induction
greatest common divisor
number theory unsolved
number theory

This is one of my problems, asked in one of the Romanian 2007 Tests (even without the finiteness requirement), see
[url]http://www.mathlinks.ro/viewtopic.php?p=815602#815602[/url]
for an incomplete discussion, which basically however identifies all solutions.

The only such finite sets are $ S \equal{} \{ n, n(n \minus{} 1) \}$, for integer $ n > 2$.
The infinite sets are $ \mathbb{N}^*$, $ \mathbb{N}^* \setminus \{1\}$, $ \mathbb{N}^* \setminus \{2\}$ and $ \mathbb{N}^* \setminus \{1,2\}$.

EDIT. In my problem, $ S$ was required to have at least two elements. Under the statement here, also all sets $ S$ with only one element trivially fulfill (since there exist no distinct $ i$ and $ j$).

In your problem, was  it required for I and j to be distinct

[quote=Konigsberg]In your problem, was it required for i and j to be distinct[/quote]

Yes (The alternative would be too ez)

*I read the problem as determine all S (infinite as well) and was meandering about for an hour :)

[b]We wish to prove that all finite sets fall under the following categories: All sets of the form $S = \boxed{(n, n(n-1))}$ for $n > 2$ and the solutions $\boxed{S = (n)}$. It can be verified that these finite sets satisfy the problem's conditions.[/b]

[b]Lemma 1[/b]
Denote f(i,j) = $\frac{i+j}{(i,j)}$. First, we prove that (i,j) is involutive iff (i,j) falls into the first category.
Note that i = i' * gcd(i,j) and j = j' * gcd(i,j). WLOG, let i < j.
For gcd(i,j) > 1, f(i,j) = i' + j' < 2j' $\leq$ j. Hence, f(i,j) = i. f(i,j) $\iff$ f(i,j) = i' + j' = i $\iff$ (i,j) = (i, i(i-1)). This follows from a gcd manipulation.
For gcd(i,j) = 1, f(i,j) = i+j > j > i. This contradicts our assumption that (i,j) is involutive.

[b]Lemma 2[/b]
If (i,j) is not involutive, at least one pair among the pairs i, f(i,j), j is not involutive.
Suppose that i, f(i,j) is involutive. Hence, i = n, $f(i,j) = n^2 - n$, and $j = n^3 - n^2 - n$ But f(i,j) and j are not involutive (follows from a simple bounding argument from Lemma 1). Contradiction.

Now suppose that f(i,j) is not involutive. We will prove that no other finite sets exist via proof by contradiction. We split the problem into the following categories:

If gcd(i,j) = 1, then the infinite set S = (i+nj), for any positive n can be generated. This is a contradiction.

Now suppose that gcd(i,j) = k > 1. From lemma 2, at least one of the pairs (i,j) , (i, f(i,j)), (f(i,j), j) is not involutive. Hence, taking the function on this pair results in a number distinct from i, j, or f(i,j). Since this number must be less than j as proved in lemma 1, repeating the procedure for the smallest three elements of S in each iteration, one obtains 2 numbers of gcd(i,j) = 1. (Else you could invoke infinite descent since S is bounded above by the largest number of such a triple.) Taking this pair gives the infinite set we found in the previous case, contradiction.

Here is a sketch of my solution I also added some motivation:

In these kind of problems where we have a lot of assumptions It is usually useful to analyize the assumption for extreme elements rather than thinking them all in a place.Consider the three small elements and analyze them if the first element is fixed the other two will be determined uniqualy but they form a contradiction.I will explain more asap.

[quote=shobber]Determine all finite nonempty sets $S$ of positive integers satisfying
\[ {i+j\over (i,j)}\qquad\mbox{is an element of S for all i,j in S}, \]
where $(i,j)$ is the greatest common divisor of $i$ and $j$.[/quote]
[Color=blue][b]My Solution[/color][/b]
Since $S$ is [u]non empty[/u], it must contain atleast 1 element, which is clearly 2, when $i=j$.
Now, we prove that this is the only [u]non-empty finite[/u] set.
Let $x\neq 2$ be the smallest number in the set.

[b][color=red]Case 1- $x$ is odd[/color][/b]

Let $x=2m+1$
Then, there also exists $\frac{2+(2m+1)}{gcd(2,2m+1)}=2m+3$ in the set.
And, repeating the process with $2m+3$, we find that $2m+5$ also exists in the set, and so on. Thus, this becomes an infinite set with all odd numbers greater than or equal to $x$, contradicting the fact that $S$ is finite.
[Color=red]
[b]Case 2-$x$ is even[/color][/b]

Let $x=2m$. Then, there also exists $\frac{2+2m}{gcd(2,2m)}=m+1$ in the set. But, this contradicts our assumption that $x$ is the least element in the set, as $2m\geq m+1\forall m\in\mathbb{N}$.

Hence, the answer is $S=\{2\}$

[quote=jin]Since $S\subset N^{\ast},S\neq\emptyset$,there exists positive integer $a\in S$
Then $\frac{a+a}{(a,a)}=2\in S$
If there exists positive integer $b\neq2$,$b\in S$
(i)b is odd,$\frac{b+2}{(b,2)}=b+2\in S$ and $b+2$ is odd,too.
so by induction,for all $k\in N,b+2k\in S$,a contradiction with S finite empty.
(ii)b is even,suppose b is the smallest even number in S except 2,(b=2m,m>2)
so $\frac{b+2}{(b,2)}=m+1<2m$,a contradiction.
So $S=\{2\}$.[/quote]

what if the problem turns into [b] [color=#f0f]for all i different from j[/color][/b] in S??

We claim that $S=\{2\}$ is the only solution.
Suppose otherwise, that there is a $2\ne a\in S$. Let $a$ be the minimal element in the set. Taking $i=j=a$, we have $2\in S$.
If $a$ is odd, taking $i=a$ and $j=2$, we have $a+2\in S$, so by induction $a+2n\in S$ for $0\le n\in\mathbb Z$, a contradiction with the finitude of $S$. Note that this means that no odd elements in $S$ exist, so $a\ge2$.
If $a$ is even, taking $i=a$ and $j=2$, we have $\frac a2+1\in S$. But:
$$\frac a2+1\le a\Leftrightarrow a\ge2,$$
which is true, so $a$ is not the minimal element in $S$. Contradiction.

[i]Solution.[/i] The answer is $\mathcal{S} = \{2\}$. This clearly works.

One can easily show $2 \in \mathcal{S}$ by putting $i=j$. If $|\mathcal{S}|=1$, then we get the claimed answer. So from here on, assume that $|\mathcal{S}|>1$. Let $a \ne 2 \in \mathcal{S}$. We will now do a case work on the parity of $a$.

If $a$ is odd, then $\gcd(a,2)=1$. Putting $i=a$ and $j=2$ we get $a+2 \in \mathcal{S}$ and $a+2$ is also odd. Simple induction yields that $a+2n \in \mathcal{S}$ for $n \in \mathbb{N}$ which is a contradiction on finiteness of $|\mathcal{S}|$. One can say pairing odd number and 2 gives infinite number of elements in $\mathcal{S}$.

If $a$ is even, then putting $i=a$ and $j=2$ we get $a/2 +1 \in \mathcal{S}$. If $a/2 +1$ is odd, then pairing it with 2 will give a contradiction as stated above. So, $a/2 +1$ must be even. Now pair $a/2+1$ with 2 to get $\frac{a+6}{4} \in \mathcal{S}$. This would give an infinite sequence of decreasing positive even numbers which is impossible. Therefore, this sequence must reach an odd number once and we are done. $\blacksquare$

[b]Remark.[/b]To be really clear which infinite sequence we are talking about in the the case where we assumed $a$ to be even. Let the the pair of 2 and $a$ give $a_1$ as a new element. Now $a_2$ is a new element after pairing 2 and $a_1$. One can now define $a_k$ easily. Induct to find $a_k$. This $a_k$ is the exact sequence we are talking about.

Take any $i=j \in S$. Then we know that $ 2 \in S$. If there is an odd number of form $2m-1 \in S$, letting $i=2m-1,j=2$ we know that $2m+1\in S$, then we can repeat this  to conclude $S$ is infinite contradiction , hence  for any $m\in \mathbb{N}$, we have $2m-1 \not\in S$.
 Choose the smallest even integer $2m \in S$. Letting $i=2m, j=2$ we have that  $m+1 \in S$. If $m+1$ is odd, we have contradiction. If it is even, we can repeat this until we get an odd integer, contradiction. Hence no other integer exists other than $2$. Therefore $S= \{ 2 \}$.

Claim: $2\in S$. Take any element $n$ of $S$. Then, $$\frac{2n}{\gcd(n,n)}=2\in S.$$

Claim 2: $S$ does not have any odd numbers. Note that any finite set of positive integers has a largest element. Suppose for the sake of contradiction that we have a maximial odd element $m$. Then, $$\frac{m+2}{\gcd(m,2)}=m+2\in S,$$ which contradicts the maximality of $M$.\\

Claim 3: If $f(n)=\frac{n}{2}+1$, and $a\geq 4$ is a even number, then there exists some integer $k\geq 1$ such that $$f^k(a)$$ is odd. Rewrite this as $$f(n)-2=\frac{n-2}{2}.$$ This is essentially saying that the distance from $a$ to 2 is halved each time. Therefore, eventually it will go between 2 and 3 and thus not be an integer. However, for it to go into a non-integer, it must have been odd right before the first time it became non-integer, hence proved.\\

Now, note that if $m\geq 4$ is even, then if $m\in S$ then $$\frac{m+2}{\gcd(m,2)}=\frac{m}{2}+1\in S.$$ However, we can keep repeating this with the new element, and by Claim 3, $S$ would then contain an odd number, contradiction, so $S$ also does not contain any even numbers $\geq4$. Hence, the answer is $$S=\{2\}$$ only, which clearly works.

We claim that $S=\{2\}$ is the only possible set.
Suppose that $S=\{a\}$ is a singleton set. Then, $\frac{a+a}{\gcd(a,a)}=2\in S$. So, $a=2$.
Now suppose that $|S|>2$. If $a\in S$, then $\frac{a+a}{\gcd(a,a)}=2\in S$. Note that if $a$ is odd, then $\frac{a+2}{\gcd(a,2)}=a+2\in S$. So, $S\supseteq\{a,a+2,a+4,\ldots\}$ which contradicts the fact that $S$ is a finite set. It follows that all elements of this set are even. Let $S=\{2\}\cap\{2a_1,2a_2,\cdots,2a_k\}$ where $a_k>a_{k-1}>\cdots>a_1>1$. However, note that $\frac{2a_1+2}{\gcd(2a_1,2)}=a_1+1\in S$. However, $2<a_1+1<2a_2$, so $a_1+1$ is a new element in $S$, which is a contradiction. So, no other sets satisfy all the given conditions.


How did you do?

Well, I did horrible, I can't divide, can't add/subtract, can't do basic logical reasoning, can't do math, can't count, can't do geometry, can't do anything!

STL

joesph- 2    his is getting too careless, better not be like this pn ur AMC, good luck next time
Andrew-4
Stephanie-7
Jim-15

or yea

congrat samson with another 20... :D

ZHOU (A): 20
WESTERDALE (A): 7
WANG (A): 2
SAVIR (A): 17
MANDELSHTAM (B): 8
JAGANNATHAN (B): 6

TOTAL: 60

Acton

(A)

Lewis: 20
Tripathy: 20
Sala: 15
Jiang: 8

(B)

Harris: 10
Wang: 10

Total: 83

I got a 14, Matt got an 8.  Other than that... no idea.

i have no clue whether any other Dallas people actually did anything week 8
but I got a 3 on round B... not my best... possible reason: did it like right before it was due, also before my mid-year exams.

Anyone speaking for  ARL

i want to see how badly STL lose :(


ratio
trigonometry
geometry
projective geometry
angle bisector
geometry solved

Let  $M $  a point on the bisector  $AD$  of a triangle $\triangle ABC$ . Lines  $BM $  and $ CM $  intersect  sides $ AC$  and  $AB $ at points   $ E , F $  respectively. Segment $ DF$  meets  $ BE $ at point  $K $  and  $DE $ meets  $CF$  at  point $ L$ . Prove that $AD$  bisects angle  $\angle KAL $

Babis

 ( it seems to need harmonic ratios ,  but how?)

I'm sure it has been posted before for at least 2 times. Anyway, this can be proved by the facts that B,K,M,E and C,L,M,F are harmonic conjugates.

Good. I hope that Darij will help us  , finding these topics.

  Babis

[b]I note [/b] $U\in BC\cap AK,\ V\in BC\cap AL, BU=x,\ UD=y,\ DV=z,\ VC=t\ \left (x+y=BD=\frac {ac}{b+c},\ z+t=CD=\frac {ab}{b+c}\right)$.

[b](Mekrazyong)[/b] $(B,K,M,E),\ (E,M,L,C)\ -\ h.d.^{*)}\Longleftrightarrow (B,U,D,C)$ $,\ (B,D,V,C)\ -\ h.d.\Longleftrightarrow \frac {UB}{UD}=\frac {CB}{CD},\ \frac {DB}{DV}=\frac {CB}{CV}$.

[b]Therefore[/b] $,\ \frac {x}{b+c}=\frac yb,\ \frac zc =\frac {t}{b+c}\Longleftrightarrow x=\frac {ac}{2b+c},\ y=\frac {abc}{(b+c)(2b+c)},\ z=\frac {abc}{(b+c)(b+2c)}$ $,\ t=\frac {ab}{b+2c}$ [b](constant values) [/b]

$\Longrightarrow\underline {\overline {\left| \frac yx +\frac zt =1\right| }} \left( \frac {UD}{UB} +\frac {VD}{VC}=1\right),\ \frac {UB}{UC}\cdot \frac {VB}{VC}=\frac {x}{a-x}\cdot \frac {a-t}{t}=\left( \frac cb \right)^2 \Longrightarrow$ $\frac {UB}{UC}\cdot \frac {VB}{VC}=\left( \frac cb \right)^2\Longrightarrow$

[b](the Steiner's theorem) The semi-lines [/b] $[AU,[AV$ [b]are isogonals in the angle [/b] $\widehat {ABC}$[b], i.e. [/b]$\angle {UAD}\equiv \angle {VAD.}$

$^{*)}$ [b][u]Definition.[/u][/b]
$ (X,Y,Z,T)$ [b]- h.d.([u]harmonic division[/u], i.e. the points Y,T are [u]harmonic[/u] [u]conjugate[/u] w.r.t. the points X,Z)[/b] $\Longleftrightarrow \frac {YX}{YZ}=\frac {TX}{TZ}.$

.

Thanks Levi!

  It could take me a long time( years!) to find this nice solution. I ' ll give the problem to some greek  engineers( students between 1965 - 75)  too , to try for something simpler , if possible.

 Babis

Indeed, this was posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=21297 .

  Darij

[size=117][color=darkred][b]Generalization :[/b] [/color][/size]\[\left\{\begin{array}{c}\triangle ABC\\\\ D\in (BC)\ ,\ M\in (AD)\\\\ E\in BM\cap AC\ ,\ F\in CM\cap AB\\\\ K\in DF\cap BE\ ,\ L\in DE\cap CF\\\\ U\in BC\cap AK\ ,\ V\in BC\cap AL\end{array}\right\|\Longrightarrow\frac{UB}{UC}\cdot\frac{VB}{VC}=\left(\frac{DB}{DC}\right)^{2}\ .\]

for the initial problem:

Denote $EF\cap BC=X$,
$B$, $C$, $D$, $X$ are in harmonic range, note lines $EB$, $EC$, $ED$, $EX$, since $BKME$ is a line, cutting those 4 lines, $B$, $K$, $M$, $E$ is in a harmonical progression, [simple with sine area formula], likewise, $C$, $L$, $M$, $F$ are harmonic

$\frac{CL}{LM}*\frac{MF}{FC}=1$
$\sin\angle CAL *\sin\angle MAF = \sin\angle LAM * \sin\angle FAC$
likewise,

$\frac{BK}{KM}*\frac{ME}{EK}=1$
$\sin\angle BAK * \sin\angle MAE=\sin\angle KAM* \sin\angle EAK$

the result follows now from a bit of algebra...i suppose angle bisector theorem could have been applied in BAE, AFC, ABC as well


Let $ n$ be a positive integer having at least two distinct prime factors. Show that there is a permutation $ (a_1,a_2,\dots a_n)$ of $ (1,2,\dots , n)$ such that

\[ \sum_{k\equal{}1}^{n} k\cos\dfrac{2\pi a_k}{n} \equal{} 0.
\]

IMO shortlist problems do not belong in [i]pre[/i]olympiad -- that's an absurd level of difficulty creep.  This was also posted (similarly inappropriately) here just this month: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=938865106&t=213714

When posting problems from such common sources as ACoPS and IMO shortlist, please do a search first.


What is the 100th digit of the decimal representation of 3/7?

[hide]3/7 is a repeating decimal (0.428571...). 100/6 = 16, remainder 4. The 4th digit of the repetition is is 5, so the answer is 5.[/hide]

[hide]
The decimal representation is $.\overline{428571}$
$\frac{100}{6}=16 r4$
So, the answer is $\boxed{5}$.[/hide]


Its been a while since i've done these problems please hlep me refresh my moemoray..thanks

$x^2+7x+12/12 * 4/x+4$

and also

$X^2 -64/x^2-16 / x+8/x+4$

[quote="ryan22"]$x^2+7x+12/12 * 4/x+4$

and also

$X^2 -64/x^2-16 / x+8/x+4$[/quote]
I think he meant $\frac{x^2+7x+12}{12}\cdot \frac{4}{x+4}$ and $\frac{\frac{x^2-64}{x^2-16}}{ \frac{x+8}{x+4}}$.
Correct me if I'm wrong.

ryan22, try to learn how to write fractions in LaTeX... just use "\frac{}{}" command...
(You can go to the LaTeX page of AoPS to learn more about LaTeX.)

Not so sure what the second one means, but for the first:[hide]$(x+4)$ is a factor of $(x^2+7x+12)$, so cross cancel to get $x+3$ and $1$. $12$ and $4$ cross cancel to get $3$ and $1$. Now we have $\frac{x+3}{3} * \frac{1}{1}$. Answer is $\frac{x+3}{3}$[/hide]

For the second one
[hide]$\frac{(x+8)(x-8)}{(x+4)(x-4)}*\frac{x+4}{x+8}$ cross cancel and you end up with $\frac{x-8}{x-4}$[/hide]


I believe there should be a flat tax, not even because it is better economically, which I dont know if it is, but purely on ethical reason. I think we should have even more free market based taxes then a flat tax. I think there should be an absolute amount for a tax rather then a fixed tax. Say everyone pays $5,000 a year or some other given amount regardless of whether they make a million or $10 a year. On ethical basis this is true because just because someone earned a lot fairly, why should they pay more then someone who is poor and the poor person get their resources through distribution, they should not. I realize that this will not work from an economic perspective because the poor would not be able to pay, in that case we will have to adopt something close to a flat tax.

What are your opinions?

I dont know why LATEX was added in the middle of the page, I dont know how to change it. (click the box that says "Disable BBCode in this post" -- Administration)

I agree entirely that people who earn more do not deserve to be taxed higher.  It makes no rational sense whatever.  Everyone always talks about giving to the poor, they need it more than us, etc. etc.,  but this is certainly not true in all cases.  From those with ability to those with need is an outright faulty statement.  One should not sacrifice anything unless he chooses to.  Anyone who disagrees with me should read Atlas Shrugged by Ayn Rand =).

Yeah, I read that book too, even though a lot of it is complicated, I was reading the wikipedia site on objectivism and I think it is a good philosophy. Yet, the problem with the tax cut being equal is that the poor people will not have enough money to pay any tax. So I guess a fixed percentage should be used, at least that is of moderate fairness.

According to Ayn Rand and her Objectivism, there should be no income tax at all (John Galt proclaims this in Atlas Shrugged).  I am not entirely sure how that works in terms of the government and its financial role for things such as the army, but I imagine that she had come up with something to make it work and it probably deserves some looking into.  There are certain economies such as in Bermuda where there is no taxation, but again I am not entirely sure how that works and if it is feasible for the U.S.

Yes, ideally that is so. But how can that be done, how can the public goods be provided. I think that public goods are not effective to produce in markets, that is wyh the work with the government, I believe only those goods should be used to spend on by the government.

I haven't spent much time thinking about this, but I imagine that a political system where huge corporations took control of say, public education, would be much more effective than with government control.  I believe we had a similar discussion with MCrawford in another thread in this forum.  The only thing that I imagine would need to be government controlled is the military.  A system with a fixed military tax for each citizen would not be so bad.  The extremely poor people who would not be able to pay this would have to be so poor that all they're doing anyway is begging on the street without a home to live in, and therefore would be served better if they would be forced to join the military after not completing this requirement.  I don't know, I'll have to think about this some more...

Ok, I will wait till you respond more. I have just one question, how exactly do you think about this problem? Like pure logic thinking or you research some information?

I just use logic to think about problems such as these.  It's just a matter of taking previous accepted ideas and building upon them as is often done with somewhat philosophical problems (e.g. using the law of causality to prove that altruism requires something mystic).  Researching some information to get previous facts to support your ideas is certainly helpful though!

Thinking about this problem requires thinking about the inventives of people to provide labour ie. work overtime in the prescence of an increasing/decreasing or flat rate tax on income or deciding to work at all. Why work overtime if I am going to be taxed so much? It also requires thinking about the extent to which the government should intervene in markets, thinking about public goods. I would like it if there were lighting along my street but if others are willing to pay for it, i won't have to (free-rider problem). I think that a progressive tax system is probably the best - the more you earn, the higher % of your income you pay in tax though higher rates should not be too high as to greatly effect overtime worked. If there is a progressive tax system, people may be less likely to work 2nd jobs, leaving them available for those who otherwise do not work.

maybe leg you are right. I base my argument solely on ethics that everyone must pay the same. Not even its economic effects which are much tougher to analyze. But I am interested in economic effects firstmost of course.

I've heard the flat tax argument for years and I'm not buying it.

First, we must consider who is served proportionally by the government.  Government helps secure property and wealth.  Those who have more of it benefit far more greatly from the government, so they should pay for the government disproportionately.

I would go so far as to say that we should decrease most taxes in favor of a wealth tax.  I know I'll hear the argument about incentives to save, but the tax would not need to be very large at all to do its job.  Besides, people on the low end of th economic scale need to save too and they would see reduced overall taxes.  Actually, probably anyone not in the upper 1% would see reduced taxes and that would mean more savings for most people.

Mathew, can you explain in more detail the savings related to taxes aspect?


[quote="Peter"]Find all functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $n\in \mathbb{N}$: \[f(f(f(n)))+f(f(n))+f(n)=3n.\][/quote]

We have that $f(f(f(1)))+f((1))+f(1)=3$,but every positive integer

number is greater or equal to $1$,therefore $f(1)=1$.

Let's suppose that we have proved that $f(k)=k$ for all $n\geq{k}$, and

let's prove that $f(n+1)=n+1$. If $f(a)=f(b)$ then $3a=f(f(f(a)))+f(f(a))+f(a)=f(f(f(b)))+f(f(b))+f(b)=3b$ so $a=b$.

We also have that $f(f(f(n+1)))+f(f(n+1))+f(n+1)=3(n+1)$,but if

$n\geq{t}=f(n+1)$ then $f(n+1)=t=f(t)$ $\Rightarrow$

$n+1=t$,contradiction. Therefore, $f(n+1)\geq(n+1)$.
 Same way if $n\geq{t}=f(f(n+1))$ $\Rightarrow$ $f(f(n+1))=t=f(t)$

$\Rightarrow$ $f(n+1)=f(t)$ $\Rightarrow$ $n+1=t$ contradiction. And

again $f(f(n+1))\geq{n+1}$. In the similar way we can prove that $f(f(f (n+1)))\geq{n+1}$. So we get that

\[f(n+1)\geq{n+1}\]
\[f(f(n+1))\geq{n+1}\]
\[f(f(f(n+1)))\geq{n+1}\]

and consequently $3(n+1)=f(f(f(n+1)))+f(f(n+1))+f(n+1)\geq3(n+1)$ $\Rightarrow$ $f(n+1)=n+1$.

$\Rightarrow$
\[f(n)=n\]
for all $n\in{N}$. :wink:


modular arithmetic
number theory unsolved
number theory

hi... i absolutely need to prove the following to complete a proof i have almost completed.... however i'm not sure it'll be true... in that case don't hesitate to tell me.... here you are

consider $h=2^\omega+1$ prove that if

$2^{h+1}+1\equiv h\pmod{2h}$ then

$2^{h+1}+1\equiv h\pmod{2(h+1)}$

thanks to everyone!!!!


geometry
parallelogram
geometry unsolved

Let $A, B$ and $C$ be three non-collinear points and $E$ ($\ne B$) an arbitrary point not in the straight line $AC$. Construct the parallelograms $ABCD$ and $AECF$. Prove that $BE \parallel DF$.

By simple angle chasing $\Delta ADF \equiv \Delta CBE$, $\Delta ABF \equiv \Delta CDE$, and $BEDF$ is a parallelogram.


calculus
integration
trigonometry
algorithm
calculus computations

Help me to evaluate the integral, thanks.
   $ \int\limits_{0}^{ \plus{} \infty }{\sin \left( {{e}^{ \minus{} x}} \right).dx}$

we can argue that converges by limit comparison test with $ \int_0^\infty e^{\minus{}x}\,dx,$ but as for finding its value by using elementary tools, don't think so.

Helo

puting $ t \equal{} e^{ \minus{} x}$ we found   $ I \equal{} \int_0^{ \plus{} \infty} \sin (e^{ \minus{} x}) dx \equal{} \int_0^1 \frac {\sin t}{t} dt \equal{} Si(1)$

OK, $ \int_0^1 \frac {\sin t}{t}\,dt$ is not an elementary number; there's no closed form representation in terms of elementary functions. Doing that integral numerically, it's about $ 0.946083070367$.

[quote="jmerry"]Doing that integral numerically, it's about ...[/quote]
Power series, Romberg, or built-in algorithm on a graphing calculator? Although I go for Romberg on many occasions, for this one, I would definitely have used the power series.

I used Romberg. Excel is the quickest tool I've got from where I post. In retrospect- yeah, it doesn't take very many terms for the power series to do that well.

If you have Internet access to make this post, then you most definitely have access to Wolfram Alpha.. In my opinion, it's way better that Excel..

I don't care whether it's better. I can implement these numerical algorithms from memory, and I like how a spreadsheet shows all of the intermediate steps. A tool that just computes suits me.


Try [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=156212#p156212]http://www.mathlinks.ro/Forum/viewtopic.php?t=24555 post #9[/url] or [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=271175#p271175]http://www.mathlinks.ro/Forum/viewtopic.php?t=41284 post #4[/url] or [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=3662#p3662]http://www.mathlinks.ro/Forum/viewtopic.php?t=1166[/url].

  darij


greatest common divisor
number theory proposed
number theory

Given is an integer  $ n > 6$. Consider all integers from interval $ (n(n \minus{} 1), n^{2})$ that are coprime to $ n(n \minus{} 1)$. Prove that the greatest common divisor of considered numbers is 1.

assume such a $ n$ exists

$ n(n \minus{} 1) \plus{} 1$ clearely is coprime to $ n(n \minus{} 1)$. since $ n(n \minus{} 1)$ is even $ gcd(n(n \minus{} 1) \plus{} 1,n(n \minus{} 1) \plus{} 3) \equal{} 1$ which means $ 3$ must devide $ n(n \minus{} 1)$ otherwise we're done. because $ gcd(n(n \minus{} 1) \plus{} 1,n(n \minus{} 1) \plus{} 5) \equal{} 1$ also $ 5$ must devide $ n(n \minus{} 1)$ so $ n\geq 15$ therefore $ 7$ must devide $ n(n \minus{} 1)$ since $ gcd(n(n \minus{} 1) \plus{} 1,n(n \minus{} 1) \plus{} 7) \equal{} gcd(n(n \minus{} 1) \plus{} 1,6) \equal{} 1$

denote $ p_k$ the k-th prime number. assume that $ p_i$ devides $ n(n \minus{} 1)$ for $ i\leq m$ for some positive integer $ m$. this means that $ n\geq 2\cdot 3\cdot 5\cdot ....\cdot p_m\geq 2^{m \plus{} 2}$ because obviously $ p_i\geq 2$ and $ 5;7$ which are definately among them are both larger than $ 2^2$

by bertrand's postulate $ p_{m \plus{} 1} < 2^{m \plus{} 2}$ so $ p_{m \plus{} 1} < n$
Now. $ gcd(n(n \minus{} 1) \plus{} 1,n(n \minus{} 1) \plus{} p_{m \plus{} 1} )$ is clearely a devisor of $ p_{m \plus{} 1} \minus{} 1$ which is a product of some $ p_i ; i\leq m$ which are all devisors of $ n(n \minus{} 1)$ and therefore coprime to $ n(n \minus{} 1) \plus{} 1$. we can conclude that $ gcd(n(n \minus{} 1) \plus{} 1,n(n \minus{} 1) \plus{} p_{m \plus{} 1} \equal{} 1$
which means that $ p_{m \plus{} 1}$ must also devide $ n(n \minus{} 1)$ otherwise we're done.

we came to a conclusion that if $ n(n \minus{} 1)$ is devisable by $ 2,3,5$ than its also by $ 7$ than also by $ 11$ by $ 13,17,19$...
since there are infinitely many primes $ n$ grows over all boundaries, meaning such a $ n$ does not exist


function
limit
induction
calculus
integration
vector
arithmetic sequence

Given:

f : R--> R is continous everywhere
\[ f ( (x\plus{}y) / 2 ) \equal{} ( f (x) \plus{} f (y) ) / 2\]

Can you find f(x)?

Have A Great Day!

[hide="Hint"]
Linear functions certainly qualify...

(*) For any three points in arithmetic progression $ x$, $ x\plus{}d$, $ x\plus{}2d$, knowing $ f$ at two of the points determines the value of $ f$ at the third.

Fix the values of $ f(0)$ and $ f(1)$. Which other values are determined by iterating the rule (*)? Can you get all natural numbers? All integers? Some rational numbers? Which rational numbers? If the values are known on a dense set, what can you infer by continuity?
[/hide]

[hide="Solution."]
Suppose we wanted to evaluate $ f(a)$. Then we can write $ f\left(\frac{a}{2}\plus{}\frac{a}{2}\right)\equal{}\frac{f\left(\frac{a}{2}\right)\plus{}f\left(\frac{a}{2}\right)}{2}\implies f(a)\equal{}f\left(\frac{a}{2}\right)$. Since $ f$ is continuous, clearly $ f$ is a constant function.
[/hide]

Hmmm. Did you mean to write

$ f(\frac{a}{2} \plus{} \frac{a}{2})  \equal{} \frac{f(a)\plus{}f(a)}{2}$?

It doesn't say a whole lot about $ f(a)$...

[quote="randomgraph"]Hmmm. Did you mean to write

$ f(\frac {a}{2} \plus{} \frac {a}{2}) \equal{} \frac {f(a) \plus{} f(a)}{2}$?

It doesn't say a whole lot about $ f(a)$...[/quote]

I meant exactly what I wrote: $ x \equal{} y \equal{} \frac {a}{2}$, and this tells me that $ f(a) \equal{} f\left(\frac {a}{2}\right)\forall a\in\mathbb{R}$. This then leads to the following conclusion:

[hide="More rigorous argument of last statement of previous post."]
Assume WLOG that $ \exists c > 0$ such that $ f(c)\neq f(0)$. Then $ f(c) \equal{} f\left(\frac {c}{2}\right) \equal{} f\left(\frac {c}{4}\right) \equal{} f\left(\frac {c}{8}\right) \equal{} f\left(\frac {c}{2^k}\right) \neq f(0)\ \forall k\in \mathbb{Z}^ \plus{}$. Thus $ \lim_{k\to \infty} f\left(\frac {c}{2^k}\right) \equal{} \lim_{x\to 0^\plus{}}f(x) \neq f(0)$, and $ f$ is not continuous, a contradiction. A similar argument is made if $ c<0$; the limit to zero would approach from the left.
[/hide]

[quote="jvenezuela716"]
I meant exactly what I wrote: $ x \equal{} y \equal{} \frac {a}{2}$, and this tells me that $ f(a) \equal{} f\left(\frac {a}{2}\right)\forall a\in\mathbb{R}$.[/quote]

You are obviously wrong. It is $ f(\frac{x\plus{}y}{2})$ on the LHS, not the $ f(x\plus{}y)$.

OK... If the hint is insufficient, here's a complete solution.

[hide="Click to reveal"]

Let $ f(x) \equal{} ax \plus{} b$ be any linear function. Then

$ f(\frac {x \plus{} y}{2}) \equal{} a \frac {x \plus{} y}{2} \plus{} b \equal{} ax/2 \plus{} ay/2 \plus{} b$

$ \frac {f(x) \plus{} f(y)}{2} \equal{} \frac {ax \plus{} b \plus{} ay \plus{} b}{2} \equal{} ax/2 \plus{} ay/2 \plus{} b$

So any linear function satisfies the condition (it is obviously continuous). Our job is to prove that only linear functions qualify.

So given any function $ f$ satisfying the condition, set $ b \equal{} f(0)$ and $ a \plus{} b \equal{} f(1)$.

Claim: for every natural $ n$, $ f(n) \equal{} an \plus{} b$.

Proof: the claim is true for $ n \equal{} 0$ and $ n \equal{} 1$ by our choice of $ a$ and $ b$. Suppose it's true up to $ n$. We have

$ f(\frac {(n \minus{} 1) \plus{} (n \plus{} 1)}{2}) \equal{} \frac {f(n \minus{} 1) \plus{} f(n \plus{} 1)}{2}$

namely

$ f(n) \equal{} \frac {f(n \minus{} 1) \plus{} f(n \plus{} 1)}{2}$

We know, by the induction hypothesis, that $ f(n) \equal{} an \plus{} b$ and $ f(n \minus{} 1) \equal{} a(n \minus{} 1) \plus{} b$. So

$ an \plus{} b \equal{} \frac {an \minus{} a \plus{} b \plus{} f(n \plus{} 1)}{2}$

and we see immediately that $ f(n \plus{} 1) \equal{} a(n \plus{} 1) \plus{} b$ as desired. QED.

In exactly the same way we can extend this to $ f(n) \equal{} an \plus{} b$ for any integral $ n$, positive or negative.

Next, for any rational number of the form $ m/2$, we can show $ f(m/2) \equal{} a(m/2) \plus{} b$. We already know this when $ m$ is even; to see this for $ m$ odd just apply the condition with $ x \equal{} (m \minus{} 1)/2$ and $ y \equal{} (m \plus{} 1)/2$. Now we can proceed to probe $ f(q) \equal{} aq \plus{} b$ for any rational $ q$ of the form $ q \equal{} m/4$, and then $ q \equal{} m/8$, etc... Finally, we have that $ f(q) \equal{} aq \plus{} b$ for any rational in which the denominator is a power of 2.

However, such rational numbers are dense in the real numbers. In other words, any real number $ x$ can be approximated by such a dyadic rational with any desired degree of accuracy. Since $ f$ is continuous, we must have $ f(x) \equal{} ax \plus{} b$ for all real $ x$.

[/hide]

jvenezula, it's not the last statement you need to make rigorous - I think you've just confused yourself on the first step. Taking $ x \equal{} y \equal{} a$ or $ x \equal{} y \equal{} a/2$ in the condition given doesn't yield your stating point. Please check again.

Incidentally, it's interesting to note that the conclusion is false if the assumption of continuity is dropped. There are many non-continuous functions which satisfy the given condition, and they can be constructed (not explicitly) by considering a Hamel basis (a basis of $ \Bbb{R}$ as a vector space over $ \Bbb{Q}$).

[quote="randomgraph"]

jvenezula, it's not the last statement you need to make rigorous - I think you've just confused yourself on the first step. Taking $ x \equal{} y \equal{} a$ or $ x \equal{} y \equal{} a/2$ in the condition given doesn't yield your stating point. Please check again.[/quote]

Yes, I'm sorry, I see my algebra error now. Sorry I didn't catch it earlier.  :(


geometry
rectangle
combinatorics unsolved
combinatorics

Let m, n be positive integers. Suppose that a given rectangle can be tiled (without
overlaps) by a combination of horizontal 1 × m strips and vertical n × 1 strips. Show
that it can be tiled using just one of the two types.

ok i'll give a hint :)
think in terms of roots of unity of both x^m=1 and x^n=1.


induction
number theory proposed
number theory

Let the sequence $a_i$ have $a_1=0$ and $a_2=1$. $a_n$ is made by writing $a_{n-1}$ and $a_{n-2}$ from left to right, for example, $a_3 = 10$, $a_4 = 101$ and $a_5 = 10110$. For which $n$ does $11$ divide $a_n$?

I admire you, it's a very beautiful problem  :thumbup:

I hope it is true

By using the induction, we can prove that when $n=3k \Longrightarrow a_n\equiv 10(mod 11) , n=3k+1 \Longrightarrow a_n\equiv 2 (mod 11)$ and $n=3k+2 \Longrightarrow a_n\equiv 1 (mod 11)$ where $k\geq 2$ so that only $a_1$ divisible by $11$ plz tell me it's true or not ???!!!

It's easy to see that the residues are periodic modulo 11, so
 $a_n \equiv 0 (mod \; 11) $ for $n=6t+1$


Euler
number theory
relatively prime
number theory theorems

Just everywhere if you mean Euler - otherwise, I've never heard about Eulerov...

You're right :blush: .

I just think for Leonhard Euler!!!!!!

There are a lot of Euler's theorems, but for number theory, I think the most famous one is the following:

Let $\varphi(m)$ be the number of numbers smaller then $m$ which are relatively prime with $m$ ($\varphi$ is also known as Euler's function). Then for every two numbers $a$ and $b$ such that $gcd(a,b)=1$ it stands that $a^{\varphi(b)}\equiv 1 \mod b$.

If $m=p_{1}^{a_{1}}p_{2}^{a_{2}}...p_{k}^{a_{k}}$ then $\varphi(m)=p_{1}^{a_{1}-1}p_{2}^{a_{2}-1}...p_{k}^{a_{k}-1}(p_{1}-1)(p_{2}-1)...(p_{k}-1)$.

Bye


Five segments have lengths such that any three of them can be sides of a - possibly degenerate - triangle. Also, the lengths of these segments are nonzero and pairwisely different. Prove that there exists at least one acute-angled triangle among these triangles.

Slight variation of http://www.mathlinks.ro/Forum/viewtopic.php?t=76893 , with the same ideas to solve.

  darij


Since the qualifying rounds have started already, I have made this thread for us to post the results at the qualifying sites. It doesn't have to complete and can be for any of the subjects. We can continue this thread until the state competition. So, if any of you have any results, feel free to post it here. I believe the Appalachian State, Catawba College, and East Carolina sites have already occurred. (results, anyone?) Also, Wake Tech is approaching. Good luck to all!

Ok.
He's what I know.

University of North Carolina At Pembroke:
Alg 2:
1st Place Xiaoyan Han (Me)

PS. Sry. I can't help that much.

Oh dude my brother won UNC pembroke alg I

I am pretty happy that he may have gotten better at math

also darn I am sad that I can't go to Wake Tech this year :( :(

I have the UNC Charlotte Rankings (the official scores seem to be a bit messed up) for comprehensive:

1.  Bryce Taylor
2.  Brendan Fletcher
3.  David Lucia
4.  Kevin Lang
5.  Elizabeth Shen
6.  Shannon Taylor
7.  Sungmin Lim
8.  Samuel Brinkley
9.  David Lee
10.  Amanda Elzey
11.  Jordan Parke
12.  Tejong Lim
13.  Alexandre Gauthier
14.  Michael Grace

For Algebra II, it was:
1.  Jack Munley
2.  Kevin Lee
3.  David Spencer
4.  Velina Kozareva
5.  Nicole Savignac
6.  Jason Yang
7.  Hayden Bader
8.  Simon Menaker

Also, I did not attend the actual awards ceremony because I was taking the APMO back home =)

[quote="jb05"]Does anyone have comprehensive results from ECU? Apparently the guy our team went with got bored and decided it would be more productive to go home early than see the results. And the results are not posted, and we have not been contacted.

UNCW is tomorrow...[/quote]

ECU never posted anything. I do not think they are able to write any good questions (too many ties). They do not have any people interested in math competitions (like UNCC, Dr. Reiter, and UNCW, Dr. Karlof).

Anyway, I predict that no one would get into top 10 (even top 20) from the ECU site in the State Comprehensive this year.

By the way, good job on a perfect AIME and good luck tomorrow.

WCU results anyone? I heard the comprehensive had some strong competitors. Also... Wingate was today too.

Here's the results for University of North Carolina at Pembroke.

Algebra I Honorable Mention
Damien Gu, Ligon Middle; Sanjay Kannan, Ligon; Peter Lup, Ligon; Rhino Nevers, 71st Classical Middle; Margaret Tobey, Ligon (not pictured  from Providence Day: Sarah Johnson, Alexander Boldt and Morgan Goodson)

Algebra I Winners
2nd place - Gordon Ma, Ligon Middle; 1st place - Allan Jiang, Ligon; 3rd place – Savannah Morgan, Providence Day

Algebra I Team 3rd Place
Ligon Middle School – Damien Gu, Gordon Ma and Peter Luo

Algebra I Team 2nd Place
Alex Boldt, Sarah Johnson, Ellis Glasco, Rachel Murray

Algebra I Team 1st Place
Ligon Middle School - Allan Jiang, Sanjay Kannan and Margaret Tobey

Geometry Honorable Mention
Jacob Troutman, West Pine Middle; Casie Jingle-West, West Pine; Hayden Rudd, West Pine; Adam Callahan, Jack Britt High; Zach Martin, West Pine; Jasper Fu, Emily Winn, Tate Krasner, Providence Day

Geometry Winners
1st place – Ben Kaiser, Pinecrest High; 2nd place - Charlie Sowers, Providence Day; Ben Weinrib, Providence Day

Geometry Team 3rd Place
West Pine Middle – Hayden Rudd, Jennie Cunningham, Julianne Gant, Zach Martin.

Geometry Team 2nd Place
Providence Day - Jasper Fu, Ben Weinrub, Andy Marx, Tucker Albert

Geometry Team 1st Place
Providence day - Tate Krasner, Emily Steward, Charlie Sowers, Emily Winn
Algebra II Honorable Mention (from left)
Michael Geneser, Jack Britt High; Garrett Berk, Pinecrest High; Abby Synder, Pinecrest; Erie Imai, Pinecrest; Katherine Rush, Providence Day

Alegebra II Winners
3rd place – Nick Kruyer, Pinecrest High; 1st place - Xiaoyan Han(Me), Jack Britt High; Nathan Bruns, Providence Day

Algebra II Team 3rd Place
Jack Britt High – Josua Pinos, Xiaoyan Han(Me), Max Umantser, Joseph Cotto

Algebra II Team 2nd Place
Providence Day – Nathan Bruns, Emily Cambern, Ntalie Burke, Jacob Dordick

Algebra II Team 1st Place
Pinecrest High – Garrett Berk, Patrick White, Nick Kruyer, Abby Synder

SPECIAL NOTE: Allan Jiang (Not_trig's bro) and Xiaoyan Han (Me) got perfect scores.

Yes, that's right, I really like to brag.... even if it's to a bunch of ppl smarter than me. lol.

UNC at Pembroke site is not so competitive. That is why people from Raleigh and Charlotte areas went there.  Also they have trouble to wrtie any good questions. They never reveal any test problems so they can use the same test every year (like ecu). I asked before why was this. They said they were too busy to write and no people care much about the test.

Interesting.... yc.
You said the raleigh ppl went there, you also said that it wasn't competitive....
the raleigh ppl going there MAKES it competitive.

lol.
Yeah. The fayetteville guy who won first in Alg II at ECU and the fayetteville girl who won second are like really good friends of mine.  They told me they went there cuz they had a guy who's doin comprehensive and they don't want to send ppl to two different schools.  Since ECU offers both.... yeah....

[quote="yc"]That is the strategy some schools use. They want to send more people to the state final so they send the second, even third team outside to compete the spots. At ECU I saw students from Wilmington and Fayetteville. Usually they keep their strongest team home.  I am not sure about this (for example, Allan Jiang may be the strongest Algebra I kid at Ligon). Anyway, I would like to see people from that site listed top 10 or 20 in the finals.
Good luck to you![/quote]

Being a student at Ligon, I know that Ligon did indeed send its second and third teams to UNC-Pembroke, but this was only because all ~12 competitors chosen from Ligon were at least moderately close.

Here's the elon math competition results. (comprehensive)

1st place: Thomas Lu/Allen Yang (meh...they couldn't break the tie)
3rd place: Andrew Hertzberg
4th place: I forgot...
5th place: Gray Symon
6th place: I forgot again...
7th place: Yimo Chen
And I forgot the rest.
Wonder if PI-Dimension remembers the other places...

omg the tiebreaker round (the first ever) was so hilarious. me & allen came up with a plan to keep the guy really frustrated, aka both hand in the same answer at the same time so the tie was unbreakable. :rotfl:

Allen and Tom are doing comprehensive?
Wow.
You guys were doing geo last year.
So you doubled up on courses?
I guess i should be glad, two less(or is it "two fewer" there's a grammer rule about it....) geniuses to make me go lower on the overall rank. lol.

PS. Tom, I'm the guy that you kept on saying "you owe me a dollar" to last year, so i'm not a stocker i actually do know u.

Tom and Allen are probably still competing for Algebra II.  They allow students who scored 120 (AMC10) or 100 (AMC 12) or more to compete in both subject (if the students are taking subjct courses) and comprehensive.

Objectively speaking, I beleive that it was in fact harder to qualify for states via Elon than via Wake Tech. Whereas Elon had at least 5 locks (ARML team members last year), Wake Tech had 4 locks at most. The main reason this happened was that all the moderately strong people (and the good people) perceived that Elon was a weaker site based on previous years, so they all flocked there.

No calculator first time for the subject tests. What is the highest scores in each region?

[list][url=http://courses.ncssm.edu/goebel/statecon/Inform/2009smc_ranklist20.pdf]Comprehensive[/url][/list]
[list][url=http://www.unca.edu/math/contest/Results/2009/MC_R_SA09.html]Algebra I[/url][/list]
[list][url=http://www.unca.edu/math/contest/Results/2009/MC_R_SG09.html]Geometry[/url][/list]
[list][url=http://www.unca.edu/math/contest/Results/2009/MC_R_SB09.html]Algebra II[/url][/list]

Eastern Region is a little behind. Xiaoyan, work hard!  :)

thanks math-4-fun.
I guess I'm pretty lucky to be in the top 10 at least. lol.

PS.  Do I know you? Because for some reason I just have the feeling that I do.......

[quote="not_trig"]darn apparently #1 alg I is a 5th grader ...[/quote]

Yes it is... my neighbor in fact.

Anybody know when they are going to post the results on the NCSSM website?

At western:

Alg 1: IDK

Alg 2: Moshe Feldmen (I think that's the wrong last name!!!)   Second was Brooke Kopelakis.

Geometry: IDK...some guy from McDowell. Second: Chris Jones

Comprehensive: Dylan Peifer Second: David Wang ... Eigth: Kim Stubbs Eleventh: Dana McAfoos

When I find out more I'll post it...That's all I can remember for right now, sorry.

I guess you could e-mail Randy Harter at [u]randy.harter@bcsemail.org[/u] if you really want to know all the results...he should have them...

They had comprehensive at Western? I thought that was only at NCSSM...

Are you talking about state...if so then no.

This was regionals that I posted.

OOOOOOO.
Yeah i was talkin bout state. sry.

@yc: me and dnkywin both took geo and alg2 last year

out of curiosity, why did you compete in geo instead of Alg 2?  Just wondering, cuz I personally suck at geo. lol.


Ok,
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam
Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam Spam

Done

Comments?

Wait till mewto comes and comments on it's Pokemon-ness...wait..I just did...

If you're still trying to beat us in spam volume, you have 3 weeks extra because everyone who posts in BA went to MOP. Hmmmm...how many Missouri people went to MOP...I wonder

[quote="math154"]it's Pokemon-ness[/quote]It is Pokemon-Ness?  Is this some new smash character combo?

Hahagooman!!!!  :rotfl: That's like a Muk!!!!  :rotfl:  :rotfl:

However, if one presses B during the evolution stage...wtf is this???

Or if Grimer is holding an Everstone  :P

Or if I've held your GameBoy/DS w/e (I hack it xD)

edit: in fact, I am rather fond of those young memories. I remember I had a lvl 100 Swampert, Groudon, Kyogre, Latios, and...um...yeah! Rayquaza. (I know my Pokemon. =P)

Oh someone's team isn't overly dependent on legendaries...Good job on geting Latios though! (For an extra challenge, get a level 100 Swords Dance/Baton Pass/Protect/X-scissor Ninjask w/ high attack EV's!!!)

Um I lost my game. Like. 5 years. Ago. (probably good for me though). And wtf are you talking about?

Probably only the bestest baton pass combo ever!!!

Erm.... the only pokemon games I ever owned were fire red and diamond(?) and I got 6 chanseys each game and they took down the elite 4 with no potions (4 killer chanseys, 2 softboilers)!!!! they were like lvl ~ 70 ish. I didnt cheat! w00t! I <3 chansey i know its like the gayest pokemon but i still love them. My secondary team was like all of eevee's evolutions (or at least water, dark, fire, grass, psychic, and one straight up beast eevee) and another lapras that I married (I <3 perish song)  (did you know there are no male chanseys or laprases :?: ).

CHANSEY EEVEE LAPRAS FTW!!!!

lol i havent talked pokemon since like last summer..... XD

[quote="shentang"]What's x-scissor? I don't like ninjask...latios rulez! Though kyogre is so much better, especially with thunder.[/quote]

X-scissor= Strong bug attack move. Physical attack, power 70? accuracy 100. Useful in combo with baton pass ninjask. Kyogre+thunder is less beastly than Groudon (earthquake,fireblast/eruption,solarbeam,fissure) which has a moveset that covers its weaknesses. Personally dynamo, I would evolve those Chanseys into Blissey.

Meh.... I evolved one of my chanseys into blisseys in diamond because at like every level they wanted to level up and I got sick of pressing b :P, and I hated it. Anyways, I sold my DS a year ago so no more Pokemon for me.

[quote="shentang"]I really don't see how groudon is so beast, every time I played a kyogre against a groudon, I win. Basically, I just use water spout or water sport, I forgot, I haven't played in a long long time, and it kills the groudon, whether drizzle or drought is in effect. And since groudon is so slow, I attack first. :D[/quote]

I WONDER WHY??? MAYBE BECAUSE KYOGRE HAS A TYPE ADVANTAGE????

And about 100% of most people have some water-combination (cause they have the most types usually) Pokemon.

lol. 100% of most people.

Mm... I never found the liking for them except Lapras.... bubble doesnt go very far.

For you people who go my school, I found out who hahagooman (the one who started this thread) is. HINT-His name starts with "a" and ends in "ojia"!!!!!!!!  :rotfl:  :rotfl:

I should've never told max that...  :mad:

I believe he means Emperor Maximilian I of Russia...

Yes. Or in Russian, da. Or in Mexican, like Emperor Maximilian of Mexico, C.


i just wondering how old aopsers at the camp will be

do you have to be a mod in order to edit a poll?

OH WHOOPS

i forgot  11th-12th for CA

srry   :oops:

Add for both.

I'll be finishing 9th, for both.

leafiness0 will get in, if he tries on the admission test (I assume he is taking C?)

I assume james4l isn't coming because of possible MOP?

No, we asked him and he said he's not coming because it's too easy for him. Sigh. I think he just pwnt all of us. Not like he didn't do so before now...

What about those Advanced USAMO classes?

Er it is called MOP and pronounced MOP, but it is officially called MOSP; therefore, it is MOP.

[quote="ak909090"]Finishing eighth for UC Santa Cruz in CA!

I'm gonna be the dumbest person there. You can recognize me by my unique eccentric and slightly lunatic look, the bloodthirsty gleam in my eyes, my incessant desire to choke people (but not to death) and my love of tap dancing and playing drums.

I'll probably be hanging around lifeisacircle, AIME15, and leafiness0 if he gets in.[/quote]
Can I apply to be in this gang?
Application:
Indian
6 ft tall
Likes Table Tennis
Plays the Viola, Violin, and Baritone
Is Gangsta
and is Freaky Freaky Fresh

YEah, I aM BaLLiN

What grade?

If you don't live in SFBA you automatically don't qualify :P


rotation
analytic geometry
search
geometry
combinatorics unsolved
combinatorics

Prove that 4 acute angle (<90) can't cover a plane

in general, $ n$ angles, $ A_iB_iC_i$, $ 1\leq i\leq n$, each of measure $ \theta_i < \frac {2\pi}{n}$ can't cover the plane. assume that the angles are labelled so that $ B_iA_i$ rotates counterclockwise through an angle of $ \theta_i$ into $ B_iC_i$. choose coordinate axes $ x$ and $ y$, and for each $ i$, define $ \alpha_i < 2\pi$ to be the angle made by $ B_iA_i$ with the positive $ x$ axis and $ \beta_i < 2\pi$ the angle made by $ B_iC_i$ with the positive $ x$ axis. since the $ n$ angles sum up to less than $ 2\pi$, the $ n$ intervals $ [\alpha_i,\beta_i]$ don't cover the interval $ [0,2\pi]$. so we may pick $ \gamma\in [0,2\pi]$ such that $ \gamma\not\in [\alpha_i,\beta_i]$ for any $ i$. now take a line $ l$ that makes an angle of $ \gamma$ with the positive $ x$ axis. $ l$ will eventually leave the interior of every angle $ A_iB_iC_i$, so $ l$ isn't totally covered.

A related problem with an affirmative answer :
http://www.mathlinks.ro/viewtopic.php?search_id=1941550359&t=1896

Pierre.

let's take a circle that the 4 vertices of 4 angle are in it. now we assume this radius is 1 and the center is O.now let's take bigger circle with the center O and radius R. Angle measure of one of these angles is a.
I want to prove that the area of intersecting zone of this angle and the circle is at most
((R+1)^2)*(a/2).
let V be the vertex of this angle. the distance between any point on the bigger circle and V is at most [(the distance between this point and O)+(the distance between O and V)]=(R+1)
Therefore area of this zone is less than the area of a part of circle with radius R+1 and arc measure is a. thus my claim is true.
If we prove that the sum of areas of such zones is less than (pi)*(R^2) proof would be cmpleted
((R+1)^2)*(A/2)<(pi)*(R^2)? [A is sum of angle measures,A<2(pi)]
(1+(1/R))^2<[2pi/A]
if R is suffeciently large this is true


trigonometry
integration
complex analysis

if $ \sum_{ \minus{} \infty}^{\infty} a_n(z \plus{} 1)^n$ is laurent expansion of $ sin(\frac {z}{z \plus{} 1})$ then what will be $ a_{ \minus{} 2}$

give me some hint
$ z \equal{} \minus{} 1$ is pole of $ \frac {z}{z \plus{} 1}$
what can we say about $ sin(\frac {z}{z \plus{} 1})$

$ \sin(z/(1\plus{}z))\equal{}\sin(1\minus{}1/(1\plus{}z)) \equal{} \sin1 \cos(1/(1\plus{}z))\minus{}\cos 1\sin(1/(1\plus{}z))\equal{}\dots$ (use the Taylor expansions of $ \sin$ and $ \cos$).

$ \ a_{n} \ \equal{} \frac {1}{2 \pi i} \int_{C} \frac { f(z).dz}{{(z \minus{} z_0)}^{n \plus{} 1}}$, will this help you?? :roll:Since the given $ \ f(z)$ is analytic throughout the annular region in which the series is defined, $ \ C$ denotes any path in the annulus region.


Let ABC be a triangle and  [AD] be median of [BC] side.Find a M point on [AD] such that
$MA^2+MB^2+MC^2$ be minumum.

[hide="the idea..."]pretty straight forward if you put it on a cooridinate plane, $A: (0,0)$, $B: (2a,2b)$, and $C: (2c,0)$, $M: (x,y)$
point $M$ is on $y=\frac{c-b}{a}*x$ ($\star$), and the sum of the distances to the verticies squared is

$[x^2+y^2]+[ (x-2a)^2+(y-2b)^2 ] + [ (x-2c)^2+y^2 ]$, now you just substitute ($\star$) and use the trivial inequality[/hide]

If the point $M$ is an any point from the plane of the triangle $ABC$, then what is the minimum of $MA^2+MB^2+MC^2\ ?$  [u]Answer:[/u] The minimum-point  is the centroid of the triangle.


integration
trigonometry
calculus
calculus computations

Prove that : $ \int_{0}^{x}{\frac {sint dt}{t \plus{} 1}} \geq 0$ for all $ x \geq 0$

Actually, that's a nonsensical expression. You can't integrate $ dx$ and also have $ x$ outside.

Once you make the obvious fix, integrate by parts- and the appropriate antiderivative for $ \sin$ is $ 1\minus{}\cos$.

Filling in the blanks:

$ I \equal{} \int\limits_0^x \frac {\sin t}{t \plus{} 1} \, dt$, $ x \geq 0$.
Call $ f(t) \equal{} \frac1{1 \plus{} t}$, $ f'(t) \equal{} \frac { \minus{} 1}{(1 \plus{} t)^2}$.
And $ g'(t) \equal{} \sin t$, $ g(t) \equal{} 1 \minus{} \cos t$.

Then:
$ I \equal{} \int\limits_0^x f(t) g'(t) dt \equal{} \left[ f(t) g(t) \right|_{0}^x \minus{} \int\limits_0^x f'(t) g(t) dt$.

That is:
$ I \equal{} \frac {1 \minus{} \cos x}{1 \plus{} x} \minus{} 1 \plus{} 1 \plus{} \int\limits_0^x \frac {1 \minus{} \cos t}{(1 \plus{} t)^2} \, dt$
$ \equal{} \frac {1 \minus{} \cos x}{1 \plus{} x} \plus{} \int\limits_0^x \frac {1 \minus{} \cos t}{(1 \plus{} t)^2} \, dt \geq 0$
Both summands are positive, as those cosines can't make the terms negative.


This is a website which is basically a giant riddle. There are 100 phases that you must complete, each one either giving you a password to gain access to a new forum containing the next phase, or else giving you directions to the next phase. There are lots of different types of codes, puzzles, and riddles on there, and I would definetly recommend that you all give it a try. I'm on phase 44 right now, so I can give you clues if you pm me and are stuck. The website is [url]http://webriddler.9.forumer.com/[/url]

I'm addicted now

too bad none of the admins are around to put me up on phase 10

Felix is the only one that I've had success with, he's on daily.

Wait, you have to log in to play?! RATS!!!


How many three-digit numbers are not multiples of 7?

[hide]$ 900 \minus{} \left \lfloor \frac{900}{7} \right \rfloor \equal{} 900 \minus{} 128 \equal{} \boxed{772}$[/hide]

[hide="different solution"]I did it a different way than the alcumus people explained it- I divided 900 (the # of 3 digit numbers) by 7 to see how many 3-digit numbers would be multiples of 7.  I got 128.57, but I didn't need the .57. That means there are 128 multiples of 7 that are 3 digit numbers.  900 minus 128 equals 772, the right answer.[/hide]

[quote="mathrocks2011"]is it a coincidence that the number of 3-digit multiples of seven is 2^7?[/quote]

Yes, it is completely coincidental.


geometry
function
algebra
domain
algorithm
3D geometry

This competition will have 5 problems. The problems are arranged in order of difficulty. No cheating!

Rules:
(1) If you only post the answer, you only get 0.5 points. You MUST post how you got the answer.
(2) No cheating, or getting help from somebody else.
(3) Post all your steps.

(1) For 1 point
Rationalize: $\frac{1}{\sqrt{a}+\sqrt{b}+\sqrt{a+b}}$

For (2), (3), and (4), if you find that there are no solutions, please prove so.
(2) For 1.5 points
Assume $x$ and $y$ are real numbers (not necessarily integers).
Find the minimum of $\sqrt{x^{2}+y^{2}}$ if $x+y=n$ ($n$ is a constant)

(3) For 3 points
Assume $x$, $y$, and $z$ are real numbers (not necessarily integers).
Find the minimum of $\sqrt{x^{2}+y^{2}+z^{2}}$ if $x+y+z=13$

(4) For 3.5 points
Find all triples of positive real numbers $(x,y,z)$ that satisfy the following system:
$\sqrt[3]x-\sqrt[3]y-\sqrt[3]z = 16$
$\sqrt[4]x-\sqrt[4]y-\sqrt[4]z = 8$
$\sqrt[6]x-\sqrt[6]y-\sqrt[6]z = 4$

(5) For 4 points
Let $t(n)$ be the maximum number of different areas that you can divide a circle into when you place $n$ points on the circumference, and draw all possible line segments connecting the points. Find a formula for $t(n)$ in terms of $n$.

(6) Tiebreaker
(a)
Pat wants to carry an antique $1.5$ meter long sword onto a train, but the conductor won't allow it as carry-on luggage. And the baggage person won't take any item that exceeds any whole dimension by $0.87$ meters. What should Pat do?

(b)
Let $f(x)$ be a function whose domain is the set of natural numbers. It satisfies $f(1)=1$, $f(2x)=x$, $f(2x+1) = f(2x)+1$. Find a nice, simple algorithm for $f(x)$.

(c)...If someone gets everything right, I will post part (c)

when are these due?

i suppose we pm you the answers

uhh... I think you should place the signups FIRST? Otherwise, t'is another marathon.

The problems are due by the end of next week... Ok, I'll place the signups. Anyone who is interested can sign up... Just say that you are interested. Anyway, what's pm?

PM=private message. If you go to the forums(if you are reading this post you are in the forums) and go up to the AoPS toolbar and click "Private Messages" you will find all of your private messages you sent and received. If you want to PM someone, go to a post made by that person and go to the very bottom of that post. Click "PM" and type the pm you wish to send. Then click send. Congratulations!(if you've done that) You've sent a pm!

For starters, copy and paste this into a pm and send it to me.

Now, also, I would suggest making individual rounds, so no one will know what all of the problems are at the first time they see them. Also, the more problems you make means more time in making them in the next rounds.

Is anybody interested? If so, please say tell me that you are.

[quote="(2)"]Find the minimum of $\sqrt{x^{2}+y^{2}}$ if $x+y=n$ (n is a constant) [/quote]

Just a thought, but you should probably give limitations to the variables, especially in problems like this (i.e. real numbers, ect).

I might be interested...

Could you post some more info, please?

Ok, I'll restrict $x$ and $y$ (I'll edit my post)
(1) pm me the answers by the end of next Thursday.
(2) Tell me whether or not you are interested by next Monday.
(3) Have fun!

why do you need sign ups, if ur interested, pm the answers to him

I'll sign up. Do we have to send you tie-breaker solutions before you notify us that there's a tie-breaker?

Yes, send me the tiebreaker solutions before the tiebreaker. Anyone who didn't do a tiebreaker won't get first if there is a tie.

Also, pm your answers to me by the end of next Thursday. (Ignore the signups part). Just pm your answers if you are interested.

Have fun!

Remember, it's not the answer that matters, it's the way you got the answer.

All right, I guess I will do it.  Seems good to have a competition based on solutions, and how you get the answer, notsomuch as the answer.   :)

Just a reminder: Last day to turn in solutions.  :D

Thanks for reminding me  :blush:

I always forget.

[quote="anirudh"]are middle schoolers allowed to participate???[/quote]
Of course. :D But the problems are due today, along with solutions. :wink:

I solved 1, 2, 3, 5, and 6b, and I only heard about this competition 4 hours ago...

I like the algebra :) Except #4  :mad:

Are we allowed to discuss the problems yet?
I thought number 5 was unnecessarily hard. Unless I did it wrong, which I probably did. :roll:

here's a joke for the first tiebreaker [hide="problem"]She asks a passenger to hold it outside a window, so she can get in without the sword, go to the passenger and ask for it back, and she carries it the whole way out of the window.[/hide]

correct?

[quote="Kurt Gödel"]Lol, number 5 was asked over here in a multiple choice round and they gave the examples $t(1) = 1$, $t(2) = 2$, $t(3) = 4$, $t(4) = 8$, $t(5) = 16$ and they asked for $t(6)$. Of course, everybody answered $t(6) = 32$  :P
[/quote]

Kurt Godel, could you please delete that site? vishalarul is giving us a few days to "fix" our solutions.

Thanks. :D

[quote="Totally Zealous"]when's the deadline?[/quote]
The deadline is Saturday.

Sorry. Really dumb mistake. Number $6b$ was supposed to have $f(2x)=f(x)$. Good thing that it won't affect the scores (no one got it right, and even if they did, only 2 people had to even go to a tiebreaker). :)

Can you post the official results and solutions?  Thanks.

Well, lotrgreengrapes7926 almost had every solution right, except for $6b$ (which no one got). So can you ask him?

Here's the answer for $6b$:
The number of $1$'s in the binary representation of the number.

I really don't know the solution... Sorry. Can someone please post that?

Lotrgreengrapes7926, could you please post your solutions (so that everyone knows how to do the problems)?

It is also possible to use RMS-AM on 2 and 3.

another way to solve 2 is to graph the line x+y=n.

the minimum value of the root is the radius of the circle that is tangent to the lind.  the radius is n rt2 /2

you can do the same for #3


There are $ 36\minus{}11$ or $ 25$ ways for this to happen.

There are $ 5$ doubles so our answer is $ \frac{25\minus{}5}{25}$ or $ \boxed{\frac45}$.

ernie, ur right, but I don't see why the answer given is 6/7.....

[quote="GameBot"]Two dice are rolled. One of the dice does not roll a 1. What is the probability that one of the dice rolled a higher number than the other?[/quote]

Because there's only one way for the "one of the dice does not roll a 1" to not be satisfied: with 2 one's.  Thus, we have $ 35\minus{}5/35\equal{}30/35\equal{}6/7$.

Ok so I was thinking about it and I was wondering how you could get 35 ways to roll the dice using constructive counting?


WLOG assume a \geq b \geq c \geq d \geq e, group the terms with (a-b) and (d-e) as factors and it's easy to show. For generalized schur for n>=3, however, it's known to be false for even n, true for n=3,5. But this method doesn't seem to apply for n>=7...

[quote=manlio]If a,b,c,d,e are all positive or all negative real numbers prove that
(a-b)(a-c)(a-d)(a-e) + (b-a)(b-c)(b-d)(b-e) + (c-a)(c-b)(c-d)(c-e) +
 +(d-a)(d-b)(d-c)(d-e) + (e-a)(e-b)(e-c)(e-d)  \geq 0
Can it be generalized?[/quote]
If $a,b,c,d,e$ are all positive or all negative real numbers prove that$$(a-b)(a-c)(a-d)(a-e) + (b-a)(b-c)(b-d)(b-e) + (c-a)(c-b)(c-d)(c-e) +(d-a)(d-b)(d-c)(d-e) + (e-a)(e-b)(e-c)(e-d)  \geq 0$$


University of Chicago
Stanford
college
Harvard
MIT
Princeton
Putnam

Hi, everyone,

I'll put out a list of criteria for colleges, and ask you please to name some colleges that you think fit those criteria. I'm sure this will be somewhat debatable, but that's all right--a lot of students are trying to figure out their application lists. I probably know the names of the "reach" colleges that are hard for almost everyone to get into, but let's also discuss colleges that may not be the very most famous, but that still offer good programs.

Here are the criteria:

1) good need-based financial aid

At the best, a college strong by this criterion should minimize the amount of debt a student will have when graduating, whatever its list price. The college should be known to be realistic and generous in judging special circumstances such as unusual family medical expenses.

2) challenging program in undergraduate mathematics, with graduate-level courses on campus

At the best, a college strong by this criterion should offer a full Ph.D. program in mathematics, and make graduate-level courses available to well prepared undergraduate students.

3) challenging program in undergraduate computer science, with graduate-level courses on campus

At the best, a college strong by this criterion should offer a full Ph.D. program in mathematics, and make graduate-level courses available to well prepared undergraduate students.

4) diverse, interesting group of smart enrolled students, who are not necessarily mostly math and computer science majors

At the best, a college strong by this criterion should offer programs in social sciences, humanities, and arts that draw students from all over the country and indeed all over the world, and be highly competitive on academic grounds for admission.

What colleges fit these criteria? All lists are welcome. You may rank your list or not as you wish.

What other criteria are worth considering when building a college application list? How does an AoPSer find a good "safety" college to build up the list from?

As I didn't apply for financial aid, I can't help you with the first.  But for the others:

2) Most major research universities offer a Ph.D. program.  In addition, many of these schools will allow qualified students to start taking graduate level courses as soon as they are ready.  Some even have BA (BS)/MA (MS) programs, that allow the undergrad to complete their BA (BS) degree and then, in the same 4 year undergraduate program, earn a Masters as well.  No separate costs, and no seperate application process

I can tell you for a fact that schools such as UChicago and Johns Hopkins both offer honors variants of many courses, and I'm pretty sure Stanford and Harvard do as well.  I'm sure there are many others, I just don't know them all.

3) Carnegie Mellon is your best bet for computer science, although admission to their program is pretty competitive.  MIT is obviously a great choice as well (of course they're great at math, but you don't me to tell you that).

Hope that helps a bit.

Carnegie Mellon is the best for computer science, but probably not as strong in math.  Stanford, Berkeley, Princeton, and Caltech are all up there in both math and computer science as well.

I know Princeton University fits (1), (2), and (4) extremely well. It is arguably the most generous University in terms of financial aid (it treats foreign students almost like domestic students in its evaluations).

I'm not so sure how reputed its Computer Science program is, though I have no doubt it is of high quality.

[quote="tokenadult"]Thanks. I am interested in the intersection of the sets {really good math colleges} and {really good computer science colleges}. MIT indeed surely belongs to that intersection, but what other colleges?[/quote]

i don't make any sort of effort to keep informed on computer science programs.  nevertheless, i know from my abortive stint as an ACM competitor that Stanford, UBC, Toronto, and Waterloo have all had top 10 putnam teams in the last 5 years and qualified for the ACM world finals.  i have no idea about the performance of east coast schools on the ACM -- it can't be that hard to look up, though.

http://cm2prod.baylor.edu/ICPCWiki/Wiki.jsp?page=Results%20World%20Finals%202008

Here's a link to this year's scores.

I know that Caltech is really good with financial aid, they give scholarships on need basis.

link
http://www.admissions.caltech.edu/affording

http://www.finaid.caltech.edu/

that is their aid office

Thanks for the further replies. I wouldn't have thought to use ACM contests as a proxy for computer science quality, so I appreciate that suggestion.


Let $ a,b,c\ge 0$. Prove the following inequality:

$ \sqrt{\frac{a}{b\plus{}c}}\plus{}\sqrt{\frac{b}{c\plus{}a}}\plus{}\sqrt{\frac{c}{a\plus{}b}} \ge 2\sqrt{1\plus{}\frac{abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}}$

[quote="Mashimaru"]Let $ a,b,c\ge 0$. Prove the following inequality:

$ \sqrt {\frac {a}{b \plus{} c}} \plus{} \sqrt {\frac {b}{c \plus{} a}} \plus{} \sqrt {\frac {c}{a \plus{} b}} \ge 2\sqrt {1 \plus{} \frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=221359

[quote="Mashimaru"]Let $ a,b,c\ge 0$. Prove the following inequality :

$ \sqrt {\frac {a}{b \plus{} c}} \plus{} \sqrt {\frac {b}{c \plus{} a}} \plus{} \sqrt {\frac {c}{a \plus{} b}} \ge 2\sqrt {1 \plus{} \frac {abc}{(b \plus{} c)(c \plus{} a)(a \plus{} b)}}.$[/quote]Without loss of generality, we assume $ a\leq b\leq c.$ Using the arithmetic mean - geometric mean inequality,

$ \sqrt {\frac {2a}{b \plus{} c}} \plus{} \frac {2a^2}{\sqrt {(a \plus{} b)^3(b \plus{} c)}}\geq\frac {3a}{\sqrt {(a \plus{} b)(b \plus{} c)}},$

$ \sqrt {\frac {2b}{c \plus{} a}} \plus{} \frac {2b^2(c \plus{} a)}{\sqrt {(a \plus{} b)^3(b \plus{} c)^3}}\geq\frac {3b}{\sqrt {(a \plus{} b)(b \plus{} c)}},$

$ \sqrt {\frac {2c}{a \plus{} b}} \plus{} \frac {2c^2}{\sqrt {(a \plus{} b)(b \plus{} c)^3}}\geq\frac {3c}{\sqrt {(a \plus{} b)(b \plus{} c)}}.$

Id est, it remains to prove that

$ \frac {3(a \plus{} b)(b \plus{} c)(a \plus{} b \plus{} c) \minus{} 2[a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b)]}{\sqrt {(a \plus{} b)^3(b \plus{} c)^3}}$

$ \geq2\sqrt {\frac {2(a \plus{} b \plus{} c)(bc \plus{} ca \plus{} ab)}{(b \plus{} c)(c \plus{} a)(a \plus{} b)}},$

which is true by difference substitution :

$ (c \plus{} a)\left[3b^3 \plus{} 4b^2(c \plus{} a) \plus{} b(c^2 \plus{} 9ca \plus{} a^2) \plus{} ca(c \plus{} a)\right]^2$

$ \minus{} 8(a \plus{} b)^2(b \plus{} c)^2(a \plus{} b \plus{} c)(bc \plus{} ca \plus{} ab)$

$ \equiv F(a,b,c)\equiv F(a,a \plus{} s,a \plus{} s \plus{} t)$

$ \equal{} 8a^5\left(s^2 \plus{} 10st \plus{} t^2\right) \plus{} 20a^4\left(s^3 \plus{} 13s^2t \plus{} 9st^2 \plus{} t^3\right)$

$ \plus{} 2a^3\left(8s^4 \plus{} 150s^3t \plus{} 193s^2t^2 \plus{} 70st^3 \plus{} 8t^4\right)$

$ \plus{} a^2(2s \plus{} t)\left(2s^4 \plus{} 71s^3t \plus{} 105s^2t^2 \plus{} 36st^3 \plus{} 4t^4\right)$

$ \plus{} 2ast(2s \plus{} t)^2\left(3s^2 \plus{} 6st \plus{} 2t^2\right) \plus{} s^2t^2(s \plus{} t)(2s \plus{} t)^2\geq0.$

prove this:
$ \sum\sqrt{\frac{a}{b\plus{}c}} \le  \frac{3}{2}\sqrt{\frac{4}{3}\plus{}\frac{4}{9}\sum\frac{a}{b\plus{}c}}$
BQ

[quote="xzlbq"]prove this:
$ \sum\sqrt {\frac {a}{b \plus{} c}} \le \frac {3}{2}\sqrt {\frac {4}{3} \plus{} \frac {4}{9}\sum\frac {a}{b \plus{} c}}$
BQ[/quote] after squaring,

$ \Longleftrightarrow 2\sum \sqrt {\frac {ab}{(b \plus{} c)(a\plus{} c)}}\le3$. By AM-GM, $ LHS\le\sum(\frac a{a \plus{} c} \plus{} \frac b{b \plus{} c}) \equal{} 3$

[quote="Mashimaru"]Let $ a,b,c\ge 0$. Prove the following inequality:

$ \sqrt {\frac {a}{b \plus{} c}} \plus{} \sqrt {\frac {b}{c \plus{} a}} \plus{} \sqrt {\frac {c}{a \plus{} b}} \ge 2\sqrt {1 \plus{} \frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}}$[/quote]

See: MATHEMATICAL EXCALIBUR 12(5) 2008  :wink:

Dear xzlbq.
if you have post solution for your ineq.
I can`t finish it. :oops:

Dear theSA:
My english is bad,You say  meaning，I do not understand，I will find someone to help me,Thank you.
BQ

[quote="xzlbq"]prove this:

$ \sum\sqrt {\frac {a}{b \plus{} c}}\geq\sqrt {9 \minus{} 3\sum\frac {b \plus{} c}{2a \plus{} b \plus{} c}}$
BQ[/quote]

1#'s ineq is stronger then your's...


I am proving that if x is an accumulation point of set S and e > 0, then there ar infinite number of points within e of x.

Proof: Suppose there is a neighborhood N of x which only contains a finite number of points of S.  Let q1, q2, ..., qn be those points of intersection of N and S, which are distinct from x, and assume r = min(1 <= m <= n) d(p,qm) to be the smallest of the numbers d(p,q1), ... d(p,qn)...

d(p,q) is defined as the distance from p to q.

  The minimum of a finite set of positive numbers is also a positive number, so that r > 0.

   The neighborhood N_r(x) contains no point q of S such that q is not equal to x, so that x is not an accumulation point of S.  This contradicts our fact that x is an accumulation point of S.

Does anybody mind if someone can show to me another proof to this because using r = min(1 <= m <= n) d(p,qm) seems uncomfortable to me?  [/code]

The result is true for T1-spaces ( Fréchet spaces, every point is closed )


algebra
function
domain
complex analysis

Prove that one can find branches of $Ln(iz+\sqrt{1-z^{2}})$ in complex plane without halfline $ix$, $x\geq 1$ and half circle $z=e^{i\phi}$, $\phi \in [0, \pi]$. Can someone explain me this?

A branch of $\text{Ln}f(x)$ exists in any simply connected domain in which $f$ doesn't have zeroes. So, it is enough to find a branch of $iz+\sqrt{1-z^{2}}$ in the given region and to show that it never vanishes. The second task is easy (just square both terms). To do the first, note again that you have a branch of $\sqrt{f}$ in any simply connected domain where $f$ doesn't have zeroes. But the only zeroes of $1-z^{2}$ are $\pm 1$ and those are on the boundary.