geometry
3D geometry
sphere
combinatorics unsolved
combinatorics

Given n non-parallel lines, with more than one intersection point, and each colored with one of 2 colors, prove that there's an intersection point that all the lines through it are colored the same way.

Is it just me, or did someone answer the question and then deleted his post? anyway, I looked in "Proofs from the book" (a very recommended reading), and I found a useful trick that (I think) solves the problem-moving the graph to a sphere and using a lemma about existence of a vertex with 2 or 0 color-changes of its edges (in a planar graph whose edges are colored in 2 colors).

[hide]
Consider the graph G whose vertices are the intersection points and the edges are the segments created by the lines. It's a planar graph. Also, all the lines through a vertex V are the same color iff all the edges connected to V are the same color - because if a line L through V doesn't have a vertice (other than V) connected to V, it implies that all the lines intersect L at V, contradiction to the fact that there's no one intersection point.

Now, move the graph to a sphere: connect each point (let's say V), with a sphere's center (call it O), and take the intersection of OV with the sphere. we will get many half circles - and we will complete them to whole circles.
Each intersection point in the new graph corresponds to an intersection point in the original graph G. The new graph, G', is planar, and simple (else, if a vertex is connected twice to himself, it means that all the lines in G pass through 1 point).

Now, we will use the following lemma: In a two-colored planar graph, there's a vertex of G with at most 2 color-changes in the cyclic order of the edges around it.
It means that there's a vertex V in G' with at most 2 color changes. It can't be 2, else all the edges of V are the same color except from one - but this one must have a "twin". So it is 0 (because it's even), and we will just take the corresponding vertex on G.
[/hide]
Still looking for another solution....

Yes, someone proposed a solution, but it was wrong. I wrote a response to that post but by the time I hit "submit" the guy retracted his solution, so I deleted my post as well.

Your solution relying on Euler's theorem and the lemma in PftB is really nice. I doubt that there is a much simpler one. I admit that initially I thought this was an easy problem, but now I don't think it really is.

Actually, I am beginning to doubt my solution - the part of completing the circles (even if it's true, it's not nice to explain). Anyway, there's a much easier solution (found by a friend) - connect all the rays in the graph that begin on a vertex and have no other vertices on them, to a new vertex, call it V'. Now we can apply the lemma (this point replaces the sphere trick) and find a vertex with 0 or 2 color changes (it most be 0, because of the new vertex), and it can't be V' (actually, if it's V', then all the lines are the same color and the problem becomes trivial).


1. Find sum of all integers attained by permuting the digits in 12345.
2. Find sum of digits of 999999999*123456789123456789123456789.
3. Each of 3 coins has been painted on its sides. Coin 1 has red on 1 side and blue on the other. Coin 2 has blue and green. Coin 3 has green and red. What's the probability that tossing 3 coins in the air, all 3 colors will appear?

Ok:

1. [hide] There are 5! possible permutations, so each digit appears 4! times in each column. So in the ones' column the sum would be (1+2+3+4+5)*4! or 180. Then for each of the other five columns would be similarly deduced, so the sums of all numbers is 180+10*180+100*180+1000*180+10,000*180= [b]1999980[/b][/hide]

3.  [hide]there is a 1/2 chance of Coin1 being red, then 1/2 C2 being blue and 1/2 C3 being green. same for if c1 isblue. in all is 1/2*1/2*1/2 times 2 (since there are only two possible ways they end up different) = [b]1/4 or .25 chance[/b][/hide]

It's nice to see some good ole math problems around here!

[quote="macaroni_28"](1+2+3+4+5)*4! or 180.[/quote]
 :huh:
[hide="hint for a quick solution"]Notice that the numbers can be paired up so that the sum of each pair is constant.[/hide]


After seeing some of the problems in Kleppner/Kolenkow's Intro to Mechanics, I kind of wished someone had told me about this book in high school. I think two of the four problems (#3 and #4) from last year's quarterfinalist test were taken directly from this book. The book also contains numerous problems that require creative thinking -- the kind that you don't normally see in a Physics C text.

Have books to recommend? Post them here!

Does anyone know of or have former tests as well? :P

former papers r available in mtg books (but difficult to find though :P )


complex analysis
function
topology
complex analysis unsolved

Assume that $ f,g$ are holomorphic functions on an open set $ U$ and $ \overline{D} (0,1)\subset U$. Prove that there exists $ z_0$ such that $ |z_0|\equal{}1$ and $ |f(z_0)|\plus{}|g(z_0)|\equal{}\text{Max}\{|f(z)|\plus{}|g(z)|: z\in \overline{D}\}$.


algebra function domain geometry LaTeX inequalities trigonometry

As I was listening to one of my classmate re-telling the plot of a story, I've got an ideea.  Maybe the class was so boring and I had a moment of brilliance :D. If I'm alowed, I would like to "create" a daily contest of problems. My ideea is:

[i][b]Problem of the day[/b][/i]
During the week (the first 5 days) I will post a different problem with a variat difficult degree. The difficulty degree ot the problems increases day bay day so Monday will be the easy one and Friday the harder. Also, in this time you can send me one of your problems so I can select one for beeing The problem of the [b][i]week[/i][/b](end).

[i][b]Problem of the week[/b][/i]
Like I said earlier, this problem (might be two problems) will be selected from the problems that I recive.

[i][b]Problem of the month[/b][/i]
At the end of the month (in the last weekend maybe), it will be a set of 3-4 problems from the main domains of mathematic.(algrbra, geometry, trigonometry...).

The score of the problems is from $1$ to $5$ for each problem of a day. This score will be cumulate for all weeks. The solution will be pm-ed to me. At this part of the contest I'm not so organised. That's why I will wait for many many suggestions to improve it from you and a accept of a moderator to do this. (There is also a poll)

Many thanks!  :)

[color=red][i][Mod Edit: One of these threads will be kept as a sticky - please post the problems and solutions in this one thread from now on. Thank you.][/i][/color]

Not bad! :D
But I hope you can post some more geometry...They are rarely seen in the marathons and other contests.

[quote="Slizzel"]Once a week I guess. It will contain all kind of problems from all categories.[/quote]
So you shall have to work to give scores to everyone everyday? That's a huge amount of work!

Yes. But since school is ending in a week i'll have a lot of time. i'm also looking for a co-partner :D anyone?

well, if you are asking for problems from people, then they can't answer that one and get points, right?  wouldn't that be bad?

Well that's bad. I thought at this. The memebers that are not participating at the contest could send. If somebody finds a way i'll be greatefull ;)

good idea :)
still got a couple of weeks exams though  :?  so if i enter, it'll be after those i guess...

Are the moderators agree  :? I hope the do :D.

If things go well, this will start Monday morning and I expect your solution in the next 24 hours since the problem is posted. I will send you a reply-post with the score that you got. Even if is much to do, i think there will be acorded partial scores.

I haven't done something like this before and if somebody, with more expericence (like towerfreak2006 or Silverfalcon) want to give me an advice, it would be nine :D.

Tomorow I will post the schedule (eg. Monday-Trigonometry... and so on). Untill the you can just keep voting. ;)

[quote="Valentin Vornicu"][quote="Slizzel"]Are the moderators agree  :? I hope the do :D. [/quote]I have absolutely nothing against something like this. In fact I like the idea. Keep it up! ;)[/quote]

Agree. :)

So when we PM you our answer, you want our full solution? Not just the answer?

If we only give the answer, is that points deducted?

If you want the solution, is it Latex only?

Sorry for all the questions heh, gotta get them out of the way eventually, and whatever way you want it, I don't mind.

For the first question: yes.
You will send PM's with a entire solution. I've made norms for each problem, but they are no standard. If a solution is correct I will score it. Each problem is noted form 1 to 5.

For the second one: yes :D
I will point with 1 point (i think) a good answer without solution. But since is a level is intemediate(or intermediate with a little $+$) don't think the problems will be so hard. :)

The last question: rather $\LaTeX$ than simple text. It's easier to read.

Some of the first week are ready thanks to the chemistry class :D (i've got one and a half to finish :blush:) and they will be after the following schedule:

Day 1 - Algebra
Day 2 - Complex Numbers
Day 3 - Geometry
Day 4 - Inequalities
Day 5 - Trigonometry

I said that their degree difficulty increases form Monday to Friday, but that could be not right :). you may consider the problems unlike I do, so this should be a problem, right :)?. That's all.

[quote="Slizzel"]. i'm also looking for a co-partner :D anyone?[/quote]
i can do it.. :)

here are two solutions. although the second didnt show that the $\gcd$ isnt greater than $576$, i think rest is easier:

[quote="K81o7"]A) Okay we can factor this to $n=(1+p)(1+q)(1-p)(1-q)=(p+1)(p-1)(q+1)(q-1)$.  We know that out of these, two are divisible by 4 and the other two by 2.  This holds for $p\equiv1\mod{4}$, giving us $0\mod{4}$ and $2\mod{4}$.  $3\mod{4}$ gives us the same numbers mod 4, so we know that each n has $2^{1+2+1+2}=2^6$ as a factor, and no greater a power of 2 as a factor.  Clearly, we have that for any prime p, one of $(p-1)$ and $(p+1)$ is divisible by 3, and one is not.  We can have $p\equiv1\mod{3}$, which gives $p-1\equiv0\mod{3}$ and $p+1\equiv2\mod{3}$, and $p\equiv2\mod{3}$ gives $p-1\equiv1\mod{3}$ and $p+1\equiv0\mod{3}$.  Thus, we have that each number in this set has $3^2$ as a factor, and no greater power of 3 as a factor.  Then, we know that for any other prime, there exists at least one such number in the set which is not divisible by that prime.  For example, let's claim that there exists a prime $x\ge5$ such that there exists no $n$ in the set with $x|n$ (or whichever way it would be written).  However, for any such prime x, we can find two primes $p$ and $q$ such that $(p+1)$, $(p-1)$, $(q+1)$ and $(q-1)$ are not divisible by x, which are, for example, $2\mod{x}$.  Thus, we know that each number in this set is divisible by $2^63^2=576$, and there exists, for each prime greater than or equal to 5, some number in this set which is not divisible by that prime.[/quote]

[quote="shyong"]a)re-write it as

$n=(p-1)(p+1)(q-1)(q+1)$ .

Consider the three consecutive integer $p-1,p,p+1$ . Since we know that for every three consececutive positive integer there will be at least one of them having $3$ as a factor . Since $p\geq 5$ and is a prime , $p$ cannot has $3$ as a fator , leaving either $p-1$ or $p+1$ is a multiple of three . Also we notice that $p$ can only be in the form $4k+1$ or $4k+3$ since it is a prime . This implies $p-1$ or $p+1$ has $4$ as a factor . But since both $p-1$ and $p+1$ are even , we have another $2$ as factor of $n$ . Hence any prime $p$ will give us $(p-1)(p+1)$ is divisible by $3\times 4\times 2=24$

So $n$ will be divisible by $24^2=\boxed{576}$[/quote]

and here are 2 solutions for b)

[quote="shyong"]b)By letting $a=\sqrt[7]{(x+127)^3}$ ; $b=\sqrt[7]{(x-127)^3}$ , we can re-write the equation into

$a^2-8b^2=7ab$ which is equivalent to $(a-8b)(a+b)=0$

For $a+b=0$ we have :arrow: $a^7+b^7=0$ :arrow: $(x+127)^3+(x-127)^3=0$
 :arrow: $x=0$ .

For $a=8b$  :arrow: $a=2^3b$ :arrow: $x+127=2^7(x-127)$
 :arrow: $x-127+254=128(x-127)$ :arrow: $254=127(x-127)$ :arrow: $x=129$

Hence the solution for $x$ is $0$ and $129$

 :)[/quote]

[quote="K81o7"]B) Let $\sqrt[7]{(x+127)^6}=a$ and $\sqrt[7]{(x-127)^6}=b$.  Then we have $a^2-8b^2=7ab \implies (a-8b)(a+b)=0$.  Thus, we either have $\sqrt[7]{(x+127)^3}=-\sqrt[7]{(x-127)^3} \implies x+127=127-x \implies x=0$ or $\sqrt[7]{(x+127)^3}=8\sqrt[7]{(x-127)^3} \implies x+127=128(x-127) \implies 127x=127*129 \implies x=129$.[/quote]

here are two solutions. btw both problems could be done by am-gm ...

[quote="kunny"]a) $(a-1)^2(a+1)(a^3+a^2+1)\geq 0\ (\because a>0)$
$\Longleftrightarrow a^6+a^3+1\geq 2a^4+a$

$\Longleftrightarrow a^3+\frac{1}{a^3}+1\geq 2a+\frac{1}{a^2}\ (\because a>0)$

$\Longleftrightarrow a^3+b^3+1\geq 2a+b^2\ (\because ab=1)\ \mathbb{Q.E.D.}$[/quote]

[quote="shyong"]
b)We have

$\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\geq 3\sqrt[3]{\frac{a^2}{bc}}=3\frac{a}{\sqrt[3]{abc}}=3a$

Similarly we have $2\times\frac{b}{c}+\frac{c}{a}\geq 3b$
and $2\times\frac{c}{a}+\frac{a}{b}\geq 3c$

Adding up all of them yields the inequality
 :)[/quote]

[quote="seamusoboyle"]I think the first solution didnt work because of how you defined a and b. Who said they were integers?[/quote]

Aren' t $a,b$ positive real numbers? :?

[quote="seamusoboyle"]I think the first solution didnt work because of how you defined a and b. Who said they were integers?[/quote]

which problem do u mean?

How is the Day 2 week 13 Problem (a) be done by AM-GM , peeta ? I cant see the am-gm way   :? Can you show ?

of course, here is my solution:

$a^3+b^3+1=\frac{a^3+1+1}{3}+\frac{b^3+b^3+1}{3}+\frac{a^3+a^3+b^3}{3}\\\ge \sqrt[3]{a^3\cdot1\cdot1}+\sqrt[3]{b^3\cdot b^3\cdot1}+\sqrt[3]{a^3\cdot a^3\cdot b^3}=a+b^2+a^2b=2a+b^2$

u can do a whole lot with am-gm :) AM-GM rocks  :coolspeak:

hi
i want to say a word how solutions should look like. i guess this is an online contest and therefore it doesnt have to look like a normal constest-style but the reader should at least be guided through the solution so he can follow (let it be through obvious transformations or a little text next to the forumals). partwise i had to look up quite some formulas of trigs, without even being told what kind it is. so please write readable solutions without leaving out important transformations. thx
example:
is it iniutive that $\frac{\tan x+1}{\tan x -1}=\frac{5}{3} \implies \tan\left(x+\frac{\pi}{4}\right)=-\frac{5}{3}$?

here are two solutions easy to read:

[quote="Gabriel(fr)"]Part a)

$y=\frac{\pi}{2} - x \implies sin\ y=sin\ (\frac{\pi}{2} - x)=cos\ x$, thus:

$\frac{sin\ x + sin\ y}{sin\ x - sin\ y}=\frac{sin\ x + cos\ x}{sin\ x - cos\ x}=\frac{5}{3}$

$\Rightarrow \frac{sin^2x + cos^2x + 2\cdot sin\ x.cos\ x}{sin^2x + cos^2x - 2\cdot sin\ x.cos\ x}={(\frac{5}{3})}^2$

$\Rightarrow \frac{ 1 + sin\ 2x}{1 - sin\ 2x}=\frac{25}{9} \implies sin\ 2x=\frac{8}{17}\implies sin\ x=\frac{4}{cos\ x.17}$ but:

$3\cdot (sin\ x + cos\ x)=5\cdot (sin\ x - cos\ x) \implies sin\ x=4\cdot cos\ x \implies cos\ x=\frac{\sqrt{17}}{17}$ in consequence:

$x=arc cos \frac{\sqrt{17}}{17}$ and $y=\frac{\pi}{2} - arc cos \frac{\sqrt{17}}{17}$

[/quote]

[quote="kunny"]
b) Since  $-1\leq \sin x\leq 1, -1\leq \cos x\leq 1$, we have $\sin ^ 2 x (\sin x-1)\leq 0\Longleftrightarrow \sin ^ 3 x\leq \sin ^ 2 x$, analogously $\cos ^ 3 x\leq \cos ^ 2 x$.
Adding these inequalities we have $\sin ^ 3 x+\cos ^ 3 x\leq 1$, the equality holds when $(\cos x,\sin x)=(-1,0),(1,0) ,(0,1),(0,-1)$,
but the pair of $(\cos x,\sin x)$ such that $\sin ^ 3 x+\cos ^ 3 x=1$ is limited to $(\cos x,\sin x)=(1,0),(0,1)$.

Thus the answer is $x=2m\pi,\frac{\pi}{2}+2n\pi\ (m,n=0,1,2,\cdots)$[/quote]
(just that in kunny's solution $m,n\in\mathbb Z$, but i chose his solution anyway because of simplicity)

peeta

here is the only complete solution.

[quote="shyong"]a)Suppose all $a_i$ are distinct (or else it wont have period of 2)
From $a_{n+1}=a_n-\frac{1}{a_n}$ , we can get

$a_na_{n+1}=(a_n)^2-1$ .  Also replace $n$ with $n+2$ we obtain
$a_{n+2}a_{n+3}=(a_{n+2})^2-1$

Substract the both to get

$a_{n+2}a_{n+3}-a_na_{n+1}=(a_{n+2})^2-(a_n)^2$

If there exist a period of $2$ for $a_n$ i.e $a_n=a_{n+2}$ , this gives us

$a_{n+2}a_{n+3}-a_na_{n+1}=0$ :arrow: $a_{n+2}a_{n+3}=a_na_{n+1}$

which is true because $a_n=a_{n+2}$

 we have $a_0a_1=(a_0)^2-1$
Also $a_1a_2=(a_1)^2-1$ :arrow: $a_0a_1=(a_1)^2-1$  ( using $a_0=a_2$)

Compare the two we get $a_0=-a_1$ (negative since all term are distinct) . Hence we get $-(a_0)^2=(a_0)^2-1$ :arrow: $a_0=\frac{1}{\sqrt 2}$ or $\frac{1}{-\sqrt 2}$

So there exist a period two sequence for $a_{n+1}=a_n-\frac{1}{a_n}$ where $a_0=\frac{1}{\sqrt 2}$ or $\frac{1}{-\sqrt 2}$ , $n=0,1,2...$[/quote]

but since this is not really the answer asked for, but more the way how to get to the answer, ill provide a solution as well

[quote="peeta"]
let $a_0$ be $\frac{1}{\sqrt2}$, then $a_1=\frac{1}{\sqrt2}-\sqrt2 =\frac{1}{\sqrt2}-\frac{2}{\sqrt2}=-\frac{1}{\sqrt2}$ and $a_2=-\frac{1}{\sqrt2}+\sqrt2= -\frac{1}{\sqrt2}+\frac{2}{\sqrt2}=\frac{1}{\sqrt2}=a_0$

the sequence $(a_n)$ with $n=0,1,2,\ldots$ has a period of $2$ iff $a_0=a_2$.
this is because $a_n$ is defined recrusivly as a function $f(a_{n-1})=a_{n-1}-\frac{1}{a_{n-1}}$. thus $f(a_k)=f(a_{k+2})$is true for all $k=0,1,2,\ldots$

therefore for $a_0=\frac{1}{\sqrt2}$ the sequence $(a_n)$ has the period $2$.
[/quote]

for the second problem there were a couple solutions by induction like this

[quote="shyong"]
b) It is easy to see that $b_2=2$ and $b_3=3$ . We shall prove by induction that $b_k=k$ for all natural $k\geq 2$

Suppose for $n=k$ ,  $b_k=k$ is true .
When $n=k+1$ ,

$b_{k+1}=\left(\frac{k}{3}+\frac{1}{k}\right)b^2_k-\frac{k^3}{3}+1$

$=\left(\frac{k}{3}+\frac{1}{k}\right)k^2-\frac{k^3}{3}+1$

$=k+1$ .

Hence it is true for all natural $k$ that $b_k=k$

So $b_{2005!}=2005!$

 :)[/quote]

here are finally the solutions of the day 5 problems.
for the first problem there were several solutions sent-in but unfortunatly there was no complete solution, so i will provide one.

[quote="Gabriel(fr)"]Part a)

By the law of cossines:

$AB^2=PA^2 + PB^2 -2.PA.PB\cdot cos\ 120^o$

$\Rightarrow AB^2=136 -2\cdot60\cdot-\frac{1}{2}=196$,     and by pythagora's:

$AC^2=AB^2 + BC^2=196 + BC^2$ then:

$BC^2 + 196=PA^2 + PC^2 -2. PA.PC. cos\ 120^0=100 + PC^2 + 10.PC$

$\implies BC^2=-96 + PC^2 + 10.PC $ and:

$BC^2=PB^2 + PC^2 -2.PB.PC.cos\ 120^0=36 + PC^2 + 6.PC$,  substituing:

$-96 + PC^2 + 10.PC = 36 + PC^2 + 6.PC \implies PC=33$[/quote]

[quote="peeta"]
(i took the solution from kalva.demon.co.uk)

b) the letters in the sketch symbolize the area and the triangles.
it is obvious that $P, U, Q, V, A$ are similar to $T$ the whole triangle.
because of the side-lengths $Q/440=P/441$ and $V/440=U/441$.
Also $T=P+U+441=Q+V+A+440\iff 441P+441U+(441)^2 = 441Q + 441V+441A+441\cdot 440 = 440P + 440U+441A+441\cdot 440$
$\iff 441 A = P+U+441=T$
Thus sides of triangle area $A$ are $1/21\times$ the sides of whole triangle. So the hypoteneuse of whole triangle is $\sqrt{440}\cdot 21$. If its height from this side is $h$, then $h - h/21 = \sqrt{440}$, so if two short sides have lengths $a, b$, we have $ab = \frac{(\sqrt{440}21)^2}{20} = 22\cdot 21^2$, and $a^2 + b^2 = 440\cdot21^2$. Hence $(a + b)^2 = 21^2(440 + 2\cdot22) = 21^222^2$, and $a + b = 21\cdot22)$
[/quote]

well there were two solutions and both totally different but nice. so i want to share them both with u:

[quote="K81o7"]First of all, there's the possibility that the people could be sitting in their original positions: that's 1.  After that, note that people can only be moved around by switching them in pairs (because they can't move more than 1 space from their places).  So, we can now count the number of ways to choose pairs of them (adjacent people only).  For 1 pair, there are 10 ways.  For 2 pairs, there are (7+7+6+5+4+3+2+1)=35 ways.  For three pairs, there are (15+15+10+6+3+1)=50 ways.  For 4 pairs, there are (10+10+4+1)=25 ways.  For 5 pairs, there are 2 ways.  Thus, the total is 2+25+50+35+10+1=123.  Note that there is also the possibility of having each person move to the right, or each person move to the left.  If we count these as well, it would be 125, but we are considering them the same as the original position.[/quote]

[quote="shyong"]
First of all let us define $S_n$ such that it is the number of way to arrange the number in the set $P_n=\{1,2,3,4,\cdots ,n\}$ with the condition that only any two consecutive integer in the set $P_n$ ,i.e $k$ and $k+1$ ( $k\in P_n$) can alternate.

It is easy to see that $S_1=1$ and $S_2=2$ .
For $S_3$ , $P_3=\{1,2,3\}$ . Now , if $1$ doesnt change it place . We can just ignore $1$ leaving the set to have only $\{2,3\}$ . But this is just $S_2=2$ . If $1$ alternate with $2$ (which is the only possible way since we define that only two consecutive integer can alternate) , we can leave out $\{2,1\}$ leaving only $\{3\}$ which is just $S_1$ .
Hence $S_3=S_2+S_1=2+1=3$

Similarly lets investigate $S_n$ . We have $P_n=\{1,2,3\cdots n\}$ . If $1$ doesnt change place , then it remains $\{2,3,4\cdots n\}$ which is just $S_{n-1}$ . If $1$ alternate with $2$ , we can leave out $\{2,1\}$ and it leave only $\{3,4,5\cdots n\}$ which is $S_{n-2}$ .

So we can conclude that $S_n=S_{n-1}+S_{n-2}$  with $S_1=1$ , $S_2=2$ (which we know it is just similar to Fibbonanci series)

Now we define $K_n$ as number of way we can arrange the number in set $P_n$ with an additional condition which is $1$ can alternate with $n$ .

Lets  investigate $K_n$ , there are three cases
(i) if $1$ doesnt alternate with $n$ then it is just $S_n$ .
Ways = $\boxed{S_n}$
(i) If every number in the set now take over the place of its left neighbour i.e $1$ change place to $2$ ; $2$ to $3$ and so on until $n$ change place to $1$ . It is another way . Similarly if all swift to the right and this gives us another way.
Ways = $\boxed{2}$
(iii)If $1$ alternate with $n$ , it remains only $n-2$ numbers in the set which is just $S_{n-2}$
Ways = $\boxed{S_{n-2}}$

Hence combine these three cases to get $K_n=S_n+S_{n-2}+2$  $(n=3,4,5\cdots )$

For special case $n=10$ , we have $K_{10}=S_{10}+S_{8}+2=89+34+2=\boxed{125}$[/quote]

[color=blue][size=150]Day 4 Solutions[/size][/color]
I received solutions for Day 2, but they're not totally clear to me, so i'm holding off until posting day 2/3 solutions until the end of today (also hoping to get some more solutions :D)

[quote="Andreas"]Week 14 Day 4

a)
$\sqrt{ab} \leq \sqrt{\dfrac{a^2 + b^2}{2}}$ hence $ab \leq \dfrac{a^2 + b^2}{2}$.
The same for $bc$ and $ca$.
We get $ab + bc + ca \leq \dfrac{a^2 + b^2}{2} + \dfrac{b^2 + c^2}{2} + \dfrac{c^2 + a^2}{2}$. So it is proved.

b)
We factorize to get $(x-1)(x+1)(x-i)(x+i) = 0$. So the solutions are $x = \pm 1$ and $x = \pm i$.
[/quote]

[color=blue][size=150]Day 3 Solutions[/size][/color]

I'll put my solutions, since I didn't get any solutions which went all the way through...

A) With the two parallel lines, we have $\angle{CBE}=\angle{ABE}=\angle{BEC}$, and thus, $\triangle{BCE}$ is isoceles, which means $BC=EC$.  Then, the angle bisector of the vertex of this triangle is also an altitude, which makes the angles at $F$ all right angles.  Then, by $ASA$ triangle congruence, $\triangle{BFC} \equiv \triangle{BFG}$.  So, we have $EC=BG$, and $EC||BG$, which is enough to tell us that this figure is a parallelogram, and the fact that $EC=BC$ tells us that it's a rhombus.  We know $CD=\dfrac{\dfrac{5}{2}}{\sin{\angle{CDF}}}=\dfrac{5}{2\sin{75^{\circ}}}$.  By the Law of Sines, $\dfrac{AC}{\sin{\angle{ABC}}}=\dfrac{BC}{\sin{\angle{BAC}}}\implies AC=5\dfrac{\sqrt{3}}{2\sin{75^{\circ}}}$.   $\therefore AD=AC-CD=\dfrac{5(\sqrt{3}-1)}{2\sin{75^{\circ}}}=\dfrac{5(\sqrt{3}-1)}{\sqrt{2+\sqrt{3}}}$, or if you like, $5(\sqrt{3}-1)(\sqrt{2-\sqrt{3}})$.

B) $S(CDI)=\dfrac{DI}{BF}S(BCF)$.  So we'll find $DI$.  We know $CD=\dfrac{5}{2\sin{75^{\circ}}}$.  By the Law of Sines, $\dfrac{DI}{\sin{30^{\circ}}}=\dfrac{CD}{\sin{45^{\circ}}} \implies DI=\dfrac{5\sin{30^{\circ}}}{2\sin{45^{\circ}}\sin{75^{\circ}}}=\dfrac{5}{\sqrt{2}\sqrt{2+\sqrt{3}}}$.  We can then multiply by $CF$ and divide by 2 to get the area to be $\dfrac{25}{4\sqrt{4+2\sqrt{3}}}$, or if you like, $\dfrac{25\sqrt{4-2\sqrt{3}}}{8}$.

[color=blue][size=150]Day 2 Solutions[/size][/color]

Once again, I got a solution sent in for Part A, but it was unclear, so I'll post my own...

A) This is the kind of problem for which you need a "moment of revelation", or whatever you'd call it: it's a USAMO problem, so..yeah.  Well It made the most sense to me that the line should go through one of the vertices, and then, to make things line up nicely, I decided to make it an angle bisector.  I claim that the angle bisector of the smallest angle (or one of the smallest if it is isoceles) in the triangle will fulfill the requirements.  But this problem is trivial with an isoceles triangle, so we'll assume the triangle is scalene.  Let this vertex be $A$, and let $B$ be the next smallest in line, while $C$ is the largest angle.  So from smallest to largest, the sides go $a$, $b$, $c$.  So we'll draw in the line, and let the line $l$ hit $BC$ at $M$.  Well, we know that the angle bisector divides the opposite side into proportional segments, so $\dfrac{MC}{b}=\dfrac{MB}{c}$.  The intersection of the two triangle has area $2S(AMC)=2S(ABC)\dfrac{b}{b+c}$.  By the triangle inequality, $a+b>c$.  Because $a<b$, we then know that $2b>c$.  Plugging this into our equation, we have that the area of the intersection is $>2S(ABC)\dfrac{1}{3}=\dfrac{2}{3}S(ABC)$.  $\mathbb{QED}$.

B) This one was actually really easy to prove.
We know that the maximum of $(1-a)(1-b)(1-c)$ occurs when they are equal, by AM-GM.  Thus, $(1-a)(1-b)(1-c)\le\dfrac{8}{27}$.  Expand it out.  $1-a-b-c+ab+bc+ac-abc\le\dfrac{8}{27}$.  We are given that $a+b+c=1 \implies 1-a-b-c=0$.  Thus, $ab+bc+ac-abc\le \dfrac{8}{27}$.

Andreas, you were the winner, by the way.  We're trying to decide whether or not people want a contest though.

[color=blue][size=150]Weekend Solutions[/size][/color]

A) Just use the pythagorean theorem.  Let $ABCD$ be a face, and $EFGH$ a face, such that $B$ and $G$ are opposite, $C$ and $F$ are opposite, and $D$ and $E$ are opposite.  You can express the lengths of the three edges of the tetrahedron $AX$, $AY$ and $AZ$ as $\sqrt{n^2+n^2+1}$ where n is the distance from $X$ to edge $BC$ ($X$ is on face $HEBC$.).  It's a bit unclear, I know but if you draw the diagram, you'll see.  Then the length of $XY$ can be expressed as $\sqrt{2(n^2-2n+1)}$.  Solving these two equations simultaneously yields that $n=\dfrac{1}{4}$, and thus, an edge of the tetrahedron has length $\sqrt{\dfrac{18}{16}}=\dfrac{3\sqrt{2}}{4}$.

B) We know that the area of this quadrilateral is $S(KEJ)-S(HEI)$.  The maximum area of $HEI$ occurs when $I$ is equidistant from the two, in the arc $FG$.  One can see that the height of the triangle would be maximized there, but if you wanted a formal proof, you can do it using the triangle area formula $\dfrac{1}{2}ab\sin{C}$ and AM-GM.  In any case, this means that the altitude of the triangle from $I$ has length $\dfrac{1}{2}+\dfrac{\sqrt{2}}{4}=\dfrac{2+\sqrt{2}}{4}$.  Thus, the area of the triangle is $\dfrac{1+\sqrt{2}}{8}$.  Now, draw the perpendicular from $I$ to $EG$, and call the foot $L$.  Then we know $IL=\dfrac{1}{2\sqrt{2}}$.  We also know that $CL=\dfrac{\sqrt{2}+1}{2\sqrt{2}}$.  So, using the fact that $CG=1$, we can use similar triangles to get that $GJ=\dfrac{IL}{CL}=\dfrac{1}{1+\sqrt{2}}$.  Adding this to $KG=1$, we get $JK=\sqrt{2}$, and thus the area of $\triangle{CJK}=\dfrac{\sqrt{2}}{2}$.  Subtract the area of $EHI$ to get $\dfrac{4\sqrt{2}}{8}-\dfrac{1+\sqrt{2}}{8}=\dfrac{3\sqrt{2}-1}{8}$.


Let $ (x_n)(n\equal{}1;2;...)$ be a bounded-above sequence such that $ x_{n\plus{}2} \geq \frac{1}{4}x_{n\plus{}1}\plus{}\frac{3}{4}x_n$ for all $ n\equal{}1;2...$
Prove that this sequence has limit.

$ x_{n}\geq C_{1}\cdot (\minus{}\frac{3}{4})^n \plus{} C_{2}$
And we have that, $ \mathop {\lim }\limits_{n \to \infty } {x_{n} } \geq C_{2}$, but we know that $ x_{n} \leq A$. We get the answer.

Is that a solution? :D Please post concrete. I don't understand.


function
calculus
integration
derivative
advanced fields
advanced fields unsolved

I am dealing with some problem involving distributions, and I am just wondering about how weak convergence works. Of course I know the definition, but I am confused by some little problem.

Suppose I have $f_{n}\to f$ weakly. $f\cdot f$ doesn't necessarly make sense -- you can't define pointwise product in general for distributions. But $f_{n}$ might be nice functions, and $f_{n}\cdot f_{n}$ might make sense for all $n$. Does that mean pointwise product destroys the convergence?

Actually, I can see why this doesn't work in general, but in my particular case my $f_{n}$ are particularly well-behaved. I mean, the Schwartz class of test functions is closed under pointwise product, right? So then I can just look at the square of the test function from before, and the inner product is the $L^{2}$ norm of what I was doing the $L^{1}$ norm of before. Can a sequence $g_{n}$ that's convergent in $L^{1}$ be divergent in $L^{2}$? Maybe this is obvious and I'm just being obtuse....

I am just having a hard time thinking of weak convergence in an intuitive way I guess. And I can't really see what to do when I write everything down. It just looks like a huge mess of integrals to me (I obtain $f_{n}$ from $f$ using a complicated smoothing operation). Maybe I should be more patient with it and do some work, but I would rather have an intuitive understanding of the reasons for this.

There's an example I like to give to explain how weak weak convergence really can be.

First the theorem: these are distributions, and we're talking about both weak limits and weak derivatives.

Suppose $f_{n}\to f.$ Then $f_{n}'\to f'.$

Coming from classical analysis, that's a startling theorem isn't it? No conditions, no exceptions: the derivative of the limit is the limit of the derivatives. The proof is just a short walk through the definitions - it looks almost silly.

Now, let $f_{n}(x)=\frac1n\sin(n^{2}x).$ We actually have $f_{n}\to 0$ in a fairly strong sense - uniformly - so it's no great surprise that $f_{n}\to0$ weakly.

Then $f_{n}'(x)=n\cos(n^{2}x).$ And by the theorem above, $n\cos(n^{2}x)$ tends weakly to zero.

This isn't a paradox or contradiction. Everything here is true. It's just that you need a little time to adjust to it being true.

--

I think Paul Dirac (who was a physicist, after all) coined the term "test function" as an analogue to the physicists' use of the idea of a "test charge" or "test mass." For instance, what is an electric field? Maybe I don't know what it "is" but if I can describe how it would act on all possible test charges. It's the same with a distribution. What "is" it? You can only answer that by asking how it acts on all possible test functions.

Thank you. Somebody else actually showed me this example as well. I think I can see why this makes sense for the particular function. I can see formally why certain things happen but the intuition doesn't come quickly. I guess that's what makes math interesting, eh?  :D


inequalities
trigonometry
geometry
incenter
geometry proposed

If $ \alpha,\beta,\gamma$ are three angles of a triangle, prove that.
$ \cos\alpha \cos\beta\cos\gamma\leq(1\minus{}\cos\alpha)(1\minus{}\cos\beta)(1\minus{}\cos\gamma)$   :?:

______________________________
Azerbaijan Land of the Fire :D

[quote="Pirkuliyev Rovsen"]If $ \alpha,\beta,\gamma$ are three angles of a triangle, prove that.
$ \cos\alpha \cos\beta\cos\gamma\leq(1 \minus{} \cos\alpha)(1 \minus{} \cos\beta)(1 \minus{} \cos\gamma)$   :?:  :D[/quote]

$ CosA.CosB.CosC \le (1 \minus{} CosA)(1 \minus{} CosB)(1 \minus{} CosC)$

$ \Leftrightarrow 8\sin^2(A/2)\sin^2(B/2)\sin^2(C/2) \ge \cos(A)\cos(B)\cos(C)$

$ \sin(A/2)\sin(B/2)\sin(C/2) \equal{} \frac {r}{4R}$

$ \cos(A)\cos(B)\cos(C) \equal{} \frac {s^2 \minus{} (2R \plus{} r)^2}{4R^2}$

So we must prove: $ \frac {r^2}{2R^2} \ge \frac {s^2 \minus{} (2R \plus{} r)^2}{4R^2} \Leftrightarrow 3r^2 \plus{} 4Rr \plus{} 4R^2 \ge s^2$ - known inequality

($ IH^2 \equal{} 3r^2 \plus{} 4Rr \plus{} 4R^2 \minus{} s^2$ ,$ I$- incenter , $ H$ - orthocenter )


Gauss induction algebra polynomial geometry

The following formula is well-know \[ 1+2+3+.....+n = \frac{n(n+1)}{2}  \]

I'm just wanted to know how many ways this can be proved.
Post your method please :)
Any advance method is permited ;)

[quote="Beginner"]The following formula is well-know $1+2+3+.....+n = \frac{n(n+1)}{2} .$ Post your method please :)  Any advance method is permited ;)[/quote]
The following identity is easy to prove
\[ \sum\limits_{k=1}^{n}A_k = \sum\limits_{k=1}^{n}A_{n+1-k}\; . \]
By considering $A_k: =k \; ,\;$ when      $S: = \sum\limits_{k=1}^{n}k$  you find    $\; \; S= n(n+1)-S \; \; \; \; \; \mbox{that is }\; \; \; \; S=\frac{n(n+1)}{2}\; .$

What flip2004 wrote is Gauss proof. Anyway, another proof is by induction. ;)

Here's what I have.  Let us have a two identical traingles in which they can lie on top of one another creating a rectangle.
one traingle will have the property $\sum_{i=1}^ni$, however, the other will be an inevrse of the other, $\sum_{i=1}^n(n+1-i)$.  The sum of $\sum_{i=1}^ni$ and $\sum_{i=1}^n(n+1-i)$ is $2S$.

Hence,
$\\2S=\sum_{i=1}^ni+\sum_{i=1}^n(n+1-i)\\=\sum_{i=1}^n(n+1)\\=n(n+1)\\\implies S=\frac{n(n+1)}{2}$

Masoud Zargar

There is a combinatorics proof if anyone is interested.

There is the normal kind of induction:
$1+2+\ldots+(n-1)+n=\frac{(n-1)(n)}{2}+n=\frac{n(n+1)}{2}$

There is this way do use induction, to make it more difficult but it's really nice :
$1+2+3+\ldots+n =$
(n is even) $1+3+5+\ldots+(n-1)\;+\;2+4+6+\ldots+n$
$=\left(\frac{n}{2}\right)^2+2\cdot(1+2+3+\ldots+n/2)$
$=\left(\frac{n}{2}\right)^2+\frac{n}{2}\cdot \frac{n+2}{2}$
$=\frac{n(n+1)}{2}$
(n is odd) about the same thing

(it uses $1+2+3+(2n-1)=n^2$)

But the quickest way (it is in fact the first induction in another order) :
$\frac{(n+1)n}{2}-\frac{n(n-1)}{2}=n$

There's a proof for the general formula of
$1^n+2^n+3^n...$
using Stirling numbers (maybe thats tetrahedr0n's proof too...)

That would be
\[ 1^x+2^x+...+n^x=\sum_{i=1}^xS(x,i)\left(\begin{array}{rr}n+1\\i+1\end{array}\right)i! \]

Masoud Zargar

[quote="jeffrey"]using Stirling numbers (maybe thats tetrahedr0n's proof too...) [/quote]

It is, and boxedexe provides the general formula.

[quote="Beginner"]The following formula is well-know \[ 1+2+3+.....+n = \frac{n(n+1)}{2}  \] [/quote]
A geometrically proof may be found on :        http://www.matharticles.com/ma_clc7e.html
Another interesting proof may be given also on the site :         http://141.51.96.22/priv/math/reihen/summenreihen.htm
Likewise, see: A.F.Beardon ,[i] Sums of Powers of Integers [/i] American Mathematical Monthly, 103 (March 1996)
and/or
http://maths.ashwyninnovations.com/BernoulliNumbers.pdf

After the formula with sterling numbers, don't forget the one with Bernoulli numbers:
$1^m+2^m+...+n^m = \frac{1}{m+1} \sum_{k=0}^m B_k \binom{m+1}{k} (n+1)^{m+1-k}$

I like to phrase that in terms of $\zeta(-k)$; those are closely related to the Bernoulli numbers.

The combinatoric proof  require Pascal's identity. we can also proof that by using Reccurence relation.
$a_n = \sum_{k=0}^n k$, so $a_{n+1} = \sum_{k=0}^{n+1} k$ doing the substraction yields
$a_{n+1} - a_{n} = n+1$.  The homogenous solution is ${a_n}^{(h)} = C$ and particular solution is ${a_n}^{(p)} = n(An+B)$. The general solution is
$a_n = n(An+B)+C$. with $a_0 = 0$ we have $C=0$. with $a_1=1$ we have $A+B =1$. with $a_2=3$ we have $A= \frac{1}{2}$ and also $B= \frac{1}{2}$ so $a_n = \frac{n}{2} (n+1)$

ZetaX use Bernoulli Polinomial $B_n(t)= \sum_{k=0}^{n} \binom{n}{k} B_k t^{n-k}$
For $(x = 2, 3, \cdots )$ , we have  \[ \sum_{k=1}^n k^x = \frac{B_{x+1}(n) - B_{x+1}}{x+1}  \]

[quote="Hokkage"]You could also use Gaussian pairing. By pairing up $n$ with $1$ you get $n+1$. Because there are $n$ numbers, there must be $\frac{n}{2}$ pairs. Multiply the two together and you get $\frac{n(n+1)}{2}$.  :D
Also, $1^2+2^2+3^2+ \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$[/quote]

That was exactly what I did in my proof.

And what already at least three persons did before you...
It would be better to stop that senseless discussion about who was first (I don't think it was mathnerd314...) and give new, interesting proves.

Define $S_n = \sum_{i=1}^{n} i$
Consider the progresion:
$S_1 , S_2 , S_3.....$
It is a second grade progresion wich satisfies:
$S_x = ax^{2} + bx + c$ for every $x$.
So calculate $S_1 , S_2$  and  $S_3$
Then we get that
$S_1= a + b +c = 1$
$S_2= 4a + 2b +c = 1+2 = 3$
$S_3=9a + 3b + c = 1+2+3=6$

Solving this system we get the values:
$a= \frac{1}{2}$
$b= \frac{1}{2}$
$c=0$

So
$S_n = \frac{1}{2} n^{2} + \frac{1}{2} n = \frac{n(n+1)}{2}$

It's also posible to use this method for $S_n= \sum_{i=1}^{n} i^{k}$ seeing that it is a  progresion with grade $k+1$ wich satisfyes:

$S_x= \sum_{i=o}^{x+1} a_{i} x^{i}$
But it is very long to do for $k > 4$

[quote="jasamper88"]Define $S_n = \sum_{i=1}^{n} i$
.....
It's also posible to use this method for $S_n= \sum_{i=1}^{n} i^{k}$ seeing that it is a  progresion with grade $k+1$ wich satisfyes:

$S_x= \sum_{i=o}^{x+1} a_{i} x^{i}$
But it is very long to do for $k > 4$[/quote]

Can we use this method to find the genral formula $\sum_{k=1}^n k^x$ ??

Well about the formula using Bernoully polynomial (above) and from ZetaX, I think we forget to state that in this case $B_1 = \frac{1}{2}$. Of course we know that $B_1 = - \frac{1}{2}$, but it will make the formula goes wrong?

I think it should be $B_1 = - \frac{1}{2}$...

[color=green]1[/color]+[color=blue]n[/color] =     n+1
[color=green]2[/color]+[color=blue]n-1[/color]=n+1
[color=green]3[/color]+[color=blue]n-2[/color]=n+1
.
.
[color=green]n-1[/color]+[color=blue]2[/color]=n+1
[color=green]n[/color]+  [color=blue]1[/color] =   n+1

The Blue Column+The Green Column=2(The wanted sum), so 1+2+..+n=n(n+1)/2

The best proof to this formula is using graphs, but i don`t know how to explain it here  :(

:idea:
1 [color=green] 1[/color] [color=blue] 1[/color] [color=violet] 1[/color]...
[color=green] 1[/color] [color=green] 1[/color] [color=blue] 1[/color] [color=violet] 1[/color]
[color=blue] 1  1  1[/color][color=violet] 1[/color]
[color=violet]1  1  1  1[/color]
.
.
.
So
1+[color=green]3[/color]+[color=blue]5[/color]+[color=violet]7[/color]+..+2n-1=$\underbrace{1+1(green)+1(blue)+1(violet)+..+1}_{n}\cdot\underbrace{1+1(green)+1(blue)+1(violet)+..+1}_{n}=n^2$

[quote="José"][quote="Cezar Lupu"]I knew that $1^{x}+2^{x}+...+n^{x}=....?$ is an open question. :?[/quote]
Is he right??????[/quote]
Regarding above comments : in case when $x$ is a positive integer , e.g. $S_{p}(n): =\sum\limits_{k=1}^n k^p$, it may be observed that there exist
real polynomials $P_m\; ,\; Q_m$ such that  for $p \ge 3$
\[ S_{p}(n)=\left\{ \begin{array}{lcr} n(n+1)(2n+1)P_{m}\left((2n+1)^2\right) \; &, & p=\mbox{even}=2m+2 \\ &&\\ n^2(n+1)^2 Q_m\left((2n+1)^2\right) \; & , & p=\mbox{odd}=2m+3 \\ \end{array} \right. .  \]
[b] Question: [/b] [color=blue]  What about polynomials $P_m$ and $Q_m$ ? [/color]

A very elementary method for this series, which I invented for this in 10th grade. :)

Dwaw a right angled triangle with both short sides equal to n, and then put n right angled triangles with short side 1 on the hypothenuse, as on the drawing. Now we have, from top to bottom: $1+2+...+n$ unit squares. And the area of the drawing is $\frac{n\cdot n}{2} + n \frac{1\cdot 1}{2} = \frac{n(n+1)}{2}$, which gives the answer. :)

Later I proved the $1+4+9+...+n^2$ in an alike way, with the volume of pyramids.

[quote="jasamper88"][quote="Beginner"][quote="jasamper88"]Define $S_n = \sum_{i=1}^{n} i$
.....
It's also posible to use this method for $S_n= \sum_{i=1}^{n} i^{k}$ seeing that it is a  progresion with grade $k+1$ wich satisfyes:

$S_x= \sum_{i=o}^{x+1} a_{i} x^{i}$
But it is very long to do for $k > 4$[/quote]

Can we use this method to find the genral formula $\sum_{k=1}^n k^x$ ??

[/quote]

Yes, because you can find  values for $k=1,2,...,x+1$. But it try to do it. For $x=4$ it is very long.[/quote]

This method wouldn't work for non-integers $x$, though, would it?

(Obviously, it works for integer $x$ because $k^x$ describes the first difference of some polynomial...)

[quote="t0rajir0u"][/quote][quote="jasamper88"][quote="Beginner"][quote="jasamper88"]Define $S_n = \sum_{i=1}^{n} i$
.....
It's also posible to use this method for $S_n= \sum_{i=1}^{n} i^{k}$ seeing that it is a  progresion with grade $k+1$ wich satisfyes:

$S_x= \sum_{i=o}^{x+1} a_{i} x^{i}$
But it is very long to do for $k > 4$[/quote]

Can we use this method to find the genral formula $\sum_{k=1}^n k^x$ ??

[/quote]

Yes, because you can find  values for $k=1,2,...,x+1$. But it try to do it. For $x=4$ it is very long.[/quote][quote="t0rajir0u"]

This method wouldn't work for non-integers $x$, though, would it?

(Obviously, it works for integer $x$ because $k^x$ describes the first difference of some polynomial...)[/quote]

I don't think it works for non integers values of $x$.


trigonometry
AMC
AIME
modular arithmetic

A sequence of numbers $ a_1,a_2,a_3,\ldots$ is defined by $ a_1\equal{}7, a_2 \equal{} \minus{}6$ and $ a_n\equal{}a_{n\minus{}1} \minus{} a_{n\minus{}2}$ for $ n > 2$. What is the sum of the first 2000 terms of the sequence?

The first few terms are...
7, -6, -13, -7, 6, 13, 7, -6, ... and we have the pattern again.
So it looks like the terms repeat in a pattern.

The first 6 add to 0, the next 6 also do, and the next and the next... until we get to $ a_{1999} \plus{} a_{2000}$. This is a new cycle, so the sum is equal to 7-6=1.

umm, dragon96?  wouldn't the third term be 13, and so the pattern is:
7,-6, 13, -19, ....

hello, we have $ a_n\equal{}\frac{\sqrt{3}}{3}\sin(\frac{1}{3}n\pi)\plus{}13\cos(\frac{1}{3}n\pi)$  and $ \sum_{i\equal{}1}^{2000}a_i\equal{}1$.
Sonnhard.

[quote="GameBot"]A sequence of numbers $ a_1,a_2,a_3,\ldots$ is defined by $ a_1 = 7, a_2 = - 6$ and $ a_n = a_{n - 1} - a_{n - 2}$ for $ n \& gt; 2$. What is the sum of the first 2000 terms of the sequence?[/quote]

Kind of coincidential that I just did an AIME problem regarding this sequence.

Let the first terms be a,b, respectively. Thus our sequence is

$ a,b,b-a,-a,-b,a-b,a$. So this repeats with period 6. The sum of the first 6 numbers is 0, so since $ 2000 \equiv 2 \pmod 6$, we need $ a+b$, which is $ 1$.

whooops, my bad.  careless mistake.  don't read my post.   :D


See here for my solution
http://www.mathlinks.ro/Forum/viewtopic.php?t=109888


given a douoble pendulum with same mass and same length, what is the period?

The penulum has one mass hanging under the other.

as in http://upload.wikimedia.org/wikipedia/commons/7/78/Double-Pendulum.svg

anyone?  Thank you

[size=150][color=blue]MOD EDIT:I directed it to the link [/color][/size]:)

actually there is a problem in using wiki links so directly use them...
it's fine now :)

i am moving this to advanced physics...this is no way an intro physics problem :)

The motion is very complicated and we can calculate period etc. only for simple cases.
I don't think you can do this without using Lagrangian formulation.

We take the Cartesian coördinates
$ x_1\equal{}L_1\cos\theta_1$
$ y_1\equal{}L_1\sin\theta_1$
$ x_2\equal{}L_1\cos\theta_1\plus{}L_2\cos\theta_2$
$ y_2\equal{}L_1\sin\theta_1\plus{}L_2\sin\theta_2$

Kinetic Energy $ T\equal{}\frac{1}{2}m_1 (\dot{x}_1^2\plus{}\dot{y}_1^2)\plus{}\frac{1}{2}m_2 (\dot{x}_2^2\plus{}\dot{y}_2^2)$

i.e. $ T\equal{}\left( \frac{1}{2}m_1 L_1^2\dot{\theta}_1^2\right) \plus{}\frac{1}{2}m_2\left(  L_1^2\dot{\theta}_1^2\plus{}L_2^2\dot{\theta}_2^2\plus{}2L_1\dot{\theta}_1L_2\dot{\theta}_2(\sin\theta_1\sin\theta_2\plus{}\cos\theta_1\cos\theta_2)\right)$

Potential Energy (w.r.t. equilibrium point) $ U\equal{}m_1g(L_1\plus{}L_2\minus{}y_1)\plus{}m_2g(L_1\plus{}L_2\minus{}y_2)$

Hence, Lagrangian $ \mathcal{L}\equal{}T\minus{}U\equal{}\left( \frac{1}{2}m_1 L_1^2\dot{\theta}_1^2\right) \plus{}\frac{1}{2}m_2\left(  L_1^2\dot{\theta}_1^2\plus{}L_2^2\dot{\theta}_2^2\plus{}2L_1\dot{\theta}_1L_2\dot{\theta}_2(\sin\theta_1\sin\theta_2\plus{}\cos\theta_1\cos\theta_2)\right)\minus{}m_1g(L_1\plus{}L_2\minus{}L_1\sin\theta_1)\plus{}m_2g(L_1\plus{}L_2\minus{}(L_1\sin\theta_1\plus{}L_2\sin\theta_2))$

We have the Lagrangian equations (coördinates $ \theta_1$ and $ \theta_2$)
$ \frac{d}{dt}\left( \frac{\partial \mathcal{L}}{\partial\dot{\theta}_1}\right) \minus{}\frac{\partial \mathcal{L}}{\partial \theta_1}\equal{}0$
and
$ \frac{d}{dt}\left( \frac{\partial \mathcal{L}}{\partial\dot{\theta}_2}\right) \minus{}\frac{\partial \mathcal{L}}{\partial \theta_2}\equal{}0$

By rote differentiation, we get the pair of equations
$ (m_1\plus{}m_2)L_1^2\ddot{\theta}_1\plus{}m_2L_1L_2\ddot{\theta}_2\cos(\theta_1\minus{}\theta_2)\equal{}\minus{}(m_1\plus{}m_2)gL_1\sin\theta_1$
and
$ m_2l_2^2\ddot{\theta_2}\plus{}m_2L_1L_2\ddot{\theta}_1\cos(\theta_1\minus{}\theta_2)\minus{}m_2L_1L_2\dot{\theta}_1\dot{\theta}_2\sin (\theta_1\minus{}\theta_2)\equal{}\minus{}m_2gL_2\sin\theta_2$
which describe the motion of the pendulum

P.S.:This is a solved example in the junior-year classical mechanics class at my local university  :)

i was catually looking for a non lagrangian solution...i remeber seein one by Djole once here...if u can search for it ..then fine


What is the greatest three-digit integer in which the product of the digits is 0?

The product of the digits will be zero if and only if at least one of the digits is a zero.  To maximize the value of the number, the zero must be in the ones place.  The 100s and 10s place should both be nines.  Therefore the number is $ \boxed{990}$.


The pedal triangle of the first isodynamic point is an equilateral triangle. (Easy if you know the fact that the first isodynamic point is the isogonal conjugate of the Fermat Point)

In the diagram, draw perpendiculars from Is1 to Lb, Lc and let the foot of the perpendiculars be D,E.

Draw an equilateral triangle DEF (with DE as a side). The line perpendicular to Is1 F and passing through F is La.

Since we're given La,Lb,Lc, we can determine the triangle.


combinatorics solved
combinatorics
games
duplicate

Two players play alternatively on an infinite square grid. The first player puts a X in an empty cell and the second player puts a O in an empty cell. The first player wins if he gets 11 adjacent Xs in a line, horizontally, vertically or diagonally. Show that the second player can always prevent the first player from winning.

[color=#cc0000][mod says: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=57339 ][/color]


Is it possible to express $ \frac {1}{2} \equal{} \frac {1}{n_1^2} \plus{} \frac {1}{n_2^2} \plus{} \frac {1}{n_3^2} \plus{} ... \plus{} \frac {1}{n_k^2}$ where $ n_1,n_2,n_3,...,n_k$ - different positive integers?

I remember having solved this problem. Indeed it is possible to do such task. I will try to think how I had done it. A key concept is that you have to use at least 2,3,4 I think, if you don´t do that you will never reach $ \frac{1}{2}$

Hope this is helpful

By the way where did you find this problem?

Daniel

I found this problem in a list of LitTMO 2001 (Lithuania Teams Mathematical Olympiad) problems.

I don't understand, why 2,3 and 4 must be used  :maybe:

polskimisiek,

How did you find that $ \frac{1}{7^2}\plus{}\frac{1}{14^2}\plus{}\frac{1}{21^2}\equal{}\frac{1}{6^2}$?  :)

I already told you that I knew this problem before. The main idea here is to work around the $ 36$ denominator
Here:
$ \frac {1}{7^{2}} \plus{} \frac {1}{14^{2}} \plus{} \frac {1}{21^{2}} \equal{} \frac {1}{7^{2}}*(1 \plus{}
\frac {1}{4} \plus{} \frac {1}{9}) \equal{} \frac {1}{7^{2}}*\frac {36 \plus{} 9 \plus{} 4}{36} \equal{} \frac {1}{36} \equal{}
\frac {1}{6^{2}}$ :wink:

It is well known that the series $ \frac{1}{n^2}$ converges
Since you can't use $ 1$ it is easy to see that if you don't use $ 2$ you will never reach the number $ \frac{1}{2}$
If you don't use neither $ 3$ nor $ 4$ you reach a contradiction... There is a posibility not to use one of $ 3$ or $ 4$ but looking the problem you realise it is not convinient

Daniel

A tougher (and I believe it's still open) problem is can $ \frac{1}{2}$ be expressed in such a form with less than 12 different numbers :smile:

[quote="ffao"]The solution is not unique. You could also try, for example

$ \frac {1}{2^{2}} \plus{} \frac {1}{3^{2}} \plus{} \frac {1}{4^{2}} \plus{} \frac {1}{5^{2}} \plus{} \frac {1}{7^{2}} \plus{} \frac {1}{12^{2}} \plus{} \frac {1}{15^{2}} \plus{} \frac {1}{20^{2}} \plus{} \frac {1}{28^{2}} \plus{} \frac {1}{35^{2}}$

EDIT: polskimisiek posted 2 minutes before me -- I believe I used only 10 different numbers, if I counted right ;)[/quote]

Well it's great then. You worked around the multipliers of 5 and 4 here...
Still, a nice question is what is the minimal number to express $ \frac{1}{2}$ in such a form (I bet it'd be some computer work needed). I knew that problem from an old book (published in 50's or 60's) so it is not hard to believe it was not so easy for the author to find an example :wink;


inequalities
real analysis
real analysis theorems

How can I prove the following:

if f(n) is O(g(n)) and k is some constant, then f(n)^k is O(g(n)^k)

I'm not sure which property to use, in order to help me prove this.  Any ideas??

Just use the definition. Write what it means that $ f(n)\equal{}O(g(n))$. Now raise the resulting inequality to the power $ k$. Then write what it means that $ f(n)^k\equal{}O(g(n)^k)$ and compare the two statements. If you have done everything right, you'll see that they are almost identical (of course, $ k$ should be positive here).


geometry
angle bisector
geometry proposed

In triangle $ABC$ we have $CA=CB$ and $CD$ is altitude. $P$ is an arbitrary point on $CD$. Let $E=AP \cap BC$ and $F=BP \cap AC$.
Assume that the incircles of triangle $ABP$ and quadrilateral $PECF$ are equal. Prove that the incircles of triangles $ADP$ and $BCP$ are equal too. :)

Triangles $\triangle ACP \cong \triangle BCP$ are congruent for any P and so are their their incircles. Let the common external tangent of the incircles of the triangle $\triangle ABP$ and kite PECF on the same side of their center line CD as B meet AB, AC, AP at X, Y, Z. Since these 2 incircles are congruent, $XY \parallel CD.$ Then the triangle pairs $\triangle ADP \sim \triangle AXZ$ are centrally similar with similarity center A and similarity coefficient $\frac{AP}{AZ}$ and so are their incircles. Likewise, the triangles $\triangle ACP \sim \triangle AYZ$ are centrally similar with the same similarity center A and the same similarity coefficient $\frac{AP}{AZ}$ and so are their incircles. Therefore, if the incircles of $\triangle AXZ,\ \triangle AYZ$ are congruent, the incircles of $\triangle ADP,\ \triangle ACP$ are also congruent. Since the point P is unique, by contradiction, the converse holds as well.

my solution is a little long...but cute
well as the problem said the incircles of $PECF$ and $ACE$ are equal. Assume that $K,L$ are centers of incircles of triangles $ABP$ and $ACE$ respectively and the prpendiculars from $K,L$ to $AE$ intersect $AE$ at $G,H$ respectively. Again as the problem say's we must have $KG=LH$.
triangles $PLH$ and $PKG$ are similar but because $KG=LH$ so triangles $PLH$ and $PKG$ are congruent and so $PL=PK$.
Assume that $I,J$ are the centers of incircles of $BPD$ and $BPC$ respectively, and the perpendiculars from $I,J$ to $CD$ intersect $CD$ at $M,N$ respectively. $IM$ and $JN$ are the inradii of triangles $BPD$ and $BPC$ so we just have to prove that $IM=JN$.
In triangle $BPK$, $PI$ is angle-bisector so: $\frac{BI}{IK}=\frac{BP}{PK}$          $(1)$
In triangle $BPL$, $PJ$ is angle-bisector so: $\frac{BJ}{JL}=\frac{BP}{PK}$          $(2)$
by $(1)$ and $(2)$ and that $PK=PL$ we have : $\frac{BJ}{JL}=\frac{BI}{IK}$ and so $IJ\|KL$ or $IJ\|MN$, so $IJNM$ is a rectangular and $IM=JN$.


show that for every naturel number n
n^2 + 3n +1 and  2n+1 are coprime

I'm not sure but..

[hide="View"]

If $n^2 + 3n + 1$ and $2n + 1$ are NOT coprime, then this means that there is a number that divides both of them. That number can be found as $n^2 + 3n + 1 - (2n + 1) \Rightarrow n^2 + n$.

But clearly, $n^2 + n$ is not factor of neither $n^2 + 3n +1$ or $2n +1$ so they must be coprime.[/hide]

a common divisor must divide also (2n+1)(n+2) which is 2n^2 + 5n + 2
IT DIVIDES ALSO THE DIFERENCE 2(n^2 + 3n +1) - (2n^2 + 5n + 1) which is n.SO a common divisor divides n and 2n+1 .But those two numbers are coprime.SO the commmon divisor is 1

[quote="Silverfalcon"][hide="View"]... this means that there is a number that divides both of them. That number can be found as $n^2 + 3n + 1 - (2n + 1) \Rightarrow n^2 + n$.[/hide][/quote]Why would the difference between the two numbers have to be the number that divides them? Also, I don't know if you care but you have a habit of using $\Rightarrow$ where you should use $=$. One statement implying another is very different from two things being equal.

[hide="my attempt"]Suppose there exists a prime $p$ such that $p\mid2n+1$ and $p\mid{n^2+3n+1}$. Then of course $p\mid{n^2+n}=n(n+1)$. Since $p\mid{n}$ would not allow $p\mid2n+1$, we have $p\mid{n+1}$.

So the assumption that such a prime exists leads to both $p\mid2n+1$ and $p\mid{n+1}$. But these are relatively prime
[hide="proof"]Suppose $(n+1,2n+1)=d$ is greater than $1$. Then for some $r$ we have $n+1=rd$, but then $2n+1=2rd-1=d\left(2r-\frac{1}{d}\right)$, which contradicts the assumption that $d>1$, which requires $d\mid{2n+1}$.[/hide]so we have a contradiction.[/hide]

:oops: Oops.

For the my attempt itself, I don't know why I choose to use Euclid's algorithrm there. Maybe because I solved one olympiad problem using that. I don't know. :P

This one is from Indian National Maths Olympiad.

Find all primes p, q and even n>2, such that
p^n + p^(n-1) + .... + p^2 + p = q^2 + q

[hide]
If p is a prime that divides n^2+3n+1 and 2n+1 then it must also divide their difference, n(n+1).  If p divides n(n+1) it must divide one and only one of n or (n+1).  Since n and n+1 are relatively prime, then any number p that divides one of them DOES NOT divdie their sum.  However, n+n+1 = 2n+1.  So if p | n(n+1), then p $\not |$ 2n+1.  So n^2+3n+1 and 2n+1 are relatively prime.[/hide]


In a $ 1 \times n$ board you have to put the integers from $ 1$ to $ n$, without two consecutive integers in two consecutive (sharing a side) squares. What's the number of posible distributions? And if instead of a board it's a circumference (the last and the first cell are consecutive)?

I hope I didn't miss calculate the final answer:
the answer to the first case is $ \sum_{k\equal{}1}^n(\minus{}1)^{k\plus{}1}(n\minus{}k\plus{}1)!(k)!$,and the second one can be done in $ \sum_{k\equal{}1}^n(\minus{}1)^{k\plus{}1}(n\minus{}k)!k!$ many ways.

[quote="BaBaK Ghalebi"]I hope I didn't miss calculate the final answer:
the answer to the first case is $ \sum_{k \equal{} 1}^n( \minus{} 1)^{k \plus{} 1}(n \minus{} k \plus{} 1)!(k)!$,and the second one can be done in $ \sum_{k \equal{} 1}^n( \minus{} 1)^{k \plus{} 1}(n \minus{} k)!k!$ many ways.[/quote]

Let's try $ n \equal{} 4$ for the first case.
Trying we see there are 2 posibilities:
$ 3142$ and $ 2413$.

With your formula (I don't know how you made it), I get:
$ ( \minus{} 1)^{1 \plus{} 1}(4 \minus{} 1 \plus{} 1)!(1)! \plus{} ( \minus{} 1)^{2 \plus{} 1}(4 \minus{} 2 \plus{} 1)!(2)! \plus{} ( \minus{} 1)^{3 \plus{} 1}(4 \minus{} 3 \plus{} 1)!(3)! \plus{} ( \minus{} 1)^{4 \plus{} 1}(4 \minus{} 4 \plus{} 1)!(4)! \equal{} 0$.

And let's try $ n \equal{} 5$ for the second case:
Trying we see there are 10 posibilities:
$ 13524$, $ 14253$, $ 24135$, $ 25314$, $ 31425$, $ 35241$, $ 41352$, $ 42531$, $ 52413$ and $ 53142$. I expect I don't miss any case.

With your formula (again, I don't know how you made it), I get:
$ ( \minus{} 1)^{1 \plus{} 1}(5 \minus{} 1)!1! \plus{} ( \minus{} 1)^{2 \plus{} 1}(5 \minus{} 2)!2! \plus{} ( \minus{} 1)^{3 \plus{} 1}(5 \minus{} 3)!3! \plus{} ( \minus{} 1)^{4 \plus{} 1}(5 \minus{} 4)!4! \plus{} ( \minus{} 1)^{5 \plus{} 1}(5 \minus{} 5)!5! \equal{} 5! > 10$

Am I missing something? Excuse me if it's that... Anyway, please tell how do you get the formula.

sorry,the way I used to deduce the formulas is totally wrong,I'll try to come up with something better


vector
induction
algebra
polynomial
invariant
linear algebra
matrix

[quote="Moubinool"]Let $ E$ be a $ \mathbb{C}$-vector space of finite dimension. Let $ u,v \in L\left(E\right)$ be such that $ uv \equal{} vu$.

Prove the equivalence of the following assertions:

[b]1/[/b] $ \mathrm{ker}\left(u\right) \cap \mathrm{ker}\left(v\right) \equal{} \left\{0\right\}$.
[b]2/[/b] $ \mathrm{Im}\left(u\right) \plus{} \mathrm{Im}\left(v\right) \equal{} E$.
[b]3/[/b] There exists some $ t\in \mathbb{C}$ such that $ u \plus{} tv$ is invertible.[/quote]

I edited your post to make things clearer (it's the "equivalence of", not the "equivalence with" - the latter could mean that $ uv \equal{} vu$ is equivalent to these three assertions).

The above problem was given to me by mathmanman some weeks(?) ago. How was it supposed to be solved? I only have something incredibly ugly, which I have no motivation at all to write up en detail. Here are the main ideas:

[color=blue][b]Lemma 1.[/b] Let $ u$ and $ v$ be two nilpotent endomorphisms of a vector space $ F$ such that $ uv \equal{} vu$. Then, the following assertions are equivalent:
[b](a)[/b] $ F \equal{} 0$.
[b](b)[/b] $ \mathrm{Im}\left(u\right) \plus{} \mathrm{Im}\left(v\right) \equal{} F$.
[b](c)[/b] $ \mathrm{ker}\left(u\right) \cap \mathrm{ker}\left(v\right) \equal{} 0$.[/color]

[i]Proof of Lemma 1.[/i] Of course, [b](a)[/b] $ \Longrightarrow$ [b](b)[/b] and [b](a)[/b] $ \Longrightarrow$ [b](c)[/b]. Proving that [b](c)[/b] $ \Longrightarrow$ [b](a)[/b] is rather easy (assume that $ F\neq 0$; if $ u \equal{} 0$, everything is clear; else, let $ x$ be a vector with $ ux\neq 0$, and let $ \left(n,m\right)$ be a "maximal" pair of nonnegative integers such that $ u^nv^mx\neq 0$ (here, "maximal" means that $ u^{n \plus{} 1}v^mx \equal{} 0$ and $ u^nv^{m \plus{} 1}x \equal{} 0$; such a "maximal" pair exists since $ u$ and $ v$ are nilpotent); then, $ u^nv^mx\in \mathrm{ker}\left(u\right) \cap \mathrm{ker}\left(v\right)$, contradicting [b](c)[/b]). Proving [b](b)[/b] $ \Longrightarrow$ [b](a)[/b] is not too hard either (just note that $ \mathrm{Im}\left(u\right) \plus{} \mathrm{Im}\left(v\right) \equal{} F$ yields $ \mathrm{Im}\left(u^k\right) \plus{} \mathrm{Im}\left(v^k\right) \equal{} F$ for every positive integer $ k$, as can be shown by induction over $ k$ and using $ uv \equal{} vu$; now apply to $ k$ large enough to have $ u^k \equal{} v^k \equal{} 0$ (remember, $ u$ and $ v$ are nilpotent), and you are done). Lemma 1 is proven.

[i]Proof of [b]1/[/b] $ \Longleftrightarrow$ [b]2/[/b].[/i] This holds for every field $ k$ instead of $ \mathbb{C}$. We can WLOG assume that $ k$ is algebraically closed or, at least, extended far enough that all eigenvalues of $ u$ and $ v$ are in $ k$. For any two elements $ \lambda$ and $ \mu$ of $ k$, denote
$ W\left(\lambda,\mu\right) \equal{} \left\{x\in E\mid ux \equal{} \lambda x\text{ and }vx \equal{} \mu x\right\}$ (the letter $ W$ stands for "weight space", which is how these $ W\left(\lambda,\mu\right)$'s are called in representation theory)
$ X\left(\lambda,\mu\right) \equal{} \left\{x\in E\mid \left(u \minus{} \lambda\mathrm{id}\right)x \equal{} \left(v \minus{} \mu\mathrm{id}\right)x \equal{} 0\right\}$ (these are some kind of extended weight spaces).
It is clear that $ W\left(\lambda,\mu\right)$ and $ X\left(\lambda,\mu\right)$ are subspaces of $ E$ with $ W\left(\lambda,\mu\right)\subseteq X\left(\lambda,\mu\right)$. Of course, both $ W\left(\lambda,\mu\right)$ and $ X\left(\lambda,\mu\right)$ are zero unless $ \lambda$ is an eigenvalue of $ u$ and $ \mu$ is an eigenvalue of $ v$ (but even if this holds, they can still be zero).

It is easy to see that $ E$ is the direct sum of all $ X\left(\lambda,\mu\right)$ over all pairs $ \left(\lambda,\mu\right)$ (only the pairs with $ \lambda$ being an eigenvalue of $ u$ and $ \mu$ being an eigenvalue of $ v$ can contribute nonzero subspaces to this sum). (To see this, write $ E$ as the direct sum of all extended eigenspaces $ U_i$ of $ u$, and then write every such $ U_i$ as the direct sum of all extended eigenspaces of $ v\mid_{U_i}$; here, it is important that $ v\mid_{U_i}$ maps $ v$ to $ v$, because $ uv \equal{} vu$.)
The decomposition of $ E$ into the subspaces $ X\left(\lambda,\mu\right)$ has the very useful property that each of $ u$ and $ v$ maps each of these subspaces $ X\left(\lambda,\mu\right)$ to itself. Hence, the maps $ u$ and $ v$ decompose into endomorphisms of these little subspaces.

Thus, assertions [b]1/[/b] and [b]2/[/b] "decompose" as well:

- Assertion [b]1/[/b] becomes equivalent to the assertion that $ \mathrm{ker}\left(u\mid_{X\left(\lambda,\mu\right)}\right) \cap \mathrm{ker}\left(v\mid_{X\left(\lambda,\mu\right)}\right) \equal{} \left\{0\right\}$, which is obviously true for all pairs $ \left(\lambda,\mu\right)\neq\left(0,0\right)$ and holds for $ \left(\lambda,\mu\right) \equal{} \left(0,0\right)$ only if $ X\left(0,0\right) \equal{} 0$ (by Lemma 1 [b](a)[/b] $ \Longleftrightarrow$ [b](c)[/b], since $ u\mid_{X\left(0,0\right)}$ and $ v\mid_{X\left(0,0\right)}$ are commuting nilpotent endomorphisms of $ X\left(0,0\right)$). Hence, assertion [b]1/[/b] is equivalent to $ X\left(0,0\right) \equal{} 0$.

- Assertion [b]2/[/b] becomes equivalent to the assertion that $ \mathrm{Im}\left(u\mid_{X\left(\lambda,\mu\right)}\right) \plus{} \mathrm{Im}\left(v\mid_{X\left(\lambda,\mu\right)}\right) \equal{} X\left(\lambda,\mu\right)$, which is true for all pairs $ \left(\lambda,\mu\right)\neq\left(0,0\right)$ (since for every such pair, we can WLOG assume that $ \lambda\neq 0$, in which case already $ \mathrm{Im}\left(u\mid_{X\left(\lambda,\mu\right)}\right)$ is the whole space $ X\left(\lambda,\mu\right)$ (because the map $ u\mid_{X\left(\lambda,\mu\right)}$ has $ \lambda$ as its only eigenvalue and thus is invertible since $ \lambda\neq 0$), and holds for $ \left(\lambda,\mu\right) \equal{} \left(0,0\right)$ only if $ X\left(0,0\right) \equal{} 0$ (by Lemma 1 [b](a)[/b] $ \Longleftrightarrow$ [b](b)[/b], since $ u\mid_{X\left(0,0\right)}$ and $ v\mid_{X\left(0,0\right)}$ are commuting nilpotent endomorphisms of $ X\left(0,0\right)$). Hence, assertion [b]2/[/b] is equivalent to $ X\left(0,0\right) \equal{} 0$.

This proves [b]1/[/b] $ \Longleftrightarrow$ [b]2/[/b] over any field $ k$. To add [b]3/[/b] into our equivalence, I fear that we need $ k$ to be infinite (at least, my proof uses it). It is very easy to see that [b]3/[/b] $ \Longrightarrow$ [b]1/[/b] and [b]3/[/b] $ \Longrightarrow$ [b]2/[/b]. I have no idea how to directly show [b]2/[/b] $ \Longrightarrow$ [b]3/[/b], but there is a

[i]Proof of [b]1/[/b] $ \Longleftrightarrow$ [b]3/[/b].[/i] WLOG say $ E \equal{} k^n$, so that $ u$ and $ v$ are $ n\times n$ matrices. We assume that [b]3/[/b] doesn't hold. Thus, $ u \plus{} tv$ is non-invertible for every $ t\in k$, so that $ \det\left(u \plus{} tv\right) \equal{} 0$ for every $ t\in k$. Since $ k$ is infinite, this yields $ \det\left(u \plus{} Xv\right) \equal{} 0$ in $ k\left(X\right)$ (where $ X$ is a polynomial variable). This, in turn, shows that there exists a nonzero vector $ p\in \left(k\left(X\right)\right)^n$ with $ \left(u \plus{} Xv\right)p \equal{} 0$. WLOG assume that $ p\in \left(k\left[X\right]\right)^n$ (otherwise, multiply $ p$ with the common denominator). Write $ p$ in the form

$ p \equal{} p_0 \plus{} p_1X \plus{} p_2X^2 \plus{} ... \plus{} p_rX^r$, where $ p_i\in k^n$ for every $ i$.

Define $ p_i \equal{} 0$ for all $ i < 0$ and for all $ i > r$. Then, $ \left(u \plus{} Xv\right)p \equal{} 0$ is equivalent to $ \left(u \plus{} Xv\right)\left(p_0 \plus{} p_1X \plus{} p_2X^2 \plus{} ... \plus{} p_rX^r\right) \equal{} 0$, what (by comparison of coefficients) is equivalent to

$ 0 \equal{} up_{i \plus{} 1} \plus{} vp_i$ for all integers $ i$.

Let $ i$ be such that $ p_i\neq p_{i \plus{} 1}$ (such an $ i$ exists, since otherwise all $ p_i$ would be equal, therefore all $ p_i$ would be $ 0$, contradicting our choice of $ p$ as a [i]nonzero[/i] vector). Let $ \rho \equal{} p_i \minus{} p_{i \plus{} 1}$; then, $ \rho\neq 0$.

Prove (by induction over $ s$) that $ u^s\left(p_i \minus{} p_{i \plus{} 1}\right) \equal{} \left( \minus{} 1\right)^sv^s\left(p_{i \minus{} s} \minus{} p_{i \minus{} s \plus{} 1}\right)$ for all integers $ s$. Conclude that $ u^s\left(p_i \minus{} p_{i \plus{} 1}\right) \equal{} 0$ for all large enough $ s$ (namely, for all $ s > i \plus{} 1$). Similarly show that $ v^s\left(p_i \minus{} p_{i \plus{} 1}\right) \equal{} 0$ for all large enough $ s$. So we have $ u^s\rho \equal{} v^s\rho \equal{} 0$ for all sufficiently large $ s$.
Now let $ \left(a,b\right)$ be a "maximal" pair of nonnegative integers such that $ u^av^b\rho\neq 0$ (here, "maximal" means that $ u^{a \plus{} 1}v^b\rho \equal{} 0$ and $ u^av^{b \plus{} 1}\rho \equal{} 0$; note that such a maximal pair exists since $ uv \equal{} vu$ and since $ u^s\rho \equal{} v^s\rho \equal{} 0$ for all sufficiently large $ s$). Then, $ u^av^b\rho\in \mathrm{ker}\left(u\right) \cap \mathrm{ker}\left(v\right)$. Thus, $ \mathrm{ker}\left(u\right) \cap \mathrm{ker}\left(v\right) \neq 0$, and [b]1/[/b] cannot hold. This proves [b]1/[/b] $ \Longleftrightarrow$ [b]3/[/b].

Did anyone understand this? I am anything but sure whether this is all correct... and since ENS oral problems can be both trivial and insanely difficult, I have no idea whether this is reasonable.

  darij

The result seems valid for any field $ K$.
We assume $ U\equal{}[u_{ij}],V\equal{}[v_{ij}]\in\mathcal{M}_n(\mathbb{K})\subset\mathcal{M}_n(\mathbb{\bar{K}})$ where $ \bar{K}$ is the algebraical closure of $ K$;
We reason by recurrence upon the dimension $ n$; if $ n\equal{}1$ then the result is obvious.
$ UV\equal{}VU$ then we may assume $ U,V$ upper triangular (why ?). Let $ (e_i)_i$ be the vectors of the basis.
$ (3)$ can be rewritten:for all $ i$,  $ v_{ii}\not\equal{}0$ or $ u_{ii}\not\equal{}0$.
$ (3)\rightarrow {(1)}$: (easy) if $ Ux\equal{}Vx\equal{}0$ then $ (U\plus{}tV)x\equal{}0$ and $ x\equal{}0$.
$ (3)\rightarrow {(2)}$: (easy) let $ y\in{E}$; there exists $ x\in{E}$ s.t. $ U(x)\plus{}V(tx)\equal{}y$.
$ (1)\rightarrow {(3)}$: if $ (3)$ is false then there exists $ i$ s.t. $ u_{ii}\equal{}v{ii}\equal{}0$. Let $ F$ be the vector space spanned by $ e_1,\cdots,e_{n\minus{}1}$. If $ i<n$ then we obtain the contradiction with $ H$ (the hypothesis at rank $ n\minus{}1$). Thus there exists $ t$ s.t. $ W\equal{}U\plus{}tV$ satisfies: if $ i<n$ then $ w_{ii}\not\equal{}0$ and $ w_{nn}\equal{}0$. $ v_{nn}\equal{}0$ implies $ Im(V)\subset{F}$. $ ker(U\plus{}tV)$ is a vectorspace of dimension $ 1$ that is $ V$-invariant because $ UV\equal{}VU$. Moreover $ ker(U\plus{}tV)\cap{F}\equal{}0$. Finally $ V_{|ker(U\plus{}tV)}$ is not onto (thus is $ 0$) and $ ker(U\plus{}tV)\subset{ker(V)}$, a contradiction.
$ (2)\rightarrow{(3)}$: if $ (3)$ is false then there exists $ i$ s.t. $ u_{ii}\equal{}v{ii}\equal{}0$. If $ i<n$ then $ H$ implies the contradiction. Thus $ i\equal{}n$ and $ Im(U)\subset{F}$, $ Im(V)\subset{F}$, a contradiction.

[quote="loup blanc"]The result seems valid for any field $ K$.
We assume $ U \equal{} [u_{ij}],V \equal{} [v_{ij}]\in\mathcal{M}_n(\mathbb{K})\subset\mathcal{M}_n(\mathbb{\bar{K}})$ where $ \bar{K}$ is the algebraical closure of $ K$;[/quote]

Please be careful here - property [b]3/[/b] doesn't behave well when extending the base field from a finite field to a larger one.

[quote="loup blanc"]We reason by recurrence upon the dimension $ n$; if $ n \equal{} 1$ then the result is obvious.
$ UV \equal{} VU$ then we may assume $ U,V$ upper triangular (why ?).[/quote]

This is nice. Any simple proof of this (simpler than reducing to the case of both $ u$ and $ v$ nilpotent (by decomposing the vector space into the $ X\left(\lambda,\mu\right)$'s) and then inducing on the dimension by replacing $ F$ by $ F\slash\left(\mathrm{Im}\left(u\right) \plus{} \mathrm{Im}\left(v\right)\right)$) ?

Anyway, your solution works. Now I know a much more beautiful (at least, than mine) solution by NoName (who claims he has no time to post it...), which roughly goes as follows:

First, let's suitably generalize the problem to all fields $ K$:

[color=blue]Let $ K$ be a field, and $ E$ a finite-dimensional $ K$-vector space. Let $ u$ and $ v$ be endomorphisms of $ E$ such that $ uv \equal{} vu$. Then, the following three relations are equivalent:

[b](1)[/b] $ \mathrm{ker}u \cap \mathrm{ker}v \equal{} 0$.
[b](2)[/b] $ \mathrm{Im}u \plus{} \mathrm{Im}v \equal{} E$.
[b](3)[/b] In the polynomial ring $ K\left[X\right]$, we have $ \det\left(u \plus{} Xv\right)\neq 0$.[/color]

It is easy to see that [b](1)[/b] $ \Longleftrightarrow$ [b](3)[/b] and [b](2)[/b] $ \Longleftrightarrow$ [b](3)[/b] are equivalent (by replacing $ u$ and $ v$ by $ u^{\ast}$ and $ v^{\ast}$), so it remains to show [b](1)[/b] $ \Longleftrightarrow$ [b](3)[/b]. I don't remember how NoName did this, but one proof was given in my post above (though I think he had a shorter one).

Finally, a question: Do we always have $ \dim\left(\mathrm{ker}u \cap \mathrm{ker}v\right) \plus{} \dim\left(\mathrm{Im}u \plus{} \mathrm{Im}v\right) \equal{} \dim E$ ? Does the rank of $ u \plus{} Xv$ has something to do with this? [EDIT: no and no, see below posts.]

  darij

[quote="darij grinberg"]Finally, a question: Do we always have $ \dim\left(\mathrm{ker}u \cap \mathrm{ker}v\right) \plus{} \dim\left(\mathrm{Im}u \plus{} \mathrm{Im}v\right) \equal{} \dim E$ ? Does the rank of $ u \plus{} Xv$ has something to do with this?
  darij[/quote]
Do you keep the asumption $ uv\equal{}vu$?

[quote="JC_math"][quote="darij grinberg"]Finally, a question: Do we always have $ \dim\left(\mathrm{ker}u \cap \mathrm{ker}v\right) \plus{} \dim\left(\mathrm{Im}u \plus{} \mathrm{Im}v\right) \equal{} \dim E$ ? Does the rank of $ u \plus{} Xv$ has something to do with this?
  darij[/quote]
Do you keep the asumption $ uv \equal{} vu$?[/quote]

Yes, of course. Otherwise, not even [b]1/[/b] $ \Longleftrightarrow$ [b]2/[/b] would hold.

If this makes your job easier, you can WLOG assume that $ u$ and $ v$ are strictly upper triangular matrices (see my post for making them nilpotent, and see loup blanc's post for making them upper triangular).

  darij

"loup blanc wrote:
The result seems valid for any field ."
"Darij wrote:
Please be careful here - property 3/ doesn't behave well when extending the base field from a finite field to a larger one. "
Hi Darij, you are right;
the following is a counter example with $ K\equal{}\mathbb{Z}/2\mathbb{Z}$ and $ n\equal{}2$.
$ U\equal{}\begin{pmatrix}1&1\\1&1\end{pmatrix},V\equal{}\begin{pmatrix}1&0\\1&1\end{pmatrix}$.
$ det(U\plus{}tV)\equal{}t^2\plus{}t\equal{}0$ for all $ t\in{K}$. But $ ker(V)\equal{}0$ !!

[quote="darij grinberg"]Finally, a question: Do we always have $ \dim\left(\mathrm{ker}u \cap \mathrm{ker}v\right) \plus{} \dim\left(\mathrm{Im}u \plus{} \mathrm{Im}v\right) \equal{} \dim E$ ?[/quote]

I see, none of you wanted to ridicule me by answering to my stupid question:

$ u \equal{} \left( \begin{array}{cccc}0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0\end{array}\right)$;
$ v \equal{} \left( \begin{array}{cccc}0 & 1 & 1 & 1 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0\end{array}\right)$

as endomorphisms of $ K^4$. Yes, they do commute. The intersection of the kernels has dimension $ 1$, the sum of the images has dimension $ 2$.

We cannot generally prove $ \dim\left(\mathrm{ker}u \cap \mathrm{ker}v\right) \plus{} \dim\left(\mathrm{Im}u \plus{} \mathrm{Im}v\right) \leq \dim E$ or $ \dim\left(\mathrm{ker}u \cap \mathrm{ker}v\right) \plus{} \dim\left(\mathrm{Im}u \plus{} \mathrm{Im}v\right) \geq \dim E$ either, because any of these inequalities would imply the opposite one by dualization.

[b]EDIT:[/b] However, $ \dim\left(\mathrm{ker}u \cap \mathrm{ker}v\right) \plus{} \dim\left(\mathrm{Im}u \plus{} \mathrm{Im}v\right) \equal{} \dim E$ at least holds in the case when $ u$ and $ v$ are two strictly upper-triangular matrices with only the first secondary diagonal being nonzero (i. e. in the case when $ u_{i,j} \equal{} v_{i,j} \equal{} 0$ for any $ i$ and $ j$ with $ j \neq u \plus{} 1$). This shows that not all pairs $ \left(u,v\right)$ of commuting nilpotent endomorphisms can be transformed into two such matrices by a similitude; in particular, this yields that not every pair of commuting endomorphisms can be transformed into their Jordan forms by one and the same similitude. Thus, to some extent, Loup Blanc's result that every pair of commuting endomorphisms can be transformed into upper triangular matrices by one and the same similitude is the best possible one unless one assumes both endomorphisms (maybe one is already enough?) to be diagonalizable.

  darij


Given $ a, b, c \ge 0$ satisfy $ a^3\plus{}b^3\plus{}c^3\plus{}abc\equal{}12$. Prove that: $ \frac{19(a^2\plus{}b^2\plus{}c^2)\plus{}6(ab\plus{}bc\plus{}ca)}{a\plus{}b\plus{}c} \ge 36$
That's a nice day for the come back with inequality!  :)

[quote="nguoivn"]Given $ a, b, c \ge 0$ satisfy $ a^3 \plus{} b^3 \plus{} c^3 \plus{} abc \equal{} 12$. Prove that: $ \frac {19(a^2 \plus{} b^2 \plus{} c^2) \plus{} 6(ab \plus{} bc \plus{} ca)}{a \plus{} b \plus{} c} \ge 36$
That's a nice day for the come back with inequality!  :)[/quote]
See here: http://canhang2007.wordpress.com/2009/11/02/inequality-33-t-q-anh/


In the process of trying to prove another inequality, I discovered the following inequality. Anybody interested in proving it?

Let   $ a,b,c>0 $. Prove that:

\[ \frac{a^2}{b^2+bc+c^2} + \frac{b^2}{c^2+ca+a^2} + \frac{c^2}{a^2+ab+b^2}\geq 1. \]

In another thread, I proposed the following inequality:

[tex]\frac{a^{2}}{b^{2}+bc+c^{2}}+\frac{b^{2}}{c^{2}+ca+a^{2}}+\frac{c^{2}}{a^{2}+ab+b^{2}}\geq \frac{\left( a^{3}+b^{3}+c^{3}\right) /2}{\left( a^{3}+b^{3}+c^{3}\right) /2+3abc}+\frac{2}{3}[/tex].

If this one turns out to be true, then it is stronger than your

[tex]\frac{a^2}{b^2+bc+c^2} + \frac{b^2}{c^2+ca+a^2} + \frac{c^2}{a^2+ab+b^2}\geq 1[/tex],

since we can easily prove

[tex]\frac{\left( a^{3}+b^{3}+c^{3}\right) /2}{\left( a^{3}+b^{3}+c^{3}\right) /2+3abc}+\frac{2}{3}\geq 1[/tex].

How do we prove this? Well, we have $a^{3}+b^{3}+c^{3}\geq 3abc$, and thus

[tex]\frac{\left( a^{3}+b^{3}+c^{3}\right) /2}{\left( a^{3}+b^{3}+c^{3}\right) /2+3abc}=1-\frac{3abc}{\left( a^{3}+b^{3}+c^{3}\right) /2+3abc}\geq 1-\frac{3abc}{3abc/2+3abc}=1-\frac{1}{1/2+1}=1-\frac{2}{3}=\frac{1}{3}[/tex],

so

[tex]\frac{\left( a^{3}+b^{3}+c^{3}\right) /2}{\left( a^{3}+b^{3}+c^{3}\right) /2+3abc}+\frac{2}{3}\geq \frac{1}{3}+\frac{2}{3}=1[/tex].

Therefore your inequality follows from mine. But is mine correct?

  Darij

I will think about your inequality, Darij. I just wanted to say that the first inequality is trivial, since it follows by $\frac{a^2}{b^2+bc+c^2}\geq \frac{2a^2}{3(b^2+c^2)} $ and Nesbit inequality. This inequality was published some years ago by Marian Ursarescu ( i really don't like his problems!) in Gazeta Matematica.

Well,if my computation is right.Your inequality:
[tex] \frac{a^2}{b^2+bc+c^2}+\frac{b^2}{a^2+ac+c^2}+\frac{c^2}{a^2+ab+b^2} \geq \frac{a^3+b^3+c^3}{a^3+b^3+c^3+6abc} +\frac23 [/tex]
is right,darij grinberg. :)
However,it is too long to write down.  :(

This inequality below is more interesting,stronger but super-long :(.
I post it as a gift for manlio  :)
Prove that:
[tex] \sum \frac{a^2}{b^2+bc+c^2} \geq \frac{a^3+b^3+c^3}{a^3+b^3+c^3+6abc}+\frac23+\sum\frac{(b-c)^2}{6(b^2+bc+c^2)} [/tex] where $a,b,c \geq 0$   :D

Your inequality is too much for me, Anh Cuong. Anyway, an ugly solution for your inequality, Darij:

  First of all, I will use Cauchy trick to find that the left-hand side minus $\frac{2}{3} $  is at least
 $\frac{ 3(a^4+b^4+c^4)+(ab-bc)^2+(ac-bc)^2+(ab-ac)^2}{6(a^2b^2+b^2c^2+c^2a^2)+3abc(a+b+c)} $ and thus at least $\frac{a^4+b^4+c^4}{2(a^2b^2+b^2c^2+c^2a^2)+abc(a+b+c)} $. Thus, it suffices to show that this last quantity is at least the right-hand side. This is not very trivial. It can be written $ (a^3+b^3+c^3)( 2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4+abc(a+b+c))\leq 6abc(a^4+b^4+c^4) $. This follows from the fact that $ 2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4$ is at most $ abc(a+b+c)$ by Schur inequality and the fact that $ (a+b+c)(a^3+b^3+c^3)$ is at most $ 3(a^4+b^4+c^4) $ by Cebasev.


AMC
AIME
AMC 12
AMC 12 B
USA(J)MO
USAMO
calculus

Hi, I took the AMC12B this year for the first time, and I scored a 105. If I understand correctly, 100 is the cutoff right? So, I barely made it. How does the AIME work? Whats the format? Also, how does the qualification for USAMO work?

I know that since I made a 105 on the AMC12B, I'm going to have to score quite high on the AIME to make USAMO, right? And how are the cutoffs determined? Are they pre-determined or determined based off student performance?

And finally, what can I do to score high on AIME? I know that I don't have the years of experience with olympiad math that most of you guys do, but I'm willing to work really hard, and I want to make USAMO.

Thanks.

As for my math background (which isn't as close to impressive as most of you guys), I took BC Calculus as a sophmore and scored a 5 on the AP exam / 800 on SATII Math Level 2.

Any suggestions and or books recommended are welcome.

Congratulations on your first AIME qualification!  :)

As for AIME cutoffs for making USAMO, qualification [i]is[/i] based on student performance, and if you want to know more, you might want to take a look [url=http://www.unl.edu/amc/e-exams/e8-usamo/usamo.shtml]here[/url] (The bottom of the page is mostly where you can find the information you're looking for.)

One of the best ways you can practice for the AIME is to try practice ones from old years. I'm pretty sure you can find those somewhere on the AMC website. ([url]http://www.unl.edu/amc[/url])

Anyway, good luck on the AIME!

What is the "general" cutoff for the AIME?

And how does the AIME scoring work?

The AIME has 15 questions, and the answers are always integers from 0 to 999, it's like the fill-in questions on the SAT.  You also have 3 hours.  TO qualify for the USAMO, you have to have one of the top number of indeces (I dunno what that number is).  USAMO index = AMC score + 10*AIME score.  Last year, you needed a 217.5 index to qualify, but it will probably be lower this year with the expansion.

Since there's like 3 weeks before the AIME, the best thing to do is take the AOPS AIME seminar, look at previous mathjams on AoPS at the AIME level, and go through AIME tests, and make sure you look a lot at the questions you missed or didn't know how to get, because you'll probably need to be able to do all the questions, to account for stupid mistakes and stuff.

AoPS II would probably also be good, but I dunno how much that would help in the amount of time you have.

Well I'm sorry to say I doubt you can make USAMO this year  :( , unless possibly you do significantly better on a longer, harder test. Anyway, here are the details:

The AIME is a 15 question, 3-hour test. All the answers are integers ranging from 0 to 999 inclusive.

USAMO qualification (for juniors and seniors) is based on the index, equal to the AMC score plus 10 times the AIME score. Essentially, the top 320 indicies out of those who took the AMC 12 get to take USAMO, so the index cutoff is determined after the AIME. Last year, the index was a 117, so depending on what the AIME is like this year, the cutoff should be around there, probably at least a bit lower. This means that you'll probably have to get an 11 on the AIME to qualify.

As for what you should do, I think the most important thing is to practice. Take a look at the AIMEs on this website, do an AIME or two at first, and see how you do. Try to solve more of the problems after the test, especially those that are a bit above your level. Also, when you get stuck on a problem, and maybe for some of the harder problems, check out the solution, so that you start to get an idea of what kinds of things people tend to think up to solve AIME problems.

As well, you might want to check out AoPS 2 a bit. I'm not sure how much you can get in a month, but there might be a subject or two where the book has some useful information that you might not have known.

Good Luck!

A good place to start is to take a practice AIME under timed conditions. That's just so you can get an idea of where you stand. Then after that you can read up on The Art of Problem Solving I and II to learn any concepts you don't know. Don't take the advice too seriously, as the best wisdom is still gained by personal experience (i.e. practicing old AIMEs).

Now if you'd really like to improve, you're going to have to work pretty hard, especially not giving up on problems that you can't do at first sight. Of course, you're allowed to look at the solution eventually, but almost no one participating in contest math for the first time is used to spending a lot of time to think.

I would estimate your maximum thinking time for a school math problem (which probably comprises most of the math you are exposed to) to be about 5 minutes. For AIME problems, be prepared to think for 30+. Try lots of different approaches--usually the solution comes from an effective first step rather than slogging through a lot of algebra.

Work on solving harder problems like this, and your overall math ability will improve so much. I can't promise you'll make USAMO because there's not much time between now and the AIME, but it should be a rewarding experience nonetheless.

P.S. AIME problems are ordered by difficulty, so #15 is generally much harder than #1. I tend to think that anyone who qualified for AIME should be able to do #1, so you should probably start there and see how far up you can go. Don't waste time attempting #15 if your expected score is a 5. Instead, attempt problems around 6-9 and gradually build up.

Hey so, since I took the AMC12B, I assume I take the AIME 2 right?

No. All qualifiers take the AIME I (March 13) unless you have a problem (i.e. school closes or you have some really important time conflict). Then you take the AIME 2.

I am gone that whole week (when the AIME1 is held) for college visits. How do I contact the AMC people and ask them if they will excuse me?

Your school has to email the AMC office (information in the teacher's manual) and request AIME 2 exams. Note that they will have to pay 25 dollars (or 50, if you have 11+ qualifiers in the school) to administer the alternate exam.

EDIT: The link is no longer dead: [url]http://www.unl.edu/amc/d-publication/d1-pubarchive/2006-7pub/07-AIME-TM/TM-%20AIME,2007.pdf[/url]

What is the date of the AIME I? AIME II?

[quote]cool! i never knew that you could take the AIME twice! My school just offers one AIME exam[/quote]
You cant take it twice, its OFFERED(but only can choose 1) twice..

[quote="DarkLinx10"]cool! i never knew that you could take the AIME twice! My school just offers one AIME exam. :)[/quote]

you can't take both of them, as far as i know... either I or II...

LOL thanks for getting my hopes up for nothing......

no offense or anything, but if you like a 3 on aime1, its not like somhow miracuosly your gonna get a 10 on aime 2

it COULD happen, just extremely unlikely.

A student can only take the AIME officially once per year. But, yes, if the AIME I date doesn't work for your school schedule, pay the extra money and clear your schedule on the AIME II date. (That's what we will have to do this year, because of a new school schedule.)

This is my first time on the AIME also.  I scored a 120 on the AMC 10 A and I am an 8th grader.  I took one practice AIME test and only got 3 right.

I realize that I have several years to "practice" for this.  However, am I better off saving some of the practice tests for the future and learning some of the math problems that are posted on AoPS or should I just take practice tests and then learn how to do those problems?

Because of the dates, I am unable to participate in the AoPS workshop offered on this.

Additionally, I am preparing for MathCounts and since it's my last year, I want to do as well as possible on that.

I am genuinely thrilled to be a participant in any math contest because I love math.

Thanks!

I would recommend taking the practice tests, and reading through the solutions very carefully until you understand it.

[quote="jnApp66"]Does anyone know where I could find the old tests online without having to buy them?[/quote]
http://www.kalva.demon.co.uk

Hi,
I'm just wondering how do you know you're qualified? Have they already sent out the letters for those qualifiers (AMC 12B)?

A few minutes ago, we found out the qualifying scores for the 10B and 12B.  :wink:

so will they start to send out the results to schools soon?

[quote="jnApp66"]Does anyone know where I could find the old tests online without having to buy them?[/quote]

http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45


geometry
vector
analytic geometry
incenter
geometry unsolved

let $ {a,b,c}$ be length of three sides of $ {\triangle ABC}$ ,with :$ {a\vec{PA} \plus{}b\vec{PB} \plus{}c\vec{PC}\equal{}\vec{0}}$
then P is the _________of $ {\triangle ABC}$.
[b]A).orthocenter     B).Centroid      C).incenter      D).circumcenter[/b]
 :) Can you give the answer with proof, waiting for your nice solutions!Dear MathLinkers

Dear Mathlinkers ,can anyone solve it ?  :)

$ P\mbox{ \minus{} incenter.}$
[b]Proof:[/b]
$ \mbox{The barycentric coordinate of incenter is: }I(a;b;c)\Rightarrow \\
\Rightarrow \overrightarrow{PI} \equal{} \frac {a.\overrightarrow{PA} \plus{} b.\overrightarrow{PB} \plus{} c.\overrightarrow{PC}}{a \plus{} b \plus{} c}. \\
\mbox{Then: }a.\overrightarrow{PA} \plus{} b.\overrightarrow{PB} \plus{} c.\overrightarrow{PC} \equal{} \overrightarrow{0}\Leftrightarrow P \equal{} I.$

Right, Nice solution !
We can also use area of the triangles to solve this problem :wink:


Prove that for all $ r,n\in [2,\plus{}\inf )$ and $ r,n\in \mathbb{N}$,

\[ r^n>\binom{n\plus{}r\minus{}1}{r\minus{}1}\].

There are more ways to put $ n$ distinguishable balls into $ r$ distinguishable boxes than there are to put $ n$ indistinguishable balls into $ r$ distinguishable boxes?

Combinatorial proofs are valid, you know.

So we are trying to prove that $ r(r \plus{} 1)(r \plus{} 2)\dots(r \plus{} n \minus{} 1) < r^n\cdot n!$, right? Well this is easy...

$ r^n \cdot 1 \cdot 2 \cdots n \equal{} (r)(2r)(3r)(4r)\dots(nr)$. Pairwise, except for the first terms (which are equal), each term of the LHS is strictly less than each term of the RHS, so we have our desired result.

Dang I only thought of the obscure solution.

[hide="obscure solution giveaway"]How many terms are there in $ (x_1\plus{}x_2\plus{}\cdots \plus{}x_r)^n$?[/hide]

That's exactly equivalent to Yongyi's first solution ;) The boxes are $ x_1, \ldots, x_r$ and the $ i$th ball goes in box $ x_j$ if and only if we choose to use $ x_j$ in the $ i$th factor when expanding out.


geometry
3D geometry
tetrahedron
integration
geometry unsolved

I have read that Volume of a Tetrahedron=$ \frac{1}{3}(Area\,\&of\,\&triangle\,\&at\,\&base)(height)=\frac{1}{6}\left[\overrightarrow{a}\,\&\overrightarrow{b}\,\&\overrightarrow{c}\right]$.

How does one account for $ \frac{1}{3}$ and $ \frac{1}{6}$ in the above formulae.

For $ z\equal{}0$ we obtain the triangle at base;let $ S$ be its area. let $ h$ be the height in the direction of $ Oz$. For $ z\equal{}u$ we obtain a triangle whose area is $ S_u\equal{}S(h\minus{}u)^2/h^2$. Then the volume $ V$ of the tetrahedron is $ V\equal{}\int_0^h{S}_udu\equal{}Sh/3$.
Here $ ABCD$ is the considered tetrahedron where $ A,B,C$ are in the plane $ z\equal{}0$. $ a\equal{}\overrightarrow{AB},b\equal{}\overrightarrow{AC},c\equal{}\overrightarrow{AD}$. $ |[a,b,c]|\equal{}|(a(crossprod){b}).c|\equal{}\parallel{}a(crossprod){b}\parallel{}h\equal{}2.area(ABC)h\equal{}2Sh$.


How do you find the product of the divisors of a certain number n? What is the logic behind the approach?

For a non-perfect square integer $ n$, you can split up the divisors into pairs whose product is $ n$. The number of these pairs is simply half the number of divisors $ n$ has. If $ t(n)$ denotes the number of divisors that an integer $ n$ has, then

$ P(n) \equal{} n^{\frac {t(n)}{2}},$

where $ P(n)$ denotes the product of the divisors of $ n$.

For perfect squares, replace $ t(n)$ with $ t(n) \minus{} 1$.

Better yet, you can make the definition:

$ P(n) \equal{} n^{\left\lfloor\frac {t(n)}{2}\right\rfloor}$.

Wait, why are you multiplying n by t(n)/2??

Thanks

Edit: Never mind. I get it now.


Pascal\u0027s Triangle
geometric sequence

Can anyone help me with thoroughly good answer to this question?

This is from Hungarian Math Competition

The array below is constructed in a similar way as Pascal's triangle. The first row in the array contains the numbers 1 and 2, every further entry is the sum of the two neighbours above.
***1  2
**1  3  2
*1  4  5  2
1  5  9  7  2
...

a) Determine the sum of the entries in 100th row.
b) What do we get if the entries in 100th row are summed with alternating signs?
c) Determine the 47th element in 100th row.

I'm not familiar with Pascal's Triangle concept. Good explanation would be appreciate.

Thanks.

You don't need to know anything about pascals triangle to solve this question - its not as if there are 'standard' ways to find out terms in these sequences.

Its just a normal problem of working out a pattern.

[hide="hint"]If you just write out what each term is in terms of 1 and 2, you should see a pattern for the first part.[/hide]

[hide="hint part 2"]
1, 2, sum is 1+2
1, 1+2, 2, sum is 1+1+2+2
1, 1+1+2, 1+2+2, 2, sum is 1+1+1+1+2+2+2+2
1, 1+1+1+2, 1+1+1+2+2+2, 1+2+2+2, 2, sum is 8*1 + 8*2.

Starting to see a pattern?
[/hide]

Try to use similar patterns to work out the other answers. Thats basically the only thing you can do with pattern-questions like these: guess at the pattern then prove it.

I'm pretty sure some of the identities in Pascal's Triangle would remain the same with this modified version, just with powers of one and two...

well, it are just the coefficients of the expansion of $(a+2b)(a+b)^n$ by induction. seeing this, everything else is straightforward.

solution to a.
[hide]you are looking for a pattern, so we start adding up the terms in each row. we see that the pattern is:
3*2^(n-1)... where n is the nth row.
so when n = 100, the sum of the terms is:
3*2^99[/hide]

The problem said it was similar to a pascals triangle.

[quote="demji"]i thought pascal's triangle was
       1
      1 1
     1 2 1
    1 3 3 1
.....
in that case is n(n+1)/2
but how would you prove it[/quote]

In pascal's triangle the sum of the terms in the nth row where the row containing "1" is the 0th row is 2^n. Pascal's triangle is the coefficients of (x+y)^n. So if x and y were both 1, we'd get the sum of the coefficients.

Oh.. :)

Let me see...

So,
the first line is 1+2
the second line is 1+2+3 or 1+2+1+2
the third line is 1+4+5+2 or 1+2+2+1+2+1+2
the fourth line is 1+5+9+7+2 or 1+2+2+1+2+2+2+2+1+2+2+2+1+2

Hmm... I was thinking about just calculating and get 3,6,12, and 24. But I don't seem to quite get "how" to get the 100th term..

Is my pattern wrong? Or am I seeing my pattern in other's post and not realizing it?

There is a pattern common to any of these problems where you start out with a line, then add elements to get the next row.

Each element in a certain row is used two times to get the next row. So the sum of the elements in one row is double the sum of the elements in the row above it.

So continuing Silverfalcon's work, if the "1  2" line was line 0, then the formula for the sum of the terms of the nth line would be 3*2^n

Silverfalcon did get a pattern: a geometric sequence with first term 3 and common quotient 2

And yes, your pattern was in other posts. Someone mentioned a pattern of 3,2*3,4*3,8*3,....


Hellos i want to prepare for mathcounts competition and im not that good at math so my goal is only state(if i get to nat's that's awesome). currently im trying to do as much sprints as possible and i have all of them. so besides doing tests, is there other things i should do? any formulas or theorems to remember? (and can somebody please explain Pick's theorem i don't quite get it)

thank you

Pick's thereom is

"IF you want to find the area of a geoboard polygon, the formula is i+b/2-1, where i is the number of points in the interior and b is the number of boundary points.

EDIT:  Oops!  Typo :blush:

Not necessarily a quadrilateral;any polygon with all of its vertices on points of the grid.

AMC10 kind of helps, but its hard compared to school/chapter. AoPS volume 1 is also good, but is has some materials not in MC as well, like complex number and some other stuff. But going through the book is quite helpful.

You might want to do a lot of the previous Mathcounts tests for State and Chapter. Since you want to get to state, then I would focus on Chapter. If you're really good at Countdown, then aim for the top 10 to give you a chance to get into Countdown and beat everyone else. If you're good at written, then practice Countdown.

You can get a lot of the previous tests by trading here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133189

In my opinion, in chapter mathcounts, AMC8 and MOEMS are probably of similar difficulty.  If you live in a difficult chapter, your focus should be to make as few careless mistakes as possible and answer as many problems on the back page as possible(for tiebraker purposes).  Well, I can't do that, but lucky for me, I don't live in a difficult chapter, so I can get by with a 25/46(around there, never received my official score) and still get first place written and cd in chapter.

And my experiences in state.........well.........I'd rather not talk about it.  :blush:

If I were you, I would probably order all the Introduction books and read over them.

Good advice Cliu.  I currently have Intro to Algebra and will get AoPS Vol. 1 soon.  I'm also going to take Advanced Mathcounts/AMC8 in the fall(currently taking Algebra I).  I probably wouldn't have failed as badly in state if I could take the test over now because I've practiced a little with AoPS.  But I've got 2 more chances, and I guess 63rd isn't that bad for a 6th grader who doesn't even understand what the sigma symbol means.

thanks for all the advices. i am working on all intro books, have aops volume 1 and 2, and taking amc 10 problem series, and intro to alg and i am currently doing a lot of past chapter and states and beginning to do some amc 10 things. thanks again

Finish the Introduction series not only helps with Mathcounts, but with AMC 10 and most of AMC 12 as well. Trust me.

[quote="jonathanchou711"]which state do you live in? or if you think that's a bit personal, do you think you live in a very competitive state? Like me, I live in a realllly easy state. Even though I'm not sure about maybach. It sounds like he/she probably lives in a realllly easy state. Because last year I got a 39/46 and i got 3rd. Hm. But it all depends pretty much on where you live. If you live in like Texas. Buckle down. You have to get to work.[/quote]

Actually, Indiana is a very difficult state.  I just live in a very easy chapter where kids don't even know how to find the number of diagonals in a convex polygon,  know how many numbers are in a sequence, or compute combinations. :P  So, jonathanchou711, you made nats!  gj

As for me, it's a little more difficult.  I probably need around 42/46.  Also, 39/46 is a very high score, but I think you typoed because I remember you earlier saying the highest score for Oklahoma was 29, not 39(39 is about what it takes to make nationals in Indiana).
Of course, even if I fail, I have 1 more shot at it before it becomes an epic failure..........

For me, I'm going to buckle down, hope for the best, and pray that I don't screw up in MC States.  But I have to worry about Mark Selkie, and I know he'll pwn me for sure, so I only have 3 slots I even have a chance at.

[quote="Mewto55555"]Wow, jonathan, that is terrible advice. Just because someone can make it to nats/state really easily doesn't mean they shouldn't work. If you want to make nats, and you are in a state where the cutoff is like 20, you should still strive to get a 40 or more on the same test. It shouldn't depend on where you live. I would not have had to work at all to make nationals this year, but that doesn't change the fact that I did. Do the best that you can regardless of your competition.[/quote]

Um.. I didn't say that you didn't have to work. I only said that if you live in a really competitive state,  you have to buckle down and get to work because you have a lot more competition. It's always better to study more.

@maybach: Oops. Didn't clarify. Sorry. I didn't make nats. At chapter, I got 39. At state I got maybe a 27? I forget. Oh. Well.

i probably live in one of the most competitive states, california, which is one reason im not aiming for nats =( the other is that im just not toooooo good at math.


For every triple $(a,b,c)$ of non-zero real numbers, form the number $\frac{a}{|a|} +\frac{b}{|b|} +\frac{c}{|c|} +\frac{abc}{|abc|}$.

The set of all numbers formed is

(A) $\{ 0\}$
(B) $\{ -4, 0, 4\}$
(C) $\{-4, -2, 0, 2, 4\}$
(D) $\{-4, -2, 2, 4\}$
(E) none of these

Sorry for this being kinda off topic, but where did you get such an old AHSME?  :?

I got this problem from 50th Anniversary AHSME, which contains one question from each AHSME.

Go to http://www.unl.edu/amc/a-activities/a7-problems/problems.html and click "50th Anniversary Problem Set".

[quote="VenDiddy"][hide]Just think of $\frac{x}{|x|}$ as yielding +1 if x is positive and -1 if x is negative. You get 1 when you divide $\frac{x}{x}$, but the +/- is retained by the x in the numerator.
[list]
[*]If all are positive then you get 1+1+1+1=4.
[*]If all are negative then you get -1-1-1-1=-4.
[*]If 2 are positive and 1 is negative then you get 1+1-1-1=0.
[*]If 2 are negative and 1 is positive then you get -1-1+1+1=0.[/list]
So I guess the answer would be $\boxed{\text{(B)}\ \{-4, 0, 4\}}$[/hide][/quote]

Oh, this is from AMC.  This was also on the 2005 MMPC.  It was #1.

But I agree with that answer.