projective geometry
combinatorics unsolved
combinatorics

assume that $X$ be set and $r$ be integer.$X_r$ is set of all subset with $r$ members of$X$.assume that $F$ is subset of $X_r$ s.t:for any$A_{i_1},A_{i_2},...,A_{i_k}\in F \Rightarrow A_{i_1}\bigcap A_{i_2}\bigcap ...\bigcap A_{i_k}\neq \phi $if :
$I(F)=min${$|T|:T\subseteq X;T\bigcap A\neq \phi, for any A\in F$}prove that $I(F)\leq \frac{r-1}{k-1}+1$

[quote="sam-n"]Let $X$ be a set. $X_r$ is set of all subsets with $r$ members of $X$. Let $F$ be a subset of $X_r$ such that for any $A_{i_1},A_{i_2},...,A_{i_k}\in F \Rightarrow A_{i_1}\cap A_{i_2}\cap\ldots\cap A_{i_k}\neq \emptyset $. Let
$I(F) =\min\{\;|T|: T\subseteq X;T\cap A\neq \emptyset$ for any $A\in F\}$. Prove that $I(F)\leq \frac{r-1}{k-1}+1$.[/quote]

To repeat that, $F$ is a set of r-element subsets of $X$ with the property that the intersection of any $k$ sets in $F$ is nonempty. Prove that if $T$ is the smallest set with the property that it has an non-empty intersection with every set in $F$, then $|T|\leq \frac{r-1}{k-1}+1$.

First, I created some examples of sets with those properties it see if they gave me any clues to this bound.

My first set was to let $X$ be a n-dimensional finite projective space with characteristic q, and $F$ is the set of $(n-1)$-dimensional subspaces. Okay, this is not an everyday example, but I did my doctoral research with such spaces, so I am very familiar with their properties. We have $r = \frac{q^{n}-1}{q-1}$. The intersection of $i$ many $(n-1)$-dimensional subspaces is a $(n-i)$-dimensional subspace, and a $0$-dimensional subspace is a non-empty point, so k = n. A suitable $T$ is any $1$-dimensional subspace in $X$, which has $q+1$ elements. The bound is $\frac{r-1}{k-1}+1 = \frac{q^n-q}{(q-1)(n-1)}+1$. For $n=2$ this bound is $q+1$, which is tight. For $n=3$ this bound is $\frac12 q(q+1)+1$, which is looser. For $n=4$, this bound is $\frac13 q(q^2+q+1)+1$, which is even looser. Tight bounds give more insight that loose bounds, so no need to look at larger n. The most interesting thing about this example is that $T$ is an intersection of $k-1$ sets in $F$. I wonder how often that will happen. The $n=2$ case is that $X$ is a finite projective plane and $F$ is the set of lines in that plane. In a projective plane, any two lines intersect, which fits with $k=2$.

Another example with a tight bound is to let $X=\{1,2,\ldots,n\}$ and let $F$ be all $(n-d)$-element subsets of $X$ for some integer $d$ between $1$ and $n-1$. Define $A_i = X\setminus\{i,i+1,\ldots,i+d-1\}$ for all $i=0,1,2,\ldots, n-d$. We can even define $A_i$ for $i\ge n-d-1$, but we will want the integers greater than $n-1$ to wrap around mod $n$. For the k-set intersection $A_1\cap A_{d+1}\cap A_{2d+1}\cap A_{(k-1)d+1}$ to be non-empty, we need $kd<n$. Let us retroactively assign $n=kd+1$. Then $r$ = $n-d$ = $kd-d+1$. A good $T$ is any $(d+1)$-element subset. The bound is $\frac{r-1}{k-1}+1$ = $\frac{kd-d+1-1}{k-1}+1$ = $d+1$. This bound is tight. And once again, $T$ can be constructed as an intersection of $k-1$ sets in $F$.

These two examples, especially the second one, gave me enough hints to construct the proof, so I will leave the rest to you young people. I wouldn't want to spoil your fun.

Erin Schram


For the positive integers $ a_{1},a_{2},...,a_{n}$ prove that $ \left(\frac {\sum^{n}_{i \equal{} 1}a^{2}_{i}}{\sum^{n}_{i \equal{} 1}a_{i}}\right)^{\frac {kn}{t}}\geq \prod^{n}_{i \equal{} 1}a_{i}$, where $ k \equal{} max\left\{ a_{1},a_{2},...,a_{n}\right\}$ and $ t \equal{} min\left\{ a_{1},a_{2},...,a_{n}\right\}$. When does the equality hold?

$ \sqrt [2]{\frac {\sum a_i^2}{n}}\ge \frac {\sum a_i}{n}$

$ \sum a_i^2 \ge \frac {(\sum a_i)^2}{n}$

$ \frac {\sum a_i^2}{\sum a_i} \ge \frac {\sum a_i}{n} \ge \sqrt [n]{\prod^{n}_{i \equal{} 1}a_{i}}$

$ (\frac {\sum a_i^2}{\sum a_i})^n \ge \prod^{n}_{i \equal{} 1}a_{i}$

Now, $ \frac {\sum a_i^2}{\sum a_i}\ge 1$ So therefore since $ \frac {k}{t}\ge 1$

$ (\frac {\sum a_i^2}{\sum a_i})^{\frac {kn}{t}}\ge(\frac {\sum a_i^2}{\sum a_i})^n$

Now, the direct application of AM-GM required that all terms are equal for equality to occur, and indeed, equality holds when all a_i are equal.


calculus
integration
logarithms
calculus computations

([b]4[/b]) Find all $ y > 1$ satisfying $ \int^y_1x\ln x\ dx \equal{} \frac {1}{4}$.


algebra
function
domain
Asymptote
irrational number

Solve in (-1,+ :inf: ):

[(x+2)^2-1]^x+(2x+4)^x=[(x+2)^2+1]^x

I wasn able to determine all the solutions...

sorry i misposted,the domain is all reals in the interval (-1, :inf: ).I edited the former post.

Well, this is mathematica talking, but I think there's only 1 solution.

By hand, you can definitely establish that x=2 is the only rational solution possible (then plug in and confirm that it works), because the a^x + b^x = c^x format can only work for x=2 according to Fermat's Last Theorem.  FLT is for integers, but for rationals (this used to say irrationals here, but I've edited it to be correct) you just multiply by the common denominator and get back to the integer form.  I'm not sure whether this can be extended to irrational number, though.

From a graph there appears to be one positive solution
Three negative solutions

And a horizontal asymptote at y=0
http://68.38.137.202/graph.JPG

Note that if all the exponents were 2 (instead of x), the equation would be an identity for all x.  Using that you can show that the only solution to the original equation for x > -1 is x = 2.

this problem was originaly discussed and solved here:

http://www.mathlinks.ro/phpBB/viewtopic.php?t=2555


It's basically more or less what you were requesting [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=229527]here[/url].

I believe there will be an option to search based on tags in the future. However, for now, if you click a tag, it'll draw up a list with the same/similar tags of other entries.

You can "search" for tags with the syntax: http://www.mathlinks.ro/weblog.php?tag=ABC,XYZ,UWV etc. (so separate the tags by commas)


function limit algebra polynomial logarithms calculus derivative

Let $ f: R \to R$ be a function such that $ f(x\plus{}y) \equal{} f(x)f(y)$ for all $ x,y \in R$ and $ f(x) \equal{} 1 \plus{} xg(x)$ , where $ \lim_{x \to 0} g(x) \equal{}1$.Then the function $ f(x)$ is :

1) $ e^x$.
2) $ 2^x$.
3) a non constant polynomial.
4) equal to $ 1$ for all $ x \in R$.

[quote="guru_bhai"]Let $ f: R \to R$ be a function such that $ f(x \plus{} y) \equal{} f(x)f(y)$ for all $ x,y \in R$  :
[/quote]

from  this we  get f(x)=a^kx

this  means  g(x)=(a^kx-1)/x

from  the given limit of g(x) , we cans ay that  log a=1  that is a=e

so answer would be option 1

actually i have a doubt. no doubt e^x satisfies this but my doubt is whether linear functions also satisfy the ans.
that is if partailly diff w.r.t x and put  y=0 in the first equation i get f'(x) = 1.which is a linear .on substituting i get f(X) = x +1 . which i see surprisingly satisfies .please help.

If it is any sort of non-constant polynomial, $ f(x\plus{}y)\neq f(x)f(y)$

[quote="vijaymenon"] i get f(X) = x +1 . which i see surprisingly satisfies .please help.[/quote]

how  do say  this  satisfies?  this obviously  doesnt satisfy the comdition.

The first condition implies that $ \log f$ is additive, and the second implies that $ f$ is differentiable (hence continuous) at $ 0$, with derivative $ 1$ at $ 0$.

From the additivity of $ \log f$ combined with the continuity at of $ f$ at $ 0$, one concludes that $ f(x) \equal{} \exp kx$ for some $ k\in\mathbb{R}$.  Since $ f'(0)\equal{}1$, $ k\equal{}1$.  This is the only possible solution.


Prove $ \sum_{k\equal{}0}^n3^kC^n_k\equal{}4^n$.

I have found a proof by combinatorial interpretation, but I hope to see some algebraic proof.

It's quite obvious by $ (1\plus{}3)^n\equal{}4^n$ . What kind of solution are you looking for?


quadratics
algebra
polynomial
Putnam
number theory unsolved
number theory

Does there exist a quadratic polynomial $ f(x)$ with integer coefficients such that when $ x$ is an integer composed of only $ 1$'s, $ f(x)$ is also composed of only $ 1$'s?

It's from MOP, yes?  It's generalized by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=177229]this[/url] Putnam problem.  (Found that by searching "repunit polynomial" (without quotes).)

The answer is yes. Take, for instance, $ f(x)\equal{}9x^2\plus{}2x$. The proof that it works is not difficult.


This is actually a math problem from contest but it's also nice question that you can do for fun.

What is the smallest number $n$ so that no matter how you place $n$ marks on a 3-by-3 tic-tae-toe board, you are assured of getting three in a row (either horizontally, vertically, or diagonally)?

Enjoy! And extra honor to whoever finds the answer with drawing. :lol:

[hide="My Answer"]
n=7

There is no where to place the seventh.
x|x|_
x|_|x
_|x|x
[/hide]

[quote="tcs09"][hide="My Answer"]
n=6

x|x|_
x|_|x
_|x|x
[/hide][/quote]
Prove it!

You posted right before I edited my when I realized that I posted the largest n that you can get an not get a three in a row. :D

OK. I explain what you did in detail.

If we place seven marks, we have a horizontal row with 3 marks(pigeonhole principle).
Now we need to prove that 7 is indeed a mininum.
6 doesn't work because of the diagram that tcs09 drew.

Therefore, 7 is the answer.

[quote="Silverfalcon"][quote="fdump"]so, isn't it 7?
anyway, eather i don't understand the problem, or it's just too simple. :)[/quote]

It is 7 and because it's too simple, it's in Puzzles & Brainteasers forum. Otherwise, I would've posted in Getting Started or Intermediate. :)[/quote]

I agree. It was too simple.


Whats your rating in the forum? Incase you don't know about the ratings. People can rate your post by clicking on the benzene shaped boxes above each of your post. I usually rate good posts with 4-5 stars and spam posts with 1 star but I don't really rate other posts. :D

To check your rating go to your profile and see.

My rating: 3.9. (was 4 yesterday, I guess somebody gave me a bad one)

Oh this sucks big time. Apparently I havent been rated yet.

(P.S - its cyclohexane shaped Shreyas :D)

Both of the above users are wrong :D its neither in aops its like a bicyclo compound (bcos of the shades)

Well mine howers really. It used to be around 1.5 bcos of a fellow called JBL whos a mod and he simply used to pick fights with me :mad: and then it increased and decreased and now if im right its around 2.5 . But i least bother as long as i can post :D

Baby, not to feel.. Go and feel at the way India is bpwling, or at the way KVKK has given his notes, but no the feel on AOPS...

all r wrong--its a hexacyclo compound :D  :rotfl:

i haven't been rated yet!!!!

[quote="Cyber-Shreyas"][quote="madness"][quote="rituraj007"]all r wrong--its a hexacyclo compound
[/quote]

And why should that be so?  :huh:   :huh:   :P[/quote]

He's right you know. Its hexacyclo, not bicyclo.[/quote]

And why do you too believe that? :huh:

[quote="Chapplikka"]Ritu, not to feel.. Go and feel at the way India is bowling, or at the way KVKK has given his notes, but no the feel on AOPS...[/quote]

See what chappli has written ritu :P :D

no,See what chappli has written
[quote="Chapplikka"]Rohit, not to feel.. Go and feel at the way India is bowling, or at the way KVKK has given his notes, but no the feel on AOPS...[/quote]

my rating sux. its 2.4 rite now. 'twas 2.6 just yesterday. :noo:  woe betide all those who've given me one or two stars. :mad:  :bruce:

my rating is 3.8
i think it's inversely proportional to the useful posts of a person :D
and shreyas, thats 6 cyclohexanes in a row. remember what the common name of that compund is called?  :D

[quote="hash_include"]
i think it's inversely proportional to the useful posts of a person
[/quote]

Yes i agree :D Its bcos the more number of useful posts you post, the more people will find mistakes and more lower your rate will get :D :D

its directly proportional to useless posts of the person :D  :rotfl:  :rotfl:

Mine is generally between 6 and 6.5, currently around 5.8.

i thought it can't  b greater than 6
how's it calculated?

It can't? I vaguely remember it being 6.3 once. Maybe I'm wrong...

hey man who's done that--my rating is 1.2
the reason 4 this shud reach hell soon :rotfl:  :rotfl:

[quote="Cyber-Shreyas"]Ashwath, there's no way your rating can be greater than 6, especially on a scale of 6.  :rotfl:

I'm going down as well. 3.7' now. I shouldn't really post in the higher mathematical forums. I get screwed there.[/quote]

Thats what i said :D :D Thell maraud you even if you make one mistake :D

I thought someone was going to go offline. Apparently not.

Okay, maybe I remember wrong. Strange, I could swear that there was a 6.1 or/and a 6.3 . Maybe it was a bug?

or again interlinking of servers :rotfl:  :rotfl:
u kno WWW is gonna shut down by 2011 due to load,videos........☻♥■

Great now my rating is 3.4.. At this rate, I'll reach 0 soon. Maybe some of my fellow Indians are rating me spam for mainly

[hide]
Chumma Jujubi[/hide]

Mine is also just 2.6. But that doesn't stop me from posting whatever I want.

rating is not even visible as long as we don't want.y care about it :D  :rotfl:  :D

mine also has shot upto 1.3 :rotfl:  :rotfl:

hey nobodys posted on my posts so wat do i do!

hey mine is given as 1.2 so wat do i do?
\


Exactly what are these, and how can they be used?

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

They can be applied to lots of problems, for instance [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79508&sid=f68b38dc2bbb1b467eafa36f81110b52]here[/url].

I downloaded a document a long time ago which listed over a hundred unique examples of problems whose solutions involved the Catalan numbers. Does anyone remember that document? I might be able to upload it if no one has the link to it.


limit
calculus
topology
real analysis
real analysis theorems

Give a formal proof of the following:

$ \lim_{n\rightarrow\infty}{\frac{1}{n}} \equal{}0$

For any $ \epsilon > 0$, if $ \delta > \frac {2}{\epsilon}$ then $ |f(\delta)| < \epsilon/2 \le \epsilon$.

If you require a proof dug from the bowels of predicate calculus, would you care to express the problem completely in quantifier form,
so that we can know your definitions of 0,1, division, and limits for a formal proof? I'll be glad to throw it into a prover for you; it's
just that translating all the machinery of limits and whatnot into symbolic logic is a bit of a pain -- and after all, the problem isn't fully
specified if it's not in symbols to begin with.  :D

That's not a valid argument; this is the Archimedean property, and you have to explicitly invoke completeness somewhere to prove it. The exact method depends on which version of the completeness axiom we're using in the definition of $ \mathbb{R}$.
If we're using the least upper bound property, here's a path:
The set $ \{1,\frac12,\frac14,\cdots,\frac1{2^n},\cdots\}$ is bounded below by zero. It therefore has a greatest lower bound $ L$. Halving all of the elements, the greatest lower bound of $ \{\frac12,\frac14,\cdots,\frac1{2^{n\plus{}1}},\cdots\}$ is $ \frac{L}{2}$. But this is just a shift of the sequence; it must have the same lower bound. That makes $ L\equal{}\frac{L}{2}$, which we can solve to find that $ L\equal{}0$.

[quote="jmerry"]That's not a valid argument; this is the Archimedean property, and you have to explicitly invoke completeness somewhere to prove it. The exact method depends on which version of the completeness axiom we're using in the definition of $ \mathbb{R}$.
If we're using the least upper bound property, here's a path:
The set $ \{1,\frac12,\frac14,\cdots,\frac1{2^n},\cdots\}$ is bounded below by zero. It therefore has a greatest lower bound $ L$. Halving all of the elements, the greatest lower bound of $ \{\frac12,\frac14,\cdots,\frac1{2^{n \plus{} 1}},\cdots\}$ is $ \frac {L}{2}$. But this is just a shift of the sequence; it must have the same lower bound. That makes $ L \equal{} \frac {L}{2}$, which we can solve to find that $ L \equal{} 0$.[/quote]

I suppose that depends on the definition of "limit", no, if $ \epsilon$, $ \delta$ isn't good enough?

Also, the rationals are Archimedean, too, but they're not complete... I don't see why completeness is necessary.

It is troublesome to make a note in English.
So I write an answer in Japanese.

任意に $ \varepsilon > 0$ をとる. アルキメデスの原理 (Archimedean property) により, $ \frac {1}{\varepsilon} < N$ を満たす $ N \in \mathbb{N}$ がある. $ \frac {1}{N} < \varepsilon$ より, $ n \ge N$ に対して,
\[ \left|\frac {1}{n}\right| \equal{} \frac {1}{n}\le\frac {1}{N} < \varepsilon \ \ \text{i.e.} \ \ \left|\frac {1}{n} \minus{} 0\right| < \varepsilon \ (n \ge N).\]
ゆえに, $ \lim_{n \to \infty} \frac {1}{n} \equal{} 0$.

There are ordered fields that aren't Archimedean; that's why completeness is necessary. In a non-Archimedean ordered field, this sequence doesn't have a limit.
I also could have used $ \frac1n$ with the map $ f(x)\equal{}\frac{x}{1\plus{}x}$ as the shift. It doesn't make much of a difference.

Of course, I didn't write all the steps out. The $ \epsilon$ and $ \delta$ come in when we apply the Archimedean property to show this particular limit.

True, but none of them have a copy of the naturals that go to infinity.

How, exactly, are you defining "limit", anyway? It would appear that even considering
a metric on $ \mathbb{R}$ circularly involves $ \mathbb{R}$ in the discussion from
the get-go. I think the fact that the codomain of $ |.|$ is $ \mathbb{R}$ makes things
formally follow without considering completeness, doesn't it?

Order induces a topology (the basic open sets are open intervals $ (a,b)$). That's how we define a topology on some other ordered field. Those fields do have copies of the rationals; it's just that they also have stuff that's bigger and stuff that's smaller.

Ah, so you're proceeding from a more topological definition rather than the standard one that would
appear in calculus?

I'm just very curious... since when does the $ \epsilon \minus{} \delta$ definition of limit for the reals
require justification? If not, what step is wrong with the (what I believe is)
standard proof I gave earlier?

Or is it not immediate that this is the definition of limit (as $ x \to \infty$ on $ \mathbb{R}$)
(as it is typically taught):

For every $ \epsilon > 0$, there is a $ \delta > 0$ such that whenever $ x > \delta$, then
$ |f(x) \minus{} L| < \epsilon$?

The $ \epsilon$/$ \delta$ definition is just a definition, which applies to any metric space, and the extension to limits at $ \infty$ is also just a definition.. No justification required.
If we assume $ n$ to be a real-valued variable, your proof is complete. If we assume it to be an integer-valued variable, we need the Archimedean property to assure the existence of a large enough integer $ \delta$. Traditionally, $ n$ denotes an integer, so the sequence version is what I assumed.

Thank you all,but the above proofs are not formal.

The problem in quantifier form is:

$ \forall\epsilon[\epsilon>0\Longrightarrow\exists n_{o}(n_{o}\in N\wedge\forall n(n\geq n_{o}\Longrightarrow |\frac{1}{n}|<\epsilon))]$ which tell us that:

Given ε>0 there exists an $ n_{o}$ belonging to the natural Nos,such that for all the natural Nos $ n\geq n_{o}$ ,$ |\frac{1}{n}|<\epsilon$

[quote="jmerry"]Thanks, but I don't care.

I don't read Japanese at all. I still find Kouichi Nakagawa's post easier to understand than something symbolic enough for you.

Besides, you haven't specified the rules. Can we take the Archimedean property as a theorem? If not, what axiom system are we working under? The real numbers are complicated enough that there are several different reasonable answers to that question, and they all lead to different proofs.[/quote]

Here is the definition of a formal proof, given by Nukular in another thread:

" I recall, a formal proof would consist of a list of statements, each of which is
either a given, an axiom, or is the resolution of two previous steps through some rule of formal logic."

[quote="τρικλινο"]
I recall, a formal proof would consist of a list of statements, each of which is
either a given, an axiom, or is the resolution of two previous steps through some rule of formal logic."
[/quote]

Firstly, that's irrelevant, since jmerry was asking what axiom system you were working from. Secondly, I don't appreciate being quoted out of context, unless you were attempting to be clever by making a personal attack, which, by the way, wasn't clever at all  :rotfl:.

Formal proofs can be written in any axiomatic system.

And Nukular i simply considered your definition correct and that is why i quoted it.

In that case, I believe jmerry was asking which one you would like a formal proof to be written in. One pins $ \mathbb{R}$ as the complete ordered field; one could proceed from its construction, or ZFC; there are many options, all essentially equivalent for purposes of this problem. Then again, it's most convenient to have as much machinery as possible developed, in which case jmerry's proof is just an English version of a formal proof, easily translatable. It's just that most theorems we manipulate sets with aren't named explicitly, and are instead numbered propositions in introductory texts, or exercises left to the reader. It would be rather cumbersome to be forced to use such a text as a crutch, or to be overencumbered by having to reprove from scratch things we already know so well.


geometry
induction
combinatorial geometry
combinatorics unsolved
combinatorics

In the square with unit side, $ n^2$ points are located so that no three points lie on one line. Prove that there exists a triangle with the vertices in these points of area not greater than $ \frac{1}{2(n\minus{}1)^2}$.

Because no three points are on the same line, we can consider arbitrary triangulation of unit square. It is easy to prove (by induction or by summing all angles) that number of triangles is $ 2(n^2 \plus{} 1)$. When we add first point - we get four triangles, after inserting next point we get exactly two more triangles. Thus, there exist triangle with area less or equal than $ \frac{1}{2(n^2\plus{}1)} < \frac{1}{2(n \minus{} 1)^2}$.

Aleksandar Ilic

Borrowing the triangulation method from the above post:

Consider the convex hull of the $n^2$ points.  Say the hull contains X points.  Then triangulating the convex hull will lead to it being divided into $x + 2(n^2-x-1) = 2n^2 -x -2$ triangles.  Since the maximum possible area of the convex hull is 1, there must be a triangle with area less than or equal to $\frac{1}{2n^2 -x -2}$.  Since $x \le n$, that value is less than $\frac{1}{2(n-1)^2}$ and we are done.

I think the fault with the alexilic's solution was that he included the vertices of the square in the triangulation.

[quote="LastPrime"][color=#FF0000]Since $x \le n$[/color], that value is less than $\frac{1}{2(n-1)^2}$ and we are done.[/quote]
And why is that true? The convex hull may contain [b]all[/b] $n^2$ points (I think the poster forgot that the number of points was $n^2$, not $n$).
The problem does not succumb so easily to just this simple triangulation idea. See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1030132&sid=460a070290d729c344fba4d6a713feec#p1030132[/url] for the correct approach.


AMC
AIME
geometry
perimeter
power of a point

A quadrilateral with perpendicular diagonals of lengths $ 39$ and $ 41$ is inscribed in a circle of area $ 1690 \pi$. Given that the perimeter of the quadrilateral is $ a \sqrt{b} \plus{} c \sqrt {d}$, where $ a$, $ b$, $ c$, and $ d$ are positive integers and neither $ b$ or $ d$ is divisible by the square of any prime, find $ a\plus{}b\plus{}c\plus{}d$.

Just a note: please only post items that are more-or-less [i]directly[/i] related to AMC/AIME/USAMO in the AMC forum.  Individual "AIME-level" problems that are not actually AIME problems should be in the appropriate high school forum.  Thanks.

I'm sorry if this is a stupid question, but is this set-up possible? I can't seem to draw a quadrilateral with perpendicular diagonals inscribed in a circle of area $ 1690\pi$...

Could somebody post a diagram?

I've just wasted a good half-hour  :|

[asy]size(200);
draw(Circle((0,0),13*10^0.5));
pair A=(-19.5,-36.19047), B=(19.5,-36.19047);
dot(A); dot(B); draw(A--B); label("\(39\)",(A+B)/2,N);
picture p = new picture;
draw(p,Circle((0,41),13*10^0.5));
clip(p,(19.5,41)--(-19.5,41)--(-19.5,-50)--(19.5,-50)--cycle);
add(p);[/asy]

Given a chord of length $ 39$ on a circle of radius $ 13\sqrt {10}$, the perpendicular diagonal would have an endpoint on the arc drawn. So the construction is impossible ..

Edit: It could be a complex quadrilateral, but I highly doubt it. I don't think the problem would change if the diagonals were $ 78 \times 82$ or something, perhaps those were the intended numbers?

[quote="azjps"]I've just wasted a good half-hour  :|

[asy]size(200);
draw(Circle((0,0),13*10^0.5));
pair A=(-19.5,-36.19047), B=(19.5,-36.19047);
dot(A); dot(B); draw(A--B); label("\(39\)",(A+B)/2,N);
picture p = new picture;
draw(p,Circle((0,41),13*10^0.5));
clip(p,(19.5,41)--(-19.5,41)--(-19.5,-50)--(19.5,-50)--cycle);
add(p);[/asy]

Given a chord of length $ 39$ on a circle of radius $ 13\sqrt {10}$, the perpendicular diagonal would have an endpoint on the arc drawn. So the construction is impossible ..[/quote]
Yea...I tried drawing a diagram like that too... haha

EDIT: Yes, th diagonals are supposed to be 78 and 82. :blush:
Sorry to all whose time I wasted.  :lol:

[hide="Solution"]The quadrilateral has perpendicular diagonals $ 78$ and $ 82$.
The circle has a radius of $ 13\sqrt {10}$.
Using power of a point:
\[ \left(13\sqrt {10} \minus{} x\right)\left(13\sqrt {10} \plus{} x\right) \equal{} 39^2
\]
Solving, we have $ x \equal{} 3$, meaning the diagonal of length $ 78$ is $ 3$ units away from the center of the circle.
Similarly, for the diagonal of length $ 82$:
\[ \left(13\sqrt {10} \minus{} x\right)\left(13\sqrt {10} \plus{} x\right) \equal{} 41^2
\]
Solving, we have $ x \equal{} 13$, meaning the diagonal of length $ 82$ is $ 13$ units away from the center of the circle.
We can split the quadrilateral into four triangles, with legs $ (36,28), (42,28), (36,54), (54,42)$.
We add up the hypotenuses to get:
\[ \sqrt {36^2 \plus{} 28^2} \plus{} \sqrt {42^2 \plus{} 28^2} \plus{} \sqrt {36^2 \plus{} 54^2} \plus{} \sqrt {54^2 \plus{} 42^2} \\
\equal{} 4\sqrt {130} \plus{} 14\sqrt {13} \plus{} 17\sqrt {13} \plus{} 6\sqrt {130} \\
\equal{} 10\sqrt {130} \plus{} 31\sqrt {13} \\
\equal{}a\sqrt{b}\plus{}c\sqrt{d}
\]
Therefore:
\[ a \plus{} b \plus{} c \plus{} d \equal{} \boxed{184}
\]
Is there an easier way to do this?[/hide]

[quote="dnkywin"]That's how I did it, and, as I said, it was a [i]hard[/i] AIME level problem. However, although your idea is right, your solution is incorrect.[/quote]

Are the lengths of the four triangles correct?


$ S_1\equal{}\sum_{k\equal{}1}^n k  cos k$
$ S_2\equal{}\sum_{k\equal{}1}^n k  sin k$
$ S_1, S_2\equal{}?$ using: $ S_1\plus{}iS_2$
pls help :(
thanx :wink:

What is $ f(x) \equal{} \sum_{k\equal{}1}^{n} e^{ikx}$?  Differentiate it.

wouldnt the above result as cos[n(n+1)/2] + i*sin[n(n+1)/2] ??

That would be the case if we were considering an infinite [i]product.[/i]  This is a sum; in particular, a sum of a geometric series.


\[ 2(\log_{3} \frac{1}{5})(\log_{5} 27)\]

i got 6/25, but dont know if thats right

Regrettably, that is incorrect. Show your work. :)

i dont even know what i did....but i obviously screwed up.

can any1 show the correct solution?

Did you remember that $ \log_x{y}\equal{}\frac{1}{\log_y{x}}$?  That makes this a bit easier.

All right. I give a hint. Because I want you to solve it by yourself.

Can you calculate $ (\log_a b)(\log_b a)?$

[hide]log base3 1/5 = log base3 1 - log base3 5

                   = - log base3 5

Change base of log base5 27:

(log base3 27) / (log base 3 5) = 3 / (log base 3 5)

So plugging everything back in we get:

2 (-log base3 5) (3 / log base3 5)

log base3 5 cancels out.

So we end up with:

(2)(-1)(3) = -6[/hide]

[quote="DaReaper"]
\[ 2(\log_{3} \frac {1}{5})(\log_{5} 27)
\]
i got 6/25, but dont know if thats right[/quote]

$ \log_3 \frac{1}{5}\equal{}\log_{27} \frac{1}{125}$

$ \log_{27} \frac{1}{125} \cdot \log_5 27 \equal{} \log_{27} \frac{1}{125} \cdot \frac{1}{\log_{27} 5}$

$ \frac{\log_{27} \frac{1}{125}}{\log_{27} 5}\equal{}\log_5 \frac{1}{125} \equal{} \minus{}3$.

I'm not that good with logs, so tell me if I'm wrong.

[quote="BOGTRO"][quote="DaReaper"]
\[ 2(\log_{3} \frac {1}{5})(\log_{5} 27)
\]
i got 6/25, but dont know if thats right[/quote]

$ \log_3 \frac {1}{5} \equal{} \log_{27} \frac {1}{125}$

$ \log_{27} \frac {1}{125} \cdot \log_5 27 \equal{} \log_{27} \frac {1}{125} \cdot \frac {1}{\log_{27} 5}$

$ \frac {\log_{27} \frac {1}{125}}{\log_{27} 5} \equal{} \log_5 \frac {1}{125} \equal{} \minus{} 3$.

I'm not that good with logs, so tell me if I'm wrong.[/quote]

I got -6 so I don't know.. check my solution.. and didn't i see you in the Mathcounts class?

[quote="BOGTRO"][quote="DaReaper"]
\[ 2(\log_{3} \frac {1}{5})(\log_{5} 27)
\]
i got 6/25, but dont know if thats right[/quote]

$ \log_3 \frac {1}{5} \equal{} \log_{27} \frac {1}{125}$

$ \log_{27} \frac {1}{125} \cdot \log_5 27 \equal{} \log_{27} \frac {1}{125} \cdot \frac {1}{\log_{27} 5}$

$ \frac {\log_{27} \frac {1}{125}}{\log_{27} 5} \equal{} \log_5 \frac {1}{125} \equal{} \minus{} 3$.

I'm not that good with logs, so tell me if I'm wrong.[/quote]

Your are right, so far, now you multiply by 2, yielding the result 6. :lol:

[quote="kunny"][quote="BOGTRO"][quote="DaReaper"]
\[ 2(\log_{3} \frac {1}{5})(\log_{5} 27)
\]
i got 6/25, but dont know if thats right[/quote]

$ \log_3 \frac {1}{5} \equal{} \log_{27} \frac {1}{125}$

$ \log_{27} \frac {1}{125} \cdot \log_5 27 \equal{} \log_{27} \frac {1}{125} \cdot \frac {1}{\log_{27} 5}$

$ \frac {\log_{27} \frac {1}{125}}{\log_{27} 5} \equal{} \log_5 \frac {1}{125} \equal{} \minus{} 3$.

I'm not that good with logs, so tell me if I'm wrong.[/quote]

Your are right, so far, now you multiply by 2, yieldinf the result 6. :lol:[/quote]

Yay

[quote="DaReaper"]why is

log base3 1/5 = log base3 1 - log base3 5

= - log base3 5[/quote]

That is one of the theorems of logs.

log basen x/y = log basen x - log basen y

And then log base3 1 = 0 so 0 - log base3 5 = -log base3 5.

$ \boxed{\log _a b^p \equal{} p\log _a b}$

Proof: Let $ \log_a b \equal{} r\Longrightarrow b \equal{} a^r.$ $ \therefore b^p \equal{} a^{rp}\Longrightarrow \log_a b^p \equal{} rp$

$ \therefore \log_a b^p \equal{} p\log_a b$.

Let $ \log_{10} 2 \equal{} a,\log_{10} 3 \equal{} b.$ Express the following expressions in terms of $ a, b.$

(1) $ \log_{10} (\sin ^ 2 60^\circ)$
(2) $ \log_{100} 4 \plus{} 2\log_{100} 27$
(3) $ \log_2 9 \plus{} \log_3 8$
(4) $ \log_{\sqrt {45}} 144$

2004 Kitasato University entrance exam/Pharmacy

[url=http://en.wikipedia.org/wiki/Kitasato_Shibasaburo]Shibasaburo Kitasato[/url]


real analysis
integration
Support
function
algebra
domain
geometry

Hello, all.

I'm not sure if this is better suited in a different location, but...

I'm suffering from a little confusion here:

Given $f(x) = e^{-x}$   and  $g(x)= x$ find the convolution of $f(x)$ and $g(x)$.

That is,

$f \ast g(x) = \int^{\infty}_{-\infty} f(x-u) g(u) du$

This one seems to be giving me a little trouble...okay, a lot of trouble.

Thanks,
Bailey

What you're probably missing is the support of the functions. Anytime we're talking about convolution, or Fourier transforms, or anything related to that, and I see someone express a function as a formula without any reference to domain or support (such as "$f(x)=e^{-x}$"), I immediately think "uh-oh, we may have a problem."

The function $f(x)=e^{-x}$ on $\mathbb{R}$ is not in $L^1(\mathbb{R})$ (is not absolutely integrable on the whole line.) Similarly the function $g(x)=x$ on $\mathbb{R}$ is not in $L^1(\mathbb{R}).$ There's no good reason for the convolution of these functions to exist.

So I'm guessing that those aren't the functions you're trying to convolve. Suppose your functions are actually this:

$f(x)=\begin{cases}e^{-x}&x\ge0\\0&x<0\end{cases}$

$g(x)=\begin{cases}x&x\ge0\\0&x<0\end{cases}$

Then, the support of $f$ is the same as the support of $g$ and both are $[0,\infty).$ The support of $f*g$ is contained in the algebraic sum of those supports, which is also $[0,\infty).$

Thus, for $x<0,\,f*g(x)=0.$ Now, suppose $x\ge 0.$

$f*g(x)=\int_{-\infty}^{\infty}f(x-y)g(y)\,dy$

$=\int_0^xe^{-(x-y)}y\,dy$ (This may the most important step right here.)

$=\ \cdots$ (Integrate it yourself - you should be able to finish it.)

Never, ever, discuss functions in Fourier transform/Laplace transform/convolution context with out being totally clear about domain, support, and piecewise definition.

Thanks, Kent.

That's very clear now.  I'm working some problems from a text (self learning) and the domains of the functions were not stated.  I guess that is where my confusion grew from.

I appreciate your help.

bailey

[quote="RaduSorici"]The convolution product is defiened on $[0,\infty)$ for piecewise continous functions.[/quote]
Two things:

1. "Piecewise continuous" is a stronger condition than is really necessary. Integrable is good enough, as are some variations on locally integrable with certain growth conditions at infinity.

2. The convolution of two functions on $[0,\infty)$ to produce a function on $[0,\infty)$ can be considered (and perhaps should be considered) to be the convolution of two functions on $\mathbb{R},$ both of which happen to be supported on $[0,\infty).$

For those not familiar with the language:

The support of $f$, written as $\text{supp}(f),$ is the closure of the set $\{x|f(x)\ne 0\}.$ (One would have to modify the definition for generalized functions - distributions - but there is a reasonable and compatible definition.)

Theorem: $\text{supp}(f*g)\subset\text{supp}(f)+\text{supp}(g).$

In general, convolution is an operation between functions defined on a (locally compact) topological group. It involves the notion of translation of functions, which makes sense on a group.


geometry
geometric transformation
reflection
rotation
symmetry
Alcumus
function

How many ways are there to arrange 6 beads of distinct colors in a $ 2 \times 3$ grid if reflections and rotations are considered the same?  (In other words, two arrangements are considered the same if I can rotate and/or reflect one arrangements to get the other.)

If rotations and reflections were considered different, there would be $ 6!\equal{}720$ ways to place the beads in the grid. But we must divide by two because we can reflect the grid and then divide by 2 again because we can rotate the grid 180 degrees. Thus, there are a total of $ \frac{720}{2\cdot{2}}\equal{}\boxed{180}$ ways to color the grid.

I know I'm resurrecting this topic, but it's because that nobody has posted on this topic saying that they agree with the alcumus answer.

I personally agree with RunpengWINS. I feel like giving a proof that you divide 6! by 4 and not what alcumus did.

[hide="Proof (I just want to find the error in this proof)"]Let $ X$ be any distribution of 1, 2, 3, 4, 5, 6 in a 2*3 grid. Let $ f(X)$ be the graph resulting from reflecting the graph vertically, let $ g(X)$ be the graph resulting from reflecting the graph horizontally, and let $ h(X)$ be the graph resulting from rotating the graph 180 degrees. It is easy to prove that $ f(g(X))\equal{}h(X)$, so if we are allowed to compose functions, then $ h(X)$ is redundant. Therefore we only need to consider reflections.

Also note that $ f(f(X))\equal{}g(g(X))\equal{}X$, and $ f(k(f(X)))\equal{}g(k(g(X)))\equal{}k(X)$, where $ k$ is some composition of $ f$ and $ g$. Therefore any composition of $ f$ and $ g$ where $ f$ or $ g$ is composed more than once is equivalent to a composition of $ f$ and $ g$ where $ f$ or $ g$ is composed at most once. This means that there are at most four possibly distinct compositions of $ f$ and $ g$: $ X$ (we don't apply any functions), $ f(X)$, $ g(X)$, and $ f(g(X))$. Note that these are all distinct. Therefore we can change a graph to any of four graphs (including itself), so if we count all distributions of 1 through 6, then we are counting $ X$, $ f(X)$, $ g(X)$, and $ f(g(X))$ exactly once for all $ X$, and we must divide by 4.[/hide]

[quote=NicksDad]The wording on this problem was misleading to me.  I read the problem to say that rotations of the arrangements were considered to be the same.  However, the problem solution considered only rotations of the 2 X 3 grid.  Thus, if the colored beads are allowed to rotate around the grid, then there are six different arrangements that are the same under rotation.  Instead of dividing 6! by 4, I divided 6! by 12, dividing by 6 to account for rotational symmetry and dividing by 2 to account for symmetry under reflection.[/quote]

Why is no-one discussing this?  :dry:  It's been 8 years now!

Anyways, I agree with you. I found that there were 4 different reflections for each board (1 reflection per side), and there were 4 rotations. $\frac{6!}{4\cdot 4}=45$. Why is this not right?  :furious:

It's 7 years, not 8.

Maybe someone could hit the "Report An Error" button...

But we can't be sure that we're right! Unless an instructor comes along now...:(

After looking at RunpengFAILS's solution, I thought that it made sense that the answer was 180? In the Alcumus solution, it only mentioned rotations, but the way I saw it was that 2 of the 4 were actually reflection.

[quote=bluehurricane]
I found that there were 4 different reflections for each board (1 reflection per side), and there were 4 rotations. $\frac{6!}{4\cdot 4}=45$. Why is this not right?[/quote]

Same!

Hi! Could someone please help me? I don't understand why I got this wrong! I thought there were 6 possible rotations and 2 reflections so I did [hide=this]$\frac{6!}{8}=90$[/hide]...

Clarifying my previous post, when I said two reflections, I meant reflecting it horizontally and reflecting it vertically. I assumed there were 6 rotations because there were 6 beads...

If it's a rectangle, shouldn't there be $4$ rotations?

No, as two of them also count as reflections. That means you would divide by 2x2, not 2x4. 6!/4 is 180, which is the answer.

I failed because I divided by $2*4$. Not $2*2$. Nooooooo........

[quote=Hypatia1728]shouldn't there be 5! ways to put the beads on the grid?[/quote]

No, the problem didn't say the shape was [i]circular[/i], and there won't be as much rotations and reflections as a circle has.

There are 6 possible rotations and 3 reflections. So, the answer should be $\tfrac{6!}{6*3} = \boxed{40}$

Why is this wrong?  :wallbash_red:

First off, there aren't 6 rotations. This is because a rotation HAS to be of the same formatting. Like you can have a rectangle standing on a corner. Also, you shouldn't be multiplying them; instead they should've been added.
So there is the original format, a 180 degree rotation, a vertical flip, and a horizontal flip. Only 1/4 of the overcomunted satisfies our case. So we multiply 6! by 1/4 to get 180.

I got this wrong around 4 months ago lol.


group theory
abstract algebra
superior algebra
superior algebra solved

Let $H$ be a subgroup of abelian group $G$.We know that we can write:
$G=H \bigcup H a_{1}\bigcup Ha_{2}\bigcup \cdots \bigcup Ha_{k}\bigcup \cdots$,and such that:$Ha_{i}\bigcap Ha_{j}=\varnothing,i \neq j$
My question is:For every $H$ and $G$,can we adjust $a_{1},a_{2},\cdots,a_{k},\cdots$ and make that $\{e,a_{1},a_{2},\cdots,a_{k},\cdots \}$ be a group.

And what about if we throw off the condition that $G$ is abelian.
For example:$G=\{e,f,f^{2},g,fg,fg^{2}\}$,$H=\{e,f,f^{2}\}$
then $G=H \bigcup Hg$
and $\{e,g\}$ is a group.

What you are asking is (or maybe is a stronger question than): Is there a subgroup of G which is isomorphic to the factor group G/H?

As the example G=Z, H=2Z shows, this is not always the case.

If G is nonabelian, the cosets of H need not form a group, in this case you cannot choose representatives which form a group.

LiMa

[quote="LiMa"]What you are asking is (or maybe is a stronger question than): Is there a subgroup of G which is isomorphic to the factor group G/H?

As the example G=Z, H=2Z shows, this is not always the case.

LiMa[/quote]
Thank you very much.
And it is wrong also $G$ is finite. :)

Yes, the smallest counterexample is G = Z/4Z, H = 2Z/4Z.
It also shows that your question is different from my question, because G/H is isomorphic to H. :)


[quote="tommy1729"]does anybody know it ? SUNDMANS SERIES FOR RESTRICTED 3 BODY PROBLEM??[/quote]
Can you tell something about it? Or just describe it...

well there you have an example of a source.
kind a strange they dont write the formula there since the site is related to physics and math  :roll:

now does anyone on this planet know this "well hidden " series expansion ???

oh cmon !

ok for the ones who know i offer beer for guys and kisses for girls  :P

or even better girls for guys and beer for girls  :D

no seriously  :|  doesnt anybody know ? :huh:  :(


calculus
integration
conics
parabola
geometry
geometric transformation
rotation

Let $a,\ b$ be positive constant numbers.Find the volume of the revolution of the region surrounded by the parabola $y=ax^{2}$ and the line $y=bx$ about  the line $y=bx$ as the axis on $xy$ plane.

Those are planar curves? or is there some rotation going on here to give a volume? or is it area you want?

Sorry for that, I have just edited.

kunny

[quote="kunny"]Let $a,\ b$ be positive constant numbers.Find the volume formed by the revolution of the region surrounded by the parabola $y=ax^{2}$ and the line $y=bx$ about  the line $y=bx$ as the axis on $xy$ plane.[/quote] \[V=\frac{\pi b^{5}}{30a^{3}\sqrt{b^{2}+1}}\]

[quote="kunny"]Regrettably your answer is incorrect. :([/quote]

fixed it: I dropped a 1/5 mid-way through my calculation.

That's right! :)  How did you solve it, benorin?

[quote="kunny"]Let $a,\ b$ be positive constant numbers.Find the volume formed by the revolution of the region surrounded by the parabola $y=ax^{2}$ and the line $y=bx$ about  the line $y=bx$ as the axis on $xy$ plane.[/quote]

I will use Pappus's senond theorem: the volume of a solid of revolutions is the area of the liminar region (the 2-d region that gets rotated) times the distance its centeroid travels during rotation (e.g. $2\pi$).

The area of the region is \[A=\int_{x=0}^{\frac{b}{a}}\int_{y=ax^{2}}^{bx}\ dydx = \frac{b^{3}}{6a^{2}}\] the coordinates of the centeroid are: \[\bar{x}=\frac{1}{A}\int_{x=0}^{\frac{b}{a}}\int_{y=ax^{2}}^{bx}x\ dydx = \frac{b}{2a}\] and \[\bar{y}=\frac{1}{A}\int_{x=0}^{\frac{b}{a}}\int_{y=ax^{2}}^{bx}y\ dydx = \frac{2b^{2}}{5a}\] hence the centeroid is $\left( \bar{x},\bar{y}\right) =\left( \frac{b}{2a},\frac{2b^{2}}{5a}\right)$

to determine the distance traveled by the centeroid, we need to first know the distance (i.e. the shortest distance) from the centeroid to the axis of rotation (namely $y=bx$), and the square of this distance is \[f(x): =d^{2}(x)=\left( x-\frac{b}{2a}\right)^{2}+\left( bx-\frac{2b^{2}}{5a}\right)^{2}\] the minimum distance occurs when \[f^{\prime}(x)=2\left( x-\frac{b}{2a}\right)+2b\left( bx-\frac{2b^{2}}{5a}\right) =0\Rightarrow x_{min}=\frac{b(5+4b^{2})}{10a(1+b^{2})}\] and $f(x_{min}) = \frac{b^{4}}{100a^{2}(1+b^{2})}$ and hence the minimum distance is \[\ell : = \sqrt{\frac{b^{4}}{100a^{2}(1+b^{2})}}= \frac{b^{2}}{10a\sqrt{1+b^{2}}}\] so the volume is then \[V=2\pi\ell A = 2\pi\cdot\frac{b^{2}}{10a\sqrt{1+b^{2}}}\cdot\frac{b^{3}}{6a^{2}}=\frac{\pi b^{5}}{30a^{3}\sqrt{b^{2}+1}}\] -Ben

[quote="kunny"]Thank you for your time and frequent replies, benorin. :)

You use Pappus-Guirdins theorem, don't you. It is very interesting to me.

This solution seems difficult for Japanese high school students.

I will post three solutions by which they might solve.

Please wait for a while.

kunny[/quote]

I tried rotating the axes so that $y=bx$ was on the $x^\prime$-axis but the resulting transformation of the parabola made the final integral a nightmare, even though the Jacobian was nice...

Solution 1.

Take a point $P(x,\ ax^{2})\ \left(0\leq x\leq \frac{b}{a}\right)$ on the parabola $y=ax^{2}$, $Q(x,\ bx)$ on the line $y=bx.$ Let the intersection point be $R\left(\frac{b}{a},\ \frac{b^{2}}{a}\right)\ (a>0,\ b>0).$

Now cut the revolution by the plane passing through the point $H,$ which is the foot of the perpendicular from the point $P,$ and perpendicular to the line $y=bx,$ we have that the sectional face is the circle with radius $\overline{PH}.$

Use the formula of the distance between a point and a line, we have $\overline{PH}=\frac{|bx-ax^{2}|}{\sqrt{b^{2}+1}},$ and by Pythagorian theorem we have

$\overline{OH}^{2}=OP^{2}-PH^{2}=\frac{(abx^{2}+x)^{2}}{b^{2}+1}\ \because a>0,\ b>0,\ x\geq 0.$

$\therefore \overline{OH}=\frac{abx^{2}+x}{\sqrt{b^{2}+1}}.$

Thus the desired volume $V$ is

${V=\int_{0}^{\overline{OR}}\pi \overline{PH}^{2}}\ d(\overline{OH})=\pi \int_{0}^{\frac{b}{a}}\overline{PH}^{2}\ \frac{d(\overline{OH})}{dx}\ dx$

$=\frac{\pi}{(b^{2}+1)^{\frac{3}{2}}}\int_{0}^{\frac{b}{a}}(ax^{2}-bx)^{2}(2abx+1)\ dx$

$=\frac{\pi b}{(b^{2}+1)^{\frac{3}{2}}}\int_{0}^{\frac{b}{a}}(ax^{2}-bx)^{2}(ax^{2}-bx)' dx+\frac{\pi}{\sqrt{b^{2}+1}}\int_{0}^{\frac{b}{a}}(ax^{2}-bx)^{2}dx$

Here the first integral is vanish and for the second one, rewrite as $\int_{0}^{\frac{b}{a}}x^{2}\left(x-\frac{b}{a}\right)^{2}dx=\frac{1}{30}\left(\frac{b}{a}\right)^{5}.$

Thus $V=\frac{\pi b^{5}}{30a^{3}\sqrt{b^{2}+1}},$ which is benorin's result.

Solition 2.

Let $\tan \theta =b,$ ....

Solution 3.

Collecting the small cones. sorry for my poor English.


inequalities
LaTeX
inequalities unsolved

Could you use LATEX to explain your inequality more clearly?

It is difficult to understand.

[quote="redi"]can you solve this please
prove that if 1≤x²-xy+y²≤2 then x to the power of 2m + y to the power of 2m ≤ 2 over 3 to the power of m
for every natyral m≥3[/quote]
Let $m$ be a positive integer greater than or equal to $3$. If $1\leq x^2-xy+y^2\leq 2$, prove that
$x^{2m}+y^{2m}\leq 3^{m}$


Let a,b,c are positive number.Prove that
$ \sum\ \frac{a^{4}} {a^{3}\plus{}b^{3}\plus{}abc}\$$ $ \geq\frac{a^{3}\plus{}b^{3}\plus{}c^{3}}{a^{2}\plus{}b^{2}\plus{}c^{2}}\$$
It is not a easy problem,i have a proof for it butit is very hard.So i hope who is here can prove it with a nice solution
If $ a,b,c$  are lenghts of a triangle  ,i had a nice proof with Moldova TST 2006,but it is not enough to solve it
SO, WHO CAN HELP ME,PLEASE

[quote="mehdi cherif"]
$ \iff$    $ \sum a^3bc \leq \sum a^3b^2$

wich is true by reordonnement inequality[/quote]
I think, the last inequality is wrong. :wink:

[quote="mehdi cherif"]let $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$

we have to prove that :

$ \sum\frac {a^4}{a^3 \plus{} b^3 \plus{} abc}\geq \frac {a^3 \plus{} b^3 \plus{} c^3}{a^2 \plus{} b^2 \plus{} c^2}$

by caushy shwartz inequality:

$ \sum\frac {a^4}{a^3 \plus{} b^3 \plus{} abc}\geq \frac {(a^3 \plus{} b^3 \plus{} c^3)^2}{\sum a^2(a^3 \plus{} b^3 \plus{} abc)}$

now we have to prove that :

$ \sum a^2(a^3 \plus{} b^3 \plus{} abc) \leq (a^3 \plus{} b^3 \plus{} c^3)(a^2 \plus{} b^2 \plus{} c^2)$

$ \iff$    $ \sum a^3bc \leq \sum a^3b^2$

wich is true by reordonnement inequality[/quote]
 If case 1 : $ \sum a^3bc \leq \sum a^3b^2$  then  proof cherif's
in case 2 : $ \sum a^3bc \geq \sum a^3b^2$
 we  need prove  that:
 $ \sum\frac {a^4}{a^3 \plus{} b^3 \plus{} abc}\geq \frac{3(a^4\plus{}b^4\plus{}c^4)^2}{3(a^7\plus{}b^7\plus{}c^7)\plus{}3abc(a^4\plus{}b^4\plus{}c^4)\plus{}abc(a^2\plus{}b^2\plus{}c^2)^2} \geq RHS$

[quote="mehdi cherif"]

$ \iff$    $ \sum a^3bc \leq \sum a^3b^2$

which is true by reordonnement inequality[/quote]
This ineq is right if $ a,b,c,$ are lenghts of a triangle ,and here is problem from Moldova TST
To Hong quy:your proof is the solution that i have said, but it is hard .
And What is $ RHS$


geometry
area of a triangle
Heron\u0027s formula

Demonstrate area of triangle whose sides are roots of $x^3+ax^2+bx+c=0 $, is $\sqrt{-\frac{a^4}{16}+\frac{a^2b}{4}+\frac{ac}{2}}$

[hide]We know that if the sides are $x$, $y$, $z$, then
$-a=x+y+z$
Therefore, using Heron's Formula, the area of the triangle is
$\sqrt{(\frac{-a}{2})(\frac{-a}{2}-x)(\frac{-a}{2}-y)(\frac{-a}{2}-z)}$
Simplify to
$\sqrt{\frac{a(a+2x)(a+2y)(a+2z)}{16}}$
Multiply it out
$\sqrt{\frac{a^4+2za^3+2xa^3+2ya^3+4xya^2+4xza^2+4yza^2+8xyza}{16}}$
$=\sqrt{\frac{a^4-2(a^4)+4ba^2-8ac}{16}}$
$=\sqrt{-\frac{a^4}{16}+\frac{a^2b}{4}-\frac{ac}{2}}$[/hide]

[quote="K81o7"][hide]We know that if the sides are $x$, $y$, $z$, then
$-a=x+y+z$
Therefore, using Heron's Formula, the area of the triangle is
$\sqrt{(\frac{-a}{2})(\frac{-a}{2}-x)(\frac{-a}{2}-y)(\frac{-a}{2}-z)}$
Simplify to
$\sqrt{\frac{a(a+2x)(a+2y)(a+2z)}{16}}$
Multiply it out
$\sqrt{\frac{a^4+2za^3+2xa^3+2ya^3+4xya^2+4xza^2+4yza^2+8xyza}{16}}$
$=\sqrt{\frac{a^4-2(a^4)+4ba^2+8c}{16}}$
$=\sqrt{-\frac{a^4}{16}+\frac{a^2b}{4}+\frac{ac}{2}}$[/hide][/quote]

Isnt it xyz = - c ?  :?  :?  :?


vector
linear algebra
linear algebra unsolved

Let $Q$ be countable set of linear maps from $\mathbb{R}^n$ to $\mathbb{R}^m$. For every $x \in \mathbb{R}^n$ denote by $Qx$ the set of vectors $Ax$, where $A \in Q$. Let $B$ be linear map from $\mathbb{R}^n$ to $\mathbb{R}^m$ such that $Bx \in Qx$ for all $x \in \mathbb{R}^n$. Is it true that $B \in Q$ ?

Sorry if I am writing non-sense, but it seems too easy. Suppose the elements of $Q$ are $ f_1, f_2,...$ and denote $V_k$ the set of vectors $x$ for which $ B(x)=f_k(x)$. These are subspaces in $R^n$, thus they are closed sets and the union is $R^n$. By Baire's theorem one of these spaces will have non-void interiour, but then this space is the whole space $R^n$ and thus $B\in Q$. What is wrong here? :?

It is ok,Harazi, nobody told it is difficult.Thanks.


Two girls and three boys are to be seated in a row of five desks. What is the probability that the students at the ends of the row are both boys? Express your answer as a common fraction.

there are 120 ways (5!) to seat the kids

in order to get the boys sitting on the outside, it can be...
1, x, x, x, 2
1, x, x, x, 3
2, x, x, x, 1
2, x, x, x, 3
3, x, x, x, 1
3, x, x, x, 2
where the x's are the other pplz and the numbers represent boy 1, boy 2, and boy 3

however, for each of those combinations, there are 3! ways to seat the pplz in between, or 6 different ways
so we would have to multiply 6 by 6 and we get 36

we put 36 over our total, 120 and we get our answer $ \frac{36} {120}$

we reduce and get $ \frac{3}{10}$

To get how many successful ways, you don't even have to list it out. There are 3 boys for the first end, times 2 boys at the second end is 6. And as vallon pointed out, there are $ 3!$ ways to arrange the remaining people, thus the probability is $ \frac{3}{10}$.


function
calculus
derivative
integration
probability
probability and stats

Let $ Z$ be a standard normal random variable $ Z$, and let $ g$ be a differentiable function with derivative $ g'$.

(a) Show that: $ E[g'(Z)] \equal{} E[Zg(Z)]$
(b) Show that: $ E[Z^{n\plus{}1}] \equal{} nE[Z^{n\minus{}1}]$
(c) Determine $ E[Z^{4}]$

(a) integration by parts gives $ \int e^{\frac{\minus{}z^2}{2}}g'(z)dz\equal{}e^{\frac{\minus{}z^2}{2}}g(z)\plus{}\int e^{\frac{\minus{}z^2}{2}}zg(z)dz$...

(b) by (a), $ E[Z^{n\plus{}1}]\equal{}E[Z\cdot Z^{n}]\equal{}E[nZ^{n\minus{}1}]\equal{}nE[Z^{n\minus{}1}]$

(c) use (b)...

Thanks a lot, I understand the problem better now.

But for (c), I get:
[hide]
$ E[Z^4] \equal{} E[Z*Z^3] \equal{} E[3*Z^2] \equal{} 3 * E[Z^2] \equal{} 3 * (Var(Z) \plus{} E[Z]^2) \equal{} 3 * ({\sigma}^2 \plus{} (0)^2) \equal{} 3(1\plus{}0) \equal{} 3$[/hide]

But I'm questioning if that makes intuitive sense.

you can also get it purely via (b): $ E[Z^4]\equal{}3E[Z^2]\equal{}3\cdot 1\cdot E[1]\equal{}3$. if u want an intuitive sense, notice that although $ E[Z]\equal{}0$, $ Z^4$ is always non-negative (and 0 with probability 0), so $ E[Z^4]$ should be positive.


geometry
3D geometry
counting
distinguishability
rotation

I got a different answer than on the solution sheet, so I'm wondering what you think the answer is.

[b]Two red, two white, and two blue faces, all unit squares, are available for building a cube. How many distinguishable cubes can be built?[/b]

Edit:
Heh, I just realized this should probably go in Intermediate problems. :D That's what I get for not reading the rules. You can move it if you want.

Here's what I get:

[hide="Solution."]
=====
Pick any face to be blue and spin the cube so it is facing you. There are two cases: the other blue square is either across the first or adjacent to it.

(i) The other blue square is across the first, in which case you can color any square red and flip it to the top. Now the other red is either adjacent to the first red or across from it. (Both adjacent cases yield the same cube up to rotation). The position of the whites are determined. Thus there are two distinct cubes here.

(ii) The other blue square is adjacent to the first, in which case you can flip it to the top. Now the color in the back is either of two colors (call it A). Then the other A is either on the bottom or on either of the two sides (the two sides yield different cubes up to rotation). The positions of the unused color up to this point are now determined. Thus there are six distinct cubes here.

So in total there are eight.
=====
[/hide]

[hide]In (ii), I think you double-counted some. Remember that you can flip the cube around so the bottom essentially becomes the back, while the two blues are still in the same positions.[/hide]

Ah, good point. So then you don't need to treat the color cases in (ii), just say it's red, so there are three colorings for that case and five in total.

Five is what they had for the solution, but I disagree.

[hide="Here is their solution."]First, suppose that there is exactly one pair of opposite faces of the same color. In this case, there is just one cube with the blue faces opposite, one with the white faces opposite, and one with the red faces opposite. There are no cubes with exactly two pairs of opposite faces of the same color. There is just one cube where all three pairs of opposite faces are the same color. [u]Finally, there is one cube where no pair of opposite faces are the same color.[/u] We get a total of five indistinguishable cubes.[/hide]

I don't agree with the underlined statement. I believe there are actually TWO cubes where no pair of opposite faces are the same color. They are mirror images, but you cannot simply rotate one to get the other. Now I think I'm right in saying that mirror images are distinguishable. (Just look at "p" and "q" :P).

I'm fairly confident that you are correct.  (One way to see this: fix the faces that are colored blue.  There are four other ways to color the remaining faces, and two symmetries, so that leaves us with two distinct colorings.)

In that case, I will personally consider my winning streak in that contest unbroken, since it all rested on that one problem. :P

How are we supposed to visualize/differentiate the rotations for the cube with no opposite same color faces?

Just fyi, this exact problem, just changing a few words, was #7 on Chapter Target last year in MathCounts.


number theory
least common multiple
floor function
induction
inequalities

the sequence $(a_{n})$ of naturals satisfy:
1) $a_{k}\mid a_{k+1}$ for all k;
2) for every $m$ exists $a_{n}$ divisible by $m$
Prove that $\sum\frac{1}{a_{n}}$ converges to an irrational number.

What if it doesn't converge? If $S_{n}=\sum_{k=1}^{n}\frac{1}{a_{n}}$, then I can define the sequence $a_{k}$ by
\[a_{1}=1,\ a_{k+1}=\text{lcm }(a_{k},\lfloor S_{k}\rfloor )\]
Such a sequence satisfies the desired properties and diverges.

[quote="knoppix"]but is the sum irrational if is finite ?

I think if the sequence would be strictly increasing the sum would always be finite, am I right ?[/quote]

1) No.
Consider the sequence formed by the succession of groups of $n!$ terms each equals to $\frac{1}{2^{n}n!}$, where $n>0$.
Clearly, each term of the sequence is an integer which divides its successor and each positive integer divides at least one term of the sequence. In another hand, the sum of each group is $\frac{1}{2^{n}}$, so that the serie converges to $1$, which is known to be rational...

2) Yes, because if $a_{n}>0$ stritly divides $a_{n+1}$, then $a_{n+1}\geq 2a_{n}$, and by an easy induction $a_{n}\geq 2^{n-1}a_{1}>0$, which forces the serie to converge.

Pierre.

what about irrationality of the sum in this case ?

Pierre;

What if one adds the condition $a_{k+1}\not = a_{k}$, so that it is strictly decreasing?

The sum will the obviously converge (each sum is at most half the previous), but what about rationality?

If we add the restriction that the $a_{n}$ are distinct, then sum is always irrational. (Note: since the sum for the particular case $a_{n}=n!$ is $e-1$, we get a simple proof that $e$ is irrational.)
[hide]
Suppose $S$ rational, where $S$ is sum. Let $S_{n}$ denote sum of first $n$ terms. If $S$ rational, then there exists positive integer $q$ such that $q\cdot S$ is an integer. There also exists integer $N$ such that $a_{N}$ is divisible by $q$. We find that
\[\epsilon=a_{N}\left( S-S_{N}\right)\]
is an integer, since $a_{N}\cdot S$ is an integer, along with $\frac{a_{N}}{a_{k}}$ for $k\leq N$. However, we also have $a_{n+1}\geq 2a_{n}$ for all $n$, so that $a_{N+k}\geq 2^{k}a_{N}$ for all $k$. In addition, equality can't occur for all $k$, since some $a_{n}$ will be divisible by a prime greater than both $q$ and 2.

Now we have
\[0<\epsilon=\sum_{k=1}^\infty \frac{a_{N}}{a_{N+k}}<\sum_{k=1}^\infty \frac{1}{2^{k}}=1\]
(where the right inequality is strict because it's a sum of inequalities, at least one of which is strict). But this means $\epsilon$ is an integer strictly between 0 and 1, a contradiction! Therefore, $S$ irrational.
[/hide]


For every natural number $ x$, let $ Q(x)$ be the sum and $ P(x)$ the product of the (decimal) digits of $ x$. Show that for each $ n \in \mathbb{N}$ there exist infinitely many values of $ x$ such that:

$ Q(Q(x))\plus{}P(Q(x))\plus{}Q(P(x))\plus{}P(P(x))\equal{}n$.

For fixed $ n$ take $ x\equal{}111...1000...0$ - there $ q\equal{}111...10$ digits '1', and any number of '0', and $ q$ has $ n$ digits '1'.
$ P(x)\equal{}0$, so $ Q(P(x))\equal{}P(P(x))\equal{}0$.
$ Q(x)\equal{}q\equal{}111...10$ - $ n$ digits '1'.
$ P(Q(x))\equal{}0$.
$ Q(Q(x))\equal{}n$.
So, $ Q(Q(x))\plus{}P(Q(x))\plus{}Q(P(x))\plus{}P(P(x)) \equal{} n\plus{}0\plus{}0\plus{}0 \equal{} n$.
We can take any number of zeros in $ x$, so, there are infinitely many $ x$.


Two circles of radius 1 overlap so that the center of each lies on the circumference of the other. What is the area of their union?

[hide="hint"]I would solve this using lunes[/hide]

[hide]

The intersection of the 2 circles is basically 2 equilateral triangles, plus 4 equal segments.

The sum of the areas of the 2 equilaterals is $ \frac {\sqrt {3}}{2}$.

The segments each have area $ \frac {\pi}{6} \minus{} \frac {\sqrt {3}}{4}$, so the sum of all 4 is $ \frac {2\pi}{3} \minus{} \sqrt {3}$.

The area of the intersection is $ \frac {2\pi}{3} \minus{} \frac {\sqrt {3}}{2}$

Since the Union is the sum of the areas of the circles minus the area of the intersection, it's equal to:

$ 2\pi \minus{} \frac {2\pi}{3} \plus{} \frac {\sqrt {3}}{2}$

$ \equal{} \frac {4\pi}{3} \plus{} \frac {\sqrt {3}}{2}$[/hide]

What the heck is lunes?

Easier solution:

The intersection of the two circles is 2 equilateral triangles of side length 1 and 4 circle segments. Each circle segment has area $ \frac {\pi}{6} \minus{} \frac {\sqrt {3}}{4}$(area of 60-degree arc minus area of equilateral triangle). We multiply that by 4 and then add to $ \frac {\sqrt {3}}{2}$(area of two equilateral triangles) and get that the total area of the intersection is $ \frac {2\pi}{3} \minus{} \frac {\sqrt {3}}{2}$. Then we subtract that value from $ 2\pi$(area of two circles) to get that the area of the union is$ \frac {4\pi}{3} \plus{} \frac {\sqrt {3}}{2}$

edit:Carbon beat me to it.

[quote="tinytim"]What the heck is lunes?

Easier solution:

The intersection is 2 equilateral triangles of side length 1 and 4 circle segment. Each circle segment has area $ \frac {\pi}{6} \minus{} \frac {\sqrt {3}}{4}$(area of 60-degree arc minus area of equilateral triangle. We multiply that by 4 and then add to $ \frac {\sqrt {3}}{2}$(area of two equilateral triangles) and get that the total area is $ \frac {2\pi}{3} \minus{} \frac {\sqrt {3}}{2}$. Then we subtract that value from $ 2\pi$(area of two circles) to get that the area of the union is$ \frac {4\pi}{3} \plus{} \frac {\sqrt {3}}{2}$

edit:Carbon beat me to it.[/quote]

What the heck ARE lunes. And lunes are a sector of a circle minus the triangle that contains the 2 points on the circumference and center. Does that make sense? I can't draw a diagram though...

Those aren't lunes. They're circle segments as mentioned in my solution and in Carbon57's solution.

No I don't think you understand. Take the area of union and draw a vertical line down the center. Then connect the circles' centers to the points of intersection of the circles.


In the "Let's Make a Deal" game, what is the probability that you will win the good prize if you switch doors after being shown one of the junky prizes?

You could be more specific, but the answer is $2/3$, isn't it? If the game I'm thinking of is the game you're thinking of.

Wut's "Let's make a deal"? :huh:

My family doesn't have cable, if you diidn't know.

I think "Let's Make a Deal" was an actual T.V. show a long time ago, but I'm not sure.

Anyway, the probability that your original guess is correct is $\frac13$.  You only have one other option, to choose the other, which must be correct with probability $\frac23$.

This is also called the Monty Hall problem.  It appeared in an "Ask Marilyn" column and has been a hot debate topic ever since.

To understand why the probability is 2/3 (and 1/3 for not changing), just look at your options:

Stay:  There are 3 doors, and 1 one of them contains the good prize.  Obviously, the probability is 1/3.  However, some people said that it was 1/2 because the host opened a door which you did not open, meaning that you have picked a door which either contains the good prize or the junky prize.  Why isn't the probability 1/2 then?  It's because he might as well not have opened the door, it doesn't affect the probability magically.  It stays the same since you chose the door independent of the host.

Switch:  If you switch, there are two options:  you switch to the good prize or you switch to the junky prize.  However, there are 3 different ways to pick the doors.  Say that the good prize is behind door 1, and the junky prizes are behind doors 2 and 3.  Then the only situation in which you would lose if you switched would be if you picked door 1.  In both other cases, you would switch to door 1.  So the probability is 2/3.

[quote="buzzer11"]Yes, it is $\frac{2}{3}$. If you read the book, "The curious incident of a dog in the night-time", this gives a good explaination why (and also some history) :).

Oh, that book is a novel, by the way, not like a math book. It's about an autistic kid whos REALLY smart, and so, there's some math involved as well.[/quote]

Ah yes, I did learn about this problem in that book. It was required reading back in 6th grade, and in fact, it's right behind me on my bookshelf.

"The curious incident of a dog in the night-time" is rather interesting.

I calculated $2^{n}$ until 42 in my head too..
It was something like OCD for him, but for me, it was just because i forgot an answer for computer science class.


Find all natural numbers $\ x,y,z$ such that the number $\frac{(x+y+z)^2}{xyz}$ is also natural.

Are you sure the solutions follow a pattern ? Maple seems to give me a lot of triplets satisfying the above condition but with no apparent connection, such as (405,25,20), (363,24,9), (36,9,9), (841,25,4), etc ...

We will give a solution to the problem in terms of a generating set and an operation that generate all the solutions.

First of all, it easy to found that if a solution verifies $x\leq y\leq z,\,\,z\leq x+y$, then $x\leq 9$, and from here we obtain that there are thirteen such solutions:
\[ (1,1,1), (1,1,2), (1,2,3), (1,4,5), (2,2,4), (2,4,6), (3,3,3), (3,6,9), (4,4,8), (5,20,25), (6,12,18), (8,8,16), (9,9,9)  \]

Now, from this thirteen solutions we can generate every other solution with the following procedure: given a solution $(x,y,z)$, we can obtain three solutions $(x,y,\frac{(x+y)^2}z)$, $(x,\frac{(z+x)^2}y,z)$, and $(\frac{(y+z)^2}x,y,z)$

As an example of this procedure, we obtain the four solutions given in the above post by julien_santini.
From $(5,20,25)$ we obtain $(20,25,405)$ directly.
From $(3,6,9)$ we obtain $(3,9,24)$, and from this we obtain $(9,24,363)$.
From $(9,9,9)$ we obtain $(9,9,36)$.
From $(1,1,1)$ we obtain $(1,1,4)$, then $(1,4,25)$ and finally $(4,25,841)$.


prove that:
[size=150]1:[/size]The sum of two even numbers is even

[size=150]2:[/size]The sum of an even number and an odd number is odd

[size=150]3:[/size]The sum of two odd numbers is even

[size=150]4:[/size]The product of two even numbers is even

1. Any even number can be expressed as $ 2a$. So when adding two even numbers,$ 2a\plus{}2k\equal{}2(a\plus{}k)$, 2 can be factored out of the sum, indicating that it is even.

2. Any odd number can be expressed as $ 2a\plus{}1$...so $ 2a\plus{}(1\plus{}2a) \equal{} 4a\plus{}1$, which is odd.

3.$ (2a\plus{}1)\plus{}(2k\plus{}1) \equal{} 2a\plus{}2k\plus{}2 \equal{} 2(a\plus{}k\plus{}1)$ Since $ 2(a\plus{}k\plus{}1)$ is divisible by two, it's even.

4. $ (2n)(2k) \equal{} 4nk$ So, in fact, any product of two even numbers is also divisible by 4.

[quote="binomial_4eva"]i thought this has something to do with mod[/quote]

You can prove them with mods, it's just easier the other way.

The two methods are basically equivalent; saying that a quantity can be written in the form $ 2(a\plus{}k)$ is more or less the same is saying that it is congruent to $ 0\bmod 2$, since all that means is that there exists some $ m$ such that the quantity can be written as $ 2\cdot m\plus{}0$.

[quote="Icy"]1. Any even number can be expressed as $ 2a$. So when adding two even numbers,$ 2a\plus{}2k \equal{} 2(a\plus{}k)$, 2 can be factored out of the sum, indicating that it is even.

2. Any odd number can be expressed as $ 2a\plus{}1$...so $ 2a\plus{}(1\plus{}2a) \equal{} 4a\plus{}1$, which is odd.

3.$ (2a\plus{}1)\plus{}(2k\plus{}1) \equal{} 2a\plus{}2k\plus{}2 \equal{} 2(a\plus{}k\plus{}1)$ Since $ 2(a\plus{}k\plus{}1)$ is divisible by two, it's even.

4. $ (2n)(2k) \equal{} 4nk$ So, in fact, any product of two even numbers is also divisible by 4.[/quote]

#2 isn't a general proof, as you assumed the numbers were consecutive.

Also, it is not trivial to me (not after PROMYS, at least) that any even number can be expressed as $ 2a$ and any odd number can be expressed as $ 2a\plus{}1$.

Yeah.

Can you elaborate on what you mean by even numbers not having form $ 2x$ and odds $ 2x\plus{}1$?

[quote="Phelpedo"]it is not trivial to me (not after PROMYS, at least) that any even number can be expressed as $ 2a$ and any odd number can be expressed as $ 2a\plus{}1$.[/quote]
Spoken like a true PROMYS alum.  :)

Now, I have a bone to pick with the first statement (since that is one way of [i]defining[/i] the even numbers), but what Phelpedo is trying to say is that it does not immediately follow from axioms that every integer which cannot be expressible in the form $ 2k$ for some integer $ k$ can be expressed in the form $ 2k\plus{}1$.  This is a consequence of lemmas which have to be proven ;)

Of course, for most official solutions I think it is safe to assume that it can be taken for granted :)


geometry
analytic geometry
inradius
function

1. A circle is inscribed in a triangle with sides of lengths 3, 4, and 5. A second circle, interior to the triangle, is tangent to the first circle and to both sides of the larger acute angle of the triangle. Find the radius of the second circle.

2. Given $\triangle{OAB}$ with $O(0,0)$, $A(8,0)$, $B(0,10)$. $C$ lies on $\overline{OB}$ and $D$ lies on $\overline{BA}$ such that $\overline{CD}\parallel\overline{OA}$. If the area of $\triangle{BCD}$ is 20, find the $x$-coordinate of point $D$.

[hide=" Hint, #1"]
Find the inradius of the triangle. Then, see what triangles you can draw with the larger and smaller radii. [/hide]

[hide=" Hint, #2"]
Look at similar triangles. How can we express the area of the smaller triangle as a function of that x-coordinate? [/hide]

Umm...neither of your answers is correct.

Ha.  Ok.  I was doing this at like midnight, so there are bound to be mistakes.

[hide="2"]
$\frac{BD}{BA}=\sqrt{\frac{20}{40}}=\frac{\sqrt{2}}{2}$.

$\overrightarrow{OD}=(0,10)+\frac{\sqrt{2}}{2}\cdot\left(8,-10\right)=(0,10)+(4\sqrt{2},-5\sqrt{2})$.

Thus the x-coordinate of point D is $4\sqrt{2}$. [/hide]

[hide="1"]$\frac{1+x}{1-x}=\sqrt 5\Longrightarrow x=\frac{3-\sqrt 5}{2}$.[/hide]


function
real analysis
real analysis unsolved

Suppose $f$ is a real continuous function on $\mathbb{R},f_{n}(t)=f(nt)$ for $n=1,2,3,...$ and $\{f_{n}\}$ is equicontinuous on $[0,1]$. What conclusion can you draw about $f$?

Just consider equicontinuity at zero: for any $\epsilon>0$, there is a $\delta>0$ such that for all $n$, $|f_{n}(x)-f_{n}(0)|< \epsilon$ whenever $0<x<\delta$.
That translates to $|f(nx)-f(0)|<\epsilon$, and thus $|f(t)-f(0)|<\epsilon$ whenever $0<t<n\delta$ for all $n$.

It's only a short step from here to conclude that $f$ is constant on $[0,\infty)$.


Sometimes I’m loud,
and viewed with distaste,
poke out my eye,
and I’ll get in your face.
What am I?

Nary an answer can be found
Over this poem or belowground.
Is this line needed to read it?
Seek the poem's best.
End your quest!

[quote="buzzer11"]is sogia's riddle answer a really mean, violent bully? :?[/quote]
lol  :rotfl:  :rotfl:  :rotfl:


Let {a_n} be an arbitrary sequence of positive numbers. Let b_n = (a_1 + a_2 + ... + a_n)(1/a_1 + 1/a_2 + ... + 1/a_n) and c_n = int[(b_n)^0.5] (int denotes the integer part of the respective term). Prove that c_1, c_2,... are all different.

Although very simple, this problem is very nice. Of course, it is enough to prove that c_n+1>=c_n+1. It is in fact sufficient to prove that sqrt(b_n+1)>=sqrt(b_n)+1. After squaring, this inequality becomes
a_n+1*(1/a_1+...+1/a_n)+1/a_n+1*(a_1+...+a_n)>=2sqrt((a_1+...+a_n)(1/a_1+...+1/a_n)), which is obvious from the Am-GM inequality.


geometry
inequalities
inequalities unsolved

we have the inqualitity
a(b+c-a)+b(a+c-b)+c(a+b-c)$ \ge$ $ \sqrt{3}$ .$ \sqrt{(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(a\plus{}c\minus{}b)(b\plus{}c\minus{}a)}$
if X=a+b-c; Y=b+c-a; Z=c+a-b;
the inqualities become :
XY+YZ+XZ$ \ge$
$ \sqrt{3XZY(X\plus{}Y\plus{}Z)}$

[quote="trannguyenquyND_VN"]we have the inqualitity
a(b+c-a)+b(a+c-b)+c(a+b-c)$ \ge$ $ \sqrt {3}$ .$ \sqrt {(a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(a \plus{} c \minus{} b)(b \plus{} c \minus{} a)}$
if X=a+b-c; Y=b+c-a; Z=c+a-b;
the inqualities become :
XY+YZ+XZ$ \ge$
$ \sqrt {3XZY(X \plus{} Y \plus{} Z)}$[/quote]
nice :)


search function MATHCOUNTS quadratics conics hyperbola algebra

Uh, anyway to make a system of equations to do this. The 4 numbers, lets say 2,2,2,2, sure they add to 8, but they don't multiply to there! I need a set of 4 numbers that go to 7.11 either way. Thanks a lot in advance!

If all the numbers are the same, let them be $x$.

Then, you have $x+x+x+x = 4x = 7.11$.

And you also have $x * x * x * x = x^{4}= 7.11$.

So then, $4x = x^{4}$, and we know that $x$ is nonzero.

Thus, $x^{3}= 4$, so $x = 4^{\frac{1}{3}}$.

But then, $4x = x^{4}= 4^{\frac{4}{3}}$, which is not $7.11$.

So, there does not exist such a number $x$.

Shoot im stupid, im sorry, all the numbers do NOT need to be the same. For example (again)

2,5,6,7 all multiply to 420, but they all add to 20. I need them to add and multiply to 420.

Get my point now, im really srry, but very good point.

this problem has been posted a million times...

Where was it posted? Can you post some links?

links would be very nice. This is 1/4 of my math grade, so it's kinda important!

see my post here:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=711&t=44345[/url]

alternately use the search function

[quote="zyryn"]Thanks Altheman, that helps a lot, but anyway I could put that into system of equations form that can be solved through substitution and graphing?[/quote]

no, there are two equations in 4 variables, you have to use the fact that 100 * the price is an integer

Right, but the thing is, that I can't prove to my math teacher using primes and that, I need proof of using a system. But since we can figure that x100 fits, I dunno...but to the point, I need a system of equations, I'm scratchin somethin onto paper now, but anyhelp would be great.

is any1 here, i really need some help to get some equations!

You plain can't solve a two equation, four variable set of equations. You pretty much have to work from the fact that one factor has to be a multiple of 79 cents upwards. It's basically logical trial and error.

of course, if anybody has a really nice way of solving this, go ahead, show it, feel smart.

[quote="Treething"]and by the way, this is a mathcounts target round problem :X.[/quote]

Are you kidding me? What year was it?

[quote="zyryn"]Right, but the thing is, that [b]I can't prove to my math teacher using primes[/b] and that, I need proof of using a system. But since we can figure that x100 fits, I dunno...but to the point, I need a system of equations, I'm scratchin somethin onto paper now, but anyhelp would be great.[/quote]

either your teacher is stupid or you are not understanding what he/she is saying...the solution is already presented...where do you get the idea that there is a magical system of equations that solves the problem??? there are only two equations that you have to solve in positive integers

[quote="zyryn"]okay, i got it now, thanks guys. Turns out that the two equations are y=7.11/x and y=-x+7.11. When you use substitution, you come out with something like -x^2+7.11x=7.11, which is a quadratic equation. Easy from there. When u graph those equations, it comes out a hyperbola and a line through it, which (luckily) graphs to two intersects

PS:Let statements
Let x=a+b
Let y=c+d

Thanks guys![/quote]

umm... it doesn't quite work that way... for instance, xy = (a+b)(c+d), not abcd


Let $ X$ be an affine algebraic variety over a field $ k$. Show that there exists a projective variety $ \bar{X}$ over $ k$ such that $ X$ is isomorphic to a dense open subscheme of $ \bar{X}$.

If $ X \subseteq \mathbb{A}^n$, take the closure of $ X$ in $ \mathbb{P}^n$ via the embedding $ \mathbb{A}^n \hookrightarrow \mathbb{P}^n$.


topology
function
real analysis
real analysis theorems

I am reading the Royden's Real Analysis book
and I have  a question for the following theorem :

Theorem (Alaoglu - page 237) :[b] The unit ball $ S^{*}\equal{}\{f: \parallel{}f\parallel{}\leq 1\}$ of $ X^{*}$ is compact in the weak* topology[/b]

Proof : if $ f\in S^{*}$ then $ |f(x)|\leq \parallel{}x\parallel{}$, and so $ f(x)\in [\minus{}\parallel{}x\parallel{},\parallel{}x\parallel{}]\equal{}I_{x}$. Then each $ f\in S^{*}$ corresponds to a point in \[ P\equal{}\prod_{x\in X}I_{x}\]
Since the later is by definition the set of all functions $ f$ on $ X$ st : $ f(x)\in I_{x}$. Thus we may think $ S^{*}$ as a subset of $ P$, [b]and the the definition of the topology for $ P$ shows that that topology $ S^{*}$ inherits as a subspace of $ P$ is the weak* topology of $ S^{*}$ [/b]

Can some one explain for me the bolded sentence above?

thanks in advance

It follows from the transitivity of induced topology.


geometry
geometric transformation
reflection
inequalities

please prove the following: sum {i=1 to n} (a_i^k/x_i) >= (sum a_i)^k/n^(k-2)*sum (x_i) for positive numbers.

edit: yeah, i think it's x_i on both sides.

Is it x^i or x_i on both left and right sides ??

BUMP, in addition to a reformating:

:Sigma:(aik/xi) :ge: ( :Sigma:(ai))k/(nk-2 * :Sigma:(xi))

where all sums are from i = 1 to n.

What types of numbers are the a's and the x's and k?

Note: Edited to reflect Ravi's comment.

Is the :Sigma:(xi) supposed to be in the denominator?  Otherwise, couldn't we make all the xi huge to violate the inequality?


geometry
circumcircle
ratio
geometry unsolved
Really old

okay this one is from Prof. Mircea Lascu from Zalau, Romaniaand Prof. V. Cartoaje from  Ploiesti, Romania. It goes like this: given being a triangle ABC for every point M inside we construct the points A[size=67]M[/size], B[size=67]M[/size], C[size=67]M[/size] on the circumcircle of the triangle ABC such that A,  A[size=67]M[/size], M are collinear and so on. Find the locus of these points M for which the area of the triangle A[size=67]M[/size] B[size=67]M[/size] C[size=67]M[/size] is bigger than the area of the triangle ABC.

so I guess no one is considering this problem to be a challenge ? :)

Well , as far as I know this is an open problem, but I haven't tried it yet either. Doesn't look like a piece of cake although.

ok i found an article witch, after tons of calculations, comes up with S'/S=

(get ready)

prod ( a^2+b^2xy+c^2y ) / (x(x+1)b^2+(x+1)c^2-a^2x)

where a,b,c the sides, and x,y,z the ratios BD/CD , with xyz=1;

gotta go
bye

PS i dont think this will help a lot, cause i guess valentin knows this
article. its gazeta matematica no 3 from 2002

yes ... i know that ... and indeed it doesn't help at all ... there is another article concerning these types of triangles in the Math Gazette 2/2003 ( just bought it today! ) - i will study  it and if I find anything intersting, or worth mentioning I will post it

The locus just looks like a monster...I'm glad I didn't try this problem before :P

[quote="Myth"]I think the question will be quite meaningless after you will see that picture I constructed using SketchPad.
It is a high-precise shape of the required locus. So it is not an any traditional line.[/quote]

The curve bordering the locus (i. e. the set of all points M such that the area of triangle $A_MB_MC_M$ equals to the area of triangle ABC) is a sextic. See [url=http://groups.yahoo.com/group/Hyacinthos/messages/6178?viscount=2]Hyacinthos messages #6178 and #6179[/url]. So, indeed, the question is meaningless. However, http://www.mathlinks.ro/Forum/viewtopic.php?t=21900 gives an assertion about a part of this locus.

  Darij

[quote="Myth"]I think the question will be quite meaningless after you will see that picture I constructed using SketchPad.
It is a high-precise shape of the required locus. So it is not an any traditional line. And I have some doubts it have a nice parametrization.[/quote]:D Thanks Mikhail for clearing this out after [b]exactly[/b] two years  :first:

[quote="Valentin Vornicu"]:D Thanks Mikhail for clearing this out after [b]exactly[/b] two years  :first:[/quote]
I think you mean two years [i]and two months[/i], don't you, Valentin? :P

[quote="Valentin Vornicu"][quote="Valiowk"]I think you mean two years [i]and two months[/i], don't you, Valentin? :P[/quote]I ment two years since the last message on the thread : http://www.mathlinks.ro/Forum/viewtopic.php?p=126#p126 :D[/quote]

Looking at the time I would say less than two years.  :D


function
number theory
algebra unsolved
algebra

Find all functions $f: \mathbb R \to \mathbb R$ such that
\[f(f(x)\plus{}f(y)\plus{}xy)\equal{}x\plus{}y\plus{}f(xy)\]
for all reals $x$ and $y$.

when I solve this function, I get to this: $ f(f(x))\equal{}x$. and I dont know how to solve it.
or the problem changes to this:
find all functions $ f: R \rightarrow R$ which $ f(f(x))\equal{}x$.

amparvardi

[quote="amparvardi"]when I solve this function, I get to this: $ f(f(x)) \equal{} x$. and I dont know how to solve it.
or the problem changes to this:
find all functions $ f: R \rightarrow R$ which $ f(f(x)) \equal{} x$.
[/quote]

1) about involutive functions :
Your equation implies $ f(f(x))\equal{}x$ but is not equivalent to.
$ f(f(x))\equal{}x$ has as solution any involutive function (it's their definition) and you have infinitely many involutive functions. One general form for them could be :

Let $ A,B,C$ any split of $ \mathbb R$ such that $ A$ and $ B$ are equinumerous and let $ h(x)$ any bijecton from $ A\to B$ :

$ \forall x\in A$ : $ f(x)\equal{}h(x)$
$ \forall x\in B$ : $ f(x)\equal{}h^{\minus{}1}(x)$
$ \forall x\in C$ : $ f(x)\equal{}x$

2) about your problem :
We have infinitely many solutions.
For example at least all involutive solutions of Cauchy's equation.
But I'm unable up to now  to prove if these are all the solutons or if some other exist.

I am sorry for my bad english

y = 0 => f(f(x)+f(0)) = x+f(0) => f is injective
y = -x => f(f(x)+f(-x)-x^2) = f(-x^2) => f(x)+f(-x)-x^2 = -x^2 => f(-x) = -f(x)

take g(x) = f(x) - x
if g is null function f(x) = x wich is solution.
otherwise f(x) injective so g(x) is too.
g(-x) = f(-x) + x = -f(x)+x = -g(x).

Replacing f(x) by g(x) in the equation we take :

g(g(x)+x+g(y)+y+xy) + g(x)+x+g(y)+y+xy = x+y +g(xy)+xy

g(g(x)+g(y)+x+y+xy) = g(xy)-g(x)-g(y) (1)

take y = 1

g(g(x)+g(1)+x+1+x)=-g(1)=g(-1)

g is injective => g(x)+2x+1+g(1) = -1 => f(x) = -x-2-g(1) (2)

take x = 1 and y = 1

g(g(1)+g(1)+3)=g(1)-g(1)-g(1)=g(-1)

2g(1)+3 = -1 => g(1) = -2 => f(x) = -x

the only solution are f(x) = x and f(x) = -x

[quote="snain"]y = 0 => f(f(x)+f(0)) = x+f(0) => f is injective
y = -x => f(f(x)+f(-x)-x^2) = f(-x^2) => f(x)+f(-x)-x^2 = -x^2 => f(-x) = -f(x)

take g(x) = f(x) - x
if g is null function f(x) = x wich is solution.
otherwise f(x) injective so g(x) is too.[/quote]
Surely not. $ f(x)$ injective does not imply $ f(x)\minus{}x$ injective !

[quote="snain"]the only solution are f(x) = x and f(x) = -x[/quote]
Surely not. As I said in my previous post, any involutive solution of Cauchy's equation is solution. And so infinitely many solutions (with axiom of choice).

[quote="snain"]...
the only solution are f(x) = x and f(x) = -x[/quote]

It has infinitely many solutions.
For Example, let $ x\equal{}y\equal{}0$,so we have: $ f(2f(0))\equal{}f(0) \Longrightarrow f(0)\equal{}0$.
Then let $ y\equal{}0$, It becomes to $ f(f(x))\equal{}x$. And a soloution for this function is this: $ f(x)\equal{}k\minus{}x$.(check it)

[quote="amparvardi"]
It has infinitely many solutions.
For Example, let $ x \equal{} y \equal{} 0$,so we have: $ f(2f(0)) \equal{} f(0) \Longrightarrow f(0) \equal{} 0$.
Then let $ y \equal{} 0$, It becomes to $ f(f(x)) \equal{} x$. And a soloution for this function is this: $ f(x) \equal{} k \minus{} x$.(check it)[/quote]
Once again, same error. Your equation implies $ f(x)$ is involutive bus [u][b]is not equivalent to [/b][/u].

All involutive functions are not solutions of your equation.
For example $ f(x)\equal{}k\minus{}x$ is involutive but is solution only if $ k\equal{}0$, so is not a good answer to snain.

[quote="pco"]
Once again, same error. Your equation implies $ f(x)$ is involutive bus [u][b]is not equivalent to [/b][/u].

All involutive functions are not solutions of your equation.
For example $ f(x) \equal{} k \minus{} x$ is involutive but is solution only if $ k \equal{} 0$, so is not a good answer to snain.[/quote]

yes, you are right ,but [size=150]snain[/size] wrote that there are only two functions  $ f(x) \equal{} x$ and $ f(x) \equal{} \minus{}x$ satisfying the problem.
it was an example I wrote,not all soloutions of the question.
((my question:why the soloutions of the $ f(f(x))\equal{}x$ aren't exactly soloutions of the original function?))

amparvardi

[quote="amparvardi"]
yes, you are right ,but [size=150]snain[/size] wrote that there are only two functions  $ f(x) \equal{} x$ and $ f(x) \equal{} \minus{} x$ satisfying the problem.
it was an example I wrote,not all soloutions of the question.
((my question:why the soloutions of the $ f(f(x)) \equal{} x$ aren't exactly soloutions of the original function?))

amparvardi[/quote]
But you gave him an example to check : $ f(x) \equal{} k \minus{} x$ and you did not yourself check it : this does not fit if $ k\ne 0$ and so your example is wrong!

And about your question :
You got $ f(f(x)) \equal{} x$ by using $ \implies$, not $ \iff$ (I said this to you already twice)
For example : $ f(x) \equal{} 1 \minus{} x$ is such that $ f(f(x)) \equal{} x$ but is not a solution of your equation.

Got it!
you're right,and I get the mistake.
thank you for your guidence!


linear algebra
matrix
algebra
polynomial
algebra unsolved
combinatorics

Let $a_{1},a_{2},...,a_{100},b_{1},b_{2},...,b_{100}$ be distinct real numbers. Consider the $100$x$100$ matrix in which the entry in the $i$th row and the $j$th column is $a_{i}+b_{j}$.

(a) Suppose the product of the entries in each column equals $1$. Prove that the product of the entries in each row equals $-1$.

(b) If the products of the entries in each column are equal, prove that so are the products of the entries in each row.

Let $ A(x)=\prod_{i=1}^{100}(x+a_{i})$ and $ B(x)=\prod_{i=1}^{100}(x+b_{i})$. The product of entries in the ith row is $ B(a_{i})$ and in the ith column is $ A(b_{i})$. If the product of the entries in each column is the same constant $ c$, then $ A(x)-c$ is a monic polynomial of degree 100 with the 100 roots $ b_{i}$, $ 1\le i\le 100$. Then it is easy to see that $ A(x)-c=B(-x)$ identically. Hence $ B(x)=A(-x)-c$, which is clearly $ -c$ at $ a_{i}$, $ 1\le i\le 100$. So the product of the entries in each row is constant at $ -c$.


[quote="Conjurer"]1) Why dont two poles of a dipole collapse on each other?

2)Why do dipoles have electric field E@ and Er and not Ex,Ey,Ez like a normal field?[/quote]

1) Dont you know... They can't. This involves simple principles of quantum electrodynamics. To put it across simple. Solve Schrodinger's equation and get the position of the two dipoles. Now apply Uncertainty principle, you will find delta x to be too small and thus you they wont collapse.

[hide]Actually the distance between the 2 dipoles is fixed. Or sometimes they are fixed at their positions. In case of molecules, thats the position of minimum Potential energy. And an ideal dipole is one with minimum distance between them.[/hide]

2) Who said they don't have normal electric field??? Differentiate Electric Potential and obtain the field.

We can easily obtain the electric field of a dipole..

Lets put the origin at the negative part:
Let the vector showing from positive part to the negative part be $ \textbf{l}$.

So the potential:
\[ \Phi(\textbf{r})\equal{}\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{|\textbf{r}\plus{}\textbf{l}|}\minus{}\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{|\textbf{r}|}\]
From here we get:
\[ \Phi(\textbf{r})\equal{}\minus{}\dfrac{1}{4\pi\varepsilon_0}\dfrac{q\textbf{l}}{r^3}\textbf{r}\]

Now the parameter we use for dipoles is $ \textbf{p}\equal{}q\textbf{l}$. Its called dipolemomentum.

The electric field of it is the negative gradient of the potential hence:
\[ \textbf{E}(\textbf{r})\equal{}\dfrac{1}{4\pi\varepsilon_0}\text{grad}\left(\dfrac{\textbf{p}\textbf{r}}{r^3}\right)\].

so from here we have:
\[ \textbf{E}(\textbf{r})\equal{}\dfrac{1}{4\pi\varepsilon_0}\dfrac{r^2\textbf{p}\minus{}3(\textbf{p}\textbf{r})\textbf{r}}{r^5}\]

Now I dont see why this is not a normal field.. Anyway what do you define as "normal field"?

I presume he wants the x component, y component and z component of the field.

ok thnx. So it basically means that if we have variables like theeta, x ,y and z in potential, so we have field E@, Ex,Ey and Ez.

And because ,should I say generally, potential of dipole has only variables theeta and R, so we have field Er and E@. Correct me if I am wrong.


i am computer stupid, so please explain in simple terms =]
im using texhshop on a mac.
how do i make the file into a pdf?

I don't have a Mac but [url=http://www.it.iastate.edu/training/videos/latexintronotes.pdf]Getting Started with LaTeX[/url] says:[quote]If you are using TeXShop, click on the Typeset button or within iTeXMac pull down under TeX ! Typeset.[/quote]You may find that document useful reading.

when i do that.
it does some crazy stuff.
maybe an error message of some sort?
it just fills up that box wiht various text.
it wont let me copy & paste.

It's probably the log file which will be kept in the same directory as the tex file with the same name. This file is important as it tells you what has happened. It will be produced even if the tex file is fine but you must read the log file carefully to see if there any errors you need to correct.

ah, i figured it out, thanks to ragnarok23.
and yes that was the issue, i screwed up some slashes around cdots.  but thanks.


Okay, this is the player list. Order matters:

cf249
??
mathelete
??
??
??
??
??

Each player has 3 grenades in their arsenal. There is no concept of turns. No consecutive turns. When you go, you have five options. Do only one of them.

1. Choose another player, and subtract points. The number of points subtracted is the post number.

2. Same as above, but add points instead.

3. Throw a grenade. Convert the post number to binary, then add up the digits. The grenade will go off that many posts later. If it does not go off in 4 days, the grenade is defused.

4. Bump the grenade to the player below you on the list. If you are at the bottom, it will go to the top.

5. Pick up a defused grenade. It goes into your arsenal.

Everybody starts off with 0 points, and they can have a negative number of points. After 75 posts, the two people with the highest number of points and the two with the lowest are on team A. The rest are on team B. The team with the higher number of total points wins. A fallen player's points does not count. When the grenade goes off, whoever it is by falls for the rest of the game. Remember, you never know whose team you will be on! And neither do I!

/in

Do we start?

If so: -2 to mathlete

cf249 0
??
mathelete -2
??
chenhsi 0
??
??
??

I'll play also.

cf249 0
??
mathelete -2
??
chenhsi 0
??
Wickedestjr 0
??

Umm, what happens when a grenade goes off?

I'll take three from mathlete.

cf249 0
??
mathelete -5
??
chenhsi 0
??
Wickedestjr 0
??

cf249 0
??
mathelete -5
??
chenhsi 0
??
Wickedestjr 0
??
ernie 4
??

ernie, alliance against mathlete?

cf249 0
??
mathelete -10
??
chenhsi 0
??
Wickedestjr 0
??
ernie 4
??

I thought I signed up by PM.

cf249 0
??
mathelete -10
??
chenhsi 0
??
Wickedestjr 6
??
ernie 4
??
aleph0 0

cf249 0
??
mathelete -10
??
chenhsi 0
??
Wickedestjr 13
??
ernie 4
??
aleph0 0

Wicked is in the lead.  :P

[quote="mathelete"]1. Choose [b]another player[/b], and subtract points. The number of points subtracted is the post number.

2. [b]Same as above[/b], but add points instead.
[/quote]

Are we allowed to heal ourselves?

cf249 0
??
mathelete -10
??
chenhsi 0
??
Wickedestjr 5
??
ernie 4
??
aleph0 0

Yes we are.

cf249 0
??
mathelete -10
??
chenhsi 0
??
Wickedestjr -4
??
ernie 4
??
aleph0 0
??

I don't think self-healing is allowed.
-10 ernie because I disagree with him.

cf249 0
mathelete -10
chenhsi 0
Wickedestjr -4
ernie -6
aleph0 0
??
??

What was up with all the question marks?

cf249 -11
mathelete -10
chenhsi 0
Wickedestjr -4
ernie -6
aleph0 0

cf249 -11
mathelete -10
chenhsi 0
Wickedestjr -4
ernie -6
aleph0 -12

In the lead!

cf249 -11
mathelete -10
chenhsi -13
Wickedestjr -4
ernie -6
aleph0 -12

-13 from chenhsi

cf249 -11
mathelete -10
chenhsi 1
Wickedestjr -4
ernie -6
aleph0 -12

cf249 -11
mathelete -10
chenhsi 1
Wickedestjr -4
ernie 9
aleph0 -12

That doesn't work.  You cannot double post.

cf249: PWNT
mathelete: -91, mushroom, 2 grenades
chenhsi: 35, 2 grenades
Wickedestjr: PWNT
ernie: 15, 2 grenades, mushroom
aleph0: -88, 2 grenades, GRENADE TO GO OFF NEXT POST

cf249: PWNT
mathelete: -91, mushroom, 2 grenades Grenade at next post
chenhsi: 35, 2 grenades
Wickedestjr: PWNT
ernie: 15, 2 grenades, mushroom
aleph0: -88, 2 grenades

Yay bump (I quadruple posted so grenade is still for next post)

cf249: PWNT
mathelete: PWNT
chenhsi: 35, 1 grenades
Wickedestjr: PWNT
ernie: 15, 2 grenades, mushroom, Grenade 50
aleph0: -88, 2 grenades

Throw Grenade at ernie:

46 = 101110 = 4

cf249: PWNT
mathelete: PWNT
chenhsi: 84, 1 grenades
Wickedestjr: PWNT
ernie: 15, 2 grenades, mushroom
aleph0: -88, 2 grenades, GRENADE TO GO OFF NEXT POST

Wickedestjr: A grenade was thrown at you, and it apparently went off.

cf249: PWNT
mathelete: PWNT
chenhsi: 134, 1 grenades
Wickedestjr: PWNT
ernie: 15, 2 grenades, mushroom
aleph0: PWNT

cf249: PWNT
mathelete: PWNT
chenhsi: 33, 1 grenade
Wickedestjr: PWNT
ernie: 15, 1 grenades, mushroom
aleph0: PWNT

Woah woah! Did ernie throw a grenade at cf249 AND give points? That is illegal! CF249, you are back in. And ernie also threw a grenade at aleph0! People do not follow directions. You cannot do two things at once. Start over. I am starting a new topic.

Sure, new game.

Ideas:

More players

Change of WC (The thing about 4 players on 1 team and everyone else is stupid, because most likely there will be <4 players left at the end)

Removal of mushrooms + stars

I don't understand the rules that much. You weren't real specific about what happens when a grenade goes off.


Find the probability that when you vote you are a member of the majority if there are n people.

a. n is even

b. n is odd

a. Obviously, utilize symmetry. The rest follows.

WEll thinking about this, b is obviously 1/2.

Uh no. Take $ n\equal{}1$. Obviously the answer is $ 1$.

BTW, am I correct on a?

Idk. No cuz u gave no answer. A will just be 1/2-nC(n/2)/2 right?

No.

$ \frac {1 \minus{} \frac{\dbinom n{\frac n2}}{2^n}}2$

Oh yeah that's what I meant. Good catch.

For b, imagine that there are $ n\minus{}1$ voters (you will vote "last").
There are several possibilities for this election:
*Let $ k\equal{}\frac1{2^{n\minus{}1}}\dbinom{n\minus{}1}{\frac{n\minus{}1}2}$.
1. It is a tie. This occurs with probability $ k$, and after this, the probability that you choose the winner is $ 1$.
2. One of the other candidates win. This occurs with probability $ 1\minus{}k$. It is clear that the probability that you choose the correct candidate is $ \frac12$, so the probability for this case is $ \frac{1\minus{}k}2$.
Add these two cases together, we find that our answer is $ k\plus{}\frac{1\minus{}k}2\equal{}\boxed{\frac{k\plus{}1}2}$.


You have triangle $ABC$, with points $D,E,F$ on sides $BC, CA, AB$ respectively, such that:

$\displaystyle\frac{BD}{DC} =\frac{CE}{EA} = \frac{AF}{FB} = 2$

Prove that the triangle made by the lines $AD$, $BE$, and $CF$ has area $\displaystyle\frac{1}{7}$ of triangle $ABC$.

[hide]

does ne1 know whether or not im allowed to for problems like these, assume best case scenario and make it a right triangle?

cuz assuming that i put it on a coordinate plane and using coordinate geometry i proved it

but yea im lookign for some better, more geometrical proofs :lol:

[/hide]

This is the fifth time I see this problem on ML!
http://www.mathlinks.ro/Forum/viewtopic.php?t=40828