vector
analytic geometry
linear algebra
matrix
invariant
combinatorics unsolved
combinatorics

Given a pair of vectors (a, b) in the plane, a move consists of choosing a nonzero integer k and changing (a,b) to either

1)     (a + 2kb, b)
or
2)     (a, b + 2ka)

A game consists of a finite sequence of moves alternating between moves of type 1 and type 2.

a.) If we start with ((1,0), (0,1)), is it possible to obtain ((1,0),  (2,1)) during a game which starts with a move of type 1?
b.) Find all pairs ((u,v),(w,x)) which can be obtained during a game with initial pair ((1,0), (0,1)), where the first move can be of any type.

The answer to the first question is negative:

If the vectors are $(x,y)$ and $(z,t)$, then we can regard the first and second transformations as being the matrices $\left(\begin{array}{cc}1&2\\0&1\end{array}\right)$ and $\left(\begin{array}{cc}1&0\\2&1\end{array}\right)$ respectively, acting on the matrices $\left(\begin{array}{cc}x&y\\z&t\end{array}\right)$ by multiplication to the left.

Basicly, the first question is whether there is a sequence of $A$'s and $B$'s s.t. $\ldots BA=B$. Assume there is such a sequence $M=AB\ldots BA=B$ or $M=BA\ldots BA=B$. We can think of the following action of the matricesd on the points of the plane: if we have a point $p=(x,y)^t$ ($x,y$ are the coordinates, but we regard the point as being a column vector), then a matrix $C$ takes this point to the point $Cp$.

Now, it's clear that $B$ fixes the point $p=(0,1)^t$, for example, but, on the other hand, any combination $\ldots BA$ takes this point strictly to the right and up, so there's no way such a combination could fix $p$. This means that no such combination can be equal to $B$.

As for the second question, I don't really know what to say. How should we classify those pairs of vectors, exactly? :? All I can say is that they are products of $A$'s and $B$'s s.t. the $A$'s and the $B$'s alternate.

The reduction of the pair of vectors modulo 2 is invariant.
The (oriented) lattice generated by the vectors is also an invariant.
There are no other invariants, I think.


Why did you put it to the Open Questions?
Finally, it is an excercise only ;)
Answer is $2^{n-1}(n+2)$, if the first row contains numbers $1,2,...,n+1$.

I think they're valid for nationals...(some of them are easy but...  :D )

Please stop posting these exercises in the Open Questions sections  ;)
These are not open aren't they?

They have to be in the Unsolved Problems section (if you don't know how to solve them) or in the Proposed and Own Problems section (if you know).

Pierre.


geometry
parallelogram
ratio
geometry solved

let ABC be a triangle with a point O in its interior.
 let BO intersect AC at M
let CO meet AB at N
        construct parallelograms OMEN and OBFC . prove
i/ the points A,E,F are collinear
ii/ AE/AF = AM*AN/AB*AC =OM*ON/OB*OC

I have a ugly soln for (i), i'm sure others we'll find better ones

Let $S$ s.t. $COMS$ is a parallelogram.

By the transerval theorem it suffices to prove that $\frac{AC \cdot SM}{SE} + \frac{CM \cdot SA}{SA} + \frac{MA \cdot SC}{SF}=0$.

Use area ratios: $\frac{CO}{CN}=\frac{K_{ABC}-K_{AOB}}{K_{ABC}}$ and $\frac{MO}{MB}=\frac{K_{AOC}}{K_{ABC}}$.

We also have $SM=CO$, $SE=CN$, $SC=MO$, $SF=MB$, so the condition is equivalent to $\frac{AM}{AC}=\frac{K_{AOB}}{K_{ABC}-K_{AOC}}$.

it is very easy to prove that $\frac{AM}{MC}=\frac{K_{AOB}}{K_{BOC}}$ :)

$\blacksquare$

I'm awaiting for a better solution (and also a confirmation for this one)

Your solution to [b](i)[/b] is correct, Perfect_radio. Here is my solution to the problem.

[color=blue][b]Problem.[/b] Let ABC be a triangle and O a point in its plane. Let the lines BO and CO intersect the lines CA and AB at the points M and N, respectively. Let the parallels to the lines CN and BM through the points M and N intersect each other at E, and let the parallels to the lines CN and BM through the points B and C intersect each other at F.

[b](i)[/b] Prove that the points A, E, F are collinear.
[b](ii)[/b] Show that $\frac{AE}{AF}=\frac{AM\cdot AN}{AB\cdot AC}=\frac{OM\cdot ON}{OB\cdot OC}$.[/color]

[i]Solution.[/i] [b](i)[/b] We work on the projective plane. Let the lines BM and CN intersect the line at infinity at the points $M_{\infty}$ and $N_{\infty}$, respectively. Then, a line through the point $M_{\infty}$ is simply a parallel to the line BM, and a line through the point $N_{\infty}$ is simply a parallel to the line CN. Thus, the point $MN_{\infty}\cap M_{\infty}N$ is the point of intersection of the parallels to the lines CN and BM through the points M and N, i. e. the point E, and the point $M_{\infty}C\cap BN_{\infty}$ is the point of intersection of the parallels to the lines BM and CN through the points C and B, i. e. the point F.

Now, applying the Pappos-Pascal theorem to the collinear point triples $\left(B;\;M;\;M_{\infty}\right)$ and $\left(C;\;N;\;N_{\infty}\right)$, we see that the points $BN\cap MC$, $MN_{\infty}\cap M_{\infty}N$ and $M_{\infty}C\cap BN_{\infty}$ are collinear. Since $BN\cap MC=A$, $MN_{\infty}\cap M_{\infty}N=E$ and $M_{\infty}C\cap BN_{\infty}=F$, it follows that the points A, E and F are collinear, and thus problem [b](i)[/b] is solved.

[b](ii)[/b] Since NE || BM and CF || BM, we have NE || CF. Thus, if the line CF intersects the line AB at a point R, then, by Thales (using the collinearity of the points A, E, F), we have $\frac{AE}{AF}=\frac{AN}{AR}$. But $\frac{AN}{AR}=\frac{AN}{AB}\cdot\frac{AB}{AR}$, and CF || BM yields by Thales $\frac{AB}{AR}=\frac{AM}{AC}$. Hence, $\frac{AE}{AF}=\frac{AN}{AR}=\frac{AN}{AB}\cdot\frac{AB}{AR}=\frac{AN}{AB}\cdot\frac{AM}{AC}=\frac{AM\cdot AN}{AB\cdot AC}$.

On the other hand, since NE || BM and ME || CN, what is equivalent to NE || OM and ME || ON, the quadrilateral OMEN is a parallelogram, and thus NE = OM. Similarly, the quadrilateral OBFC is a parallelogram, and thus BF = OC. Now, from NE || CF, we infer by Thales that $\frac{AE}{AF}=\frac{NE}{RF}$. But $\frac{NE}{RF}=\frac{NE}{BF}\cdot\frac{BF}{RF}$, and thus, using NE = OM and BF = OC, we get $\frac{AE}{AF}=\frac{NE}{RF}=\frac{NE}{BF}\cdot\frac{BF}{RF}=\frac{OM}{OC}\cdot\frac{BF}{RF}$. Now, we have RB || BN (trivial), BF || NO (this is equivalent to BF || CN) and FR || OB (this is equivalent to CF || BM). Hence, the two triangles RBF and BNO are homothetic, and thus the ratios of corresponding sides must be equal: $\frac{BF}{RF}=\frac{ON}{OB}$. Hence, $\frac{AE}{AF}=\frac{OM}{OC}\cdot\frac{BF}{RF}=\frac{OM}{OC}\cdot\frac{ON}{OB}=\frac{OM\cdot ON}{OB\cdot OC}$.

Altogether, we get $\frac{AE}{AF}=\frac{AM\cdot AN}{AB\cdot AC}=\frac{OM\cdot ON}{OB\cdot OC}$, and problem [b](ii)[/b] is solved.

  darij

I think that  this prolem is very nice and the solution is ,too


Let $n$ be a positive integer. Consider the sum $x_1y_1 + x_2y_2 +\cdots + x_ny_n$, where that values of the variables $x_1, x_2,\ldots, x_n, y_1, y_2,\ldots, y_n$ are either 0 or 1.

Let $I(n)$ be the number of 2$n$-tuples $(x_1, x_2,\ldots, x_n, y_1, y_2,\ldots, y_n)$ such that the sum of the number is odd, and let $P(n)$ be the number of 2$n$-tuples  $(x_1, x_2,\ldots, x_n, y_1, y_2,\ldots, y_n)$ such that the sum is an even number. Show that: \[ \frac{P(n)}{I(n)}=\frac{2^n+1}{2^n-1}  \]

I put a solution here:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84880[/url]

also we can solve it using recurrences $P(n)=3P(n-1)+I(n-1)$ and $I(n)=3I(n-1)+P(n-1)$ and this fact that $P(n)+I(n)=2^{2n}$.

[hide= Solution]
We proceed by induction. Call $S_n$ the sum defined above. Suppose $n=1$ so $S_1=x_1y_1$ from this, we get that the pairs $(x_1,y_1)$ that make $S_1$ even are ${(0,0), (0,1), (1,0)}$ and the one that makes it odd is $(1,1)$ thus, $P(1)=3=2^1+1$ and $I(1)=1=2^1-1$, so its true for $n=1$. Suppose it holds for $n=k$. Note that $S_{k+1}=S_k +x_{k+1}y_{k+1}$, from this we get $P(k+1)=3P(k)+I(k)$ since there are three options of $x_{k+1}y_{k+1}$ that make $ S_{k+1}$ even provided  that $S_k$ is even and one option of the pair that makes it even provided that $S_k$ is odd. Using the same argument we get $I(k+1)=P(k)+3I(k)$. From the induction hypothesis we get,
\[\frac{P(k+1)}{I(k+1)}=\frac{3P(k)+I(k)}{P(k)+3I(k)}=\frac{(3(\frac{2^k+1}{2^k-1})+1)I(k)}{(\frac{2^k+1}{2^k-1}+3)I(k)}\]

With a little bit of algebra the result follows for $k+1$ and we're done.
[\hide]


Let $f:R\longrightarrow R$ then show that:
$f(x+y+xy)=f(x)+f(y)+f(xy)$ if and only if $f(x+y)=f(x)+f(y)$.

[quote="xxreddevilzxx"]would proving the converse work?

[hide="proof attempt"]
Proof: prove that if $\displaystyle f(x+y) = f(x) + f(y)$, then
$\displaystyle f(x+y+xy) = f(x) + f(y) + f(xy)$

well... $\displaystyle f(x+y+xy)=f((x+y)+xy) = f(x+y) + f(xy) = f(x) + f(y) + f(xy)$
[/hide][/quote]

u proved the sufficiency of the condition..

Now prove the necessity of the same

that is if f(x+y + xy) = f(x) + f(y) + f(xy),

then f(x+y) = f(x) + f(y)

$f(x+y+xy)=f((x+y)+xy+0)=f(x+y)+f(xy)+f(0)$

Now $f(0+0+0)=3f(0)\Rightarrow f(0)=0\Rightarrow f(x+y+xy)=f(x+y)+f(xy)$

But then $f(x+y)+f(xy)=f(x+y+xy)=f(x)+f(y)+f(xy)\Rightarrow f(x+y)=f(x)+f(y)$

[quote="blahblahblah"]$f(x+y+xy)=f((x+y)+xy+0)=f(x+y)+f(xy)+f(0)$

Now $f(0+0+0)=3f(0)\Rightarrow f(0)=0\Rightarrow f(x+y+xy)=f(x+y)+f(xy)$

But then $f(x+y)+f(xy)=f(x+y+xy)=f(x)+f(y)=f(xy)\Rightarrow f(x+y)=f(x)+f(y)$[/quote]

Very nice solution blahblahblah.Mine is 3 times longer.
I think you have a typo in

[quote="blahblahblah"]
$...f(x+y+xy)=f(x)+f(y)=f(xy)$[/quote]

[quote="blahblahblah"]$f(x+y+xy)=f((x+y)+xy+0)=f(x+y)+f(xy)+f(0)$

Now $f(0+0+0)=3f(0)\Rightarrow f(0)=0\Rightarrow f(x+y+xy)=f(x+y)+f(xy)$

But then $f(x+y)+f(xy)=f(x+y+xy)=f(x)+f(y)=f(xy)\Rightarrow f(x+y)=f(x)+f(y)$[/quote]

why is f((x+y)+xy+0) = f(x+y)+f(xy)+f(0) ??

We only know that  f(x+y+xy) = f(x) +f(y) +f(xy)

Hmm, I think it depends on interpretation of what the functional equation actually says.

I was so busy and charmed that I didn't read the solution properly.
 :blush:

i have the same question as santosh_blr... how did you get $ f((x+y)+xy+0) = f(x+y)+f(xy)+f(0) $??

I don't really like functional equations, but I like this one :). I remember it from some preparation camp many years ago, and I was really happy when I solved it then :).

Make $y=-1$ to see that $f(x)+f(-x)=0$. After that, turn $y$ into $-y$. Adding what we get to the original functional equation, we find $f(x+a)+f(x-a)=2f(x)$, where $a=y(x+1)$. Actually, this shows that $f(x+a)+f(x-a)=2f(x)\ (*)$ for all $x,a\in\mathbb R$ (if $x=-1$, then that $a$ becomes $0$, but this is no problem, since we can look at $f(1-a)+f(1+a)=2f(1)$ and put a $-$ in front of the whole thing, thus getting $f(-1+a)+f(-1-a)=2f(-1)$). Now, making $a=x$ gives $f(2x)=2f(x)\ (**)$, and if we replace $2f(x)$ in $(*)$ according to $(**)$ we get the result.


Find the value(s) of $ x$ such that $ 8xy\minus{}12y\plus{}2x\minus{}3\equal{}0$ is true for all values of $ y$.

$ \textbf{(A)}\ \frac{2}{3} \qquad
\textbf{(B)}\ \frac{3}{2}\text{ or }\minus{}\frac{1}{4} \qquad
\textbf{(C)}\ \minus{}\frac{2}{3}\text{ or }\minus{}\frac{1}{4} \qquad
\textbf{(D)}\ \frac{3}{2} \qquad
\textbf{(E)}\ \minus{}\frac{3}{2}\text{ or }\minus{}\frac{1}{4}$

[hide]$ 8xy \minus{} 12y \plus{} 2x \minus{} 3 \equal{} (2x \minus{} 3)(4y \plus{} 1)$

It follows that:

$ x \equal{} 3/2$[/hide]


[quote="Valentin Vornicu"]Did anyone find a closed result for this sum?[/quote]

Apparently the proposers have not got a closed-form solution either, have a look [url=http://www.math.ca/Competitions/CMO/solutions/sol_2006.pdf]here (pdf).[/url]

But the problem as stated here is the "intended" version ... I've seen it (or equivalent things) solved elsewhere on the forum, but I can't find them right now.

Edit: for example, here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=467216#467216

[quote="N.T.TUAN"]Let $ f(n,k)$ be the number of ways of distributing $ k$ candies to $ n$ children so that each child receives at most $ 2$ candies. For example $ f(3,7) \equal{} 0,f(3,6) \equal{} 1,f(3,4) \equal{} 6$. Determine the value of $ f(2006,1) \plus{} f(2006,4) \plus{} \ldots \plus{} f(2006,1000) \plus{} f(2006,1003) \plus{} \ldots \plus{} f(2006,4012)$.[/quote]

can't we write it as the equation x1+x2+x3+...xn =k , the number of solutions of that is for each xi>=0 is C(n+k-1,k), each 0<=xi<=2 ,

using inclusion exclusion we can take out the solutions x1>=3,x2>=3 and subtract it from that. It is tedious.

Do you find anything wrong with that.

; We are basically looking for the sum of coefficients of $ x, x^4 $ .... x to the power of  4012  in $ (1+x+x^2) $ to the power of 2006. Can't this sum be extracted by using the complex cube roots of unity?
Correct me if I am wrong.

^Yes this problem can be solved by generating functions. Note that $f(n,k)$ is the number of solutions $(x_1,x_2,\cdots,x_n)$ to $x_1+x_2+\cdots+x_n=k$ where $x_i \in \{0,1,2\}$ for $1 \le i \le n$. Equivalently,
$f(n,k)$ is the coefficient of $x^k$ in $A(x)=(1+x+x^2)^n$. Let

$A(x)=\sum_{k=0}^{\infty}f(n,k)x^k$

Let $\omega \neq 1$ be a third root of unity. Note that

$\frac{1}{3}[A(x)+\omega^2A(\omega x)+\omega A(\omega^2 x)]=f(n,1)x+f(n,4)x^4+f(n,7)x^7+\cdots$

Plugging in $x=1$ we get

$\frac{1}{3}[A(1)+\omega^2A(\omega)+\omega A(\omega^2)]=f(n,1)+f(n,4)+f(n,7)+\cdots$

Now verify that $A(1)=3^n, A(\omega)=0, A(\omega^2)=0$ so the desired sum is simply $3^{n-1}$. The answer to the problem is then $3^{2005}$.

You could also create a bijection between the desired sum and $3^{n-1}$. Remove the last person, each remaining person has 3 choices, (0, 1 or 2). Last person is fixed in mod three (cuz number of candies is always 1 mod 3). So $3^{n-1}$ ways.

Enumerate the children $1, 2, \ldots, 2006$. Note $f(2006, 1)+f(2006, 4)+\cdots+f(2006, 4012)$ counts the number of ways to distribute any number of candies that is $1 \pmod{3}$ to 2006 children so that each child receives at most 2 candies, since the most number of candies that can be given regardless of value $\pmod{3}$ is $2 \cdot 2006 = 4012$. Let the set of all ways to do this be $A$. We claim that this is equivalent to distributing 0, 1, or 2 candies to each of the first 2005 children arbitrarily. Let the set of all ways to do this be $B$.

For any arbitrary distribution of candies to the first 2005 children with each of those children having 0, 1 or 2 candies, note there is exactly one way to give candies to the 2006-th child so that the 2006-th child receives 0, 1, or 2 candies and that the number of candies given is $1\pmod{3}$. More precisely, let $s$ be the total sum of all the candies. There is a unique value $v \in \{0, 1, 2\}$ so that $s+v \equiv 1 \pmod{3}$ since $\{0,1,2\}$ are the set of all residues $\pmod{3}$, so the number of candies given to the 2006-th child is fixed so that the total number of candies given to all the children is $1 \pmod{3}$ and all children are given 0, 1, or 2 candies. So for each element of $B$, there is exactly one corresponding element of $A$. Hence $B \subset A$.

Suppose that there exists some element of $A$ that did not correspond with some element of $B$ using the map defined above. This means there would be some way to give candies to the first 2005 children where each of those children received 0, 1, or 2 candies and the sum of the total number of candies given to all 2006 children was $1 \pmod{3}$, that was not covered in the total number of ways of distributing 0, 1, or 2 candies to each of the first 2005 children arbitrarily. This is absurd. Hence $A = B$.

Therefore $|A|=|B|$. So the answer is just $3^{2005}$; each child gets 0, 1, or 2 candies.

[hide=sol]This is the number of ways to distribute a 1 mod 3 number of candies to 2006 children such that each child receives at most 2 candies. Thus, we can represent it with the sum of the 1 mod 3 coefficients of the generating function \((1+x+x^2)^{2006}.\) This is the same as the sum of the multiple-of-3 coefficients of \(x^2(1+x+x^2)^{2006}.\) Using a 3-roots of unity filter, we get \(\frac{3^{2006} + 0 + 0}3 = \boxed{3^{2005}}.\)[/hide]


You're buying donuts for a company function. You are told to buy 100 donuts, and the choices for flavors are glazed, chocolate, and chocolate-glazed. The glazed are popular, so you think you should buy at least 20. You like the chocolate, so you want at least 1. Also, there is a company superstition that if you have an odd number of chocalate donuts in one room, bad luck will be bestowed upon the company. In how many ways can you buy the donuts?

[hide]Let a=the number of glazed, b=chocolate, c=chocolate glazed.
We know that $a \geq 20$, $b \geq 1$, b is even, $a+b+c=100$
Apply the substitution $a=d+20$, $b=2k+2$ (since we can't buy 1 b)
$d+c+2k=80$ where d, c, and k are nonnegative.
Case I: d is even, c is even.
Divide by 2, balls and urns, $\binom{42}{2}$.
Case II: d is odd, c is odd.
substitue, balls and walls, $\binom{41}{2}$
$\frac{42 \cdot 41+41 \cdot 40}{2}$
$41 \cdot 41=1681$[/hide]

Consider the following situation:

We want to distribute $m$ indistinguishable balls in $n$ distinguishable urns, $m$ does not neccessarily have to be greater than $n$. Enclose each urn by |'s, except the first and last, which we will leave open. Also, denote each ball as o. Hence, we will have something like this:

|oo|oo|,

which will mean there are four urns, 0 in the first, 2 in the second, 2 in the third, four in the last. Also, we can have something like

|||||||||||||||||||||||||||00000000||,

with 0 in a bunch, 8 in one, and 0 in each of the last two.
Getting back to the problem, we are going to have $n-1$ |'s, which enclose the $n$ urns, and $m$ balls, so $m+n-1$ balls in total. We note that the way we generate a distribution of the balls is to choose which of the $m+n-1$ "slots" are balls, the rest will be urn dividers. This makes a one-to-one correspondence.
So we are choosing where to put $m$ balls in $m+n-1$ slots, hence the number of ways to distribute the balls is just ${m+n-1\choose m}$.

[quote="vamathletes"]What are balls and urns(and balls and walls as well)?[/quote]
They're the same thing (see leet's explanation).
Did I get it right?

No, but close. You made a careless error when substituting.

I get 41C2+40C2.  Is that right?  As its a+b+2c=78, not 80


inequalities
geometry
inequalities proposed

Say x(1-y)+y(1-z)+z(1-x)=f(x)=ax+b. Then no matter which are y and z, it will get the maximum value or in 0 or in 1. So just try the values x,y,z=0,1 and se that the inequality occurs  :lol:

[quote="lasha"]Given $ (x,y,z)$ real numbers from $ [0;1]$. Prove the following inequality: $ x(1\minus{}y)\plus{}y(1\minus{}z)\plus{}z(1\minus{}x) < 1$.
P.S. I have very beautiful and short solution. I will post it later.[/quote]
I think it's wrong. Try $ x\equal{}1$ and $ y\equal{}z\equal{}0.$

Sorry I made mistake.$ x,y,z$ are from $ (0;1)$.

[quote="kunny"]$ (1\minus{}x)(1\minus{}y)(1\minus{}z) < 0$[/quote]
I think you have meant that $ (1\minus{}x)(1\minus{}y)(1\minus{}z)\plus{}xyz>0.$ :wink:

I have it's geometric solution and it's very smart and beautiful, but I know it's solution without geometry. :wink:

Take triangle $ ABC$ such that $ AB \equal{} AC \equal{} BC \equal{} 1$. Take points $ M,N$ and $ K$ on sides$ AB,BC$ and $ AC$ respectively. Let $ AM \equal{} x$,$ CK \equal{} y$ and $ BN \equal{} z$. Then, $ BM \equal{} 1\minus{}x$, $ CN \equal{} 1\minus{}z$ and $ AK \equal{} 1\minus{}y$. Obviously, the area of $ ABC$ is more then the sum of areas of triangles $ AMK,KNC$ and $ BMN$.
$ S_{ABC}\equal{}\frac{\sqrt{3}}{4}$; $ S_{AMK}\equal{}\frac{x(1\minus{}y)\sqrt{3}}{4}$; $ S_{CKN}\equal{}\frac{y(1\minus{}z)\sqrt{3}}{4}$; $ S_{BMN}\equal{}\frac{z(1\minus{}x)\sqrt{3}}{4}$;
So, by the last $ 4$ equality and fact, that $ S_{ABC}> S_{AMK}\plus{}S_{BMN}\plus{}S_{CKN}$, Dividing the both sides by $ \frac{\sqrt{3}}{4}$, we get:
                                  $ x(1\minus{}y)\plus{}y(1\minus{}z)\plus{}z(1\minus{}x) < 1$, Q.E.D. :wink:


MML team (final results): http://www.massmathleague.org/statistics/mml/200607/200607v_team.html

MML individual (final results): http://www.massmathleague.org/statistics/mml/200607/200607v_individual.html

MAML team -- WOCOMAL teams highlighted: http://www.wocomal.org/statistics/maml/2007/2007m_team.html

MAML team -- MML teams highlighted: http://www.massmathleague.org/statistics/maml/2007/2007m_team.html

MAML individual -- WOCOMAL students highlighted: http://www.wocomal.org/statistics/maml/2007/2007m_individual.html

MAML individual -- MML students highlighted: http://www.massmathleague.org/statistics/maml/2007/2007m_individual.html

The MAML statistics haven't been officially released (at least I haven't received them), but what I've put online comes from photocopies of the original scoresheets.  I doubt that there will be any substantive difference between the official data and the data that's now online.

The MML individual rankings that I've put online differ slightly from the official rankings on GeoCities; in particular, in the official rankings there were 18 instances where a student's scores were split among 2+ names.  The only effect with respect to awards is that in the official ranking, Vai of LSRHS and Khalandovsky of Andover were not in the top 50, and in my list they are; and in my list Li of ABRHS moved from 36th to 25th place.

If you see any typos, let me know.

Rick Yanco
Worcester Academy

18	M-2	Koorapati, Nipunn	11	Bedford	M	6	 	6	 	2	 	14

There's me

I screwed up the analytic geometry round badly
First round, I put down the diameter instead of the radius
third problem I completely messed up

Somehow Bedford still made it to New Englands even though we did quite poorly. We got 4th place (quite far out), but we're going! We just have to practice more.

So who are you guys?

Heh ABRHS still pulled out another math night after being in 10th place after round 3  :maybe:

And yeah Anton Li did get an 82 total.

Could you post which of the schools get to go to the
New England meet on April 27th (@Canton HS)
please?

 :)

The minimum number of teams to qualify is 4 large, 3 medium, and 2 small.  So on that basis the nine automatic qualifiers are

Large: Lexington, Canton, ABRHS, Brookline
Medium: Sharon, Westborough, Wayland
Small:  Worcester Academy, Maimonides

But there are other rules.  Any school which comes close to qualifying (because they have a score that's close to that of the last qualifier above) is usually invited.  The closest two would be Dover-Sherborn and Thayer, both just four points behind Maimonides, so I would [i]think[/i] those two would be invited before, say, Belmont, Westford, or BLS.  But I don't make that decision.

Also, league champions have an automatic invitation to NEAML.  The WOCOMAL champion was Shrewsbury, so they are invited.  The WMML champion was Deerfield, so they are invited.  The MML champion was Lexington, and they're already going.  I think either Falmouth or Dartmouth was a league champion, so one of those would go.

If/when I get the list I'll send it along, actually proofreading it this time!

Rick Yanco
Worcester Academy

I know that we (Bedford) got invited after placing 4th in the medium school division.


modular arithmetic
number theory
number theory unsolved

[u]Wilson's Theorem[/u]. Using Fermat's Little Theorem and working in $ \mathbb{Z}_{p}[X]$, we have that $ x^{p \minus{} 1} \minus{} 1 \equal{} (x \minus{} 1)\hdots(x \minus{} p \plus{} 1)$.  Setting $ x \equal{} 0$, we have $ \minus{} 1 \equal{} ( \minus{} 1)^{p \minus{} 1}(p \minus{} 1)!$. Ergo, $ (p \minus{} 1)!\equiv \minus{} 1\pmod p$ iff (Since the converse is also true) $ p$ is a prime number.

[quote="jdlkem"]Prove: (p-1)! = -1 (mod p), where p is prime.

I forgot how to type math in this forum :([/quote]

[code]$(p-1)!\equiv -1 \mod\;p$[/code]  :wink:

Another solution you can use the lemma :

For any $ k\in\;[1,p\minus{}1]$ has only $ k'$ so that $ kk'\equiv\minus{}1\mod p$


symmetry
geometry
angle bisector
geometry unsolved

In my notes, there is the following problem: "Let ABCD be a convex quadrilateral. Diagonal AC bisects <BAD. Let E be a point on side CD. Segments BE and AC intersect at F. Extend segment DF through F to intersect segment BC at G. Prove that <GAC = <EAC." I remember this was in the forum before under the title 'angle bisectors' or something but can't find it anymore, so I'll just post my question in a new topic: I tried to solve this by showing that AE, AG, AC, and the line through A perpendicular to AC are a harmonic range. Since we already have that AC, the perpendicular line, AB, and AD are a harmonic range, all we have to do is show that EG, BD, and the line through A perpendicular to AC are concurrent, and it seems like this should be the easy part of the proof. But I am not able to prove this. Is there a nice easy way to note this or is it better to solve the problem in a totally different way?

Btw, this problem seems rather powerful. Just shortly after I was reading it, I come across "Point D lies on side AC of triangle ABC such that BD = CD. Point E lies on side BC. Line [i]l[/i] passes through E and is parallel to line BD. Lines AB and [i]l[/i] meet at F. Let G denote the intersection of segments AE and BD. Prove that <BCG = <BCF" and this is of course a special case of the above.

let $P=AC\cap BD$ and the perpendicular to $AC$ at $A$ intersects line $BD$ at $Q$. You want to show that $E$,$G$ and $Q$ are collinear. By Ceva in the triangle $BCD$ with respect to the point $F$ and using the fact that $\frac{BP}{PD}=\frac{BQ}{QD}$  you have exactly what you need in menelaos theorem in $BCD$ for the points $E,G,Q$ to be collinear.

[quote="MysticTerminator"]In my notes, there is the following problem: "Let ABCD be a convex quadrilateral. Diagonal AC bisects <BAD. Let E be a point on side CD. Segments BE and AC intersect at F. Extend segment DF through F to intersect segment BC at G. Prove that <GAC = <EAC." I remember this was in the forum before under the title 'angle bisectors' or something but can't find it anymore,[/quote]

Indeed, the topic was lost during the hacker attack on 6 December 2004. Thanks for posting it again.

Here is my solution of this problem:

In http://www.mathlinks.ro/Forum/viewtopic.php?t=6082 , I showed:

[quote="darij grinberg"][b]Problem, version 1:[/b]

Let P and Q be two points in the plane of a triangle ABC such that the lines CP and CQ are symmetric with respect to the angle bisector of the angle BCA. The lines AQ and BQ meet the lines BP and AP at the points N and M, respectively. Show that the lines CM and CN are symmetric with respect to the angle bisector of the angle BCA.[/quote]

Now, in our problem, we have the triangle BDA and the points C and F in its plane. The lines AC and AF are symmetric with respect to the angle bisector of the angle DAB (since these lines simply coincide with the angle bisector of the angle DAB, because the problem requires that the diagonal AC bisects < DAB). The lines BF and DF meet the lines DC and BC at the points E and G, respectively. Hence, from my problem above, it follows that the lines AG and AE are symmetric with respect to the angle bisector of the angle DAB. Since this angle bisector is the line AC, we therefore obtain that the lines AG and AE are symmetric with respect to the line AC. In other words, < GAC = < EAC, and the problem is solved.

  Darij

By the way, if you're interested, this is the solution given in my notes:

"We apply a central projection that takes the line through A perpendicular to AC to infinity & that preserves symmetry about AC. After doing this, we know B'A', C'A', and D'A' are parallel, and that B'A' and D'A' are equidistant from C'A'. We want to prove E'A' and G'A' are equidistant from C'A'.  We can apply an affine tranformation in the direction of A'C' s.t. D''B'' is perpendicular to A''C''. Parallel lines to A'C' are preserved. But then by symmetry, it is clear E''A'' and G''A'' are equidistant from C''A''. Transforming back gives that AE and AG are symmetric about AC, so AC bisects <EAG. QED"


For $ x,y,z$ positive reals

$ \sum_{cyc}\sqrt {\frac {x}{x + y}} \leq {\frac {3}{\sqrt 2}}$

http://www.mathlinks.ro/viewtopic.php?t=158729

In that link there are two other links as well. I hope you'll find your intended proof in one of the three links.

Thank you very much for the link. I didn't remember it.


projective geometry
geometry
power of a point
radical axis
geometry proposed

Let $\mathfrak C_1,\mathfrak C_2$ be two circles .

Find the point $\mathcal A$ such that its polar Wrt the circles be a same line.

let $C_1=(O,R)$ and $C_2 = (Q,r)$

The polar of $A$ w.r.t. $C_1$ is a line perpendicular to $OA$.
The polar of $A$ w.r.t. $C_2$ is a line perpendicular to $QA$.

So $A$ must be on the line $OQ$.

Let $P$ the intersection point between the common polar line and line $OQ$.

The problem says that the points $A,P$ are inverse w.r.t. $C_1$ and inverse w.r.t. $C_2$.
($OA \cdot OP = R^2$ and $QA \cdot QP = r^2$)

so, if $A$ is in the interior $C_1$ then $P$ is out of $C_1$, and inversely.
That is, always there is one point of $A,P$ in the interior, and the other is out of $C_1$ (except the fact that $A,P$ are [u]on[/u] the circle)

Denote $w$ the circle with diameter $AP$.
Let $K$ its center and $k$ its radius.
Of course, the circle $w$ intersects $C_1$.

The power of $O$ w.r.t. $w$ is $OA \cdot OP = R^2$.
So the circles $w$ and $O$ are perpendicular.
This means that the power of $K$ with respect to $C_1$ is $k^2$.
Another conclusion is that $K$ can't be in the interior of $C_1$.

Similarly we find that the power of $K$ w.r.t. $C_2$ is  also $k^2$.

Hence, $K$ is on the radical axis of $C_1,C_2$.

$1)$ If the circles $C_1,C_2$ have two common points, then the point $K$ is in the interior of $C_1$ and $C_2$, so we can't have solution.

$2)$ If $C_1,C_2$ are tangent at point $K$, then the point $A \equiv P \equiv K$ (there is no circle centered at $K$, with non-zero radius and intersecting $C_1$ perpendicularly)

So, there are non-trivial solutions if
$3)$  the circles $C_1,C_2$ have not common points (either they have common interior points or not)

So, what we need is

- find the point $K$ in the intersection of $OQ$ and radical axis of $C_1,C_2$.
- construct a tangent segment $k$ from $K$ to one circle, say $C_1$.
- draw the circle $w=(K,k)$

The points of intersection between $w$ and line $OQ$ are the points $S_1,S_2$.

[u]1st solution[/u]: $A=S_1, P=S_2$
[u]2nd solution[/u]: $A=S_2, P=S_1$

(so we have two solutions, but there are the same points)


When $ P(x)\equal{}x^{81}\plus{}Lx^{57}\plus{}Gx^{41}\plus{}Hx^{19}\plus{}2x\plus{}1$ is divided by $ x\minus{}1$, the remainder is $ 5$; and when $ P(x)$ is divided by $ x\minus{}2$, the remainder is $ \minus{}4$. However, $ x^{81}\plus{}Lx^{57}\plus{}Gx^{41}\plus{}Hx^{19}\plus{}Kx\plus{}R$ is exactly divisible by $ (x\minus{}1)(x\minus{}2)$. If $ L$, $ G$, $ H$, $ K$, and $ R$ are real, compute the ordered air $ (K,R)$.

[hide]Let $ Q(x) \equal{} x^{81} \plus{} Lx^{57} \plus{} Gx^{41} \plus{} Hx^{19}$. Then $ P(x) \equal{} Q(x) \plus{} 2x \plus{} 1$, and let $ F(x) \equal{} Q(x) \plus{} Kx \plus{} R$. Then we know $ P(1) \equal{} Q(1) \plus{} 3 \equal{} 5\rightarrow Q(1) \equal{} 2$, and $ P(2) \equal{} Q(2) \plus{} 5 \equal{} \minus{} 4\rightarrow Q(2) \equal{} \minus{} 9$. We also have $ f(1) \equal{} 2 \plus{} K \plus{} R \equal{} 0$, and $ f(2) \equal{} \minus{} 9 \plus{} 2K \plus{} R \equal{} 0$, which leads to $ (K,R) \equal{} (11, \minus{} 13)$[/hide]

You should have stated that f(x) = x^81 + ... and blah blah blah, and also your last Q(x) should be Q(2) i think


induction
algorithm
combinatorics proposed
combinatorics

Prove the following:
Let G be a 2-connected graph, v_1 and v_2 are elements of V(G),
|V(G)|=n=n_1+n_2, where each n_i>= 1 is an integer.
Then there exists a partition (V_1, V_2) of V(G) such that v_i is an element of V_i,
|V_i|=n_i, and the subgraph of G induced by V_i is connected.

We will use induction on $n_1$.

Base $n_1=1$. We are to show $G\setminus\{v_1\}$ is connected. Indeed, since $G$ is 2-connected, for all $v\in G\setminus\{v_1\}$ there exists path from $v_2$ to $v$ which doesn't contain $v_1$.

Suppose we find pation $\{G_1,G_2\}$ s.t. $|G_1|=n_1$ and $|G_2|=n_2$, $G_i$ is connected subgraph. Let $v\in G_2$. We denote $|v_2\to v|$ the length of the [i]shortest[/i] path from $v_2$ to $v$ inside $G_2$. Let $w\in G_2$ and
\[|v_2\to w|=\max\{|v_2\to v|\ :\ v\in G_2\mbox{ and } \exists\mbox{ edge }(u,v),\mbox{ where } u\in G_1\}.\qquad (*)\]
/* corrected definition: $u\in G_1$ instead $u\in G_2$ */

We state that $G_1\cup\{w\}$ is connected subgraph and $G_2\setminus\{w\}$ is connected.
It is clear for $G_1\cup\{w\}$.
Suppose $G_2\setminus\{w\}$ isn't connected graph. Let $v$ be such vertex of $G_2\setminus\{w\}$ that there are no path from $v_2$ to $v$ inside of $G_2\setminus\{w\}$ [b](**)[/b]. Since $G$ is 2-connected, there exists path from $v_2$ to $v$ [i]which doesn't contain[/i] $w$. Let $v_2=q_1,q_2,...,q_k=v$ be that path. Due to [b](**)[/b] there is $l\in\mathbb{N}$ s.t. $q_{l-1}\in G_1$ and $q_{l},q_{l+1}..,q_k\in G_2$. Consider vertex $q_l$. Due to [b](*)[/b] [i]there exists path from $v_2$ to $q_l$ inside[/i] $G_2$. It implies that there exist path from $v_2$ to $v$ inside $G_2$ which doesn't contain $w$. Contradiction.

Since $|G_1\cup\{w\}|=n_1+1$ and $|G_2\setminus\{w\}|=n_2-1$, we are done.

That's  what I did too. Induction on n_1.

Could  you help me to prove the same theorem, but for k-connected graphs?:

Let G be a k-connected graph, v1,...,vk are elements of V(G),
|V(G)|=n=n1+,...,+nk, where each n_i>= 1 is an integer.
Then there exists a partition (V_1,..., V_k) of V(G) such that v_i is an element of V_i,
|V_i|=n_i, and the subgraph of G induced by V_i is connected.

It is harder, i have not suceed in proving it.

Is there an easier proof for the 2-connected.
Maybe using the block structure of G2..

In my opinion, it is a simple proof (for experienced persons ;) ).

Just a thing (to see if I've well understood) :
In (*) you meant $u \in G_1$ am I right? (otherwise, I do not see why $G_1 \cup \{w \}$ has to be connected).

Pierre.

Execuse!
There was misprint in definition of $w$: there is edge from $w$ to $G_1$.

do you have any ideas for proving for k-connected?

Are you sure that general theorem is correct?
I tried to prove that it is possible to partion $G$ into two subgraphs $V$ and $W$ s.t. $v_k\in W$, $v_1,v_2,...,v_{k-1}\in V$,  $|W|=n_k$, $W$ is connected and $V$ is $k-1$-connected. In this case we can apply induction on $k$.

Myth,
This theorem was proved by E. Gyori in 1976.
I don't know the reference; but i want to try to prove it before looking at his proof.  ;)
I think we have to use menger's th. to prove it.

Oops. My previous remark is nonsense. :blush:

Myth,
I read the proof of the theorem. It is very nice, it is an algorithmic proof. But the algorithm is not in any computer language or any of those ugly things  ;) .
I can post it later to discuss it.

[quote="manuel"]I read the proof of the theorem. It is very nice, it is an algorithmic proof.[/quote]
Yes, it was clear for me, because my proof for initial problem was algorithmic too. ;)


The special pages on AoPS wiki is not available.  I have used it before without error.

I have windows XP and I am using IE6.  I have a screenshot of it, but when I tried to upload the file, I got a General Error saying I cannot upload an empty file.  :maybe: I checked, and the file does exist...

Actually, this was already reported, but it took me over a dozen times to get it to load (finally).  Also, the error about not being to load an image was wierd...


a,b,c  are positive numbers prove the inequality:
$ (1\plus{}\frac{a}{b})(1\plus{}\frac{b}{c})(1\plus{}\frac{c}{a})\geq 2(1\plus{}\frac{a\plus{}b\plus{}c}{\sqrt[3]{abc}})$

Inequality equivalents to $ a^2b\plus{}ab^2\plus{}b^2c\plus{}bc^2\plus{}a^2c\plus{}ac^2 \geq 2.\sqrt[3]{a^2b^2c^2}(a\plus{}b\plus{}c)$ which is true by Muirhead.

http://www.mathlinks.ro/Forum/viewtopic.php?t=79681&search_id=768833282

Why is this in geometry?


Let p:E->B be a covering map.

Show that if B is regular, then E is also regular.

Thanks!

[quote="JRav"]See [url=http://at.yorku.ca/cgi-bin/bbqa?forum=homework_help_2001;task=show_msg;msg=0162.0001]here[/url].  It also includes the other parts of that problem.[/quote]

Actually I saw that before. But the "regularity" part is confusing. It seems that he did not finish that proof..


There are $ 5$ buttons in a row marked $ 1,2,3,4,5$ respectively. Find the no. of $ n$ button sequences such that in two consecutive moves neither the adjacent button(s) or the same button is pressed.

One can solve this via standard recursion methods (let $ a_n$ be the number of such sequences in which the last button pushed was 1 or 5; let $ b_n$ be the number in which the last button pushed was 2 or 4; and let $ c_n$ be the number in which the last button pushed was 3.  Then relate the three sequences and finish from there).  What's perhaps more interesting is that it seems that the resulting sequence is the same as sequence [url=http://www.research.att.com/~njas/sequences/A118649]A118649[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].  That sequence is the sequence of row-sums of [url=http://www.research.att.com/~njas/sequences/A106597]A106597[/url], a triangle generated by a rule similar to but more complicated than Pascal's triangle.  Can anyone prove (or disprove) that the two are the same?

Simpler variations on this problem: If we consider also 1 and 5 to be adjacent (i.e., if the buttons were on a circle, not a line) then the answer would just be $ 5\cdot 2^{n \minus{} 1}$.  If we replace "5" with "3", the resulting sequence is $ a_0 \equal{} 1, a_1 \equal{} 3, a_n \equal{} 2$ for $ n \geq 2$; if we replace "5" with "4" then the sequence is $ a_0 \equal{} 1$, $ a_n \equal{} 2 F_{n \plus{} 1}$ where $ F_0 \equal{} F_1 \equal{} 1$, $ F_{n \plus{} 2} \equal{} F_{n \plus{} 1} \plus{} F_n$ is the Fibonacci numbers.


algebra
polynomial
inequalities
trigonometry
analytic geometry
induction
LaTeX

Prove that for every interval (a,b) of real numbers and every p>=2 ,p natural ,there exist
a_1,a_2,...,a_p in (a,b) ,all IRATIONAL, such that

(a_1)^n + (a_2)^n + ... + (a_p)^n is RATIONAL for every natural number n.

Hmmm-this problem was given at the "titularization" exam for mathematics teachers in 2004.
I don't remember the solution but i do remember that the original problem had some subproblems wich lead to the solution.

Hint:try to construct a polynomial with rational coeficients,of degree p,all p roots are real and irrational,and all the roots are in the interval (a,b) and if you do that the problem is solved.

[quote="xxxxtt"]Hmmm-this problem was given at the "titularization" exam for mathematics teachers in 2004.[/quote]

What is the titularization exam?   Does this mean that some large fraction of high-school mathematics teachers in Romania can solve the above problem (it would be quite impressive, if true)?

Anyway, the point for the above problem is that if the sum of powers $s_n$ is rational for $p$ consecutive integer values of $n$ then it is true for all integer $n$, as the sequence satisfies a linear recurrence (with rational coefficients).   So one can start with an irreducible polynomial as xxxtt suggested.

Of course that isn't a very simple problem,but it had only 3 points of 100 at that exam.

Here's a way to construct such a polynomial. First, we construct a polynomial $f$ of degree $p$ with all roots real irrational and all roots in $(-1,1)$.

Let $T_p(x)$ be the $p$-th Chebyshev polynomial $T_p(x)=2^{p-1}x^p+\cdots$ where $T_p(\cos\theta)=\cos p\theta$. $T_p$ has integer coefficients. Now define $g(x)=T_p(x)-\frac13$. Multiplying the coefficients by 3, $g$ is irreducible by Eisenstein's criterion (reflected). $g$ also clearly has $p$ roots in the interval $(-1,1)$, which are distinct and irrational by the irreducibility of $g$.
Finally, we construct $f$ with a linear transformation. Choose $a',b'\in (a,b)$ rational and distinct. Let $f(x)=g\left(\frac{2x-a'-b'}{a'-b'}\right)$.

By the above work and previous posts, the roots of $f$ are the desired $a_k$.

[Edit: I should have previewed one more time after my last changes...]

[quote="fleeting_guest"]Can you give more information about the titularization exam?  Is it for all mathematics teachers (at what level?), are there exams available on Internet, etc.[/quote]

In Romania mathematics teachers,like other teachers,are of two types:stable teachers(which i call "titulars"-once such a teacher occupies a "desk",he may stay there all his carrer),and  "locum tenens" teachers-this kind of teachers must sit the "titularization" exam every year.
So,at the titularization exam,a teacher(or student who graduated the university) may become a stable teacher iff he takes 7,00 or more,and ,of course,if there are vacant desks.If he gets 5,00 to 7,00 he may occupy a desk,but only for 1 year,at the end of that year he must sit the exam again.
You can find the subjects at http://www.portal.edu.ro, but they are in Romanian.
I may say that the exam is not very simple at all,for example this year they gave 3 or 4 subproblems wich were not simple at all,like the problem above.
Of course,if you are well-prepared you can take 7,00 or more,but it is not a very simple job to take a 10.

[quote="jmerry"]Here's a way to construct such a polynomial. First, we construct a polynomial $f$ of degree $p$ with all roots real irrational and all roots in $(-1,1)$.

Let $T_p(x)$ be the $p$-th Chebyshev polynomial $T_p(x)=2^{p-1}x^p+\cdots$ where $T_p(\cos\theta)=\cos p\theta$. $T_p$ has integer coefficients. Now define $g(x)=T_p(x)-\frac13$. Multiplying the coefficients by 3, $g$ is irreducible by Eisenstein's criterion (reflected). $g$ also clearly has $p$ roots in the interval $(-1,1)$, which are distinct and irrational by the irreducibility of $g$.
Finally, we construct $f$ with a linear transformation. Choose $a',b'\in (a,b)$ rational and distinct. Let $f(x)=g\left(\frac{2x-a'-b'}{a'-b'}\right)$.

By the above work and previous posts, the roots of $f$ are the desired $a_k$.[/quote]

Why g=T_p-1/3 has also p roots in (-1,1)?!
Isn't that false?
Of course t_p has p roots in (-1,1) but how did you conclude T_p-1/3 has also?

How about a list of the roots?

Let $\alpha=\frac{\cos^{-1}\frac13}p$.
The roots are $\cos\alpha,\cos(\alpha+\frac{2\pi}p),\dots,\cos(\alpha+(p-1)\frac{2\pi}p)$

Considered in the complex plane, $T_p^{-1}[-1,1]=[-1,1]$. The values oscillate between $-1$ and $1$ $p$ times in the interval.

Another irreducible with $n$ real roots is $(X-q^2)(X-q^4)...(X-q^{2n}) + q$ for integer $q > 1$.  It is irreducible by Eisenstein's criterion for any prime divisor of $q$, and has the same sign as $\Pi (X-q^{2i})$ at $q, q^3, q^5, ... $.

[quote="fleeting_guest"]Thank you.  I found the exam, and except for the purely computational problem (verifying Pappus' theorem in coordinates) I doubt that more than 1 per cent of math teachers in the USA could pass it.[/quote]

I didn't manage to find it. Where did you find it?

Go at [url]http://portal.edu.ro/index.php/articles/c53/?letter=M[/url] and download Matematica-profesor I [subiect]

Thanks. That doesn't look toooo tough to me, although there might be some tricky parts...

Another polynom of degree p with all roots real irrational and all coefficients integers that satisfy the conclusion is constructed like this:
- if p is even, L_p = Π[i=0;(p/2 - 1)](X^2 - 3^i * 2)	with roots xi = 3^(i/2) * 2^(1/2), - 3^(i/2) * 2^(1/2) where i = 0,1,...,p/2 - 1.
The sum S = Σ[i=1;p]xi^n can be calculated for n even and odd [in this last case is 0] and gives always a rational number.

- if p = 3 take U = X^3 - 300*X + 1, U' = 3*(X^2 - 100) with roots -10 and 10, U has 3 real roots, none of them rational.
Let be Sn = Σ[i=1;3]Xi^n. By induction it's easy to prove that Sn is rational for every n.

- if p is odd at least 5, take V = L_(p-3) * U, combining roots of L_p and U we get the same result.

Now, if f is the above polynom in any case, with roots xi[i=1;p], let be m a large integer such that
(xi/m) is in (-1,1) for each i.
If a' < b' are rational in (a, b) we take :
s = (b' - a')/2,
t = (b' + a')/2,
yi = (xi/m)*s + t. yi is then in (a', b') so in (a, b) for each i.

The polynom g = f(m*(x - t)/s) has the roots yi and verify the conclusion for these roots.

S = Σ[i=1;p]yi^n = Σ[i=1;p]((xi/m)*s + t)^n using then Newton formula, switch the 2 sums and using the fact that Σ[i=1;p]Xi^n is rational for each n.

[i am new on this forum - so please double check... i will find also time for latex]


geometry
3D geometry
probability
Support
geometric sequence
geometric series

This is a problem I made up, im gonna see other people's soultions.

1.Andy, Ben, and, Colin are playing a game in which those people roll a a number cube ( 6 sides, including numbers 1,2,3,4,5,6) in that respective order and the first person to roll a 6 wins the game. What is the probability that Ben will win the game.

My solution

[hide] the probability that ben will win in the :
first try :5/6 x 1/6
2nd try : 5/6 x 5/6 x 5/6 x 5/6 x 1/6 (5/6)^4 x 1/6
3rd try : (5/6)^7 x 1/6
......

so the probability of winning is

5/6 x 1/6 x (1 + (5/6)^3 + (5/6)^6 + ...)
= (5/36) x (1+ 125/216 + ...)
=(5/36) x 1/(1-(125/216))
=(5/36) x (216/91)
=30/91
[/hide]

[hide]Since Ben is going second, The probability he will win the first time is

$\frac 56 \cdot \frac 16$

, and the second time is

$\frac 56 \cdot \frac 56 \cdot \frac 56 \cdot \frac 56 \cdot \frac 16$

because everybody else must lose, which the probability of happening is $\frac 56$. We must sum all the possibilities up and since this appears to be an infinite geometric sequence we use the formula $\frac {a}{1 -r }$ the first term being $\frac 56 \cdot \frac 16 = \frac 5{36}$ and the rate being $\left( \frac 56 \right)^3$.

Plugging in the values, the answer is $\frac {30}{91}$.[/hide]

There's really no other way to do it so I won't bother posting the obvious and only solution again

[hide]
It's an infinite series.
The chances of him winning on his first turn is: $\frac5{6}*\frac1{6}=\frac5{36}$
The chance of him winning on his second turn is $\frac5{6}*\frac1{6}*\frac5{6}*\frac5{6}*\frac1{6}=\frac5{36}*\frac{125}{216}.$
Use the formula for infinite series: $\frac{a}{1-r}$ where a is the first term and r is the constant.
$\frac{\frac5{36}}{1-\frac{125}{216}}=\frac{30}{91}$[/hide]

[hide]
The chance of Andy [b]not[/b] winning is $\frac{5}{6}$
Of this, Ben will win $\frac{1}{6}\text{ of the time, for an overall amount of }\frac{5}{36}$
If Ben does not win, the total chance of the rolling coming back around to him again is
$\frac{5^4}{6^4}$
So, the chance of him winning on two rolls is $\frac{5^4}{6^5}$
The total chance of Ben winning is the sum of an infinite geometric series:
\[\sum_{i=1}^{\infty} \frac{5}{36}\left(\frac{125}{216}\right)^i\]
\[=\frac{\frac{5}{36}}{1-\frac{125}{216}}\]
Which is:
\[\frac{\frac{5}{36}}{\frac{91}{216}}=\frac{30}{91}\]
[/hide]

That was a good problem!

[hide]Andy has a 36/91 chance
Ben has a 30/91 chance
Colin has a 25/91 chance

I think Andy has an unfair advantage.[/hide]

I think your internet browser window does not support does little pictures of fractions and numbers that people post

[quote="felmo"][quote]The chance of Andy not winning is \frac{5}{6}
Of this, Ben will win \frac{1}{6}\text{ of the time, for an overall amount of }\frac{5}{36}
If Ben does not win, the total chance of the rolling coming back around to him again is
\frac{5^4}{6^4}
So, the chance of him winning on two rolls is \frac{5^4}{6^5}
The total chance of Ben winning is the sum of an infinite geometric series:

\sum_{i=1}^{\infty} \frac{5}{36}\left(\frac{125}{216}\right)^i

=\frac{\frac{5}{36}}{1-\frac{125}{216}}

Which is:

\frac{\frac{5}{36}}{\frac{91}{216}}=\frac{30}{91}
[/quote] What does that mean?[/quote]
He probably forgot to put dollar signs.  Anyway, it's fixed now.


1.
When the sum of the first $k$ terms of the series

\[ 1^2 + 2^2 + 3^2 + \cdots + n^2 + \cdots \]
is subtracted from the sum of the first $k$ terms of the series

\[ 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1) + \cdots, \]
the result is 210.  Find the numerical value of $k$.

2.
Three coplanar circles are concentric, and the lengths of their respective radii are 10, 20, and 30.  A square is circumscribed about the smallest circle, and squares are inscribed in the other two circles.  Find the length of the shortest line segment that could contain a point on each square.

3.
Find the numerical value of $k$ for which

\[ \frac{7}{x+y} = \frac{k}{x+z} = \frac{11}{z-y}. \]

4.
Spender Sam always carried only dollars and pennies, but he never carried more than 99 pennies.  When leaving home one day, Sam noted that the number of dollars he had was 150% of the number of pennies he had.  After spending 10 dollars and 78 cents, Sam discovered that the number of dollars he had remaining was 1 less than the number of pennies he then had.  With how much money did Sam leave home that day?

5.
Two cards are randomly selected from a standard 52 card deck of playing cards, without replacement.  Find the probability that at least on king and at least one heart are selected (though not necessarily on different cards).

6.
Find the ordered pair of positive integers $(a, b)$, with $a<b$, for which
\[ \sqrt{1+\sqrt{21+12\sqrt{3}}} = \sqrt{a} + \sqrt{b}. \]

7.
The length of the base of an isosceles triangle is 20.  In the plane of the triangle, a point 4 units from this base is 10 units from each leg.  Find the area of the triangle.

8.
If Arc denotes the principle value, find all values of $x$ which satisfy

\[ \sin{(4\text{Arc}\tan{x})} = \frac{24}{25}. \]

#1:[hide]1*2+2*3+...+k(k+1) - 1^2 - 2^2 - ... - k^2 = 210
1(2-1) + 2(3-2) + ... + k(k+1-k) = 210
1 + 2 + ... + k = 210
k(k+1)/2 = 210
k^2 + k - 420 = 0
(k+21)(k-20) = 0
k = 20[/hide]

#3: [hide]Let a = x+y and b = x+z, so that the equation reduces to three variables:
$\frac{7}{a} = \frac{k}{b} = \frac{11}{b-a}$
Looking at the left and center, we have $a = \frac{7b}{k}$; substituting this in the equation $\frac{k}{b} = \frac{11}{b-a}$ yields $\frac{k}{b} = \frac{11}{b - \frac{7b}{k}}.
Cancelling b from both sides yields $k = \frac{11}{1- \frac{7}{k}$, so that
$k(1-\frac{7}{k}) = 11$
$k - 7 = 11$
k = 18.[/hide]

Other people can answer these, you know  ;)

aaaaaaaaaaaaaaaaa so much stuff to work on!!

"Must, not... reveal... hidden... content..." What I was wrong!  :mad:


MIT college Stanford Harvard Princeton Support ARML

Omg, it's tomorrow guys!

Anyone else anxious to hear their decision?

I'm not really nervous, but it would be great to get in. :)

tomorrow!!!?!!?

do you mean (1) they're sending the letters tomorrow, (2) we're getting letters tomorrow, or (3) its being posted online?

I'm quite anxious: my chances of getting in are roughly inversely proportional to the emphasis they put on grades.

And I believe it is 1 and 3.  Well, 3, definitely.  I'm not sure about 1.

[quote="joml88"]I think it's (1) and (3).[/quote]

thanks for confirming. as a side note, your quote reminds me of that one quote from henry ford (i think?) that goes "whether you think you can or not, your right"

I want to give everyone a last minute good luck :D

Hopefully we'll have many AoPSers get in EA :)

I'm not waiting on it, but my brother is extremely tense.

[quote="drunner2007"]dude I got in!!!!!!!

 :lol:[/quote]

Congratulations!

But, I got deferred. :(

Guys are you going to apply for the regular action?
I am gonna do that and hopefully we will be collegues :) .

Yes, going to college at an unusually young age is usually just called "early college," while "early admission" refers to applying at a time before the typical application deadline each year, and then getting notification of an admission result early too. Early admission takes two major forms: "early action" (what MIT does), in which the application deadline is early, notification is early, but the student still has until May to decide where to attend, and "early decision" (what Princeton does this year, but will cease doing), in which the application deadline is early, notification is early, and IF THE RESULT IS ADMISSION, the student is expected to rescind all other applications and attend that college.

Early Action is pretty great, especially when you get some nice school like MIT that lets you EA as many places are you want. From the school's perspective, I don't think it makes as much sense as a policy like Stanford's that forces you into single-choice EA. But it's great for students, especially ones here that have megahigh AMC/AIME scores and lots of other credentials, who want a fairly safe EA school (no stress over Christmas and in the spring).

Anyway, I think MIT's EA is about the greatest thing to hit college admissions since the AMC's. And the tube! And the poster! And confetti! Oh man.

Hey guys I applied for the MIT regular action too :wink:.Good luck to everyone who applied,hopefully we will be collegues :) .

i have a mit hat :wink:

how much do they look at amc scores?

-jorian

yeah, and they have one with harvard that goes

"havard... because not everyone can get into MIT"

just a question about mit...

how much do grades other than math matter...what's the lowest you could get?

-jorian

All grades including grades in math are important to MIT. An outstanding math student (a three-time USAMO qualifier and a black-MOPper) was rejected by MIT due to poor grades in history and English. Please note that his actual English level is not poor at all, which is evidenced by his 800 verbal score on SAT I.

By the way, it is unnecessary to sign your name twice.

[quote="juicybooty911"]do they still sell that "mit...because not everyone can get into Caltech" shirt?[/quote]

haha if you can get into one school, you probably can get into the other. It's just a bad prank from caltech  :roll: I actually saw people wearing these shirts

You can see some of their shirts and other stuff [url=http://store.thecoop.com/coopstore/servlet/com.bst.servlets.EStoreWebControlServlet?object=EstoreGenTreeLib&method=GetProductList&categoryuno=376&parent=Mit&refresh=true]here[/url].

I couldn't find this shirt on that site, but I've seen it in the store.  It says something like "Friends don't let other friends go to Harvard" :)

[quote="qvc122"]All grades including grades in math are important to MIT. [/quote]

Sort of.   The grades outside of math/science play different roles.  A less-than perfect grade in European history will not raise eyebrows as much as the equivalent grade in a math class.  But low grades in multiple subjects or at several times suggesting some lack of functionality or unwillingness to meet requirements, can well be a black mark for purposes of admission.   Discrepancies between high verbal test scores and low English grades are another indicator of this.

umm...it's been a while. but i guess this is the college forum, so not a big deal this time :)

jorian

Yeah, I sure hope my 2nd marking period C in English won't get me rejected now that I was already deferred...

All the rest were A's (and all A's for 1st MP, even English that time).

Congrats to all who made it early, and good luck to those who were deferred.  I am biased, but I strongly recommend that you come to MIT if accepted; it's the best place ever and we can never have too many math nerds here! :)


Find $ x$ :

  $ 5^x\minus{}2x\minus{}1\minus{}2\log_5(4x\plus{}1)\equal{}0$

[quote]Find $ x\in \mathbb {R}$ such that $ 5^x \minus{} 2x \minus{} 1 \minus{} 2\log_5(4x \plus{} 1) \equal{} 0$[/quote]
[hide="A useful hint"]Consider $ f(t)$ $ \equal{} t \plus{} 2\log_5 (t)$, $ \forall$ positive $ t$[/hide]

[quote="Duy_VN"]Find $ x$ :

  $ 5^x \minus{} 2x \minus{} 1 \minus{} 2\log_5(4x \plus{} 1) \equal{} 0$[/quote]

[b][u]solution[/u][/b]

Let $ f(x) \equal{} 5^x \minus{} 2x \minus{} 1 \minus{} 2\log_5(4x \plus{} 1)$
$ A_f \equal{} \left( \minus{} \frac {1}{4},\infty\right)$

We can easily see that $ x \equal{} 0$ and $ x \equal{} 1$ are indeed solutions of $ f(x) \equal{} 0$.

If there exists another $ x_0$ such that $ f(x_0) \equal{} 0$ then from rolle's theorem (three times)  there exists a $ x_1$ such that $ f''(x_1) \equal{} 0.$

But if i'm not mistaken in the calculations

$ f''(x) \equal{} 5^x (\ln5)^2 \plus{} \frac {32}{\ln5(4x \plus{} 1)^2} > 0$.

So $ f''(x) \not \equal{} 0$ for all $ x\in A_f$--contradiction.

So the only solutions are $ x \equal{} 0$ and $ x \equal{} 1$.

Dimitris


quadratics algebra polynomial function MATHCOUNTS AMC AIME \/closed

This fall we will offer a course on topics of advanced algebra.  We'll start off with quadratics, but quickly move to topics such as polynomials, functions, logarithmic equations, clever substitutions of variables, systems of equations, symmetric sums, advanced factoring methods, and binomial expansion.  Among the techniques taught will be those useful on (very) advanced MATHCOUNTS questions, AMC questions, and AIME questions (and some of the approaches discussed will even be useful as steps in attacking Olympiad level problems).  The bulk of the course will be AMC-AIME level difficulty. When the course is listed, we will make a pretest and posttest available to help you determine if the course is suitable for you.

I post here to ask those of you *[color=red]who might take the course[/color]* to let us know what times are best for you.  Tuesday and Sunday evenings are already taken by other courses.  Please do not respond if you are not considering taking the course.

The course will begin in early November and run through February, with some days off for winter holidays.  We know we won't be able to hit a time that works for everyone (though you might consider taking the class as a correspondence course - the transcripts for each class will be available to enrolled students throughout the course), but we'd like to reach as many of you as possible.

Thanks much for your input.

Where do we find pricing information on the courses?

On the left side of the screen, you see a menu. Click on "Enroll" under "Online Classes."

You have to wait for the pricing information to come out for "Advanced Algebra" though.

I c, can anyone send me a direct link to the page that hosts the virtual classroom software... I seem to be getting lost somewhere with the links.

Virtual Classroom is [url=http://www.artofproblemsolving.com/Classroom/AoPS_Classroom.php]here[/url].  You can only access the classroom during a class in which you are enrolled or during a Math Jam.  There's one tomorrow night (Tues) if you want to check it out.  If you have any problems, let me know - most are quickly and easily fixed.

The class information will be posted [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]here[/url] when the course is listed (there'd info there for the other two classes we have listed - Spoon, the Counting class may interest you.

I misunderstood before, I thought we had to download the virtual classroom. This method works for me. I am flying up to visit MIT early wednesday morning so tomorrow night will be hectic for me, but I will try to check it out.

Yes that counting class sounds like it would be good for me. It's frustrating that I can do almost all calculus (through multi variable) problem placed infront of me, but when it comes to many problems on this site, I just haven't a clue. I am eager to learn this type of mathematics. I will also shortly be purchasing AoPV 1 and 2, I only wish I knew about them at the BEGINNING of summer.

O well, I have to consult my parents before making any decisions but I think I would be interested in taking both of these classes(counting and advanced algebra), so if you are taking an approximate count, please include me. I will get back to you regarding these classes most likely by Sunday.

Just to let you know, Monday night's and Wednesday night's are booked for me.. I have a night class.

when eggzactly will the pretest be up, because i can solve lots of amc problems, but aim is kinda iffy. but since it does start in november i think, ill be much better by then. but still.

times good for me: not monday, any day after that after 6 pm cst :lol:

We'll list the course with its pre & post test in a week or so.

If I can't make this course, when will the next course be?

Are you referring to when will the algebra course be taught again?  If so, probably in the summer next year.  If you're asking when will more classes start, that will be in January (probably AMC Problem Series & Intro Counting).

Is what you're talking about the Algebriac solving one???

I'm referring to the Algebraic Problem Solving when I say it the algebra class will probably be offered again in summer 2004.

oh ok...thanks... and can it not be Monday??? since I have Tulsa Junior Youth Symphony then... but its okay if it can't be... just wondering.

OK - we'll put up a survey for days a few weeks before we list the class, so you can register your preference then.  We do hope to mix up the days of each course as we offer them over time to give you more opportunities to take them.


calculus
integration
trigonometry
limit
function
real analysis
real analysis unsolved

Find $ lim_{n \rightarrow \infty} (\int_0^1 sin^n x dx) ^{\frac{1}{n}}$=?

The limit of your sequence is $ \max\limits_{x\in [0,1]}\sin x\equal{}\sin 1$.

Similar problems have been discussed on this forum.

[quote="didilica"]The limit of your sequence is $ \max\limits_{x\in [0,1]}\sin x \equal{} \sin 1$.

Similar problems have been discussed on this forum.[/quote]

Could you explain the solution?

[quote="Pluraln"]Find $ lim_{n \rightarrow \infty} (\int_0^1 sin^n x dx) ^{\frac {1}{n}}$=?[/quote]

Remember if $ |a|<1$ then $ a^n\rightarrow 0$
then Use didilica's hint

[quote="1234567a"][quote="Pluraln"]Find $ lim_{n \rightarrow \infty} (\int_0^1 sin^n x dx) ^{\frac {1}{n}}$=?[/quote]

Remember if $ |a| < 1$ then $ a^n\rightarrow 0$
then Use didilica's hint[/quote]
There is no hint.
\[ \sin(1) \equal{} \lim \frac {\sin \left(1 \minus{} \frac {1}{n} \right)}{n^{1/n}} \equal{} \lim \left( \left(1 \minus{} \left(1 \minus{} \frac {1}{n} \right)\right) \sin^{n}\left( 1 \minus{} \frac {1}{n} \right) \right)^{1/n} \leq \lim \left( \int_{1 \minus{} \frac {1}{n}}^{1}\sin^{n}(x)dx\right)^{1/n} \leq \lim \left( \int_{1 \minus{} \frac {1}{n}}^{1}\sin^{n}(x)dx \plus{} \int_{0}^{1 \minus{} \frac {1}{n}}\sin^{n}(x)dx \right)^{1/n} \equal{} \lim \left( \int_{0}^{1}\sin^{n}(x)dx \right) ^{1/n} \leq \lim \left( \sin^{n}(1)\right)^{1/n} \equal{} \sin(1)
\]

More generally:
Assume that the function $ f(x)$ is continuous on $ [a,b]$ and let $ \max\limits_{x\in [a,b]}f(x) = A$. Then $ \displaystyle\lim_{n\to\infty}\left(\displaystyle\int_a^b(f(x))^n\,dx\right)^{\frac{1}{n}} = A$.


I'm trying to figure out how to transpose the following formula. It's an electrical formula and I'm not sure how to go about transpoising when you have multiple characters on each side of the dividing line:

R2=R1 A1 L2/ A2 L1

Make A1 the subject of the equation? or for that matter A2 etc.

Just in case you're interested R=resistence A=Area and L=Length.

Cheers

Transposing? Are you saying just rearranging the equation? If so, this belongs in Classroom Math...

I think what you're asking is just to solve for $A_{1}$... That's rather basic algebra. Divide both sides by $R_{1}L_{2}$ and multiply both sides by $A_{2}L_{1}$. Then it becomes:

$A_{1}= \frac{R_{2}A_{2}L_{1}}{R_{1}L_{2}}$

I'm hoping you mean something else, because this is awfully basic.

sorry, only my first post so I'll know better next time unless a forum administrator wants to move this.
Looking at it I don't even think equation rearranging is necessary because you can figure it out on inverse relationship:

L1=R1 A1 L2/A2 R2

L2=A2 R2 L1/A1 R1

Is this correct? Easier than I thought sorry.


geometry
conics
ellipse
3D geometry
sphere

The graphs of $ y^{2}\plus{}4x^{2}\plus{}9z^{2}\equal{}16$ and $ ax\plus{}by\plus{}cz\equal{}d$ intersect. What is the largest possible area of this intersection?
A) $ 4\pi$
B) $ 6\pi$
C) $ 8\pi$
D) $ 32\pi$
E) NOTA=None of the Above

[hide="lol"]
NOTA=Nerd_of_the_Ages
[/hide]
is this in 3-d geometry?
if so, the measure of the size of the rgion is volume
if it is not 3-d geometry, then is z an unknown constant?

The "graph" of $ ax \plus{} by \plus{} cz \equal{} d$ is a plane in 3-D space.
The "graph" of $ y^{2} \plus{} 4x^{2} \plus{} 9z^{2} \equal{} 16$ is the surface of an egg-like solid. It intersect the axes at $ (2,0,0)$, $ (0,4, 0)$ and $ (0, 0, 4/3)$. I don't know a name for it, but it is the 3-D equivalent of an ellipse. Its intersections with the planes x-y, y-z and x-z are ellipses. I cannot visualize what happens with a generic plane.
The question meant to ask for the area on the plane $ ax \plus{} by \plus{} cz \equal{} d$, bound by the intersection with $ y^{2} \plus{} 4x^{2} \plus{} 9z^{2} \equal{} 16$
It may be that the x-y plane gives the largest area.

ok... i get it now
so since the second equation is a plane, the intercestion is a cross section of the region defined bythe first equation.
In other words, we want the area of the largest cross section of the region inside or defined by the first equation.
rewrite the first equation so that it looks similar to the equation of an ellipse
$ \frac {x^2}{2^2} \plus{} \frac {y^2}{4^2} \plus{} \frac {z^2}{\left(\frac43\right)^2} \equal{} 1$
which is basically a sphere centered at the origin dilated in the x-axis direction by a factor of 2, in the y-axis direction by a factor of 4, and in the z-axis direction by a factor of 4/3
not too obviously, but intuitively, the largest cross section of this solid will lie in the same plane as the 2 longest axes- the x and y axes.
so we want the area of an ellipse with major axis of length 8 and minor axis of length 4, or $ \pi \cdot \frac82 \cdot \frac42 \equal{} 8\pi \implies \boxed{c}$

the area of the ellipse can be found by $ \pi a b$ where a and b are the lengths of the shortest and longest radii of the ellipse. this is because the ellipse is basically a circle dilated in one direction.

The correct problem statement would then be:

What is the largest possible area of a planar cross section of the region in space satisfying $ y^2\plus{}4x^2\plus{}9z^2 \leq 16$?

OR

What is the largest possible area of a planar cross section of the region in space bounded by the graph of $ y^2\plus{}4x^2\plus{}9z^2 \equal{} 16$?


I should solve this question,,

About an inscribed pentagonABCDE, AC//DE, and, M is the middle point of BD, if an angle AMB=angle BMC,

Show AC is bisected by BE.

Denote T is the intersection of AC and BE
A,B,C,D are concyclic    ∠BMC=∠AMB   =>  ABCD is a harmonic quadrilateral
△CMD∽△CBA      =>   MD/AB=CD/AC   => BD/AB=2CD/AC=CD/(1/2AC)
On the other hand, ∠ABE=∠ACE=∠CAD=∠CBD  ∠BAC=∠BDC
=> △BCD∽△BTA   BD/AB=CD/AT
So we have 1/2AC= AT
Thus we are done


function
inequalities proposed
inequalities

[tex]x_1,...,x_5\ge0[/tex] and
 [tex]\sum\frac{1}{1+x_1}=1[/tex];
find max:
  [tex]f=\sum\frac{x_1}{4+x_1^2}[/tex].
 GOODLUCK

This was on the 2003 CWMO: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=18378]www.mathlinks.ro/Forum/viewtopic.php?t=18378[/url].

By the way, you don't have to post things twice: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=38339]www.mathlinks.ro/Forum/viewtopic.php?t=38339[/url]

the function  $t\longmapsto \frac{t}{4+t^2}$ is cuncave then
$ S=\displaystyle\sum_{i=1}^5 \frac{x_i}{4+x_i^2} \le 5\frac{a}{4+a^2}$ where $a=\frac{x_1+...+x_5}{5}$~
By cauchy $(\displaystyle\sum_{i=1}^5 (1+x_i))(\displaystyle\sum_{i=1}^5 \frac{1}{1+x_i})\ge 5^2$ then $a \ge 4$
so $ a^2+4-5a=(a-4)(a-1) \ge 0$
then $ S \le 1$


Does exist one-to-one function $ f$ $ :$ $ R$ $ \longrightarrow$ $ R$ that for every $ x$,$ y$ $ \in$ $ R$:

$ f(x^2)$ $ \minus{}$ $ (f(x))^2$ $ \geq$ $ \frac {1}{4}$

Substitute $ x\equal{}0$ we get $ 0 \ge [2f(0)\minus{}1]^2$
So $ f(0)\equal{}\frac{1}{2}$
Similarly we get $ f(1)\equal{}\frac{1}{2}$ also. So such a function does not exist.

no because,
$ f(0)\minus{}(f(0))^2 \ge 1\div 4$ $ \Rightarrow$ $ (f(0)\minus{}1\div2)^2\le0$ $ \Rightarrow$ $ f(0)\equal{}1\div2$
$ f(1)\minus{}(f(1))^2 \ge 1\div 4$ $ \Rightarrow$ $ (f(1)\minus{}1\div2)^2\le0$ $ \Rightarrow$ $ f(1)\equal{}1\div2$
therefore $ f(0)\equal{}f(1)$ that is not one-to-one function contradiction


number theory
greatest common divisor
number theory proposed

Let $ a,b,n\in\Bbb N$, such that gcd$ (a,b)\equal{}1$. prove that

$ A\equal{}\frac1a\plus{}\frac1{a\plus{}b}\plus{}\cdots\plus{}\frac1{a\plus{}nb}$

is not an integer.

[quote="ZetaX"]Posted before. And using a much more meaningfull title would have been nice.[/quote]
Where is that's link?


limit
function
logarithms
geometric series

[hide]That's just the sum of the infinite geometric series 1, 1/2, 1/4, 1/8, etc.
So the answer is $\frac{1}{1-\frac{1}{2}}=\boxed{2}$[/hide]

Edit: scratch that, my bad

[quote="The Zuton Force"][hide]That's just the sum of the infinite geometric series 1, 1/2, 1/4, 1/8, etc.
So the answer is $\frac{1}{1-\frac{1}{2}}=\boxed{2}$[/hide][/quote]

I don't think so... for instance, $\frac{1}{8}$ shows up twice, as $\frac{1}{2} \cdot \frac{1}{4}$ and as $\frac{1}{8}$ itself.

[hide]
If you expand it out, I think you get something like

$1+\frac{1}{2}+\frac{1}{4}+2 \cdot \frac{1}{8}+2 \cdot \frac{1}{16}+3 \cdot \frac{1}{32}+3 \cdot \frac{1}{64}+\cdots$,

or

$1+\frac{1}{2}+\frac{1}{4}+\cdots = 1+1$
$\frac{1}{8}+\frac{1}{16}+\cdots = \frac{1}{4}$
$\frac{1}{32}+\frac{1}{64}+\cdots = \frac{1}{16}$
...

so we get $1+1+\frac{1}{4}+\frac{1}{16}+\cdots = 1+\frac{1}{1-\frac{1}{4}} = 1+\frac{4}{3} = \frac{7}{3}$.[/hide]

Zuton's solution is for the slightly different limit

$\large\lim_{n\to\infty}(1+\frac{1}{2})(1+\frac{1}{4})...(1+\frac{1}{2^{2^n}})$

I don't think so... the product up to $n=6$ is about $2.34736... > \frac 7 3$, and each term is slightly above 1, so the product will never decrease.

I don't think the coefficients increase in the manner you are assuming... I think they increase maybe some fibonacci thing, but I'm pretty sure it's not constant like that.  I'm just too lazy to write out too many products... up to $\left( 1 + \frac 1 {32} \right)$ wasn't fun...

Heh, let me save you some work.  The first 20 coefficients of the product up to 1/2^20 (and so of every larger term) are:

1, 1, 1, 2, 2, 3, 4, 5, 6, 8, 10, 12, 15, 18, 22, 27, 32, 38, 46, 54, 64

If you know some generating functions, by the way, you should be able to figure out what these coefficients actually represent.  Once you do, I suggest despair, as they involve [hide]partitions, and don't satisfy a nice closed formula of any sort.[/hide]

The final answer [hide]similarly doesn't have a nice closed form; to 5 decimal places, it is 2.38423.[/hide]

Wow!  That's a very interesting pattern... you can see that each coefficient is used later as the difference between two consecutive terms twice in a row, then the next coefficient is the difference between the next two pairs.  Very cool... no idea how to calculate the sum from here though  :D

Hmm, I'm not sure about that, deej21.  It's true for some coefficients, but I don't think for all.

A similar one:
Prove that: $(1+\frac{1}{3})(1+\frac{1}{3^2})(1+\frac{1}{3^3})...(1+\frac{1}{3^n})$ converges
And don't expand it like above...!!!

[hide="I can say so much at the most. Here is ... a little generalization."]\[ \boxed {\ a>1\Longrightarrow 1+\frac 1a\ <\lim_{n\to\infty}\ \prod_{k=1}^n \left( 1+\frac{1}{a^k}\right)<\ e^{\frac{1}{a-1}}\ }  \][/hide]

[quote="Virgil Nicula"][hide="Here is a generalization."]\[ \boxed {\ a>1\Longrightarrow \lim_{n\to\infty}\ \prod_{k=1}^n \left( 1+\frac{1}{a^k}\right)=e^{\frac{1}{a-1}}\ }  \][/hide][/quote]But Virgil, I find

$\prod_{k=1}^{30}\left(1+\frac1{2^k}\right)\approx 2.384231$

correct to 6 decimal places, and I tried with more than 30 iterations (50 iterations) and these 6 decimal places did not change a bit, not even the next 3 decimal places. I'm doubting your conjecture...

Yes, yes ... you are right ! Thanks. I made a little mistake in a drawing up (instead $\boxed {<}$ I put falsely $\boxed {=}$). I corrected now. I think that this limit is not calculable, maybe using the P.C. (a wished number of exact digits).

[hide]Let $f(a)=\prod_{n=1}^\infty \left( 1+\frac{1}{a^n} \right)$. Then $\ln f(a)=\sum_{n=1}^\infty \ln \left( 1+\frac{1}{a^n} \right)$.

$\ln (1+x)<x$ when $x\neq 0$ because the natural logarithm is concave down and therefore never greater than any of its tangent lines. So $\ln f(a)<\sum_{n=1}^\infty \frac{1}{a^n}=\frac{\frac 1a}{1-\frac 1a}=\frac{1}{a-1}\Rightarrow f(a)<e^\frac{1}{a-1}$.[/hide]


hullo¡¡¡ im looking for information on gaussian integers and its use in olympiad type problems. im trying to add more elememts to my problem solving arsenal. but im only interested in its use on the olympiad. also im trying look for information on roots of unity or something like that. im not sure what the exact name of it all i know is that they mention it a lot on solving some of the problems so it must be useful. anyway i know nothing about these topics but i really really want to learn so maybe someone could be kind enough to show me  all i need to know. many thanks in advance¡¡¡

The Art of Problem Solving books especially volume 2 has a wide range of topics for olympiad type problems. This includes number theory. I have only briefly looked at the number theory section but it seems to have only the important stuff you need to know for olympiad.


[quote="parinaz.sa"][quote="eLeM"][quote="doicu"]hallo,
prove for all real positive $x_1, x_2, x_3, x_4$ the ineq.
$\frac{x_1}{3x_2+x_3}+\frac{x_2}{3x_3+x_4}+\frac{x_3}{3x_4+x_1}+\frac{x_4}{3x_1+x_2} \geq 1$
 :)[/quote]

Just use Cauchy-Schwarz and then AM-GM. :)[/quote]
are you sure ??? i can't prove it with Cauchy-Schwarz and then AM-GM :oops:[/quote]

If you don't mind I will replace $x_1, x_2, x_3, x_4$ by $a, b, c, d$. It will be easier for me to write :)

So we have with Cauchy-Schwarz:
$\frac{a}{3b+c}+\frac{b}{3c+d}+\frac{c}{3d+a}+\frac{d}{3a+b}\geq\frac{(a+b+c+d)^2}{3(ab+bc+cd+da)+2(ac+bd)}$

With AM-GM we make this:
$\frac{(a+b+c+d)^2}{3(ab+bc+cd+da)+2(ac+bd)}=\frac{a^2+b^2+c^2+d^2+2ab+2bc+2cd+2da+2(ac+bd)}{3(ab+bc+cd+da)+2(ac+bd)}\geq$

$\geq\frac{3(ab+bc+cd+da)+2(ac+bd)}{3(ab+bc+cd+da)+2(ac+bd)}=1$

And so we are done :)


We divide the plane with an arbitrary number of lines and want to colour all the regions with two colours. Any two regions sharing a side should be coloured with different colours. Is this always possible?

No. You need at least four colors. Look up the "Four Color Theorem" if you'd like to know more about this sort of problem.

The famous "Four colors theorem" is for arbitrarily shaped countries. Here the countries are very specially shaped  :D

I think it can be proved by induction (it works fine for one line :roll: ) :

Let's suppose we have n+1 lines.

Let's remove one line and let's color the remaining regions with 2 colors.
We add back the removed line. and inverse the colors of all regions on one side on this line.
Now if we take 2 regions :
- either they have the line in common : they come from a region which was cut in two by the new line and it's correct
- either they don't : since their colors were inversed it's still correct.


What is the minimum value of the following expression, where a, b, and c are positive real numbers?

$\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}$?

Show your work please.

I'm not sure if this is right. I'll post if
[hide] $\sqrt 2$[/hide]
 is the answer.  :)

Well, I think it is wrong, since if a=b=c, then it sums to 1, so there.

Could someone please help?

I would appreciate it gratefully.

[quote="ch1n353ch3s54a1l"][quote="Kurt Gödel"][hide]By Cauchy in Engel form we have \[\sum_{\text{cyc}}\frac{a}{b+2c}\geq \frac{(a+b+c)^{2}}{3(ab+bc+ca)}\geq 1.\] (The last inequality is equivalent to $a^{2}+b^{2}+c^{2}\geq ab+bc+ca$, which is well-known.)[/hide][/quote]

Does the cauchy in engel form apply here..? don't numerators have to be squares?or is my mind just blank right now...[/quote]

Yes, they have to, but that's why I make them squares. Otherwise, I should've writte this: \[\sum_{\text{cyc}}\frac{a}{b+2c}\geq \frac{(a+b+c)^{2}}{3(a+b+c)}.\] But I'm multiplying each fraction (i.e. numerators and denominators) with $a$, $b$ and $c$ resp. Therefore, I obtain: \[\sum_{\text{cyc}}\frac{a}{b+2c}= \sum_{\text{cyc}}\frac{a^{2}}{ab+2ac}\geq \frac{(a+b+c)^{2}}{(ab+2ac)+(bc+2ab)+(ca+2bc)}= \frac{(a+b+c)^{2}}{3(ab+bc+ca)}.\] It's a powerfull trick!

I tried bruteforcing this and got something unexpected: \[\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}= \frac{\sum_{cyc}a(c+2a)(a+2b)}{(a+2b) (b+2c)(c+2a)}=\frac{\sum_{cyc}a(ac+2bc+2a^{2}+4ab)}{(a+2b) (b+2c)(c+2a)}= \frac{\sum_{cyc}a^{2}c+2abc+2a^{3}+4a^{2}b}{9abc+2\sum_{sym}a^{2}b}\] Let $\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\geq k$ for some real k. \[\frac{\sum_{cyc}a^{2}c+2abc+2a^{3}+4a^{2}b}{9abc+2\sum_{sym}a^{2}b}\geq k \implies \sum_{cyc}a^{2}c+6abc+2\sum_{cyc}a^{3}+4\sum_{cyc}a^{2}b\geq k(9abc+2\sum_{sym}a^{2}b) \implies 2\sum_{cyc}a^{3}+(4-2k)\sum_{cyc}a^{2}b+(1-2k)\sum_{cyc}a^{2}c\geq (9k-6)abc\] and now simple AM-GM on LHS: \[\text{L. H. S.}\geq 2\cdot 3abc+(4-2k)\cdot 3abc+(1-2k)\cdot 3abc=3abc(7-4k)\] Naturally we choose $k$ such that $3abc(7-4k)=(9k-6)abc$, $k=\frac97$. Therefore, \[\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\geq \frac97\]  \[\implies 2\sum_{cyc}a^{3}+(4-2k)\sum_{cyc}a^{2}b+(1-2k)\sum_{cyc}a^{2}c\geq (9k-6)abc\]  \[(k=\frac97) \text{By AM-GM}\] Since it is AM-GM, shouldn't this be the best minimum? But it isn't... help please!

[quote="me@home"]Naturally[/quote]

It's not natural at all.  In fact, as you have seen, it is incorrect.

Why is there any guarantee that the equality case of AM-GM and the minimal case of this expression coincide?

Edit:  [b]Here is a very simple counterexample.[/b]

I want to minimize the expression $x^{2}+1$.  By AM-GM,

$x^{2}+1 \ge 2x$

The equality case is at $x = 1$.  However, this gives $1^{2}+1 = 2$, whereas the actual minimum is obviously $0^{2}+1 = 1$.


How many integers are solutions of $ 3 \leq |n\minus{}1| < 8$?

Let's have $ n\minus{}1$ be $ x$ So, we have $ 3 \le |x| <8$. There are 5 solutions for positive x and 5 for negative x, so the answer is $ \boxed{10}$.


Let $ R$ be the region bounded by the curves $ f(x) \equal{} x^m$ and $ g(x) \equal{} x^n$ with $ x \in [0,1]$ and $ m,n$ integers where $ 0\le n < m.$ Compute the coordinates of the mass center for $ R$ and find the values of $ m$ and $ n$ so that the mass center of $ R$ stays out of the region.

The first question is easy, so the coordinates would be

$ \overline{x} \equal{} \frac {\int_{0}^{1}{x\left( x^{n} \minus{} x^{m} \right)\,dx}}{\int_{0}^{1}{\left( x^{n} \minus{} x^{m} \right)\,dx}},\,\overline{y} \equal{} \frac {1}{2}\cdot \frac {\int_{0}^{1}{\left( x^{2n} \minus{} x^{2m} \right)\,dx}}{\int_{0}^{1}{\left( x^{n} \minus{} x^{m} \right)\,dx}}.$

How to find the condition so that the mass center stays out of the region?

I feel that $ \overline{x}$ is correct but $ \overline{y}$ may not be. Please check your calculation once again.

Any way condition for mass center to lie outside the region is $ \overline{y} > g(\overline{x})$ or $ \overline{x} < g^{ \minus{} 1}(\overline{y})$. It is obvious once you draw the graphs roughly..
Both of them come down to $ \overline{y} > (\overline{x})^n$ That would give you the required condition on simplification.

[quote="savy2020"]I feel that $ \overline{x}$ is correct but $ \overline{y}$ may not be. Please check your calculation once again.[/quote]
Yes, I've edited, check it now please.

I now got your point as for the condition, thanks a lot!


I'm trying to figure out what differentiates me (not a contest winner) from contest winners. I work hard on my own, but that doesn't seem to cut it by itself. Is it background? Tons of practice? Lots of work with coaches or teachers? Raw talent? Surely combinations are in order, I'm just trying to figure out the typical profile of contest winners.

Any comments?

People who are ridiculously good at math and don't choke under pressure.

Coming from schools/areas where there is a tradition of contest excellence.
Critical mass of similarly talented peers.

it is possible that they are working harder than you, myself, i do math all the time, but it maybe that you are not taking away as much as you can from each problem, i.e. picking out specific strategies for each type of problem, also you might want to work on your test taking skills, i know careless errors kill me, also you may need to challenge yourself more

I'm kinda with Altheman on this. I make lots of careless mistakes, even though I know the right way of solving a problem. What it comes down to for me is that I need to practice test environments, but I don't know about you.

Other than genius...

I say the most important thing is challenging yourself as much as possible and developing mental stamina and determination.

Since you specified contest [i]winners[/i]:

To an extent, good luck is with them.

Avoiding careless mistakes is what i would say, because that difference would have allowed me usamo the past two years...and would have served me much much better in other competitions.
The ability to keep focus is important.


This problem was the one with the n-stars. I saw the math jam solution to this, but I don't understand why they were able to assume all points were on a circle. Couldn't a point be inside a region formed by two line segments?

http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=26

I don't understand what you mean a region formed by two line segments, but all the stars are regular n-stars, and as far as I know regular 2D figures always can be inscribed in circles.

What I meant was a point inside a region outlined by line segments...so a point trapped...sorta. But if what you say is true, is there any way to prove it?

A regular n-gon, or a regular n-star, has all of its angles and sides the same for each side.  This implies rotational symmetry, so it can be shown that all vertices (for n-stars, the outer vertices) are the same distance from the center of the n-gon or n-star (which is, therefore, the circumcenter).

I don't understand what you mean by "a point trapped."


function
trigonometry
quadratics
calculus
calculus computations

If $ f(x) \equal{} \frac{{e^x  \plus{} e^{ \minus{} x} }}{2}$
,then find $ f^{ \minus{} 1} (x)$ in terms of $ x$.

That is the definition of the hyperbolic cosine function cosh.  So it's inverse is $ \cosh^{\minus{}1}(x)\equal{}\ln(x\plus{}\sqrt{x^{2}\minus{}1})$ if $ x\geq 1$.

hello, let $ y\equal{}\frac{e^{x}\minus{}e^{\minus{}x}}{2}$, multiplying by $ 2$ we get
$ 2y\equal{}e^{x}\minus{}e^{\minus{}x}$ and multiplying by $ e^{x}$ we get
$ 2ye^{x}\equal{}e^{2x}\minus{}1 \Leftrightarrow 0\equal{}e^{2x}\minus{}2ye^{x}\minus{}1$
now you must solve this quadratic equation.
Sonnhard.


[hide="Hint"] Consider divisibility by $3$ and divisibility by $2$ separately. [/hide]

[quote="Cartman"]You write down the digits 1-6 in random order and you get a 6 digit number. Whats the probailty for that this number can be dividet by 6?[/quote]
[hide="Answer"]For it to be divisible by $6$, it has to be divisible by $3$ and $2$. The digits will add up to $21$, meaning it is divisible by $3$ no matter how you arrange it. Thus it just has to be divisible by $2$. The desired probability is $1/2$[/hide]


trigonometry
ratio
LaTeX
trig identities
algebra open
algebra

Hi, I was wondering how would you find the $\cos/\sin$ of any angle exactly without using a calculator. I picked a random number like $\cos(69^{\circ})$ and tried solving for the answer exactly by hand using the Euler's formula and other trig identities but I was unable to do so. I don't even know if such a task is possible.

It is possible, but it will get a very ugly and long term.
You could express $\sin$ and $\cos$ of $30^\circ$ and $36^\circ$, then those of $6^\circ$ and then $3^\circ$ by that. Now just add $60^\circ+6^\circ+3^\circ$, giving the result...

It is possible to express $\sin, \cos$ of all constructible angles just by square roots, but they get quite big.

[quote="ZetaX"]It is possible, but it will get a very ugly and long term.[/quote]Yes indeed, but like they say, ugly is just another name for beautiful.  :lol:

[quote="ZetaX"]You could express $\sin$ and $\cos$ of $30^\circ$ and $36^\circ$, then those of $6^\circ$ and then $3^\circ$ by that. Now just add $60^\circ+6^\circ+3^\circ$, giving the result...
It is possible to express $\sin, \cos$ of all constructible angles just by square roots, but they get quite big.[/quote]I am not quite sure what you mean... I know that you can find the $\cos/\sin$ of $30^{\circ}$ and and any other $2^{n}$ multiples or divisors of that number like 7.5, 15, 60, etc. But how do you find the $\cos/\sin$ of a number like 6 or 3?? :huh:

[quote="ZetaX"]Find them of $36^\circ$, then use the addition formulas to get $\cos, \sin$ of the difference to $30^\circ$.[/quote]I don't know, may be I am forgetting something, but how do you find the $\cos$ or $\sin$ of $36^{\circ}$?

We discussed, among others, the angle $18^\circ$, at another post. I think that some of the tricks we used there will be useful to you.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=90665]Link[/url]

Yes, thank you very much. That certainly helps a lot.

Look carefully:

Angle $18^\circ$:
$5\cdot 18^\circ=90^\circ$, so $2\cdot 18^\circ+3\cdot 18^\circ=90^\circ$ and then $\cos(2\cdot 18^\circ)=\sin(3 \cdot 18^\circ)$.
After some trigonometry equalities to put it all in terms of $x=\sin 18^\circ$, we get $4x^{2}+2x-1=0$. The positive solution is $\sin 18^\circ=\frac{\sqrt{5}-1}{4}$.

Now you have some possibilities:

$36^\circ=2\cdot 18^\circ$.

Maybe you can use the same trick as before: $5\cdot 36^\circ=180^\circ$, so $2\cdot 36^\circ+3\cdot 36^\circ=180^\circ$ and then $\sin(2\cdot 36^\circ)=\sin(3 \cdot 36^\circ)$...

I think there's a geometric interpretation from the properties of the golden ratio. ($5\cdot36^\circ=180^\circ$ appear in a pentagon.)

[quote="swapnillium"][quote="quantum leap"]If 45+60=105 and 105-36=69, why not just use the sum and difference formulas to get cos(69)?[/quote]But the problem is that we don't know the $\cos(36)$. Well, not so easily anyway. We would have to use Cardano's solution and do some really painstaking algebra like Lord Wings does in the link above. :([/quote]

I don't think it's that bad.  You know sine(18) so since $\cos(2\theta)=1-2\sin^{2}(\theta)$ you can find cos(36).  Then just use the sum/difference identities.

[quote="swapnillium"][quote="lordWings"]$5\cdot 18^\circ=90^\circ$, so $2\cdot 18^\circ+3\cdot 18^\circ=90^\circ$ and then $\cos(2\cdot 18^\circ)=\sin(3 \cdot 18^\circ)$.
After some trigonometry equalities to put it all in terms of $x=\sin 18^\circ$, we get $4x^{2}+2x-1=0$. The positive solution is $\sin 18^\circ=\frac{\sqrt{5}-1}{4}$.[/quote]Wait a minute... how did you get $4x^{2}+2x-1=0$, I keep getting $0=0$ no matter what I try![/quote]
We're looking for the solution to $\cos 2\alpha =\sin 3\alpha$.
Using repeatedly the trigonometric formulas for addition and the identity $\cos^{2}\alpha = 1-\sin^{2}\alpha$, we reach
$\cos 2\alpha =1-2\sin^{2}\alpha =1-2{}^{x}{}^{2}$,
$\sin 3\alpha =3\sin \alpha-4\sin^{3}\alpha =3x-4x^{3}$.
where $x=\sin\alpha$.
So, we're looking for the roots of $p(x)=4x^{3}-2x^{2}-3z+1=0$.
It's easy to see that $p(1)=0$, so $p(x)=(x-1)(4x^{2}+2x-1)=0$.
We'll only keep the solution with $0<x<1$ because $0^\circ<\alpha<90^\circ$. That's it.

[Edit:I added the following]
Notice that $4x^{2}+2x-1=0$ and $\cos 2\alpha =1-2x^{2}$.
Immediately,
$\cos 36^\circ=\frac{1}{2}+x=\frac{\sqrt{5}+1}{4}=\frac{\phi}{2}$,
where $\phi$ is the golden ratio.
It will be useful to remember that $\phi =2\cos 36^\circ$.

Oh, I see. My mistake was that I tried to take the cosine of both sides of $2\cdot 18^\circ+3\cdot 18^\circ=90^\circ$ and then I tried substituting $\cos(2\cdot 18^\circ)=\sin(3 \cdot 18^\circ)$ into that equation. How foolish of me. :blush:

Thanks for your explanation lordWings.  :)

Look at my previous post, I edited it.

Now, here's a direct approach to $\cos 36^\circ$.
It's almost exactly the same.
Now $2\alpha+3\alpha =180^\circ$, so
$\cos 2\alpha =-\cos 3\alpha$.
Using repeatedly the trigonometric formulas for addition and the identity $\sin^{2}\alpha = 1-\cos^{2}\alpha$, we reach
$\cos 2\alpha = 2\cos^{2}\alpha-1=2x^{2}-1$,
$\cos 3\alpha = 4\cos^{3}\alpha-3\cos\alpha=4x^{3}-3x$,
where $x=\cos\alpha$.
So, we're looking for the roots of $p(x)=4x^{3}+2x^{2}-3z-1=0$.
It's easy to see that $p(-1)=0$, so $p(x)=(x+1)(4x^{2}-2x-1)=0$.
Notice that $4x^{2}-2x-1=0$, with the change $2x=t$, becomes the famous relationship $t^{2}=t+1$, where the positive solution is the golden ratio $\phi$.
We'll only keep the solution with $0<x<1$ because $0^\circ<\alpha<90^\circ$.
So, again, $\cos 36^\circ =\frac{\phi}{2}$, or $\phi =2\cos 36^\circ$.

In conclusion,
$\phi=2\cos 36^\circ =\frac{1}{2\sin 18^\circ}$, where $\phi$ is the golden ratio.
I'm sure it's well known, but I hadn't seen it before and I think it's beautiful.
[Edit: Weird latex rendering, isn't it? It doesn't seem to be very well aligned.]


Let $ f: R \rightarrow R$ a continuous function that transforms any open interval into a closed interval. Prove that $ f$ is constant.


The symbols $ \star$ and $ \ast$ represent different operations, either +, -, $ \times$, or $ \div$, and $ x$ is a positive integer. Find $ x$ if $ 17 \star x \equal{} 54 \ast x$.

If $ \star$ is $ \times$, and $ *$ is -, the we have the equation
\[ 17\times x\equal{}54\minus{}x\implies18x\equal{}54\implies x\equal{}\boxed{3}\]
If someone could possibly explain how you could get that $ \star$ is $ \times$ and $ *$ is -, that would be great.  All I did was guess and check  :P

Here's my reasoning... not particularly elegant, but perhaps it will help someone else make a clearer statement.

You can see pretty quickly that they can't be the same operation.
x is a positive integer; the operation on the left must be one that makes 17 larger and the operation on the right must make 54 smaller.
So either it's multiplication and division; addition and subtraction; multiplication and subtraction; or addition and division.

It can't be multiplication and division because 17 isn't a factor of 54.
It can't be addition and subtraction because 17 is odd and 54 is even.
If it were to be addition and division, x would have to be 2 or 3 for 54 not to shrink too much, and neither works.

So it must be multiplication and subtraction.


How are you gonna celebrate Easter?

I'm going to my Aunts.

I had a easter egg hunt! The bunny was nice this year jk :P

easter

wtf

ok that's why they didn't have school last friday

I would go


I'm not Texan, but I have my pick of tales about Texan education. If any of you here can help make this thing better...

[quote]Wichita Falls HS, May 2009. The teacher tells him about what to expect in a TAKS test, regarding quadratic equations. Mrs. Bryce, a blonde-haired, middle-aged woman whose density is very close to 1, asks her class one thing before the class warms up.

Mrs. Bryce: Before I engage in the preparation activities for the exit-level TAKS, I need you to warm up. Please solve the following semi-linear system: $ Y\equal{}X^2\minus{}6X\plus{}8$ and $ Y\equal{}X\minus{}5$.

Ludveck: It's not true that a quadratic equation will have no roots. It will either have one or two roots. The answer is $ \frac{7}{2}\pm\,i\sqrt3$.

Mrs. Bryce: Ludveck, why do you find $ \frac{7}{2}\pm\,i\sqrt3$ where everyone else found no root? Do you imply I'm a fool that does not know anything about square roots of negative numbers?

Ludveck: Some semi-linear systems have collision (intrersection) points in the complex numbers.

Mrs. Bryce: I have always said that square roots of negative numbers don't exist. Prove me wrong. You have 24 hours to prove it and you will be able to sit TAKS like everyone else.

Ludveck: I could show you [i]everything[/i] you'll ever need to know about complex numbers.

By that Ludveck meant merely to be able to do the basics in the way of complex numbers, such as adding, substracting, multiplying and dividing complex numbers, knowing that $ \sqrt{\minus{}1}\equal{}i$ or $ i^2\equal{}\minus{}1$. It's way beyond most proficient, commended or advanced-ranked students in Grade-X TAKS. For him, the exit-level TAKS is Grade-XI TAKS and he feels that there will be no mention to complex numbers.

Mrs. Bryce: Go ahead and find me a good series of exercises about complex numbers!

What Ludveck found was an excerpt from a Quebecer-made textbook about linear algebra. He fels his teacher didn't need even adding up complex numbers, just to prove that square roots of negative numbers exist.

Ludveck: When the discriminant $ B^2\minus{}4AC$ is negative, we then find a square root of a negative number, and, when we work over the real numbers, we say that no real solution exists.

Mrs. Bryce: I win...

Ludveck: I'm not done! Of course, mathematicians want to go further and consider this solution rather than to reject it. $ \frac{\minus{}B}{2A}\pm\frac{i}{2A}\sqrt{B^2\minus{}4AC}$ if the discriminant is smaller than 0.

Mrs. Bryce: I admit defeat. But I'll list those who didn't fail and everyone else can have [i]a day off[/i] because today is dedicated to mathematics remedial for those who failed last year.[/quote]

Quite funny, but if all goes well, it can be a possibility of [i]eight[/i] days off if one had a normal school path.

[quote="CEM_College_Champ"]
Ludveck: I'm not done! Of course, mathematicians want to go further and consider this solution rather than to reject it. $ \frac { \minus{} B}{2A}\pm\frac {i}{2A}\sqrt {B^2 \minus{} 4AC}$ if the discriminant is smaller than 0.
[/quote]

[hide="Nitpickiness"]Isn't it supposed to be $ \frac{i}{2A}\sqrt{4AC\minus{}B^2}$?[/hide]

All right then...

But the teacher was fooled because she was made to forget all the math that did not pertain to TAKS-covered math (any grade) or pre-calculus. There was also a bit of Ludveck as well, in his anger towards Mrs. Bryce, he distorted the answer. Or he is a bit absent-minded...


calculus
derivative
function
real analysis
real analysis unsolved

The problem is just to find out whether it always holds, i.e., whether for any two differentiable functions $f,g : \mathbb R\to \mathbb R$ the product $f'g'$ has the intermediate value property.

[quote="t0rajir0u"]Correct me if I'm wrong, but the definition of a differentiable function $f$ only requires that $f'(x)$ be defined on $\mathbb{R}$, which means $f'g'$ is an arbitrary (and not necessarily continuous) function...?[/quote]

This much is clear, but, for example, $f'$ always has the intermediate value property even if $f'$ fails to be continuous.

One special case is easy: if $f'(a)g'(a)<0$ and $f'(b)g'(b)>0$, then $f'(x)g'(x)=0$ for some $x$ between $a$ and $b$. Indeed, either $f'$ or $g'$ changes sign between $a$ and $b$, and therefore vanishes somewhere.

I am skeptical about the general case, but do not have a counterexample.

[quote="mlok"]
I am skeptical about the general case, but do not have a counterexample.[/quote]

You are right. Here is an example. First we partially define $f(x)$ as follows:
\[f(x) = \begin{cases}0, & \text{if }x=0\\ \frac{1}{2k}, & \text{if }x\in [\frac{1}{2k},\frac{1}{2k-1}]\\-\frac{1}{2k}, & \text{if }x\in [-\frac{1}{2k-1},-\frac{1}{2k}] \end{cases}\]
It remains to define $f(x)$ on intervals $(1/(2k+1), 1/(2k))$. We do so in such a way that
[list]
a. $f(x)$ is differentiable on every interval $[1/(2k+1), 1/(2k)]$, and
b. $f(x)$ is monotone
[/list]
Similarly, we partially define $g(x)$
\[g(x) = \begin{cases}0, & \text{if }x=0\\ \frac{1}{2k}, & \text{if }x\in [\frac{1}{2k+1},\frac{1}{2k}]\\-\frac{1}{2k}, & \text{if }x\in [-\frac{1}{2k},-\frac{1}{2k+1}] \end{cases}\]
and then extend it by monotonicity.

Let us now compute $f'(x) g'(x)$. For every $x\neq 0$, either $f'(x) =0$ or $g'(x)=0$. Therefore, for every $x\neq 0$,
\[f'(x) g'(x) = 0. \]
On the other hand,
\[f'(0) g'(0) = 1 \cdot 1 = 1. \]


abstract algebra
linear algebra
matrix
superior algebra
superior algebra unsolved

Show that the group defined by generators a,b and relations a^2 = e, b^3 = e is infinite and nonabelian. (Hungerford I.9.7)

Attempted solution:

This is so intuitive: there is no relation between a and b so we can just write something like abab^2ab^2ababa and just make it as long as we want. But...how can I [i]prove[/i] that this group is NOT abelian  or that each of the elements we write are different in the quotient group F/N? Whenever, $ aba^{ \minus{} 1}b^{ \minus{} 1}$ is a relator, then we know the group IS abelian but going the other way is much different. I really don't know where to start...I just need to find any two elements and prove that they are in different cosets of the factor group...but how does one prove that two cosets are different...the number of cosets is probably infinite so you can't just like "partition" the group into the different cosets right?

Abelian implies finite here, so if you can prove that the group is infinite then it is also nonabelian.  To that end, try to prove that a particular element has infinite order.  (I might be mistaken here.)

I think one needs to find a realization in some infinite group.
Like in $ G \equal{} PSL_2(\mathbb{C}) \equal{} SL_2(\mathbb{C})/\{\pm I_2\}$ with $ Z: \equal{} \{\pm I_2\}$.
Then $ a: \equal{} \begin{pmatrix} & 1 \\
\minus{} 1 & \end{pmatrix}Z,b: \equal{} \begin{pmatrix} & \minus{} 1 \\
1 & 1\end{pmatrix}Z$ have orders $ 2,3$ in $ G$
and $ ab \equal{} \begin{pmatrix}1 & 1 \\
&1\end{pmatrix}Z$ has no finite order in $ G$,
as $ \begin{pmatrix}1 & 1 \\
&1\end{pmatrix}^n \equal{} \begin{pmatrix}1 & n \\
&1\end{pmatrix}\not\in Z$ for all $ n\in\mathbb{N}$.

Is it clear that $ a$ and $ b$ have no nontrivial relations, though?

Who said that?
It's an epimorphic image, which is all we need.

[quote="olorin"]I think one needs to find a realization in some infinite group.
Like in $ G \equal{} PSL_2(\mathbb{C}) \equal{} SL_2(\mathbb{C})/\{\pm I_2\}$ with $ Z: \equal{} \{I_2\}$.
[/quote]

Could you explain this notation? Thanks.

Sorry for the typo, I meant $ Z\equal{}\{\pm I_2\}$ the centre of $ SL_2(\mathbb{C})$.
You do know what $ SL_2(\mathbb{C})$ is, right?

[quote="olorin"]Sorry for the typo, I meant $ Z \equal{} \{\pm I_2\}$ the centre of $ SL_2(\mathbb{C})$.
You do know what $ SL_2(\mathbb{C})$ is, right?[/quote]

Yes, but what is $ PSL_2(\mathbb{C})$?

$ PSL_2(\mathbb{C})$ is the projective special linear group, which is obtained by identifying a matrix with its negative (which is what that notation means).  In other words, it is the group of $ 2 \times 2$ complex matrices with determinant $ 1$.

olorin:  Yes, I understand now.

the free product of two groups $ A$ and $ B$ (i.e. their coproduct in the category of groups) can be explicitely constructed as set of formal products whose factors are nontrivial and come alternating from $ A$ or $ B$. in this case, we have the free product of Z/3 and Z/2. from here the two properties are clear. by the way, this group is isomorphic to PSL(2,Z).