probability counting distinguishability symmetry

A gardener plants three maple trees, four oak trees, and five birch trees in a row.  He plants them in random order, each arrangement being equally likely.  Let $\frac{m}{n}$ in lowest terms be the probability that no two birch trees are next to one another.  Find $m + n$.

[hide="Read on......"]Let M denote maple, O denote oak, and B denote birch.

There are $\frac{12!}{3!4!5!}=27720$ ways to arrange the trees without any restrictions, assuming the trees are indistinguishable.

You want there there to be no birch trees next to each other.
So we have MMMOOOO in some order. We must place a B in between the M's and the O's such that none touch.

Or place B's into _ in the following diagram: _M_M_M_O_O_O_O_. There are 8 spaces to place 5 B's, or $\binom{8}{5}=56$ ways. Now we can also arrange the M's and the O's in $\binom{7}{4}=35$ ways. So we have $56*35=1960$ ways to arrange the birch tree such that no two birch trees are next to one another. So our desired probability is:

$\frac{1960}{27720}=\frac{7}{99}\Rightarrow m+n=7+99=\boxed{106}$[/hide]

Hmmm.... can someone enlighten me whether this works and if not, why?

[hide]We know that the number of ways with at least 2 birch trees together is p(2 birchs together)-p(3 birchs together)+p(4 birchs together)-p(5 birches together). We wish to find 1-this probability.

p(2 birchs together) is the same as sticking the 2 trees together into one bundle, placing the 3 maples in c(11,3) ways, placing the 4 oaks in c(8,4) ways, and then choosing one of the remaining 4 spots to be the bundle in c(4,1) ways, for a total of c(11,3)c(8,4)c(4,1) ways.

p(3 birchs together) is the same as sticking those 3 trees together into a bundle and so on and so forth for c(10,3)c(7,4)c(3,1) ways

p(4 birchs) - we stick the 4 together and so on for c(9,3)c(6,4)c(2,1)

p(5)=c(8,3)c(5,4)c(1,0)

So p(at least 2 together)=c(11,3)c(8,4)c(4,1)-c(10,3)c(7,4)c(3,1)+c(9,3)c(6,4)c(2,1)-c(8,3)c(5,4) which works out to 12740.

The total possibilities is c(12,3)c(9,4)=27720 ways.

Therefore the total ways is 91/198, which when subtracted from 1 is 107/198, for 305.[/hide]

Probably wrong; is it?

[hide="possible answer"]we are looking at permutations, mapel and oak can be the same since all they do is split the birch, there are 7, the ideal arrangement is where we have those 7 with 8 spaces, and have to choose 5 of the 8 spaces,multiply by permutations for the birch and others, over the total permutations:
$\frac{ \binom{8}{5}*7! *5!}{12!}=\frac{7}{99} \implies m+n=106$[/hide]

[hide]Look at this arrangement: MMMOOOOBBBBB. There are $\frac{12!}{3!4!5!}$ ways to make any arrangement without restrictions. Consider the following: _M_M_M_O_O_O_O_. We must put all five B's into any of the eight _'s in order to have no two birch trees next to each other. There are $\binom{8}{5}$ ways to choose which _'s to put the B's in, and $\binom{7}{3}$ ways to arrange the maple and oak trees. Thus, there are $\binom{8}{5}\cdot \binom{7}{3}$ ways for this to happen. There are then $\binom{8}{5}\cdot \binom{7}{3}=1960$ successful outcomes, so our probability is $\frac{1960}{27720}=\frac{7}{99}=\frac{m}{n}$, and $m+n=\boxed {106}$.[/hide]

Wow, it actually worked out!

[hide="solution"]There are $\frac{12}{3!4!5!}$ possible unique permutations of the trees's sequence.

The maples and oaks can be placed in $\binom{7}{3}$ ways. There are $7$ trees in this sequence and the spaces between them are where the birches will be planted. There are $8$ spaces for $5$ birches. So $\binom{8}{5}$ possible arrangements of birches flanked by different trees. For every arrangement of oak and maple, there can be $\binom{8}{5}$ arrangements of birch. Therefore $\binom{7}{3}\binom{8}{5}$ is the number of permutations of the event.

$\frac{m}{n} = {\binom{7}{3}\binom{8}{5}}{\frac{12}{3!4!5!}} = \frac{1960}{27720} = \frac{7}{99}$
$m + n = \boxed{106}$[/hide]

[hide="much faster solution"]
For every distinct way that the birch trees are placed, there will always be a same number of ways to place the rest of the trees. So we can disregard the rest of the trees and just find the number of ways to place the birch trees correctly over the total number of ways to place the birch trees:
 $ \frac{\binom{8}{5}}{\binom{12}{5}} \equal{} \frac{7}{99}$[/hide]

Could someone kindly tell me what's wrong with this? I couldn't find the flaw...

First we stipulate all the trees to be distinguishable (regardless of their kind). We first place the 5 birch trees in a rows and they can permute in $ 5!$ ways. We then place a tree in each of the 4 gaps the birch trees create, and there are $ 7!/3!4!$ ways to choose 4 trees out of 7 tress, and those 4 trees can permute in $ 4!$ ways. We are then left out with 3 trees to be placed in the 10 possible spaces created by the 9 trees. Using the similar logic to balls and urns, the 9 trees and the 3 remaining trees permute in $ 12!$ ways, but since we have already counted the permutations of the 9 trees, we should get $ 12!/9!$ instead. This is out of a total of $ 12!$ permutations so we get

$ \frac{5!*7*6*5*4*12!/9!}{12!}\equal{}\frac{7*6*5*4}{9*8*7*6}\equal{}5/18$

It doesn't work because you count "BOMMBOBOB(OM)B" at least two different times (if you switch the letters in the parentheses, it represents another permutation counted by your method through which reversing again will give that), for example.

I see... should have realized that... thanks...

[hide="What is wrong with this?"]
The total number of arrangements without restrictions is $ \dfrac {12!}{3!4!5!} = 27720 $.

Now, we count the number of ways in which there is at least one string "BB" in the arrangement of MMMOOOOBBBBB. So, consider one of these arrangements: B B _ _ _ _ _ _ _ _ _ _. Now, by symmetry, if the number of ways to arrange these is $x$, then the total number of ways where there is some string "BB" is $ \dfrac {13x}{2} $. We multiply by $13$ for the placement of the string "BB" and divide by $2$ because of non-distringuishability of the two Bs.

Now: $ x = \dfrac {10!}{3!3!4!} $, so the total number of ways with at least one BB string is \[ \dfrac {13}{2} \cdot \dfrac {10!}{3!3!4!} = 27300. \]Thus, the number of arrangements with no BBs is $ 27720 - 27300 = 420 $, so our probability is \[ \dfrac {420}{27720} = \dfrac {1}{66} \implies \boxed {067}. \]
[/hide]

[quote="ahaanomegas"]What is wrong with this?
...
We multiply by $13$ for the placement of the string "BB" and divide by $2$ because of non-distringuishability of the two Bs.[/quote]  This doesn't make sense.

Does complementary counting work for this problem?

We don't need to sort trees based on maple or oak; that is, all we need to do is find the probability then when given $12$ spots to plant a tree, that of the $5$ birches, none of them are next to each other. To find the number of ways to place the birches, we can first plant the non-birch trees next to each other (just NNNNNNN) and then insert the birch trees in between them. Since there are $8$ spots to place the birch trees and we have to choose 5 of them to place birches in, there are $\binom{8}{5}$ ways to plant the birches so that no two are touching. The total number of ways to place the birches without any constraints is simply $\binom{12}{5}$. Hence, the probability is $\frac{\binom{8}{5}}{\binom{12}{5}} = \frac{7}{99}$. $\boxed{106}$.

Hi I did complementary counting + PIE but got $\frac{11!}{3!4!3!}-\frac{10!}{3!4!2!}+\frac{9!}{3!4!1!}-\frac{8!}{3!4!} = 35840 > 27720$ which makes no sense… what happened?

It doesn't matter how the maple and oak trees are placed. There are $12\choose 5$ ways to plant the birch trees. There are $8\choose 5$ ways to plant the birch trees where they aren't next to each other, because you can imagine planting 8 trees then putting a tree in every gap between birch trees. So, the probability is $\frac{{8\choose 5}}{{12\choose 5}}$ which is $\frac{7}{99}$, so the answer is 106.

Let any non-birch be NB. Then we can look at it as _NB_NB_NB_NB_NB_NB_NB_ with $8$ blank slots. There are therefore $\binom{8}{5}$ desirable ways and $\binom{12}{5}$ total, for an answer of $\frac{\binom{8}{5}}{\binom{12}{5}} = \frac{7}{99} \implies \boxed{106}$.


[quote="aidan"]On her birthday in 1993, Robyn's age was equal to three times the sum of the digits in the year she was born. What year was that?[/quote]

[hide="solution"]It would be strange if she was born before 1900.

$ 1993\minus{}(1000\plus{}900\plus{}10c\plus{}d)\equal{}30\plus{}3(c\plus{}d)$

$ 63\minus{}10c\minus{}d\equal{}3c\plus{}3d$

$ 63\equal{}13c\plus{}4d$

The solution to that diophantine equation is $ (c,d)\equal{}(3,6)$.

We check the year $ 1936$.

$ 3(1\plus{}9\plus{}3\plus{}6)\equal{}57$

$ 1993\minus{}1936\equal{}57$.

The birthday girl was born in 1936.[/hide]


In order to pass $ B$ going $ 40$ mph on a two-lane highway, $ A$, going $ 50$ mph, must gain $ 30$ feet.  Meantime, $ C$, $ 210$ feet from $ A$, is headed toward him at $ 50$ mph.  If $ B$ and $ C$ maintain their speeds, then, in order to pass safely, $ A$ must increase his speed by:

$ \textbf{(A)}\ \text{30 mph} \qquad
\textbf{(B)}\ \text{10 mph} \qquad
\textbf{(C)}\ \text{5 mph} \qquad
\textbf{(D)}\ \text{15 mph} \qquad
\textbf{(E)}\ \text{3 mph}$

[hide="solution"]
Let $ a$ be the new speed of A.  Assume that A can make a instantaneous perpendicular lane change.  Let $ x$ be the time A has to make the change.

Distance is velocity times time, and in relation to B, A moves at $ a\minus{}40$ mph.  We are given that $ (a\minus{}40)x\equal{}30$.  However, we are also given that $ (a\plus{}50)x\equal{}210$ (since C moves towards a, their relative speeds add).  Isolating x and substitution yields $ \frac{30}{a\minus{}40}\equal{}\frac{210}{a\plus{}50}\rightarrow \frac{1}{7(a\minus{}40)}\equal{}\frac{1}{a\plus{}50}$.  Now $ a\equal{}55$, and the change in speed is $ 5$, so

$ C$[/hide]

I got so confused on this since it says 30 feet and 210 feet.. I guess we were supposed to assume it was miles? not sure.


inequalities
induction
quadratics
algebra
inequalities proposed

Yes,it's very famous problem
I think everyone has known it.

I think it's the maclaurin's symmetric inequality and this post can be moved to theorems section

I know a solution using induction . Who knows other solutions?

It can be solved using derivation.
Take $ P(x) = (x+x_1)(x+x_2)...(x+x_n)$ it has n solution.
So P^{n-2}(x) has a solution means D > 0
which gives $ n!/2x^2 + (n-1)!S_{1}x +(n-2)!S_2 = 0$
 $ D > 0 $or  $(n-1)(S_1) ^2 - 2nS_2 >0 $
for others for k  as example take n - k + 1 derivation.
take Q(x) = P(1/x)  then derivate k - 1 times.
Again R(x) = Q(1/x) then you will obtain a quadratic equation in terms
k-1, k , k+1 so inequality can be constructed inductively. q.e.d.

[quote="yptsoi"]Shouldn't it be $\geq$?[/quote]

It should...


$ a$,$ b$,$ c$,$ d$ distinct positive integers. if $ \frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c} \plus{} \frac {c}{c \plus{} d} \plus{} \frac {d}{d \plus{} a}$ is a positive integer,then prove that $ abcd$ is perfect square.

First of all, this claim is not right. Because $ a\equal{}c$ or $ b\equal{}d$ is enough to make the condition hold without $ abcd$ being square.

Assume there is one more condition that $ a,b,c,d$ are distinct. Then
$ A\equal{}\frac{a}{a\plus{}b}\minus{}1\plus{}\frac{b}{b\plus{}c}\plus{}\frac{c}{c\plus{}d}\minus{}1\plus{}\frac{d}{d\plus{}a}$ is integer.

$ A\equal{}\frac{(a\minus{}c)(b\minus{}d)(ac\minus{}bd)}{(b\plus{}c)(a\plus{}b)(d\plus{}a)(c\plus{}d)}$

We have either

$ |(a\minus{}c)(b\minus{}d)|<|ab\plus{}cd|<(b\plus{}c)(d\plus{}a)$ and $ |ac\minus{}bd|<(a\plus{}b)(c\plus{}d)$

or

$ |(a\minus{}c)(b\minus{}d)|<|cb\plus{}ad|<(a\plus{}b)(c\plus{}d)$ and $ |ac\minus{}bd|<(b\plus{}c)(d\plus{}a)$

So $ |A|<1\Rightarrow A\equal{}0\Rightarrow (a\minus{}c)(b\minus{}d)(ac\minus{}bd)\equal{}0\Rightarrow ac\equal{}bd\Rightarrow abcd\equal{}(ac)^2$.


This is from
High School Math Contest
University of South Carolina
January 29, 2005

# 8. Suppose a and b are positive integers such that (a+2b)(a-b) = 10. What
is the value of 2a - b ?

(a) 1 (b) 2 (c) 3 (d) 4 (e) 5

Anyways, I thought about and messed around with this question, but I couldn't think of a way to approach it, and I couldn't understand the answer they provided.  So if you could explain each step in solving this problem that would be great!  Also did you know how to solve this kind of problem just by looking at it?  Is there a certain way to attack these types of problems?

Thanks

EDIT: Fixed. It appears I don't know how to subtract.  :blush:

Seeing that $ a$ and $ b$ are positive integers, we know that $ a \plus{} 2b$ and $ a \minus{} b$ both divide $ 10$ with $ a \plus{} 2b > a \minus{} b$. We know that the only positive integer factors of $ 10$ are $ 1,2,5,10$. Therefore, we have two cases.

We can have

$ a \plus{} 2b \equal{} 10 \\
a \minus{} b \equal{} 1$

which yields $ (a,b) \equal{} (4,3)$. Checking the other case, we have

$ a \plus{} 2b \equal{} 5 \\
a \minus{} b \equal{} 2$

which yields no integer solutions. Therefore, $ 2a \minus{} b \equal{} 2(4) \minus{} (3) \equal{} \boxed{5\text{ (E)}}$.

Whenever you have an equation in two or more variables such that the solutions you are looking for are integers, try factoring to get an equation with a factored expression on one side and a constant on the other. Then, factor the constant and try different pairs of factors.

Thats wrong. From the first set of equations you get solutions $ (4,3)$. The second set doesn't give integer solutions.

So $ 2(4) \minus{} (3) \equal{} \boxed{5 \ (E)}$.

And you got $ 4 \ (D)$ though.

sunehra's error was in stating:

$ a\plus{}2b\equal{}5$ and $ a\minus{}b\equal{}1$.. it should be $ a\minus{}b\equal{}2$ where $ a\equal{}3$ and $ b\equal{}1$ and $ 2a\minus{}b\equal{}6\minus{}1\equal{}\boxed{5\ (E)}$ which is the same answer for $ (4,3)$


What is ACoPS??? I think you should start working on Alcumus! That's the thing that helped me a lot! Alcumus shows explanations over problems and helps you see videos that improve your skills A LOT!!!  :ninja:

ACoPS stands for Art and Craft of Problem Solving.It's a very useful book, but more at a beginning olympiad level, it appears to me. I suppose it's for people who [i]can[/i] qualify for USAMO but can't do very well there or who have [i]already qualified[/i] but have gotten somewhat of a low score.

[url]http://www.wiley.com/WileyCDA/WileyTitle/productCd-0471789011.html[/url].

Thank you for your opinion on Alcumus.

[quote="xpmath"]Art and Craft of Problem Solving[quote]
I bought this book a few days ago and i would like to hear opinions about it...
How must someone to read it???
Is it necessary to read the first chapters to go to algebra,n.t,...?

It's not strictly necessary to read the first few chapters, no, but I highly recommend it. It really helps you organize your thinking process when you're stuck on a problem.


At a party every woman dances with at least one man, and no man dances with every woman. Show that there are men M and M' and women W and W' such that M dances with W, M' dances with W', but M does not dance with W', and M' does not dance with W.

Let $M$ be one of the men who dance with the maximal number of women, $W'$ one of the women he doesn't dance with, and $M'$ one of the men $W'$ dances with. If $M'$ were to dance with every woman that $M$ dances with, then the maximality of the number of women that $M$ dances with would be contradicted, so there is a woman $W$ that dances with $M$ but not with $M'$, and we're done.

let A be the man who dances with the maximum number of girls. i.e. for any man M, the number of girls M dances with is less than or equal to the number of girls A dances with.
Think A dances with G1,G2,...Gn. No man dances with all the girls. So there is a girl(say W) such that W does not dance with A. But W have someone to dance with. Lets take him as M.
If M dances with all G1,G2,...Gn then number of girls M dances with is greater than number of girls A dances with.(Since M dances with W too). So there is a girl say G1 such that M does not dance with G1.
Now A dances with G1, A doesn't dance with W.
M dances with W, M doesn't dance with G1.
QED :)

Let A be the subset of men, and B be the subset of girls. We use a graph and divide points in two sets( draw the line between two points if they dance). Denote $b_t,t=1,2,..k$ the elements of B. Of course, #B=k. Every $b_i$ has at least one corespondent $a_j$(from A), and no $a_t$ has k corespondents in B. Denote $p_i$ the set of points from A coresponding to $b_i$. If we have $b_i$ and $b_j$ such that $b_i,b_j\in{B}$ and $p_i-\#p_i\cap p_j\ge 1$ and $p_j-\#p_j\cap p_i\ge 1$, we are done. If not, we have(we can suppose the order arbitaly) $p_1 \subset p_2 \subset ...\subset p_k$,( of course $p_i$ may be equal $p_j$), but $\#p_1\ge 1$(using hyphotesis), so there exist $a_t$ such that the man coresponding to $a_t$ dances with all the girls, imposible.


Prove that a finite number of squares with sum of their areas $ \frac 12$ fit in the unit square without overlap.

[hide="Hint"]The following algorithm, called Next Fit Decreasing, works.  First, order the squares from biggest to smallest.  Place the biggest square on the bottom-left corner of the unit square.  For each remaining square, if the square fits on top of the current column of squares, then place it there; otherwise start a new column to the right.[/hide]

I'll leave it to others to show that this algorithm works.

That is also the solution in the book from where I collected the problem. However, I have a hopefully correct, very nice and extremely short solution. Okay I won't play around anymore. Here's my solution. Sorry if it's wrong.

[hide="My solution"]Suppose that the side lengths of the squares are $ a_1,a_2,\dots,a_n$. By the given condition, we deduce that $ a_1^2\plus{}a_2^2\plus{}\dots\plus{}a_n^2\equal{}\frac 12$.

Now let us examine, exactly when two squares overlap. Suppose that the two squares with side lengths $ a_i$ and $ a_j$ overlap. Then what condition must they suffice?

[asy]size(50); pair A,B,C,D,F,G,H,I,J; A=origin; B=(0,15); C=(15,15); D=(15,0); F=(10,10); G=(10,0); H=(0,10); I=(10,15); J=(15,10); draw(A--B--C--D--cycle); draw(A--G--F--H--cycle); draw(F--J--C--I--cycle);[/asy]
[asy]size(50); pair A,B,C,D,F,G,H,I,J,K; A=origin; B=(0,15); C=(15,15); D=(15,0); K=(10,10); F=(8,8); G=(10,0); H=(0,10); I=(8,15); J=(15,8); draw(A--B--C--D--cycle); draw(A--G--K--H--cycle); draw(F--J--C--I--cycle);[/asy]

(I've tried my best with these!) From the diagram, it's clear that whenever $ a_i\plus{}a_j>1$, the squares will overlap, if they are to fit in the unit square. So now for our proof we want to show that for all $ i,j$,$ 1\le i,j\le n$ we will have $ a_j\plus{}a_j\le 1$. Assume the contrary, that for some $ i,j$, $ a_i\plus{}a_j>1$. Then we have
\[ \frac 12\equal{}a_1^2\plus{}a_2^2\plus{}\dots\plus{}a_n^2\ge a_i^2\plus{}a_j^2\ge \frac 12(a_i\plus{}a_j)^2>\frac 12\text{, contradiction!}\]
Here $ a_i^2\plus{}a_j^2\ge \frac 12(a_i\plus{}a_j)^2\iff (a_i\minus{}a_j)^2\ge 0$. So our proof is complete.

[/hide]

All you have shown is that any two such squares can be fit into a unit square.  You haven't shown that all $ n$ squares can fit.

To be more precise, you have shown that $ a_i \plus{} a_j \le 1$ for every pair of distinct $ i$ and $ j$.  But how do you proceed from there?

First I took the $ n$ squares to fit (with or without overlap) in the unit square, as in the diagram. Then I showed that they fit without overlap (I showed that no pair of squares overlap). Isn't that sufficient? Or maybe I still don't get it. :maybe:

EDIT: I think I am starting to get it now.

(Written before your edit.)

What you have shown is that, given $ n$ squares of total area $ 1/2$ with two of them distinguished, the $ n$ squares can be placed in a unit square so that the two distinguished squares are disjoint.  But that does not imply that all $ n$ squares are disjoint.

In other words, you can move any two of the squares to be disjoint.  But when you move those two, you might end up creating more overlaps.


Square $ MNOP$ is tangent to circle $ O$ at $ N$ and $ P$. How many square centimeters are in the area of the shaded region? Express your answer to the nearest whole number.

[asy]fill(Circle((0,0),10),gray(0.7));
fill((0,0)--(10,0)--(10,10)--(0,10)--cycle,white);
draw(Circle((0,0),10));
draw((0,0)--(10,0)--(10,10)--(0,10)--cycle);
label("$P$",(0,10),N);
label("$O$",(0,0),NE);
label("$N$",(10,0),E);
label("$M$",(10,10),NE);
label("4 cm",(10,5),E);[/asy]

if its a square, tehn teh radius of the circle is equal to the length of teh square, due to its tangency

its radius is 4

the shaded area is 3/4 of a full circle, and its area is therefore

$ \frac{3(4^2\pi)}{4}\equal{}12\pi$
when rounded, yields 38


When I try to change styles in prosper (specifically to Azure), it wont compile the math correctly.  I'm using MIT's Athena computers to compile it.  Does anyone know the problem?

Thanks.

You really have to give us more information than that! You need to say what goes wrong, what the errors are and best of all, give a simple file where you are having problems.

If it helps, this file I found somewhere compiles fine with Azure [b][i]provided[/i][/b] you compile via ps first as Prosper requires. Do you get any errors compiling it?
[code]% CVSId: $Id: ExampleTroisPoints.tex,v 1.1 2003/01/27 14:39:22 exupery Exp $
\documentclass[%
pdf,
%nocolorBG,
colorBG,
slideColor,
%slideBW,
%draft,
%frames
azure
%contemporain
%nuancegris
%troispoints
%lignesbleues
%darkblue
%alienglow
%autumn
]{prosper}
\usepackage{amsmath}
\begin{document}
\begin{slide}{The quest for $\pi$}
\begin{itemize}
\item The following formula computes $8$ correct digits per iteration
  (Ramanujan):
\end{itemize}
  \begin{small}
  \begin{equation*}
    \frac{1}{\pi}=\sum_{n=0}^\infty \frac{(\frac{1}{4})_n(\frac{2}{4})_n(\frac{3}{4})_n}{n!^3}\bigl(2\sqrt{2}(1103+26390n)\bigr)\frac{1}{(99^2)^{2n+1}}
  \end{equation*}
  \end{small}
\end{slide}
\begin{slide}{The quest for $\pi$}
\begin{itemize}
\item The following formula computes $8$ correct digits per iteration
  (Ramanujan):
\end{itemize}
  \begin{small}
  \begin{equation*}
    \frac{1}{\pi}=\sum_{n=0}^\infty \frac{(\frac{1}{4})_n(\frac{2}{4})_n(\frac{3}{4})_n}{n!^3}\bigl(2\sqrt{2}(1103+26390n)\bigr)\frac{1}{(99^2)^{2n+1}}
  \end{equation*}
  \end{small}
\end{slide}
\end{document}[/code]


Click on the below picture

[asy]size(300);
picture CC()
{
picture p=rotate(0)*currentpicture;
currentpicture.erase();
return p;
}

pen anglepen=green,anglefontpen=red;
pair MA (real a=0,string s='',int f=9,
         pen p=anglefontpen, pair B, pair A, pair C, real r, int m=1, pair S=nullpair, pen q=anglepen)
{
if ((A==nullpair)||(B==nullpair)||(C==nullpair)) return A;
pair U=unit(B-A), V=unit(C-A);
if ((U==(0,0))||(V==(0,0))) return A;
path P=arc((0,0),U,V);
pair D=WP(P),Q=S;
if (Q==nullpair) Q=D;
D=A+r*D;
if (m<0) m=0;
if (m<8)
 {
 int n=2*m; if (n<6) n=6; if (n>8) n=8;
 for(int k=0;k<m;++k) D(shift(A)*scale(r*(1-k/n))*P,q);
 }
if (m>7&&(!(fullbreak&&breakflag))&&pathflag) fill(A--shift(A)*scale(r)*P--cycle,q);
MP(a,s,D,f,Q,p);
return A;
}

picture CopyClean()=CC;

pair MarkAngle(real a=0,string s='',int f=9,
         pen p=anglefontpen, pair B, pair A, pair C, real r, int m=1, pair S=nullpair, pen q=anglepen)=MA;

picture p,q;

pair A=(2,3), B=(4,5), C=(2,9);
D(  D(MarkAngle("\theta",black,B,A,C,0.4,8,yellow))
  --D(MA("\gamma_3",C,B,A,0.4,3,black))
  --D(MA(-75,"\eta+\rho",6,A,C,B,0.4))--cycle );

p=CopyClean();

A=(rand()%4,rand()%4); B=(rand()%4,rand()%4); C=(rand()%5,rand()%5);
D(  D(MA("\theta^2",12,black,B,A,C,0.5,9,green))
  --D(MA("\gamma_6",C,B,A,0.5,2,black+linewidth(1)))
  --D(MA("\beta",7,A,C,B,0.5,7))--cycle );

q=CC();

add(p); add(shift((7,0))*rotate(45)*q);[/asy]

It is unclear: since these routines (with slightly different definitions) are now a part of the site cse5 version, I am rather surprised that this code does not produce an error (two similar functions with the same name). Perhaps this is because coping the code verbatim doesn't result in its execution but rather in copying the picture. I tried the same picture in the Test forum with function definitions removed and the code showed OK. This may be the root of the problem.


number theory
least common multiple
puzzles

New to the forum today. I need a math guru or puzzle solver. I have a number grid that uses seven numbers in seven rows and four columns. Each number matches with each other number two times.  It looks like this.
1236
2347
3451
4562
5673
6714
7125
I want to increase the amount of numbers.  Lets say you coached track in high school. Your season was 9 weeks long (rows)and you had 9 schools involved. Your track field was only large enough for six schools (columns) (or less) at one time. You want to meet each school the same amount of times. Do we need more weeks (rows) or a bigger field (columns)? Or can it work?  I have worked out a couple of other grids that work out even the way I want, but I have my troubles with larger numbers.  Here is one that works.
123
425
153
214
435
423
152
431
523
415
Here I have 5 numbers in 10 rows, 3 columns, each number matches with another 6 times.  I'm thinking there are other grids, patterns that would work out similarily.  Some that are larger but not unreasonable.  From the high school track example....please try other paramiters..... you could increase/decrease weeks(rows), increase/decrease columns or increase/decrease numbers (schools).  Anyway I'm sure there are other grids but I just can't get it.  Thanks.
Mkron

Here's how you need to do it:
Let us say that $s$ is the number of schools, $c$ is the number of columns, and $r$ is a the number of rows. Also, let $p$ be the number of times each school plays each other, although this will be restricted. (I haven't quite worked the restriction, see the but the case below ALWAYS works, although it may not be the most "economical". The description on how to build the table only works if $p$ is picked in such a manner.)

***
Pick a $p$ such that $(s-1)\mid p(c-1)$ ($p(c-1)$ is divisible by $(s-1)$). To do this, $p$ must be some multiple of $s-1$ and $c-1$ over $c-1$. To minimize $p$, I would suggest making it $\frac{lcm (s-1,c-1)}{c-1}$.
***

Next, you are going to need $\frac{ps(s-1)}{(c-1)c}$ (or some multiple of this) many rows. Using the example of 9 schools with a 4 column chart, we can pick $p$ to be 3 so that they play each other 3 times, and we need an 18 row chart. One example of such chart is the following:

1257
2368
3479
4581
5692
6713
7824
8935
9146
1328
2439
3541
4652
5763
6874
7985
8196
9217

Here's how I made it:
1) List the numbers 1-9 two times.
2) I listed the numbers 1-9 starting with 2, then repeated with 3.
3) I saw that 1 had been paired with 2,3,8, and 9 already. 2 had been paired with 1,3,4, and 9. I picked 5 to start with, because it hadn't been paired with either 1 or 2. After I looped around, I had made 2 1/2  columns, and every number had been paired with every number. So for the next one, I could start with an arbitrary number. I chose 2, and went down.
4) For the last column, I saw that the only 2 numbers that hadn't been paired with 1,2 or 5 twice were 7 and 8. So I started the first loop with 7 and the next with 8.

Similar methods can be used for other tables, using "loops" of number. Your second example doesn't follow my description, but my way will always build a table, even if it doesn't describe all tables.

Hope this helps!

Hey, thanks for the work.  I will try the formula in my future quests for other matching grids.
mkron


limit
real analysis
real analysis unsolved

prove that if $(a_{n})$ is non-decresing and bounded above, then it converges.

Assume the [i]Least Upper Bound axiom[/i]:

Any upper bounded, non empty subset $S$ of $\mathbb{R}$ has a least upper bound.

Let $S = \left\{ a_{n}, n \in \mathbb{N}\right\}$. Since $a_{n}$ is bounded from above, it is clear that $S$ is upper bounded.
For every $\varepsilon > 0$, there is an $n \in \mathbb{N}$ such that $\sup S-\varepsilon < a_{n}\leq \sup S$ (by definition of $\sup S$)
Since $n$ is increasing, we have $a_{n}\leq a_{n+1}\leq \ldots$ and each of the $a_{n}$ is bounded by $\sup S$.
Therefore for all $N \geq n$ we have $\sup S-\varepsilon < a_{N}\leq \sup S.$
Recapitulating, we proved that for all $\varepsilon > 0$, there is an $n$ such that for all $N \geq n$ we have \begin{eqnarray*}\sup S-\varepsilon < a_{n}\leq \sup S & \Longrightarrow & | \sup S-a_{n}| < \varepsilon \end{eqnarray*} Therefore \begin{eqnarray*}\lim a_{n}& = & \sup S \end{eqnarray*}


floor function
logarithms
function
ratio
geometric sequence

How many elements of the set $ S \equal{} \{2^0, 2^1, 2^2, 2^3, ..., 2^{2003}\}$ have a ﬁrst digit of $ 1$?

If the exponent is congruent to 0 modulo 4, then its units digit is 1. Therefore, we get:

\[ \left\lfloor\frac{2003}{4}\right\rfloor\plus{}1\equal{}\boxed{501}
\]

I think he means first digit, not units digit.

Edit: Ah, never mind. Can anyone explain why the first digit is 1 if the exponent is divisible by 4...?

If we calculate some powers of 2, we can see that first digit 1 have powers: $ 2^{10k}, 2^{10k\plus{}4}, 2^{10k\plus{}7}$ so we will form blocks of 3 elements for every possible $ k$. $ \left\lfloor\frac{2003}{10}\right\rfloor\equal{}200 \implies k_{max}\equal{}200$ so there are $ 201$ blocks totally (since $ k$ can equal $ 0$) but if we examine the last block ($ 2^{2000}, 2^{2004}, 2^{2007}$ we see that only 1 element of that block is in set $ S$ so total number of the elements is $ 201\cdot 3\minus{}2\equal{}\boxed{601}$

It's actually not quite like the others are saying... it doesn't follow a strict pattern (although the guess of 601 is pretty close).

Instead, we need to note a few things
(1) The sequence cannot "skip" #s of digits.  That is, if one element has $ s$ digits and another has $ t$ digits (where $ s\plus{}1<t$), then there is an element with $ s\plus{}1$ digits.
(2) If $ 2^i$ is the first element in the sequence with $ n$ digits, then its first digit will be $ 1$.
(3) Then $ 2^{i \plus{} 1}$ will have a first digit of $ 2$ or $ 3$.

Since we have $ 2^0 \equal{} 1$ in this sequence, there will be exactly one element whose first digit is $ 1$ for each digit of $ 2^{2003}$.  So just how many digits is that?

$ \left\lfloor log_{10}(2^{2003})\right\rfloor \plus{} 1 \equal{} \left\lfloor 2003*log_{10}(2)\right\rfloor \plus{} 1 \equal{} \boxed{603}$

How exactly did you find relation with logarithm function? Your solution seems pretty interesting but I can't fully understand it. Perhaps it's because of my limited knowledge of English.

What [b]Mathnerdmo[/b] is saying is;
If we group the powers of two in terms of the number of digits they have then we find;

$ 1,2,4,8$
$ 16,32,64$
$ 128,256,512$
$ 1024, 2048, 4096, 8192...$

Notice that every time we move up one digit we have exactly one power of two which bigins with $ 1$.
To proove this, suppose $ 2^n \equal{} a\cdot 10^k$ and $ 2^{n\plus{}1} \equal{} b\cdot 10^{k\plus{}1}$, where $ a,b < 10, a,b \in \mathbb{R}$
$ \Rightarrow \frac{2^{n\plus{}1}}{2^n} \equal{} \frac{b}{a} 10 \Rightarrow \frac{b}{a} \equal{} \frac{1}{5}$

Since $ a<10$ we must have $ b<2$ which implies $ b\cdot 10^{k\plus{}1} \longrightarrow 2^{n\plus{}1}$ has $ 1$ as its first digit.

So we have the find the number of digits of $ 2^{2003}$ in order to find the number of $ 2^n$ which have $ 1$ as a first digit.

Again let $ 2^{2003} \equal{} a\cdot 10^k$ for some intiger $ k$ and real number $ a$, then we can see that $ 2^{2003}$ has $ k \plus{}1$ digits.

$ \Rightarrow k\equal{}\log_{10} 2^{2003} \minus{} \log_{10} a$

Since $ a<10 \Rightarrow \log_{10}a <1$
$ \Rightarrow k \equal{} \log_{10} 2^{2003} \minus{} \log_{10} a \equal{} \lfloor  \log_{10} 2^{2003} \rfloor$

$ \Rightarrow k\plus{}1 \equal{}  \lfloor  \log_{10} 2^{2003} \rfloor \plus{}1$

I'm going to agree with Mathnerdmo on this one.  Another way of phrasing what he's saying:

(1) in a geometric sequence with a common ratio of 2, every term in the sequence either has the same number of digits or one more digit than the previous term.
(2) If (and only if) a term has more digits than the previous term, then it has a first digit of 1. (This obviously can't apply to the first term, because the first term doesn't have a previous term to be compared to)

then, by (2), the number of terms with a first digit of 1 is the number of terms with one more digit than the previous term, plus one for the first term, which does in fact have a first digit of 1.  because of (1), we can simply count the difference in the number of digits between the first and last terms, and add one.  Since the first term has one digit, the number we are looking for is $ (($number of digits in $ 2^{2003}) \minus{} 1 )\plus{} 1$.  In other words, we're just looking for the number of digits in $ 2^{2003}$.  hence the logarithm.

edit: haha ocha you beat me to it.

Thank you for explaining. It was helpful :)

1. How would you have figured out $ \lfloor \log_{10} 2^{2003} \rfloor$ without a calculator?

2. How about the number of elements beginning with digit $ 4$?

[hide="hint"]
For number 2, look at my post and how the powers are grouped
Notice that they are always in groups of $ 3$ or $ 4$
Every group that has four elements has exactly one power of two that has $ 4$ as its first digit...

[hide="hint 2"]
Try to set up a simultaneous equation using the number of digits in $ 2^{2003}$...
[/hide][/hide]


geometry
circumcircle
projective geometry
cyclic quadrilateral
power of a point
radical axis
geometry solved

Let $ABCD$ be a cyclic quadrilateral. Call $M$ the point of intersection of the lines $AD$ and $BC$. Let $MT$ and $MP$ be tangents to the circumcircle of the quadrilateral $ABCD$ such that the points $T$ and $P$ lie on this circumcircle. If $R$ is the point of intersection of the lines $BA$ and $CD$, then prove that the points $R$, $T$, $P$ are collinear.

[color=darkblue][b]Lemma 1 (well-known).[/b] Let $A,B,C,D$ be four collinear points for which the points $C$, $D$ are harmonical conjugate w.r.t. the pair $(A,B)$,
i.e. $\frac{CA}{CB}=\frac{DA}{DB}$. Denote the middlepoint $M$ of the segment $CD$. Then exists the relation $MD^{2}=MA\cdot MB\ .$

[b]Lemma 2.[/b] Let $d_{1}\ ,\ d_{2}$ be two fixed concurrent ( in the point $A$) lines and let $B\not\in d_{1}\cap d_{2}$ be a fixed point.
For a mobile point $M\in d_{1}$ denote the intersection $N\in BM\cap d_{2}$ and the harmonical conjugate $L$ of the point $B$ w.r.t. the pair $(M,N)$,
i.e. $(!)\ \ \frac{LM}{LN}=\frac{BM}{BN}\ .$ Then the geometrical locus of the point $L$ is a line $d$ for which $A\in d$ - [b]the polar line of the point $B$ w.r.t. the pair[/b] $(d_{1},d_{2})\ .$

[u]Proof.[/u] The parallel line from the point $L$ to the line $AB$ meets the line $d_{1}$, $d_{2}$ in the points $P$, $R$ respectively. Thus, $\frac{LR}{BA}=\frac{LN}{BN}$ and $\frac{LP}{AB}=\frac{LM}{BM}\ .$
From the relation $(!)$, i.e. $\frac{LN}{BN}=\frac{LM}{BM}$ results that $LR=LP$ and $PR\parallel AB$, i.e. the geometrical locus of the point $L$ is a line $d$ for which $A\in d\ .$

[b]Proof of the proposed problem.[/b] Denote : the circumcircle $w$ of the cyclic quadrilateral $ABCD$ ; the h-conjugate points $X$, $Y$ of the point $R$ w.r.t. the pairs $(A,B)$, $(C,D)$ respectively ; the h-conjugate points $Z$, $K$ of the point $M$ w.r.t. the pairs $(B,C)$, $(A,D)$ respectively ; the middlepoints $S$, $U$, $V$, $W$, $Q$ of the segments $[MR]$, $[MZ]$, $[MK]$, $[RX]$, $[RY]$ respectively. From the previous lemma results that $M\in XY$ and $R\in ZK$.
$UM^{2}=UB\cdot UC$ and $VM^{2}=VA\cdot VD$ $\Longrightarrow$ the line $UV$ is the radical axis between the circle $w$ and the null circle $\{M\}$.
$WR^{2}=WB\cdot WA$ and $QR^{2}=QC\cdot QD$ $\Longrightarrow$ the line $QW$ is the radical axis between the circle $w$ and the null circle $\{R\}$.
The radical axis between the null circles $\{M\}$ and $\{R\}$ is the bisector of the segment $[MR]$. Therefore : $S\in UV\cap QW$ ;  $\overline{SWQ}\parallel XY$ $\Longrightarrow$ $M\in XY$ ; $\overline{SUV}\parallel KZ$ $\Longrightarrow$ $R\in KZ$. In conclusion, and the points $P$, $T$ belong to the line $\overline{RZK}$, i.e. $R\in PT\ .$

[b]Remark.[/b] With the "high theory" (pol-point and polar-line, fascicles a.s.o., can proof this problem in ... two rows ![/color]

Ok, very good solution.
Another solution???
DARIJ, i think you have another solution!!!! Please post!!! :P

We can have another elementary proof, by Pascal and Desarques theorems as follows:

     We denote as E, the intersection point of AT, BD and as F, the intersection point of AC, DT ( in my drawing the point T, lies on the arc BC not containing the vertices A, D ).

     We consider the non convex pentagon TACBD and then, by Pascal’s theorem we have that the points M, E, F, are collinear.

     We see now that the triangles BEA, CFD, are perspective at point M and so, by Desarques’s  theorem we have that the points R, T, N, are collinear, where N, is the intersection point of diagonals AC, BD, of ABCD.

     By the same way ( considering the non convex pentagon PBDAC ), we can prove that the points R, N, P, are collinear. Hence, the points R, T, P, are collinear and the proof is completed.

     Kostas Vittas.

Very good solution, but, you can explain me the pascal theorem!!
I dont know this theorem!!
Thanks!!  :D

Yes I can. Give me please a little time, because today is a difficult day for me.

Kostas Vittas.

[quote="conejita"]DARIJ, i think you have another solution!!!![/quote]

...maybe ;)

I will use the word "polar" in the below proof, but all what I will need from the theory of polars will be the following basics:

[color=blue][b]Definition 1.[/b] If k is a circle with center O and radius r, then the [i]image of a point P under the inversion with respect to the circle k[/i] is defined as the point Q which lies on the ray OP and satisfies $ OP\cdot OQ \equal{} r^{2}$. This image is defined for $ P\neq O$ only (at least on the Euclidean plane).

[b]Definition 2.[/b] If k is a circle with center O, then the [i]polar of a point P with respect to the circle k[/i] is defined as the perpendicular to the line OP through the image of the point P under the inversion with respect to the circle k. This polar is defined for $ P\neq O$ only (at least on the Euclidean plane).

[b]Theorem 3.[/b] If k is a circle and P is a point outside of k, and the two tangents from the point P to the circle k touch k at two points U and V, then the line UV is the polar of the point P with respect to the circle k.[/color]

I suggest proving Theorem 3 using Definitions 1 and 2 as an easy exercise. Now to your problem:

We have to prove that the points R, T, P are collinear, i. e. that the point R lies on the line TP. Now, since the two tangents from the point M to the circumcircle of the quadrilateral ABCD touch this circumcircle at the points T and P, Theorem 3 yields that the line TP is the polar of the point M with respect to this circumcircle. So we have to show that the point R lies on the polar of the point M with respect to this circumcircle. In other words, we have to show that the point $ BA\cap CD$ lies on the polar of the point $ AD\cap BC$ with respect to the circumcircle of the quadrilateral ABCD.

Now, this is a fact which appears as Theorem 1 in [url=http://www.cut-the-knot.org/cgi-bin/dcforum/ctk.cgi?az=read_count&om=390&forum=DCForumID6]Yetti's [i]Quadrilateral Inversions[/i][/url] (with a proof using Pascal's theorem) and as Theorem 7 in [url=http://www.cip.ifi.lmu.de/~grinberg/Butterfly.zip]my [i]On cyclic quadrilaterals and the butterfly theorem[/i][/url] (with a proof using elementary arguments - scroll down to the proof of Theorem 9 and the first few lines below it). I don't know whether one can call it "well-known" and use it without proof on any competition, but at least it is known enough to make me wonder how a problem like the above, which is more or less a trivial consequence of this fact, could be proposed at a Chinese olympiad.

  darij


calculus
integration
function
trigonometry
calculus computations

Calculate: $ \int e^{\arccos{x}}\frac{x \plus{} 1 \plus{} \sqrt{x^{2} \plus{} 2x}}{(x\plus{}1)^{2}\sqrt{1\minus{}x^{2}}}dx$

hello, are you sure that there exists a primitive function?
Sonnhard.

I don't know I found the integral in another forum.

hello, where do you found this integral? In which forum?
Sonnhard.

Let $ x\equal{}\cos u$

$ \int e^u\frac{\cos u \plus{}1 \plus{} \sqrt{\cos^2u\plus{}2 \cos u}}{(\cos u \plus{} 1)^2 \sqrt{1\minus{}\cos^2u}}(\minus{} \sin u)du$
$ \minus{} \int e^u \frac{\cos u \plus{} 1 \plus{} \sqrt{\cos^2u \plus{} 2\cos u}}{(\cos u \plus{} 1)^2}du$

I can't seem to get any farther, are you sure it's copied correctly?

Yes, I'm sure - take a look: http://dxdy.ru/topic22986.html


There is a dartboard with two scores you can get: 4 and 9. What is the highest score you can NOT get on the dart board shown below if you may use as many darts as you wish?

What if the scores you can get are a and b, where gcd(a,b)=1? Then what's the highest score you can't get?

yeah...Max, I also found the formula when I was in mathcounts. I noticed they used that kind of problem a lot, so I tried to find a formula. I found it and proved it (and i've since been proud of that). So, for all you other peeps, how can you prove Max's formula?

 if no-one comes up with anything, i'll start posting hints

someone could actually just try out a few cases and see that the formula works...that might give you some ideas as to proving the formula...

you're supposed to be trying to find it. i've already told you how to get started: try it out on a few examples, and make sure our proposed formula works.

Im dumb. Hmm, isn't this just about the diophantine n = ax + by?

Oh. Ill have to read that chapter (along with five other ones  8-) ) in AoPS 2.

I don't think any general formula exists for the three-number problem.  Part of the reason is that there are very many things to consider other than oddness and evenness.  For example, if a|b, we're really dealing with a 2-number problem with a and c.  The formula doesn't deal with this at all, giving me the value of c - a for the function, which definitely isn't correct.


Write a fourth degree polynomial with leading coefficient 1 and with all integer coefficients such that one of its roots is $ \sqrt{2}+\sqrt{5}$.

I know what the other roots must be, they are just $ \pm{\sqrt{2}+\pm{\sqrt5}}$ but i was wondering is there an easier way to multiply the polynomial out instead of just doing long multiplication.  Vietas doesnt seem much easier.  So is that the only way to multiply the roots out or is there an easiwer way just using the orignial given root.


geometry
3D geometry
pyramid
Duke
college
geometric transformation

First, answer the person above you. If you are right, ask a question that you know the answer to that isn't too hard (like what color were george washingtons socks on the 12th of may in...)
that is also [b]about history[/b]

I'll start

What did pocohantas die from?

The most reasonable guess historians have at the moment is either pneumonia or tuberculosis. Some people think it may have been small pox.

How was Herculaneum first found (hint: it was an accident)?

a farmer was digging a well when he found chunks of Pompeii & Herculaneum

How were the Dead Sea Scrolls found?(same hint as NOTA)

Lol, I'm gonna' try to answer it without google. Lets see....
I THINK some boys were exploring, and noticed a cave. They went in and found them.

What Egyptian ruler changed the religion from polytheism so monotheism?

Akhanaten? Sp (Without google)

Which Pharaoh is buried in the Great Pyramid?

Wait!  I actually know this - um, Akhenaten or something like that.

EDIT: Darn, SnipedYou beat me to it.

Here's the question I had anyway:
No Google: What countries did Napolean force to peace in 1797?

I think SnipedYou beat you to it, miyomiyo.

I'll answer SnipedYou's question so there are no further complaints: Khufu.

[quote="miyomiyo"]
No Google[/quote]

Lol, I'm hopeless then.

Nobody here's in sophomore history?  We just covered this in my class....

France and the cuntry with Waterloo in it, i think

[quote="mz94"]France[/quote]Er, Napolean was the consul of France.

I know that napolean was a french leader and etc, but I'm in 7th grade, so I don't know french history yet.

Austria? Prussia? I'm in AP US History now so I forget a lot from last year.

Here should be an easier one:

Who was the leader of Athens during Athen's golden age? (You shouldn't need to use google if you have studied the Greeks at all)

I did Ancient Civilizations in 6th grade last year, and same here, I forget, lol. I'm guessing it's the first democratic ruler, but I forgot his name  :huh:

The answer to the one with Athens Golden Age is obviously Pericles (or whatever the spelling is).
The Napoleon question....is it Austria?
My question, how many years did the war known in history as the "100 years war" actualy last; and the years it begun and ended.

[quote="1=2"]Was it the signing of the Declaration of Independence, on July 4, 1776?[/quote] :rotfl:  :rotfl:  :rotfl:

Maybe 1=2 accidently read it as the French revolution or something... But I guess which event is most famous is kind of your own opinion.

yes, the boston tea party...the key word in my question was "sparked"
the beaches were: sword beach, juno beach, gold beach, omaha beach, utah beach, maybe there were others. Please post the complete answer (i'm curious  :wink: ).
my question:
which was the shortest war in history?

You got them all

Hmm...I remember reading somewhere about a 40 min? War. I think it was England vs. Algeria or somewhere else in Africa.

Why did Henry VIII split England from the Catholic Church?

I just studied this!
Because the Pope was refusing to allow him to divorce his wife (Catherine of Aragon).  The reason he didn't allow this was because Catherine's nephew was holding the pope captive at the time.

The Treaty of Vienna restored order in Europe after the French revolution and the Napoleanic Era.  One of the many results of it was restoring the Old Regime in France with King Louis the ----th.

16? He's really the only one I remember from World History and he's probably not the correct one either.

What island was Odysseus the king of?

The Latin Golden Age.

Which king of which country was called the "Sun king"?

Louis the XIV of France

What is the tree of life in Norse Mythology?

Yggdrasil

What is the name of the last Native American tribe to live on a reservation?

don't know

[quote="mojo_jojo"]
My question, how many years did the war known in history as the "100 years war" actualy last; and the years it begun and ended.[/quote]

It lasted 116 years ago and it began in 1337 and ended in 1453 (war between England and France).

PS: hehhehheh, I just learned this in social studies (7th grade). :rotfl:

Question:Who was our 28th vice president? No Chart to help you. Or google.

[quote="pingala"]don't know
[/quote]

Just so you know, it's the Nez Perce. haha just learned this yesterday

Thomas Marshall.

What century was the Illiad written in?

6th century bc i think

after you tel me im right or wrong, i'll post question


Why is there no Hawaii or Louisiana forum??
Could someone possibly add them..?

I think subforums take up a lot of forum space though, there probably hasn't been a lot of activity from either state for a while...

Umm... I live in Hawaii and I really don't know many people (there are a few) that  would actually use it.

That said, it would totally be cool if there was a Hawaii subforum... I would like to meet all the other island AoPSers, but, like mewto said, it's not gonna get used much.

@ mewto: :wink:

AIME_is_hard, glasgowgators95, me (Ttocs45), and mathcounts_LA all live in Louisiana. There are others, too.

Hey theres no Alaska forum! What will the Palins do?

Presumably, something better than talk to themselves about math.

Anyways, I think this TOPIC THAT ORIGINALLY WAS USELESS is dead. So leave it.

Um, I was going to post a new thread about this, but then I saw this one.  There's also no state forum for Delaware.   :?:  Could someone add one?

Seeing you live in Illionois, why would [i]you [/i]want a Delaware forum?

Hmm,  just doing the deleware people a favor....

And how do you know I live in IL?  :ninja:  :P

Because people here know how to read and look up research sources like school study skills :wink: *cough**profile location**cough*

 :D > :P

[quote="suma_milli"]Haha I was being sarcastic

:P

And okay..
:D[/quote]
i believe hawaii is more leaning towards  :D xDhhhhhhhhhhhaaaaaaahhaaa

I also believe that smaller states can group together (if they all lie in the same region) to be one forum.

That's a good idea, that would help minimize the space taken up.  Also maybe combining upstate new york and nyc would be a good thing as there's barely any activity on either forum.  Does anyone even look at topics in this forum?

uhhhh

rarely, I believe?

Well actually only the smaller states would need to group together to form one forum :D

HAHA WE'VE GOT  :D PRIDE IN THIS TOPIC

I guess, but I was saying that because barely anyone posts on the NYC or upstate new york forum?  I guess two separate forums for them would come in handy later though.

Nah not alaska and hawaii because although they're both separate from the rest of the U.S. they're not close to each other at all.  That's like combining IL and CA.

lol AK and HI are far away from the mainland US :(

Hm... even though ok is not very active anymore, we have one. no one posts there that much. prob once per 2 weeks?

The Arizona forum is dead. (seeing that I've probably posted there the most. Haha.) Should that be combined with something?

[quote="gauss1181"]Or this:

One forum for the whole state of NY

One forum for Vermont, Connecticut, West Virginia, Maine (and some small-populated New England states)

One forum for Oregon, Montana, Wyoming, Utah

Alaska and Hawaii should share the same forum, along with...

[hide] :D (JK JK JK JK)

...uh maybe Arizona and New Mexico. They're pretty inactive I guess...[/hide][/quote]

LOL, West Virginia is not a New England state.  And all I did is PM rruscyzk and ask him if he could create a WV Forum.  We only have like 5 people, but I guess that's enough.

EDIT:  OOPS, I didn;t mean to revive the old topic.  Sorry, everyone.


ARML geometry college Stanford AMC AIME email

How do I join the SF Bay Area ARML group?

Well, I can only speak from experience before I moved to Taiwan.

Basically, you should receive word about trainings in Berkeley or Stanford, which you should attend (they probably have started about now). The trainings usually consist of lectures, problem sets, and placement exams.  The placement exams determine whether you will be competing in A team, B team, or C team.

However, if you really live in the Netherlands like your profile claims, joining SFBA ARML may be quite difficult!

For more information, the person you want to make contact with is Ted Alper. He does have an AoPS user name - he's TMA - but I don't think he comes here often. Look for him in connection with the EPGY program at Stanford.

[quote="fourierseries"]
Basically, you should receive word about trainings in Berkeley or Stanford, which you should attend (they probably have started about now).
[/quote]
Last year, the meetings started after we got back our AIME scores. (And I know it didn't start yet. I didn't get an invitation yet  :P )

How do I get in contact with Ted Alper if I am interested in joining?  Does anyone know his email?

I think anyone interested can go, but they issue invitations to people with (supposedly) good scores on the AMC/AIME/other math competitions. I got mine today.

[quote="daermon"][quote="Ubemaya"]Do they cut people, or can anyone make the C team?[/quote]
i think anybody can make the C team
but i dont think you would have any chance of getting cut even if they did cut people  :P[/quote]

Wait, aren't there like 69 people, so:

${\lceil \frac{69}{15}}\rceil= 5$

$A \quad B \quad C \quad D \quad E$

Is there really going to be an E team?

No, a lot of people join just for practice, they don't actually go to ARML.

Follow ccy's link and join the Yahoo group.  Then start coming to practices--they're every Sunday from 2 to 5 (or is it 4:30?) at Stanford, building 380, room 380C.

No one gets cut from ARML.  If the number of people going is not divisible by 15, there will be some students who have to be alternates (there were 2 last year), but they still get to go to ARML.  Also, there will not be an E team.  I think if we have enough to make 4 or more teams, then there's an A team, a B team, and several C teams (like C', C*, C#, or whatever) that are divided up more or less randomly.


Should Tare graciously step down from being a F&G moderator like they do in government so they don't have to fire them?

Eh?  I don't know I have an opinion, but I don't want anyone mad at me.


Does there exist irreducible polynomial $ P(x)$ such that:
-whenever $ x$ is integer, $ P(x)$ is a composite integer (so $ P$ has rational, but not necessarily integer, coefficients)
- $ P(x)$ is irreducible
- there does not exist an integer $ n$ so that $ n | P(x)$ for all integers $ x$ with at most finitely many exceptions.

This seems to be an open probelm, known as the the Bunyakovsky conjecture.


Who here went to AMSP last year?

Will anybody be coming back?

If so, what site (CA, TX)?

i went last year.
not coming back tho T_T

have funn :D

lol suzy.
i decided to represent wisconsin. ;P lol since im stuck here for like 2 more years hopefully . >< so my initial idea for a cali beach icon didnt exactly fit lollll.

what? really. huh. i didnt know it was still on ur list of options this summer.
when do u hear back from stanford??

oh btw Ghostmaster,

they only offered it at UTD last year.
but if youre going to CA site, santa cruz is BEAUTIFUL.
well a lot of california places are [like the oc!]

michelle: ahahaha way to represent! :P
mm yeah..its off now. my app's due on the 10th..should write the essay x(
i still havent heard back from upenn tho...so odd

ahhh i see.
:O smack in mid of aps. blergh.
good luck <3

WHOA KASPER, IS THAT YOU?!?!?!?!
this is steve...so...yea...the one in ur dorm the whole time...yea...

[quote="serialk11r"][quote="sooozyy"]omg michelle, your icon's a picture of a cow
-_____- holy crap
lols

i went last year..

still deciding whether i should apply or not..and ill probaly go to ucsc one just cuz i like cali weather better,,,and its close to bay areaa so i can drop by my boyfirned's house in palo alto and come back down to irvine[home] after everything wheee[/quote]
Not sure if you're aware of this, but Palo Alto is a 30 minute drive to San Jose airport, and 2 hours to Santa Cruz :O

On a side note, interestingly I live on the same street as Jeff...[/quote]

lol ill be at jeff's house frequently over summer. im stuck on stanford campus for a while...


[quote="math92"]Ed spent 3 dollars and 40 cents for 22 stamps.  If he bought only 25 cent and 10 cent stamps, how many of each kind did he buy?[/quote]

this problem was somewhere in teh MOEMS book.

I know that the answer is 14 10¢ stamps and 8 25¢

[hide]THe number of ten cent stamps has to be four or more to compensate for the inability of twenty-five cent intervals to occupy forty cents. Let's say there are four ten-cent stamps. The remaining three dollars would be comprised of 12 twenty-five cent stamps. 4+25 is less than 22, so we have to increase the number of ten-cent stamps.We should only increase by intervals of five stamps, so the twenty-five cent stamps can fully consume the rest of $\$$3.40 .  4+5= 9 ten-cent stamps does not fulfill the requirements, but

14 ten cent stamps and
8 twenty- five cent stamps do. So ends our problem, and my horrendously long solution.  :) [/hide]

On a MOEMS test, guess and check would probably be the easiest method. :D

[hide]I guess and Checked to get 14 10-cent stamps and 8 25-cent stamps[/hide]

[hide]
suppose he only bought 10 cent stamps.
then he would spend 22*0.10 or 2.20
subtract this from 3.40 (the total amount)
to get 1.20
this is how much the # of 25 cent stamps must make up for
so, you subtract .10 from .25 to get .15
divide that into 1.20 to get  8.
subtract 8 from 22 to get 14
so, he bought 8 25-cent and 14 10-cent stamps[/hide]

this might be a better idea than guess and check, as it requires
no complex arithmetic with big numbers, and is guarateed to work the first time(as long as you make no mistakes)


Person A, B, C, and D are trying to go across a bridge at night. A, B, C, and D need at least 1, 2, 5, and 10 minutes to go across it, respectively. The bridge can only support at most 2 people, and flashlight is needed when you cross the bridge. What is the minimum number of minutes they need for all four of them to go across it?

Please don't answer this if you have seen it before.....

Can anyone beat 20 minutes?  It seems a little too simple....

Yes, 20 minutes is too long...., u can do it faster than 20 minutes

I see my flaw now.  I did that far too quickly.  I also got 17 this time.  Spoiler attached:

[hide]1 and 2 go-2 minutes

1 returns-3 minutes

5/10 go-13 minutes

2 returns-15 minutes

1 and 2 go-17 minutes

[/hide]


calculus
integration
limit
trigonometry
inequalities
function
calculus computations

if we can interchange the limit and integral, we have

$\int_{0}^{\pi}\lim_{y\to 0}\frac{\tan(y\sin x)}{y}\, dx = \int_{0}^{\pi}\sin x \, dx = 2$

then how can we prove this?

since $x < \tan x < \frac{x}{\cos x}$ for $0<x<\pi/2$,

$\sin x < \frac{\tan(y\sin x)}{y}< \frac{\sin x}{\cos(y\sin x)}< \frac{\sin x}{\cos y}$

for sufficient small $y>0$. this inequality yields

$\int_{0}^{\pi}\sin x \, dx \leq \int_{0}^{\pi}\frac{\tan(y\sin x)}{y}\, dx \leq \int_{0}^{\pi}\frac{\sin x}{\cos y}\, dx$

or

$2 \leq \int_{0}^{\pi}\frac{\tan(y\sin x)}{y}\, dx \leq \frac{2}{\cos y}$.

therefore, by squeeqing theorem, we obtain

$\lim_{y\to 0^{+}}\int_{0}^{\pi}\frac{\tan(y\sin x)}{y}\, dx = 2$

my proof is restricted to the case $y>0$, but you can easily extend this argument to prove it for $y<0$.

Extremal: that does work, but there's something you have to be careful of: that $O(y^{2})$ must include a bound that is uniform in $x.$ Otherwise, this style of argument would fail. Fortunately, in this case, there is a uniform bound there, so the argument does work.

[quote="Extremal"][quote="maokid7"]Evaluate the limit

$L=\lim_{y\to 0}\frac{1}{y}\int_{0}^{\pi}\tan(y\sin(x))dx$[/quote]
I don't understand why you can't use this
$\frac{\tan(y\sin(x))}{y}=sin(x)+O(y^{2})?$ when y->0.
and our integral -> $2$.[/quote]

how did you get that $\frac{\tan(y\sin(x))}{y}=\sin(x)+O(y^{2})$?

Since the tangent is an odd function, its Maclaurin series consists only of odd terms, and we know the first term from $\tan'(0)=1.$

$\tan u=u+O(u^{3})$ as $u\to 0.$ Hence $\tan(y\sin x)=y\sin x+O((y\sin x)^{3}).$

We note that $\sin x$ is bounded on $[0,\pi]$ and can be replaced by a constant in the estimate. Now we have

$\tan(y\sin x)=y\sin x+O(y^{3})$

$\frac{\tan(y\sin x)}{y}=\sin x+O(y^{2}).$

then the $\int O(y^{2})dx$ term will disappear right? since
$\lim_{y\to 0}xO(y^{2})=0$?

The $O(y^{2})$ part represents some continuous function $g(x,y)$ which has the property that for all small enough $y$ we have $|g(x,y)| \leq Cy^{2}$ for some fixed constant $C$. So we have that $\left| \frac{1}{y}\int_{0}^\pi g(x,y) dx \right| \leq C \pi y$ for small enough y. It's important to note here that when we say $O(y^{2})$ we mean it as a function of two variables, and that it's true for all $x$ and for small $y$. Maybe it's better to not use $O$ at all... it's unclear.


All you [hide="Kannadigas"]

fortunately, for some like me only those born to parents who have studied in Karnataka.......and others studying in Karnataka are entitled for domicile
Others.......thankfully left out :D
 [/hide]
after April 13th(IIT-JEE 2008).........

our next challenge is on 19th .......Saturday! :ninja:
So, how many of you guys taking it?!!!!!!!

Any idea about regular scores???? :huh:

P.S.: I am a Kannidiga anyway........just staying in Delhi!

me done so so in chem and phy. but math altho easy was really long. dont think neone cud have completed the paper. how've u done?

NOW WAT DO U MEAN "ME DONE [u]SO SO[/u] IN PHYSICS AND CHEM

WEL I COMPLETED MATH IT WAS QUITE EASY
BUT I'L LOSE EQUALLY IN CHEM . (4 D 1ST TIME I FELT MY LACK OF ABILITY TO MEMORISE IS GOING TO COST ME SERIOUSLY :rotfl:  :rotfl: )

u completed math? :o . no one in my centre cud complete, i dint remember the formulae in co geo, so had to leave that. ofc many problems req simple substitution.

Math was kinda shi**y.
around 3 of us here are xpecting to max the paper(atleast so far as i know. ) :D

came back from Bengaluru on Sunday evening.......

Math was ok.......... just about completed on time!(Slim chances of 'max'ing it)......although it wasn't tough yet a little lenghthy.....

Guys please, do discuss the solutions/answers when it is released by KEA on Saturday......make it sunday if you want coz of AIEEE.......

post your scores and any mistakes in the solns.......coz as per the brochure(iff u've read it carefully) [b][u]one can contact the authorities(unlike JEE) in case any answer is wrong[/u][/b] with proper documentation, substantiating the claim by 30th i think......... our papers would be evaluated only on the basis of the corrected answer key finalised by the expert committee.......

It is important, because......:
1) In 2005, students were given grace marks in two ques....
2) In 2006, grace marks in one chem and phy ques...
3) In 2007, somewhat the same!

Considering the 'consistency' of giving mistake riddled papers :rotfl: .....it is important that we make the most of this opportunity given to us...... :ninja:

Source:CET sample papers and solutions from the official website![/list]

sometimes, some wrong soln. go unnoticed......

Did I mention in a 2005 or 06 or 07.....chem paper......
ques: what happens wen R-X + AgCN......
official answer: R-CN!

but i think it should be R-NC!

Official K-CET Answer Key is out!!!!!!!!
Guys if you've checked ur score post em here!!!!!! :huh:

Those of you who haven't.......download the pdf file [url=http://kea.kar.nic.in/ANSWER%20KEYS%202008.pdf]here[/url]!!!!

Guys any idea about previous year scores and standings?????!!!!

they posted the scores and not any rank! :mad: what about u guys?


Tark scored an integer number of points on each of five tests. He then computed the following statistics for those five scores: the mean was 72; the median was 70; and the mode was 80. What was the least possible score he could have had on any test?

We begin with finding the total score.  We know that the mean is $ 72$, so we do $ 72\times 5\equal{}360$.  Having $ 5$ scores, we know that $ 70$ is the $ 3^{rd}$ number.  Also knowing we have a mode of $ 80$, we must have at least $ 2$ of them.  We may only have $ 2$ of them, because there can only be $ 2$ numbers larger than $ 70$.  So far, our total score is $ 70\plus{}80\plus{}80\equal{}230$.  This leaves us with $ 130$.  We want to minimize our lowest score, so we maximize our $ 2^{nd}$ to lowest score.  The largest it can be is $ 69$ (must be less than $ 70$, not equal to because there is only one mode).  This means that our lowest score must be $ 130\minus{}69\equal{}\boxed{61}$


geometry
circumcircle
perpendicular bisector
geometry unsolved

Let $ ABC$ be an isosceles triangle such that $ \widehat{BAC} \equal{} 100^o$
Let D be a point ([size=150]not[/size] in triangle $ ABC$) such that $ \widehat{DBC} \equal{} 35^o$ and $ \widehat{DCB} \equal{} 15^o$
$ \widehat{ADB} \equal{} ?$

Hope that I have no mistake in English.

Obviously, <BDC = 130 degs and A is the circumcenter, hence the triangles BAD, DAC are A-isosceles, with <BAD = 2.<BCD,  hence
<ADB = 150/2 = 75 degs.

Best regards,
sunken rock

[quote="sunken rock"]Obviously, <BDC = 130 degs and A is the circumcenter, hence the triangles BAD, DAC are A-isosceles, with <BAD = 2.<BCD, hence
<ADB = 150/2 = 75 degs. [/quote]
May be I'm stupid but can you tell me why $ A$ is the circumcentre of triangle $ BCD$
Please, sunken rock

The angle <BDC = 130 degs, clearly, other angles of the triangle BCD being 35 and 15 degs respectively.

1) The circumcenter of triangle BCD will be on the perpendicular bisector of BC, let it be O and, since <BDC > 90 degs, it will be in the exterior of the triangle.

2) The m(<BOC) = 360 - 2. m(<BDC) = 360 - 260 = 100 degs = m(<BAC).

Best regards,
sunken rock


When do we get to know the floor value???

(and is there any possiblity that the floor will be 6??? :| )

On page 4 of the USAMO Teachers' Manual, available on our website: "The selection of the USAMO participants will be made on before Tuesday, March 27, 2006....Please do not inquire about the USAMO participation with anyone but the AMC Director, and do not phone prior to Friday, March 31."

I will not reply to email inquiries prior to Friday, March 31 and I will not respond to questiosn on this forum prior to Friday, March 31.

At that time, the floor value will be announced.

Steve Dunbar
AMC Director

On that same note, when do we find out our AIME scores?

It says on the back that it should take about two weeks for the AIME results to come. (I think).

I don't see that anywhere; i think it should be 1 week since the USAMO qualifiers are released 1 week after the AIME II.
[quote]I will not reply to email inquiries prior to Friday, March 31 and I will not respond to questiosn on this forum prior to Friday, March 31. [/quote]

What about hints like "I can't tell you the floor score because my dog ATE/EIGHT the score forms..." or "I'm too busy playing TENnis?"

bah, theres no point in worrying too much about something thats probably out of your hands, since other people's performance affect the floor. Give the AMC people time to figure out the floor.

Go do something fun or concentrate on HW to kill time and stuff will be announced eventually.

A debate is going on about wether the floor will be 7 of 8 with a slight chance of 9. So I dont think it will be a 6.


inequalities
function
real analysis
inequalities unsolved

Because $0$ isn't root of inequation, we have $0<k\leq 1$.

$|x-1|<k\Longleftrightarrow-k+1<x<k+1$, so you only need see functtion $f(x)=\frac{3}{2x}-\frac{x}{2}-1$ on interval $(-k+1,k+1)$. Use real analysis! :)

[quote="Lena"]find $k>0$ so that if $|x-1|<k$

then this is true: $| \frac{3-x^{2}}{2x}-1|< \frac{1}{10}$[/quote]

Such a $k$ exists.

Consider, $f(x) = \frac{3-x^{2}}{2x}$ we know $f$ is continous at $x=1$.

Thus, for any $\epsilon=1/10 > 0$ we can choose $\delta > 0$ (in this case $k$ such that:
$|x-1|<k \mbox{ and }x\in D \to |f(x)-f(1)|<1/10$

But note that,
$f(x)=\frac{3-x^{2}}{2x}$
And,
$f(1)=1$.

Thus, that is exactly what you are asking hiddenly.

absolutely wrong

....

i am asking for a number between 0 and whatever

that if x-1<than that number

for every x<that number+1

the inequality will be true

my problem is, i have absolutely no clue how to find it

i've found 0<K<1/5

but i dont know if its correct


[quote="sharkman"]Rita's purse has the ratio of the numbers of quarters to the number of dimes to the number of nickels to be 3 to 4 to 2.
If the value of these coins is 8.75 dollars, how many of each coin does Rita have?

MY WORK:

3x + 4x + 2x = 8.75

9x = 8.75

x = 8.75/9

x = 0.97

I then replaced x with 0.97 for each term above and multiplied.

This is what I did:

3(0.97) = 2.91

4(0.97) = 3.88

2(0.97) = 1.94

However, I did answer the question.

How do I find how many of each coin does Rita have?

Thanks![/quote]

The first step doesnt work because the coins are not worth the same.  you have to multiply each term by its coin value

[hide]
Okay we have 3 variables, $q, d, n$ which stand for the number of quarters dimes and nickles respectively.

$0.25q+0.10d+0.05n = 8.75$ multiplying by 100 to make things easier
$25q+10d+5n = 875$

Now from the ratios given we can create 2 more equations and from that solve for d and n in terms of q.
$\frac{q}{d}= \frac{3}{4}\Rightarrow d = \frac{4q}{3}$
$\frac{q}{n}= \frac{3}{2}\Rightarrow n = \frac{2q}{3}$

so now we sub in:

$25q+\frac{40q}{3}+\frac{10q}{3}= 875$
$\frac{125q}{3}= 875$
$125q = 2625$

$\boxed{q = 21}$

now we can find the number of dimes and nickles:
$d = \frac{4q}{3}= \frac{84}{3}= \boxed{28}$

$n = \frac{2q}{3}= \frac{42}{3}= \boxed{14}$
[/hide]

[quote="GlenthemanN"]are you referring to me?[/quote]
No, to sharkman

[quote="jli"]interval was this annoying user who posted a BUNCH of really dumb problems but i think he quit or something.  at least sharkman posts his work[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=work&t=132800
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=work&t=132805
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=work&t=132803


[b]Problem[/b]
Let $ [n]\equal{}\{1,\dots,n\}$ and let $ B_k\equal{}\{A_{k1},A_{k2},\dots\}$. $ A_{ki}$ is a subset of $ [n]$ with the following restrictions.
(i) $ A_{ki}\cap A_{kj} \neq \emptyset \quad \forall i,j$
(ii)$ \cup_{i} A_{ki}\equal{}[n]$

Let $ C\equal{}\{B_1,B_2,\dots\}$. Find $ |C|$.

Let $ C_n$ denote the set $ C$ formed from $ [n]$.

Clearly, we have $ C_1 \equal{} 1$, and we will let $ C_0\equal{}1$ as well for the purpose of recursion. We can develop a recursive relation as follows:
Suppose we wish to find $ C_{n}$.To do this, we choose the elements to go with $ n$ and then find the  number of ways to partition the rest. This gives us $ \sum_{i \equal{} 0}^{n \minus{} 1}\binom{n \minus{} 1}{i}C_{n \minus{} 1 \minus{} i} \equal{} \sum_{i \equal{} 0}^{n \minus{} 1}\binom{n \minus{} 1}{i}C_{i}$. If anyone has any ideas as to evaluating this, I'm open to 'em :P...

No.  $ C_{n} \equal{} \sum_{i \equal{} 0}^{n \minus{} 1}\binom{n \minus{} 1}{i}C_{n \minus{} 1 \minus{} i}$ is the recursion for the Bell numbers (counting set partitions; sequence [url=http://www.research.att.com/~njas/sequences/A000110]A000110[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url]).  Here we have the condition that any two of the $ A_{k_i}$ must intersect, not must be disjoint.  In particular, your formula gives the wrong answer for $ n \equal{} 2$: it should be 3 (we have $ \{\{1, 2\}\}$, $ \{\{1, 2\}, \{1\}\}$ and $ \{\{1, 2\}, \{2\}\}$), not 2.

Can we say $ |B_i|\leq 2^{n\minus{}1}\quad \forall i$ ?


Given triangle $ABC$ with four Feuerbach's point $F,F_{a},F_{b},F_{c}$ prove that $ABC$ and $F_{a}F_{b}F_{c}$ are perspective with perpector $F'$ and $F,F'$ divide $I,N$ harmonically with $I$-incenter, $N$-Nagel point.

Use Monge's theorem about the condition of collinearity for 3 centers of similitudes given 3 circles.

I don't know this theorem can you post all solution!


Two people play a game in which they alternate saying numbers from 1 to n. The objective of the game is to be the first on to reach a number y by saying a number so the sum of it and all the numbers everyone has said previously is y. Basically, if there are b moves, the sum y should equal A_1+A_2+......A_b, where A_x describes the xth move. For example, the objective in this sample game is for either Mendel or Brot to be the one to make the sum 30 by saying numbers 1-6. Mendel says 3, Brot says 6, Mendel says 1, Brot says 5, Mendel says 6, Brot says 2, Mendel says 4, and Brot wins the game by saying 3, since 3+6+1+5+6+2+4+3=30. The question is if the number you want to get to is y, and you use numbers 1 through n, what number should you start out with to ensure that you'll win the game?

You have to look for killer numbers. For example in the game of 1-6 and to 30. If A picks 23, A wins, same with 16, 9 and 2. Easily, you recognize this as a modular problem, so you just notice the glaring pattern. You have the equivilant of x = y(mod(n+1)) where 0<x<(n+1).

If the desired sum is y, and the choice of numbers is 1 through n, then if the sum before your turn is y-[1, 2, ..., or n] you can win.  Thus, it is desirable to have your opponent end on y-[1 thru n].  This is possible if you end on y - (n+1), since your opponent will then have to increase the sum by some number between 1 and n.  So you want your opponent to end on y - (n+1) - [1 thru n], in which case you should end on y - 2(n+1).  We can do this over and over again, and in each case you should always end on some integer y - k(n+1) for integral k.  Thus, in order to guarantee a win, you should choose the number of the form y - k(n+1), for some integer k.  Let x be the number you should choose.  Then x = y - k(n+1), so x + k(n+1) = y, so [tex]x \equiv y mod n+1[/tex]

Applying this to the given example (y=30, n=6), we need x = 30 mod 7 = 2.


Find the coordinates of the vertices of a triangle whose circumcenter is the point $(4,3)$, circumradius is $5$, and orthocenter is the point $(-3, -1)$ given that the vertices lie on lattice points.

[hide="hint"] See link in my signature.[/hide]

$A(0,0), B(3, -1), C(0, 6)$

I don't have a solid demonstration yet, I just messed around with integer points on the circle =P  Nor can I prove that only one such triangle exists.  Alas.

Edit:  Concerning your comment about orthocenter, circumradius, and circumcenter uniquely defining a triangle, this definition fails for right triangles (where the orthocenter is a point on the circumcircle), since there are an infinite number of right triangles inscribed in a given circle with the right angle at a given point (diameters for the hypotenuse can be chosen arbitrarily).  Just thought you'd like to know  :)

Edit 2:  Wait, you weren't talking about orthocenter, circumradius, and circumcenter, were you?  Sorry  :blush:

In any case, I know that there's at least one more triangle that satisfies all your requirements (flip the triangle I gave you across the line connecting the orthocenter to the circumcenter).  I don't think it has integer coordinates though :X

I don't think that's quite correct: from my calculations, the orthocenter of your triangle is the point $(-5, -1)$... and the point $(3,-1)$ would not lie on the circumcircle described either.

On a side note, I did want a triangle with integer coordinates (edited problem statement).

Sorry... drew the diagram wrong.  Okay:

$A(0,0), B(4, -2), C(1, 7)$


In the sequence\[\frac{1}{2},\frac{5}{3},\frac{11}{8},\frac{27}{19},...\]the denominator of the $n^{th}$ term $(n>1)$ is the sum of the numerator and the denominator of the $(n-1)^{th}$ term. The numerator of the $n^{th}$ term is the sum of the denominators of the $n^{th}$ and $(n-1)^{th}$ term.
Find the limit of the sequence

the limit will equal the value of $\frac ab$, where $\frac ab$ satisfies

$\frac ab = \frac{a+2b}{a+b}$.

if we denote $\frac ab = r$, then

$r = \frac ab = \frac{a+2b }{a+b} = \frac{\frac ab + 2}{\frac ab + 1} = \frac{r+2}{r+1}$,

from which

$r^2+r=r+2$; $r = \sqrt{2}$


may be challenge for beginner section, but anyone is able to solve it if you really try.

show that for every natural $n > 1$ we have: $(n-1)^2|\ n^{n-1}-1$

I'm sorry but saying that anyone can solve it if they really try doesn't make it a getting started level problem.  If I put you in a 100 meter race against Maurice Green could you beat him if you really tried?  Ok, so those two things are a little different but the idea is still the same.  I'm going to move this up to Intermediate.

um...it's not true.  Maybe you meant (n-1)^n? But then it's just trivial.

[quote="Peter VDD"]It is almost trivial and that's why I posted it in the beginner's section.

Mister moderator, please read the problems before you move them next time![/quote]

This is *definitely* far too hard for a beginner. I wouldn't go calling these problems trivial for beginners, since they aren't going to be too happy with you :P

[quote="TripleM"]This is *definitely* far too hard for a beginner. I wouldn't go calling these problems trivial for beginners, since they aren't going to be too happy with you :P[/quote]

It's not. This may even be too easy for an exam in school. Simply apply Horner's division 2 times, and you'll get a result that is clearly integer. Now if a problem can be solved with only horner (something we learnt in 10th grade) I really can't call this too hard for there :?

i hope it's simple by bionomial expansion

[quote=blahblahblah][quote="Peter VDD"]It is almost trivial and that's why I posted it in the beginner's section.

Mister moderator, please read the problems before you move them next time![/quote]

It is not true, for n=3 we have 4|9, contradiction.[/quote]

No. We have 2^2=4 and 3^2-1=8. 4|8.

We observe that $$n^{n-1}-1=(n-1)\left[\sum_{k=0}^{n-2}n^k\right].$$
Thus it remains to prove that $(n-1)$ divides $\displaystyle\sum_{k=0}^{n-2}n^k.$

Observe that $$\sum_{k=0}^{n-2}n^k\equiv \sum_{k=0}^{n-2}1^k\equiv (n-1)\equiv 0 \pmod{n-1}$$ as $n\equiv 1\pmod{n-1}.$

Thus $n-1\mid \sum_{k=0}^{n-2}n^k\implies (n-1)^2\mid n^{n-1}-1$ and our proof is complete. $\square$


A sequence (u_n) is defined by $ u_1\equal{}u_2\equal{}1, u_{n\plus{}2}\equal{}u_{n\plus{}1}\minus{}2u_n$ for every natural numbers $ n$. Prove that $ 2^{n\plus{}2}\minus{}7u_n^2$ is a square for every natural numbers $ n$.

Let $ \alpha,\beta$ be the solutions of $ x^2 \minus{} x \plus{} 2 \equal{} 0.$

$ \alpha$ and $ \ \beta$ satisfy $ \alpha \plus{} \beta \equal{} 1$ and $ \alpha \beta \equal{} 2$ and so

$ \alpha^2 \plus{} \beta^2 \equal{} (\alpha \plus{} \beta)^2 \minus{} 2 \alpha \beta \equal{} \minus{} 3,$

$ (\alpha \minus{} \beta)^2 \equal{} (\alpha \plus{} \beta)^2 \minus{} 4 \alpha \beta \equal{} \minus{} 7.$

Then

$ u_{n \plus{} 2} \minus{} \alpha u_{n \plus{} 1} \equal{} \beta ( u_{n \plus{} 1} \minus{} \alpha u_n ) \equal{} \cdots \equal{} \beta^{n} ( u_{2} \minus{} \alpha u_1 ) \equal{} \beta^{n \plus{} 1}$

$ u_{n \plus{} 2} \minus{} \beta u_{n \plus{} 1} \equal{} \alpha ( u_{n \plus{} 1} \minus{} \beta u_n ) \equal{} \cdots \equal{} \alpha^{n} ( u_{2} \minus{} \beta u_1 ) \equal{} \alpha^{n \plus{} 1}$

$ (\beta \minus{} \alpha) u_n \equal{} \beta^n \minus{} \alpha^n$

$ (\beta \minus{} \alpha)^2 u_n^2 \equal{} \minus{} 7 u_n^2$

$ 2^{n \plus{} 2} \minus{} 7 u_n^2 \equal{} 2^{n \plus{} 2} \plus{} (\beta \minus{} \alpha)^2 u_n^2 \equal{} 4 (\alpha \beta)^n \plus{} (\beta^n \minus{} \alpha^n)^2 \equal{} (\beta^n \plus{} \alpha^n)^2,$

and $ v_n \equal{} \beta^n \plus{} \alpha^n$ satifies $ v_1 \equal{} 1,v_2 \equal{} \minus{} 3, v_{n \plus{} 2} \equal{} v_{n \plus{} 1} \minus{} 2v_n.$

So $ v_n$'s are integers for $ n\ge0,$and $ 2^{n \plus{} 2} \minus{} 7 u_n^2 \equal{} v_n^2,QED.$


prove that for positive reals,

$ \sqrt{ab}$+$ \sqrt{cd}$ $ \leq $ $ \sqrt{(a+d)(d+c)}$

t0rajir0u, can you give me a liner on this.

Looks like you made a typo, should be $ b$ instead of $ d$.

[hide="hint"]Cauchy-Schwarz[/hide]

[quote="mdk"]prove that for positive reals,

$ \sqrt{ab}$+$ \sqrt{cd}$ $ \leq$ $ \sqrt{(a+d)(b+c)}$[/quote]

I assume you meant $ (a+d)(b+c)$.

[hide="Then..."] By AM-GM, $ ab+cd+2 \sqrt{abcd}\le ab+cd+(ad+bc)$ [/hide]
Either this problem or a very similar one, though, has a really nice trigonometric solution.

you're absoluletely right, it can also be done by trig substitution.

Answer

Lets assume $ \sqrt{ab}+\sqrt{cd}> \sqrt{(a+d)(b+c)}$

$ \Longrightarrow ab+2\sqrt{abcd}+cd > ab+ac+bd+cd$
$ \Longrightarrow2\sqrt{abcd}> ac+bd$
$ \Longrightarrow 4abcd > a^{2}c^{2}+2abcd+b^{2}d^{2}$
$ \Longrightarrow a^{2}c^{2}-2abcd+b^{2}c^{2}< 0$
$ \Longrightarrow (ac-bd)^{2}< 0$

contradicion


Let $\textit{P}$ be an interior point of triangle $\textit{ABC}$ such that $\angle\textit{PBA}=\angle\textit{PCA}=\left(\angle\textit{ABC}+\angle\textit{ACB}\right)/3$. Prove that $\frac{\textit{AC}}{\textit{AB+PC}}=\frac{\textit{AB}}{\textit{AC+PB}}$.

Poland 1997 I believe, posted before (by me!), just search for it.


$ 2a^2b^2\plus{}2a^2c^2\plus{}2c^2b^2\minus{}a^4\minus{}b^4\minus{}c^4\equal{}A\cdot...\cdot B$        $ \Longrightarrow$    $ A\equal{}?,  B\equal{}?$

[quote="maksmaks24"]$ 2a^2b^2 \plus{} 2a^2c^2 \plus{} 2c^2b^2 \minus{} a^4 \minus{} b^4 \minus{} c^4 \equal{} A\cdot...\cdot B$        $ \Longrightarrow$    $ A \equal{} ?, B \equal{} ?$[/quote]

You should try to be clearer if you hope solution :

Do $ a,b$ belong to $ \mathbb N$, $ \mathbb Z$, $ \mathbb Q$, $ \mathbb R$, $ \mathbb C$, ... or something else ?

What's the meaning of $ A\cdot...\cdot B$ ? :
- one number whose first and last digits are $ A$ and $ B$ ?
- one product of numbers ?
- ....

$ a,b,c$ real numbers,  $ A\equal{}A(a,b,c),  B\equal{}B(a,b,c)$

[quote="maksmaks24"]$ 2a^2b^2 \plus{} 2a^2c^2 \plus{} 2c^2b^2 \minus{} a^4 \minus{} b^4 \minus{} c^4 \equal{} A\cdot...\cdot B$        $ \Longrightarrow$    $ A \equal{} ?, B \equal{} ?$[/quote]

Ahhhh, OK, quite more clear!

$ A(a,b,c)\equal{}2a^2b^2 \plus{} 2a^2c^2 \plus{} 2c^2b^2 \minus{} a^4 \minus{} b^4 \minus{} c^4$
$ B(a,b,c)\equal{}1$

I got it:
$ A(a,b,c)\equal{}a\plus{}b\plus{}c$
$ B(a,b,c)\equal{}{a^2}b\plus{}a{b^2}\plus{}{a^2}c\plus{}a{c^2}\plus{}{b^2}c\plus{}b{c^2}\minus{}3abc\minus{}{a^3}\minus{}{b^3}\minus{}{c^3}$.

Another one :

$ A(a,b,c)\equal{}a^6\plus{}b^4\plus{}c^2\plus{}1$

$ B(a,b,c)\equal{}\frac{2a^2b^2\plus{}2a^2c^2\plus{}2c^2b^2\minus{}a^4\minus{}b^4\minus{}c^4}{a^6\plus{}b^4\plus{}c^2\plus{}1}$

[quote="raf92"]I got it:
$ A(a,b,c) \equal{} a \plus{} b \plus{} c$
$ B(a,b,c) \equal{} {a^2}b \plus{} a{b^2} \plus{} {a^2}c \plus{} a{c^2} \plus{} {b^2}c \plus{} b{c^2} \minus{} 3abc \minus{} {a^3} \minus{} {b^3} \minus{} {c^3}$.[/quote]
Ohhh, if you want to be serious :

$ \minus{}(a\minus{}b\minus{}c)(a\plus{}b\plus{}c)(a\minus{}b\plus{}c)(a\plus{}b\minus{}c)$

But really, maths is a science of precision and rigor and such a problem statement is far from basic maths rules  :(

Ok, I noticed some vagueness in the statement too. Thanks for complete my solution.


function
number theory
prime factorization
totient function
complementary counting

How many positive integers less than 70 are relatively prime to 70 ?

[hide] The prime factorization of $ 70$ is $ 2\times{5}\times{7}$. So, we have $ 70(\frac12)(\frac45)(\frac67)\equal{}\boxed{24}$.[/hide]

You can do this by using Euler's totient function, which is what gaussintraining did.

[url]http://en.wikipedia.org/wiki/Euler's_totient_function[/url]

or by complementary counting and PIE but it'll get really messy. So I suggest using the formula.


For instance... the geometry introductory textbook. does it cover the same material as the school-level geometry courses? or does it go in deeper?

same with the counting and probablility and the number theory books.

thanks!  :)

I think it is not the better place to ask this: but can I order this books and receive them here in Portugal?

[quote="lotrgreengrapes7926"]I remember hearing the books are more in depth and cover more material. :maybe:[/quote]

so you would recommend them?

These books are definitely deeper than your regular textbooks, and they are more interesting as well.  However, they also covered most of the new material I learned in my Geometry course (for example).  So yes, I would recommend these books.


group theory
abstract algebra
superior algebra
superior algebra unsolved

$G$ a finite group whose all proper subgroups are abelian. Then $G$ is not simple.

(Under the extra hypothesis that $G$ admits at least one proper *nontrivial* subgroup, otherwise the result is false).

Assume the contrary. Then $G$ is nonabelian and simple, whence $Z(G)=\{1\}$. Consider an element $x \neq 1$ whose centralizer has maximal order (this order necessarily $\in [2,n[$), and consider a maximal subgroup $H$ of $G$ containing $x$ ($H$ exists otherwise $G$ is cyclic). It is clear that $H$ must contain at least one $p$-Sylow of $G$ whenever $p$ divides $|H|$ (Otherwise $\exists T \leq H, |T|= p$, and $T$ is included in a $p$-Sylow $S$ of $G$. But then $S \cap H \neq \{1\}$ and every element $\neq 1$ in this intersection has a centralizer whose order is greater than that of $x$, which is absurd). As a result, there must exist a prime number $p^{*}$ such that $p^{*}||G|$ but $p^{*}$ does not divide $H$. But the normalizer $N(H)=H$ (otherwise $H$ is normal), meaning $G = \cup_{a \in G} aHa^{-1}$. This is not possible since no subgroup $aHa^{-1}$ contains an element of order $p^{*}$.


vector
induction
abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Let $ V$ be a vector space of finite dimension. Let $ \phi: V \to V$ be a linear transformation. Show that there is an integer $ n$ such that $ im(\phi^n) \cap \ker(\phi^n) \equal{} \{0\}$.

Since $ \phi$ is a linear transformation, $ \phi$ is a homomorphism. So for each $ v \in V$ and any integer $ n$ we have by induction: $ \phi^n(v) \equal{} \phi(v^n)$.
If $ w \in im(\phi^n) \cap \ker(\phi^n)$,  then: $ \phi(w^n) \equal{} 0$ and there exists some $ x \in V$ such that $ w \equal{} \phi(x^n)$.

Is this even leading anywhere? The other idea I had was to look at the dim of the given intersection but then I could not conclude anything right away. Any hint will be appreciated.

[quote="akech"]So for each $ v \in V$ and any integer $ n$ we have by induction: $ \phi^n(v) \equal{} \phi(v^n)$.[/quote]
$ v^n$ is not meaningful notation in a vector space.

Think about the behavior of the sequence of vector spaces $ V, \phi(V), \phi^2(V), ...$.

You are right in saying that $ \phi$ is a homomorphism, but it is a [i]vector space[/i] homomorphism. There is no multiplicative structure on $ V$. So yeah, you have to rethink that approach.

We have:
1.  $ ...\subseteq \phi^3(V) \subseteq \phi^2(V) \subseteq \phi (V) \subseteq V$

2. $ \ker(\phi) \subseteq \ker(\phi^2) \subseteq \ker(\phi^3) \subseteq ... \subseteq V$

I still cannot feel strong about this problem. Can someone offer further hint please. I have seen a similar problem in the context of Noetherian Ring on an algebra prelim study guide.

Those two chains you gave must eventually stabilize.  Does that help?

Steve


Prove that, for all positive real numbers $a,b,c$, the following inequality holds:

$\frac{1}{b(b+c)}+\frac{1}{a(a+b)}+\frac{1}{c(a+c)}\ge \frac{27}{2(a+b+c)^2}$

Very nice problem! :)

[hide=Wrong solution, due to reversed sign][/hide]

So is it still possible to prove it with jensen?

Actually, Jensen is a generalisation of AM-GM, cauchy, AM-HM, whatever, so every inequality that can be solved with those can also be solved with Jensen :)

What I did (wrong) was Jensen on 1/x, which is just a nicer way to write AM-HM, while it should actually have been 2 AM-GM's. (so jensen on ln(x), applied twice, but as you haven't learnt logarithms yet, I don't think it would be that useful)

In fact, I'd be really interested in seeing a solution using logarithms :)
So, Peter VDD, if you have enough time, please post your sol ;)

Jensen on logarithm:

$\frac{\ln(a)+\ln(b)+\ln(c)}{3} \le \ln\left(\frac{a+b+c}{3}\right)$
$ln(\sqrt[3]{abc}) \le \ln\left(\frac{a+b+c}{3}\right)$
exp() is increasing so maintains inequality and cancels ln, so applying to both sides yields:
$\sqrt[3]{abc} \le \frac{a+b+c}{3}$, which is AM-GM. :)

And I guess it is clear how to go with AM-GM? (AM-GM the RHS then AM-GM the denominator)

[quote="Melissa"]That certainly looks interesting... of course I don't understand it because as Peter said I haven't learn logarithems yet. But I plan to do so soon!
[hide="If anybody cares to look at my simple solution, click here..."]
$\frac{2(a+b+c)^2}{27}$

$=\frac{1}{3} \cdot \frac{(a+b)+(b+c)+(c+a)}{3} \cdot \frac{a+b+c}{3}$

$\ge \frac{1}{3} \cdot \sqrt[3]{abc(a+b)(b+c)(c+a)}$      (AM-GM)

$\ge \frac{1}{\frac{1}{b(b+c)}+\frac{1}{a(a+b)}+\frac{1}{c(a+c)}}$      (GM-HM)

Raise both sides to the power of $-1$ to get the desired result. [/hide][/quote]
Indeed. It is simple enough.  :)

Source of initial inequality  : 6th Junior Balkan MO, 2002.

$a^2+b^2+c^2\leq ab+ac+bc$  :)
I don't think that this is true.

Aaaaaaah! What a nightmare, my whole solution collapses because of such a stupid reversed sign...

But anyway, my solution is to be saved, AM-HM is to be changed into GM-HM in this case, but if we work it out in an analogue way, we get exactly the same solution as Melissa's, so it's deleted

I just saw $a^2+b^2+c^2$ and $ac+bc+ac$ and immediately thought it was right without checking the $\leq$

man...

An other solution (but Melissa's is the nicest so far !)

$LHS * ((a+b) + ((b+c) + (a+c)) \geq \left( \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{a}} + \frac{1}{\sqrt{c}} \right)^2$ by Cauchy-Schwarz

Now $\frac{1}{\sqrt{x}}$ is concave so $RHS \geq \left(\frac{3}{\sqrt{(a+b+c)/3}} \right)^2$