So I was doing the combinatorics chapter in AoPS vol. 2 and came across "block walking". This style of argument is pretty foreign to me and I'm having trouble understanding how to come up with arguments like this. For example, how would you prove Vandermonde's identity with a block walking argument? Any help would be appreciated!

To arrive at block ${m+n\choose r}$, we must make make $r$ right turns out of a total of $m+n$ moves. Suppose we make $k$ right turns in the first $m$ moves. We must now make $r-k$ right turns out of the remaining $n$ moves. Thus there are ${m\choose k}{n\choose r-k}$ ways to reach the block ${m+n\choose r}$ when we divide the trip into $k$ right turns and then $r-k$ more right turns. Because $k$ can be any integer on $[0,r]$, we sum each of these cases to arrive at the identity $\sum_{k=0}^{r}{m\choose k}{n\choose r-k}={m+n\choose r}$.


function
limit
calculus
calculus computations

I need some help for one more time, if you can..
Let $ f_n(x)\equal{}nx/exp(nx)$. Does $ f_n$ tend to 0 uniformly on a)[0,1] and b)[0,infinity)?
Is there a general way to check if a sequence of functions is uniformly convergent?
Thanks.

[quote="georgech"]Is there a general way to check if a sequence of functions is uniformly convergent?[/quote]
Yes.  The definition.

Recall that if a sequence has a uniform limit, it must also be its pointwise limit, so you might try to figure out what the pointwise limit is first and then see if that limit is also uniform.

(Note that the pointwise limit of this sequence - which t0rajir0u referred to - is zero everywhere.)

What is $ f_n\left(\frac1n\right)$ and what does that tell you?

I should add that I worded my response subtly:  it is much easier to show that a sequence is [i]not[/i] uniformly convergent than to show that it is because there are many properties that a uniformly convergent sequence can be shown to have and if any of them are false you know you don't have uniform convergence.  One of those properties is that for any sequence $ x_n$ tending to $ x$, one has $ \lim_{n \to \infty} f_n(x_n) \equal{} f(x)$ where $ f$ is the uniform limit.  (Alternately, you could think of Kent Merryfield's comment in terms of the definition alone.)

$ f_n(1/n)$ = 1/e for all n.. From this, how can we deduce that $ f_n$ tends to zero, and how can we talk separately for those 2 intervals?

I think what they were trying to show you was the following:

First find the pointwise limit.
If the sequence is uniformly convergent then that must also be its uniform limit.
If the uniform limit is $ f$, we should have $ \lim_{n \to \infty} f_n(x_n) \equal{} f(x)$, where $ (x_n)$ is any sequence tending to $ x$.

Do we?

yes, but i still don't get how we will talk separately for those two intervals..what does $ f_n(1/n)\equal{}1/e$ tell us? I see that $ f_n$ is pointwise convergent on [0,infinity), but how can i use this to say if the function is uniformly convergent in these intervals separately?
Thanks

Who said it is in fact uniformly convergent?

It is pointwise convergent in [o,infinity). I don't know if it is in any of these intervals, because i don't know how to use the fact that $ f_n(1/n)\equal{}1/e$ to show that it is or that it is not unif. convergent in those intervals.
Thanks

Let's go all the way back to definitions.  A sequence of functions $ f_n$ is uniformly convergent to a function $ f$ on an interval $ I$ if for any $ \epsilon$ we can find an $ N$ such that if $ n > N$ then

$ |f_n(x) \minus{} f(x)| < \epsilon$ for [b]all[/b] $ x \in I$.

We know that if $ f$ is the uniform limit, it must be the pointwise limit, which is $ f(x) \equal{} 0$ everywhere.  Hence this sequence is uniformly convergent if and only if for any $ \epsilon$ we can find an $ N$ such that if $ n > N$ then $ |f_n(x)| < \epsilon$ for all $ x \in I$.

What happens if $ \epsilon < \frac {1}{e}$ and $ x \equal{} \frac {1}{n}$?  (These points are in both the interval $ [0, 1]$ and the interval $ [0, \infty)$.)

Okay, let's see.

The argument can go as follows.
We know that if our sequence is to have property A, then it must surely have property B.
Now fill in
A: "$ f_n \to 0$ uniformly",
B: "For all sequences $ x_n$ tending to $ x$, we have $ \lim_{n \to \infty} f_n(x_n) \equal{} f(x)$"

If it happens to be that B is not true, then A can not be true either.

Let A be "I have a blue car", B be "I have a car". Surely, if I have a blue car, then I have a car. Furthermore, if I don't have a car, then I also can't have a blue car. (Note that me having a car doesn't imply that it's blue)

Edit: guess I was a bit too late! Maybe t0rajir0u's post is a bit more clear as it refers to the definition. Mine just shows that if the series doesn't have a certain property that it should have if uniformly convergent, then it's not uniformly convergent.

Let $ g(x)$ be a continuous function on $ [0,\infty)$ that is positive on $ (0,\infty)$ such that $ g(0) \equal{} 0$ and $ \lim_{x\to\infty}g(x) \equal{} 0.$ Define $ f_n(x) \equal{} g(nx).$

a) Show that $ f_n(x)\to 0$ pointwise.

b) Show that $ f_n$ does not tend uniformly to zero.


geometry
trigonometry
inequalities
function
geometry proposed

Let triangle ABC and $ (I,r)$ is its incircle. A circle $ (I_a)$, whose radius is $ r_a$, touches AB,AC and extouches the incircle (I). The circles $ (I_b,r_b),(I_c,r_c)$ are similarly defined. Prove that $ r_a \plus{} r_b \plus{} r_c\ge r$

one can prove that $ r_a\equal{}\dfrac{r(1\minus{}\sin\frac{A}{2})}{1\plus{}\sin\frac{A}{2}}$... this can be done by solving the equation $ \dfrac{r_a}{x\minus{}r_a}\equal{}\sin\frac{A}{2}$, where $ x\equal{}AD$, and $ D$ is the closest intersection of the bisector of angle $ A$ with the incircle...

then the inequality can be rewritten as $ \sum \dfrac{1}{1\plus{}\sin\frac{A}{2}}\geq 2$, which follows from cauchy-schwarz and the well-known $ \sum \sin\frac{A}{2}\leq \dfrac{3}{2}$, and we're done  :D

[quote="thaithuan_GC"] [color=darkred]Let triangle $ ABC$ and $ (I,r)$ is its incircle. A circle $ (I_a)$, whose radius is $ r_a$, touches $ AB$ , $ AC$ and extouches the incircle $ (I)$ . The circles $ (I_b,r_b),(I_c,r_c)$ are similarly defined. Prove that $ r_a \plus{} r_b \plus{} r_c\ge r$[/color] [/quote]

[color=darkblue][b][u]Proof (similarly to Campos !).[/u][/b]

$ \blacktriangleright$ [b][u]Case 1[/u][/b] - $ I_a\in (AI)$ a.s.o. Denote $ F\in AB$ for which $ IF\perp AB$ and $ R\in IF$ so that $ I_aR\perp IF$ .

In $ \triangle II_aR$ we have $ m(\widehat {II_aR}) \equal{} \frac A2$ , $ II_a \equal{} r \plus{} r_a$ and $ IR \equal{} r \minus{} r_a$ . Thus, $ \sin\frac A2 \equal{} \frac {r \minus{} r_a}{r \plus{} r_a}$ ,

i.e. $ \frac {r_a \plus{} r}{2r} \equal{} \frac {1}{1 \plus{} \sin\frac A2}$ . Prove easily that the function $ f(x) \equal{} \frac {1}{1 \plus{} \sin x}$ is convexely on $ \left(0,\frac {\pi}{2}\right)$

because $ f''(x)\ge 0$ . Thus, $ \sum\frac {1}{1 \plus{} \sin\frac A2}\ge \frac {3}{1 \plus{} \sin\frac {\pi}{6}} \equal{} 2$ . In conclusion,

 $ \sum\frac {r_a \plus{} r}{2r} \equal{} \sum\frac {1}{1 \plus{} \sin\frac A2}\ge 2$ $ \implies$ $ 3r \plus{} \sum r_a\ge 4r$ , i.e. $ \boxed {\ \sum r_a\ge r\ }$ .

$ \blacktriangleright$ [b][u]Case 2[/u][/b] - $ I\in (AI_a)$ a.s.o. Prove analogously $ \frac {r_a \plus{} r}{2r} \equal{} \frac {1}{1 \minus{} \sin\frac A2}$ and the function $ g(x) \equal{} \frac {1}{1 \minus{} \sin x}$

is convexely on $ \left(0,\frac {\pi}{2}\right)$ because $ g''(x)\ge 0$ .Thus, $ \sum\frac {1}{1 \minus{} \sin\frac A2}\ge \frac {3}{1 \minus{} \sin\frac {\pi}{6}} \equal{} 6$ . In conclusion,

$ \sum\frac {r_a \plus{} r}{2r} \equal{} \sum\frac {1}{1 \minus{} \sin\frac A2}\ge 6$ $ \implies$ $ 3r \plus{} \sum r_a\ge 12r$ , i.e. $ \boxed {\ \sum r_a\ge 9r\ }$ .
[/color]


Prove that for any positive real numbers a,b and c
$ \sqrt {\frac {b \plus{} c}{a}}\plus{} \sqrt {\frac {c \plus{} a}{b}}\plus{}\sqrt {\frac {a \plus{} b}{c}}\geq \sqrt {6 \frac {(a \plus{} b \plus{} c)}{\sqrt[3] {abc}}}$

I think math is  very big sea, there are too many beatiful things ,too many

[quote="Enjar0"]Prove that for any positive real numbers a,b and c
$ \sqrt {\frac {b \plus{} c}{a}} \plus{} \sqrt {\frac {c \plus{} a}{b}} \plus{} \sqrt {\frac {a \plus{} b}{c}}\geq \sqrt {6 \frac {(a \plus{} b \plus{} c)}{\sqrt [3] {abc}}}$

I think math is  very big sea, there are too many beatiful things ,too many[/quote]
$ \sqrt {\frac {b \plus{} c}{a}} \plus{} \sqrt {\frac {c \plus{} a}{b}} \plus{} \sqrt {\frac {a \plus{} b}{c}}\geq \sqrt {6 \frac {(a \plus{} b \plus{} c)}{\sqrt [3] {abc}}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\sqrt{ab(a\plus{}b)}\geq\sqrt{6(a\plus{}b\plus{}c)\sqrt[3]{a^2b^2c^2}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(a^2b\plus{}a^2c\plus{}2a\sqrt{bc(a\plus{}b)(a\plus{}c)}\right)\geq6(a\plus{}b\plus{}c)\sqrt[3]{a^2b^2c^2}.$
But $ \sqrt{(a\plus{}b)(a\plus{}c)}\geq a\plus{}\sqrt{bc}.$
Hence, it remains to prove that
$ \sum_{cyc}\left(a^2b\plus{}a^2c\plus{}2a^2\sqrt{bc}\plus{}2abc\right)\geq6(a\plus{}b\plus{}c)\sqrt[3]{a^2b^2c^2},$
which true by AM-GM.


function
floor function
irrational number
algebra proposed
algebra

A little problem ,which i created :

[color=blue]Problem[/color]Let $f: \mathbb R \longrightarrow \mathbb R$ be a continous function and $a$ a rational number and $b$ an irrational number such that for $x \in \mathbb{R}$

 $f(x)=f(x+a)=f(x+b)$

Prove that $f$ is a constant function .

Note that I called the rational number $r$ and the irrational one $a$.
Also, the whole thing can be made much more rigorous but I don't want to mess with "epsilon-delta" proofs :) This is more a sketch than a proof. Anyway here it goes.

Note that $\frac{a}{r}$ is irrational. Hence by Hurwitz theorem there is infinitely many rationals $\frac{p}{q}$ such that
\begin{eqnarray} | \frac{a}{r} - \frac{p}{q} | & < & \frac{1}{\sqrt{5} \cdot q^2} \\ |q \cdot a - p \cdot r | & < & \frac{r}{\sqrt{5} \cdot q} \nonumber \end{eqnarray}
The term $\frac{r}{\sqrt{5} \cdot q}$ can be as small as we want.
(For a given $q$ there is finitely many $p$ such that $| \frac{a}{r} - \frac{p}{q} | < \frac{1}{\sqrt{5} \cdot q^2} = \varepsilon$. Therefore if $q$ is bounded there is only finitely many $\frac{p}{q}$ such that $| \frac{a}{r} - \frac{p}{q} | < \frac{1}{\sqrt{5} \cdot q^2}$. A contradiction).
Since $r$ and $a$ are both periods of $f$ it follows that their linear combination $q \cdot a - p \cdot r$ is also a period of $f.$ Therefore the period of $f$ can be as small as we want.
Let $x, y \in \mathbb{R}$ be two reals. Let $\kappa$ be a period of $f$ (to be fixed later). Then
\begin{eqnarray*} \left. f ( x + y - x \right) & = & f \left( x + \kappa \left( \lfloor \frac{y - x}{\kappa} \rfloor + \left\{ \frac{y - x}{\kappa} \right\} \right) \right)\\ f ( y ) & = & f \left( x + \kappa \cdot \left\{ \frac{y - x}{\kappa} \right\} \right) \end{eqnarray*}
We have
\[ 0 \leq \kappa \cdot \left\{ \frac{y - x}{\kappa} \right\} \leq \kappa  \]
Therefore as we choose the period $\kappa \longrightarrow 0$ we obtain
\begin{eqnarray*} \kappa \cdot \left\{ \frac{x - y}{\kappa} \right\} & \longrightarrow & 0 \end{eqnarray*}
And since $f$ is continuous, as $\kappa \longrightarrow 0$,
\begin{eqnarray*} f ( y ) = f \left( x + \kappa \cdot \left\{ \frac{y - x}{\kappa} \right\} \right) & \longrightarrow & f ( x ) \end{eqnarray*}
Therefore $f ( y ) = f ( x )$.


$abcd \times n = dcba$

How many solutions to the above equation?

$a,b,c,d$ are all different positive single digits. $n$ is a positive integer.

For example

$2178 \times 4 = 8712$

Is it possible to get more solutions ?

[hide="Conditions"] $abcd \equiv dcba \bmod 9$ (not hard to show), so either $9 | abcd$ or $n \equiv 1 \bmod 9$.  The second case is not very useful, so let's assume $9 | abcd$.

Second, $abcd \equiv-dcba \bmod 11$, so either $11 | abcd$ or $n \equiv-1 \bmod 11$.  Again, the second case is not very useful; we now know $99 | abcd$.  There are less than $89$ such four-digit multiples with four-digit reverse, so an exhaustive search is not too difficult since we only need to check half of them.   [/hide]


algebra
IMO Shortlist
functional equation
Iteration

Does there exist a sequence $ F(1), F(2), F(3), \ldots$ of non-negative integers that simultaneously satisfies the following three conditions?

[b](a)[/b] Each of the integers $ 0, 1, 2, \ldots$ occurs in the sequence.
[b](b)[/b] Each positive integer occurs in the sequence infinitely often.
[b](c)[/b] For any $ n \geq 2,$
\[ F(F(n^{163})) \equal{} F(F(n)) \plus{} F(F(361)).
\]

[hide="Solution"]Answer: Yes
Choose F(1)=0, F(361)=1, the condition (c) will become $ F(F(n^{163})) = F(F(n))$ for every $ n\ge 2$
Construct the sequence as follows:
* $ F(n) = n$ for every $ 2\le n\le 360$
* $ F(n^{163}) = F(n)$ so that the condition (b) is satisfied.
* For remain cases, let F(n) be the smallest of numbers which are other than F(k) with k<n.
There are infinitely numbers not form $ n^{163{}}$, so the condition (a) is satisfied.
We also have $ F(n^{163m}) = F(n)$ for every naturals numbers m, which implies each positive integer appears infinitely often, the condition (b) is satisfied.
The constructed sequence proves the answer "yes"[/hide]

when i click the solution it sends me to the top of the page;it's a problem with my browser?

[hide=Alternative Solution 1]Yes. Just take $F(n)$ to be the number of prime factors of $n$ (counted [i]with[/i] multiplicity).[/hide]
[hide=Alternative Solution 2]Yes. Just take $F(n)$ to be the number of prime factors of $n$ (counted [i]without[/i] multiplicity).[/hide]

Nice and easy. Here's my solution (with motivations): Note that, by condition (c), we cannot have $F(n)=0$ for $n \geq 2$. So, using condition $(a)$, we get that $F(1)=0$. To make condition (c) simpler, this motivates us to take $F(361)=1$, which gives $F(F(n^{163}))=F(F(n))$. I first thought about trying to find solutions to the equation $F(n^{163})=F(n)$, which would directly give condition (c). Even though I didn't really think that any function would work, but fortunately on trying the basic functions which I always try while working on any existential NT FE problem, the [url=https://en.wikipedia.org/wiki/Prime_omega_function]prime omega function[/url] $\omega(n)$ seemed to work (One can also think of this function by noticing the fact that $361$ has only one prime factor, which seems to coincidentally match our assumption that $F(361)=1$). And the problem's done :D.


sum of three real numbres is zero. sum of their squares is 4. then find the sum of their 4th power. i mean
a+b+c=0 ; a2 + b2+ c2 =4 ; a4 +b4 + c4=?

one thing you could do to make it even easier is

instead of expanding

$ (a^2\plus{}b^2\plus{}c^2)^2$ , you could again substitute $ (a \plus{} b \plus{} c)^2$

into $ (a^2\plus{}b^2\plus{}c^2)^2$ and you'd have very little expanding to do before finding the answer as we know that $ (a \plus{} b \plus{} c) \equal{} 0$


function
integration
real analysis
real analysis theorems

Let $f$ be a integrable function, and $g$ be a differentiable function with $g'$ continuous, positive and increasing.
Then,  $f \circ g$ and $f \circ g \cdot g'$ are integrable and:
\[ \int_a^bf(t)dt=\int_{g(a)}^{g(b)} f\circ g(t)\cdot g'(t)dt \].

In the original formula he had $f$ continuous and $g$ differentible with $g'continuous$.

I don't know yet. You can post an "anoter" variable changement, when $g'$ do not exist.


Putnam
analytic geometry
vector
induction
linear algebra
college contests

Let $Z$ denote the set of points in $\mathbb{R}^{n}$ whose coordinates are $0$ or $1.$ (Thus $Z$ has $2^{n}$ elements, which are the vertices of a unit hypercube in $\mathbb{R}^{n}$.) Given a vector subspace $V$ of $\mathbb{R}^{n},$ let $Z(V)$ denote the number of members of $Z$ that lie in $V.$ Let $k$ be given, $0\le k\le n.$ Find the maximum, over all vector subspaces $V\subseteq\mathbb{R}^{n}$ of dimension $k,$ of the number of points in $V\cap Z.$

Was this question easy or did I miss something.

I got 2^k

As did I... I proved by induction and said WLOG the space that passes through the maximum number of points in Z is spanned by unit vectors.

All right- here's a full solution.

The answer is indeed $2^{k}$, achieved by choosing our subspace to be $\{(x_{1},x_{2},\dots,x_{k},0,\dots,0): x_{i}\in\mathbb{R}\}$.

To show that $2^{k}$ is maximal, suppose $S$ is a set of more than $2^{k}$ elements of $Z$. Since every vector in $Z$ has integer coefficients and all are distinct mod 2, we can look at the elements of $Z$ as elements of $F_{2}^{n}$, where $F_{2}$ is the field of integers mod 2.
In $F_{2}^{n}$, every $k$-dimensional vector space contains $2^{k}$ points. Therefore, $S$ is not contained in any $k$-dimensional $F_{2}$-vector space, and some $k+1$ elements $v_{1},\dots,v_{k+1}$ of $S$ ale linearly independent mod 2.

Write $v_{1},\dots,v_{k+1}$ as columns of a matrix. Since these columns are linearly independent mod 2, some $(k+1)\times (k+1)$ submatrix has nonzero determinant mod 2. That submatrix thus has odd determinant, and odd integers are nonzero in $\mathbb{R}$. This means that $v_{1},\dots,v_{k+1}$ are linearly independent over $\mathbb{R}$, and $S$ is not contained in any $k$-dimensional $\mathbb{R}$-vector space.

First post heh...

So the solution is 2^k.  Obviously n independent vectors span 2^n vertices.  Also, obviously 1 vector spans at most 2 vertices(2^1).

We then show that if we have k vectors spanning x vertices for any x, we can construct k+1 vectors spanning at least 2x vertices in the following way:

Take e_i defined to be the vector with a 1 in the i-th row and 0's elsewhere.  Obviously as the set of all e_i is of dimenion n, for any k<n, there exists an e_i such that e_i is not a LC(linear combination) of the k given vectors.

Now note that for any element V of the unit hypercube the i-th entry of V is either 0 or 1.

If the i-th entry of V is 0, then V+e_i is not an element of the span of the k vectors given.  Otherwise, e_i would be in the span of the k vectors given contradicting our choice of e_i.  But V+e_i is obviously an element of span{k given vectors, e_i}.

Similar idea i-th entry of V is 1.

Thus we have constructed k+1 vectors spanning at least 2x vertices if we are given k vectors spanning x vertices.  Thus recalling we can find 1 vector spanning 2 vertices, we can find k vectors spanning at least 2^k vertices.  But if k vectors span more than 2^k vectors then it is possible to find n vectors spanning more than 2^n vertices an obvious contradiction.  Thus any k vectors span at most 2^k vertices.

I hope this was clear.  Sorry I didn't use appropriate math tools.  I thought that A4 and B4 were both almost trivial this year.  It seemed really strange.

Sorry in the above outline of a solution I meant to define V as an element of both the unit hypercube AND an element of the span of the k given vectors.

Everybody, This is my problem. :wow:  I created it 3 months ago, and now I see it as "Putnam 2006 B4". I also proved a generalization of the problem. Here you can see my posts about this problem :
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105991]Link 1[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110623]Link 2[/url]

Haha Cool Omid,  :D
They have mentioned it in the official solutions too.
(Maybe you should propose one for Shortlist  :cool: )

[quote]They have mentioned it in the official solutions too.[/quote]

According to the 2006 solutions at [url]http://www.unl.edu/amc/a-activities/a7-problems/putnamindex.shtml[/url]:

"Remark: The proposers probably did not realize that this problem appeared online about three months before the exam, at [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=105991[/url]. (It may very well have also appeared even earlier.)"

I wrote neatly 2^k on the paper with nothing else.  And I didn't get any partial credit for guessing it right?!

A right answer without a proof is definitely not worth much- in this problem, almost all of the difficulty comes in making the maximum rigorous, and the answer's pretty easy to guess even without a good reason. There's a reason I invoked linear algebra over $F_{2}$ in my post.

Putnam graders are also very tough with partial credit in general.

yeah, I heard they're very tough on partial credit.

The answer is $2^k$. I will bound only.

Assume FTSOC there exists $2^k+1$ vectors in $\{0,1\}^n$ that works. Pick $k+1$ of them, call $v_1,\cdots,v_{k+1}$ that are linearly independent over $(\mathbb{Z}_2)^n$ but they are linearly dependent over $\mathbb{R}$.

The key step is to note that the equation $c_1v_1+\cdots+c_{k+1}v_{k+1}=0$ has a nontrivial solution IN INTEGERS.  This can be proven via induction on $k$. Say we have $\gcd(c_1,\cdots,c_{k+1})=1$. Now taking mod 2 implies the $v_i$'s are linearly dependent in $(\mathbb{Z}_2)^n$. Absurd.


geometry
rectangle
ratio
induction
combinatorics unsolved
combinatorics

Consider a m*n rectangle. We place a 0 or 1 in each of the mn cells. In each 'move', we consider a full row / column and for each 0, we replace by 1, for each 1 we replace by 0. We call a positioning of 0s and 1s 'ideal' if there is no sequence of moves that decreases the number of '1's in the positioning. For each ideal positioning K , we note the number of '1's f(K). Find all values, in terms of m and n, that f(K) can assume.

$f(k)$ can assume every value between 1 and it's maximum. Now comes the real problem, which I am 99% sure it's true, but I just can't prove it. In every ideal rectangle, no matter what the sizes are, the ratio between the number of 1's in it's interior and the area of the rectangle is at most $\ds \frac 14 $. There are many figures, I think, which satisfy that the ratio is $\ds \frac 14$, without being decomposed in smaller pieces with that property, but I only managed to find these :
1x
xx

1xxx
x1xx
xx1x, or some alternatives for the $3$x$4$. With the use of these we can fill up any rectangle in which $m\cdot n=4k$. If m and n are both even it's trivial, if $m\vdots 4$ and n is odd, we can make wide bands $4k$, using one of the second type ,and $(k-3)/2$ of the first type. If they are both odd, we will use one figure of this type (or maybe even more, but I think one is enough):
1xx
xx1
xxx.
I tried to solve the "real problem" by induction, but it didn't worked.
I really want to know the solution. Will anyone help?


Is this normal?  I believe it's been about a little over 2 months since I took AMC12A.
I heard that they've already mailed the official score reports about 2 weeks ago.
When can I expect to get the report in the mail? I live in Tennessee by the way.

You should have it by now. Send the AMC an email.

It is mailed to your school, not to your home, if that's what you meant.


I know that many people do not frequent the NOML forum so I thought I would repost Cody's message here.  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47552]Here[/url] is where this quote originally came from.

[quote="mathjoker"]The National Online Math League is proud to announce the beginning of its third season of online competition. Under the direction of founder and commissioner Cody L. Patterson, NOML challenges math enthusiasts of all ages to stretch their mathematical boundaries with a series of exciting problems that begins in September and stretches into May.

In the 2005-06 season, teams from across America will square off in head-to-head matches for the right to play in Cody Bowl V, one of the biggest mathematical competition events of the year. For the first time ever, NOML is truly a league without boundaries. Middle school students will face off against college math majors on problems designed to test ingenuity rather than content knowledge. Players at the age of 13 will have the chance to collaborate with players at the age of 23 on problems that read like mathematical theorems, inviting exploration, generalization, and conjecture. All of this is due to the league's new "No Boundaries" format, which gives every player in the league access to every problem, without restrictions on roster size, individual test difficulty, and team test collaboration.

Once again, NOML seeks to create a winning formula by taking the best of the old - our two-pronged contest format - and combining it some fresh faces and some new adjustments to maximize the level of excitement throughout the season. As always, each NOML match pits two teams against each other - no free-for-alls here, either you win or you lose. Each match consists of two rounds - a fast-paced Individual Round with questions that test on-the-spot thinking and cleverness, and a mentally taxing Vertical Round with questions that invite team members to work together to explore unfamiliar problems and present their results. The "fresh faces" come in the form of a set of outstanding new problem writers, including Joel Lewis, Jeffrey Middleton, Robert Nix, Simon Rubinstein-Salzedo, and Nathan Savir. As for the "adjustments" - you'll have to wait and see!

NOML is proud to offer a math league that runs throughout the entire academic year, is on par with some of the nation's most prestigious contests in terms of problem quality, has an online format that allows students from different parts of the world to join together in team competition, and is still completely free. All you need to join is an e-mail address, and the desire to dig into some fun and engaging problems. To get in on the action, send an e-mail to codybowl@gmail.com with the following information:

* Your name,
* Your e-mail address,
* Your AIM screen name (if applicable),
* Where you're from,
* Where you go to school (if applicable),
* A few words about your previous math contest experience (if you have any - if you don't, we're happy to be your first competition!)
* And whether you're interested in owning your own team. (It's a fairly substantial responsibility, and if you're not sure you want to make such a commitment, we're happy to put you on one of our pre-existing teams.)

The fun officially begins in less than a month, with our Week 0 preseason game starting on September 1, 2005. Don't miss out on the action! E-mail today![/quote]

I don't think the statement "teams from across America" actually restricts participants to being US residents. Last year, one of the competitors lived in Shanghai, China! Assuming you're familiar with common English math terms, you should be good to go. :) Send him an e-mail, and you're on your way!

That's cool!
I have another question now: How do we orgnize the teams?

There will be a new player salary system that will be implemented for the first time this season.  This is like professional clubs having a budget cap to prevent a huge number of superstars joining one particular team. At this time, the only things you can do toward forming teams are probably notifying your potential teammates of your interest and including their names in your e-mail. Pretty soon, he'll use a formula to determine how much each player will cost and whether or not you can actually build your team.

When is the deadline?  Is it all done online?  How?

A salary system? How does he actually assign values to people he doesn't know?

Well he asked for past experience, so that might help. There might be a eval test at the beginning? iono, never taken this before either, hehe

Can someone explain how all the contests/teams or whatever works?  I'm thinking about joining, but want to know the rules.

Edit: also when do we have to sign up by?

ditto, eryaman.

i have a few friends, some in college and some in lower grades in high school. can we form a team? or are teams handled differently?

[edit] i looked at some of the problems on noml.org and im interested, but i still cant find any info on rules or scoring. the problem pdfs from year IV tell me to look on the cody bowl site for official rules, but a search on noml.org gives me nothing! can somebody please explain?

There will be a [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]Math Jam[/url] on September 6 at 7:30 PM.  Cody Patterson, NOML founder, will discuss NOML, go over a couple problems from last year, and answer your questions.

crap i just missed it  :(

is there a site with official rules and participation info?

You can check the transcript [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=123]here[/url].

I think there's a rules thing on their website now also.

I'm going to unsticky this now but keep in mind that you can still sign up for NOML and become a free agent by emailing codybowl@gmail.com


The sequences $3, 20, 37, 54, 71$ and $16, 27, 38, 49, 60, 71$ have $71$ as the common term. What is the next common term?

because the sequences are just 3 mod 17 and 16 mod 11, and we use crt to solve, 71 mod 11*17 is the common numbers, and clearly the next is 11*17+71

@ Altheman - Hide your answers & solutions please.

[hide]The first sequence is 3 mod 17; the secod is 5 mod 11. After 71, the next common term will be $\text{lcm}(11,17)+71=\boxed{258}$.[/hide]


I need help.
Which one of the following combinations of given parts does not determine the indicate dtiranlge?
A) base angle and vertex angle; isosceles triangle
b) vertex angle and the base; isosceles triangle
c) the radius of th ecircumscribed circle; equilateral tirangled
d) one arm and the radius of the inscribed icrcle; right tirangle
e) two angles and a side opposite one of them; scalene triangle

btw the answer is a) but what does c) and d) mean?

Hmm.. after deciphering the typos :D

In c), the circumscribed triangle is the triangles circumcircle, which is the circle that goes through all three vertices of the triangle.

In d), one arm would indicate one of the sides of the triangle, and the inscribed circle is the incircle which is the biggest circle that fits inside the triangle.

The grammar policeman is afraid to set foot here.  This one is too painful.

- [url=http://www.livejournal.com/community/grammarpolice/]Grammar Policeman[/url] #2059123, now driven insane by all those dtiranlges.

Actually, in d, it would really mean one leg of the right triangle -- one of the shorter sides.  If it could include the hypotenuse as well, it might not be uniquely determined.


algebra
polynomial
number theory unsolved
number theory

prove that there is not any polynomial like $ f(x)\equal{}a_nx^n\plus{}...\plus{}a_1x\plus{}a_0$ such that $ \forall n \in N : f(n)$ is a prime number.
( f(x) is not constant $ a_i \in Z$)

Take any n, suth that $ |f(n)|\equal{}m>1$. Then for any k $ m|f(n\plus{}km)$.


geometry
cyclic quadrilateral
perpendicular bisector

I made up a geometry problem, and I think it's pretty good. But keep in mind that since I made it up, there is probably a simple, easy solution that I am too blind to see. Good luck!

In cyclic quadrilateral ABCD, AB and CD intersect at E, and AD and BC intersect at F. The perpendicular bisector of AC intersects the line BD at X, and the line AC intersects the perpendicular bisector of BD at Y. Given that E, X, and Y are collinear, prove that F is also on this line. (see diagram)

Let [tex]AC\cap BD=P[/tex] and let [tex]O[/tex] be the center of the circle. It's clear that [tex]O[/tex] must be the orthocenter of [tex]XPY\Rightarrow OP\perp XY\Rightarrow OP\perp EX (*)[/tex]. On the other hand, it's well-known that [tex]O[/tex] is the orthocenter of [tex]EPF[/tex], so [tex]OP\perp EF (**)[/tex]. From [tex](*),\ (**)[/tex] we find [tex]F[/tex] to be on the line [tex]EXY[/tex].

What program did you use to draw that diagram? just curious.

[quote="grobber"]On the other hand, it's well-known that [tex]O[/tex] is the orthocenter of [tex]EPF[/tex][/quote]

Hm, I didn't know that. Is there some obvious reason why that's true that I'm just not seeing, or is it somthing you just happen to know to be true?

BTW, I used geometer's sketchpad to make the diagram.

Just use polars: [tex]EPF[/tex] is an autopolar triangle wrt the circle, and the polar of a point wrt a circle is perpendicular to the line determined by the point and the center of the circle, thus the result.


You have two jars, one with $x$ beans and other with $y$.
Each jar has a lever. when a jar has at least two beans, pressing its lever will give you a bean and will move one bean to the other jar (if a jar has less than 2 beans, the lever will not have any effect).  What is the necessary and sufficient condition on $x$ and $y$ so that you can get all of the beans, except one.

[hide="hint"]
consider x-y[/hide]

Solution:
[hide]
The necessary condition is: ${x-y\not\equiv{0}\mod{3}}$.

Proof:
At the start there are x beans and y beans. WLOG say we pull lever x. Then there are (x-2) and (y+1) beans.

Let ${x-y\equiv{k}\mod{3}}$. Then clearly ${(x-2)-(y+1)\equiv{x-y-3}\equiv{x-y}\equiv{k}\mod{3}}$

Let $x-2=x_{2}$ and $y+1=y_{2}$. Clearly no matter what lever we pull, $x_{2}-y_{2}\equiv{k}\mod{3}$ and continuing for all positive integer n, $x_{n}-y_{n}\equiv{k}\mod{3}$.

After the last lever pull, say the remaining bean is in x. Then there is 1 bean in x and no beans in y, and $1-0\equiv{1}\mod{3}$.

Since the value of $x-y\mod{3}$ never changes, for it to end with $x_{n}-y_{n}\equiv{1}\mod{3}$, it must have started that way.

Likewise, suppose we have $y-x\equiv{1}\mod{3}$; then we can do the same thing, but the last bean ends in the "y" container.

So everything works out if $x-y\equiv{1}\mod{3}$ or if $y-x\equiv{1}\mod{3}\Rightarrow x-y\equiv{-1}\equiv{2}\mod{3}$. So the only case that does not work is the only remaining case: $\boxed{x-y\not\equiv{0}\mod{3}}$ QED.
[/hide]


function
algebra
polynomial
system of equations

When f(x) is divided by x-1, the remainder is 12. when f(x) is divided by x-4 the remainder is 3. Find the remainder when f(x) is divided by (x-1)(x-4) ???

AnyOne?

Can anybody that's good at functions tell me if that's correct? :)

Sorry If I'm triple posting, but can someone tell me if I'm correct, I would really like to know if I've solved the problem right ;)

well i dont quite know what mentiser expected, but u are assuming that f is a polynomial of the second degree. then it is certainly correct, but f actually could also be $f(x)=-3x+15$ ... but what is the remainder after division (x-1)(x-4) then? also if f is of the third degree, i think it would leave a diffrent remainder, but try it urself, u only have to find a function for which f(1)=f(4)=0 ...
i think the question is not specific enough though, so u did a good job to figure out the remainder for a second degree polynomial

It is probably better to set $f(x)=Q(x)(x-1)(x-4)+R(x)$, where $Q(x)$ is the quotient and $R(x)$ is the remainder. We know that $R(x)$ must have a degree of 1 (or less) because $(x-1)(x-4)$ is a second degree polynomial, so we let $R(x)=ax+b$.

The rest is the same as Iversonfan's solution. (Just consider $f(1)$ and $f(4)$ and solve the system)


geometry
geometric transformation
reflection
analytic geometry
graphing lines
slope
MATHCOUNTS

The triangle with vertices at $A(-2,2)$, $B(-8,2)$, and $C(-8,-1)$ is reflected about the line $y=2x+1$. Express the coordinates of the reflection of $A$ as an ordered pair.

Basically, how do you reflect a point over a line?

I have been wondering that for a long time.
I believe that when a point is reflected across a line, its closest distance to the line is still the same.

[quote="236factorial"]I have been wondering that for a long time.
I believe that when a point is reflected across a line, its distance to the line is still the same.[/quote]ditto. And isn't the slope between the original point and the reflected point perpendicular to the slope of the line you're reflecting it across?

Oh yeah, why didn't I suddenly pop that out of my mind. that means the slope of the perpendicular line in this case is -1/2, right?

If A is refelected across a line y=mx+b to A', then y=mx+b perpendicular bisects
__
AA'

also,

[hide="big hint"]
If you have a point $(x,y)$ and the line $ax+by+c=0$
then the distance from the point to the line is
$|\frac{ax+by+c}{\sqrt{a^2+b^2}}|$
[/hide]

[quote="Boy Soprano"]Yes, when one reflects a point $k$ over a line $l$, its image is the other point on the line perpendicular to the $l$ that is the same distance from $l$.  (Unless I am mistaken, this is the definition.  See Coxeter's [i]Introduction to Geometry[/i].)  From this, we can construct an algorithm.  We take the line $n$ perpendicular to $l$ that passes through $k$ and find its intersection with $l$ (we can call this point $o$).  Then we can find the point that is twice as far along $n$, and this is our image of $k$.  I think I'll work more on this when I have time to find the formula, if that's simpler, and then post it unless someone else does first.
[hide]In any event, the solution in this particular case is $A'(2,0)$, I believe.[/hide][/quote]
yes that is the correct answer

I've investigated a formula.  As far as I can tell, the answer is your worst nightmare and it is much easier to use the algorithm.  The problem is much simpler, of course, if the line's slope is 1, 0, or undefined.

$A=(-2,2)$. We need to find the distance from the line $y=2x+1$. We know that the $y$-value is $2$, so $2=2x+1 \Rightarrow x=\frac{1}{2}$. The distance from $A$ to $(\frac{1}{2},2)$ is $2+\frac{1}{2}=\frac{5}{2}$.

The slope between $A$ and $A'$ is the opposite reciprocal of the slope of $y=2x+1$, so the slope is $-\frac{1}{2}$. Now we have $m=-\frac{1}{2}=\frac{y-2}{x+2}$. And from before, we have $\frac{5}{2}=\sqrt{(y-2)^2+(x-\frac{1}{2})^2}$. Then you solve that system to get $x=2, y=0 \rightarrow \fbox {(2,0)}$.

As Boy Soprano pointed out, the general solution is pretty ugly.  As is often the case with MATHCOUNTS problems, they very carefully chose the problem to give you an easier way to do it.

The slope of the line of reflection is $\frac{2}{1}$.  So we expect a "1 over, 2 up" pattern if you follow points along this line.

The slope of a perpendicular line is $-\frac{1}{2}$.  So we expect a "2 over, 1 down" pattern if you follow the points along this line.

The intersection of these two lines occurs at a lattice point, which is what gives you a reasonably easy solution.  See the diagram.

Cool diagram. That reminds me, what's the best [b]free[/b] geometry program?

[quote="rcv"]As Boy Soprano pointed out, the general solution is pretty ugly.  As is often the case with MATHCOUNTS problems, they very carefully chose the problem to give you an easier way to do it.

[/quote]

I did it with the perpendicular lines concept, and after where I found where the lines intersect (happening to be point with whole number coordinates). The answer is pretty easy from there. A diagram definetly helps.

[quote="chess64"][quote="nat mc"]$|\frac{ax+by+c}{\sqrt{a^2+b^2}}|$[/quote]

Hmmm... If $ax+by+c=0$, then substituting, we see $d=0$! I believe you mean that the distance from the point $(x_0,y_0)$ and the line $ax+by+c=0$, is $d=|\frac{ax_0+by_0+c}{\sqrt{a^2+b^2}}|$.

BTW, since when have you been taking Analytic Geometry? (That's a formula from analytic geometry!)[/quote]
ax+by+c=0 is the equation of the line.  then the point has coordinates of variables x and y.  you plug those into the equation with coeffficients.

I don't know what analytic geomoetry is, but I took regular geometry like three years ago


wait... so what's the overall score now??

We tied so we split the point in half. :P

37. Is your attention span so short that you declare a tie after playing into deadlock for 2 days?

I don't really know why they declared a tie, it's confusing. Right after cgyao and angel we're in constant battle, they declared a tie, then Izzy posted... so on and so forth. I think that these Boy vs. Girl battles are kind of spamming the fun factory. Don't get me wrong... I just saw this whole thing in a different light about 25 seconds ago, and I like this game too. But shouldn't there be one thread where all the Boy vs. Girls games are held?

yeah, maybe there should be like a boys vs. girls forum, like the mafia games forum??


factorial
algebra
polynomial
Euler
real analysis
real analysis unsolved

This is a problem me and a friend decided to tackle while at a math competition. It was rather fun and intriguing at the time and I wonder if anyone had any other comments and/or solutions pertaining to the problem.
Well here it is,
Find an explicit general expression for sums of the form
\[ \sum_{n\equal{}1}^{\infty} \frac{n^k}{x^n}\]
I have a very informal paper including my non-rigorous solution to the problem if anyone is interested.

[b]Lemma 1:[/b]  $ \sum_{n \equal{} 0}^{\infty} {n \choose k} x^n \equal{} \frac {x^k}{(1 \minus{} x)^{k \plus{} 1}}$.

[b]Lemma 2:[/b]  $ n^k \equal{} \sum_{j \equal{} 0}^{k} S(k, j) j! {n \choose j}$, where $ S(k, j)$ are the [url=http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html]Stirling numbers of the second kind[/url].  The technique I give [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=211560]here[/url] is essentially a quick way to compute the Stirling numbers for small values of $ k$.

It therefore follows that

$ \sum_{n \equal{} 0}^{\infty} n^k x^n \equal{} \sum_{j \equal{} 0}^{k} S(k, j) j! \frac {x^j}{(1 \minus{} x)^{j \plus{} 1}}$.

Edit:  An alternate form.  Define $ (n)_k \equal{} k! {n \choose k} \equal{} n(n\minus{}1)...(n\minus{}(k\minus{}1))$, the [url=http://mathworld.wolfram.com/FallingFactorial.html]falling factorial[/url].  Our two lemmas are then

$ \sum_{n\equal{}0}^{\infty} (n)_k x^n \equal{} \frac{1}{(1 \minus{} x)^{k\plus{}1}}$

and

$ n^k \equal{} \sum_{j\equal{}0}^{k} S(k, j) (n)_j$

which gives

$ \sum_{n\equal{}0}^{\infty} n^k x^n \equal{} \sum_{j\equal{}0}^{k} S(k, j) \frac{1}{(1 \minus{} x)^{j\plus{}1}}$.

Interestingly enough t0rajir0u your interpolation method used to find the Newton polynomials is much as the same method I had used to come up with coefficients of polynomials by which i named error functions. These helped me to define and explicit form and coincidently enough the sequence I found is related to the Euler triangle. any comments you have on the results i came up with would be highly appreciated.

Interesting.  The connection between the Eulerian numbers and the Stirling numbers is given [url=http://www.mathpages.com/home/kmath529/kmath529.htm]here[/url].

awesome. i will definitely look into this some more. thanks t0rajir0u.


group theory
abstract algebra
Pascal\u0027s Triangle

Prove the following well-known property:

[img]http://upload.wikimedia.org/math/9/a/f/9af27c32bff67dad09025d8d4e304b71.png[/img] (Ignore the (5))

In Pascal's triangle, each entry (the $k$th entry in the $n$th row) is the sum of the two entries just above it (one slightly to the left, one slightly to the tright).

So the sum of each row is double the sum of the previous row, and the first row has sum 2, so the statement is true by induction.

consider a group of n presidents. you want to find the number of ways of picking subgroups of k presidents who think they are the best in the world, where 0<=k<=n, which is like asking each president, do you think you are the best in the world, since there are n presidents, and each one answers yes or no, it is 2^n

A related question:

Find a nice formula for the sum

$\dbinom{n}{0}^{2}+\dbinom{n}{1}^{2}+\dbinom{n}{2}^{2}+\cdots+\dbinom{n}{n}^{2}$.

[hide="rnwang2"]$\binom{2n}{n}$?[/hide]
[hide="a proof of the original question"]Just plug in $(1+1)^{n}$ in the binomial theorem.[/hide]
[hide="a more elegant proof of the original question"]Consider constructing some arbitrary subset. Consider an arbitrary element to put in there. This element can either be in the set or not in the set. This can be said for each element. So when making a subset each element has two choices, so there are $2^{n}$ possible subsets of an n element set.[/hide]


geometry
power of a point
radical axis
geometry proposed

Let   $ABCD$  an quadrilateral  inscribed in a circle with center $O$. A circle $\omega_1$ passes through $A$  and is tangent to $OC$ at $C$.    A circle $\omega_2$ passes through $D$ and is tangent to $OB$  at $B$ . These two circles intersect in two points and let $P$  the one intersection point which is interior to  $ABCD$ .If the diagonals of $ABCD$  intersect at $E$ , prove that
   [b]i) [/b]points $O , P , E$   are collinear .
   [b]ii) [/b]$\angle APC = 90 + \angle B$  or $\angle APC = 90 + \angle D$ ( it depents on the figure !)

[u]Babis[/u]

Nice problem indeed!  :D

Part [b]i)[/b]: It's clear that $OA$ is tangent to $\omega_1$, and that $OD$ is tangent to $\omega_2$. This means that $O$ lies on the radical axis of $\omega_1$ and $\omega_2$.

Point $E$ is the intersection of two radical axes, so automaticaly, it lies on the third radical axis. Now it is clear that $O,P$ and $E$ are collinear.

Yes nice. For the first I have similar solution with Jan . For the second I think therre exist a typo or you have another figure mr Babis .
Let $S$ the center of $w_1$ .Then we have
$\angle{APC}=180-\frac{\angle{ASC}}{2}=180-\frac{180-\angle{AOC}}{2}=180-\frac{180-2\angle{D}}{2}=90+\angle{D}$


I'm considering using AOPS so I can skip precalculus (my goal is to actually skip precalculus and Calc BC if possible by using the awaited AOPS calc book that should be coming out later this year)(my school didn't let me skip Alg 2 so I've just been working through vol 1 + parts of 2 in back of class lol). (I haven't bought the books yet).

I would have to finish Intermediate Algebra and Precalculus to do so, and I'm wondering how long that would take, given I would devote 1-1.5 hours a day to going through the book (I could always just skip higher level AIME/USAMO problems and come back to them later when I'm training for AIME next year.) Would it be possible to finish both in around 5 months (and if less, how long would Calc take)? I'm already fairly well versed in Alg/Precalc, but I have some gaps in knowledge...basically, I would only need to do the following chapters (I would probably just run through all the problems in between material that I already know):

I already know most of the material in intermediate algebra, albeit to not as much depth as covered in the book, so I could probably skip a bit in here if needed (though I plan on working through it). I know basic trig for precalc, so I could skip here as well if needed.

Is the above achievable in the given timeframe (my target is around 2 month for Intermediate algebra, 2 months for precalc, then like 4 months for calc, which I have almost no background in)? Thanks. AOPS Vol 2 can probably fill me in on a good bit of stuff beforehand... Thanks.

Hi megahertz,

It really depends on the book. In your case, it might take quite a while, as Intermediate Algebra and Precalculus are the two largest books from AoPS. Because I haven't finished these, I can't give you an estimate, but with previous books about half the size of these I finished in about a month, so expect to take two or three months (a little more based on difficulty) on Intermediate Algebra and Precalculus I guess.

Have you asked your school yet about whether they will give you credit? Some schools are very picky.


The product of three different positive integers is equal to $ 7^3$.
What is the sum of the three integers?

The only possible 3 numbers are 7^2, 7, and 1.
49+7+1=57.

How do you know those are the only numbers?

Because 7 is prime, 7^3 can only have factors of 7^0, 7^1, 7^2, and 7^3.  Because the three integers are different, 7^3 can't be used, and thus the answer is 1+7+49=57


The MathCounts national competition was held last week.  New York State placed 15th out of 57 states and territories.

Andrew Ardito, a homeschooled student from upstate New York, finished 10th in the written portion (out of 228 students), qualifying for the oral countdown round.  There he was lightning quick, defeating his first three opponents before losing in the final round to Daesun Yim of New Jersey.  For finishing 2nd in the country, Andrew (as well as New York's coach, me) was invited to the White House for a picture with the president.  (Because I am not exactly a fan of Mr. Bush, I declined the invitation.)

Alex Iriza, an 8th grader from Dalton, also did well.  He finished 52nd in the country.

Dan Mendelsohn, a 7th grader from Nassau County (Long Island), finished 71st in the country.  I hope to see Dan back next year.

David Bieber, an 8th grader from upstate New York, finished 121st in the country.

Congratulations to all of them!

This year, ESPN decided not to broadcast the national competition.  MathCounts is trying to line up another channel, such as the Discovery Channel, for future years.

The MathCounts national competition was probably the most exciting math contest I have ever seen.  The MathCounts staff does an amazing job organizing the competition.

P.S.  Next year's national competition will be in Ft. Worth, Texas.

Well, congratulations to you and your students!  That's an excellent performance!

Does this mean they've switched the countdown to an elimination format?  Does New York state still use the ladder system?

Thanks!

Yes, for the last three years, the MathCounts national countdown has been a bracket (elimination) tournament among the top 12 students.  The top 4 students get a bye in the first round.  I have mixed feelings about the bracket format: the written portions should perhaps matter more than whether you buzz in half a second faster than the other person.  Even in the written portions, I feel that MathCounts places a little too much emphasis on speed.  But it's a great hook for pulling in kids, and hopefully over time those kids will get drawn to the more contemplative parts of math.

In states where countdown is official, the chapter and state countdowns are required to be in ladder format for the top 10 students.  In states where countdown is unofficial, chapters and states can choose ladder or bracket format.

In New York, where the countdown is unofficial, this year's countdown used ladder format; last year's countdown, at least in Manhattan's chapter, used bracket format.  When countdown is unofficial, I actually prefer the bracket system, since it gives all 12 students a reasonable chance to win.


This is probably very basic but... Say im typing text in Latex, and then I want a space below my text and then to start a new paragraph.  Equivilant to hitting the return key twice on the keyboard.  I know that if I hit the return key twice on the keyboard in the latex document that it will start a new paragraph for me but what If I want a space between the two paragraphs?

The short answer is don't do it - leave it to LaTeX to format your document as it will do it properly as in a book.
But I don't suppose you will be happy with that answer :) so here are a couple of ways:
1. if it's just the odd line then put a \\ at the end of the line, then in TexnicCenter, press Enter twice.
You can vary the vertical space by using \\[length] with whatever length you like.
2. if you want this always to happen between paragraphs then use
\setlength{\parskip}{\baselineskip}

Do read the [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=parskip]TeX FAQ's take on this[/url].


inequalities
limit
trigonometry
analytic geometry
logarithms
inequalities unsolved

hello,the searched maximum is $ (\sqrt{3}\minus{}1)^3$ and will be obtained for $ x\equal{}y\equal{}z\equal{}\frac{\pi}{3}$.
Sonnhard.

[quote="kuing"][quote="VIMF"]let $ \Delta ABC$. find maximum value of
$ (tanA \minus{} 1)(tanB \minus{} 1)(tanC \minus{} 1)$[/quote]

but i think this is no maximum...[/quote]

you are right

$ (tanA \minus{} 1)(tanB \minus{} 1)(tanC \minus{} 1)\le k$

The inequality does not hold for any k

$ (tanA \minus{} 1)(tanB \minus{} 1)(tanC \minus{} 1)\le (\sqrt {3} \minus{} 1)^3$

the inequality doesn't hold   for example

a=4.980985018, b=2.770509054, c=5.962791639

Let $ k > 0$ and $ A \equal{} \frac {\pi}{3}, B \equal{} \frac {\pi}{6} \minus{} k, C \equal{} \frac {\pi}{2} \plus{} k$, that

$ \lim_{k \to 0}{(\tan A \minus{} 1)(\tan B \minus{} 1)(\tan C \minus{} 1)} \equal{} \lim_{k \to 0}{(\sqrt {3} \minus{} 1)(\tan{(\frac {\pi}{6} \minus{} k)} \minus{} 1)(\tan{(\frac {\pi}{2} \plus{} k)} \minus{} 1)}$  $ \equal{} \plus{} \infty$

[quote="hedeng123"][quote="kuing"][quote="VIMF"]let $ \Delta ABC$. find maximum value of
$ (tanA \minus{} 1)(tanB \minus{} 1)(tanC \minus{} 1)$[/quote]

but i think this is no maximum...[/quote]

you are right

$ (tanA \minus{} 1)(tanB \minus{} 1)(tanC \minus{} 1)\le k$

The inequality does not hold for any k

[/quote]

what about acute triangle ?

hello, you have right, the coordinates of my point leads not to the maximum, sorry.
Sonnhard.

if $ \Delta ABC$ is an acute triangle:

WLOG, assume that $ 0< A \le B \le C <\frac{\pi}{2}$, easy to prove that $ B > \frac{\pi}{4}$, so we have $ (\tan B \minus{} 1)(\tan C\minus{}1) > 0$.

if $ A \le \frac{\pi}{4}$, that we have $ (\tan A\minus{}1)(\tan B \minus{} 1)(\tan C\minus{}1) \le 0$.

if $ A > \frac{\pi}{4}$, that we have $ A,B,C \in (\frac{\pi}{4},\frac{\pi}{2})$,
let $ f(x)\equal{}\ln{(\tan x \minus{}1)},x \in (\frac{\pi}{4},\frac{\pi}{2})$,
we can prove that $ f(x)$ is "half concave and half convex", so, if $ (\tan A\minus{}1)(\tan B \minus{} 1)(\tan C\minus{}1)$ have maximum, it must getting maximum when $ A,B,C$ have two equal, thus, we just need to find the following funtion's maximum:
$ g(x) \equal{} (\tan x\minus{}1)^2(\tan (\pi \minus{} 2x) \minus{} 1), x \in (\frac{\pi}{4},\frac{\pi}{2}).$

actually, $ g(x)\equal{}\minus{}(\tan 2x \plus{} 1)(\tan x\minus{}1)^2$, $ g'(x)\equal{}\frac{2\sin 3x}{\cos^3 x (\sin x\plus{}\cos x)^2}$, easy to prove that $ g(x)$ get maximum if and only if $ x\equal{}\frac{\pi}{3}$,

so, $ (\tan A\minus{}1)(\tan B \minus{} 1)(\tan C\minus{}1)$ get maximum if and only if $ A\equal{}B\equal{}C\equal{}\frac{\pi}{3}$, the maximum is $ (\tan \frac{\pi}{3} \minus{}1)^3\equal{}6\sqrt{3}\minus{}10$.


induction
geometry
MATHCOUNTS
FTW
topology
number theory

Combo, geo, NT, or Algebra? (in the Olympiad Sense)

My opinion: Combo is pretty beautiful because what it deals with requires no knowledge whatsoever, and normally strikes a fundamental chord in numbers. However, too many problems in my opinion are just guided induction bashes.
Number Theory is pretty great as well, for similar reasons, and in the sense that (even though Olympiad problems might not be so) the properties that resonate out of the integers are highly unintuitive and deep.

For these reasons as well, I'm inclined to dislike synthetic geometry problems.

THis is meant to be an opinionated thread, not one in which major convincing is done.

Post your thoughts!

I know I'm in the wrong area, but I've always liked Combinatorics the most. I like the fact that it is highly intuitive and that you don't need nearly as much knowledge in it as in other aspects of math.

A huge number of the formulas are intuitive enough that i've learned them on the spot. For example during my 8th grade Mathcounts year, I remember finding out balls and urns for myself while doing a problem. I thought it was a cool new technique, but when I went onto the forums, it turns out that it was called balls and urns.

[quote="pythag011"][quote="not_trig"]I like geometry due to visualblah concrete stuff, but you gotta know all these obscure and later useless theorems for it. how unfort.

I dislike combo because I suck at it. well actually it's pretty cool, but I suck at it.[/quote]

Hmm, how useful are these obscure theorems anyways? Since i'm pretty bad at geo...[/quote]

well, depends on your definition of obscure. they are pretty helpful. but Euclidean geometry is pretty much useless... :( (ok, not completely; but "pretty much")

[quote="junggi"][quote="CatalystOfNostalgia"]algebraic topology is good stuff, guys
too bad i don't actually know very much[/quote]
this[/quote]
Algebraic geo > topo


Determine $ \sqrt[6]{1061520150601}$ without a calculator.

[quote="math154"]...er...look at it in pairs: $ 1,6,15,20,15,6,1$? Look familiar?[/quote]
Yes, I noticed binomial theorem, too.
But I solved it by a general method.

Yeah using the binomial theorem (sorry cant describe it very well) you get $ (101)^6$ and the 6th root is 101.

I used the binomial theorem solution, but is there a solution without it?

You can write :
\begin{align*}
1061520150601 &= 1 \cdot 100^6 + 6 \cdot 100^5 + 15 \cdot 100^4\\
&\quad + 20 \cdot 100^3+ 15 \cdot 100^2 + 6 \cdot 100 + 1. \\
\end{align*} . . . 1, 6, 15, 20, 15, 6, 1 - As you can seem all these numbers are found in Row 6 of Pascal's Triangles!
Notice that the cofficients on powers of 100 are all binomial. In fact, we have
\begin{align*}
1061520150601 &= \binom66 \cdot 100^6 + \binom65 \cdot 100^5 + \binom64 \cdot 100^4 \\
&\quad+ \binom63 \cdot 100^3 + \binom62 \cdot 100^2 + \binom61 \cdot 100 + \binom60.\\
\end{align*}By the binomial theorem, this is equal to $(100 + 1)^6$, so its sixth root is $\boxed{101}$.

I don't know if this is a legit method but I didn't even use binomial theorem. First, I looked at the digits and found that it had 14. So I knew that it would have to be a three-digit number as 100^6 has 13 digits. I also figured out that the first digit of the three-digit number would be 1 as if it was 2 or greater than the first digit of the sqrt(1061520150601) wouldn't be 1 and would have more than 14 digits. Then I figured out the third digit of the three-digit number is 1 or 9 as a number ending with 1 or 9 taken to the 6th power would end in a digit of 1, I ended up eliminating 9 as 1061520150601 isn't divisible by three. So we discover that the number is 1_1. The last middle digit I discovered was 0 by looking at the second digit of 1061520150601 and realizing it was also 0 and that it wouldn't be possible if the middle digit wasn't 0. Therefore, I concluded the answer is 101.


Fie $\left( G, \cdot \right)$, unde $G$ e o multime si $\cdot$ e o operatie binara pe $G$. Se stie despre $\cdot$ ca:

- e asociativa;
- exista si e unic $e \in G$ astfel incat $x \cdot e=x, \, \forall x \in G$;
- oricare ar fi $x \in G$ exista $y \in G$ astfel incat $x \cdot y=e$.

Demonstrati ca $e \cdot x=x, \, \forall x \in G$ si ca $x \cdot y = e \, \, \Longrightarrow \, \, y \cdot x = e$.

Nu e prea greu exercitiul asta, dar am pierdut cam o jumatate de ora la el (sau chiar mai mult).

Nu este necesar sa se presupuna unicitatea elementului unitate(care ar trebui(!) notat cu 1 in scriere multiplicativa).

[quote="mchis68"]Nu este necesar sa se presupuna unicitatea elementului unitate.[/quote]
Ai putea sa-ti demonstrezi afirmatia? Ca eu am folosit unicitatea in demonstratia mea (la inceput am incercat s-o evit, dar nu mi-a iesit nimic).

[quote="mchis68"](care ar trebui(!) notat cu 1 in scriere multiplicativa).[/quote]
L-am notat cu $e$ pentru ca asa era notat in locul de unde l-am luat. Mai mult, acolo nu era operatia $\cdot$, ci un semn d-ala ciudat pe care nu stiu sa-l fac in $\LaTeX$.


Let $ a,b,c$ be arbitrary reals. Prove that-
$ (ab \plus{} bc \plus{} ca \minus{} 1)^2 \le (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)$

By AM-GM

$ (ab \plus{} bc \plus{} ca \minus{} 1)^2$

$ \equal{} (ab)^2 \plus{} (bc)^2 \plus{} (ca)^2 \plus{} 2abc(a \plus{} b \plus{} c) \minus{} 2(ab \plus{} bc \plus{} ca) \plus{} 1$

$ \le (ab)^2 \plus{} (bc)^2 \plus{} (ca)^2 \plus{} (abc)^2 \plus{} (a \plus{} b \plus{} c)^2 \minus{} 2(ab \plus{} bc \plus{} ca) \plus{} 1$

$ \equal{} (abc)^2 \plus{} (ab)^2 \plus{} (bc)^2 \plus{} (ca)^2 \plus{} a^2 \plus{} b^2 \plus{} c^2 \plus{} 1$

$ \equal{} (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)$

EDIT: typo fixed

correct Ocha! thats exactly my solution!
i think there is a typo in the 3rd line it is  less than or equal to.


Is it just me or all primes squared except 3 congruent to 1 mod 3?

Well, let's see...
[hide="Is this correct?"]
All primes are either $ 1 \pmod{3}$ or $ 2 \pmod{3}$.  Squaring these, we have
$ 1 \pmod{3}$ and $ 4 \pmod{3}\equiv 1 \pmod{3}$.
Thus, all primes squared other than $ 3$ are congruent to $ 1 \pmod{3}$.[/hide]

[quote="Padmavathi"]But is there any way to prove it?[/quote]

he just did with his mod proof.

Here is a more interesting fact: All primes greater than $ 3$ when squared are congruent to $ 1\bmod 24$.

[hide="Proof"]Note that all primes $ p$ greater than $ 3$ can be written in the form $ p=6k\pm 1$. So $ p^{2}=36k^{2}\pm 12k+1=12k(3k\pm 1)+1$.
Now verify that one of $ k$ and $ 3k\pm 1$ must be even, thus their product is even. hence we conclude that the right hand side leaves a remainder of $ 1$ upon division by $ 12\cdot 2=24$. [/hide]


calculus
integration
trigonometry
calculus computations

[quote="xufrog"]by the way, if integral field is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, will the result be the same?[/quote]
Yes. Use $\cos x=\frac {\sin 2x}{2\sin x}$.

[quote="xufrog"][quote="kunny"]Evaluate

\[\int_{-\pi}^{\pi} (\cos2x)(\cos 2^2x)\cdots (\cos 2^{2006}x)dx\][/quote]

let$ f(x)= (\cos2x)(\cos 2^2x)\cdots (\cos 2^{2006}x)$, then $f(x+2\pi)=f(x)$, so
\[\int_{-\pi}^{\pi} (\cos2x)(\cos 2^2x)\cdots (\cos 2^{2006}x)dx=0\]
by the way, if integral field is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, will the result be the same?[/quote]
I'm sorry, I have made foolish mistake here. We can't get the result by $f(x+2\pi)=f(x)$. For example, let $f(x)=\cos x+1$. I'll try again, soon.Sorry for your trouble.

After all, thank enescu's hint and kunny's question. A good way to test by myself.


The number 555,555 can be expressed as the product of two 3-digit numbers. Find the smaller of these two numbers.

first, let's prime factorize 555,555

$ 555,555 \equal{} 3 \times 5 \times 7 \times 11 \times 13 \times 37$

we need 2 3-digit numbers such that they have a product of 555,555

so, we use the numbers 3, 5, 7, 11, 13, and 37, to try to get 2 3-digit numbers, because we know that when we multiply them together, they will have a product of 555,555

after a few attempts, we find that $ 3\times7\times37\equal{}777$ and $ 5\times11\times13\equal{}715$, both of which are 3-digit numbers

the smaller one is 715, so that is our answer

C'mon Gamebot, you should [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1319827256&t=252116]search[/url] first!

you do know that gamebot isn't a person...
so when someone reviews and presses "ask on forum" gamebot justs posts


function
real analysis
integration
limit
calculus
real analysis unsolved

Let f be a real valued measurable function and Lebesgue integrable on the interval [0, infinity]. Show that if f is uniformly
continuous on the interval [a, infinity], for some a>=0, then  f(x) goes to zero as x goes to infinity

i dont know legesgue integration.... but maybe i can attempt it using riemann integration.
we can probably proceed by contradiction.
if $ f$ is integrable , so is $ |f|$.
if possible let $ f$ not converge to 0 as $ x \rightarrow \infty$.
then we get a sequence $ {x_n}$ going to infinity such that $ | f(x_n)| \geq \delta$.

then using this we can derive a contradiction by showing that $ |f|$ is not integrable.

does this suffice?? :maybe:

Just two hints:
1. We can suppose that our function is $ \ge 0$.

2.If it does not tend to $ 0$ when $ x\to\infty$ then there exists some $ a > 0$ and sequence $ x_1 < x_2 < x_3..... < x_n...$ for which $ f(x_n)\to a$, $ f(x_i) > \frac {a}{2}$ and $ x_n \to\infty$. Now choose some small $ \varepsilon$, by uniformly continuoty there exists $ \delta$ such that $ f(x) > \frac {a}{2} \minus{} \varepsilon$ when $ x\in [x_i \minus{} \delta ,x_i]$ :wink:

Thanks a lot . I have some questions that why do we assume that f is non- negative and how do we use the integrability of f

Just a reminder: if you relaxed the condition of uniform continuity to mere continuity on $ [0,\infty),$ the conclusion would be false. Hence any proof must explicitly invoke the uniform continuity.

My favorite example: Let $ T(x)\equal{}\max(1\minus{}|x|,0)\equal{}\begin{cases}1\minus{}x&\text{for }0\le x\le 1\\1\plus{}x&\text{for }\minus{}1\le x\le 0\\0&\text{otherwise}\end{cases}.$

Then let $ f(x)\sum_{n\equal{}1}^{\infty}nT(n^3(x\minus{}n)).$

On the one hand, $ \int_0^{\infty}f(x)\,dx\equal{}\sum_{n\equal{}1}^{\infty}\frac1{n^2}<\infty.$

On the other hand, $ f(n)\equal{}n,$ so we don't have $ \lim_{x\to\infty}f(x)\equal{}0.$ In fact, we have $ \limsup_{x\to\infty}f(x)\equal{}\infty.$

buddhu: the best way to read "Lebesgue integrable" is to read it as "absolutely integrable," and that justifies working with $ |f|.$ An improper Riemann integral might exists even though the function is not absolutely integrable. Come to think of it, that's another question that could be asked here - does convergence (possibly conditional) as an improper integral together with uniform continuity imply that the function must tend to zero?

Hello Kent ,

Could you please tell how should I go  for this problem

santug: if $ f\ge 0$ you can now estimate $ \int \limits_{[b,\infty]} f\ge \sum \int\limits_{[x_i \minus{} \delta, x_i]} f \ge \sum (\frac {a}{2} \minus{} \varepsilon)\delta \equal{} \infty$ which contradicts to the integrability of $ f$. The only thing that you should prove now is that if your function $ f$ is uniformly continuos then so is $ |f|$ :wink:


algebra
polynomial
induction
superior algebra
superior algebra solved

Hello,

First, I apologize if I have this in the wrong forum...

I am trying to prove the Vandermonde determinant through induction. That is:

[tex]V(x_0,\cdots,x_n)=\left|\begin{array}{cccc} 1 & x_0 & \cdots & x_0^n \\
1 & x_1 & \cdots & x_1^n \\ \cdots & \cdots & \cdots & \cdots \\
1 & x_n & \cdots & x_n^n \end{array} \right|=\prod_{i<j} (x_i-x_j)[/tex]

I'm sure there are a few different ways to do this. I've got the first step, but I'm a little stuck on the following:

Let [tex]V_n(x)=V(x_0,\cdots,x_{n-1},x)[/tex] where [tex]x[/tex] is a variable.  Show that [tex]V_n(x)[/tex] is a polynomial of degree [tex]n[/tex] with roots [tex]x_0,\cdots,x_{n-1}[/tex].

Can I assume for this step that if [tex]x[/tex] takes on a value from [tex]x_0,\cdots,x_{n-1}[/tex], the determinant will be zero for that value (since two rows are the same) and therefore that value is a root?  It seems a little to easy...like there should be a more complicated method.

The rest of the proof (by induction) I think I can figure out...knock on wood.

Thanks for setting me straight.

Bailey

The Vandermonde determinant [b]is[/b] that easy. The induction feels like useless clutter, since you can get all of the factors that way; just treat it as a polynomial in all of the variables at once.


inequalities
function
inequalities proposed

The problem is equivalent to \[ (a\plus{}c)(b\plus{}d)\geq abcd\cdot\sum a^{2}.\]
Denote $ \sqrt{abcd}\equal{}t$, and also $ x\equal{}a\plus{}c, y\equal{}b\plus{}d$. Then $ x\plus{}y\equal{}4$ and $ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}\equal{}x^{2}\plus{}y^{2}\minus{}2(ac\plus{}bd)\leq x^{2}\plus{}y^{2}\minus{}4t$.
Therefore it will suffice to prove that \[ xy\geq t^{2}(x^{2}\plus{}y^{2})\minus{}4t.\]But $ x\plus{}y\equal{}4$, therefore $ x^{2}\plus{}y^{2}\equal{}16\minus{}2xy$.We denote $ a\equal{}xy$ and therefore it suffices to prove that \[ t^{2}(16\minus{}2a)\minus{}4t\minus{}a\leq 0.\]
It`s discriminant is $ \triangle\equal{}8(\minus{}a^{2}\plus{}8a\plus{}2)>0$, because $ a\leq 4$. The two roots of the above equation are
\[ t_{1}\equal{}\frac{4\minus{}2\sqrt{2(\minus{}a^{2}\plus{}8a\plus{}2)}}{4(8\minus{}a)}<0\leq t\]
\[ t_{2}\equal{}\frac{4\plus{}2\sqrt{2(\minus{}a^{2}\plus{}8a\plus{}2)}}{4(8\minus{}a)}\geq 1\geq t\]

Therefore \[ t^{2}(16\minus{}2a)\minus{}4t\minus{}a\equal{}(16\minus{}2a)(t\minus{}t_{1})(t\minus{}t_{2})\leq 0,\] so the problem is solved!

The last two inequalities are easily proven, keeping in mind that $ a\leq 4.$

I hope I haven`t made any mistake :)

[quote="andyciup"]
\[ t_{2} \equal{} \frac {4 \plus{} 2\sqrt {2( \minus{} a^{2} \plus{} 8a \plus{} 2)}}{4(8 \minus{} a)}\geq 1\geq t
\][/quote]
It is wrong. See $ a\equal{}1$.

That`s right :blush:  sorry..Maybe I can fix it in some way.. Anyway nice inequality!

[quote="zaizai-hoang"]\[ f(a,b,c,d) \equal{} \left(\frac {1}{a} \plus{} \frac {1}{c}\right)\left(\frac {1}{b} \plus{} \frac {1}{d}\right) \minus{} a^2 \minus{} b^2 \minus{} c^2 \minus{} d^2
\]

\[ f(a,b,c,d) \minus{} f\left(a,\frac {b \plus{} d}{2},c,\frac {b \plus{} d}{2}\right) \equal{} (b \minus{} d)^2\left(\frac {(a \plus{} c)}{abcd(b \plus{} d)} \minus{} \frac {1}{2} \right)
\][/quote]

I do not understand  :maybe:
What do you achieve by replacing b and d by (b+d)/2?
I mean is it some standard method of proving??

I used Mixing Variable method to proving above inequality, isomorphism  :wink:

[quote="zaizai-hoang"]Using AM-GM inequality we have: $ 4 \equal{} a \plus{} b \plus{} c \plus{} d\ge 4\sqrt [4]{abcd}\to abcd\le 1$
On the other hand
\[ f(a,b,c,d) \equal{} \left(\frac {1}{a} \plus{} \frac {1}{c}\right)\left(\frac {1}{b} \plus{} \frac {1}{d}\right) \minus{} a^2 \minus{} b^2 \minus{} c^2 \minus{} d^2
\]

\[ f(a,b,c,d) \minus{} f\left(a,\frac {b \plus{} d}{2},c,\frac {b \plus{} d}{2}\right) \equal{} (b \minus{} d)^2\left(\frac {(a \plus{} c)}{abcd(b \plus{} d)} \minus{} \frac {1}{2} \right)
\]

\[ \ge (b \minus{} d)^2\left(\frac {(a \plus{} c)}{(b \plus{} d)} \minus{} \frac {1}{2} \right) \ge 0
\]
The rest part is not hard to prove:
\[ f(a,t,t,t) \equal{} \minus{} a^2 \minus{} 3t^2 \plus{} \frac {2}{at} \plus{} \frac {2}{t^2} \equal{} \frac {(t \minus{} 1)^2(9t^3 \minus{} 12t^2 \plus{} 3t \plus{} 2)}{(4 \minus{} 3t)t^2}\ge 0
\]t,
The equality hold when $ a \equal{} b \equal{} c \equal{} d \equal{} 1$[/quote]
I think you can't conclude that f(a,b,c,d)>=f(a,t,t,t) instead of f(a,b,c,d)>=f(a,t,c,t) and prove similar to conclude that f(a,b,c,d)>=f(h,h,t,t)! Anyway, your solution is fine!

[quote="ghjk"]
I think you can't conclude that f(a,b,c,d)>=f(a,t,t,t) instead of f(a,b,c,d)>=f(a,t,c,t) [/quote]
You are right.

[quote="ghjk"]
and prove similar to conclude that f(a,b,c,d)>=f(h,h,t,t)! [/quote]
I think you mean f(a,b,c,d)>=f(h,t,h,t).
We can also prove directly  f(a,b,c,d)>=f(h,t,h,t).

An elegant solution is to add the inequalities
$ ac(b \plus{} d)^4 \ge 8abcd(b^2 \plus{} d^2)$,
$ bd(a \plus{} c)^4 \ge 8abcd(a^2 \plus{} c^2)$.

of course, I have known how to use Mixing Variable method  :wink: but I am a lazy person, I think the last part is easy to prove because it has only two-variable. So I used to S.M.V for the remain part. Anyway, above inequality is really nice, quite easy and I am interested in solve it  :)
Could you solve the general case with $ n$ variable ? Maybe, (n-1)EV will kill it easily  :)

This is a generalization of a known similar one.
If $ a_1,a_2,...a_n$ are positive numbers such that $ a_1\plus{}a_2\plus{}...\plus{}a_n\equal{}n$, $ n\ge 4$, then

$ \frac 1{a_1^2}\plus{}\frac 1{a_2^2}\plus{}...\plus{}\frac 1{a_n^2} \plus{}8\minus{}n \ge \frac 8{n}(a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2)$.

[quote="Vasc"]The case $n=8$ is nice.  :wink:[/quote]
A form of the inequality in this case is nice. But the proof is the same. No?

How to give an upper bound to $\sum a^2$?? Most of the ineq.s I am concerned of, says $\sum a^2 \geq $(something).... :(

[quote=lwwwww][quote=zaizai-hoang]Using AM-GM inequality we have: $ 4 \equal{} a \plus{} b \plus{} c \plus{} d\ge 4\sqrt [4]{abcd}\to abcd\le 1$
On the other hand
\[ f(a,b,c,d) \equal{} \left(\frac {1}{a} \plus{} \frac {1}{c}\right)\left(\frac {1}{b} \plus{} \frac {1}{d}\right) \minus{} a^2 \minus{} b^2 \minus{} c^2 \minus{} d^2
\]

\[ f(a,b,c,d) \minus{} f\left(a,\frac {b \plus{} d}{2},c,\frac {b \plus{} d}{2}\right) \equal{} (b \minus{} d)^2\left(\frac {(a \plus{} c)}{abcd(b \plus{} d)} \minus{} \frac {1}{2} \right)
\]

\[ \ge (b \minus{} d)^2\left(\frac {(a \plus{} c)}{(b \plus{} d)} \minus{} \frac {1}{2} \right) \ge 0
\]
The rest part is not hard to prove:
\[ f(a,t,t,t) \equal{} \minus{} a^2 \minus{} 3t^2 \plus{} \frac {2}{at} \plus{} \frac {2}{t^2} \equal{} \frac {(t \minus{} 1)^2(9t^3 \minus{} 12t^2 \plus{} 3t \plus{} 2)}{(4 \minus{} 3t)t^2}\ge 0
\]
The equality hold when $ a \equal{} b \equal{} c \equal{} d \equal{} 1$[/quote]

$ \[ \ge (b \minus{} d)^2\left(\frac {(a \plus{} c)}{(b \plus{} d)} \minus{} \frac {1}{2} \right) \ge 0
\]$I do not think it is always true...so how could you conclude $f\ge f(a,b,a,b)$?[/quote]

Anyone understand?

[quote=andyciup]The problem is equivalent to \[ (a\plus{}c)(b\plus{}d)\geq abcd\cdot\sum a^{2}.\]
Denote $ \sqrt{abcd}\equal{}t$, and also $ x\equal{}a\plus{}c, y\equal{}b\plus{}d$. Then $ x\plus{}y\equal{}4$ and $ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}\equal{}x^{2}\plus{}y^{2}\minus{}2(ac\plus{}bd)\leq x^{2}\plus{}y^{2}\minus{}4t$.
[color=#f00]Therefore it will suffice to prove that \[ xy\geq t^{2}(x^{2}\plus{}y^{2})\minus{}4t.\][/color][/quote]

Dear  [b]andyciup[/b]
It will suffice to prove this
\[ xy\geq t^{2}(x^{2}\plus{}y^{2})\minus{}4t^3.\]

[quote=Vasc]Let $ a,b,c,d$ be positive numbers sach that $ a\plus{}b\plus{}c\plus{}d\equal{}4$. Prove that

$ \frac 1{ab}\plus{}\frac 1{bc}\plus{}\frac 1{cd}\plus{}\frac 1{da}\ge a^2\plus{}b^2\plus{}c^2\plus{}d^2$.[/quote]

BY Lagrange method it's easy to prove.


I'm doing exactly as it says to, but I can't compile it. I tried downloading MikTeX, but I don't know If it works for me. I have Adobe Reader, but the LaTeX file just won't compile. Please help me figure this thing out, because I really want to use LaTeX.

So little detail makes it impossible to tell what the problem is.
You say you downloaded MiKTeX but did you install it - it would have added entries to the Start menu?
What does "I can't compile it" mean? What did you do to compile it? Did you use a command line or an editor? What error messages did you get? Is there a log file? .... The questions are endless since none of us can see what you've done - you will have to tell us in detail.


Let $1> \beta > 0$ and the sequence $(a_{n})$  such that: $a_{1}=1, a_{n+1}=a_{n}+\frac{1}{a_{n}^\beta}$
Find all $\alpha > 0$ such that $\exists lim_{n \to \infty}\frac{a_{n}^\alpha}{n}\neq 0$ and $\neq \infty$

$a_{n}=(\alpha n)^{1/ \alpha }(1+O(\frac{ln n}{n}))$, were $\alpha=1+\beta$.

[quote="Rust"]$O(\frac{ln n}{n}))$[/quote]
what is it????? I don't understand.  :huh:

It is easy to show, that $\alpha x<(1+x)^{\alpha }-1=\alpha x+O(x^{2})$, therefore $a_{n+1}^{\alpha}-a_{n}^{\alpha }=a_{n}^{\alpha }[(1+x)^{\alpha }-1]$, were $x=\frac{1}{a_{n}^{\alpha }}, \alpha =1+\beta$.
Let $x_{n}=a_{n}^{\alpha }.$ We have $x_{n+1}-x_{n}>\alpha$, therefore $x_{n}>x_{0}+\alpha n$ and $x_{n+1}-x_{n}=\alpha+O(\frac{1}{x_{n}})$. It give $x_{n}=x_{0}+\alpha n+\frac{\alpha \beta }{2}\ln n+O(1).$


If the definition of parallel is that the lines do not intersect (on a plane), would the segment and the line in the picture be "parallel" ?(note that one is a [i]segment[/i])

I know this seems kind of obvious, but I was just wondering about segments...

a line is infinity

a segment is finity

the segments are not parallel


induction
inequalities proposed
inequalities

prove that for all positive integer $k$ we have:
                        $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}} \leq 4k^{2}(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}})$  :?  :?

we'll prove it for induction on $n$, for $n=1$ it's quite easy to show it's true... suppose it's true for $n$, we'll prove it for $n+1$

from the inductive hypothesis it remains to show that
$\frac{1}{\sqrt{k+1}}+4k^2\left(\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k(k+1)}}\right)<4(k+1)^2\left(\frac{\sqrt{k+2}-\sqrt{k+1}}{\sqrt{(k+1)(k+2)}}\right) \\\Leftrightarrow 4\sqrt{k+1}\left(\frac{k^2\sqrt{k+2}-(k+1)^2\sqrt k}{\sqrt{k(k+2)}}\right)<8k^2+8k+3$

but it's easy to notice that $LHS<0<RHS$, therefore our induction is complete and the ineq is true!


Hey Everyone, i'm taking Algebra 2/trig...for the upcoming 2008-2009 school year. I need to be prepared, so what so of things should i make sure i know, and do you know of any good websites where i could get a headstart...or books even...

Algebra 1. Also little bit of geo will always help.

ooo...yeah i'm definately good in algebra 1...but i meant like what sort of things should i like get a head start--getting to know...know what i mean?

Then maybe the [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=11]Introduction to Algebra[/url]

Also there is the two tests that they made for it. For Trig, this [url=http://en.wikipedia.org/wiki/Trigonometry]Wikipedia article[/url] will suffice for the knowledge you need.

I do not suggest getting a head start as it will make the class less interesting.


group theory
abstract algebra
number theory
relatively prime
superior algebra
superior algebra unsolved

Let $ G$ be a finite group and $ N$ a normal subgroup, such that $ N$ and $ G/N$ have relatively prime orders. Let $ \varphi$ be an automorphism of $ G$. Prove that $ \varphi(N)\equal{}N$.

[quote="CatalystOfNostalgia"]Let $ G$ be a finite group and $ N$ a normal subgroup, such that $ N$ and $ G/N$ have relatively prime orders. Let $ \varphi$ be an automorphism of $ G$. Prove that $ \varphi(N) \equal{} N$.[/quote]

In this case N is the unique normal subgroup of its order, so the conclusion is clear.

Let pi is the set of prime divisors of N.  If A and B are two normal pi-subgroups, then AB is a subgroup, and so is in fact a normal pi-subgroup.  Let M be the subgroup generated by all normal pi-subgroups. M contains N, and M is a pi-subgroup.  The index [M:N] is a divisor of the index [G:N], but [G:N] is relatively prime to every prime in pi, so in fact [M:N]=1, and M=N is the unique largest normal pi-subgroup.

Claim: It suffices to show that $ \phi(N) \not\subset G/N$ and $ \phi(G/N) \not\subset N$.

(If you are strict about it, this statement isn't correct, what i mean is, if $ x \to \overline{x}, G \to G/N, \overline{x} \neq 1$, $ \phi(x) \neq 1$.)

By lagrange's theorem, every element of $ N$ has order which divides $ N$, and every element of $ G/N$ has order which divides $ G/N$. By the definition of a homomorphism, every element of $ phi(N)$ has order which divides $ N$ and every element of $ phi(G/N)$ has order which divides $ G/N$. Since $ N$ and $ G/N$ have relatively prime orders, an element with order that divides $ G/N$ cannot also have order that divide $ N$ unless that element is $ 1$( but that's not a problem ).

The claim follows almost directly from the above paragraph(maybe cleanest by contradiction).


trigonometry
real analysis
real analysis unsolved

does there exist $x\neq 0$ and $y\neq 0$ real numbers such that

$\frac{\sin x}{x}= \frac{\sinh y}{y}$

if I have not mistake : answer is there not exist such real numbers: only it's true when $x,y\to 0$
Why? because $\frac{\sinh{y}}{y}=1+\frac{y^{2}}{3!}+\frac{y^{4}}{5!}+...>1$ when $y$ is not equal to $0$.
${\frac{\sin{x}}{x}}<1$ when $x$ is not equal to $0$.


Find the prime number $ p$ such that $ p^{2} \plus{} 11$ has exactly $ 2^{k}$ divisors where
$ k \equiv 0,1\bmod{4}$

Is this solvable ?

If $ p\equal{}2$, $ p^2\plus{}11\equal{}15$, which has $ 4$ divisors, and don't satisfies the statement. If $ p$ is odd, $ p^2\plus{}11\equiv 4 (mod8)$, so $ p^2\plus{}11\equal{}4k$, where $ k$ is a odd positive integer. But in this case the number of divisors of $ p^2\plus{}11$ is of the form $ 3\tau (k)$, and not a power of $ 2$.


E is a finite dimensional Banach space. There is a certain linear
bijection $\Phi : E\to E^\ast$. For a subspace $V$ of $E$, demote by $V^\circ$ as
its annihilator. Then orthogonal supplement of $V$ is: $V^{\perp}=\Phi^{-1}(V^\circ)$.

prove:
$(V^{\perp})^{\perp}=V$

(note: of course, $\Phi$ is a special linear bijection, but I was told we
only need it is a linear bijection here. Please help, thanks.)

sorry, read question wrong! wrong question.
we need $\Phi$ 's special property!!!


A rope with mass M and length L is held in the position shown in the attachment,with one end attached to a support.(Assume that only a negligible length of the rope starts out below the support.)The rope is released .Find the force that the support applies to the rope,as a function of time.

You got 1/4 as a coefficient,but the answer is 3/4.  :roll:

doesn't force equals the variable weight  of the part of the rope that is under the support?

The force is equal to the sum of weight of whole rope and the change of momentum:

Here is the solution.

At time t, the free end of the rope is moving at speed $gt$ , and it has fallen a distance $\frac{gt^{2}}{2}$.This distance gets 'doubled up'  below the support.So at time t,a length $\frac{gt^{2}}{4}$ is hanging at rest,and a length $L-\frac{gt^{2}}{4}$ is moving at speed $gt$.The momentum of the entire rope is therefore $P=-\rho(L-\frac{gt^{2}}{4})(gt)$,where $\rho=\frac{M}{L}$ is the mass density,and the minus sign signifies downward motion.

The forces on the entire rope are $Mg=\rho Lg$ downward, and N(the force from the support) upward.$F=\frac{dP}{dt}$ applied to the entire rope therefore gives the following:

$N=\rho Lg+\frac{d(-\rho Lgt+\frac{\rho g^{2}t^{3}}{4})}{dt}$

$N=\frac{3\rho g^{2}t^{2}}{4}$


geometry
geometric transformation
rotation
induction
search

[b]1.[/b] Let $ \Delta A_1A_2A_3$ and $ P_0$ a point.Let $ A_s$ be a point with the propreety that $ A_s\equal{}A_{s\minus{}3}, s \geq4$.
Let $ (P_n)_{n\geq0}$ with $ P_{k\plus{}1}\equal{}R_{A_{k\plus{}1},\frac{\minus{}2\pi}{3}}(P_k).$ If $ P_1986\equal{}P_0$ then $ \Delta A_1A_2A_3$ is equilateral.(We have note $ R_{A_{k\plus{}1},\frac{\minus{}2\pi}{3}}(P_k)$ a rotation of center $ A_1$ and angle $ \frac{\minus{}2\pi}{3}$ aplied to $ P_k$).

[b]2.[/b] Let P be a point in the int of the triangle ABC such as $ <PAC\equal{}<PBC$, $ PM \bot AC$, $ PL \bot BC$, and D is the midle of AB.Prove that  $ DL\equal{}DM$.

nobody? :(...as a hint use complex numbers or geometric transformations

Let $ a_1,a_2,a_3$ be the afixez of the points $ A_1,A_2,A_3$ and $ z_i$ the afixez of the points $ P_i$.
\[ \omega_{\frac {2\pi}{3}} \equal{} cos(\frac { \minus{} 2\pi}{3}) \plus{} isin(\frac {2\pi}{3}) \equal{} \bar{\epsilon} \equal{} \epsilon^2
\]

\[ z_1 \equal{} \epsilon^2 \cdot z_0 \plus{} (1 \minus{} \epsilon^2)a_2
\]

\[ z_2 \equal{} \epsilon z_0 \plus{} (\epsilon^2 \minus{} \epsilon)a_1 \plus{} (1 \minus{} \epsilon^2)a_2
\]

\[ z_3 \equal{} z_0 \plus{} (\epsilon \minus{} 1)a_1 \plus{} (\epsilon^2 \minus{} \epsilon)a_2 \plus{} (\epsilon^3 \minus{} \epsilon^2)a_3
\]
Inthe same way and with induction we will have
\[ z_{3k} \equal{} z_0 \plus{} k(\epsilon \minus{} 1)(a_1 \plus{} \epsilon a_2 \plus{} \epsilon^2 a_3)
\]
and $ z_1986 \equal{} z_0$ from where $ z_0 \equal{} (\epsilon \minus{} 1)(a_1 \plus{} \epsilon a_2 \plus{} \epsilon^2 a_3) \equal{} 0$ and the conclusion results imediatly.

The proof for the second problem I'll post it this night...now I have to go to school :D

Here is the proof for the second problem : We will proof that $ M\equal{}R_{D,2\alpha}(L)$
After all the transformation to reach the point M, we have \[ M\equal{}R_{A,\alpha}(H_{A,\frac{1}{k}}(H_{B,k}(R_{B,\alpha}(L)))\] equivalent to \[ M\equal{}(R_{A,\alpha} \circ H_{A,\frac{1}{k}} \circ H_{B,k} \circ R_{B,\alpha})(L) **\] There are very well known that $ H_{A,\frac{1}{k}} \circ H_{B,k} \equal{} T_u$ and  $ T_U \circ R \equal{}R$
From ** results that $ M\equal{}R_{C,2\alpha}$ and we will prove that $ C\equal{}D$ equiv to $ R_{C,2\alpha}(D)\equal{}D$ wich is imediatly.

#2
This is a lema in problem 6 in mathematics and youth magagine.
Let I, J is midpoint of PA, PB. Then, we prove triangeMID = triangeDJL.

Sorry! I don't have ebooks. That is a month magagine.

But you can search it in http://www.nxbgd.com.vn -> toanhoctuoitre.
It is writed by VietNamese.


In regular octagon ABCDEFGH, the distance betweeen two opposite sides is 10. What is the area of quadrilateral CAFE?

[hide]Draw the octagon and then the quadrilateral.  Then draw diagonal $\overline{CF}$.  This splits the quadrilateral into two triangles with base $\overline{CF}$.  Let $a=$ the altitude from $A$ to $\overline{CF}$.  Let $b=$ the altitude from $E$ to $\overline{CF}$.  Then $a+b=10$.  The area of the quadrilateral is:

\begin{eqnarray*}[CAFE]&=&[ACF]+[ECF]\\ &=&\frac{1}{2}(CF)(a)+\frac{1}{2}(CF)(b)\\ &=&5a+5b\\ &=&5(a+b)\\ &=&50\end{eqnarray*}[/hide]


A monument is constructed so that the floor is a regular hexagon with sides measuring 8 feet. What is the number of square feet in the area of the floor? Express your answer in simplest radical form.

The formula for a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1434514860&t=252935]regular hexagon[/url] is $ \frac{3s^2\sqrt3}{2}$.  Thus, we have $ \frac{3 \times 8^2 \times \sqrt3}{2}$, or $ 3 \times 32 \times \sqrt3$, for the answer of $ \boxed{96\sqrt3}$.


In the [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2001-ua/01usamo.html]2001 USAMO General Information[/url], it doesn't mention an AIME floor value in the selection process unlike more recent years, so I would assume that there wasn't one.  :(

I know they started the whole floor value system when I was a freshman since everyone was confused about it that year, and that was in 2002, so no 2001 did not have the floor value.

Furthermore I'd be skeptical that anything less than 200 would qualify for USAMO, but you never know.

That really sucks :( .... so that means it would be impossible for me to qualify this year since practice ones are always easier than the real ones....

Hardly. You still have a couple more months to practice, and it's not unreasonable to be able to pull yourself up on both tests in that time.

Just keep working at it.

[quote] I had a 119.5 on 2001 AMC12, and an 8 on 2001 AIME1 on practice. Did I qualify for USAMO that year?(I'm in 9th grade now but I'm not sure if they had AIME floor value for 10th grader or below back then) [/quote]

Yes, you would have qualified that year--in fact you would have qualified even as an 11th grader that year with that score.

AMC-12 was VERY hard that year--the AIME cutoff was 84.  The highest score in NY state that year was a 134.

They had a very complicated qualification scheme in the old days.  It used to be some variant of "top 120 index scores, regardless of grade", then the next 20 11th graders and below, then the next 20 10th graders and below, then the next 20 9th graders and below.

Because of ties and AMC tinkering around after the fact, it generally worked out to about 250 students.

In 2001, there was a feeling that the AMC12 was so much harder than the AMC10 (by the way, there was only one exam date that year--no A/B exams) that it was important to adjust somehow.

So, after the scores came in, the AMC applied the basic scheme I've described above to the index for all students (regardless of whether it was based on AMC12 or AMC10) and, IN ADDITION, they invited four more groups of students from the AMC-12 takers.

They originally published a very complicated table of qualifiers, explaining all the details--the table had headings like "top 120 index students", next 20 12th graders (all of whom took the AMC-12), next 20 11th graders (all of whom took the AMC-12), next 20 10th graders and below  (regardless of test), next 20 10th graders who took AMC12, next 20 9th graders and below (regardless of test), next 20 9th graders are below who took AMC12.

I remember that an 11th grader could definitely have gotten into USAMO that year with an index in the low 190s.   For a 9th grader who took the AMC12, I believe the index cutoff would have been somewhere in the 160s, if I recall correctly.

2001 AMC12 was the hardest exam in recent years.  No reason to be discouraged--your score would have been VERY competitive for USAMO qualification that year for a 9th grader.

And the AIME was tough that year too.

More generally, though, try not to focus unduly on the numbers.  Focus on having fun learning and challenging yourself!


conics ellipse factorial function hyperbola geometry parallelogram

Mathematical definitions game. Make up funny definitions for mathematical terms!

I'll start off with a few:

Acute triangle: An adorable triangle

Algebra: Undergarments made of aquatic flora

Ellipse: An online lip enhancement service

Factorial: An editorial that contains facts

Hypotenuse: But the low pot might be available

One-to-one function: An event where couples can meet

X-axis: A group of unknown, evil nations

absolute value: the value u must use
polygon: when u cant find yr parrot
hering illusion: when u suddenly see fish floating around you
product: something you buy
hyperbole: a bowl thats either high on sugar or caffeine
parallelogram: 2 units of weight that will never touch
mean: when yr angry

EDIT: changed it. happy?

[quote="mz94"]
hyperbole: a bowl thats either high on sugar, caffeine, or alcohol
[/quote]

...High on alcohol?  Is that possible?  Alcohol is a depressant...

[quote="abacadaea"]coprime. two individuals who are dating idk?[/quote]

or: a poem by a policeman

One of my favorites:

demand curve: What students often do after tests

Number Theory: The study of the causes of numbness.

[quote="bubka"]homotopy: a gay couple sporting matching T-shirts[/quote]
Alternatively, a gay hat

(for Hindi-speaking)

There's something to be said for mathematical uses of ordinary words. How about a
[url=http://en.wikipedia.org/wiki/Normal_family]normal family[/url]?

How many parents and how many children should there be in one of those?

algebra: a female zebra
covalent bond (science): the thing ur parents do when you're not around.
pascal's triangle: something resembling a cleenex box.
n factorial: number of ways of showing off in public.

matrix: what mom keeps up her sleeves  :)
$ x^{2}$: someone who used to work for a knight

Topology: Study of best ice cream toppings.
Geometry: An endangered species of a tree.
Combinatorics: Study of which combs work the best.


geometry
geometric transformation
dilation
parallelogram
geometry solved

circle $ W_{1}$ and circle $ W_{2}$ intersect at $ P$,$ Q$ draw a line $ \overline{AB}$ thought $ P$
intersect $ W_{1}$ and $ W_{2}$ at $ A$ and $ B$ respectively let $ C$,$ D$ are the midpoint
of arc($ AQ$) and arc($ BQ$) respectively if $ M$ is a midpoint of $ \overline{AB}$ prove that $ \angle$CMD =90

[hide="Angle chasing solution"]Notice that $ \angle{QPA}$ and $ \angle{QPB}$ are supplementary: therefore, $ \angle{AQC}$ and $ \angle{BQP}$ are complementary.

Dilate $ M$, $ C$ and $ D$ outwards from $ Q$ by a factor of $ 2$ to $ M'$, $ C'$ and $ D'$, respectively.  We then have that $ AC\equal{}QC\equal{}C'C\implies m\angle{C'AQ}\equal{}90^{\circ}$.  Similarly, $ m\angle{D'BQ}\equal{}90^{\circ}$.  We also have $ \angle{AQC}$ and $ \angle{BQP}$ are complementary: therefore, $ \triangle{C'AQ}\sim\triangle{QBD'}$.  Thus, $ \frac{BD'}{AQ}\equal{}\frac{BQ}{AC'}$.  But $ AB$ and $ QM'$ bisect each other: therefore, $ AM'BQ$ is a parallelogram, and so opposite sides are equal in length.  So $ \frac{BD'}{BM'}\equal{}\frac{AM'}{AC'}$.  We moreover have $ m\angle{C'AM'}\equal{}m\angle{D'BM'}$, so $ \triangle{C'AM'}\sim\triangle{M'BD'}$.  Then $ m\angle{C'M'D'}\equal{}m\angle{AM'B}\minus{}m\angle{AM'C'}\minus{}m\angle{D'M'B}$
$ \equal{}m\angle{AM'B}\minus{}m\angle{AM'C'}\minus{}m\angle{M'C'A}$
$ \equal{}m\angle{AM'B}\minus{}180^{\circ}\plus{}m\angle{M'AQ}\plus{}90^{\circ}$
$ \equal{}90^{\circ}$, as desired.[/hide]

Let $ X$ on $ CM$ such $ MC\equal{}MX$.Then $ BX\equal{}CA\equal{}QC,BD\equal{}QD$.We use directed angle. angle $ XBD\equal{}XBA\plus{}ABD\equal{}CAP\plus{}PBD\equal{}CQP\plus{}PQD\equal{}CQD$. Therefore angle $ CQD$=angle $ XBD$ or $ CQD\plus{}XBD\equal{}180$(standard angle). If $ CQD\equal{}XBD$, triangle $ CQD$ is congruent to $ XBD$. Therefore $ DX\equal{}DC$. Since $ MC\equal{}MX$, angle $ CMD\equal{}90$. I can't prove in case $ CQD\plus{}XBD\equal{}180$. Actually, this case isn't exist

Also: http://www.mathlinks.ro/Forum/viewtopic.php?t=5433

  darij


function
trigonometry
real analysis
real analysis unsolved

Let $ x_1, x_2, x_3, x_4$ be four non-zero numbers satisfying the equation:
$ \tan^{ \minus{} 1} \frac {a}{x} \plus{} \tan^{ \minus{} 1} \frac {b}{x} \plus{} \tan^{ \minus{} 1} \frac {c}{x} \plus{} \tan^{ \minus{} 1} \frac {d}{x} \equal{} \frac {\pi}{2}$.
Then,

$ A)$ $ x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \equal{} a \plus{} b \plus{} c \plus{} d$

$ B)$ $ \frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} \frac {1}{x_3} \plus{} \frac {1}{x_4} \equal{} 0$

$ C)$ $ x_1x_2x_3x_4 \equal{} abcd$

$ D)$ $ (x_2 \plus{} x_3 \plus{} x_4)(x_3 \plus{} x_4 \plus{} x_1)(x_4 \plus{} x_1 \plus{} x_2)(x_1 \plus{} x_2 \plus{} x_3) \equal{} abcd$

[hide="Note:"][b]ONE[/b] OR [b]MORE THAN ONE[/b] OPTION MAY BE CORECT.[/hide]

$ \tan ^{\minus{}1}\left(\frac{\alpha }{x}\right)\plus{}\tan ^{\minus{}1}\left(\frac{\beta }{x}\right)\plus{}\tan ^{\minus{}1}\left(\frac{\gamma }{x}\right)\plus{}\tan ^{\minus{}1}\left(\frac{\delta }{x}\right)\equal{}\frac{\pi }{2}$

We will take Cotangent of both sides (can we do this without messing up?). But first note that:

$ \cot (a\plus{}b)\equal{}\frac{1\minus{}\tan (a) \tan (b)}{\tan (a)\plus{}\tan (b)}$
$ \cot (a\plus{}b\plus{}c\plus{}d)\equal{}\frac{1\minus{}\tan (a\plus{}b) \tan (c\plus{}d)}{\tan (a\plus{}b)\plus{}\tan (c\plus{}d)}\equal{}\frac{1\minus{}\frac{(\tan (a)\plus{}\tan (b)) (\tan (c)\plus{}\tan (d))}{(\tan (a) \tan (b)\minus{}1) (\tan (c) \tan (d)\minus{}1)}}{\frac{\tan (a)\plus{}\tan (b)}{1\minus{}\tan (a) \tan (b)}\plus{}\frac{\tan (c)\plus{}\tan (d)}{1\minus{}\tan (c) \tan (d)}}$

Since RHS=0 (cot of $ \frac{\pi}{2}$ is zero):

$ (\tan (a) \tan (b)\minus{}1) (\tan (c) \tan (d)\minus{}1)\equal{}(\tan (a)\plus{}\tan (b)) (\tan (c)\plus{}\tan (d))$

Substituing.

$ \left(\frac{\alpha  \beta }{x^2}\minus{}1\right) \left(\frac{\gamma  \delta }{x^2}\minus{}1\right)\equal{}\left(\frac{\alpha }{x}\plus{}\frac{\beta }{x}\right) \left(\frac{\gamma }{x}\plus{}\frac{\delta }{x}\right)$

Expanding and collection the terms.

$ x^4\minus{}x^2 (\alpha  \beta \plus{}\alpha  \gamma \plus{}\alpha  \delta \plus{}\beta  \gamma \plus{}\beta  \delta \plus{}\gamma  \delta )\plus{}\alpha  \beta  \gamma  \delta \equal{}0$

A is false since the sum of the roots is zero.
The option C is obvious. Is true.
Option D is trick. Since $ x_1\plus{}x_2\plus{}x_3\plus{}x_4\equal{}0$ the product can be rewrited in the form of C. so D is true.

The roots can be rewrited as:
$ x_1 \equal{} \plus{}(a\plus{}b)$ , $ x_2 \equal{} \plus{}(a\minus{}b)$ , $ x_3 \equal{} \minus{}(a\plus{}b)$ , $ x_4 \equal{} \minus{}(a\minus{}b)$

So $ \frac{1}{(a\plus{}b)}\plus{}\frac{1}{(a\minus{}b)}\plus{}\frac{1}{\minus{}(a\plus{}b)}\plus{}\frac{1}{\minus{}(a\minus{}b)}\equal{}0$
And B is true.

So only A is false and the rest is true.


quadratics
number theory proposed
number theory

Let $p>3$ be a prime. Prove that
\[ p\mid 1^2 + 2^2+\ldots+(p-1)^2 \]

Either use the formula for the sum of the first $p-1$ squares or see http://www.mathlinks.ro/Forum/viewtopic.php?t=40171 with $k=2$.

[quote="justdudxit"]Nifty

[hide]Note that $\displaystyle\sum_{1}^{p-1}{i^2}= \frac{(p-1)(p)(2p-1)}{6}$.  Let $q = \frac{(p-1)(2p-1)}{6}$.  We show that q is an integer.

Case 1: $p \equiv 1 (mod 6)$.  Then $6\mid(p-1)$, therefore q is an integer.

Case 2: $p \equiv 5 (mod 6)$.  Then $2 \mid (p-1)$ and $3 \mid (2p-1)$ (since $2p-1 \equiv 3 (mod 6)$), therefore q is an integer.

Hence, the sum on the right is equal to pq, which is clearly divisible by p.  $QED$
[/hide][/quote]
I think we can make it more simple:
$6|(p-1)(2p-1)p$
$p|(p-1)(2p-1)p$
$(6,p)=1$
then$6p|(p-1)(2p-1)p$