I don't have the 2009 10B booklet on me, so I can't check my answers. Does anybody have one that they can upload so I could see how I did?

Thanks.


Awesome!

I'll be looking forward to it (That is, if I can somehow squeeze my time to take it)

Do we have to sign up ahead of time, or can we simply sit down, take it, and submit answers at the appropriate time?

You can simply sit down, take it, and submit answers at the time it is administered.

Awesome, I'll be looking forward to it. ;)


geometry
rectangle
LaTeX
real analysis
topology
advanced fields
advanced fields unsolved

Let's S be ?-algebra in set X and F  be ?-algebra in set Y . S?F  is defined as the smallest ?-algebra in set X?Y containing all measurable rectangles .
    Is it true that S?F  is equal to the set of all countable unions of measurable rectangles ?If so could you prove it ,If not what is a contradictory example ?

Could you please use only symbols that every computer can read¿ If it's mathematical, you should use $ \text{\LaTeX}$.

I guess it is $ \sigma$-algebra, $ X\times Y$ and $ S\times F$.

No, the claim is not true.

A simple example: Take $ X \equal{} Y \equal{} \mathbb{R}$ and the $ \sigma$-algebras $ S$ and $ F$ being the $ \sigma$-algebras generated by open intervals (namely the smallest $ \sigma$-algebras containing all open intervals). $ X \times Y$ is just the plane $ \mathbb{R}^2$. Consider a set $ I \in \mathbb{R}^2$ which is, say, an interval which is neither horizontal nor vertical.

Now show:

A. The set $ I$ is measurable. To see this, construct sets $ A_n$ each of which is the union of finitely many squares with side $ 1/n$ which contains $ I$. The squares should have horizontal and vertical sides, so that each one is just a product of intervals. Each $ A_n$ is definitely measurable (that is, belongs to the $ \sigma$-algebra $ S \times F$). Finally , $ I$ is the intersection of the $ A_n$, so it is itself measurable.

B. The set $ I$ is not a countable union of rectangles. This is trivial: the only rectangles contained in $ I$ are single points, and $ I$ is not a countable union of points.

All right, fedja ,these are the same you guessed . Zetax ,could you guide me to use LATEX ?
Thanks randomgraph ,I'm considering what you wrote .

$ \text{\LaTeX}$ is a large and complicated subject. Look at some of the threads in [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]this subforum[/url].

Thanks for your reply .I'm not sure to understand what you wrote precisely ,however let me review that on my own .I think you mean S is set of borl subsets of R  as well as F in your example .I is an open subset of R2 in a form unlike a rectangle ,e.g. an open disc or interior region of a triangle ,etc. I understand your proof for measurability of I ,but for example an open disc or interior region of a triangle can be regard as a countable union of open rectangles which clearly are measurable rectangles .

In the example, $ I$ is not open. It's a closed set, or perhaps would be closed if we added the endpoints. Every open set is a countable union of open rectangles and hence lies in the $ \sigma$-algebra. This $ I$ (for example, $ \{(x,x): 0\le x\le 1\}$) can easily be shown to be in the $ \sigma$-algebra, which is the Borel $ \sigma$-algebra, but $ I$ is not a countable union of aligned rectangles.

You can certainly take $ S$ and $ F$ to be the Borel sets, or Lebesgues sets. It doesn't matter much.

The set $ I$ is not an open set. Take $ I$ to be $ \{(x, x) | x \in [0,1] \}$, say. Or take $ I$ to be a circle, or any curve, it doesn't matter. The point is that any such curve is measurable, but it doesn't contain any rectangles other than single points or finite sets of points. Therefore it is never a countable union of rectangles.

Thanks to all .I understood what you intended ,good luck .


limit
logarithms
linear algebra
matrix
vector
invariant
complex numbers

Let $\{x\}$ be a sequence of positive reals $x_1, x_2, \ldots, x_n$, defined by: $x_1 = 1, x_2 = 9, x_3=9, x_4=1$. And for $n \geq 1$ we have:

\[x_{n+4} = \sqrt[4]{x_{n} \cdot x_{n+1} \cdot x_{n+2} \cdot x_{n+3}}.\]

Show that this sequence has a finite limit. Determine this limit.

It is strange this is the third problem from Vietnam TST. Usually they are much harder.  ;)

I'm having a hard time justifying that the limit exists.  Can anyone help?

[quote="Rafal"]Prove that $\{x_n\}$ is bounded (from the fact that at least one of the four elements of the seqence before the $n$-th element is not smaller and at least one not bigger than $x_n$) and that $limsupx_n=liminfx_n$ using the fact  (what's easy to prove looking at the value $\frac{x_{n+1}}{x_n}$) that ${\lim_{n->\infty}\frac{x_{n+1}}{x_n}}=1$.[/quote]

Of course $(x_n)$ is bounded, but do you care to elaborate on the last part?  I don't think it's that easy, and as they say, the devil is in the details.

By the way, \limsup, \liminf, and \rightarrow give $\limsup$, $\liminf$, and $\rightarrow$.

There is direct proof using circulant matrices for this type of problems. Of course, otherwise it is a little bit difficult to explain all details.

Suppose that there exist two distinct subsequences $\{a_n\},\{b_n\}$ of the sequence $\{x_n\}$, such that
$\lim_{n->\infty}a_n>\lim_{n->\infty}b_n$ (if the sequence $\{x_n\}$ is not convergent we can obviously choose such two).
So $\lim_{n->\infty}\frac{a_n}{b_n}=c>1$.
Thus there exist such $n_0$ that $\frac{a_n}{b_n}>c_1>1$ for all $n>n_0$.
Thus putting $a_n=x_i$ $b_n=x_j$ we obtain supposing $i>j$
$\frac{a_n}{b_n}=\frac{x_i}{x_j}=\frac{x_i}{x_{i-1}}\frac{x_{i-1}}{x_{i-2}}...\frac{x_{j+1}}{x_{j}}=
\sqrt[4]{\frac{x_{i-1}}{x_{i-2}}\frac{x_{i-2}}{x_{i-3}}...\frac{x_{j}}{x_{j-1}}}=\sqrt[4]{\frac{x_{i-1}}{x_{j-1}}}=...$(analogically)$=\sqrt[4^{j-1}]{\frac{x_{i-j+1}}{x_1}}=\sqrt[4^{j-1}]{x_{i-j+1}}$.
Because $0<m<x_n<M<\infty$ ($\{x_n\}$ is bounded), so $\sqrt[4^{j-1}]{m}<\frac{a_n}{b_n}=\sqrt[4^{j-1}]{x_{i-j+1}}<\sqrt[4^{j-1}]{M}$, so for sufficient, big $j$ so sufficeint big $n$ (thus $n>n_0$ it's $|\frac{a_n}{b_n}-1|<c_1-1$ - contradiction.
The same is when we suppose $j>i$ and then taking sufficient, big $i$ (then $\sqrt[4^{i-1}]{\frac{1}{M}}<\frac{a_n}{b_n}=\sqrt[4^{i-1}]{\frac{1}{x_{j-i+1}}}<\sqrt[4^{i-1}]{\frac{1}{m}}$).

I don't see any devil.

[Edit] there is one stupid mistake above so it should be corrected.

[Edit] [2] ok, that one was stupid...I think linear algebra will help, I don't have time to specify my new idea but I'll just wright an introduction:
first we construct a new sequence $\{y_n\}$, such that $y_n=\log_3x_n$, so we have
a recurential sequence $y_{n+4}=\frac{y_n+y_{n+1}+y_{n+2}+y_{n+3}}{4}$ with given $y_1=0,y_2=2,y_3=2,y_4=0$.
Let's define in standard $R^4$ a linear transformation given $f(x_1,x_2,x_3,x_4)=(x_2,x_3,x_4,\frac{x_1+x_2+x_3+x_4}{4})$. We can wright easily the matrix of this endomorphism $End(R^4)$, and find the Jordan matrix and other stupid but easy things in order to easily compute $A^n$ where $A$ is the matrix of $f$ defined in vectors $(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)$, so we can find the last coefficient (which equals $x_{n+4}$) in the vector $f^n((0,2,2,0))$, sorry but I'm not a specialist in English when talking about linear algebra :).. this must work...and of course we will obtain the limit of the sequence $\{y_n\}$ computing this way.

[Edit] [3]  I suppose harazi might mean something similar what I noticed reading previous posts...by the way, what's that method with circulate matrices ?

It is practically the same method as the last one you presented, Rafal.

I didn't read the Rafal's solution carefully,but I think he didn't answer the what is the limit.
I think the limit shoulde be 1.but My solution very long and I don't intends to post it.who can have a shorter solution?

_________________

Let $y_n=\log _3{x_n}$. We have $y_1=0,y_2=y_3=2,y_4=0,y_n=\frac 14 (y_{n-1}+y_{n-2}+y_{n-3}+y_{n-4})$. We have characteristic equation: $4x^4=x^3+x^2+x+1$. One is roots is 1, for other roots we have $4x^3+3x^2+2x+1=0$, it give |x|<1. Used beginner means it is easy to solve recurent equation. We have $y_{n+1}+2y_{n+2}+3y_{n+3}+4y_{n+4}=y_n+2y_{n+1}+3y_{n+2}+4y_{n+3}=...=0+1*2+2*2+3*0=6.$ Therefore \[ \lim_{n\to \infty } y_n=3/5,\lim_n x_n=3^{3/5}  \].

[quote="Hawk Tiger"]Hi ,rust.
I use a similar method with yours ,but more long(may be sth useless)
but you maybe made a mistake
$y_1+2y_2+3y_3+4y_4=10$but not $6$
so the limit should be 1[/quote]
But we have to prove that exits limit of $y_n$. Then how do you do this?

Read the Rust's post.
$y_n=\lambda+\mu_1 r{_1}^n+\mu_2 r{_2}^n+\mu_3 r{_3}^n$ where the ($r_i)_i$ are complex numbers with modulus $<1$.
Thus $\lim(y_n)=\lambda$. Considering $y_1,y_2,y_3,y_4$, we obtain $\lambda=1$ and $\lim(x_n)=3$.


Find all function $f(x): \mathbb{R}^* \rightarrow \mathbb{R}^*$ satisfying:  \[ f(f(x))+f(x)=6x \forall x>0 \]

Sorry the correct problem is $f: R^+ \rightarrow R^+$
[u][b]Help me please![/b][/u] :omighty:

Solution with missing steps:

We have $f(f(x)) = 6x - f(x)$

Put $f(x)$ in place of $x$ we see that $f(f(f(x))) = 6f(x) - f(f(x)) = 7f(x) - 6x$

Continuing this way we see that $f^k(x) = a_kx + b_kf(x)$ where $f^k(x) = f(f^{k-1}(x))$

$a_k$ and $b_k$ satisfy a simple second order recurrence relation (of the form $c_{n+2} = Ac_{n+1} +Bc_{n}$) which can be solved easily.

Notice that $a_k$ and $b_k$ are alternately positive and negative. We can use this and the fact that $f^k(x) > 0$ to bound $f(x)/x$ between $a_{2k}/b_{2k}$ and $a_{2k-1}/b_{2k-1}$. Taking limits as $k \to \infty$ we see that $f(x) = 2x$


trigonometry calculus derivative geometry circumcircle function conics

Let $ a$ and $ b$ be positive real numbers with $ a\ge b$. Let $ \rho$ be the maximum possible value of $ \frac{a}{b}$ for which the system of equations
\[ a^2\plus{}y^2\equal{}b^2\plus{}x^2\equal{}(a\minus{}x)^2\plus{}(b\minus{}y)^2\]has a solution in $ (x,y)$ satisfying $ 0\le x<a$ and $ 0\le y<b$. Then $ \rho^2$ can be expressed as a fraction $ \frac{m}{n}$, where $ m$ and $ n$ are relatively prime positive integers. Find $ m\plus{}n$.

[hide]Consider a cyclic quadrilateral ABCD with
$ \angle B \equal{} \angle D \equal{} 90$, and $ AB \equal{} y, BC \equal{} a, CD \equal{} b, AD \equal{} x$.

$ AC^2 \equal{} a^2 \plus{} y^2 \equal{} b^2 \plus{} x^2$

From Ptolemy's Theorem $ ax \plus{} by \equal{} AC(BD)$

$ AC^2 \equal{} (a \minus{} x)^2 \plus{} (b \minus{} y)^2 \equal{} a^2 \plus{} y^2 \plus{} b^2 \plus{} x^2 \minus{} 2(ax \plus{} by) \equal{} 2AC^2 \minus{} 2AC*BD$
Simplify that we have $ BD \equal{} AC/2$

Note the circumcircle of ABCD has radius AC/2, hence BD = radius and has an arc of 60 degrees, so
$ \angle C \equal{} 30$. Let $ \angle BDC \equal{} \theta$.

$ a/b \equal{} BC/CD \equal{} \sin \theta / \sin(150 \minus{} \theta)$, where both $ \theta$ and $ 150 \minus{} \theta$ are $ \leq 90$ since
triangle BCD must be acute. Since $ \sin$ is an increasing function over (0, 90), $ \sin \theta / (\sin(150 \minus{} \theta)$ is also increasing function over (60, 90).

$ a/b$ maximizes at $ \theta \equal{} 90$ => $ a/b$ maximizes at $ 2/\sqrt {3}$[/hide]

These solutions are very nice. I was thinking geometric during the test too, but stuck to normal conic section curves and such.

[hide="Non-geometric..."]
Notice that the given equation implies
\[ a^2\plus{}y^2\equal{}b^2\plus{}x^2\equal{}2(ax\plus{}by)\]
We have $ 2by \ge y^2$, and so $ 2ax \le a^2 \implies x \le \frac{a}{2}$.  So then notice $ b^2\plus{}x^2\equal{}a^2\plus{}y^2 \le a^2$, so $ b^2 \ge \frac{3}{4}a^2 \implies \rho^2 \ge \frac{4}{3}$.  And the solution $ (a, b, x, y)\equal{}(1, \frac{\sqrt{3}}{2}, \frac{1}{2}, 0)$ satisfies the equation.  Hence $ \rho^2\equal{}\frac{4}{3}$, yielding 007.
[/hide]

i was going through the AOPS wiki and I noticed that K81o7's solution has a small typing error in it. He actually  says that the minimum value is 4/3, not the maximum. This doesn't affect the validity of his solution however i do want it to be corrected.

[quote=calc rulz][hide]Consider the points $ (a,y)$ and $ (x,b)$. They form an equilateral triangle with the origin. We let the side length be $ 1$, so $ a \equal{} \cos{\theta}$ and $ b \equal{} \sin{\left(\theta \plus{} \frac {\pi}{3}\right)}$. Thus $ f(\theta) \equal{} \frac {a}{b} \equal{} \frac {\cos{\theta}}{\sin{\left(\theta \plus{} \frac {\pi}{3}\right)}}$ and we need to maximize this for $ 0 \le \theta \le \frac {\pi}{6}$. A quick differentiation shows that $ f'(\theta) \equal{} \frac {\cos{\frac {\pi}{3}}}{\sin^2{\left(\theta \plus{} \frac {\pi}{3}\right)}} \ge 0$, so the maximum is at the endpoint $ \theta \equal{} \frac {\pi}{6}$. We then get
\[ \rho \equal{} \frac {\cos{\frac {\pi}{6}}}{\sin{\frac {\pi}{2}}} \equal{} \frac {\sqrt {3}}{2}
\]
so $ \rho^2 \equal{} \frac {3}{4}$ and the answer is $ 3 \plus{} 4 \equal{} \boxed{007}$.

[/hide]

EDIT: Hmm I think I switched a and b, but the answer is still the same.[/quote]

Sorry for the bump, but wouldn't in this solution $f(\theta)$ be maximized at $\theta=0$? This gives a value of $p=\dfrac{2}{\sqrt{3}}$ which matches the other solutions...

He switched $a,b$, which changes the problem a bit.

For a non-calculus way to maximize the function in calc rulz's post:
[hide]
I will use degrees. Let $f(x)=\frac{\cos x}{\sin(x+60)}$. We need to maximize $f$ on $[0,30]$.

Suppose $k$ is an upper bound of $f$ on this range; in other words, assume $f(x)\le k$ for all $x$ in this range. Then: \[\cos x\le k\sin(x+60)=k\cdot\left(\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x\right)\]
\[\rightarrow 0\le \left(\frac{\sqrt{3}}{2}-1\right)\cos x+\frac{k}{2}\sin x\rightarrow 0\le (\sqrt{3}k-2)\cos x+k\sin x\]
\[\rightarrow (2-\sqrt{3}k)\cos x\le k\sin x\rightarrow \frac{2-\sqrt{3}k}{k}\le \tan x,\]
for all $x$ in $[0,30]$. In particular, for $x=0$, $\frac{2-\sqrt{3}k}{k}$ must be less than or equal to $0$, so $k\ge \frac{2}{\sqrt{3}}$.

The least possible upper bound of $f$ on this interval is $k=\frac{2}{\sqrt{3}}$. This inequality must hold by the above logic, and in fact, the inequality reaches equality when $x=0$. Thus, $f(x)$ attains a maximum of $\frac{2}{\sqrt{3}}$ on the interval.
[/hide]

This method also applies to maximizing/minimizing functions like $f(x)=\frac{x^2+4x+13}{3x^2+2x+3}$ for $x$ in the reals. See #48 here: http://holbrook.bergen.org/oldcomp/2015/Grade8.pdf
The solution is here: http://holbrook.bergen.org/oldcomp/2015/Grade8sol.pdf

[hide=Non-Calculus Solution]
Consider the points $O(0,0)$, $B(x,b)$ and $A(a,y)$ in that order (clockwise). The problem statement gives that these three points form an equilateral triangle. Let $l_1$ be the line parallel to the x-axis going through $B$ and let $l_2$ be the line parallel to the y-axis going through $A$. Let $l_1$ hit the y-axis at $S$ and $l_2$ the x-axis at $P$. Let $l_1 \cap l_2 = T$. Since $OSTP$ has length $a$ and width $b$, finding the rectangle with $a/b$ maximized reduces to finding the triangle such that $OSTP$ has the longest length. This clearly happens when $A$ lies on the x-axis in which case we get that the ratio of $a$ to $b$ is the ratio of the side of an equilateral triangle to it's height which is $\rho = \frac{2}{\sqrt{3}}$ giving us an answer of $\boxed{007}$.
[/hide]


A pair of linear operators $ \sigma , \tau \in \mathcal{L(V)}$ are simultaneously diagonalizable if there is an ordered basis $ \mathcal{B}$ for $ V$ for which [$ \tau$]$ _{B}$ and [$ \sigma$]$ _{B}$ are both diagonal. Prove that two diagonalizable operators $ \sigma$ and $ \tau$ are simultaneously diagonalizable if and only if $ \sigma\tau \equal{} \tau\sigma$.

 "$ \rightarrow$"

$ \sigma\tau$ =$ P^{ \minus{} 1}[ \tau]_{B} P \cdot P^{ \minus{} 1} [\sigma]_{B}P$.

which are two diagonal matrices, with eigenvalues down the diagonal. can i simply switch $ [ \tau]_{B}$ and $ [\sigma]_{B}$ and conclude $ \sigma\tau \equal{} \tau\sigma$? or does there need to be more work?

also, how do i go in the other direction ( assuming $ \sigma\tau \equal{} \tau\sigma$)?

thanks.

Diagonalize one, keeping equal eigenvalues in blocks, and look at the entries of the other with respect to that basis. It still commutes, so it's block diagonal. Diagonalize each of those blocks, and we're done.

how would i go in the other direction though? (where $ \sigma, \tau$ are diagonalizable operators and $ \sigma\tau \equal{} \tau\sigma$)

the problem gives a hint , that if $ \sigma\tau \equal{} \tau\sigma$ then the eigenspaces of $ \tau$ are invariant under $ \sigma$. My problem, is I m not so sure of where invariance will take me.

help is greatly appreciated.

thanks

What I wrote [b]is[/b] the other direction.

This is pretty much what jmerry did (he did in matrix representation, i do in basis), hope this helps.

Decompose $ V$ into a direct sum of eigenspaces of $ \tau$. Restrict $ \sigma$ to each eigenspaces, which makes sense because it's invariant. Decompose each eigenspace into direct sum of eigenspaces of $ \sigma$. Pick an eigenbasis for each eigenspace, we are done.

just one question:

you picked an eigenbasis for each eigenspace by virtue of the fact that we're dealing with eigenspaces right?

also, what's the difference between a basis and an eigenbasis?

thanks.

when you pick an eigenbasis for the eigenspace, do you just simply write elements $ v \in$ eigenspace as a linear combination of what?

( the general question is just how you would write the bases picked for each eigenspace)

thanks

Oops, I am sorry, I meant to say pick a basis in each eigenspace.


Find all $ (x,y)$ such that $ sin(x \plus{} y) \equal{} sinx \plus{} siny$ & $ |x| \plus{} |y| \equal{} 1$

sin(x+y) = sinx + siny

<=> 2sin((x+y)/2) cos((x+y)/2) = 2 sin((x+y)/2) cos((x-y)/2)

<=> cos((x+y)/2) = cos((x-y)/2)

or cos((x+y)/2) - cos((x-y)/2) = 0

<=> 2 sin(x/2) sin(y/2) = 0

=> sin(x/2) = 0   or  sin(y/2) = 0

           x = 2n*pi  or  y = 2n*pi ,  where n Є Z

lxl = 2n*pi ,   n Є N
lyl = 2n*pi  ,  n Є N.

which means that

lxl = 0, 2pi, 4pi,......

lyl = 0, 2pi, 4pi, ......

but I don't see how lxl + lyl = 1 :noo:

Could you please post the complete solution? Thanks.

From Attila the Hun's solution, we have $ x\equal{}0$ or $ y\equal{}0$
If $ x\equal{}0$ then $ y\equal{}\pm 1$
If $ y\equal{}0$ then $ x\equal{}\pm 1$

[quote="apratimdefermat"]Attila's solution is incomplete ......But what Attila has done also suggests that $ sin(x \plus{} y)/2 \equal{} 0$ . He evidently should not have canceled it...And kyoshiro's answers are also not complete

Hint:And proceeding that way we have to find the conditions which satisfy $ |x| \plus{} |y| \equal{} 1$[/quote]
Oh you're right! :oops:
I don't take notice of that.
If  $ \sin\frac {x \plus{} y}{2} \equal{} 0$ then $ x \plus{} y \equal{} 2k\pi$ for $ k\in \mathbb{Z}$
Since $ |x \plus{} y|\le |x| \plus{} |y| \equal{} 1$ then $ x \plus{} y \equal{} 0\implies x \equal{} 0.5,y \equal{} \minus{} 0.5$ or $ x \equal{} \minus{} 0.5,y \equal{} 0.5$

oh, sorry guys.  :blush: you cannot cancel out sin(x+y/2)


We define \[ X \equal{} \left\{ {t\varphi_1 \left( x \right): t \in \mathbb{R}} \right\}
\]where $ \varphi_1$ is the first eigenfunction of operator $ \minus{}\Delta$ in bounded smooth domain. Is it true that $ X$ is convex and compact in $ H_0^1$?

$ X$ is just a line, isn't it? A line is convex, but is not compact.


On the figure, the quadrilateral $ ABCD$ is a rectangle, $ P$ lies on $ AD$ and $ Q$ on $ AB.$ The triangles $ PAQ, QBC,$ and $ PCD$ all have the same areas, and $ BQ \equal{} 2.$ How long is $ AQ$?

[img]http://i250.photobucket.com/albums/gg265/geometry101/NielsHenrikAbel1995Number2.jpg[/img]

A. 7/2
B. $ \sqrt{7}$
C. $ 2 \sqrt{3}$
D. $ 1 \plus{} \sqrt{5}$
E. Not uniquely determined

Let$ \left\{\begin{array}{ccc}AQ & \equal{} & a \\ AP & \equal{} & b \\ PD & \equal{} & c \end{array}\right.$
So,$ \left\{\begin{array}{ccc}\triangle PAQ & \equal{} & \frac{ab}{2} \\ \triangle QBC & \equal{} & b\plus{}c \\ \triangle PCD & \equal{} & \frac{ac}{2}\plus{}c \end{array}\right.$
Than the assumption of the problem
$ \left\{\begin{array}{ccc} \frac{ab}{2} & \equal{} & b\plus{}c \\ b\plus{}c & \equal{} & \frac{ac}{2}\plus{}c \\ \frac{ac}{2}\plus{}c & \equal{} & \frac{ab}{2} \end{array}\right.$
$ \iff \left\{\begin{array}{ccc} b & \equal{} & \frac{ac}{2} \\ \frac{ac}{2}\plus{}c & \equal{} & \frac{ab}{2} \end{array}\right.$
$ \iff \frac{ac}{2}\plus{}c \equal{} \frac{a\frac{ac}{2}}{2} \equal{} \frac{a^2c}{4}$
$ \iff a^2 \minus{} 2a \minus{} 4 \equal{} 0$
$ \iff a\equal{}1\pm\sqrt{(\minus{}1)^2\minus{}1\cdot(\minus{}4)}\equal{}1\pm\sqrt{5}$
Thus $ a\equal{}1\plus{}\sqrt{5} \ \ (\because a>0)$

[quote="Silverfalcon"]On the figure, the quadrilateral $ ABCD$ is a rectangle, $ P$ lies on $ AD$ and $ Q$ on $ AB.$ The triangles $ PAQ, QBC,$ and $ PCD$ all have the same areas, and $ BQ \equal{} 2.$ How long is $ AQ$?

[img]http://i250.photobucket.com/albums/gg265/geometry101/NielsHenrikAbel1995Number2.jpg[/img]

A. 7/2
B. $ \sqrt {7}$
C. $ 2 \sqrt {3}$
D. $ 1 \plus{} \sqrt {5}$
E. Not uniquely determined[/quote]
[hide]Set the length of $ BC$ as x. So the area of triangle $ BQC$ is x. Since the areas of $ PAQ$, $ QBC$, and $ PCD$ are the same, they all have areas of x. Set the length of $ AQ$ to y. Since the area of triangle $ PAQ$ is x, the side length of $ PA$ is $ \frac{2x}{y}$. Since the length of $ AD$ is x, the length of $ DP$ is $ \frac{xy\minus{}2x}{y}$. And since the length of $ CD$ is $ y\plus{}2$, we now have an equation.
$ \frac{xy\minus{}2x}{y}\cdot\frac{y\plus{}2}{2}\equal{}x$
Simplifying this we now have $ xy^{2}\minus{}4x\equal{}2xy$.
Divide by x to have a quadratic for y.
$ y^{2}\minus{}2y\minus{}4\equal{}0$.
$ \frac{2\pm\sqrt{20}}{2}$
This simplifies to our final answer, $ 1\pm\sqrt{5}$. Since the lengths cannot be negative, the final answer is $ \boxed{1\plus{}\sqrt{5}}$[/hide]


The diagonals of a rhombus have lengths of 18 and 24. Find the area of a circle inscribed in this rhombus.

[hide] The area of this rhombus is (18)(24)/2=9(24) since the area of the rhombus is half the product of the diagonals.

Now, we know that the diagonals of a rhombus are perpindicular and they bisect each other.  So, call the rhombus ABCD and AB=24 and BD=18.  Say that the diagonals intersect at O.  Thus, we have that OA=12 and OB=9.  By the Pythogorean Theorem, we have that AB=15=BC=CD=AD.

Note that a rhombus is also a parrelelogram.  So if we make AX perpindicuar to CD at X, we can say that the area of the rhombus is (AX)(CD).  Since we know that CD=15, we can say that the area of the rhombus is AX(15).  Now, since the area is (18)(24), we can say that AX(15)=9(24), so AX=72/5.  This is going to be the diameter of our circle.  Thus, the radius is 36/5=7.2.  Finally, we can say that the area is $(7.2)^{2}\pi=\boxed{51.84\pi}$ [/hide]


superior algebra
superior algebra solved

$A$ is a ring without zero divisors. Show that all elements in $(A,+)$ have the same order.

Consider $0\neq a,b\in A$. Then we have $(na)b=a(nb)$ for all $n\in\Bbb N$.  This implies $na=0\Leftrightarrow nb=0$ so $a$ and $b$ have the same order. The same holds for all pairs of non-zero elements of $(A,+)$, q.e.d.

Misha


inequalities
trigonometry
inequalities unsolved

Suppose  $\theta_{i}\in(-\frac{\pi}{2},\frac{\pi}{2}), i = 1,2,3,4$. Prove that, there exist $x\in \mathbb{R}$, satisfying two inequalities \begin{eqnarray*} \cos^2\theta_1\cos^2\theta_2-(\sin\theta\sin\theta_2-x)^2 &\geq& 0, \\ \cos^2\theta_3\cos^2\theta_4-(\sin\theta_3\sin\theta_4-x)^2 & \geq & 0 \end{eqnarray*}
if and only if \[ \sum^4_{i=1}\sin^2\theta_i\leq2(1+\prod^4_{i=1}\sin\theta_i + \prod^4_{i=1}\cos\theta_i). \]

Is it just me, or is Chinese Olympiad getting easier?

This one is straight forward manipulation, using sum-to-product and product-to-sum rules repeatedly. I'll sketch the idea, but I'll leave out the details.

Let $p = \cos(\theta_1 + \theta_2)$, $q = \cos(\theta_1 - \theta_2)$, $r = \cos(\theta_3 + \theta_4)$, $s = \cos(\theta_3 - \theta_4)$.

The first inequality is equivalent to $-p \leq x \leq q$ and the second inequality is equivalent to $-r \leq x \leq s$. So $x$ exist iff $\max(-p, -r) \leq \min(q, s)$, or $\min(p, r) + \min(q,s) \geq 0$. Since $|\theta_i | \leq \pi / 2$, we have $p + q \geq 0$, and $ r + s \geq 0$. So the first condition is equivalent to $p + s \geq 0$ and $q + r \geq 0$.

The second condition rearranges to $(p+s)(q+r) \geq 0$. The only if part is clear. The other way is also easy, since if both $p+s$ and $q+r$ are nonpositive (with at least one negative), then $p+q+r+s < 0 $, contradicting the $p + q \geq 0$ and $r + s \geq 0$ requirement.

I think q1 of this year CMO is quite easy if comparing it with the first one of last year's.

It got easier because you guys improved skills over the year and therefore it seems easier. It's a good sign, you know. :P  :D

[quote="billzhao"]Is it just me, or is Chinese Olympiad getting easier?[/quote]Maybe it's just you getting better over the year :D rather than the level going down ;)

i think it's the standard going down cos i managed to solve it too.  was quite shocked initially.

it is the easiest problem of the 2005 Chinese MO,
but at that time I nearly spent an hour to solve it.


Two regular square pyramids have all edges 12 cm in length. The pyramids have parallel bases, and each has a vertex at the center of the other's pyramid's base. What is the total number of cubic centimeters in the volume of the solid of intersection of the two pyramids? The answer is in the form $y\sqrt2$. Find $y$.

The intersection is two smaller similar pyramids with all sides length 6.


Compute
$\sum_{k=1}^{\infty}\frac{\binom{2k-1}{k}}{2^{2k-1}}$

Edit: had to edit the power of two on the bottom

Don't the terms increase without bound?  In that case the sum is $\infty$.
When I use a symbolic operator (Mathcad 12), it gave me $\frac{-1-i}{2}$ for the sum.  (What?!?)

Normally, one wouldn't say "compute" when [hide]the sum diverges.[/hide]

I wonder why it diverges? I got this sum from considering the expected length of a path. Imagine the real number line, with movement constrained to the integers. Starting from the origin, with an equal probability of going to the left or right 1 unit, what is the expected length of the paths that reach the integer $+1$?

Every path to $+1$ has an odd length. If $2k-1$ denotes the number of movements, exacly $k$ of those moves must be to the right. Using reasoning like this you get the sum I had, right?

In your counting, among the length 3 paths are paths that begin by moving one step to the right. Similarly for longer paths. I don't think you intend to overcount like this.
Also, you're simply summing the individual probabilities, not calculating the expected value. You have to multiply the probability by its corresponding path length before summing.

yeah ur right. then maybe it would be $\sum_{k=1}^{\infty}\frac{(2k-1)(\binom{2k-1}{k}-\binom{2k-2}{k-1})}{2^{2k-1}}$

[quote="JBL"]Let's take the path ending at -1, because I like that better.  Then the number of such paths of length 2k + 1 is the number of paths of length 2k among the positive integers which end at 0 (and then you step down to -1).  Calculating this number is a classical problem.  The answer is given by the Catalan numbers, $C_{n}= \frac1n{2n\choose n}$ (prove this, it's fun!), a sequence that shows up in a suprisingly wide variety of contexts.  (Someone should go make a Wiki page about them.)  I think it will still diverge, for the second reason that Bictor noted -- multiplying by the path length adds a big size boost.  (The best tool for telling whether things like this could possibly converge is Stirling's approximation of the factorial, which gives exceedingly good approximations of things like $2n\choose n$.)[/quote]
IMHO, calculating the Catalan number from its recurrence relation using generating functions is a very formidable task indeed!


There are 6 people- W,H,M,C,G,F.they  are

Murderer,Victim,Judge,Police,Witness,Hangman.The Murderer Was later Hanged.
1)M knew both Murderer and the Victim.
2)Judge asked  C to Discribe the Murder incident.
3) W was Last to see F alive.
4) police Found G at the Murder site.
5) H and W have never met.

Find who is who?

I don't understand this... :|  :huh:  :maybe:

[quote="lotrgreengrapes7926"]This should be moved to Games and Fun Factory or something.[/quote]

Or "Puzzles and Brainteasers"...

Where did you get this problem?


People that have qualified for USAMO in the past, please poll your scores above. If you haven't I ask that you do not answer on this thread. After some discussion, it seems that this year's AIME was particularly easy and uncommonly susceptible to irrational and petty calculation or interpretation errors. Of course, USAMOers are not above making such mistakes. Generally, it can be assumed that a good majority of past USAMOers will be the ones making the highest scores on subsequent AIMEs. I'd like you guys you poll your scores so we can compare them to what results are on the official poll by Mr. Crawford.

:D

you probly want to change it to 7, 8 and not 6-8

If a lot of past USAMOers score 6 - 8, sure. But that's not the case. :P

[quote="peter"]I messed up so bad... only got 6, and got 7 last year, and it was easier this year, and...[/quote]

It seems a lot of us scored a lot less than we'd expected...

[quote="avolfson"]So what does this mean about USAMO-ers?

(And thus the test?)[/quote]
Implication: They are better ;)

Three scores of 5 or less?  Are you sure that only past USAMOers are voting?

The statistics are all muddled now!

I'd like to trust the members of this site to only poll their score if they've qualified for the USAMO in the past. Although it does seem odd that USAMOers might get 5 or below, this is not impossible. A myriad of unintended mistakes could sink just about anyone.

[quote]A myriad of unintended mistakes could sink just about anyone.[/quote]

Indeed.   There are several cases of students who got honorable mention on USAMO (meaning that they not only qualified for USAMO, but also wrote a very good exam placing in the top 25) in one or more years, but then got very low scores (1 to 4) on AIME in a subsequent year.

a 12, but i should have gotten more haha

because... 2^2, contrary to what you might expect, does NOT equal 8 (referring of course to problem 8)
i ended up with 3/111=1/37 instead of 2/111

[quote="dareonion"]i should have gotten more haha
[/quote]

Hehe...I think just about everyone polling here believes that. :D


conics
hyperbola
trigonometry
analytic geometry
graphing lines
slope

$P$ is arbitrary point on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ with $x>0$ . If the tangent line at $P$ passes through the directix $x=\frac{a}{e}$ at $T$ and the focus point of the hyperbola is $F(ae,0)$ . Prove that $\angle PFT=90^\circ$

[quote="shyong"]$P$ is arbitrary point on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ with $x>0$ . If the tangent line at $P$ passes through the directrix $x=\frac{a}{e}$ at $T$ and the focus point of the hyperbola is $F(ae,0)$ . Prove that $\angle PFT=90^\circ$[/quote]

We first parameterize the hyperbola, with the choice

$P(t) = (x(t),y(t)) = (a \sec t, b \tan t)$, $t \in [0,2\pi)$.

Since we have $e = c/a$, and $c^{2}= a^{2}+b^{2}$, it follows that

$b = a \sqrt{e^{2}-1}, \qquad (b > 0)$.

The slope of the tangent line at $P(t)$ is simply

$m = \frac{dy}{dt}\Big/\frac{dx}{dt}= \frac{b \sec^{2}t}{a \sec t \,\tan t}= \frac{\sqrt{e^{2}-1}}{\sin t}$.

Therefore the tangent line through $P(t)$ has the form $y-b \tan t = m(x-a \sec t)$, and its intersection with the directrix is

$T(t) = \left(\frac{a}{e}, \frac{a}{e}\sqrt{e^{2}-1}(\csc t-e \cot t)\right)$.

From here, it is a straightforward (but tedious) computation to evaluate the dot product $\overrightarrow{TF}\cdot \overrightarrow{PF}$ and verify it is zero.


Let $ x,y,z>0$. Prove that
$ \frac{xyz}{(1\plus{}3x)(x\plus{}8y)(y\plus{}9z)(z\plus{}6)} \leq \frac{1}{7^4}$

Could you do it without using [b]Huygens Inequlaity[/b]?

can you kindly describe how you got this??i mean what was the total process? :(

This is the inequality
$ (a_1\plus{}b_1)(a_2\plus{}b_2)...(a_n\plus{}b_n)\geq (\sqrt[n]{a_1a_2...a_n}\plus{}\sqrt[n]{b_1b_2...b_n})^n$ for nonnegative numbers. The proof is very simple (use $ GM(AM)\geq AM(GM)$).

[quote="limes123"]This is the inequality
$ (a_1 \plus{} b_1)(a_2 \plus{} b_2)...(a_n \plus{} b_n)\geq (\sqrt [n]{a_1a_2...a_n} \plus{} \sqrt [n]{b_1b_2...b_n})^n$ for nonnegative numbers. The proof is very simple (use $ GM(AM)\geq AM(GM)$).[/quote]
This is Holder'Inequality


Joey plans to go to Mary's house, but he lost her house number. He knows she lives on Green Street, and the houses on that street are numbered consecutively from 310 to 325 inclusive. If he randomly selects a door on which to knock, what is the probability that he'll knock on the correct door? Express your answer as a common fraction.

There are a total of $ 325\minus{}310\plus{}1\equal{}16$ houses on the street, and only one is Mary's house. So our probability is $ \boxed{\frac{1}{16}}$.


Does a formula replace for  $ \sum_{k\equal{}1}^n k^4$ , in terms of $ n$?

Yes, the formula is $ \sum_{k\equal{}1}^{n}k^{4}\equal{}\frac{1}{5}n^{5}\plus{}\frac{1}{2}n^{4}\plus{}\frac{1}{3}n^{3}\minus{}\frac{1}{30}n$.  There actually is a pattern for higher powers involving the Bernoulli numbers.

One way to derive this formula is to simplify $ (2r\plus{}1)^{5} \minus{} (2r\minus{}1)^{5}$, sum from $ 1$ to $ n$ and it telescopes nicely.


Find the maximum value of product of no. of positive divisors of 1st n positive divisors of a number. Choose 1st n according to ascending order.

Let $ M_n\in\mathbb{R}\cup\infty$ denote the maximum depending upon $ n$, then $ M_1\equal{}1$ and for $ n>1$ we have $ M_n\equal{}\infty$.

Proof: $ M_1\equal{}1$ is trivially fulfilled and for $ n>1$ note, that the product of the first $ n$ smallest divisors of $ p^{n\minus{}1}$ is given by $ p^{\frac{n(n\minus{}1)}{2}}$ which goes to infinity for $ p\rightarrow\infty$ in the set of primes

P.S. If I have misread your problem statement, pls state it in mathematical form

[quote="hjbrasch"]Let $ M_n\in\mathbb{R}\cup\infty$ denote the maximum depending upon $ n$, then $ M_1 \equal{} 1$ and for $ n > 1$ we have $ M_n \equal{} \infty$.

Proof: $ M_1 \equal{} 1$ is trivially fulfilled and for $ n > 1$ note, that the product of the first $ n$ smallest divisors of $ p^{n \minus{} 1}$ is given by $ p^{\frac {n(n \minus{} 1)}{2}}$ which goes to infinity for $ p\rightarrow\infty$ in the set of primes

P.S. If I have misread your problem statement, pls state it in mathematical form[/quote]

I dont mean product of divisors but product of No. of divisors

its simple. The answer is n!. Try to find out the proof.


integration
calculus
derivative
function
algebra
domain
calculus computations

${f\left(x_{n+2}\right)=-\int_{f\left(x_{n}\right)}^{f\left(x_{n+1}\right)}f(t) \, dt}$
${x_{n+2}=x_{n+1}+x_{n}}$

OH my god, it has a solution?!

Just an idea:
$f(x_{n+2})=-\int_{f(x_{n})}^{f(x_{n+1}}f(t)dt$
be g(x) the anti-derivative of f, so:
$f(x_{n+2})=g(f(x_{n}))-g(f(x_{n+1)}) = f(x_{n}+x_{n+1})$
so: $g(g'(x))-g(g'(y))=g'(x+y)$ if  y=0 and g'(x)=u
$g(u)-K=u$ hence: g(x)=x+K so f must be a constant function, what contradicts the first integral ( since it would imply that f(x)=0) so this has no solutions. If 0 is not in the domain of f... so we need another stuff.

its wrong separete  into  x  and y

because if u put y ->  0    so x will not be variable, will be constant.

it´s just a sequence starting in x_0=0 and x_1= any real .But this is not so usefull at all....


so thats like 96%? I'm too lazy to figure it out :(

[quote="ccy"]A circle has a radius of 50 meters. The radius is increased by 40%. By what percent is the area increased?[/quote]

[hide] Radius $1$=$50$.
Area $1$=$(pi)*50^2$.

Radius $2$=$70$.
Area $2$=$(pi)*70^2$.

Then, the area of the circle $2$ is $196\%$ of the area of the circle $1$. Then, the answer is $\boxed {96\%}$.[/hide]

[color=darkred][size=75][Latex fixing][/size][/color]

[hide]I don't know why the answer didn't appear. However, the answer is 96%.[/hide]

[hide]the answer is $\frac{7}{5}^2-1\times 100$, which is 96%[/hide]


algebra
polynomial
quadratics
algorithm
algebra unsolved

Given a polynomial $P(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_2x^2+a_1x+a_0 ~(a_n \neq 0),$ where $a_n,a_{n-1},\ldots ,a_1,a_0 \in \mathbb{Z}$, prove that if $P(a+\sqrt{b})=0$ then $P(a-\sqrt{b})=0$.

Note: $a+\sqrt{b}$ and $a-\sqrt{b}$ are irrationals.

P has to be divisble by the quadratic that has both a+-root(b) as roots

use euclidean division algorithm

remainder is linear, say its cx+d

c(a+root(b))+d=0; you get that c=0, d=0, so the remainder is zero

haha got itt

[hide]

it suffices to show that $P(x)$ is divisible by $(x - (a + \sqrt{b}))(x - (a - \sqrt{b})) = x^2 - 2ax + (a^2 + b^2)$

dividing by this will leave a degree 1 remainder, hence, for some rational $c,d$, we let the remainder be $cx + d = 0$

so we know that:

$c(a+ \sqrt{b}) + d = 0$

$ac + d + c\sqrt{b} = 0$

since we know that $c\sqrt{b} = 0$ because of how we match the rationals with the irrationals, and since $b \not = 0$ due to the condition that $a \pm \sqrt{b}$ is irrational, we know that $c=0$

hence we have $ac + d = 0\implies d = 0$ and our remainder is thus $0$

this implies that our quadratic is a factor of $P(x)$ and that $a-\sqrt{b}$ is a root as well

[/hide]

Put $\overline{a+\sqrt{b}}=a-\sqrt{b}$.  It's not hard to prove that, if $x,y,z$ are of the desired form and $\alpha$ is a real scalar, then
\begin{eqnarray*} \overline{x\pm y}&=&\overline{x}\pm\overline{y} \\ \overline{z}^n&=&\overline{z^n}\\ \overline{\alpha z}&=&\alpha\overline{z} \\ \end{eqnarray*}
The above imply that
\[ \overline{P(z)}=P(\overline{z})  \]
for $z$ of the desired form.  But that is exactly what we want.

[quote="gotztahbeazn"]
since we know that $c\sqrt{b} = 0$ because of how we match the rationals with the irrationals[/quote]

$a$ doesn't have to be rational. The roots of $x^4-10x^2+1=0$ are $\pm\sqrt{2}\pm\sqrt{3}$

[quote="blahblahblah"]$\overline{z}^n=\overline{z^n}$[/quote]

Please, can you post the proof of that for irrational $a$?

[quote="blahblahblah"] It's not hard to prove that, if $x,y,z$ are of the desired form.[/quote]

You mean $x=a$ and $y=\sqrt{b}$?

My airplane continues flying very high.

[quote="Farenhajt"]

[quote="blahblahblah"]$\overline{z}^n=\overline{z^n}$[/quote]

Please, can you post the proof of that for irrational $a$?[/quote]

I think $a$ and $b$ are required to be rational.

[quote="blahblahblah"]I think $a$ and $b$ are required to be rational.[/quote]

The statement of the problem says nothing about $a$, just that $a\pm\sqrt{b}$ is irrational, which can be satisfied even with irrational $a$ (take $\sqrt{2}\pm\sqrt{3}$ as an example).

[quote="scorpius119"]For $\overline{z^n}=\bar{z}^n$ it is enough to prove $\overline{xy}=\bar{x}\bar{y}$ and the power rule follows from induction.

This works because $\left(x_1-y_1\sqrt{b}\right)\left(x_2-y_2\sqrt{b}\right)=(x_2y_2+by_1y_2)-(x_1y_2+x_2y_1)\sqrt{b}$ and $(x_2y_2+by_1y_2)+(x_1y_2+x_2y_1)\sqrt{b}$ factors into $\left(x_1+y_1\sqrt{b}\right)\left(x_2+y_2\sqrt{b}\right)$.[/quote]

Given that it's nowhere said that $a$ must be rational, in $\overline{xy}=\bar{x}\bar{y}$ how would you define $\overline{(\sqrt{2}+\sqrt{3})(\sqrt{2}+\sqrt{5})}$ without resorting to $(\overline{\sqrt{2}+\sqrt{3}})(\overline{\sqrt{2}+\sqrt{5}})$? And the statement holds even for irrational $a$: if $\sqrt{2}+\sqrt{3}$ is a root, then $\sqrt{2}-\sqrt{3}$ is also a root.

I'm sure this is not so hard. But i still can't find the correct approach. Someone could help?

Thanks in advance.

The statement is wrong for $a$ not rational.
Take $a= \sqrt 2$ and $b=4$.

But something more general is correct:
When considering $K=\mathbb{Q}[a,b]$ (or also any field $K$) we see that when  $\sqrt b \not\in K$ then the desired implication is correct since $f: \mathbb{Q}[a,\sqrt b] \to \mathbb{Q}[a,\sqrt b] ; f(x+y \sqrt b) = f(x-y \sqrt b)$ ($x,y \in \mathbb{Q}[a,b]$) is an automorphism of $\mathbb{Q}[a,\sqrt b]$ fixing $\mathbb{Q}[a, b]$.

Now i think it's clear [b]ZetaX[/b]. Thank you.

So [b]blahblahblah[/b] approach is correct, since $a$ and $b$ are rationals.

[quote="M4RI0"]this isn't an olympiad problem[/quote]
Explain...
I think it's theoretically possible to be one!

[quote="M4RI0"]this isn't an olympiad problem[/quote]

I don't know if it is or not.

This problem was posted originally in Pre-Olympiad forum. And since nobody give a complete solution i asked a moderator to move it. And ... well, till now it's keep without a complete solution using high school arguments.  :)


function
calculus
derivative
real analysis
real analysis solved

[quote="harazi"]If f is a real derivable function then the set $ f(A)$ (where A is the set of points where the derivative is 0) is neglijable.[/quote]
What does it mean "neglijable"? Countable, zero mesuare or something else?

Sorry, Myth, the definition I knew was that it can be covered with intervals having the sum of their lenghts no matter how small.

$f(A)$ has only one element in that case, so it sure has measure zero..

Anyone, any idea for this problem? Moubi, do you have any references?

Oh... I have too many problems to solve... Maye I will find time for this one in the nearest future...

It seems I am in great form these days...  :blush:

Let $f:[a,b]\to \mathbb{R}$. Consider for each $x\in A$ decreasing sequence $h_1(x)>h_2(x)>...>0$ of numbers s.t.
\[\forall n\in\mathbb{N}\ \forall \delta\in\mathbb{R},\ |\delta|<h_n(x)\ :\ |f(x+\delta)-f(x)|\leq \frac{h_n(x)}{n}.\]

Fix $\epsilon>0$. Let $n$ be such number that ${\frac{4}{n(b-a)}<\epsilon}$. Denote $I_n(x)=\big(x-h_n(x),x+h_n(x)\big)$, $x\in A$. Note [b](!)[/b] that $|f(I_n(x))|\leq \frac{2}{n}|I_n(x)|$.

Consider set $A_n=\bigcup_{x\in A}I_n(x)$.

Exercise 1 (general statement, not related to the current problem).
[i]There is at most countable set $\{x_k\}\subset A$ s.t. $A_n=\bigcup_{x_k}I_n(x_k)$.[/i]

Exercise 2.
[i]For each $m$ it is possible to delete some elements from $\{I_n(x_k)\}_{k=1}^{m}$ in such way that $\bigcup_{k=1}^{m} I_n(x_k)$ will not change and each point from $[a,b]$ will be covered by at most two intervals from remainings.[/i]

Note that if system of intervals from $[a,b]$ covers each point at most two times then total length of these intervals is at most $2(b-a)$.

The rest is clear: $f(A)\subset f(A_n)$, and we will show that $|f(A_n)|\leq \epsilon$.
Let $A_n^m=\bigcup_{k=1}^{m}I_n(x_k)$, then $A_n^{m}\subset A_n^{m+1}$ and $A_n=\bigcup_{m}A_n^m$. So it is sufficient to prove $|f(A_n^m)|\leq\epsilon$.

Indeed, due to Exercise 2 we can express $A_n^m$ as union of $I_n(x_{k_i})$ ($1\leq k_i\leq m$, $i=1..s_m$) and each point from $[a,b]$ is covered at most two times. Thus
\[|f(A_n^m)|=\left|\bigcup_{i=1}^{s_m}f(I_n(x_{k_i}))\right|\leq\] \[\sum_{i=1}^{s_m}\left|f(I_n(x_{k_i}))\right|\leq \frac{2\sum_{i=1}^{s_m} |I_n(x_{k_i})|}{n}\leq \frac{4(b-a)}{n}\leq\epsilon.\]

That's all.

There was small bug in my solution, so I slightly revised it. :P


linear algebra
matrix
pigeonhole principle
linear algebra unsolved

But even if the generalization is nice, it is obvious. Now, something that really does not seem obvious: which are the invertible matrices $nxn$ such that the set of all elements of all powers of $A$ is finite?

Well .. if the set is finite than clearly $\exists m \in \mathcal l N^*$ such that $A^m=I_n$.. so we have to solve the matrix equation $A^m=I_n$ if $(\lambda_1,\lambda_2,..,\lambda_n)$ are the eigenvalues of the matrix $A$ then $(\lambda_1^m,\lambda_2^m,..,\lambda_n^m)$ are the eigenvalues of the matrix $A^m$ so we have the system $\lambda_i^m=1, \forall i \in \{ 1,2,...,n\}$ so the answer should be all the matrices that have the eigenvalues belonging to the set of the roots of the equation $x^m=1$ ? hope i didn`t do something wrong....

Ah, sorry, I said non-sense. I was thinking about matrices as in the problem, that is with values in ${-1,1,0}$.

[quote="spix"]Let $A\in\mathcal l M_2(\Re)$ so that $\det(A)\neq 0$ and $\forall k \in \mathcal l N^*, A^k$ has all of it`s elements from $\{-1,0,1\}$.
1.Prove that $\exists m \in \mathcal l N^*$ such that $A^m=I_2$ (let $m_A$ be the smallest $m \in \mathcal l N^*$ such that $A^m=I_2$.
2.Give examples of $A,B\in \mathcal l M_2(\Re)$ and $C \in \mathcal l M_4(\Re)$ such that $m_A=3, m_B=2, m_C=6$

That was the text of the problem from the contest. In the contest i simply studied the values that $tr(A)$ and $\det(A)$ could take and afterwards using Hamilton-Cayley i got max :D. I think that the generalisation is much more beautifull:
Let $A\in\mathcal l M_n(\Re)$ so that $\det(A)\neq 0$ and $\forall k \in \mathcal l N^*, A^k$ has all of it`s elements from $\{a_1,a_2,...,a_p\}$ and there exist $i,j$ such that $a_i=0,a_j=1$.
Prove that $\exists m \in \mathcal l N^*$ such that $A^m=I_n$[/quote]

Solution:
1) Notice that there are maximum $3^4$ matrices that can be formed using only $-1,0,1$ so in the set $A,A^2,A^3,......$  at least two matrices repeat themselves so $A^t=A^s$ since $\det(A)\neq 0$we have that $A^{|t-s|}=I_2$.
2)$B=-I_2,A=\left ( \begin{array} {cc} -1&1\\-1&0 \end{array} \right )$ we can put $C=\left ( \begin{array} {cc} A&O_2\\O_2&B \end{array} \right )$.

[quote="spix"]
I think that the generalisation is much more beautifull:
Let $A\in\mathcal l M_n(\Re)$ so that $\det(A)\neq 0$ and $\forall k \in \mathcal l N^*, A^k$ has all of it`s elements from $\{a_1,a_2,...,a_p\}$ .
Prove that $\exists m \in \mathcal l N^*$ such that $A^m=I_n$[/quote]
The Idea of generalisation is nice but notice that i removed the condition (and there exist $i,j$ such that $a_i=0,a_j=1$) we can easily deduce that from the pigeonhole principle ,there are $p^{n^2}$ matrices that the matrices  of the set $\{A^k |k \in N^*\}$ might be equal to, let's consider the set $S=\{A^k | 1\leq k\leq p^{n^2}+1\}$ according to the pigeonhole principle there exists $r$ and $q$ integers $\in [1,p^{n^2}+1]$ such that $A^r=A^q$ and $q\geq r$.
Thus ${A^r(A^{q-r}-I_n})=0_{M_n(R)}$  ,i think $q-r$ is the $m$ you talked about .
Thisalso leads to the fact that $0,1\in \{a_1,a_2,...,a_p\}$:).

I think i spoiled it :D i didn't notice that you mentioned the idea of pigeonhole while talking about the case of $3^4$ sorry :blush:

[quote="spix"]Could please state the pigeonhole principle?  :oops:[/quote]

In romanian it is "Principiul cutiei": If you put at least $n+1$ pigeons in $n$ pigeonholes, then you'll have $2$ pigeons in the same pigeonhole.

thanks :!: .... Isn`t it also called Dirichlet`s principle?


We have 3 Eriksons here at mathcamp: Lars Erikson, Waffle Erikson, and Soda Erikson. They all play bughouse.

The probability that Lars will play bughouse on any given day is 1. The probability that Waffle will play is 0.3. The probability that all of them will play is 0.1 and the probability of any given pair of them  except soda-waffle playing is 0.2. What is the probability of Soda playing with Waffle on any given day?

Watch out - these events are not independent so multiplying is not valid.

[hide="hint"]
Make a venn diagram with the probabilities.
[/hide]

im confused  :huh:  :huh:  :huh:
don't some of these not make sense if you compare them individually?  is there something special about[hide="hint"]venn diagrams[/hide] that I don't understand?

The solution actually works. You don't have to use the hint if you don't want to, but there's no contradiction within the problem.

Let me give you an example. There is a 50% chance it will rain today and a 30% chance that it will be cloudy. The probability that it will be rainy AND cloudy is not necessarily 15% because they are not independent.


If x and y are real and i = $\sqrt{-1}$,

find the value of x + y when:

7+(x+y)i = 3x+4y+7i

[hide] answer: 7 [/hide]

How to solve this problem?

This should go to High School Basics, or Intermediate at highest.

The imaginary parts have to be equal, so $(x+y)i = 7i \Rightarrow x+y = 7$.


[quote="poonprang"]$ 2^x\minus{}3^y\equal{}9$ Find (x,y) positive integer[/quote]
none, since $ y>0$ implies $ 3|(3^y\plus{}9)$, so $ 3|2^x$, which is impossible

Now first we know that x and y are positive integers so both of them should be greater than 0. Now x=1 and x=2 yield no solution. So we may assume that x is greater than 2. Now we know that 2^x (for x greater than 2) is divisible by 4. Now we say that -3^y = 3(mod 4) and 3^y = 1 (mod 4). Now y is even. Now checking on modulo 3 shows that 2^x=1 (mod 3) so x is an even number , too. Put x=2a y=2b. (2^a - 3^b)(2^a + 3^b) = 7. So 2^a - 3^b = 1 and 2^a + 3^b=7. Now it is easy to see that a=2 and b=1. So there exists a unique solution x,y which is (4,2) :)


given    t = F[A + B/X + (D/X^2) + (E/X^3)]

rearrange for X = ?

7 engineers at work have tries and failed at my work-  I nned help urgently[/code][/list]

For $F\neq 0$, Rewrite the given equation as $\frac{E}{X^{3}}+\frac{D}{X^{2}}+\frac{B}{X}+A-\frac{t}{F}=0$.

Let $Y=\frac{1}{X}$, we have $EY^{3}+DY^{2}+BY+A-\frac{t}{F}=0$. You can use Tartalia or Cardano's formula with respect to a cubic equation.

I hope this help you.

This is the point I got up to.
can you please resolve this for me- [/hide]

[url=http://planetmath.org/encyclopedia/CardanosFormulae.html]Please Take a look[/url]

kunny


AMC
AIME
floor function
algebra
polynomial
quadratics
geometry

I'll start it off, but can't do much today..... need to finish USAMTS.

[hide="#1"]The points that are within one radius away from a particular point on the circle are on the 120 degree arc of the circle, 60 degrees on each side. So there are $2 \left\lfloor \frac{2006}{6} \right\rfloor = 2*334= \fbox{668}$ points one radii away.[/hide]

Why is it rounded down? I didn't understand that. I thought it rounded up because it included the point you were starting from.

Oh, I thought it meant $n\equiv 1\bmod{2}$. That's real nice. :dry:

#15
Using tangent of sum of angle formula we get tan(CDB) = (3+7/24)/(1-21/24) = 79/3.
Note that triangle HIG similar to AGD similar to DFG similar to AFB similar to BDC. Let

DF = a.
FH = 3a
FG = 79a/3
AF= 3a/79.
HG = FG - FH = 70a/3
AG = FG + AF = 6250a/(79*3)

BC=24(79/2)= 632 = AD.

HG/AG = HI/632 = (70a/3)/(6250a/(79*3)) = (79*7)/625
so HI = (79^2*8*7)/625.

IG = HI*79/3 = (79^3*8*7)/(625*3).  Add all the prime factors out there: 2+3+5+7+79=96. This problem is a bit too easy for #15,  it's just write everything interms of a ratio.

#14: Meh factor that thing into (P(x)-2)((P(x))^2 + P(x) + 3)) = 0

So P(x)=2 if P(x) is a real polynomial, for x= 1, 2, ... , n.  Let Q(x)=  P(x) - 2.

Q(x) is monic, with roots 1, 2, ,..., n, and degree n. Hence Q(x) = (x-1)(x-2)...(x-n)
P(x) = (x-1)(x-2)...(x-n) + 2

so P(0) = (-1)^n n! + 2, (P(1)+P(2)+P(3)...+P(n))^3 = (2n)^3 = 8n^3. Now P(0)>8n^3 at least n is 8. When  n=8 we have 8!+2=P(0). Remainder 322

[hide="#12"]The tangent line will have equation $y=mx$, where $m$ is slope. Substitute into circle equation to get $(x-2)^2+(mx-2006)^2=1\Rightarrow (m^2+1)x^2-(4012m+4)x+(2006^2+3)=0$.
Since the line is tangent, only 1 solution, so set the discriminant equal to 0. $(4012m+4)^2-4(2006^2+3)(m^2+1)=0$. $4012^2=4*2006^2$, so these $m^2$ terms cancel, leaving $-12m^2+8*4012m-4(2006^2+3)=0$. The sum of the solutions is equal to $\frac{8*4012}{12}=\frac{2*4012}{3}$, so the answer is $\left\lfloor \frac{4012}{15} \right\rfloor =267$.[/hide]

Aaaah, scorpius.....I got down to the quadratic like you, but actually [b]solved [/b](and made a mistake) for the two solutions and added them. I feel so dumb. :oops:

[quote="furious"][hide="9"]9.
$f(2)+f(3)+...+f(n) \ge \frac{2006}{1337} - \frac{101}{n}+\frac{99}{n+1}$
the left side becomes $1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n-2}-\frac{1}{n}+\frac{1}{n-1}-\frac{1}{n+1}$
which simplifeis to $\frac{3}{2}-\frac{1}{n}-\frac{1}{n+1}$ so,
$\frac{3}{2}-\frac{1}{n}-\frac{1}{n+1} \ge \frac{2006}{1337} - \frac{101}{n}+\frac{99}{n+1}$
$\frac{100}{n}-\frac{100}{n+1} \ge \frac{1}{2674}$
$\frac{1}{n(n+1)} \ge \frac{1}{267400}$
$n(n+1) \le 267400$
from here i just did some guestimation and obtained the value rather quickly. i'm sure there is a much better way to solve from here.
$n=516$[/hide][/quote]

Well you can approximate the value of sqrt(267400)

Made a dumb mistake on this one... missed the middle group.

[hide="5"]$16\rightarrow 80$, $243\rightarrow 255$, $2187\rightarrow 2499$, so we have $65+13+313=\boxed{391}$.[/hide]

Wow, #15 is really easy once I got the fact that tanBDE = 3...I probably would have gotten on the test had i had that.

I had a completely wrong definition for Monic polynomial. Great fun doing that problem.

I stilll think #7 should be 448, given the definition on the test.

For whoever asked about #10 --

Completing the square on the first equation, you get $16(x-10)^2+(y-30)^2=40000$, or $\frac{(x-10)^2}{50^2}+\frac{(y-30)^2}{40^2}=1$. Doing the same for the other one yields $\frac{(x-30)^2}{40^2}+\frac{(y-10)^2}{50^2}=1$. Solving for c (distance from center to foci), we get $c=\sqrt{a^2-b^2}=30$. Thus, the foci of the first one are $(10\pm 30, 30)=(-20,30)$ and $(40,30)$. Doing the same for the second one, we get $(30,10\pm 30)=(30,40)$ and $(30, -20)$. Using the Shoelace Theorem, we find the area to be 1800, and A/2 = 900.

The center of an ellipse $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ is the point $(h,k)$. $c$ is the distance from the center to either of the foci.

The problem I have with that solution of #7 is that it states $f(n)$ is the number of [i]divisors[/i] of $n$, not the number of [i]positive[/i] divisors of $n$. Thus, $f(2006^{6002})$ would be twice what you got, and $f(f(2006^{6002}))$ would be twice more and thus four times the given answer.

[quote="dakyru"]http://mathworld.wolfram.com/Divisor.html

It is commonly accepted that "divisor" refers to only the positive ones.[/quote]

THAT IS NOT TRUE!!!

At PROMYS I learned divisors refer to negative ones!

Either way  that problem should be tossed out.

[quote="Elemennop"]That's true, I've seen many questions on tests where there is a clear difference between simply divisors and positive divisors.

The question itself should have stated positive to get rid of ambiguity, but it didn't...I think both answers should be accepted.[/quote]

I see what you mean. Sorry for the ambiguity; if you want me to go back and rescore tests, I can, but if people are just satisfied with themselves knowing they got the "right" answer, I won't bother.

Full solutions to problems #14 and #15 are located in my blog (see link in my profile - feel free to click the ads while you're there :P ).

Did it simply state "divisors of the number" or did it actually say something more, or imply more directly that the negative divisors were involved? I have seen problems that use negative divisors as well, but not worded in the way that this problem was. I think that #7 and #15 were not worded properly, but I'm not going to complain about my answers there. It's not like this test even matters in the long run.

No, those problems simply state 'number of divisors of n', and almost always imply both negative and positive integers.

As I said, I don't really mind, since this test doesn't really count for anything. It was nice to see that I got a 7 in 1 hour 50 minutes, with at least two dumb mistakes on my part, and the heaviest topic was the one I'm weakest at (Geometry).

Elemennop, those problems are the ones that should be written better. They should specifically state both positive and negative divisors, as well-known and reliable sites like [url=http://mathworld.wolfram.com/]Mathworld[/url] give a different definition of a [url=http://mathworld.wolfram.com/Divisor.html]divisor[/url] than that.

[quote="paladin8"][quote="JesusFreak197"]I think that #7 and #15 were not worded properly, but I'm not going to complain about my answers there. It's not like this test even matters in the long run.[/quote]

 :( If I spent more time on it and had more people proofread it it would've turned out better. My apologies for poor wording that may have thrown you off.[/quote]

I thought it was a great test even with the mistakes :)

a very nice test Jeffrey.
i just finished it and got a 8.
the first question was almost like one of the questions i had seen two weeks ago in a local competition. :D

i missed#2 due to a computational mistake. but i got 3-9.

again on #7, it reminded me of a problem which i saw on the same competition about the number 667, i missed it on the test, because i thought it was a prime, but on this i instantly knew that it was 23*29

i also answered #10,11. kinda guessed on them, due to time but got them both wrong.

overall a very nice test. :)

[quote="Elemennop"][quote="JesusFreak197"][quote="Elemennop"]The problem I have with that solution of #7 is that it states $f(n)$ is the number of [i]divisors[/i] of $n$, not the number of [i]positive[/i] divisors of $n$. Thus, $f(2006^{6002})$ would be twice what you got, and $f(f(2006^{6002}))$ would be twice more and thus four times the given answer.[/quote]

Frankly, while you're right that it didn't specify, that's a rather stupid argument. You're being unnecessarily picky. When was the last time that a problem of that kind was referring to negative as well as positive divisors?[/quote]

Multiple times in Southeast tournaments here I've seen an answer asking for divisors of a number n taking into account negative divisors as well as positive.[/quote]

Argh, yeah, these southeast tournaments are often try to trick you by being ambiguous, and you have to basically guess their definition of divisors so that you don't get the question wrong.

Despite a few small errors, this test was quite good, and I liked the geometry questions mostly.  The analytic questions... were too easy quite frankly.  The foci of the ellipse one was a regular Mu Alpha Theta question (really easy competition  :P ).  I did like the telescoping number 9.   :D

Overall, quite reasonable difficulty for this early in the year  :lol: Fun test

i thought it was a good test, the problems were at a reasonable difficulty, but complaining about the divisors is just being picky, it does not matter if you do a little better on this test or not

[quote="1234567890"]What's the shoelace thm?

Also, anyone have a easy way to do #13 without brute force? But I'd like to see any solution, brute force or elegant.[/quote]
If you use vector geometry, you can derive a formula for the area of a triangle given three points in two dimensions (so $z = 0$ for all three points).  Take the absolute value of the cross product of the two vectors (where one vector is $P_1$ to $P_2$, for example, and the second vector is $P1$ to $P_3$), divided by two.  If I remember correctly, the formula ends up being something like

$A = \frac{|(x_2 - x_1)(y_3 - y_1) - (y_2 - y_1)(x_3 - x_1)|}{2}$

$= \frac{|(x_2 y_3 - x_2 y_1 - x_1 y_3) - (y_2 x_3 - y_2 x_1 - y_1 x_3)|}{2}$

, which is sometimes known as the Shoelace Theorem (though I honestly don't know where that name came from, since no reference that I've seen refers to it that way).  Somebody correct me if I'm wrong, because I tried to take the cross product in my head there.

It's called the shoelace theorem because it isn't simply limited to 3 points, but to any number of points that define a convex polygon. Given the points $(x_1,y_1), (x_2,y_2),\ldots(x_n,y_n)$ in clockwise or counterclockwise order, the area of that convex polygon will be $\frac{1}{2}|(x_1y_2+x_2y_3+\ldots+x_ny_1)-(y_1x_2+y_2x_3+\ldots+y_nx_1)|$. The 'shoelace' name comes from the fact that you can arrange the points in two columns, one of x and one of y, with the first row being $x_1, y_1$, second being $x_2,y_2$, and so on until you use up all the points, and the last row being $x_1, y_1$ again. You can then do the products of all the diagonals, and them all up, and take one half the positive difference in the sums. The diagonal multiplication gives an effect of a shoelace when drawn.

#13 -- Consider a number of tails $n$. The tails split the sequence into $n+1$ regions. Anytime a tails occurs, the region changes ownership; this is a result from the fact that whenever a person flips a tail, the other person gets to flip and thus heads in the sequence after a tail belong to a different person than the heads right before the tail in the sequence. For example, with two tails, the first region is Al's, the second is Bob's, and the third is Al's again. You can take the remaining heads (8), and allot them in ordered pairs of $(A,B)=(8,0),(7,1),(6,2),(5,3),(4,4)$. Then, you use the same formula for 'marbles in urns' as for this; that is, given R regions for a person and H heads to allot into them, the number of ways you can allot them is $\binom{R+H-1}{R-1}$. Then, you simply have 11 cases for 11 different number of tails, and then just manually compute all the possible combinations such that $A\ge B$.

What I've found with positive/negative divisors is that when the problem seems more algebraic ($x, y, z$, etc.), then "divisors" also refers to negative divisors, whereas when the problem seems more number-theoretic ($k, m, n$, etc.), then it's more likely that the problem assumes a restriction to $\mathbb{N}$.

Fun test, I got #14 and #15 pretty quickly but Jeffrey showed them to me (not their solutions) before, so maybe that doesn't count...?  I kept telling him they were too easy, but evidently not :lol:

heh yea nice job with the test

how do u think this mock compares to a real AIME difficulty wise?

I think it was a bit easy...not excessively, but a bit on the easy side.  Probably similar or a bit harder than last year's AIME...

Most people thought it was a bit easier than a "real" one, as not well-defined as that is. I'll probably make a more difficult one later in the year - the topics will likely be based on stuff from my blog; I might do some monthly problem sets related to my blog posts also if anyone might be interested (or even if no one is...).


limit
calculus
integration
logarithms
calculus computations

Hint: Try to write it as a Riemann sum- what integral does this represent?

I know what the answer is but I just want to know if you know it (lol)

[quote="arbelos08"]${ \lim_{n\to\infty} ~\frac {1}{n}(\frac {1}{1 + \frac {1}{n}} + \frac {1}{1 + \frac {2}{n}} + \dots \frac {1}{1 + \frac {n}{n}} })$[/quote]

This is equal to

$ \lim_{n \to \infty} ~\frac{1}{n}\sum_{i=1}^{n} \frac{1}{1+\frac{i}{n}}$

which equals

$ \int_1^2 \frac{1}{x}~dx = [\ln{|x|}]_1^2 = \ln{2}$

I hope this work is correct.


Write this equation in function notation.

20   -30-  40- 50- 60-70
175-166-157-148-139-130

Its supposed to be a chart

If you mean $ f(20)\equal{}175$, etc., then the only relevant polynomial function is $ f(10x)\equal{}193\minus{}9x$ for all integer $ x$.

There are an infinite number of solutions for $ f(n)$ if $ n\ne10x$.


please find positive integer solutions to the following equation.

a^2+b^2+c^2=a^2．b^2

please help me!  thz

[quote="wailam010"]please find positive integer solutions to the following equation.

$ a^{2}+b^{2}+c^{2}=a^{2}\times b^{2}$

please help me!  thz[/quote]
(i texed ur prob)
it is a classical problem which can be found in many books.
[hide="hint"]
use modular arithmetic with the cases. $ mod 8$
a)all 3 are odd
b)2 even,1 odd(one of the a,b even)
c)all even.
u can find that only all even case can be true.

then use infinite descent.
hope that the problem is soved  :wink:
so no +ve int solution.
[/hide]

[hide="maybe..."]There are no non-trivial solutions. The equation is equivalent to $ (a^{2}-1)(b^{2}-1)=c^{2}+1$. If one of $ a,b$ is odd, then $ c^{2}+1\equiv 0\pmod 4$, absurd. So both of them must be even, and so must $ c$. Now all prime factors of $ c^{2}+1$ are of the form $ 4k+1$. So the prime factors of $ a^{2}-1$ and $ b^{2}-1$ must be of this form as well, which is a contradiction.[/hide]

Btw, this isn't the right place to post these problems. you should post them in the intermediate or pre-olympiad forum.

EDIT: i thought the parity check was easy so i skipped it. but as t0rajir0u points out, i included it.

I will answer the question that got locked for some reason

Question

Use Chinese remainder theorem, or otherwise, find the last four digits of 1x3x5x7x...x2007.

Answer

Let $ x = 1\cdot 3\cdot 5\cdot \ldots \cdot 2007$

$ 10000 = 2^{4}5^{4}$

$ 1\cdot 3\cdot 5\cdot 7 \equiv 105 \equiv 9\,(mod\,16)$

$ 9\cdot 11\cdot 13\cdot 15 \equiv (-7)(-5)(-3)(-1) \equiv 9\,(mod\,16)$

$ \Rightarrow 1\cdot 3\cdot 5\cdot \ldots \cdot 15 \equiv 81 \equiv 1\,(mod\,16)$

But $ \frac{2000}{16}= 125$

$ \Rightarrow 1 \cdot 3\cdot 5\cdot \ldots \cdot 1999 \equiv 1^{125}\equiv 1\,(mod\,16)$

$ 2001\cdot 2003\cdot 2005 \cdot 2007 \equiv 1\cdot 3\cdot 5\cdot 7\cdot \equiv 9\,(mod\,16)$

$ \Rightarrow x \equiv 9\,(mod\,16)$

We know $ 5\,|\,x,\,\,15\,|\,x,\,\,25\,|\,x,\,\,35\,|\,x \Rightarrow 5^{4}\,|\,x$

$ \Rightarrow x \equiv 0\,(mod\,625)$

We now have a pair of simultaneous congruences which can be solved easily by the Chinese Remainder Theorem.

$ x \equiv a_{1}N_{1}x_{1}+a_{2}N_{2}x_{2}\, (mod\,n)$

$ n = 16\cdot 625 = 10000$

$ a_{1}= 9$ and $ a_{2}= 0$ (so we don't have to find $ N_{2}$ and $ x_{2}$)

$ N_{1}= \frac{10000}{16}= 625$

$ \Rightarrow 625x_{1}\equiv 1\,(mod\,16)$

$ \Rightarrow x_{1}= 1$ is the solution within the range $ 0 \leq x_{1}\leq 15$

$ \Rightarrow x \equiv 9\cdot 625\cdot 1+0 \equiv 5625\, (mod\,10000)$

Hence last four digits are $ 5625$.

[quote="nayel"]all prime factors of $ c^{2}+1$ are of the form $ 4k+1$. So the prime factors of $ a^{2}-1$ and $ b^{2}-1$ must be of this form as well, which is a contradiction.[/quote]

This point deserves more explanation, since I don't think it's at all obvious, especially if any of $ a, b, c$ are odd.

[hide="Discussion"] The same logic applies to $ a$ as $ b$.

[b]Case:[/b]  $ a$ is even.  Then $ a^{2}-1 \equiv 3 \bmod 4$, so it has at least one prime factor of that form.

[b]Case:[/b]  $ a$ is odd.  Then $ 8 | a^{2}-1 \implies 8 | c^{2}+1 \implies c^{2}\equiv 7 \bmod 8$, contradiction. [/hide]

[hide="infinite descent"] Using modulo 4, we find that $ a,b,c$ must be even, so we let $ a=2a_{1}$, and define $ b_{1},c_{1}$ analogously, substitute and divide out by 4 to get ${ a_{1}^{2}+b_{1}^{[}2}+c_{1}^{2}=a_{1}b_{1}$, apply the same process, etc., by infinite descent there are no solutions other than $ (0,0,0)$.[/hide]


trigonometry
geometry
rectangle
inequalities unsolved
inequalities

In any convex quadrilateral

$ 4(\sqrt{2}-1) \leq \sum \tan \frac{\hat A}{4} <2 $

But according to [url=http://www.mathlinks.ro/viewtopic.php?t=14059]this[/url] post, $\sum\sin \frac A2>2$, and it's well-known that $\sin x<\tan x$, so maybe there's something wrong?

Besides, if the quadrilateral is a rectangle, the sum of the tangents is $4>2$.

Thank you very much Grobber to have pointed out at my misprint.

It is in fact $ \tan \frac{\hat A}{4} $  :blush:


[quote="mathmatatics"]I got a 31 at chapter, and now I just did 2003 state competition sprint and target (timed) and got 39 *sadface*[/quote]

yeah I did bad at chapter (cuz I bombed the sprint) but when I took the state for last year I did better. How much did you get on the sprint for chapter?

[quote=".:Logic:."]Please help me by answering this:

How much would I need to get on that 2003 State Sprint and Target if I wanted to get in the top 4 in my state. (Which is [b]Ohio[/b])

I'm just trying to judge how close I am to my goal :)[/quote]

Well to be on the safe side assuming you do not bomb the Target and etc. you would need at least 25 on the Sprint to guarentee a good place ment?

Hmm... well.. a week-and-a-half ago, I took it and got a 27 on the sprint and 7 on the target.... (is 7 considered Bomb?? :huh: )

I've actually improved a lot since then (kinda weird, in such little time, but true), so I think I'd get a....

Bah forget it...

Can you please just tell me what you think I should get on 2006 State to have a 60-70% chance for top 4? Same thing?  :maybe:
[hide]
I worry and speculate too much, don't I?
[/hide]

[quote="mathmatatics"]I got a 31 at chapter, and now I just did 2003 state competition sprint and target (timed) and got 39 *sadface*[/quote]
probably because you are under pressure during the actual test so you are like paranoid about making sure your answer is right and in the process you waste alot of time. i know from experience :( . that and also the 2007 chapter test was really hard.

[quote="mathmatatics"]I got a 31 at chapter, and now I just did 2003 state competition sprint and target (timed) and got 39 *sadface*[/quote]
Eh, that's weird.
Last year before states I was getting 20/5 on sprint/target on average, then at states I got a 23/6.
Weird.

At chapter, I got 23/4 (Yes, I got four because I did all the last four wrong because I was being retarded)

On the 2003 State, I got 27/7 just like Logic

I'm not talking about points, just number correct

LOL.  At Chapter I got 13/8...as in 21 points...and i'm still going to nats...  don't take chapter as a good indication...  you might have a bad day...

does that... sort of? give away anything about this year's state test?

I don't think it does... but it might be wise to just edit it.

Yeah... I bombed the chapter test... It might be waking up at 5:00 when I'm used to around 7:00....and my lack of coffee...

I edited the previous post to be more precise...sorry

I have not taken the state and I personally think that that doesn't have the slightest inkling of even possibly revealing something. In fact, I have no clue what your post is talking about. Who exactly should edit what?

You guys are all lucky-- i was on the B team, and still don't know my score.

ubemaya, footballrocks already edited it.

[quote="indianamath"]last year before states i averaged 18/12 on old states and i got 17/10.
[/quote]

Shawn Qian told me that that score was still the best on Clay.. :rotfl: ..pretty sad..considering Mrs. Arnold said that Clay was going to win-what team place did they get last year anyways at state?

EDIT: Apparently he got a 21 last year...

[quote="Ubemaya"][quote="bpms"]Eh, my average score-my real score=-3 usually on mc.
Before nats bpms was averaging 30s on old nats, then he got a 33 at nats.[/quote]

There's a HUGE difference between 30 on an old Nat and a 33 on the 2006 Nats.[/quote]
Yeah, luck.

[quote="mathmatatics"]At chapter, I got 23/4 (Yes, I got four because I did all the last four wrong because I was being retarded)
[/quote]

Me too, except I got 22 on the sprint, AND I wasn't being retarded, I was trying my best.  :huh:
I suck at this.

[quote="indianamath"]lol my band teacher's favorite phrase:

"every expert was once a beginner" :dry:  :sleeping:  :no:[/quote]

MY band teachers favorite phrase:

from the first day of 6th grade:

"Treat your instrument like you treat your gun"

me: "sir, i dont have a gun...."

 :|

I don't need to worry about it, because I already took my chapters, (I got States coming up on MArch 24th)


[panic mode]The SAT is in two weeks, and basically, I've never even done a day of SAT preparation or even touched a prep book ever! Should I stop math for two weeks and do SAT instead? What's the best way to study for the SAT in two weeks? I am getting 1400 on W+CR based on my PSAT (800 M), and want to get that to at least a 1500+. Is it worth it to do vocab for 2 weeks (getting a lot there)? What's the best way, overall, to study writing and reading for SAT in a short period of time? I know you can't cram and improve immensely, but any help would be appreciated. [\panic mode]

Don't panic. What makes you think a few points on top of an already high score will mean much of anything? You don't need to improve immensely.

You should at least do some practice tests; knowing what the questions look like is always useful.

Well you're SAT's is here already, but i'll tell everyone else that's in panic mode. You should get the vocab down by setting goals of how much words to memorize in a day. Study the roots word too, it's a big help on the vocab. Don't try to memorize every word unless you want to get like a 1600 on your SAT. (Study 15 words and 10 roots word a day for example) Read some classic books before you go to sleep and review the tips for writing essays. For math, get some example problems and slash through them one by one. Learn some tatics for the SAT's too, such as guessing and crossing out "distrators" I can't guarantee that you'll get like a 1400 or something on your SAT, but you'll get a pretty decent score.

[quote="bluejay206"]Well you're SAT's is here already, but i'll tell everyone else that's in panic mode. You should get the vocab down by setting goals of how much words to memorize in a day. Study the roots word too, it's a big help on the vocab. [b]Don't try to memorize every word unless you want to get like a 1600 on your SAT.[/b] (Study 15 words and 10 roots word a day for example) Read some classic books before you go to sleep and review the tips for writing essays. For math, get some example problems and slash through them one by one. Learn some tatics for the SAT's too, such as guessing and crossing out "distrators" I can't guarantee that you'll get like a 1400 or something on your SAT, but you'll get a pretty decent score.[/quote]

You're a new member here so you probably don't know this but...I think just about everyone on this forum would like to get a 1600.  There are a bunch of people on this forum who have achieved 1600's (even 2400's on the new SAT).

Anyways, I agree that you don't need to memorize every word in the vocab books(not even to get a 2400).

1234567890, is it necessary to go obsessive compulsive over every test you take (AIME, SAT, AIME, etc.)?

As a sidenote, I suspect that a large number of (american national) AoPSers would not consider 1400 (resp. 2100?) a decent score.  Certainly I would feel a lot more comfortable applying to elite colleges if I had a 2250, or better yet 2300+.

I'm taking it on January 28th also for the second time...hoping for 800M/650V/750W (got 780M/560V/560W first time and 80M/60V/60W on PSAT). I think taking practice tests is really the only way to do very well on the SAT, with about a 30-40 minute practice each day for a month should raise your score by atleast 200 pts.

I agree with zoogies...reading a lot and staying calm'll do you the most good. The hardest thing about the SAT is being able to read the questions clearly...mistakes come up when you're so jittery that you don't read the whole question. It's probably a good idea to take a look at test format and a few old SATs before you take it for real - it makes it that much easier to walk in and ace the thing, at least in my experience. (Never touched a prep book or did vocab books, but took the new SAT for the first time this year after a few years of the old one for CTY and popped out a 2400 first go 'round) Oh, and don't worry if/when you don't know all the words on the test - process of elimination can get you at least a 7-something, I think, with pretty basic vocab knowledge and some familiarity with the way they write the test. Go with your gut on words - most of the time just knowing "good" "bad" "verb" or "noun" is enough. For the Writing section...just make your intro paragraph and your conclusion nice and fancy sounding, (They don't take off for using dead-wrong statistics, and they love bombastic BS) and read the grammar out loud under your breath. (wrong answers will [i]sound [/i]wrong)

Assuming you can perfect the math, learn the format of the test by taking a few practice tests, and just stay calm when you take the real thing.  The people who go nuts over it tend to make stupid mistakes.

[quote="zoogies"][quote]
You're a new member here so you probably don't know this but...I think just about everyone on this forum would like to get a 1600. There are a bunch of people on this forum who have achieved 1600's (even 2400's on the new SAT).
[/quote]

HAHAHA yes! AoPS is quite probably the geekiest forum on the planet. It definitely transcends collegeconfidential, anyway.

But while it's probably true that most people on AoPS want a 2400, I still think memorizing vocabulary is just an incredibly wasteful use of your time. Read books. If you read enough, you'll be well off on the critical reading - and after all, the section is now called "Critical Reading." If you don't know a word, there are no analogies that spell sudden death on a question. If you have a general sense of the language and it use, you can probably do a fair job of inferring words you don't know in the sentence completions.

I realize the Dec test date is already past, but...panicking is probably the worst thing you can do for yourself on an SAT :)[/quote]

I disagree with you about reading is only way to do well on the tests. This is true for americans who are born here and use English for entire life.

But if you think about immigrants or exchange students, this is not true. You do not have enough time to read certain books to suddenly become a better reader. Moreover, critical reading section, although it is called [i]reading[/i], is different than reading. The questions may be similar to what you saw before but this section contains essays in politics, history, or other things that are not "that" fun as books you normally read. You have time limit and I definitely do not agree that reading book is the best way to prepare for this. I mean, you should keep reading but reading solely wouldn't get EVERYONE perfect score on this section. I highly doubt this (not saying that this is not possible but it's not very likely).

I think you'll find that by and large immigrant and foreign students score better on English tests.

Some do.  But in a sense it's a rather meaningless comment, as there are no particular qualities common to immigrant and foreign students.  Some come from English speaking countries with a higher standard of education, and some speak little to no English and come from places with a lower standard of education.  However, it's probably true that immigrant and foreign applicants score lower than average when applying to elite schools, which is a statistic relevant to this thread.

The reason immigrants from non-English-speaking countries have to learn English when they arrive in the United States (as most of my ancestors did) is simply that English is the majority language of the country, the only language with official status in most states for most purposes, and the language most likely to be known as a second language by Americans whose native language is some language other than English. So the reason I say READ, READ, READ, and READ in English books and magazines that you like is not only because that is a great way to boost your college admission test scores, but also because that is a great way to become better acquainted with the country your parents choose to move to or that you are choosing to study in. Limiting yourself to cramming for tests will cause you to miss out on many delights of American life.

I am a rare American who has spent six years of my adult life living in a non-English-speaking country and whose children are bilingual. There is no doubt that it is more difficult to take a college entrance test in a second language than in a native language, more difficult to take it in a language used mostly at school than in a language used primarily at home. But the one and only language for college entrance tests in the United States is English, and some international students from countries where no Indo-European language is spoken get above 700 on the verbal section of the SAT, and quite a few first-generation immigrant children from a variety of countries ace the SAT or the ACT. Learning to take a test in a new language is difficult, but it is possible. I admire the people who take on that challenge.


If $xy = a, xz =b,$ and $yz = c$, and none of these quantities is zero, then $x^2+y^2+z^2$ equals:

$ \textbf{(A)}\ \frac{ab+ac+bc}{abc}\qquad\textbf{(B)}\ \frac{a^2+b^2+c^2}{abc}\qquad\textbf{(C)}\ \frac{(a+b+c)^2}{abc}\qquad\textbf{(D)}\ \frac{(ab+ac+bc)^2}{abc}\qquad\textbf{(E)}\ \frac{(ab)^2+(ac)^2+(bc)^2}{abc} $

[i]Solution put into hidden text by moderator--tokenadult[/i]

[hide]Since $(xyz)^2=xy\cdot yz\cdot zx=abc$, we have $xyz=\pm\sqrt{abc}$ so $x=\pm\sqrt{\frac{ab}{c}}$ ,similarly $y=\pm\sqrt{\frac{ca}{b}},z=\pm\sqrt{\frac{bc}{a}}$.

Therefore $x^2+y^2+z^2=\frac{ab}{c}+\frac{ca}{b}+\frac{bc}{a}=\frac{(ab)^2+(bc)^2+(ca)^2}{abc}$.[/hide]

[quote="DPopov"][hide]
First we note that $x^2=\frac{ab}{yz}$, $y^2=\frac{ac}{xz}$, and $z^2=\frac{bc}{xy}$. Substituting for the denominators we get

\[x^2=\frac{ab}{c}\]
\[y^2=\frac{ac}{b}\]
\[z^2=\frac{bc}{a}\]

Adding these three we obtain

\[\frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}=\frac{(ab)^2+(ac)^2+(bc)^2}{abc}\]

Thus the answer is $E$

[/hide][/quote]

Oh! you have already posted,DPopov.

I didn't notice your posting,sorry.

kunny

no problem, you are free to post your solution as well just use a spolier. I actually like yours a little better although it is basically the same.

[quote="DPopov"]use spoiler please...[/quote]

DPopov,How can I use a spoiler?

kunny

To hide your text (put it in a spoiler) put [ Hide ] before it and [ /Hide ] after it (without the spaces).  Right above where you type your messages there is a button that can do this for you.  You can just highlight the desired text and click Hide.  Here's an example:

[code]
[Hide]If $x+2=3$, then $x=1$[/Hide]
[/code]

will give [hide]If $x+2=3$, then $x=1$[/hide]

If you want you can change the text displayed on the link by doing this:

[code][Hide=My Solution]If $x+2=3$, then $x=1$.[/Hide][/code]

[hide="My Solution"]If $x+2=3$, then $x=1$.[/hide]

Yes. That's right answer.

I just hope this problem would be solved by more beginners than advancers.

It is very kind of you to explain in detail, joml88.
I could understand.

kunny


function
calculus
derivative
real analysis
real analysis unsolved

Prove there exists a nonnegative continuous function on $ [0,1]$ such that $ f(0)\equal{}f(1)\equal{}0$, the derivative exists a.e. and where it exists it is strictly positive.

Just subtract the Cantor staircase from $ x$.

Ah. I'm starting to get frustrated with how obvious most of these old prelim problems are, yet I'm not seeing them. I only have less than two weeks left, and I'm starting to feel hopeless.  :mad:

Fedja, could you please explain this Cantor staircase ? :)

There are many equivalent ways to think about the construction. The most intuitive, I think, is to do the first couple of steps and just say repeat this process.

It is defined on to be constant on the closure of the intervals you remove in the Cantor set in the following way. On the interval $ [1/3, 2/3]$ we let $ c(x) \equal{} \frac {1}{2}$. On the intervals $ [1/9,2/9]$ and $ [7/9,8/9]$ you let $ c(x) \equal{} \frac {1}{4}$ and $ c(x) \equal{} \frac {3}{4}$ respectively. Then on the $ \frac {\cdot}{27}$ intervals you let it be $ \frac {\cdot}{8}$ values (see if you can fill these in). So basically on the $ 3^n$ denominator intervals, you have values with denominator $ 2^n$ such that they increase.

So each point not in the Cantor set has a neighborhood that is constant, and since the Cantor set has measure zero, the derivative is zero almost everywhere. Also, $ c$ is monotone increasing on all of $ [0,1]$ and is onto, so it has no discontinuities and is thus continuous.

This is sometimes also known as the devil's staircase. Here's the graph (of course, highly simplified...)

http://www.cs.cmu.edu/~cdm/pics/Devils-staircase.jpg

Look at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22230&start=20]this old topic.[/url] You'll find an exposition of the Cantor function at post 41, but look also at posts 25, 32, and 46-51 for a different example that could also have been used for the problem that started this topic.


inequalities
trigonometry
complex numbers
inequalities theorems

I don't know where to post this. If this is not the correct place, please moderators move it.

What is the importance in the inequalities that $(x,y,z) \mathbb \in{R}$, $x,y,z>0$,  $xyz=1$, or $x+y+z=1$ ??.

I mean, what can I do when I know any of this? Because this 4 conditions are always in the problems proposed here in MathLinks.

Thanks in advance.

Maybe I am not the best person to answer such a question, since I just started learning inequalities a couple weeks ago, but I sort of figured these out myself, and I'll try to share my thoughts about it.

Knowing what sort of numbers are $x, y, z$ (or whatever) tells you what you're allowed to do. If you know they're positive, you can do AM-GM, for example. Make sure you know the conditions for each inequality. Some inequalities hold for complex numbers in addition to reals, but you need to be careful of that as well -- not all do!

Statements like $xyz=1$ or $x+y+z=1$ can be used to homogenize an inequality. Note that the first one is degree 3 = degree 0 and the second is degree 1 = degree 0. So if someone gives an inequality that is not homogeneous, you might be able to use the given condition to homogenize the inequality (and thus make it easier to solve -- some inequalities work specifically on homogeneous expressions, like Muirhead).

Also, sometimes such a condition suggests a substitution you might make, like $a=\frac{x}{y}, b = \frac{y}{z}, c = \frac{z}{x}$ when you're given $abc=1$.

The best way to really learn these is by reading people's solutions and trying to solve problems on your own. I hope this helps. :)

I think the main reason for giving often conditions like $xyz=1$ or $x+y+z=1$ is to write the ineqalities in a neater form. Like for example if a homogenous ineq contains at several places the sum $xy+xz+yz$, this looks just awkward. Others can be made looking symmetrical in $x,y,z$ vs. $\frac1x, \frac1y,\frac1z$ by putting $xyz=1$ etc. Esthetical... And IMHO the nicest solutions are often [i]not[/i] those that start with homogenizing the thing :lol:
It could be added that non-homogenous conditions like $a+b+c=abc$ can be hints towards a trigonometrical substitution (like $a= \tan\alpha$) with maybe a geometrical interpretation.
My 2 pennys' worth.


Oh that is too easy ;)

We know that $\sum _{i=1}^nx_i^2 \times\sum_{i=1}^2y_i^2\times \sum_ {i=1}^n z_i^2\geq \sum_{i=1}^nx_iy_i\times \sum_{i=1}^n y_iz_i\times \sum_{i=1}^nz_ix_i$
(where $x_i,y_i,z_i\in\mathbb{R}$ and $i=1,2,...,n$).
Let put the $z_i=\sqrt {x_iy_i}$ in above inequality,then attention to the new inequality
 put it $x_i=a_i^{\frac m2}$ and $y_i=a_i^\frac {m+4}{2}$(where $1\leq i\leq n)$ so that the inequality holds. :D


ARML
summer program
Mathcamp
USAMTS
probability
geometry
calculus

I scored six.  Alas, I forget which two problems I missed.

[quote="Kent Merryfield"][quote="JesusFreak197"]I think one on the Southern California A team got a 5.[/quote]
We had two freshmen with 5's - Macky and Yolanda. I'm sure if you look nationally, you'll turn up a several 6's - there are a lot of teams, after all.[/quote]
I know someone who went to mathcamp and i think he mentioned there was a guy named Macky there who held the world record in rubik's cubes. Same one?

Now, enough on 2005 ARML, let's get to 2006. I got a 4 (2,3,4,7)... The IGP problem blew it...instead of 68, I put 34...what sheer carelessness...no idea on 5 and 8. Had no time to do 1 because I ran out of space and I messed my paper up trying to solve 2. (Why the heck is it 91/128, anyway on 8?)

I got a 6 (1, 2, 3, 4, 5, and 7).  I brute-forced 8 (incorrectly due to a minor (=5/512) probability error somewhere and my lack of multiplication by 5, both of which were caused by limited time) before getting 7.  Then, I thought I'd messed up on 7 and almost erased my answer in the last 15 seconds, but decided to leave it.  I later realized I'd been right.  Did any other 8th graders get 6 (or more)?

What did you get?

I got 6 - I was being a complete idiot and changed my correct answer on #1 to something incorrect about 15 seconds before time was called and thence I couldn't go to tiebreakers.  :mad:

Five correct (1,2,3,4,7) (freshman)
TJ had three freshmen with 5's, and one with a 6.

I got 2. I think most of the people in our team had at least 3 (I'm talking about Lehigh Valley Ice, not Fire) and I could've too but I don't know how to take square root on #2. Anyway, my goal was to score a natural number so I'm happy. Good job!

[quote="Kent Merryfield"]Yes, that's him. His name is actually Shotaro Makisumi. He does post on AoPS, and you'll also see his name on the USAMTS leader board.[/quote]

That's me!

I got a 6 this year: (2,3,4,5,6,7). I misinterpreted #1, as did many people, thinking that n had to have the same number of digits in base 3, 5, AND 10. That trivialized the problem (clearly n=1, 2), but I didn't think that was a problem for #1. I still feel it was worded badly, though.

Why don't we open up a new poll/thread and post there? I wasted time reading the results from last year.

I got a 3. I am sort of mad at myself for missing easy ones. I got 4, 5, and 7. I think i shouldve gotten the two geometry ones correct.

[quote="litoxe2718281828"]Copying my post from the other thread:

I got a 6 this year: (2,3,4,5,6,7). I misinterpreted #1, as did many people, thinking that n had to have the same number of digits in base 3, 5, AND 10. That trivialized the problem (clearly n=1, 2), but I didn't think that was a problem for #1. I still feel it was worded badly, though.[/quote]

Oh...that actually is a valid way to look at it, when I think about it.  They should reconsider for that one...

I got a 7 (1 2 3 4 5 7 8)...ended up getting 2:26 on the tiebreaker...I first panicked a little, spaced out, and then got focus and got it.

I got 6 (1, 2, 3, 4, 6, 7). I actually had the right answer for #5, but I somehow thought that it was wrong... Oh well :roll:

I think 9 or 10 people in our team (Chicago A) got 6+, including 2 7's.

He frt, what chair were you in? We were right next to you guys. I was in 12.

Omg, I got a 6 (1,2,4,5,6,7) (9th grade). I can't believe it.  I missed #3! I had x and x-36, but i read the wrong angle! Should have gotten a 7, maybe 8 lol.

#8 was actually not that bad, especially since you had like 8 minutes to work on it since 7 was fairly easy(just needed to see it, which was hard) (EDIT: Nice job K81o7 for getting 8! One of two people,wow):

[hide]Define $\ f ( x )$ to be the # of ways to make sure all $x$ comps are linked.  First off, $f( 1 )= 1$, $f ( 2)=1$, $f ( 3 )= 4$.  Now we find $f (4 )$ and $f(5 )$ recursively.  For 4, we find the number of ways you cant do it by splitting the links up (example 3 comps linked, 1 not), so we look at partitions of 4 not including 4.
$(3,1)$ gives us $\binom41\cdot f(3)\cdot f(1)=16$
$(2,2)$ gives us $\frac12\cdot\binom42\cdot f(2)\cdot f(2)=3$ (we need the $\frac12$ because choosing AB is the same as choosing CD)
$(2,1,1)$ gives us $\binom42\cdot f(2)\cdot f(1)\cdot f(1)=6$
$(1,1,1,1)$ gives us $f(1)\cdot f(1)\cdot f(1)\cdot f(1)=1$ so
$f(4)=2^{\binom42}-16-3-6-1=38$

Again, for 5, we find the # of ways yo cant link them with partitions.
$(4,1)$ gives us $\binom54\cdot f(4)\cdot f(1)=190$
$(3,2)$ gives us $\binom53\cdot f(3)\cdot f(2)=40$
$(3,1,1)$ gives us $\binom53\cdot f(3)\cdot f(1)\cdot f(1)=40$
$(2,2,1)$ gives us $\frac12\binom52\binom32\cdot f(2)\cdot f(2)\cdot f(1)=15$
$(2,1,1,1)$ gives us $\binom52\cdot f(2)\cdot f(1)\cdot f(1)\cdot f(1)=10$
$(1,1,1,1,1)$ gives us $f(1)\cdot f(1)\cdot f(1)\cdot f(1)\cdot f(1)=1$ so
$f(5)=2^{\binom52}-190-40-40-15-10-1=728$ then
$\frac{728}{1024}=\boxed{\frac{91}{128}}$[/hide]

[quote="Xantos C. Guin"]Also, since there are only 8 questions, an individual's score depends more on what type of problems are on the individual round rather than his overall skill level. But skill level is still important.[/quote]I don't think I would put it quite like that.  I think most people with highish scores would do well no matter what questions were on the test.  But it definitely is fluky, because one out-of-the-ordinary question or a single stupid mistake can obviously make an enormous difference for the individual, much more so than any other competition I know (except ones where the questions are so easy that there are a few people who could score perfects and the winners are the ones who don't make any stupid mistakes).

If you walk around in New York enough in early January (as I did this year), the ink in your pen actually [i]does[/i] start to freeze.

Hmm, I didn't know Ryan Ko won last year.

i got a six(1,2,3,4,5,7) as a freshman
I missed #6 because I was so stupid. I wrote the answer as the 4th degree function instead of 2nd degree function, which kills me..
& #8, I didn't exclude the ones doesnt work when 5 or 6 links are connected

oh wells, this is my first year at ARML

[quote="Kent Merryfield"]A number of posts on this topic have had the form, "I got a ___ and I'm only in the ___ grade." The clear implication is that the posters expect to do better when they return to ARML in the future.

I've coached an ARML team for five years now, and our organazation has had three teams in each of the last three years, so a number of students have had the opportunity to come more than once. In that time, 16 different individuals have been three-time ARML participants. Here are the average scores of those 16 students:

First year: 3.00
Second year: 2.94
Third year: 4.38

We haven't had any four-year participants yet, although we're reasonably likely to have several in 2007.[/quote]

Hey, I feel special! I can improve! :P This is definitely true, though, even if I suspect that such statistics may be influenced by the difficulty of the tests. In any case, I find that if I make a stupid mistake in the first four it seems to have an effect on my attitude during the last four, which usually isn't a good thing. So like maybe instead of improving I just got lucky. :o

I didn't get the chance to go, but I got 6 when taking it as a practice (1, 2, 3, 4, 5, 7), beating the majority of people on the MA A team.

wow impressive, I have herad of arml but never have tried any practice or anything

You should find out about our (Alabama) ARML tryouts. I don't know how good you are, but it will be practice at least.

ARML is fun, and a good place to practice your math skills, like Cicatriz said. You should come to tryouts and practice math problems just to gain some experience if nothing else.

I think this is my first legitimate post on AOPS.

Um... how did [b][i]3[/i][/b] people get 8??  :rotfl:

This shows how people lie so much on polls.

yeah, it does, doesn't it?

i didn't really think that 8 was that hard ... i tried it, realized my method was wrong, tried again, got another wrong method, and then tried again and got the right answer about 15 seconds after time was called  :ewpu:

The time limit is what makes it hard, especially considering that there's so much computation.

[quote="samath"]How do 5 people now have 8??
Like, the fact that the most people said they have a 6.
Well, that could just show that AoPS people are, in general, smarter. :lol:
But somehow i still doubt it.[/quote]
I don't think the sixes are lies, if someone wants to lie they'd vote for 8 probably.

[quote="mchoi815"]i don't think that the sixes are lies.  i was a seventh grader last year and got a six, making the team high scorer.  however, the arml website says everyone got a seven or below except for three people who got number 8 but possibly missed something else.[/quote]

Only 1 person got a score of 8. A lot of people got 7, though.

Yeah and he was from our team! Go Indiana Gold!


function
calculus
derivative
algebra
system of equations
calculus computations

I am confused about this problem, so if anyone could explain how it is done, I would really appreciate it!  Thanks!

For the surface defined by the function

F(x,y)=2(x^2)y - (y^2) - (4x^2) + (3y), find and classify all critical points.

From this, I've gotten that

The derivative with respect to x is 4xy-8x.
The derivative with respect to y is 2x^2 - 2y + 3.

However, I'm not sure how to find the critical points.

Set both derivatives equal to zero and solve the system.

[quote="Kent Merryfield"]And the system of equations is nonlinear - the interesting cases usually are. Algebraic manipulation is more of an art then a technique, so I'd be very reluctant to say "do it this way." There's often room for and insight that lets you try a particular thing with a particular system. The most straightforward, plodding thing you can do is to try to solve one equation for one variable in terms of the other, and then to substitute that into the other equation to get some kind of equation in one variable. You could do that in this case and it would work, although it's not the most elegant thing to do.[/quote]

Ok, thanks!  Well, I solved for y for the first equation and got y = 2.

Then, I solved for x in the second equation and got + or - the square root of .5.

Would this mean that my critical points are...

(sqrt[.5], 2)
and (-sqrt[.5], 2) where (sqrt[.5],2) is a maximum and (-sqrt[.5],2) is a minimum?

[quote="JBL"]You missed a possibility: given the equation $4xy-8x = 0$, you can't divide by $x$ unless $x \neq 0$.[/quote]

But can't you do it like this....

4xy - 8x = 0

Add 8x to both sides, getting:

4xy = 8x

Then divide both sides by 4x to get:

y = 2

You could do that, if you knew that $4x \neq 0$.

Or to put what JBL is trying to say a little more bluntly, [i]factor[/i] that equation:

$4x(y-2)=0$

So [i]either[/i] $4x=0$ or $y-2=0.$

That's the thing about nonlinear equations - they're allowed to have multiple roots.

So there are three critical points then?
(-sqrt[.5], 2) ,
(-sqrt[.5],2) ,
and (0, 2)?

Yes. Now, about the classification - what about second derivatives?


$ \frac{2}{1\minus{}\frac{2}{3}}\equal{}$

\[ \textbf{(A)}\ \minus{}3 \qquad
\textbf{(B)}\ \minus{}\frac{4}{3} \qquad
\textbf{(C)}\ \frac{2}{3} \qquad
\textbf{(D)}\ 2 \qquad
\textbf{(E)}\ 6
\]

[hide="Click for solution"]
$ \frac{2}{1\minus{}\frac{2}{3}}\equal{}\frac{2}{\frac{1}{3}}\equal{}2 \times 3\equal{}6$, or answer choice $ \boxed{\textbf{(E)}}$.
[/hide]