Suppose that give us the following information about the signal $ x (n)$

[b] i) [/b] $ x(n)$ is a real signal and torque.
[b] ii) [/b] $ x (n)$ has period $ N \equal{} 10$ and Fourier coefficients $ a_k$.
[b] iii) [/b] $ a_ {11} \equal{} 5$.
[b] iv) [/b] $ \displaystyle\frac{1}{10} \displaystyle \sum_ {n \equal{} 0} ^ 9{| x (n) | ^ 2} \equal{} 50$.

Show that $ x (n) \equal{} A cos (Bn \plus{} C)$, and specify the numerical values of the constants $ A$ $ B$ and $ C$.

Greetings :D

Conventions for the definition of the Fourier transform vary, notably in the placement of $ 2\pi.$ Can you give us the definitions of Fourier transform and inverse Fourier transform that you are working under?

And, for at least one of the problems you've posted, can you show us a partial computation? What have you done so far?


geometry Mafia search Gauss email articles Support

Does anyone know anything about gang activity in America such as how they are started, some examples of gangs, why kids join them, etc. Im trying to do a report and Im serching for info but Ive been unable to find very much besides what I already have.

I already got that. I was looking to see if anyone else had any info on local gangs in the area. We already know that if there is a gang here than we dont know about it. Its to small of a town to keep everything hidden forever so I guess we dont have any here in the city we live in.

Want me to get info from my parents about gangs in like New York and places?  They are still in contact with people and can tell me stuff for you.

I heard about that. That was saddening. I might have to look into that some more though because I dont really know all that much about that issue. Its a neat thing to study in all actuality. Its neat to see how many gangs orginally spun off the KKK which is now considered a cult. But do you know anything about some of the gangs around your area?

There are gangs here where we live aalindsey theres one in my neghborhood they steal stuff out of our front lawns and break a lot of things and are basically really destrustive, but they haven't hurt anyone that I know of.

Are you sure its a gang or is it just hoodlems being a nusence? A gang is like the Crips and the Bloods or the Mexican Mafia or the Bloody Red Dragons. There is a difference between a group of roudy friends and a gang. Gangs are groups of organized criminals. They have symbols, colors, and rules that they live by in order to stay true to that asossiation.

It's increasing in the Bay Area (cali)

A kid was killed (stabbed) near this high school. I'm not sure if it's gang related or not. They think that two boys got into a fight (probably from different gangs) and one stabbed the other to death

[quote="gauss202"][quote="BHorseMath"]Gangs in America have taken a increase over the past few years and the kids who join them are tipically either ones who do not get any attention at home or the ones who are so overloaded they are triny to get away.  Does that help at all?[/quote]

I don't believe that's true. I actually think that gang activity has gone down over the last decade. It was particularly high in the late 80s and early 90s, but I think it has been on the decline since then. If it has risen anywhere, it is most in suburban areas of all places. I've seen some television reports on this, but I can't remember which ones to cite them. I'm sure if you do a search you can find tons of information on the internet. You just have to be patient enough to wade through all the search results.[/quote]

I've got to agree with Gauss here.  New York has hardly anything reported as "gang-related" any more, although in the past (before my time, largely) gangs were a serious problem.

I would recommend you do a search on specific gangs -- pick one or two and look them up.  Be ready to make distinctions between real information and garbage, because there will probably be lots of garbage.

Actually from the info Ive found, gang numbers are large at the moment. The reason you dont see them in New York as much now is because alot of them are now run out of less populated areas and prisons of all places. The state with the highest recorded amount of gangs are Illinois and California actually. Gangs now try to stay anonymous by keeping it quiet. In larger cities. Its a new tactic and because of this gangs are becomeing more dangerous. I know here in Texas on North Side a well known gang is based called the Mexican Mafia. An Asian gang based out of Boston called the Bloody Red Dragons is spreading to other parts of the U.S. I think the most well known as rivals and large gangs are the Crips and then the Bloods. They are two gangs which I think almost every state has. I know this much but I was hoping someone would know more about some gangs located near them.

Is this a project? Why are you interested in finding these gangs? What can you do to stop them? How can discussing it here on the math forum solve the problem? What is the chance someone here is involved in a serious gang?

Wait! I know the answer to my last question. 0%.

You better believe no one is involved in a gang that posts or even lurks here if my parents get the slightest idea that there may be someone like that on here they will pull me from this site so fast that I won't even get to tell you people!

Okay I thought since most of you live in a place other than I do I thought that maybe yall might beable to tell me some stuff about kids who are in gangs at your school school. It is a project and Im sorry I bothered you I was just seeking info from those who might beable to help me actually. Nevermind.

the crips and the bloods are pretty famous.  You'll find good things on them.

[quote="Tare"]You could also say the best because they have a strict code of honor and chivalry...there are corrupted yakuzas though[/quote]
There is no "best" when it comes to organized crime and violence.

my brother forwarded me an email saying that gangs (in Boston area, i think) as an initiation ritual, have a person drive around at night with their headlights off, and they have to chase after and shoot the first person who signals to them to turn their lights on.  There are websites saying it's a hoax, though, but just be aware.

I know that it's probably well after you're done with the paper or whatever you needed the information for, but I just read a really good article in the New Yorker about the Aryan Brotherhood (also known as the Brand), a gang that started and is based entirely in prison.  Actually, there is a lot of gang activity in prison entirely separate from what goes on outside -- if you're still interested in the topic,  you might try to get a look at the article.

On NPR radio today, I heard that most kids start joining gangs and stuff because they dont have a father.  The father usually gives kids support and teaches them right from wrong.

they think they aren't cool so they try to roll with the gangstas

Yo the gangstas don't live in the ATL from what I've seen! Funny Usher lives here.
OK other than that did you guys realise that the death eaters are the wizard's KKK???

Yeah, the fact that they went around the World Cup wearing masks and hurting people that they didn't like because of their physical characteristics kinda gave it away for me :P

well my mum was buying us DE costumes when she asked didn't it seem like the KKK and wouldn't we outrage people trick or treating...but my brother still got one and hopefully i will

Local Gangs:
The VILLE (2X)
The TRE (3X)
The HILL (4X)

The Tribe - Too insignificant

National Gangs:
Crips
Bloods

Neither of them are around here.

[quote="moogra"]isn't gangs made up of balcks and mexicans? thats what some people say... (not being racist)and there are asian gangs and american kkk gangs[/quote]

It's obvious you know NOTHING about gangs. In fact, I am dying of laughter at this moment.

My sister dated an ex-gang member several years ago (he was a good guy by the time she met him though). He explained to me that drug trafficking and gangs are still big, at least in certain parts of Atlanta where he used to live.

Also, I've heard of kids nearby trying to become crips and bloods, but the gangs around the city i live in are too small to be a really big public nuisance or hazard.

gangs in the ATL!!!??? *funcia officially dies*
Well not in Buckhead!! Or I think is is N Atlanta

I am sure gangs are everywhere. There are friendly gangs and mean gangs, depending on the definition of the word gang.


permutation
inequalities
combinatorics
IMO
IMO 1997

Let $ x_1$, $ x_2$, $ \ldots$, $ x_n$  be real numbers satisfying the conditions:
\[ \left\{\begin{array}{cccc} |x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n | & \equal{} & 1 & \ \\
|x_i| & \leq & \displaystyle \frac {n \plus{} 1}{2} & \ \textrm{ for }i \equal{} 1, 2, \ldots , n. \end{array} \right.
\]
Show that there exists a permutation   $ y_1$, $ y_2$, $ \ldots$, $ y_n$  of $ x_1$,  $ x_2$, $ \ldots$, $ x_n$  such that
\[ | y_1 \plus{} 2 y_2 \plus{} \cdots \plus{} n y_n | \leq \frac {n \plus{} 1}{2}.
\]

$|(y_1+...+ny_n)+(ny_1+...+y_n)|=(n+1)(1)=n+1$ --> not to fulfil our conditon, $|y_1+...+ny_n|$ and $|ny_1+...+y_n|$ must differ by more $n+1$.

Now if we switch 2 neighbouring elements, then $|y_1+...+ny_n|$ changes by $k.(y_k-y_{k-1})+(k-1)(y_{k-1}-y_k) = 1(y_{k}-y_{k-1})$, which is at most $n+1$.

And this is a contradiction, since there would be no way to cross the gap, but both ends can be attained by switching.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14981]The idea in this problem helps a lot in finding the solution.[/url]

I did the hard way by adding all possible permutations together to get their average, which is $ (n\plus{}1)/2$.  And since they are all equal, or at least one of the permutations is greater than average, which means there exist a permutation less than average.  Am I right?

I thought this problem is pretty trivial, maybe I missed something

So obviously if $x_1=x_2=...=x_n$, then the statement is true, and we actually have a equality.

The obvious answer to the general solution of course is to arrange the $n$ elements in increasing order, let $y_1$ be the largest of the elements, and $y_n$ be the smallest, i.e.
$y_n\le y_{n-1}\le ... \le y_1$

WLOG assume $x_1+x_2+...+x_n=1$
observe now that the average of the elements is
$x_1+x_2+...+x_n=\frac{1}{n}$, so let $y_1=\frac{1}{n}+q$ for some some value $q$.  This $y_1$ must exist since all $n$ elements cannot be all less than $1/n$ which is the average.  And let us for the sake of argument put the extra $q$ at any place other than the first position, say , let $y_n=(1/n)-q$

So the equation now becomes

$(q+\frac{1}{n})+2(\frac{1}{n})+3(\frac{1}{n})+...+n(\frac{1}{n}-q)$

$=\frac{n+1}{2}+(1-n)q \le \frac{n+1}{2}$

So apply similar arguments to a more general case, so i will stretch out my arms before I attempt it..., Ok here it is:

let $y_1=\frac{1}{n}+q_1$
$y_2=\frac{1}{n}+q_2$
...
$y_k=\frac{1}{n}+q_k$
$y_{k+1}=\frac{1}{n}-q_{k+1}$
....
$y_n=\frac{1}{n}-q_n$

where
$q_1 \ge q_2 \ge q_3 \ge ... \ge q_k \ge 0 \ge -q_{k+1}\ge....\ge-q_n$

$q_i \le \frac{n+1}{2}-\frac{1}{n}$ for $i=1,2...,k$, and $-q_i\ge -(\frac{n+1}{2}-\frac{1}{n})$ for $i=k+1,...,n$

Observe $q_1+q_2+q_3+...+q_k-q_{k+1}-....-q_n=0$

Now, $y_1+2y_2+...+ny_n=$

$=\frac{n(n+1)}{2}n+q_1+2q_2+3q_3+...+kq_k-(k+1)q_{k+1}-....-nq_n$

Now, observe
$\frac{n+1}{2}-\frac{1}{n}\ge q_1\ge q_2 \ge...\ge q_k\ge0$

And,

$\frac{n+1}{2}-\frac{1}{n}\ge q_n\ge q_{n-1}\ge...\ge q_{k+1}\ge0$

But what is $q_1+2q_2+3q_3+...+kq_k-(k+1)q_{k+1}-....-nq_n$?  Well it is the following equations added together:

$q_1+q_2+q_3+...+q_k-q_{k+1}-....-q_n=0$

+____$q_2+q_3+...+q_k-q_{k+1}-....-q_n=-q_1$

+________$q_3+...+q_k-q_{k+1}-....-q_n=-q_1-q_2$

                                                              .....

+_______________$q_k-q_{k+1}-...-q_n=-q_1-q_2-.....-q_{k-1}$

+__________________$-q_{k+1}-...-q_n=-q_{k+1}-...-q_n$

                                                              .....

+____________________________$-q_n=-q_n$

So, $q_1+2q_2+3q_3+...+kq_k-(k+1)q_{k+1}-....-nq_n=-(k-1)q_1-(k-2)q_2-....-q_{k_1}-(k+1)q_{k+1}-...-nq_n \le 0$

Therefore, $q_1+2q_2+3q_3+...+kq_k-(k+1)q_{k+1}-....-nq_n\le 0$

So I have proved--or, at least I think I "proved"--that given $x_1+x_2+...+x_n=1$
and, $|x_i| \le (n+1)/2$ then the permutation of {$x_1,...,x_n$}, i.e., $y_n\le y_{n-1}\le ... \le y_1$ satisfies $y_1+2y_2+....+ny_n \le (n+1)/2$

Now I want to show this permutation is $\ge -(n+1)/2$

[quote=solver6]Solution :

First case : There exists two permutations $y_1, y_2, \ldots, y_n$, $z_1, z_2, \ldots, z_n$ of set $x_1, x_2, \ldots, x_n$, such that $a:=y_1 + 2y_2 +\ldots + ny_n\geq 0$, $b:=z_1 + 2z_2 +\ldots + nz_n\leq 0$.

In this case we can use some elementary transpositions $\sigma_i$ which switches 2 neighbouring elements and sends set $y_1, \ldots, y_n$ to set $z_1, \ldots, z_n$. Note that for every transposition $\tau : (y_i, y_{i+1})\to (y_{i+1}, y_i)$, $|(y_1 + 2y_2 +\ldots + ny_n) - (y_{\sigma(1)} + 2y_{\sigma(2)} +\ldots + ny_{\sigma(n)})|\leq |y_i - y_{i+1}|\leq n+1$. So from continuous principle we know that $a\geq 0$, $b\leq 0$ and so by some permutations we can get to sequence $|t_1 + 2t_2 +\ldots + nt_n|\leq (n+1)/2$.

Second case : For every permutation $y_1, y_2, \ldots, y_n$, number $y_1 + 2y_2 +\ldots + ny_n$ is positive. In this case consider cyclic permutation $\sigma : (1, 2, \ldots, n)\to (2, 3, \ldots, 1)$ and if $y_{\sigma^i(1)} + 2y_{\sigma^i(2)} +\ldots + ny_{\sigma^i(n)}> (n+1)/2$ for all $i$, then we can get $\sum_{i=1}^n i = \sum_{i=1}^n(y_{\sigma^i(1)} + 2y_{\sigma^i(2)} +\ldots + ny_{\sigma^i(n)})> n(n+1)/2$. Contadiction. $\Box$[/quote]

What is continuous principle?

Beautiful problem! Here's my solution: Let $\sigma$ denote some permutation $(y_1,y_2, \dots ,y_n)$ of $(x_1,x_2, \dots ,x_n)$, and let $f(\sigma)$ denote the sum $y_1+2y_2 +\dots +ny_n$. WLOG take $x_1+x_2+ \dots +x_n=1$ (if its $-1$, then replace $x_i$ by $-x_i$). Also, let $x_1 \leq x_2 \leq \dots \leq x_n$. Finally, FTSOC assume that $|f(\sigma)|>\frac{n+1}{2}$ for all possible permutations $\sigma$.

Let $\sigma_r$ denote the permutation formed by interchanging $y_r$ and $y_{r+1}$ in $\sigma$, for some $r \in [n-1]$. We call such a move (of interchanging adjacent elements) a [i]Pro[/i] move. Note that $f(\sigma_r)-f(\sigma)=y_r-y_{r+1}$. Then, by triangle inequality, we get that $$||f(\sigma_r)|-|f(\sigma)|| \leq |f(\sigma_r)-f(\sigma)|=|y_r-y_{r+1}| \leq |y_r|+|y_{r+1}| \leq n+1$$
where the last inequality follows from the condition that $|y_i| \leq \frac{n+1}{2}$ for all $i \in [n]$. Our main claim is that $f(\sigma_r)$ and $f(\sigma)$ must have the same sign. Suppose not, and WLOG take $f(\sigma) \geq 0$ and $f(\sigma_r) \leq 0$. By our original assumption, we get $f(\sigma)>\frac{n+1}{2}$ and $f(\sigma_r)<-\frac{n+1}{2}$. Thus, $f(\sigma)-f(\sigma_r)>n+1$, which contradicts the above inequality. Hence, our assumption was wrong, and so $f(\sigma)$ and $f(\sigma_r)$ must have the same sign.

Now, let $\sigma_1=(x_1,x_2, \dots ,x_n)$ and $\sigma_2=(x_n,x_{n-1}, \dots ,x_1)$. It's easy to see that there exists a sequence of [i]Pro[/i] moves taking $\sigma_1$ to $\sigma_2$ (First take $x_1$ to the end, then take $x_2$ to the end, and so on). By the above para, this means that $f(\sigma_1)$ and $f(\sigma_2)$ have the same sign. Since $x_1 \leq x_2 \leq \dots \leq x_n$, and $x_1+x_2+ \dots +x_n>0$, so we must have $f(\sigma_1)>0$. This means that $f(\sigma_2)$ must also be positive. Also, note that $$f(\sigma_1)+f(\sigma_2)=(n+1)(x_1+x_2+ \dots +x_n)=n+1$$
But, using our assumption, we have $f(\sigma_1),f(\sigma_2)>\frac{n+1}{2}$, and so their sum must be greater than $n+1$, a contradiction. Hence, done. $\blacksquare$

[b]REMARKS:[/b] The solution was found in the exact opposite order. First the condition to be proved was shown to be equivalent to proving that $f(\sigma_1)$ and $f(\sigma_2)$ have same sign. Then this was shown by defining the [i]Pro[/i] moves, and characterizing what happens to $f(\sigma)$ on undergoing a [i]Pro[/i] move.

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

WLOG $\sum x_i = 1$ (by negating $x_i$) and $x_1 \le x_2 \le \dots \le x_n$. Notice that [list] [*]The largest possible value of the sum in question is \[ A = x_1 + 2x_2 + 3x_3 + \dots + nx_n. \] while the smallest value is \[ B = nx_1 + (n-1)x_2 + \dots + x_n. \] [*]Meanwhile, the [i]average[/i] value across all permutations is \[ 1 \cdot \frac{n+1}{2} + 2 \cdot \frac{n+1}{2} + \dots + n \cdot \frac{n+1}{2} = \frac{n+1}{2}. \] [/list]
Now imagine we transform the sum $A$ to the sum $B$, one step at a time, by swapping adjacent elements. Every time we do a swap of two neighboring $u < v$, the sum decreases by \[ (iu + (i+1)v) - (iv + (i+1)u) = v-u < n+1. \]
We want to prove we land in the interval \[ I = \left[ -\frac{n+1}{2}, \frac{n+1}{2} \right] \] at some point during this transformation. But since $B \le \frac{n+1}{2} \le A$ (since $\frac{n+1}{2}$ was the average) and our step sizes were at most the length of the interval $I$, this is clear.

It is equivalent to show there is a permutation such that \[ y_1 + 2y_2 + \cdots + ny_n \in \left[-\frac{n+1}{2}, \frac{n+1}{2}\right] \]
WLOG let $x_1 < x_2 < \dots < x_n$. Then \[ x_1 + 2x_2 + 3x_3 + \dots + nx_n \] is maximal and \[ x_n + 2x_{n-1} + 3x_{n-2} + \dots + x_1 \] is minimal.
Since they sum to $(n + 1) \cdot (x_1 + x_2 + \dots + x_n) = \pm (n + 1)$, either one of them lies in the interval, or they lie on opposite sides of the interval.

[b][color=pink]Claim:[/color][/b] Exchanging adjacent $y_i$ and $y_{i+1}$ changes the sum by at most $n+1$.
[i]Proof.[/i] $((i+1)y_i + iy_{i+1}) - (iy_i + (i+1)y_{i+1}) = y_i - y_{i+1}$ so this follows by the bound. $\blacksquare$
Thus, taking discrete continuity from the smallest permutation to the largest finishes.

Nice AC hybrid.

Let $\sigma = (y_1, y_2, \dots, y_n)$ be such a permutation for which the absolute value is minimized, and assume FTSOC it is greater than $\frac{n+1}2$. We can also assume WLOG that the sum in question is greater than $\frac{n+1}2$ (in the other case, repeat the same argument with orderings and directions flipped.)

First, I claim that $\sigma$ is inversely sorted. Otherwise, there exist indices $y_i < y_{i+1}$, and by swapping $y_i$ and $y_{i+1}$, the sum $S$ decreases by $$\Delta S = y_{i+1} - y_i \leq n+1.$$ However, this means that the newer permutation $\sigma'$ yields a smaller value of $S$, contradiction.

Hence $\sigma$ is inversely sorted. On the other hand, by Chebyshev we have $$y_1+2y_2+\cdots+ny_n \leq \frac{n+1}2 (x_1+x_2+\cdots+x_n).$$ Notice that we originally assumed that the sum was greater than $\frac{n+1}2$ and any reduction past that value will be a contradiction. This implies that $x_1+x_2+\cdots+x_n = 1$ precisely, so we have the result.


Compute the least prime p such that p-1 equals the difference of the squares of two positive multiples of 4

Yes.  Can you prove that your answer is the least prime that meets the condition?

Check all smaller cases.
$8^2-4^2+1=49$, $12^2-8^2+1=81$, both of which composite. Lastly, $12^2-4^2+1=129$ which is too big.

I got that one too

[hide="My explanation"]$p$ must be of the form $16k+1$. So if $(4a)^2-(4b)^2=p-1$, $a^2-b^2=k$.
$k=1\Rightarrow p=16+1=17$ prime but $b>0$ so no
$k=2\Rightarrow p=33$ not prime
$k=3\Rightarrow p=49$ not prime
$k=4\Rightarrow p=65$ not prime
$k=5\Rightarrow p=81$ not prime
$k=6\Rightarrow p=97$ is prime but 6 cannot be a difference of two squares
$k=7\Rightarrow p=113$ is prime and $7=2\cdot3+1=4^2-3^2$

Answer: $p=113$
[/hide]


Solution relies upon tangent double and triple angle formulae and also some standard angles (30, 45, 60) which I assume.

tan30 = 2tan15/(1 - tan^2(15)) giving 0 < tan15 = 2 - sqrt(3).

Let x = tan5.

tan15 = (3x - x^3)/(1 - 3x^2) = 2 - sqrt(3) giving:

x^3 - 3(2 - sqrt(3)x^2 - 3x + 2 - sqrt(3) = 0.

Multiplying through by sqrt(3) + 1,
(sqrt(3) + 1)x^3 + (3 - 3sqrt(3))x^2 - (3sqrt(3) + 3)x + sqrt(3) - 1 = 0

This can be rearranged to give:
[3 - x^2][sqrt(3) + 1]    =       1 / x    =  tan85  =
--------------------------
[1 - 3x^2][sqrt(3) - 1]

[sqrt(3) + x][sqrt(3) - x][1 + 1/sqrt(3)]        =
-----------------------------------------------
[1 - sqrt(3)x][1 + sqrt(3)x][1 - 1/sqrt(3)]

tan(60 - arctan(x))tan(60 + arctan(x))tan(45 + 30) =

tan(55)tan(65)tan(75) as required.

[N.B. There are possibly three solutions here of which x = tan5 is certainly one.  I haven't tried to find the others.]


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^3\plus{}b^3\plus{}c^3\equal{}3.$ prove that
\[ (a\plus{}b^2c^2)(b\plus{}a^2c^2)(c\plus{}a^2b^2)\geq8a^2b^2c^2\]

[quote="bunhiacovski"][quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3.$ prove that
\[ (a \plus{} b^2c^2)(b \plus{} a^2c^2)(c \plus{} a^2b^2)\geq8a^2b^2c^2
\]
[/quote]
[b]Lemma[/b](Pham Kim Hung)
$ a\ge b \ge c \Rightarrow (a \plus{} b^2)(b \plus{} c^2)(c \plus{} a^2)\ge (a^2 \plus{} b)(b^2 \plus{} c)(c^2 \plus{} a)$

thus,we only need prove the inequality with case : $ a\ge b \ge c$
we have

$ (a \plus{} b^2c^2)(b \plus{} a^2c^2)(c \plus{} a^2b^2)\ge (a^2 \plus{} bc)(b^2 \plus{} ac)(c^2 \plus{} ab)$

we need prove that $ (a^2 \plus{} bc)(b^2 \plus{} ac)(c^2 \plus{} ab)\ge 8a^2b^2c^2$

$ \Leftrightarrow \minus{} 3r^2 \plus{} 3r \plus{} q^3 \minus{} 3pqr\ge 0$ ($ p \equal{} \sum a, q \equal{} \sum ab$ and $ r \equal{} abc$)

$ pqr\le \frac {q^3}{3}\Rightarrow \minus{} 3r^2 \plus{} 3r \plus{} q^3 \minus{} 3pqr\ge \minus{} 3r^2 \plus{} 3r \plus{} q^3 \minus{} q^3 \equal{} 3r(1 \minus{} r)$

$ 3 \equal{} a^3 \plus{} b^3 \plus{} c^3 \ge 3abc \Rightarrow abc\le 1\Leftrightarrow r\le 1$

$ \Rightarrow 3r(1 \minus{} r)\ge 0$  :wink:

we have done ~> :P  :wink:[/quote]

\[ \left( {a^2  \plus{} bc} \right)\left( {b^2  \plus{} ca} \right)\left( {c^2  \plus{} ab} \right) \ge 8a^2 b^2 c^2
\]

which true by AM-GM.  :huh:

[quote="bunhiacovski"]
we have

$ (a \plus{} b^2c^2)(b \plus{} a^2c^2)(c \plus{} a^2b^2)\ge (a^2 \plus{} bc)(b^2 \plus{} ac)(c^2 \plus{} ab)$

[/quote]
Are you sure? :wink:
By my opinion it is equivalent to $ a^4b^4c^4\plus{}a^2b^2c^2\geq2abc,$ which obviously wrong.

[quote="Honey_S"]And the lemma must be
\[ \left( {a^2 \plus{} b} \right)\left( {b^2 \plus{} c} \right)\left( {c^2 \plus{} a} \right) \ge \left( {a \plus{} b^2 } \right)\left( {b \plus{} c^2 } \right)\left( {c \plus{} a^2 } \right)
\]
with $ a \ge b \ge c$.[/quote]
   I think you are not right . You can check again  :rotfl:

[quote="thanhak5"]I think you are not right . You can check again  :rotfl: [/quote]
I agree with thanhak5. Try $ a\equal{}2, b\equal{}1, c\equal{}0$  :wink:
Will the inequality be flipped then??  :maybe:

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3.$ prove that
\[ (a \plus{} b^2c^2)(b \plus{} a^2c^2)(c \plus{} a^2b^2)\geq8a^2b^2c^2
\]
[/quote]

Assume $ a \ge b \ge c$

By expandinng we get the inequality is equalivent
\[ a^3 b^3 \plus{} b^3 c^3 \plus{} c^3 a^3 \plus{} abc \plus{} a^4 b^4 c^4 \ge 5a^2 b^2 c^2
\]
Bescause of $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3$ so
\[ a^3 b^3 \plus{} b^3 c^3 \plus{} c^3 a^3 \plus{} abc \plus{} a^4 b^4 c^4 \ge 5a^2 b^2 c^2
\]

\[ \Leftrightarrow a^3 b^3 \plus{} b^3 c^3 \plus{} c^3 a^3 \plus{} \frac {{abc\left( {a^3 \plus{} b^3 \plus{} c^3 } \right)}}{3} \plus{} \frac {{9a^4 b^4 c^4 }}{{\left( {a^3 \plus{} b^3 \plus{} c^3 } \right)^2 }} \minus{} 5a^2 b^2 c^2 \ge 0
\]

\[ \Leftrightarrow \left( {a^3 b^3 \plus{} b^3 c^3 \plus{} c^3 a^3 \minus{} 3a^2 b^2 c^2 } \right) \plus{} \left( {\frac {{abc\left( {a^3 \plus{} b^3 \plus{} c^3 } \right)}}{3} \minus{} a^2 b^2 c^2 } \right) \plus{} \left( {\frac {{9a^4 b^4 c^4 }}{{\left( {a^3 \plus{} b^3 \plus{} c^3 } \right)^2 }} \minus{} a^2 b^2 c^2 } \right) \ge 0
\]

\[ \Leftrightarrow \frac {{\left( {\sum {ab} } \right)\left( {\sum {a^2 \left( {b \minus{} c} \right)^2 } } \right)}}{2} \plus{} \frac {{abc\left( {\sum a } \right)\left( {\sum {\left( {b \minus{} c} \right)^2 } } \right)}}{6} \minus{} \frac {{a^2 b^2 c^2 \left( {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc} \right)\left( {\sum a } \right)\left( {\sum {\left( {b \minus{} c} \right)^2 } } \right)}}{{2\left( {a^3 \plus{} b^3 \plus{} c^3 } \right)^2 }} \ge 0
\]

\[ \Leftrightarrow S_a \left( {b \minus{} c} \right)^2 \plus{} S_b \left( {c \minus{} a} \right)^2 \plus{} S_c \left( {a \minus{} b} \right)^2 \ge 0
\]

\[ S_a \equal{} \frac {{a^2 \left( {\sum {ab} } \right)}}{2} \plus{} \frac {{abc\left( {\sum a } \right)}}{6} \minus{} \frac {{a^2 b^2 c^2 \left( {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc} \right)\left( {\sum a } \right)}}{{2\left( {a^3 \plus{} b^3 \plus{} c^3 } \right)^2 }}
\]

\[ S_b \equal{} \frac {{b^2 \left( {\sum {ab} } \right)}}{2} \plus{} \frac {{abc\left( {\sum a } \right)}}{6} \minus{} \frac {{a^2 b^2 c^2 \left( {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc} \right)\left( {\sum a } \right)}}{{2\left( {a^3 \plus{} b^3 \plus{} c^3 } \right)^2 }}
\]

\[ S_c \equal{} \frac {{c^2 \left( {\sum {ab} } \right)}}{2} \plus{} \frac {{abc\left( {\sum a } \right)}}{6} \minus{} \frac {{a^2 b^2 c^2 \left( {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc} \right)\left( {\sum a } \right)}}{{2\left( {a^3 \plus{} b^3 \plus{} c^3 } \right)^2 }}
\]
We have
\[ \frac {{abc\left( {\sum a } \right)}}{6} \ge \frac {{a^2 b^2 c^2 \left( {a^3 \plus{} b^3 \plus{} c^3 } \right)\left( {\sum a } \right)}}{{2\left( {a^3 \plus{} b^3 \plus{} c^3 } \right)^2 }}
\]

\[ \Leftrightarrow a^3 \plus{} b^3 \plus{} c^3 \ge 3abc
\]
So it is enough for us to show the inequality after resetting
\[ S_a \equal{} \frac {{a^2 \left( {\sum {ab} } \right)}}{2} \minus{} \frac {{3a^3 b^3 c^3 \left( {\sum a } \right)}}{{2\left( {a^3 \plus{} b^3 \plus{} c^3 } \right)^2 }}
\]

\[ S_b \equal{} \frac {{b^2 \left( {\sum {ab} } \right)}}{2} \minus{} \frac {{3a^3 b^3 c^3 \left( {\sum a } \right)}}{{2\left( {a^3 \plus{} b^3 \plus{} c^3 } \right)^2 }}
\]

\[ S_c \equal{} \frac {{c^2 \left( {\sum {ab} } \right)}}{2} \minus{} \frac {{3a^3 b^3 c^3 \left( {\sum a } \right)}}{{2\left( {a^3 \plus{} b^3 \plus{} c^3 } \right)^2 }}
\]
We will show that $ S_a \ge 0$
\[ S_a \ge 0
\]

\[ \Leftrightarrow \frac {{a^2 \left( {\sum {ab} } \right)}}{2} \ge \frac {{3a^3 b^3 c^3 \left( {\sum a } \right)}}{{2\left( {a^3 \plus{} b^3 \plus{} c^3 } \right)^2 }}
\]

\[ \Leftrightarrow \left( {a^3 \plus{} b^3 \plus{} c^3 } \right)^2 \left( {\sum {ab} } \right) \ge 3ab^3 c^3 \left( {\sum a } \right)
\]
By Holder inequality we have $ 9\left( {a^3 \plus{} b^3 \plus{} c^3 } \right) \ge \left( {\sum a } \right)^3$. Therefor it enough to prove that
\[ \left( {a^3 \plus{} b^3 \plus{} c^3 } \right)\left( {\sum a } \right)^2 \left( {\sum {ab} } \right) \ge 27ab^3 c^3
\]
But $ \left( {a^3 \plus{} b^3 \plus{} c^3 } \right)\left( {\sum a } \right)^2 \left( {\sum {ab} } \right) \ge 27a^2 b^2 c^2 \left( {\sum a } \right)$ so we just need to prove
\[ 27a^2 b^2 c^2 \left( {\sum a } \right) \ge 27ab^3 c^3
\]

\[ \Leftrightarrow a\left( {\sum a } \right) \ge bc
\]
which obviously true.

By the same way we can also prove that $ S_b \ge 0$.
\[ S_b \ge 0
\]

\[ \Leftrightarrow b\left( {\sum a } \right) \ge ca
\]
Moreover we can also prove $ S_b \plus{} S_c \ge 0$ by the same way too.
\[ S_b \plus{} S_c \ge 0
\]
\[ \Leftrightarrow \left( {b^2  \plus{} c^2 } \right)\left( {\sum {ab} } \right)\left( {\sum {a^3 } } \right)^2  \ge 6a^3 b^3 c^3 \left( {\sum a } \right)
\]
We have $ b^2  \plus{} c^2  \ge 2bc$ so we only need to prove
\[ \left( {\sum {ab} } \right)\left( {\sum {a^3 } } \right)^2  \ge 3a^3 b^2 c^2 \left( {\sum a } \right)
\]
By the same way of Holder we get it is
\[ a \plus{} b \plus{} c \ge a
\]
which obviously true. We are done.

[quote="arqady"]I think, the inequality  $ a^3b^3 \plus{} a^3c^3 \plus{} b^3c^3\geq3(abc)^{\frac {5}{3}}$ gives an easier proof. :wink:[/quote]

I have edited my solution. Thanks arqady!

But I prefer my solution to yours.

Do you have any mistake in typing since the inequalty is inhomogenous.

It is too late now (in VietNam). So see you later.

[quote="Honey_S"]

Do you have any mistake in typing since the inequalty is inhomogenous.

[/quote]
We have the condition: $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3.$ :wink:
Good night!


limit
real analysis
real analysis unsolved

is true that $\sum_{n=0}^{+\infty} |a_n|$ is convergente if and only if $\lim_{n\to\infty}n*a_n=0$?

in the case of affermative answer, I'd like to see the demonstration. Thanks

[quote="(¼§¡!)³»x"]is true that $\sum_{n=0}^{+\infty} a_n$ is convergente if and only if $\lim_{n\to\infty}n*a_n=0$?

in the case of affermative answer, I'd like to see the demonstration. Thanks[/quote]

No, it is not. Consider for example the $a_n=\frac{1}{n log n}$, here we have that $\lim_{n \to \infty}na_n=0$ but the series $\sum_n |a_n|$ doesn't converge.

[quote="eugene"]To give a counterexample to the other equvalence consider, say, $a_n=\frac{1}{n %Error. "logn" is a bad command.
}$, here we have that $\lim_{n \to \infty}na_n$ but the series $\sum_n a_n$ doesn't converge.[/quote]

uhmmm... but if $a_n=1/n$ then $na_n=1$ so limit $\lim_{n \to \infty}na_n=1$ not 0, the other equivalence maybe is true...

I edited it,it was a typo. To show that the other implication is also not true, consider the series $\sum_n |a_n|$, where $a_n=0$ if for all natural $p$ $n \neq 2^{2p}$ and $a_n=2^{-p}$ if there exist natural $p$ such that $n=2^{2p}$. You may notice that the series converge but $\lim_{p \to \infty}2^{2p}a_{2^{2p}}= +\infty$.

yes, you are right, however i have not made any restriction on the succession, the example that you show is that of a succession defined for cases and therefore in that example the limit of $na_n$ doesn't exist being $a_n=0$ almost anywhere, therefore if the case of succession defined for cases is excluded by the possible $a_n$ at least an implication should be true, the case $a_n=1/(nln(n))$ excludes 1 of two implications


modular arithmetic
number theory unsolved
number theory

Let $ k$ be a positive integer and $ f(n)\equal{}k2^n\plus{}1$ for all positive integer $ n$. Do there exist positive integers $ i$ and $ j$ st for all positive integers $ m$, $ m\equiv i\pmod j$, $ f(m)$ is not a prime?

The answer is yes.

Let $ f(a)\equal{}k2^a\plus{}1\equal{}h$, then we have $ gcd(2,h)\equal{}1$.
Now choose $ j\equal{}2\varphi(h)$, $ i\equal{}\varphi(h)\plus{}a$, we have $ k2^m\equal{}k2^a\equal{}\minus{}1(mod \;\; h)$ and $ k2^m\plus{}1>h$ for all $ m\equal{}i(mod \;\;j)$
So $ h|f(m)$ and $ f(m) \neq h$ for all $ m\equal{}i(mod \;\; j)$, done.


The product of three consecutive integers is 2184 . What is the sum of the integers ?

(A)34 (B) 33 (C) 36 (D) 39 (E) noneofthese


A cylindrical closed toxic waste container eighteen inches high with an outer diameter of 12 inches is made of stainless steel with a uniform thickness of $ \frac14$ inch. Assuming the cylinder is full, to the nearest percent what is the volume of the liquid to the volume of the entire cylinder?

The liquid takes up $ (\frac {23}{4})^2\pi$ inches of space, or $ 33 \frac {1}{16} \pi$.  The whole cylinder is $ 36\pi$.  The answer is $ \frac {33.0625}{36}$, or $ \boxed{92\%}$.

Shouldn't it be 33 1/16 pi instead of 33 1/12 pi?

I got this problem today and when asking on forum, it redirected me here. The correct answer is 89%, idk how.


Tinky Winky sleeps fifteen hours a day, while Dipsy is asleep only 40% of the time. If a person picks a random time to look at Tinky Winky and Dipsy, what are the [i]odds[/i] that they are both asleep?

[hide]
Whichever one you said first sleeps 15 hours, which is 5/8 of the day. Whichever one you said second sleeps 40% of the day, which is 2/5. 2/5*5/8=1/4.
[/hide]

[hide="taking answer from last guy"]
note that the question asks for odds. Thus, a probability of $\frac14$ is $\frac13$ odds (read 1 to 3).
[/hide]

oops i put the probability instead of the odds.
THIS IS MY PROBLEM. I DO NOT READ QUESTIONS.
ON NATS, WHEN IT ASKED FOR THE 3 DIGIT NUMBER, I WROTE THE 2 DIGIT NUMBER
ITS NOT EVEN FUNNY.
 :rotfl:
SHUT UP.

Hey, don't feel bad... at Nats, when it asked for A,BCD,EFG, I put down A+B+C+D+E+F+G :blush: :oops_sign:

[hide]
The probability is 3/8*2/5 or 3/20, so the odds are 3:17
[/hide]

Edit: I can't cancel fractions properly... the right answer is:

[hide]
The probability is [b]5/8[/b]*2/5 or 1/4, so the odds are 1:3
Hehehehehe...
[/hide]

[quote="mathnerd314"]Hey, don't feel bad... at Nats, when it asked for A,BCD,EFG, I put down A+B+C+D+E+F+G :blush: :oops_sign:

[hide]
The probability is 3/8*2/5 or 3/20, so the odds are 3:17
[/hide][/quote]
Uh...that's wrong. And the odds are [hide] $1: 3$ [/hide]

Ha, I do stuff like that all the time...and my math teacher hates me for it XD. She's always telling me to check my work...but of course, I don't have the attention span for it, so I don't.

...I lose interest in things far too quickly.

[hide]And yep, I agree...the answer's 1:3[/hide]

[quote="binonunquineist"]Not really since you have to consider the small probability that the day is a daylight savings day with only 23 or 25 hours. Thus this is a bad question.

If you do not consider that then yes, it is 1:3.[/quote]

Either way it evens out if you average all of the days :p .

Actually if you take into account daylight savings and leap years and all that good stuff I got the odds as $840083/2520217$ which is still kind of close to $1/3$.

Even though I kind of symphatize binonunquineist's first answer, I will answer with 5/8*2/5(5/8)=1/4... ;)

[quote="Treething"][quote="Ravi B"]More importantly, all the solutions so far have assumed that whether Tinky Winky is asleep is independent of whether Dipsy is asleep.  (Otherwise, you can't apply the multiplication principle.)  But that seems to be an unlikely assumption: People (and teletubbies?) tend to sleep at night, so their sleep times are correlated.

I would say that the problem can't be solved without extra information.[/quote]

Have you seen the Teletubbies? Po is ALWAYS asleep.

Assume that they sleep at random intervals.[/quote]
Maybe if I watch the Teletubbies, I'll be a Nats champion :rotfl:


calculus
integration
real analysis
advanced fields
advanced fields unsolved

m is a measure on the line,for every borel subset of R ,m(A)=integral of f on A (lebesgue integral),f continuous.Suppose that for some B, m(B)>0.Then B has some interior points.

That doesn't work, unless I misundertand the problem.

Let $f(x)=1.$ That's certainly continuous. Let $B$ be the set of irrational numbers in $[0,1].$ Then $m(B)=\int_Bf=1$ but $B$ has no interior points.


The following is my answer and reasoning for the famous Chicken and Egg Question:

The egg came first, because the characteristics of a chicken is that it lays eggs and is, itself, hatched from an egg. Something other than a chicken could have layed an egg which borne the first chicken.

[quote="mcalderbank"]Er, If you beleive in evolution, it would make sense for a chicken to evolve from a lesser organism.[/quote]

I do not see a contradiction. A chicken evolves from a [i]lesser[/i] organism (one that layed eggs, as opposed to one that gave "live-birth"), which layed an egg from which the first chicken was hatched.

[quote="filletwho"]Mutations. But what defines a chicken? I mean, the modern chicken evolved gradually over time, but at what point is it considered a chicken? That is, what definins feature did it have that made it so that it is a chicken?[/quote]

I made the initial assumption that the person who came up with the problem used the word "chicken" for essentially the same organism we consider today to be a "chicken". But actually this assumption is not really needed, [i]as long as[/i] one of the assumptions about the characteristics of a chicken is the fact that it was born from an egg, it doesn't matter what other characteristics the "chicken" has, for evolution (oh, yeah; you've got to assume the existence of evolution as well) will have divided time into "pre" and "post" existence of a chicken.

[quote="peter"]Er....dinosaurs laid eggs, so the egg came first![/quote]

No, that would prove that the chicken came first.

It's: "Dinosaurs hatched from eggs, so the egg came first!"

Assuming there are levels of illogical-ness.

[quote="furious"]what nobody knows is that a dinosaur laid an egg with a mutation. this mutation caused the egg to hatch the modern-chicken. so, the egg came first.[/quote]
Technically, that's not right...it was a long series of mutations + chickens with bad mutations being killed off over millions of years...so I have to say the same thing as filletwho. At what point along the series of mutations is the chicken considered a chicken?

My answer would still be the same as well: it doesn't matter, because as long as we agree that this organism (which is undergoing so many mutations) was hatched from an egg, then the egg came first, no matter what your definition of a chicken is.

The reason that one would be compelled to agree is due to the fact that live birth is more "sophisticated" than hatching from an egg, so as long as we believe that the chicken did not descend from an organism which gave birth live, then we can safely rest that the egg came first - no matter what each person's definition of a chicken is.

[quote="sam3.14"]If we rephrase the question as, "Which came first, the chicken or the chicken egg?", then my reasoning is as follows:

Lets say that a chicken is an organism some given precise DNA code. Let a chicken egg be an egg from which a chicken (by the above definition) will hatch. Since mutations occur before the formation of the egg, the first chicken egg had the precise DNA of a chicken in it, and the egg came first.[/quote]

Yep, very similar to my solution - I concur.

[quote="Scrambled"]i believe that, since without the chicken, most eggs wouldnt survive
therefore the chicken came first[/quote]

But you see, eggs containing chickens (I'll assume that what you meant) does not need specifically a chicken to survive. It just needs something which gives heat and protection. Something which is genetically (slightly) different would not cause an egg to perish.

[quote="filletwho"]Unless u reason that a chicken egg must be layed by a chicken.[/quote]

Well, I would use the argument that a the definition of a "chicken" is quite loose, but we know for a fact that the DNA within the egg is identical to the DNA of the chicken which will hatch from the egg, and so if we consider each chicken to be distinct in their DNA (which is indeed the case, unless they happen to be identical twins), then the egg will always come before the first chicken.[/quote]

i believe the cow came first, that's why the cows want you to go to Chick-fil-A

Prokaryotes that reproduce asexually(binary fission) came first.


[quote="Jutaro"]The Central American Olympiad is an annual competition. The ninth Olympiad is held in 2007. Find all the positive integers $n$ such that $n$ divides the number of the year in which the $n$-th Olympiad takes place.[/quote]
If the 9th Olympiad was held in 2007 it means the 1st Olympiad was held in 1999. And in general the $n$-th Olympiad is held in $1998+n$.

We ask for what $n$ does $n | (1998+n)$ which is true if and only if $n|1998$.

So all positiver integer $1998+n$ such that $n|1998$ are all the year the Olympiads that you seek.

Since the 9th one was held on 2007, in general, the nth olympiad is held in the year 1998 + n.

Therefore we need to find all n that satisfy 1998 + n = 0 (mod n ). Therefore 1998 = 0 (mod n).

Therefore all n we find are  such that n/1998. In other words, the divisors of 1998. :)

Using the theory of prime factors, I can highlight that 1998 acts as 0, which could say that if 1998 has a prime factor then that prime factor will appear again by adding that amount of the prime factor. For example

2 divides 1998 so 2 divides 1998+2

3 divides 1998 so 3 divides 1998+3

and so progressively.

So we have that 1998 has 16 divisors so if we add those 16 to 1998 we will then have 16 answers.

Answers:
1|1999
2|2000
3|2001
6|2004
9|2007
18|2016
27|2025
37|2035
54|2052
74|2072
111|2109
222|2220
333|2331
666|2664
999|2997
1998|3996


The flag that is posted next to "Carolinas" [url=http://www.artofproblemsolving.com/Forum/index.php?f=249]here[/url] is the NC flag.  It should be a neutral flag, or else a Gamecock flag.  Go Steve Spurrier.  Also, I have a question:  How well will Steve Spurrier do in his first season at USC?  Will they go to a bowl (barring another brawl)?  Will they beat Clemson?  Who will start at QB?  Did anyone see the Spring Game on ESPN or at Williams-Brice?

I didn't think of anywhere else to post this without bothering all of the other, non SC people, so I posted it here.

Congrats to Duke on losing to Michigan State!!:rotfl:

I think that they will go to all the way

I like the NC flag! They could put 1 flag on top of the other.


Let $n,k$ two positive integrers with $n>k$. Prove that $\frac{1}{n+1} \cdot \frac{n^n}{k^k(n-k)^{n-k}} < \frac{n!}{k! (n-k)!} < \frac{n^n}{k^k(n-k)^{n-k}}$

Don't repost problems especially if you know that they already in resources !  :mad:
http://www.mathlinks.ro/Forum/viewtopic.php?p=469045#p469045


It says that the deadline for when the aplicant will be informed of the desicion was 4/22/09 but what if the application form takes longer to get there?

 im from NJ and they said it would take 3 days to reach them so i dont think it should be a problem for me
but i was just wondering how it would affect other people.

Yea... Are they going to receive it in time? It wont or shouldnt count as late though... Since most people are already getting their acceptance now..

im not too worried i guess because its postmarked so it should be fine.

I just scanned my sheets, put the images in a word document, and emailed a huge file to them. heh.


trigonometry geometry circumcircle inradius

Prove that, in a triangle $ ABC$,

$ \sin^{2}\frac{A}{2} \plus{}\sin^{2}\frac{B}{2} \plus{} \sin^{2}\frac{C}{2} \plus{} 2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \equal{} 1$.

hello, using the formulas
$ \sin\left(\frac{\alpha}{2}\right)\equal{}\sqrt{\frac{(s\minus{}b)(s\minus{}c)}{bc}}$
$ \sin\left(\frac{\beta}{2}\right)\equal{}\sqrt{\frac{(s\minus{}a)(s\minus{}c)}{ac}}$
$ \sin\left(\frac{\beta}{2}\right)\equal{}\sqrt{\frac{(s\minus{}a)(s\minus{}c)}{ac}}$
and plugging these in the left-hand side of your equaton we get
$ {\frac { \left( s\minus{}b \right)  \left( s\minus{}c \right) }{bc}}\plus{}{\frac {
 \left( s\minus{}a \right)  \left( s\minus{}c \right) }{ac}}\plus{}{\frac { \left( s\minus{}a
 \right)  \left( s\minus{}b \right) }{ab}}\plus{}2\,{\frac { \left( s\minus{}a \right)
 \left( s\minus{}b \right)  \left( s\minus{}c \right) }{abc}}$
with $ s\equal{}\frac{a\plus{}b\plus{}c}{2}$
now setting
$ a\equal{}y\plus{}z,b\equal{}x\plus{}z,c\equal{}x\plus{}y$ so we have
$ {\frac {yz}{ \left( x\plus{}z \right)  \left( x\plus{}y \right) }}\plus{}{\frac {xz}{
 \left( y\plus{}z \right)  \left( x\plus{}y \right) }}\plus{}{\frac {xy}{ \left( y\plus{}z
 \right)  \left( x\plus{}z \right) }}\plus{}2\,{\frac {xyz}{ \left( y\plus{}z \right)
 \left( x\plus{}z \right)  \left( x\plus{}y \right) }}$
and after a short computation we get $ 1$ as the searched result.
Sonnhard.

Thanks once again, Sonnhard.

Is there an easier/quicker way?


Mornin'
Let X be a partial ordered set. Is there a set of sets Y (partial ordered by subset' s relation), isomorphic with X?

I mean, that there exists f from X onto Y (bijection) and ordered preserved.

I don' t know if there is a solution.

Bye

yes, there is. for each $ x\in X$, define $ S(x) \equal{} \{y\in X|y\leq x\}$, and define $ (Y,\subseteq)$ to be the poset whose order is set inclusion, "$ \subseteq$," and whose elements are the sets $ S(x)$, for $ x\in X$. now we show that the two posets are isomorphic. the isomorphism should naturally be given by $ f: X\to Y$, $ f(x) \equal{} S(x)$. first, this map is bijective because if $ x\neq y$, then $ S(x)\neq S(y)$ since otherwise we'd have $ x\leq y$ and $ y\leq x$, which would imply $ x \equal{} y$. the map preserves order because if $ x\leq y$, and $ r\leq x$ (i.e., $ r\in S(x)$), then $ r\leq y$ (by transitivity of the partial order), implying $ r\in S(y)$.


Melvyn has two passengers in his car as he drives slowly through Edgware - one is in the front and the other in the back. When they arrive at their destination, they meet up with 4 others.

As they all get out, it starts to rain heavily. Six of them start walking fast but still get wet. However, the passenger from the back of Melvyn’s car remains completely dry, even though he makes no attempt to hurry and is wearing no hat, coat or umbrella.

How is this possible?

i dont know...does he stop time?
-jorian :P

[hide]he had someone drive him into the house[/hide]

uhh..probably not[hide]the guy standing next to him is so tall that he blocks the rain.[/hide]

[quote="rgrgrg"][hide] There are 7 guys (Melvyn+2+4), but only 6 move.  The guy who remains dry does not walk through the rain, but rather remains under something, possibly a garage or maybe the car, as to not get wet. [/hide][/quote]If he was under the car he would get squashed like a bug.

I hope this is right:
[hide]The "passenger from the back" is in a coffin. It's his funeral, and the other six are rushing because they're carrying the coffin.[/hide]

[quote="7h3.D3m0n.117"][quote="rgrgrg"]I hope my funeral has more than 6 people at it.[/quote]

 :rotfl:  :rotfl:  :rotfl:  lol the funeral answer i think is the best[/quote]

I was just gonna say that, ive heard that question b4.

***Spoiler Ahead ***
>
>
>
>
>
.
.
>
>
>
There is nothing to say that there is a dead person in a coffin.
In fact, here is a case where the solution can be David Blaine
in the coffin, and he will be buried alive (using shallow breathing and perhaps a hidden oxygen breathing device inside the walls of the coffin) as a stunt for Target stores in order to raise money for charity.


calculus
integration
trigonometry
calculus computations

Find the minimum value of $ \int_0^{\frac{\pi}{2}} |x\sin t\minus{}\cos t|\ dt\ (x>0).$

$ \int_{0}^{\pi /2}\left| x\sin t\minus{}\cos t\right| dt$

$ \equal{}\sqrt{x^{2}\plus{}1}\left[ \int_{0}^{\tan ^{\minus{}1}\frac{1}{x}}\sin (\tan ^{\minus{}1}\frac{1}{x}\minus{}t)dt\plus{}\int_{\tan ^{\minus{}1}\frac{1}{x}}^{\pi /2}\sin (t\minus{}\tan ^{\minus{}1}\frac{1}{x})dt\right]$

$ \equal{}\allowbreak 2\sqrt{1\plus{}x^{2}}\minus{}x\minus{}1\allowbreak$

At $ \ x\equal{}\frac{1}{\sqrt{3}},$ $ \left( \int_{0}^{\pi /2}\left| x\sin t\minus{}\cos t\right| dt\right) _{\min }\equal{}\sqrt{3}\minus{}1$

[hide="Hint:"]Let $I:= \int_{0}^{\frac{\pi}{2}}|m\sin x-\cos x|\ dx;$($m>0$)
We have:
$m\sin x-\cos x=\left (\frac{m}{\sqrt{m^2+1}}\sin x-\frac{1}{\sqrt{m^2+1}}\cos x  \right )\sqrt{m^2+1}
 = \sqrt{m^2+1}.\sin(x-\alpha);$
$\Rightarrow \frac{I}{\sqrt{m^2+1}}=\int_{0}^{\frac{\pi}{2}}\left |\sin(x-\alpha)  \right |\ d(x-\alpha)$
$ \overset{(x-\alpha)=t}{\rightarrow} \frac{I}{\sqrt{m^2+1}}=\int_{-\alpha}^{-\alpha+\frac{\pi}{2}} \left |\sin t  \right |\ dt$;
 where: $\sin \alpha=\frac{1}{\sqrt{m^2+1}}$[/hide]


can some one tell me if dna is water or fat soluable?? please explain... cause i need to knwo what will happen when dna is mixed with water,..

..and also there is one more question if u can answer ......"What happened when the isopropanol was added to the liquid that passed through the filter? What does this tell you about the behaviour of DNA?"------ i knwo when we mix it the dna begins to appear and floats to the surface... but what is its behavious?

thank you-------
- titan

DNA is soluble in water. Also, isopropanol causes the DNA to precipitate.  :)

Any hydrophobic organic substance is a lipid. as DNA is a nucleotide polymer, it is water hydrophilic, and thus water soluble.


OK, not everyone has AoPS volume 2, but I'm pretty sure most everybody here has it, so here's the question:

On example 6-12, the question is "Find the polynomial whose roots are twice those of x4 -3x 2 +x-9."

Well, the answer is x4 -12x 2 +8x-144.

I understand how you have to multiply the constant by 24 because its xyza, which is 2*2*2*2. And I understand why you have to multiply the x 2  coefficient by 2 2 , because the coefficient is the 2nd symmetric sum, which is xy+xa+..... and it has 2*2 factors in each term.

What I DON'T get is why you multiple the x coefficient by 23 because isn't the coefficient supposed to be the kth symmetric sum? In this case the 1st.  And so, wouldn't it be x+y+z+a, which would be times 2 if the roots were 2 times?  So I'm confused about why you times 8.  Could someone help me clear it up?

Mr.Ocax

Did you read the explanation at the beginning of the solution?  I'm a bit confused to what you're asking because it didn't mention symmetric sums at all.

well, yes, I should have specified that I understood the explanation and everthing, but if you actually think about it, it doesn't make sense.  b/c the coeffecient of the x is the sum of the roots, which, after you subsitutue in twice the roots, you would get 2x+2y+2a+2x=2(x+y+a+z)

Mr.Ocax

No. You got it backwards. the 1st symmetric sum in this case is:

- coefficient of x^3/coefficient of x^4.

-coefficient of x/x^4 is the third symmetric sum:
abc + abd + acd + bcd =  -1
thus we had 1 as the coefficient of the x term of the old equation

since it is looking for:

8abc + 8abd + 8acd + 8bcd = ???

multiply the former by 8, then we have 8 as coefficient of the x term in the new equation.

Nearly all discussions of AoPS Volume 2 material belongs in the Intermediate forum.  I'll move this there.

Consider the polynomial $(x-r_1)(x-r_2)(x-r_3)(x-r_4)$.
The coefficients are $1, -\sum_{cyc} r_1, \sum_{cyc} r_1r_2, -\sum_{cyc} r_1r_2r_3, r_1r_2r_3r_4$ upon expansion.

Now consider the polynomial $(x-2r_1)(x-2r_2)(x-2r_3)(x-2r_4)$.
The coefficients are $1, -\sum_{cyc} 2r_1, \sum_{cyc} 4r_1r_2, -\sum_{cyc} 8r_1r_2r_3, 16r_1r_2r_3r_4$ upon expansion.

So the coefficient of $x^n$ is multiplied by a factor of $2^{4-n}$.


trigonometry
function
real analysis
real analysis theorems

prove using the ε- δ definition that the limit of :

$ lim_{n\to 0} \frac{sinx}{x} \equal{} 1$

What is the point of wanting to use the raw definition here? There's a reason why we prove theorems (the Squeeze Theorem, for instance): so we can use theorems in useful cases rather than always being stuck with the raw definition.

that still does not answer my question

Can, or can we  not prove the above limit by using [b]only the ε-δ definition[/b]??

Ah, this should be an entertaining thread.   [hide=":roll:"]Yes, yes, I know. :spam: [/hide]

An $ \varepsilon \minus{} \delta$ proof (I hope you will find it satisfactory).

[u]Lemma [/u] If $ f(x)\to L$ and $ g(x)\to L$ for $ x\to0,$ and $ f(x) \le h(x) \le g(x)$
for $ x\in ( \minus{} r,r),$ $ r > 0,$ then $ h(x)\to L$ for $ x\to0.$

We need the proof. The hypotheses are

1) $ \forall\ \varepsilon > 0\ \exists\ \delta_\varepsilon\,\colon\ x\in( \minus{} \delta_\varepsilon, \delta_\varepsilon)\ \Longrightarrow\ L \minus{} \varepsilon < f(x) < L \plus{} \varepsilon$

2) $ \forall\ \varepsilon > 0\ \exists\ \delta'_\varepsilon\,\colon\ x\in( \minus{} \delta'_\varepsilon, \delta'_\varepsilon)\ \Longrightarrow\ L \minus{} \varepsilon < g(x) < L \plus{} \varepsilon$

It follows
$ \minus{} \varepsilon < h(x) \minus{} L < \varepsilon$ for $ \vert x \vert < \tilde\delta_\varepsilon$
where $ \tilde \delta_\varepsilon$ is the smaller between $ \delta_\varepsilon$
and $ \delta'_\varepsilon$ (End of the proof)

We apply the lemma to $ \cos x < {\sin x \over x} < 1$ (standard trigonometry).

In this case $ \delta'_\varepsilon$ may be any positive number due to the constancy of the
function $ g(x)\equiv1.$ We only need to estimate $ \delta_\varepsilon.$

Now (let's take $ \varepsilon < 1$) $ 1 \minus{} \varepsilon < \cos x < 1$  is equivalent to
$ \sqrt {1 \minus{} (1 \minus{} \varepsilon)^2} > \sin x > \minus{} \sqrt {1 \minus{} (1 \minus{} \varepsilon)^2}.$ Since
$ \vert \sin x \vert \le \vert x \vert$ we take $ \delta_\varepsilon \equal{} \sqrt {1 \minus{} (1 \minus{} \varepsilon)^2}$ and this the $ \delta$ you are asking (I hope).

The real question to ask is not what methods you're allowed to use: it's what definition of $ \sin$ you're using.

A [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=120492]previous discussion[/url] of this limit.


A chord $\overline{ST}$ of constant length slides around a semicircle with diameter $\overline{AB}.$ $M$ is the midpoint of $\overline{ST}$ and $P$ is the foot of the perpendicular from $S$ to $\overline{AB}$. Prove that angle $SPM$ is constant for all positions of $\overline{ST}.$

Hmm, not as hard as it seemed when i first read it. Appearances can be deceiving. :D

[hide="Solution"]Let the center of the semicircle be $O.$ It follows that $\overline{OM}$ is the perpendicular bisector of $\overline{ST}.$

Hence, $\angle SPO = \angle SMO =90^\circ$ and quadrilateral $MOPS$ is cyclic. It follows that
\[2 \angle SPM = 2 \angle SOM= \angle SOT, \]
or the measure of the arc intercepted bt chord $\overline{ST}.$ Since the length of $\overline{ST}$ is constant, $\angle SPM$ is constant. $\mathbb{QED}$[/hide]


Let $a_{1}, a_{2}, ...a_{n}$ be a permutation of the set $S_{n}={1,2,3...n}$. An element $i$ in $S_{n}$ is called a fixed point of this permutation if $a_{i}=i$. Let $g_{n}$ be the number of derangements of $S_{n}$. Let $f_{n}$ be the number of permutations of $S_{n}$ with exactly one fixed point. Show that $|f_{n}-g_{n}|=1$.

[hide]$f_{n}$=$n(g_{n-1})$  This can be seen if one tries to visualize it.  (There's n ways to make one fixed point and since everything else is different, it's $g_{n-1}$).
$g_{n}$=$(n-1)(g_{n-1}+g_{n-2})$.  I will prove this later.
$n(g_{n-1})-(n-1)(g_{n-1}+g_{n-2})=g_{n-1}-(n-1)(g_{n-2})=g_{n-1}-f_{n-1}$  Since these are in absolute value signs, the multiplication by -1 won't matter.
$|f_{n}-g_{n}|=|f_{n-1}-g_{n-1}|=...=|f_{1}-g_{1}|=1$
Now, the proof that I held off.  Recall the other formula for derangement, $n!/0!-n!/1!+n!/2!-...+\frac{(-1)^{n}n!}{n!}$.  When one applies this formula to the three gs and multiplies through, $n!/0!-n!/1!+n!/2!-...+\frac{(-1)^{n}n!}{n!}=(n-1)(n-1)!/0!-(n-1)(n-1)!/1!+(n-1)(n-1)!/2!-...+\frac{(-1)^{n-1}(n-1)(n-1)!}{(n-1)!}+(n-1)!/0!-(n-1)!/1!+(n-1)!/2!-...+\frac{(-1)^{n-2}(n-1)!}{(n-2)!}$.  When all the equal terms on the left and right sides are subtracted off, $\frac{(-1)^{n-1}(n)!}{(n-1)!}+\frac{(-1)^{n}n!}{n!}=\frac{(-1)^{n-1}(n-1)(n-1)!}{(n-1)!}$ or $n-1=n-1$ or $1-n=1-n$.[/hide]


Five distinct points $ A, B, C, D$ and $ E$ lie on a line, but not
necessarily in that order. Use the information below to determine
the number of units in the length of segment $ DC$.

- $ E$ is the midpoint of segment $ AB$.

- $ D$ is the midpoint of segment $ AE$.

- Both $ C$ and $ E$ are the same distance from $ B$.

- The distance from $ D$ to $ B$ is 9 units.

15. mem.

So you set up the points as shown [geogebra]b9916e84c316021e8cc3cdcb50344021688d437d[/geogebra].

As you can see, $ DB\equal{}\frac12EB\plus{}EB\equal{}9$
$ EB\equal{}6$
$ EB\equal{}BC\equal{}6$
$ DC\equal{}DB\plus{}BC\equal{}9\plus{}6\equal{}15$

how does 1/2EB+EB=9and how does BC equal 6


calculus
integration
logarithms
algebra
partial fractions
calculus computations

Can one solve this integral using substitution only?
$\int{\frac{x^{8}}{(x^{4}-1)^{3}}\,dx}$

The quick answer is yes.  If interested, your solution will look like this:

[hide]

$\frac{1}{128}\left(-\frac{16 x}{(x^{4}-1)^{2}}-\frac{36 x}{x^{4}-1}-10 \arctan x+5 \log (x-1)-5 \log (1+x) \right)$

[/hide]

(and yes, I cheated).

I know the solution, this is the integral of rational function. But I want to know if  you can get that solution using only substitution.

The expression being integrated is equal to
\[\frac{1}{x^{4}-1}+\frac{2}{(x^{4}-1)^{2}}+\frac{1}{(x^{4}-1)^{3}}\]
The substitutions $\frac{x}{(x^{4}-1)^{2}}$ and $\frac{x}{x^{4}-1}$ are good choices for getting rid of the terms with denominators $(x^{4}-1)^{3}$ and $(x^{4}-1)^{2}$, respectively. You'll be left with the integral
\[\int \frac{1}{x^{4}-1}\ dx\]
which is easy enough by partial fractions, but the answer will involve $\ln |x+1|$, $\ln |x-1|$, and $\arctan x$, so don't expect a nice substitution without already solving the integral.


We know that the relative error defined:

$ \delta\tilde{x} \equal{} \frac {\Delta\tilde{x}}{x} \equal{} \frac {x \minus{} \tilde{x}}{x} \newline$
$ \delta\tilde{y} \equal{} \frac {\Delta\tilde{y}}{y} \equal{} \frac {y \minus{} \tilde{y}}{y}$

and so:

$ x \equal{} \tilde{x}(1 \plus{} \delta{x}) \newline$
and
$ y \equal{} \tilde{y}(1 \plus{} \delta{y}) \newline$

So to prove that in multiplication, the relative error sums:

$ xy \equal{} \tilde{x}\tilde{y}(1 \plus{} \delta\tilde{x})(1 \plus{} \delta\tilde{y}) \equal{} \tilde{x}\tilde{y}(1 \plus{} \delta\tilde{x} \plus{} \delta\tilde{y} \plus{} \delta\tilde{x}\delta\tilde{y} )$

Can someone explain why $ \delta\tilde{x} \plus{} \delta\tilde{y} \plus{} \delta\tilde{x}\delta\tilde{y} \approx \delta\tilde{x}\tilde{y}$ ?

By that, the proof is complete, but I dont understand that last part.

Thanks alot.

This is wrong. Should be: $ \delta\tilde{x} \plus{} \delta\tilde{y} \plus{} \delta\tilde{x}\delta\tilde{y} \approx \delta\tilde{x} \plus{} \delta\tilde{y}$ because $ \delta\tilde{x}\delta\tilde{y} < < \delta\tilde{x}$ and $ \delta\tilde{x}\delta\tilde{y} < < \delta\tilde{y}$ ;)

$ \boxed{xy\approx \tilde{x}\tilde{y}(1\plus{}\delta\tilde{x} \plus{} \delta\tilde{y})}$


analytic geometry
real analysis
real analysis unsolved

Suppose $k \geq 3$, $x,y \in \mathbb{R}^{k}$, $|x-y|=d>0$ and $r>0$. Show that:

(a) If $2r>d$, then there exist infinitely many $z \in \mathbb{R}^{k}$ such that $|z-x| = |z-y| = r$.

Hint: the set of such $z$ forms a $k-2$ dimensional sphere.

Yes I can see that from a geometrical point of view. But I have problems writing it down in a precise way, from the analytical point of view.

Introduce a new coordinate system s.t. the point $x$ has coordinates $(d/2,0,\dots, 0)$; the point $y$ has coordinates $(-d/2,0,\dots,0)$. Write the equations $|z-x| = r$ and $|z-y| = r$ in terms of coordinates $z_{i}$.


geometry
circumcircle
geometric transformation
reflection
parallelogram
power of a point
radical axis

[quote="Ulanbek_Kyzylorda KTL"]In an acute triangle ABC, let H be intersection of its heights and D be midpoint of AC. Show that the line Dh passes through an intersection point of the circumcircle of ABC with the circle for which BH is a diameter[/quote]
Reflect $ H$ with respect to $ AC$. Suppose $ H \to H'$, so $ \angle{AH'C}\equal{}\angle{AHC}\equal{}180^o\minus{}\angle{ABC}$. So $ H'$ is on the circumcircle of $ \triangle{ABC}$.

Let $ BO$ (O is the circumcenter of triangle ABC) meets the circumcircle at $ E$, so $ \angle{AH'E}\equal{}90^o$. Hence $ H'E \parallel AC$. Since $ H'$ and $ H$ are symmetric WRT $ AC$, so we get that $ AC$ passes through the midpoint of $ HE$. Since $ AH \perp BC$ and $ EC \perp BC$, so $ AH \parallel EC$. Hence $ EH$ also passes through the midpoint of $ AC$, which is $ D$. So let the other intersection of $ DH$ with the circumcircle of $ \triangle{ABC}$ be $ F$, we get that $ \angle{BFH}\equal{}90^o$. Hence $ F$ is also on the circle with $ AH$ being the diameter.

[quote="Ulanbek_Kyzylorda KTL"]In an acute triangle ABC, let H be intersection of its heights and D be midpoint of AC. Show that the line Dh passes through an intersection point of the circumcircle of ABC with the circle for which BH is a diameter[/quote]
We have $ OD\equal{}O'H$ and also $ OD \parallel OH$. So, $ OO'HD$ is a parallelogram and $ OO'\parallel DH$. Let $ X\equal{}(O)\cap(O')$ and $ BX$ is the radical axis of $ (O)$. So, $ (O')$. Also, $ OO' \perp BX \Longrightarrow DH \perp BX$. Let $ BX \cap DH\equal{}X'$, $ HX \perp BX$ (As $ BXA'H$ concyclic) and $ HX' \perp BX \Longrightarrow X \equiv X'$ and we are done.

Let $ X,Y,Z$ be the feet of the altitudes issuing from $A,B,C.$ Inversion with pole $H$ and power $\overline{AH} \cdot \overline{HX}$ takes 9-point circle passing through $ Y, D$ into the circumcircle $(O)$ of $ ABC$ and sideline $BC$ into the circle with diameter $ \overline{BH}$ $\Longrightarrow$ Second intersection $P$ of $(O)$ with  the circle with diameter $\overline{BH}$ is the inverse of $ D$ under the referred inversion $\Longrightarrow$ Points $D,H$ and $ P$ are collinear.

Actually the main idea behind my solution is also a transformation, that is, to translate the dotted circle to $ (O')$ :)
BTW I don't think that solutions that use inversion are quite elementary. :wink:


We have two decks of cards. Each with $n$ cards. There existis a matching between the decks (for every card in the first deck there is an identical one in the second, and vicecersa..; a bijection). We shuffle one of the decks and place it on top of the second non-shuffled deck. For each card on the top shuffled deck we count the number of cards between it and its corresponding card in the second deck. What is the sum of these numbers?

err not sure if this is right but

[hide]

i thought of it as for the first deck, since the cards aren't in order i just labeled different "spots"

so the first card of the first deck(regardless of what it is) is one card above the 2nd deck and is in the first "spot"

and so on until the very top card of the first deck is in the 52nd "spot"

and i denoted these spots as $n_1$

then since the 2nd deck is unshuffled, every spot corresponds with the value of the card, so the last card on the bottom is the 52nd card, and i denoted these values as $n_2$

So for example, say the 4 of diamonds was 5 cards above the 2nd deck, and the corresponding 4 of diamonds would be 4 cards into the 2nd deck, so the number of cards inbetween is:

$(5-1) + (4-1) = 7$

thus, this will be true in all cases and you get:

$(n_1 - 1) + (n_2 - 1) =$ number of spots in between

So you are looking for:

$\displaystyle\sum_{1}^{52} (n_1 + n_2 - 2) = \sum_{n_1 = 1}^{52} n_1 + \sum_{n_2=1}^{52} n_2 - \sum_{1}^{52} 2$

This is just asking for the sum of all the possible values of $n_1$ and $n_2$ and subtract $(52)(2)$

So you get:

$2\displaystyle\frac{(53)(52)}{2} - 52(2) = (53)(52) - (2)(52) = 51(52) = 2652$

[/hide]

[quote]We shuffle one of the decks and place it on top of the second non-shuffled deck[/quote]
You might, if you please, shuffle the second deck too .. answer is not  to going  change...:)

[hide]Start with  unshuffled decks, if you  interchange any two cards in the top   (or bottom , for that matter) the sum is not going to change... and if you can keep interchanging ..you can reach any permutation..:).. so you might as well  do the calculation when the decks are unsuffled ..:)  [/hide]

When any two cards in one deck are switched, one distance increases by some number, and the other decreases by the same number.  Thus, we can consider any case, and the sum will be the same.  Consider them unshuffled, arranged the same way.  Then each distance will be $n-1$.  There are $n$ of these distances, so the final answer is $n(n-1)$.

yep.. my aprroach yields $n^2$ (because in the distance i counted the first card too) here it goes:
Let $a_i$ be the position of the top card $i$ counting cards from the bottom and let $b_i$ the position of the bottom card $i$ counting it from the bottom too.
so this implies that the sum must be $ \sum_{i=1}^{n}(a_i-b_i)$ and this it is easy to see that it is $\sum_{i=1}^na_i-\sum_{i=1}^{n}b_i =n^2$


geometry
circumcircle
rectangle
geometry unsolved

Given a circle k with the diameter AB, let P be an arbitrary point of the circle different from A and B. The projections of the point P to AB is Q. The circle with center P and radius PQ intersects k at C and D. Let E be the intersection of CD and PQ. Let F be the midpoint of AQ, and G is the foot of perpendicular from F to CD. Prove that EP=EQ=EG and that A, G and P are colinear.

Let $ X,Y$ be the othogonal projections of $ Q$ onto $ PB,PC.$ The inversion with center $ P$ and radius $ PQ$ takes the circumcircle $ k'$ of the rectangle $ PXQY$ into the line $ AB$ and the line $ CD$ into $ k.$ Since $ AB$ is orthogonal to $ k,$ then $ k'$ is othogonal to $ CD$ $ \Longrightarrow$ $ CD$ passes through $ X,Y$ and the midpoint $ E$ of $ PQ.$ Circles with diameters $ AQ$ and $ PQ$ are orthogonal to each other,  thus $ FY \perp EY$ $ \Longrightarrow G \equiv Y.$ Then $ P,G,A$ are collinear and $ EP \equal{} EQ \equal{} EG.$

We have $ \mathcal {P}_{E/(k;R)} \equal{} CE.ED \equal{} R^2 \minus{} OE^2 \equal{} \mathcal {P}_{E/(P;PQ)} \equal{} PQ^2 \minus{} PE^2$.Since we have $ R^2 \minus{} OE^2 \equal{} PQ^2 \minus{} PE^2$=> $ EQ^2 \plus{} QO^2 \plus{} PQ^2 \equal{} R^2 \plus{} PE^2$=> $ PE \equal{} EQ$ (QED) :)

We may use twice the inversion, at $P$ and at $A$, easy!

Best regards,
sunken rock


I just thought that no one was starting new marathons nowadays, so I decided to start one. but first, a few rules:
1) please, PLEASE don't post anything that'll get this locked.
2) keep it within classroom math range.

with that out of the way, I'll start:

[b]Problem #1[/b]

x squared - x = 42

[hide]
$ x^2 \minus{} x \equal{} 42 \implies x^2 \minus{} x \minus{} 42 \equal{} 0 \implies (x \minus{} 7)(x \plus{} 6) \equal{} 0 \implies \boxed{x \equal{} 7\ \text{or}\ x \equal{} \minus{} 6}$
[/hide]

Note: There already exists a classroom math marathon in this thread: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194761[/url]. For future problems and solutions, post in that thread.

It never said solve for x. Maybe it means write in latex form $ x^2\minus{}x\equal{}42$. Maybe it means write in the form ax^2+bx+c: $ x^2\minus{}x\minus{}42\equal{}0$

I think that he meant that you should solve for x ;)


modular arithmetic
geometry
5th edition
geometric transformation
reflection
homothety
ratio

Could someone give me a hint? :? About 3.2.

It is clear $n$ is odd. My idea was to consider a number $m=\frac{n-1}{2}$. Then we know that $m\pmod{p}\leq \frac{p-3}{2}$ for all odd primes $p\leq m$.

Thank you,Myth.
But problem 3.2 is geometry,isn't it?

Ops... I saw number 2...  :blush:

There you need to prove that circles, constructed on $AA'$ and $BB'$, are perpendicular to the circumcircle.

Problem #3:

Let $ABC$ be a triangle and $A' \in BC$, $B' \in CA$ and $C' \in AB$ be three collinear points.
(a) Prove that that each pair of circles of diameters $AA', BB'$ and $CC'$ has the same radical axis
(b) Prove that the circumcenter of the triangle formed by the intersections of the lines $AA', BB'$ and $CC'$ lies on the common radical axis found above.

For part (a), see Coxeter & Greitzer, Geometry Revisited, chapter 2, paraghraph 4, theorem 2.43 (or 2.47, some editions do not have the theorems numbered correctly). A different proof is similar to the solution of problem  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31399]"Prove 6 points lie on a circle"[/url].

[color=blue][b](a)[/b][/color] Let $K, L, M$ be the midpoints of the segments $AA', BB', CC'$, i.e., centers of the circles $(K), (L), (M)$ with diameters $AA', BB', CC'$. Let $D, E, F$ be the intersections of the circles $(K), (L), (M)$ with the sides $BC, CA, AB$ different from the points $A', B', C'$. Since the angles $\measuredangle ADA', \measuredangle BEB', \measuredangle CFC'$ are all right, the points $D, E, F$ are the feet of altitudes of the triangle $\triangle ABC$ and these altitudes intersect at the triangle orthocenter $H$. The intersections $D', E', F'$ of the altitudes $AD, BE, CF$ with the circumcircle $(O)$ of the triangle $\triangle ABC$ are reflections of the orthocenter $H$ in the triangle sides $BC, CA, AB$. The power of the orthocenter $H$ to the circumcircle $(O)$ is

$HA \cdot HD' = HB \cdot HE' = HC \cdot HF'$.

Since $HD' = 2HD, HE' = 2HE, HC' = 2HC$, we also get

$HA \cdot HD = HB \cdot HE = HC \cdot HF$.

But these are powers of the orthocenter $H$ to the circles $(K), (L), (M)$. Thus the orthocenter is either the radical center of the circles $(K), (L), (M)$ or these 3 circles belong to the same pencil and the orthocenter is on the common radical axis of this pencil.

Let $A_1, B_1, C_1$ be the midpoints of the sides $BC, CA, AB$, forming the triangle $\triangle A_1B_1C_1$. From similarity of the triangle $\triangle ABC$ and its medial triangle $\triangle A_1B_1C_1$ with homothety center $G$ (their common centroid) and homothety coefficient -2, it follows that the midpoints $K, L, M$ (i.e., the centers of the circles $(K), (L), (M)$) are on the side lines $B_1C_1, C_1A_1, A_1B_1$ of the medial triangle. Since the given points $A', B', C'$ are collinear, by Menelaus' theorem,

$\frac{AC'}{BC'} \cdot \frac{BA'}{CA'} \cdot \frac{CB'}{AB'} = 1$

From similarity of the triangle $\triangle ABC$ and its medial triangle $\triangle A_1B_1C_1$,

$\frac{AC'}{BC'} = \frac{B_1M}{A_1M},\ \ \frac{BA'}{CA'} = \frac{C_1K}{B_1K},\ \ \frac{CB'}{AB'} = \frac{A_1L}{C_1L}$

$\frac{B_1M}{A_1M} \cdot \frac{C_1K}{B_1K} \cdot \frac{A_1L}{C_1L} = \frac{AC'}{BC'} \cdot \frac{BA'}{CA'} \cdot \frac{CB'}{AB'} = 1$

Again, by Menelaus' theorem, the points $K, L, M$ are collinear. 3 circles with collinear centers have their pairwise radical axes parallel. Hence, a finite point with the same power to all 3 circles exists only if these 3 radical axes are identical, i.e., if the 3 circles belong to the same pencil. Since the circles $(K), (L), (M)$ obviously intersect (the orthocenter on their common radical axis is inside of each), they form an elliptic pencil intersecting at with 2 base points $X, Y$. This concludes the proof of part (a).

[color=blue][b](b)[/b][/color] WLOG, assume that the points $A', B'$ are inside of the segments $BC, CA$ and the point $C'$ on the line $AB$ is outside of the segment $AB$. (The case, when the line $A'B'C'$ intersects neither of these segments, is handled in exactly the same way.) Let $P \equiv BB' \cap CC'$, $Q \equiv CC' \cap AA'$, $R \equiv AA' \cap BB'$ be the vertices of the triangle $\triangle PQR$ formed by the lines $AA', BB', CC'$. This triangle is cut by 4 lines $C'AB$, $BA'C$, $CB'A$, $A'B'C'$ and we can apply Menelaus' theorem to each:

the points $C', A, B$ are on the side lines $C' \in PQ,\ A \in QR,\ B \in RP$,
the points $C, A', B$ are on the side lines $C \in PQ,\ A' \in QR,\ B \in RP$
the points $C, A, B'$ are on the side lines $C \in PQ,\ A \in QR,\ B' \in RP$
the points $C', A', B'$ are on the side lines $C' \in PQ,\ A' \in QR,\ B' \in RP$

$(1)\ \ \ \frac{PC'}{QC'} \cdot \frac{QA}{RA} \cdot \frac{RB}{PB} = 1$

$(2)\ \ \ \frac{PC}{QC} \cdot \frac{QA'}{RA'} \cdot \frac{RB}{PB} = 1$

$(3)\ \ \ \frac{PC}{QC} \cdot \frac{QA}{RA} \cdot \frac{RB'}{PB'} = 1$

$(4)\ \ \ \frac{PC'}{QC'} \cdot \frac{QA'}{RA'} \cdot \frac{RB'}{PB'} = 1$

For example, dividing equation (1) by (4) and equation (2) by (3), we get

$\frac{QA}{RA} \cdot \frac{RB}{PB} = \frac{QA'}{RA'} \cdot \frac{RB'}{PB'}$

$\frac{QA'}{RA'} \cdot \frac{RB}{PB} = \frac{QA}{RA} \cdot \frac{RB'}{PB'}$

Multiplying the results leads to

$\frac{QA}{RA} \cdot \frac{QA'}{RA'} \cdot \left(\frac{RB}{PB}\right)^2 = \frac{QA'}{RA'} \cdot \frac{QA}{RA} \cdot \left(\frac{RB'}{PB'}\right)^2$

$\frac{RB}{PB} = \frac{RB'}{PB'}$

In a similar way, we can find $\frac{PC}{QC} = \frac{PC'}{QC'}$ and $\frac{QA}{RA} = \frac{QA'}{RA'}$. This means that the circle $(M)$ centered on the line $PQ$ and passing through the points $C, C'$ on this line is the locus of points with a constant ratio of distances $\frac{PC}{QC} = \frac{PC'}{QC'}$ from the points $P, Q$. Similarly, the circle $(K)$ centered on the line $QR$ and passing through the points $A, A'$ on this line is the locus of points with a constant ratio of distances $\frac{QA}{RA} = \frac{QA'}{RA'}$ from the points $Q, R$. Finally, the circle $(L)$ centered on the line $RP$ and passing through the points $B, B'$ on this line is the locus of points with a constant ratio of distances $\frac{RB}{PB} = \frac{RB'}{PB'}$ from the points $R, P$. Consequently,

$\frac{PX}{QX} = \frac{PY}{QY},\ \ \frac{QX}{RX} = \frac{QY}{RY},\ \ \frac{RX}{PX} = \frac{RY}{PY}$

$\frac{PX}{PY} = \frac{QX}{QY} = \frac{RX}{RY}$

Invert the figure in a circle $(X)$ centered at the base point $X$ of the elliptic pencil of the circles $(K)-(L)-(M)$ and passing through the other base point $Y$ of this pencil, i.e., with radius $r = XY$. The vertices $P, Q, R$ of the triangle $\triangle PQR$ are carried into points $P', Q', R'$ and the circumcircle $(O)$ of the triangle $\triangle PQR$ into the circumcircle of the triangle $\triangle P'Q'R'$. The point $Y$ is on the inversion circle and as such, it stays in place. The distances $P'Y, Q'Y, R'Y$ of the inverted points $P', Q', R'$ from the fixed point $Y$ are equal to

$P'Y = PY\ \frac{r^2}{XP \cdot XY} = r\ \frac{PY}{PX}$

$Q'Y = QY\ \frac{r^2}{XQ \cdot XY} = r\ \frac{QY}{QX}$

$R'Y = RY\ \frac{r^2}{XR \cdot XY} = r\ \frac{RY}{RX}$

i.e., they are the same. This means that the point $Y$ is the circumcenter of the triangle $\triangle P'Q'R'$. But the elliptic pencil of the circles $(K)-(L)-(M)$ is carried in this inversion into a pencil of lines $k'-l'-m'$ concurrent at the point $Y$. Hence, the circumcircle $(Y)$ of the triangle $\triangle P'Q'R'$ is perpendicular to every line of the inverted pencil. Since inversion preserves angles between curves, the circumcircle $(O)$ of the original triangle $\triangle PQR$ is perpendicular to all circles of the original pencil $(K)-(L)-(M)$. As a result, the center of the circle $(O)$ is on the radical axis of this pencil.

mecrazywong used menelaus to finish 3.2. Maybe you can ask him for hte solution. (It's quite short)

OK, here is my solution, but I didn't use Menelaus.
Actually proving the 4 points are harmonic conjugate use Menelaus, but as it is well known, so I didn't mention the proof.

what is an elliptic pencil?
how do we find the points $X$ and $Y$?

mecrazywong, i don't understand your notation. who is $L$?

I hope this is different from the other posted solutions:

Let $D,E,F$ be the feet of the perpendiculars from $A,B,C$ respectively (in $ABC$). Also, let $H$ be the orthocenter of $A,B,C$. The circles $(AA'),(BB'),(CC')$ pass through $D,E,F$ respectively, so the powers of $H$ wrt these three circles are $HA\cdot HD,HB\cdot HE,HC\cdot HF$, which are equal. In the same way we can prove that the powers of the orthocenter of $AB'C'$ wrt the three circles are equal, and the same goes for the triangles $A'B'C,A'BC'$. This means that the orthocenters must lie on a common axis of the circles.

[quote="Myth"]In order to prove it you need to prove the following lemma.

Lemma.
Suppose cevians $AY$, $BB'$, $CC'$ intersects in the point $A'$. Then $A''X\cdot A''Y=A''A'^2$, where $A''$ is the midpoint of $AA'$ (see the figure below).

I don't know how other people did it, but my way is described above.[/quote]

i can't prove it
can anyone help me?

edit: nevermind

[quote="Pascual2005"]I dont know what that hapened to me, but instead of problem 2, i sent a solution of another problem in the forum!, crazy, ah? :D[/quote]I am sorry :( I knew that but I could not warn you unfortunately. There were many students that sent wrong solutions to that problem and had corrections afterwards :)

[quote="perfect_radio"]
mecrazywong, i don't understand your notation. who is $L$?[/quote]
L is the circumcenter.

[quote="perfect_radio"]what is an elliptic pencil?
how do we find the points $X$ and $Y$?
[/quote]

A pencil of tangent circles is called parabolic, a pencil of circles intersecting at 2 points is called elliptic, and a pencil of circles with no common points is called hyperbolic. The pencil of circles $(K)-(L)-(M)$ is elliptic, because the orthocenter $H$, which lies on the radical axis of this pencil, is inside of each of the 3 circles. The 2 intersection points $X, Y$ of an elliptic pencil of circles are called the base points of the pencil.

this my solution:
a)
first i want nameing some points and circles.let $\omega_1,\omega_2,\omega_3$ be circles with diameters $AA',BB',CC'$ repectively and $H_1,H_2,H_3,H_4$ be orthcenters of triangles $\triangle ABC, \triangle BA'C', \triangle B'C'A,\triangle B'CA'$
respectively and  let $AH_3,C'H_3,B'H_3$ meet $B'C',AB',AC'$ at $A'',B'',C''$
respectively ,we have :
$\angle C'C''C=90$ thus $C''\in \omega_3$ and similary
$B''\in \omega_2$ ,$A'' \in \omega_1\\$ but the power of $H_3$ into $\omega_1,\omega_2,\omega_3$ is
$H_3A''.H_3A$ and $H_3B''.H_3B'$ and $H_3C'.H_3C''$ respetively \\and $H_3$ is the orthoceter of $\triangle B'C'A$ thus:
$H_3B''.H_3B'=H_3A''.H_3A=H_3C'.H_3C''\\$ thus the power of $H_3$ into $\omega_1,\omega_2,\omega_3$ is equal, similary we can show that the power of
$H_1,H_2,H_4$ into$\omega_1,\omega_2,\omega_3$ are equal too ,so the centers of $\omega_1,\omega_2,\omega_3$
are coliner and they have the same radical axis and the solution is complete.
b)
let$O_1,O_2,O_3$ be centers of $\omega_1,\omega_2,\omega_3$ thus $O_1,O_2,O_3$ are the midpoints of $AA',BB',CC'$ repectively and let $D_1=BB'\cap CC',D_2=AA'\cap BB',D_3=AA'\cap CC'$ and $O$ be circumcenter of $\triangle {D_1}{D_2}{D_3}$ and $R$ circumradius of $\triangle {D_1}{D_2}{D_3}$ so we have :
$(BB'D_1D_2)=-1 \Leftrightarrow   (newton) {O_2} {D_1}.{O_2}{D_2}={O_2{B}}^2$
and by power of $O_2$ into cicumcircle of $\triangle {D_1}{D_2}{D_3}$we have:
${O_2}{D_1}.{O_2}{D_2}={O_2 {O}}^2-R^2$ thus :
${OO_2}^2-R^2=O_2B^2$     $(I)$
but the power of $O$ into $\omega_2$ is :
${OO_2}^2 - {O_2B}^2\Leftrightarrow$ by using $(I)$we have ${OO_2}^2-{O_2B}^2=R^2$
similary we know that $(CC'D_1D_3)=-1,(AA'D_2D_3)=-1$ and we can show that
the power of $O$ into $\omega 1,\omega 3$ is $R^2$ too and thus
$O$ lies on the radical axis and the solution is complete.

[quote="zhaobin"]$payman_pm$,can I know what do you mean by $(BB'D_1D_2)=-1 \Leftrightarrow (newton) {O_2} {D_1}.{O_2}{D_2}={O_2{B}}^2 $.
I don't know the meaning of  $(BB'D_1D_2)=-1$  :([/quote]
it's name  is harmonic ratio (BB'D_1D_2)=-1 and that means $\frac{D_2{B}}{D_2{B'}}=\frac{D_1{B}}{D_1{B'}}$
you can simply show that in a harmonic ratio like$(BB'D_1D_2)=-1$  we have :
${O_2} {D_1}.{O_2}{D_2}={O_2{B}}^2$
there is a lot of theorems about harmonic ratio that i using one of thgem here! :D

where can we discuss about the rest of round_1 problems?

If anyone is interested, here are my solutions. Does anybody know a better way to do p2?

Find all integers $n \ge 5$ such that the residue of $n$ when divided by each prime number smaller then $\frac{n}{2}$ is odd.

------

Obviously, $n$ must be odd.

Also no composite $n$ can fulfill the statement: Let $n = p_{min} \cdot k$ with $p_{min}$ as the smallest prime dividing $n$. With $n \ge 6$ we have $p_{min} \le k < \frac{p_{min}}{2} \cdot k = \frac{n}{2}$.

No prime $n > 31$ can fulfill the statement: For $n \ge 34$ there is always a prime number with $\frac{1}{4}n < p < \frac{1}{3}n$. This means there is a $p$ with $n = 3p + r$ and $0 \le r < p$. Since $n$ and $3p$ are odd, $r$ is even.

The only solutions are therefore $5$, $7$, $13$ and $31$.

Any questions? :ninja:

I solved the first problem at this way:
Of course $x=y$ is always a solution.Suposse there are another solution with $x<y$
1-I proved that $x \equiv y\equiv 0 (mod 5)$ and that $x \equiv y \equiv 0(mod 401)$.So $y>x>2005$

After that I proved that $y<2005$,so we have a contradition, finishing the solution.


Richard--
I'm one of the gangstaz in the Saturday Problem Solving classes and I was just wondering if you could set up a little board for me and my homies.
You know, a place where we could discus problems that we encounter in class, or topics that we encounter in class that may be difficult.

Keep on chillaxin'
The Idiot without a village.

Great minds.  Yesterday I sent an email to Mathew suggesting we do this.

Will do soon.

(For the rest of the board - Idiot is referring to a saturday weekend class we hold at UCSD in San Diego.  If any of you are anywhere near San Diego and want to join us, let us know.)


geometry
cyclic quadrilateral
geometry proposed

$ 1.\ \blacktriangleright$ Let $ ABCDE$ be a convex pentagon inscribed in a given circle and for which $ AC\perp BD$ . Ascertain the its maximum area $ [ABCDE]$ .

$ 2.\ \blacktriangleright$ Let $ ABCDEF$ be a convex hexagon inscribed in a given circle and for which $ \|\begin{array}{c} AC\perp BD \\
\ EF\parallel AD\end{array}$  . Ascertain the its maximum area $ [ABCDEF]$ .

I consider that these extremum problems are very interesting.
The first was given at a districtual olympiad in Romania (author, Marcel Chirita).
I"ll come back soon with my proofs. Succesfully !

[quote="Virgil Nicula"] [color=darkred]$ 1.\blacktriangleright$ Let $ ABCDE$ be a convex pentagon inscribed in a given circle $ C(O,R)$

and for which $ AC\perp BD$ . Ascertain the its maximum area $ [ABCDE]$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Suppose w.l.o.g. $ DE \equal{} AE$ . Denote $ \left\|\ \begin{array}{ccc} AB \equal{} u & ; & CD \equal{} v \\
 \\
BC \equal{} x & ; & AD \equal{} y\end{array}\ \right\|$ and the distance $ \delta_d(X)$

of the point $ X$ to the line $ d$ . Since $ AC\perp BD$ obtain $ x^2 \plus{} y^2 \equal{} u^2 \plus{} v^2 \equal{} 4R^2\ (*)$ and $ \delta_{AD}(O) \equal{} \frac x2$ .

Apply the [b]Ptolemy's theorem[/b] in the cyclic quadrilateral $ ABCD\ \ : \ \ AC\cdot BD \equal{} xy \plus{} uv$ . Thus,

$ [ABCDE] \equal{} [ABCD] \plus{} [AED] \equal{}$ $ \frac 12\cdot AC\cdot BD \plus{} \frac 12\cdot AD\cdot \delta_{AD}(E) \equal{}$ $ \frac {xy \plus{} uv}{2} \plus{} \frac y2\cdot\left(R \minus{} \frac x2\right)$ .

Therefore, the area $ [ABCDE]$ is  maximum $ \Longleftrightarrow$ $ \left[2uv \plus{} y(2R \plus{} x)\right]$ is maximum, where exists $ (*)$ .

$ \blacktriangleright\ \ \ uv\mathrm {\ \minus{} \ max.}$ $ \Longleftrightarrow$ $ \left\|\begin{array}{c} u^2\cdot v^2\mathrm {\ \minus{} \ max.} \\
 \\
u^2 \plus{} v^2 \equal{} 4R^2\mathrm{\ \minus{} \ constant}\end{array}\right\|$  $ \Longleftrightarrow$ $ u^2 \equal{} v^2 \equal{} 2R^2$ $ \Longleftrightarrow$ $ u \equal{} v \equal{} R\sqrt 2$ .

$ \blacktriangleright\ \ \ y(2R \plus{} x)\mathrm {\ \minus{} \ max.}$ $ \Longleftrightarrow$ $ y^2\cdot (2R \plus{} x)^2\mathrm {\ \minus{} \ max.}$ $ \Longleftrightarrow$ $ \left(4R^2 \minus{} x^2\right)\left(2R \plus{} x\right)^2\mathrm {\ \minus{} \ max.}\ \Longleftrightarrow$

$ \left\|\begin{array}{c} (2R \minus{} x)\cdot\left(\frac {2R \plus{} x}{3}\right)^3\ \mathrm {\ \minus{} \ max.} \\
 \\
(2R \minus{} x) \plus{} 3\cdot\frac {2R \plus{} x}{3} \equal{} 4R\mathrm{\ \minus{} \ constant}\end{array}\right\|$ $ \Longleftrightarrow$ $ 2R \minus{} x \equal{} \frac {2R \plus{} x}{3} \equal{} R$ $ \Longleftrightarrow$ $ x \equal{} R$ si $ y \equal{} R\sqrt 3$ .

In conclusion, $ AB \equal{} CD \equal{} R\sqrt 2$ , $ BC \equal{} R$ and $ AD \equal{} R\sqrt 3$ $ \Longrightarrow$ $ \boxed {\ \begin{array}{c} AB \equal{} CD \equal{} R\sqrt 2 \\
 \\
BC \equal{} DE \equal{} EA \equal{} R\end{array}\ }$ .[/color]


logarithms
number theory
number theory unsolved

Is it true that for any prime $p$ and any positive integer $a$ such that $p>a$, there are infinitely many primes $q$ such that $q \equiv a (mod p)$ ?

It is Dirichle theorem: Let (a,m)=1. There are infinetely many primes p, suth that p=a(mod m) o. It proved by analitic method, consider $L(s,\lambda )=\prod_{p}(1-\frac{\lambda (p)}{p^{s}})^{-1}.$

And there is an elementary proof for the cases $a \equiv \pm 1 \mod p$. For all others, you seem to need the L-series Rust gave.

That looks very hard....
What's that exactly ??
Is that some kind of modern number theory ???
Thx for giving me information about that...

Thats "analytic number theory" (an about 100 years old branch of number theory).
This part uses complex/real analysis to get behaviors of such series etc.
For example, Dirichlet proved $\sum_{p \equiv a \mod m }\frac{1}{p}= \infty$ (the sum runs over the primes $\equiv a \mod m$), thus there are infinity many primes $p \equiv a \mod m$. More exactly, he has shown $\sum_{p \equiv a \mod m }\frac 1{p^{s}}\sim \frac{-\log(s-1)}{\phi(m)}$.
Some time later, a more  general theorem occured: Chebotarev's density theorem.
Another example is the famous prime number theorem, which can be proved by using the Zetafunction $\zeta(s) : = \sum_{n=1}^\infty \frac{1}{n^{s}}$. There are also generalisations into the residue classes or for more prime divisors.

You can take a look at http://www.maths.ex.ac.uk/~rjc/etc/dirichlet.pdf to see the methods, the used series and the proof itself.


induction
combinatorics proposed
combinatorics

Given a set $X$ with $n$ elements and a collection $\mathcal F$ of $\ge n+1$ distinct $3$-element subsets of $X$, show that there are at least two sets in $\mathcal F$ which have exactly one element in common.

P.S. Yes, yes, I know it's rather easy, but I thought it was cute and worth having here :).

Hmm... Have you a more elegant solution than the following?

In all the following 'subset' means '$3$-element subset from $X$'.

First, the result does not hold for $n < 5$ since for such values there are no more than $n$ subsets.

For $n=5$, let $X = \{a,b,c,d,e \}$.
Since there are exactly $10$ subset, it follows that we have to choose all of them except at most $4$. (1)
Wlog, we may assume that $\{ a,b,c \}$ is one of the chosen sets.
Suppose for a contradiction, that there is no two chosen subsets with exactly one common element. (2)
Since $n=5$, it is clear that any two of the chosen subsets have a non-empty intersection. Thus, we may assume that $\{ a,b,d \}$ is another chosen subset. From (2), it then follows that each of the following subsets $\{a,d,e \}, \{b,d,e \}, \{c,d,e \}, \{ a,c,e \}, \{b,c,e \}$ is not a chosen one, contradicting (1).
Thus, the result holds for $n=5$.

Now, we prove the result by induction on $n$. We then assume that it holds for each  $5 \leq p \leq n-1$.
First note that if some element from $X$, say $a$, appears at most once in the chosen subsets then, we may use the induction hypothesis with $X' = X - \{a \}$ and with the same subsets excpet at most one (the one which contains $a$ if any).
Thus, any element of $X$ belongs to at least two of the chosen sets.

Assume, for a contradiction, that the desired conclusion does not holds. (3)

Let $a,b$ two distinct elements which belong to a same subset.
Assume that this appears in at least three of the subsets, say $\{a,b,c \}, \{a,b,d \}, \{a,b,e \}$.
From above, we know that $c$ belongs to at least one other subset so that, from (3), it appears with $a$ or $b$, say $a$ in $\{a,c,x \}$, where $x \neq b$.
If $x = d$ then the chosen subsets $\{a,b,e \}$ and $\{a,c,d \}$ contradict (3).
If $x \neq d$ then $\{a,b,d \}$ and $\{a,c,x \}$ contradict (3).

It follows that any pair which belongs to one of the subsets belongs to at most two of them.

Assume that $a,b$ belong to only one common subset, say $F = \{a,b,c \}$.
We then consider that '$a=b$' and omit $F$. This leads to a new set $X'$ with $n-1$ elements, and to at least $n$ chosen subsets from $X'$ (no other subset than $F$ contains both $a$ and $b$ thus they still are $3$-element subsets from $X'$). From the induction hypothesis, we know that there are two subsets from $X'$, say $F'_1$ and $F'_2$, with exactly one common element, say $x$ (4).
If $x \neq a$ (in $X'$) then the corresponding subsets $F_1$ and $F_2$ from $X$ have the same property, contradictiing (3). Then, $x = a$ in $X'$. To avoid the same contradiction, it follows that $a \in F_1$ and $b \in F_2$. But, from (3) and using $F$, we then have $c \in F_1$ and $c \in F_2$, which implies that $F'_1 \cap F'_2 = \{a,c \}$, contradicting (4).

Then, we know that each of the pair which belongs to one of the subsets belongs to exactly two of them. (5)

Let $\{ a,b,c \}$ be one of the subsets. Then $\{ a,b,d \}$ is another one.
Moreover, $b,c$ must belong to another common subset, say $\{b,c,x \}$, where $x \neq a$.
If $x \neq d$ then the subsets $\{a,b,d \}$ and $\{b,c,x \}$ contradict (3). Thus $x = d$ and $\{b,c,d \}$ is a chosen subset too.
Similarly, we deduce that $\{ a,c,d \}$ is a chosen subset.
Moreover, if $a$ belongs to another subset it is not with $b,c,d$ otherwise it would contradicts (5). Therefore, if $a$ belongs to another subset it is $\{ a,x,y \}$ where $x,y \notin \{b,c,d \}$, and the subsets $\{a,x,y \}$ and $\{ a,b,c \}$ contradict (3).
Similarly, it follows that none of the elements $a,b,c,d$ appears in another subset. Note that the other subsets than the 4 above only contain elements from $Y = X- \{a,b,c,d \}$ and that they are at least $n-3$. It follows that $|Y| = p$ with $ 5 \leq p \leq n-1$, and the induction hypothesis ensures that at least two of them have exactly one common element, contradicting (3) again.

Thus, in any case, we have the desired contradiction. The induction step follows, and we are done.

Pierre.

I didn't read all of it  :blush:.

First of all, you're right: I forgot to mention $n\ge 5$. Terribly sorry about that.

I think the following is a valid proof, but you be the judge of that :).

Consider the sets as elements $x_i,\ i\in\overline{1,n+1}$ in $\mathbb Z_2^n$. There are $n+1$ of them, so there is a zero sum set, because they cannot be independent. Let $\sum_{i\in I}x_i=0$. Assume WLOG $1\in I$. We then have $x_1+\sum_{i\in I\setminus \{1\}}x_i=0$ (all of this is computed in $\mathbb Z_2^n$). Now define $<x_i,x_j>$ with values in $\mathbb Z_2$ in the usual manner, i.e. some sort of pseudo-scalar product. It's not a real scalar product because we can have $<v,v>=0$ if $v$ has an even number of $1$'s.

We have $<x_1,x_1>+\sum_{i\in I\setminus\{1\}}<x_1,x_j>=0$, but if we assume the conclusion is wrong, we get $<x_i,x_j>=\delta_{ij}$, so the computation above would yield $1$ instead of $0$, a contradiction.

Is it correct?

It is clearly correct, and I'm completely dumb, because I wrote a paper on it last year to prove in the same way that the maximal number of odd sets with pairwise even intersection is $n$ and other related results... :maybe:

My Alzheimer is beginning...

Pierre.

You are hero, Pierre!
I also have induction solution. I affraided that I should write it. You saved me!   :)

I really wouldn't worry about that (being dumb) if I  were you.. :).

[b](Moderator Remark: you = Pierre)[/b]


hmm i was thinking about this and i wanted to know if it was true or not

if you have a triangle, and you take the midpoints of two sides and connect them, then inscribe a circle in the smaller triangle, and in the quadrilateral formed

is the midpoint of the two tangency points that lie on the midline, also the midpoint of the midline itself?

I find this really interesting. However, I won't respond because it's too closely related to Problem 5 of the USAMTS. If you don't believe me, take a close look at the problem.

To those of you who know the answer to this, please don't respond until Mr. Rucszyk gives the okay to discuss USAMTS problems.

Thanks for your cooperation.  :)

I agree.

This is actually EXACTLY what number 5 is (although I didn't get to solve this problem due to I didn't have much time).

Mod, can you lock this topic?

[quote="chess64"]gotztahbeazn, what were you thinking? Almost ALL high school AoPSers are working on USAMTS! :rotfl:[/quote]What are you suggesting? Conspiracy? :roll:


Colleges get to see what scores you made each time on the SAT, but is that also true for all the SAT 2's?  :huh:  Also, if you didn't want to give them a specific SAT 2, will they still be able to receive it?  Thanks!

I think your College Board record of testing shows results of all SAT I and SAT II tests taken after eighth grade, all on one big list. That's the kind of record we see when we look up the tests taken here online. I think that's what colleges see too when the score reports go out to them.

I think the AP scores go on the separate "AP score report";  I don't see my ap scores when i log onto collegeboard.

One more question:
Do the colleges get all my tests scores including APs, SAT IIs?
Or do i get to select which tests to show??

Thanks,

Colleges get all the scores in your SAT record, which from the answers above appears to be all of your SAT I scores (after eighth grade) and all of your SAT II scores. You do not get to choose, college by college, which colleges see which scores.


i think this is the right place for this.

Evaluate the infinite product.

$\displaystyle\prod_{n=2}^{\infty}\frac{n^3-1}{n^3+1}$

$\displaystyle\prod_{n=2}^{\infty}\frac{n^3-1}{n^3+1}$

[hide="solution"]
$\prod_{n=2}^{\infty} \frac{n^3-1}{n^3+1} = \prod_{n=2}^{\infty} \frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)} =\prod_{n=2}^{\infty} \frac{n-1}{n+1} \cdot \prod_{n=2}^{\infty} \frac{n^2+n+1}{n^2-n+1}$

$\prod_{n=2}^{\infty} \frac{n-1}{n+1}$ Notice that $(n+2) - 1 = n+1$
Thus the product telescopes to $2$.

$\prod_{n=2}^{\infty} \frac{n^2+n+1}{n^2-n+1}$ Notice that $(n+1)^2-(n+1)+1 = n^2+n+1$
Thus the product telescopes to $\frac{1}{3}$.

The answer is then $\frac{2}{3}$.[/hide]

i like paladin8's solution better, however, i myself solved it more like FMako.
but on the last step i did someting different, but still got the same answer. saddly i dont remember what i did, and i cant find the paper that i wrote it down on.  :(
i think i might have written out the first few terms to see what it telescoped to.

I did the problem myself and got the right solution, I was just wondering what the term "telescope" means, since it seems to be a pretty common word you all throw around. Does it simply mean that a product cancels such that it reduces to a number?

boxedexe, that is not a solution... you need to show you how went from there to the answer, that was the bulk of the problem.  You wouldn't get many points if you put that in a competition.

Arjunlandes, yes, that is what telescoping is.  Although many sums cancel out in a similar way too.

In Putnam, it'd get 0/10. Well, maybe 1 for the right answer if you're lucky. They're harsh.


Well, I can say that MOEMS is much easier than MathCounts.

[quote="BOGTRO"]MathCounts and MOEMS are somewhat different competitions.

MathCounts gives you problems that most people can do, but its very fast paced at 30 problems, 40 minutes for the sprint, 6 minutes for 2 problems on the Target, and Countdown is even quicker (most of the time).

MOEMS on the other hand, gives you problems that most people can do, but gives you much more time to solve it (I think the average is 4-5 minutes, as opposed to 10-80 seconds on MC).

Therefore, I'd suggest doing Mathcounts for a 6th grader, assuming you want a challenge.

However, if you are just getting started, and are looking for a competition to get some fairly easy problems that you are given 4,5 minutes for, to get an idea of Math Competitions, try MOEMS, and then move on to MathCounts in 7th or 8th grade.[/quote]

From the other thread.  This shoudl explain nearly all.

AMC 8, in my opinion, is between these two in difficulty.  Mathcounts is hardest.

[quote="dzo"]Is there any quick comparison of the Mathcounts and Math Olympiad (MOEMS) programs? Obviously one can look at the websites, but does anyone have any experience with both such that they can offer some idea about the advantages and disadvantages of both for a 6th grader? TIA.[/quote]

Mathcounts is a lot harder than MOEMS. You have a better chance of doing good in MOEMS, then in MC. MOEMS, is shorter I think. MC lasts a couple months.

Which one is more popular: MOEMS or MC for middle school kids?

Generally, most schools who do competitions have Mathcounts, while MOEMS is usually only in some.

Yeah, my mathcounts coach didn't even know what MOEMS was...

I hadn't even heard of MOEMS, until I got on this website, and even then I was like "what is this?".

Yeah, same here.

MathCounts is the well-known competition, so your principal may have heard of this, but I don't think many know of MOEMS.

I learned about MathCounts on the first day of middle school.

mathcounts has progressive levels.
moems on the other hand, is a competition where four or five kids from a school often get a perfect score.

moems, amc8, mathcounts, in list of increasing commonity  :?: lol weird word.

uhhh thats about it. moems is also a lot easier than mc.

your math team might be having MOEMS as well as MC. that's how it is in our school, anyways.

Usually if a school is doing AMC, then they are also doing MathCounts, and if they are doing MOEMS, then they are probably doing MathCounts and maybe AMC also.

Math Olympiads for Elementary and Middle Schools


hey could someone please help me with this question?

A light is at the top of a pole 20 m high. From a point on  the ground 8m from the base of the pole a ball is thrown upwards with an inital velocity of 13m/s. At what rate is the shadow of the ball moving aloing the grouind 1 s later.

Yeah, and this definitely looks like homework...my class just finished related rates. :)

[quote]you should post calculus problems in the calculus section[/quote]
And please don't post the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60015]same problem[/url] in more than one place.


analytic geometry
graphing lines
slope
calculus
calculus computations

(i) Given that $ 14x^{2} \minus{}7xy \plus{}y^{2} \equal{}2$, show that $ \frac{dy}{dx} \equal{} \frac{28x\minus{}7y}{7x\minus{}2y}$.

(ii) The points $ L$ and $ M$ on the curve $ 14x^{2} \minus{}7xy \plus{} y^{2} \equal{} 2$ each have $ x$-coordinate $ 1$.  The tangents to the curve at $ L$ and $ M$ meet at $ N$. Find the coordinates of $ N$.

Differentiate the 1st one

$ 28x \minus{} 7( x\frac{dy}{dx} \plus{} y ) \plus{} 2y\frac{dy}{dx} \equal{} 0$

$ \frac{dy}{dx} ( \minus{} 7x \plus{} 2y ) \equal{} 7y \minus{} 28x$

$ \Rightarrow \ \frac{dy}{dx} \equal{} \frac {28x \minus{} 7y}{7x \minus{} 2y}$

This will be the slope of the tangent at any point on this curve.

[b]x = 1[/b] gives the points [b]L[/b] and [b]M[/b] on the curve.

$ 14 \minus{} 7y \plus{} y^2 \equal{} 2 \\ y^2 \minus{} 7y \plus{} 12 \equal{} 0$

$ (y \minus{} 4)(y \minus{} 3) \equal{} 0$

Thus, $ L \equiv ( 1,3 ) , \ M \equiv ( 1, 4)$

 the slope of the tangents at L and M will respectively be 7 and 0.

Equation of tangent at L :

$ y \minus{} 3 \equal{} 7 (x \minus{} 1) \\  y \equal{} 7x \minus{} 4$

Tangent at M :

$ y \minus{} 4 \equal{} 0 ( x \minus{} 1) \\ y \equal{} 4$

Find their point of intersection ... that will be y = 4 and 8/7

$ N \equiv ( \frac{8}{7}, 4 )$


function
real analysis
real analysis solved

$X=[-1,1] \cap Q$ (={Natural Number})
$Y=\{y=\sqrt{1-x^2}|x \in X\}$
Function $F: X \rightarrow Y$
Let $x=(x_1,x_2), y=(y_1,y_2)$
Prove that: when $x_1,x_2 \in B_\delta \cap X$,
$||x-y||<\epsilon$

It did make sense!
I'm sorry maybe my poor expression make it nonsense ... :(