Euler
combinatorics theorems
combinatorics

Prove that a graph G that has an Euler path has either two vertices of odd degree or no vertices of odd degree.

That's trivial, the more interesting statement is that the converse also holds.

here goes the proof of both statements:
[b]theorem:a nonempty connected graph is eulerian if and only if it has no vertices of odd degree.[/b]
[i]proof:[/i]let $ G$ be eulerian,and let $ C$ be an euler tour of $ G$ with origin (and terminus) $ u$.each time a vertex $ v$ occurs as an internal vertex of $ C$,two of the edges incident with $ v$ are accounted for.since an euler tour contains every edge of $ G$,$ d(v)$ is even for all $ v \not= u$.similarly since $ C$ starts and ends at $ u$,$ d(u)$ is also even.thus $ G$ has no vertices of odd degree.
[u]conversely[/u],suppose that $ G$ is a noneulerian connected graph with at least one edge and no vertices of odd degree.choose such a graph $ G$ with as few edges as possible.since each vertex of $ G$ has degree at least two,$ G$ contains a closed trail.let $ C$ be a closed trail of maximum possible length in $ G$.by assumption,$ C$ is not an euler tour of $ G$ and so $ G-E(C)$ has some component $ G'$ with $ \epsilon (G')>0$,where $ \epsilon$ denotes the number of edges in a graph.
since $ C$ is itself eulerian,it has no vertices of odd degree;thus the connected graph $ G'$ also has no vertices of odd degree.since $ \epsilon (G')<\epsilon (G)$,it follows from the choice of $ G$ that $ G'$ has an euler tour $ C'$.now,because $ G$ is connected,there is a vertex $ v$ in $ V(C) \cap V(C')$,and we may assume,without loss of generality,that $ v$is the origin and terminus of both $ C$ and $ C'$.but then $ CC'$ is a closed trail of $ G$ with $ \epsilon (CC')>\epsilon (C)$,contradicting the choice of $ C$.

as a corollary of the above theorem we get the following statement:
[b]corollary:a connected graph has an euler trail iff it has at most two vertices of odd degree.[/b]
[i]proof:[/i]if $ G$ has an euler trail then,as in the proof of the above theorem,each vertex other than the origin and terminus of this trail has even degree.
[u]conversely[/u],suppose that $ G$ is a nontrivial connected graph with at most two vertices of odd degree.if $ G$ has no such vertices then,by the theorem above $ G$ has a closed euler trail.otherwise,$ G$ has exactly two vertices ,$ u,v$ of odd degree.in this case let $ G+e$ denote the graph obtained from $ G$ by the eddition of a new edge $ e$ joining $ u,v$.clearly,each vertex of $ G+e$ has even degree and so,by the above theorem,$ G+e$ has an euler tour $ C=v_{0}e_{1}v_{1}...e_{\epsilon+_{1}}V_{\epsilon+1}$ where $ e_{1}=e$.the trail $ v_{1}e_{2}v_{2}...e_{\epsilon+1}v_{\epsilon+1}$ is an euler trail of $ G$.


I'll probably taking Intermediate Algebra class this year but I want to ask something.

I'm thinking of studying through Mathematics of Choice by Ivan Niven instead of AoPS class for AMC and AIME.

Is that good book and not much hard?  (I bought Geometry Revisited thinking it was easy) :?

Okay, for Geometry Revisited I must say you had plenty of warning. :P I personally think Mathematics of Choice more of an advanced-side intermediate/elementary olympiad rather than introduction/beginning-to-mid intermediate level.

I do highly recommend the Intermediate Counting (it might have a different name now) course offered here though. It was one of the first ones I took and one of the ones I remember the best and apply the most.

So, wait.

If I buy that book, do you mean that it won't be useful now?

I think it's 21.95 dollars.. I don't want to waste money if I can't use that book...

I'm not saying it won't be useful. It's a good book no doubt, but I'm not familiar with your level in combinatorics.

the material in mathematics of choice is not that complicated, i think you can use it now and get a lot from it.  you'll see problems in there similar to those on the AMC, maybe AIME.  i think the aops counting class is probablly better though, i did the aops class and then read the book, and the book then seemed real easy.

Are these books you speak of, in AoPS book list?

Yes, they are on the list of books found when you click the Resources link on the left.

personally, i enjoyed Mathematics of Choice...i read it more out of interest rather than for competition purposes, but i'm sure if you thoroughly understand the material covered, you'll be able to see it's applications in AMC and AIME oriented problems. there is no harm done if you read it anyway

mathematics of choice is the best math book ive ever read... but maybe i have a disposition because i love combinatorics  :)

I'm halfway through it, and so far it seems to be at the level of intro counting, judging from the do you need this? test. I'm doing Mathematics of Choice so I can be ready for Intermediate Counting in fall.

Does Mathematics of Choice cover everything done in intro counting on AoPS?  I don't have time for 2 classes this summer since I'm in an independent study (with intermediate algebra), but I'd like to do intermediate counting at its next offering rather than intro since i'm able to do approximately half the "Do You Need This Test" with little difficulty, yet I'm clueless as to the half I can't do.  Would I be able to fill in the gaps and go to Intermediate Counting by using Mathematics of Choice?

If you read Mathematics of Choice and do many of the problems, you'll be ready for Intermediate Counting.


linear algebra
matrix
vector
linear algebra unsolved

Let A be an n by n matrix with real entries and let W be the set of all vectors w of R^n such that Aw=0.

Prove that W intersects the row space of A at the zero vector only

We want to prove that $ ker(A)\cap{I}m(A^T)\equal{}\{0\}$. Let $ u$ s.t. $ Au\equal{}0,u\equal{}A^Tx$. Then $ AA^Tx\equal{}0$ and $ x^TAA^Tx\equal{}\parallel{}A^Tx\parallel{}^2\equal{}0$ where $ \parallel{} \parallel{}$ is the euclidean norm. Thus $ A^Tx\equal{}u\equal{}0$.  Moreover these vector spaces span $ K^n$ because $ dim(ker(A))\plus{}dim(Im(A^T))\equal{}n$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

find all $n$ such that $A_n$ has a subgroup of order $n$.

For odd numbers $n,\ A_n$ has a cyclic subgroup of order $n$ generated by an $n$-cycle. For $n=4k,\ A_n$ has a subgroup isomorphic to $\mathbb Z_2\times\mathbb Z_{2k}$, as described in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=man&t=46589]this topic[/url], in post #3. It's easy to see that that group is contained in $A_n$. For $n=2k$ with $k\ge 5$ odd we can consider the subgroup of $A_n$ generated by $(1,2)(3,4)(5,6,\ldots,k+4)$.

Since $n=2$ clearly doesn't work, this leaves the case $n=6$ to investigate. $n=6$ works as well, since we can embed $S_3$ in $A_6$: if a certain permutation of $1,2,3$ is even, then we leave it as it is, and regard it as a permutation of $1,2,3,4,5,6$. If, on the other hand, it's odd, then we multiply it with $(4,5)$. The set of all these permutations is an isomorphic copy of $S_3$ in $A_6$.

I guess this means the answer is: $n>2$? :?


A small kingdom consists of five square miles. The king, who is not very good at math, wants to
divide the kingdom among his 9 sons. He tells each son to mark out a region of 1 square mile.
Prove that there are two sons whose regions overlap by at least 1/9 square mile.

This is[b] official Solution[/b]: There are 9 regions inside 5 square miles, so there must be at least 4
square miles of overlaps. There are (9 × 8)/(2 × 1) = 36 pairs of regions, so some overlap for a
pair of regions must be at least 4/36 = 1/9 square miles.

I don't understand where "36 pairs of regions" come from. Because 1+2+3+...+8=9.8/2, so I assume that they meant every 2 regions drawn by any 2 sons will overlap. But this statement is clearly not true because we can't infer it from the given hypothesis. Therefore, can anybody help me by explaining this one more detailed?Thank you in advance!

The solution is using the pigeonhole principle. Basiclly we have to find the minimum that any of the two fields overlap.

Since there are 9 sons, there are at most $ \binom{9}{2}\equal{}36$ locations where two fields overlap.

Since there is an excess of $ 4$ square miles. the minimum that two fields overlap is $ \frac{4}{36}\equal{}\frac{1}{9}$

[quote="ocha"]The solution is using the pigeonhole principle. Basiclly we have to find the minimum that any of the two fields overlap.
[/quote]
Thank you for your explanation! I didn't think of finding the minimum for [b]any[/b] of the two fields, which means we can assume that all of the 9 regions intersect each other. Your proof is even clearer than the official one(at least to me it is). :)

[quote="ghjk"] Your proof is even clearer than the official one(at least to me it is). :)[/quote]

Thankyou  :P .

Here is another solution. It states:

Suppose by way of contradiction that the overlapping area between any
2 rectangles is less than 1/9 square mile. Let A denote the total nonoverlapping
area covered by the 9 small rectangles. So
A > 1 +8/9+7/9+6/9+5/9+4/9+2/9+2/9+1/9= 5:
This contradicts the fact that the 9 small rectangles is totally enclosed in
the larger rectangles. Hence our assumption is incorrect. That is, there
must exists two rectangles whose overlapping area is greater than or equal
to 1/9 square miles.

i kinda don't understand why we have  8/9, 7/9, 6/9 etc... I know it's 1-1/9 etc.. but i just can't wrap my hand around it. Explain please.

This is how I did it. I assumed contrariwise. So ten we have an overlapping area greater than 8/9. But we have 9 sons. so 9(8/9)=8>5. Contradiction. Does this work? How can I ameliorate it?

[quote="Arrange your tan"][quote="ghjk"]A small kingdom consists of five square miles. The king, who is not very good at math, wants to
divide the kingdom among his 9 sons. He tells each son to mark out a region of 1 square mile.
Prove that there are two sons whose regions overlap by at least 1/9 square mile.

[/quote]
To all users to whom it may concern:

There is not a set rectangular area of five square miles, nor is
there a kingdom made up of a definite set of five one-square
mile regions, each one mile along on each side.  All the
problem states is that the area consists of "five square miles."
If suppose we stick to connected shapes, that could be any shape
that constitutes that area.  What each son is to mark out is not a
set rectangular shape of one square mile.  Each region for a son
is to be "1 square mile," so, if we were to stick to connected
shapes, each son's region can be any shape that constitutes
one square mile.

With this knowledge, consider whether this problem is true.
If it is still true, then perhaps then solution needs tweaking.

And if you meant something DIFFERENT in the English, THEN
you have to adjust the WORDS in the statement of the problem.
And I suspect what those changes, if at all, would be.[/quote]
That's the original problem! I didn't change any words. :wink:

[quote="Arrange your tan"]What I am hinting at is if someone wanted to have
the problem to necessarily be worked with square
regions of area, THEN the wording would have to be
different, along the lines of:

"A small kingdom consists of five miles square.  [ . . . ]
He tells each son to mark out a region of 1 mile square.
And so on."

1 square mile has the same area as 1 mile square, but
1 mile square is specifically a square shape with 1 mile
on each side, whereas 1 square mile is NONSPECIFIC
in shape.[/quote]Umm, nowhere in the solution says also that they have to be square regions... they are also of arbitrary regions...  :roll:  Perhaps you are referring to bigman's solution [u]only[/u], which involves with rectangles. If not convinced, how about you try and draw weird shapes totalling 5 sq miles of area and divide it among 9 sons...? Watch the overlaps and count them perhaps... and if you can find a counterexample and provide us a glimpse to a scanned pic, your efforts will be merited  :lol:

I had a problem just about exactly like this, but the wording used rectangles. So will my soltion be considered legitimate?

How about for the general area? instead of saying i am wrong, how about you give me a solution or show me how to do it?

@ ocha: where does the excess 4 come from? i don't get it

[quote="Arrange your tan"][quote="10000th User"][/quote]Umm, nowhere in the solution says also that they have to be square regions... they are also of arbitrary regions...  :roll:  Perhaps you are referring to bigman's solution [u]only[/u], which involves with rectangles. If not convinced, how about you try and draw weird shapes totalling 5 sq miles of area and divide it among 9 sons...? Watch the overlaps and count them perhaps... and if you can find a counterexample and provide us a glimpse to a scanned pic, your efforts will be merited  :lol:[/quote]

Do not type the sarcastic "Umm" expression.  It is as pointless as
typing any of these: (very) easy, (very) obvious, trivial, difficult, etc.
These are subjective and don't belong in mathematics.  It's immature
writing and poor style to use "umm."  Just make your statement.

My phrase "To all users to whom it may concern:" covered bigman.

bigman has a wrong/incomplete solution, because he did not show
it for a 5 square mile area of general shape.[/quote]I must have missed the reply here...
@Arrange your tan: if you are making this personal, please PM me.

@bigman: the "excess" of 4 sq miles comes from the fact that there are 9 sons and each is going to take 1 sq mile; but then, there are only 5 sq miles available, so 9-5=4 sq miles is the total area of the regions that overlap within that same 5 sq miles.


On a football team, the founders are Coach Yellsalot, Coach Grunt, and Coach Unmotivator. the assistants are Coach Touchdown, Tackle, Donthurtme, and Faster. The "Novice" coaches are Coach Timeout, Benchwarmer, Getoutofhere, Goodbye, Idontdomornings, and Whyamidoingthis.

If 2 founders, 3 assistants, and 4  "Novice" coaches are selected to got to a football meeting, how many combinations are there?

[hide]
is the answer 180?[/hide]

[hide]$ \binom{3}{2}*\binom{4}{3}*\binom{6}{4}$=180, I think.[/hide]
EDIT:No clue how that happened, musta been asleep.

[hide="uhhh"]

So, of the 3 founders, we need to choose to.
C(3,2) = C(3,1) = 3.

Of the 4 assistants, we need to choose 3.
C(4, 3) = C(4, 1) = 4

Of the 6 novices, we need to choose 4.
C(6, 4) = C(6, 2) = (6 * 5)/2 = 3 * 5 = 15

3 * 4 * 15 = 12 * 15 = 180

So yeah, you got it right. Or I'm pretty sure  :maybe: [/hide]

edit: you beat me to the post, but I think you mad a latex error

Umm... just in case you want it in Latex, because xpmath messed his up.  He'll probably go back in and edit it, but whatever.

There are $ 3$ founders, $ 4$ assistants, and $ 6$ novices.

[hide="Solution"]To find the total combinations, we simply multiply the ways to choose $ 2$ founders, $ 3 assistants$, and $ 4$ novices.

This is merely $ \binom {3} {2}*\binom {4} {3}*\binom {6} {4}\equal{}$$ 3*4*15\equal{}180$.
[/hide]


In a triangle $ ABC$, let $ P$ be a point such that $ PA \equal{} PB \plus{} PC$. $ R$ is the midpoint of the chord $ AB$ (the one that contains $ P$) of the circumcircle of the triangle $ ABP$ and $ S$ is the midpoint of the chord $ AC$ (containing $ P$) of the circumcircle of the triangle $ ACP$. Prove that circumcircles of the triangles $ BPS$ and $ CPR$ touch each other.

I think you mean the midpoints of the arcs $ AB,\ AC$, instead of the corresponded cords.

You are welcome in this Forum, Kostas Vittas.

[quote="livetolove212"]To mod: Please lock this topic because this problem was in Vietnam [b]Mathematics and Youth[/b] Magazine last month.
To [b]Zaq[/b]: I'm so disappointed  :(
Sorry dear Mr. [b]Vittasko[/b].[/quote]
Oh, I think the problem has expired to send the solution for magazines, so I just want to discuss. I'm sorry.


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

suppose we have a polynomial P, $ deg(P) \equal{} n$ ... Prove that the $ n\plus{}1$ polynomials $ P(x), P(x\plus{}1),...,P(x\plus{}n)$ are l.i.

Take the Wronskian and use some row operations to bring it to a Vandermonde matrix ;)

The finite difference polynomial $ \Delta P(x) : \equal{} P(x\plus{}1) \minus{} P(x)$ has degree $ n\minus{}1$. Inductively, the $ k^{\textrm{th}}$ iterate $ \Delta^k P(x)$ has degree $ n\minus{}k$. One may even obtain the formula
\[ \Delta^k (x) \equal{} \sum_{i\equal{}0}^k (\minus{}1)^i \genfrac {(} {)} {0pt} {} {k} {i} P(x\plus{}k\minus{}i).\]
Do this until $ \Delta^n P(x)$, of degree $ 0$. Since $ P(x), \Delta P(x), \ldots, \Delta^n P(x)$ are linearly independent, so are $ P(x), P(x\plus{}1), \ldots, P(x\plus{}n)$.

P.S. By the way, we better work over a field of characteristic $ 0$, otherwise the result is no longer true.


Czy ktoś z was wie może skąd można dostać książkę : "Wstęp do geometrii dawnej i nowej"- Coxeter :?: Byłbym wdzięczny za informację

Jak ktoś chce, mam w formie cyfrowej (DjVu)

Ja tez, dzisiaj zastanawialem sie gdzie to znalezc. Tak samo "Geometrię Trojkąta".

http://www.lo3.wroc.pl/~seminarium/download/geometry_revisited.djvu
http://www.lo3.wroc.pl/~seminarium/download/introduction_to_geometry.djvu

Ma ktoś może nierówności Hardyego, Polya, Littlewooda (przynajmniej po angielsku, ja mam tylko po rosyjsku)

MMMM

sorry a te pliki djvu to czym sie otwiera jesli mozna spytac?

to jakis plugin, ale czym to sie je? w jakim programie to otworzyc bo acdsee to jakos nie idzie

Zainstaluj i kliknij na plik, który chcesz otworzyć

[quote="Michal Marcinkowski"]Ma ktoś może nierówności Hardyego, Polya, Littlewooda (przynajmniej po angielsku, ja mam tylko po rosyjsku)[/quote]

Ja mam, ale nie na kompie niestety...

@Michal Marcinkowski moglbys zapodac wersje po rosyjsku ? Bylbym wdzieczny ;-)

thx Michal Marcinkowski;) również skorzystałem z twych linków i jestem bardzo wdzięczny;)


function
search
absolute value
real analysis
real analysis unsolved

let f be a continuous function from R to R bounded from below,show there exists a real no. y s.t.  f(y)-f(x)<=[y-x],for all real x ,where [g]=absolute value of g. :?:

Hint: Consider $y$ where the function $f(y)+|y|$ attains its minimum.

yury,how to show f(y)+mod y attains and has minimum? :?: if we do that
 we are home

the function $g(x) : = f(x)+|x|$ is continuous, and goes to infinity as $|x|$ goes to infinity. so there are two constants $M,N$ such that $g(x)\le N$ for all $|x|\le M$. so you can search for its minimum in $[-M,M]$.
but since it's a continuous function defined on a compact set, it attains a minimum (bolzano-weierstrass theorem? is it called like this?).

[quote="ma_go"]you can search for its minimum in $[-M,M]$.
but since it's a continuous function defined on a compact set, it attains a minimum (bolzano-weierstrass theorem? is it called like this?).[/quote]
Yes, but then you have only showed the obvious fact that $g$ has a minimum on a closed and bounded interval, not that it has a global minimum..

uhm.. what i actually meant (which is totally different from what i wrote :oops: ) is that $g(x)>N: =g(0)$ for $|x|>M$... sorry! :blush:

(i guess now it should be clear: $\min g \le g(0) = N$, so $\min g$ is taken in $[-M,M]$...)


Is there any way, in LaTeX, to put information in the source file that won't show up in the compiled document, like comments in programming languages? I'm asking because I'm typing up a long list with \begin{enumerate} and \end{enumerate}, and I want to keep track of which 'item' corresponds to which number by putting the number as a 'comment'.

If you put a percent sign before it it wont appear in the compiled version.  e.g.
%[put comment here]

Thanks. Now I have another question. Why is it that when I do something like
[quote]Stuff here

\includegraphics*[viewport 0 0 180 130]{beggeomiddia}

More stuff here[/quote]
The picture gets its own page, even though it's not that big? And I don't get what the viewport thing is, the output seems to be the same for any choice of the coordinates...

You probably want to put a \centerline around it. A method I find useful for error-testing of the dimensions is to put a \fbox around it as well.

You're missing an equals sign after the viewport command:

[code]
\includegraphics*[viewport=0 0 180 130]{beggeomiddia}
[/code]

The numbers define what part of your image will be included.  In the example above, it will include the part of the image with lower left corner (0,0) and upper right corner (180,130).  I believe the units are 72 = 1 inch, so the above command should produce an image roughly 2.5 inches wide and 2 inches high.

Thanks. It works now. But, another question: how do you typeset an arc? Like an upside-down version of the 'breve' thingy?

No, he actually means the opposite of that, something like $\widehat { }$ but instead of begin composed of two segments, he wants an arc :)

I really don't know how to do that  ...

[quote="DPatrick"]So you mean like a curved circumflex?

If you use the tipa package (necessitating a \usepackage{tipa} declaration at the top), the command is \textroundcap

You can also do it, using the phonetic package, using \ohill[/quote]Yep, that's what he wants.

[quote="LynnelleYe"]*ahem* [i]she[/i]. But thanks...[/quote]Oops, well didn't notice :) ...

an arc is $\stackrel{\frown}{something}$

So I guess we would make the arc $AC$ in this way: $ \stackrel{\frown}{AC}$ :)

[quote="ComplexZeta"]No, don't use a stackrel. Use Dave's suggestion.[/quote]That would require tipa package installed on the forum which is not (I think :) ).

[quote="LynnelleYe"]What's wrong with \stackrel? (It's what I ended up using)[/quote]Well ... basically Simon wants to use available macros, rather than creating his own :) So, as this is sort of like creating your own macro, he advises agaisnt it.

I usually use \widehat{AC} for comodity ... if I say "the arc $\widehat{AC}$" everyone understands what I am talking about.

It's not about that. The stackrel doesn't look good for something longer than one (or maybe two) characters because it won't stretch the $\frown$. But if we can't use the tipa package, then we need to do something else. I like the widehat better. But if you have only one or two characters, the stackrel is okay.


Let be $ a,b,c,d>0$ such that $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}16$ . Prove that :

$ (a\plus{}1)(b\plus{}1)(c\plus{}1)(d\plus{}1)\le 81$

[hide]
This is equivalent to $ (w\minus{}1)^2 \plus{} (x\minus{}1)^2\plus{}(y\minus{}1)^2\plus{}(z\minus{}1)^2\equal{}16 \Longrightarrow wxyz \le 81$

The condition gives
$ 12 \plus{} 2(w\plus{}x\plus{}y\plus{}z) \equal{} w^2\plus{}x^2\plus{}y^2\plus{}z^2 \ge \frac{1}{4}(w\plus{}x\plus{}y\plus{}z)^2$

Subbing $ w\plus{}x\plus{}y\plus{}z\equal{}k$ gives $ k^2\minus{}8k\minus{}48 \le 0 \Longrightarrow (k\minus{}4)^2\le 64 \Rightarrow k\le 12$

$ \therefore 12 \ge w\plus{}x\plus{}y\plus{}z \ge 4\sqrt[4]{wxyz} \Longrightarrow wyxz \le 81$[/hide]

$ 4(a^2\plus{}b^2\plus{}c^2\plus{}d^2) \ge (a\plus{}b\plus{}c\plus{}d)^2 \Longleftrightarrow a\plus{}b\plus{}c\plus{}d \le 8$

$ (a\plus{}1)(b\plus{}1)(c\plus{}1)(d\plus{}1) \le \left(\frac{a\plus{}b\plus{}c\plus{}d\plus{}4}{4}\right)^4 \le \left(\frac{12}{4}\right)^4\equal{}81$

Done


topology
real analysis
real analysis unsolved

Can anyone help me show directly that [a,b] is covering compact? By directly, I meant that I am not allowed to show that [a,b] is sequentially compact which implies covering compact, by a theorem we learned.

Hint:
If we let U be an open covering of [a,b] and let C={ x in [a,b]: finitely many members of U cover [a,x]}. Try to show that b is in C by the least upper bound property.

I really have trouble with the covering compact concept. We have the definition of covering compact as : M is a metric space, A in M is covering compact if every covering U of A has a finite subcovering.

There's the classical proof that if $ [a,b]$ can not be covered by finitely many of the open sets, then one of the sets $ [a, M]$ and $ [M,b]$ can't be covered either, say $ [a,M]$. But then you can divide that into two parts as well, one of which cannot be covered. And continue dividing like this until the sets get small enough so that they must be covered, a contradiction.

Let's follow the hint. It should be clear that $ a\in C$: indeed, just one open set is enough to cover $ [a,a]$. Let $ d$ be the least upper bound for $ C$. We want to show that $ d\equal{}b$. Suppose not; that is, $ d<b$. There is an open set $ V\in U$ such that $ d\in V$. The openness of $ V$ implies $ [d\minus{}\epsilon,d\plus{}\epsilon]\subset V$ for some $ \epsilon>0$. By definition of $ d$, the interval $ [a,d\minus{}\epsilon]$ can be covered by finitely many members of $ U$. ... I'll let you do the rest.  :)

By the way, sequentially compact does not imply covering compact in general (you need an additional property of the space, like second countable).

[quote="mlok"]By the way, sequentially compact does not imply covering compact in general (you need an additional property of the space, like second countable).[/quote]

By "in general" do you mean in any topological space? I thought that in a metric space, covering and sequential compactness are equivalent.


let ABC be a triangle and O be center of it's circumcircle .let M and N be two points on AB and AC such that MAN = MON .
prove that AM + AN + MN > BC .

here is my solution:
let T and K be two points on the circumcircle of ABC such that: AM=MT and AN=NK. We have:
AM+AN+MN=MT+NK+MN. We have:
MT+NK+MN>TK(or=TK). So if I prove that TK>BC(or=BC), the problem will be solved.
MOA=MOT(triangles) so MOA=MOT(angles) and NOA=NOK(triangles) so NOA=NOK. So TOK=2MON(angles) MON=MAN so TOK=2MAN on the other hand BOC=2MAN so BOC=TOK so TK=BC :D


When showing that

\[ \sin n\theta \equal{} \sum_{k\equal{}0}^n \binom{n}{k} \cos^k \theta\,\sin^{n\minus{}k} \theta\,\sin\left(\frac{1}{2}(n\minus{}k)\pi\right)
\]

is true for all $ n$, how do we sum when $ n\equal{}\frac{1}{2}$?

The sum is infinite.  Are you familiar with the generalized binomial theorem?


A pyramid has a rectangular base 10 cm by 25 cm and four triangular faces; two opposing faces both with altitude 13 cm and the other two faces with altitudes of 20 cm and 15 cm.  What is the volume of the pyramid?

[hide="Answer"]1000[/hide]

[hide]Take the two cross-sections of the triangle perpendicular to the sides, and you get a triangle with top two sides $ 13$ and $ 13$, and one with $ 20$ and $ 15$. They share heights, so obviously, the one with 20 and 15 must have the larger base $ = 25$, while the smaller one has base $ 10$. Therefore, the triangles have height $ 12$. Using the formula for volume of a pyramid, $ A= \frac{25*10*12}{3}=1000 cm^{3}$[/hide]

Thanks - Did you use the Pythagorean Theorem to find that the height was 12 ?

Somewhat, you should already know that 5-12-13 are the side lengths of a right triangle because it comes up so often, but if not, then yeah, you use Pythagoreans to get that.


Hi I'm a 10th grader at my highschool and I was considering taking the AMC 10 this year instead of the AMC 12.  Unfortunately, my school only administers the AMC 12.  Is there any way I can still take the AMC 10 (at least one of the testing dates)?

Also which test gives me a better chance to make the USAMO?  Normally I score around 100-120 on the AMC12, whereas i just took an AMC 10 and scored 144.5.  Which test would you recommend taking?

AMC 12 is given more consideration than the AMC 10 for AIME/USAMO qualification. If you can score 100-120 on the AMC 12, you have a pretty good chance of making it given that you get a decent AIME score.

I think that the AMC 10 is better for 10th graders, because then you only have to qualify through a floor score, and even if there is an index, it is lower. For example, last year, the qualification for 10th graders and below was 6+ on AIME and 202.5+ on index, while for 11th and up, it was 204+ on index, and it is definitely easier to get a higher score on AMC 10 than AMC 12.

Regarding your question about taking the AMC 10, if you go to the AMC website at unl.edu/amc, you'll find a list of testing sites where the B test is offered (usually at local universities, colleges, etc.). If you find one near you, then you should go take it.

If you can pull off that high for AMC 10, I say you should see if you can get AMC 10. But if you can't, don't worry too much and focus more on AIME because it sounds like you will qualify to AIME either way so might as well focus on AIME more.


geometry
geometric transformation
vector
parallelogram
induction
linear algebra
linear algebra unsolved

$f: \mathbb R^{n}\longrightarrow\mathbb R^{n}$ is a bijective map, that Image of every $n-1$-dimensional affine space is a $n-1$-dimensional affine space.
1) Prove that Image of every line is a line.
2) Prove that $f$ is an affine map. (i.e. $f=goh$ that $g$ is a translation and $h$ is a linear map.)

Common people, this is a good problem. :D  Think on it !!

Actually, 1) implies again 2). Here is a proof:

   By composing with a translation one can assume that $f$ has a fixed point $O$. Put $g(v)=f(O+v)$. We will prove that $g$ is linear by induction.
   Let us do first for $n=2$. Because $f$  is injective, two parallel lines are taken into two parallel lines. This shows that $g(rv)$ is collinear to $g(v)$ $g(rv)=h(r)g(v)$ for some real $h(r)$. Now, take $u$ not collinear to $v$ . Using Thales it follows that $g(ru)=h(r)g(u)$. It is immediate to prove that this result holds for $u$ collinear to $v$ (pass by some other vector which is not collinear to any of them). Next, using this property, it is clear that $h$ is multiplicative. Now, take $u,v$ not collinear. By looking at the parallelogram $OACB$ with $A=O+u$, $B=O+v$, $C=O+u+v$ it follows that $g(u)+g(v)=g(u+v)$. It is not difficult (simple algebraic manipulation) to prove that the result holds for collinear $u,v$: take $w$ not collinear to them and write $g(u+v)=g(u+(v/2+w)+(v/2-w))=g(u+(v/2+w))+g(v/2-w)=...=g(u)+g(v)$.
    Thus $h$ is additive and multiplicative thus it is the identity. Thus $g$ is linear and conclusion for $n=2$.
   Now, pass to general case. Take any hyperplan $H$, then $g$ restricted to $H$ is linear by induction and by composing with a linear map we may assume that $f(H)=H$. By considering a hyperplan that contains $u,v$ and using this observation, we deduce that $g(au+bv)=ag(u)+bg(v)$. This shows that $g$ is linear and finishes the proof.
   Unfortunately, for a) my solution is quite cumbersome and I prefer to see some easier proof.

by induction ?
the image of any $n-2$ dimensional subspace is a $n-2$ dimensional subspace: take $U$ such a $n-2$ dimensional subspace, $U=H_{1}\cap H_{2}$ where $H_{i}$ are hyperplanes. Because $f$ is a bijection, $f(U) = f(H_{1}) \cap f(H_{2})$, where $f(H_{i})$ are distinct hyperplanes. Continue like this and I think the result will follow (I've not checked)


abstract algebra
superior algebra
superior algebra unsolved

$R=\{X, X \subseteq\mathbb{Z}\}$

$X+Y=X\triangle Y=(X-Y)\cup(Y-X)$

$X.Y=X\cap Y$

$I=\{X\mid X\subseteq\mathbb{Z}, X is finite\}$

$R$ is a ring and $I$ is a ideal of$R$, so$R$is a $R$ module and $I$is a
submodule of $R$.

prove that $I$ is not finitely generated

If $X\in\mathrm{span}(X_{1},\dots,X_{n})$, then $X\subseteq\bigcup_{i=1}^{n}X_{n}$.

For any finite collection of $X_{1},\dots,X_{n}\in I$ there are elements of $I$ that are not subsets of $\bigcup_{i=1}^{n}X_{n}$.

Moreover we have $R.(X_{1},\dots,X_{n}) = R.(\bigcup_{i=1}^{n}X_{n})$ for any finite collection of $X_{i}\in R$, so the ring is locally cyclic.

LiMa


We define rectangular arrangement of vertices on a rectangular grid of $ x$ by $ y$ connected. A 1 x 1 square has 2x2 vertices. An ant must travel along the non-diagonal gridlines connecting the vertices and get from one corner to the opposite corner, touching every vertex exactly once. [b]Which $ x$'s and $ y$'s satisfy the above requirements? Justify your answer with proof. [/b]

do you need help, or is this a challenge?
i figured out *i think* that either x or y must be even but im not too good at writing proofs
if both x and y are even, you can achieve this by going along the vertices horizontally and snake back to the right until you reach the last corner
if one of x and y is even, then you move along horizontally if there is an odd number of vertices in a column, or vertically if there is an odd number of vertices in a row. then you snake back and forth until you reach the last corner

if both x and y are odd, we can prove that you cant go to each vertex and arrive at the last corner as follows:
color the first vertex black and color the other vertices in a checkerboard fashion
the opposite vertex in the same column will be white because the vertices will alternate in color an odd number of times, and by the same logic, the opposite corner will be black
since there is an even number of columns, we can consider each alternating column starting with the leftmost one. every vertex in the aforementioned columns will be different in color than the one to its right. therefore,  there will be the same number of black and white vertices.
because we must go to every vertex, and every tie you go to another vertex you change colors, the last vertex traveled to must be white and therefore cannot be the opposite vertex

not Q. E. D.

I think that your proof is rigorous.

Isn't it?


I have a nice solution for the first .By Cauchy Swartz
We $ 2\sum\frac{1}{3\plus{}a}\equal{}\sum\frac{1}{3\plus{}a}\plus{}\sum\frac{1}{3\minus{}a}\geq \frac{36}{3\cdot 6}\equal{}2$ so the desired . :)

Now I found a nice solution also to te second (by your method arqady  :D )

If one of the numbers is bigger than $ \frac {9}{4}$  (let $ x\geq \frac {9}{4}$)
then it suffices to prove that

$ \frac {1}{4 \minus{} b} \plus{} \frac {1}{4 \minus{} c}\geq\frac {5}{28}$ or equivalently

$ \frac {144 \minus{} 8b \minus{} 8c \minus{} 5bc}{28(4 \minus{} b)(4 \minus{} c)}\geq 0$ (1) but $ 8b \plus{} 8c \plus{} 5bc < 24 \plus{} 24 \plus{} 45 < 144$ so (1) is true .

Now let all the varibles $ \leq \frac {9}{4}$

We have to prove that
$ \sum\frac {1}{4 \minus{} a}\geq{3}{4}$ or equivalently

$ \sum (\frac {1}{4 \minus{} a} \minus{} \frac {9}{32}(\frac {1}{3 \minus{} a} \minus{} \frac {1}{3 \plus{} a}) \minus{} \frac {1}{4})\geq 0$   or

$ \sum (\frac {a^2(9 \minus{} 4a)}{16(4 \minus{} x)(3 \minus{} x)(3 \plus{} x)})\geq 0$  which is true and we are done . :)

Arqady which is your solution ?

Your proof for the first inequality is geniusly easy.  :lol:
My proof for the second is same, but if $ a\geq\frac {9}{4}$, I think the following  reasoning is more simple:
$ \sum_{cyc}\frac {1}{4 \minus{} a}\geq\frac {1}{4 \minus{} \frac {9}{4}} \plus{} \frac {2}{4 \minus{} ( \minus{} 3)} \equal{} \frac {6}{7} > \frac {3}{4}.$
The following problem is harder ( but still true ):
Let $ \{a,b,c\}\subset( \minus{} 3,3)$ and $ \frac {1}{3 \plus{} a} \plus{} \frac {1}{3 \plus{} b} \plus{} \frac {1}{3 \plus{} c} \equal{} \frac {1}{3 \minus{} a} \plus{} \frac {1}{3 \minus{} b} \plus{} \frac {1}{3 \minus{} c}.$
Prove that:
\[ \frac {1}{9 \minus{} a} \plus{} \frac {1}{9 \minus{} b} \plus{} \frac {1}{9 \minus{} c}\geq\frac {1}{3}
\]


number theory
prime numbers
number theory unsolved

For which $ n \geq 2, n \in \mathbb{N}$ are there positive integers $ A_1, A_2, \ldots, A_n$ which are not the same pairwise and have the property that the product $ \prod^n_{i \equal{} 1} (A_i \plus{} k)$ is a power for each natural number $ k.$

There cannot exist a natural number $ n$ satisfying the condition.

Assume that there exist integers $ A_1$, $ A_2$, .., $ A_n$ such that $ \prod^{n}_{i \equal{} 1} (A_i \plus{} k)$ is a power for all natural $ k$.

We may assume without loss of generality that $ A_n$ is the maximum among $ A_i$.

Due to the infinitude of prime numbers, there exists a prime $ p > A_n$.
Let $ k \equal{} p \minus{} A_n$, and observe that since $ A_i$ are pairwise distinct, we obtain that $ p| \prod^{n}_{i \equal{} 1} (A_i \plus{} k)$
but $ p^2$ cannot divide $ \prod^{n}_{i \equal{} 1} (A_i \plus{} k)$. This is a contradiction to the fact that $ \prod^{n}_{i \equal{} 1} (A_i \plus{} k)$ is a power.

From All-Russian Olympiad a year prior: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=209458


I think that $ f$ is $ F$, am I wrong? :wink:

Pierre.

you posted the same problem here [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=172054[/url] (in persian),I gave a solution there and [b]ehsan2004[/b] gave a link [url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=cauchy&t=51585[/url] (in english)...

now you're posting it again in another forum,why!? :huh:


Let [tex] x,y,z \in [a,b] [/tex]
Find the greatest value of $ P $ where
[tex] P=\frac{xy}{xz+yz}+\frac{xz}{xy+yz}+\frac{yz}{xy+xz} [/tex]


search
algebra
binomial theorem
absolute value

Question regarding when series converge in the binomial theorem (with fractional and/or negative exponents).

This is through reading on section 16.6 of Art Of Problem Solving Volume 2, specifically bottom of p.190 to top of p.191.

I'm too lazy to retype what exactly they said, so hopefully you have a book if necessary.

Basically, the book states that way out in the expansion of the terms of $ (x \plus{} y)^n$, we have, after a little bit of rearranging,

$ \binom{n}{large n} \cdot x^n \cdot \frac{y^{large n}}{x^{large n}}$

$ x^n$ is fixed, and the combination tends to a finite limit. Therefore, we care about the last part tending to zero. This occurs when the absolute value of x is greater than that of y.

I agree with this, and understand where it comes from. However, there is something I am confused about. What happens when we switch x and y and expand $ (y \plus{} x)^n$ ? We then get that the absolute value of y must be greater than that of x. So basically, does this mean that the first term of  (x + y), the x, must be greater in absolute value for the sequence to converge?

This doesn't make sense to me. For example,

$ (1 \plus{} .2)^{\minus{}1/3} \equal{} (.2 \plus{} 1)^{\minus{}1/3}$

However when expand with the binomial theorem rules, the former converges to about 0.9422, while the latter does not. This is exactly what AoPS' proof said: the first term must be larger in absolute value. $ 1 > .2$.

But I don't understand- these two should be equal, but one converges and the other does not. This, of course, is just a form of the proof, just with numbers.

Help/clarification? I really appreciate it, sorry this post is so long.

This very same issue was discussed not quite one month ago:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=17541833&t=314330
I found it with a search for "binomial theorem," but many other similar searches would have worked just as well (if not better).

Oops, sorry. I'll make sure to search it first next time. Thanks!


real analysis
calculus
calculus computations

My book is presenting a theorem in a chapter on multivariable analysis. It says:

Let $ T: V\rightarrow W$ be a linear transformation from one normed space to another. The following are equivalent:

(a) $ \parallel{}T\parallel{} < \infty$
(b) $ T$ is uniformly continuous
(c) $ T$ is continuous
(d) $ T$ is continuous at the origin

For a proof, it says to assume (a). For all $ v, v' \in V$, linearity of $ T$ implies that
$ |Tv\minus{}Tv'|\le \parallel{}T\parallel{} |v\minus{}v'|$
which gives (b). Then it goes on to show how (b) gives (c) and implies (d) ( it literally just states it). Then the book uses an $ \epsilon \minus{}\delta$ argument to show how (d) "verifies" (a).

My question is how does (a) imply (b), shouldn't the definition of uniform continuity at least be shown from the linearity that was demonstrated? I just don't see how the author can use linearity to show that (a) "gives" (b) unless we introduced for every $ \epsilon> 0, |v\minus{}v'|< \delta$ and etc. ...

You can take $ \delta \equal{} \epsilon/\|T\|$, if you really want.

Another point is that $ V$ should be a complete space, which is never assumed. (Maybe, it is in your definition of a normed space?)


algorithm
algebra
polynomial
modular arithmetic
algebra unsolved

Sequences $ (x_n)$ and $ (y_n)$ are constructed as follows: $ x_0 \equal{} 365$, $ x_{n\plus{}1} \equal{} x_n\left(x^{1986} \plus{} 1\right) \plus{} 1622$, and $ y_0 \equal{} 16$, $ y_{n\plus{}1} \equal{} y_n\left(y^3 \plus{} 1\right) \minus{} 1952$, for all $ n \ge 0$. Prove that $ \left|x_n\minus{} y_k\right|\neq 0$ for any positive integers $ n$, $ k$.

[quote="April"]Sequences $ (x_n)$ and $ (y_n)$ are constructed as follows: $ x_0 \equal{} 365$, $ x_{n \plus{} 1} \equal{} x_n\left(x^{1986} \plus{} 1\right) \plus{} 1622$, and $ y_0 \equal{} 16$, $ y_{n \plus{} 1} \equal{} y_n\left(y^3 \plus{} 1\right) \minus{} 1952$, for all $ n \ge 0$. Prove that $ \left|x_n \minus{} y_k\right|\neq 0$ for any positive integers $ n$, $ k$.[/quote]
I think I got it: Prove that $ x_n \equiv 365 \bmod 1987$ and $ y_n \equiv 16 \bmod 1987$.

Shouldn't it be "$ x_{n \plus{} 1} \equal{} x_n\left({x_n}^{1986} \plus{} 1\right) \plus{} 1622$, and $ y_0 \equal{} 16$, $ y_{n \plus{} 1} \equal{} y_n\left({y_n}^3 \plus{} 1\right) \minus{} 1952$"?

[quote="Erken"][quote="bambaman"]Nice solution! Is there any trick to show that 1987 is a prime?[/quote]

yes, it is called "writeaprogrammonC++". :wink:

seriously, I doubt the existence of such a trick.[/quote]
You're probably right... and also, it was proven that there's an algorithm with polynomial run time to check primality only a few years ago...  XD
There it a 'trick' to calculate $ 16^4 \pmod {1987}$ though:
$ 1987 \equal{} 2^{11} \minus{} 61 \implies 16^4 \equal{}$
$ 2^{16} \equal{} (2^{11} \minus{} 61)2^{5} \plus{} 61 \cdot 2^{5} \equiv 61*32 \equal{} 1952 \pmod {1987}$

[quote="bambaman"]Nice solution! Is there any trick to show that 1987 is a prime?[/quote]
It's very easy in fact to check it. Just check all primes less than $ \sqrt{1987}$ and see if the divide $ 1987$...


In $\triangle ABC$, let $D$, $E$ and $F$ be points on $BC$, $CA$ and $AB$, respectively. The intersection of $AD$ and $BE$ is $P$, the intersection of $BE$ and $CF$ is $Q$, and the intersection of $CF$ and $AD$ is $R$.

If $AF: FB=BD: DC=CE: EA=1: 2$, what is the area ratio of $\triangle PQR: \triangle ABC$?


superior algebra
superior algebra unsolved

Let $ F$ be a field of characteristic $ p \neq 0$. Let $ K \equal{} F(t)$ be the field of rational functions.

What is the fixed field of the automorphism $ \sigma$ of $ K$ that maps $ t$ to $ t\plus{}1$?

Let $ P(t)/Q(t)$ be an element of $ F(t)$. Let $ n$ be a point at which $ Q \neq 0$. Then $ P(n)/Q(n) \equal{} k$ for some $ k$. Thus $ P(n)\minus{}kQ(n) \equal{} 0$ for all $ n$. It follows that $ P(t)\minus{}kQ(t) \equal{} R(t)(x^p\minus{}x)$, so $ P(t)/Q(t) \equal{} R(t)/Q(t)(t^p\minus{}t)\plus{}k$. Since there was a point at which $ Q \neq 0$, we have $ Q(t)$ is not a divisor of $ t^p\minus{}t$, so $ P(t)/Q(t)$ must be some constant plus a multiple of $ t^p\minus{}t$. It is easy to see that these all work. Now, suppose that $ Q(t) \equal{} 0$ for all $ t$. Then $ t^p\minus{}t | Q(t)$. Thus $ Q(t) \equal{} Q(t\plus{}1)$. Now on the other hand we need the numerator to be the same. Thus $ P(t) \equal{} P(t\plus{}1)$ for all $ t$, but this means that $ t^p\minus{}t | P(t)\minus{}P(0)$. Thus $ P(t) \equal{} R(t)(t^p\minus{}t)\plus{}k$ for some $ k$, and as we noted $ Q(t) \equal{} S(t)(t^p\minus{}t)$ for some $ S$. Thus $ P(t)/Q(t) \equal{} R(t)/S(t)\plus{}k/S(t)(t^p\minus{}t) \equal{} [R(t)\plus{}k/(t^p\minus{}t)]/S(t)$. If $ k \neq 0$ then this doesn't make much sense, so we have $ P(t)/Q(t) \equal{} R(t)/S(t)$ after factoring out $ t^p\minus{}t$ from the top and bottom. We can keep doing this until $ t^p\minus{}t$ doesn't divide $ Q$, and then proceed as we did at the beginning.


$ ABCD$ and $ EFGH$ are congruent parallelograms. $ AD \equal{} 10 \ \text{cm}$, $ MC \equal{} 8 \ \text{cm}$, and the area of $ ABCD$ is $ 112\ \text{cm}^2$.  What is the number of centimeters in $ EN$? Express your answer as a decimal to the nearest tenth.

[asy]size(200); import olympiad; unitsize(20,20);
picture p,q; real x = 2; real h = 6;
draw(p,(0,0)--(8,0)--(8+x, h)--(x,h)--cycle);
label("10",(x/2,h/2),W); label("8",(5,0),N);

label(p,"$D$",(0,0),W); label(p,"$M$",(x,0),S); label(p,"$C$",(8,0),E); label(p,"$B$",(8+x,h),E); label(p,"$A$",(x,h),W);

pair n = intersectionpoint((x,h)--((x,h) + 10*(h,-x)), (8,0)--(8+x,h));

draw(q,(0,0)--(8,0)--(8 + x, h)--(x,h)--cycle);
label(q,"$H$",(0,0),W); label(q,"$N$",n,E); label(q,"$G$",(8,0),E); label(q,"$F$",(8+x,h),E); label(q,"$E$",(x,h),W);

draw(p,(x,h)--(x,0),linetype("8 8")); draw(p,rightanglemark((x,h),(x,0),(8+x,0),16));
draw(q,(x,h)--n,linetype("8 8"));
draw(q,rightanglemark((x,h),n,(8+x,h),16));

add(shift(14*right)*q);
add(p);[/asy]

um since they are congruent then, the side of the other one would be ten and the are is 112 so just divide by 10 to get 11.2.


function conics parabola quadratics LaTeX

Assume that AoPS is growing in terms of member at this rate:

1 new member on Day 1.

7 new members on Day 2.

17 new members on Day 3.

31 new members on Day 4.

49 new members on Day 5.

Assuming that AoPS had no members at the beginning of Day 1, at what day will AoPS have 10,000 new members join?   :)

first we have to remember that it's a function

this function is a parabola meaning it's quadratic

if we set the value x as the number of days

and y as the number of people

using the first 3 given statements, we know that

$17=(3)^2a+(3)b+c$
the third one
$7=(2)^2a+(2)b+c$
the second one
$1=(1)^2a+(1)b+c$

I punched these in to my calculator...(using matrices)

so my solutions turned out

$a=2, b=0, c=-1$

therefore

the function we have been desperately looking for is...

$y=2x^2-1$

now I think I gave everyone enough hints ;)

good luck to rest of you guys

could you hint at what the function is?  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:

function....

when there is some kind of graph drawn in a graph paper

and when you draw any vertical lines, and it intersects only once, it's a function

for instance

a circle drawn in a graph paper is not a function since it intersects twice if you drew imaginary verticals through the graph....

we know it's a function since there is only growth...

Or, in a function y=something involving x, when there is a value given for x, only one value can come up for y.

oh... I finally remember it

the vertical thing I mentioned earlier is called Vertical Line test

[hide]$2x^2-1=10,000\\
2x^2=10,001\\
x^2=5,000.5\\
x \approx 70.7\ldots \\
\text {So the } 70^{\text {th}} \text { day.}$[/hide]

:D

[color=darkred][size=75][Edited for latex][/size][/color]


Support FTW factorial AMC AIME Columbia Mafia

[color=blue][b][size=200]Survivor: U.S.A.[/size][/b][/color]

[size=150]Format[/size]

I am not going to make a real long explanation about what the format is...I feel that is somewhat unnecessary...

Anyways, the game will consist of [color=red]tribes, challenges, voting off, and even more surprises.[/color]

The number of players will most likely be somewhere between 12 and 20, depending on interest.

One thing: If you are not in this game, please do not post here...thank you.

Everyone who is planning on playing must pm me their answers to the quiz I have bolded (and you can't play if you don't submit this) - but until the game starts, you may also say whether or not you're playing in-thread.

[b]Registration Quiz:

What time zone are you in?
What's your favorite:
baseball team?
basketball team?
football team?
pizza topping?
country?
number?
color?
How many AoPS games have you played in (if it's over 3, I don't need an exact number)?
How much knowledge do you have on the reality show?
Do you have a strategy for the game?
What time slot (in EST) works best for you?[/b]

Player List:
[color=red]6 submissions thus far[/color]
_______________________________

Wow, there seems to be a good amount of interest...so

the deadline for all applications will be the end of this weekend...assuming that people are applying as fast as they are now (I'm surprised no one's posted on here though)

thanks

[color=red]9 people have signed up so far[/color]

that means there's at least a good 7 spots or so, give or take a few (if there's well over the number that I'm allowing for play, then some people might not be able to play, but for the most part, the first 8-12 to sign up will have a guaranteed spot pretty much)

once again, you can post here if you're playing (though you still have to send me registration)

13 or so applications, but I would really like to get to 16 before I start this

of course...but I'd like to get 16 people to play first...12 might do...

yeah, I'm just waiting for a good weekend (either this one or next) where I can get this started...once it's started, it's downhill from there...so keep posted!

I'm in!  WOOOT!

(the message was too short)

are we allowd to edit posts?

when will we start?

ok, yeah...sooooo

[size=200][color=red][b]Meets and Greets, part 1[/b][/color][/size]

[color=blue]Ok everyone...this is your first phase of survivor usa...I hope everyone had a very safe and fun voyage...however, this is when the real fun begins!

Please feel free to talk with each other for the entirety of this task in this thread about personal stuff or w/e...it may help you later on ;) ;) ;) ;) ;) ;)...after this task, this thread will be mostly all business :)

To begin off with, I'll need someone who considers themselves [size=150]A DEMOCRAT, A REPUBLICAN, AND ANOTHER PERSON WHO CONSIDERS THEMSELVES A CENTRIST
[/size]
a couple things:
- please post this in bold (multiple people can go for the same position, just not multiple positions for one person)
- the first person to get 4 supporters (you can support multiple leader people...so like I could support all of mccain obama and barr if they were running for positions in this game...i would count as 1 supporter for all of them) will get the position (please just say if you support someone in thread by saying "support x for y position" in bold)

thanks, and good luck[/color]

Okay, so everyone else is [b]29[/b], including me.

Alright, well, just to show that I'm active;

[b]29[/b]

that means everyone is here

except for the mod  :roll:

Well, tasks have been posted on the quicktopics, so START.

Ok, now if anyone can post on the quicktopic how to reach the number with the numbers provided, it counts - so only one person needs to post this, and time is of the essence.

ha, this explains why no one's been posting, I thought I had posted the results...

anyways, team [b]13[/b] wins

that means that ProtestanT, Wickedestjr, Garyzx, cf249, and Brut3Forc3 all need to send me one vote and two sentences on why. I am very sorry, I thought I had posted this.

hmmm, we need to spark some energy into this game.

i hope it's just the fact that jhredsox forgot...

If somebody is inactive, then I am willing to replace.


limit
real analysis
real analysis unsolved

Let $ a$ $ \in$ $ (0,1)$ and consider the set
 $ A_a$={$ x$ $ \in$ $ [0,1]$ : $ \lim_{n\to{\infty}}$ $ \frac{\sum^{n}_{i=1}a_{i}}_{n}$ $ =a$, $ where$ $ x=0.a_{1}a_{2}a_{3}...$ is the binary expansion of $ x$}
Prove that $ A_a$ is a Borel set of type $ G_{\delta\sigma\delta}$ or simpler.

Let $ S_n(x) \equal{} \sum_{k \equal{} 1}^n x_k$ for $ x\equal{}0.x_1x_2\dots$

We have $ x\in A_a$ iff $ \forall k > 0\ \exists N > 0\ \forall n > N\quad x\in A_{a,n,k}$, where $ A_{a,n,k} \equal{} \{x: |S_{n}(x)/n \minus{} a| < 1/k\}$. It is easy to see that $ A_{a,n,k}$ is $ G_\delta$, qed.


which is bigger?
$log_{4}9$ and $log_{9}25$
prove your answer, do you?

[quote="hien"] hence $\log_{3}2>1$
[/quote]
I cant believe :?:

Yeah, he flipped the inequality signs.  The answer makes sense intuitively too, because 25 is a much smaller fraction of 81 than 9 is of 16, so $\log_{4}9$ should be closer to 2...


Hello.

How can I prove that there are infinite natural numbers that can be represented as:

$ n^{2}$ for natural $ n$,
and
 $ \frac {q(q \plus{} 1)}{2}$ for a natrual $ q$.

Thanks!

$ q(q \plus{} 1) \equal{} 2n^2$. Hence $ q \equal{} 2b^2, q \plus{} 1 \equal{} a^2$ or $ q \equal{} b^2$ and $ q \plus{} 1 \equal{} 2a^2$ $ \Longrightarrow$ $ a^2 \minus{} 2b^2 \equal{} 1$ or $ 2a^2 \minus{} b^2 \equal{} 1$ which are [url=http://mathworld.wolfram.com/PellEquation.html]Pell's equations[/url]. First has a trivial solution $ (3,2)$ and the second $ (1,1)$.

[b]A much simpler approach:[/b]

If

$ \frac {q(q \plus{} 1)}{2}$

is a perfect square then

$ \frac {4q(q \plus{} 1)(4q(q \plus{} 1) \plus{} 1)}{2} \equal{} 4 \cdot \frac {q(q \plus{} 1)}{2}(4q^{2} \plus{} 4q \plus{} 1) \equal{} 4 \cdot \frac {q(q \plus{} 1)}{2}(2q \plus{} 1)^{2}$

is a number that satisfies the two conditions that you are imposing.

Since $ 1^{2} \equal{} 1 \equal{} \frac {1 \cdot 2}{2}$ satisfies both constraints, we are done.  :wink:

realtime please post this in the pre olympiad section

[url=http://www.mathlinks.ro/viewtopic.php?t=254064][color=red]Here[/color][/url] you can find a possibly related exercise...  :wink:


Well, I'm panicking... I turned in the solutions on Sunday, and there's still no check. Because of holidays?

Did you send it with email, snail-mail, or fax? If you used email, they haven't logged them in yet. Otherwise, you should PM or email an admin.

I used snail-mail, I turned them in Monday.

I emailed it... thanks for the heads-up. (And how exactly did you know?)

RRuscyk said it in another post.

Seriously, I don't even pay any attention to whether I get a check mark yet. Maybe if when everybody else got their scores and mine was blank I would begin to worry, but there's no point in worrying now - they always take a while after the deadline to file them.

Still no check for me and I sent them on the 18th via email.  I guess I won't worry at least for a few more days?

Yeah, still no check mark. I'll hope it's up tomorrow. (Living on the East Coast has its problems)

Yeah...I snail mailed mine and I still don't have the check mark, hopefully mail's just slow....omg if the usps lost it then i'm gonna freaking kill them, i didn't have time to make a copy  :censored:

I don't have a little green check mark either and I emailed mine, so I daresay it's not just your problem.

W00t!

I got a green check mark for the solutions that I e-mailed 2 hours before the deadline.

dude, moral of story: don't panic unless it's like next year and you haven't gotten results yet

I would have freaked out and like go crazy if that happened

and I do have the check mark  :o


Well, it's my first year, and I have two questions about the competition.

1. If you make it to State round because you are in one of the top teams from Chapter, do you also do the individual rounds as well as the team round in State?

2. I may be mistaken, but what's the point of team round in State if it doesn't do anything to determine your final ranking and whether you move on to National or not?

Heh. Good question(s).
For the first, I believe that it's up to your coaches who's on the team and individual.
And the second....I have no idea.  :huh:

Well I sent a letter to the coordinator of my county. He said:

25% of the participants (the top 25% of the participating Teams plus the
highest ranking Individuals who are not on one of the winning teams) will go
on to the Connecticut State Competition on Saturday March 8, 2008 at the
University of Hartford.

What I mean by the first question is, if you make it to the state championship because you are in the top 25% of participating teams, do you do both the team and the individual round(s)?

1. If you make it as a team, then yes, the whole team does the individual and team rounds. If you go as an individual, though, you don't do the team round.

2. Well, in Pennsylvania, the coach of the winning team coaches the state team for the national competition.

the team round affects the overall team score, so I think it determines who coaches the national team.  Also, it's pretty important at nationals.

Is it me, or is the MC ranking system a bit complicated? :|


complex numbers
Harm Derksen
algebra
convex polygon
Perfect Squares
IMO
IMO 1990

Prove that there exists a convex 1990-gon with the following two properties :

[b]a.)[/b] All angles are equal.
[b]b.)[/b] The lengths of the 1990 sides are the numbers $ 1^2$, $ 2^2$, $ 3^2$, $ \cdots$, $ 1990^2$ in some order.

I think a nice idea is to consider the 1990-th roots of unity...and after that consider the vectors with sizes $1^2,2^2,\ldots,1990^2$ and multiples of the roots of unity. Drawing it would be easier to show the idea...=]

[quote="Ilthigore"]Is there an official solution that doesn't use complex numbers? Were they 'IMO syllabus' back in 1990? :S[/quote]
Actually,there is an unofficial solution that doesn't use complex numbers,but it is much longer and a bit ugly.

[quote="Erken"]Actually,there is an unofficial solution that doesn't use complex numbers,but it is much longer and a bit ugly.[/quote]

How about writing it down here please?  :)

Use Complex numbers.
Let $\{a_1,a_2,\cdots,a_{1990}\}=\{1,2,\cdots,1990\}$ and $\theta=\frac{\pi}{995}$.
Then the problem is equivalent to that there exists a way to assign $a_1,a_2,\cdots,a_{1990}$ such that $\sum_{i=1}^{1990}a_i^2e^{i\theta}=0$.
Note that $e^{i\theta}+e^{i\theta+\pi}=0$, then if we can find a sequence $\{b_n\}$ such that $b_i=a_p^2-a_q^2, (p\not=q)$ and $\sum_{i=1}^{995}b_ie^{i\theta}=0$, the problem will be solved.
Note that $2^2-1^2=3,4^2-3^2=7,\cdots,1990^2-1989^2=3979$, then $3,7,\cdots,3979$ can be written as a sum from two elements in sets $A=\{3,3+4\times199,3+4\times398,3+4\times597,3+4\times796\}$ and $B=\{0,4\times1,4\times2,\cdots,4\times 198\}$.
If we assign the elements in $A$ in the way that $l_{a1}=3,l_{a2}=3+4\times199,l_{a3}=3+4\times398,l_{a4}=3+4\times597,l_{a5}=3+4\times796$, then clearly $\sum_{i=1}^{199}l_{aj}e^{i\frac{2\pi}{199}}=0, (j\in\{1,2,3,4,5\})$
Similarly, we could assign elements in $B$ in that way ($l_{b1}=0,l_{b2}=4,\cdots$) to $e^{i\frac{2\pi}{5}}$.
Then we make $b_i$ according to the previous steps. Let $i\equiv j$ (mod 5), $i\equiv k$ (mod 199), and $b_i=l_{aj}+l_{bk}$, then each $b_i$ will be some $a_ p^2-a_q^2$ and $\sum_{i=1}^{995}b_ie^{i\theta}=\sum_{j=1}^5\sum_{i=1}^{199}l_{aj}e^{i\frac{2\pi}{199}}+\sum_{k=1}^{199}\sum_{i=1}^{5}l_{bk}e^{i\frac{2\pi}{5}}=0 $
And we are done.

Set f(x)=x*(x-1)*(x^10-1)/(x^2-ε(5))*(x^1990-1)/(x^10-ε(199)).
Here, ε(n) denotes e^(2πi/n).

We can claim that:
f(x) has 1990 terms with degree 1 through 1990.
The coefficients are a permutation of all 1990-th roots of unity.
f(x) is divisible by (x-1)^3.

Therefore, f(1)=f''(1)=0, summing them yields the answer.

Generalization: if n has at least s+1 distinct prime factors, then such a polygon with sides 1^s, 2^s, ..., n^s exists.

Solution from [i]Twitch Solves ISL[/i]:

Throughout this solution, $\omega$ denotes a primitive $995$th root of unity.

We first commit to placing $1^2$ and $2^2$ on opposite sides, $3^2$ and $4^2$ on opposite sides, etc. Since $2^2-1^2=3$, $4^2-3^2=7$, $6^2-5^2=11$, etc., this means the desired conclusion is equivalent to \[ 0 = \sum_{n=0}^{994} c_n \omega^n \] being true for some permutation $(c_0, \dots, c_{994})$ of $(3, 7, 11, \dots, 3981)$.

Define $z = 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796}$. Then notice that \begin{align*} 	z &= 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796} \\ 	\omega^5 z &= 23 \omega^5 + 27 \omega^{204} + 31\omega^{403} + 35 \omega^{602} + 39 \omega^{801} \\ 	\omega^{10} z &= 43 \omega^{10} + 47 \omega^{209} + 51\omega^{408} + 55 \omega^{607} + 59 \omega^{806} \\ 	&\vdotswithin{=} \end{align*} and so summing yields the desired conclusion, as the left-hand side becomes \[ (1+\omega^5+\omega^{10}+\cdots+\omega^{990})z = 0 \] and the right-hand side is the desired expression.

I like the first part of this answer:
[quote=v_Enhance]Solution from [i]Twitch Solves ISL[/i]:
Throughout this solution, $\omega$ denotes a primitive $995$th root of unity.

We first commit to placing $1^2$ and $2^2$ on opposite sides, $3^2$ and $4^2$ on opposite sides, etc. Since $2^2-1^2=3$, $4^2-3^2=7$, $6^2-5^2=11$, etc., this means the desired conclusion is equivalent to \[ 0 = \sum_{n=0}^{994} c_n \omega^n \] being true for some permutation $(c_0, \dots, c_{994})$ of $(3, 7, 11, \dots, 3981)$.
[/quote]
But I can't follow the 2nd part, especially how we can get
$\omega^5 z = 23 \omega^5 + 27 \omega^{204} + 31\omega^{403} + 35 \omega^{602} + 39 \omega^{801}$
from
$z = 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796}$

So this is my version for 2nd part:

We have to prove that $\sum_{k=0}^{994} (4d_k-1) \omega^k = 0$ where $d_k \to \{1,2,3,4...,995\}$ and $\omega=\cos \frac{2\pi}{995} + i \sin \frac{2\pi}{995}$

(Define $\omega$ as a primitive 995th root of unity is also good, but I avoid it because there are many primitive 995th roots of unity)

Because $\omega^k=(\cos \frac{2\pi}{995} + i \sin \frac{2\pi}{995})^k=\cos \frac{2k\pi}{995} + i \sin \frac{2k\pi}{995}$
We can show that $\omega^{995}=1$ and $\sum_{k=0}^{994}\omega^k = 0$
Because $995=5 \cdot 199$ we can also show that $\sum_{k=0}^{4}\omega^{199k} = 0$ and $\sum_{k=0}^{198}\omega^{5k} = 0$

Let's make our equation simpler:
$\sum_{k=0}^{994} (4d_k-1) \omega^k = 4 \sum_{k=0}^{994}d_k \omega^k -  \sum_{k=0}^{994} \omega^k=0 \iff  \sum_{k=0}^{994}d_k \omega^k=0$

Now we split the sum into 5 smaller sums:
$\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}d_{5k} \omega^{5k} + \sum_{k=0}^{198}d_{5k+1} \omega^{5k+1} + \sum_{k=0}^{198}d_{5k+2} \omega^{5k+2} + \sum_{k=0}^{198}d_{5k+3} \omega^{5k+3} + \sum_{k=0}^{198}d_{5k+4} \omega^{5k+4}$

Here is the trick: we define that $d_{k+995j}=d_{k}$ (just remember that $d_k$s are the sides of 995-got) and because $\gcd(5,199)=1$, we can rewrite our group of sums to this:
$\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}d_{5k} \omega^{5k} + \sum_{k=0}^{198}d_{5k+199} \omega^{5k+199} + \sum_{k=0}^{198}d_{5k+2\cdot 199} \omega^{5k+2 \cdot 199} + \sum_{k=0}^{198}d_{5k+3 \cdot 199} \omega^{5k+3 \cdot 199} + \sum_{k=0}^{198}d_{5k+4 \cdot 199} \omega^{5k+4 \cdot 199}$

So:
$\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}(d_{5k} \omega^{5k} + d_{5k+199} \omega^{5k+199} + d_{5k+2\cdot 199} \omega^{5k+2 \cdot 199} + d_{5k+3 \cdot 199} \omega^{5k+3 \cdot 199} +d_{5k+4 \cdot 199} \omega^{5k+4 \cdot 199})$
$\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}(d_{5k}+ d_{5k+199} \omega^{199} + d_{5k+2\cdot 199} \omega^{2 \cdot 199} + d_{5k+3 \cdot 199} \omega^{3 \cdot 199} +d_{5k+4 \cdot 199} \omega^{4 \cdot 199}) \omega^{5k} $

Now if we choose :
$d_{5k} \to \{1,6,11,...,198\cdot 5+1\}$
$d_{5k+199}=d_{5k}+1$
$d_{5k+2\cdot 199}=d_{5k}+2$
$d_{5k+3\cdot 199}=d_{5k}+3$
$d_{5k+4\cdot 199}=d_{5k}+4$

So:
$\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}(d_{5k}+ (d_{5k}+1) \omega^{199} + (d_{5k}+2) \omega^{2 \cdot 199} + (d_{5k}+3) \omega^{3 \cdot 199} +(d_{5k}+4) \omega^{4 \cdot 199}) \omega^{5k} $
$\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}(d_{5k}(1+\omega^{199} + \omega^{2 \cdot 199} + \omega^{3 \cdot 199} + \omega^{4 \cdot 199})+(\omega^{199}+2\omega^{2\cdot 199}+3 \omega^{3 \cdot 199}+4\omega^{4 \cdot 199})) \omega^{5k} $
$\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}(d_{5k}\cdot 0+(\omega^{199}+2\omega^{2\cdot 199}+3 \omega^{3 \cdot 199}+4\omega^{4 \cdot 199})) \omega^{5k} $
$\sum_{k=0}^{994}d_k \omega^k=(\omega^{199}+2\omega^{2\cdot 199}+3 \omega^{3 \cdot 199}+4\omega^{4 \cdot 199})\sum_{k=0}^{198} \omega^{5k} $
$\sum_{k=0}^{994}d_k \omega^k=(\omega^{199}+2\omega^{2\cdot 199}+3 \omega^{3 \cdot 199}+4\omega^{4 \cdot 199})\cdot 0 $
$\sum_{k=0}^{994}d_k \omega^k= 0 $ Q.E.D

[i][size=75]Solved with ARCH hints and Evan's stream[/size][/i]

We use complex vectors to represent the sides of the 1990-gon. Rather than utilizing the 1990th roots of unity, since 1990 is 2 mod 4, we can add up the vectors of opposite sides in jumps of 2 and use the 995th roots of unity.

Let $w$ be a 995th root of unity. Pairing consecutive perfect squares $(2k-1)^2$ and $(2k)^2$ as opposite sides, we require
\[\sum_{k=1}^{995} a_n \cdot w^k = 0\]

where the $a_n$ are of the form $(2m)^2 - (2m-1)^2 = 4m-1$. We finish using a clever arrangement of $a_n$. $\blacksquare$

[quote=andib2n]I like the first part of this answer:
[quote=v_Enhance]Solution from [i]Twitch Solves ISL[/i]:
Throughout this solution, $\omega$ denotes a primitive $995$th root of unity.

We first commit to placing $1^2$ and $2^2$ on opposite sides, $3^2$ and $4^2$ on opposite sides, etc. Since $2^2-1^2=3$, $4^2-3^2=7$, $6^2-5^2=11$, etc., this means the desired conclusion is equivalent to \[ 0 = \sum_{n=0}^{994} c_n \omega^n \] being true for some permutation $(c_0, \dots, c_{994})$ of $(3, 7, 11, \dots, 3981)$.
[/quote]
But I can't follow the 2nd part, especially how we can get
$\omega^5 z = 23 \omega^5 + 27 \omega^{204} + 31\omega^{403} + 35 \omega^{602} + 39 \omega^{801}$
from
$z = 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796}$
[/quote]

The expression for $\omega^5 z$ follows from $20\omega^{5} + 20\omega^{204} + 20\omega^{403} + 20\omega^{602} + 20\omega^{801}$ being $20\omega^5$ times the sum of the fifth roots of unity, or 0. The remaining expressions follow analogously.

In order to construct the wanted polygon, we'll deal with complex vectors for each of the sides of the polygon, where each pair of opposite sides is of the form $(4i^2,(2i-1)^2)$.

Let $z$ be a $995-$th root of unity and let $d_i =  4i^2 = 4i^2 - (2i-1)^2 = 4i-1 $. We want to find a permutation of the $s_i$'s for which the following holds :
$$\sum_{i=0}^{994} s_i \cdot z^i = 0$$

Since $995 = 5 \times 199$, then $z^{199}$ is a $5-$th root of unity, thus $20 z^{5i} (z^0 + z^{199}+z^{398}+z^{597}+z^{796} )=0 $ for all $0\geq i \geq 198 $    $...(**)$.
\\

Now by recursively combining property $(**)$ with the fact that $t = 3z^0 + 7z^{199} + 11z^{398} + 15z^{597} + 19z^{796}$, we can compute all of the $z^{5i}t$ to prove that the desired sum is no more than
$$t\times \left(\sum_{i=0}^{198}z^{5i}\right) = 0$$

$$\mathbb{Q.E.D.}$$


Let $(F_n)$ the Fibonacci sequence (with $F_1=F_2=1$).
If $p$ is prime, prove that $p|F_{p-1}$ or $p|F_{p+1}$  ;)

You are an joker!  :lol:  :lol:  :lol:

It doesn't hold for $p=5$ ;)
Note that $F_{p-1}F_{p+1}=F_p^2-1$ and use the result from http://www.mathlinks.ro/viewtopic.php?p=157824 .

You are right   :rotfl:
(Sorry for $p=5$...)


Al and Bert are fellow con artists who deliver coded messages to each other to communicate. Recently Bert was put in jail for stealing a rare and expensive diamond. Only a few days after this, Al sent him a friendly letter asking him how he was. On the inside of the envelope of the letter, he hid a code.

Yesterday, Bert escaped and left the envelope and the letter inside the jail cell. The police did some research and found the code on the inside of the envelope, but they haven't been able to crack it. Could you help the police find out what the message is?

The code is as follows:

llwatchawtfeclocklnisksundialcirbetimersool

Remove all the time related words, and consider the reverse of the message.

llwatchawtfeclocklnisksundialcirbetimersool [ original code ]

IIawtfelniskcirbesool [ words pertaining to time removed ]

loose bricks in left wall [ reversed letters ]

Nice.

I had this pretty cool code book. So someone was sent to jail, and his friend sent him a paragraph-long letter. Of course, this was checked by the guards, and nothing was fishy in the letter (unlike this one). For example, the letter said stuff like I'll miss you, I hope you don't die, and stuff like that. So the guards delivered it to the prisoner. But it turned out to be a code that you read 3 letters after all punctuation. So in the end, the prisoner escaped, and they lived happily ever after.


probability AMC AIME USA(J)MO USAMO search

I've always found probability rather difficult. It seems that even the simplest questions in terms of practicality can turn into seemingly impossible probability questions.

Does anyone have any difficult or neat probability questions that they can share? I would appreciate the opportunity to see more questions and hopefully improve.

If you're looking for probability problems at the Intermediate level (hard AMC, all AIME, easy USAMO level), you might consider taking the Intermediate Counting/Probability course here at AoPS.  That course includes dozens and dozens of problems of a great variety.

How would I go about doing that? I've heard of the classes before, but didn't know where to begin.

[quote="MCrawford"]On the left side panel, you should see "Online School" and "Enroll" buttons.  Click those to learn more about AoPS classes.  You might also ask advice from students familiar with those classes.  Plenty of discussion of the classes can be found if you use the search feature on the forum panel (at the top of this page).[/quote]
Is it free? if not, how do i pay?

Classes do require an enrollment fee.

You can pay using credit card online, call in using a credit card, or mail a check.  Instructions are in the enrollment page information.


algebra
polynomial
inequalities
real analysis
real analysis unsolved

Let $h$ a real trigonometric polynomial of degree $n$.Show that $||h'||_{\infty}\leq n||h||_{\infty}$.

that is  :)  what is your solution eugene ?


ratio
geometry
circumcircle
inradius
area of a triangle
geometry open

prove that the ratio of the area of a triangle to the area of the triangle determined by the points of the tangency of the incircle equals the ratio of twice the length of the circumradius to the length of the inradius of the given triangle.

Well, by Euler's theorem on pedal triangles, if $ S$ and $ S'$ are the areas of $ ABC$ and of the pedal triangle of $ P$, the following holds:
\[ S'\equal{}\frac{R^2\minus{}OP^2}{4R^2}S\]
where $ O$ is the circumcenter.
So, if $ P\equal{}I$, we have
\[ S'\equal{}\frac{R^2\minus{}R^2\plus{}2Rr}{4R^2}S\]
that is
\[ \frac{S'}{S}\equal{}\frac{r}{2R}\]


geometry
probability
function
trigonometry
ratio
Pythagorean Theorem
angle bisector

I know this sounds kinda rude, but can y'all post a few more geometry problems in this section, cuz I haven't done one in a while on this site, most of these problems are like probability and functions, etc. which I think are fun, but I'm looking for a little bit more geometry please  :blush: (that don't have to use Trigonometry much cuz I'm bad at Trig)

2.)  a,b,c are the sides and x,y,z are the interior angles of of the triangle ABC then
prove pi/3 =< (a.x+b.y+c.z)/(a+b+c) < pi/2

[quote="matematikator"]1. If two corresponding angles of a quadrilateral are obtuse then the diagonal connecting to these angles is less than the other diagonal.[/quote]

Yeah.... :?
That's fairly obvious but it's not a problem  :(

3. If the angles of a convex polygon form an arithmetical progression then prove that each of the angles is bigger than 360.

4. There is a line segment MN in a sector AOB of a circle with radius R=AO=BO. Prove that  MN<=R or MN<=AB (Take <AOB<180)

5. Prove that 0.4<r/h<0.5, where r is the radius of inscribed circle and h is the altitude of right angle.

I know I'm being a little picky, but can you also add in problems too other than proofs please! I can't do proofs at all.... :(

[quote="matematikator"]:D    It's a problem:      :D

Two altitudes of the triangle ABC are 12 and 20. Prove that the third is less than 30.[/quote]

Please, a problem as in find the area of this or find the length, etc. Not proofs please!

Point E is selected on side AB of triangle ABC in such a way
that AE : EB = 1 : 3 and point D is selected on side BC so that CD : DB = 1 : 2.
The point of intersection of AD and CE is F. Find EF/FC + AF/FD.

[quote="matematikator"]:D    It's a problem:      :D

Two altitudes of the triangle ABC are 12 and 20. Prove that the third is less than 30.[/quote]

Okay, well, I'm just going to treat it as a regular problem and not a proof, you can tell that the third side is going to be less than 30 because you know that this is probably a right triangle with the altitudes 12 and 20, doing the pythagorean theorem we find that 12^2+20^2=c^2, where C=544^1/2=4(34^1/2)=23.32, so that is less than 30. Well, is that kind of what you wanted with these proofs?

In triangle ABC, angle CBA = 72, E is the midpoint of side AC and D is a point on side BC such that 2BD = DC; AD and BE intersect at F. Find the ratio of the area of triangle BDF to the area of quadrilateral FDCE.

Let triangle ABC be a right triangle with a 30 degrees angle at B and a 60 degrees angle at A. Let CD be the altitude to the hypotenuse and let the angle bisector at B intersect AC at F and CD at E.
Show that BE : EF = (3 + 2.[3]^0.5) : 1 and CE : ED = 2 : [3]^0.5.

In a triangle ABC, the intersection point of AK, BL and CM is P. If AP=a,AK=d,BP=b,PL=d
and CP=c,PM=d such that a+b+c=43 and d=3 then find a.b.c

Let ABCD be a unit square and M,N be interior points on the sides AB,BC respectively, such that AM/MB =7 and CN/NB=2. Let P be the intersection point of the lines CM and DN.
Find AP.

In an isosceles triangle ABC(AB=AC, <BAC=30) points Q and P are marked on the side AB and the median AD, respectively, so that PC=PQ (Q is different from B). Find <PQC.

Hey thanx Matimatikor, I'm gonna work on them right now  :)

since Iversonfan2005 is working on the latter ones, i'll try the proof problems  :)


counting
derangement
pigeonhole principle
combinatorics proposed
combinatorics

How many ways to fill the board $ 4\times 4$ by nonnegative integers, such that sum of the numbers of each row and each column is 3?

[quote="Litlle 1000t"]The answer is 2008 :lol:[/quote]
wow :w00tb: Please post the proof.

[hide=" incomplete"]set 1 {0,0,0 3} then we can arrange it to form 4.3.2.1=24 arrays
set 2{0,0,1,2}
set 3{ 0,1,1,1} then we can again arrange it 2.3.4=24 ways

let us take a detailed look at set 2
the first row can be arranged in 4!/2! ways =12 ways
second row can be arranged in 12 ways, but we have to put some constraints .. need to look at these[/hide]

set A={0,0,0 3}
set B={0,0,1,2}
set C={0,1,1,1}

There are some cases for the four rows in some order:
AAAA, AABB,ABBB,ABBC, ABCC, ACCC
BBBB, BBBC,BBCC,BCCC
CCCC

AAAA=CCCC total 24 possibilities each

I seem to be getting $ 2152$.  How do you come up with $ 2008$?

Edit: Scratch that, double counted one class of solutions; 2008 seems right.  Would love to see a clean proof of it.

[quote="implex"]set 1 {0,0,0 3} then we can arrange it to form 4.3.2.1=24 arrays
set 2{0,0,1,2}
set 3{ 0,1,1,1} then we can again arrange it 2.3.4=24 ways[/quote]
I think nothing can be done like this. For one row it is known(quite common) that number of nonnegetive integer sum solution is $ \binom{s \plus{} n \minus{} 1}{n \minus{} 1}$ where $ s$ is the sum and $ n$ is number of numbers. I proved it with a nice solution with partitions. But its $ 2d$. And thats the problem. I too would love to see a clean solution.

Here are the details; they are not very interesting.

If four non-negative integers are to sum to $3$, they are either $\pi(3, 0, 0, 0)$, $\pi(2, 1, 0, 0)$, or $\pi(1, 1, 1, 0)$. Let $a, b, c$ denote the number of rows that are of each of the three forms, respectively. We can now do casework on the value of $a$. (We use the term makeshift row/column; this refers, in an arrangement with at least one $3$, to the entries of a row that are not in the column containing a $3$, or similarly to the entries of a column that are not in the row containing a $3$.)

[b]Case 1 : $a = 4$[/b] - then there are clearly $4! = 24$ arrangements.

[b]Case 2 : $a = 3$[/b] - clearly any row or column with a $3$ in it cannot have any other entries. Then any additional entries cannot lie in either of the three rows or columns containing one of the three $3$s. This leaves but a single square, which must be filled with a $3$, so $a = 4$ contradiction. Thus no valid arrangements here.

[b]Case 3 : $a = 2$[/b] - there are $\binom{4}{2} = 6$ ways to choose the two rows containing the $3$s and $4 \times 3 = 12$ ways to choose the columns. Again, no entries can lie in either of the two rows and two columns containing one of the $3$s. Then we are left to fill a makeshift $2$x$2$ square so that each row and column sums to $3$, but no row or column is $\pi(3, 0)$. The only possibility is $\pi(2, 1)$ which we can permute in $2$ ways. This case then gives $6 \times 12 \times 2 = 144$ arrangements.

[b]Case 4 : $a = 1$[/b] - there are first $4$ ways to choose the row to contain the $3$ and $4$ choices for the column it is in. This leaves us to fill a makeshift $3$x$3$ square with rows either $\pi(2, 1, 0)$ (corresponding to $b$) and $(1, 1, 1)$ (corresponding to $c$; all other entries in the row and column containing the $3$ must be $0$). We can now do casework on the value of $b$.

[i]
Subcase 4.1 : $b = 3$[/i] - we can pick the permutation of the first makeshift row in $3! = 6$ ways. We easily get $2$ possible permutations for the second makeshift row. Finally, the last makeshift row is fixed. In all, this is $12$ sub-arrangements.

[i]Subcase 4.2 : $b = 2$[/i] - then $c = 1$. We have $3$ choices for which makeshift row is $(1, 1, 1)$. The next makeshift row can be any of the $3! = 6$ permutations of $(2, 1, 0)$. Finally, the third makeshift row is fixed. Here we have $18$ sub-arrangements.

[i]Subcase 4.3 : $b = 1$[/i] - this is impossible; we would need some makeshift column to sum to $2 + 1 + 1 = 4 > 3$.

[i]Subcase 4.4 : $b = 0$[/i] - all entries in the sub-square are $1$s, so we have just a single possibility here.

In total, this case gives $16 \times (18 + 12 + 1) = 496$.

[b]Case 5 : $a = 0$[/b]

[i]Subcase 5.1 : $b = 4$[/i] - that is, each row is a permutation of $(2, 1, 0, 0)$. There are $24$ ways to position the $2$ in each row, since no two $2$s can lie in the same column. It only remains to position the ones; this is easily just $D_4 = 9$, the number of derangements of $4$ objects. Thus we have $24 \times 9 = 216$ arrangements here.

[i]Subcase 5.2 : $b = 3$[/i] - we have two fundamentally different arrangements here. In one, every $(2, 1, 0, 0)$ row has its $1$ in the same column and the $1$s in the $(1, 1, 1, 0)$ row line up with the $2$s. There are $4$ choices for the $(1, 1, 1, 0)$ row and $4$ choices for how we permute it. The $2$s in the three $b$-rows can be permuted in $6$ ways. This gives $4 \times 4 \times 6 = 96$ arrangements.

The other is a little more complicated. Again, we have $4$ choices for the $(1, 1, 1, 0)$ row and $4$ choices for its ordering. This time, there is at least one $2$ in the column containing the $0$ of the $(1, 1, 1, 0)$ row; we have three choices for which row it comes from and $3$ choices for which column to put the $1$ from that row in. We can now choose the position of the other two $2$s in $2$ ways and the positions of the other two $1$s in $2$ ways. This contributes another $4 \times 4 \times 3 \times 3 \times 2 \times 2 = 576$ arrangements.

[i]Subcase 5.3 : $b = 2$[/i] - our two $(1, 1, 1, 0)$ rows cannot line up, since otherwise by pigeonhole either we have a column summing to at least $2 + 2 = 4 > 3$ or $1 + 1 + 2 = 4 > 3$. Then they are not ordered the same way, so there are $\binom{4}{2} = 6$ ways to choose their positions and $4 \times 3 = 12$ ways to order them. Here, we can choose the position of the two $2$s in $2$ ways and the position of the two $1$s in $2$ ways, for a total of $6 \times 12 \times 4 = 288$ arrangements here.

[i]Subcase 5.4 : $b = 1$[/i] - we have $4$ choices for the $(2, 1, 0, 0)$ row and $12$ ways to permute it. The columns with the $0$s from this row must contain $1$s in the next three columns, so there are $\binom{3}{1} = 3$ ways to finish. Here we get $4 \times 12 \times 3= 144$ arrangements.

[i]Subcase 5.5 : $b = 0$[/i] - clearly there are $4! = 24$ arrangements here.

----------------------------------------------------------------------------------------------------------------------------------

Adding these up, we get \[24 + 144 + 496 + 216 + 96 + 576 + 288 + 144 + 24 = \boxed{2008}\]


calculus
derivative
analytic geometry
graphing lines
slope
function
inequalities

most of these can be solved by expanding and using AMGM.
 $\frac{x+y}{2}\geq{\sqrt{xy}}$

Or you can substitute x=-y+1 and take the derivative :D

But "no calculus in this forum" so too bad...

[quote="Ray"]If X+Y = 1 then what is the minimum and maximium of:

(1+X3)(1+Y3)  where X3 means X cubed
(1+1/X)(1+1/Y)
(1+X2)(1+Y2)
(1+SQRT(X))(1+SQRT(Y))[/quote]

$(1+x^{3})(1+y^{3})$ has no minimum, its maximum however is $\frac{1+\sqrt{5}}{2}$

likewise, $(1+x^{2})(1+y^{2})$ has no maximum, but its minimum is $\frac12$

too lazy to do any more

Given: $X+Y = 1$

Arithmetic Average of X and Y
$\frac{X+Y}{2}= 0.5$
Geometric average of X and Y
$\sqrt{xy}$
By AM-GM
$0.5 \geq \sqrt{xy}$

See if that helps at all.

I'll post how to do this with derivatives when I get home from school.
Basically, you can think of a derivative as the instantaneous slope of a graph at a point. Minimums and maximums are defined by each having slope zero. So we take the derivative of a function f(x) and set it equal to zero, then we can find what the x-value of the maximums and minimums are. Then, we substitute that into the original function and we get our y-values.
Note that you have to manually check whether a point given by this process is a maximum OR a minimum. It treats both as the same thing. Make sure you put the function into a graphing calculator and manually check.

Also, using the first definition ($X+Y=1$),
set $y = 1-x$ and substitute.

[quote="maokid7"]question. did you mean for both $x$ and $y$ to be positive?[/quote]

Yes X and Y are positive. I also used the inequality to solve the problems. I am thinking maybe these problems were too simple to be posted.

Ray

[quote="Treething"][quote="Ray"]If X+Y = 1 then what is the minimum and maximium of:

(1+X3)(1+Y3)  where X3 means X cubed
(1+1/X)(1+1/Y)
(1+X2)(1+Y2)
(1+SQRT(X))(1+SQRT(Y))[/quote]

$(1+x^{3})(1+y^{3})$ has no minimum, its maximum however is $\frac{1+\sqrt{5}}{2}$

likewise, $(1+x^{2})(1+y^{2})$ has no maximum, but its minimum is $\frac12$

Except that:$\frac{1+\sqrt{5}}{2}$ is equal to 1.618033989 and if x=0 and y=1 then $(1+x^{3})(1+y^{3})$ is equal to 2 and the minimum is given when X=Y=0.5. But I wasn't clear about the X and Y being in the range of (0,1). If X and Y can be anything like X = 1000 and Y=-999 then there are no maximums or minimums.
Ray

too lazy to do any more[/quote]

As promised, the way to do it with a bit of simple calculus.
I'll do one as an example.
$f(x) = (1+\frac{1}{x})(1+\frac{1}{y})$
We want the fraction to be entirely in terms of X.
Using $y = 1-x$ from above, we can substitute and simplify.
I'll leave the algebra up to you.
The resultant function looks like:
$f(x) = \frac{-x^{2}+x+2}{-x^{2}+x}$
Look up derivatives if you are interested in how this works. For now, I'll simply tell you that it *does* work.

The derivative of this function works out to be

$f'(x) = \frac{4x-2}{x^{2}(x^{2}-2x+1)}$
When f'x is zero (hint: use the quadratic formula), you will have a local minimum or local maximum.

In this case, the original function has vertical asymptotes at x = 1 and x = 0, and the local minimum works out to be $\boxed{(0.5,9)}$

[color=darkred][b][u]DiscreetFourierTransform[/u][/b] (what long name !), in the my opinion, the your speech is intricacy and wrong for ... the most simple exercise from all.[/color]

$\boxed{\ x>0\ ,\ y>0\ ,\ x+y=1\Longrightarrow \frac{25}{16}\le \left(1+x^{2}\right)\left(1+y^{2}\right)<2\ }\ .$

[color=darkblue][b]Proof.[/b] $1=(x+y)^{2}=x^{2}+y^{2}+2xy$ $\Longrightarrow$ $\left(1+x^{2}\right)\left(1+y^{2}\right)=(xy)^{2}-2(xy)+2=t^{2}-2t+2=f(t)\ ,$
where $xy=t\in \left(0,\frac{1}{4}\right]\subset (-\infty ,1]$ and the function $f$ is decreasing, i.e. $\frac{25}{16}=f\left(\frac{1}{4}\right)\le f(t)<f(0)=2\ .$

$\boxed{\ x>0\ ,\ y>0\ ,\ x+y=1\Longrightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\ge 9\ }\ .$

[b]Proof.[/b] $\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=$ $\frac{(1+x)(1+y)}{xy}=$ $\frac{1+(x+y)+xy}{xy}=$ $\frac{xy+2}{xy}=$ $1+\frac{2}{t}=f(t)\ ,$
where $xy=t\in \left(0,\frac{1}{4}\right]$ and the function $f$ is decresing, i.e. for any $0<t\le \frac{1}{4}\ ,\ f(t)\ge f\left(\frac{1}{4}\right)=9\ .$

$\boxed{\ x>0\ ,\ y>0\ ,\ x+y=1\Longrightarrow 2<\left(1+\sqrt x\right)\left(1+\sqrt y\right)\le \frac{3+2\sqrt 2}{2}\ }\ .$

[b]Proof.[/b] $\left(1+\sqrt x\right)\left(1+\sqrt y\right)=$ $1+\sqrt{xy}+\sqrt{1+2\sqrt{xy}}=$ $1+t+\sqrt{1+2t}=f(t)\ ,$ where $\sqrt{xy}=t\in\left(0,\frac{1}{2}\right]$
and the function $f$ is increasing, i.e. for any $0<t\le\frac{1}{2}\ ,\ 2=f(0)<f(t)\le f\left(\frac{1}{2}\right)=\frac{3+2\sqrt 2}{2}\ .$[/color]


analytic geometry
geometry
geometric transformation
reflection

The coordinates of $ A,B$ and $ C$ are $ (5,5),(2,1)$ and $ (0,k)$ respectively. The value of $ k$ that makes $ \overline{AC}\plus{}\overline{BC}$ as small as possible is:
$ \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 4\frac{1}{2} \qquad\textbf{(C)}\ 3\frac{6}{7} \qquad\textbf{(D)}\ 4\frac{5}{6} \qquad\textbf{(E)}\ 2\frac{1}{7}$

[hide="Cheap"]So $ \overline{AC} \equal{} \sqrt {5^2 \minus{} (5 \minus{} k)^2} \equal{} \sqrt {(5 \minus{} 5 \plus{} k)(5 \plus{} 5 \minus{} k)} \equal{} \sqrt {k(10 \minus{} k)}$.
And $ \overline{BC} \equal{} \sqrt {2^2 \minus{} (1 \minus{} k)^2} \equal{} \sqrt {(2 \minus{} 1 \plus{} k)(2 \plus{} 1 \minus{} k)} \equal{} \sqrt {(1 \plus{} k)(3 \minus{} k}$.

Therefore, $ \overline{AC} \plus{} \overline{BC} \equal{} \sqrt {k(10 \minus{} k)} \plus{} \sqrt {(1 \plus{} k)(3 \minus{} k)}$.

At this point, I didn't feel like actually thinking, so I brute forced it  :D

Plugging in answer choices, we get:
$ \textbf{(A)} \equal{} 21$
$ \textbf{(B)}\approx 16$
$ \textbf{(C)}\approx 19$
$ \boxed{\textbf{(D)}}\approx 14$
$ \textbf{(E)}\approx 19$

Yes, yes, I know, there's a better way.[/hide]

Edit: Wow, screwed up the distance formula  :| I'm really out of it today.

[hide="grn_trtle:"]Unfortunately, that answer is wrong.  :huh:
Reflect $ B$ over the y-axis to get $ B'\equal{}(\minus{}2,1)$. We see that $ \overline{BC}\equal{}\overline{B'C}$, so we want the minimum value of $ \overline{AC}\plus{}\overline{CB'}$. This is shortest when $ C$ lies on the line $ AB'$. So we solve for the y-intercept of the line going through $ A$ and $ B'$... Using $ y\equal{}mx\plus{}k$ and plugging in the 2 points gives $ k\equal{}2\frac{1}{7}$, or E.[/hide]


There is a frog jumping on a $ 2k \times 2k$ chessboard, composed of unit squares. The frog's jumps are $ \sqrt{1 \plus{} k^2}$ long and they carry the frog from the center of a square to the center of another square. Some $ m$ squares of the board are marked with an $ \times$, and all the squares into which the frog can jump from an $ \times$'d square (whether they carry an $ \times$ or not) are marked with an $ \circ$. There are $ n$ $ \circ$'d squares. Prove that $ n \ge m$.

[hide]Label the unit squares by pairs of integers $ (a,b)$ with $ 1\le a,b\le2k$. Consider the $ \times$'d squares $ (a,b)$. If $ a$ is even and $ b\le k$, the squares $ (a\minus{}1,b\plus{}k)$ is $ \circ$'d. If $ a$ is odd and $ b\le k$, the squares $ (a\plus{}1,b\plus{}k)$ is $ \circ$'d. If $ a$ is even and $ b>k$, the squares $ (a\minus{}1,b\minus{}k)$ is $ \circ$'d. If $ a$ is odd and $ b>k$, the squares $ (a\plus{}1,b\minus{}k)$ is $ \circ$'d. Clearly, these $ \circ$'d squares are pairwise distinct. Hence the number of $ \circ$'d squares is at least $ m$.[/hide]

[quote="Xell"]gwen01 : but the frog can jump from different x'd squares to the same o'd square !![/quote]Sure, I am aware of that. That's why I correspond one x'd square with one o'd square (bijection).

[hide=" Horribly phrased solution."]
Note that moving $k$ units to one direction and 1 unit to its perpendicular direction produces a path of length $k^2+1.$ From x'd square, the frog can go to at least two other cells (either move $k$ units up, 1 unit right; $k$ units down, 1 unit right; $k$ units left, 1 unit up; or $k$ units right, 1 unit left) depending on its position (for example, it cannot move $k$ units left if it's in the leftmost column). Furthermore, the original x'd square gets uniquely determined by those two squares. So we have a bijection, giving $n \geq m.$ $\blacksquare$ [/hide]

Wrong. A frog on the top-left corner can only move to one square that is mentioned by you (namely $k$ right, $1$ down).

Oops, I see. The argument still holds though... the original position of the frog is uniquely determined by the set of squares it can travel to.

Sure, but what then? You still haven't established the bijection.

Yeah, that's why I called it a "horribly phrased solution". :P Well, the sets of squares the frog can move to from a x'd square are all mutually disjoint, so given a set of x'd squares $\{\mathcal{P}_1, \mathcal{P}_2, \cdots , \mathcal{P}_x\},$ the frog can find a set $\{\mathcal{Q}_1, \mathcal{Q}_2, \cdots , \mathcal{Q}_x\}$ such that the frog can move from $\mathcal{P}_i$ to $\mathcal{Q}_i$ for all $i.$ I'm not very sure how to explain this in words.


[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=296[/img][hide=":-)"]please help me answer the question. I've been looking for an answer forever!!!!!!![/hide]

[hide]Dont see the ans,its too easy[/hide]
[hide]Take linear pair of both angles and then.....[/hide]


my school doesn't invigilate AIME during march break, and i cannot take the aime II because i write CMO that day, i need help. any suggestions?

lol...thats a problem lots of ppl r facing...cant help on this one...sry...

Write both! Write one contest in the morning and one in the afternoon.

I have the same problem.
APMO on Mar 15, CMO on Mar 28 and AIME which i may write will be on the same day. Can I write both AIME and CMO on the same day?
Just to make sure

Yes you can.

Btw APMO is March 12 not 15.

[quote="Kileel"]Jack, possibly you could write the AIME I at my school: in Fredericton, NB.  If you are taking the APMO, you may consider taking both in Fredericton (APMO is one day before AIME I).

This likely is a bad idea if you don't have any family in the area with which you could stay.  Then you'd have to write AIME II and CMO on the same day I guess...[/quote]

Im not taking apmo, so i could just write aime i. and btw, how long does it take to travel from halifax to fredericton?
aime II and cmo is on the same day, so it takes about 6 hours a day, which is not a good idea i think, cuz my head might explode.
i'll talk to other teachers about it, and see if there's a teacher that could invigilate me.


Find the general equatio of the parabola whose focus is $ {(2,3)}$ and directrix $ {x\minus{}4y\plus{}3}\equal{}0$

[quote="21 7 15"]Find the general equatio of the parabola whose focus is $ {(2,3)}$ and directrix $ {x \minus{} 4y \plus{} 3} \equal{} 0$[/quote]

Apply the definition $ \ \frac {PS}{PM} \equal{} 1 \implies PS \equal{} PM$

$ \implies \sqrt {(x \minus{} 2)^2 \plus{} (y \minus{} 3)^2} \equal{} \frac {|x \minus{} 4y \plus{} 3|}{\sqrt {1^2\plus{}4^2}}$

$ \implies (x \minus{} 2)^2 \plus{} (y \minus{} 3)^2 \equal{} \left(\frac {x \minus{} 2y \plus{} 3}{17}\right)^2$

Sorry i edited my post

Firstly its $ {x\minus{}4y\plus{}3}$ not $ {x\minus{}2y\plus{}3}$.
Secondly i want to know how you got $ \sqrt{5}$ as denominator.

I got the same answer as makar and I squared everything and simplified to get $ 16x^2\plus{}y^2\plus{}8xy\minus{}76x\minus{}78y\plus{}212\equal{}0$
In a parabola the distance from any point on the parabola i the equidistant from the focus and the directrix.

The perpendicular distance from a point $ (x_1, y_1)$ to a line $ ax\plus{}by\plus{}c\equal{}0$ is given by:

$ d \equal{} \frac{|ax_1 \plus{} by_1 \plus{} c|}{\sqrt{a^2 \plus{} b^2}}$

The equation [b]makar[/b] has basically says:
"Distance from (x,y) to focus = Distance from (x,y) to x-4y+3=0"
which is subsequently squared.

[b]Elefont[/b] expands the resulting equation and presents it in standard conic form.


The sum of two prime numbers is 20.  What is the greatest possible product of these two primes?

[hide="Hint"]
Think about what the prime means. This should give you only two possible choices.[/hide]

[hide="BIG_HINT"]
Two choices are {3,17} and {13,7}. Which of the product is larger? :-D [/hide]

[hide="ANSWER"]
You should have it by now. 91 is the answer.[/hide]

[hide]the numbers are 13 and 7 becuase they are close together

13*7=[b]91[/b][/hide]

[hide][color=indigo]The only primes that work are 3+17 and 7+13.
The bigger product is[color=green] 91 [/color](which is 7*13.)  [/color][/hide]

[color=green]sorry, the colors didn't work out[/color]

[quote="LadyKn1ght"][hide][color=indigo]The only primes that work are 3+17 and 7+13.
The bigger product is[/color][color=green] 91 [/color][color=indigo](which is 7*13.)  [/color][/hide][/quote]

It does now. Click quote to see why.

[hide]91[/hide]
Can you tell me where i can get some informations about basic number theory online?


in how many ways can 5 identical rooks be placed on a 8 by 8 board such that 4 of them form a rectangle?

[hide]Identify each square on the chessboard by an ordered pair of integers $ (x,y)$ where $ 1\leq x,y\leq 8$.
Suppose the vertices of the rectangle are $ (a_1,b_1),(a_1,b_2),(a_2,b_1),(a_2,b_2)$, and WLOG let $ a_1<a_2$ and $ b_1<b_2$.  There are $ 7\plus{}6\plus{}5\plus{}\ldots\plus{}0\equal{}28$ ways to choose the values of $ a_1$ and $ a_2$.  This is so because each term in the sum corresponds to how many possibilities there are for $ a_2$ when $ a_1\equal{}1,2,3,\ldots,8$.  Similarly, there are 28 ways to choose $ b_1$ and $ b_2$.  Then there are 60 empty squares left to place the 5th rook.  We don't have to worry about overcounting because of our assumptions and because the fifth rook can't create a new rectangle.  So the answer is $ 28^2\cdot 60\equal{}\boxed{47040}$.[/hide]

[b]@ OP:[/b] Do you mean for the sides of the rectangle to be parallel to the sides of the board, as was assumed above?

[hide="solution"]The number of rectangles is $ {8\choose 2}^2$.  There are 60 spaces left over so $ {8\choose 2}^2\cdot 60$[/hide]


[img]http://img175.imageshack.us/img175/8490/geometria25fz3.png[/img]

$ FB \equal{} BG \equal{} AB\plus{}AC \equal{} EB\plus{}ED$

$ FA \equal{} AB\frac{\sin30}{\sin54}\equal{} AB\frac{\sqrt{5}\minus{}1}{2}$

$ FB \equal{} AB\frac{\sqrt{5}\plus{}1}{2}\equal{} AB\cdot 2\cos36 \equal{} EB\frac{2\sin36\cos36}{\sin24}\equal{} EB\frac{\sin72}{\sin24}$

$ EG \equal{} EB\frac{\sin72\minus{}\sin24}{\sin24}\equal{} EB\frac{2\sin24\cos48}{\sin24}\equal{} EB\cdot 2\cos48$

$ \frac{AE}{EC}\equal{}\frac{\sin12\sin54}{\sin24\sin18}\equal{}\frac{\frac{1}{2}\sin48\minus{}\frac{1}{2}\sin24}{\sin24\sin18}\equal{}\frac{2\cos24\minus{}1}{2\sin18}\equal{}$

$ \equal{}\frac{2\cos48(2\cos24\minus{}1)}{4\cos48\sin18}\equal{}\frac{2\cos48(2\cos24\minus{}1)}{4\sin42\sin18}\equal{}$

$ \equal{}\frac{2\cos48(2\cos24\minus{}1)}{2\cos24\minus{}1}\equal{} 2\cos48 \equal{}\frac{GE}{EB}$

so the triangles AGE and CEB are similar and $ \angle AGE \equal{} 18\Longleftrightarrow\angle DEB \equal{} 36$


what are the steps needed to take in order to solve this problem by taking square roots? Thanks alot for your help.

(x-1)2=36  The x-1 is squared i didn't know how to make a small 2

take the square roots of both sides sqrt(x-1)^2=x-1
sqrt(36)=6 OR -6

Case 1.
x-1=6
x=7

Case 2.
x-1=-6
x=-5

$(x-1)^{2}= 36$

$x-1 = \pm\sqrt{36}$ find the square root of each side

$x = 1\pm6$ bring the one over to isolate x

and there you have it.


[quote="Stargirl240"]You do mean the maximum ways you can arrange them, right? :huh:[/quote]

Yes, maximum way!

[quote="moldlee"]We put 6 different color balls in a row.

If two balls were next to each other in an arrangement, then

they cannot be next to each other in the following arrangements.

How many times can we rearrange the balls this way?[/quote]

[hide]
5 times[/hide]

[quote="moldlee"][hide]solution : 5

(why?)
hint : 3 ball => 2 ways
4 ball => 3 ways
5 ball => 4 ways
n ball => n-1 ways
by induction.... good problem!!![/hide]

[size=75][color=darkred]Please hide solutions/answers when posting in response to a problem. Thanks.
-nebula42[/color][/size][/quote]

[hide="question about your solution"]
How would you induct?
[/hide]

I don't think there is a simple induction, so I think he might have meant pattern.

3 balls A, B, C;  2ways (ABC, ACB)
4 balls A, B, C, D; 3ways (ABCD, ACBD, BADC)
....

[quote="SplashD"][quote="moldlee"][hide]solution : 5

(why?)
hint : 3 ball => 2 ways
4 ball => 3 ways
5 ball => 4 ways
n ball => n-1 ways
by induction.... good problem!!![/hide]

[size=75][color=darkred]Please hide solutions/answers when posting in response to a problem. Thanks.
-nebula42[/color][/size][/quote]

how did you get 2 ways for 3 balls?

You have ABC and that's it. B can't be next to any other ball in the next arrangement since it was next to A and C in the first arrangement.[/quote]

[quote="moldlee"]If two balls were next to each other in an arrangement, then

they cannot be next to each other in the following arrangements.
[/quote]

[quote]3 balls A, B, C; 2ways (ABC, ACB) [/quote]

in the first arrangement, B was next to C and in the second arrangement, B is also next to C, which contradicts the info given in your problem.