linear algebra
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algebra
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abstract algebra
group theory
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there are no simple groups of order 112 and 120
(120 is tricky i guess because it contains a simple nonabelian group)

also in general, can two simple groups of same order be nonisomorphic?

I don't know all of the simple groups off the top of my head, so I'm not sure about your first statement/question.

For your second, I have an example: There are two simple groups of order 168.

For the first, take the group of $3\times 3$ invertible matrices over the field with two elements. There are 168 of these, and they form a simple group under multiplication.
For the second, take the group of $2\times 2$ matrices of determinant 1 over the field with 7 elements. Take the quotient of this group by its center, $\pm I$. This leaves a different simple group of order 168.

Both groups belong to the "PSL" family of simple groups.

:D well my group theory professor said that there  is a gag for group theory professors going on retirement : i am now as old as the order of the smallest nonabelian group, hopefully i don't reach that of the next one: and he meant A5 and L2( 7 )
and it is possible for a lot of orders like for instance 111,123,...
but what i meant was, is there a fast trick by using sylow groups and permutation representations on them etc.. to show that there aren't any of order 112 and 120
are they really nonisomorphic, those two groups of order 168?
i know that A5, L2(4) and L2(5 ) are all simple , but one can easily show that they all are isomorphic

thx for the help however

A proof that the two groups of order 168 are not isomorphic:

The first has no elements of order 8: in the base field $(x^8-1)=(x-1)^8$, so a matrix of order dividing 8 has a power of $x-1$ as its minimal polynomial, and therefore has $(x-1)^3$ as its characteristic polynomial. Since $(x-1)^3|x^4-1$, the matrix has order dividing 4.

In the second group, the equivalence class of $\begin{pmatrix}2&4 \\ 3&2 \end{pmatrix}$ has order 8:
$\begin{pmatrix}2 & 4 \\ 3& 2 \end{pmatrix}^2=\begin{pmatrix}2&2 \\ 5&2 \end{pmatrix}$
$\begin{pmatrix}2&4 \\ 3&2 \end{pmatrix}^4=\begin{pmatrix}0&1 \\ 6&0 \end{pmatrix}$
$\begin{pmatrix}2&4 \\ 3&2 \end{pmatrix}^8=\begin{pmatrix}1&0 \\ 0&1 \end{pmatrix}$

A really quick way to show that there are no simple non-abelian groups of order $112$ would be employing Burnside's Theorem: There are no simple non-abelian finite groups with order having less than $3$ distinct prime factors. Since $112=2^4\cdot 7$, that's that.

I don't know how to prove the theorem, of course :).

;)
thank you jmerry for that explanation
grobber : that is an interesting theorem that gives me certainty about 112
( i hadn't heard about it yet  :blush: )
of course if my professor would have allowed use of that, half of his problems
could have been discarded :D

so are there any people here who have a clue for 112

hi

i'd like to come back to this topic

i see no hole at all in jmerry's proof

but on the other hand i have found a course claiming PSL(2,7) is isomorphic to PSL(3,2)

and on so many sites, like http://www.mathreference.com/grp-fin,g168.html

they speak of the simple group of order 168, as if it is unique up to isomorphism?

so can anyone explain what is going on, is there only one? or is the second one often forgotten?

thx

fred

They're wrong if they say it's the only one or that the two I described are isomorphic. The group described at the site you linked to is PSL(3,2); it has no elements of order 8 because it can be described as a permutation group on 7 elements. It's an oversight, and I had no idea it was common.
I guess people like to believe that groups must be isomorphic if you match a few simple properties (in this case, the number of Sylow subgroups of each order).

Here's a cute exercise to show that isomorphisms aren't always there when you think they should be: Find two non-isomorphic groups of order 16 which have the same number of elements of each order.

:?

okay so the overall tone here is that these groups are both simple with 168 elements but are not isomorphic

however i also asked this on 'ask an algebraist' (
http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=1182.0001
)

this person claims it is actually a theorem in a book that they are isomorphic??

so what is going on, here people are certain that they aren't, and on other forums and in books the opposite is claimed?

sorry for insisting on this but i'm really interested

frederic

I was about to post my proof there, but you did already. He's wrong, and the book being quoted is wrong; this short proof of nonisomorphism invalidates any proof of isomorphism, however long.

:?

hi

on http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=1182.0001.0001.0001

an argument against jmerry's proof is now given

the determinant of your representant is -1 which is no square modulo seven
it is, when we see PSL(2,7) as SL(2,7)*SC(2,7)/SC(2,7) , thus a subgroup of PGL(2,7)
only an element of PGL(2,7) and not of PSL(2,7)

so does this mean it is true that their is only one simple group of order 168

or can you fix your proof?

plz help,

fred

Now that I've had some time to think about it, it all falls apart. There are no elements of order 8 in either group; both have 2-Sylow subgroups isomorphic to the dihedral group.

In $PSL_3(F_2)$, the upper triangular matrices form a 2-Sylow subgroup, which is easily seen to be the dihedral group.
In $PSL_2(F_7)$, the "rotations" and "reflections" of the form $\begin{pmatrix}a&b\\-b&a\end{pmatrix}$ or $\begin{pmatrix}a&b\\b&-a\end{pmatrix}$ form a 2-Sylow subgroup. $\begin{pmatrix}2&2\\-2&2\end{pmatrix}$ and $\begin{pmatrix}2&-2\\2&2\end{pmatrix}$ are the elements of order 4, $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ is the non-identity element in the center, and the "reflections" $\begin{pmatrix}3&2\\2&-3\end{pmatrix}, \begin{pmatrix}3&5\\5&-3\end{pmatrix}, \begin{pmatrix}2&3\\3&-2\end{pmatrix}, \begin{pmatrix}2&4\\4&-2\end{pmatrix}$ are the other elements of order 2.

thanks for the remarks

i found it all very interesting

on this site i found a theorem :
http://www.maths.qmul.ac.uk/~pjc/class_gps/ch2.pdf

theorem 2.12 states that the PSL(2,p) with p odd prime are always unique simple groups of their order

so that implies PSL(2,7) being isomorphic with PSL(3,2)

thx

fred

[quote="vess"][b][i]A simple proof that no group of order $120$ is simple: [/i][/b] Let $G$ be any group of order $120$. By the Sylow theorems, the number of $5$-Sylow subgroups of $G$ is either $1$ or $6$; in the first case, there is nothing to show because the unique Sylow $5$-subgroup has to be normal. Thus, we may assume that $G$ has precisely $6$ Sylow $5$-subgroups. Consider the action by conjugation of $G$ on the Sylow $5$-subgroups; this action yields an embedding of $G$ into $S_6$. As $G$ is simple, we must have $G \leq A_6$, and $[A_6:G] = 3$. This is a contradiction, because $A_6$ is simple and therefore cannot act on a three element set. Q.E.D.[/quote]

This approach works even faster when dealing with groups of order $112$ (what I wrote when I mentioned Burnside's Theorem is not a proof :)).

There are $7$ Sylow-$2$ subgroups, so we have a homomorphism from $G$ to $S_7$. It's injective, because otherwise the kernel would be a normal non-trivial subgroup of $G$. This means that $G$ is embedded in $S_7$, and thus in $A_7$, because it has no subgroups of index $2$. "Unfortunately", $|A_7|$ is not divisible by $112$ :).


Compute :

\[ C_{2n}^n \plus{} 2C_{2n\minus{}1}^n \plus{} 4C_{2n\minus{}2}^n \plus{} ...\plus{}2^n \cdot C_{n}^n \equal{} ?\]

\[ ( \text{where , } C_n^k \equal{} \frac{n!}{k! \cdot (n\minus{}k)!} ) .\]

[hide]I can't prove it [yet], but I think $ \sum_{k \equal{} 0}^{n}2^k\binom{2n \minus{} k}{n} \equal{} 4^n$.[/hide]

$ 2^k\binom{2n\minus{}k}{n}$ is the number of subsets of $ \{ 1,2,\cdots ,2n\plus{}1\}$ with at least $ n\plus{}1$ elements such that the $ (n\plus{}1)$st element is equal to $ 2n\minus{}k\plus{}1$.

So the sum (as it ranges through all possible values of the $ (n\plus{}1)$st element) counts the number of subsets of $ \{ 1,2,\cdots ,2n\plus{}1\}$ with at least $ n\plus{}1$ elements. But these subsets can be paired with subsets of $ \{1,2,\cdots ,2n\}$ (complement the subsets containing $ 2n\plus{}1$ and leave the subsets not containing $ 2n\plus{}1$ alone). So the answer is in fact $ \boxed{2^{2n}}$.

has somebody an algebraic solution for this sum ?

The counting argument is a lot nicer, but here you go:

[hide="Non-counting Method"]We use induction.  The base case $ n \equal{} 1$ gives $ 1 \equal{} 1$, which is true.  Assume that
$ \sum_{k \equal{} 0}^{n}2^k\binom{2n \minus{} k}{n} \equal{} 2^{2n}$

Rewrite the sum to get
$ \sum_{k \equal{} 0}^{n}\binom{2n \minus{} k}{n} \plus{} \sum_{k \equal{} 1}^{n}\binom{2n \minus{} k}{n} \plus{} 2\sum_{k \equal{} 2}^{n}\binom{2n \minus{} k}{n} \plus{} \cdots \plus{} 2^{n \minus{} 1}\sum_{k \equal{} n}^{n}\binom{2n \minus{} k}{n} \equal{} 2^{2n}$
(This is the tricky part, to see why this works notice that each binomial $ \binom{2n\minus{}k}{n}$ appears $ 2^{k}$ times in our new expression)

Now use Hockey Stick to get
$ \binom{2n \plus{} 1}{n \plus{} 1} \plus{} \binom{2n}{n \plus{} 1} \plus{} 2\binom{2n \minus{} 1}{n \plus{} 1} \plus{} \cdots \plus{} 2^{n \minus{} 1}\binom{n \plus{} 1}{n \plus{} 1} \equal{} 2^{2n}$

Multiplying both sides by 4,
$ 4\binom{2n \plus{} 1}{n \plus{} 1} \plus{} 4\binom{2n}{n \plus{} 1} \plus{} 8\binom{2n \minus{} 1}{n \plus{} 1} \plus{} \cdots \plus{} 2^{n \plus{} 1}\binom{n \plus{} 1}{n \plus{} 1} \equal{} 2^{2(n \plus{} 1)}$

We're almost done, we just have to do some stuff with that first term to get what we want.  We have
$ 4\binom{2n \plus{} 1}{n \plus{} 1} \equal{} 2\binom{2n \plus{} 1}{n \plus{} 1} \plus{} \binom{2n \plus{} 1}{n} \plus{} \binom{2n \plus{} 1}{n \plus{} 1} \equal{} 2\binom{2n \plus{} 1}{n \plus{} 1} \plus{} \binom{2n \plus{} 2}{n \plus{} 1}$

Substituting this into our equation, we get
$ \binom{2n \plus{} 2}{n \plus{} 1} \plus{} 2\binom{2n \plus{} 1}{n \plus{} 1} \plus{} 4\binom{2n}{n \plus{} 1} \plus{} \cdots \plus{} 2^{n \plus{} 1}\binom{n \plus{} 1}{n \plus{} 1} \equal{} 2^{2(n \plus{} 1)}$

The induction is complete, so we're done.[/hide]
Actually it turned out pretty nice.  :lol:


logarithms
function
calculus
integration
calculus computations

Consider the equation $ \frac{dx}{dt}\equal{}x^{a}$ where $ a>0$.

(i) Find the solution of this equation satisfying $ x(0)\equal{}1$
(ii) For which values of $ a$ do the solutions of this equation become infinite in finite time?

[quote="Carcul"][hide="Click to reveal hidden content"](i) Your differential equation is equivalent to $ x^{\minus{}a}dx \equal{} dt$. Now integrate both sides:

$ \frac{x^{1\minus{}a}}{1\minus{}a}\equal{} t\plus{}C\Rightarrow x(t) \equal{} [(1\minus{}a)(t\plus{}C)]^{\frac{1}{1\minus{}a}}$.

Making x(0) = 1 we get $ C \equal{}\frac{1}{1\minus{}a}$.[/hide][/quote]

You forgot a specific solution when $ a\equal{}1$

[hide]
For $ a\neq 1$, let $ f(t)\equal{}t\plus{}\frac{a\minus{}1}{x^{a\minus{}1}}$ so $ f^{\prime}(t)\equal{}0$ and $ f$ is constant, equal to $ a\minus{}1$. This gives
\[ t\plus{}\frac{a\minus{}1}{x^{a\minus{}1}}\equal{}a\minus{}1\Rightarrow x\equal{}\sqrt[a\minus{}1]{\frac{a\minus{}1}{a\minus{}1\minus{}t}}\]
When $ a>1$, this approaches $ \infty$ as $ t\rightarrow (a\minus{}1)^{\minus{}}$. (Actually, one might question the possibility of $ x$ being equal to 0 somewhere when $ a>1$. I'm not sure if there's any immediately obvious and rigorous way to deal with this case. I posted the case where $ a\equal{}2$ yesterday, and it was a bit annoying.)

For $ a\equal{}1$, let $ f(t)\equal{}xe^{\minus{}t}$ so again $ f^{\prime}(t)\equal{}0$ and $ f$ is constant, equal to 1. This gives $ x\equal{}e^{t}$.
[/hide]

[quote="countdownmath"]Scorpius, I'm not sure how you came up with the $ f(t)$'s in your solution.
Could you please be more specific?[/quote]

The usual method for solving separable differential equations such as this involves rewriting in the form
\[ A(x)\ dx \equal{}B(t)\ dt\]
and integrating both sides. In this case, the equation becomes
\[ \frac{1}{x^{a}}\ dx \equal{}1\ dt\Rightarrow\minus{}\frac{a\minus{}1}{x^{a\minus{}1}}\equal{}t\plus{}C\]
for some constant $ C$ if $ a\neq 1$.

If $ a\equal{}1$, then it becomes
\[ \frac{1}{x}\ dx \equal{}1\ dt\Rightarrow\ln x \equal{}t\plus{}C\Rightarrow x\equal{}C_{2}e^{t}\]
for constants $ C,C_{2}$. Therefore, $ t\plus{}\frac{a\minus{}1}{x^{a\minus{}1}}$ (if $ a\neq 1$) and $ xe^{\minus{}t}$ (if $ a\equal{}1$) should be constant. However, the substitution is usually a better method of proof because dividing by a function which might be 0 on an interval might miss solutions. These substitutions deal with that issue when $ a\leq 1$, but not when $ a>1$. Whether it is acceptable to claim that we must have
\[ x^{\prime}\equal{}x^{a},\ x(0)\equal{}1\Rightarrow x\equal{}\sqrt[a\minus{}1]{\frac{a\minus{}1}{a\minus{}1\minus{}t}}\]
when $ a>1$ from the above integration or substitution probably depends on the class. I think the substitution would be acceptable in most cases, and either method would work in an AP class. But given my limited calc background, you should ask someone else.


The number is irrational because it contains arbitrary long sequences of e.g. zeros.

The key is to prove that any rational is a repeating decimal.  That follows if we prove that any $\frac{1}{n}$ is a repeating decimal.  But that is true, because if $n$ is divisible by 2 or 5, then we have $\frac{1}{n}=\frac{1}{2^{a}}\frac{1}{5^{b}}\frac{1}{k}$, where neither 2 nor 5 divides $k$.  Then $k|10^{\phi(k)}-1$.  So then, if we take the result of that division, and multiply it by $10^{-\phi(k)}+10^{-2\phi(k)}+...$, then that decimal multiplied by $k$ yields $1$, and is thus a representation of $\frac{1}{k}$.  Then the repeating decimal we get can be multiplied by $\frac{1}{2^{a}}\frac{1}{5^{b}}$, because that is not a repeating decimal.

Perhaps it's easier to see that a rational has repeating decimal representation just by making the long division:
if you divide a by b, there are only b possible remainders, since by euclidean division $a=qb+r$ with $0\leq r<b$. So after at most b steps we get the same remainder which will start a new cycle of digits in your quotient (the same digits in the quotient from the last time we had that remainder will reappear.)

I'll explain my previous post in more detail. Suppose the number is rational. Then it has a repeating decimal represention. Let $k$ be the length of the longest sequence of consecutive zeros in this repeating part. (if there is a longer string of zeros before the repeating part starts, take that length for k instead). But that gives a contradiction, because somewhere there will be 'pasted' the number $10^{k+1}$ which has k+1 consecutive zeros...


algebra
polynomial
number theory
relatively prime
divisibility tests

Hmm..
Since we're not given that a+b, b+c, c+a are relatively prime, consider the following solution:

Let N = f(a,b,c)

Obviously f(a,-a,c) = f(a,b,-b) = f(-c, b, c), so a=b, b=c, c=a are all solutions to the polynomial.
so (a+b)(b+c)(c+a) divides f.

as for the 3 part, brute forcing with fermat's little theorem should give the answer, I think.

[quote="jpark"]
as for the 3 part, brute forcing with fermat's little theorem should give the answer, I think.[/quote]

This does not appear to be working for me...

As for the $3$ part, every coefficient of $C^333_n$ is divisible by 3, I think.  Or I could be wrong.  I'll come back to this after the math jam is over.

[quote="Fierytycoon"]Show that [tex]N = (a+b+c)^{333}-(a^{333}+b^{333}+c^{333})[/tex] is divisible by [tex]3(a+b)(b+c)(c+a)[/tex].

The book says that all we must do is show that [tex]a+b|N[/tex], [tex]b+c|N[/tex], and [tex]c+a|N[/tex], but why does this imply that [tex](a+b)(b+c)(c+a)|N[/tex]?

Thanks.[/quote]

They mean divisibility of polynomials in variables $a,b,c$ with integer coefficients.
$a+b$ and $b+c$ are relatively prime as polynomials. Alternatively you can think of $b,c$ as integers and the whole expression as a polynomial in $a$, which has (say) $(-b)$ as a root.  You might need to work more to show the divisibility by 3, or to prove first the two-variable version (Fermat's little theorem or similar, also understood as a statement about polynomials).

I accidentally put (a-b)(b-c)(c-a), apparently.
It is now corrected.
Give me a minute to think about Fermat's Little Theorem approach

I think it works. It's just a lot of tedious casework, which I'm not really willing to do.

Anyways, here's a suggestion how you might start with the FLT.

a^2=1 mod 3 for all a unless a=3k (so consider this case separately)
so a^333=a mod 3 etc.
Similarly (a+b+c)^333=a+b+c mod 3 if all a,b,c are not congruent to 0 mod 3.
So (given expression) = 0 mod 3 if all a,b,c are not congruent to 0 mod 3.

If not... I guess it would be easier. It's basically the same.  :D

Binomail expansion would be nice too.

[quote]a^2=1 mod 3 for all a [b]unless[/b] a=3k (so consider this case separately)
so a^333=a mod 3 etc. [/quote]

Of course, you do  notice that  $a^3 = a \bmod 3$  for all values  of  $a$ including when  a=3k)  :)

I think it easily comes out from: (a+b+c)^3-a^3-b^3-c^3=3(a+b)(a+c)(b+c)

[quote]Of course, you do notice that  for all values of  including when a=3k) [/quote]

That's why i said that it'd be easier when a,b,c are multiples of 3...  ;)


inequalities
Cauchy Inequality
inequalities proposed

For $ a,b,c$ positive real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{}1$ prove that

$ \sum {\sqrt{1\minus{}ab}}{\sqrt{1\minus{}bc}} \geq 2$

Here is my solution for this very nice problem

We have that $ 1\minus{}ab\equal{}\frac{c^2\plus{}1}{2}\plus{}\frac{(a\minus{}b)^2}{2}$

So by Cauchy Swartz we have
$ \sum\sqrt{1\minus{}ab}\sqrt{1\minus{}bc}\geq \sum\frac{\left|(a\minus{}b)(b\minus{}c)\right|}{2}\plus{}\sum\frac{ca\plus{}1}{2}\geq\frac{a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca}{2}\plus{}\sum\frac{ca\plus{}1}{2}\equal{}2$

qed  :D

[quote="silouan"]Here is my solution for this very nice problem

We have that $ 1 \minus{} ab \equal{} \frac {c^2 \plus{} 1}{2} \plus{} \frac {(a \minus{} b)^2}{2}$

So by Cauchy Swartz we have
$ \sum\sqrt {1 \minus{} ab}\sqrt {1 \minus{} bc}\geq \sum\frac {\left|(a \minus{} b)(b \minus{} c)\right|}{2} \plus{} \sum\frac {ca \plus{} 1}{2}\geq\frac {a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca}{2} \plus{} \sum\frac {ca \plus{} 1}{2} \equal{} 2$

qed  :D[/quote]
Yes, that's right Silouan. Congratulations!

No chance for others to try.  :lol:

[quote="silouan"]Here is my solution for this very nice problem

We have that $ 1 \minus{} ab \equal{} \frac {c^2 \plus{} 1}{2} \plus{} \frac {(a \minus{} b)^2}{2}$

So by Cauchy Swartz we have
$ \sum\sqrt {1 \minus{} ab}\sqrt {1 \minus{} bc}\geq \sum\frac {\left|(a \minus{} b)(b \minus{} c)\right|}{2} \plus{} \sum\frac {ca \plus{} 1}{2}\geq\frac {a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca}{2} \plus{} \sum\frac {ca \plus{} 1}{2} \equal{} 2$

qed  :D[/quote]

Very nice solution, Silouan. Congratulations!! :)


If you have an equilateral triangle inscribed in a circle, and the area of the equilateral triangle is 1, then what is the diameter of the circle?

If you draw a line from the center of the equilateral triangle to the circle, you get the radius of the circle. Since the area of the triangle is 1, a side of the triangle is $ \sqrt {\frac {4\sqrt {3}}{3}}$. Then you multiply that by $ \frac {\sqrt {3}}{3}$ to get $ \frac {2\sqrt [4]{3}}{3}$. Since that is the radius, multiply that by 2 to get $ \frac {4\sqrt [4]{3}}{3}$.

Is that right?

What's that picture supposed to mean? That vertical line you drew is an altitude of the triangle, not the diameter of the circle.

[geogebra]fdaec6f4e7b29067963292e9ed870340d18c0eec[/geogebra]

That segment inside the triangle is the radius.  The longer one (altitude/median/angle bisector/perpendicular bisector) line you were showing is not the radius.

EDIT: dang it, circumcenter i an invalid command

I know where you got that question from!

First you set the equilateral triangle formula for area equal to 1. Solve this for the side length. Then, multiply by sqrt3 for height of triangle. Through vertx/median relationships, we know to multiply by 2/3 for the radius. Then, multiply by 2 for diameter.

And yes, I got that part right on the competition. (We are NOT discussing Mathcounts, so I think it is legal)


Let $x$ and $y$ be positive integers such that: $3x^{2}+x=4y^{2}+y$. Show that $x-y$,$3x+3y+1$ and $4x+4y+1$ are squares of integers.


algebra
polynomial
Rational Root Theorem
difference of squares
special factorizations

factorise: x^6-6x+5.
check for( x-1)^2.
i am unable to go further .

[quote]
We have:
$ (x^3)^2 \minus{} 6x \plus{} 5 \\
\equal{} (x^3 \minus{} 5)(x^3 \minus{} 1) \\
\equal{} \boxed{(x^3 \minus{} 5)(x \minus{} 1)(x^2 \plus{} x \plus{} 1)}$[/quote]

this is not corect because

 $ (x^3 \minus{} 5)(x^3 \minus{} 1) \equal{} x^6 \minus{} 6x^3 \plus{} 5$ not $ x^6 \minus{} 6x \plus{} 5$

Never mind then. Using the rational roots test, you get to:
[hide]\[ {\left(x\minus{}1\right)}^2 \cdot \left(x^4\plus{}2x^3\plus{}3x^2\plus{}4x\plus{}5\right)\][/hide]

just a question is there a general method how to factorize stuff?  or should i post a seperate topic for this?

There are methods such as the Rational Root Theorem, Difference of Squares, and Sum and Difference of Cubes.  Those are the basic ones.  Of course long/synthetic division too.

yes and decartes' rule of signs can be very usefull!!! if you dont know what this is just googel it.  From that we can use long/sythetic division to find the zeroes of the equation and then we know that for a zero well call z (x - z) is always a factor of the original equation!!!(very very very very usefull ---- i <3 remiander theorum!)

anyway i find factorizing more art than science ...
thank you for the answers :lol:

if by art you meen that you dont follow a strick procedure and kinda go with the flow then i agree.  Thats how my mom describes driving!!! :)

:lol: (driving)

yes it was meant in that way .... but that is the most beautiful thing about math  :D

yes my mom is kinda teaching me to drive(just kinda listening to her so that in a few years i will be ready!!!) and when we went around  a bend i asked why shee accelerated instead of decelerated because thats what i had learned and she said that its an art becuase you dont always have to go by the rules you have to *cough=corny* be one with the road! :(


linear algebra
matrix
analytic geometry
limit
linear algebra unsolved

Let $E$ be an euclidian space and $u$ a linear map of norm at most 1. Prove that $ Im(u-I)$ and $Ker(u-I)$ are ortogonal, where $I$ is the identity function.

I'll replace $u$ with $A$ (which will also denote the matrix of $A$ in a convenient basis) :).

We may as well assume that $x\in \ker (I-A)$ is $e_1$, for simplicity. We must now show that $<e_1,y>=<e_1,Ay>,\ \forall y\in\mathbb R^n$ (I assumed $E=\mathbb R^n$), which is equivalent to saying that the first coordinate of $y$ is the same as the first coordinate of $Ay$.

Since we assumed that $Ae_1=e_1$, it means that $A(1,1)=1$, so we have to prove that given a matrix $A$ with $1$ in position $(1,1)$ s.t. the associated transformation has norm $1$, $A$ has first row $(1,0,\ldots,0)$. This follows quickly from the fact that a matrix whose associated transformation has norm $1$, has all rows of norm at most $1$.

It is a classical of X / Ensae / Mines-Ponts / Centrale

$E$ euclidean space dimension n
$u\in L(E)$  suppose for any x, $||u(x)||\leq ||x||$

Prove
1/ for any x, $||u^{*}(x)||\leq ||x||$  here $u^{*}$ adjoint of $u$
2/ $u(x)=x$ is equivalent to $u^{*}(x)=x$
3/ $E=ker(u-I)\oplus^{\bot}Im(u-I)$

4/ Find $\lim_{p\rightarrow +\infty}\frac {1}{p}\sum_{k=0}^{p-1}u^{k}$

another approach:
$M=I-A^*A$ is positive. Suppose that $x \in Ker(A-I)$. Then
$<x, My>=<x, y-A^*Ay>=<x,y>-<Ax, Ay>=<x,y-Ay>$. But because
$M$ is positive and $|Mx|=0$ we conclude that : $<x,My>=0$ for all $y$.
Conclusion follows.

Moubi, this is Von Newmann ergodic lemma. And you know, it is not so easy to find all the ideas, especially when the examinator doesn't  give you those hints.  ;) . To Alekk: great solution!

[quote="harazi"]...it is not so easy to find all the ideas, especially when the examinator doesn't  give you those hint...[/quote]

At the end of MPSI it is normal that students not solve it without hints.
At the end of math-spé (3/2)  there are a lot of of students solve it without
hints. About  (5/2) students they are very confortable on such problems.


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

Hello! Can ou help me to evaluate this integral:
$\int_{-1}^{1} \cos\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right)dx$?

Hint: if you see a symmetric interval (such as $[-1,1]$) it's at least worth asking whether this function is even, odd, or neither.

I can't evaluate it anyway. :(  Mr. Merryfield can you help me?

An even function composed with an odd function is even, which doesn't help that much. Make the substitution $u=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\tanh x.$ Then $dx = \frac{du}{1-u^2}.$ The integral becomes:

$2\int_0^{\tanh 1}\frac{\cos u}{1-u^2}\,du$

Seeing it in that form leaves me with the sense that this is very unlikely to have an elementary closed-form solution.

Thanks anyway. Maybe there is mistake in the book.


I don't know where else to put this, so here goes...

I got this email from a guy named Alin/Ioan Serdean, from Romania, apparently. He says he got my email from artofproblemsolving/mathlinks, which is entirely possible. But he says he wants to exchange olympiad problems, solutions, etc., and so he wants my address and email. Why is he asking for my email in an email to me??? Someone I know got an email exactly like this too... has anyone else here gotten one?

Here's the text of the message:

[quote]Dear Sir,
Forgive my daring to approach you. Please give me just a few seconds.
           I am a Romanian passionate about Mathematics.
  My address is:
Serdean Ioan, str. Pandurilor, No 1B, 335700,
Orastie, Hunedoara, Romania, Europa.
e-mail: alin_serdean@yahoo.com
I obtained your e-mail from diverse forums: Mathlinks,
Art of Problem Solving etc.
I am very interested in an informational
Exchange, useful to both of us I hope, on various themes:
a)    Olympiads from your country, but ONLY WITH SOLUTIONS
b)      Other contests, but ONLY WITH SOLUTIONS
c)      Training problems, math themes etc.
d)      University Olympiads for Students.
e)      Magazines, Journals, Math books, all related to math Olympiads and math contests.
I can read English, French( I read quite good) Italian, Spanish, Portuguese (only mathematic text)
If you could be so kind to tell me your postal address for us to carry out correspondence and your E-MAIL address(ESPECIALLY).
I also offer the things presented at the points above from Romania and other areas.
I have some materials electronically formatted (files, e-books etc., DVDs etc.) and some are photocopied or books. As much as possible I offer free what I have.
If you want something special, which I can’t offer you free, but I’ll have to buy them I shall tell you the prices before buying them.
If you have friends from neighbor counties, or from other areas who are interested in this kind of exchange on math themes which I propose, I would be very grateful if would communicate my address to them.
              Thank you for your time.
                                                                            All my best regards,
                                                                              Ioan Serdean

P.S: If you would be so kind to tell me the country from which are you.
[/quote]

What are your thoughts on this?

He used to send it as a mass PM but, after some discussion in Moderator's Lounge, Valentin said that he "would deal with serdean", which, probably, meant some kind of banning or restriction of privileges. At any rate, it seems that there were no more PMs after that. Asking for an e-mail address in an e-mail is, indeed, a weird idea but it may be just a leftover from the time when he used PMs. I do not know what to think of the guy. At least, he seems to be real and traceable: googling his name produces a few sites like [url]http://rgmia.vu.edu.au/members/Serdean.htm[/url] or [url]http://bibliophil.bibliotecamm.ro/?s=catalog&f=3&ca%5Bc1%5D=4&ca%5Bc1t%5D=3&ca%5Bc1v%5D=Serdean,%20Vasile[/url]. The approach he chose is, indeed, sort of unconventional but how would you go about this if you were in his shoes? So, decide yourself what to do. As long as he doesn't ask your credit card number, I do not see any real danger in communicating with him and it is quite possible that he is really interested in what he offers. :)

Do you know (of) him? Have you answered his e-mail and run into any trouble? Where do those !!!!!!!!!!! come from? :? My point was that there is a real person with this name who is a mathematics professor in his 50's or 60's (judging from the picture on [url]http://www.colegiulavlaicu.ro/engleza/index.html[/url]), published a few articles and books about teaching mathematics and problem solving, etc. I do not give a 100% warranty that the letter came from him and not from some impostor but you can always choose to answer directly to his institution address secretariat@colegiulavlaicu.ro (this is the address of the college secretary but putting "for Ioan Serdean" or something like that in the subject line will most likely ensure that the e-mail will be delivered to the right addressee) and ask whether it was he who wrote the letter. Probably, I'll try to do exactly this myself if nobody does it in the next few days or gives a valid reason why this might be a bad idea.

The question is, of course, whether you are interested in his offer at all. Anyway, I do not intend to promote him on AoPS in any way. I just wanted to make a suggestion that we give opinions based on some evidence. The best thing would be an opinion of somebody who knows a person who knows a person.... who knows (real) Ioan Serdean personally (the shorter the chain, the better) but, lacking this, I think we can at least rely on institutional websites and similar things.

[quote="fedja"]The question is, of course, whether you are interested in his offer at all. Anyway, I do not intend to promote him on AoPS in any way. I just wanted to make a suggestion that we give opinions based on some evidence. The best thing would be an opinion of somebody who knows a person who knows a person.... who knows (real) Ioan Serdean personally (the shorter the chain, the better) but, lacking this, I think we can at least rely on institutional websites and similar things.[/quote]I know him personally, but that doesn't excuse this kind of behavior (even after I specifically stressed out that what he is doing is called spamming, he still continued to do it - which is, I believe, unacceptable).

The user and his IP have been banned.  We'll continue taking that action as he pops back up, as well.

Oh cool! I got the same mail as well. Now I replied and sent him some Indian MO links. I hope it'll be okay,...

Hi,
My name is Daniel Popa and I'm teacher at Colegiul National Aurel Vlaicu Orastie.
Serdean Ioan is a math teacher and Alin Serdean is his son.
Mr. Serdean Ioan is second principal and he collect math problems from around the world.

He could have just looked around AoPS... like in the contest section, or even in the Wiki. Speaking of which, I think I'll go do some AMC 12s.

I just got a much shorter email from him that goes like this:

____________________

Dear friend,

I am a pasionate by math. I would like to  begin an informational exchange on e-mail regarding different math topics. We can exchange materials, e-books. I belive that this exchange is useful to both of us. Hoping this invitation will interest you, I send you my best regards.

Yours, I.S.

P.s.I’m interesting in ….(regards attach)

____________________

I replied to him asking who he is and he replied, attaching more materials, but without telling me that he got my email from aops.  So I only knew he got it from here after searching his name.  Given that he looks harmless but has sent this to a number of people, should i continue to reply him?

He wants to trade math ebooks. If it's not breaking the law where you live, and you want math ebooks, I don't see why you shouldn't continue to correspond with him. That's between you and him. But be aware that he might be giving you and he might be expecting from you things which are copyrighted.

I don't know if he was told to stop, but I got it again today.

[quote]Dear friend,
1.      I am passionate roumanian thachers  about mathematics.
2.      I have your address from different friends and mathematics forums you collaborate.
3.      It would interest me an exchange of materials which will be benefic to both sides: I send you some books, articles, magazines, notes, training materials, etc. I wait the same thing from you.
4.      If you are interested please tell me.
                                                                                Ioan Serdean[/quote]

Yeah he's been told, in two languages actually. There's really nothing else that I can do at this point :)

Guys.
I receive the same e-mail today but I think I may understand him. Make it between you and him.

I got those emails from serdean too.

[quote="serdean"]
Dear friend,

1.      I am passionate roumanian thachers  about mathematics.

2.      I have your address from different friends and mathematics forums you collaborate.

3.      It would interest me an exchange of materials which will be benefic to both sides: I send you some books, articles, magazines, notes, training materials, etc. I wait the same thing from you.

4.      If you are interested please tell me.

                                                                                Ioan Serdean[/quote]

With a lot of attachments, which are useful. I did not know his identity so I did reply , but without any attachments  :blush:

I think we have to make an understanding. :|

Yeah, he sent some messages to me, too. I said that I would want to colloborate but I don't think that I have materials which can be useful to him. After every message I sent, he sent a really irrelevant message with some really useful documents. My question is that doesn't he know English much? Or is he just pretending like he didn't understand. Initially, it seemed like a good cooperation, but now it started to be disturbing.

he wrote [quote]
Dear friend,

1. I am passionate roumanian thachers about mathematics.

2. I have your address from different friends and mathematics forums you collaborate.

3. It would interest me an exchange of materials which will be benefic to both sides: I send you some books, articles, magazines, notes, training materials, etc. I wait the same thing from you.

4. If you are interested please tell me.

Ioan Serdean [/quote]

i'm confused[/quote]

Thank goodness,
Actually, I just got one, too, and I'm very nervous about what to or what to not do.
Thank to Google.  :D , Googling his name brought me here and made me understand what to do now.
Thank to all of you.  :lol:

[quote]Dear friend,
1.      I am passionate roumanian thachers  about mathematics.
2.      I have your address from different friends and mathematics forums you collaborate.
3.      It would interest me an exchange of materials which will be benefic to both sides: I send you some books, articles, magazines, notes, training materials, etc.
4. I wait the same thing from you.
5.      If you are interested please tell me.
                                                                                Ioan Serdean[/quote]

I got the similar email too....:maybe:

I replied him and he send me some ebooks but I haven't got any ebooks so I just send some problems to him :lol:

[quote]
Dear friend,

1. I am a teacher with a passion for the mathematics in contests and the specialties about it: algebra, geometry, inequalities, analysis, numbers, combinatorics.                                                                                2.I am from a small town in Romania and I am working at the high school from this town.

3. If you are interested I am proposing you an exchange of information which I hope to be useful for both sides. This exchange consists in sending without costs: problems with solutions, books, articles, magazines, training materials.

4. You are probably wondering how come I have your address. I have found it on different math forums and blogs and I have decided to write to you because I saw you have a passion for mathematics as me.

5. For being easier I propose to send one to another by e-mail the titles that we want.

6.  I would like the following from you and I have placed it in the “books” file.

7. Of course you may not have anything of what I asked for. Our friendship and correspondence will continue because it is impossible for me to have all that you have.

8. I am waiting for you to tell me what you are interested in.

9. Hoping that I have caught your attention and that we will collaborate in the future please tell me if you are interested in this exchange.

10. I guarantee that what you will be sending me will be used only by me and I won’t be sending it to anyone. I am asking for the same thing.

11. I am also interested in blogs and links about mathematics.

12. Once again I am looking forward to your answer.

                                                                             With respect,i.s.
[/quote]

[quote]
Dear Mr. Professor,
1) Thanks a lot for you answer me!
2) SORRY about my english too, but i think that we are to understand one other.
3) I am interested and articles, notes, mathematical problems of training.
4) I am interested in problems that I need and  solutions.Especially problems of the following countries: China, Macao, Hong Kong, Korea, Singapore, Thailand, Japan, Taiwan,
Mongolia, Vietnam.
5) I can not handle me and I can read only in Spanish, Portuguese, English, French.
6) You can 't help if you send my email friends and passionate teachers of mathematics.
7) What to send?
8) Waiting for news!
         Sincerely, i.s.
[/quote]

[quote="pacman2812"]I got emails from both guys, and they somewhat got to my hotmail account (my "official" email account for aops is from yahoo, and I got it really full of creepy Serdean mails). Weird.
It's definitely not funny, as, if this might have been with not bad intentions, it might have been a spammer or even a spambot spreading. Also, it's very very unusual that he might have been sending PMs to such lots of people on AoPS (I don't know how he got my Hotmail account, as I don't know why he emailed me with such insistence).[/quote]
I am not really sure if I would call him a "spam", because he has sent me quite a lot of books, and I have also exchanged many ebooks with him.
[quote]

Dear friend,
1) I'm glad we will work!
2) To be forgiven for that my English is not excellent!
3) I am interested in books from the list below.
4) I am also interested but have problems and solutions especially in the following countries: India, Macao, Korea, Singapore, Philipinnes, Vietnam, Thailand, Japan,
Hong-Kong, Taiwan, China, Mongolia.
5) Waiting for news!
6) I have many books and therefore I would like you to send as much as I possibly those books that I do not have them.
7) I do not understand how I could use Mathlinks.
8) I am on this forum.
9) Please back the specifications!
10) Waiting for news!
                  Sincerely, i.s.[/quote]
Again I suggest we should do something so that I.S and us(/me) can be friendly and useful to each other. :maybe:
PS: He does not know my age! :oops:

I don't think this is spam. Cause if he's sharing with you for free already, and he knows you have the perfect chance just to take his stuff, then why would he even PM you? I think he's actually wanting to trade! Not just spam.

this guy suddenly creeps up after some months of silence tellin sthng abt bad health and all.. again talks abt some collaboration... again as usual gives a huge chunk of e-books(unfortunately alluring as they r quite useful) ... what am i to do??? i am kinda scared cuz is it legally wrong or sthng for trading books like he is doing so? as if so ... wont i be questioned unnecessarily too? :(
and i want a sincere advice.. i dint reply to him those months back and still he did send me a mail recently... Is this guy wrong buy any legal law of any country?
and since i am from india.. is it wrong to trade books with him?
i see that one of the book he gave in his recent e-mail is copyright stuff that was published in 2009!! ..... and unrelatedly it is awesome.... now.. CAN I OR NOT USE IT! DO  INDIAN LAW PROHIBIT THIS????
i want some expert comments...
(since when did maths become sthng to be scared of... i really wonder.... all everybody shud see is its beauty not this kinda things so d guy is definitely irritating as hell..)
Valentin what should i do??? please help me...

i kno i am computer savy or anything.. is it wrong if some1 sends him a virus or sthng.. and valentin sincehtmu kno him personally can u tell me why in this link..
http://www.staff.vu.edu.au/RGMIA/members/Serdean.
there is no details?

I really appreciate him but if you are annoyed simply mark him as a spam in your inbox and we are done. No need for the posts

Also you could hide your email to prevent further spam from other people who saw your email on AoPS.


I'm in Spanish half this year :P . Seriously, my school has a really weird program... Ok, I'll start.

Hola! Como te llamas? Me llamo Alina. Escriben aqui, por favor.

I'm not sure if that made any sense... :maybe:

Hola, me llamo Taylor. (upsidedown ?) Como estas?

Hey, no fair. What does feo(a) mean?

Estoy mal. :( Y tu, como estas?

[quote="junggi"]uhhhh wtf
Tu eres feo tambien.[/quote]

Y TU MAMA TAMBIEN.

"feo/a" significa "ugly".

estoy muy bien gracias. Que hiciste ayer?(I know its a dumb question)

ummmmm... What does that mean?  :(  Only now I realize how much Spanish I don't know... :(

I said "what did you do yesterday?"..
o well I'll stop with the past tense stuff.
lets start again

Que vas a hacer esta semana?

uhhh ignore me.
a mi me gusta el chocolate

Mmmmmmmmmm.... A Me gusta mucho comer. La comida es muuuuuuuy sabrosa.  :P

moosing???

At this rate we'll beat all the other language things... :lol:

Hola! Mi comida favorita es chocolate. Y tu? Que te gusta comer?

No como nada porque no me gusta comer. :o

No me digas! :o Tu eres un robot?  :lol:

:rotfl:  :rotfl:

Tu eres gracioso, pero you soy mas graciosa. :P

$\frac{six*n*x}{cos(x)}=tan(x)$
so
$\frac{i*n*x}{co}=tan(x)$
and
$\frac{i}{co}=ta$
Hence,
$i=taco$!!!

Yo soy uno taco! Tu eres un taco muy sabroso...  :icecream:  :rotfl:

[quote="Ignite168"][quote="Dave Vather"][quote]Hola! Yo tengo que ir, pero yo (will) escribir mas (later). Que es tu comida favorita? (I don't know what else to ask  )

You no tengo un idea que las personas hablan...
In case that didn't make any sense, which I'm almost sure of, I don't have any idea what the people before me said...  [/quote]

"Yo tengo que ir" doesnt makes much sense... maybe you wanted to say "Me tengo que ir" (ir = to go, if you make it reflexive, irse = to leave)
"Escribiré más después" = I will write more later.
"No tengo idea de qué dijeron las personas antes que yo"
= "I don't have any idea what the people before me said"

If you wish to practice your spanish, send me a PM  :wink:[/quote]

I READ YOUR NAME AS DARTH VADER[/quote]
YO TAMBIEN

[quote="Dave Vather"][quote]Hola! Yo tengo que ir, pero yo (will) escribir mas (later). Que es tu comida favorita? (I don't know what else to ask  )

You no tengo un idea que las personas hablan...
In case that didn't make any sense, which I'm almost sure of, I don't have any idea what the people before me said...  [/quote]

"Yo tengo que ir" doesnt makes much sense... maybe you wanted to say "Me tengo que ir" (ir = to go, if you make it reflexive, irse = to leave)
"Escribiré más después" = I will write more later.
"No tengo idea de qué dijeron las personas antes que yo"
= "I don't have any idea what the people before me said"

If you wish to practice your spanish, send me a PM  :wink:[/quote]

QUE?!?!?!?! Pero mi profesora ensena que is correcto! (dont' know if that made sense...)

She says tengo+que+infinitivo=I have to.... So tengo que ir would mean I  have to go.... right? Or not... Why would you put me rather than yo????

Ok, so no one is replying or answering mi pregunta. :mad:

Ustedes deben hablar aqui! Por favor???

This topic is dying....

Hola! Que es tu clase favorita? Mi clase favorita es matematicas, pero no me gusta el profesor del clase.

If you say "Tengo que ir" you are saying that you have to go somewere, but not exactly saying that you are leaving from here now, for example:
"No quiero ir a la escuela hoy, pero tengo un examen y tengo que ir" Means "I don't want to go today to the school, but i have an exam and i have to go"
However if you want to say you have to leave and that you can't post anything right now, you would say:
"[b]Me[/b] tengo que ir, no puedo escribir en este momento" or something like that. In spanish you can't say just "I have to go" ("Tengo que ir") Unless you say it whit reflexive form ("Me tengo que ir") But your teacher is right: Tengo que + verb in infinitive = I have to + verb in infinitive. The only thing is that sometimes, we use the reflexive form of the verb (As Irse, or even in other verbs that can't be used unless you make them reflexibe for example desayunar[b]se[/b] (to breakfast))

That kinda makes sense... I don't think we've gotten to that yet.... By reflexive verb do you mean one where you have to use Me or Te or Le or one the other ones with? Like gustar?

I think its the form of a verb that's like BLAHBLAH oneself.
like washes oneself, or pleases oneself(which is gustar i think)

gustar is actually different. Gustar is a transitive verb with the thing that's nice, or pleasing as the subject.

So when you say Me gusta la clase, you're saying the class pleases me, not I like the classes (when you translate literally).

Reflexive verbs are like BLAHBLAH oneself, and end in "se". Gustar is not reflexive; I guess gustarse would work in a way though, if you say Me me gusta or something weird like that meaning I please myself, or he pleases himself, or something similar.

That's... really weird... when in the spanish curriculum is that taught?

No me gusta hacer la tarea :mad:


inequalities
ratio
LaTeX
geometry
geometric transformation
reflection
parallelogram

My motives: I'm thinking about giving a talk to my school math club about geometric constructions. I'd like to learn some more constructions before I do so.

Your motives: Whatever you want them to be! :yup:

[b]How this thread works:[/b] Let's compile a list of constructions for fun (compass and straightedge only)! This will be similar to a marathon. Someone posts a construction, somebody gives a way to do it. The constructions you post should have reasonable difficulty, something we can all figure out and not have to look up on the Internet. For example, "construct a $60^\circ$ angle is acceptable, but construct a regular 17-gon is definately not. Without further adieu.... let the fun begin! (As a sidenote, please let there be no spam. If you spam, I will delete your post and attempt to put you under "quarantine" or something along the lines of that. Not pretty.  :wink:)

[b]Construction#1[/b]
Describe how to bisect an angle.

[hide="First solution!"] Draw a circle with an arbitrary radius and centered at the vertex of the angle. Then, note the two points where the circle intersects the angle. Call these A and B. Construct the circle with radius AB and center A, then the circle with radius BA and center B. Let the intersection be C. Then, connect the vertex and C and you're done.[/hide]

Do I have to prove it too? If I do, then it involves congruent triangles and I'm too lazy to get into it now.

Construction #2:
Given three segments (drawn for you) that satisfy the triangle inequality, construct the triangle with sidelengths equivalent to the lengths of the three given segments.

For your problem, just take one segment, and draw two circles with radii of the lengths of the other segments, centered at the endpoints of the first segment, the circles' intersection is the desired point

---

here are a few...

3)Perpendicular to a given line through a given point

4)Parallel to a given line through a given point

5)Circumcenter, 6)incenter, 7)orthocenter, 8)centroid, 9)9-pt center, of a given triangle

construct a triangle given:
10)
a) the lengths of the the medians
b) the length of two sides of the triangle and the median drawn to the third side
c)given two medians and the side they are not drawn to

11) lengths of two sides and their internal angle bisector

12) given two intersecting lines, and a point, A (not on the lines), construct a triangle with A as a vertex, and whose angle bisectors (extended) are the two lines

13)divide a segment into a ratio fraction of itself
14)geometric mean
15)given a circle, and a point, construct the inversive image of that point with respect to the circle
16)given two circles, construct the inversive image of one circle over another

17)circle tangent to a given line and passing through two given points

18)given two lines that intersect [but you cannot see the intersection, it is too far away], construct a line passing through the intersection of the first two lines

19) a right triangle is drawn with the line containing the altitude to the hypotenuse, and the 9 point circle is drawn, after erasing everything except for the circle and the line containing the altitude, reconstruct the triangle

I hate you. You're so unloyal to Euclid. I'm only doing the Euclid ones. (Breaks out copy) :lol:

And I'm assuming you can bisect a line segment.

[hide="3"]Draw a circle centered at the given point so that it intersects the line, then take the line segment determined by the points of intersection. Bisect it and connect it to the given point.[/hide]

[hide="4"]Pick two arbitrary points on the line. Let them be A and B, then let the point off the line be P. Draw a circle with center B and radius equal to AP. Draw a circle with center P and radius equal to AB. Let the intersection of the circles be Q, PQ is the parallel.[/hide]

ahhh i think 17 is euclid too, but I forgot how to do it

And for 5-9 just construct the angle bisectors/medians, etc. by bisecting lines, bisecting angles, and dropping perpindiculars (for altitudes, etc.)

Wow, that's a nice list ther Altheman. I'll give these some thought. :)

For the geometric mean, there's a way to multiply segments, and there's a way to square root. I forget exactly how to multiply, but for square rooting, one way is to bisect then to construct the square from the half-segment then draw the diagonal.

[hide="Geometric mean"]Let the two numbers be $m$ and $n$.  Draw line segment $\overline{AB}$ with $AB=m$.  Extend $\overline{AB}$ to a point $C$ such that $BC=n$.  Construct the midpoint of $\overline{AC}$ (call it $O$) and draw circle $O$ with radius $\overline{AO}$.  There exists a point $D$ on $\widehat{AB}$ such that $\overline{BD}\perp\overline{AC}$.  It can easily be proven that $BD=\sqrt{mn}$.[/hide]

How do you show arcs in LaTeX?

[quote="lingomaniac88"]$BD=\sqrt{mn}$.[/quote]
It can? Maybe it's just my lack of geometrical knowledge, but I don't think it's that easy...

And what is the construction for multiplication?

Here's one for the Perpendicular Bisector:

[img]6665[/img]

[quote="Phelpedo"]It follows from your standard similar triangle ratios when given the altitude to the hypotenuse.[/quote]
Oh right. :blush:

I swear, I knew that once...

Here are three involving tangents:

20) Given a circle and an exterior point, construct both lines from the point tangent to the circle.

21) Given a circle and a point on the circumference of the circle, construct the line tangent to the circle at that point.

22) Given two circles, draw their comon tangent lines. (No part on one circle is contained in the other)

Altheman, I know this sounds silly but what's the 9-point circle? I don't know much about advanced geometric concepts, as I haven't read Geo Revisited and that other good stuff yet.  :D

[quote="4everwise"]
Altheman, I know this sounds silly but what's the 9-point circle? I don't know much about advanced geometric concepts, as I haven't read Geo Revisited and that other good stuff yet.  :D[/quote]
I know something about geometry tht 4everwise doesn't!?

Please let me savor this moment...

See: http://en.wikipedia.org/wiki/Nine_point_circle

Hehe, thanks for the link.  :P

For my 3 tangent problems, i solved the last 2 intuitively, so there might be faster methods. I'll post them in case they're different from the accepted methods. (I thought 21 was cool :yup:)

[hide="21"]You're given a circle and a point on that circle. Using the compass, construct another circle whose center is at the given point on the circumference of the circle, and whose radius is equal to the radius of the given circle

Draw the diameter of this new circle (line containing center of old circle and the given point). The perpendicular bisector of this diameter is the desired tangent.[/hide]

[hide="22"]Draw the line connecting the centers of the two circles. Next, draw another line that passes through the interiors of both circles and intersects the line you drew just before. (Similar triangles!) By similar triangles, that intsersection point is the point where both common tangents will intersect. Finish the problem with construction 20.[/hide]

[hide="20"]Call the center of the circle O and the point outside P. Construct a circle with diameter OP (Draw OP, find its midpoint M, set the radius equal to MP and make the center M). Call the points of intersection between the two circles A and B. PA and PB are tangent to circle O.

Proof. OAP and OBP intercept straight angles and are thus right angles. Since radius OA is perpendicular to AP, AP is tangent to the circle. Same for OB.[/hide]

I don't think your 22 solution works.
[hide]The second line you draw needs more restraints. One way to do it would be to construct diameters of the circles perpendicular to the first line and then draw lines containing corresponding diameter endpoints. (How you pair the points up will give you either the external or internal tangents.)[/hide]

Here's another one for geometric mean, i think.  :)

[hide="geometric mean, 14"]Let the lengths of two segments be $m$ and $n$ and let $x=\sqrt{mn}.$ It follows that $(x)(x)=(m)(n),$ which can be constructed with Power of a Point.

It's easy to construct a segment with length $m+n.$ Start with segment $AB=m$ and add $bc=n$ (all the points collinear). Next, draw a circle (diameter must be greater than $m+n.$ :P) and construct the segment within the the circle. Note that the chord with length $2x$ must be bisected at point $B.$

To do this, I drew the line (radius) containing the center of the circle and $B.$ The part of the perpendicular to that radius at $B$ contained in the circle has length 2x, so half of it has length $x.$

Hope that's understandable.  :blush: [/hide]

23. Construct a regular pentagon with a given side length.

24. Construct a segment with side length $\sqrt{k}, \ k \in \mathbb{N}$ given a segment with length $1$.

(edited:  :ninja: )

25 and 26 have already been posted by Altheman. Also, the first 24 is known to be impossible (you can cheat and do a marked ruler construction though).

And there are two 24s.

Anyway, for the second 24, construct the isosceles right triangle with $\sqrt{k}$ as the hypotenuse, double the leg. Then, apply the division construction (draw another line coming from one of the vertices of the line, mark off $k$ equidistant posts, connect the last points, then draw parallels through the points that you marked), then you're done.

What's the nice way to construct the golden ratio? I know how to do it algebraically, we all know that it's equal to $\frac{1+\sqrt{5}}{2}$ which is easy to construct. Just draw 1, then draw a right triangle with sides 1 and 2, copy the hypotenuse and add it to the 1, then bisect it. Then, subtract 1 and we're done.

But there's a pure geometric way that's simpler, it's in book 1 of Euclid, but I forgot how. Does anyone know?

[quote="Altheman"][quote="cincodemayo5590"]23. Construct a regular pentagon with a given side length.

24. Construct a regular heptagon with a given side length.

24. Construct a segment with side length $\sqrt{k}, \ k \in \mathbb{N}$ given a segment with length $1$.

25. Given a triangle, find the centroid.

26. Given a triangle, construct the incircle, circumcircle, orthocenter.[/quote]

lol nice to post like 4 things that i already posted...also #24 is impossible...cos 2pi/7 is a solution to a cubic...

23....it hinges on the golden ratio...which i am too lazy to do, but yea...

--

i suggest to look back at 16,17,18,19 again, they are some of my favorites...

22. is a pretty nice problem...think of how you calculate the common internal and external tangents[/quote]

Oh, rofl, i wasn't paying attention. Yay! I'll get rid of them.

[quote="13375P34K43V312"]
Anyway, for the second 24, construct the isosceles right triangle with $\sqrt{k}$ as the hypotenuse, double the leg. Then, apply the division construction (draw another line coming from one of the vertices of the line, mark off $k$ equidistant posts, connect the last points, then draw parallels through the points that you marked), then you're done.[/quote]

Huh? You start with just a segment of length $1$, and not all numbers in the form $\sqrt{k}$ can be hypotenuses of an isosceles triangle with integer sides.... unless I'm not understanding you, which is probably more likely.  :P

[hide="Anyway, my idea was..."]
To construct the segment of length $k+1$ from the original segment and to draw the GM of the the segments with length $1$ and $k$.[/hide]

Oh, another question... how do you take the square of a number in constructions? (with segment defined as of length $1$, which I think would be necessary)

you can use power of a point to do 24 as well.

Here's an odd one...

25. Given three segments labeled $a,b, \text{and }c$, find an algebraic and a purely geometric (i'm not sure about this one... does it exist?) method to find the roots of $P(x)=ax^{2}+bx+c$.

solution to #10 (!!!)

(the solution for (a) really needs a diagram. i should learn metapost or something...)
[hide="a"]
We first prove a lemma: given any triangle, it is possible to construct a second triangle with lengths equal to the medians of the first triangle, and if a third triangle with lengths equal to the medians of the second triangle is constructed, the third triangle will be similar to the first with side ratio 3/4.

Suppose we are given a triangle $\triangle ABC$. Let $D$ be the midpoint of $\overline{BC}$, $E$ be the midpoint of $\overline{AC}$, and $F$ be the midpoint of $\overline{AB}$. Reflect $F$ over point $E$ to $C_{1}$. Then $\overline{BD}$ and $\overline{EC_{1}}$ are congruent and parallel, so $BDC_{1}E$ is a parallelogram and $BE=DC_{1}$. Also, $AE=EC$ and $FE=EC_{1}$, making $AFCC_{1}$ a parallelogram and $CF=C_{1}A$. Therefore, $\triangle ADC_{1}$ has side lengths equal to the medians of $\triangle ABC$.

Now let $F_{1}$ be the intersection of $\overline{AD}$ and $\overline{FC_{1}}$ and $D_{1}$ be the intersection of $\overline{AC}$ and $\overline{DC_{1}}$. We have $AF_{1}=F_{1}D$ and $C_{1}E=EF=2EF_{1}$, so $E$ is the centroid of $\triangle ADC_{1}$!!! We can now compute the ratio of the areas between $\triangle ADC_{1}$ and $\triangle ABC$:
\[[ADC_{1}]=2[ADD_{1}]=\frac{3}{2}[ADC]=\frac{3}{4}[ABC]\]

Finally, reflect $F_{1}$ over point $E_{1}$ to $C_{2}$, so that $\triangle ADC_{2}$ is the "third triangle" we need in our proof. Since $[AD_{1}C_{2}]=\frac{9}{16}[ABC]$, it only remains to show that the two triangles are similar. But this follows from $\overline{AB}\parallel \overline{DE}\parallel \overline{D_{1}C_{2}}$, so $\angle AD_{1}C_{2}=\angle BAC$ and $\angle AC_{2}D_{1}=\angle ABC$, making $\triangle D_{1}C_{2}A$ and $\triangle ABC$ similar.

With our lemma, we can construct a triangle with our three given lengths and then form a second triangle with the median lengths of the first. To find the desired triangle, just scale the second one by 4/3.
[/hide]

[hide="b"]
Given the side lengths $x,y$ and the median length $m$, let $A$ and $D$ be two points spaced $m$ apart and reflect $A$ over point $D$ to $A_{1}$. Construct a point $B$ such that $AB=x$ and $A_{1}B=y$ by intersecting two appropriate circles, then reflect $B$ over point $D$ to $C$. $ACA_{1}B$ is a parallelogram because the diagonals bisect each other, so $AC=y$ making $\triangle ABC$ the desired triangle.
[/hide]

[hide="c"]
This is basically a combination of the first two: given a side $x$ and medians $m,n$, first construct a triangle with sides $m,n$ and included median $x$. Then construct a triangle with lengths equal to the medians of the first one and scale the second triangle by 4/3.
[/hide]

26. Given two intersecting lines and a point not on either line, construct both circles passing through the given point and tangent to each of the given lines.

[b]Re. Number 25:[/b]

Whoa... let's revert to a simpler problem? I'm not even sure that you can divide segment lengths by eachother geometrically...  :maybe:

[quote="cincodemayo5590"][quote="A M Y"][b]Re. Number 25:[/b]

Whoa... let's revert to a simpler problem? I'm not even sure that you can divide segment lengths by each other geometrically...  :maybe:[/quote]

Lol, ok. I'm pretty sure you can... but its fine. I'll post that as a separate question in the pre-olympiad forum. :P

25. Given three pairwise externally tangent circles (i.e. they form what looks like a triangle), construct a fourth circle that is tangent to all three.[/quote]

[url]http://www.ics.uci.edu/~eppstein/junkyard/tangencies/apollonian.html[/url]

how is this easier?

[quote]12) given two intersecting lines, and a point, A (not on the lines), construct a triangle with A as a vertex, and whose angle bisectors (extended) are the two lines

15)given a circle, and a point, construct the inversive image of that point with respect to the circle

17)circle tangent to a given line and passing through two given points

18)given two lines that intersect [but you cannot see the intersection, it is too far away], construct a line passing through the intersection of the first two lines

19) a right triangle is drawn with the line containing the altitude to the hypotenuse, and the 9 point circle is drawn, after erasing everything except for the circle and the line containing the altitude, reconstruct the triangle[/quote]

[hide="12"]Reflect $A$ over $l_{1}$
Reflect $A$ over $l_{2}$
Join the two obtained points to obtain line $m$
Then
$m \cap l_{1}=B$
$m \cap l_{2}=C$[/hide]
[hide="15"]If the point is in the exterior of the circle, call it $P$.
Draw the tangent segments $PA,PB$
Let $Q\in OP \cap AB$. Then $Q$ is the inverse of $P$

If the given point is inside the circle, call it $Q$.
Join the points $O,Q$
Bring a perpendicular line to $OQ$ passing through $Q$. This line intersects the circle at the points $A,B$. Bring the tangent at $A$. This tangent will intersect line $OQ$ at $P$.
Then $P$ is the inverse of $Q$.
Of course, if the given point is on the circle then it maps to itself, while the center has no image (infinity)[/hide]
[hide="17"]Let $A,B$ the two points and $m$ the line.
Of course $A,B$ must be on the same side of $m$, or else it can't be a tangent.
Let $C\in m$ be the touch point

a) if $AB\parallel m$ then bring the perpendicular bisector of $AB$. This will intersect $m$ at $C$ we are done.

b) if $AB \not\parallel m$, let $P\in AB \cap m$.
From the power of point we get $PA\cdot PB = PC^{2}$, so it's enough to find the geometric mean of $PA,PB$[/hide]
[hide="18"]Bring two parallel lines intersecting the given lines, to construct a trapezoid $ABCD$.
Join the midpoints of $AB,CD$[/hide]
[hide="19"]Assume that $\angle BAC$ is the right angle. Then the circumcenter $O$ is the midpoint of $BC$.
Let $AD$ be the altitude corresponding to the hypotenuse.
The nine-point circle passes through the midpoints of the sides, and through $A$ and $D$.
So the points $A$ and $D$ are known (the intersection between the circle and the line).
Bring a perpendicular line to $AD$ passing through $D$. Call this line $m$. This line intersects again the nine-point circle at $O$
(if it was a 90-45-45 triangle, then the points $D,O$ coincide but there's no problem)
Finally draw the circumcircle $(O,OA)$ and intersect it with the line $m$ to find the points $B,C$.[/hide]


You are all stuck in the Suburbs of Malton. It has had a severe zombie outbreak which has lead it to complete lock down.
Now after a long time the 12 of you have gathered in one building.
they are:

- Temperal
- perfect628
- 142857
- miyomiyo
- hunter34
- Peter
- pianoforte
- BoesFX
- CheeseIsGood
- Hamster1800
- abacadaea
- toadoncart

Everything seems well as you have a generator and radio.
What more can you ask for.
The Generator allows you to see well(who cares) and also powers the radio transmitter and helps you search in this building.
The radio allows you to communicate to other survivors (and tells you when you are in lynch or lose).
There is one slight problem, you barricaded the building but the only person competent enough to rebuild them was thrown out.
That means you have some serious zombie sympathizer among you.
You had a report from a neighboring building that he worked with 2 people one destroyed the radio and the other the generator.
In addition to that you hear a report form a neighboring building...
"Hello this the the *static* police department, we have a person on the loose, he has killed at least 3 people, his name is
*static* no please don't I didn't mean to no *static* (the distinct sound of blood trickling)...
Finally you know that the free running lanes to neighboring buildings have been destroyed
So you are all stuck here
this is here so that I know you read the entire post please say zed in you first post otherwise I am not sure what to do
(You know that amongst the 12 of you there is a 3 person mafia and a sk)
It is time for some serious mob lynch action
As it is true that as soon as someone dies they will become a zombie you will be unable to determine roles of the dead.
Or alignments for that matter.
Or the fact that their dead.
Well no if they start stinking you know their dead.
Unless they haven't washed.
Which all of haven't.
No really if they haven't moved they're probably dead.
Or asleep.
Nah dead.

Well good luck and don't complain.

Rules:
Each person has 1 vote they may change it at any time.
The vote must be in the form to count:
[b]Vote: Name [/b]
No communication outside of the thread unless your role allows it.
No Self Voting
I will make a deadline if I don't see progress.
If you fail to submit a night choice I will chose the worst possible choice for you.
(most will make you die and have nothing to do with your role)
That is unless you tell me you will be inactive for that time period
in that case it will be random.
I leave myself the right to change the rules whenever needed

The Game Begins with day
Kill someone already.

slight error on the player list-i replaced ajai. zed.

Sounds fun!  Zed.  Better go check on 26....

Z, not zed!

Anyway, I'm wondering - what's "sk"? As in there are 3 mafia and a sk.

sk=vigilante except not on either side. they win if everyone else is dead.

perfect what are you talking about I see no error
sk is short for serial killer.
+what perfect said

Me, uh... CheeseIsGood and, uh... toadoncart are all mafia. yeah. That's right.

EDIT: Zed.

Wait, but I'm mafia....

[b]Vote: CheeseIsGood[/b]

You are?

I think Temperal is probably the SK, since he seems to be trying to get me and Toadoncart killed. Either that, or he is mafia and trying to take two(one?) innocent(s) with him.

Zee....

Zombies on the Suburbs?

[b]Vote Peter[/b], because I can't do that in the previous game.

So will we be noticed that who is dead
Zed!

Yeah, so I'm just going to claim that CheeseIsGood, toadoncart, and I are mafia and that Temperal was trying to get us killed...

Oh also I claim that Temperal is the sk just because.

Therefore, the most obvious choice for me is to [b]unvote: CheeseIsGood[/b] and [b]Vote: Temperal[/b]

:rotfl:

[b]Vote: everybody[/b] because [i]I'm[/i] mafia and I just want to kill everybody.  But then I [b]Unvote: everybody[/b] because that would be too conspicuous.  Then I [b]Vote: miyomiyo[/b] because he just claimed mafia.  Then I [b]Unvote: miyomiyo[/b] because he's me.  Then I [b]Vote: miyomiyo[/b] for acting weird.  Then I [b]Unvote: miyomiyo[/b] because he's still me.

Hm.

And I also did not lie about my role. I was a former army soldier and indeed my role pm said that abacadaea was the group medic. Osht, he was out of first aid kits? lol.

Funny how I trusted 1=2 after 142857's role claim, then everyone turns on him on the basis of our contradicting claims :)

And I was a vigilante.  Surprise surprise.

As I suspected, Hamster was telling the truth the whole time...I was actually considering reviving abacadaea, b/c he was a doctor. Good thing I didn't...People would have been pissed with him back. (no offense)

Roles

Peter
[hide]you are one of many survivors currently located in Malton.
The entire place has been infested with zombies and under complete lockdown. Your natural instinct as a human is to fight them.
There is one slight problem. You ran off from an insane assylum to which you were sentenced for mass murder.
You really aren't insane, so you in this chaos go around killing off your fellow survivors.
Lucky for you no one in this building knows about your past.
For more flavor and details see my first post.

(You are a serial killer, each night you may send me someone you want to kill)[/hide]

CheeseisGood
[hide]you are one of many survivors currently located in Malton.
The entire place has been infested with zombies and under complete lockdown. Your natural instinct as a human is to fight them.
There is one slight problem. You are a zombie sympathiser, you with Temperal and Toadoncart are working together to help the zombies take over this building.
Each night you as a group may try to kill 1 person.
In addition to that you may communicate with these zombies.
(You are mafia with Temperal and Toadoncart, each night one of you will have to perform the kill while the 2 others may perform their night actions
Your action is to get 1 dead persons role)[/hide]

Temperal
[hide]you are one of many survivors currently located in Malton.
The entire place has been infested with zombies and under complete lockdown. Your natural instinct as a human is to fight them.
There is one slight problem. You get a very nice paycheck from your boss whenever you destroy a generator.
You and CheeseisGood and Toadoncart are working together to help rid the survivors in this building so you can destroy the generator.
Each night you as a group may try to kill 1 person.
In addition to this you want to destroy the generator to win.
(You are mafia with Temperal and Toadoncart, each night one of you will have to perform the kill while the 2 others may perform their night actions
Your action is to try and damage the generator)[/hide]

Toadoncart
[hide]you are one of many survivors currently located in Malton.
The entire place has been infested with zombies and under complete lockdown. Your natural instinct as a human is to fight them.
There is one slight problem. You get a very nice high whenever you destroy a radio.
You and CheeseisGood and Temperal are working together to help rid the survivors in this building so you can destroy the radio.
Each night you as a group may try to kill 1 person.
In addition to this you want to destroy the radio to win.
(You are mafia with Temperal and CheeseisGood, each night one of you will have to perform the kill while the 2 others may perform their night actions
Your action is to try and damage the radio)[/hide]

Perfect628
[hide]you are one of many survivors currently located in Malton.
The entire place has been infested with zombies and under complete lockdown. Your natural instinct as a human is to fight them.
There is one slight problem. You were a necrotech employee before this all happened and really can't do much to fight them.
you do have a syringe though.
(You may at one point in the game during the night revive 1 of the dead.)[/hide]

142857
[hide]you are one of many survivors currently located in Malton.
The entire place has been infested with zombies and under complete lockdown. Your natural instinct as a human is to fight them.
There is one slight problem. You are a cop. You are used to eating donuts in your car while watching who comes to a person's house
Each night you may watch one person and see everyone who visited them that night
(You are a watcher)[/hide]

Miyomiyo
[hide]you are one of many survivors currently located in Malton.
The entire place has been infested with zombies and under complete lockdown. Your natural instinct as a human is to fight them.
There is one slight problem. You are a loner. You don't do your work with others.
Each night you will kill off someone you think is guilty on your own.
(you are a vigilante)[/hide]

Hunter34
[hide]you are one of many survivors currently located in Malton.
The entire place has been infested with zombies and under complete lockdown. Your natural instinct as a human is to fight them.
There is one slight problem. You weren't any good with weapons. Instead you went to med school, where you became a doctor.
Each night you may protect 1 person from getting killed.
(you are a doctor)[/hide]

Pianoforte
[hide]you are one of many survivors currently located in Malton.
The entire place has been infested with zombies and under complete lockdown. Your natural instinct as a human is to fight them.
There is one slight problem. You were a firefighter before this outbreak. You may try to fix either the generator or radio or search for things
(Each night you may search for something or repair the generator or radio, your choice)[/hide]

BoesFX
[hide]you are one of many survivors currently located in Malton.
The entire place has been infested with zombies and under complete lockdown. Your natural instinct as a human is to fight them.
There is one slight problem. You were in the army as a scout before this madness. As a result you became good at tracking people
Each night you may choose 1 person to track who they go to.
(You are a tracker)[/hide]

Hamster18400
[hide]you are one of many survivors currently located in Malton.
The entire place has been infested with zombies and under complete lockdown. Your natural instinct as a human is to fight them.
There is one slight problem. You served in the infatry in the army. But by now you are out of ammo.
You do know that your group medic abacadea is innocent.
(You are masons with Abacadea)[/hide]

Abacadea
[hide]you are one of many survivors currently located in Malton.
The entire place has been infested with zombies and under complete lockdown. Your natural instinct as a human is to fight them.
There is one slight problem. You served as a medic in the army. But you ran out of first aid kits.
You do remember Hamster was in your unit.
(You are masons with Hamster) [/hide]

My Analysis on the game

Peter was screwed because of his popularity
he had killed Abacadea

The mafia had a huge advantage going into day 2
all 3 were alive and only 142857 had any clue about them.
Then the game became inactive and the mafia just died off leaving Donkeykong
I tried to help him by giving him info about the theme

Miyomiyo was a vig who actually couldn't kill, this is determined by the theme

142857 had chosen the target getting killed by the mafia the 1st 2 nights
also a watcher can't see if one targets themself (1=2 did that)

BoesFx  tracked the Sk n1, the mafia extra night 2 and then the useless townie and finally 142857

Piano's role was originally created to balance the game if it became mafia favored but due to inactivity became useless
he could find items that would allow him to do stuff

and the doctor was a selfish son of a gun (almost always them self)
and the reason I allowed this is no one would know who is alive

and the mafia had another way to win
destroy generator...

and the 2 masons were just masons

anyways I believe that the town deserved the win
the power roles were well played while the mafia played there roles badly

in addition the theme was Urbandead
I was hoping DonkeyKong would have forced a name claim once I gave him the site but that didn't work out
he didn't know about the roles having special names
Cheese if he survived would have been able to figure that out I think

the radio was there to tell you when it was Lynch or Lose
due to the town play that was not needed.
The generator affected night actions, and if both were eliminated mafia wins

That basically means if mafia can get a deadlock
 with the town they can instead of killing someone
and providing evidence
win the other way
it just takes to long
and you need those specific people to be alive

[quote="perfect628"]Woah, this game was way more thought out on the mod's part than the actual play would suggest...Thanks, Ars, this was a very interesting game.[/quote]
Yes, I agree. I am glad that I participated this game.

@Ars: Sorry, I thought the site you gave me was a game site, and it looked like it. I don't play online games..

[quote="Hamster1800"]ROFL!!!

[quote="Hamster1800 on page 1"]
Yeah, so I'm just going to claim that CheeseIsGood, toadoncart, and I are mafia and that Temperal was trying to get us killed...

Oh also I claim that Temperal is the sk just because.
[/quote]

The three mafia, right there :P[/quote]

Yep, that's why I got pissed at Temperal.. a mistake on my part.  :P

And I did recognize the theme, and would have made a flavored claim if I wasn't lynched before I had a chance to defend myself.

The most interesting thing is that there is a tracher alongwith a watcher, and a doctor along with a group madic.

It makes people don't trut me!

Why do I always post using my dad's account?! :(

really interesting game, it's a shame the medics and the doctors were killed off... :P

You guys do know that the medic had no power, right?

Anyways, great setup.  However, I'm not sure what role the radio played.

and donkeykong I gave you the site with the starting roles/desciptions
so that's all you needed to see
you didn't have to play
and if both were destroyed mafia wins

radio tells the town if its lynch or lose
and I tried to make so the role names would imply a different role then the actual role
cop-watcher
scout-tracker
medic-mason
firefighter-townie

and that's why I kind of gave up once the game became inactive
I thought it was to complicated

[quote="CheeseIsGood"]If the either the radio or generator was destroyed by toadoncart/Temperal, then I think the Mafia would have won.[/quote]
I don't know why, but I really haven't gave a hard thought on that.. My bad.
It's mainly my fault that towns won.. The scums might have had the upper hand all the way through if we destroy the generator, which I was capable of.. Sorry.


50 with two distinctions (problems 4 and 5 of round 2)

wait...they're not due yet, correct?

-interesting_move

The third one isn't, no. It's due on January 4th.

I didnt get my round 2 back, but my friend did, so i know its not something to do with the region. Do they take longer when the solutions are faxed in?

i don't think so, maybe just messed up in the mail carrier or something, you will eventually get it, if you dont, they have it, (you get the copy) so you can ask for it..

i faxed mine and got the results a couple days ago.

hmm, is anyone going to give their scores or answer the poll?

the graders are so careless, they don't carefully read your solution at all.

http://68.38.137.202/round2.pdf

Look at question #5.
All they did was see the 13 at the end (which was not even my answer to the question) and gave a 4/5, saying the correct answer was 14, with the comment "You missed an edge. The answer is 14."

Unbelievable. Even when I explicitly stated in the first sentence, we show that the minimum is FOURTEEN. And I provided one of the two possible correct polyhedrons (up to an isomorphism in the polyhedral graph) WITH FOURTEEN edges.

Now, I guess this is forgiveable since the graders are not paid, but that is NO excuse to be negligent, even in volunteer work!

24 + 23 = 47 with one commendation.

At least they only took off one point.  I got a 3/5 on the last one  (because I got it incorrect...)

yeah. this might wreck your chances for getting AIME, fairyofwind...;)

the graders are tottaly careless, "why can't x be this" i clearly explained it on my second page, oh well.

Also my friend for round 1 sent in a wrong circle (for the triangular number one) and got a 5/5.

insaneness... i totally b.s-ed number 5 and got a commended on it... wow...
my first round is a little screwed up tho (19/25)... owel, i guess a 43/50 for a frosh is good enough...

25 + 23 = 48 total, 2 + 1 = 3 commended(#3 on round 2)


function
trigonometry
analytic geometry
graphing lines
slope

Let x be a positive real number. Find the maximum possible value of
$ \frac{x^{2}+2-\sqrt{x^{4}+4}}{x}$

$ \frac{x^{2}+2-\sqrt{x^{4}+4}}{x}=x+\frac{2}{x}-\sqrt{x^{2}+\frac{4}{x^{2}}}$, let $ x+\frac{2}{x}=t$, by A.M.-G.M., $ t\geq 2\sqrt{2}$.

$ \therefore \frac{x^{2}+2-\sqrt{x^{4}+4}}{x}=t-\sqrt{t^{2}-4}=\frac{4}{t+\sqrt{t^{2}-4}}$, since $ t$ and $ t+\sqrt{t^{2}-4}$ are increasing functions for $ t\geq2\sqrt{2}$, the maximum value is $ 2(\sqrt{2}-1)$ when $ t=2\sqrt{2}\Longleftrightarrow x=\sqrt{2}$.

Certainly, you meant to write that the maximum is $ 2(\sqrt{2}-1)$.

Alternative solution:

From previous my post, let $ t=\frac{2}{\cos \theta}$, $ t\geq 2\sqrt{2}\Longleftrightarrow 0<\cos \theta \leq \frac{1}{\sqrt{2}}$. It is sufficient to consider the case of $ 0<\theta <\frac{\pi}{2}$, so $ \frac{\pi}{4}\leq \theta < \frac{\pi}{2}$.

Thus $ t-\sqrt{t^{2}-4}=\frac{2}{\cos \theta}-\sqrt{\tan^{2}\theta}=-2\cdot \frac{\sin \theta-1}{\cos \theta-0}\ \because \frac{\pi}{4}\leq \theta < \frac{\pi}{2}$. Now let the points $ A(0,\ 1),\ P(\cos \theta ,\ \sin \theta)$ for $ \frac{\pi}{4}\leq \theta \leq \frac{\pi}{2}$,

$ \frac{\sin \theta-1}{\cos \theta-0}$ means the slope of the line $ AP$. Since the slope is increasing for $ \frac{\pi}{4}\leq \theta < \frac{\pi}{2}$,

we have the minimum value when $ \theta =\frac{\pi}{4}$. Thus desired maximum value is $ -2\cdot \frac{\sin \frac{\pi}{4}-1}{\cos \frac{\pi}{4}}=2(\sqrt{2}-1)$, when $ \theta =\frac{\pi}{4}$.


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umm just wondering,
what does this have to do with the AMC?
 :huh:

It doesn't. That's why it's getting locked.


floor function
ceiling function
number theory unsolved
number theory

Prove that for any positive integer n, there exists an integer k which can be expressed as a sum of two
squares and such that $n \leq k \leq n+3n^{\frac{1}{4}}$


Express 0.045(repeating) [I don't know how to make the repeating bar thing on this] as a common fraction.

How would you do this?

Lets say $ 0.0\overline{45} \equal{} x$
Then $ 45.\overline{045} \equal{} 1000x$
Subtract both equations and you get $ 999x \equal{} 45$
So $ x \equal{} \frac {45}{999} \equal{} \frac {5}{111}$

If you wanted only $ 45$ repeating its similar

Lets say $ 0.\overline{045} \equal{} y$
Then $ 0.\overline{45} \equal{} 10y$
And $ 45.\overline{45} \equal{} 1000y$
Now subtracting last 2 you get $ 990y \equal{} 45$
And fraction is $ \frac {1}{22}$

Oh! I get it. Thanks.

(And by the way, I like your quote.)


geometry
perimeter
analytic geometry
integration
calculus
trigonometry
function

There is an enclosed field with radius R. A dog is tied with a rope of length L to a point on the perimeter of the field. Given R, find a closed form for L, such that the dog can reach exactly one half of the field.

I'm putting this in algebra because the geometry part of it is pretty trivial.

this should still be in geometry! center the circle at the origin $ O$, and let the rope be tied to $ A(R,0)$. you know how to find the $ x$ coordinate of the point of intersection in the first quadrant of circles $ C(A;L)$ and $ C(O;R)$. (geometry of it is trivial ;)) denote it by $ a$. then $ L$ will be the solution to

$ \int_{R\minus{}L}^{a}\sqrt{L^2\minus{}(x\minus{}R)^2} dx \plus{}\int_{a}^{R}\sqrt{R^2\minus{}x^2} dx \equal{} \frac{\pi R^2}{4}$

ugly, eh?

Why was this moved away from olympiad? I believe there is a non-calculus solution. Can a moderator please move it back to olympiad?

i'd love to see a non-calculus solution, if you could ever show me one

[hide="The area expression is elementary."] In particular, let $ C$ be the center of the circle, $ P$ be the point at which the dog is tied, and $ A, B$ be points on the circle such that $ PA \equal{} PB \equal{} L$.  Divide the area that the dog can reach by the line $ AB$, and suppose that $ P$ is above $ AB$.

The area above $ AB$ is given by $ \frac{1}{2} R^2 (4\theta) \minus{} [ABC]$, where $ \theta \equal{} \frac{1}{2} \angle ACP \equal{} \arccos \frac{L}{2R}$ and $ [ABC] \equal{} \frac{1}{2} R^2 \sin 2 \theta$.

The area below $ AB$ is given by $ \frac{1}{2} L^2 (2(\frac{\pi}{2} \minus{} \theta)) \minus{} [ABP]$, where $ [ABP] \equal{} \frac{1}{2} L^2 \sin (\pi \minus{}2 \theta)$. [/hide]

as I said, there are non-calc ways to do this problem. The problem is expressing them into a closed form for L. Mods, please move this back to olympiad section

What makes you think there is a closed form? Transcendental equations like what t0rajir0u wrote almost never have closed form solutions.

[quote="ml2"]Because many of the variables he had in there can be expressed in terms of L and R. I got and expression just based on those, but couldn't exactly simplify it.[/quote]

All of the "variables" in the expression can be expressed in terms of $ L$ and $ R$.  The problem is that for a given $ R$ there is no expression in terms of elementary functions for $ L$.

For example, the equation $ xe^x \equal{} 3$ has no solution in terms of elementary functions.


This problem involves areas of math not generally used in competition (advanced combinatorial game theory).  Familiarity with the game Nim will be helpful.

There are 2 players.  There is a pile of 100 counters, of which the first player to move may take up to and including 50 on the first turn.  The winner is the player who takes the last counter.  The number of counter which can be taken on a given turn is twice the number the opponent took on the previous move.  (That is, if I took 3, you could take 6.  If you chose to take only 1, I would be limited to 2, etc.)

The problem is multipart, each with a different level of difficulty.
[list]
(1) Determine, with proof, the optimal opening move, or set of such moves, if several are equivalent.
(2) Determine a general method for determining the optimal move in any situation.
(3) Generalize the method found in the previous part to any integer scalar.  (That is, any multiplier for determining the number of counters allowed to be taken given the number taken on the previous turn.)
[/list]

You may solve any or all, though I do not forsee anyone being able to solve the third part without solving the second, or the second without solving the first.  Note also that even the first part is quite difficult (barring brute force, which is forbidden), and the second and third are exceptionally hard.  Finally, yes, I know that the solution to this problem is available online, but please to not look at it.  The problem is very interesting to solve (though it may take some time).

To clarify:

the MAXIMUM number of counters a player can take is double the number taken by the other player on the previous move, but the player can take any lesser number (though not 0)

[color=cyan]Well, if they could take 0, it would lead to a rather interesting situation . . .[/color]

I have a solution to parts 1 and 2, the gist of which follows in spoiler:

[hide]

We can represent any position in this game as an ordered pair (x, k) such that x is a nonnegative integer representing the number of counters left and k is a positive even integer representing the maximum number of coins that may be taken at the next move.  We define winning and losing positions as usual (you want to move TO winning positions), so that (0, k) is a winning position for any k (hereafter "any k" will mean "any positive even k", as those are the only possible k's), (1, k) and (2, k) are all losing positions, and so on.  I claim the following algorithm determines if a position is winning or losing:

Algorithm:

If k>=x, then (x, k) is losing.

If x>k and x is a Fibonacci number, then (x, k) is winning.

If x>k and x is not a Fibonacci number:

  Let f be the largest Fibonacci number less than x; then (x, k) and (x-f, k) are simultaneously winning or losing.

This, if true, would of course allow us to determine the 'best' move from any given position (a, b): simply check all of the b possibilities of moves and see if any are winning positions.

To prove this, we need to define the following:

Call a number s 'special' if (s, k) is winning for all k<s.

Say that a number n 'acts as a zero' for a M of numbers if for any m in M, (m, k) is winning or losing with (m-n, k).

We then prove the following:

Every special number s acts as a zero for all integers x such that s<x<3s/2.

Now we wish to prove that the set of special numbers is the same as the set of (positive) Fibonacci numbers, and that every special number F_n (the nth Fibonacci number) acts as a zero for all x such that F_n < x < F_(n+1).

We verify this up to a certain point (say n=8); ie, {special numbers} = {Fibonacci numbers} for all numbers <=8 and that special numbers act as a zero for the required set for all numbers <8.  The idea is to then show by induction that the the current highest-known special number acts as a zero for the required set and that the next special number is also the next Fibonacci number.

Letting the largest special number known be F_n, we know from above that F_n acts as a zero for (F_n + F_(n-2)) since F_(n-2)<(F_n)/2.  Furthermore, we can argue that since from the hypothesis F_(n-2) acts as a zero for x such that F_(n-2)<x<F_(n-1) we also have that (F_n + F(n-2)) acts as a zero for all x such that (F_n + F_(n-2))<x<F_(n+1).   (This is a bit tricky but can be done.)  Then we show that (given all proven so far) F_(n+1) is the next special number, and we're done.

Yes, I know I have not proven all of my assertions but I hope that the proofs are all either apparent or at least make intuitive sense with some work.  I have not proven them here because the proofs (as I now have them stored in my head) would take a LONG time to write out.

I would be very interested in a shorter or nicer solution than this, or even some quick proofs of the statements I have left not fully proven here.  This post is quite long enough as it is...

[/hide]

I laude you, setofallsets.  you might want to try making a completely rigorous proof, as it has a little unexpected difficulty.

[hide]

now that you know the fibbonacci relationship, look at identity nim (pretty easy to analyze), then try to generalize to any integer scalar.

[/hide]

if you are able to properly extend the pattern in identity and doubling nim to tripling nim, you will likely have found the general form.  beware, however, that there is a way to generate the series for tripling nim that IS NOT a generalization of the previous rule, do not fall into that trap.

look at it some more, in a few days i will post a link to the preprint on this problem, then a full explanation of the underlying concept (which is unpublished).

Dr. Reiter's preprint on this problem (beware, it is dense):
[url]http://www.math.uncc.edu/~hbreiter/dynamic.pdf[/url]

His paper doesnt include the generalization onto any base, but it does discuss the general concept of representing the number of counters as a number in a semibinary base, and decreasing the number of 1s.   I will give the full explanation as soon as I have some spare time.

I have looked through the paper, and I was rather disappointed to find that it did not clear up my major questions about our specific cases: namely, how to generate the initial string of what I called 'special numbers' before the recursion takes over (in terms of n (n being our integer scalar)), whether we can write that s_i = s_(i-1) + s_(i-k) for some fixed k which is a function of n, and if so, how to compute k in terms of n.  I was able to (using essentially the same ideas as before) show that the 'special numbers' satisfy the property that s_i = s_(i-1) + s_m, where s_m is the minimal special number such that n(s_m) > s_i, but couldn't get much farther.
I am guessing that there is such a k that is fixed given n, but I could figure out no formula for k in terms of n.  Nor could I solve the larger problem of finding the initial sequence of special numbers before the recursion kicks in.  (I did find an algorithm simpler than just checking positions one by one, but I could not get a way to find them in terms of n.)

I'd be quite interested to know the answers to these questions.


Inequality
Sequence
inequality system
algebra
IMO Shortlist

Suppose that $ n \geq 2$ and $ x_1, x_2, \ldots, x_n$ are real numbers between 0 and 1 (inclusive). Prove that for some index $ i$ between $ 1$ and $ n \minus{} 1$ the
inequality

\[ x_i (1 \minus{} x_{i\plus{}1}) \geq \frac{1}{4} x_1 (1 \minus{} x_{n})\]

If
Suppose that $ x_{i}(1\minus{}x_{i\plus{}1})<\frac{1}{4}x_{1}(1\minus{}x_{n})\leq \frac{1}{4}$ is true for all $ i\equal{}1,n\minus{}1$

Then for $ i\equal{}1$ we get:

$ x_2>\frac{3}{4}>\frac{1}{2}$

Else, is obvious that if $ x_i>\frac{1}{2}$ then $ x_{i\plus{}1}>\frac{1}{2}$

because $ x_{i}(1\minus{}x_{i\plus{}1})<\frac{1}{4}$.

Then $ x_{n\minus{}1}>\frac{1}{2}$

(If $ x_n\equal{}1$ the problem is solved alone)

So $ x_{n\minus{}1}(1\minus{}x_n)<\frac{1}{4}x_{1}(1\minus{}x_{n})$

$ \implies  x_{n\minus{}1}<\frac{1}{4}$    That is a contradiction.


Ok, so I have a project for my programming class that requires the use of an ellipse.  However, I need it in polar form.

Everyone knows that the equation of an ellipse in cartesians is:
$\frac{(x - h)^{2}}{a^{2}} + \frac{(y - k)^{2}}{b^{2}} = 1$
Where $(h,k)$ is the center of the ellipse, $a$ is $\frac{1}{2}$ the length of the major axis, and $b$ is $\frac{1}{2}$ the length of the minor axis.

Upon converting this to polars I ran into much difficulty and a [b]very[/b] ugly answer:

For simplification, let $M = \cos^{2}{\theta}*b^{2} + \sin^{2}{\theta}*a^{2}$
                            let $N = -2b^{2}h\cos{\theta} - 2a^{2}k\sin{\theta}$
                            let $P = b^{2}h^{2} + a^{2}k^{2} - a^{2}b^{2}$

$r = \frac{-N \pm \sqrt{N^{2} - 4MP}}{2M}$

Not only is this a very ugly and cumbersome result, but I can't figure out how to solve it for $\theta$, which is precisely why I need to use polar graphing.

Isn't there an easier formula that [b]can[/b] be solved for $\theta$?

Thanks.

i can tell you this:
the polar equation for the eillips at $(0,0)$, where the equation in cartesian is: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is :
$r^2=\frac{b^2}{1-e^2\cos^2{\theta}}$ where $e$ is the ecentrisity of the ellipse,
i can prove it if you want to...


$ a_1,\dots,a_n > 0$ satisfy $ \sum\frac {1}{a_i} \equal{} 1$
prove that for $ z_1,\dots,z_n\in\mathbb{C}$ we have $ |z_1 \plus{} \dots \plus{} z_n|^2 \le a_1|z_1|^2 \plus{} \dots \plus{} a_n|z_n|^2$


inequalities
function
logarithms
real analysis
real analysis unsolved

Prove that:

$ a^{a}\plus{}b^{b}>\equal{}a^{b}\plus{}b^{a},a>\equal{}b>0.$

more difficult case is if :
$ 1 > a\geq b$
we have
$ \frac {b^{a}}{a^{b} \plus{} b^{a}}(\frac {a}{b})^{a} \plus{} \frac {a^{b}}{a^{b} \plus{} b^{a}}(\frac {b}{a})^{b} \geq 1$
use AM-GM (if $ \alpha_{1} \plus{} \alpha_{2} \equal{} \alpha$ then next inequality is true :$ \frac {\alpha_{1}x_{1} \plus{} \alpha_{2}x_{2}}{\alpha}$
$ \geq$$ (x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}})^{\frac {1}{\alpha}}$)
therefore we have
$ \frac {b^{a}}{a^{b} \plus{} b^{a}}(\frac {a}{b})^{a} \plus{} \frac {a^{b}}{a^{b} \plus{} b^{a}}(\frac {b}{a})^{b}\geq (\frac {a}{b})^{ab^{a} \minus{} ba^{b}}\geq 1$
or
$ ab^{a} \minus{} ba^{b}\geq 0$
I note that the function $ f(x) \equal{} \frac {\ln x}{1 \minus{} x}$ is increasing in interval $ x\in (0,1)$ therefore we have
$ \frac {\ln a}{1 \minus{} a}\geq \frac {\ln b}{1 \minus{} b}$or
$ ab^{a} \minus{} ba^{b}\geq 0$.
Done


geometry
incenter
geometric transformation
homothety
ratio
geometry unsolved

Let $ ABC$ be a triangle with 4 Feurbach points $ F, F_{a}, F_{b}, F_{c}$. Then the Steiner line of $ F$ wrt $ F_{a}F_{b}F_{c}$ passes through the incenter of similitude of the incircle and the nine-point circle.

Note: The Steiner line of $ F$ wrt $ F_aF_bF_c$ is the image of the Simson line under the homothety centered at $ F$ with ratio 2.

This problem is pretty hard! :)  (hopefully, someone has already had some germ of ideas).