pigeonhole principle
search
combinatorics proposed
combinatorics

$ 49$ students join in contest with $ 3$ problems. The piont for each problem is $ 0\rightarrow 7$. Prove that there exist $ 2$ students $ A ,B$ satisfy the piont for each problem of $ A$ always $ \geq$ the point for each problem or $ B$.

well, pigeonhole is crying out desperately to me...

It well may, but the reasoning is arduous.

Theoretical framework. Denote
$ P : \equal{} \{(x,y,z) \ \mid \ x,y,z \in \{0,1,2,3,4,5,6,7\}\}$
the set of all posible triplets of scores on the three problems ($ |P| \equal{} 8^3 \equal{} 512$). The relation
$ (a,b,c) \preceq (x,y,z) \ \textrm{if and only if} \ (a\leq x) \land (b\leq y) \land (c\leq z)$
is a partial order relation on $ P$. Define $ P_k : \equal{} \{ (x,y,z) \in P \ \mid \ x\plus{}y\plus{}z\equal{}k\}$, for $ 0\leq k\leq 21$ (all $ P_k$ are antichains, i.e. made of elements mutually not comparable). We have $ |P_0| \equal{} |P_{21}| \equal{} 1$, $ |P_1| \equal{} |P_{20}| \equal{} 3$, ..., up to $ |P_{10}| \equal{} |P_{11}| \equal{} 48$.

If we could partition $ P$ into $ 48$ disjoint sets, each a  chain (i.e. with any pair of elements comparable), as we have $ 49$ students, by pigeon-hole principle, two will be in the same chain, hence the required result about their scores. This would also show $ 49$ is the lowest number available, since an example of $ 48$ students without that property is given by the maximal antichain $ P_{10}$. (This is not a coincidence; Dilworth's theorem states that the cardinality of the largest antichain is equal to the minimal number of chains in a partition by chains of $ P$).

The above is correct. I will soon post the references to how this problem is settled in all its generality.

A graded poset is said to "have the Sperner property" or to "be Sperner" if there is an antichain of maximal size consisting of all points of a particular rank.  The sought-after result is that a product of chains is Sperner; I'm moderately confident that I've seen a proof of this statement in general, but 2 minutes with a search engine wasn't enough for me to find a reference.


Prove that if $a,b,c\geq0$ then
\[ \frac a{\sqrt{a^2+bc}}+\frac b{\sqrt{b^2+ca}}+\frac c{\sqrt{c^2+ab}}\geq2. \]

This seems to be wrong. I thought about such inequality some day ago, but I think a found a counterexample.

You are right, Lovasz. $c\rightarrow+\infty$

if  $a,b,c \ge 0$then
（1）$\frac{a}{{\sqrt{a^2 + \lambda bc} }} + \frac{b}{{\sqrt{b^2 + \lambda ca} }} + \frac{c}{{\sqrt{c^2 + \lambda ab} }} \ge \frac{3}{{\sqrt{1 + \lambda } }}(\lambda \ge 8);$

（2）$\frac{a}{{\sqrt{a^2 + \lambda bc} }} + \frac{b}{{\sqrt{b^2 + \lambda ca} }} + \frac{c}{{\sqrt{c^2 + \lambda ab} }} \ge 1(0 < \lambda \le 8)$

[quote]Prove that if $a,b,c\geq0$ then
\[ \frac a{\sqrt{a^2+bc}}+\frac b{\sqrt{b^2+ca}}+\frac c{\sqrt{c^2+ab}}\geq c.  \][/quote]
Which will be the best constant?
The number $1$ is not really familiar...


Are easy questions supposed to have more points?  Like I had question of the probability of not getting an even prime when rolling a dice, which is easy, yet it was worth 4 points.

I'm pretty sure it's supposed to go $ 1\minus{}4$ points in order of difficulty (least to most)...but at the beginning, there's a lot of fluctuations.

Then i got another ez question with what is the probability when rolling a die you get a prime number.  that question was worth 1 point....


topology
complex analysis
complex analysis unsolved

Consider $U=\mathbb{C}-\{a,b\}$ where $a \neq b$. Show that there is a closed path $\gamma$ in $U$ such that $Ind_{a}(\gamma)=Ind_{b}(\gamma)=0$ but $\gamma$ is not null-homotopic.

Well, I found the path, and graphically it is obvious that it is not null-homotopic. Having no remembrance of algebraic topology, how to prove this fact formally ?

uhm..
the fundamental group of $U$ is $F_{2}$... you can take two generators $\alpha$ and $\beta$, and since there's no relation, $\alpha\beta\alpha^{-1}\beta^{-1}\neq e$...


Ring Theory
algebra
function
domain
number theory
relatively prime
superior algebra

Let K be a number field with ring of integers O and class number h. Suppose that for every ideal I in O, there exists some integer r, relatively prime to h, such that I^r is principal. Why does this imply that O is a Unique factorization Domain?

Because $ x \mapsto x^r$ is an injective map of the class group to itself (by recntly posted stuff, but it is not hard to show at all), sending $ (1)$ to $ (1)$. Thus every ideal is principal.


I have a strange favor for someone up for he challenge.

I want to find a solution for the following.

An employee works 80 hours per pay period at $18.50 per hour. after 80hrs the pay goes up to pay and a half$27.75 per hour.

I want to know what I have to change the rate to every pay period so that each pay period $200.00 are subtracted from the total amount paid.
Now I can only change the regular hour rate because whatever the regular hour rate the Overtime must be that X 1.5
ie
Regular hrs.        80 hrs @ $18.5              =  $1,480.00
Over Time          11.96 hrs @ $27.75         =  $331.89
Total:$1,811.89

Now I have to get $1,811.89 down exactly$200.00

I am in heavy need for a formula which I can use to figure this out every pay period.

Thank you any help is great. :blush:

im confused on the question. im probably wrong too

ummmm
$ xr_1\plus{}yr_2\minus{}200$ $ r_1$ and $ r_2$ are the normal and overtime rates and x and y are the hours worked.

[quote="kaptainyui"]Now I have to get $ 1,811.89 down exactly$200.00[/quote]
I don't understand why you want this, or where you want me to take the money from (i.e. if you just don't want $ \$200$ then subtract $ \$200$), or is there some sort of scalar involved.  :?:

Well its not as simple. See It cant simply bee subtracted because of tax problems. But if its subtracted using a diffrent pay rate then it has nothing to do with taxes.

Well I am guessing its a bit hard to explain so thanks for your help.


integration
calculus
trigonometry
function
real analysis
real analysis unsolved

\[ \int\limits_0^{\plus{}\infty}e^{\minus{}x^2}\sin(ax) dx
\equal{}
\frac{1}{2}\sum_{n\equal{}0}^{\plus{}\infty}\frac{(\minus{}1)^nn!}{(2n\plus{}1)!}a^{2n\plus{}1}\]

could anyone prove it ?

Can you compute $ \int_0^{\infty}x^{2k\plus{}1}e^{\minus{}x^2}\,dx\ ?$

Once you do that, then try expanding $ \sin(ax)$ as a power series.

[quote="knoppix"]
\[ \int\limits_0^{ \plus{} \infty}e^{ \minus{} x^2}\sin(ax) dx \equal{} \frac {1}{2}\sum_{n \equal{} 0}^{ \plus{} \infty}\frac {( \minus{} 1)^nn!}{(2n \plus{} 1)!}a^{2n \plus{} 1}
\]
could anyone prove it ?[/quote]

i am not really sure why you call it a [i]mystery[/i] formula ? where is the mystery in it ?  :?:

$ \int_0^{\infty}\; e^{ \minus{} x^2}\; \sin(ax) \;dx \equal{} \int_0^{\infty}\; e^{ \minus{} x^2}\;\left( \sum_{n \equal{} 0}^\infty \; \frac {( \minus{} 1)^n}{(2n \plus{} 1)!} \; (ax)^{2n \plus{} 1} \right) \;dx$

$ \equal{} \sum_{n \equal{} 0}^\infty \; \frac {( \minus{} 1)^n}{(2n \plus{} 1)!} \; a^{2n \plus{} 1} \;\;\; \underbrace{ \int_0^{\infty}\;x^{2n \plus{} 1}\; e^{ \minus{} x^2}\; dx } _ { \equal{} \; \frac {1}{2}\; n! }$

$ \equal{} \frac {1}{2}\; \sum_{n \equal{} 0}^\infty \; \frac {( \minus{} 1)^n\; n!}{(2n \plus{} 1)!} \; a^{2n \plus{} 1} \; \qquad ( \; \text{where}\qquad n > \minus{} 1\;, \quad a\in \mathbb R )$

$ \equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}$

to get the above integral, do

$ \int_0^{\infty}\;x^{2n \plus{} 1}\; e^{ \minus{} x^2}\; dx \equal{} \minus{} \frac {1}{2}\; \int_0^{\infty}\;x^{2n} \;e^{ \minus{} x^2}\; d( \minus{} x^2)\qquad \qquad \ldots \qquad (1)$

$ \equal{} \minus{} \frac {1}{2}\;\left[ \underbrace { x^{2n} \; e^{ \minus{} x^2}\; \bigg |_0^\infty} _ { \equal{} \quad 0} \; \minus{} \;\int_0^{\infty}\;\frac {d}{dx}( x^{2n} ) \;\;\; e^{ \minus{} x^2}\; dx\right]$

$ \equal{} \frac {1}{2}\;\left[ 2n\; \int_0^{\infty}\; x^{2n \minus{} 1} \; e^{ \minus{} x^2}\; dx\right] \equal{} \frac {1}{2}\;\left[ 2n\;\; \left( \minus{} \frac {1}{2}\right) \;\int_0^{\infty}\; x^{2n \minus{} 2} \;e^{ \minus{} x^2}\; d( \minus{} x^2) \right]$

keep going in a similar fashion to $ (1)$, and you will see that you eventually get

$ \frac {1}{2}\;n \; (n \minus{} 1) \; (n \minus{} 2)\cdot \; \; \ldots \; \; \ldots \cdot \; \;3\cdot 2 \cdot 1 \; \equal{} \; \frac {1}{2}\;n!$

now - nothing :) Thanks !

I have the next one ;)
I'm not starting new topic because it is similar:

prove:
\[ \int_0^{\plus{}\infty}\frac{\sin(tx)}{e^t\minus{}1} dt\equal{}\sum_{n\equal{}0}^{\infty}\frac{x}{x^2\plus{}n^2}\]
for all real $ x$.

I tried with power series for sine (as you did above), but then I would need to compute
\[ \int_0^{\plus{}\infty}\frac{t^{2n\plus{}1}}{e^t\minus{}1}\]
As Mathematica said, $ \equal{}\Gamma(2n\plus{}2)\zeta(2n\plus{}2)$, but how can I obtain it ?

Ah, but the trick is a little different. Any time you see a simple two-term denominator, consider the possibility of representing that as a geometric series.

$ \frac{1}{e^t\minus{}1}\equal{}\frac{e^{\minus{}t}}{1\minus{}e^{\minus{}t}}\equal{}\sum_{n\equal{}1}^{\infty}e^{\minus{}nt}$

Try interchanging that sum with the integral in question.

By the way, in your formula, the sum should start with $ n\equal{}1,$ not $ n\equal{}0.$ It's incorrect as written, and you can convince yourself of that by taking the limit as $ x\to0.$

[quote="knoppix"]
prove:
\[ \int_0^{ \plus{} \infty}\frac {\sin(tx)}{e^t \minus{} 1} dt \equal{} \sum_{n \equal{} 0}^{\infty}\frac {x}{x^2 \plus{} n^2}
\]
for all real $ x$.[/quote]

use what kent said.

$ \int_0^{\infty}\; \frac {\sin\; (tx)}{e^t \minus{} 1} \; dt \equal{} \int_0^{\infty}\;\sin\; (tx) \; \sum_{n\equal{}1}^{\infty}e^{\minus{}nt} \; dt \equal{} \sum_{n\equal{}1}^{\infty}  \int_0^{\infty}\; e^{\minus{}nt} \; \sin\; (tx) \;dt$

$ \equal{}  \sum_{n\equal{}1}^{\infty} \frac{x}{n^2\plus{}x^2}$

make sure to fix your starting index to $ n\equal{}1$

$ \equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}$

to show that $ \int_0^{\infty}\; e^{\minus{}nt} \; \sin\; (tx) \;dt\;\equal{}\; \frac{x}{n^2\plus{}x^2}$

you need to do integration by parts twice. see the basic idea here:

http://en.wikipedia.org/wiki/Integration_by_parts

[quote="knoppix"]

As Mathematica said, $ \equal{} \Gamma(2n \plus{} 2)\zeta(2n \plus{} 2)$, but how can I obtain it ? [/quote]

i don't know what that means ? are you trying to get at a particular sum , and if so, what is the value of $ x$ you are using ?

Thank you very much for this !!

[quote="misan"]
[quote="knoppix"]As Mathematica said, $ \equal{} \Gamma(2n \plus{} 2)\zeta(2n \plus{} 2)$, but how can I obtain it ? [/quote]
i don't know what that means ? are you trying to get at a particular sum , and if so, what is the value of $ x$ you are using ?[/quote]

when put \[ \int_0^{ \plus{} \infty}\frac {t^{2n \plus{} 1}}{e^t \minus{} 1}dt
\]
into Mathematica, the program replies $ \Gamma(2n \plus{} 2)\zeta(2n \plus{} 2)$.

Hence I was interested in how to prove that for any $ n\in\mathbb{N}$
\[ \int_0^{ \plus{} \infty}\frac {t^{2n \plus{} 1}}{e^t \minus{} 1}dt \equal{}\Gamma(2n \plus{} 2)\zeta(2n \plus{} 2)
\]

($ \Gamma$ is obviously Euler's Gamma function, $ \zeta$ is Riemann Zeta)

[quote="misan"]
to show that $ \int_0^{\infty}\; e^{ \minus{} nt} \; \sin\; (tx) \;dt\; \equal{} \; \frac {x}{n^2 \plus{} x^2}$

you need to do integration by parts twice. see the basic idea here:
[/quote]
Complex numbers is a faster way too.


parameterization
probability
probability and stats

In class, I was given a short, simple problem something along the lines of :

$ X$ and $ Y$ are independent Binomial random variables with respective parameters $ (n, p )$ and $ (m, p)$, calculate the conditional distribution of $ X$ given that $ X\plus{}Y \equal{} n$. I noticed that when it's calculated, it kind of looks like the form for the hypergeometric random variable for $ P(X \equal{} i)$. Is there a combinatorial argument for why this is, or is it just one of a coincidence that they look alike?

It can't be hypergeometric. A hypergeometric random variable is unbounded, while this variable is bounded by $ n$.

If you had something else in mind, please find a way to clearly ask that question.

No, the [url=http://en.wikipedia.org/wiki/Hypergeometric_distribution]hypergeometric distribution[/url] is bounded.

For the problem in question: the probability that $ X\equal{}k$ given $ X\plus{}Y\equal{}r$ is given by

$ \frac{\binom{n}{k}p^k(1\minus{}p)^{n\minus{}k}\binom{m}{r\minus{}k}p^{r\minus{}k}(1\minus{}p)^{m\minus{}r\plus{}k}}{\binom{m\plus{}n}{r}p^r(1\minus{}p)^{m\plus{}n\minus{}r}}\equal{}\frac{\binom{n}{k}\binom{m}{r\minus{}k}}{\binom{m\plus{}n}{r}}$

Yes, that is hypergeometric.

The explanation is something like this. We do our Bernoulli trial $ n\plus{}m$ times - $ n$ of them are trials of the first kind and $ m$ are trials of the second kind. We know, overall, that we have $ r$ successes. So think of it this way: we throw the $ n\plus{}m$ trials into a bag, and pick out $ r$ of them to be the $ r$ successes. How many of those $ r$ are of the first kind? That's a setup for a hypergeometric distribution.

Oops. I got the terminology mixed up with the negative binomial.


trigonometry
limit
real analysis
real analysis solved

$x_n = \sin x_{n-1}$
Show that $\lim_{n\to\infty} n x_n = 3$?

or it is a similar problem because the one above seems not correct.

i think the limit is $\sqrt{3}$ but i could be wrong...i remember that it was posed to me at IMO but i didn't think about it.

Peter

If $x_0>0$
$x_n \sim \sqrt{\frac{3}{n}}$
With Cesaro

actually, it is divergent, but $\lim_{n\to\infty} nx_n^2$ exists(the problem was posed to me with $\lim_{n\to\infty} \sqrt{n}x_n$, that's why i said something with $\sqrt{3}$).
consider $a_n=\frac{1}{x_n}$. we have the following: $a_n$ is strictly increasing, positive and unbounded(if it was bounded, it had a limit, contradiction). thus we can apply the cesaro stolz theorem( http://planetmath.org/encyclopedia/StolzCesaroTheorem.html ):
\[\lim_{n\to\infty} nx_n^2=\lim_{n\to\infty} \frac{1}{a_{n+1}^2-a_n^2}\].
since $a_{n+1}=a_n+\frac 16a_n^{-1}+O(a_n^{-3})$, we have $a_{n+1}^2-a_n^2=\frac 13+O(a_n^{-2})$, and the right side equals 3(remember $a_n$ is unbounded).

Peter


calculus
integration
trigonometry
calculus computations

Evaluate $\int {\frac{{d\theta }}{{\cos ^3 \theta - \sin ^3 \theta }}}$.

My solution is rather long, involving repeated changes of variables. Hope that someone will find a shorter one.

It's

$\frac{1}{\cos ^3 \theta - \sin ^3 \theta }= \frac{1}{3} (\frac{2}{\cos \theta - \sin \theta }+ \frac{\cos \theta - \sin \theta }{1+\cos \theta \sin \theta })$

Now

$\frac{\cos \theta - \sin \theta }{1+\cos \theta \sin \theta }=\frac{2(\cos \theta - \sin \theta) }{2+2 \cos \theta \sin \theta }=\frac{2(\cos \theta - \sin \theta) }{1+(\cos \theta + \sin \theta)^2}$

and you make the change of variable $x=(\cos \theta + \sin \theta)$. :)

For the other you make the change of variable $x=tan\frac{\theta}{2}$  :D

and you have $\cos \theta =\frac{1-x^2}{1+x^2}$ , $\sin \theta =\frac{2x}{1+x^2}$ and $d\theta =\frac{2dx}{1+x^2}$

Then you can easily evaluate $\int \frac{2dx}{1-x^2-2x}$ :lol:
I hope this will help you  :) (and i hope it has no typos  :D )

However, the first integral (by letting x=cos(theta) + sin(theta) ) will still lead to lengthy evaluation. Enlighten me if I missed something  :)

What you are missing is that $(arctanf(x))'=\frac{f'(x)}{1+f^{2}(x)}$ .  :)

If you make the change of variable $x=(\cos \theta +\sin \theta)$ you have $dx = (\cos \theta -\sin \theta)d\theta$

so $\int \frac{2(\cos \theta - \sin \theta){d\theta }}{1+(\cos \theta + \sin \theta)^2}$ becomes $\int \frac{2dx}{1+x^2}$.

But $\int \frac{2dx}{1+x^2} =2\cdot arctan(x)$ or $\int \frac{2(\cos \theta - \sin \theta){d\theta }}{1+(\cos \theta + \sin \theta)^2} =2 \cdot arctan(\cos \theta +\sin \theta)$.   ;)

For $\int \frac{2dx}{1-x^{2}-2x}$ you write $1-x^{2}-2x=1-(\frac{x+1}{\sqrt2})^2$ and if you put $u=\frac{x+1}{\sqrt2}$

you take the integral $\int \frac{\sqrt{2}du}{1-u^2} =\sqrt{2}\cdot arctanh(u)$ or $\int \frac{2d\theta}{\cos \theta - \sin \theta }=\sqrt{2}\cdot arctanh(\frac{1+Tan(\frac{\theta}{2})}{\sqrt2})$.

Thanks for your more elegant solution  :)

What I did was to express the combinations of $Sin{x}$ and $Cos{x}$ in terms of a single trigo function.
That went well, but was a tad too tedious.


geometry
trapezoid
ratio
trigonometry
complex numbers
geometry unsolved

Let $ ABCD$ is a trapezoid with $ BC//AD$ and $ AD>BC$ and let $ \angle BAD \equal{} \angle CDA \equal{} 60^0$. Let $ \angle C_1 \equal{}\angle C_2$ as picture. Let $ M$ is midpoint of $ BC$, let $ d$ is a line pass $ M$ and $ d \perp BC$, let $ d$ meet $ Cx$ at $ E$. Let $ N$ is a point on $ AC$ so that $ EN \perp DN$. Prove that $ N$ is midpoint of $ AC$.

Very easy ?
Denote N be the midpoint of AC. MN=AB/2=CD/2, ME=CE/2, EMN=ECD=150,
$ \rightarrow\triangle EMN\sim triangle ECD\rightarrow\triangle ECM\sim\triangle EDN\rightarrow\angle END\equal{}90^o$

[quote="mr.danh"]Very easy ?
Denote N be the midpoint of AC. MN=AB/2=CD/2, ME=CE/2, EMN=ECD=150ú
$ \rightarrow\triangle EMN\sim triangle ECD\rightarrow\triangle ECM\sim\triangle EDN\rightarrow\angle END \equal{} 90^o$[/quote]
I don't think so!
You must from $ \angle DNE \equal{}90^0 \longrightarrow$ N is midpoint of AC.

[quote="thanhnam2902"]
You must from $ \angle DNE \equal{} 90^0 \longrightarrow$ N is midpoint of AC.[/quote]
????????
My post implies that the midpoint N of AC lie on the circle whose diameter is DE, it means that N is the intersection of that circle with AC, two expressions of N are concide.

Rotating the triangle ECD around E 60º clockwise and contracting to half, C comes to M and D to the middle of AC (the angle between CD and midline of CBA is 60º). That is, the triangle DEN having an angle of 60º (angle DEN) and the ratio of its adjacent sides ½, is rightangled; as the circle of diameter DE intersects |AC| in a single point, the conclusion follows.

Best regards,
sunken rock

[quote="thanhnam2902"]Let $ ABCD$ is a trapezoid with $ BC//AD$ and $ AD > BC$ and let $ \angle BAD = \angle CDA = 60^0$. Let $ \angle C_1 = \angle C_2$ as picture. Let $ M$ is midpoint of $ BC$, let $ d$ is a line pass $ M$ and $ d \perp BC$, let $ d$ meet $ Cx$ at $ E$. Let $ N$ is a point on $ AC$ so that $ EN \perp DN$. Prove that $ N$ is midpoint of $ AC$.[/quote]
Consider the complex plane with $ M$ as center.
Let $ a,b,c,d,e,n$ be the complex numbers corresponding to the points $ A,B,C,D,E,N$.
Let $ b = -x$, $ c = x$, where $ x \in \mathbb{R}_+$, and let $ P(p)$ be the projection of $ M$ on $ AD$. Then $ p = -iy$ for $ y \in \mathbb{R}_+$.
Then we obviously have $ a = -x-\frac{y}{\sqrt{3}}-iy$, $ d = x+\frac{y}{\sqrt{3}}-iy$, since $ \tan 30 ^\circ = \frac{1}{\sqrt{3}}$.
We also have $ \angle MCE = 30 ^\circ$, so $ E = i\frac{x}{\sqrt{3}}$. Then let $ N'(n')$ be the midpoint of $ AC$. Then $ n' = \frac{a+c}{2} = \frac{-\frac{y}{\sqrt{3}}-iy}{2}$.
And $ EN' \perp DN' \iff i \cdot \frac{e-n'}{d-n'} \in \mathbb{R}$. But $ \frac{e-n'}{d-n'} = \frac{i}{\sqrt{3}}$ so $ EN' \perp DN'$, and thus it follows that $ N = N' \iff n=n'$. Hence $ N$ is the midpoint of $ AC$. QED :)


inequalities
logarithms
calculus
integration
function
Cauchy Inequality

Here's a problem that "came" to my mind when I was doing history :P.

[b]Problem.[/b] Prove, without the use of a calculator (or approximation), that
\[ \frac {\pi \minus{} e}{\pi \plus{} e} < \frac {3}{2}(\ln\pi \minus{} 1).
\]

Consider $ f(x)\equal{}3(x\plus{}1)\ln x \minus{} 2x$.
We have $ f(1)\equal{}\minus{}2$ and $ f'(x)\equal{}3\ln x \plus{} 1 \plus{} \frac{3}{x} > 0 \,$ for $ x > 1$
hence $ f(x)>\minus{}2$ for $ x>1$. Take here $ x\equal{}\frac{\pi}{e}$ and we are done.
Of course I hope you will belief me that I can prove the inequality $ \frac{\pi}{e}>1$ without the use of a calculator.  :lol:

I simply proved Cauchy's Inequality in Integral form: $ \left(\int_a^bfg\right)^2\leq\int_a^bf^2\int_a^bg^2$.  Then, I set $ (f(x),g(x)) \equal{} \left(\frac {1}{\sqrt {x}},\sqrt {x}\right)$ and integrated over the interval $ [e,\pi]$.

[size=117][color=darkred][b]Very nice the your proof, [u]boxedexe[/u] ![/b][/color][/size] [color=darkblue]Here is and an another (similar to [b]Digger[/b]'s) natural proof.[/color]

[quote="boxedexe"] [color=darkred]Prove that (without the use of a calculator or approximation) $ \frac {\pi \minus{} e}{\pi \plus{} e} < \frac {3}{2}(\ln\pi \minus{} 1)$ .[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Consider the function $ f: (0,\infty )\rightarrow\mathcal R$, where $ f(x) \equal{} \frac 32\cdot\ln x \minus{} \frac {x \minus{} 1}{x \plus{} 1}$ . Observe that $ f(1) \equal{} 0$ and $ f'(x) \equal{} \frac {3x^2 \plus{} 2x \plus{} 3}{2x(x \plus{} 1)^2} > 0$ for any $ x > 0$ . Therefore, the function $ f$ is strict increasing and particularly, $ \pi > 3 > e$ $ \Longrightarrow$ $ \frac {\pi}{e} > 1$ $ \Longleftrightarrow$ $ f\left(\frac {\pi}{e}\right) > f(1)$ $ \Longleftrightarrow$ $ f\left(\frac {\pi}{e}\right) > 0$ $ \Longleftrightarrow$ $ \frac {3}{2}(\ln\pi \minus{} 1) > \frac {\pi \minus{} e}{\pi \plus{} e}$ .[/color]

[quote="Virgil Nicula"][size=117][color=darkred][b]Very nice the your proof, [u]boxedexe[/u] ![/b][/color][/size] [/color]

[quote="boxedexe"] [color=darkred]Prove that (without the use of a calculator or approximation) $ \frac {\pi \minus{} e}{\pi \plus{} e} < \frac {3}{2}(\ln\pi \minus{} 1)$ .[/color] [/quote]
[/quote]

Yes, nice. But it gives something more:
$ \frac {\pi \minus{} e}{\pi \plus{} e} < \frac {1}{2}(\ln\pi \minus{} 1)$  :lol:

In this case yor proof is preserved whereas those of mine would demand additional observation of the second derivation.

[quote="Altheman"]Can somebody give a more precise formulation of the "Cauchy Inequality in Integral form?" Are f,g any integrable functions [on that interval of course]? What are the conditions for equality?[/quote]

The inequality is:
Suppose $ f$ and $ g$ are integrable functions on $ (a,b)$.  Then $ \int_a^b f^2(x)\mathrm dx\cdot\int_a^b g^2(x)\mathrm dx \ge \left(\int_a^b f(x)g(x)\mathrm dx\right)^2.$

[hide="Proof"]Set $ A\equal{}\int_a^b f^2(x)\mathrm dx,B\equal{}\int_a^b g^2(x)\mathrm dx,C\equal{}\int_a^b f(x)g(x)\mathrm dx$ and $ h(\gamma)\equal{}\int_a^b \left(f(x)\minus{}\gamma g(x)\right)\mathrm dx.$ Then $ 0\le h(\gamma)\equal{}B\gamma^2\minus{}2C\gamma \plus{}A$ and $ h(\gamma)$ has a nonpositive discriminant which implies $ C^2\le AB.$ Equality holds if there is a $ \gamma$ with $ f(x)\minus{}\gamma g(x)\equal{}0\;\forall x\in(a,b)\iff f$ and $ g$ are linearly dependent.[/hide]


trigonometry
calculus
integration
complex numbers

[hide="Solution"]$ \cos (n\arctan 2\sqrt 2)\equal{}\text{Re}[(\cos\arctan 2\sqrt 2\plus{}i\sin\arctan 2\sqrt 2)^n]$ by DeMoivre.

Drawing a right triangle with legs $ 4$ and $ \sqrt 2$, we find the hypotenuse to be $ 3\sqrt 2$, and therefore $ \cos\theta\equal{}\frac{1}{3}$ and $ \sin\theta\equal{}\frac{4}{3\sqrt 2}$

So we have $ \text{Re}\left[\left(\frac{1}{3}\plus{}\frac{4i}{3\sqrt 2}\right)^n\right]\equal{}\text{Re}\left[\left(\frac{\sqrt2 \plus{} 4i}{3\sqrt 2}\right)^n\right]\equal{}\frac{1}{3^n}\cdot\text{Re}\left[(1 \plus{} 2\sqrt 2 i)^n\right]$.

$ (1\plus{}2\sqrt 2i)^n\equal{}\sum_{k\equal{}0}^{n}\binom nk (2\sqrt 2i)^n$. These terms are only real when $ n\equiv 0,2\bmod 4$.
In both cases, $ n$ is even, which makes the square root integral, so the real part of this whole thing is some integer $ j$.

Therefore, the answer is $ \frac{j}{3^n}$ for some integer $ j$, which is a rational number. $ \Box$[/hide]

[hide="Aliter"]
Let $ a_n \equal{} \cos n \theta$ where $ \theta \equal{} \tan^{ \minus{} 1} 2 \sqrt 2$

We have the recurrence $ a_{n \plus{} 2} \equal{} 2a_{n \plus{} 1} a_1 \minus{} a_n$

$ a_1 \equal{} \cos \theta \equal{} \frac {1}{3} \in \mathbb{Q}$ and $ a_2 \equal{} \cos 2 \theta \equal{} 2 \cos^2 \theta \minus{} 1 \equal{} \minus{} \frac {7}{9} \in \mathbb{Q}$

The above recurrence tells us that whenever $ a_n$ and $ a_{n \plus{} 1}$ are rational so is $ a_{n \plus{} 2}$

Since $ a_1$ and $ a_2$ are rational, this result is true for all $ a_n$[/hide]


You want $\lim_{x\to2^+}e^\frac{3}{2-x}$ right? The answer is $0$.

I think that spix wanted to say that the answer is $e^{- \infty} = 0$.

Let's be more careful here! Check the one-sided limits.

$\lim_{x\to2^-}e^{\frac3{2-x}}=\infty$

$\lim_{x\to2^+}e^{\frac3{2-x}}=0$


22.  Objects A and B move simultaneously in the coordinate plane via a sequence of steps, each of length one.  Object A starts at (0,0) and each of its steps is either right or up, both equally likely.  Object B starts at (5,7) and each of its steps is either left or down, both equally likely.  Which of the following is closest to the probability that the objects meet: 0.10, 0.15, 0.20, 0.25, 0.30

How do you do this problem?

[hide="Hint"]At which points can the objects meet?  From here, you can use casework.[/hide]

[hide="Solution"]Let us refer to the (0,0) object as A and the (5,7) object as B.

Since there are 12 steps from (0,0) to (5,7) or from (5,7) to (0,0), both objects must move six steps before meeting.  Thus they can meet at (5,1), (4,2), (3,3), (2,4), (1,5), or (0,6).  For (5,1), the probability that A will go there is $\frac{\binom{6}{1}}{2^6}$, since out of the 2^6 possible paths, only $\binom{6}{1}$ (out of 6 total steps, 1 up) will get to (5,1).  Similarly, the probability that B will go there is $\frac{\binom{6}{0}}{2^6}$.  So the total probability is $\frac{\binom{6}{1} \binom{6}{0}}{2^{12}}$.

For (4,2), the prob. of A going there is $\frac{\binom{6}{2}}{2^6}$ and the prob. of B is $\frac{\binom{6}{1}}{2^6}$.

Doing the same for the others, we see that the total probability is
\[\frac{1}{2^{12}} \left( \binom{6}{0}\binom{6}{1} + \binom{6}{1} \binom{6}{2} + \binom{6}{2} \binom{6}{3} + \binom{6}{3} \binom{6}{4} + \binom{6}{4} \binom{6}{5} +  \binom{6}{5} \binom{6}{6} \right)\]
\[= \frac{1}{2^{11}} \left( \binom{6}{0}\binom{6}{1} + \binom{6}{1} \binom{6}{2} + \binom{6}{2} \binom{6}{3} \right)\]
(since the first three terms in the parentheses are the same as the last three)
\[= \frac{1}{2^{11}} (396) \approx .193\]

So 0.2, (C).[/hide]

I don't really know probability or combinatorics, so this is probably wrong
[Hide/]
Consider the paths of particle A and particle B; consider only the paths where A and B end up meeting. Draw these paths in. Notice that since the two meet, they form one continuous path, from (0,0) to (5,7). Notice also that the motions possible alloted to B are the exact opposite of A. This problem then reduces to this problem:

Given a particle A starting at the origin, moving either 1 unit directly upwards or 1 unit directly to the right, find the probability that it hits (5,7).

This is obviously C(12,5)*1/2^12, which is about .1933. So it is closest to .20. Thus, the answer is [b].20[/b]. It is nice that this is consistent with mathfanatics


inequalities
function
LaTeX
inequalities solved

Prove that : abc+bcd+cda+dab<= (1 + 176.abcd)/27
for all positive real number a,b,c,d satisfy: a+b+c+d=1
Who can solved???
I'll help you

now this is another problems that has been troubling me.i'd luv to see a soln now.

I think I have a solution. Here it is:

The inequality was abc+bcd+cda+dab<=(1+176abcd)/27

a+b+c+d=1>=4*(abcd)^1/4 that means that
1>= 256 abcd

the inequality becomes
abc+bcd+cda+dab<=432abcd/27
abc+bcd+cda+dab<=16abcd
We divide with abcd

1/a+1/b+1/c+1/d<=16

Let there be a concave function f(x)=1/x
Using Jensen:
f(a)+f(b)+f(c)+f(d)<=4f((a+b+c+d)/4)

1/a+1/b+1/c+1/d<=4*1/(1/4)=16

Hope it's the corect solution

Hey.Your solution is wrong.
Applying AM-GM, we get
1/a + 1/b +1 /c +1/d >= 16/(a+b+c+d) =16 !!

Try to solv!

[quote="hxtung"]For all positive real number $ a,b,c,d$ satisfy $ a \plus{} b \plus{} c \plus{} d \equal{} 1$, prove that

$ bcd \plus{} cda \plus{} dab \plus{} abc\leq\frac {1 \plus{} 176abcd}{27}$ .[/quote]$ (a \plus{} b \plus{} c \plus{} d)^4 \plus{} 176abcd \minus{} 27(a \plus{} b \plus{} c \plus{} d)(bcd \plus{} cda \plus{} dab \plus{} abc)$

$ \equiv F(a,b,c,d) \equal{} F(a,a \plus{} x,a \plus{} x \plus{} y,a \plus{} x \plus{} y \plus{} z)$

$ \equal{} 5\left(3x^2 \plus{} 4y^2 \plus{} 3z^2 \plus{} 4yz \plus{} 2zx \plus{} 4xy\right)a^2$

$ \plus{} 2\left(7x^3 \plus{} 10y^3 \plus{} 8z^3 \plus{} 14x^2y \plus{} 7x^2z15y^2z \plus{} 25y^2x \plus{} 18z^2x \plus{} \plus{} 21z^2y \plus{} 25xyz\right)a$

$ \plus{} (3x \plus{} 2y \plus{} z)\left(8y^3 \plus{} z^3 \plus{} 12y^2z \plus{} 9y^2x \plus{} 9z^2x \plus{} 6z^2y \plus{} 9xyz\right)\geq0$,

which is clearly true for $ a\leq b\leq c\leq d$.

So who can edit these sollution ?

It's difficult to see so add to Latex is well :)

It's problem from IMO Shortlist 1993. See IMO Shortlist Section of Forum.  :)


Let  ABC be a triangle.Find the minumum value of  $x\geq 0$  such  that
x+sin A/2 , x+sin B/2,  x+sin C/2  form the sides of a triangle.

max of sin C/2 is greater than 1, min of sin A/2 and sin B/2 is 1.Because the 3 results are lengths of sides of a triangle, have :
2x >= x + 1
x >= 1.
so it is 1.

It was the problem 816 of College Math Journal january 2006. Why did you posted it here?

[quote="rem"]max of sin C/2 is greater than 1...[/quote]
Sorry, what do you mean?
I have a quite long solution
[hide="here"]WLOG, $A$ is the largest angle, so we have $\frac{\pi}{3}\leq A<\pi$. We claim that for $\alpha \in \mathbb {R}$, $0\leq \alpha <1$, there exist angle $A$ s.t
$\alpha +sin(\frac{A}{2})=2(\alpha+sin(\frac{\pi -A}{4}))$. Our claim is equivalent to the existence of $A$ s.t
$sin(\frac{A}{2})-\sqrt{2}(cos(\frac{A}{4})-sin (\frac{A}{4}))=\alpha$. But consider $f(A)=sin(\frac{A}{2})-\sqrt{2}(cos(\frac{A}{4})-sin (\frac{A}{4}))$. We have $f(\pi)=1$ and $f(\frac{\pi}{3})=-0.5$. By the continuity of $f$, it follows that our claim is true.
Now it's easy to see that the required value of $\alpha$ is at least 1. Next we will prove that 1 is good. It means, for any triangle $ABC$,
$1+sin(\frac{A}{2})$, $1+sin(\frac{B}{2})$, $1+sin(\frac{C}{2})$ form the sides of a triangle. Because $g(A)=sin(\frac{A}{2})$ is monotonic increasing in $(0,\pi)$, it's enough to show that $1+sin(\frac{A}{2})<(1+sin(\frac{B}{2}))+(1+sin(\frac{C}{2}))$, or equivalently, $sin(\frac{A}{2})<1+sin(\frac{B}{2})+sin(\frac{C}{2})$. But it's obvious since $sin(\frac{B}{2})$ and $sin(\frac{C}{2})$ are positive, while $sin(\frac{A}{2})<1$[/hide]


I would really like to attend Mathcamp, but I just don't have the time to take 5 whole weeks out of summer.  Does anyone know if there is a way to make arrangements to go for just 2 or 3 weeks or so?

yeah a couple people did that last year.

I would only suggest that if you have some other summer program or event with a fixed date that you must go to. Every second of Mathcamp is worth it so I do suggest you stay as long as you can.

But it is definitely possible. Some people came late, some people left early, in fact one person in '04 could only come for the last week.

Yeah. A girl stayed for a week, a bunch of people came a week late an a bunch of people left early. I think that a guy went to RSI and came back for the last two weeks.

I am an international student, and 5 weeks sure seemed long, but once you see what mathcamp is like, you won't want to leave. Also, no matter what you do, don't miss week 5. It's the best with the all nighter and all the extremely cool classes you get to choose.

Officially this is not allowed: the number of students at camp is limited (by dorm space, number of staff members, etc.) and if someone comes for less than the whole came, then that prevents someone else from coming for 5 weeks.

Unofficially, if a good reason arises that someone has to miss perhaps a week of camp, we'll usually allow it.  (Typical examples of good reasons: if MOSP and Mathcamp have a week of overlap; conflict between the end of Mathcamp and the start of school; a family reunion that your parents insist you not miss.)

But if you can only come for two or three weeks, we will almost certainly ask you to apply to Mathcamp again some summer where you are less busy and can commit for the full duration of the camp.  We did have a situation last year where a new student wanted to come for the first three weeks and a 3-time alumnus wanted to come for the last two weeks, so that worked out for everyone, but you can't count on that happening.   I can only think of one other time in recent years when we have allowed a new student to come for just three weeks, and this was because the student was competing at the IMO for the first two weeks of camp.

I had a few more questions about this partial term issue, if you don't mind.

I've heard a lot of awesome things about MathCamp so I'd like to go this year. Unfortunately, I'll be attending a national leadership and debate competition that may last until the 8th of July. (Final date confirmation still pending.) So it could be the 10th before I could get to MathCamp. Should I still apply?

Also, do you have a special tuition rate for students (like me) who would be missing part of the camp?

As far as I know, you should still apply, because I think your competition will qualify as a "good reason" for missing a week of camp. I doubt they have a reduced tuition rate in your case, since you are only missing a week.

To make sure, though, you should either email Mathcamp, Mira, or Dave, or wait until Dave responds here.

Yeah, a few kids came a week late last year, so you will probably be fine (didn't Zhou arrive a little late last year?)

You should certainly still apply -- that's a perfectly valid reason for missing the beginning of Mathcamp.  (If your conference ends on July 8th, though, we would encourage you to arrive at MC by July 9th if at all possible!)

As Yasha anticipates, there won't be any sort of discount if you come late: we wouldn't have any trouble finding someone else who would come (and pay) for the full five weeks, the college wouldn't give us a discount, etc.

After all, the next best thing to 5 weeks at mathcamp is 4 weeks at mathcamp(well, accually 34 days at matchcamp is the next best thing, but 4 weeks is still up there)  :D

*stabs sirclucky*
If you really want to get technical, 34.999999 (etc. etc. etc.). But that happens to equal 35. AAAAAAAaaaaaaaaaaaaAAAAAAAAAAAAAaaaaaaaaaaaaAAAAAAAAAaaaaah.

*stabs drspyder*

stop brining other forum personaitles onto this forum :P(I know treething on another forum... but there hes known as drspyder and im know as sirclucky  :D )

And I consider half a day at mathcamp still a day of mathcamp. 1/2 of infinty is still infinity.

can i sttend for three weeks?
because i cannot attend July 3-July 8 because of EPGY(Stanford) summer program which took my hard work to get in... (although MathCamp was my priority, since EPGY was earlier in deadline, i had to prioritize that)
I cannot attend August 1- August 7, because I am attending Mu Alpha Theta Nationals Competition... im shooting for a national championship this year.... hehe :D

please post ASAP. thanks in advance.

reddevil, regarding your request to come to Mathcamp for three weeks, please see my first post in this topic.

what are the requirements to attend the math camp?

[url]http://www.mathcamp.org/apply.php[/url]
The due date is April 27.

would: 4weeks(because of Mu aLpha Theta Nationals..) be a good reason?

i have to really go to MAT because i have pre-paid and prepared... i wouldnt have registered if i knew about MathCamp a long time ago.


abstract algebra
superior algebra
superior algebra unsolved

This is a problem from Noncommutative Algebra by Farb and Dennis, Springer, 1993 (pg26) which I cannot solve.
Prove that the ring End(M) of R endomorphisms of an R-module M is Artinian if M  is both Artinian and Noetherian.
I can do the special case when M is not indecomposable.

Ha, I just had this problem assigned as homework.

  If M is Noetherian, then it is finitely generated (M Noetherian <=> any submodule of M is finitely generated, also an exercise in Farb and Dennis).

   Let [code] a_{1},....a_{n} [/code] be a set of generators of M then any endomorphism f is completely described by its action on [code] a_{1},....a_{n} [/code].

   Set T:[code]End_{R}(M) -> M^{n}[/code] (where [code]M^{n}[/code] is the R-module given by the direct sum of n copies of M)

      [code]T(f)= (f(a_{1}),....,f(a_{n}))[/code].

  Then it is easy to check that T is a module homomorphism, thus its image, [code]T(End_{R}(M))[/code] is a submodule of [code]M^{n}[/code] .

  Furthermore T(f)=T(g) => f and g agree on the generators hence they are the same => T injective. Then [code]End_{R}(M)[/code] is isomorphic to a submodule of [code]M^{n}[/code] .

    Now from a previous exercise you know that if you have the short exact sequence [code]0-> A^{'} -> A -> A^{"}->0[/code] then [code]A Artinian <=> A^{'}, A^{"} Artinian[/code].

    Taking [code]A=A^{'}+A^{"}[/code] (here + signifies direct sum) you find that the direct sum of two Artinian modules is Artinian, hence [code]M^{n}[/code] is Artinian.

    Finally, taking [code]A^{'}[/code] to be a submodule of A and [code]A^{"}=A/A^{'}[/code], we get a short exact sequence and thus any submodule of an Artinian is an Artinian.

    Combining the last 2 results with the fact that [code]End_{R}(M)[/code] is isomorphic to a submodule of [code]M^{n}[/code] we get that [code]End_{R}(M)[/code] is Artinian.[/code]


geometry
3D geometry
function
number theory proposed
number theory

[color=darkblue] Find all natural numbers, not divisible by $ 10$, which are perfect cubes, and after deleting the last three digits, they are also nonzero perfect cubes.
[/color]

Suppose $ n$ works, $ n \equal{} 1000a \plus{} b$, $ 0 < b < 1000$. (First we have $ 0\leq b$, but if $ b \equal{} 0$, then 10 divides $ n$, so $ 0 < b$)
Then
\[ 1000a \plus{} b \equal{} k^3
\]

\[ a \equal{} j^3
\]

\[ b \equal{} k^3 \minus{} 1000j^3 \equal{} k^3 \minus{} (10j)^3 \equal{} (k \minus{} 10j)(k^2 \plus{} 10kj \plus{} 100j^2)
\]
Since this is more than 0 , $ k^3 > 1000j^3$, or $ k > 10j$. So both terms of are the product are positive, and integers. Since if $ j > 3$, the second term is more than 1000, $ j\leq3$.

Case 1: $ j \equal{} 3$. Then $ k > 10j \equal{} 30$ and $ (k \minus{} 30)(k^2 \plus{} 30k \plus{} 900) < 1000$, obviously no solutions. (the function is increasing in $ k$, and is already too big when $ k \equal{} 21$.
Case 2: $ j \equal{} 2$. Then $ k > 10j \equal{} 20$ and $ (k \minus{} 20)(k^2 \plus{} 20k \plus{} 400) < 1000$, again no solutions.
Case 3: $ j \equal{} 1$. Then $ k > 10j \equal{} 10$ and $ (k \minus{} 10)(k^2 \plus{} 10k \plus{} 100) < 1000$. 11 works, so does 12, but 13 doesn't so again since it is increasing, no solutions above 12.

Summarizing, $ n \equal{} 11^3 \equal{} 1331$ or $ n \equal{} 12^3 \equal{} 1728$.

Let the natural numbers required be $ n\equal{}1000a\plus{}b$, where $ 1 \le b \le 999$
Then $ a\equal{}x^3$ for some natural numbers $ x \Rightarrow n\equal{}(10x\plus{}p)^3$ for some natural number $ p$ and $ 1000x^3 <(10x\plus{}p)^3 \le 1000x^3\plus{}999$
Because $ (10x\plus{}p)^3\equal{}1000x^3\plus{}300x^2p\plus{}30xp^2\plus{}p^3$, we have $ 0<300x^2p<999 \Rightarrow x\equal{}1, p\equal{}1$ or $ 2$ or $ 3$
After checking , $ n\equal{}11^3\equal{}1331$ or $ n\equal{}12^3\equal{}1728$ are the solutions


linear algebra
matrix
superior algebra
superior algebra solved

This one is solved on the forum, posted by Pcalin and solved by Moubinool

cheers!

Lagrangia has a good memory

Here my trick for the memory: some cod-liver oil, this is good for your brain before µath exam

http://www.mathlinks.ro/viewtopic.php?p=4570&highlight=#4570

PiDeltaPhi where did find this nice problem ?

Wich forum ?

Thx for the link.
I found it on another forum and it was unsolved.


Prove that, if three primes(greater than 3) are in A.P. then their common difference is divisible by 2 and 3 both.

we suppose that $ p_i\equal{}p\plus{}ih>3$ (where i\in\{0,1,2\}) are primes.
$ 2|p_2\minus{}p_1\equal{}h$ (because $ p_1,p_2$ are odd).
we have $ p_1,p_2,p_3\equiv \pm 1(mod\ 6)$
then $ 6|(p_2\minus{}p_1)(p_3\minus{}p_2)(p_3\minus{}p_1)\equal{}2h^3$ so $ 3|h$


calculus
derivative
LaTeX
real analysis
real analysis unsolved

Hi,
I'm not sure if this is the right place to post this, but I haven't been able to solve it, and was hoping for some help.  The problem is as stated:

Show that
$\frac{1} {M_x + N_y}$, where $M_x+N_y$ is not identically zero, is an integrating factor of the homogeneous equation $M(x, y)dx+N(x, y)dy=0$ of degree n.

[hide="Hint"]Use it and show that $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$.  We need $M=\Psi_{x}$ and $N=\Psi_{y}$ in order to solve the equation (think of the chain rule).  Therefore $\Psi$ must be $C^{2}$ so its mixed partial derivatives are equal.  That is,
\[ \frac{\partial^{2}\Psi}{\partial x\partial y}=\frac{\partial^{2}\Psi}{\partial y\partial x}. \][/hide]

I know, but it hasn't been turning up exact.  While playing around yesterday though, I found that if we tried setting the integrating factor as $\frac{1}{Mx+ Ny}$ instead, it would turn out to be an exact equation.  What do you think; could this be a typo on the writer's behalf?

Isn't that what you originally posted?  If it works, then it would be an integrating factor (I trust that you took the right steps).

no; in the way it was originally written, $x$ and $y$ were subscripts, denoting the partial derivatives; in this case, the way I found it to work, $M$ and $N$ were multiplied by $x$ and $y$

Oh yeah, sorry, I thought you just forgot to make them subscripts in $\LaTeX$.  Could you post what you did with the first case?


Mr. Jones has 6 children. Assuming that the gender of each child is determined independently and with equal likelihood of male and female, what is that probability that Mr. Jones has more sons than daughters or more daughters than sons?

This is $ 1\minus{}\text{probability that there are 3 male and 3 female}$.  This probability is $ \binom{6}{3}\left(\frac{1}{2}\right)^6\equal{}20 \times \frac{1}{64}\equal{}\frac{20}{64}\equal{}\frac{5}{16}$.  So our answer is $ 1\minus{}\frac{5}{16}\equal{}\boxed{\frac{11}{16}}$.

how is their sol correct? they're assuming that order matters, which has not been specified in the question and usually is not taken into consideration for such kinds of problems...

[hide=Sol]The only other case other than the case that Mr. Jones has more sons than daughters or more daughters than sons is that there are 3 sons and 3 daughters. Ans = 1-(6c3)/2^6 = 11/16[/hide]


trigonometry
logarithms
AMC
AIME
AMC 10
MATHCOUNTS
HMMT

So I'm currently writing a mock AMC 10.

If you're interested in [b]editing/reviewing it[/b] [i]or[/i] [b]taking it[/b] leave a comment below.

[quote="anirudh"]I'll do it as long as it has a tex file.[/quote]

Uh...why do you need a tex file?

[quote="anirudh"]To fix up minor errors instead of creating an errata.[/quote]

you could always get a mod to fix up a minor error.

Besides questions should be edited before posted online.

[quote="rnwang2"]Ok. It's completed. I'm just double checking all the questions and answer choices. I'll probably post it some time later this afternoon.[/quote]
Did you get someone in >10th grade to review it? Methinks that would be a good idea (just in case)... How hard do you think this is going to be for an AMC 10?

I think this may be leaning toward the difficult side for an AMC 10. The reason I did not make it an AMC 12 is because there is no trigonometry or logarithms involved.

Ok here it is. Just PM me the answers. Everyone is welcome to take it.
[b]EDIT: The deadline has been moved to Sunday, January 7.[/b]

[b]EDIT: I have attached a new version with #14 fixed.[/b]

[b]EDIT: I have attached a new version with a clarification of #4 (please download the [i]first[/i] one).[/b]

Wow I just realized how nice #25 really is.

[quote="rnwang2"][quote="SplashD"]the answers to #11 are a decent amount off of what I got as an answer.

Questions you might want to check: #12, #14(the wording with solid and water), #21[/quote]

Thanks, I have fixed #14.

#11, #12, and #21 have no problems.[/quote]

There is definitely something wrong with #12. Let me PM you...

EDIT: Oh nvm, my mistake.

EDIT2: Wait, #17 is confusing. It says the men do not wish to be on the same committee... does that mean we are choosing 2 of them? One with 2 women and 3 men, and one with 4 women and 5 men??

You are picking 3 men from the 8 men. However, 2 of the 8 men do not wish to serve on the same committee.

Yeah, #17 is worded weirdly, but it's technically written correctly.  It took me like 3 readings to get it... :wink:

EDIT: (Didn't wanna spam up the threads so edited) Yea Adeel, that got me too.  I was like WTF.

I see. (The confusion arises from the term "same committee", which implies that there is more than one committee.)

i'm confused....
so are you going to count #24 and #14 or not????

Well, I think #24 is being counted (the problem definitely makes sense, and it is not hard if you have seen the idea before).

[b]Please PM me the answers before noon (central time) tomorrow. [/b]
Then we can start posting solutions to the problems   :) . I will make a separate thread for that.

[quote="bigboy82892"]Ok, it's noon in central time now, so, who got $16,19,20,23,24$? :roll:[/quote]
[hide="16"] $\text{digital root}\equiv\mod9$
$(9^{6130}+2)^{4875}\pmod{9}\equiv2^{4875}\equiv2^{4875\mod6}\equiv2^{3}=\boxed{8}$ [/hide]
[hide="19"] Hard to explain.  :| [/hide]
[hide="20"] Find the coordinates of Bob and Joe in terms of $t$. Find the distance between them by the distance formula; it will be the square root of a quadratic. Find the minimum of the quadratic and square root it. [/hide]
[hide="23"] It's actually a fairly easy problem. Draw a radius to where one of the 6 cm lines intersect the circle. Right triangle with legs $\frac{x}{2}$ and $6$, hypotenuse of length $4+\frac{x}{2}$. This is because the distance from the intersection of the 4 cm line and circle, to the center of the star is a radius. [/hide]
[hide="24"] Didn't get it, but you could plug it into a graphing calculator, unless it's too big to be fractionized. :D [/hide]

I have created a separate thread for solutions.

PLEASE ASK FOR/POST SOLUTIONS [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127421]HERE[/url]

EDIT: lol, lotrgreengrapes, we posted the same stuff in different threads.

Sorry  :blush: Anyway, my answers above were more outlines than solutions.


calculus
integration
real analysis
rudin
real analysis unsolved

How do I calculate this

$ S \equal{} \int_{\mathbb{R}}\sum_{n\in \mathbb{N}} n\cdot \chi_{\left[n,n \plus{} \frac {1}{2^n} \right] }(t) d\lambda$

where $ \lambda$ is the lebesgue measure.
Is there a theorem to make the sum commute with the integral ?

P.S.

I've reached that it should be equal to [url=http://www.mathlinks.ro/viewtopic.php?t=207367]this[/url] but
I'm not entirely sure that I knew what I was doing.

Also why is the lebesgue measure needed here ?
Wouldn't a more "simple" kind of measure do ?
I find Lebesgue measure hard to grasp.

You're probably confusing yourself unnecessarily by thinking about Lebesgue measure.

I don't know what you mean by a simpler measure, but here the Lebesgue integral is equal to the Riemann integral, and there is very little to worry about. (if you want to make this [i]very[/i] rigorous, you need to prove what I just said, of course, but you should review the basic principles of Lebesgue integration first)

The integral you want is $ \int_{-\infty}^\infty \sum_{n\in \mathbb{N}} n\cdot \chi_{\left[n,n + \frac {1}{2^n} \right] }(t) dt$.  I hope that the following steps are each clear, but if they're not, feel free to ask.

$ = \sum_{k=-\infty}^{\infty} \int_{k}^{k+1} \sum_{n\in \mathbb{N}} n\cdot \chi_{\left[n,n + \frac {1}{2^n} \right] }(t) dt$

$ = \sum_{k=1}^{\infty} \int_{k}^{k+\frac{1}{2^k}} \sum_{n\in \mathbb{N}} n\cdot \chi_{\left[n,n + \frac {1}{2^n} \right] }(t) dt$

${ = \sum_{k=1}^{\infty} \int_{k}^{k+\frac{1}{2^k}} k } dt$

${ = \sum_{k=1}^{\infty} \frac{n}{2^n} } dt$

the problem is I've never found an example where it would really matter
that the measure was lebesgue measure and not any other measure.
I mean,an example in which having lebesgue measure instead of any other
measure would make a very big difference.

You keep saying "any other measure," but can you name some?  Are you talking about the Borel measure?

If you are talking about the Lebesgue [i]integral[/i], a related but different concept from Lebesgue measure, then it is a fairly simple to say the difference with Riemann integration.  Whenever a Riemann integral exists, it is the same as the Lebesgue integral, but sometimes the Riemann integral does not exist, while the Lebesgue integral does.  This is precisely why Lebesgue integration was invented.

Here is an example: Take $ f(x)\equal{}\chi_{\mathbb{Q}}$, that is, $ f(x)$ is $ 1$ on the rationals and $ 0$ on the irrationals.  The Riemann integral $ \int_{\minus{}\infty}^{\infty} f(x)dx$ is not defined.  However, the Lebesgue integral is defined and equal to just what you think it should be: $ 0$.

The best place to read about these things is in an analysis text that has a focus on Lebesgue measure.  The last chapter of Baby Rudin, for example, is a thorough introduction.

You can commute the integral and the sum because the summands are all positive; just apply monotone convergence.


Each term of a sequence is generated by adding the two previous terms. What is the sum of the values of the four missing terms in the following sequence?
\\
\[ \minus{}12, \underline{\hspace{0.2in}},  \minus{}5, \underline{\hspace{0.2in}},  \minus{}3,  \underline{\hspace{0.2in}},   \minus{}4,  \underline{\hspace{0.2in}},  \minus{}9, \minus{}14, \minus{}23\]

\[ \minus{}12, \underline{a}, \minus{}5, \underline{b}, \minus{}3, \underline{c}, \minus{}4, \underline{d}, \minus{}9, \minus{}14, \minus{}23\]
We have
\[ \minus{}12\plus{}a\equal{}\minus{}5 \Rightarrow a\equal{}7 \\
\minus{}5\plus{}b\equal{}\minus{}3 \Rightarrow b\equal{}2 \\
\minus{}3\plus{}c\equal{}\minus{}4 \Rightarrow c\equal{}\minus{}1 \\
\minus{}4\plus{}d\equal{}\minus{}9 \Rightarrow d\equal{}\minus{}5\]
The sum is $ 7\plus{}2\minus{}1\minus{}5\equal{}\boxed{3}$


If the average stride of a horse walking is $ 2\frac{3}{4}$ feet and that of a man is $ 2\frac{1}{2}$ feet, how many man strides would be equal to 40 horse strides?


If you can solve equations form $f(x)=0$, where $f$ is a polynomials of degree $4$ then this problem is easy.

[quote="Svejk"]Yes but b=2 isn't the only root  :([/quote]
$b+(7-b^{2})^{2}-11=(b-2)f(b)$. That is all!

[quote="kg2100"]Very easy!! :)

[img]http://img011.photo.21cn.com/photos/user/24454028/20070331231229/o/8339237.jpg[/img][/quote]
I don't understand. What are you doing?  :huh:


Solve for $ x$

$ \sin ^4 x \minus{} \sin ^4 \frac{\pi }{4} \equal{} \sin ^4 (x \minus{} \frac{\pi }{4})$  :D

[quote="addikaye03"][quote="TRAN THAI HUNG"]Solve for $ x$

$ \sin ^4 x \minus{} \sin ^4 \frac {\pi }{4} \equal{} \sin ^4 (x \minus{} \frac {\pi }{4})$  :D[/quote]

LHS: =(sin^2(x)-sin^2(pi/4))(sin^2(x)+sin^2(pi/4))
=(sinx-sin(pi/4))(sinx+sin(pi/4))(sin^2(x)-i^2sin^2(pi/4))
=(sinx-sin(pi/4))(sinx+sin(pi/4))(sin(x)+isin(pi/4))(sinx-isin(pi/4))
therefore x=+-pi/4,arcsin(+-isin(pi/4))

i think i went very wrong actually.. could u bring the RHS to the LHS and let x=sinx, y=sinpi/4, and z=sin(x-pi/4), then expand (x^4-y^4-z^4)...?[/quote]
Just a spammer  :mad:
Does anyone has solution ??? Please help me  :huh:
I don't know how to solve
P.s : I don't spam. I post this just because i want this topic to be on top again  :)

hello ,applying the addition formulas we get $ \sin(x)^4\minus{}\frac{1}{4}\equal{}\minus{}\cos(x)^4\minus{}\sin(x)\cos(x)\plus{}\cos(x)^2\plus{}\frac{1}{4}$, simplifying this we get the equation
$ \frac{1}{2}\plus{}2\cos(x)^4\plus{}\sin(x)\cos(x)\minus{}3\cos(x)^4\equal{}0$.Converting this in $ \tan(\frac{x}{2})$ we get $ \frac{1}{2}\plus{}\frac{2(1\minus{}\tan(\frac{x}{2})^2)^4}{(1\plus{}\tan(\frac{x}{2})^2)^4}\plus{}\frac{2\tan(\frac{x}{2})(1\minus{}\tan(\frac{x}{2})^2)}{(1\plus{}\tan(\frac{x}{2})^2)^2}\minus{}\frac{3(1\minus{}\tan(\frac{x}{2})^2)^2}{(1\plus{}\tan(\frac{x}{2})^2)^2}\equal{}0$. By substituting $ \tan(\frac{x}{2})\equal{}t$and simplifying we have to solve the following equation in $ t$
$ (t^2\plus{}2t\minus{}1)(t^6\plus{}2t^5\plus{}9t^4\minus{}12t^3\minus{}9t^2\plus{}2t\minus{}1)\equal{}0$.
Sonnhard.

[quote="SHIRO_DHKT"][quote="TRAN THAI HUNG"]Solve for $ x$

$ \sin ^4 x \minus{} \sin ^4 \frac {\pi }{4} \equal{} \sin ^4 (x \minus{} \frac {\pi }{4})$  :D[/quote]

$ \sin x \minus{} \sin \frac {\pi }{4} \equal{} \sin (x \minus{} \frac {\pi }{4}) \\
\Leftrightarrow 2\sin \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right)\cos \left( {\frac {{x \plus{} \frac {\pi }{4}}}{2}} \right) \equal{} 2\sin \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right)\cos \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right) \\
\Leftrightarrow \left[ \begin{array}{l} \sin \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right) \equal{} 0 \\
\cos \left( {\frac {{x \plus{} \frac {\pi }{4}}}{2}} \right) \equal{} \cos \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right) \\
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \frac {{x \minus{} \frac {\pi }{4}}}{2} \equal{} k\pi (k \in Z) \\
\cos \frac {x}{2}\cos \frac {\pi }{8} \minus{} \sin \frac {x}{2}\sin \frac {\pi }{8} \equal{} \cos \frac {x}{2}\sin \frac {\pi }{8} \plus{} \sin \frac {x}{2}\sin \frac {\pi }{8} \\
\end{array} \right. \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi (k \in Z) \\
\minus{} \sin \frac {x}{2}\sin \frac {\pi }{8} \equal{} \sin \frac {x}{2}\sin \frac {\pi }{8} \\
\end{array} \right. \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi (k \in Z) \\
\sin \frac {x}{2}\sin \frac {\pi }{8} \equal{} 0 \\
\end{array} \right. \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi (k \in Z) \\
\sin \frac {x}{2} \equal{} 0 \\
\end{array} \right. \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi \\
\frac {x}{2} \equal{} l\pi \\
\end{array} \right.(k,l \in Z) \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi \\
x \equal{} l2\pi \\
\end{array} \right.(k,l \in Z) \minus{} \minus{} \minus{} \minus{} > Bingo!!$
 :roll: So it no need to be pi/4
I prove that i'm not a spammer  :rotfl:[/quote]
I have no idea what problem did you prove??? :huh:

[quote="SHIRO_DHKT"][quote="TRAN THAI HUNG"]Solve for $ x$

$ \sin ^4 x \minus{} \sin ^4 \frac {\pi }{4} \equal{} \sin ^4 (x \minus{} \frac {\pi }{4})$  :D[/quote]

$ \sin x \minus{} \sin \frac {\pi }{4} \equal{} \sin (x \minus{} \frac {\pi }{4}) \\
\Leftrightarrow 2\sin \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right)\cos \left( {\frac {{x \plus{} \frac {\pi }{4}}}{2}} \right) \equal{} 2\sin \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right)\cos \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right) \\
\Leftrightarrow \left[ \begin{array}{l} \sin \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right) \equal{} 0 \\
\cos \left( {\frac {{x \plus{} \frac {\pi }{4}}}{2}} \right) \equal{} \cos \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right) \\
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \frac {{x \minus{} \frac {\pi }{4}}}{2} \equal{} k\pi (k \in Z) \\
\cos \frac {x}{2}\cos \frac {\pi }{8} \minus{} \sin \frac {x}{2}\sin \frac {\pi }{8} \equal{} \cos \frac {x}{2}\sin \frac {\pi }{8} \plus{} \sin \frac {x}{2}\sin \frac {\pi }{8} \\
\end{array} \right. \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi (k \in Z) \\
\minus{} \sin \frac {x}{2}\sin \frac {\pi }{8} \equal{} \sin \frac {x}{2}\sin \frac {\pi }{8} \\
\end{array} \right. \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi (k \in Z) \\
\sin \frac {x}{2}\sin \frac {\pi }{8} \equal{} 0 \\
\end{array} \right. \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi (k \in Z) \\
\sin \frac {x}{2} \equal{} 0 \\
\end{array} \right. \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi \\
\frac {x}{2} \equal{} l\pi \\
\end{array} \right.(k,l \in Z) \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi \\
x \equal{} l2\pi \\
\end{array} \right.(k,l \in Z) \minus{} \minus{} \minus{} \minus{} > Bingo!!$
 :roll: So it no need to be pi/4
I prove that i'm not a spammer  :rotfl:[/quote]

I guess ur just a spammer after all fuckwit. My solution above was an attempt. Im new to AOPS, didnt know if i was right/wrong, now that i see i was wrong im glad i can now see the error of my way, i think u should do the same.

[quote="addikaye03"][quote="SHIRO_DHKT"][quote="TRAN THAI HUNG"]Solve for $ x$

$ \sin ^4 x \minus{} \sin ^4 \frac {\pi }{4} \equal{} \sin ^4 (x \minus{} \frac {\pi }{4})$  :D[/quote]

$ \sin x \minus{} \sin \frac {\pi }{4} \equal{} \sin (x \minus{} \frac {\pi }{4}) \\
\Leftrightarrow 2\sin \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right)\cos \left( {\frac {{x \plus{} \frac {\pi }{4}}}{2}} \right) \equal{} 2\sin \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right)\cos \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right) \\
\Leftrightarrow \left[ \begin{array}{l} \sin \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right) \equal{} 0 \\
\cos \left( {\frac {{x \plus{} \frac {\pi }{4}}}{2}} \right) \equal{} \cos \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right) \\
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \frac {{x \minus{} \frac {\pi }{4}}}{2} \equal{} k\pi (k \in Z) \\
\cos \frac {x}{2}\cos \frac {\pi }{8} \minus{} \sin \frac {x}{2}\sin \frac {\pi }{8} \equal{} \cos \frac {x}{2}\sin \frac {\pi }{8} \plus{} \sin \frac {x}{2}\sin \frac {\pi }{8} \\
\end{array} \right. \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi (k \in Z) \\
\minus{} \sin \frac {x}{2}\sin \frac {\pi }{8} \equal{} \sin \frac {x}{2}\sin \frac {\pi }{8} \\
\end{array} \right. \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi (k \in Z) \\
\sin \frac {x}{2}\sin \frac {\pi }{8} \equal{} 0 \\
\end{array} \right. \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi (k \in Z) \\
\sin \frac {x}{2} \equal{} 0 \\
\end{array} \right. \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi \\
\frac {x}{2} \equal{} l\pi \\
\end{array} \right.(k,l \in Z) \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi \\
x \equal{} l2\pi \\
\end{array} \right.(k,l \in Z) \minus{} \minus{} \minus{} \minus{} > Bingo!!$
 :roll: So it no need to be pi/4
I prove that i'm not a spammer  :rotfl:[/quote]
I guess ur just a spammer after all -. My solution above was an attempt. Im new to AOPS, didnt know if i was right/wrong, now that i see i was wrong im glad i can now see the error of my way, i think u should do the same.[/quote]
Nonsense,
That is my solution. The one who just post a stupid message is a spammer. I post my solution, and that solution is right, at least in my opinion. You are wrong,it 's just your problem. What were you thinking when you said that i spammed.
I just joked after I had posted my solution  :mad:
I'm not angry. If you found something wrong I my solution just post, do not said that someone spammed as a spammer. :mad:

[quote="SHIRO_DHKT"][quote="TRAN THAI HUNG"]Solve for $ x$

$ \sin ^4 x \minus{} \sin ^4 \frac {\pi }{4} \equal{} \sin ^4 (x \minus{} \frac {\pi }{4})$  :D[/quote]

$ \sin x \minus{} \sin \frac {\pi }{4} \equal{} \sin (x \minus{} \frac {\pi }{4}) \\
\Leftrightarrow 2\sin \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right)\cos \left( {\frac {{x \plus{} \frac {\pi }{4}}}{2}} \right) \equal{} 2\sin \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right)\cos \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right) \\
\Leftrightarrow \left[ \begin{array}{l} \sin \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right) \equal{} 0 \\
\cos \left( {\frac {{x \plus{} \frac {\pi }{4}}}{2}} \right) \equal{} \cos \left( {\frac {{x \minus{} \frac {\pi }{4}}}{2}} \right) \\
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \frac {{x \minus{} \frac {\pi }{4}}}{2} \equal{} k\pi (k \in Z) \\
\cos \frac {x}{2}\cos \frac {\pi }{8} \minus{} \sin \frac {x}{2}\sin \frac {\pi }{8} \equal{} \cos \frac {x}{2}\sin \frac {\pi }{8} \plus{} \sin \frac {x}{2}\sin \frac {\pi }{8} \\
\end{array} \right. \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi (k \in Z) \\
\minus{} \sin \frac {x}{2}\sin \frac {\pi }{8} \equal{} \sin \frac {x}{2}\sin \frac {\pi }{8} \\
\end{array} \right. \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi (k \in Z) \\
\sin \frac {x}{2}\sin \frac {\pi }{8} \equal{} 0 \\
\end{array} \right. \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi (k \in Z) \\
\sin \frac {x}{2} \equal{} 0 \\
\end{array} \right. \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi \\
\frac {x}{2} \equal{} l\pi \\
\end{array} \right.(k,l \in Z) \\
\Leftrightarrow \left[ \begin{array}{l} x \equal{} \frac {\pi }{4} \plus{} k2\pi \\
x \equal{} l2\pi \\
\end{array} \right.(k,l \in Z) \minus{} \minus{} \minus{} \minus{} > Bingo!!$
 :roll: So it no need to be pi/4
I prove that i'm not a spammer  :rotfl:[/quote]
Why do you have $ \sin x \minus{} \sin \frac {\pi }{4} \equal{} \sin (x \minus{} \frac {\pi }{4})$ ? My equation is $ \sin ^4 x \minus{} \sin ^4 \frac{\pi }{4} \equal{} \sin ^4 (x \minus{} \frac{\pi }{4})$
 :) Let take it easy guys,It's not neccessary to argue right? :)

Pardon my double post, but I made a mistake in my observation of Dr Sonnhard Graubner's solution, as tan 22.5 solves this equation, which is half pi/4.

[quote="gurmies"]Hey there, by inspection it's evident that x = π/4 is a solution to the equation.  This means that t = 1 has to be a solution to the polynomial expression provided above by Dr Sonnhard Graubner.  However it's not, and unless i'm mistaken here, Dr Sonnhard Graubner may have made an error.

My attempt is as follows:

sin^4(x) - sin^4(π/4) = sin^4(x - π/4)

                              = [sin(x).cos(π/4) - cos(x).sin(π/4)]^4

                              = 1/4[sin(x) - cos(x)]^4

So,

sin^4(x) - 1/4 = 1/4[sin(x) - cos(x)]^4

4sin^4(x) - 1 = sin^4(x) - 4sin^3(x)cos(x) + 6sin^2(x)cos^2(x) - 4sin(x)cos^3(x) + cos^4(x)

4sin^4(x) - [sin^2(x) + cos^2(x)]^2  = sin^4(x) - 4sin^3(x)cos(x) + 6sin^2(x)cos^2(x) - 4sin(x)cos^3(x) + cos^4(x)

4sin^4(x) - sin^4(x) - 2sin^2(x)cos^2(x) - cos^4(x) = sin^4(x) - 4sin^3(x)cos(x) + 6sin^2(x)cos^2(x) - 4sin(x)cos^3(x) + cos^4(x)

2sin^4(x)  + 4sin^3(x)cos(x) - 8sin^2(x)cos^2(x) + 4sin(x)cos^3(x) - 2cos^4(x) = 0

Now dividing by 2cos^4(x) produces:

tan^4(x) + 2tan^3(x) - 4tan^2(x) + 2tan(x) - 1 = 0

y^4 + 2y^3 - 4y^2 + 2y - 1 = 0

If P(y) = y^4 + 2y^3 - 4y^2 + 2y - 1

P(1) = 0 ======> As per initial inspection

Hence, y - 1 is a factor and after performing long division,

(y-1)(y^3 + 3y^2 - y + 1) = 0

The problem lies right there, the expression y^3 + 3y^2 - y + 1 definitely has one real root, which I cannot find, but can merely approximate.  So i'm not sure whether my method is completely wrong, that's for you judge.  Also worthy of mention is the almost symmetrical arrangement of coefficients of y^4 + 2y^3 - 4y^2 + 2y - 1, perhaps it can lead to a solution.  Any help appreciated![/quote]
With Cardano, you can easily find the root of that equation.  :|

hello, note that i have made the substitution $ t\equal{}\tan(\frac{x}{2})$ and solving $ t^2\plus{}2t\minus{}1\equal{}0$ gives us
$ \tan(\frac{x}{2})\equal{}\minus{}1\plus{}\sqrt{2}$ or $ x\equal{}\frac{\pi}{4}\plus{}2k_1\pi$, with $ k_1 \in \mathbb{Z}$
$ \tan(\frac{x}{2})\equal{}\minus{}1\minus{}\sqrt{2}$ or $ x\equal{}\minus{}\frac{3\pi}{4}\plus{}2k_2\pi$, with $ k_2 \in \mathbb{Z}$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, note that i have made the substitution $ t \equal{} \tan(\frac {x}{2})$ and solving $ t^2 \plus{} 2t \minus{} 1 \equal{} 0$ gives us
$ \tan(\frac {x}{2}) \equal{} \minus{} 1 \plus{} \sqrt {2}$ or $ x \equal{} \frac {\pi}{4} \plus{} 2k_1\pi$, with $ k_1 \in \mathbb{Z}$
$ \tan(\frac {x}{2}) \equal{} \minus{} 1 \minus{} \sqrt {2}$ or $ x \equal{} \minus{} \frac {3\pi}{4} \plus{} 2k_2\pi$, with $ k_2 \in \mathbb{Z}$
Sonnhard.[/quote]
I can't understand you Dr.


Did anyone watch House last night? Because I didn't quite understand the ending. What was the deal with the "you're in for an all-nighter" thing? Was it the season finale? Because it was either extremely anti-climactic or I'm really missing something substantial. Also what happened to the kid?

I'm confused...

House figured out that the kid had too much iron in his blood or something, and that he would have to get it drained once in a while for the rest of his life.

After that, House didn't tell Foreman that the kid had already been diagnosed, and he lets Foreman keep doing the tests. I still haven't figured out why he did that, though... Maybe House was mad at Foreman for accusing him of canceling the interview?

That wasn't the finale; their will be at least one more episode.


Solve for x E [0,2pi]: sin(3x) - sin(x) = cos(2x)

let's expand this

sin(3x) - sin(x) = cos(2x)
=> sin(2x)cos(x) + sin(x)cos(2x) - sin(x) = cos^2(x) - sin^2(x)
=> [2sin(x)cos(x)]cos(x) + sin(x)[cos^2(x) - sin^2(x)] - sin(x) = cos^2(x) - sin^2(x)
=> 2sin(x)cos^2(x) + sin(x)[cos^2(x) - sin^2(x)] - sin(x) = cos^2(x) - sin^2(x)

we have a lone sin(x), so let's change everything to sin(x)

=> 2sin(x)(1 - sin^2(x)) + sin(x)[1 - sin^2(x) - sin^2(x)] - sin(x) = 1 - sin^2(x) - sin^2(x)
=> 2sin(x) - 2sin^3(x) + sin(x) - 2sin^3(x) - sin(x) = 1 - 2sin^2(x)
=> 2sin(x) - 2sin^3(x) + sin(x) - 2sin^3(x) - sin(x) - 1 + 2sin^2(x) = 0
now this is ugly, let's try and simplify this a bit

=> -4sin^3(x) + 2sin^2(x) + 2sin(x) - 1 = 0
this is a cubic in sin(x), to cut down on typing for the mean time, let's write sin(x) as y, we get:

-4y^3 + 2y^2 + 2y - 1 = 0 ........ah, that looks a little better
=> -2y^2(2y - 1) + (2y - 1) = 0
=> (2y - 1)(1 - 2y^2) = 0
=> y = 1/2 or y = +/- sqrt(1/2)

but, y = sin(x)
=> sin(x) = 1/2 or sin(x) = +/- sqrt(1/2) = +/- 1/sqrt(2) = +/- sqrt(2)/2
=> x = pi/6, 5pi/6 or x = pi/4, 3pi/4, 5pi/4, 7pi/4

The problem is that when i graph this to check, the only valid answers are pi/6 and 5pi/6. Could someone check this for me.

THANKS

I plugged in the values wrong into my calculator. They are correct. Sorry for the confusion.

[hide="alternatively..."]
\[\sin 3x=\sin (2x+x)=\sin 2x\cos x+\cos 2x\sin x\]
\[\sin x=\sin (2x-x)=\sin 2x\cos x-\cos 2x\sin x\]
so that the equation becomes
\[2\cos 2x\sin x=\cos 2x\]
Either $\cos 2x=0$ or $\sin x=\frac{1}{2}$. You should be able to figure out the solutions from here...
[/hide]

[hide]$\implies 3\sin x-4\sin^{3}x-\sin x=\cos 2x$
Let $\sin x=y$
$\implies 4y^{3}-2y^{2}-2y+1=0$.
Solve: Using Rational Roots Thm, 1/2 is a root, and the other two roots are $\pm\sqrt{2}/2$.

Since $\sin x=1/2, \pm\sqrt{2}/2$, the solutions in $[0,2\pi]$ are $\pi/4, \pi/3, 2\pi/3, 3\pi/4, 5\pi/4, 7\pi/4$.[/hide]


[quote="jacob8888888"]An arithmetic series is called a concatenation series if the sum of the serries is represented by the concatenation of the first and last terms. For example, [b]17[/b]+19+21.....+[b]85[/b]=[b]1785[/b]. Find a concatenation series with 41 consecutive intergers and a four digit sum. What is the sum of the integers of this series?

Thanks[/b][/quote]

[hide]
Let:
$a=$ the first number in the series
$b=a+40$ (the last number in the series)
$s=$ the sum of the numbers in the series
Using the formula for the sum of an arithmetic series,
$41\frac{(a+b)}{2}=100a+b$
$41(a+20)=101a+40$
$41a+820=101a+40$
$780=60a$
$a=13$
So the sum of the numbers in the series is 1353.
[/hide]


[b](1) If 2 water tap fill a tank in 75/8 hours. If a water tap of larger diameter has fill the tank 10 hour less than the first tap.
then find their seperate time to fill the tank.[/b]

let  the  volume  coming out of the  small tap be  x  lt per hour  and let the volume of water coming out of bigger tap be  y lt per hour.

               75/8 (x+y) = capacity of the tank.

le the  small tap take   `s`  hours to fill the tank. then   big tap  taes  s-10 hours to fill the tank.

                    xs= 75/8(x+y)    and  y(s-10)=75/8(x+y)

            now u have 3 equations  and 3 unknowns.  solve  them  to obtain the required answers


geometry
function
search
blogs
angle bisector

Hi everyone,

Do you know any program that can be used to draw diagrams in Geometry (with the functions of locating midpoints, midlines, angle bisector) that is free? (got from Internet.) or having a "good" trial time or or cheap one?

Thanks in advance.

I've seen this question on AoPS/ML at least 3 or 4 times.  Try a search of the forum.

rrusczyk was talking about a new program, [url]http://www.geometer.org/geometer[/url], recently on his blog, and it seems pretty good.  Geometer is free, as is KSEG ([url]http://www.mit.edu/~ibaran/kseg.html[/url]), which is also good, though somewhat simpler and less powerful.

I suggest buying Geometer's Sketchpad.  It is so much easier to use than anything else out there.  I was using KSEG until my friend gave me Geometer's Sketchpad.  It is SO much better, even though it is similar.

I have heard of Sketchpad, but I dont have money to buy it, and I dont know anyone having it (except for the teachers, but they wont let me borroe it.)I'll just try the KSEG then. Thank you for all replies.

You can download Geometers sketch pad(trial) in its site.

http://mathsrv.ku-eichstaett.de/MGF/homes/grothmann/java/zirkel/doc_en/
I think this program is great and its FREE :D

Yeah, C.a.R it's my favourite, it's very useful. I think it's better than Cabri and many others.  :D


geometry
3D geometry
combinatorics proposed
combinatorics

divided a square into any smaller square prove that there is at least two equal square .

I wouldn't bet on that:

[url]http://mathworld.wolfram.com/PerfectSquareDissection.html[/url]

Did you conjecture this?

but i think a similar thing with cubes would be a correct statement..... :?


What'd you get at chapter?

Just trying to estimate my position at states this Sat.

EDIT: please reply with your chapter location, thanks!

EDIT 2: I got a 31 and got second, while first in my chapter was 33. I think our chapter is very weak.

Using my various contacts :lol: I know that one of the chapters was won with a 35.

Our chapter won with a 45  :wink:

Are you toledo? I think David Li won in toledo. This is cleveland.

Mr. Frost Sir: Which one? Cincinnati?

I got 13th in Cincinnati (my first time participating), but am not sure of my score: It's between 27 and 29.

I've improved a bit since then.. hoping to make Nationals some how :P

Sycamore is Godly :(

isn't sycamore a national private school?

Whichever chapter Sycamore is in, they had top 6 scores and 7 of top 8 overall.

mr. frost, I just hope that the contact you used wasn't my coach.

sycamore definitely is a private school, there's one in Indianapolis, IN. I think mky dad told me that sycamore was national though. It makes sense, since it costs alot and the level of the teachers.

All I know is the best MathCounts school in Ohio = Sycamore :P (It's Public in Cincinnati)
==========================================
Hmm.... could someone from Ohio, who participated last year, do the [b]2003[/b] State Target and Sprint, and tell me:

1) What they got on it.
2) How much I should be able to get if I stand an ~okay~ chance (60-70%) of making it to the nationals??

I hope this isn't too much :)

I went to the Akron University chapter competition. I'm from Copley Fairlawn Middle School. We and Revere had the top 10 places in our chapter.

[quote="hoshattack"]i am not sure which chapter I am in for Ohio, but I know I got in the 20's  :(  I still made it to state.  Unfortunately this is my first and last year of mathcounts.  :mad:[/quote]
Same.

I got 27~29 and barely made it to state from the Cincinnati chapter.
Again, my 1st and last year. :(

=( for both of you. This is my 2nd year (im an 8th grader now)
Last year I got 1st at chapter with a 39 overall. At state I got 62nd

[quote="frost13"]Official word : Cut off last year in Ohio was a 37(23/14).[/quote]
Cut-off for the nationals??

YAY!!  :w00t:

I just got that exact score on last years State, meaning I have a chance to go to the nationals :lol:

Thanks!

[quote="yhh3002"]dude, our chapter is much more competitive than yours...

the scores were

1=45
2=45
3=44
4=43
5=40
6=39
7=38
8=37
9=36
10=35[/quote]

Um, that is a little bit off.

1=45, 2=43, 3=41, and the scores go from there.  10 was indeed a 35.  This is the hardest chapter in Michigan, and usually the national qualifiers come from here.

Ummm...lol

You guys should move to South Carolina:

Our Chapter:

1st (me): 33
2nd: 32
3rd: 29
4th: 25
5th: 24...

You guys get the point lol

Too good..  :|

One of the best Chapters in the nation, from those scores...

Yah, but our chapter kinda stinks because our people only know how to do chapter competitions. They can't do state levle :(

is that chapter in ohio? if so, where/which one?

i just noticed, but isn't there already a thread asking for how ppl did in chapter?

well thats good, i'll probably bomb it and get like 200th place...
or I may get lucky and go to nats! :D

oh, and mwpl11, this is specifically for Ohio, because not many ppl acutally visit their state forum.

last year I got 16/3

for a grand total!!!!! 22 points lol

I was like 200 something place

This year I'm going for nats!

[quote="hoshattack"]
oh, and mwpl11, this is specifically for Ohio, because not many ppl acutally visit their state forum.[/quote]

oh, oops :oops: .

See you guys tomorrow!!

Good Luck to Everyone!!  :thumbup:


algebra
polynomial
group theory
superior algebra
superior algebra unsolved

thank you!

it was clear that the proof consists of simple algebraic manipulations, but how to find them ...  :maybe: is there a sort of topological motivation?

[quote="-oo-"] how to find them?[/quote]
I have heard that Gröbner bases is a method to find some kind of relations in algebraic structures. However, I'm not very familiar with those bases. Gröbner bases are discussed for example in Eisenbud's book "Commutative algebra with a view toward algebraic geometry".

Gröbner bases are there to find out if a polynomial lies in an ideal generated by some other polynomials... over a commutative ring, or more likely, over a field. Is this really of any help for a (noncommutative) group theory problem? I cannot name sources, but I think I heard such problems can even be undecidable...

  darij


From a set 2n+1 weights(where n is natural number), if any one weight is excluded, then the remaining 2n weights can be divided into 2 sets of n weights that balance each other. Prove that all the weights are equal

This question is often phrased using sports.  I searched for "football" and found the most recent version that I've commented on:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=275557
(Following links from there should lead to at least one solution.)
Searching for "japan 1999 weight" works pretty well, too:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=27219


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

I can show that for a local noetherian ring $ (A, \mathfrak{m})$ that $ \hat{A}$ is flat over $ A$. How to show that it's faithfully flat? What for non-local rings?

Does the following make any sense?

[quote]
It suffices to show that $ \mathfrak{m}\hat{A} \neq \hat{A}$. Assume that $ \mathfrak{m}\hat{A} \equal{} \hat{A}$ for $ \hat{A} \neq 0$, then according to NAK there is an $ x \in \hat{A}$ with $ x \equiv 1 \ mod (\mathfrak{m} \hat{A})$ such that $ x\hat{A} \equal{} 0$. But $ x \equal{} 1 \plus{} m$ is a unit in $ \hat{A}$ for the $ \mathfrak{m}$-adic completion, so $ x\hat{A} \equal{} \hat{A} \equal{} 0$.
[/quote]

Still not sure about that...  :blush:


Putnam
number theory proposed
number theory

Find the last digit of this numbre
$\left[\frac{10^{20000}}{10^{100}+3}\right]$.

This is the exact same problem as Putnam 1986 A2.


Let f(n) denote the sum of (all) digits of natural number n. Example: f(2004)=2+0+0+4=6. Prove that for each natural n we can choose convenient value of natural parameter p such that the equation f(npx)=f(x) has solution in natural numbers x that doesn't contain any "9" in its notation (of course in base 10).

any ideas about this one.. i have no clue.
i think this is advanced but if its easy u can move it..

[edit]  Thanks!

Here's one thought -- every natural has multiples made up exclusively of 0's and 1's.  You could construct your 0's and 1's in such a way as to force the existence of a root, perhaps.  Multiples of 9 seem like promising numbers to be converted in this way.


trigonometry
Putnam
function
algebra
polynomial
integration
algebra unsolved

Prove that:

$cos(x) = \frac{e^{ix} + e^{-ix}}{2}$

using de Moivre's theorem: $cos(x) + isinx = e^{ix}$

This was a [b]Putnam[/b] problem?   :huh:

$e^{-ix} = \cos (-x) + i \sin (-x) = \cos x - i \sin x$
$e^{ix} + e^{-ix} = 2 \cos x$

Alternately, if you want to be smart about it,

$e^{-ix} = \frac{1}{\cos x + i \sin x} = \frac{\cos x - i \sin x}{\cos^2 x + \sin^2 x} = \cos x - i \sin x$

Edit:  [b]Remark:[/b]  The result above, while simple, is useful in allowing us to quickly define trig and arctrig functions on complex values.

Furthermore, we can set any arbitrary complex number $z = e^{ix}$ (without adding an $r$ term, because we can have $x$ complex) and arrive at the conclusion

$2 \cos x = z + \frac{1}{z} \Rightarrow 2 \cos nx = z^n + \frac{1}{z^n}$,

An easy proof of the result that $z^n + \frac{1}{z^n}$ can be written as a polynomial in $z + \frac{1}{z}$.

Agg, it's too simple.  Thanks why didn't I see it before  :rotfl:

It wasn't a Putnam problem at all. The real Putnam question is the following one:

For which $n \in \left\{1,2,\,\cdots,10\right\}$ is $\int_{0}^{2\pi} \cos x \cos 2x\,\cdots \cos nx \,dx \neq 0?$


For Algebra2/Trig I am having difficulty doing the homework, I don't understand how to solve it.

Does anyone know of a site that gives tutorials for this subject?
Or anyone who is willing to try and bare with me to help me out on this?

I need help with Recursive and Explicit solving by the way...

Thanks
    -totally_stupid? (aka, Kira)

Can you post a specific question?
The AOPS wiki (link on the left) has tutorial-like pages on lots of things

ya kno, your teacher is an efficient resource...


LaTeX
combinatorics unsolved
combinatorics

Junior Balkan MO 2008, Problem 4

A [b][i]4*4[/i][/b] table is divided into [i]16[/i] white unit square cells. Two cells are called neighbors if they share a common side. A move consists in choosing a cell and the colors of neighbors from white to black or from black to white. After exactly [i]n [/i]moves all the  cells were black. Find all possible values of[i] n[/i]. :?:

Posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=211773  (I searched for "JBMO 2008" (without quotation marks) to find this thread.)

obviously n can be any number such that $ n>\equal{}4$(there is an easy choice of 4 squares that makes the whole square black).
now all we have to do is prove that with 3 moves we cannot achieve a entirely black square.
but this is obvious, since each move changes at most 5 squares(we then have that using 3 moves we change at most 15 squares,absurd). Done. :D

[quote="v235711"]obviously n can be any number such that $ n > \equal{} 4$ (there is an easy choice of 4 squares that makes the whole square black).[/quote]  In LaTeX, \geq gives $ \geq$.  I'd be curious to see your solution with $ n \equal{} 5$  :roll:

oops, sorry , I have a mistake in my solution. it is not for all n>=4, it is just for n even>=4...
I still have to find out whether there is not a sequence of moves using an odd number of moves...
sorry for my mistake :oops:
but can we color the table black using 5 moves?what is the least odd n required? :maybe:

If you read the link I posted, it is shown that no odd $ n$ is possible: note that every time you flip any square, you flip exactly one of the squares in positions (1, 2), (2, 4), (3, 1) and (4, 3).


What is the purpose of a math jam?  What are they used for?  How often do we have math jams?

[quote="Nerd_of_the_Ages"]math jams are for various purposes. For example, the math jam that's going on right now while I type is about the upcoming "Introduction to Number Theory" class. They're usually used to discuss math problems or upcoming classes. Math jams are pretty rare, although recently there have been a lot, because of all the upcoming classes.[/quote]

Nice description, But they are also about theories or special things, like HSSiM(Pardon me if I misstyped it)

whoops, meant to say intro to algebra, not intro to number theory


bob drives to work on the highway. he averages 45 mph for the trip. when he drives the back route he averages 36 mph. the trip on the back road is 3 miles shorter and takes 5 more minutes than the highway
route. how many miles long is the highway route??

[hide] Using $ d=rt$, we can set up a system of equations.

$ d=45t$
$ d-3=36(t+\frac{1}{12})$

Solving the second equation,

$ d-3=36t+3 \implies d=36t+6$.

Substituting,

$ 45t=36t+6 \implies t=\frac23$.

Thus,

$ d=45(\frac23)\implies d=\boxed{30}$.[/hide]

[b][size=150]Bubba the bicyclist goes for a long distance ride. three hours later, bubbete decides to bring him lunch, so she hops in her car and zooms down the road at a blistering 42 mph.
If it takes her an hour and a half to catch Bubba, how fast was she riding?[/size][/b]

[hide]
Bubba distance = bubbete distance.

x*4.5 = 42*(1.5)

x=14.
[/hide]

[color=green][size=200][b]THX I UNDERSTAND IT NOW. :)[/b][/size][/color][/b]

[b][size=150]Two trains are 300 miles apart and traveling toward each other on parrallel tracks. One travels at 35 mph and the other 45mph. At the front of the faster train is a bee that flies exactly 72mph. The bee flies to the front of the slower train, then instantaneously turns around and flies to the front of the faster train. The bee continues flying from the front of one train to the other until the trains pass. At the exact moment the trains passed, what was the total distance the bee flown[/size]?[/b]

Don't be posting homework for other people to solve!
I go to the same school as you guys and have the same Teacher!
Do your own homework, Just like I did!

audilover, do not post Lomes problems or AIME15 will stalk you. I go o your school HPMS

BTW, are you guys both 7th graders(audilover&sheeprule), cause only 7th graders could be stupid enough to post homework! :rotfl:

[b]I[size=150] apologize. Except, I wanted to share this amazing problem with people. I can't do that? What if I already know how to solve it? I can't post it becuase it's a problem from my homework? Anyways, I am an eigth grader so you just got PWNED. Who is this anyway? Nikil? or somebody? and who is AIME15? and ALSO! WHo posts things at 8 in the morning? That is truly sick.[/size][/b]

Turn down the font size.

Also, DO NOT post homework unless you have tried it, and SHOW that you have tried it.

Maybe you should also learn to spell your teacher's name (@mathlearner2012) :wink:

[b]WELL, just to let you know, I post things in big font for the people who have trouble reading small font. SInce most people that go on this website have glasses and bad eyesight, I post it bigger so that more people have the opportunity to be enlightened. AND, if you are AIME15 then who in the world is AIME15USAMO?!! You both have the same picture and same location. Its either somebody is a poser or you guys are married or related.
And now let's end all this and focus on math again. :)[/b]

Are you blind? AIME15 and AIME15USAMO have completely different locations and pictures. And I don't think big font will enlighten people anymore than AOPS already does. And its weird how you live in China, but have get the same math problems as I do.  :wink:  :wink:


This is my first post here, salutes to everybody :).

Well, how could i obtain "alpha" from this equation:

[color=blue]Rho = (A * tan [ (90- alpha) * (PI/ (2*180)) ] ) / ( [(1-n)/(1+n)]^(Ec/2) )
[/color]

Where "n" is:[color=blue]n = Ec * sin( alpha * (PI/180) )[/color]

And:[color=blue]
"A" and "Rho" are known values.
"Ec" is a constant value: Ec = 0.08181.[/color]

Some help would be pretty much appreciated, i'm being stucked with this for two days :?

Edit: Hm... i'm not sure if this post should've been posted in this thread, sorry if not.  In such case, please, somebody tell me where i should post it, or move it. :blush:

Well,  this problem seems to be quite difficult, at least for me.
I've could not dedicate to this problem many time, i've been very busy with a lot of other more urgent things.

The only advance i've been able to do in the little time i've had available for this (this evening concretely ;) ) is shown below, but first i have to apologize, because the equation i posted originally was not totally correct, so i sorry indeed if someone has been trying to solve it.

Here i go with the original, and correctly wrote this time, equation:
[color=blue]
Rho = A * [ tan [ (90- alpha) * (PI/ (2*180)) ]  /  [(1-n)/(1+n)](Ec/2) ] N
[/color]
Where "n" is: [color=blue]n = Ec * sin( alpha * (PI/180) )[/color]

And:[color=blue]
"A", "Rho" and "N" are known values.
"Ec" is a constant value: Ec = 0.08181.
[/color]
And here are the steps i've done today:
[color=blue]
1.-  Rho/A =  [ tan [ (90- alpha) * (PI/ (2*180)) ]  / [(1-n)/(1+n)](Ec/2)   ] N

2.-  [Rho/A]1/N =  tan [ (90- alpha) * (PI/ (2*180)) ]  /  [(1-n)/(1+n)](Ec/2)

3.- [Rho/A]1/N =  tan [ (45- alpha/2) * (PI/ 180) ]  /  [(1-n)/(1+n)](Ec/2)

4.- [/color] (PI/180) is there to convert the angle into Radians, so i can get it out of there:[color=blue]

[Rho/A]1/N =  tan [ (45- alpha/2)  ]  /  [(1-n)/(1+n)](Ec/2)

5.- [/color]We have that :[color=blue]
tan(45- alpha/2) = ( tan (45) - tan (alpha/2) ) / ( 1 + tan(45)*tan(alpha/2) ) = (1 - tan(alpha/2) ) / ( 1 + tan(alpha/2) )  , [/color]aplying this result we have:
[color=blue]
[Rho/A]1/N =  [(1 - tan(alpha/2) ) / ( 1 + tan(alpha/2) )]  /  [(1-n)/(1+n)](Ec/2)

6.- [/color] I'm going to substitute "n" getting out (PI/180), so now[color=blue]  n = Ec * sin( alpha )[/color]
And then:
[color=blue]
[Rho/A]1/N =  [(1 - tan(alpha/2) ) / ( 1 + tan(alpha/2) )]  /  [(1 - Ec*sin( alpha ))/(1+ Ec*sin( alpha ))](Ec/2)  [/color]

This final equation still looks menacing :mad: , i'm not sure about persisting in this way, but perhaps somebody could see something in it  i cannot at this moment, i hope that :).

hi, i am quite sure that you posted in a wrong forum. May any moderator move this topic.


This is in a book. I dont know how to organize the information, and also i dont know how to use the info about a loss in the share price, please if you can help me.

An investor has 3 shares, Delta Airlines, Hilton Hotels and Mcdonalds. He says that 2 days ago the shares went down 350 dollars and yesterday they went up 600 dollars. His stock broker remembers that 2 days ago Delta lost 1 dollar per share, Hilton lost 1.5 dollars, and Mc gained 0.5 dollars. He also remembered that yesterday Delta gained 1.5 dollars, Hilton lost again 0.5 dollars and Mc gained 1 dollar. Is it possible to know with this info how many shares of each company has the investor?
If the investor says he has 200 shares of Mc, how many does he have of the other 2?

Thanks to anyone who can give me a hand!

Note that all that information amounts to two equations and three variables.  you have an equation of the form (a1)x + (b1)y + (c1)z = -350
and an equation of the form (a2)x + (b2)y + (c2)z = 600
So with this information alone, it is not possible.  If you are given one of x, y, and z, it is of course possible to solve for the other two.  (unless the equations happen to degenerate, which I doubt happens)
Hope this helps.
But I don't think this is olympiad level.


When three lines intersect a circular region, what is the largest number of regions that can be formed within the circle?

To ensure the greatest number of regions with 3 lines, we should draw construct a triangle using those 3 lines. If you draw the figure out, there are 3 regions with angles vertical to the triangles' angles, 3 regions between the other 3 regions, and 1 triangle. That totals to 7 regions.

I made a picture of it. Sorry if you have to download it, and i hope it works!


in the cyclic pentagon $ ABCDE$ we have  $ AC\parallel DE$ and $ <AMB\equal{}<BMC$ where $ M$ is the midpoint of $ BD$.prove that the line $ BE$ cuts the segment $ AC$ into two equal parts

let CM meet the circumcircle at P, so we have $ \angle PMD \equal{} \angle BMC \equal{} \angle AMB$, then $ PM \equal{} MA$, then $ DP \equal{} AB$, then $ \angle PCD \equal{} \angle ABC$, then CA is the C-symmedian of $ \triangle BCD$, then ABCD is an harmonic quadrilateral and so BD is the B-symmedian of $ \triangle ABC$ and BE (its B-isogonal) the B-median of $ \triangle ABC$.


Alcumus
videos
geometry
number theory
Support

Previously, I asked whether and when new videos would be available.  Now, I am running a poll to see which topic would win the "popular vote".  Also, it would be a big help if a moderator or administrator could provide some input about what's next.  Personally, in my opinion, an algebra or number theory video section would work nicely, or videos regarding AoPS Volume 1.  Please post your opinions here.  :D

[quote="FAQ"]Will there be other subjects included in Alcumus?

Yes. We are developing material for Alcumus that supplements the rest of our texts, beginning with Introduction to Algebra. There is currently no available timeframe estimate as to when the Algebra problems will be ready. [/quote]

Reading the FAQ is useful :wink:

Blargh I'd like it if some Intermediate-level stuff was in Alcumus.

Thanks AIME15.  I kinda knew they were starting on Algebra, but wanted to know if they'd finish anytime soon.  And it seems like not, unfortunately  :( .  I also wanted to see the opinion of other people and if there would be any other plans(I could use a good number theory set of videos).  But the way it seems, Alcumus will be standing still and not much will happen for a while(unfortunately).

It wasn't a complaint, I was just saying.  I just hope there'll be new problems before 2010(plus videos!)

As of today at 12:00 P.M., Problem Solving Techniques and Algebra I/II are in the lead with 2 votes each.  Everything except, "Algebra III", "Contest Specific Advice", and "Other" has received at least 1 vote.

Now number theory has 2 (because of me). I think number theory is good AFTER Algebra. This is because it is the order that is recommended with the introduction series. Next, Geometry would be good.

With 5 and 4 votes respectively, the current leaders are "Algebra I/II" and "Problem Solving Techniques".  If your looking at this and have not voted, PLEASE VOTE NOW. Thank you!  :lol: