.[/hide]


Find all positive integers k, l, m, n such that
(2^k+2^l) 2  = 2^m + 2^n

$2^{2k}+2^{2l}+2^{k+l+1}=2^m+2^n$

suppose WLOG 2k is the smallest. WLOG n must be equal to this. (cuz of mod $2^k$)
We have $\bmod 2^{2k}$: $2^{2(l-k)}+2^{l-k+1}=2^{m-n}$.
At this instance we know $2^{2(l-k)}=2^{l-k+1}$ so $l-k=1$ OR $m=n$

So all solutions are those of the form: $(2^{n+1}+2^n)^2=2^{2n+3}+2^{2n}$
Or: $(2^n+2^n)^2=2^{2n+1}+2^{2n+1}$

Peter

Indeed. Or to fix the hole in my proof: when m=n.

There is another solution thus: $(2^n+2^n)^2=2^{2n+1}+2^{2n+1}$


1) Let $n$ be an integer greater than four, and let $P_{1}P_{2}\cdots P_{n}$ be a convex $n$-sided polygon. Stephanie wants to draw $n-3$ diagonals that partition the region enclosed by the polygon into $n-2$ triangular regions and that may intersect only at the vertices of the polygon. In addition, he wants each triangluar region to have at least one side that is also a side of the polygon. In how many different ways can Stephanie do this ?

These two don't really belong in the same post.  Also, they may be a little easy for the Olympiad forums.  (Lots of stuff of this difficulty ends up in Pre-olympiad.)

The second looks like no fun.  For the first,
[hide="outline"]For $n \geq 4$, we have exactly two triangles with two sides as sides of the polygon.  We can choose two such edges, $P_{i-1}P_{i}$ and $P_{i}P_{i+1}$ in $n$ ways, and we draw chord $P_{i-1}P_{i+1}$.  It's clear that we now need to include exactly one of the chords $P_{i-2}P_{i+1}$ or $P_{i-1}P_{i+2}$; then, after making that choice, we have exactly two choices for the next chord, and so on.  There are $n-4$ total choices to be made, giving us a preliminary count of $n\cdot 2^{n-4}$; however, for each such triangulation we can start an either end, so we must divide this answer by 2 to get the final result, $n \cdot 2^{n-5}$.[/hide]

First, I don't consider the second problem as an easy one, since my computation needs too much cases division and thus, has a great risk to make a miscalculation.. So here I'm curious whether anyone having a simpler way to count it..

Second, why must we include exactly one of $P_{i-1}P_{i+2}$ or $P_{i-2}P_{i+1}$ ? For example, we have a convex hexagon $ABCDEF$. WLOG, let $FA$ and $AB$ be the first two consecutive edges we choose. Then, it's not certain that we must take $EB$ or $AC$ to finish the partition, since we can just draw $BD$ and $DF$.

EDIT : Sorry, my reasoning was wrong..  :blush:

These two problems have really nothing together, thus it's total nonsense to post them together.

For 1), see problem 1 in http://www.mathematik.uni-muenchen.de/~yarotsky/jarnik/Jarnik_daniel_harrer.pdf (LMU selection round for Vojtech Jarnik International Mathematical Competition 2007).

Okay, I have posted the second problem in another topic..


geometry
inequalities unsolved
inequalities

The problem is equivalent to find the minimum value of  PA*PC+PD*PB where P is a point  inside the unit   square  ABCD.
Construct a triangle DCQ congruent to BAP.Then
  area (CPDQ)=area(APB)+area(CPD)=1/2=1/2sinPCQ*PC*CQ+1/2sinPDQ*DP*DQ=<
=<1/2(PC*CQ+DP*DQ) .But PA=CQ,PB=DQ.

Another geometric proof:

Construct point $R$ in the exterior of the square $ABCD$ such that $\triangle ABP \equiv \triangle DCR$.

Using Ptolemy's Theorem, $PA \cdot PC + PB \cdot PD = RD \cdot PC + RC \cdot PD \geq CD \cdot PR = 1$.


trigonometry
function
geometry
perpendicular bisector
geometry proposed

:!:   :idea:   Consider the triangle $ABC$ and the point $P$ such that $PC=PB$ and $PA=AB$. Prove that $\boxed {\sin(B-C)\mbox{ = 2}\sin C \cos(B-2sx)}$,  where we noted $x=m(\widehat {PBC})$ and $s=1$ or $s=-1$ if the line $BC$ don't separate or separates the points $P$ and $A$.
[u]Remark[/u]: If $m(\widehat {PAB})=2\cdot m(\widehat {PAC})$  we obtain a problem from $IMO,\ 1994\ .$

Here is my approach:
I will prove it for the case:the line [tex]BC[/tex] separates[tex] P[/tex] and [tex]A[/tex] or[tex] s=-1[/tex].The rest case is similarly to as.we have [tex]P[/tex] lies in perpendicular bisector of [tex]BC[/tex].Suppose that this line cuts [tex]AC[/tex] at[tex] T[/tex].we have [tex]\widehat{ABT}=B-C[/tex].And by sin function theorem we have :[tex]sin(B-C)=sin2C.\frac{AT}{AB}=2sinCcosC\frac{AT}{AB}.[/tex]Since [tex]AB=AP[/tex],we have [tex]\frac{AT}{AB}=\frac{AT}{AP}=\frac{sin\widehat{APT}}{sin\widehat{ATP}}=\frac{cos(B+2x}{cosC}[/tex](By [tex]\widehat{APT}=2\pi-\widehat{BPC}-\widehat{APB}-\widehat{TPC}=2\pi-(\pi-2x)-(B-x)-(\frac{\pi}{2}+x)=\frac{\pi}{2}-B-2x-->sin APT=cos (B+2x) [/tex]and [tex]\widehat{ATP}=90+C-->sin\widehat{ATP}=cosC[/tex] and we are done

See and http://www.mathlinks.ro/Forum/viewtopic.php?t=54461 I will present here the my solution to-morrow, 2/14.

I have solved this problem in Spring 2005,a long time!As I think it's really easy.
  And today is 2/14-happy valentine day!Love for maths and... :D


calculus
integration
function
geometry
calculus computations

Integrate the following:

(X+1)(X+5)dx     x+3 = u

In simple words, what is integration?

well its sort of like finding the antiderivitive
for definite integrals of continuous functions you can think of it as finding the area under a curve

[quote="sharkman"]In simple words, what is integration?[/quote]
Go read a calculus textbook. There's a reason we devote basically a whole course to integration.


Let $a,b,c$ be the positive real numbers such that $ab+bc+ca=1$.
Find the minimum value of
$S=\sum\frac{\sqrt{3a^2+b^2}}{ab}$

\begin{eqnarray*} S=\sum_{cyc} \frac{\sqrt{3a^2+b^2}}{ab} &\ge& \frac12 \sum_{cyc} \left(\frac{3a+b}{ab}\right) = 2\left(\frac1{a}+\frac{1}{b}+\frac1{c} \right) \\ &\ge& \frac{6}{\sqrt[3]{abc}}=\frac{6}{\sqrt[6]{ab.bc.ca}} \ge \frac{6}{\sqrt{\frac13 (ab+bc+ca)}} = 6\sqrt{3} \end{eqnarray*}


Solve the following simultaneous equations:

$x+y=x^{2}+2xy+y^{2}$,

$x-y=x^{2}-2xy+y^{2}$.

[hide="Solution"]

Using factoring for both equations, the Zero Product Property, and pairing factors results in four real solutions: $\boxed{(x,y)=(0,0),(1,0),(.5,-.5),(.5,.5)}$[/hide]

Nice but too easy for Pre-Olympiad.

Just sum up, you get that $(x,y)=(\frac{1}{2}, \pm \frac{1}{2})$
or $y=0$, giving the solutions $(0,0)$ and $(1,0)$.


calculus
integration
geometry
rectangle
calculus computations

[b]Q.  [/b]Given $\int_{-2}^{4}(|3-|3x\parallel )dx$, first sketch the graph $y=f(x)$ on the given interval.

Then find the integral of f using your knowledge of area formulas for rectangles, triangles and circles.

[b]A.  [/b]The region represented by the integral consists of three triangles above the x-axis, one with base 1 and height 3, the second with base 2 and height 3, and the third with base 3 and height 9.  So the value of the integral is $\frac{27}{3}+\frac{6}{3}+\frac{3}{3}=\frac{36}{3}=\frac{12}{1}=12$

You almost got it. The only mistake: area of a triangle is $\frac{1}{2}\times{\rm base}\times{\rm height}$.

I understand what I did wrong, thanks for your input.

$\frac{27}{2}+\frac{6}{2}+\frac{3}{2}=\frac{36}{2}=\frac{18}{1}=18$


algebra
polynomial
number theory proposed
number theory

[quote="IWW"]Find the nonzere second degree trinomials $ P_1,P_2,P_3,P_4$ have integer coefficients such that
$ (P_1(x))^3 \plus{} (P_2(x))^3 \plus{} (P_3(x))^3 \equal{} (P_4(x))^3$[/quote]

I am not sure what a second degree trinomial is.
If it is a polynomial of second degree, than one possible solution is
$ P_1(x)\equal{}12x^2$
$ P_2(x)\equal{}\minus{}10x^2$
$ P_3(x)\equal{}\minus{}9x^2$
$ P_4(x)\equal{}x^2$.


group theory
abstract algebra
number theory
least common multiple
superior algebra
superior algebra unsolved

Prove that a group $ G$ is a cycle group iff for any subgroup $ H$ of $ G$, there exists a positive integer $ k$ such that

  $ H \equal{} \left\{ a^k: \ a\in G \right\}$

I'm sure there is a faster way to do it than this:

=>: Let $ G$ be cyclic. The subgroup of a cyclic group is cyclic so we know $ H$ is cyclic. Consider $ G/H$, let the order of this group be $ k$ (its finite because a cyclic group is either isomorphic to $ \mathbb{Z}$ or $ \mathbb{Z}/n\mathbb{Z}$ for some $ n$). Now, this means there are $ k$ distinct cosets of the form $ g\plus{}H$ (i'm going to write additively because cyclic means abelian). Now if $ 1$ generates $ G$, then it descends to a generator of $ G/H$. That means the $ k$ distinct cosets can be represented as $ 0\plus{}H$, ..., $ k\minus{}1\plus{}H$. Therefore the coset contains of $ k$ multiples.

Here I'm assuming the condition on the subgroup means that taking ANY element $ a\in G$, taking it to the $ kth$ power sticks it into $ H$ (if this isnt true the above direction still holds)

<=: Suppose all subgroups of $ H$ of $ G$ are $ k$ powers of elements in $ G$. Suppose now any generating set contains at least two elements say $ a$ and $ b$. Then if we pick the subgroup $ <a>$ we get that $ <b>$ should also be an element in it hence $ b\equal{}a^k$ contradiction.

for "<=" , I don't understand how you got the result :?:
And I've heard that it's a difficult problem.

yeah I think I'm misunderstanding what the condition on the subgroup for <= is because if its what I think you mean, then the problem doesn't seem terribly bad.

My impression of what you meant is:

all subgroups $ H$ have the property given any $ a$ in $ G$, there exists $ k$ s.t. $ a^k$ is in $ H$

Let the cyclic group  $ G$ be $ \{a^1,a^2,\dots ,a^n\}$ with $ a^n\equal{}1$ (identity element) of finite order $ n$. Let the orders of the elements in order be $ \{r_1,\dots,r_n\}$. Let the non trivial subgroup $ S$ (which is cyclic again) be $ \{a^r,a^{2r},\dots ,a^{kr}\equal{}1\}$ where $ r|n$ and their orders being $ \{O(a^r),\dots ,O(a^{kr})\}$ respectively. Consider $ \ell\equal{}LCM [\{r_1,\dots,r_n\}\backslash \{O(a^r),\dots ,O(a^{kr})\}]$. Obviously, $ a^\ell \notin G\backslash S \quad \forall a\in G$. Secondly, $ r \nmid \ell$. We are done.

No one solved this problem?  :?:
If for any subgroup $ H$ of $ G$, we have $ H \equal{} G^k \equal{} \left\{g^k: g\in G\right\}$ for some integer $ k$, how to prove $ G$ is cycle?
I saw a solution in the book, confused by the last step of it.

[hide="SOLUTION:"]For arbitrary $ a\in G$, and assume that $ \langle a \rangle \equal{} G^k$. So there exists $ x\in G$ such that $ x^k \equal{} a$.
Assume that $ \langle x\rangle \equal{} G^j$, then there exists $ y\in G$ such that $ y^j \equal{} x$.
If $ j \equal{} 1$, we are done. If $ j > 1$, assume that $ y^k \equal{} a^i \equal{} y^{jik}$, and we obtain $ y^{k(ji\minus{}1)} \equal{} e$,
which implies $ o(y)$ (the order of $ y$ in $ G$) is finite, and hence $ o(a)$ is finite. Moreover, for any $ g\in G$,

$ g^{ko(a)} \equal{} (a^s)^{o(a)}\equal{}e$

Now take a cycle subgroup $ \langle b\rangle$ of $ G$ with the greatest order in $ G$, and assume that $ \langle b\rangle \equal{} G^m$ and $ o(b) \equal{} n$.
Since $ \langle b\rangle$ has the greatest order, it's not hard to prove that $ \gcd(m,n)\equal{}1$.
Moreover, we can prove $ G\equal{} \langle b\rangle$.[/hide]

I don't understand the last step in the solution. (i.e. how do we get that $ G \equal{}\langle b\rangle$?)


algebra
function
domain
real analysis
real analysis unsolved

If f(x) is exist and not the infinite , so what's the value range of x and f(x) ?
[img]http://img012.photo.21cn.com/photos/user/24454028/20070404193846/o/8391269.jpg[/img]

Has been discussed several times.

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=tower+power&t=77942

I have knowed that the domain of $x$ is $[\frac{1}{e^{e}}, e^{ \frac{1}{e}}]$ , and the range of $f(x)$ is $[\frac{1}{e}, e]$

But I can't prove it alone  , I need you help .

eager for your proof   ,   Thanks !

[quote="t0rajir0u"]Has been discussed several times.

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=tower+power&t=77942[/quote]
I could not find my answer in that post

And I will continue waiting till a nice guy help me

The question has, indeed, been posted several times. You may want to look at [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=89869[/url] first. See if you still have any questions after reading that thread.


In the $ 3\times4$ grid shown, the points are one unit apart horizontally and vertically. Given that two points are randomly selected from the grid, what is the probability, expressed as a common fraction, that the distance between them is $ \sqrt2$?
[asy]dot((0,0));
dot((1,0));
dot((2,0));
dot((3,0));
dot((0,1));
dot((1,1));
dot((2,1));
dot((3,1));
dot((0,2));
dot((1,2));
dot((2,2));
dot((3,2));[/asy]

there are 12 points, so we can choose two of those points in 12 choose 2 ways, or $ \frac{12*11}{2}\equal{}66$

for each of the little squares inside (there are 6) there are two lines that are exactly $ \sqrt{2}$ units long, the one connecting the top right hand corner to the bottom left, and the top left corner to the bottom right
since there are 6 of these squares, and each square has 2 of these lines that are $ \sqrt{2}$ units long, there are 12 such lines

so our answer is just $ \frac{12}{66}\equal{}\frac{2}{11}$

could you send an easier solution please? One that makes more sense?


Guys,

I'm learning more about the AMS package and I've had some difficulty with a particular environment: [b]alignat[/b].  I've read the ams manual (http://www.ams.org/tex/amslatex.html) and checked out a website (http://www-texdev.mpce.mq.edu.au/l2h/mathdocs/testmath/node39.html) but it still isn't clear the purpose of the [b]&&[/b]'s, and also the argument after the \begin{alignat}[b]{#}[/b] --- is it the no. of columns? How do I specify the horizontal spacing as it is supposed to do?

Thank you.

Here's an example of alignat:

\begin{alignat}{2}
L_1 &= R_1 & \quad L_2 &= R_2 \\
L_3 &= R_3 & \quad L_3 &= R_4
\end{alignat}

I don't use this much, but I think the point is to align multiple elements of equations.  If you write \begin{alignat}{n}, then you are aligning 2n elements in each line, so you use 2n - 1 ampersand characters (&).

Not sure where the && appears.


limit
logarithms
inequalities
number theory unsolved
number theory

Maybe you meant "$k_n$ is the smallest ..."? :?  ;)

[quote="Kapi"]Let a sequence $(u_n)$ with $u_n>0$.
$k_n$ is a largest positive integer satisfies $\sum\limits_{i=1}^{k_n}{\frac{1}{i}} \geq \sum\limits_{i=1}^n{u_i}$
prove that:$\exists \lim\limits_{n \to \infty}{\frac{k_{n+1}}{k_n}}$ if only if $\exists \lim\limits_{n \to \infty}{u_n}$[/quote]
Not sure to understand the election of  $k_n$, is the minimal satisfying the property?. because if $\sum_{i=1}^\infty u_n$ is bounded (and hence  $u_n \to 0$) I can take, for  every pair $(n,n+1)$, another pair $(k_n,k_{n+1})$ such that the limit  $k_{n+1}/k_n$ soen't exists because the harmonic series diverges.  Then I can consider if $n$ is even $k_{n+1}=nk_{n}$  and if $n$ is odd $k_{n+1}=L$ cosntant  such that  $\sum_{i=1}^L \frac 1 i> \sum_{i=1}^\infty u_n$.

Best Regards.
RobertuX

[quote="grobber"]I'm pretty sure I've seen this before on the forum. Posted by Pigfly, maybe? :?[/quote]
And there was a solution?

In any case, it was for the particular case $u_n=1$. And this is much more beautiful and seems to be more difficult also.

I can't remember (I was answering Myth :); I didn't see Harazi's post), but it cannot be that hard, just sort of boring to write rigorously.

The case when $\sum u_n<\infty$ is clear, so we may assume $k_n\to\infty$. The condition that $\frac{k_{n+1}}{k_n}$ has a limit is equivalent to the fact that $\ln k_{n+1}-\ln k_n$ has a limit, i.e. $\frac 1{k_n}+\frac 1{k_n+1}+\ldots+\frac 1{k_{n+1}}$ has a limit. It should follow rather quickly from here, if I am not mistaken.

Then maybe you will have the time to complete the proof. Proofs of easy problems cannot be difficult to write, right?  ;)

Actually, come to think of it, it looks like there's more to it :). I'll give it more thought, you have made your point :D.

Then again, no, it's not hard. :)

We have $H_{k_n}-S_n<\frac 1{k_n}$ and $H_{k_{n+1}}-S_{n+1}<\frac 1{k_{n+1}}$, meaning that $H_{k_{n+1}}-H_{k_n}-(S_{n+1}-S_n)\to 0$ as $n\to\infty$, where $H_n$ is the $n$'th harmonic number, and $S_n$ is the $n$'th partial sum of the series $\sum u_i$. Since $S_{n+1}-S_n=u_{n+1}$, it means that $u_n$ has a limit iff $H_{k_{n+1}}-H_{k_n}$ does, which has a limit iff $\frac{k_{n+1}}{k_n}$ does.

Ok, man, spell it out: what's wrong? :D No, I'm never sure of my solutions, as I keep mentioning. You're torturing me here!!! :D

I just wanted to check if you are sure.  :D . You're quite nervous, grobber, what is happening?

Ohhhh, so it was but a test.. You're evil, man :) :D.

Yes, I am. I remember I posted another problem some time ago, which was of course ignored:
  In the case $ u_n=1$ prove that the sequence $ k_{n+1}-ek_n$ is bounded. What do you think about this?

I apologize for every single one of your beautiful problems that I have ever ignored. Really :).

We get $e^{n+1-\gamma-\varepsilon_{n+1}}\le k_{n+1}\le e^{n+1-\gamma-\varepsilon_{n+1}+\frac 1{k_{n+1}}}$, where $\gamma$ is Euler's constant, and $\varepsilon_n\to 0$. A similar inequality holds for $k_n$, where every $n+1$ is replaced by $n$.

We have, for instance, $k_{n+1}-ek_n\le e^{n+1-\gamma-\varepsilon_{n+1}+\frac 1{k_{n+1}}}-e^{n+1-\gamma-\varepsilon_n}$, which is equal to $e^{n+1-\gamma}\cdot(e^{\varepsilon_n-\varepsilon_{n+1}+\frac 1{k_{n+1}}}-1)$. We can show that this is bounded by using the fact that $e^x-1$ is equivalent to $x$ as $x\to 0$, so we can apply this to $x=\varepsilon_n-\varepsilon_{n+1}+\frac 1{k_{n+1}}$, replacing $e^{\varepsilon_n-\varepsilon_{n+1}+\frac 1{k_{n+1}}}$ with $\varepsilon_n-\varepsilon_{n+1}+\frac 1{k_{n+1}}\ (*)$. It's easy to show this is bounded, but we need a stronger estimate for our $\varepsilon_n$. Since $H_n=\ln n+\gamma+O(\frac 1n)$, it means that $\varepsilon_n=O(\frac 1{k_n})$, so we may freely assume that we only have $\frac 1{k_{n+1}}$ in $(*)$. Now the conclusion follows, since $k_n$ is equivalent to $e^{n-\gamma}$.

Now of this, I really am not sure.  ;)

Nice, grobber, you're fast. No problem, anyway they are not interesting nor beautiful.

[quote]Let a sequence $(u_n)$ with $u_n>0$.
$k_n$ is a smallest positive integer satisfies $\sum\limits_{i=1}^{k_n}{\frac{1}{i}} \geq \sum\limits_{i=1}^n{u_i}$
prove that:$\exists \lim\limits_{n \to \infty}{\frac{k_{n+1}}{k_n}}$ if only if $\exists \lim\limits_{n \to \infty}{u_n}$[/quote]
This problem was used in Vietnam TST 1999,if I'm not wrong it was problem 4(the easiest problem in Second Day).This is my first post in Forum! :)


Consider the sequence of numbers: 4, 7, 1, 8, 9, 7, 6, ....  For n>2, the nth term of the sequence is the units digit of the sum of the two previous terms.  Let $ S_n$ denote the sum of the first $ n$ terms of this sequence. The smallest value of n for which $ S_n$>10000 is

(a) 1992
(b) 1999
(c) 2001
(d) 2002
(e) 2004

[hide]
Write the sequence out, and you find its periodic. The sequence is 4,7,1,8,9,7,6,3,9,2,1,3 and then it repeats. So it has period 12. The sum of the first 12 terms is 60. Therefore, every 12 terms, S_n increases by 60. Therefore, $ S_{166(12)} \equal{} S_{1992}$ = 9960. Then,

S_1993 = 9964
S_1994 = 9971
S_1995 = 9972
S_1996 = 9980
S_1997 = 9989
S_1998 = 9996
S_1999 = 10002.

So the answer is (b).[/hide]


Let $1=d_1<d_2<....<d_k=n$ be all different divisors of positive integer n written in ascending order. Determine all n such that:
\[d_{d_4-1}=(d_1 + d_2 + d_4)d_8 ; \ \ k=12\]

The answer is 416 and 1989.

Notice that $ n$ is not a square and for every $ 1\leq k\leq 12$, we have $ d_k\cdot d_{13\minus{}k}\equal{}n$.
Since $ (d_1\plus{}d_2\plus{}d_4)d_8$ is a divisor of $ n$,
\[ n \equal{} d_5\cdot d_8 \geq (d_1\plus{}d_2\plus{}d_4)d_8 > d_4 d_8.\]
Therefore, $ d_1\plus{}d_2\plus{}d_4\equal{}d_5$ and $ d_4\minus{}1 \equal{} 12$, i.e., $ d_4\equal{}13$ and $ d_5\equal{}14\plus{}d_2$.

Since the number of divisors of $ n$ is 12, $ n$ has one of the forms: $ pqr^2$, $ pq^5$, $ p^2q^3$, or $ p^{11}$, where $ p, q, r$ are distinct primes, one of which is $ 13$.

It is clear that the case $ n\equal{}p^{11}\equal{}13^{11}$ is not a solution.

Notice that $ d_2$ is the smallest prime factor of $ n$, and thus $ d_2\leq 13$. The smallest prime factor of $ d_5\equal{}14\plus{}d_2$ must be not smaller than $ d_2$, leaving only the cases $ d_2\equal{}2,\ 3$ or $ 5$.

The case $ d_2\equal{}2$ implies a solution $ n\equal{}13\cdot 2^5\equal{}416$.
The case $ d_2\equal{}3$ implies $ d_5 \equal{} 17$, and $ n\equal{}13\cdot 17\cdot 3^2\equal{}1989$.
The case $ d_2\equal{}5$ implies no solutions.

[color=#f00][b]Claim:-[/b][/color]only number $1989$ is the unique solution

[color=#00f][b]Proof:-[/b][/color]  we Notice $d_{j}d_{13-j}=n$, also $(d_{1}+d_{2}+d_{4})d_{8}=d_{d_{4}-1}$ which gives that $d_{1}+d_{2}+d_{4}=d_{5}$ which gives $d_{4}-1=12 \implies d_{4}=13$ and $d_{5}=14+d_{2}$\\

now since $d_{2}$ is the smallest prime factor of $n$ we have $d_{2} \in \{2,3,5,7,11\}$.  Also, the number $n$ can have at max three distinct prime factors so:\\

 [color=#000][b]Case 1:-[/b][/color]$d_{2}=2$ then we have $d_{15}=16$ which gives $4,8$ are also in divisor pair list but that contradicts $d_{4}=13$ \\

[color=#000][b]Case 2:-[/b][/color]$d_{2}=3$\\

this case forces $d_{3}=3,d_{5}=17$ and hence only $n=1989$ can work in this\\

[color=#000][b]Case 3:-[/b][/color] $d_{2}=5$\\

this case gives $d_{5}=19$ which is a contradiction\\

now it's not hard to check that $7$ and $11$ also don't works and hence we only  have $n=1989$ as the unique number satisfying this property $\square$


calculus
AMC
AMC 12
derivative
integration
percent
geometry

Same questions as zabelman on the AP Calculus BC exam...

"(If reading this thread later) How'd you do?

(I must remind you however that we cannot discuss any specific questions on the exam)"

How many right do you have to get (as a percentage) on the exam to get a 5? 4? 3? 2? 1? Is the grading similar to other AP exams, where you an 80%+ is equivalent to a 5? 60%+ is equivalent to a 4? Or is it more strict, since Calculus is supposedly one of the easiest exams if you know the material well.

I suppose it was overall not too bad, but I wonder about two free response questions.

Without mentioning anything specifically at all, and I'll thank others not to as well (I would think test admin. is done in all time zones now anyway), is it normal for Logistic Equations to be covered in a BC curriculum? I know Legrange Error Analysis is, which is a shame, because with the words "I can't teach this," it did not appear in my curriculum. And when I asked Simon online, he didn't know, so I figured "it does not matter much."

I undoubtedly failed. With a 2-- my most likely score-- I bring shame upon AoPS and all the smart people here. I skipped about 10 multiple choice, and some of the sections of the free response were extremely hard-- for me. I should probably cancel my score-- that is how difficult I thought it was. If there were some sort of meter of difficulty, I would place the AP Calculus BC exam between the last 10 questions on the AMC12 and the first 7 one the AIME.

Did anyone else here study for the BC exam independently, i.e. take the AB class but the BC exam? How did you guy s do? I totally blanked out with the formulas that I needed to know. I figured that if there was a difficult formula, then they would provide it-- sort of like the AP Physics exam. I also did not take a practice BC exam, which was a big mistake, because I was too busy frantically reviewing infinite series, which I thought was the most difficult BC topic. What tripped me up were actually AB topics like derivatives-- I don't know why though.

mathnovice that is what I did. I did pretty awesome except for one blasted question that I ran out of time on.

For the first free response and the all of the multiple choice I answered all of them, most with 100% certainty that they were correct. A VERY low estimate of my combined first score of the first two sections would be like 85%, depending upon how they grade my free response. I mean for some I just wrote down an integral because It seemed pretty obvious if you know the general form. I'd say a more reasonable estimate would be around 95%.

But on the last free-response I totally nailed the first question, and the last question, except for the something that rymes with Megrange Air I didn't have time to totally the interval, thereby putting them into the equation, so I just did the thing with the alternating series, maybe they will give me some points?

But the second one I really ran out of time on and simply jotted some partial fractions stuff and a little bit about what the constant was equal to, then I ran out of time. I could have easily gotten 90%+ overall had I not ran out of time, now i'm going to be lucky to get a 5 on the BC. Or am I? What is the cut off?

[quote="scm007"]What is the cut off?[/quote]
Around a 68%, but I doubt you did precisely as well as you claimed, even if you did very well.

[quote="mathnovice"]I also did not take a practice BC exam, which was a big mistake. What tripped me up were actually AB topics like derivatives-- I don't know why though.[/quote]

I'm not a big fan of test prep for the sake of test scores, but I am a big fan of learning the material through multiple sources, and then taking ACTUAL past tests to gauge knowledge of the material. I would definitely advise NEVER taking an AP test without taking as much of old actual tests as is made available, under the best simulation of actual testing conditions you can arrange. Maybe only do that once or twice for each exam, but definitely know that much about the test before taking it for keeps.

Good luck next time. You've still got more time to digest more calculus.

OK, whatever you say. I'll I know is that I have taken many practice tests, and many of the questions were identical to the ones I have seen before, and know how to do. I'm almost 100% postive that all of the free response questions that I did answer were right, just how many points I get is determined by the graders.

So even if I only get around 50% of the last section, all I need to average is 77% of the first 2 sections, and I'm fairly confident I did at least that, regardless of what you say. I've gotten a 5 on all of the practice tests I have taken, many over 90%.

What really pisses me off is that i couldn't even get to the second problem because I kept screwing up the last problem. If I hadn't redone the last one 2 times I could have finished the whole thing. Then I would have been a no-brainer that I would get a 5. Now I might have to take hte whole damn thing over again.

Moderator's note:  there is no reason to make three consecutive posts.  Edit to add information.

It's sounds like the whole test was harder than you thought it would be( except for scm007).  That kind of scares me because I was thinking about independent studying for BC in an AB class next year.  Do you think that I should still take the BC?  Also how much calc. experience did you guys have before this year.  I have two calculus books at home(a big one and a small one, size wise)  so I start looking at calculus over the summer to get a head start.  What do you recommend?

Yeah definitely still take the BC. Just manage your time well and don't make stupid mistakes that require you to waste your time fixing :)

The BC stuff is not much harder at all. If you get a firm grasp of the AB stuff it should come naturally. I'm still pissed off about how stupid that was, GOD!!!!

BC was nice except 2 free responses i didnt completely fully(yes u guessed it the last one is one of them)

I think we need something 65% from the total possible score to get a 5.  heard that the BC curve is very generous ,same like in math2c.

what i found funny was that i did much much better in non calculator section (multiple choice)than the calculator one ....

If the curve is 65%, then you have a 5, scm-- even if you underestimate. There's no need for you to worry at all.

Was this exam harder or easier than previous BC exams? I am hoping that the curve will run as low as possible.

Eh, people always do horribly, but I believe they try to make all the exams have the same difficulty, and do so with no more than a few percent variation in the cutoffs.

If any of you do multivariable next year I really recommend the book [u]Vector Calculus[/u] by Marsden and Tromba. Even if your course doesn't use it, try to find it from the library or University bookstore. It's the best book to learn multivariable calc that I know of. Also the Schaum's Outline [u]Advanced Calculus[/u] is a good supplement.

I'll take BC Calc and Stats next year (sophomore year).  I don't know what stats is like though, can anybody fill me in?  Should I take it?  If I had my way, I wouldn't ever take it, but maybe it's not all bad.

stat's really easy, you can learn it in less than a week and get a 5. my brother just took it. i'm probably gonna be taking it next year too.

For anyone who is interested, the free response questions from this year are available on the collegeboard website.  Here are some handy links:

[url=http://apcentral.collegeboard.com/repository/ap04_frq_calculus_ab_36102.pdf]2004 FRQ Calc AB Form A[/url]
[url=http://apcentral.collegeboard.com/repository/ap04_frq_calculus_ab__36107.pdf]2004 FRQ Calc AB Form B[/url]
[url=http://apcentral.collegeboard.com/repository/ap04_frq_calculus_bc_36108.pdf]2004 FRQ Calc BC Form A[/url]
[url=http://apcentral.collegeboard.com/repository/ap04_frq_calculus_bc__36109.pdf]2004 FRQ Calc BC Form B[/url]

For the others, go to http://apcentral.collegeboard.com/repository/ and scroll down to ap04_frq_.  It just takes a while to populate the entire list.

[quote="cats..."]http://users.adelphia.net/~sismondo/AP2004.html

i dont know how helpful that is though, cause it's difficult to remember your answers.[/quote]

Cool, I think I got 1, 2, 4, 6, and parts of 3 and 5.  Thanks for the link.

And here ([url=http://clem.mscd.edu/~talmanl/AB2004]AB Form A[/url], [url=http://clem.mscd.edu/~talmanl/AB2004B]AB Form B[/url], [url=http://clem.mscd.edu/~talmanl/BC2004/]BC Form A[/url], [url=http://clem.mscd.edu/~talmanl/BC2004B]BC Form B[/url]) is another set of solutions.  It seems to be fairly popular for professors to solve the problems after they go up on the College Board website.

I never look in this forum and I may be late, but whatever...

I thought the BC exam was ridiculously easy... I was quite disappointed actually. The section I multiple choice only took me about 20 minutes, and the section II took me 22. I thought too many of the problems could be done without writing anything on the paper. There was only one problem from polar equations, and it laughably easy... I don't understand what the AP writers were thinking this year.

As for the response portion, I think the questions were fair, not in difficulty, but fair to the test takers. They had mixed difficulty, and I think it was nicely done. Logistics Diff EQ is no different than any other separation of variables, and even though I had no looked a an AP curriculum, I'd say there is nothing wrong with them placing the question on the test.

I never took Calc BC, no teacher in my school even knows calc BC, lol... I come from a school where the average calc ab class size is 10. I studied BC on my own.

I am of the opinion that Calc AB and BC should be one class, because there is little matterial covered in calc ab.

I have checked through various sources, which I will not name of course, but I got every multiple choice question right, and as soon as I find the time, I will look over some posted answers to the extended response, which I quite confident I got 100% right.

ROFL!!! NOOO!

lol I'm so stupid, I just looked at the extended response questions and I got everything right down to 6. d!!!

haha the last question, and it's really easy.... wow!

I left in the f(0), f'(0), f''(0)... instead of substituting f(x) for the given question... what a dumbass mistake on my part, oh well... so much for a perfect score. :(

I took the AP Calc BC exam last year through self-study. It took me a whole year, but I got a 5.

Even Calc is self-studiable. After all, Calc isn't really math :-p

Stat in two days was rather difficult though. I missed 5 on the MC of Stat b/c I didn't know a few definitions. I also skipped one FR completely

[quote]I also skipped one FR completely[/quote]

Which one?

And still another one!

http://www.jamesrahn.com/CalculusI/Ap/2004/2004_AP.HTM

Whee.

Yeah I got everything right that I answered. All of 1,2,3,4, and 6. None of 5.

DO you guys think that his answers would warrant full credit? If they do then I wasted a LOT and I mean a LOT of time on the free-response questions then.

Got 1, 2, 3, and 4 completely. Although I did screw up on 5a, and 6c. However, I got the right answer on 5a, although I did some really wierd stuff.

[quote="masamune2387"]Even Calc is self-studiable. After all, Calc isn't really math[/quote]
You have a fascinating outlook on life.


function
induction
algebra
domain
geometry
3D geometry
absolute value

find all the functions
\[ f : N \cup {\{0}\} \longrightarrow N \cup{\{0}\}
\]

\[ f(m^2 \plus{} n^2) \equal{} f(m)^2 \plus{} f(n)^2 ,  \ f(1) > 0
\]

[quote="harpeng"]:D It is clear that there f(x)=x or 1/2
I'm not ready for better solution. :P
 Wait a moment please... :wink:[/quote]

seems to me that $ \frac{1}{2} \not \in \mathbb{N}_0$ :rotfl:

Take $ m \equal{} n \equal{} 0$, thus have $ f(0) \equal{} 2f(0)^{2} \longrightarrow f(0) \equal{} 0$
Take $ m \equal{} 1$, $ n \equal{} 0$, thus have $ f(1) \equal{} f(1)^{2} \plus{} f(0) ^ {2} \longrightarrow f(1) \equal{} f(1) ^ {2} \longrightarrow f(1) \equal{} 1$.

Got no clue in getting f(2)... :(

[quote="KenHungKK"]
Got no clue in getting f(2)... :([/quote]

Take $ m \equal{} n \equal{} 1$, thus have $ f(2) \equal{} 2f(1)^{2} \longrightarrow f(2) \equal{} 2$

The problem is quite famous and not hard.
First of all,we prove that
$ f(n)\equal{}n$,for all $ n\leq 10$.
Then with induction on $ m$,we prove that
$ f(n)\equal{}n$,for all $ 5m< n\leq 5(m\plus{}1)$.
For $ m\equal{}0,1$,it was proved before.
Now if we prove for all $ m\leq k\minus{}1$,we must prove for $ m\equal{}k$.
But it is obviously follows from the following conditions:
$ (5k\plus{}1)^2\plus{}2^2\equal{}(4k\plus{}2)^2\plus{}(3k\minus{}1)^2$,
$ (5k\plus{}2)^2\plus{}1^2\equal{}(4k\plus{}1)^2\plus{}(3k\plus{}2)^2$,
$ (5k\plus{}3)^2\plus{}1^2\equal{}(4k\plus{}3)^2\plus{}(3k\plus{}1)^2$,
$ (5k\plus{}4)^2\plus{}2^2\equal{}(4k\plus{}2)^2\plus{}(3k\plus{}4)^2$,
$ (5k\plus{}5)^2\equal{}(4k\plus{}4)^2\plus{}(3k\plus{}3)^2$

Well, we need to show
for n=0,1,2,3,4 f(n)=n (or not? I'm not sure)

 Anyway, we already got f(0)=0, f(1)=1, f(2)=2.  f(4)=f(2^2+0^2)=4
 f(5)=f(4+1)=f(2)^2+f(1)^2=5
 f(25)=f(5^2+0^2)=25
 f(3^2+4^2)=f(3)^2+f(4)^2
 25=f(3)^2+16
 f(3)^2=9
 f(3)=3 sincef(3)>0

 And we can use mathmatical induction by Erken.

 I think we can extend this stuff like this-> f(m^3+n^3)=f(m)^3+f(n)^3
 Solution may be simmilar to this.

I think it's hard to prove those .  :starwars:
although it will not help but I solved
\[ f : \mathbb{Z} \longrightarrow \mathbb{Z}
\]

\[ f(a^3 \plus{} b^3 \plus{} c^3) \equal{} f(a)^3 \plus{} f(b)^3 \plus{} f(c)^3
\]
it can be proved by induction and using the equation  $ (2k \plus{} 1)^3 \equal{} (2k \minus{} 1)^3 \plus{} (k \plus{} 4)^3 \plus{} (4 \minus{} k)^3 \plus{} ( \minus{} 5)^3 \plus{} ( \minus{} 1)^3$  if $ a$ is even write it for $ \frac a2$,
so we can write $ a^3\plus{}5^3\plus{}1^3$as the sum of three cubes with smaller absolute value ........... :!:

hasn't any body solved $ f(m^3\plus{}n^3)\equal{}f(m)^3\plus{}f(n)^3$  :?:

[quote="KenHungKK"][quote="harpeng"]
 I think we can extend this stuff like this-> $ f(m^{3} \plus{} n^{3}) \equal{} f(m)^{3} \plus{} f(n)^{3}$
 Solution may be simmilar to this.[/quote]

I don't think so,

in your MI initial case proof, you've used $ 3^{2} \plus{} 4^{2} \equal{} 5 ^ {2}$,

but by Fermat's Last Theorem,

If an integer $ n$ is greater than $ 2$, then the equation $ a^{n} \plus{} b^{n} \equal{} c^{n}$ has no solutions in non-zero integers $ a$, $ b$, and $ c$.

Then how to prove the case 3??

But I think the general form of this problem worths consideration, also how about the domain and range is in non-negative real number??

Can somebody give help to solve this:
[list]
$ f(m^{3} \plus{} n^{3}) \equal{} f(m)^{3} \plus{} f(n)^{3}$ over non-negative integer (domain and codomain)
$ f(m^{2} \plus{} n^{2}) \equal{} f(m)^{2} \plus{} f(n)^{2}$ over non-negative real (domain and codomain)
[/list][/quote]

 Well, of course I use a different skills. :huh:


analytic geometry
graphing lines
slope
trigonometry
ratio
calculus
derivative

Find the slope of the tangent line to the polar curve $r = sin{3\theta}$ at $x = \frac{\pi}{4}$

I think you meant $\theta=\pi/4$, not $x$. One approach is to write the curve in parametric form: $x=r\cos\theta =\sin 3\theta\,\cos\theta$, $y=r\sin\theta =\sin 3\theta\,\sin\theta$. The slope is the ratio of derivatives $\frac{dy/d\theta}{dx/d\theta}$ evaluated at $\theta=\pi/4$.

Oops.  :blush: Yeah, I did mean $\theta = \frac{\pi}{4}$


Let $ t_{k} \left(n\right)$ show the sum of $ k^{th}$ power of digits of positive number $ n$. For which $ k$, the condition that $ t_{k} \left(n\right)$ is a multiple of 3 does not imply the condition that $ n$ is a multiple of 3?

$\textbf{(A)}\ 3  \qquad\textbf{(B)}\ 6  \qquad\textbf{(C)}\ 9  \qquad\textbf{(D)}\ 15  \qquad\textbf{(E)}\ \text{None}$

[hide="Solution"]The answer is $ B$.

Observe $ a^3 \equiv a (mod 3)$.
In general $ a^{2k\plus{}1} \equiv a (mod 3)$ where $ k$ is non-negative integer.

$ (a_{n\minus{}1}a_{n\minus{}2}...a_{0}) \equiv a_{n\minus{}1} \plus{} a_{n\minus{}2} \plus{} ... \plus{} a_{0} \equiv  a^{2k\plus{}1}_{n\minus{}1} \plus{} a_{n\minus{}2}^{2k\plus{}1} \plus{} ... \plus{} a_{0}^{2k\plus{}1}  (mod 3)$

The only even number is 6.
Now look at $ t_{6}\left(221\right) \equal{} 129 \equiv 0 (mod 3)$ but $ 221 \equiv 2 (mod 3)$.
[/hide]


Hi all,

I am currently wrinting my master thesis regarding filametous fungi who's names should always be writting in italic. I have chosen to use the standard font for the section (and sub- subsubsection) titles.

To write in italic in selected section titles i have tried to use the sectsty package
[code]\usepackage{sectsty}
\subsubsectionfont{\normalfont\itshape}[/code]

and to \renewcommand{subsubsection}
[code]\makeatletter
\renewcommand{\subsubsection}{\@startsection{subsubsection}{3}{0pt}%
                {-3.25ex plus -1ex minus -.2ex}%
                {1.5ex plus .2ex}%
                {\normalfont\normalsize\itshape}}
 \makeatother[/code]
but both of these "solutions" changes the apperance of the section-number as well.

Does anyone know a way to change the font of only the section title, leaving the section number looking like all the other section titles?

I am, by the waym using the scrartcl document class

Any help is greatly appreciated :)
Kenneth

You need to tell LaTeX that it's only the title that you want to change so you need to bring in the subsubsection parameter as in [code]\let\oldsubsubsection\subsubsection
\renewcommand{\subsubsection}[1]{%
  \oldsubsubsection[#1]{\normalfont\itshape #1}} [/code]You might also want to have a look at the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=titlesec]titlesec package[/url] if you want radical customisation of section commands.

Hi mate,

Thank you very much for the help :)


Sa se dem ca daca $f: \mathbb R\to  \mathbb R$ este o functie strict convexa ,atunci nu se pot gasi patru puncte pe graficul functiei care sa fie varfurile unui paralelogram.

banuiesc ca ai si solutia.mie mi se pare simpla, insa cred ca e vorba de redactare..

Evident consuideri un reper cartezian XOY cu abscisa OX. Se construeiste o perpendiculara d pe OX in punctul unde functia f ia valoarea minima. astfel, graficul functiei este separat in doua parti ca sa zic asa. oricum am plasa 2 puncte in partea dreapta pe functie , dreapta determinata de ele nu poate fi paralela cu perpendiculara d, deoarece functia este STRICT convexa. deci acea dreapta va intersecta pe d. Analog, oricum am alege o dreapta in partea stanga ea va intersecta pe d. deci cele doua drepte se vor intersecta si ele. in concluzie nu putem gasi un paralelogram, deoarece laturile lui trebuie sa fie paralele, ceea ce nu se poate.

Evident nu se poate gasi un paralelogram nici pe graficul unei functii strict concave

Eu am gasit o rezolvare diferita de a ta folosind numere complexe si inegalitatea lui Jensen.Nu am vazut totusi rezolvarea oficiala.

[quote="Zamfirmihai"]Evident consideri un reper cartezian XOY cu abscisa OX. Se construieste o perpendiculara d pe OX in punctul unde functia f ia valoarea minima.[/quote]
Problema nu-i prea grea, dar vezi că ţi-a scăpat un caz (gândeşte-te la $e^{x}$).

probabil ca asta se face pe clasa a XI-a sau a XII-a. Eu sunt doar a X-a. ma refer la cazul care mi-a scapat..


How many ways are there to assign each of 6 friends to either the chemistry class or the biology class if one of these six, Manoj, refuses to be in a class without any of his friends?

[quote=AIME15]Each one has $ 2$ choices, so $ 2^6\equal{}64$.  However, we must subtract two off because there are two cases where Manoj is alone.  Thus, the answer is $ 64\minus{}2\equal{}\boxed{62}$.[/quote]

Could someone explain why it's $2^6$ and not $6^2$ because I'm confused. I thought because each one has 2 choices and because there are $6$ kids, it would be $6^2$. I'm sorry but could someone explain why it's $2^6$ and not $6^2?$ Thank you! :)

[quote=rnatog337]@above The first kid has 2 choices, the second kid has 2 choices, and everybody up to the sixth kid has 2 choices.

Therefore it is $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 2^6$

If it were $6^2$ then there would only be two kids and each kid has 6 choices to go to.[/quote]

Thank you very much! I understand now. :D


integration
trigonometry
logarithms
calculus
calculus computations


In Wikipedia, a coversing space is defined as follows:

[quote]A covering map is a continuous surjective function p from a topological space, C, to a topological space, X, such that each point in X has a neighbourhood evenly covered by p. This means that for each point in X, there is associated an ordered pair, (K, U), where U is a neighborhood of x and where K is a collection of disjoint open sets in C, each of which gets mapped homeomorphically, via p, to U (as shown in the image). In particular, this means that every covering map is necessarily a local homeomorphism. Under this definition, C is called the covering space of X.[/quote]

Now define a map from the real line to a unit circle.

[quote]p: R -> S^1 with p(t) = (cos(t), sin(t))[/quote]

Obviously R is a covering space of S^1. My question is: Isn't the interval (-2*pi, 2*pi) also a covering space of S^1? (Any open arc in S^1 has a preimage of 1 or 2 open intervals in (-2*pi, 2*pi), which are mapped back to open arcs on the circle...) Why Wikipedia says:

[quote]In fact, S^1 and R are the only connected covering spaces of the circle[/quote]

Thanks

Thanks, man. I see the point.

[quote="FibonacciFan"]The most natural covering map $ p: ( \minus{} 2\pi, 2\pi) \to S^1$ would be $ p(x) \equal{} e^{ix}$. But consider $ p^{ \minus{} 1}$ of any neighborhood of the point $ (1,0)$. A neighborhood of the form $ U \equal{} \{\cos \theta \plus{} i \sin \theta | \theta \in (\epsilon,\epsilon)\}$ would have an inverse image of

$ p^{ \minus{} 1}(U) \equal{} ( \minus{} 2\pi, \minus{} 2 \pi \plus{} \epsilon) \cup (\epsilon, \epsilon) \cup (2 \pi \minus{} \epsilon, 2\pi)$

The middle interval maps homeomorphically onto $ U$, but the outer neighborhoods don't since they don't even map onto $ U$. For example $ p(( \minus{} 2\pi, \minus{} 2 \pi \plus{} \epsilon)) \equal{} \{\cos \theta \plus{} i \sin \theta | \theta \in (0,\epsilon)\} \neq U$. If you try to combine the two outer intervals, that won't work either.

If you think about, there would be similar problems any open interval in the real line that has non-infinite endpoints.[/quote]

At first, I thought "mapped homeomorphically onto $ U$" means that "mapped onto something that is homeomorphic to $ U$". Now I understand what it really means is "mapped onto $ U$ by a homeomorphism".


That's a problem from short-list 1995. The key idea is:
$\frac{a_i^2+a_{i+1}^2-a_{i+2}^2}{a_{i}+a_{i+1}-a_{i+2}} \leq{a_i+a_{i+1}-2}$ and the rest is obvious


Have you seen these movies? I for one think that The Prestige was a little more science fiction than The Illusionist but you never know.
Do you know any other good movies which involve magicians?

I haven't watched The Illusionist. I think the plot of The Prestige is really interesting (i like the idea of killing yourself every night). But it is impossible to duplicate an object in those days (according to Einstein's formula, 4493775 Tera-Joule is needed to create a 50-kg man). If Tesla can, why dont he just duplicate banknotes or loads of AC generators to bury Edison alive? :D  :D  :D

That cloning device seems totally out of reach for that period.
The Illusionist is worth seeing and I think it is much more interesting.

[quote="modeler"]If Tesla can, why dont he just duplicate banknotes or loads of AC generators to bury Edison alive? :D  :D  :D[/quote]

I don't think Tesla realized that his machine duplicated things, whats-his-face just realized that accidentally when he stumbled across all of the top hats.

Geez, haven't you guys ever heard of spoiler alerts?

<SPOILER ALERT>

Luckily for me, I've already seen both movies, and I can say without a doubt that The Prestige is a much better film.  Better story, better twists, better characters, just better in every way.  I don't really care whether it's realistic, since it's a movie and all.

The Illusionist bothered me in that it the illusions shown were completely ridiculous for the day's technology.  Seemed like a copout.  (This is not an issue of realism.  It's an issue of authenticity.)  But more importantly, I don't understand why we should be happy that the "evil" guy dies despite not having killed anyone.  I also guessed the ending but I was hoping that it wouldn't end that way since it seemed like a terrible ending.  (In contrast, I loved the ending of The Prestige.)  Even brushing those criticisms aside, it's just not as good as The Prestige.  And the whole "rich aristocrat falls for poor boy, and then plot to run away together" plot?  If you're going to have a cliched plot, you need to execute it very well or very originally.  Anyway, it's not that bad of a movie, just much worse tha Prestige.


MATHCOUNTS geometry AMC USA(J)MO USAMO Euler

Even though the team round at state doesn't really affect anything, I was wondering how people did.

what? there are still 3 weeks until states

Funtwo, maybe for you there is. The states can hold their competitions anytime in the month of March.

[quote="gauss1181"]My school team didn't make it to states, so I went as an individual.[/quote]

Same here(except change the verb "went" to future tense.) I wonder for how many AoPSers this is true?

[quote="AIME15"]Not in all states :)

Hm, I posted EPIC FAIL because that's what our team will get.  Harker, Miller, (insert some other school here) will beat us, so we won't place in top 3.[/quote]

AIME15: Harker didn't make State this year. Neither did Redwood. :O

OMG

Wow, we could place top 3 this year.

Did hopkins qualify?

Our discussion should be moved to bay area forum :)

if we have a good day,we may actually do very well at states

In Kansas, our team got first.  Once they said, "2nd place goes to Lakewood", our team did yelled yes so loud everyone was looking at us cause we knew we got first.

Our team got 1st in Oklahoma but I failed written...
WAAAAY tooooo much pressure!

Edit: Pressure as in "last year of mathcounts, must do good or i'll beat myself"

haha, OK, pressure, think about being in CA, TX, MA, or VA...

"OMG

Wow, we could place top 3 this year.

Did hopkins qualify?

Our discussion should be moved to bay area forum Smile."

Uh, yeah..
First was Miller (duh).
Second was Hopkins.
Third was Kennedy.
Fourth was us (Challenger Sunnyvale).

PM me if you want specific details (eg people).

[quote="AIME15"]Not in all states :)

Hm, I posted EPIC FAIL because that's what our team will get.  Harker, Miller, (insert some other school here) will beat us, so we won't place in top 3.[/quote]

Harker, not so much... they aren't even in States!

Miller, yes. (insert some other school here) is probably Hopkins  :D
That is, if some of the team members (cough) decide to actually come to the meetings.

[quote="AIME15"]I think we have a semi-decent chance to beat you guys...

(semi-decent being about 15%)

Does pythag011 have a team for mathcounts?[/quote]

Of course not! My school doesn't even want me to tell anybody about mc because they are afraid that someone will want to start a math club... so I'm participating as an individual

And no, please do not put this is bay area forum. Bay area forum is not about mc, its about olympiad math... fine you can put results in but nothing else.

Do not expect me to win unless all the probs are usamo probs...

Heh, it would be nice if I made nats... then I could tell everyone to do usamo instead of mc at nats.

[quote="AIME15"]Not in all states :)

Hm, I posted EPIC FAIL because that's what our team will get.  Harker, Miller, (windermere) will beat us, so we won't place in top 3.[/quote]

Folks, I've got to emphasize something:

Harker DID NOT make states; in fact NONE of their team members qualified.

They placed seventh (tied) in the chapter (five teams go), and their best team member (who missed one, by the way) was eleventh - tenth place was the cutoff.

And really, my team quite fails, as it seems as if random people keep disappearing at spontaneous moments during the team meeting.

hmm... how do you people know that Miller/Hopkins are good teams?
never heard of them
AMIE15:
yep our team will epic fail
we should do better next year though. Almost all of the people in our team are in 7th grade.

[quote="theprodigy"]our team failed :oops:
jk i was the only one from my school
team round was EASY[/quote]

Are you talking about state or chapter?

[quote="AIME15"][quote="jerry125"]hmm... how do you people know that Miller/Hopkins are good teams? [/quote]

Because I know people that you don't.  Miller has Eulers_Apprentice, miller4math, EcstaticPotter, and darksigma.  They will win states by about...10 points? :D

[quote="jerry125"]we should do better next year though. Almost all of the people in our team are in 7th grade.[/quote]

However, nobody on our team is very good :(

Also, spell my username correctly :D[/quote]

Well, at least you can beat my team. (Which doesn't exist.)

I'm desperately hoping for either:

i) A ridiculously hard test.
ii) A ridiculously easy test and IMO#2-level tiebrakers (also, the tiebrakers must be combinatorics or nt.)

My team is a complete n00b. Heh yeah. The only times when we really beasted was when we had full teams when I was in 6th and 8th grade. 7th and 6th at chapter for team...epic fail. In fact, my school never got anything above that. :(

In addition to that, only two kids in the history of my school's MATHCOUNTS participation, me and some other kid who graduated in 2007, made it to chapter countdown, which takes top 10 participants. That is epic fail. Plus, that kid and I were the only kids to make it to states, and I was the only one (and the first one) to qualify for Nats.

But there's also one more n00bish thing...there were only three years in my school's MATHCOUNTS history that lasted 20+ years, where we had full four-person teams. :mad:

i wasn't on my schools team, cuz its beast, but we got 1st in states and chapter. All 4 individuals from our school(including me :D) made it to states. then, i epic failed the written round. nervousness???

[quote="gauss1181"]My team is a complete n00b. Heh yeah. The only times when we really beasted was when we had full teams when I was in 6th and 8th grade. 7th and 6th at chapter for team...epic fail. In fact, my school never got anything above that. :([/quote]

Sounds like you need to change schools....

My school has been pretty good but not any more. We won states for 5 years in a row, sent 3 or 4 people to national CD, and this is in Michigan. In fact, all people from Michigan to make National CD are from my school... I think.

My school placed 1st at state...
I wasn't on the school team but I made nats :D
The school team members placed 1,2,3,5
The individuals got 4 (me), 7, and the other didn't place in the top ten, but I think he got in the top 30...
My school pwns :P  :P  :P

geez, I wish I was in a school like yours...
It's so stupid that of the entire history of my school's MATHCOUNTS participation, only two kids (including me) placed in top 10 at chapter.

Keep in mind that this is Missouri, so results may be somewhat...inflated.

im actually glad my school doesn't care about math. that means i dont have to go to hours of practice so i get more free time

Same as funtwo. I plan on going to a bad high school for that reason...

And I just realized that Bay Area mc date is different from SoCal mc date.

[quote="the.bookblogger"]hm.
our team (we're in pa) placed fourth (fail) at states, even though individual scores were 19, 25, 26(me), and someone not in the top 30.
woow.[/quote]

I am on that team.  4th at PA isn't very bad, considering the competition we had.  Although, if we did a little better, we could've gotten a trophy  :P .  And, to clarify, those aren't our scores, but our placing.  Also, I apparently got moved from 19th at state to 20th, but I don't know why...

I'm on the Puerto Rican team. My school got 1st and I got 1st as individual W000!

Darn...... Our school has never let go of either 1st or 2nd in state, but unfortunately, we came in kind of short this year. :(   But I'm pretty sure we're going to have a pretty solid team next year, though.  :D

Our team got 2nd in both chapter and state.  :wink: Our school is so bad... it placed 7th last year in CHAPTERS. I'm sure it wasn't even in the top 10 before that, and this was the first time we even made it to states.

Did the same team beat you? (at chapter/state)

Yes, it's a magnet school, and we are regular...public... They are like best in NC, and we are worst (or 2nd to worst). 1st place for us is impossible.

Ohh...NC :D

Ligon Middle School is the 1st place team?

Yes... it got 1st last year too, and twice or more before that...


[quote="perfect628"]Define SoD(x) to mean the sum of the digits of x. Evaluate SoD(SoD(SoD(4444^4444)))[/quote]
[hide]
$SoD(x)=$(sumation, with [[log10x]] at top and n=0 at bottom)$[[\frac{x mod 10^{n+1}}{10^{n}}]]$

therefore the answer to the problem is $3$, or the SoD of the answer is $3$, or the SoD of the SoD of the answer is $3$, or if SoD'ed enough until it is one digit long, the answer will come out to be $3$

edit--nvm...well testing up to the 7th power, it never gets to 3, but the SoD of the SoD's until its a single digit is always odd...
edit #2--well...the 8th power came out to an even sod: 1+5+2+1+2+2+1+3+1+5+4+7+2+7+9+6+2+3+7+2+6+2+8+0+2+7+8+0+1+6=112, 1+1+2=4
[/hide]

Is this mathcounts level?...because i certainly don't get it :huh:

[quote="perfect628"]Define SoD(x) to mean the sum of the digits of x. Evaluate SoD(SoD(SoD(4444^4444)))[/quote]

Is there supposed to be a trick to this or something?

The only way I am seeing it is brute force which wouldn't be nice...

This is like, a problem from the how to write proofs (I think)

Does anyone know the source please?
If this was a MathCounts Practice, I would love to see it, even though I am done with it.

[hide]This has to do with $\equiv 9$.

Notice that number of digits in $4444^{4444}$ is at most 4*4444=17776, so this means that the digit sum is at most 17776*9=159984.

Now, notice that the maximum sum of this occurs at 149999, which has a sum of 31.

Now, notice that this has a sum of 4.

So the operation is always a single-digit number, from this case, even with an upper bound.

And also, when you keep on summing the digits until it reaches a single-digit number, you see that this number is the remainder when divided by 9.

So, the remainder when divided by 9 of $4444^{4444}$, with some patterns is

7, 4, 1, 7, 4, 1, etc.

So the answer is 7, I believe.

But I don't know

$7$[/hide]

[hide="hint"]A number mod 9 is equal to the sum of its digits.[/hide]
[hide="a hint that expands on the previous one"][/hide]
[hide]$4444^{4444}$ mod 9 is
$7^{4444}$ mod 9.
$7^{6}=1$ mod 9
$7^{4}$ mod 9
$7$
[/hide]
EDIT: Neil beat me to it by like one minute

[quote="mathgeniuse^ln(x)"][quote="bpms"][hide="hint"]A number mod 9 is equal to the sum of its digits.[/hide]
[hide="a hint that expands on the previous one"][/hide]
[hide]$4444^{4444}$ mod 9 is
$7^{4444}$ mod 9.
$7^{6}=1$ mod 9
$7^{4}$ mod 9
$7$
[/hide][/quote]

You have to show that indeed the sum will end up being a single digit number.

For example, how do we know that the sum of the sum of the sum of the digits can't be 16, or 25, etc.[/quote]
You proved that, so I'm fine :D. I could just copy and paste your solution :P

Why is the number of digits in $4444^{4444}$ at most 17776?

[quote="Ubemaya"]Why is the number of digits in $4444^{4}444$ at most 17776?[/quote]

Because

$10000^{4}>4444^{4}$, I am doing this without a calculator.

So this means that the number of digits is less than 17777, and since it is less than that it has at most 17776.

[quote="Ubemaya"]Doesn't $10000^{4}$ have 17 digits?[/quote]

I meant $10000^{4444}$ and since it has 17777 digits, then $4444^{4444}$ has at most 17776 digits.

And likewise  I meant $4444^{4444}$ not $4444^{4}$

[quote="mathgeniuse^ln(x)"]Does anyone know the source please?[/quote] http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWriteWrite.php

Actually, I got this somewhere else, but w/e.

What was the source from where you got it?

A book i can't remember the name of...i only remembered the problem. i'll look for the title.


A person has 15 dollars. She has twice as many dimes as fifty-cent pieces, and the same number of quarters as dimes. She has twice as many nickels as dimes and a dollar's worth of pennies. how many of each coin does

The question is ambiguous, since it doesn't state a limit on pennies. The way I understand it though:

Since she has a dollar's worth of pennies, the question simplifies to $ 14$ dollars in quarters, dimes, and nickels, with the conditions stated. We have:

$ 4N\equal{}2D\equal{}2Q\equal{}H$, where N=number of nickels, D=number of dimes, Q=number of quarters, and H=number of 50-cent coins (half dollars).

Since there are 10 cents in a dime, 5 cents in a nickel, etc, this simplifies to $ 20x\plus{}20x\plus{}50x\plus{}50x\equal{}1400$, where x is the number of half-dollars. This becomes $ 140x\equal{}1400$, or $ x\equal{}10$. Therefore, there are 10 half-dollars (5 dollars), 20 quarters (5 dollars), 20 dimes (2 dollars), and 40 nickels (2 dollars), in addition to the 100 pennies mentioned in the beginning of the problem.

[b][size=150][color=darkblue]Cookie has 97 coins in her collection-all either nickels, dimes , quarters. If 77 of her coins are quarters and dimes and the total value of her coins is $ \$$13.95, how many quarters does she have[/color]?[/size][/b]

Since out of 97 coins, 77 are quarters or dimes, there must be 20 nickels. Thus, there are $ 13.95\minus{}5 \cdot 20\equal{}12.95$ dollars left.

Now, we have $ q\plus{}d\equal{}77$ and $ 25q\plus{}10d\equal{}1295 \implies 5q\plus{}2d\equal{}259$.

We substitute $ d\equal{}77\minus{}q$ into $ 5q\plus{}2d\equal{}259$ to get $ 5q\plus{}2(77\minus{}q)\equal{}259$, so $ 3q\plus{}154\equal{}259$. Solving, we have $ 3q\equal{}105$, so $ q\equal{}\boxed{35}$.

[b][size=200]WOW I CANT BELIEVE IM THAT DUMB. O YEAH, CAUZ ITS QUARTERS AND DIMES!!!! OMG. THANK U![/[/size]b]

[quote="Nlakkamraju"]Don't Post homework on AOPS!
Do it yourself!
I got the same problems for Homework![/quote]

Nlakkamraju You make spam here :spam:
I like the interesting problems from Homework , from books , from contests. Here is the good  collection of problems with  different explanations  or with  links to nice site.
The knowledge how to solve problems is much  more then the grades or the class ranks.

I don't Spam!

That person was posting homework on the Forum so they wouldn't have to do it at all. They could just copy other people's answers.

:mad:

I wasn't spamming, just pointing out the obvious( to me).

You are NOT supposed to post homework on the forum.

[b]wei shen me? wo zhi dao! wo shi yi ge bun dan. Ni bu ke yi gou shu wo, ni zhi dao ma?  :mad:  and is ilovepink really going to be at the competiton?!!! WOW! cant wait to see her beat u. Wats her name? I need to meet her!!![/b]


Suppose

\[ n(n\plus{}1)a_{n\plus{}1} \equal{} n(n\minus{}1)a_n \minus{} (n\minus{}2)a_{n\minus{}1}\]
for every positive integer $ n \geq 1$.

Given that $ a_0 \equal{} 1, a_1 \equal{} 2,$ find

\[ \frac{a_0}{a_1} \plus{} \frac{a_1}{a_2} \plus{} \frac{a_2}{a_3} \plus{} \cdots \plus{} \frac{a_{50}}{a_{51}}\]

I think you could prove by induction (or by finding the pattern  :wink:  ) that for every

positive integer n $ \geq 2$,  $ a_{n} \equal{} \frac {1}{n!}$

So the sum will be $ \frac {1}{2} \plus{} 4 \plus{} ( 3 \plus{} 4 \plus{} \cdots \plus{} 51 )$

$ \implies$ S = $ \frac {1}{2} \plus{} 4 \plus{} \frac {49}{2} \cdot (2 \cdot 3 \plus{} 48 \cdot 1)$

S = 4.5 + 1323 = $ \boxed{1327.5}$

 :)


geometry
circumcircle
inequalities
email
analytic geometry
induction
trigonometry

For the past 3-4 months or so we have been discussing relatively boring things, nearly all of them were math contests.*Boring*.

     So, now that all the contests are over, we need a break, I think. This is called math battle, and it is a 2-team math competition that is really fun.

     Here are the rules. We are going to have 2 teams of 6-10 people each. After everyone confirms, I will post 8 problems. You will have 5 days to solve them(by real rules, it's 2 hours, but here, it's different). So, you need to make sure that you will be participating in this seriously. Each team must select a captain, and you need to tell me who is the captain before I post the problems.

     During the 5 days, you can pm each other about the solutions and this way discuss them.

     At the end of the 5 days,e each captain must pm me the solutions that his/her team came up with. The captains don't have ot write up the whole solution, but need to give me an outline of their solution or something like that.

     After I receive the solutions from both captains, you will begin presenting them by posting the solutions in this topic. AND you can't change your solutions once the captains sent the solutions to me.
     REMEMBER: You need to clearly explain your solutions.

     In the end of 5 days, I will randomly choose one team. Say it is team 1. The captain from team 1 will post if his team will present a solution or it will be the other team to present a solution. The captain of one team announces what problem the other team must present(the decision should be not just the captain's but the whole team must participate in the decision). The teams will alternate by deciding what solutions the other team will present.

     If a team needs to present a problem, one of the team members must post the solution of that problem. If they don't have a solutoin, they must post "Pass" and the other team will present the solution of that problem. After a solution is posted, one of the team members from other team must post questions regarding the solution. No more than 3 questions in total; one question at a time. After the "questioning person" asks all the questions, and the presenter answers them, I can ask the presenter some questions, too. Every team member can be the questionning person or the presenter at most 2 times.

     Now, the scoring. If a solution is completely correct and the presenter answers all the questions correctly, his team gets 12 pts. If the solution has errrors, the presenter loses some of the 12 pts. If the questioning person spots an error in the solution and posts what it is, his team gains points. If a team decides to not present a solution and to not "Pass", the other team automatically gets 6 pts. A team cannot pass after another team already "passed" the solution to the first team.

     After all problems are presented, the team with more pts wins. I will also announce the MVP.

     Now, everybody can sign up and ask questions. I would also like one more person to be a ref.

I think you guys start reading the rules and leave thinking this is boring.

ANYBODY CAN PLAY SO FEEL FREE TO SIGN UP. WE NEED PEOPLE FOR THIS!!!!!!!!!!!!!!!!!!!!!! :!:

sounds funny.
i'll sign up.
(i wish i don't mess everything up for my team :blush: )

[quote="soapbubble"]i'm in..haha..soundz fun...been doing too much eng/hist proj these dayz..need sum math...XDXD...and..just a thought..can we do a girl/guy mix team? for example, say each team must have a certain number of each gender at least...[/quote]
yes, we'll do that. but this depends on how many ppl sign up.
PLEASE SIGN UP FASTER. THIS THING WILL TAKE A LONG TIME AND  IT SHOULD BE FINISHED BY JUNE 2006 :!:

I was too lazy to read the rules.  :read:   Whatever this thing is, I'll just sign up. :D

Sign me up too, although I might not be that good but I might learn something from it. :P

I'll play.
Does MVP stand for Most Valuable Player?
And when are we to confirm?

We need 5 more pepople. If you sign up you automatically confirm

I posted here yesterday, and it disappeared!! :o
Yufei, did you delete it?  :?

[quote="rem"]In my opinion, at least 4 of the problems have very nice interesting solutions. If you solve the geometry problems with coordinate geometry and ineq with a Muirhead bash and word problems with looking at all possible cases, the problems might seem boring. I consider all these problems really nice. Also very interesting problems either take a month to be solved or are ridiculously easy.so stop whining. we'll see how interesting your problems are in math battle 2.[/quote]
Most of the problems are really nice, I must agree.

Here is my solution to #6.

1)
Showing one example is sufficient.
Suppose that there are three elves, who knows each other. On day 1, one of them gets sick and one gets immunity.
 :stretcher:  :whistling:  :(
Suppose that immunity only lasts for one day for everyone. On day 2, we get this:
 :whistling:    :(   :stretcher:

The elf that was sick now has immunity. The elf with immunity now lost his/her immunity. The elf that was normal is now sick. This is the exactly same situation as the day 1. Therefore, this epidermic lasts forever.

2)
Let's start by defining some groups.
Let $G_0$ be the set of elves that got sick on day 1. Let $G_1$ be the set of elves that do not belong in $G_0$ and has a friend in $G_0$. In a similar way, we define $G_n$ be the set of elves that do no belong in any $G_m$(m<n) and has a friend in $G_{n-1}$(n is a natural number). Everyone belongs in exactly one $G_m$(m is any non-negative integer), even though there could be some elves that are not friend with anyone in $G_m$(m is any non-negative integer), but we do not need to consider them in this problem at all. (basically this whole thing is like the Erdos Number)

First, note that two elves are friends to each other only if the index of the group they belong in is differ by 1 or 0.  Finding two elves that are friends to each other yet have the index of the group differ by at least 2 gives us contradiction.

Consider day 1. Only $G_0$ is sick and no-one has immunity.
Consider day 2. Only $G_1$ is sick and only $G_0$ has immunity.
Suppose that on day n, only $G_{n-1}$ is sick, $G_{n-2}$ has immunity and no $G_m$($m \geq n$) has immunity.(n is at least 2)
Then, in day n+1,
Only $G_n$ is sick (because the flu can only spread to $G_{n-2}$ and $G_{n}$. $G_{n-2}$ has immunity. $G_n$ has no immunity in day n and all elves in this set has a friend in $G_{n-1}$.)
$G_{n-1}$ has immunity (because $G_{n-1}$ was sick in day n)
no $G_m$($m \geq n+1$) has immunity (because they were never sick).
By induction, $G_{n-1}$ is sick, $G_{n-2}$ has immunity and no $G_m$($m \geq n$) has immunity on day n.

Suppose that $G_a$ is an empty set(such set exists unless there are infinitely many elves :D ). On day a, $G_{a-1}$ is sick, $G_{a-2}$ has immunity. The next day, the epidemic ends(because the flu can only spread to $G_{a-2}$ and $G_a$. $G_{a-2}$ has immunity. $G_a$ has no immunity but it is empty.) QED

Ok. Great job. Now team 2 asks questions about the solution. You can [b]post the questions within the next 48 hrs[/b]. Please don't ask stupid questions like "Why does $2 + 2 = 4$?". Questions must realate only to the solution. Obviously if you find any errors post here about them.

No questions? or is team 2 not participating? I'll give team2 24 more hrs then I will decide on how many pts team 1 will get. REMEMBER: If you ask good questions and find an error your team will get points. ;)

[quote="rem"]OK. no questions.
Great job lightrhee :omighty: . The solution is completely correct with no errors.
So, [b]team 1 gets 12 pts[/b].
Now team 1 must post the solution to which problem team 2 will present. Please reply asap.[/quote]
i dont get this.
can you verify the members of the two teams?
i thought I was on the opposite team as lightrhee.

It's a messed up game that's what it is.

Should we continue then? It seems that not many people from my team are participating. And I haven't been here for a while either. (BTW, I'm on Rhee's team.)

Congrats to Rhee for winning CMO.

Edit: Woops. My bad. Sorry for the inaccuarate info.

[quote="otaku"][quote="lightrhee"][quote="startogalaxy"]Congrats to Rhee for winning Eulcid and CMO.[/quote]
 :huh: I didn't win Euclid this year. I'm in group 2.[/quote]

clearly he said Eulcid  :rotfl:
jk[/quote]
 :blush:  Oops
____________________
EDITED: Oh. That wasn't me. :D

Just to add one more post to my post count. The game is dead.

[quote="M4RI0"]Could anyone post solutions to problems number 2,5 and 7 :?:[/quote]

#2
Consider a right triangle ABC, so that AB=5, BC=4, and CA=3. Put a point P inside ABC, so that angle APB=150, angle BPC=120 and angle CPA=90. Let $\frac{y}{\sqrt{3}}$,x, and z be PA, PB, and PC. Then this x, y and z satisfy the conditions.
To find the value, use sine law to calculate triangle APB, BPC, and CPA.

#5
I think I had a solution after this game 'died'. However, I completely forgot how it went.

#7
Move some terms and we get \[(\frac1{2a}+\frac1{2b})(\frac1{2b}+\frac1{2c})(\frac1{2c}+\frac1{2d})(\frac1{2d}+\frac1{2a})\geq(\frac1{ab}+\frac1{cd})^{2}(\frac1{bc}+\frac1{da})^{2}\] , \[\ln (\frac1{2a}+\frac1{2b})+\ln (\frac1{2b}+\frac1{2c})+\ln (\frac1{2c}+\frac1{2d})+\ln (\frac1{2d}+\frac1{2a})\geq 2\ln (\frac1{ab}+\frac1{cd})+2\ln(\frac1{bc}+\frac1{da})\] .

Let $f(x,y)=\ln (\frac1x+\frac1y)$. f(x,y) is convex(I think), so using two-variable Jensen, we are done.
I have seen a better solution, but I don't remember how it's done.


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Let $A \in M_{n}(\mathbf C)$ and $C(A)$ be a subspace of $M_{n}(\mathbf C)$ consisting of matrices commuting with $A$. Prove that $\dim C(A) \ge n$.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=73310[/url] states that if $AB=BA$ then $B$ is a polynomial of $A$ and since $A\in M_{n}(C)$ there are $a_{1},a_{2},...,a_{n}$ such that $A^{n}+a_{n}A^{n-1}+...+a_{1}=O_{n}$; from here the conclusion is visible.

That problem has an extra hypothesis! We have to consider the other case, too.

We may assume that $A=diag(\lambda_{1}{I}+N_{1},\cdots,\lambda_{k}{I}+N_{k})$ where $N_{i}$ is a $(n_{i},n_{i})$ Jordan nilpotent matrix ($N_{i}^{n_{i}-1}\not=0$). $\mathcal{C}(N_{i})=\mathbb{C}[N_{i}]$ is of dimension $n_{i}$. Thus $dim(\mathcal{C}(A))\ge n_{1}+\cdots+n_{k}=n$.
  More generally if $A$ is a $(n,n)$ nilpotent matrix ($A^{k}=0$) and if $dim(ker(A^{i}))=\alpha_{i}$, $i=1,\cdots,k$,  then $dim(\mathcal{C}(A))=\alpha_{1}^{2}+(\alpha_{2}-\alpha_{1})^{2}+\cdots+(\alpha_{k}-\alpha_{k-1})^{2}$.
Conclusion: if $A\in\mathcal{M}_{n}(\mathbb{C})$ then $\exists{u_{1}},\cdots,u_{k}$ s.t. $u_{1}+\cdots+u_{k}=n$ and $dim(\mathcal{C}(A))=u_{1}^{2}+\cdots+u_{k}^{2}$.

The last post invoked Jordan form- here's an arbitrary-field version.

Given $A\in M_{n}(F)$, decompose $A$ into rational canonical form: $A=S^{-1}XS$, $X=\begin{bmatrix}B_{1}&0&\cdots&0\\ 0&B_{2}&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&B_{r}\end{bmatrix}$, with each $B_{j}$ a companion matrix.
Matrices of the form $S^{-1}\begin{bmatrix}p_{1}(B_{1})&0&\cdots&0\\ 0&p_{2}(B_{2})&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&p_{r}(B_{r})\end{bmatrix}S$ (each $p_{j}$ is a polynomial) commute with $A$. If $B_{j}$ is $m_{j}\times m_{j}$, the space of matrices of this form has dimension $m_{1}+m_{2}+\cdots+m_{r}=n$.

There may be other matrices that commute with $A$; we get these if the characteristic polynomials of the $B_{j}$ have nontrivial common factors, which makes it possible to place nonzero off-diagonal blocks.


Oh, sorry. Looks like ProtestanT already did this for the smaller numbers. (But he got slightly different numbers. Hmm.)

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1125948#1125948[/url]

ya pretty graph ftw :)

When I did mine I might have undercounted by 1 in some places, and I think you might also have overcounted by 1 in some places. I think it's the 2nd choice (you overcounted by 1 in some places), because I don't remember undercounting by 1. Let's check it again :)

I'm an island! No one else scored my score or the scores either one more or one less.

Still, that's kind of weird. Did one of my solutions give me a 4, or did I manage to get four sixes?

Interesting how the peaks this year aren't the same as last years at all.

[quote="Math Geek"]Wow, so many failures.  Now I don't feel as bad for getting a 2.  I still feel pretty bad, though. :ninja:[/quote]

Hey, I actually beat someone with my 3...

[quote="apple pi"]
27 -   3 - ***
28 -   3 - ***
[/quote]

I count four 27s and two 28s.  9/4 and 7/3 of my score, respectively.

That's because the PDF has apparently been updated with one 28 being changed to a 27.

I got the "rarest" score below 17 :P

Means nothing though, except that I got a weird amount of credit on a problem, probably #1.


$a \equiv b (mod c )$. I'm sure that there is something like $a\equiv bw (mod cw)$, $w \in N$. can somebody help me to find real formula.

not that, i;m not so dump. Maybe not $a \equiv f(b) \mod cw$

Well, there will be an $d$ with $a \equiv b+cd \mod cw$, but that $d$ will depend on $a,b,c$.

ok that is what i need. How $d$ depends?

You have $a=b+cd$ for some $d$, so you can even get equality as integers.
Formula: $d=\frac{a-b}{c}$