MATHCOUNTS
ratio
geometry
probability
3D geometry
tetrahedron
calculus

[size=150][b]First Round Matchups[/b][/size]

Kool_Dudy      vs.      RC-7th

Yifman12         vs.      Mlee06

theone853      vs.     white_horse_king88

bob123           vs.     Genius1601

h_s_potter      vs.      Lucky707

Everyone esle has a bye.

The time limit for round 1 problems will be 25 minutes.  Those ppl who are involved in this round(see above) need to pm me with a time that they are available.

Btw,  I want to post an attachment for the bracket but it is a word document.  Does anyone have any ideas for how I can make it accessible to everyone?

[edit]  Here is the list of competitors:

lonelytraveler8
Dune
Kool_dudy
RC-7th
EFuzzy
nr1337
yifman12
mlee06
tandjsnell
Peter VDD
SingtheSorrow
h_s_potter2002
Lucky707
white_horse_king88
theone853
LynelleYe
HashBrownie
bob123
Genius1601
Mat_88
soccerplayer

Convert the word file to a pdf file.
I'm not sure how to do it on a windows machine but if by some chance you have a mac just start like your going to print it and before clicking print click save to pdf instead. Unfortunately not very many people use macs.

Ok, I did the Go2pdf thing and here is the bracket... thanks to those who told me to use go2pdf.

[edit] I'll post scores when everyone has done there questions.

Joml, Can you get the bracket on one page so it's easier to read?

[quote="h_s_potter2002"]Can we have a contest to see who can fill out the bracket the most closest to how everything works out?[/quote]

Let's see...

On the first round that I'm in, I'll lose.

In the second round I'm in, I'll lose again and be out of the tournament! :lol:

Are the questions going to be PMed to us?

[quote="RC-7th"]After you PM Joml88 to tell him when you're avaliable[/quote]

So I suppose I should PM Joml88 to say when I'm available now?  (sorry - I haven't been on for a few days) (I'm not in this round though)

Does that mean you are going to post the problems after the round is over?  If so then I'll just wait till then, since I'm not in the first round.

[quote="joml88"]The way I am doing this is that I will appoint a time with everyone who is in the round.  At this time I will send them the questions and they will get a certain amount of time(25 min. for round 1 questions).  For the first round I want to let all the people participating in the round do the questions before I let other people see the questions.  I will be a little leniant.  So if you want to see the questions but aren't in the round pm me and I'll pm you the questions just as long as you don't talk about them or pm someone else about them etc.  I trust you wont.

Btw,  if those in the first round do not allot a time and answer the questions before Friday they will be disqualified.  I am available most of the time since it is the summer now.  I want to try to get this tournament finished before my school year starts.[/quote]

OK I should be available for round 2 on weekdays between 8:00 AM EST and 7:00 PM EST.

Moderator edit by tokenadult:

[i]I am going to be ruthless in editing out flames on this forum. Politeness counts, and I will back up my fellow moderators to the fullest, with gratitude for their contributions to the forum.[/i]

[quote="genius1601"][quote="joml88"][edit] I'll post scores when everyone has done [size=200]there [/size]questions.[/quote]
wrong use of THEIR buddy.[/quote]

I don't think you'd want to get on the bad side of the tournament coordinator.  ;)

Never mind. Obviously, this was the one I got wrong so I only got 4 on the third set of tiebreakers. Sorry everybody.

Ya, I was thinking about the n-2 part...but then I got confused. Hehe...

$\frac{180(n-2)}{n}$  must be an integer.  If we expand the top and break the fraction apart we get \[180-\frac{360}{n}\] So n must be a factor of 360.  $360=2^3\cdot3^2\cdot5$  so it has 24 factors.  2 of these(n=1 or n=2) wont work because the angle has to be positive.  Hence there are 22.

[quote="joml88"] If we expand the top and break the fraction apart[/quote]

This is what I noticed afterwards, however I didn't post it because I didn't want to and I didn't edit my previous post because I didn't think anyone would see it.

[quote="white_horse_king88"]so n can also be divisble by 360, but can't be 360[/quote]

Why can't n=360?  It worked in the example I posted before.  And each angle was 1 less than a straight line.

Tiebreaker 3, #4 (I like this question!):

[hide]We look at the equation (mod 8). All squares are  :equiv: 0, 1, or 4 (mod 8), so the LHS is the sum of two of {0, 1, 4}. The RHS is clearly 6 (mod 8). Since neither 4+1, 1+0, nor 4+0 = 6, the equation has no integer solutions.[/hide]

[quote="JSRosen3"]Tiebreaker 3, #4 (I like this question!):

[spoiler]We look at the equation (mod 8). All squares are  :equiv: 0, 1, or 4 (mod 8), so the LHS is the sum of two of {0, 1, 4}. The RHS is clearly 6 (mod 8). Since neither 4+1, 1+0, nor 4+0 = 6, the equation has no integer solutions.[/spoiler][/quote]

This was the hardest tiebreaker in my opinion.  I don't think that either white_horse_king88 or theone853 came up with this solution.  In 25 minutes with 4 other problems no one can expect them to!  I gave them both credit for it however.  The source of this problem for those who were wondering is the CanMO 1969.

I don't think that anyone has done the last one yet

[quote]5. How many paths from (0,0,0) go to (3,3,3) in exactly 9 moves without passing through (1,1,1) or (2,2,2)?
(A move is defined as increasing either the x, y, or z value by 1 e.g. possible moves from (0,0,0) are (0,0,1), (0,1,0), or (1,0,0))[/quote]

I don't know. I think this was harder for me than number 4 because I particularly stink at combinatorics and 3-D block walking problems.

[quote="RC-7th"]For this i think you can use the Principle of Inclusion and Exclusion, The number of paths to (3,3,3)-Paths through (1,1,1)-Paths through (2,2,2)+Paths through(1,1,1)and(2,2,2). It's easy enough to get the 4 numbers. Remember the number of paths to (x,y,z) is analogous to finding the number of ways to organize x indistinguisiable items, y items, and z items[/quote]

 I don't get it.  Aren't you supossed to find out how many paths there are to (3, 3, 3)?  If you're supossed to find out that for the problem, then why do you need to go through all those extra steps using that number?

[quote="RC-7th"]For this i think you can use the Principle of Inclusion and Exclusion, The number of paths to (3,3,3)-Paths through (1,1,1)-Paths through (2,2,2)+Paths through(1,1,1)and(2,2,2). [b]It's easy enough to get the 4 numbers. [/b]Remember the number of paths to (x,y,z) is analogous to finding the number of ways to organize x indistinguisiable items, y items, and z items[/quote]

Can you elaborate by showing the method because the trouble I ran into was that once either x, y, or z reached 3, the possibilities were got thrown off for me, so if there's a way to do this, I would love to see it.

I don't feel confident about this answer so could someone post the real answer!
To go from (0,0,0) to (3,3,3) you must add 3 xs, 3 ys and 3 zs. So, the total number of ways to get from (0,0,0) to (3,3,3) is just the number of ways you can order the word: xxxyyyzzz which is $\displaystyle\frac{9!}{3!3!3!}$. To find the number of ways that dont go through (1,1,1) or (2,2,2) we find the number of paths that do go through (1,1,1) or (2,2,2) and subtract from our other number.
By P.I.E, #Paths through (1,1,1) or (2,2,2) = #Paths through (1,1,1) + #Paths through (2,2,2) - #Paths through (1,1,1) and (2,2,2)
So, essentially N = (# of ways for xyz)*( # of ways for xxyyzz) + (xxyyzz)(xyz)  (xyz)(xyz)(xyz) = $\displaystyle\frac{3!}{1!1!1!}\displaystyle\frac{6!}{2!2!2!}+ \displaystyle\frac{6!}{2!2!2!}\displaystyle\frac{3!}{1!1!1!}- \displaystyle(3!)(3!)(3!)$ = 864.
So the number of paths that do not go through (1,1,1) or (2,2,2) are $\displaystyle\frac{9!}{3!3!3!}-864=816$.

Edit: This is an elaboration of RC-7th's method. This is, I believe, a question from an AMC-series contest and I believe this is the published method of solving the problem. I do know I've seen it before. Its a pretty cool solution though.

PorkChop's solution looks correct.  I made this problem myself.  In the intro. to counting/prob.  we did problems a lot like this one.  So I really got the idea from that.  I ve never seen this on an AMC but it is probably on at least one of them.

If anyone is confused as to how to get here try starting in two dimensions without restrictions.  e.g.  how many routes are there from (0,0) to (3,3) if you can only go in the positve x or positive y direction?

I actually thought about ordering xxxyyyzzz by first placing the x's, then placing the y's,and lastly placing the z's to get C(9,3)C(6,3)C(3,3)or 9!/3!3!3!


A graph has $ 2002$ points. Given any three distinct points $ A, B, C$ there is a path from $ A$ to $ B$ that does not involve $ C$. A move is to take any cycle (a set of distinct points $ P_1, P_2, ... , P_n$ such that $ P_1$ is joined to $ P_2$, $ P_2$ is joined to $ P_3$, ... , $ P_{n\minus{}1}$ is joined to $ P_n$, and $ P_n$ is joined to $ P_1$) remove its edges and add a new point $ X$and join it to each point of the cycle. After a series of moves the graph has no cycles. Show that at least $ 2002$ points have only one edge.

[quote="nguyentrang"]Let $ G$ be a graph with 2002 vertexes.It is known that if we delete some vertex then graph is still connected,i.e there is a path between any two vertexes. A move is to take any simple cycle and remove its edges and add a new vertex and join it to each vertex of the cycle. After a series of moves the graph has no simple cycles. Show that at least $ 2002$ vertexes have only one edge.[/quote]
We will prove that after series of moves,vertexes of the original graph $ G$ will have only one edge.
We will call graph strongly connected, if after deleting any vertex from the original graph $ G$,graph is still connected.From the statement $ G$ is strongly connected.
At first we will prove that after a move graph will stay strongly connected.
Indeed if $ G_1$ is some strongly connected graph and $ v$ is one of the vertexes of $ G$.Consider an arbitrary path $ l$ in $ G_1$, that does not contain $ v$.After a move we obtain graph $ G_2$.Denote the new vertex as $ u$.Some edges in $ l$ could be deleted, but the ends of the deleted edges now are connected with $ u$.Consider minimal segment of $ l$,that contains all deleted edges, and its ends are connected with $ u$.So we obtain a new path that doesn't contain $ v$ and the graph without $ v$ is still connected.
So after all moves our graph is still strongly connected.Suppose that there is still vertexes from original graph $ G$ with more than 1 edge.
Let $ A\in G$ and $ A$ is connected with $ B$ and $ C$.But then after deleting $ A$ there is a path that connects $ B$ and $ C$.Let it be $ p$.
But then $ p$ with edges $ \overline{AB}$ and $ \overline{AC}$ forms a simple cycle,contradiction.
So we have at least 2002 vertexes with unit degree.

Very nice solution, Erken  :lol:

[Moderator edit: see also http://www.mathlinks.ro/Forum/viewtopic.php?t=98552 ]


algebra
polynomial
integration
calculus
function
linear algebra

Hey all,
this is an assignment, so no solution please. Just a nudge in the right direction, as I am having trouble starting this one.

define the inner product of two polynomials p(t) and q(t) of degree less than or equal to 2 as follows:

$ < p,q > \equal{} \int_{ \minus{} 1}^1t^2p(t)q(t)dt$

Prove that $ < p,p > \equal{} 0 <\equal{}\equal{}> p \equal{} 0$

one of those implications is trivial, so I just need a nudge  for the first one. I can see that this is true intuitively, since <p,p> is the integral of an even function, but I am having trouble expressing it as a proof. Any help?

Not only is that function $ t^2p(t)^2$ even (which by the way it's irrelevant), but it is non-negative and continuous.

Right, I am just stuck on how to begin proving that the integral of a nonnegative function is also nonnegative

An integral is basically a sum. Whatever integral you're using, it's defined as a limit of sums of nonnegative bits (function values times set measures).

Now, showing that it's positive when $ p$ is not identically zero requires the continuity. Find somewhere where $ t^2p^2$ is positive, and stick a box under the graph.

[hide]The function $ f(t) \equal{} t^2p^2(t)$ is continuous in any interval [a,b], as f(t) is a polynomial. Since $ f(t) \geq 0$, if f(t) is not identically zero on [a,b] there is a $ t_0$ such that $ a \leq t_0 \leq b$ and $ f(t_0) > 0$. But since f(t) is continuous there exists an $ \epsilon > 0$ such that $ t \in [t_0\minus{} \epsilon, t_0 \plus{} \epsilon] \equal{}> f(t) > 0$. Hence,

$ \int_a^b f(t)\,dt \equal{} \underbrace{\int_a^{t_0\minus{} \epsilon}f(t)\,dt}_{\geq 0} \plus{} \underbrace{\int_{t_0\minus{} \epsilon}^{t_0\plus{} \epsilon}f(t)\,dt}_{> 0} \plus{} \underbrace{\int_{t_0\plus{} \epsilon}^{b}f(t)\,dt}_{\geq 0} > 0$.[/hide]


analytic geometry
geometry
geometric transformation

The points $ B(1, 1)$, $ I(2, 4)$ and $ G(5, 1)$ are plotted in the standard
rectangular coordinate system to form triangle $ BIG$. Triangle $ BIG$
is translated five units to the left and two units upward to triangle
$ B'I'G'$, in such a way that $ B'$ is the image of $ B$, $ I'$ is the image of $ I$,
and $ G'$ is the image of $ G$. What is the midpoint of segment $ B'G'$?
Express your answer as an ordered pair.

We want to find the midpoint of $ B'G'$.
We can first find the midpoint of $ BG$, which is $ (\frac{(1\plus{}5)}{2},\frac{(1\plus{}1)}{2}) \equal{} (3,1)$
The translations move the point to the left 5 units and upwards 2 units.
Thus, the midpoint of $ B'G'$ is $ (3\minus{}5,1\plus{}2) \equal{} \boxed{(\minus{}2,3)}$.


geometry
inequalities
probability
IMO
IMO 2006

Are there any news about Germany (the problems 2 and 6, I mean)?

Thanks

okay, I think you are Darij's mother yeah?? Did I guess  :)  if yes please say me ...
no the results are not known us for now ... if they will put them I will let you know

Sincerely
Davron Latipov

She's from Karlsruhe, and Darij is from Karlsruhe too, so... :)

I am quite confident Darij will get a gold medal.  :first:

Yes, it would seem that Darij, Peter, Zetax and Noname will each receive a gold medal, as predicted  :P  :winner_first:

I don't think all four of them will get a gold medal :huh:

First, I doubt whether gold will be on 28 (of course it's possible, let's say the chances are 50 50).

Second, are you sure you're not mixing up the names?

[quote="ZetaX"]Well, when valentin is right, gold begins at 17 points  :P
(I've 17...)[/quote]Ouch  :blush:  Sorry ... I had no knowledge of names, just DEU1, etc. and I knew who is Darij and who is Peter from Christian (Reiher) but I forgot to ask about the other two (possibly) gold medalists so I just filled in with the known MathLinks usernames :D

Arne, it is yet not clear if the cut-off will be 28 or 29, but 29 is certainly gold. I strongly belive that 28 will also be gold.

[quote="mom"][quote="Valentin Vornicu"]29 is certainly gold.[/quote]

It means, my son will finally get gold! I'm so happy and congratulate him and all his friends.

Natalia Grinberg,
former number 1 in the 1981 USSR team, who was not allowed (in the last minute) to quite the beloved motherland to participate at IMO in Washington.[/quote]

I am really sorry for you! But now we know why Darij is so pretty good at IMO. Our beloved professor of geometry and inequalities!  :D

Don't be sad ZetaX.  :)  Big congrats to the German team. You pwned the IMO. OK. Let's Darij do the chat-speak.  :first:

PS: Natalia, I told you Darij is going to score gold!  :lol:

Hallo,
I wanted to ask a question,
I just began studying CS in Karlsruhe and i wanted to know what does Darij have in mind to study then...ob he will stay in Karlsruhe too.? :-)
See ya

[quote="Searius"]I just began studying CS in Karlsruhe and i wanted to know what does Darij have in mind to study then...ob he will stay in Karlsruhe too.? :-)[/quote]

Hallo, no, sorry, I'm intending to study mathematics at the university of Munich (LMU to be precise) with computer science as a secondary subject.

Thanks to all for the congrats...

  Darij

[quote="José"]How many points did Peter Scholze score?[/quote]
35 or so.

[quote="José"]Strange. He got 42 in the last IMO. What happened to him?[/quote]
Who knows...There is also a Chinese who got 42 last year and didn't manage to solve all this year......

So, I have to conjecture that this IMO was more difficult thean last years' ... Am I correct?

Well, of course this IMO was MUCH harder. Only 10 people solved problem 6, for example.

In fact, problems 2 and 5 were harder than usual, and problems 3 and 6 were extremely hard.

[quote="andyciup"][quote="Bluesea"]to me it's so surprise that peter didn't solve problem 3 :o .However he did a great job;congratulations[/quote]

Why is that?[/quote]

Maybe because he got 42 in México, so you may think that, in the time period between México and Slovenia, he may have improved in math, so he had a high probability of scoring 42 again.

[quote="rem"]So Peter got perfect on IMO in grade 10... amazing! :) He will probably own next year, too.
Germany did really well this year. Congrats![/quote]
He will be 13th grade next year and is 12th grade now and did perfect score in 11th grade and participated first in 10th grade.

[quote="conceive"][quote="ZetaX"][quote="shobber"]Is it you last IMO, darij? And same question to ZetaX.[/quote]

It is the last IMO for the whole team except Peter, so lets see what happens next year...[/quote]

Well, do you still be with us ZetaX and Darij? Or are any of u going to leave the forum...?

conceive[/quote]

I will not, and darij will neither (I just asked him and he said: "I'm currently posting something").

congrats to the german team. you did much better than i had expected (although peter messed up another exam :D )  :winner_first:
dg

[quote="pavel kozlov"][quote="shobber"][quote="Soarer"]this isn't really true since difficulty of each imo varies. For example JPN2's score drops from 42 to 26.[/quote]
And CHN4 drops from 42 to 34......[/quote]
But RUS4 rises from 41 to 42.[/quote]

and Vladimir Serbinenko from Switzerland went from HM to gold  :lol:

[quote="Arne"][/quote][quote="pavel kozlov"][quote="shobber"][quote="Soarer"]this isn't really true since difficulty of each imo varies. For example JPN2's score drops from 42 to 26.[/quote]
And CHN4 drops from 42 to 34......[/quote]
But RUS4 rises from 41 to 42.[/quote][quote="Arne"]

and Vladimir Serbinenko from Switzerland went from HM to gold  :lol:[/quote]
Incredible.. :ninja:
from 030700 to 777610
What a huge boost!!

[quote="José"]How many points did ZetaX got?[/quote]

DEU 2	Daniel Harrer	711710:17   Bronze medal

[quote="Soarer"]
Incredible.. :ninja:
from 030700 to 777610
What a huge boost!![/quote]
That's perfect on day 1!!!! That's amazing! :o

Hey Valentin,
do u casually have the results of BMO 2002..
i partecipated there but i havent found it written noware??
U would do me a favour.
(perhaps somebody else can find it easily)

PS : Guys..the guy IURIE BOREICO.. perfect again!

[quote="Arne"][/quote][quote="pavel kozlov"][quote="shobber"][quote="Soarer"]this isn't really true since difficulty of each imo varies. For example JPN2's score drops from 42 to 26.[/quote]
And CHN4 drops from 42 to 34......[/quote]
But RUS4 rises from 41 to 42.[/quote][quote="Arne"]

and Vladimir Serbinenko from Switzerland went from HM to gold  :lol:[/quote]

Preved for Alexander Magazinov and Vladimir Serbinenko, my congratilations.


calculus
integration
invariant
modular arithmetic

Hi all, I was studying the invariance principle and I found a few problems that I'm stuck on. Could you please help?

Start with the integers 1,2...4n-1. In one move you may replace any two integers by their difference. Prove that an even integer will be left after 4n-2 moves.

A sequence is defined over non-negative integral indexes in the following way: a_0=a_1=1 and a_(n+1)a_(n-1)=a^2_n +2007. Find the invariant for this sequence.

The number 2^29 has 9 digits all different. Which digit is missing? Don't google the number or do other naughty things! Show your reasoning!

Thanks,

SuperMathBoy

I can't understand ques no 2
[hide="hint to 1"]
the number of odd integers is an invariance here. :) [/hide]

[hide="hint to 3"]A number is congruent to the sum of its digits modulo
9. [/hide]

[hide="soln to 3"]Lemma:A number is congruent to the sum of its digits modulo
9.
proof:
let the no. be $ x_1 \plus{} 10x_2 \plus{} 100x_3 \plus{} 1000x_4 \plus{} ..........$
sum of digits is $ x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \plus{} .............$
difference of number and sum of its digits is $ (x_1 \minus{} x_1) \plus{} x_2(10 \minus{} 1) \plus{} x_3(100 \minus{} 1) \plus{} x_4(1000 \minus{} 1) \plus{} .....$
so,difference=$ 9x_2 \plus{} 99x_3 \plus{} 999x_4 \plus{} ....... \equiv 0 \pmod{9}$
so,$ x_1 \plus{} 10x_2 \plus{} 100x_3 \plus{} ....... \equiv x_1 \plus{} x_2 \plus{} x_3 \plus{} ..... \pmod{9}$
so,lemma proved

here,$ 0 \plus{} 1 \plus{} 2 \plus{} ......... \equal{} 45$ (it is divisible by 9)

again,$ 2^{29}\equiv \minus{} 4 \pmod{6}$
so,the missing digit is $ 4$

I suppose it is rigorous enough[/hide]

I have basically the same thing for #3, but I'll say it a bit differently.

[hide="#3, basically same idea"]
Again, we prove that a number is congruent to the sum of its digits (mod 9).

[hide="proof"] Consider some number $ x_1x_2x_3...x_n$.

It is clear that the number is $ 10^{n\minus{}1}{x_1}\plus{}10^{n\minus{}2}{x_2}\plus{}...\plus{}10x_{n\minus{}1}\plus{}x_n$. Considering this in mod 9, the number is $ x_1\plus{}x_2\plus{}...\plus{}x_n$. However, that is just the sum of the digits, so our claim is proven.[/hide]

Now, we have $ 2^6\equiv1\pmod{9}$, so $ 2^{29}\equiv{2^5}\equiv5\pmod{9}$. As the sum of the integers 0 through 9 is 45, the number missing must be $ \boxed{4}$.[/hide]

[hide="For 2"]Consider the equation for $ n$ and $ n\plus{}1$.[/hide]


Let $ABC$ be an acute angled triangle and point $M$ chosen differently from $A,B,C$. Prove that $M$ is the orthocenter of triangle $ABC$ if and only if \[\frac{BC}{MA}\vec{MA}+\frac{CA}{MB}\vec{MB}+\frac{AB}{MC}\vec{MC}= \vec{0}\]

Trivial. If $M$ is orthocenter then equality is TRUE (well know). If we have equality then by $BC\cdot BM=\frac{BC^{2}+BM^{2}-CM^{2}}{2}$ and $CB\cdot CM=\frac{BC^{2}+CM^{2}-MB^{2}}{2}$ we have $AM\cdot BC=0$ similary $BM\cdot CA=0$. Done!

Solution of my students //http://trungtuan.wordpress.com/2007/09/24/solution-of-problem-2-grade-9-rno-2007/

[quote="TTsphn"]Good solution .
But $ S[AMB]$ is direct area so you must use direct angle.[/quote]
I mean that, it is useful area .

Actually, I haven't seen Grades 7 and 8 of this competition posted here. Have they been cancelled or what?


abstract algebra
group theory
analytic geometry
function
superior algebra
superior algebra unsolved

I don't understand this proof, specifically the part in red, I don't understand. Please help me understand this step in the proof. Thanks!

Let Tors be the category whose objects are torsion abelian
groups; if $ A$ and $ B$ are torsion abelian groups, we deﬁne $ \text{Mor}_{\text{\textbf{Tors}}}(A, B)$ to
be the set of all (group) homomorphisms $ \phi : A \rightarrow B$. Prove that direct
products exist in Tors; that is, show that given any indexed family $ {A_i}_{i \in I}$
where each $ A_i$ is a torsion abelian group, there exists a torsion abelian group
which serves as a direct product for this family in Tors.

Proof- Let $ T$ be the torsion subgroup (that is, the subgroup of elements of ﬁnite
order) of $ P \equal{} \prod_{i \in I} A_i \equal{} \{f : I \rightarrow A_i : f(i) \in A_i \text{ } \forall i\}$; here of course
$ P$ is the direct product of $ \{A_i\}_{i \in I}$ in the category Ab. Let $ j : T \rightarrow P$ be
the inclusion and for each $ i \in I$, let $ \pi_i : P \rightarrow A_i$ denote the usual projection
map; that is, $ \pi_i(f) \equal{} f(i)$. (In coordinate notation, $ \pi_i(a_0, a_1, \ldots) \equal{} a_i$.)
For each $ i \in I$ deﬁne $ \tau_i : T \rightarrow A_i$ by $ \tau_i \equal{} \pi_i * j$. I claim that the group
$ T$ together with the maps $ \{\tau_i\}_{i \in I}$ constitute a direct product for $ A_{i \in I}$ in
Tors. Well, given a torsion group $ S$ and maps $ \sigma_i : S \rightarrow A_i$ for each $ i \in I$,
one deﬁnes $ h : S \rightarrow T$ as follows: given $ s \in S$, let $ h(s) \in T$ be the function
deﬁned by $ \color{red}{[h(s)](i) \equal{} \sigma_i(s)}$. Then clearly $ \tau_i * h \equal{} \sigma_i \text{ } \forall i \in I$. Moreover,
if $ h' : S \rightarrow T$ is any other map such that $ \tau_i * h' \equal{} \sigma_i$, then for any $ s \in S$ and
$ i \in I$, $ \color{red}{[h(s)](i) }\equal{} (\tau_i * h)(s) \equal{} \sigma_i(s) \equal{} (\tau_i * h')(s) \equal{} \color{red}{[h'(s)](i)}$, so $ h \equal{} h'$.

I don't understand what $ \color{red}{[h(s)](i) \equal{} \sigma_i(s)}$ is.

Function-valued functions can be confusing; I've always found that stepping through the notation in painstaking detail is very helpful.

By definition, h : S --> T
Therefore, $ h(s) \in T$
By definition of T, h(s) is a function $ I \to A_i$
h(s)(i) is, therefore, evaluating h(s) at i. (And $ h(s)(i) \in A_i$)


The smaller angle between the hands of a clock at $ 12: 25$ p.m. is:

$ \textbf{(A)}\ 132^\circ 30' \qquad
\textbf{(B)}\ 137^\circ 30' \qquad
\textbf{(C)}\ 150^\circ \qquad
\textbf{(D)}\ 137^\circ 32' \qquad
\textbf{(E)}\ 137^\circ$

[hide="Solution"]A clock can be divided into $ 12$ sectors, each worth $ 5$ minutes. Each of these is $ \frac {360}{12} \equal{} 30$ degrees, and there are $ 5$ sectors, so we have $ 5\cdot 30 \equal{} \boxed{\textbf{(C)}\ 150^\circ}$.[/hide]

Ah, forgot that the hours hand moves slightly   :mad:
To be fair, they might have been using analog clocks in 1955, but it's all digital now ;)

while the minuts hand moves $ x$  hours hand moves $ \frac {x}{12}$. A hour is equal to $ 60$ minuts, and the circle has $ 360 \circ$  ,so $ 1$ minut represent $ \ 6 ^{\circ}$. In our case the angle is : $ 25 \cdot 6 \minus{} \frac {25}{12}\cdot 6 \equal{} 150 \minus{} 12,5 \equal{} 137^{\circ}30'$ or $ \boxed B$


geometry
probability
inequalities
symmetry
summer program
Mathcamp
geometry proposed

The first problem I saw somewhere on a math teacher's web(http://www.denizyuret.com/math/problems.html)(problem number 50),
Here we go
1- Looking at a line segment of an unknown length what is the probability when two points are chosen randomly on the line that  the resultant line segments will create a triangle? and then What is the probability that when 3 lengths are chosen randomly on the line that the resultant segments would constitute a triangle?

Let's try the first problem. Suppose that the line segment is defined by points A and D. Then we are going to pick two points B and C on line segment AD.

The first problem is to define random. Here we want to pick the points uniformly from AD. This means that each interval of the same length has the same chance of having the point fall into it.

The point chosen closest to A will be B and the one farthest from A will be C. That makes it simple, because the points will be in the order A, B, C, and D. A triangle can only be formed from segments AB, BC, and CD if C if more than half way along AD. Otherwise, the other 2 sides of the triangle could 'reach' all the way back to D.

Let's call the midpoint of AD point E. Thus |AE| = |ED|. One of the two points has to be chosen from the segment ED to form a triangle.

Since we clarified the problem that choosing a random point is to choose points uniformly random. The probability of choosing a point that lies in AE is the same as the probability that the point lies in ED, since the segments are the same length. The probability is 1/2.

The probability both points are chosen from AE is (1/2)*(1/2) = 1/4. When that happens no triangle is formed. Therefore the probability of forming a triangle is 1-1/4, which is 3/4.

I tried my hand on the second one but i don't have a complete solution and i am not certian that my steps are correct so please complete mine if you can or come up with your own.

The second problem is simply chossing 3 lengths on a line segment of unknown length, which is equivelant of saying that we will choose 3 points randomly on the line, and figure out what the probablity is that the absloute value of the 3 lengths would constitute a triangle.
Let the line segment have a length X. So the line begins at 0 and ends at +X.
Also let the 3 points choosen be A, B, C.
Since we are talking triangles
B + C > A  , A + B > C , and A + C > B.  The probablity of  B + C > A   occuring is P(B+C>A given A,B) , Then the complement of that probability would be ( B < A , C < A, and B + C < A) all occuring at the same time. The probability of B<A is .5( I think) and the probablity of C < A is also .5 and the probability of         B + C <  A  is .25 and therefore the P(complement)=(.5)(.5)(.25)=.0625
Thus P(B+C>A given A,B)=1-.0625= .9375, but this condition must apply for all 3 inequalities created by the sides of the triangle and therefore the probablity is (.9375)^3.
Yet it is not over yet because there I might have over counted some triangles so i use the PIE
let B+C>A be set Z        and   C+A>B   be set Y      and  B + A > C be set W
now using the PIE the P(3 lengths(points) making a trinagle) = (.9375)^3 - 3( intersection between any 2 of the 3 sets(Z,Y,W) due to symmetry) + ( intersection of all 3 sets X,Y,Z)

And There I stop because I don't know what to do from there; that is if these steps are somewhere in the vicinity of the correct solution.

I made a mistake by putting this here because i don't have a solution yet for the second part of the problem.

I am working on the solution to be sure I've got it right. I believe the answer will be 1/2. Remember that it is important how the word random is defined. We often think of choosing uniformly random, but that was not defined in the problem. We could just as well have stated that there are 6 fixed points and you choose some of these with some probabilities.

The introduction Geometrical Probability by Kendall and Moran has an interesting discussion on how defining random leads to 3 answers: 1/2, 1/3, and 1/4.

If it is not a bother could you define random for the 3 situations and solve each situation. I am trying to find the book you wrote about, hopefully i can find it.

I found the answer, and i believe it is the same no matter what definition of random you choose. I have solved it, but there is that summer mathcamp qualifying quiz thing on the internet and the application deadline is april 25th so i will post it 26th of april just to be safe.

by the way Stereologist, the answer is not 1/2.

sorry, it is 1/2, i just was to happy that i found my awsome solution that i rushed and didn't complete it.But now it is complete.


If you set up the equation:

$ \zeta(s) \equal{} a\zeta(s\plus{}b)$, (A ∩ B)∈ℝ

where zeta is the riemann zeta function

Is it possible to find da/db?

Should not be too hard, since the only choice is $ a\equal{}1$ and $ b\equal{}0$.  :D

[quote="mlok"]Should not be too hard, since the only choice is $ a \equal{} 1$ and $ b \equal{} 0$.  :D[/quote]

Why would that be?

ζ(s) = ζ(s) for all s

ζ(s) = aζ(s + b) for some relation a to b

Even if it's not a closed form, I want to find da/db.


The mean of three test scores is 74. What must a fourth test score be to increase the mean to 78?

Let's say that we want the test scores of 78,78,78,78 but we only have 74,74,74.
Since the 1st test is 4 less than our target, and our 2nd test is 4 off the target, and the 3rd is 4 off, we our 12 points off our test goal.  We make up those 12 points with 78+12=90

Also, 78*4=312 points needed
74*3=222 points so far
312-222=90 points

The sum of the $ 3$ test scores is $ 74 \times 3$ or $ 222$.  If the average needs to be a $ 78$, the sum of the test scores would be $ 78 \times 4$, or $ 312$.

$ 312 \minus{} 222 \equal{} \boxed{90}$

EDIT: Beaten.

Also using weighted averages, 3 time it is 4 below, so the other time it must be 12 above, so 78+12=90


ratio
limit
calculus
calculus computations

Determine the convergence of divergence of the series:
$ \sum_{n\equal{}1}^{\infty} \left( \frac{1}{n^2} \minus{} \frac{1}{2^n} \right)$

hello, your series is convergent and we get $ \sum_{n\equal{}1}^{\infty}\left(\frac{1}{n^2}\minus{}\frac{1}{2^n}\right)\equal{}\frac{\pi^2}{6}\minus{}1$
Sonnhard.

Do you just take the sum of each and subtract?

hello, using the generalized ratio test we get
$ \lim_{n \to \infty}\frac{a_{n\plus{}1}}{a_n}\equal{}1/2<1$ so our series converges.
Sonnhard.

A more amusing and intuitive way to think about it :
Draw a square with side length 1.
Then draw a line from the midpoints of two parallel lines. (So draw a line cutting the square in half, but not a diagonal)
Then draw another line cutting one of the halves in half. And continue this way. You'll see that the sum of all these halves is the unit square itself, ie. 1^1 = 1 ^^
A thing to note I guess : Changing $ 2$ to any value greater than $ 2$ will converge. And this makes sense since all the terms will be smaller anyways.

Determine the convergence/divergence of the series:
$ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^n \sqrt{n}}{n\plus{}1}$

Would the alternating series test work here?

Size matters.

For large $ n,$ how big is $ \frac{\sqrt{n}}{n\plus{}1}\ ?$ The first thing you should do is "de-clutter" that and say that it looks like $ \frac1{\sqrt{n}}.$

Does that tend to zero (without oscillating)? Yes, it does. So the alternating series test applies.

But before we even get to the question about the alternating series test, we should decide whether this converges absolutely. Does this go to zero fast enough for absolute convergence? That's asking whether $ \sum\frac1{\sqrt{n}}$ converges, and you should know that immediately.

(math92, I need an answer from you before saying anything more.)

[color=green][Please do not quote the preceding post. It's redundant.][/color]

Thanks for this help.

I'm unsure of how you "de-cluttered" to get $ \frac {1}{\sqrt {n}}$.
$ \sum\frac1{\sqrt {n}}$ does not converge because of the p-series test.  $ p \leq 1$ so it diverges.

The "de-cluttering" is the essence of the limit comparison test.

We see that in the sum $ n \plus{} 1,$ the $ n$ is much, much more important than the $ 1.$

$ \lim_{n\to\infty}\frac {\frac {\sqrt {n}}{n \plus{} 1}}{\frac {\sqrt {n}}{n}} \equal{} \lim_{n\to\infty}\frac {n}{n \plus{} 1} \equal{} 1.$

That means that $ \frac {\sqrt {n}}{n \plus{} 1}$ equals the term $ \frac {\sqrt {n}}{n} \equal{} \frac1{\sqrt {n}}$ plus something that is much smaller than that term as $ n\to\infty.$ These two terms are essentially the same size, and series containing them will behave in essentially the same way.

Now: you have denied the absolute convergence of the original series by limit comparison to $ \sum\frac1{\sqrt {n}}$ but affirmed its convergence by the alternating series test. Therefore, the original series converges conditionally.

Thanks.  The limit comparison makes a lot more sense to me now.
I don't see where you affirmed its covnergence by the alternating series test though.

[quote="Dr Sonnhard Graubner"]hello, using the generalized ratio test we get
$ \lim_{n \to \infty}\frac {a_{n \plus{} 1}}{a_n} \equal{} 1/2 < 1$ so our series converges.
Sonnhard.[/quote]

- We don't need the ratio test at all there, esteemed S. G.
- @math92: I guess that K. M. inferred from one of your previous posts that you already had a proof of the conditional convergence of the series. In any case, this part amounts to showing that $ \sqrt {n}/(n \plus{} 1)$ is a decreasing sequence.


[quote="Iversonfan2005"][quote="MCrawford"][quote="joml88"]Hopefully I can get better at math before ARML...ever since USAMO I have let my math skills deteriorate....the problem solving part at least.[/quote]

If Michael Jordan had played every minute of every game, he would have been tired by the beginning of each finals.[/quote]

He did play close to every minute a game. His career average was like 43 minutes a game for the Bulls, and he was never tired...not even when he had the flu in 97 and 98.  :)[/quote]
well playing 43 minutes does not equal playing 48 minutes, so mcrawford's point is still valid

[quote="furious"][/quote][quote="Iversonfan2005"][quote="MCrawford"][quote="joml88"]Hopefully I can get better at math before ARML...ever since USAMO I have let my math skills deteriorate....the problem solving part at least.[/quote]

If Michael Jordan had played every minute of every game, he would have been tired by the beginning of each finals.[/quote]

He did play close to every minute a game. His career average was like 43 minutes a game for the Bulls, and he was never tired...not even when he had the flu in 97 and 98.  :)[/quote][quote="furious"]
well playing 43 minutes does not equal playing 48 minutes, so mcrawford's point is still valid[/quote]

 :oops:  :rotfl:

[quote="jb05"]A game is 48 minutes? Why?

Excuse my ignorance; I don't watch/play sports.[/quote]

That's just the way it is. 12 minutes per quarter, that's just the playing time, see with free throws, fouls, commercials, timeouts, etc. it takes about two and a half hours to watch a whole game.

I see. It just seems like it should be 45 minutes.

11 minutes and 15 seconds a quarter seems a little odd to me  :D  :P

[quote="furious"]it could also be 3 quarters of 15 minutes with 2 intermissions like hockey.[/quote]

I don't understand hockey :ninja: Why only 3???? It makes no sense

Well how was I to know it had 4 quarters (I actually didn't; that's how sportsy I am :) )? And what's wrong with three quarters? Only problem is that it should be three thirds, but I assume that was just a typo.

It's always fun to go back to the title and wonder how exactly "Some collected geometry problems for the Carolinians" became "Why is a basketball 48 minutes" ;) .

Well, they are actually periods.  A hockey game has three periods of 20 minutes.

Heh, this is quite a bit off topic, I'll split the thread ;)

Does anyone actually miss hockey???? I don't know a single player in the NHL nor do I know who won the championship last year (okay, I do, but I had to think about it).

i missed the hockey playoffs cuz that's the only time i watched hockey. go lightning!!! and actually, i usually only watch the nba and baseball at playoff time as well cuz the seasons are so long.

I don't watch much TV at all and I haven't kept up with sports in a long time.  I usually keep up with the playoffs and such though.  I don't miss hockey at all, though ;)

I also don't watch much tv. Just the news, and occasionally we'll watch American Idol or a good sit-com. Only sports I watch are golf and tennis, and since we only get 4 channels (no cable), we only do that in hotels.

None of y'all watch South Park! :o How can y'all live without USA and TBS?

i really don't like south park but i will occasionally watch it when nothing else is on. i did get a kick out of the golden psp episode. as for usa and tbs, i never watch them.
(and once again, this is getting off-topic)

amazing how this conversation changed so drastically :o

[quote="furious"](and once again, this is getting off-topic)[/quote]

Well, splitting the topic again probably won't work...so I'll just change the name of the topic  :)

This is like an official random thread or something :rotfl:

Proud to be a part of the first official random thread in the Carolinas :) .

and while we've off topic i mine as well say that i'm getting very frustrated trying to find a set of custom car wheels with the following specs: 16X7.5, 45(mm i think) offset, 5x114.3 bolt pattern. it's like they don't even make em. looks like i'll be leaving the factory wheels on there  :(
edit: well i found some adaptors but they don't seem to fit tuner wheels.

I can't wait till I get a car.... :lol:  Hmmm...should I get an SUV?

get whatever you like. i personally don't like SUV's but that's just me.
i have a 1990 nissan 300zx (sports car for those of you who don't know alot about cars).

and joe, are you gonna have a car to drive next year?


Write a program that reverses a string s using pointers but without using strlen or getline.(in C)

you can make a strlen function by doing:

[code]int mystrlen (char* str) // use char* because you don't know how long it is
{
    int i;
    while( str[i++] ); // equivalent to while(str[i++]) {}
    return i-1; // go back 1
}
[/code]

and then you can do:

[code]void reverse (char* str)
{
    int i, len = mystrlen(str); // only calculate strlen once
    char t;
    for( i=0 ; i*2 + 1 < len ; i++ ) // NEVER write i*2 + 1 < strlen(str)
    {
        t = str[i]; // store prev value of str[i]
        str[i] = str[len-i-1];
        str[len-i-1] = t;
    }
}
[/code]


For all positive integers $m,n$ define $f(m,n) = m^{3^{4n}+6} - m^{3^{4n}+4} - m^5 + m^3$. Find all numbers $n$ with the property that $f(m, n)$ is divisible by 1992 for every $m$.

[i]Bulgaria[/i]

$1992=8*3*83$

We will prove that for arbitrary $m, n$ $f(m, n)$ is always divisible by $3$ and $8$. Well, it's almost obvious because if $(m, 3) > 1$ or $(m, 2) > 1$ all terms are simply divisible by $3$ or $8$ respectively, and if $(m, 3) = 1$ or $(m, 2) = 1$ we see that this divisiblity is consequence of the fact that $m^2 = 1 \mod{3}$ or $\mod{8}$.
So now, let's try the divisiblity by $83$.
If $83|m$ then it is obvious that divisiblity holds, so let's assume that $83 \not | m$. Also, we see that $m \equiv 1 \mod{83}$ or $m \equiv -1 \mod{83}$ implies that $83|f(m, n)$ so suppose that $83 \not | m^2-1$.
Ok, so we want to have $f(m, n) \equiv 0 \mod{83}$, or $m^{3^{4n} + 3} - m^{3^{4n} + 1} \equiv m^2 - 1 \mod{83}$ or $m^{3^{4n} + 1} \equiv 1 \mod{83}$
Because $83$ is prime there exist a primitive root in $\mathbb{Z}_{83}$, call it $g$. Taking $m=g$ we obtain that
$g^{3^{4n}+1} \equiv 1 \mod{83}$ and because g is the primitive root we must have $82|3^{4n}+1$ which means that
$3^{4n} \equiv -1 \mod{82}$ or $(-1)^{n} \equiv -1 \mod{82}$
From this we obtain that $n$ is odd. Ofcourse we can reverse all steps to see that this condition is also sufficient.
So the answer is: every odd $n$.


What is the largest prime factor of   (  30^2 -  7^2 ) ?

We can factor $ 30^2\minus{}7^2$ as $ (30\minus{}7)(30\plus{}7)$. Thus the largest prime factor is $ \boxed{37}$.


only apx 500 students qualify. 160 by AMC 10. you need something like a 200 for an index. it doesnt seem like that few students would have an index. of 200+. maybe its just me.

yeah idk hopefully ill make it nxt year!

err... is there a point to this topic??
discussion??
most people screw up on either amc 10/12 or aime, so they don't get 200+
maybe people here are just too beastly


A particular deck of cards has 20 cards. 7 cards are black on both sides, 6 cards are black on one side and red on the other, and 7 cards are red on both sides. You randomly choose one card, and you see that one of the sides is red. What is the probability that the other side of the card is also red?

[hide]
Since there are 20 cards, there are 40 sides.  Yes, this does matter.
However, you cannot include the sides of cards that are black on both sides, because you can't see a red side if you pick one of those.  That leaves 26 sides.
There is a $\frac{14}{26}$ chance that you pick a side of one of the red/red cards.  So the answer in simplest form is $\frac{7}{13}$ [/hide]

Hope everything makes sense!

umm, i got something different, but i suck at this sort of probability, so im not very sure:
[hide]
The only way that you could be looking at a red side, but have not-red on the other is if you picked a black-red card. Therefore, find the probability of that happened, [b] but with the condition of already have a red side [/b] in other words, the total probability should be the number of red sides. The number of red sides is 6(1) + 7(2) = 20, while the number of black-red cards = 6, so prob. of getting black-red card = 6/20, so by complementary counting, the prob. of getting a red side = 1-6/20 = 14/20 = 7/10.
uh, CircleSquared, im not sure why you included the black sides of the black-red cards in your denominator. However, once again, im not very good at this subject, so please excuse me if im just acting stupid[/hide]
btw: Did you post this problem as a continuation of fanatic's "card probability"? The questions are very similar. If not, i suggest you look at that topic,too.

[quote="CircleSquared"]Umm, given the condition you get a card with a red face, there is twice the probability that it's red-red as red-black, because red-red has 2 red sides.

So we need to include the number of sides, not the number of cards.[/quote]
im not sure if i know what you mean... I agree with you on the part about counting the sides (thats what i did in my solution), but i think we should only count the red sides, since it is given that we already have one. I also agree that red-red has twice the probability of occuring under these circumstances...
... so were you agreeing with me, or disagreeing, then? im confused  :maybe:

[hide]There are 20 red faces total - 14 of which are from red/red cards and 6 of which are from red/black cards. I could have seen one of them.

Thus there is a $14/20 = \boxed{7/10}$ possibility that it is the red face from the red/red card.[/hide]

Silly me, I know what I did wrong.
Yes, I now agree that
[hide]
$\frac{7}{10}$
[/hide]
is the correct answer.
 :oops:


group theory
abstract algebra
linear algebra
linear algebra unsolved

let $ K$ be a finit corps
show that $ K^*$ is cyclic and find all field morphismes from $ GL_n(K)$ to $ K^*$

bellahsayn,
From French to English a "corps fini" is a called a finite field.
You probably could find the answer to your question on this forum, otherwise, an answer in French is at http://pagesperso-orange.fr/megamaths/ccof0001.pdf

More generally, any finite multiplicative subgroup of a field is cyclic. There are some proofs [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=27050]here[/url].


No,it is not strong than cauchy,
it is strong than the cauchy's special case n=10 ;)


geometry
blogs
3D geometry
pyramid
counting
distinguishability
probability

I was bored, so I decided to create a really easy 2-part mini-tournament! [b]You have only 15 minutes to solve as many problems as you can without the use of calculators. Please PM me your answers. The answers are due by July 4th.[/b] The person with the most cumulative points after round 2 gets a really sweet* prize! Because of the semi-strict time limit, the total number of points needed to advance to round 2 is $ 4$.
*sweet=kinda stupid

[hide="Click When Ready (No Calculators)"]
1. For how many pairs $ (x, y)$- with x and y being positive integers- is it true that $ 3x \plus{} 2y \equal{} 100$.
Points:1

2. $ x$ and $ y$ are integers that are non-negative with the property that $ 3^{x \plus{} 6}6^{y \plus{} 2}$ is a perfect cube. Find the least possible product of $ xy$.
Points:1

3. Find and express in simplest terms:
$ \frac {2 \plus{} 4 \plus{} 6 \plus{} ..... \plus{} 4,002 \plus{} 4,004}{1 \plus{} 3 \plus{} 5 \plus{} ..... \plus{} 4,001 \plus{} 4,003}$
Points:1

4. Jeeves is thinking of a number between 1 and 100. He says, " My number is a prime number with tens digit $ x$ and units digit $ y$ such that $ 5x \equal{} y \plus{} 1$. Find the value of $ (x \plus{} y)(x^2 \minus{} xy \plus{} y^2)$.
Points:1

5. $ i$ integers are written out and then added together. Carl adds 100 to the sum of the first $ i$ integers and gets the sum of the next $ i$ integers. Find $ i$
Points:2

6. If $ (d \plus{} a \plus{} f \plus{} j \plus{} b \plus{} o \plus{} m \plus{} s \plus{} u \plus{} n)^2$ is expanded and then simplified, what will the coefficients sum to?
Points:2

7. Jo constructs a right triangle with sides $ 5, 12,$ and $ 13$, and then inscribes a perfect circle in it. What is the area of the inscribed circle?
Points:2

8. There are integers $ w, l, o, g$ with the property $ g \equal{} 5w \plus{} 3l \plus{} 5o$ and $ g \equal{} 4w \plus{} 5l \plus{} 4o$. Suppose $ 131 \le g \le 150.$ What is the value of $ w \plus{} l \plus{} o$?
Points:3[/hide]

You guys can discuss the difficulty (or lack thereof) of the test, just dont post your answers here

I did it in ten minutes, 30 is just way too much.

EDIT: Oh, I forgot to send them in.  :oops:

Alright, edited the time to 15 minutes (how I chose 15 I have no idea). for everyone who took the test before I edited it, you got lucky and had 15 more minutes, though I doubt you needed it. The qualifying score is so low (intentionally) that 15 minutes wont make a difference.

[hide="Scores"]Yay, I already got 8 scores back. Thats 8 more than I expected!
1. blackhole: 12/13
1.arb95 12/13
3. ffao:11/13
4.PerpetualMotion:10/13
4. RoFlLoLcOpT:10/13
6. yaofan:6/13
7. alex960532:4/13
8. waterbird: 3/13 [/hide]

[hide="Scorers that are Advancing to Round 2"]
1. blackhole: 12/13
1.arb95 12/13
1. Shentang:12/13
4. ffao:11/13
4. Bacteria: 11/13
6.PerpetualMotion:10/13
6. BOGTRO:10/13
6. RoFlLoLcOpT:10/13
6. James4l:10/13
10. yaofan:6/13
11. alex960532:4/13
11. Mathboy101:4/13
[/hide]
there are 12 people advancing to round 2 so far out of 13 people that have taken the test. Thats 13 more people who have taken the test than I had expected!

[quote="Thunder365"][b] Please PM me your answers. The answers are due by July 4th.[/b]
[/quote]
So Round II will begin shortly after July 4th or on the same day?

Probably the same day. If I happen to be way too busy, then definitely the next day.

WOW thunder you created a tournament...
Hmmm blackhole got a 12/13. Pretty shocking. I would expect him to get a 13...

yes I did and its pretty easy.

*waits for fermatsmymom to get the first perfect*

[hide="Scorers that are Advancing to Round 2"]

1. blackhole: 12/13
1.arb95 12/13
1. Shentang:12/13
1. AIME15: 12/13
5. abcak:11/13
5. urc: 11/13
5. ffao:11/13
5. Bacteria: 11/13
9.PerpetualMotion:10/13
9. BOGTRO:10/13
9. RoFlLoLcOpT:10/13
9. James4l:10/13
13. yaofan:6/13
14. SuperNerd123:5/13
14. Dojo:5/13
16. alex960532:4/13
16. Mathboy101:4/13[/hide]

there are 17 people advancing to round 2 so far out of 18 people that have taken the test. Thats 18 more people who have taken the test than I had expected!

*note: round 2 will be a lot more....beastier (not really, but whatever)

Also, Im kinda.... disappointed.... that nobody got a perfect so far. I was sure that AIME15 would.

(yes, I triple posted :P )

So seeing as I have probably hit a peak with the number of people who are doing this tournament, congrats to the following for making it to round 2 [b](however, round 1 isnt over yet, so if you have not taken it, please do!)[/b]:
1. blackhole
1.arb95
1. Shentang
1. AIME15
5. abcak
5. urc
5. ffao
5. Bacteria
9.PerpetualMotion
9. BOGTRO
9. RoFlLoLcOpT
9. James4l
10. PowerofPi
14. yaofan
15. SuperNerd123
15. Dojo
17. alex960532
17. Mathboy101

Although I must say that Round 1 wasnt all that challenging. Round 2 will have much better questions for sure. Also, thanks to athunder for helping with round 2 a lot. Join his tournament!!! [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?p=1546590#1546590]athunder's Pwnage Tourney[/url]

[u][i][b]REMEMBER, ROUND 2 WILL BE POSTED ON JULY 4TH!![/b] [/i][/u]I will try to PM everyone a reminder, and I really hope everyone who advanced to round 2 takes the test. Even if you do not plan on taking round 1, you can still take round 2 just for fun.

[quote="Thunder365"][u][i][b]REMEMBER, ROUND 2 WILL BE POSTED ON JULY 4TH!![/b] [/i][/u][/quote]
Since the first round was due in a week, I'm guessing this is due by July 21st?

[quote="Thunder365"]Actually Round 2, which is being held on July 4th, has problems that have a lot more points, so basically whoever does well on round 2 is the winner.[/quote]
@AIME15, above is what Thunder365 commented on his blog about Round II. So I think it is transferred over to our Round I score, but it's worth much more points in the scoring system.

[quote="AIME15"]Do scores from Round 1 carry over to Round 2?[/quote]

Yes. In the first post I said that the person with the most cumulative points after round 2 gets a really sweet prize. However, the questions in round 2 are weighted a lot more.

ROUND 2 IS HERE!!! Please read the following directions:

Remember, [b]no calculator usage is allowed whatsoever[/b]. PM me your answers when you finish. [b][i][u]You only get 25 minutes. Please respect the test writers by following the time limit. [/b][/i][/u]

This will be the final and deciding round for the grand prize! Entries are due next Saunday, July 12th. However, I am going to california so I will not grade any test sent after this Tuesday until probably Monday the 13th.

NOTE: I took off the PDF file because AIME15 made me realize that I had a lot of typos in it.

[hide="CLICK WHEN READY (NO CALCULATORS)"]1.	How many triangles does a pyramid with $ 12$ edges have? (2pts.)

2.	Jose is trying to find the sum of all the numbers in the tenth row of Pascal’s Triangle, which starts at row zero, shown below. What is that sum? (To create Pascal’s triangle, shown below, start with a 1 at row zero. To construct the elements of following rows, add the number directly above and to the left with the number directly above and to the right to find the new value) (2pts.)
…………………....…$ 1$--Row Zero
………………….…..$ 1$ $ 1$--Row One
…………………....$ 1$ $ 2$ $ 1$--Row Two
…………………...$ 1$ $ 3$ $ 3$ $ 1$--Row Three
…………………………$ \vdots$

3. Mary is babysitting $ 3$ indistinguishable babies. To make them stop crying, she wants to distribute $ 7$ distinguishable lollipops among the $ 3$ indistinguishable babies. How many ways can she do that?(Babies can get any number of lollipops) (2pts.)

4. John has $ 10$ sticks with distinct length, each with integer length, with the property that no three of these sticks can make a triangle. What is the least possible length of the longest stick he could have? (2pts.)

5. Let $ A$ be the sum of the roots of the equation $ X^2 \plus{} 3X \plus{} 9 \equal{} 0$ and let $ B$ be the product of the roots of that equation. What is $ A \plus{} B$? (2pts.)

6. I have $ N$ positive integers, not necessarily distinct, that have a sum of $ 47$. What is the greatest possible product of my $ N$ numbers? (Express in exponential form) (2pts.)

7. On planet Jizz, the only currency is in $ 5$ and $ 17$ cent coins. Jo walks into a store to buy a toy worth $ X$ cents, but realizes there is no possible way he can pay for it. What is the greatest possible value of $ X$? (Assume that Jo has an endless supply of $ 5$ and $ 17$ cent coins) (2pts.)

8. How many digits are in the following: $ 4^{70} \cdot 125^{45}$? (2pts.)

9. What is the value of the following: $ \sqrt {42\sqrt {42\sqrt {42\sqrt {42 \ldots}}}}$? (2pts.)

10. When a printer is numbering all the pages of a book, 1653 digits are used. How many pages does the book have? (2pts.)

11. Simplify: $ \sqrt {84 \plus{} 30\sqrt {3}}$.(3pts.)

12. Find the sum of the infinite sequence: $ 1 \minus{} \frac {1}{2} \minus{} \frac {1}{4} \plus{} \frac {1}{8} \minus{} \frac {1}{16} \minus{} \frac {1}{32} \plus{} \frac {1}{64} \minus{} \frac {1}{128} \minus{} \frac {1}{256}\ldots$ (3pts.)

13.	An equiangular hexagon has sides with length $ 9, 11, 10, 6, X,$ and $ Y$ in clockwise order. What is $ X \plus{} Y$? (3pts.)

14.	$ 3$ points, $ X, Y,$ and $ Z$ are selected on the circumference of a circle. What is the probability that all $ 3$ points lie on a semi-circle? (4pts.)

       15. Cubes with side lengths of $ 3$,$ 4$,$ 5$,$ 6$,...$ 28$ are completely painted with the color red. Then every cube is taken apart into unit cubes. Find the total number of cubes that have no paint on them? (4pts.)
[/hide]

So ya, I got 8 submissions so far (wont reveal scores yet to keep the suspense).

But what happened to the other 10 people who took the test?

I don't have time right now, but I will take it before the 13th.

CONGRATULATIONS to everyone who took both tests! The scores were very high. It was also very tight. Second and third place was determined by only one point, as was first and second place. I would like to thank athunder for helping me out with the questions and making a pdf of the test (even though I didnt use it because I realized I made too many typo's in the original version so there were lots of typo's in the pdf)

 Without further ado, here is your top 5.....
[hide="5th place"] ffao! :thumbup:
[hide="4th place"]Arb95!! :clap:

Now heres where it get really close...
[hide="3rd place"]RoFlLoLcOpT!!! :winner_third:
AIME15!!!! :winner_second:
  SHENTANG!!!!!:winner_first:  :clap2: Hmmmm ya didnt think this one through. Um I'll think what to give you. So far your prize is the pride of being the winner of my tournament :D.

[/hide][/hide]
[/hide]
Again, good job to everyone who participated. Im hoping to have another tournament soon!

Also, the solutions probably wont come out :( (really busy this summer). PM me to get the solution to whichever questions you want.

:rotfl: My two misreadings and more fails apparently didn't matter that much :rotfl:


geometry
circumcircle
trapezoid
geometric transformation
reflection

In $ \triangle{ABC},\hat{ABC}\equal{}60^{\circ}$ and $ \hat{ACB}\equal{}70^{\circ}$.
Point $ D$ is on the line segment $ BC$ such that $ \hat{BAD}\equal{}20^{\circ}$.Prove that $ AB\plus{}BD\equal{}AD\plus{}DC$

Hint: $O$, the circumcenter of $\triangle ABC$ lays on $AD$, and the circle $\odot BOC$ intersects $AD$ at a point $P$ so that $BPCO$ is an isosceles trapezoid; the reflection $Q$ of $D$ in $BP$ is such as $AQP$ is an isosceles triangle.

Best regards,
sunken rock


I have two mathematical question which is really want to ask and know how to do

Date: 10/22/2006 at 06:23:27
From: laafox@gmail.com (Kelvin)
To: dr.math@mathforum.org
Subject: Mathematical Induction

[Question]
Prove by induction that, for all positive integers m and n,
the smallest integer greater than (sqrt(3)+1)2n is divisible by 2n-m

[Difficulty]
I dont know whether my working is right or not. But i feel that it
seems now right.

[Thoughts]
For m = 1, n = 1
BASIC: L.H.S.: (sqrt(3)+1)2(1) = 2(sqrt(3)+1)
R.H.S.: 2(1) - 1 = 1
2(sqrt(3)+1) is divisible by 1
Therefore, It is true for n = 1, m = 1

Assume n = k, m = c where n = 1, m = 1, That is:
(sqrt(3) + 1) 2k is divisible by 2k - c

For n = k + 1, m = c + 1.
STEP: (sqrt(3)+1)(2(k+1)) = (sqrt(3)+1)(2k+2)
>(sqrt(3)+1)(2k-c) which is divisible of (2k-c)

It is true for n = k+1, m = c+1, where n = k, m = c, by Mathematical
Induction. It is true for the smallest integer greater than (sqrt(3)+1)2n
is divisible by 2n-m, for all positive integer m, n

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

[Question]
Prove by M.I that, for all integers n >= 2,

x^n - nxy^(n-1) + (n-1)y^n is divisible by (x-y)^2

[Difficulty]
I cant work out the final step when n = k + 1

[Thoughts]
For n = 2
BASIC: L.H.S.: x^2 - 2xy + y^2 = (x-y)^2
is divisible by (x-y)^2
Therefore, It is true for n = 2,

Assume n = k, where n = 1, That is:
x^k - kxy^(k-1) + (k-1)y^k = N(x-y)^2 where N is positive integer

For n = k + 1.
x^(k+1) - (k+1)xy^k + ky^(k+1)

then i dont know what to do with this. i dont know what to do.

[hide]Both polynomials $x^{k}-kxy^{k-1}+(k-1)y^{k}$ and $(x-y)^{2}$ are homogenous so your question is equivalent to show that $\left( \frac{x}{y}\right)^{k}-k\left( \frac{x}{y}\right)^{k-1}+k-1$ is divisible by $\left( \frac{x}{y}-1\right)^{2}$ as when $y=0$ this is evident. Let $z= \frac{x}{y}$ we have to prove that $z^{k}-kz+k-1$ is divisible by $(z-1)^{2}$ when $k \ge 2$. For $k=2$ this is true (you did this).
Assume that (*) $z^{n}-nz+n-1$ is divisible by $(z-1)^{2}$.
Let's prove that (**)$z^{n+1}-(n+1)z+n$ is divisible by$(z-1)^{2}$.
$**-*= z^{n+1}-z^{n}-z+1=(z^{n}-1)(z-1)$ and this obviously is divisible by $(z-1)^{2}$ . As * is is divisible by$(z-1)^{2}$ it follows that the same is true for ** .
Advice: Try to learn LaTex (this is simple) :)
[/hide]


방학은 인간을 망치는 무시무시한 시간입니다.

마음먹었던 공부들... '그거 먹는건가요 우걱우걱' 수준이 되어버렸으니... (피토한다)

음... 그냥 그렇다고요 -ㅁ-;;

하도 게시판이 썰렁해서 그냥 주절거려봤습니다. 아래는 그냥 사소한 공식 하나

$ \prod_{n\equal{}1}^{\infty} \left[ 1 \plus{} \frac{4x^2}{(4n\plus{}1)^2}\right] \equal{} \frac{\cosh (\pi x)}{16 \, \Gamma \left( \frac{3}{4} \right)^2} \, \Gamma \left( \minus{}\tfrac{1}{4} \plus{} \tfrac{ix}{2}\right) \Gamma \left( \minus{}\tfrac{1}{4} \minus{} \tfrac{ix}{2}\right)$

여기서 선배님을 뵙게 될줄은..ㅋ

방학에도 공부하시는군요 굳

제가 몇가지 문제를 내 드릴께요 ㅋㅋ

1.방학때 뭘해야 할까요??ㅜㅜ
2.적분을 잘 할려면 어떻게 해야하나요??ㅜㅜ
3.전 누구일까요??ㅋㅋ(hint:08입니다)

what does that formula mean?

i agree breaks SUC-k

방학이 재미있지요!!!
언제나 공부만 하는게 재미있어요?
아니면 놀러갈데 없나요?

@jonathanchou711 Dude, I'm indian... how do you think i feel?

wow 하하

anyways, i don't understand why people come to a different language forum and go like WTF lol xD
(no offense)

it's like some Korean goes into the Chinese/Indian forum and going like UHH...

@remy
aren't you the guy who got second in state NC mathcounts?
i got 12th. :(

Yeah, that's me. Wow, you still "remember"! :P

12th place is still really good, in my opinion. You only barely missed the countdown cutoff. Of course, compared to how I did in 7th grade, when I totally failed at chapters.
Yeah, I was really lucky this time.

Man I need to make state... My first year!

[quote="thkim1011"][quote="sos440"]방학은 인간을 망치는 무시무시한 시간입니다.[/quote]
I disagree.. that's when I get most time to do math :P[/quote]

Well, if you don't wanna do math during breaks then it could be a scary time...
Or do what I do, have a 24 hour game marathon and get dumb :P

Hey remy, you cane 2nd? Wow, thats awesome. I came 7th in the NC state competition, and I totally bombed the countdown round. :(

The fact that ARML is in June tells it all

[quote="jonathanchou711"]I'm chinese and I feel left out. :([/quote]

Why are you feeling left out in the Korean forum lol

[quote="thkim1011"][quote="sos440"]방학은 인간을 망치는 무시무시한 시간입니다.[/quote]
I disagree.. that's when I get most time to do math :P[/quote]

Do you live in Korea? >.< It would be scary in Korea

I live in the us and i can speak korean fluently!

안녕하세요. 저는 이 AOPS에 다닌지 8개월이 됐습니다. 참 재미있어요. 안녕히계세요 여러분!

and penguin 25 is right  why do yo feel left out in a korean forum if you are a chinese>?

그냥 가! Lol I once did that to a 1 year old

Uhh... lol you feel left out in a Korean forum. GO VISIT THE CHINESE FORUM!!!

I mean really, breaks kinda suck. You expect it to be nice and it's kinda boring unless you get to go somewhere or something. If you don't 99% all of your other friends are gone somewhere else. So bah.

이것의 뜼은 무엏입니까?....

[quote="sos440 - 오오, 역시 휴가란 좋은 것이구나!"]일병 말봉 달고 1차정기 나왔습니다.

9박 10일동안 즐겁고 신나는 날들을 보내다가, 재입대하는 기분으로 다시 끌려가겠죠.

엏엏엏...[/quote]

Not to be mean and offensive but you are bad at Korean... ALL OF YOU...

It means that

I passed my first year in the army and came out for a vacation

I spent 9 nights and 10 days happily and when it was over, it felt like I was re-entering my first year of the army.

CRYING... :(

너는 미국으로 언제 이사 했어? 어떤 AOPSER들은 미국에 태어난 는데도 한국어를 배우기 위해 노력했다는것은 대단한것이야. 너는 한국에서 다른 사람보다는 조금더 오랫동안 공부했어 너가 당연히 한국말을 더 잘하겠지.

Sorry for being offensive everyone... :(

I was born in Korea, then moved to France, then moved back to Korea and came to US :)

Well I only have the korean skill of a 10 year old man... :(


[quote="fireforce"]How do you do this problem?
Triangle ACB is similar to triangle DEF. If AB=14, BC=9, AC=12, and DE=8, then what is the perimeter of triangle DEF?
A)$\frac{3 \sqrt{89}}{2}\\\\$ B)$\sqrt{119}\\\\$
C)$6\sqrt{3}\\\\$ D)$\frac{4 \sqrt{13}}{3}\\\\$
E)$NOTA$[/quote]

[hide]the perimeter of ACB is 35, and AC~DE so $\frac{8}{12}\times35=\frac{70}{3}$

How come all the answers except E have square roots in them..?[/hide][/hide]

[quote="ch1n353ch3s54a1l"]How come all the answers except E have square roots in them..?[/quote]

Alas, I have contemplated this fact too and  yet the answers is [hide]I know the answer but i do not know why [hide] also this is a region 1 competition question [hide] But, if you really want know its "somehow" D.[/hide][/hide][/hide]

r u sure its perimeter and not area?  because i dont see any way to get rads in perimeter...


Please join!!!

Dojo's game died out because he was inactive. I want to continue it. Basically, you pm me your move and I'll make matchups, and do score. It will be all sudden death, best out of 1. Double-elimination, random draw-out-of-a-hat pairings. Hope to get 8-10 ppl!!!

[quote="pinkmuskrat"]
Dojo's game died out because he was inactive. [/quote]
Well, actually I got 2 applications.

bump...

I hope I can get some more applications, but this games will not start until after mc nats. I'm going to be very busy doing work and going places. (Not having access to aops) If we get more people, I'll pm all players a reminder before it begins.

If this works, 48 hours after each start of round will equal an inactive forfeit.

since yaofan joined.... the kid who beat me in first round cd XD...
ill be competitve and join

[quote="jonathanchou711"]Is there any more open spotsies?[/quote]

well considering only like 2 people joined i'm pretty sure there are more spots...

Yes, please join! There are tons of open spots. I'm hoping for atleast 8 people, but as many people as want to can join. Deadline and start of game is after MathCounts nationals. Since this is in a while, I'll pm all participants prior to the start.

Applications so far- yaofan, weird888, jonathonchou711, hedgehog12
Hope for a few more

@jonchou- Am I correctly assuming that you join?

Cool. Remember it does not start until after MC Nats, so 10-13 ish of May. One more player necessary to make 6, but as many as possible hopefully.

/in
The message is too small. Please make the message longer before submitting.

hedgehog, i have submitted rock,
please play sciccors

when i post my tricks, there is a pattern that will let you win.
hope u can find it out:)

i pmed my move a while ago. btw, remember my message, pinkmuskrat, as you will regret the day you try to mess with me, MWAHAHAHAHAHAHAHAHAHAHAHAHA!!!!!!!! *laughs maniacally for at least 10 minutes*

Round 1
Joe10112 def. weird888
jonathanchou711 def. AIME15
splatyango def. connaissance, tiebraker
eli140 def. yaofan, tiebraker
hedgehog12 def. louiethechewy123, inactivity

Round 2- number of losses in parentheses
These will be concluded w/ inactivity at 10:00 ET.
splatyango(0)- Joe10112(0) (Joe will win, unless splatyango pms quick)
eli140(0)- louiethechewy(1) (eli140 will win)
hedgehog12(0)- weird888(1) (weird888 will win)

FOR SURE
connaissance(1) def. AIME15(2)
yaofan(1) def. jonathanchou711(1)

If these people don not pm in time, then
louiethechewy123- 10th Place
AIME15- 9th Place

And, 3rd Round- Deadline is Tuesday, 5/19 10:00 ET
splatyango vs. jonathanchou711
eli140 vs. Joe10112
yaofan vs. weird888
hedgehog12 vs. connaissance

GL!

yaofan...buddy...
well, i have pmed to pink that i would play sciccors, so i beg of you to play tissue or paper or w/e they call it, thankyou

I spoke too soon! hedgehog12 sent his move in, so there are 2 ppl who will be considered inactive.
weird888 def. hedgehog12
Next round matchups are still the same- Thanks all!

At this point, two players have been eliminated. Everybody else except for eli140 and Joe10112 have 1 loss, with those two being undefeated.

And, 3rd Round- Deadline is Tuesday, 5/19 10:00 ET
splatyango vs. jonathanchou711
eli140 vs. Joe10112
yaofan vs. weird888
hedgehog12 vs. connaissance

10th- louiethechewy gets 10000 imaginary dollars
9th- AIME15 gets 15000 imaginary dollars

And, 3rd Round- Deadline is Tuesday, 5/19 10:00 ET
splatyango vs. jonathanchou711
Joe10112 def. eli140
weird888 def. yaofan
hedgehog12 vs. connaissance

As of right now- connaissance, splatyango, jonathanchou711 would forfeit. 3 is not enough for me to delay deadline. BTW i have tiebreaks for the placing

10th- louiethechewy gets 10000 imaginary dollars
9th- AIME15 gets 15000 imaginary dollars
8th- splatyango gets 20000 imaginary dollars
7th- connaissance gets 30000 imaginary dollars
6th- jonathanchou711 gets 50000 imaginary dollars
5th- yaofan gets 75000 imaginary dollars

Joe10112 is undefeated, rest have 1 loss.
weird888- eli140
Joe10112- hedgehog12

GL :)

Deadline is Thurs. 5/21, 10:00 ET AM
Round 4
Joe10112 is undefeated, rest have 1 loss.
weird888- eli140
Joe10112- hedgehog12

sorry about this inactivity, i didn't come to AOPS in a long time, then completely forgot about this. Can I  resign?

We had some inactivity again.  :(

All here have 1 loss.
This is now round robin for the championship.

Joe10112
weird888
hedgehog12

Each player plays agianst everybody else.
PM me your choice against each of the two opponents. Thanks! No deadline, but do ASAP.

This game has now phailed.

Let's just thank everyone for playing and declare weird888 the champion, because he pmed a move and nobody else did.


Let a,b,c be positive numbers such that abc=1. Prove that

  (a+b)(b+c)(c+a)  \leq ((a+b+c)/2)^6

By AM-GM inequality:
(a+b)(b+c)(c+a)  \leq (8/27)(a+b+c) 3
abc  \leq (1/27)(a+b+c) 3

=>

(a+b)(b+c)(c+a)abc  \leq (8/729)(a+b+c) 6 < (1/64)(a+b+c) 6

So is not very strong inequality.


Two variable circles $\ C_1 $ and  $\ C_2 $ intersect at  the fixed points  $A $  and  $B$ . Let  $AC $ and  $AD $ diametres in these circles. The lines $AC$  and $AD $ meet circles $\ C_2 $ and $\ C_1 $ at  $ E $ and  $ F $ .Prove that the circumcenter of  $\triangle  AEF $  belongs to a fixed line.

Babis

It seems that $B$, $E$, $F$ are collinear, how can I prove it?

[quote="stergiu"]Do you mean $ C, B, D$ ?

Babis[/quote]
Yes.
Sorry I made some typos.

I' m afraid that you have not a good figure!

Angles $\angle ABC $ and $\angle ABD $ are right , since we have diametres , so.....  [b]A hint:[/b] somewhre you must see an  orthocenter.Do not give it up!

Babis

Well , I ' ll write  a goog hint (solution ! )

 a)    Points  $C ,  B , D$ are colinear

 b) angles $\angle CFA  $and $\angle  DEA $ are right

 c) The point $R  $of inersection of the lines $ CF , ED $ is orthocenter in triangle  $ACD$ , so$ A , R , B $ are on the same line( on altitude AB).( in my figure  $ R $is an interior point of triangle $\triangle ACD $ . The center of each  circle is  interior to the  other))

 d) Quadrilateral $ AERF $  is inscribed  and $ AR $ is a diameter , so the circumcenter of triangle  $\triangle AEF  $is on the fix line $AB$ .

[u]Babis[/u]

Uhm ... you can make it shorter : the circumcenter $O$ of $FEA$ is the intersection of the perpendicular bisectors of $FA$ and $EA$; these two lines pass through the centers of C1 and C2, let me call them $O_1, O_2$. Now, $A$ is the orthocenter of $OO_1O_2$, so $AB$ passes through $O$.


Can someone please tell me what this famous Jenson's inequality is that everyone uses?

First of all, it's spelled Jensen.

Secondly it basically states that if a function $f$ is concave up, the weighted average of $f$ over several $x_{i}$ is $\geq f(x_{avg})$, where $x_{avg}$ is the weighted average of all of the $x_{i}$'s.

More formally, $\frac{\Sigma a_{i}f(x_{i})}{\Sigma a_{i}}\geq f(\frac{\Sigma a_{i}x_{i}}{\Sigma a_{i}})$.

And of course, if the function is concave down, the inequality is reversed.


Denote by $\mathcal{S}_{n}^{++}(\mathbb{Z})$, the set of symmetrical positive definite matrices of $\mbox{GL}_{n}(\mathbb{Z})$.
Find out the set of $n \in \mathbb{N}$ s.t. the following statement holds true:

$\forall A \in \mathcal{S}_{n}^{++}(\mathbb{Z}), \exists P \in \mbox{GL}_{n}(\mathbb{Z})~|~ A= P^{T}P$.

Using Minkowski's theorem, one can prove it up to 4. Using Hermite basis, one can prove it also for 5. The first counter-example that I saw in some papers was to n=8. So, what about the other values?

That's a good start harazi.
$n=6$ or $n=7$ holds true as well.


The surface area of sphere $A$ is $96\%$ more than the surface area of sphere $B$. The volume of sphere $A$ is $r\%$ more than the volume of sphere $B$. What is the value of $r$? Express your answer to the nearest whole number.

174

The message is too small. Please make the message longer before submitting.

[hide]
the suface area is 96% more, or 196% of the suface area, which means the radius is 40% more.
to get the volume you would cube 14(10+ 40% of 10), so 2744, then subtract the cube of 10, which is 1000, so 1744. then conver to a percent to get 174%[/hide]

then you have to divide by 10^3 and you get your answer as a decimal.

ooops forgot that part...EDITED...thanks  :oops:
see... im really dumb


Two trains start moving in opposite directions at the same time, from cities A and B which are separated by a distance of 300 km. 4 hours after the two trains met each other, the train traveling form B to A reaches city A. 9 hours after they met each other, the train traveling from A to B reaches city B.Which are the velocities of the trains?

I´m having problems. I started like this: Let $ t$ be the time when they both met, $ x$ the speed of the train traveling from B to A, and $ y$ the speed of the train traveling from A to B, then:

$ (t)(x \plus{} y) \equal{} 300$
$ (t \plus{} 4)(x) \equal{} 300$
$ (t \plus{} 9)(y) \equal{} 300$

then $ t \equal{} 300/(x \plus{} y)$

substituding $ t$ in both equations and equaling them i obtain

$ 4x^2 \plus{} 300x \plus{} 4xy \equal{} 9y^2 \plus{} 300y \plus{} 9xy$ which i am not able to solve.

Please try to solve it yourself and post the solution step-by-step, so i can see other ways to solve it. Thanks!

Assuming your system of equations is set up right, you can solve as follows:

[hide="Outline"]
We have

$ tx \plus{} ty \equal{} 300 \\
tx \plus{} 4x \equal{} 300 \\
ty \plus{} 9y \equal{} 300.$

Notice that viewing it like this shows that it would be simplest to add the last two equations together to obtain:

$ (tx \plus{} ty) \plus{} 4x \plus{} 9y \equal{} 2(300)\implies 300 \plus{} 4x \plus{} 9y \equal{} 2(300)\implies 4x \plus{} 9y \equal{} 300.$

From here, you can solve for $ y$ and substitute in equation 3 to obtain two systems in $ x$ and $ t$. After you compute $ t$, $ y$ follows.
[/hide]


conics
ellipse
geometry unsolved
geometry

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14794[/img]

 :(

[color=darkblue]
   Let $ \omega_1$ be the circle. Suppose $ MC\le MD$.

   Let $ E$ on the line $ CD$ such that $ M$ is midpoint of $ ED$. Let $ \omega_2$ be a variable circle, which passes through points $ E$ and $ D$.

   In the circle $ \omega_2$ draw chords $ QX$ and $ RY$, such that line $ QX$ passes through $ A$ and $ M$ and line $ RY$ through $ B$ and $ M$. Let $ \left\| \begin{array}{cc} K \in QY \cap ED \\
L \in RX \cap ED \end{array} \right\|$.

   By [b][u]Butterfly's theorem[/u][/b] $ M$ is midpoint of $ KL$.

   Then $ \boxed{P \in BK \cap AL}$.[/color]

[hide="Image"][img]http://i058.radikal.ru/0806/5e/55f45d6a73a5.png[/img][/hide]

P isn't on circle

 :(

[url]http://mathworld.wolfram.com/ButterflyTheorem.html[/url]

 :(

$ E$ is the midpoint of $ AB$ (one ellipse focus). Parallel to $ CD$ through $ E$ cuts the given circle $ (O)$ circle at $ U, V.$ $ G$ is the midpoint of $ UV$ (ellipse center) Pick arbitrary point on the circle different from $ A, B,$ say $ C.$ $ CA, CB$ cut the line $ UV$ at $ X, Y.$ $ Z$ is the midpoint of $ XY.$ $ H$ is foot of the perpendicular from $ E$ to $ CZ$ (an ellipse tangent). $ H$ is on the ellipse pedal circle $ (G)$ with center $ G$ and radius $ GH.$ Construct the ellipse tangents from $ M.$ Circle with diameter $ ME$ cuts the pedal circle $ (G)$ at $ X, Y.$ The ellipse tangent $ MX$ cuts the given circle $ (O)$ at 2 points $ P, P'$ and the ellipse tangent $ MY$ at $ Q, Q'.$ $ PA, PB$ cut the line $ CD$ at $ K, L.$ If $ M$ is inside the segment $ KL,$ then $ MK \equal{} ML$ and $ P$ solves the problem. Similarly for $ P', Q, Q'.$ Generally, only one interection of $ (O)$ with each ellipse tangent satisfies the condition.

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14811[/img]

 :thumbup:


search
linear algebra
linear algebra theorems

Can someone give me the proof of Cayley-Hamilton's theorem?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=127728168&t=98156]This topic[/url] would be a good place to start.

This site does have a search function. Now, a phrase like "Cayley-Hamilton" will get a lot of hits, but you should still be able to sort through them to find something useful.


$ a,b,c > 0$ and $ a\plus{}b\plus{}c \equal{} 3$

$ (a \plus{} b^2)(b \plus{} c^2)(c \plus{} a^2)\leq13 \plus{} \frac {abc}{2}$

[quote="Tomekk"]Try $ a \equal{} b \equal{} c \equal{} 1$.[/quote]

My Friend 8 < 13

[quote="spanferkel"]then try $ a \equal{} b \equal{} c \equal{} 2$.  :rotfl:[/quote]

$ a\plus{}b\plus{}c \equal{} 3$  :wink:


I think that the old  method used by Peter in the following topic works also here  :lol:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66936

Of course, It's even solved by Tchebyshev, I think.

How did you solved by Chebychev ?with using yur leema and after using Chebychev ?

[quote="silouan"][quote="hungkhtn"]No, directly.[/quote]

How. Could you show  it please ?(Sorry if I  am annoying)[/quote]
SEE [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=61915]HERE[/url]
hope help :)

KMN said a solution  ;) .It is difficult to write it on details.


limit
logarithms
real analysis
real analysis unsolved

Let $ D_n(p)=\left(1+\frac{1}{n}\right)^{n+\frac{1}{p}} - e \; , \; n=1,2,... .$  Find (if exists) all positive integers  $ p $ for which  $ \lim\limits_{n\to \infty}n^2\cdot D_n(p) $ exists.

$p=2$ and the limit in this case $-\frac{1}{4}e$
am i right?

$p=2$ is correct but the limit is $+\frac{e}{12}.$

$\ln\left(1+\frac1n\right)=\frac1n-\frac1{2n^2}+\frac1{3n^3}+\cdots.$

I looks like eugene overlooked the third term of this series.


Let $ n$ be a positive integer.  Factor

$ 3^{3^n}(3^{3^n}\plus{}1) \plus{} 3^{3^n \plus{} 1} \minus{} 1$

Source: Mathematical Olympiad Treasures, Section 1.1, Problem 11

Does this factorization have to have rational coefficients?  If not, we can treat it as a regular quadratic, although we end up with some sqrt5.

I think so.  There should be some clever solution based on the factorization of $ x^3\plus{}y^3\plus{}z^3 \minus{} 3xyz$

[quote="mysmartmouth"]I think so.  There should be some clever solution based on the factorization of $ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz$[/quote]

Triiiicky.
I know what x,y,and z are.  But I don't know the factorization.  And yes this factorization does exist.

$ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz \equal{} (x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} zx)$.

[hide]
$ x \equal{} 3^{2 \cdot 3^{n \minus{} 1}}$, $ y \equal{} 3^{3^{n \minus{} 1}}$, $ z \equal{} \minus{} 1$ should do it.[/hide]


Prove that any composite three-digit number must have a prime factor less than or equal to $ 31$

[hide="solution"] Suppose the opposite. Then, as the number is composite, it has at least two prime factors, which are both greater than $ 31$. Thus, the number is at least $ 37^2\equal{}1369$, a contradiction.[/hide]


induction
number theory unsolved
number theory

Find all positive integers $ n\geq 3$, so that, for any two positive integers $ m, r$, both $ <n \minus{} 1$, there exist $ m$ distinct elements of the set $ \{1,2,...,n \minus{} 1\}$ whose sum is congruent to $ r$ when considered in modulo $ n$

[quote="praneeth"]Use induction,
for m=1, it is obviously true.
Let it be true m=k and prove it for m=k+1[/quote]

Show it...  :lol:


Suppose $n\ge 2$ and    $0< a_1\le a_2\le \ldots \le a_n$. [b] Prove/or disprove [/b]
\[ \begin{array}{|c|} \hline \\ \displaystyle \frac{\sum\limits_{k=1}^n a_k^2+\left(\sum\limits_{k=1}^{n}a_k\right)^2} {\sum\limits_{k=1}^{n-1} a_k^2+\left(\sum\limits_{k=1}^{n-1}a_k\right)^2} \ge \displaystyle \frac{n+1}{n-1}\\ \\ \hline \end{array}\; .  \]

it's clear that
$\sum_{i=1}^{n-1}a_i^2+(\sum_{i=1}^{n-1}a_i)^2=2(\sum_{i=1}^{n-1}a_i^2+\sum_{1\leq i<j\leq n-1}a_ia_j)=2S$
and
$\sum_{i=1}^na_i^2+(\sum_{i=1}^na_i)^2=2(\sum_{i=1}^na_i^2+\sum_{1\leq i<j\leq n}a_ia_j)=2(S+a_n(a_1+a_2+...+a_n))$
then, we must prove that
$\frac{S+a_n(a_1+a_2+...+a_n)}{S}\geq \frac{n+1}{n-1} \\\Leftrightarrow nS+na_n(a_1+a_2+...+a_n)-S-a_n(a_1+a_2+...+a_n)\geq nS+S \\\Leftrightarrow a_n(n-1)(a_1+a_2+...+a_n)\geq 2S$

now, from cauchy-schwarz and the condition we have that
$(n-1)\sum_{i=1}^{n-1}a_i^2\geq (\sum_{i=1}^{n-1}a_i)^2 \\\Rightarrow (n-2)\sum_{i=1}^{n-1}a_i^2\geq 2\sum_{1\leq i<j\leq n-1}a_ia_j) \\\Rightarrow n\sum_{i=1}^{n-1}a_i^2\geq 2S \\\Rightarrow na_n\sum_{i=1}^{n-1}a_i\geq 2S$

now, we must show
$a_n(n-1)(a_1+a_2+...+a_n)\geq na_n\sum_{i=1}^{n-1}a_i=na_n(a_1+a_2+...+a_{n-1}) \\\Leftrightarrow na_n(a_1+a_2+...+a_n-1)+na_n^2-a_n(a_1+a_2+...+a_n)\geq na_n(a_1+a_2+...+a_{n-1}) \\\Leftrightarrow na_n^2\geq a_n(a_1+a_2+...+a_n)$
but this clearly follows from the condition, and we're done!!!  :D

btw nice inequality!

Clearly the LHS is increasing in $a_{n}$, hence it suffices to assume that $a_n=a_{n-1}$.  The rest is not difficult, since $a_{n-1}$ is at least as large as the rest of the variables, and hence larger than the mean of the remaining variables, or the root mean square of the remaining variables, etc.  From there, some manipulations with the power mean inequality or something similar gives the result.


vector
linear algebra
linear algebra unsolved

Let $ V$ be a finite dim vectorspace of characteristic $ 0$ and $ A \in L(V,V)$.  Let $ T_A$ be the linear operator on $ L(V,V)$ given by $ T_A(B)\equal{}AB\minus{}BA$. If $ B$ is a characteristic vector of $ T_A$ with non-zero characteristic value, then show that $ B$ is nilpotent.

[url=http://www.mathlinks.ro/viewtopic.php?t=203645]this[/url] would be helpful.


articles ratio geometry function inradius perimeter algebra

I came across in an article that read:

It is well known that primitive Pythagorean triples are given by $ (m^2 \minus{} n^2, 2mn, m^2 \plus{}n^2)$, where $ m$ an $ n$ are relatively prime and at least one of them even, up to switching the first two terms. Interestingly, all these triangles have unique ratios $ A/s^2$.

NOte: A = area, S = semiperimeter

1.) What does he mean by "up to switching the first two terms"?
2.) No matter how I solve, how do I know that they have unique ratios? Aren't there infinite number of Pythagorean triples?

Maybe I don't understand the English or did I miss something? Please help.

1)  The triples $ (3, 4, 5)$ and $ (4, 3, 5)$ are really the same triple except only one of them fits the formula because the way the formula is stated the even leg is always the second one listed.

2)  $ A \equal{} rs$, so the ratio we're looking at is really $ \frac {r}{s}$ where $ r$ is the inradius.  Can you express $ r, s$ in terms of $ m, n$ and show that the resulting function is an injection?  Alternately, what can you say about two Pythagorean triangles with the same inradius and semiperimeter?

Please tell me if my I am correct.
Expressing in terms of m and n, we have

$ A \equal{} \frac{n(m\minus{}n)}{m(m\plus{}n)}$ and $ s\equal{}m(m\plus{}n)$ so $ r\equal{}\frac{A}{s}\equal{}\frac{n(m\minus{}n)}{m(m\plus{}n)}(\frac{1}{m(m\plus{}n)})\equal{}\frac{n(m\minus{}n)}{m^2(m\plus{}n)^2}$.

Now we are looking for $ r/s$ so,
$ \frac{r}{s}\equal{}\frac{\frac{n(m\minus{}n)}{m^2(m\plus{}n)^2}}{m(m\plus{}n)}\equal{}\frac{n(m\minus{}n)}{m^3(m\plus{}n)^3}$.

Then, I have to show that $ \frac{n(m\minus{}n)}{m^3(m\plus{}n)^3}$ is an injection. Am I right? Sorry but I am not a math major.

You seem to have gotten mixed up.  What we have is $ A \equal{} mn(m^2 \minus{} n^2), r \equal{} n(m\minus{}n), s \equal{} m(m\plus{}n)$.  There are a few approaches you can take from here:  you can show that $ r$ and $ s$ are relatively prime (or nearly so), or you can write everything in terms of $ \frac{m}{n}$.

ok, i got it.

so i have to show that $ f(m,n) \equal{} \frac{mn(m^2 \minus{} n^2)}{m(m\plus{}n)}$ is an injective function. As far as I can remember, to show that a function is injective, $ f(x_1) \equal{} f(x_2)$ whenever $ x_1 \equal{} x_2$. But there are two variables. How do I go about it?

Why do I have to show that $ r$ and $ s$ are relatively prime? Does it have to do with being injective. Sorry, I'm so dumb at this, but I really am willing to learn.

[quote="guile"]As far as I can remember, to show that a function is injective, $ f(x_1) \equal{} f(x_2)$ whenever $ x_1 \equal{} x_2$. [/quote]
Other way around.  An injective function maps distinct arguments to distinct images.  (Every function has the property you describe.)

[quote="guile"] But there are two variables. How do I go about it?[/quote]
One way, again, is to write everything in terms of $ \frac{m}{n}$, and then it becomes a problem in one (rational) variable.  Note that the restriction that $ m, n$ are relatively prime and that one of them is even is key, otherwise this reduction is not unique.

So... eh... let's see here...

$ \frac{mn(m^2\minus{}n^2)}{m(m\plus{}n)}$
$ \equal{}\frac{n(m\plus{}n)(m\minus{}n)}{m\plus{}n}$
$ \equal{}n(m\minus{}n)$

From there on out idk how to continue on with how to prove that it is injective.

That's still not the function we're looking at.  $ \frac{A}{s^2} \equal{} \frac{r}{s} \equal{} \frac{n(m\minus{}n)}{m(m\plus{}n)} \equal{} \frac{t \minus{} 1}{t^2 \plus{} t}$ where $ t \equal{} \frac{m}{n}$ is a rational number already in lowest terms, hence in bijection with the relevant ordered pairs $ (m, n)$.  Can you show that this function $ f(t)$ in one variable is injective?

Also, again, consider the geometric argument about two Pythagorean triangles with the same perimeter and inradius.

i guess, they have the same area.
by the way, torajirou, are you a teacher? you sound like one. (not giving the answer but hints. if not, you can really be a good teacher.  :) )

Yes, but that's true of any two such triangles.  In this case, the statement "two Pythagorean triangles with the same perimeter and inradius are the same triangle" is equivalent (or nearly; I'm not entirely sure) to the statement that the value of $ \frac{r}{s}$ is unique.  Try proving the statement in this form.

And no, I'm not, but thanks :)

t0jajir0u:

I have discovered where did I get $ \frac {n(m \minus{} n)}{m^3(m \plus{} n)^3}$. It is in the rescaling. The article says, it has to be rescaled such that $ s \equal{} 1$. So we divide everything by $ s \equal{} m(m \plus{} n)$. If I computed correctly, then $ r/s \equal{} f(m,n) \equal{} \frac {n(m \minus{} n)}{m^3(m \plus{} n)^3}$. Is there a way that we could turn this into a function of two variables just like you did previously. I tried to use your technique but I did not get anywhere.

[quote="guile"]The article says, it has to be rescaled such that $ s \equal{} 1$. So we divide everything by $ s \equal{} m(m \plus{} n)$. [/quote]
I'm not sure what you mean; that's already covered by the first division by $ s$ and you don't need to divide by $ s$ again.  Could you link to the article?

Also, have you worked on the geometric argument?  It might be more intuitive.

[quote="t0rajir0u"]

I'm not sure what you mean; that's already covered by the first division by $ s$ and you don't need to divide by $ s$ again.  Could you link to the article?

Also, have you worked on the geometric argument?  It might be more intuitive.[/quote]

The article has no diagram but here it is. Our professor told us to understand the article for reporting so I had to expand a bit.

The website:
http://www.focusonmath.org/FOM/resources/publications/

The title is Heron's Triangles and Moduli Spaces

The point of setting $ s \equal{} 1$ is to make everything scale-free, i.e. to quotient out by similarity.  Since $ \frac{A}{s^2} \equal{} \frac{r}{s}$ is already scale-free, you definitely don't need to divide again; use my expression.

Anyway, it occurs to me that the condition that at least one of $ m, n$ be even is actually harder to work with when you set $ t \equal{} \frac{m}{n}$, so that may not be the best approach.  How much of the discussion about moduli spaces did you understand?

[quote="t0rajir0u"]The point of setting $ s \equal{} 1$ is to make everything scale-free, i.e. to quotient out by similarity.  Since $ \frac {A}{s^2} \equal{} \frac {r}{s}$ is already scale-free, you definitely don't need to divide again; use my expression.

Anyway, it occurs to me that the condition that at least one of $ m, n$ be even is actually harder to work with when you set $ t \equal{} \frac {m}{n}$, so that may not be the best approach.  How much of the discussion about moduli spaces did you understand?[/quote]

not much really, not to mention that i do not understand some of the English in the article. i have answered most of the exercises though. moduli spaces i think is in higher mathematics. i don't have those subjects.

if  you can see graph though, in the sketchpad, the domain of the functions where $ s \equal{} 1$is from $ (0,1)$, does that mean that $ f(t)$ is injective in that domain?

thanks for your help. i really appreciate it.  :)

"Moduli space" is a high-level word, but the object it describes is not necessarily complicated.  (One must distinguish between complicated and high-level mathematics.)  In the most concrete sense, a moduli space is just a way to parameterize a family of geometric objects (here, triangles).


A line containing the points $ (9, 1)$ and $ (5, 5)$ intersects the $ x$ -axis at what point?

The first thing we want to do is write our equation. To do this, we need to find the slope. The formula is:

$ \frac{y_2 \minus{} y_1}{x_2 \minus{} x_1}$

Now we substitute our points.

$ \frac{5 \minus{} 1}{5 \minus{} 9} \implies \frac{4}{\minus{}4} \implies \minus{}1$

Our slope is $ \minus{}1$. So our equation looks something like this:

$ y \equal{} \minus{}x \plus{} b$

What we need now is $ b$, the $ y$-coordinate of the point that intersects the $ y$-axis. To find this, we substitute one of our points in.

$ y \equal{} \minus{}x \plus{} b$

$ 5 \equal{} \minus{}5 \plus{} b$

$ 10 \equal{} b$

$ b$ is equal to 10. Our equation is:

$ y \equal{} \minus{}x \plus{} 10$

Now we need to find our $ x$-coordinate. Where it intersects the $ x$-axis, the $ y$-coordinate is $ 0$. So let's substitute this.

$ y \equal{} \minus{}x \plus{} 10$

$ 0 \equal{} \minus{}x \plus{} 10$

$ \minus{}10 \equal{} \minus{}x$

$ 10 \equal{} x$

Our $ x$-coordinate is $ 10$, and the point is $ (10, 0)$.

[quote="isabella2296"]The first thing we want to do is write our equation. To do this, we need to find the slope. The formula is:

$ \frac {y_2 \minus{} y_1}{x_2 \minus{} x_1}$

Now we substitute our points.

$ \frac {5 \minus{} 1}{5 \minus{} 9} \implies \frac {4}{ \minus{} 4} \implies \minus{} 1$

Our slope is $ \minus{} 1$. So our equation looks something like this:

$ y \equal{} \minus{} x \plus{} b$

What we need now is $ b$, the $ y$-coordinate of the point that intersects the $ y$-axis. To find this, we substitute one of our points in.

$ y \equal{} \minus{} x \plus{} b$

$ 5 \equal{} \minus{} 5 \plus{} b$

$ 10 \equal{} b$

$ b$ is equal to 10. Our equation is:

$ y \equal{} \minus{} x \plus{} 10$

Now we need to find our $ x$-coordinate. Where it intersects the $ x$-axis, the $ y$-coordinate is $ 0$. So let's substitute this.

$ y \equal{} \minus{} x \plus{} 10$

$ 0 \equal{} \minus{} x \plus{} 10$

$ \minus{} 10 \equal{} \minus{} x$

$ 10 \equal{} x$

Our $ x$-coordinate is $ 10$, and the point is $ (10, 0)$.[/quote]

an easier way, since u r given (9,1) and (5,5), you can just presume that when u add "x", u must subtract "y"

thus add 1 to the x-coordinate and subtract 1 from the y-coordinate, giving the coordinate of (10,0)

hence, the answer is still [b]10[/b]

Gee, I wrote all that for nothing... :maybe:

Or notice sum of coordinates is 10. y is 0 so x is $ 10$


Need Help, If you get the answer explain why and how.

Six polititions, three red and three blue, are training for a relay race and have put on identitcal outfits to disguise their political affiliation.  Either all of the red polititions tell the truth and all blue ones lie or vice versa.  One of the candidates holds a baton as they run together around a circular track in a clockwise direction.  Candidate B is in front of A, C is in front of B,etc., and A is also in front of F as they are spread out evenly around the circle.  They were all asked the same two questions as they passed by a stationary reporter: "Does the candidate in front of you tell the truth?" and "Does the candidate behind you hold the baton?" Their replies are as follows"

 Candidate A  "Yes, No"
 Candidate B  "No, Yes"
 Candidate C  "Yes, No"
 Candidate D  "No, Yes"
 Candidate E  "No, Yes"
 Candidate F  "No, No"

Which one is actually holding the baton?
               Answer: Candidate____

Explain your work and reasoning!!!

There are two cases:
Case 1
[code]
 Candidate A  "Yes, No" Truth
 Candidate B  "No, Yes" Lie
 Candidate C  "Yes, No" Lie
 Candidate D  "No, Yes" Truth
 Candidate E  "No, Yes" Lie
 Candidate F  "No, No" Truth
[/code]
Case 2
[code]
 Candidate A  "Yes, No" Lie
 Candidate B  "No, Yes" Truth
 Candidate C  "Yes, No" Truth
 Candidate D  "No, Yes" Lie
 Candidate E  "No, Yes" Truth
 Candidate F  "No, No" Lie
[/code]
However, in Case 1, there will be one person with the baton, while in case 2, there will be two, so case one is the real case and $ \boxed{\mbox{Candidate E}}$ is holding the Baton.

[quote="ianw"]Surely this question has been stated incorrectly/has no answer?

The reasoning, given the following facts:

- There is one baton
- Three politicians always tell the true
- Three always lie

Case 1 - the baton is behind someone who always tells the truth. In which case they say 'yes'. The baton then cannot be behind the candidates who lie, but as they lie they also must all say yes. We have 4 'yes' and 2 'no' responses

Case 2 - the baton is behind someone who always lies. So they must say 'no' when asked this question. As must the 3 who tell the truth, as the baton isn't behind them. So we have 4 'no' and 2 'yes'.

These are the only two cases possible and show that there CANNOT be 3 'yes' and 3 'no' responses to the second question, as is given... which means the question is wrong in some way.[/quote]

You are correct!!! Your the only 1!

[quote="fishythefish"]You can also use trial and error easily. test the statements for the following cases:

A holds the baton
B holds the baton
C holds the...
etc.

See what happens![/quote]

That helped soo much!

[quote="dreambold22"]int there some sort of riddle similar to this that HAS a real answer? something like "dude A always lies, dude B always tells the truth. u ask them both the same question. what do you ask them?"

or something like that. i forgot. anyone know the question and answer?[/quote]

Ask them a math question (that you know the answer to)! Then you would know who the liar is.
There are too many ways to approach this. "Is the person next to you thinking that the person next to him think that the..."

The problem you are referring to follows:

You are a prisoner, with two doors in the cell, one leading to death and the other to freedom. There are also two guards. One tells the truth and the other lies. What do you ask them to ensure your freedom?

There are also several other variations of this riddle, changing the situation or who tells the truth and when.

Also, don't revive old threads like this.

What do you mean, don't revive old threads like this? I posted that a long time ago...

Unless you're setting yourself as an example of what not to do...

Oh wait, am I reviving an old thread too now?

I'm talking to dreambold. Someone start a new thread if you want to continue this.


Solve for $ x$: $ \frac{x\plus{}3}{4} \equal{} \minus{}x\minus{}3$.

We have $ x\plus{}3\equal{}\minus{}4x\minus{}12$. Thus $ 5x\equal{}\minus{}15$, and $ x\equal{}\boxed{\minus{}3}$.

Simply incorporate the standard manipulations, starting with multiplying both sides by 4 to eliminate the existence of the fraction on the left hand side.

$ \frac{x\plus{}3}{4}\equal{}\minus{}x\minus{}3$
$ x\plus{}3\equal{}4(\minus{}x\minus{}3)$
$ x\plus{}3\equal{}\minus{}4x\minus{}12$
$ 5x\equal{}\minus{}15$
$ x\equal{}\boxed{\minus{}3}$


geometry
trigonometry
area of a triangle
Heron\u0027s formula
trig identities
Law of Cosines

could anyone give me a hint on where to start when trying to prove that the area of a quadrilateral is equal to
A = sqrt((S-a)(S-b)(S-c)(S-d))?  I have fiqured out a proof of Heron's Formula located in this pdf file.

First of all, I believe that formula is only for cylic quads.

I believe joml88 is right.  It's called Barahmgupta's formula or something.  Perhaps it makes use of the law of cosines?  Only I thought the guy who came up with it was from centuries ago with only very basic knowledge...


Suppose you start with 10 dollars. Every turn*, you flip a fair coin: if it's heads, you double your money, and if it's tales, you lose everything. Once you lose, that's the "end" of the round.

What is the expected "maximum" value you would achieve in many rounds of play? Keep in mind if you lose immediately, your maximum for that round is 10, because that's what you started with.

(I don't know the answer-- I'm asking.)

*edited from "round" for clarity.

[hide="Solution (edited)"] Flipping $k$ heads in a row puts your maximum at $10 \cdot 2^{k}$, and the probability of doing so is $\frac{1}{2^{k+1}}$ (since you need to end with tails).  Hence the expected value is

$\sum_{k=0}^{\infty}\frac{10 \cdot 2^{k}}{2^{k+1}}= \sum_{k=0}^{\infty}5 = \boxed{ \infty }$. [/hide]

I don't understand why you're summing $k$ to infinity. :maybe:

[hide="This is the way I interpreted the problem"]Presumably, to make money on this game you'd have to stop at some point because the probability of getting a tails approaches 1 as you keep playing. So you might say that you'll flip $k$ times before stopping or losing. Then, as you showed, the expected money you'll make is 10 dollars, no matter what $k$ is.[/hide]

[quote="MellowMelon"]I don't understand why you're summing $k$ to infinity. :maybe: [/quote]

(My solution has been edited slightly.)

[b]Definition:[/b]  Given an event that gives outcomes of values $v_{0}, v_{1}, ... v_{k}$ with corresponding probabilities $p_{0}, p_{1}, ... p_{k}$, the expected value is given by

$E = \sum_{i=0}^{k}v_{i}p_{i}$

The probability $p_{i}$ of flipping [b]exactly[/b] $i$ heads in a row (and then losing by flipping tails) is $\frac{1}{2^{i+1}}$.  The value $v_{i}$ associated with this outcome is $10 \cdot 2^{i}$.  Hence the expected value for a round that goes up to $k$ flips is

$E = \sum_{i=0}^{k}\frac{10 \cdot 2^{i}}{2^{i+1}}= \sum_{i=0}^{k}5 = \boxed{ 5(k+1) }$

In the finite case.  For $k = 1$ this is $10$ (breaking even), but for larger $k$ the expected value increases to infinity.

So does that expected value imply that after many rounds of play, the "maximum" money held could not be expected to converge on any particular value? My calculator trials would seem to confirm that idea.

It all depends what you define as the "value" of a particular outcome.  So yes, if that value is the maximum number of dollars, then the expected value is infinite.