invariant
induction
combinatorial geometry
combinatorics unsolved
combinatorics

$5$ points are given in the plane, any three non-collinear and any four non-concyclic. If three points determine a circle that has one of the remaining points inside it and the other one outside it, then the circle is said to be [i]good[/i]. Let the number of good circles be $n$; find all possible values of $n$.

More generally, we may prove the following result :
Let $S$ be a set of $2k+1$ points in the plane, no three collinear, no four concyclic. A circle is said to be an halving circle if it has three points of $S$ on its circumference and exactly $k-1$ points of $S$ in its interior (and thus exactly $k-1$ in its exterior).
Thus, the number of halving circles is $k^2$.

Reference :
F. Ardila, The number of halving circles, A.M.M. 2004, p.586-591.

Pierre.

Please give a solution here, either for the extended version or the problem at hand.  :)

Yeah! Federico is our Team leader! he has shown us many times the proof of this beautiful result!

anyway it is to hard to write it down, i just give some hints:

Draw all circles and show that moving a point to another region which is a neighboor of it starting region doesnt change the number of halving circles! (the point may only pass trough one boundary)

Conclude this number is invariant

Use induction in $n$, to prove for $2n+1$ points it is $n^2$.

if you don't have access to the american mathematical monthly,
you can read the article "the number of halving circles" here:
  http://www.math.washington.edu/~federico
it is available as a pdf file.
.f.

[quote="orl"]5 points are given in the plane. Any three of them are non-collinear. Any four are non-cyclic. If three points determine a circle that has one of the remaining points inside it and the other one outside it, then the circle is said to be good. Let the number of good circles be $ n,$ find all possible values of $ n.$[/quote]
I think that $ n\equal{}3,4,5$  :)
(using convex hull prove least 3 circle good ,least 5 circle not good)
Am I right? :?:

[quote="math10"][quote="orl"]5 points are given in the plane. Any three of them are non-collinear. Any four are non-cyclic. If three points determine a circle that has one of the remaining points inside it and the other one outside it, then the circle is said to be good. Let the number of good circles be $ n,$ find all possible values of $ n.$[/quote]
I think that $ n \equal{} 3,4,5$  :)
(using convex hull prove least 3 circle good ,least 5 circle not good)
Am I right? :?:[/quote]
I had wrong.
$ n\equal{}4$ (not equal 3 and 5).

[quote="JBL"]If you're going to post a solution, post a solution.  Posting answers (right or wrong) without any proof is just pointless filling of space (especially when someone else has already stated the answer to a more general version of the problem).

Federico Ardila's webpage is now here: http://math.sfsu.edu/federico/[/quote]
please check link?(I can't open it  :( )

It works fine for me.  The article itself is at http://math.sfsu.edu/federico/Articles/circles.pdf


MATHCOUNTS
analytic geometry
graphing lines
slope

Do not discuss any problems from the MATHCOUNTS Chapter Competition until further notice.  Do not discuss your score.  Do not solicit others to discuss the Chapter Competition.  (Chapter Competitions happen on different dates around the country.)

[quote="Last year, MCrawford"]Students who discuss any information including scores and difficulty level of the test will be put in the AoPS penalty box and not be allowed to post for a considerable period of time.[/quote]
Furthermore, during the competition windows, [b]please identify the source of any problem you post[/b].  Such as "2002 Chapter Target #4".

Until such time as MATHCOUNTS releases the Chapter Competitions, I will relentlessly delete all offending posts, and I will refer the matter to AoPS Administrators and/or the National MATHCOUNTS organization.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=69442]The previous admonition against discussing the School Competition still holds.[/url]  Do not discuss the School Competition until after February 26.

[quote="biffanddoc"]but we can say whether or not we made states or not and our rank, right?[/quote]
Yes, you can say if you are advancing to State.  Beyond that, it's a slippery slope, so be very, very careful.

If you do not exceed what is released to the "general public and the press" you will be OK.  I am told that the instructions to the Chapter Coordinator contains the following statement:
[quote="Instructions to Coordinators"]A final report consisting of final rankings -- [b]not scores[/b] -- of the top 25 percent of individuals and top 40 percent of teams should be posted at the finish of the competition.  [b]This report is the only listing that should be released to the general public and the press.[/b][/quote]

So we're allowed to discuss the competition [b]after[/b] the competition season--are we allowed to discuss our scores after that time window? Just checking.

after the competition window, we can say anything abotu the competition or problems

Yeah. The point of this is just not to give an unfair advantage to those who have later competition dates.

Unlike AMC, MATHCOUNTS does not take precautions against that, not that any of you would cheat would you?  :roll:

When does the Chapter competition window end?

we can discuss scores. Not specific problems because there may still be people taking the test

[quote="samath"]Why can't we discuss scores?  I don't see how that helps people still taking the test in any way.[/quote]Scores might indicate the level of difficulty of a certain test and may influence (either in bad or in good) a person's behaviour on the test (even without you or that person wanting to do so).

today? then my chapter was one of lasdts to end. It was yesterday

A number of people have posted in violation of this Announcement.  Read the first post of this thread.
[quote="rcv"]Do not discuss any problems from the MATHCOUNTS Chapter Competition until further notice.  Do not discuss your score.  Do not solicit others to discuss the Chapter Competition.[/quote]
A moderator will advise when discussion of the Chapter Competition may begin.

Meanwhile, I'm sorry I have to yell to get your attention.  But...  [b][color=red][size=150]DO NOT DISCUSS ANY ASPECT OF THE CHAPTER COMPETITION UNTIL FURTHER NOTICE!!![/size][/color][/b]


calculus
integration
function
trigonometry
calculus computations

For $ x\geq 0,$ define a function $ f(x)\equal{}\sin \left(\frac{n\pi}{4}\right)\sin x\ (n\pi \leq x<(n\plus{}1)\pi )\ (n\equal{}0,\ 1,\ 2,\ \cdots)$.

Evaluate $ \int_0^{100\pi } f(x)\ dx.$

is there an x missing in the argument of $ \sin(\frac {n\pi}{4})$?


graph theory
combinatorics solved
combinatorics

we say the edge$\varrho$ in graph G is critical if $\alpha (G-\varrho)> \alpha(G)$.now assume that xy,xz are critical ,y is not connected to z,prove that G contains a cycle with odd vertices as a subgraph.

Maybe this sounds stupid, but what's $\alpha(G)$?

maximum number of vertices which are not connected to each other

That is the maximum size of an independent set of vertices (which is called the independance number of the graph).

Pierre.

I found this notation in a book. It means the independence number of $G$ in that book, i.e. the greatest $r$ s.t. $\overline{K^r}\subseteq G$, where, in general, $K^r$ is the complete graph on $r$ vertices and $\overline X$ is the complement of $X$.

Are you referring to the same thing?

I'm terribly sorry.. I guess you guys posted the answers while I was searching and typing my last message.

Thanks!

I don't know that much about graph theory, but I think Konig's theorem states that a graph has an odd cycle iff it's not bipartite, so we should assume that it's bipartite and obtain a contradiction.

I'm sure you've tried this, so maybe it doesn't yield a quick result. Anyway, I have to go now, so I can't think about it for a while (I'm off to a party :D).

I don't know if this is correct or not, but it seems correct to me right now :).

Let's assume $G$ is bipartite. Take $V_1,V_2$ to be the two sets of points. There exist two sets, $S_1=A_1^1\cup A_2^1,\ S_2=A_1^2\cup A_2^2,\ A_i^j\subseteq V_i$, such that $x\in S_i$ (assume $x\in V_1$; maybe we'll need it, I'm not sure right now) and such that $S_i=\overline{K^{\alpha(G)}}$ and the only edge in $S_1\cup\{y\}$ is $xy$ and the only edge in $S_2\cup \{z\}$ is $xz$. It's clear that $y,z\nin S_1\cup S_2$.

We have $|A_1^1|+|A_2^1|=|A_1^2|+|A_2^2|=\alpha(G)$. From the inclusion/exclusion principle we get $|A_1^1\cup A_1^2|=|A_1^1|+|A_1^2|-|A_1^1\cap A_1^2|,\ |A_2^1\cup A_2^2|=|A_2^1|+|A_2^2|-|A_2^1\cap A_2^2|$. By adding these two we find that $(|A_1^1\cup A_1^2|+|A_2^1\cap A_2^2|)+(|A_2^1\cup A_2^2|+|A_1^1\cap A_1^2|)=2\alpha(G)$.

Let's assume that one of $|A_1^1\cup A_1^2|+|A_2^1\cap A_2^2|,\ |A_2^1\cup A_2^2|+|A_1^1\cap A_1^2|$ isn't $\alpha(G)$. In this case one of them is $>\alpha(G)$, and this is false because if we assume, for example, that $|A_1^1\cup A_1^2|+|A_2^1\cap A_2^2|>\alpha(G)$, then the set $(A_1^1\cup A_1^2)\cup (A_2^1\cap A_2^2)$ would be a graph with more than $\alpha(G)$ independent vertices, and that's not possible. This means that $|A_1^1\cup A_1^2|+|A_2^1\cap A_2^2|=|A_2^1\cup A_2^2|+|A_1^1\cap A_1^2|=\alpha(G)$.

Now consider the set $((A_2^1\cup A_2^2)\cup (A_1^1\cap A_1^2)-\{x\})\cup\{y,z\}$. It obviously yields a graph with $\alpha(G)+1$ independent vertices, so we've got a contradiction.

I apologize in advance for any mistakes I might have made :).

Well, i will not use the word 'mistakes' because the general idea seems to be the same as mine ( :D ) but, at least for me, there are some unclear points :
a) I do not see why it is obvious that $y,z \in S_1 \cup S_2$
b) I do not see why $(A_1^1 \cup A_1^2) \cup (A_2^1 \cap A_2^2) $ will give a independant set of points.

Pierre.

a) I'm sorry about this. I meant to say that it's obvious that $y,z$ are [b]not[/b] in $S_1\cup S_2$, I used \nin (thinking that it comes from "[b]n[/b]ot [b]in[/b]) and hoped that this was a valid Tex instruction :D. In the last few lines I used the fact that $y,z$ aren't in $S_1\cup S_2$ (in order to show that the cardinal of that nasty-to-write set is $\alpha(G)+1$, and not $\alpha(G)$ or $\alpha(G)-1$).

b) No point from $A_1^1$ is connected to a point of $A_2^1$ because $S_1=A_1^1\cup A_2^1$ is graph with no edges (we defined it to be like that). This means that no point of $A_1^1$ is connected to any point of $A_2^1\cap A_2^2$. We do the same for $A_1^2$ and we find that no point of $A_1^1\cup A_1^2$ is connected to a point of $A_2^1\cap A_2^2$ (because $G$ is bipartite, and no point from $V_i$ is connected to another point of $V_i$, and $A_1^1,A_1^2\subset V_1,\ A_2^1,A_2^2\subset V_2)$.

Now I give mine :
Let $Y$ be a maximum set of independent points in $G \setminus xy $
Let $Z$ be a maximum set of independent points in $G \setminus xz$

- First, we prove that $x,y \in Y$ and $x,z \in Z$.
Suppose that $x \notin Y$ and $y \notin Y$ then clearly, $Y$ is also a set of independent point in $G$, which contradicts that $|Y| > \alpha (G)$. Thus at least one of $x,y$ is in $Y$.
But, if $x \in Y$ and $y \notin Y$, then $x$ is non adjacent to each of the other vertiex in $Y$ and $y \notin Y$ so that the edge $xy$ let the vertex set $Y$ to be an independent set of points for $G$ itself, which leads to the contradiction above. Thus if $x \in Y$ then $y \in Y$ too. We use a similar reasoning to prove that if $y \in Y$ then $x \in Y$ too. From above it follows that $x,y \in Y$.
Similarly $x,z \in Z$.

It follows that $y \notin Z$ and $z \notin Y$.

Let $G'$ be the restriction of $G$ to the set of vertices $(Y \triangle Z) \cup \{ x \} $, where $ \triangle $ denotes the symmetric difference.
note that $y,z \in Y \triangle Z$ but $x \notin Y \triangle Z$ and $y,z \notin Y \cap Z$.

Let's assume, for a contradiction, that $G'$ is bipartite, with vertex set partition $A,B$.
Then $G'$ has an independent set of points $S$ such that
$|S| \geq max (|A|,|B|) \geq \frac {1} {2} (|(Y \triangle Z) \cup \{x \} | = \frac {1} {2} (|Y \triangle Z| + 1) $. (1)

Now, we prove that the set $T = S \cup ((Y \cap Z) \setminus \{x \}) $ is an independent set of points for $G$ :
Let $a,b$ be in $T$.
- If $a,b \in S$ they are nonadjacent since $S$ is an independent set of points for $G'$ and that $G'$ contains vertices $x,y,z$, and therefore the edges $xy$ and $xz$.
- If $a,b \in (Y \cap Z) \setminus \{ x \} )$ then they are not $x$ and they both belong to $Y$ which is an independent set of points for $G \setminus xy$. Thus the edge $xy$ leaves them nonadjacents.
- If $a \in S$ and $b \in (Y \cap Z) \setminus \{ x \}$ : Since $S \subset (Y \triangle Z) \cup \{ x \}) \subset Y \triangle Z$ it follows that for example $a \in Y$. In another hand, $b \notin \{ x, y \}$ and $b \in Y$. Thus, they are nonadjacents for $G \setminus xy$, and adding the edge $xy$ leaves $a,b$ nonadjacents.

But :
$|T| = |S \cup (Y \cap Z) \setminus \{ x \} )| = |S| + |(Y \cap Z) \setminus \{ x \} )| = |S| + |Y \cap Z| - 1 $
Thus, from (1) :
$|T| \geq \frac {1} {2} (|Y \triangle Z| + 2|Y \cap Z| - 1) = \frac {1} {2} (|Y| + |Z| - 1) \geq \frac {1} {2} (\alpha (G) + 1 + \alpha (G) + 1 - 1)$.
It follows that $|T| \geq \alpha (G) + \frac {1} {2} > \alpha (G)$.
A contradiction.

Thus, $G'$ is not bipartite. From König's theorem (as quoted by Grobber), it follows that $G'$ has an odd cycle, and so does $G$.

Pierre.

Hartman is mathematician, this problem has a hint maybe it leads to shorter solution,as pbornsztein said take $H=G[Y\Delta Z]$ and then use the theorem (I don't know its name )which tell us if G is bipartite then $\alpha(G)=\beta'(G)$
$\beta (G)$ means the minimum number of edges in $E(G)$ such that cover all vertices of G.

The theorem you are reffering to is also due to König (with the extra hypothesis that there is no isolated vertex).

Pierre.


The sides of a triangle have lengths $ 11,15,k$ . For how many integral values of $ k$ is the triangle obtuse?

edit:$ k$ is an integer

I think k should be the integer because if not ,there are a lot of values of k. :wink:  :)


calculus
integration
limit
function
calculus open

How can it be shown that the following limit exists, and is not zero? Numerical evidence suggests that the limit is approximately 2.17.

$\lim_{n\rightarrow\infty}\sqrt{n}\int_{-1}^{1}\left( \frac{2+x}3\right)^{n}(1-x^{2})^{-1/2}\, dx$

Substituting $t=1-x$, the integral becomes $\sqrt{n}\int_{0}^{2}\left(1-\frac{t}{3}\right)^{n}\sqrt{t}^{-1}\sqrt{2-t}^{-1}\,dt$
Now set $y=nt$; the integral becomes $\int_{0}^{2n}\left(1-\frac y{3n}\right)^{n}\cdot\frac1{\sqrt{y}\sqrt{2-\frac{y}{n}}}\,dt$.

As $n\to\infty$, the integrand tends to $\frac1{\sqrt{2}}e^{-\frac{y}{3}}y^{-\frac12}$; the limit is $\int_{0}^{\infty}\frac1{\sqrt{2}}y^{-\frac12}e^{-\frac{y}{3}}\,dy$.

This can be evaluated exactly; the substitution $u=\frac{y}{3}$ makes this into a multiple $\frac{\sqrt{3}}{\sqrt{2}}\int_{0}^{\infty}u^{-\frac12}e^{-u}\,du$ of the standard Gamma function integral. The limit is exactly $\sqrt{\frac{3\pi}{2}}$.


number theory
greatest common divisor
number theory unsolved

Which are all $ (a,b,n) \in \mathbb{N}^3$ such that $ a^{n\minus{}1}\plus{}b^{n\minus{}1}|a^n\plus{}b^n$?

We can assume WLOG that $ gcd(a, b) \equal{} 1$.  Then

$ \text{gcd} (a^{n \minus{} 1} \plus{} b^{n \minus{} 1}, a^n \plus{} b^n) \equal{} \text{gcd} (a^{n \minus{} 1} \plus{} b^{n \minus{} 1}, b^n \minus{} ab^{n \minus{} 1} ) \equal{} \text{gcd} (a^{n \minus{} 1} \plus{} b^{n \minus{} 1}, a \minus{} b) \equal{} \text{gcd}( 2b^{n \minus{} 1}, a \minus{} b) \equal{} 1, 2$

Hence $ a^{n \minus{} 1} \plus{} b^{n \minus{} 1} \equal{} 2$.  This occurs if $ n \equal{} 1 \implies a \equiv b \bmod 2$ or $ a \equal{} b \equal{} 1 \implies n \in \mathbb{N}$.  Furthermore, if $ (a, b, n)$ is a solution then so is $ (ka, kb, n)$, so we also have the more general solution set $ a \equal{} b, n \in \mathbb{N}$.

[quote="t0rajir0u"]We can assume WLOG that $ gcd(a, b) = 1$.  Then

$ \text{gcd} (a^{n - 1} + b^{n - 1}, a^n + b^n) = \text{gcd} (a^{n - 1} + b^{n - 1}, b^n - ab^{n - 1} ) = \text{gcd} (a^{n - 1} + b^{n - 1}, a - b) = \text{gcd}( 2b^{n - 1}, a - b) = 1, 2$

Hence $ a^{n - 1} + b^{n - 1} = 2$.  This occurs if $ n = 1 \implies a \equiv b \bmod 2$ or $ a = b = 1 \implies n \in \mathbb{N}$.  Furthermore, if $ (a, b, n)$ is a solution then so is $ (ka, kb, n)$, so we also have the more general solution set $ a = b, n \in \mathbb{N}$.[/quote]
I don't think that if $ (a,b,n)$ is solution then $ (ka,kb,n)$ can find out all solution.Do you agree?
I have a other solution
$ d = \gcd(a,b)$
$ a = dx,b = dy$ where $ \gcd(x,y) = 1$
$ d^{n - 1} (x^{n - 1} + y^{n - 1})|d^n(x^n + y^n)$
But from ${ \gcd(x^{n - 1} + y^{n - 1},x^n + y^n)\in \{1,2}$
If $ \gcd (x^{n - 1} + y^{n - 1},x^n + y^n) = 1$ then
$ \leftrightarrow x^{n - 1} + y^{n - 1}|d$
So for each $ (x,y)\in N$ and $ \gcd(x,y) = 1$ then exist $ d\in N$ such that
$ x^{n - 1} + y^{n - 1}|d$ (1)
So solution is $ (dx,dy,n)$ satisfy (1)
Case 2:$ \gcd (x^{n - 1} + y^{n - 1},x^n + y^n) = 2$
$ \leftrightarrow x^{n - 1} + y^{n - 1}|2d$
Same about.

Fll solutions gtt from $ a\equal{}(x^{n\minus{}1}\plus{}y^{n\minus{}1})x,b\equal{}(x^{n\minus{}1}\plus{}y^{n\minus{}1})y,n\in N$ were $ (x,y)\equal{}1.$

[quote="Rust"]Fll solutions gtt from $ a \equal{} (x^{n \minus{} 1} \plus{} y^{n \minus{} 1})x,b \equal{} (x^{n \minus{} 1} \plus{} y^{n \minus{} 1})y,n\in N$ were $ (x,y) \equal{} 1.$[/quote]
Why are these all the solutions?

If I'm not wrong, you can't get $ (1,2k\plus{}1,1)$ with that form..

Let $ d\equal{}(a,b), x\equal{}a/d, y\equal{}b/d$, because $ (x^{n\minus{}1}\plus{}y^{n\minus{}1},x^n\plus{}y^n)\equal{}t\le 2$ from $ a^{n\minus{}1}\plus{}b^{n\minus{}1}|a^n\plus{}b^n$ we get $ x^{n\minus{}1}\plus{}y^{n\minus{}1}|d(x^n\plus{}y^n)$. Therefore $ x^{n\minus{}1}\plus{}y^{n\minus{}1}|2d$. And all solutions are
$ (xd,yd,n)$ were $ x^{n\minus{}1}\plus{}y^{n\minus{}1}|2d, \ (x,y)\equal{}1$.


A guidance counselor is planning schedules for 30 students. Sixteen students say they want to take French, 16 want to take Spanish, and 11 want to take Latin. Five say they want to take both French and Latin, and of these, 3 wanted to take Spanish as well. Five want only Latin, and 8 want only Spanish. How many students want French only?

HOW CAN SOLVE THIS QUESTION USING VENN DIAGRAMS?

[hide] Draw three intersecting Venn Diagrams, one for French, one for Spanish, and one for Latin (for refrence).  There are 5 People who take Spanish and Latin, and 3 people who take Spanish, Latin , and French.  This means that only 5-3=2 people take only French and Latin.  There are 5 people who take only Latin, and 8 people who take only French.  We also know that there are 11 students who take Latin total, so there are 11-5-3-2=1 people who take Spanish and Latin only.  Since there are 8 people who take only Spanish and 16 people who take Spanish in all, there are 16-1-3=4 people who take only Spanish and French.  Therefore, there are 4 people who take only Spanish and French, 3 people who take all, and 2 people who take only Latin and French.  There are also 16 people total in the French Class.  Therefore, there are 16-4-3-2=$7$ people take only French.  [/hide]

This is exactly what I seek.  I needed guidance through the venn diagram part.

Thanks!


I think everyone should start playing...

Umm could you at least post a link, and some basic strategies/instructions?
Thanks.

haha it's an interesting game.. only beat the easy level so far.

[url]http://handdrawngames.com/desktopTD/Default.asp?cc=1[/url]

Oohh, I played this a long time ago. Never deigned to play past easy, though. *Goes and plays normal mode*

Going cold turkey is when you cut yourself off from a source of addiction in hopes of quitting.
I have no idea where the idiom came from, though.

[quote="daermon"]dude i got addicted to that game
bleh
i beat hard a few days ago, and am trying to go cold turkey[/quote]

No, don't! I need someone to beat my crappy scores.

aren't there cheats for that game? There was an old thread a while ago

strategy, make a maze of pellet towers so it takes longer for the creepo guys to get through.

That works for a while, but once you get to the harder ones, pellet towers do practically nothing.

I do it once in a while on the 10K gold mode, where you can really build a lot of stuff.  The later levels get insane, though.  I prefer to build quake towers within the maze of pellet towers; quake towers will kill just about anything.  You still have to deal with the flying ones, though, and whatever I do, I tend to die around level 80.

hacks ftw and you'll beat the high scores... lol

hmm...
ive seen like 200x of these kind of games
they're actually pretty generic *gets kicked*
but they can be fun  :D

Lol
As long as something's on the forum page there isn't really a difference, even if the thread's like 3 months old.
I came across this a few days ago. It wasted a good few hours...

yup.. just came across this thing..
interesting.. i remember playing one of these before!!
oh wait.. that was the one i died on lvl 12... :blush:

BUMP *spam*

onslaught.playr.co.uk/

got to wave 426 in easy mode

Wow, that game has crazy changed since I last played it. I think my high score was a couple billion or something, using nukes. My internet was really slow so I lagged out like crazy toward the end. I also tried a lot of laser chain strategies, but I kept lagging out at wave 100.

What are those frickin huge starting towers? And why don't you have any other towers? And what combo is that using?

wave 426, [b]no[/b] basic towers, those are railguns that you get at 1500 kills. HUGE SPLASH DAMAGE YAY

(also, I never use combos)

eh...i was on 380 something on easy and i decided to send 20 peeps at a time...and i died...yeah...coulda made it farther  :D

hey posts here actually count!! YES I CAN BOOST MY POST COUNT
ok just so that this post doesnt count as spam...i couldnt figure out how to do anything on here

Dang, I tried onslaught for the first time and only got to wave 357... Wow, lewis pwns me in math AND random flash games.

I remember Desktop tower defence. my top score on normal was 8124, I have a screenshot, but i'm posting this at home.

http://www.freewebarcade.com/game/anti-td/

an interesting way to see tower defense games

instead of being the one who makes towers you send out creeps

am on level 5 now

[quote="james4l"]www.freewebarcade.com/game/anti-td/

an interesting way to see tower defense games

instead of being the one who makes towers you send out creeps

am on level 5 now[/quote]

Sorry, this game is innovative, but that doesn't make it fun. I still prefer tower def over creep invade.

[quote="mewto55555"][quote="moogra"][quote="Nerd_of_the_Ages"]Actually, tower defense games are really common now. In one version, they actually found a build where you can beat all 50 levels using only the most basic weapon. You have to literally cover up every single building tower, and you can barely make it.[/quote]

Would that be Bloon Tower Defense 2? I did that once.
The games are useless and all can be hacked.[/quote]

Unless I'm crazy and/or mistaken (being from MO i may be both!), you can't beat Bloons Tower Defense 2 with only the basic monkeys because of lead balloons. You can beat Bllons Tower Defense (1) that way, but it is hard and you need to place the monkeys (monkies?) expertly.[/quote]

You're just mistaken. Theres things called road spikes.

I can and have beat the first one with only normal monkeys. I'll post my set up when I get home from school.

Hm, I just tried bloon tower defense last week, and I have to say, that's a pretty hard td game.

YO I GOT TO LEVEL 111 IN FLASH CIRCLE TD!!!
http://www.candystand.com/play.do?id=17995
Check out the high scores, I'm trigalg693, 15th place all time w00t. If I built some towers as a last stand I would've gotten 13th but whatever. I had 13 level 2 damage towers, about 50 level 1 splash, 20 level 5 air...but the air was f***ing ridiculous, like 30000 hp so I died.

It's retarded how your monthly ranking is higher than your weekly ranking

[quote="Brut3Forc3"][quote="moogra"][quote="mewto55555"]Right but road spikes can't pop lead balloons. Only cannons can.[/quote]

Again, the game is easily hacked and this is easy.[/quote]
I think the point of a game is [b]not[/b] to hack it, but to actually have fun instead...
However, I do agree that Bloons TD 1 and 2 are both really easy... :D[/quote]
I'm feeling nooby.

[quote="serialk11r"]So I figured out a better strategy, only building damage towers, and got to level 389 :D 12th place all time :)
If I built more air defense I think I could've gone a lot farther, trying that right now...[/quote]

Sadly, your weekly ranking is lower than your monthly ranking


For #1, it says that give 4 out of 5 for a minor gap or a really poorly written solution.

Define poorly written solution. Does it mean just a solution that's unclear or missing major steps or hard to read?

[quote="worthawholebean"]Any of the above. If I understand the rules correctly, graders can arbitrarily add/subtract a point for well/badly written solutions.[/quote]
Wouldn't that just be more of a subjective opinion, making your score depend on whoever grades your paper because people have different opinions about what's good and what's bad about a proof?

[quote="gauss1181"]Wouldn't that just be more of a subjective opinion, making your score depend on whoever grades your paper because people have different opinions about what's good and what's bad about a proof?[/quote]Yes.

Proofs will not be graded with 100% consistency. It just isn't realistic.

Read our article [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php]"How to Write a Solution"[/url], particularly sections 2 and 3.

People who write solutions that look like the "How Not to Write the Solutions" examples are the people who are likely to get a point deducted.


I am a high school senior and currently taking AP Java (my school only offers A)
tho I have no previous experience in programming, I developed a strong interest in computer science.

I plan to major in mathematics in MIT, and i thought it would be enjoyable to have cs as second major (or minor) in addition to math. but I'm worried as I see people who went to IOI and/or have plenty of experience of programming.

What is it like to major cs in college? is it most likely problemsolving and programming design? how hard is it to fill out those gaps?

and also, can you recommend some summerreading for beginners?

Thanks.

Well i can just advise for starting :D

Best site to practise programming (contest programming like IOI or ACM wrold final) is USACO and ACM.

And for book, coreman's book is best. though u can look for Sahni's book as well!

Thanks.
It seems like too hard for me yet, but I'll try.

Don't worry! The great programmers were beginner once! :D

USACO training pages will teach you a ton, pretty much everything you need to know to program algorithms efficiently -- http://ace.delos.com/usacogate.  Algo programming is a lot of fun, but I've heard of a lot of people changing their majors just because of the tediousness of programming large projects.

Yup! I also practise there. by the way in which level u r? I am in 4.3.

I am stuck on cryptcowgraphy. So far only 7 test cases pass and the 8th one runs in some 50 seconds. :(  How many test cases are there by the way? Am I close?


If it takes 4 hours to graduate 63 kids at Wintergreen Interdistrict Magnet School and only 2 hours to graduate 492 kids at nearby Hamden Middle School, approximately how many more kids, to the nearest ten-thousandth, graduate from Hamden Middle per minute? You may not use a calculator and assume the rates are constant.

[hide]It takes 4 hours to graduate 984 kids at the middle school so the answer is $\frac{984-63}{4\cdot 60}=\frac{921}{240}=\frac{307}{80}\approx \boxed{4}$[/hide]

Woops, problem should say to the nearest ten-thousandth, not hundredth.

[hide]At Hamden Middle you have $\frac{492}{120}=4.1$ kids per minute. At Wintergreen you have $\frac{63}{240}=.2625$ kids per minute. The difference is 3.8375 kids per minute.[/hide]
hmm...I wonder how you graduate part of a kid. :)

[hide]At Hamden Middle, it will take $\frac{492}{120}=4.1$ kids for every minute.

At Wintergree it will take $\frac{63}{240}=0.2625$ kids for every minute.  The difference hence is $3.8375$ kids per minute.[/hide]


For the linear expressions $A(x),\ B(x),\ C(x),$ assume that  $\{A(x)\}^2+\{B(x)\}^2=\{C(x)\}^2$ holds, then
Prove that both $A(x)$ and $B(x)$ can be expressed the constant times $C(x) .$

I'll take linear expressions to mean $f(x) = mx + b$.

$A(x) = ax + p; B(x) = bx + q; C(x) = cx + r$

Then $A(x)^2 + B(x)^2 = (a^2 + b^2)x^2 + 2(ap + bq)x + (p^2 + q^2)$ and $C(x)^2 = (c^2)x^2 + 2(cr)x + (r^2)$, giving

$a^2 + b^2 = c^2$
$ap + bq = cr$
$p^2 + q^2 = r^2$

Squaring the second relation,

$a^2 p^2 + 2abpq + b^2 q^2 = c^2 r^2 = (a^2 + b^2)(p^2 + q^2) = a^2 p^2 + b^2 p^2 + a^2 q^2 + b^2 q^2 \implies 2abpq = b^2 p^2 + a^2 q^2$

Factoring, this gives us $(aq - bp)^2 = 0 \implies aq = bp \implies \frac{a}{b} = \frac{p}{q}$, so clearly $A(x)$ and $B(x)$ can be expressed as multiples of the other, and the result follows.

(Some of these university entrance problems are a little easy.  :P )

Your answer is correct, t0rajir0u. :)

As you say, Japanese university entrance exam has been declining the difficulty compared with 10 years ago except

a part of universities. :o I will post  these problems.

kunny


$ \overline{AB}$ when $ AB$ is a $ 2$-digit integer means that $ \overline{AB} \equal{} 10a \plus{} b$, right? :maybe:


geometry
3D geometry
algebra unsolved
algebra

a and b are two distinct positive integers.  How many (a,b) pairs which makes  2a.(a^2+3.b^2)  a cube...

$ 2a^{2}\plus{}6ab^{2}\equal{}(a\minus{}b)^{3}\plus{}(a\plus{}b)^{3}$ ,$ a\plus{}b\equal{}y,a\minus{}b\equal{}x$ $ \implies$
$ x^{3}\plus{}y^{3}\equal{}z^{3}$  (fermat's last theorem)
only solution $ x\equal{}0$  $ \implies$  $ a\equal{}b$  :wink:


group theory
abstract algebra
superior algebra
superior algebra solved

let $N\triangleleft G$ and $H$ be a subgroup of $G$. Show $N \cap H \triangleleft H$

A subgroup $A$ is normal in $B$ if and only if $bab^{-1}\in A$ for all $a\in A, b\in B.$ If $n\in N, n\in H, h\in H$, we have clearly that $hnh^{-1}\in H,$ and since $gng^{-1}\in N$ for all $g\in G,$ $hnh^{-1}\in N.$

This is really quite trivial from the definitions.

We could also apply the fact that a group is normal if and only if it is the kernel of some homomorphism. Take such a homomorphism from $G$ for $N$; the restriction of this homomorphism to $H$ clearly has kernel $N\cap H.$


logarithms trigonometry geometry rectangle integration calculus function

The region bounded by the curve $ y\equal{}\ln(\cos x)$, the $ x$-axis and the line $ x \equal{} 1.5$ has area $ A$. The region is divided into $ 5$ strips, each of width $ 0.3$. By considering sets of rectangles, find upper and lower bounds for $ A$.

This is just a computation of upper and lower Riemann sums.  See [url=http://en.wikipedia.org/wiki/Riemann_sums#Methods]here[/url] for an explanation on how they're done.

Thanks, JRav. What bothered me about the question, which I've paraphrased, is that it specified $ 5$ rectangles for the lower bound for $ A$.

Can the left-most rectangle have zero height?

[quote="AndrewTom"]
Can the left-most rectangle have zero height?[/quote]

Yes, a rectangle in a Riemann Sum can can have a "height" of zero.

But the left-most rectangle for the Lower Sum in this instance should not have zero height. It should have a height of $ \ln(\cos(.3))$, since that value is less than zero (remember you want a [i]lower[/i] sum).

Hi gauss202. Shouldn't the left-most rectangle you describe be for the upper bound for $ A$ as the curve is monotonically decreasing and is below the $ x$-axis?

P.S. How do you get (a bit of) someone else's post to appear in a white rectangle?

This one is bothering me as I get a lower bound of $ 0.518466765$ and an upper bound of $ 1.313101862$ which seem quite far apart.

I'm also bothered about the stipulation of $ 5$ suitable rectangles for the lower bound.

I guess I could check it again andd then use more rectangles to get better and better approximations.

Is is possible to get an exact answet for $ \int_{0} ^{1.5} \ln (\cos x)\, dx$?

Why are you so troubled about using 5 rectangles?  There's nothing wrong with using 5 partitions.

If you don't have many partitions, your upper and lower sums will be somewhat far apart.  I'm not sure if that integral has a closed form but if you stick it in a calculator, it should give you 7 or so decimals.
[quote="AndrewTom"]P.S. How do you get (a bit of) someone else's post to appear in a white rectangle?[/quote]
Copy the part of the post you want to quote.  Follow the syntax below.
[code][quote="The person you are quoting"]Paste quote here.[/quote][/code]

Thanks for the tip, JRav. I'll try it, but I'm already using CTRL C for the backslash character which doesn't work on my computer.

I'm troubled because (unless I'm mistaken) one of the $ 5$ "suitable rectangles" must have zero height.

Also, I'd like to know if my values for the upper and lower bounds are correct.

P.S. Any thoughts on my Reduction formula question, especially tangents at the pole and sketching the curve $ r \equal{} \sin^{3}\theta$? I think I know what it looks like (a single loop symmetrical about $ \theta \equal{} \frac {\pi}{2}$) but I don't know how to post graphs.

P.P.S. Is taking more rectangles the only way of reducing the difference between the upper and lower bounds for $ A$?

[quote="An0maly"]Press the quote button in the bottom right of a post to quote their reply. As for your problem, I got the same values. I'm pretty sure the left most rectangle for the lower bound will have zero height. It should have height $ ln(cos(.3))$ for the upper bound.

Yes, the more rectangles you use, the lower the difference between upper and lower bounds should be.[/quote

Thanks An0maly.

Have you checked my values in the Newton-Raphson question? I'd like to know whether I've got the right answers or not.

In case you would like to know, the exact value of the integral is (to a few decimals) .8305 (the actual integral is negative but since you want area, we just need the magnitude).

Thanks, JRAv.

How did you get it? By taking more and more rectangles, via a computer?

Is it possible to do the integration?

I just stuck it in a calculator.  That function does not have an elementary antiderivative so you can't just plug in numbers.  If there is a way to do that integral, it will require some sort of algebraic manipulations and identities.

Thanks again JRav.

I wonder if anyone can do this integral.

$ \frac{\pi}{2}$ is approximately $ 1.570796327$ which is a little more than $ 1.5$ and I know that $ \int_{0}^\frac{\pi}{2} \ln(\cos x)\,dx$ = $ \int_{0}^\frac{\pi}{2} \ln(\sin x)\, dx \equal{} \frac{\pi}{2}\ln\frac{1}{2}$ which is approximately $ 1.088793045$. But that's not much help.

Perhaps kunny or farenhajt or Dr Sonnhard Graubner or ... might look at it.

We had a brief discussion about this integral in 'random' bounds here:

http://www.mathlinks.ro/viewtopic.php?t=209391

and the conclusion was that apart from the bounds $ <0,\frac{\pi}{2}>$ there's no closed form. Btw,

http://www23.wolframalpha.com/input/?i=int_{0}^{1.5}log(cos(x))dx

Thanks hsiljak. That's very useful.

I guess you're right. But I have a lot of faith in people like kunny, farenhajt, Sonnhard etc and so I haven't given up hope yet.

Using the trapezium rule, I get $ 0.915784313$, an over-estimate of about $ 0.085$ (I thought it would be a lot more).

Using the mid-point rule (that is, taking rectangles whose heights are the values of the function at the mid-points of the intervals), I get $ 0.79356784$, an underestimate of abpot $ 0.0369$ (closer!).

I haven't used Simpson's rule as we have an odd number of intervals (but we could use smaller intervals).

Are there other ways of evaluating this integral approximately?

Would it be possible to use series such as

$ \ln x \equal{} (x\minus{}1) \minus{} \frac{1}{2} (x\minus{}1)^{2} \plus{} \frac{1}{3} (x\minus{}1)^{3} \minus{} ...\, (0 < x \le 2)$

and $ \cos x \equal{} 1 \minus{} \frac{x^{2}}{2!} \plus{} \frac{x^{4}}{4!} \minus{} \frac{x^{6}}{6!} \plus{} ...$

or $ \ln\cos x \equal{} \minus{}\frac{x^{2}}{2} \minus{} \frac{x^{4}}{12} \minus{} \frac{x^{6}}{45} \minus{} \frac{17x^{8}}{2520} \minus{}... \minus{} \frac{2^{2n\minus{}1}(2^{2n}\minus{}1) B_{n}}{n(2n)!} x^{2n} \minus{} ...\, (x^{2} < \frac{\pi^{2}}{4})$

where $ B_{n}$ are the Bernoulli numbers?

Don't go crazy.  If you want a really good way to approximate the integral, go with [url=http://en.wikipedia.org/wiki/Simpson%27s_rule]Simpson's rule[/url].

hello, or use this method
http://math.fullerton.edu/mathews/n2003/RombergMod.html
Sonnhard.

Thanks, JRav. Halving each interval, Simpson's rule gives me $ 0.834306664$ and over-estimate of about $ 0.0038$

Thanks, Sonnhard: I clicked on the weblink but can't get any response from "Download this Mathematica Notebook Romberg integration". Perhaps because I haven't got Mathematica?

I've found $ \int \ln(\cos x)\,dx$ on an integrator site as

$ \frac{ix^{2}}{2} - x \ln (1+e^{2ix}) + x\ln(cos x) + \frac{1}{2} Li_2(-e^{2ix)}$,

where $ Li_2$ is the polylogarithmic function given by $ Li_2 (z) = \sum_{k=1}^{\infty} \frac{z^{k}}{k^{2}}$.

The only site which claims to do definite integrals, unfortunately gives the value of the definite integral as $ -9.$ something.

hello, yes, this is the reason, you will need Mathematica, but i think you can download a trial version.
Sonnhard.

Thanks Sonnhard.

I've edited my previous entry.


Hi!  I wrote my solutions to USAMTS R1/1 by hand, and my mom scanned it.  However, she accidentally submitted only one page.  She then emailed Mr. Patrick asking him if I could resubmit my solutions by US mail, and received a reply which said I could.

My solutions were postmarked on the deadline.  However, (a month later) when I checked my profile and clicked on the magnifying glass for Round 1, only Page 1 of my solutions showed.

I was kind of wondering if USAMTS received the solutions I mailed...  Thanks in advance!

USAMTS ID: 13534

Methinks emailing USAMTS directly is a better approach.  Hope that helps. :)

What shows up under the "magnifying glass" icon is always what you electronically submitted, so even if you mailed a replacement later, it won't get updated on the website.

I can't specifically check for your paper until we get them back from the graders next week.

p.s.: yes, for a specific issue like this, it's always best to email usamts@usamts.org, and always include your ID#.


Can anybody tell me what a "square-free" number is? I see it quite often but I don't know what it is.

$n$ is a square free number if and only if for every $p$ prime such that $p|n$ , $p^2 \not|n$

Look here : http://planetmath.org/encyclopedia/SquareFreeNumber.html

Or, if you prefer mathworld : http://mathworld.wolfram.com/Squarefree.html


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

Let R be a integral domain and a,b be elements of R.  If a^m=b^m and a^n=b^n and (m,n)=1, show that a=b.

$m\wedge n =1\Longrightarrow \exists (u,v)\in \mathbb{Z}^2 ~|~ mu+nv=1$ (Bezout's theorem).

${\left\{ \begin{array}{ll} a^m=b^m \\{ a^n=b^n} \end{array} \right.\Longrightarrow\left\{ \begin{array}{ll} a^{mu}=b^{mu}\\ a^{nv}=b^{nv} \end{array} \right.\Longrightarrow a^{mu+nv}=b^{mu+nv}\Longrightarrow a=b$

leibniz should specify he is working in the field of fractions of $R$ because I assume that Alope wanted to say $m,n\in\mathbb N$ otherwise $a^n$ does not always make sense in $R$. And this is not just because $a$ can be $0$.

There is a solution without the use of the field of fractions:
We assume that $ab\neq 0$, the $0$ case is trivial.
If $m,n\geq 0$ and $gcd(m,n)=1$, then there exist $u,v\in\mathbb N$ such that $mu-nv=\pm1$.
Assume that $mu-nv=1$. Then $a^m=b^m\Rightarrow a^{mu}=b^{mu}\Rightarrow a^{nv+1}=b^{nv+1}$. But $a^{nv}=b^{nv}$ and since they are nonzero by assumption and since we are working in an integral domain, we have $a^{nv+1}=b^{nv+1}=b^{nv}b=a^{nv}b\Rightarrow a=b$.

If $mu-nv=-1$, then $nv-mu=1$ and we can follow the steps of the previous case.

The case $a=0$ is trivial. So, we can assume that $a\neq 0$ , so $b\neq 0$ too.

$m$ and $n$ can be in $\mathbb{Z}$ because $R$ is integral, so it had a fraction's field. And that's why they give it.

[quote="amfulger"]leibniz should specify he is working in the field of fractions of $R$ because I assume that Alope wanted to say $m,n\in\mathbb N$ otherwise $a^n$ does not always make sense in $R$. And this is not just because $a$ can be $0$.

There is a solution without the use of the field of fractions:
We assume that $ab\neq 0$, the $0$ case is trivial.
[color=red][b]If $m,n\geq 0$ and $gcd(m,n)=1$, then there exist $u,v\in\mathbb N$ such that $mu-nv=\pm1$[/b][/color].
Assume that $mu-nv=1$. Then $a^m=b^m\Rightarrow a^{mu}=b^{mu}\Rightarrow a^{nv+1}=b^{nv+1}$. But $a^{nv}=b^{nv}$ and since they are nonzero by assumption and since we are working in an integral domain, we have $a^{nv+1}=b^{nv+1}=b^{nv}b=a^{nv}b\Rightarrow a=b$.

If $mu-nv=-1$, then $nv-mu=1$ and we can follow the steps of the previous case.[/quote]
Last remark, what you used here is a strong result of Bezout's theorem!

[quote="leibniz"]Last remark, what you used here is a strong result of Bezout's theorem![/quote]

Call it what you want. It follows from the existence of $u,v\in\mathbb Z$ such that $mu+nv=1$ by case study of the signs of $u$ and $v$. Not hard at all.


[quote="probability1.01"]Find all functions $f : \mathbb{R}\to \mathbb{R}$ with

$f(f(x)y+x/y) = xyf(x^{2}+y^{2}),$
$f(1) = 1.$[/quote]

Maybe you should have written $f: R-{0}\to{R}$ insted of $f: R\to{R}$?

Let's suppose you have done that :lol: .

If there is no such an $x_{0}$ such that $f(x_{0})=0$ then let's find an $y$

such that $f(x)y+x/y=x^{2}+y^{2}$. We can always do that because
\[f(x)y+x/y=x^{2}+y^{2}\]
<=>
\[y^{3}-y^{2}f(x)+y(x^{2})-x=0\]

and the latter is a cubic equation for $y$ which always has a nonzero real

solution $y_{0}$.Now we have that $f(x)y_{0}+\frac{x}{y_{0}}=x^{2}+y_{0}^{2}$=> $f(f(x)y_{0}+\frac{x}{y_{0}})=f(x^{2}+y_{0}^{2})$=>$xy_{0}=1$ =>$y_{0}=\frac{1}{x}$ => $f(x)=\frac{1}{x}$ for all nonzero real $x$.

The other case that is left is when there is such an $x_{0}$ that $f(x_{0})=0$ which gives us that $f(x)=0$ for all $x\in{R-{0}}$,but this seems easier to achieve, doesn't it?

Sorry, I forgot to mention that y is not zero. With that correction, it can still be R-->R. So how exactly did you handle the case when there is $x_{0}$ with $f(x_{0}) = 0$? I'm afraid I don't understand what you did (unless you haven't done that part yet).

The right side is symmetric, so if we switch the $x$s and $y$s on the right side, we dont change equality. Thus,

$f(f(x)y+x/y)=f(f(y)x+y/x)$

Assuming the function is 1-1,

$f(x)y-y/x=f(y)x-x/y$
$f(x)xy-y=f(y)xy-x$
$f(x)-f(y)=(y-x)/(xy)=(k-1/x)-(k-1/y)$
So $f(x)=k-1/x$, but $f(1)=k-1/1=1$, so $k=2$

$\therefore f(x)=2-1/x$

If the mapping isnt bijective i dont know what to do

gopherhole112, you mixed up the signs somewhere in your calculation. But anyways I don't think you can easily prove the function is bijective. Anyone want to complete Tiks' proof?


Given $a,b,c \geq 0$ . Prove that

$a^3+b^3+c^3-3abc
 \geq max{(a-b)^2 a, (a-b)^2 b, (b-c)^2 b, (b-c)^2 c, (c-a)^2 c, (c-a)^2 a}$

good luck  :D

Suppose that $ a \geq b \geq c $ then $a(a-c)^2 $is max
We have:
[tex] a^3+b^3+c^3-3abc -a(a-c)^2=2a^2c-a(3bc+c^2)+b^3+c^3 [/tex]
Let:
[tex] f(a)=2a^2c-a(3bc+c^2)+b^3+c^3 [/tex] and we need to prove that
$ f(a) \geq 0$ where $a \geq b \geq c $.
This is not too difficult,we have
$ f(b)= b^3+c^3-b^2c-bc^2 \geq 0 $ and $ (a1+a2)/2= (3b+c)/4 \leq b$
So $ f(a)\geq 0 $ with every $ a \geq b $
 :)


geometry
3D geometry
tetrahedron
geometry unsolved

Let $ A_1A_2A_3A_4$ be a tetrahedron. Denote by $ B_i,\ i \equal{} 1,2,3,4$ the feet of the altitude from a given point $ M$ onto $ A_iA_{i \plus{} 1}$ (where we consider $ A_5$ as $ A_1$). find the smallest value of $ \sum_{1\le i\le 4}A_iA_{i \plus{} 1}A_iB_i$

This is a problem on M&Y in Vietnam. Pleases lock it!


Hello,

Prove that :

\[ \displaystyle \forall n\in \mathbb{N}^{*},\;\lfloor\sqrt{n}\plus{}\sqrt{n\plus{}1}\rfloor \equal{}\lfloor\sqrt{4n\plus{}2}\rfloor.\]

Let $ k\leqslant\sqrt{4n+2}<k+1\implies {k^2-2\over 4}\leqslant n<{k^2+2k-1\over 4}$

[b]Case 1.[/b] $ n=2s$.

Then $ {k^2-2\over 4}=s^2-{1\over 2}$ and $ {k^2+2k-1\over 4}=s^2+s-{1\over 4}$, therefore

$ s^2\leqslant n\leqslant s^2+s-1$

Then

$ s\leqslant\sqrt{n}\leqslant{s^2+s-1}$

$ \sqrt{s^2+1}\leqslant\sqrt{n+1}\leqslant{s^2+s}$

Summing those two up, we get

\begin{eqnarray*}2s=s+s<s+\sqrt{s^2+1}\leqslant\sqrt{n}+\sqrt{n+1}\leqslant\sqrt{s^2+s-1}+\sqrt{s^2+s}<\sqrt{s^2+s+{1\over 4}}+\sqrt{s^2+s+{1\over 4}}=2s+1\end{eqnarray*}

Hence $ [\sqrt{n}+\sqrt{n+1}]=2s=k=[\sqrt{4n+2}]$

[b]Case 2.[/b] $ k=2s+1$.

Then $ {k^2-2\over 4}=s^2+s-{1\over 4}$ and $ {k^2+2k-1\over 4}=s^2+2s+{1\over 2}$, therefore

$ s^2+s\leqslant n\leqslant s^2+2s$

Then

$ \sqrt{s^2+s}\leqslant\sqrt{n}\leqslant\sqrt{s^2+2s}\qquad(1)$

$ \sqrt{s^2+s+1}\leqslant\sqrt{n+1}\leqslant s+1\qquad(2)$

Now we'll prove $ 2s+1\leqslant\sqrt{s^2+s}+\sqrt{s^2+s+1}$. Squaring it, we get $ 4s^2+4s+1\leqslant 2s^2+2s+1+2\sqrt{(s^2+s)(s^2+s+1)}$, which is equivalent to $ s^2+s\leqslant\sqrt{(s^2+s)(s^2+s+1)}$, and that yields $ \sqrt{s^2+s}(\sqrt{s^2+s+1}-\sqrt{s^2+s})\geqslant 0$, which obviously holds.

Hence, with apparent $ \sqrt{s^2+2s}<s+1$, by summing $ (1)$ and $ (2)$, we get

$ 2s+1<\sqrt{n}+\sqrt{n+1}<2s+2\iff[\sqrt{n}+\sqrt{n+1}]=2s+1=k=[\sqrt{4n+2}]$


Adding on to that, you usually have summations that involve expressions like $ \frac{1}{k\minus{}1}$ or $ \frac{1}{k(k\minus{}1)}$ and you have to represent them as a difference of two fractions which usually cancel out. The process of "decomposing" the original fraction is known as partial fraction decomposition. It is also used in calculus to make it easier to integrate rational expressions. Click [url=http://en.wikipedia.org/wiki/Partial_fraction]here[/url] for more information.

Would the series $ (1\minus{}\frac{1}{2})(1\minus{}\frac{1}{3})\dots(1\minus{}\frac{1}{10})$ be considered telescoping b/c nearly everything cancels out?  BTW, the product is $ \frac{1}{10}$.

[quote="Yongyi781"]Telescoping is when you have a sum $ \sum_{i \equal{} k}^n f(i)$ (finite or infinite) which is composed of two or more terms, where one term from $ f(i)$ "cancels" with the other term from $ f(i \plus{} 1)$, such as:
\[ \sum_{i \equal{} 1}^\infty \frac 1{i} \minus{} \frac 1{i \plus{} 1}
\]
In this case, when you expand it, you get $ \frac 11 \minus{} \frac 12 \plus{} \frac 12 \minus{} \frac 13 \plus{} \frac 13 \minus{} \dots$. You probably notice here that everything after $ \frac 11$ cancels, so the final sum is simply $ \frac 11 \equal{} 1$.[/quote]

eh im not sure but i dont think so notice how it is the sum not product. just pointing it out

Telescoping, in the strictest sense, is defined for sum.

[quote="Wikipedia"]In mathematics, a telescoping series is an informal expression referring to a series whose sum can be found by exploiting the circumstance that nearly every term cancels with either a succeeding or preceding term. Such a technique is also known as the method of differences.[/quote]

However, we could also use the term for products in which fractions "cancel out".

Think about why this statement is true: [i]You can use telescoping whenever you can write your sum in the following format:

$ \sum c_0 \cdot f(n) \plus{} c_1 \cdot f(n\minus{}1) \plus{} c_2 \cdot f(n\minus{}2) \ldots \plus{} c_k \cdot f(n\minus{}k)$

such that $ c_0 \plus{} c_1 \plus{} \ldots \plus{} c_k \equal{} 0$[/i]

Try writing out several terms in arbitrary examples such as:

$ \sum_{n\equal{}0}^{\infty} \frac{1}{n} \minus{} \frac{2}{n\minus{}1} \plus{} \frac{3}{n\minus{}2} \minus{} \frac{2}{n\minus{}3}$

Write it out to, say, 10 terms (you can arrange them in a grid so that it looks neat). Don't simplify at all (except subtracting in the denominators) until you've written out all the terms. What cancels? What stays?

What would happen if you were to sum that to infinity? What would cancel, what would stay?

[b]Regarding "telescoping" products:[/b] One way of thinking about these is to take the logarithm of the product, and then change that into a sum of logarithms of individual terms using properties of logs. Once the telescoping is done, you can undo the logarithm. Sometimes the sum of logarithms will decrease without bound as you use more and more terms. In this case, the original product is just $ 0$.

Hi!
Sorry If I am out of topic but this subforum is high scholl basics and you use series...(I don't disagree but I wan't to ask sth)
Are the series included in the high school syllabus (in USA).If yes how do you define them?
I am aking that,because in hig shcool (Greece) I have not taught series.I saw series for first time at the second semester of my university...


a is a real number and a # 0 Find all function $ f: R\minus{}>R$ such that : $ f(a\plus{}x)\equal{}f(x)\minus{}x$

Consider $ g(x) \equal{} f(x) \plus{} \frac {2}{a}.x^2 \minus{} 2x$
Replace to the equation we have :
$ g(x \plus{} a) \equal{} g(x)$
So solution is :
$ f(x) \equal{} g(x) \minus{} \frac {2}{a}x^2 \plus{} 2x$ where g is a periodic function with $ g(x) \equal{} g(x \plus{} a)$

He has defined $ g(x)$ in the first line.

Exactly $ g(x) \equal{} f(x) \plus{} \frac{x^2}{2a} \minus{} \frac {x}{2}.$

[hide][quote="Harry Potter"][quote="Rust"]Exactly $ g(x) \equal{} f(x) \plus{} \frac {x^2}{2a} \minus{} \frac {x}{2}.$[/quote]

Oh My god  :rotfl:
Do you know that we must find $ f(x)$   ?
You  :spam: ?
I realy don't like your post  :sleeping:[/quote]
You are mad. :mad: . Check things more carefully. You shouldn't expect solutions here so that every single step is sooo clarified that an one-year old can understand. What do you think? {Let everyone else do the hard part, why should I use my head?}. Don't you even understand that if one finds $ g$ then one finds $ f$ trivially by the relation $ f(x)\equal{}g(x)\minus{}\frac{x^2}{2a}\plus{}\frac{x}{2}$. It's ok if one don't understand. In that case this is not how he should reply. Ask politely to clarify. I don't understand many things here(I'm only a ninth grader), But I don't reply arrogantly like you. And you noted that two more people have posted here and many others have visited this topic. What do you think? Everyone is stupid and you are the smartest?[/hide]

It is not hard to chek, that $ f_0(x)\equal{}\frac x2 \minus{}\frac{x^2}{2a}$ is solution.
If $ f_1(x),f_2(x)$ are solution, then $ g(x)\equal{}f_2(x)\minus{}f_1(x)$ is periodic $ g(x\plus{}a)\equal{}g(x)$. Therefore all solutions give by $ f(x)\equal{}g(x)\plus{}f_0(x)$ with some periodic $ g(x)$.

[hide]@Harry Potter: Hnm, cậu làm thế chẳng hay ho gì đâu? Đề bài có bắt phải trình bày tại sao để tìm ra cách đặt $ g$ đâu? Xin lỗi nhưng mình nghĩ cậu đã hơi nông cạn:).[/hide]

harry potter i am not surprised from the nickname that u have chosen.... :D  :rotfl:


The two circles have even radius and are cut in $A$ and $B$. A line passing by $A$ recuts the first circle in $M$, the second in $N$. prove $BN=BM$

Even radius means equal radius, right?
In that case, have angle AMB equal to angle ANB. So their sines are equal. Angles MAB and BAN add up to 180 degees, so their sines are equal.Use Sine Law in MAB and BAN, get:
NB = AB * sinBAN / sinANB
=AB * sinMAB / sinAMB
=MB, QED

now; two circles  are cut in $A$ and $B$. $(D)$ is a variable line passing by $A$  recuts the first circle in $M$, the second in $N$ . prove that the medians of segment $MN$ are passing  by a constant point.

sorry.
 :blush:  :blush:  medians of segment $MN$

How do we get medians of a line segment? :?


vector
analytic geometry
geometry
geometric transformation

[quote="Prime factorization"]The derivation requires a lot of algebra so I'll just explain the basic idea.

The shortest distance between two parallel lines is a line that intersects both parallel lines at 90 degrees. The slope of this line is the negative reciprocal of the slope of the parallel lines.

You then use a bunch of algebra to find the two intersection points between the 2 parallel lines and the new line, then you apply the shortest distance formula.

If that's way too confusing, search for my "work over the summer thread" and check the geometry section. I have a derivation there that should be a little clearer.[/quote]

As a note, we can generalize this in a slightly simpler way, using Cauchy's Inequality.

[quote="b-flat"]I noticed that most of you assumed that $ \sqrt{14\plus{}8\sqrt3}\equal{} a\plus{}b\sqrt3$.  However, this is not always the case; we only know that $ \sqrt{14\plus{}8\sqrt3}\equal{}\sqrt{a}\plus{}\sqrt{b}$, if it can be simplified.

[hide="to simplify"]We know that $ a\plus{}b \equal{} 14$ and $ 2\sqrt{ab}\equal{} 8\sqrt3$, so thus, we have:

$ a\plus{}b \equal{} 14$

$ ab \equal{} 48\Rightarrow a \equal{}\frac{48}{b}$

Substituting, we have:

$ b\plus{}\frac{48}{b}\equal{} 14$
$ b^{2}\minus{}14b\plus{}48 \equal{} 0$
$ (b\minus{}6)(b\minus{}8) \equal{} 0$

Thus, $ b \equal{} 8$ or $ b \equal{} 6$ and we have that $ \sqrt{14\plus{}8\sqrt3}\equal{}\sqrt8\plus{}\sqrt6 \equal{} 2\sqrt2\plus{}\sqrt6$[/hide][/quote]

wow, i never thought of it that way. but it makes perfect sense. thats something new to put into my tool chest. :)

[quote="Prime factorization"]1337, please share with us this generalization of distance formulas.[/quote]

Say we wanted to find the distance from $ (x_{0},y_{0})$ to the line $ Ax\plus{}By\plus{}C\equal{}0$. We can prove by vectors or massive coordinate geometry that the desired distance is $ \frac{|Ax_{0}\plus{}By_{0}\plus{}C|}{\sqrt{A^{2}\plus{}B^{2}}}$. However, there is a slightly slightly slightly SLIGHTLY nicer way to do this, we can translate $ (x_{0},y_{0})$ to $ (0,0)$ and apply an appropriate translation to $ Ax\plus{}By\plus{}C\equal{}0$ so that the distance is the same, and work as follows.

(I'll leave the details of the proof to you, but here's a discrete example.)

So we want the distance from $ (0,0)$ to $ 5x\plus{}12y\equal{}60$. If we have a point $ (x,y)$ in space, on the line $ 5x\plus{}12y\equal{}60$, the distance between the points is $ \sqrt{x^{2}\plus{}y^{2}}$, so we wish to minimize $ \sqrt{x^{2}\plus{}y^{2}}$, because the minimum distance between the origin and a point on the line is really what the distance from a point to a line is (to prove this rigorously, think Pythagoras). Anyway, by Cauchy, $ \sqrt{x^{2}\plus{}y^{2}}\sqrt{5^{2}\plus{}12^{2}}\ge5x\plus{}12y\equal{}60$, so the minimum is 60/169.

[quote="diophantient"][quote="13375P34K43V312"]
(I'll leave the details of the proof to you, but here's a discrete example.)[/quote]
You mean concrete right? :D[/quote]

whooops


1. Determine $ x$, $ y$, and $ z$, given:
\begin{align*} x^3 + y & = 3x + 4 \\
2y^3 + z & = 6y + 6 \\
3z^3 + x & = 9z + 8 \end{align*}
2. Prove $ 2\left(a^5 + b^5 + c^5\right) = 5\left(a + b\right)\left(b + c\right)\left(c + a\right)\ldots$[color=red]I don't know what the "Z" means...[/color]
3. Given that $ x,\ y,\ z,\ w$ are real numbers, and $ x + y + z + w = x^3 + y^3 + z^3 + w^3$, determine $ w\left(w + x\right)\left(w + y\right)$.

123456789: I believe you made a typo in your rewriting of the problems.

The original 3rd problem stated to determine $ w(w\plus{}x)(w\plus{}y)$, not $ x(w\plus{}x)(w\plus{}y)$.


function
limit
induction
algebra unsolved
algebra

Find all continuous functions $f: \mathbb R \to \mathbb R$ satisfying
\[ 12f(f(x)) \equal{} 4f(x) \plus{} 3x\] for all $ x \in \mathbb R$.

[quote="ll931110"]I think this problem is familiar:

Find all continuous functions $ f: R \rightarrow R$ satisfying
$ 12f(f(x)) \equal{} 4f(x) \plus{} 3x$ for all $ x \in R$[/quote]

Yes, this is a rather classical problem.

1) $ f(x)$ is a bijection
$ f(a)\equal{}f(b)$ $ \implies$ $ 12f(f(a))\minus{}4f(a)\equal{}12f(f(b))\minus{}4f(b)$ $ \implies$ $ a\equal{}b$ and $ f(x)$ is injective
$ f(x)$ injective and continuous implies monotonous.
$ \lim_{x\to \plus{}\infty}f(x)\equal{}\pm\infty$ else, if $ \lim_{x\to \plus{}\infty}f(x)\equal{}L$ $ \lim_{x\to \plus{}\infty}12f(f(x))\equal{}12f(L)$ while $ \lim_{x\to \plus{}\infty}(4f(x)\plus{}3x)\equal{}\plus{}\infty$
Same, $ \lim_{x\to \minus{}\infty}f(x)\equal{}\pm\infty$
And so $ f(x)$ is surjective, hence bijective.

2) The only continuous decreasing solution is $ f(x)\equal{}\frac{1\minus{}\sqrt {10}}{6}x$
Let $ f(a)\equal{}b$ and the sequence : $ u_0\equal{}a$, $ u_1\equal{}b$ and $ 12u_{n\plus{}2}\equal{}4u_{n\plus{}1}\plus{}3u_n$
It's easy to see with induction that $ u_{n\plus{}1}\equal{}f(u_n)$
And $ u_n\equal{}\frac{ar_2\minus{}b}{r_2\minus{}r_1}r_1^n\plus{}\frac{b\minus{}ar_1}{r_2\minus{}r_1}r_2^n$ with $ r_1\equal{}\frac{1\plus{}\sqrt {10}}{6}$ and $ r_2\equal{}\frac{1\minus{}\sqrt {10}}{6}$ the two roots of $ 12x^2\equal{}4x\plus{}3$

So $ u_{n\plus{}1}\minus{}u_n\equal{}\frac{ar_2\minus{}b}{r_2\minus{}r_1}(r_1\minus{}1)r_1^n\plus{}\frac{b\minus{}ar_1}{r_2\minus{}r_1}(r_2\minus{}1)r_2^n$
Then, if $ ar_2\minus{}b\neq 0$, $ \lim_{n\to \plus{}\infty}\frac{u_{n\plus{}2}\minus{}u_{n\plus{}1}}{u_{n\plus{}1}\minus{}u_n}$ $ \equal{}\lim_{n\to \plus{}\infty}\frac{f(u_{n\plus{}1})\minus{}f(u_{n})}{u_{n\plus{}1}\minus{}u_n}\equal{}r_1>0$ and $ f(x)$ is increasing.
So the only decreasing continuous solution could be such that $ b\equal{}ar_2$ and so $ f(x)\equal{}\frac{1\minus{}\sqrt {10}}{6}x$
And this indeed is a solution.

3) The only continuous increasing solution is $ f(x)\equal{}\frac{1\plus{}\sqrt {10}}{6}x$
Let $ f(a)\equal{}b$ and the sequence : $ v_0\equal{}b$, $ v_1\equal{}a$ and $ v_{n\plus{}2}\equal{}\minus{}\frac 43v_{n\plus{}1}\plus{}4v_n$

It's easy to see with induction that $ f(v_{n\plus{}1})\equal{}v_n$  (notice we need to prove first that $ f(x)$ is a bijection, and we did in part 1)
And $ v_n\equal{}\frac{ar_2\minus{}b}{r_2\minus{}r_1}r_1^{1\minus{}n}\plus{}\frac{b\minus{}ar_1}{r_2\minus{}r_1}r_2^{1\minus{}n}$ with the same $ r_1$ and $ r_2$ than in point 2 above.

So $ v_{n\plus{}1}\minus{}v_n\equal{}\frac{ar_2\minus{}b}{r_2\minus{}r_1}(\frac{1}{r_1}\minus{}1)r_1^{1\minus{}n}\plus{}\frac{b\minus{}ar_1}{r_2\minus{}r_1}(\frac{1}{r_2}\minus{}1)r_2^{1\minus{}n}$
Then, if $ ar_1\minus{}b\neq 0$, $ \lim_{n\to \plus{}\infty}\frac{v_{n\plus{}1}\minus{}v_{n}}{v_{n\plus{}2}\minus{}v_{n\plus{}1}}$ $ \equal{}\lim_{n\to \plus{}\infty}\frac{f(v_{n\plus{}2})\minus{}f(v_{n\plus{}1})}{v_{n\plus{}2}\minus{}v_{n\plus{}1}}\equal{}r_2<0$ and $ f(x)$ is decreasing.
So the only increasing continuous solution could be such that $ b\equal{}ar_1$ and so $ f(x)\equal{}\frac{1\plus{}\sqrt {10}}{6}x$
And this indeed is a solution.

4) synthesis.
We just have two continuous solutions to this equation :
$ f(x)\equal{}\frac{1\minus{}\sqrt {10}}{6}x$
$ f(x)\equal{}\frac{1\plus{}\sqrt {10}}{6}x$


Let f(x) be geometrically concave on I, and a,b be contained within I.  How do you show that the value of f(x) at the average of a & b is >= the avereage of f(a)&f(b)?

[quote="Alope"]Let f(x) be [color=blue]geometrically concave [/color]on I[/quote]
What is this? i.e?

Sorry, I should have included that.

Let f(x) be defined on any type of interval I. For any subinterval [a,b] in I, we let P: (a,f(a)) and q: (b,f(b)) be two points of the graph lying over the endpoints of the interval. We say f(x) is geometrically convex on I of the graph of f(x) lies on or below the chord PQ, for all [a,b] in I. An equivalent analytic formulation of this is (f(x)-f(a))/(x-a) <= (f(b)-f(a))/(b-a) for any triple a<x<b in I.


I'm trying to figure out how to use bibtex in texniccenter (so I don't have to type out the bibliography entries manually), but I keep getting the same error. There is probably something simple I'm not doing right.

[code]\documentclass[12pt]{article}
\begin{document}
MIGHTY!
\bibliographystyle{plain}
\bibliography{bibtexttest2}
\end{document}[/code]

I copied and pasted the bibfile from a tutorial:

[code]@string{jgr = "J.~Geophys.~Res."}

@MISC{primes,
   author = "Charles Louis Xavier Joseph de la Vall{\'e}e Poussin",
   note = "A strong form of the prime number theorem, 19th century",
   year = 1879
   }[/code]

Here is the output log

[code]This is pdfTeX, Version 3.1415926-1.40.10 (MiKTeX 2.8) (preloaded format=pdflatex 2009.11.10)  11 JAN 2010 14:07
entering extended mode
**E:/RSI*2009/Paper/Capstone/test.tex
("E:/RSI 2009/Paper/Capstone/test.tex"
LaTeX2e <2009/09/24>
Babel <v3.8l> and hyphenation patterns for english, dumylang, nohyphenation, ge
rman, ngerman, german-x-2009-06-19, ngerman-x-2009-06-19, french, loaded.
(E:\miktex-portable\tex\latex\base\article.cls
Document Class: article 2007/10/19 v1.4h Standard LaTeX document class
(E:\miktex-portable\tex\latex\base\size12.clo
File: size12.clo 2007/10/19 v1.4h Standard LaTeX file (size option)
)
\c@part=\count79
\c@section=\count80
\c@subsection=\count81
\c@subsubsection=\count82
\c@paragraph=\count83
\c@subparagraph=\count84
\c@figure=\count85
\c@table=\count86
\abovecaptionskip=\skip41
\belowcaptionskip=\skip42
\bibindent=\dimen102
)
("E:\RSI 2009\Paper\Capstone\test.aux")
LaTeX Font Info:    Checking defaults for OML/cmm/m/it on input line 2.
LaTeX Font Info:    ... okay on input line 2.
LaTeX Font Info:    Checking defaults for T1/cmr/m/n on input line 2.
LaTeX Font Info:    ... okay on input line 2.
LaTeX Font Info:    Checking defaults for OT1/cmr/m/n on input line 2.
LaTeX Font Info:    ... okay on input line 2.
LaTeX Font Info:    Checking defaults for OMS/cmsy/m/n on input line 2.
LaTeX Font Info:    ... okay on input line 2.
LaTeX Font Info:    Checking defaults for OMX/cmex/m/n on input line 2.
LaTeX Font Info:    ... okay on input line 2.
LaTeX Font Info:    Checking defaults for U/cmr/m/n on input line 2.
LaTeX Font Info:    ... okay on input line 2.
 ("E:\RSI 2009\Paper\Capstone\test.bbl"

[b]LaTeX Warning: Empty `thebibliography' environment on input line 3.[/b]
) [1

{E:/miktex-portable/pdftex/config/pdftex.map}][/code]

When I run it, it gives me

"MIGHTY!

References"

but it doesn't list the references.

Thanks for the help!

The bibliography lists texts that were referenced in the document. So add something like
[code]See \cite{primes}[/code]to the document, compile up to 3 times and you will see the reference.
If you want to include a reference without citing it then use
[code]\nocite{primes}[/code]


In my book aops 1, I am on p. 44, where it first introduces adding and multiplying by mods.

However, it goes from 12+13 is congruent to 7+8 (mod 5), then 12^2 is congruent to 7^2 mod 5 (which I get) to quickly saying that->

 since 9899 is congruent to 4 (mod 5), and 7677 is congruent to 2 (mod 5), you can say 9899 x 7677 is conguent to 3 (mod 5), rather than multiplying 9899 and 7677 and modding out the product.

Can someone explain why the latter can be done.

1) I have no idea why you can just multiply like the above.

2) I do not understand what it means to "mod out the product."

thanks!

So you have $ 5k \equal{} 9899 \minus{} 4$ and you have $ 5l \equal{} 7677 \minus{} 2$

And you want to know $ 5j \equal{} 9899 * 7677 \minus{} x$ Well then $ 5j \equal{} (5k \plus{} 4)(5l \plus{} 2) \minus{} x$ and $ 5j \equal{}  5^2kl \plus{} 20l \plus{} 10k \plus{} 8 \minus{} x \equal{} 5j \equal{} 5(5kl \plus{} 4l \plus{} 2k) \plus{} 8 \minus{} x$. Clearly $ 5 | 8 \minus{}x$. so $ x \equal{} 3$.

"mod out the product" just means take the modulus of the product. You can think of modulus as a remainder operaton

so 5 mod 4 is 5 remainder four. Other definitions of mod $ x \equiv y \pmod{p}$ when $ p | (x \minus{} y)$. or alternatively when $ pk \equal{} x \minus{} y$ for some integer k.

The fundamental concept here is that equivalence $ \bmod n$ is an [b]equivalence relation[/b], which means it satisfies all the normal properties of the $ \equal{}$ sign.  In particular, if

$ a \equiv b \bmod n$
$ c \equiv d \bmod n$

then

$ ac \equiv bd \bmod n$.

[hide="Proof"] Write $ a \equal{} b \plus{} pn$ and $ c \equal{} d \plus{} qn$.  Then

$ ac \equal{} (b \plus{} pn)(d \plus{} qn) \equal{} bd \plus{} n(pd \plus{} qb \plus{} pqn) \equiv bd \bmod n$. [/hide]

[quote="t0rajir0u"]If

$ a \equiv b \bmod n$
$ c \equiv d \bmod n$

then

$ ac \equiv bd \bmod n$.[/quote]
Are there any restrictions on the variables?

no anyway they will come down to the same thing right?  :)

In this discussion it's understood that $ a, b, c, d$ are integers and $ n$ is a positive integer greater than $ 1$.  However, the notion of a [url=http://en.wikipedia.org/wiki/Congruence_relation]congruence relation[/url] can be generalized significantly.

[quote="t0rajir0u"]equivalence $ \bmod n$ is an [b]equivalence relation[/b], which means it satisfies all the normal properties of the $ \equal{}$ sign.[/quote]
That equivalence mod $ n$ is an equivalence relation gives you reflexivity ($ a \equiv a$), symmetry (if $ a \equiv b$ then $ b \equiv a$) and transitivity (if $ a \equiv b$ and $ b \equiv c$ then $ a \equiv c$), nothing more or less.  The fact that equivalence mod $ n$ "behaves properly" with respect to addition and multiplication (and, as a consequence, exponentiation) has to do with the fact that the integers modulo $ n$ are the quotient of a ring (the integers) by an [i]ideal[/i] (the multiples of $ n$), [i]not[/i] just an equivalence relation.

My apologies.  I was really thinking about congruence relations, then.   :oops:


Can someone tell me what they like about The Art and Craft of Problem Solving by Paul Zeitz? Many people seem to use it, but it seems to lack depth. For such a thin book, I don't see how it can cover everything from problem solving strategies to algebra, number theory, combinatorics, and calculus. I have had the book for about a year now, and I have never used it mainly because there are so many other books that can do a much better job than it does, especially in its coverage of specific areas. I think it is best suited for people who already have all the background needed for USAMO and are only fine-tuning their skills rather than for beginners. However, people keep recommending this book, and I am curious to know why.

What I like about The Art and Craft of Problem Solving by Zeitz is its ATTITUDE. I think he has an infectious enthusiasm about problem-solving. I also have Engel, Problem-Solving Strategies; Lozansky and Rousseau, Winning Solutions; Edward Barbeau et all, 500 Mathematical Challenges; and a whole lot of other problem-solving books, but Zeitz's book continues to be one of the most enjoyable to pick up and page through late at night.

Just throwing my two cents in:
It really depends on what you want.
I borrowed a friend's copy of Zeitz, and I found that I didnt like it. Its impression to me was that it started off really babying, but then took off like a rocket into much harder material.
Engel, on the other hand, was a book that I bought without any prior experience, but I had some people advise it as a great book. Kudos to them. He makes no bones about the material, and is very straightforward. He lays down the groundwork for the chapter, presents some very interesting problems and solutions, and then includes about 100 (i guess on avg this is correcT) problems for the reader to do. Most of them have solutions, but a couple I found say "Do it yourself"... which really shows where this book lies. Its not about the infectious enthusiasm, but rather teaching by problem-solving and investigating.

Personally, I think that Art and Craft is a simply beatiful book.  However, I got it at the best possible time (a few months after my first MOP) when I had the background to tackle it, but still had much to learn from it.  I think it is great on the philosophy, and it has all sorts of cool stuff in it.  I find it inspiring.


Self explanitory poll.

Also, what were some problems you got wrong?  I missed target # 6,8.

Go here [url]http://mathcounts.org/NETCOMMUNITY/Page.aspx?pid=295&srcid=296[/url] for the answers.

Wow this year's middle schoolers in Alabama stink if 1st place was only a 31..... I made like a 26ish last year and probably wasn't even in top 15

You would've been close. 1st was 34, 3rd was 31, but the rest could've had ties  :wink:

don't make fun @ me...... jk! just because alabama didn't score so high this year doesn't mean that it sucks. our team next year is going to be beast because we have dynamo, me, that randolpher, and a beast kid from my school(pizitz) that beat me at chapter.

did you take the AMC lifesapian?

:rotfl:

(at me being beast)

 :rotfl:

"That randolpher".... gah . He can be so annoying at regular mt tournaments. Every person that gets called out thats not him, he cheers. and then when 2nd place gets called out and he knows he's won, he jumps into the air. (Mavis does this too, but hes not as bad.)

@lifesapain: If you were on the team at state last year, you may remember will richardson or whatever calling me "bootyliscious" . Because of that he got 10 demerits or something and didnt get to go on the class trip to chicago!

HAHAHAHAHHAHAHh rofl @ dynamo. I remember that so well its still in my mind. You were the one Will called bootylicious? It was the only funny thing to happen there. No he didn't get in trouble for that. Our coach just talked to him and the rest of us laughed. Who on earth told you about the class trip thing? This year was Chicago, last years was Nantahala. And he didn't get to go to Nantahala because he called a teacher's shorts gay and got his 30th? demerit.This was 3 minutes before he could have been safe.

I love "that randolpher" hes Zaki Ahmed. Yes I did take the AMC 10 this year. Dont ask for my score it was in the low 100's. So dynamo and chestnut are 7th graders in Algebra maybe? thats good. Zaki should make nats next year and so could evil_ed's brother.

Gosh, thanks.

My teacher told me this year he was banned from the chicago trip the day we were talking about it, I thought she meant for the "bootyliscious" comment. BTW, I can totally see him saying that. I was SO glad he didnt even make countdown. (mwahahahaha!)

No, no. You have it all wrong. Next years nats team:

Botong (if this is who you are talking about, chestnut)
Mavis (ash)
Chestnut
Zaki

I will fail.

What are you talking about dynamo? Will R did make countdown, he was 8th written....

[quote="chestnut"]HAHAHAHAH! lifesapian, i beat you @ the AMC 10!
121.5(yes i did take the AIME and got a 4!)

Somehow, i take pre-algerbra  :rotfl:

yes, botong is beast

yes, Zaki is sooooooooooo annoying! At the VHHS toun. it wuz me and him when they were announcing second and first, he asked me for my score, and after i said my score...... he jumped in the air? rofl!!![/quote]

We dont do AMC or AIME or whatever bcuz it cost waay too much.

Gah. I am super hoping Zaki doesnt read the aops forums. We are being a bit mean.....
LOLZ DID U BELIEVE ME??? I HOPE HE SEES THIS!!!! MWAHAHAHAHA I AM EVIL ON AOPS!!!!!

I heard Botong talking to some guy telling him something and then he said: "Oh, yes, I bet you'll do good next year!"

@ lifesapain: well, I am getting that thing... oh whats it called.... its where you forget stuff... you know, that thing...

EDIT: also, I believe he must have then been pwnd by that beast kid with the squeaky voice....

EDIT AGAIN: we dont go to VHHS at challenger but they do at grissom. I want to take one of their tests, thats the reason we dont go is cuz my coach says its too hard....

Well that's a silly reason to not go. The test IS deathly hard, but that means its going to be hard for everyone, not just Challenger. So it'll be just like any other tourney, but with lower scores. Like you can be in the top 10 with a score of 35/100. At least thats how it was in Geometry.

[quote="chestnut"]i FAILED SO HARD!!!!!

i get a 21 on the sprint, and then bomb the target. A 10!  :wallbash_red:[/quote]

What targets did you miss??

[quote="chestnut"]4, 5, and 6

don't ask me how i got 8 right. i got it when the guy in our room called 15 seconds :lol:[/quote]

Oh yeah, didnt you just LOVE the engineering talks in between breaks? I personally was FASCINATED and AWESTRUCK.

Gah. Everyone in my family wants me to be an engineer, once I started math team. NONE of my 3 older siblings were on math team...... feel weird in my family...


Source: Austria-Poland Olympiad 1998

Consider all pairs (a,b) of natural numbers such that the product a^a * b^b written in decimal form ends in [i]exactly[/i] 98 zeroes.  Find the pair (a,b) for which ab is the smallest.

IMPORTANT EDIT:  Oops, I used a + instead of a * before.  Sorry.

IMPORTANT EDIT #2: Previously, the problem was missing the word "exactly".

just a guess (i don't feel like writing up a real solution now...and i might be wrong anyways lol): 800?

oops...it said pair. My guess is (32, 125)..and their product is 4000...

I doubt it. I think it goes deeper. The pair (1, 100) also words, and their product is 100.

Well, a=2^w*5^x, and b=2^y*5^z, with w, x, y, z, integers less than or equal to 5.

The main point I'm trying to make is a and b do not have to be relatively prime.

(1,100) doesn't work.  1^1 * 100^100 = 100^100, which would have 200 zeroes, not 98.  Unless I'm missing something...

Do you disagree that a number that ends in 200 zeros ends in 98 zeros? Or did you mean [i]exactly[/i] 98 zeros?

Oh.  I guess something that ends in 200 zeroes [i]would[/i] end in 98 zeroes.  But it seems too easy...I can't think of an easy way to beat 100.  I'll have to check.

Uh oh.  The site I got them from didn't have answers.

Oh, actually, the problem [i]does[/i] say exactly 98 zeroes.  So there's your answer...(1,100) does not work.

I have no clue if this is right, but the number of factors of 5 will need to be a multiple of 5, and the number of factors of 2 will need to be a multiple of 2. Thus it won't be possible to get exactly 98 5s. Thus we must get 98 2s. That leaves us with 2 possibilities: either use a for 5s and b for 2s or use a for 5s and 2s and b for just 2s. Let's try the first: We get a=75 or at least 105 and b=98. If we use the second possibility, we get a=50, but then what is b? We need 48 factors of 2 from b^b, and how do we do that? If b=12, then we get 24 factors of 2, and if b=24, we get 72. So there aren't any possibilities for that case. Thus the best solution we can get is (75, 98).

After re-reading and re-working the problem, I got the same solution as you, ComplexZeta. Yay!


I tried to write an essay in English. I copied the beginning of a Finnish document and edited it here is the code
[code]
\documentclass[12pt]{article}
\usepackage[T1]{fontenc}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
\newcommand{\F}{\mathbb{F}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newtheorem{defn}{Definition}[section]
\newtheorem{nota}{Notation}[section]
\newtheorem{lem}{Lemma}[section]
\newtheorem{cor}{Corollary}[section]
\newtheorem{prop}{Proposition}[section]
\newtheorem{theo}{Theorem}[section]
\title{Document}
\author{Puuhikki}
\maketitle
ffff
\end{document}
[/code]
Latex gave an error: ! Package babel Error: You haven't defined the language finnish yet.

What am I doing wrong?

Your code compiles without any errors on my computer. Also, the example you gave cannot be a minimal one: I'm ready to bet 100 to 1 that if the code you posted gives an error, you can remove quite a few lines from it and still have the same error (the reason for the suggestion of making a [b]truly[/b] minimal example is that, firstly, in 90% of cases, you will be able to see the problem yourself, and, secondly, in the remaining 10% of cases, we'll have a very clear idea of where to look).

My wild guess so far would be that you are using some Finnish version of $ \text{\LaTeX}$ and that it requires to specify some settings concerning some pecularities of Finnish language in every document but I can be completely off track here. What was the original Finnish heading?

The Finnish version starts like this (from my master's thesis)
[code]\documentclass[a4paper]{article}
\usepackage[T1]{fontenc}
\usepackage[latin1]{inputenc}
\usepackage[finnish]{babel}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{stmaryrd}
\usepackage{euscript}
\begin{document}
\newcommand{\F}{\mathbb{F}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\syt}{\operatorname{syt}}
\newcommand{\card}{\operatorname{card}}
\newcommand{\Hom}{\operatorname{Hom}}
\newtheorem{lau}{Lause}[section]
\newtheorem{defn}{M\"a\"aritelm\"a}[section]
\newtheorem{merk}{Merkint\"a}[section]
\newtheorem{lem}{Lemma}[section]
\newtheorem{esim}{Esimerkki}[section]
\newtheorem{cor}{Korollaari}[section]
\newtheorem{prop}{Propositio}[section]
\newtheorem{teor}{Teoreema}[section]
\newtheorem{huom}{Huomaa}[section]
\renewcommand{\refname}{\thesection\ Viitteet}
\renewcommand{\bibname}{\thesection\ Viitteet}
\begin{titlepage}
[/code]
An I'm using the standard teTeX system which I have installed on my Ubuntu using standard repositories and apt-get.

/etc/apt/sources.list

# deb cdrom:[Ubuntu 7.10 _Gutsy Gibbon_ - Release i386 (20071016)]/ gutsy main restricted
# See http://help.ubuntu.com/community/UpgradeNotes for how to upgrade to
# newer versions of the distribution.

deb http://fi.archive.ubuntu.com/ubuntu/ hardy main restricted
deb-src http://fi.archive.ubuntu.com/ubuntu/ hardy main restricted

## Major bug fix updates produced after the final release of the
## distribution.
deb http://fi.archive.ubuntu.com/ubuntu/ hardy-updates main restricted
deb-src http://fi.archive.ubuntu.com/ubuntu/ hardy-updates main restricted

## N.B. software from this repository is ENTIRELY UNSUPPORTED by the Ubuntu
## team, and may not be under a free licence. Please satisfy yourself as to
## your rights to use the software. Also, please note that software in
## universe WILL NOT receive any review or updates from the Ubuntu security
## team.
deb http://fi.archive.ubuntu.com/ubuntu/ hardy universe
deb-src http://fi.archive.ubuntu.com/ubuntu/ hardy universe
deb http://fi.archive.ubuntu.com/ubuntu/ hardy-updates universe
deb-src http://fi.archive.ubuntu.com/ubuntu/ hardy-updates universe

## N.B. software from this repository is ENTIRELY UNSUPPORTED by the Ubuntu
## team, and may not be under a free licence. Please satisfy yourself as to
## your rights to use the software. Also, please note that software in
## multiverse WILL NOT receive any review or updates from the Ubuntu
## security team.
deb http://fi.archive.ubuntu.com/ubuntu/ hardy multiverse
deb-src http://fi.archive.ubuntu.com/ubuntu/ hardy multiverse
deb http://fi.archive.ubuntu.com/ubuntu/ hardy-updates multiverse
deb-src http://fi.archive.ubuntu.com/ubuntu/ hardy-updates multiverse

## Uncomment the following two lines to add software from the 'backports'
## repository.
## N.B. software from this repository may not have been tested as
## extensively as that contained in the main release, although it includes
## newer versions of some applications which may provide useful features.
## Also, please note that software in backports WILL NOT receive any review
## or updates from the Ubuntu security team.
# deb http://fi.archive.ubuntu.com/ubuntu/ gutsy-backports main restricted universe multiverse
# deb-src http://fi.archive.ubuntu.com/ubuntu/ gutsy-backports main restricted universe multiverse

## Uncomment the following two lines to add software from Canonical's
## 'partner' repository. This software is not part of Ubuntu, but is
## offered by Canonical and the respective vendors as a service to Ubuntu
## users.
# deb http://archive.canonical.com/ubuntu gutsy partner
# deb-src http://archive.canonical.com/ubuntu gutsy partner

deb http://security.ubuntu.com/ubuntu hardy-security main restricted
deb-src http://security.ubuntu.com/ubuntu hardy-security main restricted
deb http://security.ubuntu.com/ubuntu hardy-security universe
deb-src http://security.ubuntu.com/ubuntu hardy-security universe
deb http://security.ubuntu.com/ubuntu hardy-security multiverse
deb-src http://security.ubuntu.com/ubuntu hardy-security multiverse

As fedja has already said, please give a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]complete minimal working example[/url] that exhibits the error. Neither of the codes you gave exhibit the error.
Start with the complete document then cut out as much as possible so the resulting document is small but still has the error.

This is the minimal example.
[code]
\documentclass[a4paper,12pt]{article}
\begin{document}
a
\end{document}[/code]
I also tried to reinstall tetex via synaptic but that does not resolve the error.

Thanks, but that shouldn't give any error at all. Can you post the log that you get?

screen:

This is pdfTeXk, Version 3.141592-1.40.3 (Web2C 7.5.6)
 %&-line parsing enabled.
entering extended mode
(./essee.tex
LaTeX2e <2005/12/01>
Babel <v3.8h> and hyphenation patterns for english, usenglishmax, dumylang, noh
yphenation, croatian, ukrainian, russian, bulgarian, czech, slovak, danish, dut
ch, finnish, basque, french, german, ngerman, ibycus, greek, monogreek, ancient
greek, hungarian, italian, latin, mongolian, norsk, icelandic, interlingua, tur
kish, coptic, romanian, welsh, serbian, slovenian, estonian, esperanto, upperso
rbian, indonesian, polish, portuguese, spanish, catalan, galician, swedish, loa
ded.
(/usr/share/texmf-texlive/tex/latex/base/article.cls
Document Class: article 2005/09/16 v1.4f Standard LaTeX document class
(/usr/share/texmf-texlive/tex/latex/base/size12.clo)) (./essee.aux

! Package babel Error: You haven't loaded the option finnish yet.

See the babel package documentation for explanation.
Type  H <return>  for immediate help.
 ...

l.3 \select@language{finnish}

?

log-file:
This is pdfTeXk, Version 3.141592-1.40.3 (Web2C 7.5.6) (format=latex 2008.10.5)  5 OCT 2008 23:33
entering extended mode
 %&-line parsing enabled.
**essee.tex
(./essee.tex
LaTeX2e <2005/12/01>
Babel <v3.8h> and hyphenation patterns for english, usenglishmax, dumylang, noh
yphenation, croatian, ukrainian, russian, bulgarian, czech, slovak, danish, dut
ch, finnish, basque, french, german, ngerman, ibycus, greek, monogreek, ancient
greek, hungarian, italian, latin, mongolian, norsk, icelandic, interlingua, tur
kish, coptic, romanian, welsh, serbian, slovenian, estonian, esperanto, upperso
rbian, indonesian, polish, portuguese, spanish, catalan, galician, swedish, loa
ded.
(/usr/share/texmf-texlive/tex/latex/base/article.cls
Document Class: article 2005/09/16 v1.4f Standard LaTeX document class
(/usr/share/texmf-texlive/tex/latex/base/size12.clo
File: size12.clo 2005/09/16 v1.4f Standard LaTeX file (size option)
)
\c@part=\count79
\c@section=\count80
\c@subsection=\count81
\c@subsubsection=\count82
\c@paragraph=\count83
\c@subparagraph=\count84
\c@figure=\count85
\c@table=\count86
\abovecaptionskip=\skip41
\belowcaptionskip=\skip42
\bibindent=\dimen102
) (./essee.aux

! Package babel Error: You haven't loaded the option finnish yet.

See the babel package documentation for explanation.
Type  H <return>  for immediate help.
 ...

l.3 \select@language{finnish}

? x

Here is how much of TeX's memory you used:
 201 strings out of 94102
 2267 string characters out of 1165833
 47703 words of memory out of 1500000
 3567 multiletter control sequences out of 10000+50000
 3938 words of font info for 15 fonts, out of 1200000 for 2000
 637 hyphenation exceptions out of 8191
 22i,0n,19p,137b,37s stack positions out of 5000i,500n,6000p,200000b,5000s
No pages of output.

Also please delete the aux file before compiling as that can cause problems from a previous compilation.

Thanks! Deleting the aux-files solved the problem.


linear algebra
matrix
vector
combinatorics unsolved
combinatorics

Since it's a real symmetric matrix of rank at most $n-1$ (because the sum of the entries of each row is zero), this is equivalent to showing that it's positive semidefinite, or, in other words, that all its eigenvalues (which we know are real because the matrix is symmetric) must be non-negative.

Assume $\lambda<0$ is an eigenvalue. Then ($L$ is the Laplacian matrix) $L-\lambda I$ would be singular, but it's been shown before on the forum that if an $n\times n$ complex matrix $(a_{ij})_{i,j}$ satisfies, for all $|a_{ii}|>\sum_{j=1,j\ne i}^n|a_{ij}|,\ \forall i\ (*)$ then it's non-singular, so we have a contradiction (it's obvious that $L-\lambda I$ satisfies $(*)$).

[quote="grobber"]Since it's a real symmetric matrix of rank at most $n-1$ (because the sum of the entries of each row is zero), this is equivalent to showing that it's positive semidefinite, or, in other words, that all its eigenvalues (which we know are real because the matrix is symmetric) must be non-negative.
[/quote]
Well, but how do you prove the existence of $U$?

[quote="liyi"]
Well, but how do you prove the existence of $U$?[/quote]

Didn't want to do [b]any[/b] of the work yourself, huh? :)

Let's prove this: given a real symmetric semipositive definite $n\times n$ matrix $A$ of rank $1\le r\le n$, we can find a real $n\times (n-r)$ matrix $U$ s.t. $UU^t=A$.

Since $A$ is real symmetric, it's orthogonally diagonalizable. This means that we can find an orthogonal matrix $K$ s.t. $A=KDK^t$, where $D=\ddag(\lambda_1,\ldots,\lambda_n)$ is a diagonal matrix with non-negative entries on the main diagonal, at least $r$ of which are $0$. We can assume WLOG that these are the last $r$, i.e. $\lambda_{n-r+1}=\ldots=\lambda_n=0$. Now take $v_i$ to be the row vector of length $n-r$ having $\sqrt{\lambda_i}$ as its $i$'th entry and $0$ everywhere else (for all $1\le i\le n$). Let $T$ be the $n\times(n-r)$ matrix having the vector $v_i$ as its $i$'th line for $1\le i\le n$, and put $U=KT$. Then, since $D=TT^t$, we have $A=KTT^tK^t=(KT)(KT)^t=UU^t$, as desired.

[quote="liyi"][quote="guru_ji"][quote="liyi"]Let $G$ be a connected graph on $n$ vertices and $L$ its Laplacian. Prove that there is $U \in \mathbb R^{n\times (n-1)}$ such that $L = UU^T$.[/quote]What is the definition of the Laplacian of a graph?[/quote]
http://mathworld.wolfram.com/LaplacianMatrix.html[/quote]
Thank you. And for the question.


What is the number of ordered pairs of positive integers that satisfy the equation 2x + 3y = 763?

[quote="236factorial"]What is the number of ordered pairs that satisfy the equation 2x + 3y = 763?[/quote]

well , that seems infinity pair to me since theres no restriction in x and y ....

[quote="236factorial"]sorry,  :blush: , positive integers.[/quote]

if so then it is just diophantine equation ....

a [i]what[/i]?

Hey that does sound familiar  :D

[hide]
Since 763 is odd, and 3(253)=759, y has to be an odd integer from 1 to 253. For every such y, there is exactly one x that satisfies the equation. Therefore, there are 127 possible ordered pairs. [/hide]

So it's just the greatest integer smaller that 763/6 (because LCM(2,3) = 6)? I think I've got it better now.


1. A fair six sided die is rolled 3 times. The positive difference of the first two rolls is equal to the third roll. What is th probability that al least one 3 is rolled?

I keep getting 5/12 but the answer is 7/15...

2. A four digit number is created using each of the digits 4,5,8, and 9 exactly once. What is the probability the number will be a multiple of 4?

I got 1/12 but answer is 1/6

3. In a science fair, there were six projects, each entered by a pair of students. During the fair some student visited other projects in the room ans shook hands with some of the other project entrants. No students shook hands with their own partners. After the fair, the students dscussed the projects they visited. Hannad learned that no two of the other eleven participants had shaken the same number f hands. How many hands did Hannah's partner shake?

Don't get it...

4. In a string of numbers, two adjacent digits are considered as a two digit number. For instance 11012 contains 10, 11, and 12. What is the number of digits in the smallest string that contains all of the two digit numbers from 10-99 inclusive?

[quote="john alabbad"]
2. A four digit number is created using each of the digits 4,5,8, and 9 exactly once. What is the probability the number will be a multiple of 4?[/quote]
[hide="Click for Solution"]

For a number to be divisible by $ 4$, the last two digits must be divisible by $ 4$.  The possibilities we have for that are:
\[ (1)48
\]

\[ (2)84
\]
With each one of these, the two digits can be in any order, or $ 2! \equal{} 2$ orders.  Thus our numerator is $ 2 \times 2 \equal{} 4$.  The denominator is simply $ 4! \equal{} 24$ because you want to put tose $ 4$ digits in any order.  Our answer is thus $ \frac {4}{24} \equal{} \boxed{\frac {1}{6}}$
[/hide]

[quote]3. In a science fair, there were six projects, each entered by a pair of students. During the fair some student visited other projects in the room ans shook hands with some of the other project entrants. No students shook hands with their own partners. After the fair, the students dscussed the projects they visited. Hannad learned that no two of the other eleven participants had shaken the same number f hands. How many hands did Hannah's partner shake?

[/quote]

[hide="Click for Solution"]

The first two partners can not shake hands with each other.  They both can shake hands with the remaining $ 10$ people though, meaning those two shake hands $ 10 \times 2 \equal{} 20$ times.  The next pair cannot shake hands with each other or the group before, so they can shake hands with $ 8$ people each.  Thus $ 8 \times 2 \equal{} 16$ handshakes.  The next two can shake hands with $ 6$ people for a total of $ 12$, the next two $ 4$ for a total of $ 8$ and the last two $ 2$ for a total of $ 4$.  THe last group has shaked hands with everyone so they don't need to.  Thus there were $ 20 \plus{} 16 \plus{} 12 \plus{} 8 \plus{} 4 \equal{} 60$ handshakes.  Since there were $ 12$ people, each one shoke hands with $ \boxed{5}$ people, which is our answer.[/hide]

[hide="2"] the only two endings that work are $ 48$ and $ 84$, which can be achieved in $ 2$ ways each, for a total of $ 4$ multiples of $ 4$. this is $ \frac{4}{4\cdot3\cdot2}\equal{}\frac16$ [/hide]

i have to go, i'll post the rest later.

argh. beaten.

[quote="john alabbad"]The correct answer to the number string one is 99.[/quote]

I just realized I was wrong.  By the way, what is the source?

These are from OPLET.

Does anyone know how to do the dice one?

[hide="1"]
For total possibilities, note that a sequence of three rolls can be made as long as the first two rolls are not equal, so there are $ 6\cdot5 \equal{} 30$ total possibilities.

For the possible sequences, either the third roll is a $ 3$ or one of the first two rolls is a three or both, giving $ 14$ possibilities: $ (1,4,3),(4,1,3),(2,5,3),(5,2,3),(3,6,3),(6,3,3);(1,3,2),(3,1,2),(2,3,1),(3,2,1),(3,4,1),(4,3,1),(3,5,2),(5,3,2)$.

Thus, the probability is $ \frac {14}{30} \equal{} \boxed{\frac7{15}}$.
[/hide]
[hide="3"]
I don't think I 007math's logic is correct even though he got the right answer...for example, he skipped the condition that all the handshake numbers must be different for the people other than Hannah.

Note that the least handshakes that anybody can make is $ 0$, and the most that anybody can make is $ 10$. Since the $ 11$ people other than Hannah all have different amounts of handshakes, and there are only $ 11$ possibilities, exactly one person must have $ n$ handshakes for $ 0\le n\le10$. Let $ a_n$ denote the person, not including Hannah, that has $ n$ handshakes.

Assume that $ a_{10}$ and $ a_0$ are not together. Then $ a_{10}$ has to shake hands with $ 10$ other people, which is impossible if he doesn't shake $ a_0$'s hand, contradiction. So $ a_{10}$ and $ a_0$ are partners.

Now, assume that $ a_9$ and $ a_1$ are not together. Since $ a_1$ only shook one hand, that hand must have been $ a_{10}$'s hand, since $ a_{10}$ shook everybody's hand except for $ a_0$'s. But if $ a_9$ was to have shaken $ 9$ hands, he must have shaken at least one of $ a_0$ or $ a_1$'s hands, or otherwise he won't have enough hands to shake. But we've already seen that neither of these hands can be shaken anymore, contradiction. So $ a_9$ and $ a_1$ are also together.

We can similarly proceed to show that $ a_8$ must be with $ a_2$, $ a_7$ must be with $ a_3$, and $ a_6$ must be with $ a_4$.

$ a_{\boxed5}$ is the only person left; he must be with Hannah (who also shook $ 5$ hands).
[/hide]
[hide="4"]
The optimal case will have every two digit number's last digit start a new number; we need $ 91$ digits for this. However, if the units digit of one of these two digit numbers is $ 0$, we cannot continue with another two digit number. Thus, we must add on one to each of the times we have a $ 0$, so we now are at $ 100$ digits. But if we end with a $ 0$ in our string of numbers, we can reduce this to $ \boxed{99}$ digits.

Now, to be rigorous, we have to show that such a sequence of digits is constructable.

Consider the following "one" sequence: $ 1121314151617181910$, which takes out all numbers with a $ 1$ in them. This takes $ 19$ digits.

Now, add the "two" sequence to this: $ 22324252627282920$, which takes out all numbers with a $ 2$ in them. This takes $ 17$ digits.

We can similarly continue for "$ n$" sequences to get a sum of $ 19 \plus{} 17 \plus{} \cdots \plus{} 5 \plus{} 3 \equal{} 99$ (where there are $ 9$ addends for each "$ n$" sequence!).
[/hide]


If $ w$ is a non-real cube root of unity, show that $ 1+w+2^{2}= 0$, and deduce that $ (1+w)^{3}(1+2w+2w^{2})$=1

(i)

$ \omega^{3}= 1$
$ \omega^{3}-1 = 0$
$ (\omega-1) (\omega^{2}+\omega+1) = 0$
$ \omega-1 = 0 \implies \omega = 1$ or $ \omega^{2}+\omega+1 = 0$

Since the first is clearly false, the second must be true.

(ii)

$ (\omega+1)^{2}= \omega^{2}+2 \omega+1 = \omega \implies \omega+1 = \omega^{1/2}$

$ (1+\omega)^{3}(1+2\omega+2\omega^{2})= (\omega)^{3/2}[2(1+\omega+\omega^{2})-1]= (-1) (-1)=1$


Regular hexagon $ ABCDEF$ is inscribed in a circle and $ P$ is a point on arc $ BC$ .Prove that :
\[ PE \plus{}PF\equal{}PA \plus{}PB \plus{}PC \plus{}PD\]


I found this problem interesting although I'm not sure whether my solutions include all possible ones.

Anyway, this problem is level of Target Round. I'm not sure for which level though. Either State or National, I guess.

In how many different ways can the integer 10 be written as the sum of four positive integers in which the order of the term is irrelevant? [That is, 4+2+2+2 is regarded same as 2+4+2+2.]

[hide="Hmm"]

I personall don't like this kind of problem but here are the ones I found:

Perhaps I missed some but I hope I didn't. :)

{1,1,1,7}
{1,1,2,6}
{1,1,3,5}
{1,1,4,4}
{1,2,2,5}
{1,2,3,4}
{2,3,3,2}
{2,4,2,2}
{2,5,1,2}
{2,6,1,1}
[/hide]

What do you think? :alien:

:blush: Haha..

Knew I missed it.

Hey, I remember seeing similar problem like this. Anyone knows neat way to figure out how many ways are there? :)

We still don't have the correct answer.  I think the way to solve this kind of problem is simply to be very well-organized.  I'll elaborate after somebody gets the correct answer.

I think you double counted a few times; To avoid this, i think it helps to write the numbers in [b] numerical order [/b].. to avoid miscounts.... (i think they were 1126 and 1225, i don't exactly remember)[/b]


Let ABC be a triangle with bisector AD, median CM. Assume AD=2CM, calculate $ \hat{C}$

Can you explain me what is the definition of bisector and median ? I am not sure with the definition that I have, really.

Median-A cevian (line) from a vertex of a triangle that bisects the opposite side of the triangle of that vertex.
Bisector-In the problem I think it's angle bisector.It's a cevian that divides the angle into 2 equal angles(bisects). :D