Professor McMillan and his wife Arne organize a party for four more couples. Introductions are made and a certain number of handshakes take place; no one shakes hand with his/her partner, and no one shakes hands more than once with the same person. At the end of the party the professor asks everyone else how many people they have shaken hands with, and recieves answers that are all different . How many people shook hands with Arne????

a fairly well known problem.

nobody could have shaken hands with 9 people because that would imply someone shook hands with his/'her problem (sorry, *partner*). so the 9 different answers were 0, 1, ..., 8. since the person who shook hands with 8 people must have shaken hands with everyone else except 0, 0 must be 8's partner. similarly, 7's partner is 1, 6's partner is 2, 5's partner is 3, and 4's partner must then be mr. macmillan. so arne shook hands with 4 people.


Two infinite, nonconducting sheets of charge are parallel to each other. The sheet on the left has a uniform surface charge density $ \rho$, and the one on the right has a uniform charge density $ -\rho$. Calculate the value of the electric field at points (a) to the left of, (b) in between, (c) to the right of the two sheets?

left right 0
in between it is $ \frac{\rho}{\epsilon_{0}}$ toward the plate with negative charge

Why are the answers as such and how can the formula $ \mathbf{E}= \frac{\rho}{2{\epsilon_{0}}}$ be applied?

yes u got the formula right.by superposition principle the presence of one charged body doesn't affect the field due to other hence just go ahead and apply the law and the formulae


Compute the value of the summation with the Hockeystick Identity:

$ \sum_{i=1}^{100}\sum_{j=1}^{i}\sum_{k=1}^{j}\sum_{m=1}^{k}1$.


trigonometry
inequalities
function
real analysis
real analysis unsolved

$ f(x)\equal{}\sum^\infty_{k\equal{}1} \frac{x^{4k}}{(4k)!}$
Find $ x\in\mathbb{R}\ni f(x)<1$ :!:  :roll:

$ x\equal{}0$.

Or did you mean to find [i]all[/i] $ x$?

Yup, all x in $ \mathbb{R}$ in other word i mean the interval of x
help me how to find it... :lol:

You could use that $ f$ is the unique solution of the inhomogeneous ODE \[ f^{(4)}(x)\equal{}f(x)\plus{}1\] with the initial conditions $ f(0)\equal{}f'(0)\equal{}f''(0)\equal{}f'''(0)\equal{}0$ to get a representation of $ f$ in terms of elementary functions. To reduce complexity note that $ f$ is an even function.

[hide]
At first, one can prove that
\[ f(x) \equal{} \frac {1}{2}(\cosh{x} \plus{} \cos{x}\minus{}2)\]
Then we have that $ \forall x > 0$:
\[ f'(x) \equal{} \frac {1}{2}(\sinh{x} \minus{} \sin{x}) > 0\]
because $ \forall x > 0: \sinh{x} > \sin{x}$. Similarly, $ \forall x < 0$:
\[ f'(x) \equal{} \frac {1}{2}(\sinh{x} \minus{} \sin{x}) < 0\]
$ f'(0) \equal{} 0$ - and in $ x \equal{} 0$ we have a minimum:$ f(0) \equal{} 0$ and $ \forall x : f(x)\ge 0$
So, $ f(x)<1$ is true on the interval $ (\minus{}x_0, x_0)$, where $ x_0$ - is the positive root of the equatuion $ \frac {1}{2}(\cosh{x} \plus{} \cos{x}\minus{}2)\equal{}1$ or equivalently:
\[ \cosh{x} \plus{} \cos{x}\equal{}4\]
[/hide]

[quote="Third Edition"]You could use that $ f$ is the unique solution of the inhomogeneous ODE
\[ f^{(4)}(x) \equal{} f(x) \plus{} 1\]
with the initial conditions $ f(0) \equal{} f'(0) \equal{} f''(0) \equal{} f'''(0) \equal{} 0$ to get a representation of $ f$ in terms of elementary functions. To reduce complexity note that $ f$ is an even function.[/quote]

Ma, ma, ma... I didn't get it at all....
Please explain to me what is inhomogeneous ODE  :maybe:
im a newbie, he579xx.. :blush:

Just go through greenvert's posting. He did all the details by giving the solution of the ODE and developing a criterion for the inequality $ f(x)<1$. ... but I haven't checked it I should admit.

But.. But.. But...
Well then... Hmmfh... I admit.. I myself didn't know what's $ sinh$ and $ cosh$
hehehehe579xx...
(Am i this fool :fool: )
Come on explain.. explain.. :jump:

Hm, $ \sinh{x}$ and $ \cosh{x}$ are [url=http://en.wikipedia.org/wiki/Hyperbolic_function]hyperbolic functions[/url].


integration
Sequence
inequalities
algebra
Summation
calculus
IMO Shortlist

We can use induction. For $n=1$ this is trivial. Now, if it's true for $m\leq n$, from $(n+1)a_{n}+\frac{n+1}{2}a_{n-1}+\ldots+a_{0}=0$ and $(n+2)a_{n+1}+\frac{n+2}{2}a_{n}+\ldots+a_{0}=0$, we get $(n+2)a_{n+1}=\sum_{i=0}^{n}a_{i}\left(\frac{n+1}{n+1-i}-\frac{n+2}{n+2-i}\right)=\sum_{i=0}^{n}a_{i}\frac{i}{(n+1-i)(n+2-i)}$ which is positive.
I really think that it's not OK that you posted this problem.

I rather think it is not OK that this problem was used in the Polish MO...

  Darij

[quote="Goblin"]I really think that it's not OK that you posted this problem.[/quote]

And I think it's not ok you know that this problem is not ok  :D

[hide="Is this right?"]
We shall prove the assertion by induction.  It is obvious that $ a_1\equal{}\frac{1}{2}>0$.
Let $ f_n(x)\equal{}a_0x^n\plus{}a_1x^{n\minus{}1}\plus{}\cdots\plus{}a_n$ and assume that $ a_1,a_2,...,a_n>0$.  Notice that $ \int_0^1f_n(x)dx\equal{}\sum^n_{k\equal{}0} \frac{a_{n\minus{}k}}{k\plus{}1}\equal{}0$.

We have $ f_{n\plus{}1}(x)\equal{}xf_n(x)\plus{}a_{n\plus{}1}$ and we are given, again, that $ \int_0^1f_{n\plus{}1}(x)dx\equal{}0$.  So, we have

$ \int_0^1xf_{n}(x)\plus{}a_{n\plus{}1}dx\equal{}0$

$ \left[a_{n\plus{}1}x\plus{}\int xf_{n}(x)dx \right]_0^1\equal{}0$

$ \left[a_{n\plus{}1}x\plus{}x\int f(x)dx\minus{}\int \int f_{n}(t)dtdx\right]_0^1\equal{}0$

$ a_{n\plus{}1}\equal{}\int_0^1 \int f_{n}(t)dtdx$

We have $ \int_0^1 \int f_{n}(t)dtdx\equal{}\int_0^1 \sum^n_{k\equal{}0} \frac{a_{n\minus{}k}}{k\plus{}1}x^{k\plus{}1}dx$.  This is equal to

$ \sum^n_{k\equal{}0} \frac{a_{n\minus{}k}}{(k\plus{}1)(k\plus{}2)}$

But it is clear that $ \frac{\minus{}a_0}{(n\plus{}1)(n\plus{}2)}<\sum^{n\minus{}1}_{k\equal{}0} \frac{a_{n\minus{}k}}{(k\plus{}1)(k\plus{}2)}$ since all $ a_{n\minus{}k}$ are positive and since $ \frac{\minus{}a_0}{n\plus{}1}\equal{}\sum^{n\minus{}1}_{k\equal{}0} \frac{a_{n\minus{}k}}{k\plus{}1}$.  Thus, $ a_{n\plus{}1}>0$.

[/hide]

[quote=ISL 2006]
Students familiar with the technique of generating functions will immediately recognise $\sum a_nx^n$ as the power series expansion of $x/ \log(1-x)$ (with value $-1$ at $0$). But this can be a trap; attempts along these lines lead to unpleasant differential equations and integrals hard to handle. Using only tools from real analysis (e.g. computing the coefficients from the derivatives) seems very difficult.
[/quote]

It isn't that difficult! There exists a solution with only real analysis, and a pretty tasteful integral formula for the coefficients.

[hide=Problem]
The sequence of real numbers $a_0,a_1,a_2,\ldots$ is defined recursively by
\[a_0=-1,\qquad\sum_{k=0}^n\dfrac{a_{n-k}}{k+1}=0\quad\text{for}\quad n\geq 1.\]
Show that $ a_{n} > 0$ for all $ n\geq 1$.
[/hide]

[hide=Formula]
\[a_n=\int_0^1 \frac{t(1-t)(2-t)\dots (n-1-t)}{n!} dt,\quad n\ge 1\]
[/hide]

[hide=Solution]
We rather define the function $G(x)=\sum_{k\ge 0} -a_kx^k$, because then the $x^0$ coefficient is $1$. With this notation,
\[
G(x)\cdot \left(\sum_{k\ge 0} \frac{x^k}{k+1}\right)=1,
\]
because the $x^n$ coefficient on the LHS for $n\ge 1$ is exactly $0$ due to the recursive equation. There is a unique $G(x)$ which satisfies this formal equation, and if we construct another $G^*(x)$ which satisfies this same equation for all $x\in I$ for some interval $I$, then [url=http://math.stackexchange.com/questions/83951/uniqueness-of-power-series]it follows[/url] that it satisfies the formal equation as well and thus $G^*\equiv G$.

It is [url=https://en.wikipedia.org/wiki/Taylor_series#Examples]clear[/url] that
\[
\sum_{k\ge 0} \frac{x^{k+1}}{k+1}+C=\int \left(\sum_{k\ge 0} x^k\right)dx=\int \frac{1}{1-x}dx=-\log(1-x)+C
\]
for all $|x|<1$, and hence $\sum \frac{x^k}{k+1}=\frac{-\log(1-x)}{x}$ (this follows from equating $C$'s for $x=0$). Therefore, we shall proceed to find the Maclaurin series of
\[
G^*(x)=\frac{-x}{\log(1-x)}.
\]

The first issue that comes up is that $G^*$ isn't actually defined for $x=0$ (albeit $\lim_{x\to 0} G^*(x)$ exists due to L'Hopital's rule), yet we're looking for the derivatives of $G^*$ at $0$. The big idea which eliminates this gap is to give another definition for $G^*$, which tamely accepts us plugging in $x=0$. One can connect this to the [url=https://en.wikipedia.org/wiki/Logarithmic_mean]integral formula for the logarithmic mean[/url], gaining the formula
\[
G^*(1-x)=F(x)=\int_0^1 x^t dt=\begin{cases} \frac{x-1}{\log x} \text{ if } x\in(0;\infty)\setminus\{1\} \\ 1 \text{ if } x=1 \end{cases}
\]
Even better, we can [url=https://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign]differentiate under the integral sign[/url] (thanks to mssmath for pointing this out) to obtain
\[
F^{(n)}(x)=\int_0^1 t(t-1)\dots(t-n+1) x^{t-n} dt,
\]
where the integrand has the sign $(-1)^{n-1}$ (or $0$) for all $t\in[0;1]$.

Now from [url=https://en.wikipedia.org/wiki/Taylor's_theorem#Taylor.27s_theorem_in_one_real_variable]Taylor's theorem[/url],
\[
F(x)=\sum_{j=0}^{n-1} \frac{F^{(j)}(1)}{j!}(x-1)^j + \frac{F^{(n)}(\xi)}{n!}(x-1)^n,
\]
where $\xi$ is between $1$ and $x$. Since the formulas above imply $|F^{(n)}(\xi)|\le n!\cdot F(\xi)$, it follows that the remainder term converges to $0$ for $|x-1|<1$, and hence
\[
F(x)=\sum_{j\ge 0} \frac{(-1)^j F^{(j)}(1)}{j!}(1-x)^j
\]
Since $G^*(x)=F(1-x)$ for all $x$ in some interval $I$, it follows that $G^*\equiv G$, and hence
\[
a_n=-\frac{(-1)^n F^{(n)}(1)}{n!}=\int_0^1 \frac{t(1-t)(2-t)\dots (n-1-t)}{n!} dt>0.
\]
[/hide]

[quote=malinger]The sequence of real numbers $a_0,a_1,a_2,\ldots$ is defined recursively by \[a_0=-1,\qquad\sum_{k=0}^n\dfrac{a_{n-k}}{k+1}=0\quad\text{for}\quad n\geq 1.\]Show that $ a_{n} > 0$ for all $ n\geq 1$.

[i]Proposed by Mariusz Skalba, Poland[/i][/quote]

[b]Here is my solution,[/b] we prove the problem by induction on $n\ge 1.$ When $n=1$ we have, $0=a_1+\frac{a_0}{2},$ which means that $a_1=-\frac{a_0}{2}=\frac{1}{2}>0.$ Assume that the numbers $a_1,a_2,\ldots,a_n$ are greater then zero for some positive integer $n.$ Since $x_0=-1<0,$ we can see that we will have some problems when we keep it and try to prove that $a_{n+1}>0,$ so we try to remove it. Because, $\sum_{k=0}^{n+1} \frac{a_{n+1-k}}{k+1}=\sum_{k=0}^{n}\frac{n-k}{k+1}=0,$ we have  \begin{align*} a_{n+1}(n+2)&=(n+1)\left(a_n+\frac{a_{n-1}}{2}+\cdots+\frac{a_0}{n+1}\right)-(n+2)\left(\frac{a_n}{2}+\frac{a_{n-1}}{3}+\cdots+\frac{a_0}{n+2}\right)\\&=(n+1)\sum_{i=0}^{n-1}\frac{a_{n-i}}{i+1}+(n+2)\sum_{i=0}^{n-1}\frac{a_{n-i}}{i+2}\\&=\sum_{i=0}^{n-1}\frac{(n-i)a_{n-i}}{(i+1)(i+2)}>0,\end{align*}
which means that $a_{n+1}>0.$ This ends the induction and solves the problem.

We will use induction. Note that the base case is trivial, so assume $a_{n-1},\ldots, a_1>0$. We have that $\sum_{k=0}^{n-2}\frac{a_{n-1-k}}{k+1}=\frac{1}{n}$, so $$\sum_{k=0}^{n-2}\frac{n-1}{n}\cdot\frac{a_{n-1-k}}{k+1}=\frac{n-1}{n^2}\implies\sum_{k=1}^{n-1}\frac{a_{n-k}}{k+1}<\frac{n-1}{n^2}\implies a_n=-\sum_{k=1}^n\frac{a_{n-k}}{k+1}>\frac{1}{n+1}-\frac{n-1}{n^2}=\frac{1}{(n+1)n^2}>0$$ as desired.

Solution from [i]Twitch Solves ISL[/i]:

In fact, it suffices to take the two consecutive equations \begin{align*} 	0 &= \frac{a_n}{1} + \frac{a_{n-1}}{2} + \frac{a_{n-2}}{3} 		+ \dots + \frac{a_0}{n+1} \\ 	0 &= \frac{a_{n+1}}{1} + \frac{a_n}{2} + \frac{a_{n-1}}{3} + 		\frac{a_{n-2}}{4} 		+ \dots + \frac{a_0}{n+2} \end{align*} and subtract $\frac{n+1}{n+2}$ times the first equation from the second (eliminating the $a_0$) to obtain a new recursion \begin{align*} 	a_{n+1} &= \left( \frac{n+1}{n+2} \cdot \frac11 - \frac12 \right) a_n 	+ \left( \frac{n+1}{n+2} \cdot \frac12 - \frac13 \right) a_{n-1} \\ 	&\qquad+ \left( \frac{n+1}{n+12} \cdot \frac13 - \frac14 \right) a_{n-2} + \dots 	+ \left( \frac{n+1}{n+2} \cdot \frac1n - \frac1{n+1} \right) a_1. \end{align*} This recursion has the property that all the coefficients in parentheses are positive. Since $a_1 = 1/2 > 0$, this implies the result by induction.

@above
 Hi Evan,to eliminate $a_0$ we need to subtract $\frac{n+1}{n+2}$ of the 1st equation from the 2nd one.

I really messed up all the indices, huh :(

Fixed, thanks for pointing it out.

Induct on $n$, the case $n=1$ trivial since $a_1=1/2$. So we can assume $a_1,\ldots,a_{n-1}>0$; this makes all the inequalities easier to deal with. Note that
\[ \frac{a_1}{n-1} + \frac{a_2}{n-2} + \cdots + \frac{a_{n-1}}{1} = \frac{1}{n} \implies \frac{1}{n}a_1 + \frac{n-1}{n(n-2)}a_2 + \cdots + \frac{n-1}{n} a_{n-1} = \frac{n-1}{n^2}. \]
That is,
\[ \sum_{k=1}^{n-1} \frac{n-1}{n(n-k)}a_k = \frac{n-1}{n^2}.\]
In order to show $a_n>0$, we want to show
\[\frac{a_1}{n} + \frac{a_2}{n-1} + \cdots + \frac{a_{n-2}}{3} + \frac{a_{n-1}}{2}  = \sum_{k=1}^{n-1} \frac{a_k}{n+1-k} < \frac{1}{n+1}.\]
We see that $\tfrac{n-1}{n(n-k)} \ge \tfrac{1}{n+1-k} \iff (n-1)(n-k+1) \ge n(n-k) \iff k\ge1$, which is always true. And it is easy to see that $\tfrac{1}{n+1} > \tfrac{n-1}{n^2}$. Hence,
\[ \frac{1}{n+1} > \frac{n-1}{n^2} = \sum_{k=1}^{n-1} \frac{n(n-1)}{n(n-k)}a_k \ge \sum_{k=1}^{n-1} \frac{a_k}{n+1-k} .\]
This completes the proof.

[hide=Remarks]
Initially, I was trying to do stuff with the expanded $a_{n-1}$, but actually the induction just tells us that all the numbers are positive, so the inequalities can work without having to flip signs or whatever. Anyways, the key was to modify the equality by multiplying by $\frac{n-1}{n}$ to get an inequality on the coefficients of the $a_i$'s.

Also, it may be worth noting that I thought $\frac{n-1}{n(n-2)} < \frac{1}{n-1}$ for some time. This made debugging annoying.
[/hide]

It suffices to show that $\tfrac{a_1}{n} + \tfrac{a_2}{n-1} + \ldots + \tfrac{a_{n-1}}{2}< \tfrac{1}{n+1}$ for all positive integer $n \geq 2$. We strong induct on $n$.

The base cases are clear; We can easily calculate $a_2 = \tfrac12$ and $a_2 = \tfrac{1}{12}$. Note that $\tfrac14 < \tfrac13$ and $\tfrac{5}{24} < \tfrac14$ hence $n = 2$ and $n = 3$ hold as desired.

In the inductive step, we are allowed to assume that the result holds for all $n \leq k$. Hence by the inductive hypothesis we may assume that the terms $a_0, a_1, \ldots , a_k$ are all positive and $\tfrac{a_1}{k} + \tfrac{a_2}{k-1} + \ldots + \tfrac{a_{k-1}}{2}< \tfrac{1}{k+1}$. We are motivated to relate the LHS to something we can find. More specifically, we can actually derive from the problem recursion that\[\frac{a_1}{k} + \frac{a_2}{k-1} + \ldots + \frac{a_{k}}{1} = \frac{1}{k+1}.\]Multiplying both sides by $\tfrac{k}{k+1}$ yields\[\frac{a_1(k)}{(k)(k+1)} + \frac{a_2(k)}{(k-1)(k+1)} + \ldots + \frac{a_{k}(k)}{(1)(k+1)} = \frac{k}{(k+1)^2}.\]In fact, we can actually check that $\tfrac{1}{k+2 - i} \leq \tfrac{k}{(k+1-i)(k+1)} \implies k^2 + 2k \geq (k+1)^2 - i$ is always true hence by our inductive hypothesis, the expression we seek, $S$, can be bounded:\[S \leq \frac{k}{(k+1)^2} < \frac{1}{k+2}\]as desired. $\blacksquare$

[color=#00f][b]Remark:[/b][/color] Induction was intuitive. I also wrote everything in non-summation notation because maybe it would be more clear to see the pattern? Also note that our entire comparison of sequences is correct because all previous terms are positive by inductive hypothesis.

[hide]We induct on $n$, with the base case $n=1$ being easy to check. Now we will prove that $n+1$ works, and note that it suffices to prove \[\frac{1}{n+1} > \sum_{i=1}^{n-1} \frac{a_i}{n+1-i}\]. To do this, first note that by induction, we have for $n$: \[\frac{1}{n}=\sum_{i=1}^{n-1} \frac{a_i}{n-i}\] Now we do the obvious thing to try to force the second equation to be kinda like the first - multiply both sides by $\frac{n}{n+1}$ \[\frac{1}{n+1}=\sum_{i=1}^{n-1} \frac{n}{n+1} \cdot \frac{a_i}{n-i}\]

From this, it suffices to show that we always have \[\frac{n}{(n-k)(n+1)} >\frac{1}{n-k+1}\] for $1 \leq k \leq n-1$, but this inequality after expanding is equivalent to $0>-k$ and so we are done. [/hide]

Note that
\[(n+1) \left(\frac{a_0}{n+1} + \frac{a_1}{n} + \cdots  + \frac{a_{n-1}}{2}+\frac{a_n}{1}\right)=0=n\left(\frac{a_0}{n} + \frac{a_1}{n-1}+\cdots + \frac{a_{n-1}}{1}\right)\]
Thus,
\[(n+1)a_n = \sum_{i=1}^{n-1} \left(\frac{n}{n-i}-\frac{n+1}{n-i+1}\right) a_{i}=\sum_{i=1}^{n-1} \left(\frac{i}{(n-i)(n-i+1)}\right) a_{i}\]

Using this we claim by induction that for all $n\geq 1$, we have $a_n>0.$ Clearly this is true for $a_1 = \frac12 >0$. Then, by the previous equation the inductive step is clear, so we're done. $\blacksquare$.

A2?

We use strong induction; the base case $n=1$ is trivial. We have that
\[\frac{a_n}{1}+\frac{a_{n-1}}{2}+\dots+\frac{a_1}{n}=\frac{1}{n+1}\]
and since
\[\frac{a_n}{2}+\frac{a_{n-1}}{3}+\dots+\frac{a_1}{n+1}<\frac{a_n}{1\cdot \tfrac{n+2}{n+1}}+\frac{a_{n-1}}{2\cdot \frac{n+2}{n+1}}+\dots+\frac{a_1}{n\cdot \tfrac{n+2}{n+1}}=\frac{1}{n+2}\]
we get
\[\frac{a_{n+1}}{1}=\frac{1}{n+2}-\left(\frac{a_n}{2}+\frac{a_{n-1}}{3}+\dots+\frac{a_1}{n+1}\right)>0\]
so we are done. $\blacksquare$

Let $P(n)=\sum_{k=0}^n\dfrac{a_{n-k}}{k+1}$, then $0=nP(n-1)-(n+1)P(n).$ Thus, \[0=-(n+1)a_{n}+\left(\frac{n+1}{2}-n\right)a{n-1}+\left(\frac{n+1}{3}-\frac{n}{2}\right)a{n-2}+\dots\] and $\frac{n+1}{k}-\frac{n}{k-1}$ here is always positive, unless $k=n+1$ where it's zero, so $(n+1)a_n>0$ implying result.

Induction on $n$, with $n=1$ trivial. Now, suppose that for some $n-1$, we have $a_{n-1},\ldots,a_1>0$. The idea is to separate the sum
$$\boxed{\frac{a_{n-1}}{1}+\frac{a_{n-2}}{2}+\cdots+\frac{a_1}{n-1}}+\boxed{\frac{a_0}{n}}$$
into positive and negative parts as shown. Now, going from the above equation to
$$\frac{a_{n-1}}{2}+\frac{a_{n-2}}{3}+\cdots+\frac{a_1}{n}+\frac{a_0}{n+1}:=S,$$
we multiply the positive part of the equation by at most $\tfrac{n-1}{n}$ (by considering each term), while we multiply the negative part of the equation by $\tfrac{n}{n+1}>\tfrac{n-1}{n}$, hence $S<0$. Since $S+\tfrac{a_n}{1}=0$, it follows that $a_n>0$, finishing the induction. $\blacksquare$

We use induction, and the base case that $a_1=\frac12>0$ is easy to check. Now assume that it is true for $n=m-1$. Then for $n=m$ we know that
		\[a_{m}+\frac{a_{m-1}}{2}+\frac{a_{m-2}}{3}+\ldots+\frac{a_0}{m+1}=0\]
		and also that
		\[a_{m-1}+\frac{a_{m-2}}{2}+\frac{a_{m-3}}{3}+\ldots+\frac{a_0}{m}=0\]
		Multiplying both equations by $m+1$ and $m$ respectively, and then subtracting the second from the first, we get
		\[a_m+\left(\frac{(m+1)a_{m-1}}{2}-ma_{m-1}\right)+\left(\frac{(m+1)a_{m-2}}{3}-\frac{ma_{m-2}}{2}\right)+\ldots+0=0\]
		Notice that $\frac{j+1}{i+1}-\frac{j}{i}<0$ as long as $j>i>0$, which is always the case here for $m+1$ and the denominator of each fraction. As all the $a_i$ are greater than $0$ by the induction hypothesis, each term in paranthesis is less than $0$, making $a_m>0$, which finishes the proof.

We will proceed strong inductively. Consider the two equations
\begin{align*}
\frac{a_0}n + \frac{a_1}{n-1} + \frac{a_2}{n-2} + \cdots + a_n &= 0 \\
\frac{a_0}{n+1} + \frac{a_1}n + \frac{a_2}{n-1} +\cdots + \frac{a_n}2+a_{n+1} &=0.
\end{align*}
Subtracting $n$ times the first from $n+1$ times the second, $$(n+1)a_{n+1} - a_n - \sum_{i=1}^n \frac{a_{n-i}}{i(i+1)} = 0.$$ But all the terms subtracted are positive by the hypothesis, so this implies $a_{n+1} > 0$ as well.

Induct on $n>0.$ We trivially deduce that $a_1=\frac 12,$ providing the base case $n=1.$ Next, the inductive step follows from
$$(n+2)a_{n+1}=\sum_{k=0}^n \left( \frac{n+1}{n-k+1}-\frac{n+2}{n-k+2} \right)a_k=\sum_{k=1}^n \frac{ka_k}{(n-k+1)(n-k+2)}>0\quad \blacksquare$$

Not sure why this was even in shortlist...
Base case is done with $a_1=1/2$. Inductive step: We know $$\frac{a_{n-1}}{1}+\cdots+\frac{a_1}{n-1}=-\frac{a_0}{n} \text{  (1)},$$ and want to prove $$\frac{a_{n-1}}{2}+\cdots+\frac{a_1}{n}<-\frac{a_0}{n+1} \text{  (2)}$$ because it would follow that $a_n$ would need to be positive s.t. the LHS is added with a positive number to maintain equality. Indeed, note that each term of LHS of (1) is multiplied by at most (n-1)/n to get LHS (2), while RHS of (1) is multiplied by n/(n+1). Put rigorously, RHS(2)=RHS(1)n/(n+1)$>$RHS(1)(n-1)/n=LHS(n-1)/n$>$LHS(2), as desired. $\blacksquare$

solved with grents

First $a_1=\frac12>0.$ Consider $n\sum_{k=0}^{n-1}\frac{a_{n-k-1}}{k+1}-(n+1)\sum_{k=0}^n \frac{a_{n-k}}{k+1}=0.$ This rearranges to $\sum_{k=1}^{n-1} a_k\left(\frac n{n-k}-\frac{n+1}{n+1-k}\right)=a_n(n+1)$ and we easily see that $\frac n{n-k}>\frac {n+1}{n+1-k}$ so we finish by induction.

bruh

This problem is easier than A1 in my opinion...


Assuming those are mixed numbers:

Rewrite each of them as the sum of $ 1$ and the fraction:

$ 1\plus{}\frac{1}{17}\plus{}1\plus{}\frac{2}{17}\plus{}1\plus{}\frac{3}{17}\plus{}\ldots\plus{}1\plus{}\frac{16}{17}$

Grouping yields:

$ 16(1)\plus{}\frac{1}{17}\plus{}\frac{2}{17}\plus{}\frac{3}{17}\plus{}\ldots\plus{}\frac{16}{17}$

Combining the fractions:

$ 16\plus{}\frac{1\plus{}2\plus{}3\plus{}\ldots\plus{}16}{17}$

Using the formula for the sum of the first $ n$ positive integers yields:

$ 16\plus{}\frac{136}{17}$

Which equals

$ 16\plus{}8\equal{}\boxed{24}$.

As the main concern here is speed, I might mention that $ \frac {1 \plus{} 2 \plus{} 3 \plus{} \ldots \plus{} 16}{17}\equal{}\frac{\frac12\cdot16\cdot17}{17}\equal{}\frac12\cdot16\equal{}8$ is faster than first evaluating to $ \frac{136}{17}$.

Agh, sorry, I was being dumb.  Canceling the 17's makes it much faster.


Let $ABCD$ be a convex quadrilateral. We choose 4 points like $P,M,N$ & $Q$ on $AB,BC,CD$ & $DA$. Prove that $PMNQ$ is also convex.

let me see, can you proove it with the definition???

[hide]If $P \equiv B$, $M \equiv C$, $N \equiv D$, $Q \equiv A$ it's obvious.

Then consider $P, M, N, Q$ as distinct points.

$\angle PBM < 180^o$ also $\angle BPM > 0$ and $\angle BMP > 0$.

The same for $\Delta PAQ$, $\Delta QDN$, $\Delta NCM$.

Hence $\angle QPM < 180^o$.

Repeat the same for others angles. (There are also the cases when one, two or three vertex are coincident).

So $PMNQ$ is convex.[/hide]

its clear that none of the angles$QPM,PMN,MNQ,NQP$ are more than $180$.

convex quadrilaterall is a quadrilateral inwich none of its angles are more than $180$.  ;)


[quote="arqady"]Let $ \{a,b,c,d\}\subset[1,2].$ Prove that
\[ 16(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} d^2)(d^2 \plus{} a^2)\leq25(ac \plus{} bd)^4
\]
[/quote]
Very nice, arqady. :)
Since $ a,b,c,d \in [1,2]$, we have $ a^2 \plus{} c^2 \le \frac {5}{2}ac, b^2 \plus{} d^2 \le \frac {5}{2}bd$, therefore
\[ (b^2 \plus{} a^2)(b^2 \plus{} c^2) \equal{} b^4 \plus{} (a^2 \plus{} c^2)b^2 \plus{} a^2c^2 \le b^4 \plus{} \frac {5}{2}b^2ac \plus{} a^2c^2 \equal{} \frac {1}{2} (2ac \plus{} b^2)(ac \plus{} 2b^2)
\]
And
\[ (d^2 \plus{} a^2)(d^2 \plus{} c^2) \le \frac {1}{2}(2ac \plus{} d^2)(ac \plus{}2 d^2)
\]
Therefore
\[ LHS \le 4(2ac \plus{} b^2)(2ac \plus{} d^2)(ac \plus{} 2b^2)(ac \plus{} 2d^2)
\]
By using the same estimation, we have
\[ (2ac \plus{} b^2)(2ac \plus{} d^2) \equal{} 4a^2c^2 \plus{} 2(b^2 \plus{} d^2)ac \plus{} b^2d^2 \le 4a^2c^2 \plus{} 5abcd \plus{} b^2d^2 \equal{} (ac \plus{} bd)(4ac \plus{} bd)
\]

\[ (ac \plus{} 2b^2)(ac \plus{} 2d^2) \equal{} a^2c^2 \plus{} 2(b^2 \plus{} d^2)ac \plus{} 4b^2d^2 \le a^2c^2 \plus{} 5abcd \plus{} 4b^2d^2 \equal{} (ac \plus{} bd)(ac \plus{} 4bd)
\]
which gives us
\[ LHS \le 4(ac \plus{} bd)^2(4ac \plus{} bd)(ac \plus{} 4bd) \le (ac \plus{} bd)^2(4ac \plus{} bd \plus{} ac \plus{} 4bd)^2 \equal{} 25(ac \plus{} bd)^4
\]
:)


calculus
integration
function
real analysis
real analysis unsolved

Suppose that $ f: [a,b] \rightarrow \mathbb{R}$ is Riemann integrable on $ [a,b]$ and $ \frac{1}{f}$ is bounded on $ [a,b]$. Prove that $ \frac{1}{f}$ is Riemann integrable on $ [a,b]$.
Relevant Theorems/Definitions
Definition - Riemann integrable - if upper integral of $ f(x)dx$= lower integral of $ f(x)dx$.
Theorem - Riemann integrable iff for each $ \epsilon > 0 \exists$ a partition $ P$ of $ [a,b]$ such that $ U(P,f)\minus{}L(P,f) \leq \epsilon$.
Theorem - Riemann integrable iff $ \exists A \in \mathbb{R}$ such that $ \forall \epsilon >0 \exists$ a partition $ P$ of $ [a,b]$ such that $ \forall$ marked refinements $ Q$ of $ P$, $ |S(Q, f) \minus{} A | \leq \epsilon$, where $ A\equal{}\int^a_b f(x)dx$.
Theorem - Let $ f: [a,b] \rightarrow \mathbb{R}$ be a continuous function on $ [a,b]$. Then $ f$ is Riemann integrable on $ [a,b]$.
etc.

I don't know how to start this problem. I need a head start on where to go with this one. Thanks in advance.

Here's a generalization: If $ f$ is Riemann integrable on $ I$ and $ g$ is Lipschitz on the range of $ f$ with constant $ M$, $ g\circ f$ is Riemann integrable on $ I$.

Given that Lipschitz constant $ M$, what can you say about $ |U(P,g\circ f)\minus{}L(P,g\circ f)|$ in terms of $ |U(P,f)\minus{}L(P,f)|$?


Try to find Rubik's cubes that are the most expensive, best looking, or best in some other category. They don't have to be normal 3x3x3 Rubik's cubes; variations such as smaller and larger cubes and mutations like the Siamese cube are also welcome. :)

I'll start off with the [url=http://www.zontikgames.com/parlour_games/rubiks_cube]Zontik Rubik's Cube[/url], which is made of solid chrome, with inlaid leather sides. :D

[img]http://www.zontikgames.com/images/items/rubiks_cube/detail_43.jpg[/img]

Solid Chromium Steel...under high cubing speed that would fall out of your hands.
Anyone seen the NIB magnet clear cube?

Eastsheen cubes are said to be really good. The new V-Cubes that are coming out are good too, maybe even better.

The dice Rubik's Cube would be easy blindfolded.

It would be cool if there was a Sudoku Rubik's Cube.

[quote="D3m0n Shad0w"]It would be cool if there was a Sudoku Rubik's Cube.[/quote]
[img]http://www.bookofjoe.com/images/whitecube_1.png[/img]

I actually have a sudoku rubik's cube where each square has a 3x3 grid on it, so when you complete it, each face makes a full, 9x9 sudoku puzzle.

can't find a picture right now

[quote="1=2"]DON'T BUY THE ICE RUBIK'S CUBE!!! IT HAS FALLEN APART ON ME 5 TIMES, AND IT HAS BEEN 2 WEEKS!!![/quote]

An ice Rubik's Cube? Won't it melt?  :P

That Sudoku Rubik's Cube is cool! I want one, now.

Theres an ice 2x2 cube (not made of ice, it's just clear). All non-eastsheen 2x2 fails. I have a rubik's one and it is dead.

My father, brother and I went on Amazon to check out the Original 3x3 Rubik's Cubes a while ago, but all the new ones do is flash with lights and cost more than it really should.
We couldn't find a really good genuine regular one.

Go to rubiks.com to order some overpriced ones.
You can also go to your local walmart or target stores to get a cube.

Agh! The Pain!

I think someone posted a link to a 4D Rubik's Cube somewhere.

[quote="D3m0n Shad0w"]The local Wal-Mart doesn't have a very good one; Target doesn't have any.[/quote]

Order some from walmart.com or something then. Hmmmm look on the internet I guess.

[quote="Carbon57"]Um...FYI, amazon does have rubik's cubes...

[url]http://www.amazon.com/Winning-Moves-5025-Rubiks-Cube/dp/B00081RYNC/ref=pd_bbs_sr_1?ie=UTF8&s=toys-and-games&qid=1213065697&sr=8-1[/url]

You can also go to a Barnes and Nobles...I'm pretty sure they sell them, and they're usually in stock.w[/quote]

Whoever is turning that cube sucks at cubing. I see a 1 move a second turn with palms, including a double turn taking more than enough time. Not the mention the fact that the person is making L2 and R2 turns only.

To stay on topic:
I think any cube+silicone spray added correctly can turn to an epic cube. Of course the stickers may need replacing as well.
Tip: If your 4x4 cube sucks, don't try turning it too hard. I broke the centers 4 times already.
Tip #2: If your 5x5 cube is TOO good, as in it POPS WAY TOO MUCH, don't use it. Mine pops at least 1 time a solve and it's not even lubed.

It's obviously not a very good cube...but he said wanted one of the normal ones...and it's not too bad for a beginner, or someone who's learning.

No I mean they're using that as advertisement. Actually beginners can turn that fast as well.

[quote="pacman2812"][img]http://mononcqc.files.wordpress.com/2007/05/cube.png[/img]
5D rubik "hyperhypercube" :o[/quote]

 :wow: :huh:

How does that work? (And HOW IN THE WORLD are you supposed to solve it?)

I don't know... The 4D hypercube is not too weird, but I don't understand its movements :wacko:


what are the last two digits of 2^222?

calculator allowed

it was asking for last 2 digits, not the last digit.

LoL, rep

u did the wrong thing, but got the right answer

indeed, but show the solution anyways for the last 2 digits if you can

Ok. SOlution:

[hide]since we are looking for last 2 digits, let us look at powers of 2 mod100.

now, it just so happens that 2^22=4(mod100).

Meaning 2^220=4^10(mod100).

This is small enough on ur calc..

2^220=76(mod100)

so 2^222=4*76(mod100)=4(mod100)

SO the answer is 04.[/hide]

[hide]the ending digits are: 02, 04, 08, 16, 32, 64, 28, 56, 68, 76, 52, 04 and then it repeats without the 02 anymore. integer divide 222 by 17 and subtract one from the remainder because of the first 02 and the remainder is then zero. that means that the two ending digits are 04.[/hide]

Wow!!! I actually got a question correct!!!

Don't you feel proud! :)
I mean, the first time I said something smart in Intermediate (actually, the [b]only[/b] time i said something smart :) ), I was so happy I ran around the house while playing "Eye of the Tiger" on my comp, which ended up badly, obviously :lol:


A circle $ \omega$ is outscribed over an acute triangle $ ABC$. $ AN$ and $ CK$ are altitudes. The median $ BM$ crosses $ \omega$ at $ P$. Point $ Q$ is chosen on $ BM$ such that $ MQ\equal{}MP$. Prove that $ B,K,Q,N$ are cyclic.

Let $ H$ be orthocenter of triangle $ ABC$. We have to show $ \angle BQH \equal{} 90$ or equivalently, $ \angle HQP \equal{} 90$. Since $ \angle AHC \equal{} \minus{} \angle ABC \equal{} \minus{} \angle APC \equal{}  \angle AQC$, $ A,H,Q,C$ are cyclic. $ \angle HQP \equal{} \angle HQA \plus{} \angle AQP \equal{} \angle HCA \plus{} \angle CPQ \equal{} \angle HCA \plus{} \angle CAB \equal{} 90$. So we are done.

Following problem is more general problem.
Let $ P$ be a arbitrary point on circumcircle of triangle $ ABC$ with orthocenter $ H$ and $ Q$ be a point such $ QAPC$ is parallelogram.  Then $ BP$ is perpendicular to $ HQ$.


Please help me   :maybe:  :lol:
Find the last three digits of the number $ 2003^{2002^{2001}}$

Note $ \phi(10^{3})=400$ and Euler's Theorem says if $ \gcd(a,n)=1$ then $ a^{\phi(n)}\equiv 1 \pmod{n}$.  Now, $ 2002^{2001}\equiv 2^{2001}\equiv x \pmod{400}$.  Use Chinese Remainder Theorem or some other technique to finish this (that is get this as small as possible).  Say you get $ x$.  Then, via Euler's Theorem we have that

$ 2003^{2002^{2001}}\equiv 3^{2002^{2001}}\equiv 3^{x}\pmod{10^{3}}$

See if this helps you get started.

[quote="baz"]I got 241 after some monotonous arithmetic.[/quote]

can you explain what you did..

I got $ 3^{{2}^{81}}\mod 1000$..but i kinda forgot what to do..$ \phi(1000)=400$,$ \phi(400)=160$

[hide="Solution"]
\[ \varphi(1000)=400 \implies 2003^{ 2002^{2001}}\equiv 3^{ 2002^{2001}}\bmod 1000 \equiv 3^{ 2^{2001}}\bmod 1000 \]
Now, we look at $ 2^{2001}$ in $ \bmod 400$. By Euler's, the Chinese Remainder Theorem and a bit of testing, we find:
\[ 2^{20}\equiv 1 \bmod 25 , \ 2^{20}\equiv 0 \bmod 16 \implies 2^{20}\equiv 176 \bmod 400 \]
Now, by the Chinese Remainder Theorem again, we can see that $ \left( 2^{20}\right)^{n}\equiv 2^{20}\bmod 400$, so:
\[ 2^{2001}\equiv \left(2^{20}\right)^{100}\cdot 2 \equiv 176 \cdot 2 \equiv 352 \bmod 4000 \]
Thus, we have that:
\[ 3^{2^{2001}}\equiv 3^{352}\equiv \left( 3^{-1}\right)^{48}\equiv (-333)^{48}\equiv 333^{48}\bmod 1000 \]
Time for some arithmetic: Note that we only have to calculate the last three digits of the product. This will save some time; I was able to fit all this neatly on one sheet of paper.
\[ 333^{48}\equiv 889^{24}\equiv 111^{24}\equiv 321^{12}\equiv \ldots \equiv 681^{3}\equiv \boxed{ 241}\bmod 1000 \]
[/hide]


Hi all--

I'm writing a puzzle hunt for bay area high schools to participate in.  In the hunt, teams of up to 10 people (including up to 1 teacher) from local high schools will race to solve a set of math, science, or logic puzzles.  The hunt will take place on Saturday, April 11th,  on Stanford University campus, lasting most of the day.  It will be in the style of the MIT mystery hunt and the Microsoft college puzzle hunt (you can find descriptions on wikipedia, and puzzles at http://www.mit.edu/~puzzle/hunthistory.html and http://www.collegepuzzlechallenge.com).

If you are a high school student in the area and you would think your school would be interested in fielding a team, or want to hear more about the hunt, email me at 11235813sam@gmail.com (or reply here).  Also, feel free to alert whomever might be interested in this.

I've attached an FAQ that gives more details about how the hunt will work.

Thanks
Sam Bankman-Fried
Crystal Springs Uplands School, Junior

Cool!  If you need someone on campus to help out, I might be available that day.


Your source is scarcely informative so that I should note that the problem is taken from Titu Andreescu's & Co "360 Problems For Matematical Contests".


AMC
AMC 8
AMC 10
AIME
USA(J)MO
USAJMO
AMC 12

What are you best?  Worst?  Average?  Tomorrow is the AMC 8 and I started to wonder how you guys did.

My best 25/25 was 9:51, my average is about 13-15:00 and my worst is 20:00.

Sometimes  instead of going back and checking my answers, I just check it after I'm done, hoping to examine my accuracy.

You're saying you finished the test in 9 minutes and 51 seconds, getting a perfect? My average over 20 tests is about 24.2, all of the scores being between 23-25, if that is what you are asking.

Cool.  I probably should take my time.  I've gotten 8/9 25/25s today, and I'd hate to get a 24 on the real test tomorrow.

Especially because being in the 8th grade, it's my last year to take it.

[quote="CubeX"]Unless you get a 25 on the AMC 8, i don't think you can tske the AMC 10.[/quote]

No, you don't need to qualify for any of the AMC exams.  (Unless I'm mistaken.)

You can take the AMC 8, 10, and 12 without qualifying. Here's the qualification system:

AMC8 --> nothing

AMC10 --> AIME --> USAJMO --> maybe Red MOP

AMC12 --> AIME --> USAMO --> MOP --> IMO

Also, perfect in 10 mins is really good. Good job.

You have to take the AMC 12 to get to USAMO (and all of the other MOPs and also IMO)???????

I didn't know that...

But thanks for telling me :)

Darnit I was hoping to get there through AMC 10 lol

This year is the first year they are splitting up like that.

I wish the AMC 8 could take you to some other test.  It seems so pointless.  Oh well, I'm hoping for a 25.  :)

[quote="PowerOfPi"]You can take the AMC 8, 10, and 12 without qualifying. [/quote]

I know for certain that at my school, you have to get at least a 21/25 on the AMC 8 in order to qualify for AMC 10.
Why is that? I mean, why does our school require us to qualify while most other places you don't have to? :(

[quote="jksong"]
I know for certain that at my school, you have to get at least a 21/25 on the AMC 8 in order to qualify for AMC 10.
Why is that? I mean, why does our school require us to qualify while most other places you don't have to? :([/quote]

At my school we (I) set a qualifying score after results come in because of space restrictions ... we try to take about 30 kids so they can all take it at the same time in the same room.  The cutoff is usually between 18 and 21 depending on the year.

[quote="qvc196"][quote]You have to take the AMC 12 to get to USAMO (and all of the other MOPs and also IMO)???????[/quote]

You don't. You can take AMC10. However, you do have to get 11 or higher on AIME to qualify for USAMO. If you get less than 11 on AIME, you can still qualify for USAJMO through AMC10/AIME index.[/quote]

Yeah, but in AMC 10, you can't make MOP unless you are in 9th grade and qualify for Red MOP

[quote="Mewto55555"]Trying to time yourself on the AMC 8 seems...unfortunate. You should maximize the time spent checking because once you are good enough, a perfect on an AMC 8 becomes a question of not making stupid mistakes.

Also you do NOT need a 25 on the AMC 8 to qualify for the 10/12. You just have to find a school hosting it (your local high school) and badger them until they let you take it too.[/quote]

Good advice, that makes a lot of sense.  I'll definitely keep that in mind.

[quote="CubeX"]What does AMC 12 cover?[/quote]
Most things covered in a regular high school math cirriculum up to 12th grade + discrete math and number theory


[color=darkred]Let $ABC$ be a triangle. Prove that for any $\{x,y\}\subset \mathbb R$,

$\boxed{\ x+y=A\Longrightarrow\frac{c\cdot\sin (B+x)+b\cdot\sin (C+y)}{c\cdot\sin x+b\cdot\sin y}=\frac{b+c}{a}\mathrm{\ (constant)\ }\ }$.

Study some particular cases.[/color]

Let there be $M$ on $(BC)$ so that $\angle{BAM}=x$ and $\angle{CAM}=y$.

now we have $c \cdot \sin(B+x)+b \cdot \sin(C+y)= c \cdot \sin{\angle{AMB}}+\b \cdot \sin{\angle{AMC}}= \sin{\angle{AMB}}\cdot (b+c)$

and now the denominator : $c \cdot \sin{x}+b \cdot \sin{y}=BM \cdot \sin{\angle{AMB}}+CM \cdot \sin{\angle{AMC}}$ (from the sin law) $\Longleftrightarrow c \cdot \sin{x}+b \cdot \sin{y}= \sin{\angle{AMB}}\cdot (BM+CM) = \sin{\angle{AMB}}\cdot a$, which ends it.


Prove that: $cos^{n}x.cosnx = \frac{1}{2^n}\sum\limits_{i=0}^{n}\large{n\choose i} cos2ix$, for all $n\in N*$ and $x\in R$.

See also : http://www.mathlinks.ro/Forum/viewtopic.php?t=78118 :)


Triangle $A_1 A_2 A_3$ has a right angle at $A_3$. A sequence of points is now defined by the following iterative process, where $n$ is a positive integer. From $A_n$ ($n \geq 3$), a perpendicular line is drawn to meet $A_{n-2}A_{n-1}$ at $A_{n+1}$.
(a) Prove that if this process is continued indefinitely, then one and only one point $P$ is interior to every triangle $A_{n-2} A_{n-1} A_{n}$, $n \geq 3$.
(b) Let $A_1$ and $A_3$ be fixed points. By considering all possible locations of $A_2$ on the plane, find the locus of $P$.

I don't know why all of the links of http://www.kalva.demon.co.uk I can't open.

(a) It can be proved just by Sperner Theorem

For a triangle $ABC$ right angled at $C$, define its "skew-center" as the second intersection of the circles with diameters $AC$ and $BD$, where $D$ is the foot of altitude of $ABC$ from $C$. Note that the skew-center of $ABC$ is not necessarily the same as the skew-center of $BAC$.

[b]Lemma[/b]:  The skew center of a right triangle lies inside the triangle.
[b]Proof[/b]: Trivial by angle chasing.

[b]Part a)[/b] Let $P$ be the skew-center of $A_1A_2A_3$. We shall show that $P$ satisfies the requirements.  We do so by proving that $P$ is the skew center of $A_{n-2}A_{n-1}A_{n}$ for all $n\ge 3$. We shall use induction.

The base case $n=1$ is obvious by definition. Suppose $P$ is the skew center of $A_{k-2}A_{k-1}A_k$.  By definition, $P\in (A_{k-2}A_k)$ and $P\in (A_{k-1}A_{k+1})$ (Here $(XY)$ refers to the circle with diameter $XY$). Clearly $A_{k+2}\in (A_{k-1}A_{k+1})$. Now we have $$\angle A_kPA_{k+2}=2\pi-\angle A_{k+1}PA_{k+2}-\angle A_kPA_{k+1}=(\pi-\angle A_{k+1}PA_{k+2})+(\pi- \angle A_kPA_{k+1})=\angle  A_{k+1}A_k+\angle A_{k+1}A_{k-1}A_{k-2}=\frac{\pi}{2}$$So $P\in (A_kA_{k+2})$. But also $P\in (A_{k-1}A_{k+1})$, so $P$ is the skew center of $A_{k-1}A_kA_{k+1}$. This completes the induction, so we conclude that $P$ is the skew-center of all triangles $A_{n-2}A_{n-1}A_n$ . This together with [b]Lemma[/b] implies that  $P$ is interior to every triangle $A_{n-2} A_{n-1} A_{n}$, $n \geq 3$, proving our claim.

[asy]
import graph; size(7.726755742739088cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-0.7829171322827649,xmax=6.943838610456323,ymin=-0.06350701568125115,ymax=4.534562378445503;
pair A_k=(0.44,1.3), P=(1.4338043821210804,1.706699099133097);
D(A_k--(6.04,1.32)); D((0.4297853343707351,4.160106376194181)--(6.04,1.32)); D((0.4297853343707351,4.160106376194181)--A_k); D(A_k--(1.5904450112061543,3.5725358204778406)); D(shift((0.43489266718536757,2.7300531880970906))*xscale(1.4300623083051773)*yscale(1.4300623083051773)*arc((0,0),1,-89.79537308603418,90.20462691396585)); D(shift((3.815222505603077,2.4462679102389204))*xscale(2.49361470664756)*yscale(2.49361470664756)*arc((0,0),1,153.14971347367387,333.1497134736739)); D((1.5904450112061543,3.5725358204778406)--(1.5985464331870287,1.304137665832811)); D(shift((1.0192732165935143,1.3020688329164054))*xscale(0.5792769109262276)*yscale(0.5792769109262276)*arc((0,0),1,0.20462691396587018,180.20462691396588),linetype("2 2")); D((1.5904450112061543,3.5725358204778406)--P,linetype("2 2")); D(A_k--P,linetype("2 2")); D(P--(1.5985464331870287,1.304137665832811),linetype("2 2"));
D(A_k); MP("A_k",(-0.1431857383173036,1.345901211648906),NE*lsf); D((6.04,1.32)); MP("A_{k-1}",(6.084199549815234,1.4158718328638784),NE*lsf); D((0.4297853343707351,4.160106376194181)); MP("A_{k-2}",(0.4665582465560267,4.264675696616324),NE*lsf); D((1.5904450112061543,3.5725358204778406),linewidth(3.pt)); MP("A_{k+1}",(1.6260713981184254,3.6349401056815727),NE*lsf); D(P,linewidth(3.pt)); MP("P",(1.0962938374907776,1.675762711662347),NE*lsf); D((1.5985464331870287,1.304137665832811),linewidth(3.pt)); MP("A_{k+2}",(1.0263232162758054,1.046027120727596),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]

To see that this is only possible $P$, simply observe that the sidelengths of the triangle $A_{n-2}A_{n-1}A_n$ tend to zero as $n\to\infty$. So if there were another point $Q$, the distance $PQ$ would eventually exceed the largest side of $A_{n-2} A_{n-1} A_{n}$, and the triangle will not be able to accommodate both $P$ and $Q$.

[b]Part b)[/b] By the definition of skew-center, the required locus is clearly a subset of the circle with diameter $A_1A_3$, excluding $A_1,A_3$.  But not all points on this circle lie on the locus.
[asy]
import graph; size(7.682604816586902cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-2.2490914484553635,xmax=5.433513368131538,ymin=-0.2146804943158581,ymax=4.357115386188629;
pen qqwuqq=rgb(0.,0.39215686274509803,0.);
pair A_1=(-0.46,4.3), A_3=(-0.4,0.76), A_2=(4.994373056994819,0.8514300518134715), A_4=(1.1818594352315481,3.2619215538942914), A_5=(1.2237984106033264,0.7875220069593784), A_6=(0.0761667036350339,1.5131211133793963), A_7=(1.2111740628187921,1.5323585262469175), P=(0.9311024018547345,1.398098833084234);
D(arc(A_1,0.2981606009024671,-89.02897806892084,-64.38804884260522)--(-0.46,4.3)--cycle,qqwuqq); D(arc(A_3,0.2981606009024671,0.9710219310791641,25.61195115739478)--(-0.4,0.76)--cycle,qqwuqq); D(arc(A_5,0.2981606009024671,90.97102193107915,115.61195115739476)--(1.2237984106033264,0.7875220069593784)--cycle,qqwuqq); D(arc(A_2,0.2981606009024671,147.69655948789924,172.33748871421486)--(4.994373056994819,0.8514300518134715)--cycle,qqwuqq);
D(A_1--A_3); D(A_1--A_2); D(A_3--A_2); D(A_4--A_5); D(A_3--A_4); D(A_5--A_6); D(A_6--A_7); D(shift((-0.43,2.53))*xscale(1.7702542190318316)*yscale(1.7702542190318316)*arc((0,0),1,-89.02897806892084,90.97102193107915),dotted); D(shift((3.0881162461131835,2.0566758028538814))*xscale(2.255312029284104)*yscale(2.255312029284104)*arc((0,0),1,147.69655948789924,327.69655948789926),dotted); D(P--A_3,linetype("2 2")); D(P--A_1,linetype("2 2")); D(P--A_5,linetype("2 2")); D(P--A_2,linetype("2 2"));
D(A_1); MP("A_1",(-0.8974300576975125,3.99),NE*lsf); D(A_3); MP("A_3",(-0.8676139976072658,0.7593108019655326),NE*lsf); D(A_2); MP("A_2",(5.035965900261582,0.9481458492037614),NE*lsf); D(A_4,linewidth(3.pt)); MP("A_4",(1.2195102087100036,3.323491969726745),NE*lsf); D(A_5,linewidth(3.pt)); MP("A_5",(1.269203642193748,0.9183297891135147),NE*lsf); D(A_6,linewidth(3.pt)); MP("A_6",(-0.19178330222834047,1.6537926046729323),NE*lsf); D(A_7,linewidth(3.pt)); MP("A_7",(1.2493262688002502,1.5941604844924389),NE*lsf); D(P,linewidth(3.pt)); MP("P",(0.6231890069050695,1.3059385702867212),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]
Observe that by similarity of triangles $A_1A_2A_3$ and $A_3A_4A_5$ and by the fact that $P$ is the skew center of both of them, we have $\angle PA_1A_3=\angle PA_3A_5=\angle PA_3A_2$. Analogously, from similar triangles $A_3A_4A_5$ and $A_5A_6A_7$, we have $\angle PA_3A_5=\angle PA_5A_7$. But $PA_5A_2A_4$ is cyclic, giving   $\angle PA_5A_7=\angle PA_2A_1$. Thus we conclude that $\angle PA_1A_3=\angle PA_3A_2=\angle PA_2A_1=\omega$, the Brocard angle of $A_1A_2A_3$. But, $\cot\omega=\cot A_3+\cot A_1+\cot A_2=0+\cot A_1+\frac{1}{\cot A_1}\ge 2\implies\tan \omega \le \frac 12\implies \omega\le \arctan {\frac 12}$. Clearly equality can be achieved. Also $A_2$ can lie on either side of $A_1A_3$. Thus the desired locus is an arc of $(A_1A_3)$ subtending an angle of $2\arctan {\frac 12}$ at $A_1$ and having midpoint $A_3$, excluding the point $A_3$.
[asy]
import graph; size(7.682604816586901cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-2.222555154975045,xmax=5.460049661611856,ymin=0.10534521731945252,ymax=4.677141097823941;
pen qqwuqq=rgb(0.,0.39215686274509803,0.);
pair A_1=(-0.46,4.3), A_3=(-0.4,0.76), A_2=(4.994373056994819,0.8514300518134715);
D(arc(A_1,0.29816060090246704,-115.59402924599883,-62.463926891842824)--(-0.46,4.3)--cycle,qqwuqq);
D(CR((-0.4,0.76),0.17638888888888887)); D(A_1--A_3); D(A_1--A_2); D(A_3--A_2); D(A_1--(1.004,1.492),linetype("2 2")); D(A_1--(-1.828,1.444),linetype("2 2")); D(shift((-0.43,2.53))*xscale(1.7702542190318316)*yscale(1.7702542190318316)*arc((0,0),1,217.84091957692317,324.10112428523513));
D(A_1); MP("A_1",(-0.9007098243074412,4.259716256560488),NE*lsf); MP("A_3",(-0.920587197700939,0.42338319161541743),NE*lsf); D(A_2); MP("A_2",(5.032686133651653,0.9501335865431084),NE*lsf); label("Locus of P",(-0.17518569544477133,1.168784693871584),NE*lsf); label("$2\arctan{\frac 12}$",(-1.0000966912749303,3.4646213208205765),NE*lsf,fp+qqwuqq);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]

Generalisation:

In triangle $A_1A_2A_3$, angle at $A_3$ is strictly larger than other angles. A sequence of points is now defined by the following iterative process, where $n$ is a positive integer. For $n\ge 3$, a point $A_n$ is taken on segment $A_{n-2}A_{n-3}$ such that $\angle A_{n-2}A_nA_{n-1}=\angle A_1A_3A_2$. Prove that if this process is continued indefinitely, then one and only one point $P$ is interior to every triangle $A_{n-2} A_{n-1} A_{n}$, $n \geq 3$.

Also identify the point $P$, if possible.


On the sides $ AB, BC, CD$ and $ DA$ of the rhombus $ ABCD$, respectively, are chosen points $ E, F, G$ and $ H$ so, that $ EF$ and $ GH$ touch the incircle of the rhombus. Prove that the lines $ EH$ and $ FG$ are parallel.


If $a ,b ,c$ are angles of a triangle, Prove:

1) $cos a + cos b + cos c > 1$

2) $(cot a)(cot b)(cot c) \le \frac{\sqrt3}{9}$

Uh....i don't know whether it's appropriate for intermediate, but it looks a bit too easy for the more advanced forums....    :huh:

Shyong means $\cos A+\cos B+\cos C=1+ 4\sin A/2\sin B/2\sin C/2$

[quote="shyong"]1)write it as

$\cos A+\cos B+\cos C=1-4\sin A/2\sin B/2\sin C/2$

2)I thought of Jensen . Maybe more elementary way ? :)[/quote]

can you show how to use jensen for #2?
i thought of it too....but can't get it
thanks  :)

consider $f(x)=\ln\cot x$ over $(0,\pi)$


$a_n$ is sequence of positive numbers satisfying \[a_n < a_{n+1} + a_{n^2}\] for all $n \in N$. Prove that \[\sum_{n=1}^{\infty} {a_n}\] diverges.

I just loved this problem when I saw it. The idea is just beautiful: if we write $ a_2\leq a_3+a_4\leq a_4+a_9+a_5+a_{16}\leq...$ we can prove that at each stage we don't have repetitions of the same term in the sum and this contradicts Cauchy's criteria, thus the series diverges.

[quote="harazi"]we can prove that at each stage we don't have repetitions of the same term in the sum [/quote] Could you post your proof, Harazi? This statement is by no means obvious to me (or, maybe, I'm missing something?  :? )

No, in fact that was the nice part of the problem. Define the sets of indices of the terms appearing in the different sums as $A_1,A_2,...$. Thus $A_1$ contains 2, $A_2$ contains $3,4$ and so on.  Assume that we have in $A_p$ a certain repetition. The number which is repeated must be of the form $n^2$ and was produced by adding 1 to $n^2-1\in A_{p-1}$ and squarig $n\in A_{p-1}$. Since $n^2-1$ is not a perfect square, it could not be obtained from $A_{p-2}$ by squaring, thus it was obtained by adding 1 to $n^2-2\in A_{p-2}$. In the same way we can prove that $ n^2-i\in A_{p-i}$ for $i=1,2,...,2n-2$ It means that $p\geq 2n-1$. On the other hand, since $n\in A_p$, it is clear that $n\geq p$ which is impossible.

Can't we do this: $a_n<a_{n+1+a_{n^2}<a_{n+2}+a_{(n+1)^2}+a_{n^2}<a_{n+3}+a_{(n+2)^2}+a_{(n+1)^2}+a_{n^2}<...<a_{n+T}+a_{(n+T-1)^2}+a_{(n+T-2)^2}+...+a_{n^2}}$, now if we fix $n$ and take $T$ arbitrary large we can leave out the case when numbers $n+T,(n+T-1)^2,...,n^2$ aren't distinct and we can see that the partial sum between terms of the series from $n+T$'th term to $(n+T-1)^2$ is at least $a_n$-fixed number-contradiction with Cauchy criterion

Unfortunately, Eugene's approach contains a mistake: for large $T$ the smallest index is $n^2$, not $n+T$.


function
geometry
rectangle
floor function
analytic geometry
number theory theorems
number theory

I'm looking for a proof of following theorem: for any positive coprime integers m,n we have $ \sum_{i \equal{} 0}^{m \minus{} 1} [\frac {ni}{m}] \equal{} \frac {(n \minus{} 1)(m \minus{} 1)}{2}$, or $ \sum_{i \equal{} 0}^{m \minus{} 1} [\frac {ni}{m}] \equal{} \sum_{i \equal{} 0}^{n \minus{} 1} [\frac {mi}{n}]$, becauce I have already proved that $ \sum_{i\equal{}0}^{m\minus{}1} [\frac{ni}{m}]\plus{}\sum_{i\equal{}0}^{n\minus{}1} [\frac{mi}{n}]\equal{}(n\minus{}1)(m\minus{}1)$. Wikipedia says that a proof of it's generalization is placed in Concrete Mathematics. A Foundation of Computer Science by Ronald L. Graham, Donald E. Knuth, Oren Patashnik, page 95. I would be grateful if someone could prove it or tell me how the generalization is going. Unfortunately when suddenly I need Concrete Mathematics it is no more in school library.

consider the rectangle with vertices $ (0,0),(m,0),(m,n),(0,n)$. the diagonal lies on the line with equation $ y\equal{}\frac{n}{m}x$. since $ m,n$ are coprime, the only lattice points on the diagonal are the endpoints. $ \lfloor \frac{ni}{m}$ counts the number of lattice points with $ x$ coordinate $ i$ and lying on or under the diagonal. thus the sum in question counts the number of lattice points in the interior of the rectangle and under the diagonal. since there are no interior points on the diagonal, exactly half of them are under the diagonal (the other half above). since the total number of interior points is $ (m\minus{}1)(n\minus{}1)$, the result follows.


inequalities
trigonometry
inequalities unsolved

Prove the following inequality for all positive reals $ a,b,c$ :

$ \sqrt{a^{2}\plus{}b^{2}\minus{}ab}\plus{}\sqrt{b^{2}\plus{}c^{2}\minus{}bc} \ge \sqrt{a^{2}\plus{}c^{2}\plus{}ac}$

:)

It's an old inequality...

Let $ ABC$ be a triangle such that $ \angle BAC\equal{}60^{\circ}$, $ AB\equal{}a,BC\equal{}b$. Let $ ACD$ be a triangle such that $ \angle CAD\equal{}60^{\circ}$, $ AC\equal{}b$, $ AD\equal{}c$. Consider triangle $ BCD$, we have $ BC\plus{}CD\ge BD$, we get $ \sqrt{a^{2}\plus{}b^{2}\minus{}ab}\plus{}\sqrt{b^{2}\plus{}c^{2}\minus{}bc}\ge\sqrt{a^{2}\plus{}c^{2}\plus{}ac}$.

Q.E.D. :D

[quote="b.s.o"]Prove the following inequality for all positive reals $ a,b,c$ :

$ \sqrt {a^{2} \plus{} b^{2} \minus{} ab} \plus{} \sqrt {b^{2} \plus{} c^{2} \minus{} bc} \ge \sqrt {a^{2} \plus{} c^{2} \plus{} ac}$

:)[/quote]
Consider $ \triangle APB$  such that $ \angle APB \equal{} \frac{\pi}{3}, AP \equal{} a, BP \equal{} b$. Then by $ AB \equal{} \sqrt{a^2\plus{}b^2\minus{}2ab \cdot \cos \frac{\pi}{3}} \equal{} \sqrt{a^2\plus{}b^2\minus{}ab}$. Construct outside of $ \triangle APB$ $ \triangle PBC$ such that $ PC \equal{} c, \angle PBC \equal{} \frac{\pi}{3}$, then $ BC \equal{} \sqrt {b^{2} \plus{} c^{2} \minus{} 2bc \cdot \cos \frac{\pi}{3}} \equal{} \sqrt {b^{2} \plus{} c^{2} \minus{} bc}$ and $ AC \equal{} \sqrt{a^2\plus{}c^2 \minus{} 2ac \cdot \cos \frac{2\pi}{3}} \equal{} \sqrt {a^{2} \plus{} c^{2} \plus{} ac}$.
So $ AB\plus{}BC \ge CA \iff \sqrt{a^2\plus{}b^2\minus{}ab}\plus{}\sqrt{b^2\plus{}c^2\minus{}bc} \ge \sqrt{a^2\plus{}c^2\plus{}ac}$

[quote="b.s.o"]Prove the following inequality for all positive reals $ a,b,c$ :

$ \sqrt {a^{2} \plus{} b^{2} \minus{} ab} \plus{} \sqrt {b^{2} \plus{} c^{2} \minus{} bc} \ge \sqrt {a^{2} \plus{} c^{2} \plus{} ac}$

:)[/quote]
Another solution by Minkowski:
$ \sqrt {a^{2} \plus{} b^{2} \minus{} ab} \plus{} \sqrt {b^{2} \plus{} c^{2} \minus{} bc} \equal{}$ $ \sqrt{\frac{1}{2}} ( \sqrt {a^{2} \plus{} b^{2} \plus{} (a\minus{}b)^2} \plus{}\sqrt {c^{2} \plus{} (c\minus{}b)^{2} \plus{} b^2}) \ge$ $ \sqrt{\frac{1}{2}}\sqrt{(a\plus{}c)^2\plus{}c^2\plus{}a^2} \equal{} \sqrt {a^{2} \plus{} c^{2} \plus{} ac}$

[quote="b.s.o"]Prove the following inequality for all positive reals $ a,b,c$ :

$ \sqrt {a^{2} \plus{} b^{2} \minus{} ab} \plus{} \sqrt {b^{2} \plus{} c^{2} \minus{} bc} \ge \sqrt {a^{2} \plus{} c^{2} \plus{} ac}$

:)[/quote]

this ineq holds for any reals $ a,b,c$...

see also: http://www.mathlinks.ro/Forum/viewtopic.php?p=1643379#1643379


algebra
polynomial
inequalities proposed
inequalities

$ a,b,c,d,e$ are real numbers such that $ a,b>0$ and $ ad^2\plus{}be^2\minus{}4ace<0$.prove that  $ ax^4\plus{}bx^3\plus{}cx^2\plus{}dx\plus{}e$ dont have real roots.

[quote="MJ GEO"]$ a,b,c,d,e$ are real numbers such that $ a,b > 0$ and $ ad^2 \plus{} be^2 \minus{} 4ace < 0$.prove that  $ ax^4 \plus{} bx^3 \plus{} cx^2 \plus{} dx \plus{} e$ dont have real roots.[/quote]
Try polynomial $ x^4\plus{}x^3\minus{}2x^2\plus{}x\minus{}1$ which satisfies your condition but has a root $ x\equal{}1$.


I hear you guys talking about like 60+ team scores but the second team in my state got EDITED OUT (but it was bad)???

[quote="biffanddoc"]I hear you guys talking about like 60+ team scores but the second team in my state got EDITED???[/quote]

My coach was too embarresed to even tell us our score....One of my teammates got a 0! <- NOT KIDDING

Are you serious???? I still don't know our score, but we had 3 in the top ten and didnt even get third?? That's wrong

[quote="LuCy4EvA"]what is the total that you can get on a team round? I think we got everything right... but our second team only missed one.... so it made us feel really bad....

wait, are we talking about state or chapter???[/quote]

im talking about chapter

Utah is pretty bad.....well...maybe just my region.  The only reason we got first is because of my score(that was a pretty arrogant remark)...which isn't a perfectone).  But second place only got way less points....and my teammates aren't exactly the brightest crayons in the..crayon box.   Biffandoc, what state are you from?

[size=75]
[color=orange]<Edited by solafidefarms. Please, no talking about scores.>[/color][/size]

im from delaware and I am talking about state. I don't know my team score yet

ah, that might makes a difference... I was talking about chapter.... our state is this Saturday... hope I'll do okay... at least the top 50..... :lol:

YOU CAN NOT TALK ABOUT STATE SCORES!!!!!

You may say your rank and your team's rank, but no score of any kind, team or individual!!!

I second what frost said... if you can, delete out anything that mentions your scores... and I assume a moderator will come in and do it if you don't... if you're talking about chapters, then pls state in your post that you're talking about chapters....

haha, yea, there's only four days left before state..... ah... we have practice every single day this week.... i miss regular seventh grade math team.... it's way easier, but I still have to do the homework.... feeling very behind with all the eighth graders...... wish me luck! :?  :D  :?  :(  :(  :lambda

4 days for my team too, good luck Lucy et al

Thank you for editing out your scores... even though I'm not a mod, we still have to keep it fair (even if I haven't competed yet)... so ya it'll be cool when we can talk about states and AIMEs, but just wait :(

Yes. PLEASE don't talk about scores. If you see that happening, please PM a moderator so they can edit it.

Thank you
Billy

Why did you edit mine??? I was talking about chapter

Well scores range quite a bit between states.  I've heard of scores of around <snip> qualifying for nationals in some places, while the top four scores for Illinois state last year were on average 43+.

[color=darkblue][size=75][Edited by moderator rcv.  Please do not discuss scores until after April 2.][/size][/color]

Moderator--err... I was talking about from previous years.  Sorry for being unclear  :blush: .

[quote="biffanddoc"]Good luck. How many compete (we had 97)??[/quote]

We had 180, I believe compete, from Colorado.  :D

It was much better than the chapter, since the auditorium we took the test in was really jampacked with people and got really stuffy during the chapter (the chapter had around 200 some people, maybe more.)

[quote="grizzlydude_10"]I don't know if our pure 6 grade team is really good or if the rest of the state is just plain bad our team placed 2 in utah.  we were the only team with every member in the top 20 tho no one placed in the top 10. we had 11 12 15 and 20. most of the other teams had 1 really good star and the rest were bad.[/quote]

I got 2nd in Utah (CountDown round) and our team didn't place in the top 7 even! :( Also, you guys had a good coach (My nationals coach now) and it was your school's first year paticipating

Our state had 200 people last year but only 166 came this year.

Ohio had a good 250 people at states this year I think (and I was second!) :D

Wow, that's a lot - what state is that? California?

Hmmm...I think I saw you at Georgetown last year and a State last year, so I know what you look like.  But I never actually talked to you in person, and anyways Im' too old for mathcounts now.

what school did you go to?
whats your name?

Oooh Georgetown. I got 3rd this year with a 44 at Georgetown. Mark got a 45 and Alex Ryan got a 46! The coolest mini-MC competition ever, mostly because all of the best Texas schools are there.

I think I got ninth at Georgetown last year
I was an idiot then
I didn't go this year because of Orchestra


For a real number $ x$ in the interval $ (0,1)$ with decimal representation

\[ 0.a_1(x)a_2(x)...a_n(x)...,\]

denote by $ n(x)$ the smallest nonnegative integer such that

\[ \overline{a_{n(x)\plus{}1}a_{n(x)\plus{}2}a_{n(x)\plus{}3}a_{n(x)\plus{}4}}\equal{}1966 .\]

Determine $ \int_0^1n(x)dx$. ($ \overline{abcd}$ denotes the decimal number with digits $ a,b,c,d .$)

[i]A. Renyi[/i]

I wonder that with an arbitrary number in the interval $(0,1)$, it is not sure that there exist the number $n(x)$ satisfy that condition.

sorry. Is it a math olympiad question? I've never seen such in competitions.

This is actually a standard exercise, hardly disguised. Indeed, we are asked to compute $\mathbb{E}[n(X)]$ where $X$ is uniformly distributed on the interval $(0,1)$. But since $X$ can also be represented as $X = \sum_{k \geq 1} U_k \, {10}^{-k}$ where $\{U_k\}_{k=1}^{\infty}$ is an i.i.d sequence of discrete random variables uniformly distributed on $\{0,1, \ldots, 9\}$, the problems only ask how long we are expected to wait before observing $(U_{t},U_{t+1},U_{t+2},U_{t+3})=(1,9,6,6)$. A similar exercise might even have already been discussed on this forum. This can for example be solved by the usual [url=http://goo.gl/evja8]first step[/url] analysis.


So far I have downloaded the free

-Kedlaya inequalities notes
-Mildorf inequalities notes
-Kedlaya's Geometry Unbound

Are there any other packets out there?

I have some number theory notes from Reid Barton, but he doesn't explain stuff very well.


Two non-intersecting circles, not lying inside each other, are drawn in the plane.
Two lines pass through a point P which lies outside each circle. The first line intersects the first circle at A and A′ and the second circle at B and B′; here A and B are closer to P than A′ and B′, respectively, and P lies on segment AB. Analogously, the second line intersects the first circle at C and C′ and the second circle at D and D′. Prove that the points A, B, C, D are concyclic if and only if the points A′, B′, C′, D′ are concyclic.

ABCD is cyclic iff $ \overline{PA} \cdot \overline{PB} \equal{} \overline{PC} \cdot \overline{PD}$. For the power of P wrt the first and the second circle we obtain $ \overline{PA} \cdot \overline{PA'} \equal{} \overline{PC} \cdot \overline{PC'}$ and $ \overline{PB} \cdot \overline{PB'} \equal{} \overline{PD} \cdot \overline{PD'}$; sobstituing these on $ \overline{PA} \cdot \overline{PC} \equal{} \overline{PB} \cdot \overline{PD}$ we obtain $ \overline{PA'} \cdot \overline{PB'} \equal{} \overline{PC'} \cdot \overline{PD'}$ that is true iff A'B'C'D'  is cyclic.

Two non-intersecting circles, not lying inside each other, are drawn in the plane. Two lines pass through a point $P$ which lies outside each circle. The first line intersects the first circle at $A$ and $A'$ and the second circle at $B$ and $B'$, here $A$ and $B$ are closer to $P$ than $A'$ and $B'$, respectively, and $P$ lies on segment $AB$. Analogously, the second line intersects the first circle at $C$ and $C'$ and the second circle at $D$ and $D'$. Prove that the points $A, B, C, D$ are concyclic if and only if the points $A', B', C', D'$ are concyclic.


[quote="jeez123"]Every $x$ digits , in a base-$Y$ number, can be grouped to one digit in base $Y^{x}$.

Can someone explain this for me more please? proof? Thanks![/quote]
Think about it.  For digit in the last x'th digit place, the value of the x-digit is Y^x.  That's how bases work.
So, in base Y^x, the x-th digit added to all the other digits will be in the first digit place.  Why?  Because if the sum of all the x digits in base Y was greater than Y^x, then there would be x+1 digits in the base Y.

Can you show me some examples instead of the variables? I'm still confused  :( .

Actually that was a bad explanation.
Let's look at a four digit number in base ten.
$3856$
the 3 is in the 4th digit place, so it has value $3\cdot10^{4}$.
But in base 10000, this will only be one digit.  Because, in base 10000, the first digit will be the ones place, and the next digit will be the 10000's place.  3856 is obviously less than 10000, so it will be confined to the ones place; the first digit.

What if our initial number was greater than 10000?  That would mean the number would have 5 digits, so in base $10^{5}$, it would still never reach the second digit, the $10000$'s place.


$N$ is the largest number less than $1000$ that has exactly $20$ divisors, including 1 and itself. Find $N$.

[quote="anyone"][hide]To find the number of divisors a number has, you add one to each exponent in its prime factorization and multiply them together. There are a few ways to do this such that the number has 20 divisors, but the best way will be a number with prime factorization in the form $xyz^{4}$. Its easy to find from this point that the maximum value less than 1000 that you can get is $11*5*2^{4}=\boxed{880}$[/hide][/quote]

Almost.  :)

[hide]I get 912 from brute force- 3*19*2^4, I got 880 first but after looking at what you wrote..i reconsidered and found that it has to be 2^4*x where x is a number slightly bigger than 11*5=55, and 19*3 works[/hide]


[hide="Solution"]You just need to find positive integer values of $ x$ and $ y$ such that $ 13x \plus{} 23y \equal{} 160$. You can use some algebra or guess-and-check to find that $ x \equal{} 7$ and $ y \equal{} 3$. This answer means that Melanie has 91 mushrooms, of which $ \frac{8}{13}$ are white. Melanie has 56 white mushrooms. Romina has 69 mushrooms, of which $ \frac{11}{23}$ are white. Romina has 33 white mushrooms. $ 56 \plus{} 33 \equal{} \boxed{89}$ white mushrooms.[/hide]

Is there a solution to this that doesn't involve guess and check?

Well, solving a Diophantine equation [url=http://mathforum.org/library/drmath/sets/select/dm_diophantine.html]isn't quite guess and check.[/url]

[quote="orangefronted"]Melanie and Romina had 160 mushrooms altogether. The mushrooms are either brown or white. 5/13 of Melanie's mushrooms are brown and 11/23 of Romina's mushrooms are white. How many white mushrooms are there?[/quote]

160/13 x 5 =100 160-100= 60 there are 60 white mushrooms


AMC
AIME
geometry
3D geometry
pyramid
ratio
USA(J)MO

I brute forced a division and ended up with 69 instead of 224.

I thought $15/60$ is not equal to $35/140$

$1+1+1+1+1+1=7$

I did the summation problem to $1000$ instead of to $2007$.

Multiplied wrong on the pyramid problem and on the two circles problem.

 :mad: bummer

[rant=careless mistakes]
my answer : right answer
#1 $81$ : $83$ i said $82^{2}= 7xxx$
#2 $52$ : $54$ i didn't see the "beside the walkway" and then said $12 = 14$
#5 $538$ : $539$ i did $1000-32$ instead of $1000-31$
#7 $774$ : $477$ when i subtract, i do right to left, so i read the number out as $7, 7, 4$.
#11 $755$ : $955$ i did my addition wrong somewhere...
[\rant]

It seems that I have gotten worse at math and this year's AMC and AIME confirm my suspicion. I've gotten A LOT more careless over the year. On this year's AIME, I got 6 out of the 12 problems I solved wrong because of careless mistakes(while some mistakes are stupider than others, I could've gotten all these questions easily if I had been more careful).

I had a similar ratio of right to wrong (3 right out of 7 that should have been), but it's still better than last year...

w00t I only made one. I'm proud of myself for keeping the number low. I thought it would be like 4 or 5.

But unproud of myself for making the number existant at all.

1 out of 8. My other two mistakes weren't really stupid mistakes, I just went about them the wrong way.

2. Beside the walkway =/= on the walkway

Messed up on 5 i should've gotten.
I could've made USAMO, considering i answered 9

Made careless errors on 5 of them, but caught four of the mistakes.  :lol:

omfg i got 3 :oops:  with 3 stupid errors.

2) misread beside the walkway as on the walkway.

5) laziness made me not brute force this, so I didn't get it right. <_<

Well I wasn't getting anywhere on the last few so with about a half hour left I went to checking, which I usually never do on AIMEs.  Like I check as I go, but not afterwards.  And good thing I did because I found 2 of 3 mistakes, so just 1 here.  And that one happened to be the one I didn't have time to check.  A lesson learned?  Too bad this is my last AIME.

1000 divided by 12 is not 86.66.. - problem 1
forgot to subtract 1 as a power of 2 - problem 7
other mistakes not quite so annoyingly stupid.

this might be the dumbest thread ever

you had 3 hours, plenty of time to check over for mistakes...and what are careless mistakes but just doing the math wrong?

As we've said, this thread is all about raising people's self-esteem.  :P

[quote="Pineappleperson"]hey u said u were gonna hurt ur self if u didn't make top 50 in mmpc[/quote]
A wallbash or two doesn't count as hurting yourself ^_^.  Getting carried off in an ambulance does.

This thread is so people can vent somewhere other than real threads.

I would have appreciated a diagram or two on the geometry questions.

For people who got 210 instead of 105 on #4, it is NOT an ABSOLUTELY STUPID mistake.

An ABSOLUTELY STUPID mistake $\rightarrow 10*2 = 12$

or $546/2 = 255$

or you misread one part of the question.

[quote="Miu"][quote="kyyuanmathcount"]Hurting yourself is not the answer Mo Zhou  :D .[/quote]

There's some1 named Mo Zhou??
That's weird lol[/quote]

...
It's not that weird, if you're Chinese.

[quote="Miu"]So Mo where ru from? And since I assume you're Chinese, where are you originally from (which part of China)?

btw this year's AIME is pretty tricky, I made 2 careless mistakes[/quote]

lol....sorry to reply back so late... :lol:

i'm in MA now...

[quote="abacadaea"][quote="1two3"]Example of stupid mistakes
1. 3+4=6
2. sqrt(225)=25
3. 2007-1980=37

You get the idea.

Mod Edit: I'm making this the 'official' thread for complaints about the AIME I. Try to refrain from making pointless posts, even if you list out your mistakes, maybe comment on how you might've avoided it, or just simply dismiss as "absolutely insane."[/quote]
whats the difference between AIME I and AIME II. why cant this just be the complaint thread for all AIME?

thought the area of a circle was $ \pi r$
need to find the integers like 987 to 45 or something to get my answer. i got 942, forgot to add 1.[/quote]

u took AIME?
i thought you hadn't since ur goal is 2+?  :lol:

oh, does it mean the actual thing? no never have taken AIME. yes my goal is prety pathetic(that doesnt mean you can laugh :mad: ).i meant on practice. i thinks that one was 2004 or something


hi,

prove that :

$a> b , f(a)=f(b) = k ,f$ is derivable in $k \Rightarrow f$ is

derivable in $[a,b]$ :blush:

First: this should be in the calculus computations and tutorials section.  Second: I don't think the question makes sense, and in the best sensible interpretation I can give it, it isn't true.


I wanted to paste this problem into my failed article, I think it is not hard and not look very awful:
Given a $ABCD$, the midpoints of $BC$ and $AD$ are $E$ and $F$ respectively. $BD$ and $AC$ intersect $EF$ at points $G$ and $H$ respectively. $M$ is a point on $EF$ situated on rays $GF$ and $HF$ . Prove that $S{}_A{}_B{}_M+S{}_A{}_C{}_M=S{}_D{}_B{}_M+S{}_C{}_D{}_M$ .

May be....Ray GF and [b]HE[/b], i.e. $M\in (HG)$ !

Nooooooooooo, all is correct, the problem is right, man. :)

I understood. Equivalencily, if I note $I\in AC\cap BD,$ then $M\in {Int(\widehat {AID})}\cap {EF}.$ It is O.K. ?

You know your solution is quiet confusing from checking viewpoint, I've got solution very easy, but before reading it check out this problem:
Problem $X$:
We got hexagon $ABCDEF$. Prove that if the lines connecting the midpoints of opposite sides of hexagon intersect in one point then  $S{}_A{}_C{}_E=S{}_B{}_D{}_F$.(opposite is also right)
Solution of the problem:
Let's make some assignments: $\vec{MA}=\vec{a}$, $\vec{MB}=\vec{b}$, $\vec{MC}=\vec{c}$, $\vec{MD}=\vec{d}$. So $ME=\displaystyle\frac{\vec{c}+\vec{b}}{2}$ and $MF=\displaystyle\frac{\vec{a}+\vec{d}}{2}$, as the $M$ lies on line $EF$, then $\vec{ME}\times\vec{MF}=0$ the same as $\displaystyle\frac{\vec{c}+\vec{b}}{2}\times\displaystyle\frac{\vec{a}+\vec{d}}{2}=0$ so $\vec{c}\times\vec{a}+\vec{c}\times\vec{d}+\vec{b}\times\vec{a}+\vec{b}\times\vec{d}=0$ let write vectoral products as areas $S{}_C{}_M{}_A-S{}_C{}_M{}_D+S{}_B{}_M{}_A-S{}_B{}_M{}_D=0$ so $S{}_C{}_M{}_A+S{}_B{}_M{}_A=S{}_C{}_M{}_D+S{}_C{}_M{}_D$. :) Was the solution beautiful?

How is in generally, when $M\in (EF)$, with respect to the first problem ?
See:
 http://www.mathlinks.ro/Forum/viewtopic.php?t=45995
 http://www.mathlinks.ro/Forum/viewtopic.php?t=46011

You can solve genral problem by using vectoral product. :)

I know use the vectorial method, but I was not interesting to aplly here. Your solution is nicely and naturally.


algebra
polynomial
linear algebra
linear algebra theorems

There is a well-known identity for any $A \in \mathcal{M}_2 \left( \mathbb{C} \right)$, with the elements $a,b,c,d$: $A^2-(a+d)A+\det A \cdot I_2 =0_2$.

My teacher said there are similar identities for $A \in \mathcal{M}_n \left( \mathbb{C} \right)$. Do you know how would that go?

The theorem says:

If \[ A = \left( {\begin{array}{*{20}c} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \ldots & a_{nn} \\ \end{array}} \right)  \] and \[ \left| {\begin{array}{*{20}c} a_{11}-x & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22}-x & \ldots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \ldots & a_{nn}-x \\ \end{array}} \right| = x^n+c_{n-1} x^{n-1}+\ldots+c_1 x + c_0  \] then \[ A^n+c_{n-1} A^{n-1}+\ldots+c_1 A+ c_0 I_n = 0_n . \]

The theorem looks really nice, but is there any nice proof of it?

[quote="cezar lupu"]Of course there is a proof. :D[/quote]

do you happen to know one? does it involve complicated stuff ?

No, it doesn't. Now i'm going to sleep, but don't worry, tommorow i'll post it. ;)

well, will you post it?

mean time, i have found a (short) proof of it, but it involved subspaces and other stuff which I really don't understand

The solutions are found here: http://planetmath.org/encyclopedia/CayleyHamiltonTheorem.html (attachments)

[quote="cezar lupu"]Ok, here it goes. Let us consider $(X I_{n} _A)^{*}$ the conjugate of $XI{}_{n}{}_A\in M_{n} (\mathbb{C}).$
 Then we will have $p_{A} (X) I_{n} =(I_{n} X-A)(I_{n} X-A)^{*}.$ First we observe that the coeficient of of the main diagonal if $I_{n}X-A$ is a monic polynomial of degree $n-1.$
 So, we deduce that $(I_{n} X-A)^{*}=X^{n-1}+C_{n-2}X^{n-2}+...+c_{1} X +C_{0}$ and from here it is clear how to proceed. ;)

P.S. Can you post in detail the solution with subspaces? :?[/quote]

What do you mean by $I_{n} _A$? I don't think I can follow your proof  :blush:


Is it true that if $X\oplus Z = Y \oplus Z$, then $X=Y$?

I've thought about such example:
Let $X\oplus Z = Y \oplus Z =$space of all convergent sequences of real numbers,

Z = all sequences convergent to 0,
X = constant sequences
Y = sequences of form (0, c, c, ...)

Is it good? :)

I suppose "=" indicates an isomorphism between normed vector spaces. If $X,Y,Z$ are finite-dimensional, then the statement follows by counting their dimensions. For infinite-dimensional spaces there are many counterexamples. Yours does not seem to work, though, because $X$ and $Y$ are isomorphic.

Consider $X=\{0\}$, $Y=Z=\ell^{2}$ (square-summable sequences). The map $((y_{i}),(z_{i}))\mapsto (y_{1},z_{1},y_{2},z_{2},\dots)$ is an isomorphism $\ell^{2}\oplus \ell^{2}\to\ell^{2}$. Also works with other sequence spaces.

I suppose when = is normal equality, then my example is good?

If "=" means that two sets have the same elements, then there are no such examples, because $X\oplus Z=Y\oplus Z$ implies $X=Y$. Indeed, $X\oplus Z$ is the set of all ordered pairs $(x,z)$ with $x\in X$ and $z\in Z$. If there is an element $x\in X\setminus Y$, then $(x,z)\in (X\oplus Z)\setminus (Y\oplus Z)$ for any $z\in Z$.

Hmm?

I talk about direct sum from linear algebra, i.e.:

Let $U,W \subseteq V$ be subspaces of vector space U, then $U+W = \{u+w | u\in U, w\in W\}$. If $U \cap W = \{0\}$ (or equivalently each vector from U+W has unique representation in form $u+w$, where $u\in U, w\in W$) then we say that it's direct sum of subspaces U, W and write $U \oplus W$.

Oh. I thought of [b]external[/b] direct sums.

For internal sums your example works. And you don't even need infinite-dimensional spaces: $\mathbb R^{2}$ can be represented as a direct sum of two lines in a lot of ways. For example, $X=\{(t,0)\colon t\in\mathbb R\}$, $Y=\{(t,t)\colon t\in\mathbb R\}$, $Z=\{(0,t)\colon t\in\mathbb R\}$.


A point in space $ (x,y,z)$ is randomly selected so that $ \minus{}1\le x\le 1$,$ \minus{}1\le y\le 1$,$ \minus{}1\le z\le 1$. What is the probability that $ x^{2}\plus{}y^{2}\plus{}z^{2}\le 1$?

If you look at this as a geometry problem the shape of your restrictions is a cube with side length 2
The X^2+Y^2+Z^2 <= 1 is a sphere with radius 1
THe volume of the cube is 8 and the volume of the sphere is 4pi/3
the answer is the volume of the sphere over the volume of the cube which is pi/6


geometry incenter parallelogram rectangle

Triangle $ABC$ has an incenter $I$ l its incircle touches the side $BC$ at $T$. The line through $T$ parallel to $IA$ meets the incircle again at $S$ and the tangent to the incircle at $S$ meets $AB , AC$ at points $C' , B'$ respectively. Prove that triangle $AB'C'$ is similar to triangle $ABC$.

[quote="Ali.Kh"]We show $\angle{AC'B'}=\angle{C}$. For this we show that angle between $BC,B'C'$ is $\angle{B}-\angle{C}$. We show $\angle {SIM}=180-\angle{B}+\angle{C}$. This is easy because angle between $AI,IM$ is $180-\dfrac{\angle{B}-\angle{C}}{2}$[/quote]

please specify point M.

[quote="madhusudan kale"]

please specify point M.[/quote]
$M=T$

Let us denote the incircle of the triangle as $\omega.$ Let $D,E$ be the points where $\omega$ touches $AC,AB$ respectively. Let $M=SS\cap BC$ and let the midpoints of $TE,ES,SD,DT$ be $U,V,W,X$ respectively. Obviously $M$ is the pole of $ST$ wrt $\omega,$ and it is given that $ST\parallel AI.$ Since the polar of $A$ wrt $\omega$(which is the line $DE$) is perpendicular to $AI,$ so we obtain $ST\perp DE.$ Therefore the quadrilateral $UVWX,$ apart from being a parallelogram, also has $WS\parallel DE$ and $WX\parallel ST$, implying $\angle SWX=\frac{\pi}{2}.$ Thus $UVWX$ is a rectangle, and therefore cyclic. Note that an inversion about $\omega$ (ie, centred at $I$ and radius $IT$) maps $V$ to $C',\ W$ to $B',\ X$ to $C,$ and $U$ to $B.$ Therefore the new quadrilateral $BCB'C'$ is also cyclic. So, $\angle AB'C'=\angle ABC$ and therefore $ABC\sim AB'C'.$
[img]http://i48.tinypic.com/282ovlt.jpg[/img]

$\Box$
Edit: 1800th post.

Let $A'=BC\cap B'C'$. $AI$ is the bisector of $\angle BAC$, $A'I$ is the bisector of $\angle BA'C'$, $A'I\bot ST\parallel AI$, done.

Best regards,
sunken rock


integration
inequalities unsolved
inequalities

g=1/sqrt(1+2006^2)+1/sqrt(2+2006^2)+...+1/sqrt(2006+2006^2)
find [2006*g], where [x] is the greatest integer <=x.

or see attachement.


symmetry
trigonometry
geometry
angle bisector
congruent triangles
trig identities
Law of Cosines

hi,
our teacher gave us this assignment and i gave it a try but im not sure...i proved that the base angles of isosceles triangles are congruent, and then i am trying to prove that equilateral triangles are equiangular...can i do this:

triangle ABC is equilateral

given triangle ABC with segment AC congruent to BC | given
triangle ABC is isosceles | definition of isosceles
angle A and angle B are base angles | definition of base angles
angle A congruent Angle B | proved that base angles are congruent

given triangle ABC with segment AB congruent to AC | given
triangle ABC is isosceles | definition of isosceles
angle B n angle C are base angles | def of base angles
angle B congruent angle C | proved

angle A congruent Angle C| transitive theorem of congruence
triangle ABC is equiangular | def of equiangular

can you please check if I do it correctly?
thx for the help you guys are great for keeping these forums up
also im in middle school but my other thread i think i posted it in highschool but i remember that i posted it in middle school sorry for posting in highschool.

I think the general outline of your proof is good.

However, you might want to be a little more rigorous and prove that $ \overline{AC}\cong\overline{BC}\text{ in }\triangle ABC\implies\angle A\cong\angle B$, which is simply [url=http://en.wikipedia.org/wiki/Isosceles_triangle_theorem#Proof]Pons Asinorum, or the Isosceles Triangle Theorem[/url].

[hide="A Cleaner Proof"]
[geogebra]af8547301340b4a8aca17fe26c147d6c86ff3c7f[/geogebra]
$ \triangle ABC$ has congruent legs $ \overline{AC}$ and $ \overline{BC}$ | Given
$ \overline{AD}$ is the angle bisector of $ \angle BAC$ | Constructed
$ \angle BAD\cong\angle CAD$ | Definition of angle bisector
$ \overline{AD}\cong\overline{AD}$ | Obvious
$ \triangle BAD\cong\triangle CAD$ | SAS Congruence
$ \angle ABC\cong\angle ACB$ | Congruent Parts of Congruent Triangles are Congruent (CPCTC)
$ \angle ACB\cong\angle BAC$ | Symmetry among the sides (we could've started with the lines $ \overline{AB},\overline{BC}$, or $ \overline{AB},\overline{AC}$ as well)
$ \angle ABC\cong\angle ACB\cong\angle BAC$ | Transitive Theorem of Congruence
$ \triangle ABC$ is equiangular | Definition of equiangular
[/hide]

hi,
thx for looking over it.

i have looked at your link and proof but i already said that i proved that base angles are congruent... i did not do it your way tho, my way was like:
Given triangle ABC with AC conguent to BC
consider the correspondence ABC <--> BAC
1. AC congruent to BC |given
2. BC congruent to AC|symmetric theorem
3. angle C congruent to angle | reflexive theorem
4. triangle ABC is congruent to triangle BAC | SAS
5. Angle A = Angle B | CPCTC

it looks very similar but i dont use angle bisector its a good idea i forgot about it tho thanks for giving the link to that proof i have more ideas how to write proofs now thanks...

thank you and God bless....

[quote="math154"][hide="A Cleaner Proof"]
[geogebra]af8547301340b4a8aca17fe26c147d6c86ff3c7f[/geogebra]
$ \triangle ABC$ has congruent legs $ \overline{AC}$ and $ \overline{BC}$ | Given
$ \overline{AD}$ is the angle bisector of $ \angle BAC$ | Constructed
$ \angle BAD\cong\angle CAD$ | Definition of angle bisector
$ \overline{AD}\cong\overline{AD}$ | [b]Reflexive Property[/b]
$ \triangle BAD\cong\triangle CAD$ | SAS Congruence
$ \angle ABC\cong\angle ACB$ | Congruent Parts of Congruent Triangles are Congruent (CPCTC)
$ \angle ACB\cong\angle BAC$ | Symmetry among the sides (we could've started with the lines $ \overline{AB},\overline{BC}$, or $ \overline{AB},\overline{AC}$ as well)
$ \angle ABC\cong\angle ACB\cong\angle BAC$ | Transitive Theorem of Congruence
$ \triangle ABC$ is equiangular | Definition of equiangular
[/hide][/quote]

One tiny error I found.

Hmm... I have a different proof:

 [geogebra]53f2de530846c2fb2289ab05223cd050f767222b[/geogebra]

$ \sin y\equal{}\frac{\sqrt{3}}{2}$.

$ \sin 90\minus{}y\equal{}\frac{1}{2}$.

Solving, from the first equation:

$ 60$, or more then 360, which can't be right.

Thus, $ y\equal{}60$.

Do the same with the other sides to prove.

Why do we know the center line is both a perpendicular bisector and a median?

[quote="BOGTRO"]Hmm... I have a different proof:

 [geogebra]53f2de530846c2fb2289ab05223cd050f767222b[/geogebra]

$ \sin y \equal{} \frac {\sqrt {3}}{2}$.

$ \sin 90 \minus{} y \equal{} \frac {1}{2}$.

Solving, from the first equation:

$ 60$, or more then 360, which can't be right.

Thus, $ y \equal{} 60$.

Do the same with the other sides to prove.[/quote]

BOGTRO, when proving things like this, it is usually best to find the [b]simplest[/b] method.

One could accuse your proof of circular logic, because it involves trigonometry (and other things), which is based off of many of these elementary facts.

For example, you can't use the Law of Cosines to prove SAS Congruence, since the Law of Cosines depends upon the fact that SAS Congruence holds (do you see why?).

Then what is?

anyway, isn't this true by definition?

[url=see this]http://en.wikipedia.org/wiki/Equilateral_triangle[/url]

...that the simpler the method, the better. :wink:

In your proof, you assume that the median is also the perpendicular bisector and also that $ x\in(0,\pi),\sin x \equal{} \frac {\sqrt3}2\implies x \equal{} 60^\circ$ (justify this!).

And it's not true by definition, because equilateral simply means that the sides are the same. It easily follows, but it's not part of the meaning.


I want to learn Asymptote for USAMTS, and I was wondering where I could get started. I looked around the site but couldn't find a tutorial? Sorry if I'm missing something....  :maybe:

Try [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote_%28Vector_Graphics_Language%29]this[/url]. From AoPS Wiki's main page, clicking LaTeX graphics gets you there.

Thank you! When I run Asymptote now, I get something like a command prompt. And the whole using Asymptote as a tool in TeXnicCenter is not working....

@buzzer11: the same thing happens to me, i have been trying to learn asymptote and latex for 2 months, asymptote has keeped me stumped.

[quote="buzzer11"]:( does anyone have any advice?[/quote]It would help if you gave much more information
[quote="buzzer11"]When I run Asymptote now, I get something like a command prompt. [/quote]Yes, that's how it should be. Asymptote is a DOS program so will open a DOS window.
[quote="buzzer11"]And the whole using Asymptote as a tool in TeXnicCenter is not working....[/quote]That tells us absolutely nothing. You must be much more specific and tell us what you've done and exactly what happens.

[quote="buzzer11"]I followed the instructions [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Getting_Started]here[/url].[/quote]And what happened? You mustn't be so shy because none of us can tell what has gone wrong unless [b]you[/b] tell us. Those instructions usually work so how can we know what is happening on your computer without you saying [b]in detail[/b]?
You don't even say if you've tried Asymptote as detailed [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Getting_Started/Windows/Interactive_Mode]here[/url] or the 3 step compile 1) LaTex=>PDF, 2) Asymptote 3) LaTex=>PDF as detailed [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration#Using_Asymptote_in_LaTeX]here[/url] and if you have, what happened.


One of the things I noticed about the fibonacci numbers was that:
$f^{2}_{n}= f_{n-1}f_{n+1}\pm 1$

Does this always hold? (Can someone prove it?) And when is the sign on the $1$ positive or negative?

"just noticed" as in this was posted a week or two ago

rofl, i didnt see the other thread though... I was just doing our math hw with induction and saw this.


function
inequalities
LaTeX
algebra proposed
algebra

Here is my solution for Problem 2 of 4th mathlinks contest round 3
Determine all functions $f: \mathbb {R} \rightarrow \mathbb{R}$ that $f(x) \geq 0$ for all positive real $x$,$f(0)=0$ and for all reals $x,y$: \[ f(x+y-xy)=f(x)+f(y)-f(xy)\]
_____________________________________________________________
Proof:We have:
\[ f(x)+f(y)=f(x+y-xy)=f(x)\]
So for all real $x,y$ that $y<>0$(Instead of $x$ put $x+y-xy$ and instead of $y$ put $\frac {y}{y-1}$):
\[ f(x+y-xy)+f(\frac {y}{y-1})=f(x)+f(\frac {y^2}{y-1}-x)\]
Adding the 2 equations :
\[ f(xy)+f(\frac {y^2}{y-1}-xy)=f(\frac{y}{y-1})+f(y)\]
Now suppose $y<>0$(Put $x/y$ instead of $x$)
\[ f(x)+f(\frac {y^2}{y-1}-x)=f(\frac{y}{y-1})+f(y)\]

Now putting $x=0$ in the last equation we have $f(\frac{y}{y-1})+f(y)=f(\frac {y^2}{y-1})$ therefore:
\[ f(x)+f(\frac {y^2}{y-1}-x)=f(\frac {y^2}{y-1})\]
Now we can see taht for all $u<0$ there exist a $y$ that $\frac {y^2}{y-1}=u$ so:
\[ f(x)+f(y)=f(x+y) \forall  x,y : x+y < 0 \]
Now suppose $x,y$ are 2 numbers that $x+y \geq 0$ there exist real $k>max \{x,y \}$
$f(x)+f(-k)=f(x-k),f(y)+f(-k)=f(y-k) \Rightarrow f(x)+f(y)+f(-2k)=f(x-k)+f(y-k)=f(x+y-2k)=f(x+y)+f(-2k)$
So :
\[ f(x+y)=f(x)+f(y) \forall x,y \in \mathbb {R}\]
Now from condition $f(x)\geq 0$ for $x \geq 0$ we have $f(x)=ax, a \geq 0$

At first, some typo corrections to your very nice solution, Omid:

[quote="Omid Hatami"]Here is my solution for Problem 2 of 4th mathlinks contest round 3
Determine all functions $f: \mathbb {R} \rightarrow \mathbb{R}$ that $f(x) \geq 0$ for all positive real $x$,$f(0)=0$ and for all reals $x,y$: \[ f(x+y-xy)=f(x)+f(y)-f(xy)\]
_____________________________________________________________
Proof:We have:
\[ f(x)+f(y)=f(x+y-xy)=f(x)\][/quote]

This should be: $f\left(x\right)+f\left(y\right)=f\left(x+y-xy\right)+f\left(xy\right)$.

[quote="Omid Hatami"]So for all real $x,y$ that $y<>0$(Instead of $x$ put $x+y-xy$ and instead of $y$ put $\frac {y}{y-1}$):[/quote]

The $y<>0$ should be replaced by $y\neq 1$.

[quote="Omid Hatami"]\[ f(x+y-xy)+f(\frac {y}{y-1})=f(x)+f(\frac {y^2}{y-1}-x)\]
Adding the 2 equations :
\[ f(xy)+f(\frac {y^2}{y-1}-xy)=f(\frac{y}{y-1})+f(y)\]
Now suppose $y<>0$(Put $x/y$ instead of $x$)[/quote]

This time, $y<>0$ should mean $y\neq 0$. (Note that the inequality sign in LaTeX is "\neq", not "<>").

[quote="Omid Hatami"]\[ f(x)+f(\frac {y^2}{y-1}-x)=f(\frac{y}{y-1})+f(y)\]

Now putting $x=0$ in the last equation we have $f(\frac{y}{y-1})+f(y)=f(\frac {y^2}{y-1})$ therefore:
\[ f(x)+f(\frac {y^2}{y-1}-x)=f(\frac {y^2}{y-1})\]
Now we can see taht for all $u<0$ there exist a $y$ that $\frac {y^2}{y-1}=u$ so:
\[ f(x)+f(y)=f(x+y) \forall  x,y : x+y < 0 \]
Now suppose $x,y$ are 2 numbers that $x+y \geq 0$ there exist real $k>max \{x,y \}$
$f(x)+f(-k)=f(x-k),f(y)+f(-k)=f(y-k) \Rightarrow f(x)+f(y)+f(-2k)=f(x-k)+f(y-k)=f(x+y-2k)=f(x+y)+f(-2k)$[/quote]

LaTeX code correction:

$f\left(x\right)+f\left(-k\right)=f\left(x-k\right);\ \ \ \ f\left(y\right)+f\left(-k\right)=f\left(y-k\right)$
$\Rightarrow\ \ \ \ f\left(x\right)+f\left(y\right)+f\left(-2k\right)=f\left(x-k\right)+f\left(y-k\right)=f\left(x+y-2k\right)=f\left(x+y\right)+f\left(-2k\right)$.

[quote="Omid Hatami"]So :
\[ f(x+y)=f(x)+f(y) \forall x,y \in \mathbb {R}\]
Now from condition $f(x)\geq 0$ for $x \geq 0$ we have $f(x)=ax, a \geq 0$[/quote]

As I've said, that's a beautiful solution. Since you asked whether somebody has another solution, I am posting a solution by Fiachra, which I just found in my archive (I believe it was lost during the hacker attack on MathLinks). I, personally, don't understand the solution since it is written in a very compressed way, but anyway, here is it:

___________________________________________________________________

[u][b]Fiachra's solution:[/b][/u]

(1) Put $y=-1$, $x=x$ and $x=-x$. Add the equations to get $f\left(2x-1\right)+f\left(-2x-1\right)=2f\left(-1\right)$, or $f\left(x\right)+f\left(-x-2\right)=2f\left(-1\right)$.
(2) Use this to modify the original equation to get $f\left(xy-x-y-2\right)=f\left(xy\right)-f\left(x\right)-f\left(y\right)+2f\left(-1\right)$. Also put $x=-x-2$, $y=-y-2$ and modify similarly to get $f\left(xy+3x+3y+6\right)=f\left(xy+2x+2y+4\right)-f\left(x\right)-f\left(y\right)-2f\left(-1\right)$. Add these two and sub in $p=xy$, $q=x+y+2$ to get $f\left(p-q\right)+f\left(p+3q\right)=f\left(p\right)+f\left(p+2q\right)$.
(3) This implies that $f\left(p-3q\right)+f\left(p+3q\right)=2f\left(p\right)$. (You have to worry about positives and negatives here; there are restrictions on $p$ and $q$. But it works out O.K.) Now we let $y=\frac{x}{x-1}$ so that $x+y-xy=0$, and use this in the original equation (and also our last result) to get $2f\left(x\right)=f\left(2x\right)$. Now $f\left(x\right)+f\left(y\right)=f\left(x+y\right)$; we solve for $x\in\mathbb{Q}$ as usual and use $f\left(x\right)\geq 0$ for $x>0$ to generalise to $x\in\mathbb{R}$.
___________________________________________________________________

  Darij

This was especial case of: 'Hoszue' Functional Equations , A very nice Approach of solving it based on : Inserting New variable like z , in equation.


linear algebra
matrix
complex analysis
real analysis
real analysis unsolved

New topic: new structure for the College forums to come soon. Please make suggestions below in this topic.

Old message was: I am curious if you would be intersted in a forum for Complex Analysis ..

Couldn't people just post stuff about complex analysis in the Analysis forum?

I'm with grobber on this one.

I also think that a Complex Analysis forum would not have any significant attendance on this site for the moment.

It would be great!!

the similitude of complex analisis and real analisis is only the name , they are quite different....

PS. how can I vote?

It would be great to have a forum on Complex Analysis!

So what is the decision??
 And how can I vote?

Yeah, it would be fine to have a subforum for complex mathematics... especially now I'm learning that one :)

At the moment however (that is as a first-year student) I feel most need to have the algebra split in matrix- and non-matrix stuff (the thing with rings and structures and such... I thought that's a 3rd-year subject for us? I haven't even understood a single problem of those. And I guess I can't be the only one not having this in early years?)

Yes, I most definitely agree.  People take Linear Algebra much earlier than Abstract Algebra, and would be able to participate better in the Algebra forum if they knew they could stick to a sub-forum where they understand everything.

Alex

Hmm... well one thing you might consider is making a college-level combinatorics forum. we don't have that at the moment.

That could feature combinatorial things not in the high school program, which are now posted in the olympiad combinatorics forum (and from which I hear people from pre-olympiad scared away by it, because they don't know any theorems for graphs, or other combinatorial things :)

[ well, it's not my idea in fact but I'm certainly for it ] :P

Greetz, Peter

Combinatorics can be combined with IMO stuff. PreOlympiad people have the PreOlympiad forums, so everyone is happy :P :)

Yes that is right. :) But I didn't know if you counted graphs as IMO stuff. They are handy indeed. But they're officially not in the high school curriculum ;)

The discussion about crazy Europeans has been moved here: http://www.mathlinks.ro/Forum/viewtopic.php?p=247089#p247089
Please continue on topic in this topic. That is, make suggestions about the a new structure of the College Forums.

[quote="Valentin Vornicu"]The College forums will be reorganized sometime this June ;) I am awaiting suggestions (besides the Linear Algebra thing that is going to happen then :) ).[/quote] :arrow: I just notice complex analysis isn't in there? :) or is that a typo?

Anyway a suggestion of mine would be to disable non-moderators to post "New Topic"s directly into the "Solved Problems" forums (if they still exist after the change - which I guess). Well not much of my worries :) but I think it isn't meant like that.


[quote="linzk3"]At a small party, each person greets each other person by shaking hands.  How many handshakes take place if the number of people is:
a. 3
b. 4
c. 5
d. 6
e. 101
?[/quote]
[hide] Say that there are $n$ people.  We know that each person shakes each of the $n-1$ other people, so there are $n(n-1)$ handshakes.  However, for every handshake, two people are involved, so we overcounted by a factor of two.  Thus, the formula is $\frac{n(n-1)}{2}$.  Plugging in our numbers, we get:
\[A=3 \]

\[B=6 \]

\[C=10 \]

\[D=15 \]

\[E=5050 \]
[/hide]

For n people who all shake hands with each other, the formula for the total number of handshakes is
[hide]
$(n-1) nCr 2$ or $\frac{n^{2}-n}{2}$
[/hide]
Therefore, the answers are:
[hide]
a)3
b)6
c)10
d)15
e)5050 (I think... I did that in my head)
[/hide]


linear algebra
matrix
complex numbers
number theory proposed
number theory

A quickie: show that if $a,b,x,y \in \mathbb{C} \setminus \{0\}$ are non-zero complex numbers such that $ax^n+by^n \in \mathbb{Z}$ is an integer for all integers $n \in \mathbb{Z}$, then $|xy|=1$.  :)

What about $a=b=0$  :P  a real quickie  :D

Pierre.

:D  :D  :D

Oh, yes, of course $abxy \neq 0$.  :lol: I've edited the post above.  :P

Is it really a problem in number theory? Is it hard?
 :(

It could also happen that $x=y,\ a+b=0$, but if we assume this is not the case, we can even show that $xy=\pm 1$:

Let $u_{n}=ax^{n}+by^{n}$. If $x=y$ then, by assumption, all $u_{n}=(a+b)x^{n},\ n\in\mathbb Z$ are non-zero integers, and it follows that $x=\pm 1$ . If $x\ne y$, all determinants $\begin{vmatrix}u_{n-1}&u_{n}\\u_{n}&u_{n+1}\end{vmatrix}=ab(xy)^{n-1}(x-y)^{2},\ n\in\mathbb Z$ are non-zero integers, so $xy=\pm 1$.

[quote="Rust"][quote="vess"]A quickie: show that if $a,b,x,y \in \mathbb{C}\setminus \{0\}$ are non-zero complex numbers such that $ax^{n}+by^{n}\in \mathbb{Z}$ is an integer for all integers $n \in \mathbb{Z}$, then $|xy|=1$.  :)[/quote]
Let $u_{n+1}=cu_{n}+du_{n-1},c,d\in Z , \ x,y-$ roots $z^{2}-cz-d=0$ ($x+y=c,xy=-d$.
If $u_{0}=a+b, u_{1}=ax+by$ are integer, then $u_{n}=ax^{n}+by^{n}\ \forall n\in Z_{+}$ are integer. But $xy=-d$ can be all integer.[/quote]

But $u_{n}$ should be integer for $n\in\mathbb{Z}$ not only $n\in\mathbb{N^{*}}$

And I think grobber is right :wink:


here's the problem: In the world of cheapville:

a donkey costs 1/4 of a dollar
a horse costs  3/4 of a dollar
an elephant costs 5 dollars

The king of cheapville gives you 20 dollars and tells you to buy 20 animals [b]and[/b] use up all the 20 dollars. How many of each animal must you buy?

I solved this problem in 3-5 minutes using algebra, inequalities, and some common sense, but I would like to know if there is a faster solution.

hmmm... this method took me about 1-1.5 minutes, but this might be the same method you used.

let d=donkeys h=horses e=elephants(since they are so cheap Im assuming they are stuffed animals :P )

We know:
1/4d+3/4h+5e=20
d+e+h=20

1st equation: multiply by 4 to get d+3h+20e=80
subtract the second equation from this new one to get
2h+19e=60.
Therefore e must be 0 or 2. If e=0, we need 30 horses (too much). If e=2, we need 11 horses and 7 donkeys. Plugging this into the first equation we see that d=7, h=11, e=2 works.


when you go from

a(a+b-c)=-b(a+b-c)
to
a=-b
you divide by zero since
a+b-c=4+2-6=0

and dividing by zero always causes problems...

[quote="xxxyyyy"][quote="aarabhi"]let a=4 , b= 2 and c = 6
ie, a+b=c
*ing both sides by a+b
(a+b) * (a+b) = c(a+b)
a*a + ab + ba + b*b = ca+cb
a*a + ab-ca = cb-ba-b*b
a(a+b-c)= -b(-c+a+b)
a = -b
a+b=0
ie 4+2=0

pls tell me where the error is[/quote][/quote]

between the steps $a(a+b-c)= -b(-c+a+b)$ and $a = -b$,
you have divided by 0 ($a+b-c$ = $0$), and division by $0$ will always land you in problems...

do you get it?[/quote

pls give more details. (a+b-c) is there in both sides. so we can cancel (a+b-c) from both sides? then will get a=-b....

[quote="aarabhi"][quote="xxxyyyy"][quote="aarabhi"]let a=4 , b= 2 and c = 6
ie, a+b=c
*ing both sides by a+b
(a+b) * (a+b) = c(a+b)
a*a + ab + ba + b*b = ca+cb
a*a + ab-ca = cb-ba-b*b
a(a+b-c)= -b(-c+a+b)
a = -b
a+b=0
ie 4+2=0

pls tell me where the error is[/quote]

between the steps $a(a+b-c)= -b(-c+a+b)$ and $a = -b$,
you have divided by 0 ($a+b-c$ = $0$), and division by $0$ will always land you in problems...

do you get it?[/quote

pls give more details. (a+b-c) is there in both sides. so we can cancel (a+b-c) from both sides? then will get a=-b....[/quote][/quote]

I think if you wish to cancel out a common factor form two sides of an equality, then the factor should not evaluate to $0$ which in thiis case does.

like the following is true:

0*2=0*4

since both sides are clearly equal to 0
but fairly obviously, we can't cancel the 0.
In your example, the 0 was just (a+b-c) but apart from that it was roughly the same

There is no mathematical operation as 'canceling'. When you cancel a factor, you are actually just dividing it out of the equation. This is all well and good, unless that factor happens to be zero. That's exactly what is happening here.