group theory
abstract algebra
linear algebra
matrix
superior algebra
superior algebra unsolved

Let $ G\equal{} \mathbb{Z}/ n \mathbb{Z}$.

If $ n\equal{}3$, I can find a non trivial morphism $ \varphi : G \to Aut(G\times G)$, hence there is a semi-direct product $ (G\times G)\rtimes G$ (which is not direct).

In general, I am wondering for which $ n>3$ we can define a semi-direct product $ (G\times G)\rtimes G$ (which is not direct), or even better $ (G\times G \times \cdots \times G)\rtimes G$

[quote="Fleury"]Let $ G \equal{} \mathbb{Z}/ n \mathbb{Z}$.

If $ n \equal{} 3$, I can find a non trivial morphism $ \varphi : G \to Aut(G\times G)$, hence there is a semi-direct product $ (G\times G)\rtimes G$ (which is not direct).

In general, I am wondering for which $ n > 3$ we can define a semi-direct product $ (G\times G)\rtimes G$ (which is not direct), or even better $ (G\times G \times \cdots \times G)\rtimes G$[/quote]

This works for all integers n > 1.  Aut(Z/nZ × Z/nZ) basically has Aut(Z/pZ × Z/pZ) as a subgroup for any prime p dividing n, and it has an element [1,1;0,1] of order p, so there is always a non-trivial homomorphism from Z/pZ into Aut(Z/pZ × Z/pZ), and so also a non-trivial (but not necessarily faithful) homomorphism from Z/nZ into Aut(Z/nZ × Z/nZ).

Similarly Aut(Z/nZ × ... × Z/nZ) has Aut(Z/nZ × Z/nZ) as a subgroup, as long as the first group has at least two factors.

A model for that group: matrices of the form $ \begin{bmatrix}1 & a & b \\
0 & 1 & c \\
0 & 0 & 1\end{bmatrix}$ over $ \mathbb{Z}/n\mathbb{Z}$. The subgroup with $ c \equal{} 0$ is normal, and conjugation by $ \begin{bmatrix}1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1\end{bmatrix}$ acts as that shear on it.

Thank you. I understand that
[quote="JackSchmidt"]This works for all integers n > 1.  Aut(Z/nZ × Z/nZ) basically has Aut(Z/pZ × Z/pZ) as a subgroup for any prime p dividing n, and it has an element [1,1;0,1] of order p,
[/quote]
But not this
[quote="JackSchmidt"]
so there is always a non-trivial homomorphism from Z/pZ into Aut(Z/pZ × Z/pZ), and so also a non-trivial...
[/quote]

jmerry gave a very good answer to this already.  In the group {[1,a,b;0,1,c;0,0,1]:a,b,c in Z/nZ}, the subgroup {[1,a,b;0,1,0;0,0,1]:a,b in Z/nZ} is isomorphic to Z/nZ × Z/nZ, is normal, and is complemented by {[1,0,0;0,1,c;0,0,1]:c in Z/nZ}.  The action of conjugation by C=[1,0,0;0,1,1;0,0,1] on Z/nZ × Z/nZ is the action I suggested in the case n was prime: [1,1;0,1], as in it fixes B=[1,0,1;0,1,0;0,0,1] and sends A=[1,1,0;0,1,0;0,0,1] to A*B.  The group has the presentation <A,B,C:A^n=B^n=C^n=1,AC=CAB, BA=AB, BC=CB>.

To more directly answer your question about the homomorphism:

Notice that Aut(Z/nZ × Z/nZ) is isomorphic to the group of invertible 2 × 2 matrices over Z/nZ, and that it has a cyclic subgroup of order n generated by [1,1;0,1].  The homomorphism from Z/nZ to <[1,1;0,1]> given by i+nZ↦[1,i;0,1] is non-trivial (even injective) and so works to answer your question.

Got it, thank you very much JackSchmidt.


Let $p$ be a prime with $p>5$, and let $S=\{p-n^2 \vert n \in \mathbb{N}, {n}^{2}<p \}$. Prove that $S$ contains two elements $a$ and $b$ such that $a \vert b$ and $1<a<b$.

We show that the smallest possible $a$ works.

Let $n= \lfloor \sqrt p \rfloor$.
If $n^2+1=p$, then we set $a=p-(n-1)^2=n^2+1-n^2+2n-1=2n$ and $b=p-1^2=n^2$. This works, because surely $p$ is odd, so $2|n$ giving $a=2n|n^2=b$.

In all other cases we can set $a=p-n^2>1$. There are three cases:
a) $n^2+n > p$: Then we set $c=n^2+n-p>0$. Additionally $n^2 < p$ gives $c=n^2-p+n<n$.
b) $n^2+n=p$: this implies $n|p$, impossible.
c) $n^2+n < p$: Then we set $c=p-n^2-n>0$. Additionally $p < (n+1)^2$ gives $c=p-n^2-n=p-(n+1)^2+n+1<n+1$. If $c=n$, then we would have $n=p-n^2-n$ giving $n|p$, impossible. So again $c<n$.

In all possible cases, we now set $b=p-c^2$ and get $b=p-c^2 \equiv p- (\pm n)^2 \equiv p-n^2 \equiv 0 \mod (p-n^2)$, being the result.

Let $[\sqrt p]=k$. Now let $p=k^{2}+x$ where $x\leq 2k+1$.Hence $x \in S$. Now consider the number $|k-x|$.So $|k-x|<\sqrt p$. Now let $b=p^{2}-|k-x|^{2}$. We see that $x|b$.Hence we are done.

Excuse me for my bad english.


Let a fibonacci sequence:$F_1=F_2=1$ and $F_{n+2}=F_{n+1}+F_n$
Prove that:
$\sum\limits_{i=1}^\infty{\frac{1}{F_i}}\geq \frac{10}{3}$

I think you meant $\sum_{i=1}^\infty$.
Anyway $\sum_{i=1}^{11}\frac{1}{F_i}>\frac{10}{3}$ :)


[quote="grobber"]Discussed quite a few times on the forum. Use a particular case of Turan's Theorem: a graph on $n$ vertices with no complete quadrilaterals has at most $\frac{n^2}3$ edges. If we join to points iff the distance between them is $>\sqrt 2$ we get such a graph.[/quote]

How to have an idea as the Turan's theorem, is there a simple proof or idea to understand it would be true for that theorem, or know a normal problem-solver that ?

Also, is it possible to find a proof without that?

Interpret the question as a graph, the points are the vertices and they are connected iif their length is $>\sqrt{2}$. Now, it is obvious that there are no 4-cycles, let alone 4-complete graph, by turan, there are at most $(1-1/3)n^2/2 = n^2/3$

A beautiful problem indeed :)
[quote=Megus]Given any $n$ points on a unit circle show that at most $\frac{n^2}{3}$ of the segments joining two points have length $> \sqrt{2}$.[/quote]

Consider the problem in graph terminology. The main claim is that $G$ is $K_4 - \text{free}$. Now this is obviously true since otherwise we will have four points $A,B,C,D$ such that $2\cdot \pi = \angle AOB + \angle BOC + \angle COD + \angle DOA < 4 \cdot \frac{\pi}{2} = 2\cdot \pi$, a contradiction. Hence we are done by Turan's theorem.

Join two points by an edge ,iff their distance is greater than $\sqrt{2}$.
Obviously ,by dividing the circle into four quarter circles,it is evident that ,there cannot be 4-cycles in our graph.
Clearly ,we achieve the desired bound as it is asked for a $K_{4}$ due to Turan.
But the bound for the 4-cycle is well known ,too so the bound in the problem is not sharp..


geometry
trigonometry
trig identities
Law of Sines

Archimedes' Book of lemma, proposition 15.

[img]http://agutie.homestead.com/files/ArchBooLem15.gif[/img]

Wow!!! Thats really cool. I could use this on a lot of my geometry homework (assuming that  it really works) :!:

you can do some angle chasing, then use the law of sines on CDE, and do a little trig to get the side interms of the radius...and you are done


Kyle has 112 unit cubes. If no unit cube may be cut into smaller
pieces, what is the minimum number of additional unit cubes he
needs in order to build a solid cube using all the unit cubes?

We want the smallest $ n$, an integer, where $ n^3$ is greater than 112. This is $ 5^3 \equal{} 125$.

We can build a cube with this many units, so we need $ 125 \minus{} 112 \equal{} 13$ additional units.


It's not likely much people in my school would join it...
so I might have to seek for people in the outside.
Anyone in long island who needs a team member?
or inside NY...
or anywhere in US...

Are inter state or inter school teams allowed?

just a bump up see if anyone interested
maybe it's too early to form a team  :P

[quote="PUMaC_PR"]Yes, teams may be composed of students from multiple schools or states, as long as they compete in Division A.[/quote]
Wouldn't that make the "initial registration limit of two teams per school" useless?


Mary takes the train home from work each day, arriving at the station at 6:00, at which time her butler, Jeeves, meets her and drives her home. Jeeves always drives at the same constant speed, and always arrives at the station at exactly 6:00. One day Mary takes an earlier train which arrives at the station at 5:00. She immediately begins walking home. On the way, she meets Jeeves, who is on his way to meet the 6:00train. Mary gets into the car, and they drive home, arriving 20 minues earlier than usual. How many minutes had Mary been walking?

This seems like a problem from the Harker Invitational?
[hide="Solution"]
Say that Mary has been walking for $ x$ minutes.  Also, say that they reach home $ t$ minutes after $ 6: 00$ on every usual day.  Thus, the butler starts driving to the train station $ t$ minutes before $ 6: 00$ every day.  So, the butler left at $ 5: (60 \minus{} t)$ and saw Mary at $ 5: x$.  Thus, the butler has been driving for $ x \plus{} t \minus{} 60$ minutes before seeing Mary.  It therefore takes him another $ x \plus{} t \minus{} 60$ minutes to get back to Mary's house.  So, he drove for a total of $ 2(x \plus{} t \minus{} 60)$ minutes.  Thus, he reached Mary's house at $ 5: (60 \minus{} t \plus{} 2x \plus{} 2t \minus{} 120) \equal{} 5: (t \plus{} 2x \minus{} 60)$.  They normally reach Mary's house at $ t$ minutes after $ 6: 00$, which is $ 5: (60 \plus{} t)$, but they reached 20 minutes early today, so they reached Mary's house at $ 5: (60 \plus{} t \minus{} 20) \equal{} 5: (40 \plus{} t)$.  Thus, $ 2x \plus{} t \minus{} 60 \equal{} 40 \plus{} t\implies x \equal{} 50$.
[/hide]


search
combinatorics unsolved
combinatorics

[img]http://i220.photobucket.com/albums/dd160/jrb599/poset.gif[/img]

Actually I think it's 3!*3! +12, so 48.

The search on this forum is text-based, so it would be much preferable if you could type in your posts rather than using images.

Your question is only semi-comprehensible.  Are you asking for the number of linear extensions of the boolean algebra $ B_3$, i.e. the set of subsets of a three-element set ordered by inclusion?

If so, the solution offered by the image is bogus (you can't move one of the two-element subsets before one of the one-element subsets unless they are disjoint from each other), and your answer is correct ($ 3! \cdot 3!$ counts the number of linear extensions with all of the singletons preceding all of the two-element sets, and there are $ 12$ linear extensions in which one two-element set precedes one singleton).

The associated sequence is [url=http://www.research.att.com/~njas/sequences/A046873]A046873[/url] in the [url=http://www.research.att.com/~njas/sequences/]OEIS[/url] -- the numbers get very big very quickly.


Let 'a' be an odd positive integer greater than 17 such that (3a-2) is a perfect square. Show that there exist distinct positive integers 'b' and 'c' such that (a+b), (b+c), (c+a) and (a+b+c)are four perfect squares.

is this problem such bad that, none wants to reply :|


Hello, I'm new here. My brother has been trying to solve Goldbach's conjecture and I'm 100% sure his using the wrong approach. Since I want to disprove him, I need your help :P What he does is taking a very large number (214124124124), and he adds the singles to get a new number and then, he continues from there. For example, if I have 1234, then the "new" number would be 10. See what I'm saying? I've been trying to tell him that adding the singles doesn't mean anything (or does it? I'm 95% with the background I have of math, it doesn't mean anything). Can anyone help me out here?

You mean digit's sum, right¿
Well, why should they even have anything to do with Goldbach... and as it seems that it's in base $10$ which is a rather random base. But what do you want from us¿

he wants an affirmation that this approach is wrong

[quote="ZetaX"]Well... that would mean to show that Goldbach was wrong or it's unprovable ;)
(If there is a proof $P$, call his approach $A$ and you get a proof using $A$ by writing down $A$ and then forgetting what you wrote and writing down $P$ ;) )[/quote]
Would you mind explaining that again? XD

Digit sum has no relationship between prime number nor sum of two numbers.

For example, both 37 and 28 have digit sum 10, but the first one is prime while the second one is composite. (Clearly you cannot prove a number is prime only because its digit sum is prime)

The sum of digit sums of two numbers does not necessarily equals to the sum of digit of the sum of two numbers. For example, 19 has digit sum 10, 22 has digit sum 4, 19+22 = 41 has digit sum 5. 10 + 4 does not equal 5, get it?

However, Goldbach conjecture concerns the sum of two primes!  :D  So the sum of digits is nothing related to it.

The sum of digit reminds me about some fortune telling stuff :rotfl:


parameterization
integration
calculus
function
conics
parabola
calculus computations

Question:Calculate the lengths of the following curve
$ y^2 \equal{} 2px, x\in [0, x_0]$ for some parameters $ p, x_0.$

The general formula is
$ \int \sqrt {(\frac {dx}{dt})^2 \plus{} (\frac {dy}{dt})^2}dt$

If we put our parameters I get the integral

$ \int \sqrt {1 \plus{} \frac {p}{2x}}dx$
I even don't know how to solve this integration.

why $ dt$? why you make it complicate? when $ y$ is the function of $ x$ we use this formula : $ \int \sqrt{1\plus{}(\frac {dy}{dx})^2} dx$.

Ooohh..yes,it's just typo.I couldn't solve

$ \int\sqrt{1\plus{}\frac{p}{2x}}dx$

[quote="jmerry"]Well, that's the wrong integral.[/quote]

\[ \begin{gathered}
  x\left( y \right) \equal{} \frac{{y^2 }}
{{2p}} \hfill \\
  x'\left( y \right) \equal{} \frac{y}
{p} \hfill \\
  y'\left( x \right) \equal{} \frac{1}
{{x'\left( y \right)}} \equal{} \frac{1}
{{\frac{y}
{p}}} \equal{} \frac{p}
{y} \hfill \\
  \left( {y'\left( x \right)} \right)^2  \equal{} \frac{{p^2 }}
{{y^2 }} \equal{} \frac{{p^2 }}
{{2px}} \equal{} \frac{p}
{{2x}} \hfill \\
  \int_0^{x_0 } {\sqrt {1 \plus{} \left( {y'\left( x \right)} \right)^2 } } dx \equal{} \int_x^{x_0 } {\sqrt {1 \plus{} \frac{p}
{{2x}}} } dx. \hfill \\
\end{gathered}
\]

OK, my mistake. I was thinking of the parabola written the other way, which gives a $ \sqrt{1\plus{}4c^2x^2}$.

$ \int\sqrt{1\plus{}\frac{p}{2x}}\,dx\equal{} \int\sqrt{1\plus{}\frac{p}{2x}\plus{}\frac{p^2}{16x^2}\minus{}\frac{p^2}{16x^2}}\,dx \equal{}\int\sqrt{\left(1\plus{}\frac{p}{4x}\right)^2\minus{}\left(\frac{p}{4x}\right)^2}$

You have a choice here: trigonometric substitution or hyperbolic substitution. Both are complicated.
Alternately, substitute $ y\equal{}\sqrt{2px}$ to convert to the other form, which is more direct.


Hey!!Great idea!
Which ones did you like??

....and where you a guide?I was the guide of Romania.

 :)

So I guess that means that there weren't any hot guys,right?
Well...I would vote for a Mexican guy (don't know his name but he was very hot!)
 ;)

Girls,where are you???

Zoe was a guide also, if she's who I think she is ;) Anyway, we are still waiting here for your first post with "words" :D

but may be there are some other candidates.... who can be?

Thanks for the comments...  :maybe: I can't vote for the hottest guy, but i think that VinChester is the greatest babnik(girl "hunter") of the IMO... ;-)

Hehe  :lol: but U know I didn't even make friends with girls there  :( so I don't think I can be nominated


Prove that any prime number $2^{2^{n}}-1$ cannot be represented as the difference of two 5th powers of integers. n is a positive integers.

Sorry, there was a mistake, edited now. :D


I'm too dumb for my education. Here's another problem that I can't even start solving:

Show that if
\[z = (\frac{1 + i\sqrt{3}}{2})(10 + \sqrt{108})^{1/3} + (\frac{1 - i\sqrt{3}}{2})(10 - \sqrt{108})^{1/3}\]
then $z^3 = - 20 - 6z$. Use this to decide the value of $z$.

As I said, I can't even show the first step, so all help is appreciated.

(a+b)^3=a^3 + b^3 + 3ab(a+b) where a+b=z
Notice after that -2 is a root for z^3 = -20 -6z.

I'm just not seeing that $z^3 = -20 - 6z$, even using what you said. Can you show me?

For the second part, I can factorize the polynomial to get $(z + 2)(z - (1 + 3i))(z - (1 - 3i)) = 0$, but I don't know what root to choose.

a^3=-10-sqrt108
b^3=-10+sqrt108
ab=3(1,pi/3)(1,-pi/3)cubicrt(10+sqrt108)(10-sqrt108)z=3x1xcubicrt(-8)xz=-6z
Write z as k+ip in the firt formula and compare the argument to the argument of the solutions.


[color=darkred]Let $ABC$ be a triangle. The its incircle $C(I)$ touches the lines $AB$, $AC$ in the points $E$, $F$. Denote the middlepoint $M$ of the side $[BC]$. Prove that the $A$- altitude and the lines $EF$ and $MI$ are concurrently if and only if $r_{a}=2R$.

[b]Remark.[/b]

$\boxed{r_{a}=2R\Longleftrightarrow\cos B+\cos C=1+\cos A\Longleftrightarrow\cos A=\frac{p-a}{p}}$.[/color]

it's not difficult at all, just use that if $X \in MI \cap AD$, then $AX = r$, is probably here on the forum, but it is in fact simple to prove, using thales ($NI \| RD$).

now $M,I,R$ are collinear $\Longleftrightarrow AR=r$, therefore from the transversal theorem $\frac{p-b}{c}\cdot b \cos{C}+\frac{p-c}{b}\cdot c \cos{B}= \frac{h_{a}-r}{h_{a}}\cdot a$, and expanding this stuf, we will simply get $\cos{B}+\cos{C}= 1+\cos{A}$.


What is the smallest prime that is 10 less than a perfect square?

A prime can't be even so the perfect square must be odd. We can quickly go through these:

25 - 10 = 15, nope

49 - 10 = 39, nope

81 - 10 = 71, yay

Therefore, we have 71.

Or you can see that it can't end in $2, 4, 5, 6, 7, 8, $ or $0$, which leaves you with $1, 3, $ and $9$. The only perfect squares with these endings (under $100$) are: $1, 9, 49, $ and $81$. We know $1$ or $9$ can't be the answer, since $1-10<0$ and $9-10<0$. We try $49$ to find that $49-10=39=3\cdot13$. Our only option is $81$, which gives us $71$. Therefore, our answer is $\boxed{71}$


Find all natural numbrers such that:
$ 7^{3x} \plus{} 13^{y} \equal{} 2^{3z}$

$ 13^y \equal{} (2^z\minus{}7^x)\left ( (2^z\minus{}7^x)^2 \plus{} 3\cdot2^z7^x\right )$, so we need $ 2^z\minus{}7^x\equal{}1$.
The solution $ z\equal{}1$, $ x\equal{}0$ does not work.
The only other solution, according with Preda Mihailescu's proof of Catalan's conjecture, is $ z\equal{}3$, $ x\equal{}1$, leading to $ y\equal{}2$.

If not, we have $ 2^z\minus{}7^x\equal{}13^t$, with $ t\geq 1$, and then we need $ 13 \mid 3\cdot 2^z7^x$, impossible.


Let $ (a_n)_{n}$ be a sequence ,with $ a_1\in\mathbb{N^*}$ given, and satisfying $ a_{n\plus{}1}\equal{}\mu(a_n)\plus{}6$ ,where $ \mu(x)$ represents the smallest natural number $ n$ such that $ x| n!$.Prove that
\[ s_n\equal{}\left(\sum_{k\equal{}1}^n 6n \frac{a_k}{k} \right)\minus{}t\ln(t)\]
is convergent for a unique value of $ t$.

Something is fishy with the statement: as written, the limit either exists for all $ t$ or doesn't exist for all $ t$. Please, post the correct statement of the problem.

The statement is the one above.I had also thought of what you pointed out .because if $ t$ was a constant then the nature of the series $ s_n$ wouldn't be affected by the value of $ t$.My guess would be that $ t$ should be determined in terms of $ n$ ? I'm not sure.That's why I posted. I thought that I was missing something.

So, let's state it as "find a good asymptotic formula for $ \sum_{k \equal{} 1}^n \frac {a_k}{k}$"
(If we can do that, we'll probably be able to figure out what was meant; as written, it is a complete nonsense).


superior algebra
superior algebra unsolved

Let $ G$ be finite group and $ p\mid |G|$, $ p$ prime. The map $ \phi(x) \equal{} x^p$ is a homomorphism. Prove that $ G$ has a nontrivial center.
Some facts that i have already found:
1) G has a unique(and hence normal) p-Sylow subgroup.
2) If $ N \equal{} \{x^{p^n}, \ x\in G\}$, $ p^n$ - order of p-Sylow, then $ G \equal{} PN$.
3) $ x^{p \minus{} 1}y^p \equal{} y^p x^{p \minus{} 1}\ \forall x,y\in G$.

hint 1: [hide]how does $ \phi$ behave on $ N$?[/hide] hint 2: [hide]how does the center of $ P$ behave with respect to $ N$?[/hide]

Oh i forgot that $ P$ must have a non-trivial center..thanks for reminding me [b]ma-go[/b]  :blush:

So, i finished my solution by proving that $ Z(P)$ is not trivial and then taking $ t\in Z(P): \ t\equal{}y^{p\minus{}1}$.
Ma-go, can you post your solution? I didn't understand what you mean.

he probably pointed you to the following simple facts using (3) and (2):

Let $ z\in Z(P),h\in N$ then (2) gives $ h\equal{}g^{p^n}$ for some $ g\in G$ and because of $ n>0$ we can set $ x\equal{}g^{p^{n\minus{}1}}$, then from (3) $ z^{p\minus{}1}x^p\equal{}x^pz^{p\minus{}1}\Longleftrightarrow z^{p\minus{}1}h\equal{}hz^{p\minus{}1}$, i.e. $ z^{p\minus{}1}$ commutes with $ P$ and $ N$ (the latter since $ h\in N$ was arbitrary). Applying (2) again, we see that $ z^{p\minus{}1}$ commutes with $ G$, i.e. $ 1\not\equal{}z^{p\minus{}1}\in Z(G)$.

P.S. As you have already pointed out $ (z\mapsto z^{p\minus{}1})\in Aut(Z(P))$, ie. we actually have $ Z(P)\subseteq Z(G)$


geometry
inequalities
combinatorics solved
combinatorics

Let [tex]A_i[/tex], [tex]1 \leq i \leq 5[/tex], be five regions on the plane, such that

(1) [tex]\left|{\bigcup_{i=1}^5 A_i}\right| \leq 1[/tex]

(2) [tex]|A_i| \geq 1/2[/tex] for each i.

Show that there exists two distinct indices i, j, such that [tex]|A_i \cap A_j| \geq 1/5[/tex].

(| | denotes area)

I think this is a real simple problem.
The only thing you must know to solve it is the formula of inclusion-exclusion.

The solution looks like :

1>=|UA(1=<i=<5)|>=|A1|+|A2|+|A3|+|A4|+|A5| - (sum over i, j  1<=i<j<=5 |Ai intersect Aj|) + {intersection of taken three of subsets and so on} >=|A1|+|A2|+|A3|+|A4|+|A5| - (sum over i, j  1<=i<j<=5 |Ai intersect Aj|) <=>
(sum over i, j  1<=i<j<=5 |Ai intersect Aj|) >= 3/2 <=> there exists
such i and j for what |Ai intersect Aj|>=3/10>1/5.
and we are done.

[quote="vetali"]
(sum over i, j  1<=i<j<=5 |Ai intersect Aj|) >= 3/2 < = > there exists
such i and j for what |Ai intersect Aj|>=3/10>1/5.
and we are done.[/quote]

But (sum over i, j  1<=i<j<=5 |Ai intersect Aj|) has $\binom{5}{2}=10$ summands in it. Which only gets us to the existance of i, j, with $|A_i \cap A_j| \geq \frac{3}{20}$, which doesn't help.

Vetali's solution is correct...
He just used an inequality that comes from the inclusion-exclusion principle...
I've talked with him about this problem today and his solution is very reasonable...
He said it would be possible to prove that 1/5 is the best aproximative value for this problem.. I don't want to interfere so that he will explain the solution again himself... :):):)

______________________________
"Our days are never coming back..." (Highway Song-> System of a down)

Sorry ,  but I havn,t the chance to write the solution until now.

So let's go on.

A(i,j) - region that is the intersection of A(i) and A(j)

So by the principle of inclusion and exclusion we get that
 A - Sum(A(i), 1<=i<=5) + Sum(A(i, j), 1<=i<j<=5)-Sum(A(i, j, k), 1<=i<j<k<=5) + Sum(A(i, j, k, l), 1<=i<j<k<l<=5)-A(1,2,3,4,5)>=0;

A it is the "piece of cloth" of area 1 on which we put another litle pieces of cloth -   A(i)'s.

Aplying the same reasoning for the A(1) covered (partially or not ) by A(1,2),       A(1, 3), A(1, 4), A(1, 5)
 we get that :
A(1)-Sum(A(1, i), 2<=i<=5)+Sum(A(1,i,j), 2<=i<j<=5) -                                    -Sum(A(1,i,j,k), 1<=i<j<k<=5)+A(1,2,3,4,5)>=0

obtain the same relations for A(2), ..., A(5) summing them get :
Sum(A(i), 1<=i<=5) - 2*Sum(A(i, j), 1<=i<j<=5)+3*Sum(A(i, j, k), 1<=i<j<k<=5) - 4*Sum(A(i, j, k, l), 1<=i<j<k<l<=5)+5*A(1,2,3,4,5)>=0;
multiply this relation with 1/3 and add with the first one we get :
A- 2/3*Sum(A(i), 1<=i<=5) + 1/3Sum(A(i, j), 1<=i<j<=5)-1/3* Sum(A(i, j, k, l), 1<=i<j<k<l<=5)+ 2/3*A(1,2,3,4,5)>=0;
but as A(1,2,3,4,5) appears in each A(i, j, k ,l) so Sum(A(i, j, k, l), 1<=i<j<k<l<=5)>= 5*A(1,2,3,4,5) so -1/3* Sum(A(i, j, k, l), 1<=i<j<k<l<=5)+ 2/3*A(1,2,3,4,5)<=0

and we finally get that Sum(A(i, j), 1<=i<j<=5)>=2*Sum(A(i), 1<=i<=5)- 3*A>=
2(5*1/2)-3=2;
whece the conclusion follows.

I have found an example provin g that this aproximation can not be more improved :
an unit square :
    12   13    14    15      23
    24   25    34    35       45
    123 124 125   134     135
    145 234 235   245    345
    where the numbers indicate region. ex 12-region of A1 intersect A2

so we are done.


Given a right triangle with integer leg lengths $a,\ b$. Find the values of $a,\ b$ such that the inradius of the triangle is 5.

[hide="almost a solution"]
Any right triangle with integer side lengths can have these lengths expressed as $k(m^{2}-n^{2})$, $2mnk$, and $k(m^{2}+n^{2})$, where $k$, $m$ and $n$ are positive integers and $m>n$. So:
\begin{eqnarray*}5(\frac{k(m^{2}-n^{2}+2mn+m^{2}+n^{2})}{2})&=&\frac{k^{2}(m^{2}-n^{2})(2mn)}{2}\\ 10m(m+n)&=&k(m^{2}-n^{2})(2mn)\\ 5(m+n)&=&k(n)(m-n)(m+n)\\ 5\cdot1\cdot1&=&k(n)(m-n) \end{eqnarray*}
If $k=5$, $n=1$ and $m=2$, so $(15,20)$ is a solution, and if $k=1$, then $n=1$ and $m=6$, so $(12,35)$ is also a solution.[/hide]

edit: fixed hide tag
edit2: changed "solution" to "almost a solution" and edited solution slightly, but I am still having trouble with the integral hypotenuse, which may be why I only get two solutions.

Actually, the problem doesn't say that the hypotenuse is an integer, so you can't immediately assume that they are of that form (although you can prove that this must be the case). Also, you don't know that the triangle is primitive, so you have to scale all of them by $k$. This doesn't end up mattering in the end either, but you should be careful about making such assumptions.

[hide]
It can be proven that the inradius of a right triangle with legs $a$ and $b$, and hypotenuse $c$, is $\frac{a+b-c}{2}$.

Thus we have $a+b-\sqrt{a^{2}+b^{2}}=10$.

After some simplification, we get

$ab-10a-10b+50=0\implies (a-10)(b-10)=50$.

This yields the solutions $(11, 60)$, $(12, 35)$, and $(15, 20)$. [/hide]

That's right. :)

The inradius can be also expressed as $\frac{ab}{a+b+\sqrt{a^{2}+b^{2}}},$ but in this case rnwang2's result for the inradius is important.


Here's a creation of my own.  I'll be interested to see if there are any short easy solutions I overlooked.  Mine isn't too ugly but it's kind of drawn out.
If $a+b+c=1$
Prove $\displaystyle \frac{a}{1+\frac{1}{b}+\frac{1}{c}}+\frac{b}{1+\frac{1}{c}+\frac{1}{a}}+\frac{c}{1+\frac{1}{a}+\frac{1}{b}}\leq\frac{1}{7}$

Well lets see :)
$a^2+b^2+c^2 \geq \frac{1}{3}$ and we know that $\frac{1}{b}+\frac{1}{c} \geq \frac{4}{b+c}$ So we
have that $LHS \leq \sum\frac{a}{1+\frac{4}{b+c}}=\sum \frac{a(b+c)}{4+b+c}=\sum \frac{a(1-a)}{5-a}
=1-\sum \frac{4a}{5-a}$ so we have to prove that $\sum \frac{4a}{5-a}\geq \frac{6}{7}$
Multiply all by $\sum 4a(5-a)$ and using Cauchy Buniakovski get that
$(4a+4b+4c)^2\geq\frac{6}{7}(20-4(a^2+b^2+c^2))$ or
$16\geq\frac{6}{7}(20-\frac{4}{3})$ which is true so this solves the ineq.

I got so messed up with the signs :) I edited though ;)


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ k$ be a field such that $ \mathbb{C}$ has infinite transcendence degree over $ k$.
If $ L$ is an extension field of $ k$ of finite transcendence degree, then $ L$ is isomorphic to a subfield of $ \mathbb{C}$.
Why??

Think: How would you construct $ L$ within $ \mathbb{C}$?

[hide]Suppose $ L$ has transcendence degree $ n$ over $ k$. Choose $ x_1,\cdots,x_n \in \mathbb{C}$ that are algebraically independent over $ k$. Then $ L$ can be realized as an algebraic extension of $ k(x_1,\cdots,x_n)$. This can be embedded in $ \mathbb{C}$ since every polynomial with coefficients in $ \mathbb{C}$ has a root, by the Fundamental Theorem of Algebra.

[hide="More specifically"]Choose a basis $ \{e_\alpha\}_{\alpha \in I}$ for an index set $ I$ for $ L$ over $ k(x_1,\cdots,x_n)$. For each $ e_\alpha$, let $ p_{\alpha}$ be its minimal polynomial. All the roots of each $ p_{\alpha}$ lie in $ \mathbb{C}$ by the Fundamental Theorem of Algebra, so we can look at the field $ M$ generated by all the roots of all these polynomials. Then $ L$ may be embedded in $ M$.[/hide][/hide]


Let $c$ be a fixed positive integer, and let $a_{1},a_{2},...$ be a sequence of positive integers such that $a_{n}< a_{n+1}< a_{n}+c$ for every positive integer $n$. Let $s$ denote the infinite string of digits obtained by writing the terms in the sequence consecutively from left to right, starting from the first term. For every positive integer $k$, let $s_{k}$ denote the number whose decimal representation is identical to the $k$ most left digits of $s$. Prove that for every positive integer $m$ there exists a positive integer $k$ such that $s_{k}$ is divisible by $m.$

Pick $k$ such that $c,m<10^{k}$.  For $r=1,\ldots,m-1$, pick a $k$-digit number $n_{r}\equiv-10^{rk}\mod m$, and let

$N=\sum_{i=1}^{m-1}n_{i}10^{(m-i)k}=\overline{n_{1}n_{2}\ldots n_{m-1}}\cdot 10^{k}$.

Now pick $n$ such that $a_{n-1}< N\leq a_{n}< a_{n-1}+c < N+10^{k}$.  Let $s$ be the string of digits up through the last digit of $a_{n-1}$.

If $s\equiv r \mod m$ for some $1\leq r<m$, then $s$, together with the first $rk$ digits of $a_{n}$, is

$10^{rk}s+10^{(r-1)k}n_{1}+10^{(r-2)k}n_{2}+\ldots+n_{r}\equiv 0 \mod m$.


calculus
derivative
function
calculus computations

$ f(x) \equal{} f_1(x) 1_{(\minus{}\infty, \minus{}2]}\plus{}f_2(x)1_{(\minus{}2,0)}\plus{}f_3(x)1_{[0,\infty)}$

$ f_1(x)\equal{}x^2\plus{}10x\plus{}8$, $ f_2(x)\equal{}ax^2\plus{}bx\plus{}c$, $ f_3(x)\equal{}x^2\plus{}2x$

$ f_1'(x)\equal{}2x\plus{}10$, $ f_2'(x)\equal{}2ax\plus{}b$, $ f_3'(x)\equal{}2x\plus{}2$

let $ y\equal{}mx\plus{}d$ be our tangent line that touches at three places..

we notice since each derivative is linear, and therefore 1-to-1, we must touch each function once.

$ m\equal{}2x_1\plus{}10\equal{}2ax_2\plus{}b\equal{}2x_3\plus{}2$ so $ x_3\minus{}x_1\equal{}4$

also $ mx_i\plus{}d\equal{}f_i(x_i)$, so $ 4m\equal{}m(x_3\minus{}x_1)\equal{}f_3(x_3)\minus{}f_1(x_1)\equal{}x_3^2\plus{}2x_3\minus{}x_1^2\minus{}10x_1\minus{}8$

lets write m now as $ m\equal{}\left (\frac{x_3}{2}\right)^2\minus{}\left(\frac{x_1}{2}\right)^2\plus{}\frac{x_3}{2}\minus{}5\frac{x_1}{2}\minus{}2\equal{}$

$ \left(\frac{x_3}{2}\minus{}\frac{x_1}{2}\right)\left(\frac{x_3}{2}\plus{}\frac{x_1}{2}\right)\plus{}\frac{x_3}{2}\minus{}5\frac{x_1}{2}\minus{}2\equal{}$

$ x_3\plus{}x_1\plus{}\frac{x_3}{2}\minus{}5\frac{x_1}{2}\minus{}2\equal{}\frac{3x_3}{2}\minus{}\frac{3x_1}{2}\minus{}2\equal{}4$

so $ x_1\equal{}\minus{}3$, $ x_3\equal{}1$, $ f_1(x_1)\equal{}\minus{}13$, $ f_3(x_3)\equal{}3$, $ d\equal{}\minus{}1$

now we just have to solve for $ f_2(x_2)\equal{}4x_2\minus{}1$, and $ f_2'(x_2)\equal{}4$

$ 4\equal{}2ax_2\plus{}b \to x_2\equal{}\frac{4\minus{}b}{2a}\in (\minus{}2,0)$,
$ ax_2^2\plus{}bx_2\plus{}c\equal{}4x_2\minus{}1 \to a(\frac{4\minus{}b}{2a})^2\plus{}(b\minus{}4)\frac{4\minus{}b}{2a}\plus{}c\plus{}1\equal{}0$

$ \to c\plus{}1\equal{}\frac{(4\minus{}b)^2}{4a} \to 4a(c\plus{}1)\equal{}(4\minus{}b)^2$

now we use continuity to set the endpoints equal

$ f_2(0)\equal{}f_3(0)\equal{}0 \to c\equal{}0$
$ f_1(\minus{}2)\equal{}f_2(\minus{}2) \to \minus{}8\equal{}4a\minus{}2b \to 2a\equal{}b\minus{}4$

so we get $ 4a\equal{}4a^2 \to a\equal{}1 \to b\equal{}6$

so$ f_2(x)\equal{}x^2\plus{}6x$


Can someone help me simplify:

$\frac{ \sin 54}{2\sin^{2}72+\sin 54}$

[hide="A once heavy solution on LaTeX but not anymore since my TeX crashed :( "]Use the half-angle theorem and the subtraction theorem.

and start with the fact that $x=36^\circ$.

\begin{eqnarray*}\frac{\sin 54^\circ}{2\sin^{2}72^\circ+\sin 54^\circ}&=&\frac{\cos x}{2\sin^{2}2x+\cos x}\\\\&=&\frac{\cos x}{8\sin^{2}x \cos^{2}x+\cos x}\\\\&=&\frac{1}{8\sin^{2}x \cos x+1}\\\\&=&\frac{1}{8\sin^{2}36^\circ\cos 36^\circ+1}\end{eqnarray*}

So, then, use the facts derivable from other calculations that I am not posting here because my latex doesn't work...

Anyway, the final result is that $\cos 36^\circ=\frac{1+\sqrt{5}}{4}$. Also, a varied version of $\sin 36^\circ$ is that $\sin 36^\circ=\sqrt{\frac{5-\sqrt{5}}{8}}$, so that means that $\sin^{2}36^\circ=\sqrt{5-\sqrt{5}}{8}$.

Then, the other part.

\begin{eqnarray*}\frac{1}{8\frac{\sqrt{5}}{8}+1}&=&\frac{1}{\sqrt{5}+1}\\\\&=&\boxed{\frac{\sqrt{5}-1}{4}}\end{eqnarray*}[/hide]

[hide="hint"]this expression is equal to sin 18[/hide]

hmm i was looking for a way to show that that simplifies down to $\sin 18$ without having to use anything like $\cos 36 = \frac{1+\sqrt{5}}{4}$


geometry
perimeter
rectangle
geometry proposed

prove that given a convex polygon of perimeter $ P $ and area $ S $, one can place inside of it a circle of radius $ \frac SP $

This has been posted and solved before.

[Moderator edit: indeed: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=51137 ]

hmm.  i guessed that it might have been.  i just couldn't find it...

A famous problem from old USSR olympiad.
Construct a rectangle of height $S/P$ on each side of the polygon. All these rectangles cannot cover the whole polygon...


calculus
integration
algebra unsolved
algebra

solve equation  with   number positive intergal :
$ a \plus{} 2b \plus{} 5c \plus{} 10d \plus{} 20e \equal{} 300$
  can you help me ?
  or  question  only how many root ?
who  can write languge  Pascal  or C++  for this problem


real analysis
calculus
integration
function
Functional Analysis
advanced fields
advanced fields unsolved

Let m_k be a sequence of absolutely continuous with respect to n-dimensional Lebesgue measure L^n,radon  measures on R^n such that  m_k(A)= integral on A of f_k dL^n for every borel set A.
Show that if 1 <p <=oo and sup{ ||f_k||p : k natural number} <oo and if m_k-->m weakly then m <<L^n.
Show tha the above is false if p=1.[/list]

Let's try this in $\text{\LaTeX}.$ Tell me if anything is wrong. The problem is (I think):

Let $m_k$ be a sequence of absolutely continuous Radon measures on $\mathbb{R}^n$ such that $m_k(A)=\int\limits_Af_k dx$ for every Borel set $A.$

(Note: the absolute continuity is with respect to $n$-dimensional Lebesgue measure, and the "$dx$" in that integral refers to $n$-dimensional Lebesgue measure.)

Show that if $1<p<\infty$ and $\sup\{||f_k||_p: k\in\mathbb{N}\}<\infty$ and $m_k\to m$ weakly, then $m$ is absolutely continuous with respect to Lebesgue measure.

Show that the above is false for $p=1.$

That it's false for $p=1$ is easy.

Let $B_n$ be the ball of radius $\dfrac1n$ centered at the origin.

Let $f_n(x)=\dfrac1{m(B_n)}\chi_{B_n}(x)$

These are uniformly bounded in the $L^1$ norm, but they converge weakly as measures to the Dirac delta (the point mass at the origin), which is not absolutely continous.

I don't have time to post the main answer, but as a hint: Consider $L^q,$ where $q$ is the dual exponent to $p.$

I'm not particularly happy about the following argument. I think there's a better argument out there somewhere, but here goes anyway.

We have that $m_k\to m$ weakly as measures, and that $\int\varphi\, dm_k=\int\varphi\cdot f_k\,dx$ with $||f_k||_p$ bounded.

$L^p$ is the dual space of $L^q.$ A closed ball in a dual Banach space is compact in the weak* topology. Hence there is a subsequence $f_{k_j}$ that converges weakly to $f\in L^p.$ That is, for all $g\in L^q$, $\int f_{k_j}\cdot g\,dx\to\int f\cdot g\mdx.$

Now suppose $\varphi\in C_0\cap L^q.$ (Such functions are dense in both $C_0$ and $L^q.$)

On the one hand, $\int\varphi\,dm_{k_j}\to\int\varphi\,dm.$

On the other hand, $\int\varphi\cdot f_{k_j}\,dx\to\int\varphi\cdot f\,dx.$

Thus, $\int\varphi\,dm=\int\varphi\cdot f\,dx.$ By the density of our test functions, that determines $m$ as the absolutely continuous measure given by integration with the locally integrable function $f.$

The argument doesn't work for $p=1$ because $L^1$ is not a dual space.


Maybe I am wrong, but I saw a similar bloody inequalities. I solved one of them here. But the solution was rather ugly, it used calculus.

it isn't the USA Mo,
we have $a^5-a^2+3 \ geq a^3+2$
and then $\cdots$ :)

no, I think this one has a better solution that calculus.
May be we should try to minorate the Left side of the inequality. :lol:

This has already been discussed and several solutions were given after the USAMO. I really like this problem, especially because the solution is so smart.

The problem involved a very hard to see (although trivial) first step ($a^5-a^2+3 \geq a^3+2$). Most people who saw this step probably succeeded to solve the problem, while those who didn't would have had a really hard time.

Oh, really! Great! WoW!
We can write $(a^3+1+1)(1+b^3+1)(1+1+c^3)\geq (a\cdot 1\cdot 1+1\cdot b\cdot 1+1\cdot 1\cdot c)^3=(a+b+c)^3$ by Holder!

[quote="Reyes"]My problem :$a,b,c,k>0$ then :
$(a^3+k)(b^3+k)(c^3+k)\geq (a+b+c+k-2)^3$.[/quote] The direct application of the same idea with Holder (see my post above).

[quote="erdos"]Let $a$, $b$ and $c$ be positive real numbers. Prove that :
$\left( a^{5}-a^{2}+3\right) \left( b^{5}-b^{2}+3\right) \left( c^{5}-c^{2}+3\right) \geq \left( a+b+c\right) ^{3}$[/quote]

Although this has been solved by a lot of different methods now, I am posting the links:

http://www.mathlinks.ro/Forum/viewtopic.php?t=24286
http://www.mathlinks.ro/Forum/viewtopic.php?t=5383
http://www.mathlinks.ro/Forum/viewtopic.php?t=5726
http://www.mathlinks.ro/Forum/viewtopic.php?t=12202

  darij


we can solve this problem just like i solved it on http://www.artofproblemsolving.com/Forum/viewtopic.php?t=149358

this is more difficult.

but after we have all point's is easy to calculate the area. the area is just the determinat of the points dividided by $2$.

so i will not make all the calculations, because it is need a lot of time :(

sorry

I am trying on a geometric solution.I don't have answer.:)


Well as Valentin pointed out, it is not really american geography :D but, here goes (quiz courtesy of Soroban, hope he doesnt mind)
Hide your answers if you are going to reply.  Thanks.  Yeah, please dont use a map or google, it spoils the fun :D  if you dont know, just answer dont know or just guess

1. what is the first non-us country directly south of detroit?
2. Of the original 48 states, which is the furthest north?

3. Is the capital of Kentucky prounounced LOO-IS-ville or LOO-EE-ville?

4. If you go directly south from Tallahassee, Florida,
what is the first South American country you will enter?

5. Going through the Panama Canal from the Atlantic Ocean
to the Pacific Ocean, how many miles west will you travel?

i didnt look up anything so ill prob be wrong on some

[hide]
1.Domican Republic
2.Maine
3.Loo-is-ville
4. Surinam
5. 2[/hide]

[hide]
1) A guess: Panama
2) Maine
3)LOO- EE-ville
4) venezuela
5)Very few, the canal is almost completely N-S[/hide]

1. [hide]Cuba... Dominican Republic is further East (I lived there)[/hide]

I never get tired of hearing that number 3...   [hide]Neither.  It's pronounced "Frankfort". :wacko:  :D [/hide]

I might as well get into this:

[hide]1. Canada (Windsor, Ontario)
2. Minnesota (the "Northwest Angle" in the Lake of the Woods)
3. well, I had relatives who lived there a couple of decades ago, and I remember it as approximately LOO-uh-vuhl with no stress on either of the last two syllables.

[Edit - oops, forgot to read the question carefully. Yeah, Frankfort.]

4. The Atlantic-to-Pacific direction of the Panama Canal is southeast; the "number of miles to the west" would be a negative number.
5. The line straight south from Tallahassee passes through the Pacific Ocean west of the mainland of South America.[/hide]

[hide]1. No clue, America? 2. Michigan! 3. Latter(<-sound smart there) 4. Venezuelaish 5. log(tan(don't know any more functions(12)))[/hide]

I got a really mean one for all of ya :) What's the capital of Romania's district Prahova? :D :D :D

[quote="DPatrick"][quote="bugzpodder"]hehe  :D[/quote]You're mean!  :)

I've got one for you: without looking at a map, is the captial of New Brunswick St. John's or Saint John?[/quote]

i am pretti sure St. John's in in Newfoundland and Labourdor.  I'll take this as a trick question so I'll say neither :P

oops, stupid me
i know capital is frankfurt
didnt pay attention lol,  :blush:
[hide]
how is canda south of detroit?
[/hide]

[hide]
Windsor, ON is directly below Detroit.
[/hide]

[quote="bugzpodder"][quote="DPatrick"]
I've got one for you: without looking at a map, is the captial of New Brunswick St. John's or Saint John?[/quote]

i am pretti sure St. John's in in Newfoundland and Labourdor.  I'll take this as a trick question so I'll say neither :P[/quote]Fredericton is the capital of New Brunswick (one of my best friends lives there), though Saint John is NB's largest city.

BTW, the drive from Fredericton to Saint John is amusing: most the drive is smack through the middle of a military base (yes, Canada has a military) for about 50 km, and there are all these scary-looking warning signs not to leave the main road because you might get shot or bombed or something.

[quote="DPatrick"][quote="bugzpodder"]hehe  :D[/quote]You're mean!  :)

I've got one for you: without looking at a map, is the captial of New Brunswick St. John's or Saint John?[/quote]

Frederickton(sp?), if you're looking for the city in New Brunswick, then it's Saint John

An old classic question (No tricks but answer may surprize some):
In  US which state is
A:   North most
B  South  most
C, East most
D West most

If you have not heard it before try to answer it without looking at a  map...

[quote="themonster"][hide]
A: Minnesota, but I just saw this on this thread
B: Hawaii
C: Maine
D: Alaska
[/hide][/quote]

Two right two wrong. :?

[hide]I think C and D could both be Hawaii depending on how you look at it...[/hide]

[hide]
NOrth=Alaska
South=Hawaii
West=Alaska
East=Alaska[/hide]


Hi!

I have a question: suppose you have two square symmetric, positive definite matrices A and B.

When can I say that the dominant eigenvalue of A is bigger than the dominant eigenvalue of B?

If needed, I can assume they share the same diagonal.

bye!
Alessio

if the maximum of $\frac{x^T A x}{x^T x}$ for all nonzero x is bigger than
 $\frac{x^T Bx}{x^T x}$ for all nonzero x

is that something you might be looking for?

Well, actually I am looking for a condition one can easily check...

I have one: let $|A_i|$ be the sum of the absolute values of the $i$-th row of $A$. Let $A$ be symmetric and positive definite.

Each eigenvalue of A, and in particular the largest $\lambda_1$, is such that:
$\min_i |A_i| \leq \lambda_1 \leq \max_i |A_i|$.

So if  $\min_i |A_i| \geq \max_i |B_i|$, the largest eigenvalue of $A$ needs to be bigger than the largest eigenvalue of $B$.  Still this condition is very very conservative, since it is true that smallest eigenvalue of $A$ is bigger than the largest of $B$...

I was kind of hoping for a sharper condition... but thanks for the reply :)

Alessio


When you watch a video and submit a comment, you go to a page where there is no link to go back. You will have to do it with your browser, or click "Get Problem" and navigate through the tabs. It is such a small thing, it is a nice feature that could be added to Alcumus. My question is, why isn't there one of those buttons?

Maybe they think that with the browsers' built in "Back" button, a second button with the same outcome would be redundant?

Yeah, but with some browsers (Google Chrome) the back button will redo your comment too.

You have to click the back button twice, and if you accidently click the play button, you lose 5 karma. I am just saying it would be nice feature. The same thing for subject reports too, there is no link back to the complete report.

Yeah that is true..
I've always thought about that..

But when you submit your comment, shouldn't it immediately be stored on the server? So clicking the back button won't matter?

When I post a comment it stays on the video screen. No back button needed.


I want to install Linux but also leave Windows XP on my computer. I want to partition my hard drive so that I do not lose my data. I found several freeware programs, such as fips, that can do this, but only with FAT16 or FAT32 file systems, not the NTFS which Windows XP uses. Some web sites recommend PartitionMagic, but this program is not free. Do you know any free programs which can resize NTFS partitions?

I believe there is a tool that comes with windows. WinDiskFS? I'm not sure the name, and I dont remember if it does NTFS or not, but u should take a look into it.

I had the same situation and I misplaced my PartitionMagic disc.

Try making a floppy that contains ntfsresize.

(http://mlf.linux.rulez.org/mlf/ezaz/ntfsresize.html)
I used BG-Rescue because it was the fastest download.

Follow the instructions closely (at the ntfsresize site, if you are using the command line).

Then (in Windows XP), go to Control Panel>Administrative Tools>Computer Management.

On the left hand side, click Disk management.  Click on your the unpartitioned space (gray) on the drive.  Click Create Partition.

It worked for me 2 weeks ago installing linux :)

[quote="rookworm"]several distributions include this in their installer, such as Suse and (I believe) Fedora, and several commercial distros.[/quote]

Yes.  My second thought after mispacing my PartitionMagic CD was to go find my old Fedora 3 cd to partition the hard drive.  But I then realized that I misplaced that too... :huh:

I have Debian, which comes with a partitioning tool (I forget which one) but I read that when you use it you lose all your data. I will look into the programs you suggested.

If nothing works, you can always burn important files on a DVD and then reinstall XP and Linux...which i'm sure you don't want to do because its a big waste of time

[quote="rookworm"][quote="jmadsen"]The other thing I don't know is, how big should I make each partiton? I plan on using 4 partitions: 1 for Linux, 1 for Windows, one for documents (shared), and a swap space. I don't know what needs to go on the partitions for each OS.[/quote]

The swap partition should be twice as large in your ram, generally; it should also be the first partition, so that it may be accessed more quickly. Give Linux at least 10 Gbs for all the programs you might want to try out, and give the lion's share of the space to a FAT32 partition for shared documents between the two OSs; mount this as /home/ on Linux. The amount you give to Windows really depends on how many programs you wish to install.[/quote]

Well, my ram is 777,712 K  :P

I'm not that skilled at Linux so I just gave Linux 20G, and 60G for Windows XP.

Do you think 10 GB is enough? I'm already using ~10 GB for Windows.

Wait...when I right-click "C:" and choose "Properties" it says that I am using 13.7 GB (about 2 GB of which is documents which would end up in the shared partition). But when I highlight all the files and folders in C:, right click and choose "Properties," it says that the total file size is 10.8 GB. I have "show hidden files" turned on. Why is there such a large discrepancy?

[quote="jmadsen"]*bump*

also should i use fat32 for the shared partition?[/quote]

Well, yeah, I suppose you should.  Linux isn't really known for good NTFS support (corrupted drives, anyone?), though I've heard that support recently has become more stable...and Windows doen't do very well at supporting file systems that Linux might use...

FAT is a standard, and isn't as priporatory as NTFS.

In short, yes.

[quote="jmadsen"]Okay, I downloaded BG-Rescue Linux and wrote the image to floppies. When I boot the floppy it takes forever to load and finally says "insert root floppy." Now, I have two floppies, so I assume this means to put the other one in. But when I do that, there is an error.[/quote]

Ah.  It took me 3 reboots to get this right.  Put in the first disk in and boot.  Dont thuch anything after you select your resolution.  When it stops for about 20 seconds, put the 2nd disk in.  Press enter.  Then it should start uncompressing the disk and all that stuff.  Hope you got the install from the command prompt page handy.

Disclaimer: I am not liable for your actions :)

After I select the resolution, it says "Loading...", then after a while when it is finished loading a bunch of stuff scrolls by. When it stops scrolling it says
[code]/dev/scsi/host0/bus0/target0/lum0: p1 p2 p3 p4[/code]
then it seems to wait for me to enter something. But I don't know what to type, and it doesn't seem to matter anyway. Then it says
[code]VFS: Insert root floppy and press ENTER[/code]
so I put in the other floppy and press enter and it says
[code]VFS: Cannot open root device "fd0" or 02:00
Please append a correct "root=" boot option
Kernel panic: VFS: Unable to mount root fs on 02:00[/code]

Sorry if you already said what to do, but I don't understand.


$1)$  In how many ways can the number $199$ be written as the sum of distinct Fibonacci numbers?
$2)$  What about $200$?
$3)$  Describe all positive integers that can be written as the sum of distinct Fibonacci numbers.
$4)$  Describe all positive integers that can be written as the difference of two Fibonacci numbers.

(Say $0$ is not a Fibonacci number and $1$ is twice.)

Fibonacci.  Also, you should make it clear where you think the Fibonacci numbers start, because the way I use them, 0 is a Fibonacci number and 1 is a Fibonacci number twice.

[quote="miyomiyo"]$1)$  In how many ways can the number $199$ be written as the sum of distinct Fibonacci numbers?[/quote][hide="Hint"]$199 = 144+55$[/hide]

JBL, I think his point was that you shouldn't be like, $1=1+0=1+0+0=1+0+0+0=...$ and say there are an infinite number of ways of doing it, but that you can use the number $1$ up to two times.

For number 1:
[hide]
First write out all the numbers we can potentially use:
$1,1,2,3,5,8,13,21,34,55,89,144$
Note that $199=144+55=(89+55)+(34+21)$

Note that $\sum_{j=1}^{n}F_{j}= F_{n+2}-1$, so $1+1+2+...+55=144-1=143$, so this sum cannot be expressed without a number $\geq 89$.

Also, $1+1+2+...+34+89=89-1+89=177$, so if we split again, the $55$ is also required. This leaves us to make another $55$.

$1+1+2+3+...+21=55-1$, so if we split, the $34$ is also required, leaving another $21$ to make. $21=13+8$. $1+1+2+3+5+8=20$, so the $13$ needs to be there if we want to split again.

etc.

We thus have:
$199=144+55=89+55+34+21=89+55+34+13+8=89+55+34+13+5+3=89+55+34+13+5+2+1$

5 ways.
[/hide]

So for number 2, we experience the same limitations, and find that the breakdowns are the same, except there is an additional $1$ appended to each.

For number 3:
[hide="Procedure"]
Fibonacci numbers larger than two are always less than twice the previos Fibonacci number. Hence, starting from any given integer $n$, if the greatest Fibonacci number not exceeding that number is $F_{k}$, then $0 \leq n-F_{k}< F_{k}$. Each time, this procedure is repeated, the resulting number is between $0$ (inclusive) and the original number (exclusive). This leaves a finite number of options. However, if the number is not yet at 0, it must decrease on the next iteration of the procedure. By Pigeonhole, this procedure must reach 0; that is, it must be possible to write the number as the sum of distinct Fibonacci numbers.
[/hide]


Prove that for natural numbers $ p>3$ , if $ p$  and $ 2p\plus{}1$  are prime numbers than $ 4p\plus{}1$ is not a prime number.

[quote="ridgers"]Prove that for natural numbers $ p > 3$ , if $ p$  and $ 2p \plus{} 1$  are prime numbers than $ 4p \plus{} 1$ is not a prime number.[/quote]
Divisibility with $ 3$.. If $ p \equiv 0,1 \bmod 3$ then either $ 3 \mid p$ or $ 3 \mid 2p\plus{}1$. Hence $ p \equiv 2 \mod 3$ and $ 3 \mid 4p\plus{}1$

[quote="ridgers"]Prove that for natural numbers $ p > 3$ , if $ p$  and $ 2p \plus{} 1$  are prime numbers than $ 4p \plus{} 1$ is not a prime number.[/quote]
Note that $ 4p \plus{} 1 \equal{} (2p \plus{} 1)^2 \minus{} 4p^2$.Since $ p$ and $ 2p \plus{} 1$ are primes,we have $ (2p \plus{} 1)^2 \equiv 1\mod 3$ and $ 4p^2 \equiv 1\mod 3 \Rightarrow 3\mid 4p \plus{} 1 \Rightarrow$ Q.E.D  .


Find all functions f:Q->R with the property:
f(x+y)=f(x)+f(y)+3xy(x+y+6) -8 for every x from Q and f(-8)=0.


group theory
superior algebra
superior algebra unsolved

[quote="Mathstunned"]Please explain my answere.. How many homomorphisms are there of Z onto Z? There are 2. Why though?? help?

Note the following fact:

if G is a cyclic group with generator g, and if f : G -> H is an onto group homomorphism, then f(g) is a generator for H.

Can you prove this? If so, then you can use it to answer your question. That is, if

f : Z -> Z

is an onto homomorphism, then since 1 is a generator of Z, then f(1) also has to be a generator of Z. What are the only two generators of Z?

This is all I have, but I need help with the generators!  Thanks!!![/quote]

The generators of $ \mathbb{Z}$ are $ 1$ and $ \minus{}1$...

[quote="Mathstunned"][quote]The generators are 1 and -1[/quote]

How do you come up with the generators though??[/quote]

Well, $ \minus{} 1 \equal{} \minus{}(1)$ I mean is the inverse of 1 in $ \mathbb{Z}$

Every multiple of a number is multiple of the inverse of that number so every number is a multiple of $ \minus{} 1$


Is there a formula to find the area if you know and only know the medians?

srry bad wording
 given nothing but the length of the mdians
if that's any better

yes, but it is messy, first be able to solve given the medians, find the sides, then use herons

use the median forumlas, $2a^2+2b^2=c^2+4m_c^2$, etc to solve for the sides in terms of the medians, then plug them into the form of heron with squares, $16A^2=2\sum_{cyc}a^2b^2-\sum_{cyc}a^4$ and simplify, because of the simplicity of the forumla, there should be a geometric interpretation too, perhaps using the medians dividing the triangle into 6 triangles with equal area

[quote="Elemennop"]The area of a triangle is $4/3$rds the area of the triangle formed by it's medians. Thus, just find the area of the triangle formed by the medians and multiply accordingly.

Ah, now the hard part, is proving this fact.[/quote]

Proving this is not at all tough.
Let AD be the median in triangle ABC. O be the centeroid.
Extend AD to H such that AO = HO. And we are done :-)
As OBD and HCD are congruent.
We can express each side in HCD as 2/3rd of the triangle sides.

[quote="frt"]Aarav, I understand your proof except for the last line...
Do you mean "we can express each side of $\triangle HCO$ (not $\triangle HCD$ as $\frac23$ of the [i]medians[/i] ($m_1$, $m_2$ and $m_3$)"?

Once we have that, the rest is quick.
Since each side length of $\triangle HCO$ is $\frac23$ of each median, the area of $\triangle HCO$ is $\left(\frac23\right)^2$ of the area of the triangle formed by $m_1$, $m_2$ and $m_3$.
We also know that $\triangle HCO=\frac13 \triangle ABC$, so the area of $\triangle ABC$ is $\frac43$ of the triangle formed by the medians.[/quote]

Sorry, coming to this thread after a long time. You are right it's triangle HCO which has sides as 2/3 of the length of the medians. That was a typo error :-)

[quote="frt"]Actually, there's a simple formula:

Let $m_1$, $m_2$ and $m_3$ be the lengths of the 3 medians of the triangle.

Then $\text{Area}=\frac43 \sqrt{s_m (s_m-m_1)(s_m-m_2)(s_m-m_3)}$, where $s_m=\frac12(m_1+m_2+m_3)$[/quote]

Really??? Nice formula!!!!


the problem is wrong.
maybe you should check the question once more.

[quote="dondigo"]Prove the identity:
\[ 2\cos{\frac{2\pi}{7}}=4\cos{\frac{2\pi}{7}}\cdot\cos{\frac{\pi}{7}}-1  \][/quote]

[hide="proof"]
Letting $A=\frac{\pi}{7}$ , hence

${4\cos 2A\cdot\cos A-1\\\\=\frac{\sin A}{\sin A}(4\cos 2A\cdot\cos A-1)\\\\=\frac{\sin 4A-\sin A}{\sin A}\\\\=\frac{\sin 3A-\sin A}{\sin A}\\\\=\frac{2\cos 2A\sin A}{\sin A}\\\\=2\cos 2A}$

[hide="proof by complex number"]
consider $x^7=1$ then $x=\cos \frac{2k\pi}{7}+i\sin \frac{2k\pi}{7}$

where $k=-3,-2,-1,0,1,2,3$

Also from vieta formula we know that $\sum^3_{k=-3}\cos \frac{2k\pi}{7}+i\sin \frac{2k\pi}{7}=0$

Hence $\sum^3_{k=-3}\cos \frac{2k\pi}{7}=0\implies 2(\cos \frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7})+1=0\\\implies 2\cos \frac{2\pi}{7}+4\cos\frac{\pi}{7}\cdot\cos\frac{5\pi}{7}+1=0\\\implies 2\cos\frac{2\pi}{7}=4\cos\frac{2\pi}{7}\cdot\cos\frac{\pi}{7}-1$[/hide][/hide]


I tried to ask this question before, but no one seemed to notice...How does a school register for the Canadian ARML in london?

[quote="jpark"]In Waterloo, CMC is sponsoring 12 people from the region. (I thought you had to register by region?)

It might be easier to talk to your local university..
Sarah? Maybe Dr. Sands and U of C would sponsor you to come to london..  :D[/quote]

Aww I'd love nothing more!  lol...miss my ontario buddies!  hehehehe.  We could say that...er....I'm an exchange student from Alberta...and thus qualified...lol.

Mr. Tom Griffiths from London decided to organise a Canadian team contest after deciding that ARML was too difficult / discouraging for the average math student.

The Canadian ARML's official name (at this time) is McTic, for Math Challenge Team Invitational Contest.  McTic is, hence, invitational.  However, if you'd like to be invited, Mr. Griffiths' phone number and e-mail are posted on his [url=http://www.math.uwo.ca/math-challenge/]Math Challenge Website[/url] at the bottom of the page - the area code is 519.  (I'm not sure if I'm allowed to post phone numbers directly on AoPS?)  Technically, it's a regional contest, I think, as Jenn said.  McTic isn't a highly official contest with ultra-strict entry rules (as far as I know) - if you wanted to form a team from your school or city, Mr. Griffiths would probably be willing to consider it.

McTic will be a lot of fun!  I'm on the London team.

Don't ask why we didn't call it CARML. You might be surprised to know that they actually went through quite a few names (They were seriously considering calling it OI for Ontario Invitational) before deciding on McTic.  I think they liked the Scottish(?) flavour.  And Mr. Griffiths said you could call it O'McTic (O for Ontario) to make it Irish(?).

There will be team, power, relay, and individual contests, as well as an individual mental arithmetic contest where all you're allowed to do is write the answer on the page, and you have <10 seconds per question to do so.  Questions like what is 42^2-38^2, or what is 45^2, or what is 99^2, or what is 38*42?

There might also be a song contest and super relay.  And there will be a banquet of some sorts.  If you'd like to see anything else at the contest, just post it here and I'll put in a word with Mr. Griffiths to add it.  (No promises on it actually happening, of course!)

[quote="geoffreynham"]

Don't ask why we didn't call it CARML. [/quote]

lol, maybe its a good thing its not called CARML...it reminds me too much of candy and sugar....yum...

[quote="jpark"]So why not seriously consider entering Sarah?  :D  :D  :D[/quote]

Hmm...Alberta seems a little far to be in the regions eh....

I talked with Mr. Griffiths today, and he said he was more than enthusiastic about having many teams participate.  If you can get a team from Alberta organised to come, that would be very exciting.  We can arrange to pick you up from London airport.  Anyone at all in Canada or US (or even China, for that matter) who is interested in forming a team just needs to call or e-mail Mr. Griffiths.  Hope to see you in May!  (The day after my birthday, by the way!!  Don't forget the belated birthday presents! :lol: )

Hmm...interesting offer....I don't know how we'd manage to form a team though, I'd need to talk with both Dr. Sands and Dr. Liu and find some sort of criteria to form a team...Yiyi, think its possible?

[quote="YY"]Well...I'm not sure...although I'd be really glad if that happens. It sounds like fun.
Maybe we can ask about it?

And BTW: I can't help but realize that a lot people have birthdays close to contests...[/quote]

Hahaha, mine's 6 days before CMO, not that special.  Anyone else with special birthdays?

I mentioned ARML to Dr. Sands a year and a half ago I think, but disregarded it cuz of money issues and stuff, maybe I should bring it up again

[quote="Sunny"]
Hahaha, mine's 6 days before CMO, not that special.  Anyone else with special birthdays?
[/quote]
Mine's 7 days before CMO  :?  :D

Wow, Sarah, I think you should really consider it!!

EVIL TWINS MEET AGAIN!!!  :D  :D

Well when do teams have to be in?  April is an awful month for me but I may have time to ask around...

There's not an official stated deadline, I think, but Mr. Griffiths has to book accomodations at Huron College sooner rather than later.  E-mail him! :) Hope to see you in a couple months!

Update: Mr. Griffiths hasn't heard anything from Mr. Crippin.  So I e-mailed Mr. Crippin.

Hmm... Jenn and Elyot and Will and Mike... want to make it a day trip to London instead?  Could you get a parent to drive you down to London just for the Saturday?  I'm sure something could be worked out with Mr. Griffiths - maybe you'd have to pay a bit of money for food, but he'd probably let you take part in the contest unofficially... Can you talk with each other about it?  (via phone or e-mail?  AoPS isn't fast enough for this sort of communication since not all of you check it regularly.)  Reply here if you're interested.

idunno about this anymore... if it all get's worked out, then I'll go... but I have some major work I've gotta do next weekend and I'm not sure how much time I'll be able to devote to this thing.  I haven't got anything back from Crippin... I've been gone all weekend playing chess at provincials, so I haven't learned anything new.  I probably won't come though... I guess we'll see.

I think it's worth emailing Crippin one last time, to see why Griffith hasn't received anything when Crippin says he emailed.  Because Crippin did say "yes, i just gotta talk to griffiths".

I talked to certain people tentatively about whether they're free Victoria Day weekend, and I'm sure we'll be able to pull together a team, so no worries about getting a team/drivers.

However, if sponsorship doesn't work out and we're not gonna go officially, then might as well not go at all; that would simply be too confusing :(

Are you sure Crippin e-mailed?  Is Mr. Crippin's e-mail pwcrippin@math.uwaterloo.ca as is published on the UWateroo website?  Does he check that e-mail address?

Probably Crippin just forgot about emailing :(

Yes that is his email; Elyot got a response.  Well since we already asked, can you email him again to ask about sponsorship?  He's agreed twice, but he just keeps forgetting

ok.  we're going to MCTIC!

griffiths emailed me today and said:

Dear Elyot,
I heard from Geoffrey at the weekend that you definitely had a team coming from Waterloo.  If that is so Prof Crippin has said that he will provide the funding, but you will need to contact him to get the money, i.e. $1000.. I will also need to know almost immediately, today or tomorrow, as I have to inform the college of numbers for rooms and food. If you are coming remember that there is a banquet on Saturday night and you will need dress pants and shirt and equivalent for the girls..
Tom Griffiths

So I told him we have people.  We have to get this finalized ASAP!

Yay!  Ok we have people and rides.

I have no idea how this sponsorship thing's gonna work though...like do we have to go get something from Crippin, or will he deal with it with Dr. Griffiths?  Or does one of us has to throw in $1000 first and wait to get it back?

OK... Crippin and Griffiths have the payment worked out.  Out last hurdle is DRIVERS!!!  We've gotta find somebody who can help drive us to London, or we're gonna end up having to take a greyhound!

Oh my goodness!!!  I'm Excited!!!  With a capital E, even!  Hrm, *thinks*... Elyot always spells his name with a capital E... oh well.  Great job everyone who has organised this so far!  You're so close!  As for drivers... how many people can Mike's dad drive?

Wow.  Who is going to be on the Waterloo team?

By the way, I have to say... you guys are really really lucky... you don't have to pay anything!  Massey and London players have to pay $50 each for McTic (the rest is sponsored by their respective school/city)  Mind you, you might have to pay for Greyhound... but even then, it's a deal for you.

Lol haha well I'm sure we wouldn't have been able to go without your help.  :)

So yeah, thanks a lot for helping us work out all the stuff to be able to come, and look forward to seeing you guys Friday!

Haha... I've posted so much that now I'm at Hodge Conjecture!!  lol...

Looking forward to meeting you, for sure.  Oh, and I'll post some updated info here tomorrow... stuff you should bring, things you should know, etc.

Updated Stuff:

Team Question will be 50 minutes long.
Relays will be 8 minutes each, with 6 points for 4 minutes and 4 points for 8 minutes.
Individual questions will be a set of 8 questions for which you have one hour to solve as  many as possible.
Mental math questions will be 5 seconds each.  Each question will be read twice.  You will be permitted to think about any question at any given time, and write down any answer whenever you want to (i.e. question 3 when question 7 is being read out), but you may not use the paper for rough work or for recording the question.

What to bring:
Something reasonably formal for the banquet (like National Camp)
Clothes etc. for two nights
Pencil (I wouldn't bet on them supplying one)
Something fun for the evening (chess, etc.)
The usual stuff

Any other questions?

Wow... That was fun!!   :lol:  :lol:

It was great to see you all!  I haven't had so much fun in a long while.  Lots of good memories for everyone, I'm sure.

I was really surprised when they said London won.  I thought Massey would win for sure.  I guess that the fact that it's a new contest and that London had a slightly better idea of what was coming helped us.  But, of course, the competition is secondary to the fun times and the friendships.  :)

On another note... unfortunately, this may be the last time I post of AoPS for a while... I've been spending far too much time here recently, trying to get a Waterloo team organised.  I'll come back and visit every so often!

Oh - and my signature wasn't just random.  If you rearrange the letters in "I'm at school," you get "Math is cool!"

YES!

MCTIC was so ownage!  The competition was good, and we had some great fun too.  Nice to see Geoff, Taotao, Jimmy, Shaun, Francis, and everyone else.  Nothing beats playing SMB and Dr. Mario on a huge TV.  That was amazing.  Thanks to everyone for this experience!

Yep, thanks to everyone who helped to make it happen, Elyot, Geoff, Mike, Thomas...

Hopefully next year we'll be more organized (we didn't find our last team member until the Friday of, lol), and hope to see you guys again next year!


You have infinitely many boxes, and you randomly put 3 balls into them. The boxes are labeled 1, 2, ... Each ball has probability 1/2^n of being put into box n. The balls are placed independently of each other. What is the probability that some box will contain at least 2 balls?

Well, I have done the first part. Does this look right? (spoiler)

[hide]For a box k, the probability p_k = ((1/2)^k)^3 + 3(((1/2)^k)^2)(1-(1/2)^k). After some simplification, we have p_k = 3/(4^k) - 2/(8^k). So to find the probability that some box will contain at least 2 balls, we have to evaluate Sum[k=1->n] p_k = Sum[k=1->n] 3/(4^k) - 2/(8^k). Now, looking at the form and knowing ComplexZeta's tricky mind, the sum is probably telescopic in some way...[/hide]

I didn't write this problem. But what you have done so far looks correct.

did you pull that off a tournament of the towns competition?

No, it was from Harvard MIT Math Tournament.

Doh! That's the second time tonight that I've missed seeing an easy geometric summation.

has anyone here actually done the contest?

Yeah, I am from MA, so I did it just this year for the first time. It was fun, particularly the Guts Round. A bit unorganized though...

We went last year.

Yeah the GUTS round was disorganized...

we were originally in about 20th place.... at least those were the 'real-time' standings DURING the contest.

Afterwards we're shocked as it turns out we (AAST) ended up winning the thing.

Incredibly sloppy though. I think they originally gave our points to one of our other teams or something... how chaotic.

But, the whole thing is at least a fun experience to try out your math skills.

The team round question last year was also quite interesting. It turns out the entire thing relates almost directly to graph theory. Looking at it that way simplifies it so much.

i was hoping you were from MA nukular.  Im looking for people from MA so we can discuss MAML Olympiad questions after the test.

Why don't you post them here? We would be happy to work on them.

yeah but after we take the test my teacher doesnt let us keep it, so i need to get the questions from someone else online.


[hide]It would be the number divided by 3 since the smallest divisor not 1 is 3[/hide]

[hide]2^2 congruent to 1 (mod3)

(2^2)^7 congruent to (1)^7 = 1 (mod 3)

1 - 1  = 0 (mod3)[/hide]

[quote=Syntax Error]What is the largest integral divisor of 2^14-1 that is less than 2^14-1?[/quote]

Yes, $3$ is a factor of the expression, and then you can just divide.

Wow, I didn't realize AoPS forums existed in 2003  :rotfl:

wow that was a humongous bump

@above me too :rotfl:

This was, if I remember correctly when randomly looking at old posts, the oldest post in MSM

[hide=Solution]Since $2^{14}-1$ is odd, $2$ is not a factor. Since $2\equiv -1\pmod 3$, we have $2^{14}-1\equiv (-1)^{14}-1\equiv 0\pmod 3$ which means $2^{14}-1$ is a multiple of $3$. Thus, our answer is $\frac{2^{14}-1}{3}=5461$.$\blacksquare$[/hide]

[quote=Math2k06]Apparently some sneaky person deleted their post to make it seem like Edud_looC bumped it. Sneaky. But the aops admins know who you are...[/quote]

Oh that's depressing. :( At least the post count gives me evidence to say I didn't bump it. Take a look at the post numbers. Where is Post #7?

[quote=RetroTurtle][quote=aopsuser305]the answer is 5461[/quote]

17 year bump  :rotfl:
And I also got 5461[/quote]

[quote=Math2k06]Apparently some sneaky person deleted their post to make it seem like Edud_looC bumped it. Sneaky. But the aops admins know who you are...[/quote]

Their post is preserved, so we know eduD_looC didnt bump it.
$2^{14}-1=16383$, and divide by $3$ to get 5461

[quote=MathJams]^above I just realized your name was Cool_Dude backward XD[/quote]

AHAHAHAHAHAHA lol the captitalization in my username hints to flip the username backwards

[hide=Solution]FIrst of all, it is best to divide it into a bunch of numbers and stuff: $2^{14}-1=(2^7+1)(2^7-1)$ From here, we get that the factorization would be $(129)(127)$. We see that $127$ is prime and $129=3 \cdot 43$. From this, we just take out the smallest integral divisor from the factorization to get $(43)(127)=5461$, the answer many have gotten!

[quote=dori_vinnie][quote=aopsuser305]the answer is 5461[/quote]

That is the biggest bump I've ever seen on AoPS lol (17 YEARS!)
[size=50][url=https://play.unity.com/mg/fps/ablaze-so]If you're really that bored click here or something[/url][/size]
Also, we know that aopsuser305 bumped it, and not eduD_looC, so you can't get away with it by simply deleting your post[/quote]

please do not advertise. reported.

[quote=ATGY][quote=Syntax Error]What is the largest integral divisor of 2^14-1 that is less than 2^14-1?[/quote]

I have an alternate solution

We can factor $2^14 - 1$ to $(2^7 - 1)(2^7 + 1) = 127\cdot129 = 127\cdot3\cdot43$.

so 3 can divide it, 2 can't divide it though, so our answer is $(2^{14} - 1)/3$[/quote]

Or $127\cdot 43$


function
real analysis
calculus
integration
topology
real analysis theorems

Is L1 metric space complete?
L1 is a space of  functions asolutely integrable on a compact.
I'm not very familiar with Math-english under compact I meant [a;b]. In other words a piece of a line.

The answer is a qualified "yes" under two stipulations. The first is that the functions be absolutely integrable in the Lebesgue sense. The second is that the elements of the metric space are not functions but rather equivalence classes of functons under the equivalence relation $f\sim g \iff f(x)=g(x)$ almost everywhere.

The proof of this theorem (which is called the Riesz-Fischer Theorem) uses the Dominated Convergence Theorem and the Monotone Convergence Theorem - in other words, the central tools of the Lebesgue integration theory.

Without the Lebesgue integral, there's no way to have this theorem. In fact, I would argue that having this theorem is one of the most important motivations for developing the Lebesgue theory.

By contrast, let's say we take $C[-1,1]$ (the continuous functions on $[-1,1]$) and consider it to be a metric space in the $L^1$ norm.

Consider the sequence of functions given by

$f_n(x)=\begin{cases}0&x<0\\ nx&0\le x\le \frac1n \\ 1&x>\frac1n\end{cases}$

Then $(f_n)$ is a Cauchy seqence but there is no function in $C[-1,1]$ for it to converge to.

[quote="Kent Merryfield"]The answer is a qualified "yes" under two stipulations.
1) The functions be absolutely integrable in the Lebesgue sense.
2) The elements of the metric space are not functions but rather equivalence classes of functons under the equivalence relation $f\sim g \iff f(x)=g(x)$ almost everywhere.
...
Without the Lebesgue integral, there's no way to have this theorem.
...
By contrast, let's say we take $C[-1,1]$ (the continuous functions on $[-1,1]$) and consider it to be a metric space in the $L^1$ norm.

Consider the sequence of functions given by

$f_n(x)=\begin{cases}0&x<0\\ nx&0\le x\le \frac1n \\ 1&x>\frac1n\end{cases}$

Then $(f_n)$ is a Cauchy seqence but there is no function in $C[-1,1]$ for it to converge to.[/quote]

1) No. Riemann sence in my case.
2) No. Simple functions are meant not classes.

This example is known.
The thing is to find an example to show that L1 itself isn't full.
Or to prove that a Cauchy sequence convreges to a function from L1.

Here's an example of one thing that can go wrong with Riemann-integrable functions:
$f_n(x)=\begin{cases}\frac1{\sqrt{x}}&\frac1n\le x\le 1\\ \sqrt{n}&0\le x\le\frac1n\end{cases}$
This sequence is Cauchy in the $L^1$ norm, but the limit function is not Riemann-integrable because it is unbounded.

I would add that the set of all Riemann integrable functions on $[a,b]$ with the $L^1$ norm is not a metric space because it fails the positive definite property of a metric. Consider the function $f$ which is identically zero and $g$ which is zero except at one point, at which it is 1. Both functions are Riemann integrable. However $d(f,g)=\|f-g\|_1=0$ but $f$ and $g$ are not the same function.

In other words, when you said

[quote]2) No. Simple functions are meant not classes. [/quote]
you were asking for something which you cannot have. If you want this to be a metric space, you're going to have to stop regarding the $f$ and $g$ of my example as being different.

You don't have this problem with $C([a,b])$ with the $L^1$ norm. That is a metric space - just not a complete one. Note also that jmerry's example was of functions in $C([a,b]).$

As a piece of conventional nomenclature, $L^1$ most commonly means the set of (equivalence classes of) Lebesgue integrable functions. What do you think "L" stands for, anyway?

[quote="Kent Merryfield"]The proof of this theorem (which is called the Riesz-Fischer Theorem) uses the Dominated Convergence Theorem and the Monotone Convergence Theorem - in other words, the central tools of the Lebesgue integration theory.

Without the Lebesgue integral, there's no way to have this theorem. In fact, I would argue that having this theorem is one of the most important motivations for developing the Lebesgue theory.[/quote]

I just wanted to say I found this post helpful. I had not made this connection.


Show that the irrational numbers are dense in R, using the fact that sqrt(2) is irrational

If you can prove that $ \forall x, y\in\mathbb{R}$ so that $ x < y,\exists q\in\mathbb{Q}: x < q < y$ then the result you want follows. I guess you're taking a beginning real analysis course? I'm in one myself, I have a similar problem to prove that the set  $ \{m\plus{}\sqrt2 n : m, n\in\mathbb{N}\}$ is dense in $ \mathbb{R}$.

Actually, I'm sure this is just a typo, but to clarify, you want $ \forall x<y\in\mathbb{R},\exists q\in\mathbb{R}\minus{}\mathbb{Q}: x < q < y$, remembering that $ \mathbb{I}\equal{}$ the irrationals $ \equal{}\mathbb{R}\minus{}\mathbb{Q}$.


I have been trying to do these problems all weekend but I dont know how to do them and I dont knoe anyone who is good in math so I was hoping I could get some help on here any help will be greatly appreciated.
3. Find the length of arc S, if r= 10 and theta = pi over 6.

4. Find the area A of the sector if theta = pie over 4 and r = 6

5. If the latitude of Chicago is about 42 degrees and the radius of the earth is about 4000 miles, how far is Chicago from the eaquator? (use pie = 3.1416)

please can some one help me its a matter of life and death lol.

The length of an arc is given by $L = r \theta$ and the area of a sector is given by $A = \frac{1}{2} r^2 \theta$.  That should take care of these.

[hide="Proof"]
You know that there are $2\pi$ radians in a circle, and that the circumference of a circle is $2\pir$.  If you're only considering $\theta$ radians (in an arc), then the fraction of the circumference that is on the arc is $\frac{\theta}{2\pi}$, so the length of the sector is $\frac{\theta}{2\pi} (2\pi r) = r\theta$.

As for the area, make a similar argument using the fact that the area of a $2\pi$ sector (the entire circle) is $\pi r^2 = \frac{1}{2} 2\pi r^2$.
[/hide]


On the sprint, #15 did not specify whether or not CD>AB, so there were two acceptable answers, 18 or 30.  No one put 18 though. =P.  30 was the desired answer.

Target Round: Basic rules apply.  Submit within 1-2 days.

Honor system again.

Good luck!

Edit: OH. Right. Yeah. Thanks for cleaning it up.  BTW, results will be posted on Thursday since we have one person unable to submit answers until tomorrow (ajain).

Only two of the nine or 10 who did the sprint have submitted their target answers.


Since everyone has a favorite number (or almost everyone), how many people here have a favorite word? Mine's lexicographer. I also like ambidextrous.

one of my favorite words is oxymoron, just cause it sounds weird...

defenestration: the act of chucking something out a window

btw this has topic definitely been posted before

this might be innappopriate, but I remember how there was this thing explaining how a certain bad word is the most useful word in the english language.

[quote="filletwho"] a certain bad word is the most useful word in the english language.[/quote]

I agree

supercalafradgalisticexpilidoscious, excuse the spelling  :D

[quote="Danbert"]Syzygy

         1a. Either of two points in the orbit of a celestial body where the body is in opposition to or in conjunction with the sun.
         1b. Either of two points in the orbit of the moon when the moon lies in a straight line with the sun and Earth.
         1c. The configuration of the sun, the moon, and Earth lying in a straight line.
   2. The combining of two feet into a single metrical unit in classical prosody.[/quote]

WOAH!!!! A word with no vowels! Awesome!

[quote="tarquin"]The full name of the tryptophan synthetase A protein, an enzyme with 267 amino acids, extends to 1,913 letters. However, I will not post it because it would take way too much space.[/quote]

methionylglutaminylarginyltyrosylglutamylserylleucyl phenylalanylalanylglutaminylleucyllysylglutamylarginyl lysylglutamylglycylalanylphenylalanylvalylprolylphenyl alanylvalylthreonylleucylglycylaspartylprolylglycylisol eucylglutamylglutaminylserylleucyllysylisoleucylaspartyl threonylleucylisoleucylglutamylalanylglycylalanylaspartyl alanylleucylglutamylleucylglycylisoleucylprolylphenyl alanylserylaspartylprolylleucylalanylaspartylglycylprolyl threonylisoleucylglutaminylasparaginylalanylthreonylleucyl arginylalanylphenylalanylalanylalanylglycylvalylthreonyl prolylalanylglutaminylcysteinylphenylalanylglutamyl methionylleucylalanylleucylisoleucylarginylglutaminyllysyl histidylprolylthreonylisoleucylprolylisoleucylglycylleucyl leucylmethionyltyrosylalanylasparaginylleucylvalylphenyl alanylasparaginyllysylglycylisoleucylaspartylglutamylphenyl alanyltyrosylalanylglutaminylcysteinylglutamyllysylvalyl glycylvalylaspartylserylvalylleucylvalylalanylaspartylvalyl prolylvalylglutaminylglutamylserylalanylprolylphenylalanyl arginylglutaminylalanylalanylleucylarginylhistidylasparaginyl valylalanylprolylisoleucylphenylalanylisoleucylcysteinyl prolylprolylaspartylalanylaspartylaspartylaspartylleucyl leucylarginylglutaminylisoleucylalanylseryltyrosylglycyl arginylglycyltyrosylthreonyltyrosylleucylleucylserylarginyl alanylglycylvalylthreonylglycylalanylglutamylasparaginyl arginylalanylalanylleucylprolylleucylasparaginylhistidyl leucylvalylalanyllysylleucyllysylglutamyltyrosylasparaginyl alanylalanylprolylprolylleucylglutaminylglycylphenylalanyl glycylisoleucylserylalanylprolylaspartylglutaminylvalyllysyl alanylalanylisoleucylaspartylalanylglycylalanylalanylglycyl alanylisoleucylserylglycylserylalanylisoleucylvalyllysylisol eucylisoleucylglutamylglutaminylhistidylasparaginylisoleucyl glutamylprolylglutamyllysylmethionylleucylalanylalanylleucyl lysylvalylphenylalanylvalylglutaminylprolylmethionyllysyl alanylalanylthreonylarginylserine.

[quote="baseballfreak2"]Is that true?[/quote]

Depends on your definition of a word, but that is an acronym for a antidisestablishmentarianist. I wouldn't consider it a word.

My favourite word, since the very first time I heard it, is a Russian one:
олимпийский
It's an adjective, which means "olympic" in English. I don't know any other word, which would be so melodious.

[quote="djimenez"][quote="09husbri"]Misunderstimate :)[/quote] I have looked in three dictionaries I have at home and in two that are online, and that word is not there! The meaning is understood, but I really don't think there would be a real situation where you would use it.[/quote]

Misunderestimate is a word that George W. Bush "created" by accident.

[quote="WindSlicer"][b]Ennui[/b][/quote]

Actually, that word was why I met my current best friend. We both missed it.

"Misled" is a cool word too.  It makes it look like someone can "misle" you, which sounds kind of bad :).

[b]Humuhumunukunukuapua'a[/b]

The Hawaiian state fish.

omphaloskepsis (pardon the spelling) means the contemplation of your belly bottom :rotfl:

I like totoleetarian. Sorry but that's how they spell it in our school :P yes I know i'm weird.

zenzizenzizenzic- a number raised to the 8th power (I'm not lying)

i always liked penultimate (second last)

:rotfl: LOL- words with no vowels? oy gavolt!  My, try sky, fly, zyzyva -- all these simple words have no vowels in them.

My favorite is-  :ninja: floccinaucinihilipilification (the categorizing of something as worthless or trivial)

[quote="Zyloch"]My favorite word is definitely [b]pneumonoultramicroscopicsilicovolcanoconiosis[/b]

Hope I didn't have a typo in there :)[/quote]

I don't know if this has been mentioned before, but it's pneumonoultramicroscopicsilivolcano[b]k[/b]oniosis.

i hav ah speling be onn therzdaye

As far as [b]I[/b] rmember its [b]c[/b]oniosis

slacker

 :o I tried to say just slacker but it was to small so I'm adding this. What's up with that? *question mark smilie here*(it says that I have 3 smilies)


limit
logarithms
calculus
calculus computations

Caculate limit of following sequence: $ n(1\minus{}\sqrt[r]{1\minus{}\frac{1}{n}})$ where $ r$ is fixed positive integer.

hello, we have $ \lim_{n\to \infty}\frac{1\minus{}\sqrt[r]{1\minus{}\frac{1}{n}}}{\frac{1}{n}}\equal{}\lim_{n\to \infty}\minus{}\frac{\frac{1}{r}\left(1\minus{}\frac{1}{n}\right)^{\frac{1}{r}\minus{}1}}{\frac{1}{n^2}}\equal{}\frac{1}{r}$.
Sonnhard.

$ \begin{aligned*}\lim_{n\to\infty}n\left(1 - \left(1 - \frac {1}{n}\right)^{\frac {1}{r}}\right) & = \lim_{n\to\infty}n\left(1 - e^{\frac {\ln\left(1 - 1/n\right}{r}}\right) \\
& = \lim_{n\to\infty}n\left(1 - e^{\frac {\frac { - 1}{n} - \cdots}{r}}\right) \\
& \sim\lim_{n\to\infty}n\left(1 - e^{\frac { - 1}{rn}}\right) \\
& \sim\lim_{n\to\infty}n\left(1 - 1 + \frac {1}{rn}\right) \\
& = \frac {1}{r} \end{aligned*}$


calculus
integration
trigonometry
calculus computations

Find the volume of the solid formed by the revolution of the curve $ x \equal{} a(1 \plus{} \cos \theta )\cos \theta ,\ y \equal{} a(1 \plus{} \cos \theta)\sin \theta\ (a > 0,\ \minus{} \pi < \theta \leq \pi)$ about the $ x$ axis.

:(

you mean the part $ 0 \le \theta \le \pi$ revolves about the x-axis, right ?

$ \textbf V_{required} \;\equal{}\; \bigg| \; \int_0^{\pi} \; \pi\;y^2 \; dx \; \bigg|$

$ \equal{}\;  \bigg| \; \pi a^3 \int_0^{\pi} \; (1\plus{}\cos\theta)^2\;\sin^2\theta\cdot \left[ (1\plus{}\cos\theta)(\minus{}\sin\theta) \plus{} (\minus{}\sin\theta)(\cos\theta)\right] \; d\theta \; \bigg|$

$ \equal{}\;  \bigg| \; \pi a^3 \int_0^{\pi} \; (1\plus{}\cos\theta)^2(1\minus{}\cos^2\theta)(1\plus{}2\cos\theta)\cdot (\minus{}\sin\theta\;d\theta) \; \bigg|$

$ \equal{}\;  \bigg| \; \pi a^3 \int_{1}^{\minus{}1} \; (1\plus{}u)^2(1\minus{}u^2)(1\plus{}2u)\;du\; \bigg|  \qquad ( \text{where} \quad u\;\equal{}\;\cos\theta )$

$ \equal{}\;  \bigg| \;  \pi a^3 \left( \minus{} \dfrac {1}{6} \minus{}  \dfrac{15}{6} \right) \; \bigg|$

$ \equal{}\; \boxed {\dfrac {8}{3} \pi a^3 }$

You have solved it to the point. Your answer is correct. :)

This parameterized curve is Cardioid, so $ V\equal{}\pi \left(\int_{\minus{}\frac{a}{4}}^{2a} y^2dx\minus{}\int_{\minus{}\frac{a}{4}}^0 y^2dx\right).$


find the remainder when $ |(i\plus{}1)^2|\plus{}|(i\plus{}2)^2|\plus{}|(i\plus{}3)^2|...(i\plus{}100)^2|$ is divided by 1000


"The incircle of triangle ABC touches BC, CA, AB at A', B', C'. Prove that
$ AB'^2 \plus{} BC'^2 \plus{} CA'^2 \geq AB'.B'C' \plus{} BC'.C'A' \plus{} CA'.A'B' \geq B'C'^2 \plus{} C'A'^2 \plus{} A'B'^2$"
I think that it's too a nice problem! :P

To mod: Please lock this topic because it is a contest problem in Mathematics And Youth Magazine of Viet Nam. The dealine is 15-04-2008!