Functional Analysis
real analysis
real analysis unsolved

I proved that $ C_0, \ell^\infty, \ell^1$ are Banach spaces using the fact ( Abs Con. of Serie $ \Rightarrow$ Con. of series In $ X$ iff X is Banach space)

I need to prove that $ \ell^p , 1 < p < \infty$ is a Banach space .

Pf :

Lef $ f_n \in \ell^p$ such that $ \sum\limits_{n \equal{} 1}^{\infty} \parallel{}f_n\parallel{}_p < \infty$ . let $ f \equal{} \sum\limits_{n \equal{} 1}^{\infty} f_n$

My technique for previous spaces is to do the following steps :

1) Prove $ f \in \ell^p$ .

2) $ f(i)$ is defined $ \forall i \in \mathbb N$.

3) $ \parallel{} f \minus{} S_k\parallel{}_p \rightarrow 0$.

I did the following for (1) :

I need to show $ \sum\limits_{i \equal{} 1}^{\infty} |f(i)|^p \equal{} \sum\limits_{i \equal{} 1}^{\infty}|\sum\limits_{n \equal{} 1}^{\infty} f_n(i)|^p$ .

Now , for $ p \equal{} 1$ , I can prove that sum $ < \infty$

How could I complete ?

Give me a hint  :)


calculus
integration
function
real analysis
real analysis unsolved

Suppose that $ f: \mathbb R^2\rightarrow \mathbb R$ is a non-negative and continuous function that $ \iint_{\mathbb R^2}f(x,y)dxdy\equal{}1$. Prove that there is a closed disc $ D$ with the least radius possible such that $ \iint_D f(x,y)dxdy\equal{}\frac12$.

I don't understand the question. Of course there are minimal disks with half the integral, by simple continuity of the integral. Expand from the center, and you will find a disk with that integral. Vary the centers and radii, and the disks are a compact set, with those containing half the integral a closed subset.


algebra
polynomial
superior algebra
superior algebra unsolved

Show that a finite field of size p^n is isomorphic to F[x]/(c(x)), where c(x) is an irreducible polynomial of degree n in F[x].

(Here, F[x] denotes the finite field of polynomials and order of F[x]=p).

Does it have something to do with finite implies perfect implies separable implies simple extension?

[quote="rawkerrxx"]Does it have something to do with finite implies perfect implies separable implies simple extension?[/quote]

Maybe.  Can you prove that if $ F\subset K$ is any simple field extension, the same result holds?


calculus
derivative
function
algorithm
Gauss
linear algebra
calculus computations

What is an equation of the form,

$ \displaystyle\sum_{i=0}^{n} c_i f^{(i)}(x)=0$

where we know the function and its derivatives and we must solve for the non-trivial constants $ c_i$ called?

$ f^{(i)}(x)$ denotes the i-th derivative

It wouldn't be a differential equation would it? Since we already know the function, we just need the constants

If we know the values of the derivatives at certain points, such as $ f(0)\equal{}0$, $ f'(0)\equal{}1$, and so on, it's an initial value problem.

It's a linear algebra problem: you have a bunch of functions (and that they're derivatives of a given function is irrelevant); presumably they're linearly dependent (otherwise the only choice is $ c_i \equal{} 0$ for all $ i$), and you're trying to find a zero linear combination.  I don't know a better name for this process in general (although there are lots of names for particular algorithms you might try to use to carry it out, like Gauss-Jordan elimination).

Thanks. The problem seems to be very similar to a linear algebra problem like you said. I noticed that I had to solve a particular equation in this form by re-writing the second dervative as a linear combination of the first derivative and the original function. I don't know if there is a specific name for this.


geometry
circumcircle
parallelogram
geometry proposed

Let a triangle ABC and its circumcircle with center O.The tangents to the circle at B and C intersect at P. Let ANBM be a parallelogram (with N,M belong to BC and PB respectively) and Q the intersection of MN and AP.
[b]If L is the intersection of the tangent to the circle at A with BC, then prove that L,Q and O are collinear.[/b]

Very nice problem  :) (Στη Σλοβενία το έφτιαξες ?)
I will give the hint of my solution
[hide]$\widehat{LQA}=\widehat{AQO}=90$  or AQ is the height of the right-angled triangle $ALO$  [/hide]

[color=darkblue][b][u]Lemma 1.[/u][/b] Given are $\triangle ABC$ and two points $M,N$ which belong to the side $[BC]$ so that $\widehat{BAM}\equiv\widehat{CAN}$.
Denote the second intersections $X,Y$ of the lines $AM$, $AN$ respectively with the circumcircle $w=C(O,R)$ of the triangle $ABC$. Then there are the relations $\boxed{AM\cdot AY=AN\cdot AX=AB\cdot AC}\ .$ [b]Indication.[/b] $\triangle ABN\sim\triangle AXC$ and $\triangle ACM\sim\triangle AYB\ .$ [b]Some particular cases :[/b]

$\blacktriangleright 1.\ K\in BC\ ,\ AK\perp BC\ ,\ AK=h_{a}\Longrightarrow \boxed{\ 2Rh_{a}=bc\ }\ ;\ 2S=ah_{a}$ $\Longrightarrow \boxed{\ 4RS=abc\ }\ .$
$\blacktriangleright 2.$ The bisector $[AD$ of the angle $\widehat{BAC}$, where $D\in BC$, meets again the circle $w$ in the point $E$.
Then $AD\cdot AE=bc$ $\Longrightarrow$ $AD\cdot (AD+DE)=bc$ $\Longrightarrow$ $AD^{2}=bc-DA\cdot DE$ $\Longrightarrow$ $\boxed{AD^{2}=bc-DB\cdot DC}\ .$
$\blacktriangleright 3.$ If the point $M$ is the middlepoint of the side $[BC]$, the point $S\in (BC)$ so that $\widehat{BAS}\equiv\widehat{CAM}$
and the point $T$ is the second intersection of the line $AS$ with the circle $w$, then $\boxed{\ AT\cdot m_{a}=bc\ }\ .$
$\blacktriangleright 4.\ L\in BC\cap AA\ ;\ E\in w\ ,\ A\not\equiv E\ ,\ AE\parallel BC\Longrightarrow AL\cdot AE=bc\ .$ If $b\ne c$, then $\boxed{\ AL=\frac{abc}{\left|b^{2}-c^{2}\right|}\ }\ .$
If the point $M$ is the middlepoint of the side $[BC]$, $L\in BC\cap AA$ and the point $K$ is the projection of the point $A$ on the sideline $BC$, then $\widehat{ALO}\equiv\widehat{MAK}\ .$

[b][u]Lemma 2 (well-known).[/u][/b] Given are $\triangle ABC$, the middlepoint $M$ of the side $[BC]$ and the points $P\in BB\cap CC$, $S\in BC\cap AP$. Then $\widehat{SAB}\equiv\widehat{MAC}$.

[b][u]A proof for the preolympiad level.[/u][/b] Denote the circumcircle $w=C(O,R)$, the second intersection $T$ of the line $AP$ with the circle $w$, the intersection $S\in BC\cap AP$ and the middlepoint $Q'$ of the segment $AS$. Prove easily that $\boxed{\frac{SB}{SC}=\left(\frac{c}{b}\right)^{2}}$ (from the [u]lemma 2[/u] and the [u]Steiner's theorem[/u]), ${\boxed{AS=\frac{2bc}{b^{2}+c^{2}}\cdot m_{a}}}$, $L\in TT$ and $\boxed{\ Q'A=\frac{bc}{2m_{a}}\ }$ (from the [u]lemma 1[/u]). Therefore, $\frac{SB}{c^{2}}=\frac{SC}{b^{2}}=\frac{a}{b^{2}+c^{2}}$ $\Longrightarrow$ $SB=\frac{ac^{2}}{b^{2}+c^{2}}\ ;$ $\widehat{ANB}\equiv\widehat{NBP}\equiv \widehat{BAC}$ $\Longrightarrow$ $\triangle ABN\equiv\triangle CBA$ $\Longrightarrow$ $\frac{c}{a}=\frac{NB}{c}=\frac{AN}{b}$ $\Longrightarrow$ $NB=\frac{c^{2}}{a}$, $AN=\frac{bc}{a}$, $NS=NB-SB=\frac{c^{2}}{a}-\frac{ac^{2}}{b^{2}+c^{2}}=\frac{c^{2}(b^{2}+c^{2}-a^{2})}{a(b^{2}+c^{2})}\ .$ Thus, $\frac{QA}{QS}=\frac{AM}{NS}=\frac{NB}{NS}$ $\Longrightarrow$ $\frac{QA}{QS}=\frac{b^{2}+c^{2}}{b^{2}+c^{2}-a^{2}}$ $\Longrightarrow$ $\frac{QA}{AS}=\frac{b^{2}+c^{2}}{4m^{2}_{a}}$ $\Longrightarrow$ $QA=\frac{2bc}{b^{2}+c^{2}}m_{a}\cdot \frac{b^{2}+c^{2}}{4m^{2}_{a}}$ $\Longrightarrow$ $\boxed{\ QA=\frac{bc}{2m_{a}}\ }$ $\Longrightarrow$ $QA=Q'A$ $\Longrightarrow$ $Q\equiv Q'$ $\Longrightarrow$ $Q\in OL\ .$

[b][u]A proof for the olympiad level.[/u][/b] The point $L$ belongs to the polar axis of the point $P$ $\Longrightarrow$
The point $P$ belongs to the polar axis of the point $L$ (the point $A$ belongs to the polar axis of the point $L$) $\Longrightarrow$
The line $AP$ is the polar axis of the point $L$ $\Longrightarrow$ $LO\perp AP$ $\Longrightarrow$ $\boxed{\ Q\in LO\ }.$[/color]

[quote="Virgil Nicula"]  [b][u]A proof for the olympiad level.[/u][/b] The point $L$ belongs to the polar axis of the point $P$ $\Longrightarrow$
The point $P$ belongs to the polar axis of the point $L$ (the point $A$ belongs to the polar axis of the point $L$) $\Longrightarrow$
The line $AP$ is the polar axis of the point $L$ $\Longrightarrow$ $LO\perp AP$ $\Longrightarrow$ $\boxed{\ Q\in LO\ }.$[/color][/quote]

That was also my solution  :wink:  . Thank you Virgil for the nice lemmas and their represantation here   :)


modular arithmetic
function
superior algebra
superior algebra solved

We say that a ring $A$ has the property $( \mathcal P )$ iff $\forall x \in A$, $\exists n_x \in \mathbb N$, $n_x \equiv 1 \pmod 2$, such that $x^{n_x} \in \{ 0,1 \}$.

(a) Prove that if $A$ has $( \mathcal P )$, then it is a field.

(b) All finite fields that satisfy $1+1=0$ have $( \mathcal P )$.

(c) Find an infinite field which doesn't have $(\mathcal P)$ and which fulfills the condition $1+1=0$.

Could you please check the statement again? Maybe $n_x$ is in fact supposed to be even instead of odd? Because as it is stated, it's not true that all finite fields have $(\mathcal P)$: if $p$ is any odd prime, then $\mathbb F_p$ (the field with $p$ elements) does not have the property, since a primitive root $g$ modulo $p$ satisfies $g^n=1$ iff $p-1|n$, which means that $n$ must be even.

Oh, Ok. Thanks for correcting it.

(b) and (c) are more or less obvious: for (b) just notice that since the multiplicative group of the field with $2^k$ elements has order $2^k-1$, every non-zero element of the field satisfies $x^{2^k-1}=1$, and for (c) take any transcendental extension of $\mathbb F_2$, the field with $2$ elements (the field of rational functions in one indeterminate with coefficients in $\mathbb F_2$, for example).

(a) First notice that $-1=1$, since there is an odd $n$ for which $-1=(-1)^n=1$. Suppose $n>1$ is the minimal positive integer for which $x^n=0$. Then $1+x^{n-1}$ is invertible (because $x^{n-1}$ is nilpotent), so there is an odd $m$ such that $(1+x^{n-1})^m=1\Rightarrow x^{n-1}=mx^{n-1}=0$, contradicting the minimality of $n$. This means that the only nilpotent (which here is equivalent to being non-invertible) element of $A$ is $0$, and we're done.


integration
real analysis
real analysis unsolved

[color=darkblue]Let sequence $ (u_n)$ so that: $ \begin {cases} u_n \ge 0 \\
u_n \minus{} 2u_{n \plus{} 1} \plus{} u_{n \plus{} 2} \ge 0 \\
u_1 \plus{} u_2 \plus{} ... \plus{} u_n \le 1 \end {cases}$, $ \forall n \in N^*$.

Prove that $ u_{n \plus{} 1} \le u_n, \forall n \in N^*$.[/color]

Define $ f: [1,\infty) \to \mathbb{R}$ as follows:
$ f(n) = u_n$ for all $ n \in \mathbb{N}$
$ f$ is linear in $ [n,n+1]$ for every positive integer $ n$.

Now $ u_n\ge 0$ says that $ f\ge 0$,
and $ u_n-2u_{n+1}+u_{n+2}\ge 0$ says that $ f$ is convex,
and from $ u_1+\cdots+u_n \le 1$ we can conclude that $ \int_1^{\infty}f(x)\ dx <1$.
That means that $ f(x) \to 0$ for $ x \to \infty$, and f is descending.
So in particular ${ u_{n+1}\le u_n}$ for all $ n$.

[quote="PeterP"]Define $ f: [1,\infty) \to \mathbb{R}$ as follows:
$ f(n) = u_n$ for all $ n \in \mathbb{N}$
$ f$ is linear in $ [n,n + 1]$ for every positive integer $ n$.

Now $ u_n\ge 0$ says that $ f\ge 0$,
and $ u_n - 2u_{n + 1} + u_{n + 2}\ge 0$ says that $ f$ is convex,
and from $ u_1 + \cdots + u_n \le 1$ we can conclude that $ \int_1^{\infty}f(x)\ dx < 1$.
That means that $ f(x) \to 0$ for $ x \to \infty$, and f is descending.
So in particular ${ u_{n + 1}\le u_n}$ for all $ n$.[/quote]

Thank PeterP very much.


[quote="gollywog"]Let $f: R\to R$ be a continous function such that $f(x) \in Q \iff f(x+1) \notin Q$. Prove there is no such function.[/quote]

 [b]HINT[/b]

Since $f$ is continous in IR , $f$ must be constant , because if not , then it's range is an interval , which of course contains rational and irrational numbers. The rest is simple.

[quote="stergiu"]Since $f$ is continous in IR , $f$ must be constant , because if not , then it's range is an interval , which of course contains rational and irrational numbers. The rest is simple.[/quote]
I don't understand this. Please clear it up.
Anyway here is my proof.

$g(x): =f(x)-f(x+1)\not\in\mathbb{Q}$ and $h(x): =f(x)+f(x+1)\not\in\mathbb{Q}$ for all $x\in\mathbb{R}$.
But $g$ and $h$ are continuous, and between $2$ different real numbers lies always a rational number.
So by the intermediate value theorem $g$ and $h$ are constant, and so is $f(x)={g(x)+h(x)\over 2}$.
But now $f(x)=f(x+1)$. Contradiction!


group theory
abstract algebra
Ring Theory
number theory
greatest common divisor
superior algebra
superior algebra solved

Define $R=\{\frac{m}{n}|m, n\neq0\in \mathbb{Z},\; p_{1}, p_{2},\ldots, p_{k}\nmid n\}$, where $p_{i}$ is a prime number. Prove that $R$ has exactly $k$ maximal ideals.$m, n$ are coprime.[added in edit.]
[hide]Consider $(p_{1}), \dots, (p_{k})$.[/hide]

Let $I_{i}= \{ \frac{p_{i}}{n}| n \in \mathbb{Z},\ n \neq 0 \}$, with $1 \leq  i \leq  k$.

Then, $\forall x, y \in I_{i}$, $x = \frac{a*p_{i}}{b}$, $y = \frac{c*p_{i}}{d}$, so $x-y = \frac{(ad-bc)*p_{i}}{bd}\in I_{i}$.
Therefore, $(I_{i},+)$ is a subgroup of $(R,+)$.

Clearly, it is closed under multiplication, so $I_{i}$ is a subring of $R$.

And, $\forall z \in R$, $z = \frac{e}{f}$, so $z*x = x*z = \frac{ae*p_{i}}{bf}\in I_{i}$.  Thus, $I_{i}$ is an ideal in $R$.

Now, assume $\exists J \subset R$, such that $I_{i}\subset J$, and $J$ is an ideal in $R$.

Then, $\exists w \in J$, such that $w \notin I_{i}$.

Let $w = \frac{s}{t}$, with $gcd(s, t) = 1$.
Since $w \notin I_{i}$, $s$ is not divisible by $p_{i}$.

Thus, $gcd(s, t*p_{i}) = 1$.
Therefore, $\exists u, v \in \mathbb{Z}$, such that $us+v(t*p_{i}) = 1$.

We know that $\frac{v*p_{i}}{u}\in I_{i}$, therefore, it is in $J$.

Thus, $w+\frac{v*p_{i}}{u}= \frac{us+vt*p_{i}}{ut}= \frac{1}{ut}\in J$.

We know that $\frac{ut}{1}\in R$.
Therefore, $(\frac{ut}{1})*(\frac{1}{ut}) = 1 \in J$, since $J$ is an ideal in R.

But now, $\forall x \in R$, $x*1 = x \in J$.  Therefore, $J = R$, a contradiction.

Therefore, $I_{i}$ is maximal.

Since the $i$ used was arbitrary, this holds $\forall p_{i}$, with $1 \leq  i \leq  k$.

Therefore, $R$ has $k$ maximal ideals.

QED

[quote="PTynan89"]Therefore, $R$ has $k$ maximal ideals.
QED[/quote]
Therefore, $R$ has at least $k$ maximal ideals. :wink:

[quote="qleak"]http://www.sosmath.com/CBB/viewtopic.php?t=25598&highlight=

It would be in your best interest to show that this ring is a PID, the result follows easily enough from that.

cheers,

-Q[/quote]Why bother proving that? We have proved that $(p_{1}),\dots ,(p_{k})$ are maximal ideals,  let $I$ be an ideal of $R$ and $\frac{a}{b}\in I,\; p_{1},\dots, p_{k}\nmid a$(otherwise $I$ would be either one of $(p_{i})$ or not a maximal ideal), then $\frac{b}{a}\in R$ hence $1\in I$.  And we're done.
In the case you're suggesting a proof totally different from PTynan89, please let me know.

Yeah, I suppose showing it is a PID is overkill. Your proof is quite nice.

Here's what I had in mind (details ommitted).

PID => maximal ideals are exactly the prime ideals.

Easy to show $(p_{k})$ are prime ideals.

Any non-invertible element $x$ must be contained in one of the $(p_{k})$ Hence $(x) \subseteq (p_{k})$ for some k for all non invertible x. Since in a PID this is all the proper ideals.

cheers,

-Q


If the graph of the equation $ y \equal{} x^2 \minus{} 4x \plus{} 4$ intersects the $ x$-axis at $ (a, b)$, what is the value of $ a \plus{} b$?

hello, we have $ x^2 \minus{} 4x \plus{} 4 \equal{} (x \minus{} 2)^2$ so the intersection point with the  $ x$-axsis is $ P( 2,0)$, we have
$ a \equal{} 2$ and $ b \equal{} 0$ and their sum is $ a \plus{} b \equal{} 2$.
Sonnhard.

Actually iy will intersect at (2,0) for a sum of 2.

hello, yes, it's correct, i have edited it.
Thank you,Sonnhard.

Very interesting problem but of course the answer is 2.


[quote="Fermat -Euler"]find an integer n satisfaies:

$\sqrt{1+5^n+6^n+11^n}$ is an integer[/quote]

[hide]

What we need to do is just to investigate the last digit of the expression or we just take $\mod 10$

Notice that  we will have $1+5+6+1\equiv 3 \mod 10$ and this shows that no integer can be formed by any $n$ because a perfect square never ends in $3$ .

 :)
[/hide]

So, n should be 0. And this should be the only integer.

yes zhaoli

can any one find another integer

No, Shyong proved that there is NO other such integer. Please read the posts above!

What about negative exponents? They could still add up to a integer

No, they can't since for $n \leq -1$ we have $1 < 1 + 5^n + 6^n + 11^n < 2$.

[quote="Fermat -Euler"]sorry but there exist an integer satisfaies this relation  ;)[/quote]

Opps , I think i take $n$ as natural number for granted  :P  . Anyway , I doubt there is another $n$ satisfying it , unless you are talking about $n\rightarrow -\infty$


I'm getting this pop up window every now and then, is this normal?

i.e.

If it looks like this:

http://images.ientrymail.com/webpronews/yourdocumenthasbeensaved.gif

, it's a JavasScript laert, and if it's like this:

http://isc.sans.org/diaryimages/ff.PNG

it's a window.

Plus, what does it say? (Both caption and content)

Is it AoPS related or just you web-surfing?

The only pop-up we have on the site is the one announcing you of new private messages in your inbox. That would explain the "now and then". You can turn that off in your [url=http://www.mathlinks.ro/Forum/usercp.php?mode=prefs&sub=privacy]preferences page[/url].

[quote="rrusczyk"]Shouldn't be AoPS-related.  We don't have those annoying pop-up ads on our site.[/quote]

[img]http://i38.photobucket.com/albums/e119/moogra4u/asdf-1.png[/img]

[quote="i_like_pie"]If you click on a $ \text{\LaTeX}$ expression that contains more than 700 characters (I think?), then you get an empty window like one that would contain Asymptote code. This is a reported bug, and I think it's currently being looked into by Mr. Vornicu.[/quote]

Actually it's 30 characters lol.

The list of people who won the Nobel Prize is quite more than 30 characters.

I know 30+ characters not including the dollar signs


alot of the questions i have gotten are only worth one point and they come from subject areas which i have mastered. i dont recall getting a question from topics i havnt mastered(recently)

is this a problem?

Well, I think the question set goes in the same order for everyone, so you just gotta go through those.


Let $ n,\ell$ be positive integers, and let $ A_1,\ldots,A_k \subset \left\{1,2,...,n\right\}$ be such that, for every $ i_1,i_2,\ldots,i_{2\ell}$, we have $ \left|A_{i_1} \cup A_{i_2} \cup \cdots \cup A_{i_{2\ell}}\right| \leq n \minus{} \ell$. Show that $ k \leq 2^{n \minus{} \ell}$.

Sorry for my spam.It has been posted yet :blush:
See also [url=http://www.mathlinks.ro/viewtopic.php?t=4400]here[/url]

[Mod edit: replaced your problem by the generalization, which hasn't got a separate topic yet.]


[color=darkblue]Fie patrulaterul convex $ABCD$ pentru care $AB\parallel CD$. Notam mijlocul $M$ al laturii $[AB]$. Sa se arate ca daca exista un punct $N$ pe dreapta $BC$ astfel incat punctul $P\in CM\cap DN$ sa verifice relatia $\frac{ND}{NP}=2\cdot \frac{CM}{CP}\ \ (*)$, atunci patrulaterul $ABCD$ este paralelogram si in acest caz orice punct $N$ al dreptei $BC$ verifica relatia $(*)$.[/color]

Analitic iese [i]destul[/i] de uşor.

Fie $C(c,0), \, D(d,0), P(a,b)$ şi $\frac{NP}{ND}= k$. Rezultă că $\frac{CP}{CM}= 2k$. Făcând nişte calcule ne iese că
\[N \left( \frac{a-d}{1-k}+d, \frac{b}{1-k}\right), \, M \left( \frac{a-c}{2k}+c, \frac{b}{2k}\right) . \]
Punctul $B$ trebuie să se afle pe $CN$ si să aibă ordonata egală cu a lui $M$. Ecuaţia lui $CN$ este \[\frac{x-c}{y}= \frac{(a-d)+(d-c)(1-k)}{b},\] iar $y_{M}= \frac{b}{2k}$. Introducând acest $y$ in ecuaţie, rezultă că \[x_{B}= \frac{(a-d)+(c-d)(k-1)}{2k}+c .\] Diferenţa $x_{M}-x_{B}$ este egala cu
\[\frac{(a-c)+(d-a)+(d-c)(k-1)}{2k}= \frac{d-c}2 , \]
deci $BM = \frac{DC}2$, i.e. $ABCD$ e paralelogram.

Observam că orice pereche $(P,k)$ determină un anumit paralelogram. Dar dacă alegem numai perechile $(P,k)$ care determină un punct fix $M$, atunci paralelogramul pe care-l obţinem nu va depinde de perechea aleasă. Există o bijecţie intre $(P,k)$ şi $N$, deci orice punct $N$ al dreptei $BC$ verifică relaţia $\left( \star \right)$.

[hide="solutie sintetica"]Fie $S$ simetricul lui $C$ fata de $M$. Rezulta imediat ca $CBSA$ este paralelogram, de unde obtinem $SA\| BC$. Totodata, relatia din ipoteza se scrie echivalent $\frac{ND}{NP}=\frac{CS}{CP}$, de unde rezulta ca $SD\| BC$. In final, prin axioma paralelelor, deducem ca $AD\| BC$, deci $ABCD$ este paralelogram si ultima cerinta a problemei este evidenta.[/hide]

Solutia 3.
Fie $DM\cap BC=\{R\}$. Aplicand Teorema lui Menelaus pentru triunghiul $DPM$ si transversala $R-N-C$  rezulta ca \[ \frac{DR}{RM}\cdot\frac{MC}{PC}\cdot\frac{NP}{ND}=1,\eqno{(\star)}\] iar din relatia din enunt :$\frac{ND}{NP}=2\cdot\frac{CM}{CP}$ si din $(\star)$, rezulta ca $\frac{DR}{RM}=2$, de unde $DM=RM$. Deci $M$ este mijlocul segmentului $DR$.Dar in triunghiul $DRC$, $B\in (CR)$ astfel incat $MB\parallel DC$, de unde $MB$ este linie mijlocie. De aici avem $2\cdot MB=CD\Longrightarrow AB=CD\Longrightarrow ABCD$ paralelogram!


geometry
incenter
circumcircle
geometry proposed

Given an acute triangle $ ABC$, consider a circle $ 1$ passing through $ B,C$. Consider another circle $ 2$ internally tangent to $ 1$ at $ T$ and also tangent to the sides $ AB,AC$ at $ P,Q$ respectively. Let $ M$ be the midpoint of the arc $ BC$ of $ 1$ containing $ T$. Prove that $ BC, MT,PQ$ are concurrent.

[quote]Given an acute triangle $ ABC$, consider a circle $ (O)$ passing through $ B,C$. Consider another circle $ (\mathcal {K})$ internally tangent to $ (O)$ at $ T$ and also tangent to the sides $ AB,AC$ at $ P,Q$ respectively. Let $ M$ be the midpoint of the arc $ BC$ of $ (O)$ containing $ T$. Prove that $ BC, MT,PQ$ are concurrent.[/quote]
Let $ A'$ be the second intersection of $ AB$ with $ (O)$ and $ A''$ be the midpoint of arc $ A'B$ of $ (O)$ that doesn't containing $ C$. Then according to the [i]Sawayama and Thébault's theorem[/i], we may conclude that the incenter $ I$ of $ \triangle A'BC$ lies on $ PQ$. Also note that $ A'',$ $ P$ and $ T$ are collinear. Then by the [i]Pascal theorem[/i] for inscribed hexagon $ A'MTA''CB$, we conclude that $ MT,$ $ BC$ and $ IP$ are concurrent, which leads to that fact of the problem. Our proof is completed then. $ \square$

Let $ K$ be the intersection of $ PQ$ and $ BC$.Applying menelaus Theorem for $ \triangle ABC$ we'll get
$ \dfrac{KB}{KC}.\dfrac{AP}{PB}.\dfrac{CQ}{QA}\equal{}1$ => $ \dfrac {KB}{KC}\equal{}\dfrac{PB}{QC} (1)$
 Extended $ TM$ meet $ BC$ at $ K'$. It's easy to see $ TK'$ is the external bisector of $ \angle BTC$ => $ \dfrac{KB}{KC}\equal{}\dfrac{TB}{TC}$. Applying Theorem generalized ptolemaic we have $ \dfrac{BP}{QC}\equal{}\dfrac{TB}{TC} (2)$. Since $ (1), (2)$ => QED

[quote="Agr_94_Math"]mathvnpro, could you explain your solution more? I do not see how Sawayma Theorem has been applied. Also, do not reason out how $ A",P,T$ are collinear. Please elaborate.[/quote]
Dear [i]Agr_94_Math[/i],

$ \bullet$ First, the [i]Sawayama and Thébault's theorem[/i] states that:
[color=darkblue]
[i]"For $ D$ is an arbitrary point on $ BC$, then the line through the tangency points $ P,$ $ Q$ of the circle that tangents to $ AD$, $ BC$  respectively and internally tangents with the circumcircle $ (O)$ of $ \triangle ABC$ will contains the incenter $ I$ of $ \triangle ABC$."[/i][/color]

Proof of this property can be seen at the file I attach below.

$ \bullet$ Second, the reason that $ A'',$ $ P$ and $ T$ are collinear based on a very well- known result, which is:

[color=darkblue][i]"Let $ (I)$ be the circle that internally tangents to $ (O)$ at $ K$. Let $ BC$ be an arbitrary chord of $ (O)$ that tangents to $ (I)$ at $ A$. Then $ AK$ goes through the midpoint of arc $ BC$ that doesn't contain $ K$.[/i][/color]

In your problem, just take $ (I)$ as $ (\mathcal {K})$, $ (O)$ $ \equiv$ $ (O)$, take $ K$ as $ T$ and $ BC$ as $ BA''$, then the result is lead as follow.

Best regard,
[i]mathVNpro[/i]

Dear Agr_94_Math.Based on my solution and MathVnpro's solution,I think the word 'internally tangent' is unnecessary :)


Let $a<b$ be relativly prime natural numbers. Prove there is at least $one$ prime number with this form $a+kb$, $k \in N$ :yup:

If you can prove this, the full Dirichlet-theorem follows.

Really? Does it follow in an elementary way?

If you have found one prime $a+bk$, then you find a another one of that type by setting $a'=a+b(k+1)$, giving a prime of type $a'+bk' = a+b(k+k'+1)$. And then continue this...

I am sorry,I think with $a<b$ should work

If $ a \plus{} bk$ is prime ($ a < b$), then take $ b \equal{} b(ak \plus{} 1)$ this time: you will find a prime of type $ a \plus{} k^\prime(b(ak \plus{} 1)) \equal{} a \plus{} b(k^\prime(ak \plus{} 1))$, again of desired type. Now proceed again...

[quote="ZetaX"]this time: you will find a prime of type $ a \plus{} k' (b(k \plus{} 1)) \equal{} a \plus{} b(k'(k \plus{} 1))$, again of desired type.[/quote]
Could you elaborate on how this works? We can only create a prime of the form $ a \plus{} k'(b(k\plus{}1))$ if $ a$ and $ b(k\plus{}1)$ are relatively prime. All we know is that $ a$ and $ b$ are relatively prime.

Right, that $ k$ should be an $ ak$.
Alternatively, take $ b\equal{}b^s$ for big enough $ s$.

Edited now.


Solve the ecuation $ x! \plus{} y! \plus{} z! \equal{} 2^{u!}$

You can find the solution , but in roumain..If you cannot understand it I will rewrite it in english.
http://www.mathlinks.ro/viewtopic.php?p=1039556#1039556


Let $ k \geq 2$ be a positive integer for which consider the set $ \mathcal{K} \equal{} \left\{1, 2, 3, \ldots, 2k\right\}$. Let $ p$ be the number of [i]disjoint[/i] partitions of $ \mathcal{K}$ into $ k$ pairs such that the sum of the two elements from a pair is a prime number. Prove that $ p \geq 2$.

The fact that $p\geq 1$ is a cute theorem by Greenfield and Greenfield, see [url]http://nd.edu/~dgalvin1/pdf/bertrand.pdf[/url].
Since the proof is based on Bertrand's postulate (Tchebysheff's theorem), it will produce as many different $p$ such partitions as there are primes between $2k$ and $4k$. By a result of Ramanujan, see [url]http://en.wikipedia.org/wiki/Ramanujan_prime[/url], we have $\pi(x) - \pi(x/2) \geq 1,2,3,4,5,\ldots$ for all $x \geq 2,11,17,29,41,\ldots$. Thus for $k\geq 4$ we will have $p\geq 2$. For the small values, we have $(1,2),(3,4)$ and $(1,4),(2,3)$ for $k=2$, and $(1,6),(2,5),(3,4)$ and $(1,2),(3,4),(5,6)$ and $(1,4),(2,3),(5,6)$ for $k=3$.

You produce multiple partitions by this method, but is it immediately clear that the partitions are disjoint?

What is the definition of [i]disjoint[/i] partitions?

Let $X=\bigcup_{i=1}^s A_i = \bigcup_{j=1}^t B_j$ represent two partitions of $X$ (so $A_i \cap A_{i'} = B_j \cap B_{j'} = \emptyset$ for all $i\neq i'$ and $j\neq j'$). Clearly we cannot have $A_i \cap B_j = \emptyset$ for all $i,j$, so I welcome a definition ... maybe $A_i \neq B_j$ for all $i,j$ ... is that it?

My reading was that the word "disjoint" was just redundant, asking that the partitions be into disjoint pairs.

So was mine ... let us wait for Cosmin to clarify that  :)

Indeed, I meant $A_i \neq B_j$ for all $i,j$. Thanks for the really neat proof for $p \geq 1$. I had a similar thing in mind that involved Bertrand.


inequalities
symmetry
inequalities proposed

Let $ a,b,c$ be positive reals such that $ [max(a,b,c)\plus{}min(a,b,c)]/2<\equal{}median(a,b,c)$

Prove:

$ (a\plus{}b)/c \plus{} (b\plus{}c)/a \plus{} (c\plus{}a)/b >\equal{} (2/3)*(a^2/bc \plus{} b^2/ca \plus{} c^2/ab) \plus{} 4$

[quote="Protonios"]Let $ a,b,c$ be positive reals such that $ [max(a,b,c) \plus{} min(a,b,c)]/2 < \equal{} median(a,b,c)$

Prove:

$ (a \plus{} b)/c \plus{} (b \plus{} c)/a \plus{} (c \plus{} a)/b > \equal{} (2/3)*(a^2/bc \plus{} b^2/ca \plus{} c^2/ab) \plus{} 4$[/quote]
Let $ a\geq b\geq c.$ Then $ 2b\geq a \plus{} c$ and $ b\minus{}c\geq a\minus{}b.$
We obtain: $ \sum_{cyc}\frac {a \plus{} b}{c}\geq\sum_{cyc}\frac {a^2}{bc} \plus{} 4\Leftrightarrow\sum_{cyc}(a \minus{} b)^2(2c \minus{} a \minus{} b)\geq0.$
But $ \sum_{cyc}(a \minus{} b)^2(2c \minus{} a \minus{} b)\geq$
$ \geq(a \minus{} b)^2(2c \minus{} a \minus{} b) \plus{} (a \minus{} c)^2(2b \minus{} a \minus{} c) \plus{} (a \minus{} b)^2(2a \minus{} b \minus{} c) \equal{}$
$ \equal{} (2b \minus{} a \minus{} c)(\left((a \minus{} c)^2 \minus{} (a \minus{} b)^2\right)\geq0.$

Here is the solution i posted in the greek forum .

Due to the symmetry wlog assume that $ a\geq b\geq c$
Now

$ \frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {a \plus{} c}{b} \minus{} \frac {2}{3}(\frac {a^2}{bc} \plus{} \frac {b^2}{ca} \plus{} \frac {c^2}{ab}) \minus{} 4 \equal{} \\
\frac {(a \plus{} b \minus{} 2c)(2a \minus{} b \minus{} c)(2b \minus{} c \minus{} a)}{3abc}\geq 0$


How many four-digit positive integers have at least one digit that is a 2 or a 3?

$ \textbf{(A) } 2439 \qquad \textbf{(B) } 4096 \qquad \textbf{(C) } 4903 \qquad \textbf{(D) } 4904 \qquad \textbf{(E) } 5416$

[hide]
it is same as to find all possiblity - those without 2,3 .Hence

first digit = 7 choices ( not including 0,2,3)
second,third,fourth digit = 8 choices (not including 2,3)

Hence the total way is 9000 - 7x8x8x8 = 5416

$\boxed{E}$
[/hide]


algebra
polynomial
search
linear algebra
matrix
analytic geometry
linear algebra unsolved

I have been given a set of data $ (x_i,y_i)_{1\leq i\leq n}\in \mathbb{R}^2$, $ x_i\ne x_j$ as $ i\ne j$. Let $ \epsilon > 0$ be given. Determine the polynomial $ P: \mathbb{R}\to\mathbb{R}$ with minimal degree and real coefficients such that $ |P(x_i) \minus{} y_i| < \epsilon$ for all $ 1\leq i\leq n$. I know already Lagrange's interpolation polynomial but this is not the same problem.

It probably makes more sense to turn this on its head: given a fixed maximum degree, what's the best approximation we can get?

If we were using the Euclidean norm, this would be a least-squares optimization, and those are well understood. With the uniform norm, there's probably no nice way to do it in general.

OK, what can we do anyway? First, look at the properties of a "best possible" approximation of degree $ k$, assuming that this approximation is not the interpolating polynomial. There are some points at the maximum error $ \epsilon$. If there are $ k\plus{}1$ or fewer of these, we can add a small multiple of a polynomial that decreases the error at all of those points, and thus push the error below $ \epsilon$ everywhere. This is impossible, so there are at least $ k\plus{}2$ such points, and no polynomial of degree $ \le k$ has the same sign as the errors at all of those points. We enforce this by having the errors alternate sign as the $ x_i$ increase, changing sign at least $ k\plus{}1$ times.

This reduces the problem to a finite search. Choose $ k\plus{}2$ of the $ x_i$, labeled for convenience as $ w_0,w_1,\cdots,w_{k\plus{}1}$ in increasing order. Interpolate a linear combination of $ 1,x,x^2,\cdots,x^k,q(x)$ on this set and the corresponding $ y$ values, where $ q(w_i)\equal{}(\minus{}1)^i$. The polynomial part of this is the best approximation on the $ w_i$ subset, and the absolute coefficient of $ q$ is the error of this approximation. Repeat this procedure for all subsets of size $ k\plus{}2$, and note the largest $ q$ coefficient found this way. That maximum is the minimum overall error, and the corresponding polynomial is the best approximation.

Computationally, this is a very slow process- $ \binom{n}{k\plus{}2}$ is often very large. There may be a way to improve the search, likely using linear programming ideas.

As an example, let's take some random data and try a fit- in this case, 4 points to fit a line. I chose 2,3,5,7 as the $ x$-values, and my RNG (0-100 uniform, rounded down) gave me 23,37,71,80 respectively. I want to fit the best possible line $ y \equal{} ax \plus{} b$ in the $ \infty$-norm to the four points $ (2,23),(3,37),(5,71),$ and $ (7,80)$.

For each triple of points, we interpolate as described in my previous post; this amounts to solving a system of linear equations. For 2,3,5, this system is $ \begin{bmatrix}1 & 2 & 1 \\
1 & 3 & \minus{} 1 \\
1 & 5 & 1\end{bmatrix}\cdot\begin{bmatrix}a \\
b \\
c\end{bmatrix} \equal{} \begin{bmatrix}23 \\
37 \\
71\end{bmatrix}$. The solution of that system is $ a \equal{} \minus{} 10,b \equal{} 16,c \equal{} 1$, meaning that $ y \equal{} \minus{} 10 \plus{} 16x$ is the best possible line on these three points, with absolute error $ 1$ at each of them. Repeating for the other triples, we get $ (a,b,c) \equal{} (1.5,11.4, \minus{} 1.3)$ at $ (2,3,7)$, $ (a,b,c) \equal{} (7.1,11.4, \minus{} 6.9)$ at $ (2,5,7)$, and $ (a,b,c) \equal{} (11,10.75, \minus{} 6.25)$ at $ (3,5,7)$. The largest $ c$ value is the $ \minus{} 6.9$, so we use that triple. The best fit line is $ y \equal{} 7.1 \plus{} 11.4x$, with maximum error $ 6.9$ on our points. Three of the four points have error $ \pm 6.9$, while the fourth is off by the smaller value $ 4.3$.
The other triple all lead to large error at the point not used in interpolation. $ y \equal{} \minus{} 10 \plus{} 16x$, for example, is off by $ 22$ at $ x \equal{} 7$.

The primes among the random values chosen were pure coincidence, as was their increasing order. The order does make things nicer, so I got lucky.

Try the [url=http://en.wikipedia.org/wiki/Bernstein_polynomial]Bernstein polynomials[/url].

Erm... What exactly would that mean in this particular context? :? (note that the $ x$ data are not even on $ [0,1]$...)

Hm.  On closer consideration, that would only work if the $ x$ coordinates were in an arithmetic progression.  Never mind.


[quote="orl"]Miklos Schweitzer 2001:[/quote]

is it my problem that I can't see anything after 2001 or there is nothing there to see?

[quote="goodar2006"]is it my problem that I can't see anything after 2001 or there is nothing there to see?[/quote]

nothing!

so where can I find them and their solutions?

Here are the problems. You may post them to get some solutions. :)

[quote="goodar2006"]thanks Amir, and just see how  :censored: Iranians are......problem 1 of this competition is the same as problem 7 of the Iranian 3rd round final exam 2010.  :mad: and this is not the only case......[/quote]

and very very similar to this problem:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=848051&sid=d2e7dd3c6fdb0beb2c8c8c129d72241a#p848051


Can a square of side length 2.1 be covered by six squares of side length 1?

No.

For the corners of the 2.1 side lengthed square to be covered, there must be a square in each of the corners:

[img]http://i169.photobucket.com/albums/u236/Mozza314/mathmisc/21Square.gif[/img]

This means a + 2.1 units * 2.1 units is leftover, and the maximum span of the 2 remaining squares is only 2.


A 98 x 98 chess board has the squares colored alternately black and white in the usual way. A move consists of selecting a rectangular subset of the squares (with boundary parallel to the sides of the board) and changing their color. What is the smallest number of moves required to make all the squares black?

according to kalva,

Answer: 98.

There are 4·97 adjacent pairs of squares in the border and each pair has one black and one white square. Each move can fix at most 4 pairs, so we need at least 97 moves. However, we start with two corners one color and two another, so at least one rectangle must include a corner square. But such a rectangle can only fix two pairs, so at least 98 moves are needed.

It is easy to see that 98 suffice: take 49 1x98 rectangles (alternate rows), and 49 98x1 rectangles (alternate columns).

I don't really understand this solution... D: can someone explain it to me? Thank you.

Also, there are ~48 freshmen qualifiers this year. What USAMO score do you think I'll need to make it to MOP?

The solution begins by considering only those squares along the border.  There are $4 \times (98 - 1)$ such squares, and each square bijects to a pair of such squares, one of which is black and the other of which is white.  (Number the squares in sequence; for each square, pick the square immediately after in sequence, and there is your pair.)

When a pair is "fixed," the white member of the pair becomes black.  A rectangle that consists only of a single white square fixes two pairs, and a rectangle that contains two white squares diametrically opposite to each other fixes four pairs, and no other members can be added to the rectangle without reversing some number of black squares on the border, meaning at least $97$ moves are required.

However, two of the corners are white, and a rectangle that contains a corner can only fix two pairs (since going to the opposite side would require going through the border, which switches black squares on the border unnecessarily).  Hence, at least $98$ moves are required.

The $98$ moves in question come from alternating rows and then alternating columns.  Consider the much smaller case of $4$, where - is white and + is black.

- + - +
+ - + -
- + - +
+ - + -

Select rows $1$ and $3$ only.

+ - + -
+ - + -
+ - + -
+ - + -

Now select columns $2$ and $4$.

+ + + +
+ + + +
+ + + +
+ + + +

[quote="AnnieS1527"]What do you mean by diametrically opposite? If it's opposite horizontally or vertically, I don't think you would fix 4 pairs. Since black is always opposite white and vice versa, you would fix 2 pairs and mess up 2 pairs.[/quote]

Only opposite either horizontally or vertically.  I suppose "diametrically" was a bad way of putting it.  For example:

+ + + +
- + + -
+ + + +
+ + + +

The two minus signs are opposite in the way I intended.  That way, they can both be contained inside a rectangle of height $1$.


[quote="Timothy"]The sum of 9 consecutive integers is 9.  What is the least of these 9 integers?[/quote]

[hide]Set the lowest of those integers equal to $x$

Then:

$9x+36=9$

$x+4=1$

$x=-3$[/hide]

[quote="pianoforte"][quote="Drunken_Math"]Taking this to the extreme:

If the sum of 100 consecutive integers is 9900, what is the fifth smallest of those integers?[/quote]

[hide]
$x+x+1+x+2...x+99=9900$
$100x+\frac{99\cdot100}{2}=9900$
$200x+9000=1800$
$200x=900$
$x=4.5?$
$x+4=9.5?$ :?
did I do something wrong?[/hide][/quote]

$9900*2\not=1800$
$99*100\not=9000$
$1800-9000\not=900$

 ;)


trigonometry geometry circumcircle symmetry

Cyclic pentagon $ ABCDE$ has side lengths $ AB\equal{}BC\equal{}5$, $ CD\equal{}DE\equal{}12$, and $ AE \equal{} 14$. Determine the radius of its circumcircle.

After a bunch of trig, I got $ 7r\equal{}(5\sqrt{r^2\minus{}36}\plus{}12\sqrt{r^2\minus{}6.25})(\sqrt{(r^2\minus{}36)(r^2\minus{}6.25)}\minus{}15)$. My calc gives 6.0421811 as the solution, I don't know what that is.

[quote="worthawholebean"]Cyclic pentagon $ ABCDE$ has side lengths $ AB \equal{} BC \equal{} 5$, $ CD \equal{} DE \equal{} 12$, and $ AE \equal{} 14$. Determine the radius of its circumcircle.[/quote]

Possible Hints, but not sure.

[hide]We need to make the figure so that there is some sort of symmetry.  The best choice we have is probably going to be introducing a new point C' such that BC'=12 and C'D=5, on minor arc BCD.  Notice that the radius is still the same.

Now we have the ability to use Ptolemy's Theorem on Various Quadrilaterals within ABC'DE; we should end up with three equations and three variables.

Then, we can solve this equation, and notice that we can get a triangle, with two sides and a diagonal, using the formula $ R\equal{}\frac{abc}{4K}$, we would be done.[/hide]

You get a cubic when you do that. Let x be the length of the three diagonals that are the same, we have 14x+60=yz, 12x+70=xy, 5x+168=xz, resulting in $ x^2(14x \plus{} 60) \equal{} (12x \plus{} 70)(5x \plus{} 168)$, x=15 works in this (but you'd have to just guess). But AC'E is isoceles, so it makes it easier to find the answer, which I get is $ 225/2\sqrt{176}$.

I tried a bunch of approaches with no success, but I know that you can rearrange the sides of the pentagon and retain the circumcircle radius.

Well, one method, as said before, is to use symmetry and rearranging the pentagon. The best thing would probably be to make the side lengths 5, 12, 12, 5, 14, in that order. There would be an axis of symmetry at this point, which goes through the midpoint of the side with length 14 and through the vertex with side lengths 12 on each side.


Can any1 provide me with the name of any introductory book on transformational geometry?

Can any1 provide me with any introduction on transformational geometry.  :P

once some 1 provides me with the name of the book,u better check it out as well.Hope it will help u to find out the ans to ur gr8 ques. :D

Hey!!
 :blush: What's transformational geometry?
I'vent heard of it!!

[i]Geometry Revisited[/i] has a chapter on transformations. I think it's a pretty good introduction.

'Geometrical Transformation' Vol 1 & 2 by Yaglom are quite exhaustive.

Notes on Transformation Geometry from New Zealand MO site.


Three balls marked 1,2, and 3, are placed in an urn. One ball is drawn, its number is recorded, then the ball is returned to the urn. This process is repeated and then repeated once more, and each ball is equally likely to be drawn on each occasion. If the sum of the numbers recorded is 6, what is the probability that the ball numbered 2 was drawn all three times?

$\displaystyle \text{(A)} \ \frac{1}{27} \qquad \text{(B)} \ \frac{1}{8} \qquad \text{(C)} \ \frac{1}{7} \qquad \text{(D)} \ \frac{1}{6} \qquad \text{(E)} \ \frac{1}{3}$

[hide="Solution"]He picked 3 times, the following outcomes add up to 6
1, 2, 3
1, 3, 2
2, 3, 1
2, 1, 3
2, 2, 2
3, 1, 2
3, 2, 1

Therefore there is 1/7 chance.[/hide]

[hide="Answer"]We have the following possibilities for drawing a ball:

$(1,2,3)$
$(1,3,2)$
$(2,1,3)$
$(2,2,2)$
$(2,3,1)$
$(3,1,2)$
$(3,2,1)$

Therefore, we have $7$ possibilities and one gives what we want, so the answer is $\frac{1}{7}\Rightarrow \boxed{C}$.[/hide]

This is a #15? :huh:

[hide]He can pick (1,2,3) or (2,2,2). THere are 3!+1=7 ways total, with one way favoring the (2,2,2). THus the probability is 1/7. [/hide]


What is the sum of the first 13 positive odd integers?

sum of first n positive odd numbers = $ n^2$

$ 13^2\equal{}169$

answer : 169


[hide="Solution"] \begin{eqnarray*}\sin x &=& \frac{1}{2}\sin \frac{x}{2}\cos \frac{x}{2}\\ &=& \frac{1}{4}\sin \frac{x}{4}\cos \frac{x}{4}\cos \frac{x}{2}\\ &=& ... \\ &=& \frac{1}{2^{n}}\sin \frac{x}{2^{n}}\cos \frac{x}{2^{n}}... \cos \frac{x}{2}\\ \end{eqnarray*}

A bit of calculus is enough to see that

${\lim_{n \to \infty}\frac{ \sin \frac{x}{2^{n}}}{2^{n}}}= x$

Which produces our desired result. [/hide]

[hide="Erratum"] Of course, I made repeated use of the identity

$\sin 2 \theta = 2 \sin \theta \cos \theta$

And so that should be

$\sin x = 2^{n}\sin \frac{x}{2^{n}}\cos \frac{x}{2^{n}}... \cos \frac{x}{2}$

And the limit should be

$\lim_{n \to \infty}2^{n}\sin \frac{x}{2^{n}}= x$ [/hide]

Well, look at it this way, lim sinx/x = 1 from a formula I learned in
                                 x-0
math class. when the left side of the eq is taken lim (x-0), cos (x/2n) is 1 no matter what. so it results in (1)(1)(1)....(1). Therefore 1=1

The fact that the limit as $x$ approaches $0$ is the same for both sides doesn't mean that they're the same for all $x$.  Furthermore, you can't assume that because each factor goes to $1$, the entire product goes to $1$; there are an infinite number of factors, so in fact it might go to $0, e, \infty$ (it doesn't in this case, but you haven't shown to the contrary.)

Example:  $\lim_{n \to \infty}\left( 1+\frac{1}{n}\right)^{n}= e \neq 1$ even though each individual factor $1+\frac{1}{n}$ tends to $1$.

[quote="t0rajir0u"]The fact that the limit as both sides approach $0$ is the same doesn't mean that they're the same for all $x$.  Furthermore, you can't assume that because each factor goes to $1$, the entire product goes to $1$; there are an infinite number of factors, so in fact it might go to $0, e, \infty$ (it doesn't in this case, but you haven't shown to the contrary.)

Example:  $\lim_{n \to \infty}\left( 1+\frac{1}{n}\right)^{n}= e \neq 1$ even though each individual factor $1+\frac{1}{n}$ tends to $1$.[/quote]

Huh? 0 divided by anything is 0. R u saying that I have to write a whole proof on that too?

Sorry, a small typo in the first sentence.  But I wasn't talking about dividing by $0$.  I was talking about limits.

You said that when both sides are taken $\lim_{x \to 0}$ they are both $1$, so they must be equal.

But consider $x$ and $2x$.  When I take $\lim_{x \to 0}$ they both go to $0$.  Does that mean that they are equal?  [b]Of course not![/b]

Secondly, you did not actually show that

$\lim_{x \to 0}\cos \frac{x}{2}\cos \frac{x}{4}... = 1$

It's not enough that each factor goes to $1$ because [i]there are an infinite number of factors.[/i]


A man stands on weighing scales with his hands by his sides with a mass in each hand.Now,what happens to the reading on the scales if he $quickly$ moves his hands:

a)Up towards his head

b)down towards his legs

My first obvious reaction is that the weight

a) increases
b) decreases

Is there some tricky part to this problem?

And on what basis di you give your solutions...those are my thoughts too but i'm totally confused on this sum...there are like so many factors i'm thinking of..i'm just confused. :?

For the 1st oneI think the weight would decrease first then increase
And for the 2nd one opposite

1st one: This is because.....
When you move your hands upward, they exert a force on your body (momentum) in the upward direction so then that would cause the body to move upward slightly, Hence causing the weight to decrease

Then after that the body has to return to its usual position, so the weight would increase.

I only used Newton's laws. Your center of mass moves upwards so there has to be some force that caused this. The same for the other case.

[quote="riddler"]When you move your hands upward, they exert a force on your body (momentum) in the upward direction so then that would cause the body to move upward slightly, Hence causing the weight to decrease[/quote]Do you really think you could lift yourself if you moved your hands up very quickly? :)

[quote="Bus"]Do you really think you could lift yourself if you moved your hands up very quickly? [/quote]

well, if you [b]really[/b] want to... maybe you will! :lol:

[quote="bus"]That's right! By moving your hands upwards, you actually push the rest of your body downwards! That's the "opposite reaction" from Newton's third law. If you wanted to lift yourself, you'd have to move your hands [i]down[/i] instead.[/quote]

Yep. Your shoulders act as the pivot and the weight of the arms going upwards pushed the body down the question a.

The answer in the first case is increase,i guess...though it is given decrease in my book..

So the answer of the first one is increase, so that is due to centre of mass? :o

[quote="Rushil"]Actually, IMPOSSIBLE!!!! [/quote]
[quote="Bus"]That's right! By moving your hands upwards, you actually push the rest of your body downwards! That's the "opposite reaction" from Newton's third law. If you wanted to lift yourself, you'd have to move your hands down instead.[/quote]

I was kidding...

[quote="bus"]That's right! By moving your hands upwards, you actually push the rest of your body downwards! That's the "opposite reaction" from Newton's third law. If you wanted to lift yourself, you'd have to move your hands [i]down[/i] instead.[/quote]

This is the answer riddler.

And for the second case it is the opposite.

If we use the formula for gravitational attraction between 2 bodies,the the distance decreases in the second case and hence attraction increases,so the weight will be more?I think the change would be negligible.

Interesting question.

Masses in hands up, weight increases then comes back to what it was before. (Except for the negible decrease because the center of mass is farther from the center of the earth.  Even with an inverse square rule, the distance to the center of the earth is thousands of miles, a decrese of part of a person's height is negible with the strength of the human body and the mass of the earth.  If we put in the mass of the sun, the human body would be torn apart by the "tidal" forces (differences between the acceleration due to gravity), even in free fall.)

The reason for this is the scale measures the force you apply towards the earth.  Moving your center of gravity up, requires a push down.  (Just like jumping.)  Reversing the hand weight movement is just like crouching, reversed scale measurement.

If you don't believe this, try it.  You can get a measurable effect with just your hands.  If you don't have a scale try standing on a location where you are balanced (moving towards and back from a fulcrum).   Aahh the joy of experiemential confirmation of theory.

Incidentially, they found hand weights that were used for long jumpers in some of the orginial Olympics.  We don't know how they worked, about 5 kg each, one for each hand.  Supposedly their jumps were much longer than our current records, they ran down a curving track out into a flat area, jumping and aparently abandoning the weights.  The force on landing was sufficient for several to have broken both legs in competition.

Anyway, you get the idea.


Hai folks, I would like you to view your replies on these queries..

Query 1

I have been told that the so-called "prodigies", exceptionally gifted people synthesize information in an unusual manner,and that there's nothing particularly "special" about them genetically, etc from any normal bright person, is this true?
Please answer this query with appropriate elaboration, I mean answer this query with a succinct reply.

Query 2

I would also like to know what are the characteristics, traits of exceptionally gifted people, please don't say that they possess a "wildly" superior intellect, I strongly disagree with that particular notion , I feel their perception of a problem, etc, is vastly different, is my understanding correct?

I feel there are many such exceptionally gifted, etc people here in this forum, I would like them( for that matter, anyone)  to answer these queries. I thank you all.........  :)
[/quote]

1)  What do you mean by "special"?  Most components of personality have some degree of heritability, so it stands to reason that giftedness is one of them.

2)  [url=http://www.ri.net/gifted_talented/character.html]This[/url] isn't a bad resource to start with, or [url=http://www.us.mensa.org/Content/AML/NavigationMenu/Programs/GiftedChildren/Factsonbeinggifted/BeingGifted.htm]this[/url].

[quote="Gen8"]I would also like to know what are the characteristics, traits of exceptionally gifted people, please don't say that they possess a "wildly" superior intellect, I strongly disagree with that particular notion [/quote]

so basically you are only interested in answers which fit into your preconceived notions of giftedness?

if you read [url=http://www.jstor.org/pss/3482178]this[/url] article (about terence tao as a child), i don't see how you can deny that he has a 'wildly superior intellect' to just about every human being on the planet.  it's likely that some of what we classify under giftedness is merely perceiving things in a different way, but a lot of it is just vastly superior ability.


Find the sum of all integers between 80 and 90 which divide $3^{32}-1$.

As always, leave this challenge for less experienced problem solvers if you find this one easy.

[hide]Note that a^2-b^2=(a+b)(a-b), and that, since 32 is an even power, 3^32 is a perfect square, as is 1.

3^32-1 = (3^16+1)*(3^16-1)

3^32-1 = (3^16+1)*(3^8+1)*(3^8-1)

...

3^32-1 = (3^16+1)*(3^8+1)*(3^4+1)*(3^2+1)*(3^1+1)*(3^1-1)

Since 3^[even number] is odd, adding 1 makes it even, so you can factor a 2 out of each term, which leaves us with

3^32-1 = 2^6*1*2*5*17*41*193*21523361

Simplified, this becomes

3^32-1 = 2^7*5*17*41*193*21523361

And then factor all of the "candidate numbers:"

81=3^4

82=2*41

83=83 (prime)

84=2^2*3*7

85=5*17

86=2*43

87=3*29

88=2^3*11

89=89 (prime)

The numbers that have all of their factors included in the factorization of 3^32-1 are divisors of 3^32-1, so  82 and 85 (in bold) are factors of 3^32-1.  Their sum is 167.[/hide]


I did... I'm sure there are a lot of other people who did as well... Why do you ask?

Oh, can anyone tell me how to do it?
Thanks a lot


Find all integers $x,y,z$ satisfying $x^2+y^2+z^2=2xyz$.

I think I've seen it before around here.

At least one number has to be even. If only one is even, the LHS is $M_4+2$, while the RHS is $M_4$, so all of them must be even. Replace $x$ by $x'=\frac x2$ (and the same for $y,z$) to get $x'^2+y'^2+z'^2=4x'y'z'$. The same reasoning shows $x''^2+y''^2+z''^2=8x''y''z''$ (where $x''=\frac{x'}2$), and so on. We find that $x^2+y^2+z^2$ is divisible by $2^n$ for any $n$, so $x^+y^2+z^2=0$, iff $x=y=z=0$.

[quote]I think I've seen it before around here. [/quote]

This is a special case of Markov equation, namely :
\[a^2+b^2+c^2 = nabc\]
where $a,b,c,n$ are positive integers.

It is well-known (and even maybe already seen here) that this equation has solution if and only if $n=1$ or $n=3$.

Pierre.


let $ x,y$ be real numbers, such that $ x^3 \minus{}3xy^2 \equal{} 11$ and $ y^3\minus{}3x^2y\equal{}2$, find the value of$ x^2\plus{}y^2$.

[quote="waelalghamdi"]let $ x,y$ be real numbers, such that $ x^3 \minus{} 3xy^2 \equal{} 11$ and $ y^3 \minus{} 3x^2y \equal{} 2$, find the value of$ x^2 \plus{} y^2$.[/quote]
[hide="hint"]
$ 11(x^3\minus{}3xy^2)\minus{}2(y^3\minus{}3x^2y)\equal{}0$
Let $ a\equal{}\frac{x}{y}$ then
$ 11a^3\minus{}3a^2\minus{}33a\minus{}2\equal{}0$
[/hide]

Squaring the equalities we get

$ x^6\minus{}6x^4y^2\plus{}9x^2y^4\equal{}121$

$ y^6\minus{}6x^2y^4\plus{}9x^4y^2\equal{}4$

Adding those up we get

$ x^6\plus{}3x^4y^2\plus{}3x^2y^4\plus{}y^6\equal{}125\iff (x^2\plus{}y^2)^3\equal{}125$

Hence $ x^2\plus{}y^2\equal{}5$

FOLLOW UP:

The above result can be used to solve the system:

$ x^3 \minus{} 3x(5 \minus{} x^2) \equal{} 11\iff 4x^3 \minus{} 15x \minus{} 11 \equal{} 0$

$ x \equal{} \minus{} 1$ is an obvious root, so the factorization gives $ (x \plus{} 1)(4x^2 \minus{} 4x \minus{} 11) \equal{} 0$, therefore $ x_1 \equal{} \minus{} 1, x_{2,3} \equal{} {1\over 2}\pm\sqrt {3}$

Now the second equation gives $ y(5 \minus{} x^2 \minus{} 3x^2) \equal{} 2\iff y \equal{} {2\over 5 \minus{} 4x^2}$, so the solutions are

$ ( \minus{} 1,2),\left({1 \plus{} 2\sqrt {3}\over 2},{ \minus{} 2 \plus{} \sqrt {3}\over 2}\right),\left({1 \minus{} 2\sqrt {3}\over 2},{ \minus{} 2 \minus{} \sqrt {3}\over 2}\right)$

COROLLARY: Multiplying the second equation by $ \minus{} i$ and adding to the first, we get $ (x \plus{} yi)^3 \equal{} 11 \minus{} 2i$. Therefore, the above points in the complex plane are the values of $ \sqrt [3]{11 \minus{} 2i}$. Also, that fact can be used thus: $ |x \plus{} yi|^3 \equal{} |11 \minus{} 2i| \equal{} 5^{3/2}\implies x^2 \plus{} y^2 \equal{} |x \plus{} yi|^2 \equal{} 5$


[quote="tranthanhnam"]prove that :
 $ab+bc+ca \geq 8(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2)$ with a,b,c positive real[/quote]
This is a problem in ussue of recent Math & Magazine.So you don't permit to post and discuss problem now. Please Admin locks this topic.


Find all n s.t. n divides $ 2^{n}\plus{}1$ . Im sorry if it had been posted before

A weaker version of this was used in 1990 IMO [url]http://www.mathlinks.ro/viewtopic.php?p=366466#366466[/url]
Edit: It must be weaker instead of stronger. Sorry for the mistake. The explanation is useless now.

That's actually a weaker version since the conditions are more specific.

You're welcome  :) .
There is actually an infinite amount of numbers that satisfy the condition for this problem, the first few are given here:[url]http://www.research.att.com/~njas/sequences/A006521[/url]
A paper on it is here:[url]http://www.maths.ed.ac.uk/~chris/n_divides_2to_nplus1.pdf[/url]

There is also another nice result, that for any k, we can find an n such that n has exactly k prime divisors and n divides $ 2^{n}\plus{}1$.
I read a proof of this which inductively created such n. I would love to see another proof


A coin and a die are tossed and rolled, alternating, starting with the coin, until either the coin shows a head or the die (6-sided) shows a 2. What is the probability that the die shows a 2 first?

[hide]1/7 right??  You can get H,T1,T2,T3,T4,T5,or T6 and only one of them works which is T2 so it should be 1/7 but im not sure[/hide]

You can get the 6 numbers on the die, the head on the coin, and the tail on any given toss and roll cycle, so there are 8 possibilities on each cycle, but that seems easy, doesn't it?

The probability of getting any number that will stop the cycle is 2/8 (Head or 2) or 1/4. The chance of getting a 2 after not getting heads on the first flip (because it is flipped before rolling the dice) is 1/12. So 1/12 divided by 1/4 is 1/3 and that is the answer???? I think it should be smaller for some reason.

[hide="this is what I got the second time around"]I got 1/7. You have 1/12+5/12(1/2)+(5/12)^2(1/2)... giving you (1/12)/(1-5/12). Simplifying that, I got 1/7.[/hide]

wouldnt it be this...
[hide]
the prob you get a two on the first roll is 1/2*1/6=1/12
the prob on the second roll is 1/2*5/6*1/2*1/6=5/144
if u do it one more time, you will notice that it is a geometric sequence with 5/12 as the common product ( i dont know the exact term).

so you use the formula to find sum of infinite geometric sequence:
which is the first term/ 1-common product:
(1/12)/1-5/12=[b]1/7[/b][/hide]

edit: go braves, i think u got it right the first time, i just dont know how u did it

[hide]I agree with 1/7. There is 1/2 a chance you would get tails, and 1/12 a chance that you would roll a 2 [i]first[/i]. This is because once you roll a two, there is a 1/2 chance you would also flip tails and would not have rolled a 2 [i]first[/i]. So you add 1/2 and 1/12 to get 7/12, and 1 of those 7/12s is the dice. 1/7[/hide]

[quote="hockeyadi23"]Go braves got the answer but I don't agree completely with his solution... since the probability of ending with heads is 1/2, the probability of t5 is 1/12.. and so forth.. unless there's some trick here I don't know..[/quote]

um....7 possible outcomes...one that works=1/7


1.  The positive integers $x_1, x_2, ... , x_7$ satisfy $x_6 = 144$, $x_{n+3} = x_{n+2}(x_{n+1}+x_n)$ for $n = 1, 2, 3, 4$. Find $x_7$.
2.  Find all real solutions to:
\begin{eqnarray*} 3(x^2 + y^2 + z^2) = 1 \\ x^2y^2 + y^2z^2 + z^2x^2 = xyz(x + y + z)^3 \end{eqnarray*}

3.  $ABCD$ is a tetrahedron. $DE$, $DF$, $DG$ are medians of triangles $DBC$, $DCA$, $DAB$. The angles between $DE$ and $BC$, between $DF$ and $CA$, and between $DG$ and $AB$ are equal. Show that: area $DBC$ $\leq$ area $DCA$ + area $DAB$.

4.  The sequence $a_1, a_2, a_3, ...$ is defined by $a_1 = 0$, $a_n = a_{[n/2]} + (-1)^{n(n+1)/2}$. Show that for any positive integer $k$ we can find $n$ in the range $2k \leq n < 2k+1$ such that $a_n = 0$.
5.  $ABCDE$ is a convex pentagon such that $DC = DE$ and $\angle C = \angle E = 90^{\cdot}$. $F$ is a point on the side $AB$ such that $\frac{AF}{BF} = \frac{AE}{BC}$. Show that $\angle FCE = \angle FDE$ and $\angle FEC = \angle BDC$.

6.  Given any $n$ points on a unit circle show that at most $\frac{n^2}{3}$ of the segments joining two points have length $> \sqrt{2}$.

1. http://www.mathlinks.ro/Forum/viewtopic.php?p=366254#p366254
2. http://www.mathlinks.ro/Forum/viewtopic.php?p=366256#p366256
3. http://www.mathlinks.ro/Forum/viewtopic.php?p=292958#p292958
4. http://www.mathlinks.ro/Forum/viewtopic.php?p=366259#p366259
5. http://www.mathlinks.ro/Forum/viewtopic.php?p=366260#p366260
6. http://www.mathlinks.ro/Forum/viewtopic.php?p=366261#p366261


email Putnam function Princeton college MIT Harvard

Many of you on this site know me as Richard's co-author on the Art of Problem Solving books.  That couldn't be farther from the truth--in fact, Richard was [i]my [/i]co-author.  Either way, we wrote those books because the problem solving skills we honed through study, competition, and the Math Olympiad Program (MOP) were vital in preparing us and others we know for careers in a wide variety of fields.

For example, I work at [url=http://www.janestcapital.com]Jane Street Capital[/url], a financial trading firm in downtown New York City, using my problem solving capabilities to analyze markets and find profitable trading strategies.  This requires not only math and computer science, but also game-playing strategy, because I don't work in a vacuum.  There are many players in every market, so I have to figure out not just how to value securities, but also how to best adjust my tactics to the motivations and actions of the other players.  This combination of quantitative and strategic thinking has allowed Jane Street to grow from a five-person shop six years ago to one hundred and twenty employees today, and we're still as aggressive and intellectually curious as ever.

The abilities I developed pursuing goals like making the MOP have been so useful in my work that now I'm looking for people with similar backgrounds to join me at Jane Street.  As a student, I had no idea a career like mine even existed, much less that it would be open to someone like me.  Therefore, my company is sponsoring this [url=http://www.artofproblemsolving.com/Forum/index.php?f=338]Careers in Mathematics[/url] Forum in the hope of helping brilliant problem solvers learn early in life about the avenues open to them.

We also sponsor the forum, of course, to find our next great trader, researcher, or programmer.  We are constantly seeking new people with problem solving prowess and a strategic mindset.  If you're an avid card player, or a master of Diplomacy or Settlers of Catan, you're probably one of us.  Contact me by email at hr@janestcapital.com, or write c/o Jane Street, One New York Plaza 33rd Floor, New York, NY 10004.

Sandor Lehoczky

i've lately been thinking about careers concerning mathematics and which college program i should aim for in order to do this.  what kind of education do you have to have to be an employee at your place?  Is a bachelors in math enough or do you want something more like an MBA or even a PhD?

I certainly don't intend to answer for Sandor, but I think the smarter firms on Wall Street are far more interested in what you can do as an employee than what papers you have behind your name.  Interviews often consist of problem solving exercises relating directly to the job.

Getting an interview to begin with sometimes has to do with the degrees you've earned, but the Information Age is quickly helping employers find workers who match the skillset they need without pouring through an increasingly large pile of decreasingly meaningful resumes.  In fact, that's a big part of the purpose of the new forum.

Mat is basically on target here.  It's more who you are and how you think than what you know or what degrees you have.  We have successful people from many different backgrounds, but the problem-solving mindset that goes with study in math and/or science is a very appealing thing for a candidate to have.

Sandor

Hi Sandor,

I am currently a senior in high school trying to decide what university to attend next year (one week left for me to think about it  :) ). Anyways, I have recently become interested in the financial market. I read the career descriptions on your group's website, but am still curious about what exactly traders and quants at your group do. Maybe you could provide a brief description of what a typical week in the life of a quant working in your group would be like. If so, I would really appreciate it.

Also, after doing some research on the net, it seems to me that most financial companies are simply looking for good math and problem-solving skills in candidates who wish to become quants. Currently, one of my college options is going to UPenn and getting a Bioengineering degree as well as a degree at Wharton (finance, perhaps). Note that this plan has no mention of mathematics. Nevertheless, would this plan allow me to go out after college and get a good job as a quant? Maybe scoring well on the Putnam exam would look good on my resume as an indication of good problem-solving skills?

Any insight you (or anyone else here) could provide would be very helpful. Thanks.

A day-in-the-life is really hard, because each person at Jane Street finds their way into a particular niche that suits their own skills.  We have people who are relatively pure traders (focus almost exclusively on optimally implementing strategies and outgaming other market participants) and relatively pure quants (focus on quantitative analysis and research), but also people who fall somewhere in between, combining implementation with experimentation and research.  Even among those groups, we have traders that stand on a trading floor, traders that work off data feeds upstairs, traders that trade primarily over the phone; traders for whom speed of execution is the key to the game, traders for whom careful thought and risk assessment are paramount.  Among quants, too, some focus on scientific-style research, others on improving existing strategies, others on writing infrastructure code.

Another problem with a day-in-the-life approach is that trading fits the Edisonian "1% inspiration, 99% perspiration" model.  The specific details of coding, spreadsheet work, analysis, discussion are individually boring unless one understands the underlying spirit.  Ideas are very important, but the bullheadedness to follow them through in the face of bad data, incomplete information, low signal-to-noise, etc. is perhaps more so.

Bioengineering + Wharton seems like an excellent course of study to me for trading and many other possible careers.  Make sure there's some computer science/coding in there, either as part of your major or electively.  Doing well on the Putnam would of course look good, but don't pursue it unless you really are interested, as it's a lot of work and can really distract from other things.

Sandor

Sandor,

Thanks for your reply. About your comment in regards to the Putnam...

I do indeed plan on studying for the Putnam because I find math competitions very interesting. However, from poking around, I get the impression that the good firms on Wall Street are essentially just hiring random really smart people and training them to be quants themselves. Of course, there are a lot of good students at the top colleges who get high GPAs. Thus, how can one distinguish oneself as a smart person from these masses? You seemed to somewhat lower the importance of doing well on the Putnam (in terms of getting a good job), while I had thought that having a line saying "Top x scorer on the Putnam" would really make one's resume stand out to top Wall Street firms as an indication of someone who is really good at problem-solving.

Richard can attest to this: only fools would hire "random" smart people.  There are hordes of smart people out there, but not many people with the spectrum of qualities that make a good trader/quant.

That said, of course a good score on the Putnam would be a plus, possibly a big plus with some firms.  For me, there would be lots of other things that would be more important, if the candidate was obviously intelligent and mathematically strong.

Sandor

Your resume is only the first step.  No matter how brilliant it is, if you go into the interview and don't perform well on the quantitative and problem-solving type questions which are asked there, you probably will not be hired.  If you do very well in your collegiate studies and are at a well-known, good school, you should at least get past this resume filter.  At that point, all the Putnam points in the world will be only a small help, if at all.

I also highly recommend doing internships during college if you can.  Jane Street, for example, hired a number of college interns this year.  This gives you a chance to impress firms that would hire you for full-time jobs after college and lets you know if a job in the financial world is right for you.  Admittedly, there aren't many firms hiring summer interns like this, but Jane Street does, and that's a great opportunity.

I see, so you are saying that in most cases, its really the interview that makes or breaks a person's opportunity to work at a particular firm (you can tell that I'm rather naive about the job application process).

I have one more quick question if you don't mind...

Like I said, I may decide to double major in Bioengineering and in some degree from Wharton. If I decide to go into finance as a trader/quant, do you think that a degree in Finance would really help me? Or would a degree in, say, Entrepreneurship or Management, serve me just as well in the terms of the knowledge that I gain? I am asking because it appears that most Wall Street firms train their traders/quants themselves anyways, so it may be rather pointless for me to major in Finance. Instead, I could major in a field like Entrepreneurship (or something else, like Applied Mathematics or Computer Science) and gain knowledge that I would not gain anyways by going through a training program at a good Wall Street firm. Furthermore, since I do not know at the moment what I would like to do with my life, I want to keep as many doors open as possible. Thus, if having a degree in Finance is not critical to enter the financial world, then I may opt to major in something else do that more job opportunities would be open to be after I complete my four years of college.

Thanks again.

I don't mean to say that the interview is all that matters so you don't have to worry about your skills or college performance - I'm saying that 'make the list on the Putnam' is not as important as 'do well in your interviews'.

As for your major, it depends on what you want to do.  If you want an interesting quantitative job in trading, majoring in Finance is not the best way to do it.  Major in something technical, learn how to code, and do summer internships if you can.  Most of the places that are hiring smart math folks figure they can teach you the finance.  Moreover, most of them won't trust college to teach you the finance.

Hmm...so how competitive is the job market for students who want interesting quantitative jobs in finance? I know that you and Sandor were both successful with your experiences in the financial world, but you were both top-level math competitors in high school. Is it difficult for a good technical student who has very good (but not genius-level) problem-solving skills and a lot of the other qualities of promising traders/quants to get a job at a good Wall Street firm? I don't know how many technical students aim to go into the financial world after graduating from college and how many good job positions are out there, so maybe you could provide a rough sense of how competitive things are out there.

It's definitely competitive, particularly to go to a selective shop where the problem solvers do a lot of the most important work (and receive a lot of the reward), like Jane Street.  (This is not true everywhere in finance - there are a lot of finance jobs that are about relationship building and sales.)

However, you don't need to be a MOPper to be an excellent problem solver, and you don't need to be a MOPper to get an excellent job in trading.  Sandor and I both worked with a number of very smart people who weren't MOP-like.

I will echo Sandor's comment about coding - learn your way around a computer.

[quote="ffdbzathf"]do traders and/or quants often or even ever have to travel?  i know that investment bankers are constantly being sent by plain or train or however to meet with clients.  also, if it's appropriate for this forum, whats the approximate beginning salary for a trader or quant, and what sort of bonus and salary increases are there?[/quote]

Travel: Depends on the job.  I went to Russia, Argentina, and Brazil as part of my trading job.  Others in my group went to Japan and London.

Salary: I'll defer to Sandor on starting salary, but I'll note that if expected value of year five-ten salary is what's most important to you, you will not find a better profession anywhere than going into trading, provided you have the requisite skills and desire.  If you want to be a multi-billionaire, you'll have to start your own Google or Dell, but if you'll be happy with at least a mid-6-digit salary after being in a career for several years, then finance/trading is a good bet.  It's no guarantee this will happen, to be sure, but if you get with a good company and do good work, you won't be in a position to complain about the pay.

[quote="MCrawford"][quote="maokid7"]as stated before the level of competition for internship positions is very high.
I assume that after u send in a resume w/ you initial application they then select a few to conduct interviews with, how hard is it to get an interview if you don't go to a school like Harvard, MIT, and other schools with that kind of reputation.[/quote]

Nearly impossible if you don't lose the chatspeak.

If you've got the talent, the interests, and credentials backing up the two, it's always possible to get an interview.[/quote]
thanks for the answer. yes I am aware that there are times when I need to proof read what I have typed/written, I also understand that many things used in chat rooms are not acceptable on a job/internship application.

This all looks really interesting. I'm thinking about learning something about finance and programming as I major in math, so that perhaps one day I can work for a big firm like this.

Is there a big competition between mathematicians that what to work in finance?For the posts I have read here ,I got the impression that there are few jobs in finance that are available for mathematicians.
This is opposite for what is happening in my country- all the biggest (and the best  :lol: ) companys in finance cant wait for mathematicians to finish their studys at the university so they could work for them,espacialy the ones that major in finance and business.
But,also,only about 60 mathematicians per year finish their studys,including the ones that major in computing science,theoretical math and so..., not to mention that U.S. has about 50 times more people...
I would like to work in U.S. so I am want to know more about the labor market...

Regards

How does University of Waterloo stand in terms of having enough clout to be considered for an interview for a quant internship/job as compared to the big schools in the states?

In addition, how much programming do you need to know, and for what language?

Also, would an accounting background be of any help? I am planning to double major in accounting and math.

Thanks!

[quote="rrusczyk"]
Salary: I'll defer to Sandor on starting salary, but I'll note that if expected value of year five-ten salary is what's most important to you, you will not find a better profession anywhere than going into trading, provided you have the requisite skills and desire.  If you want to be a multi-billionaire, you'll have to start your own Google or Dell, but if you'll be happy with at least a mid-6-digit salary after being in a career for several years, then finance/trading is a good bet.  It's no guarantee this will happen, to be sure, but if you get with a good company and do good work, you won't be in a position to complain about the pay.[/quote]

This is a very well said and good advice for many.  While I understand compensation is not the strongest factor to many, the folks that are looking to become multi-billionaires should weigh the risks and consider if being a multimillionaire works for them instead.

This has been said before also, but even folks that are not simply financially driven should consider the space for the problems and work they would be doing alone.  Throw in the compensation and it makes it an attractive field for those that can get in.

[quote="ffdbzathf"]how exactly would one gain programming ability/experience through college without working in comp sci?[/quote]

i'm not well qualified to give advice in quant related matters but perhaps i can suggest that one can gain some programming experience through research projects to complement the experience they get through work experience and school.

I coded primarily in Fortran and Matlab in research projects as a grad student. My current job is mostly about SAS and spreadsheets. Recently, the motivation to land a quant job promted me to get some familiarity with Cpp and OOD. After picking up a Cpp book(a free online book from the For Dummies series) and learning the basics, i'm feeling comfortable enough to begin collaboration in some research projects involving Cpp coding with a former professor of mine.

Of course my goal is to have a shot at a quant job. I think being able to put some Cpp programming experience down on your resume won't hurt. Regardless if the source is work experience, formal education or research projects.

This is just my 2 cents worth. I'd welcome any input or advice from the more qualified members in the forum. Thanks.

By the way, programming is a subject that's became more interesting to me as I learn more about it. So much that I now describe my interests as physics, math and programming.

Cheers,
Mike.

I am currently a freshman at MIT exploring different major and career choices. My original intention was to declare Computer Science, a major I enjoy very much and offers what I believe would be flexible and varied career options (software engineering, product management in Silicon Valley all the way to consulting, finance, etc). I am currently considering a second major to double in, and I'm having a hard time deciding between Economics and Applied Math. Yes, I realize MIT has the Sloan school, but there are quite a lot of "fuzzy" classes in Sloan's Management Science degree track (Management Psychology, etc).

Assuming I went with CS and Applied Math, I'd still be taking a few Sloan classes geared towards preparing for a career as a quant, such as whatever class I can find that would get me started in stochastic calculus, etc.

Anyway, I guess my real question is, for a second major, which of the following would be most useful to a student seeking a quantitative analyst-type job?

-economics
-management science (MIT Sloan)
-applied math

Thanks in advance!

It depends on what exactly you want to do. If you want to work for a place like DE Shaw or Jane Street, it doesn't matter too much exactly what classes you take. You just need to be focused, have good analytical skills, work hard, etc. Even when I interviewed for Morgan Stanley, a big bank, the guy told me that it didn't really matter what you learned in classes, because what goes on in the finance world is leaps and bounds ahead of what's taught in undergraduate finance courses.

I'd say you should take what interests you. If you have a choice still, then go for something more quantitative, especially something that involves any kind of modeling or probability theory. If you don't mind sucking it up and taking a horrendously boring class, then go ahead and take some corporate finance classes or something. But it's not the number of finance classes you've taken that will make or break your applications to these companies. I think it's better if you do what interests you and therefore excel and show your true strengths. That will make more of an impression on recruiters than a long list of worthless and boring classes about binomial option pricing models and CAPM.

DEShaw are in the relatively small minority of firms who are largely indifferent to your topics of study.
I'm glad aznphatso managed to dodge the fuzzy courses, since they typically add bulk, not strength.

In another place (Slashdot.com) I became embroiled in a flame war when I pointed out that as a quant headhunter I was becoming increasingly suspicious of CS qualifications, and that a couple of mildly well known places have CS  courses that we don't count as degrees of any kind. Some managers at top tier banks have flatly told us that they don't want to see CVs from CS people at all.
Since you enjoy CS, you might as well make the best of it and learn C++. Some people will suggest Java or another quiche language, ignore them, perhaps laughing gently as you do so.  Signal processing should be on your shopping list.

For quant finance generally, applied maths are more useful, though if you take to signal processing some of the time series analysis which typically is to be found in economics is useful.

I'm generally with Xevarion that you should do courses you find interesting, with a bias towards those in which you feel you are stronger than your peers, though of course that correlates.
If you've got the right maths, then not only should you whizz through the binomial tree, but quickly realise its limitations in this field. CAPM is on many finance courses, though the most useful part of that is the insight into the many ways it goes wrong.

A good path is to get better at maths in general and then work through of Wilmott's big book of banking
[url]http://www.amazon.com/Paul-Wilmott-Quantitative-Finance-Set/dp/0470018704/ref=pd_bbs_sr_1?ie=UTF8&s=books&qid=1202926834&sr=8-1[/url]

From the little I know of your skills, I think some form of algorithmic trading might be good fun, since it's high end s/w development (and occasionally h/w as well),  and quite sexy at the moment.

Hi, I'm currently a freshman, but I'm very interested in trying to become a trader.

On the Jane Street website, I saw that the ideal applicant would have to come from a top school.  Would double majoring in computational and applied mathematics and mathematical economic analysis from Rice University be considered "top"?

Getting an interview to begin with sometimes has to do with the degrees you've earned, but the Information Age is quickly helping employers find workers who match the skillset they need without pouring through an increasingly large pile of decreasingly meaningful resumes. In fact, that's a big part of the purpose of the new forum.

[url=http://www.math123xyz.com]Interactive Math Resources[/url]

Hi,
This is sort of a weird question but since there is a recruiter from in the industry, I would like to put it out.
I've been in a Financial Engineering PhD program at a tier 1 US university for about a year now.  The mathematics is very interesting and is a good intersection of PDE, Probability, and CS.  However, whenever I see the numerical application of my work in Bloomberg, I can't help but feel that all I'm doing is some how help a fund get better at gambling on the stock market.  Then when this fund gets better at gambling, the other firms have to hire better quants to improve their models.  It's like a "modeling arms race".  When I look back at history in the longer run, I can' explain to myself how has the advances in quantitative finance in the last 25 years made the world a better place.

Sorry if this seems rude, but it's been seriously bothering me for quite a while now.  My alternative would be to pursue something in Operations Research or switch into Mechanical or Electrical Engineering.  My understanding is that high level Mechanics is quite math heavy too?

John

No need to post the same thing twice John: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204998

Hello
 Im currently studying my second year of university (my "major" is math). Recently ive been  thinking about what  i want to do with my  life once college is done. A couple of days ago  i found this article on from the American Mathematical Society called "A view on the transition from academia to finance" by Catherine Oneil. In this article she describes what her job is like as a quantitative analyst at D. E. Shaw and i must say i found it very appealing to me. From the work enviorment she describes to the excitement of living in NYC, i think being a "quant" is what im looking for in a job. Anyway studying in a university in Mexico which is probably completely unkown to any  firm  in the US and not being an american citizen makes getting this job look pretty hard ; still there must  be a way (legal obviously) to get a shot, isnt there?
 I lived in the states for quite some time (7 years) while my dad was studying to get his phd  from Miami University. So maybe that helps in some way to get my citizenship or at least to be able to work in the US legally. I would definitely have to get a Phd or a masters degree at a top college right? I've been looking at the mathmatical finance programs from Columbia and Chicago University. They seem  very interesting. Anyway my question is (finally!!  :) ) :
 Is it possible to get a job(particularly as a "quant") in the US if youre initially not a citizen and in wich ways can this be done?

 Anyway many thanks in advance for your feedback.  :)

[quote="Dominic Connor"]

Since you enjoy CS, you might as well make the best of it and learn C++. Some people will suggest Java or another quiche language, ignore them, perhaps laughing gently as you do so.  Signal processing should be on your shopping list.

For quant finance generally, applied maths are more useful, though if you take to signal processing some of the time series analysis which typically is to be found in economics is useful.

[/quote]

I also do not know about the future of statistics, signal processing, machine learning, data mining, operations research (or whatever you want to call it) in the finance sector but I wanted to point out that D.E. started to become a [url=http://nips.cc/Sponsors/]NIPS sponsor.[/url] NIPS and ICML are the two major machine learning conferences in this field.


One (over-complicated) [hide="solution"]Let $f(x) = (x - a)(x - b)(x-c)(x-d)$.  If $abcd = p_4$ and $ab + ac + ad + bc + bd + cd = p_2$ then we have from the given that $abc + abd + acd + bcd = S\cdot p_4$ so that $f(x) = x^4 - Sx^3 + p_2x^2 -Sp_4x + p_4$.  We have that we can deduce from this that $\frac{f'(1)}{f(1)} = S$.  If $\frac{f'(1)}{f(1)} = T$ then substituting and expanding we have $T(1 - S + p_2 - Sp_4 + p_4) = 4 - 3S + 2p_2 - Sp_4$ must have a unique solution for $T$ and that this solution must in fact equal $S$.  So when we put $T = S$ into this equation, we must have only a single root.  However, $p_2$ and $p_4$ can take on any values we wish (you have to know a little modern algebra to know that their values are independent of the given values in the question, or else this wouldn't work) and so all those terms must disappear -- the only time this happens is when we have $S = 2$.  I do think the question should contain the additional condition that none of $a, b, c, d$ be equal to 1, however, or else even 2 does not in fact work (since we cannot deduce simply from $a + b + c + d = 2$ and $\frac 1 a + \frac 1 b + \frac 1 c + \frac 1 d = 2$ even that the $\frac 1 {1 - a}$ expression exists.[/hide]

Can you explain me why $p_2 , p_4$ can take on any values ? Why can you deduce $S=2$ is the only solution from the relatioship : $T(1 - S + p_2 - Sp_4 + p_4) = 4 - 3S + 2p_2 - Sp_4$ ? Thanks

What is Aro?

They can take on any values because they are the coefficients of a polynomial with roots $a, b, c, d$.  Since any polynomial has roots, whatever values I choose for $p_2$, $p_4$ will give me a polynomial with roots (perhaps complex) and they will represent a set of four numbers which give the desired value for their sum, reciprocal sum, product and $p_2$-sum.

The problem says that we can deduce from that equation that $T = S$.  Thus, we must have that $T = S$ is the only possible solution, so replacing $T$ with $S$ must give us an equation which is only satisfied by one value of $S$.  Since $p_2$ and $p_4$ can be anything, we need to eliminate them.  They only die when $S = 2$.

Thanks JBL !
 ARO = All Russian Olympiad problems
 Indeed , we can solve problem "clearly" (based on your ideas) :
 $S(1-S+p_2-Sp_4+p_4)=4-3S+2p_2=Sp_4$
$\iff (S-2)(S-2+Sp_4-p_2)=0$

Neat problem!  Here's the official solution.

If the result holds for the real numbers $a$, $b$, $c$, $d$, then it must also hold for the real numbers $1/a$, $1/b$, $1/c$, $1/d$, so
\[ \sum \frac{1}{1 - 1/a} = \sum \frac{a}{a - 1} = S, \]
which implies that
\[ \sum \frac{a}{1 - a} = -S. \]

Then
\[ S - (-S) = \sum \frac{1}{1 - a} - \sum \frac{a}{1 - a} = 4, \]
so $S = 2$.

Now to prove that $S = 2$ works.  From the given conditions, $\sum a = 2$ and $\sum abc = 2abcd$.  Then
\begin{eqnarray*} \sum \frac{1}{1 - a} &=& \frac{4 - 3 \sum a + 2 \sum ab - \sum abc}{1 - \sum a + \sum ab - \sum abc + abcd} \\ &=& \frac{4 - 6 + 2 \sum ab - 2abcd}{1 - 2 + \sum ab - 2abcd + abcd} \\ &=& \frac{-2 + 2 \sum ab - 2abcd}{-1 + \sum ab - abcd} \\ &=& 2. \end{eqnarray*}
Hence, $S = 2$ works.

What a nice solution ! :)
 Dear nsato , could you please tell me where you got the solution of this problem ?

Like I said, it's the official solution.  I happen to have the solution booklet with me.


Let $ABC$ be a triangle .Prove that:
1)$cos\frac{A}{3}+cos\frac{B}{3}+cos\frac{C}{3}>\frac{5}{2}$
2)$sin^{3}A+sin^{3}B+sin^{3}C <2$

[color=purple]Do the following inequalities true chienthan? [/color]
  $sin^{3}A+sin^{3}B+sin^{3}C\leq \frac{9\sqrt{3}}{8}$

Chien than  : It is not strong  :D
@ chien than : Chú sôi nổi nhỉ :D


calculus
derivative
function
parameterization
calculus computations

Let [b]f[/b]([i]t[/i])=([i]x(t),y(t)[/i]) be a C2 parametrization of the curve with [b]f[/b]'([i]a[/i])=/=[b]0[/b].
Prove that [i]y'(x)= dy/dx = (dy/dt)/(dx/dt)[/i]
and [i]d(^2)y/d(x^2) = (dy'(x)/dt)/(dx/dt)[/i]  in the neighborhood of [i]a[/i].

To me seems that the condition of [b]f[/b]'([i]a[/i])=/=[b]0[/b] is not enough...I think it requires that the first component of [b]f[/b]'([i]a[/i]), which is [i]x'(a)[/i]=/=0 in order to use the implicit function theorem ????

Or am I in the wrong track???

Yes, that's correct. While there is a tangent line if the first component is zero and the second isn't, that tangent is vertical and the derivative fails to exist. The right condition for all theorems in this family is that whatever derivative you need to invert is invertible.


let f:A->A with A=[a,b](a<b) satisfy:
       |f(x)-f(y)|<|x-y|,x,y in A.
Let c in A.
Put a(1)=c;
      a(n+1)=f(a(n)),n>=1;
Prove that:  {a(n)} have a limit when n->infinity.

Take a look at this one: http://www.mathlinks.ro/Forum/viewtopic.php?t=18269


number theory
prime numbers
number theory proposed

Find all $n$ such that for any distinct prime divisor of $n$ ,$a$ & $b$ also $a+b-1$ is divisor of $n$. :roll:

This problem is from Russian Math.Olympyad 2001, grade 9, №8.

Actually, problem 9.8 asked for the [i]odd[/i] numbers $n$ with that property, and this one is problem 10.8.

Let's solve the one where $n$ is odd first:

Suppose $3\le p_1<\ldots<p_k$ are the prime factors of $n$, and $n=p_1^{a_1}\cdot\ldots\cdot p_k^{a_k}$. Now, since $p_1\ge 3$, for all $j\ge 2$ we have $p_j\not\|p_1^2-1$, so for all $u_j\le a_j,\ j=\overline{2,k}$ the numbers $p_1+p_2^{u_2}\cdot\ldots\cdot p_k^{u_k}-1$ and $p_1^2+p_2^{u_2}\cdot\ldots\cdot p_k^{u_k}-1$ are powers of $p_1$ (if $p_1^2|n$; otherwise just the numbers of the first type are powers of $p_1$). In the case $p_1^2|n$, their difference is $p_1^2-p_1$, and this is only possible if the first one is $p_1$ and the second one is $p_1^2$, and from $p_1+p_2^{u_2}\cdot\ldots\cdot p_k^{u_k}-1=p_1$ we find $k=1$, i.e. $n$ is a power of $p_1$. When $p_1^2$ does not divide $n$, again, we get $p_1+p_2^{u_2}\cdot\ldots\cdot p_k^{u_k}-1=p_1$, and $n=p_1$. Either way, $n$ turns out to be a power of a prime. Conversely, all powers of prime numbers work, so this is the answer.

In the general case (i.e. when we allow $n$ to be even) the trick above still works if we restrict our attention to odd divisors, because whenever $a,b$ are odd, $a+b-1$ is also odd. The above then proves that $n$ has one of the forms $q^a$ for some prime $q\ge 2$ or $2^ap^b$ for some odd prime $p$. Let's look at $n=2^ap^b$. If $a=1$, then $p^b+1$ is a divisor of $n$ which is larger than $\frac n2$, and this is impossible, so $a\ge 2$. If $p>3$ then both $p+1$ and $p+3$ must be powers of $2$ and we get a contradiction again. This means that $p=3$. Also, $b=1$ because otherwise $10=9+1$ would be a divisor of $n$, but $5$ does not divide $n$, and $a\le 2$ because otherwise $10=2^3+3-1$ would divide $n$, and we get the same contradiction. This means that $n=12$, and these are all the solutions to the problem: $n$ is either a power of a prime or $n=12$.


search
algebra
system of equations
algebra unsolved

Solve the system of equations:
$ \left\{ \begin{array}{l}
 50{x^3} \plus{} y \equal{} 37x \plus{} 14 \\
 25{y^3} \plus{} z \equal{} 18y \plus{} 8 \\
 75{z^3} \plus{} x \equal{} 56z \plus{} 20 \\
 \end{array} \right.$

[quote="litbon"]Solve the system of equations:
$ \left\{\begin{array}{l} 50{x^3} \plus{} y \equal{} 37x \plus{} 14 \\
25{y^3} \plus{} z \equal{} 18y \plus{} 8 \\
75{z^3} \plus{} x \equal{} 56z \plus{} 20 \\
\end{array} \right.$[/quote]
The same method in the post #2 works!.
It seems the problem in USATST this year is not new! http://www.mathlinks.ro/viewtopic.php?search_id=1114260908&t=289587


geometry
circumcircle
geometric transformation
homothety
ratio

So anyways, this problem was created while I was playing around with Tom Davis' [url=http://www.geometer.org/geometer/Problems/index.html]Geometer[/url] program.  I am not sure how difficult it is.

Let $ X$ be a variable point on the circumcircle of triangle $ ABC$, and let $ D, E,F$ be the projections of $ X$ onto the altitudes of $ \triangle ABC$.  As $ X$ varies, describe the locus of the circumcenter of $ \triangle DEF$.

Clearly, $ D$, $ E$, and $ F$ lie on the circle diameter $ XH$. Let $ O_1$ be the circumcenter of $ \triangle DEF$. $ O_1$ is the midpoint of $ XH$. A homothety center $ H$ ratio $ \frac {1}{2}$ maps $ X$ to $ O_1$, the locus of $ X$ is a circle, the locus of $ O_1$ is then a circle. In particular, this homothety maps $ A$, $ B$, $ C$ to the midpoints of $ AH$, $ BH$, and $ CH$. The circumcircle of this triangle is well-known; it is the 9 point circle.

Nice problem!

Thank you, Altheman!
It looks like this method generalizes to cyclic $ 2k \plus{} 1$-gons with concurrent "altitudes."


ABCDEF is a hexagon inscribed in a circle of radius r.

AB = CD  = EF = r.

Prove that the midpoints of BC, DE, FA are the vertices of an equilateral triangle.

A picture for reference:
[img]http://www2.hkedcity.net/sch_files/a/lpl/lpl-lsk/public_html/propeller.GIF[/img]
Hint:
[hide]
Let $z_1,z_2,z_3$ be complex numbers.
Then $z_1 z_2 z_3$ forms an equilateral triangle in the complex plane if and only if \[z_1+{\omega}z_2+{\omega}^2z_3=0\]
where $\omega=cos{\frac\pi3}+ {i}sin{\frac\pi3}$.
[/hide]
Proof:
[hide]
Consider constructions of points and lines in the complex plane.
Let $z_1,z_2,z_3,z_4,z_5,z_6$ be the complex numbers representing the points A,B,C,D,E,F in the complex plane.
Put the centre of the circle at the origin O.
Then
$z_1+{\omega}z_2=0---(1)$
$z_3+{\omega}z_4=0---(2)$
$z_5+{\omega}z_6=0---(3)$
where $\omega=cos{\frac\pi3}+ {i}sin{\frac\pi3}$
In order to show the mid-points form an equilateral triangle, it suffices to prove that
$\frac{z_2+z_3}2+{\omega}\frac{z_4+z_5}2+{\omega}^2\frac{z_6+z_1}2=0---(*)$
$(1)\times\omega^2:\omega^2{z_1}+z_2=0---(4)$
$(2)\times1:                   z_3+{\omega}z_4=0---(5)$
$(3)\times\omega:    \omega{z_5}+\omega^2{z_6}=0---(6)$
$((4)+(5)+(6))\div 2\Rightarrow (*). Q.E.D.$
[/hide]

Remark: The necessary and sufficient condition for three complex numbers to represent an equilateral triangle is not shown because it can be found in most textbooks.
I am seeking a purely geometric proof. If you know how to prove the statement by other methods( gemoetry, trigonometry, co-geom, and so forth), please let me know.
$Thanks ! $


a) Prove that there exist infinitely prime number $ p$ such that $ p^2\not|2^{p \minus{} 1} \minus{} 1$
b) Let $ x\ne y$ be two coprime positive integers.
Prove that there exist infinitely prime number $ p$ such that the exponent of $ p$ in factorization of $ x^{p \minus{} 1} \minus{} y^{p \minus{} 1}$ is odd.

[quote="hana1122"]a) Prove that there exist infinitely prime number $ p$ such that $ p\not|2^{p \minus{} 1} \minus{} 1$
b) Let $ x\ne y$ be two coprime positive integers.
Prove that there exist infinitely prime number $ p$ such that the exponent of $ p$ in factorization of $ x^{p \minus{} 1} \minus{} y^{p \minus{} 1}$ is odd.[/quote]
Are you sure for the first???
What about fermat little theorem my friend??? :wink:

oh, here are items my post.
(a) link: http://www.mathlinks.ro/viewtopic.php?p=1203091#1203091

It is an open question if there are infinitely many non-Wieferich-primes:

http://www.elmath.org/index.php?id=display_topic&topic=1


Solve the equation: \[ 2008^x\plus{}2009^x\equal{}2015x\plus{}2\]

[hide="Solution"]We want the roots of $ f(x) \equal{} 2008^{x} \plus{} 2009^{x} \minus{} 2015x \minus{} 2$.  By inspection $ f(x)$ has a root at $ x\equal{}0$.  Taking the derivative, we get $ f'(x) \equal{} 2008^{x}\ln 2008 \plus{} 2009^{x}\ln 2009 \minus{} 2015$.  This is clearly strictly increasing, so it can only have one root.

If $ f(x)$ had more than two roots, say it had roots $ a,b,c$ such that $ a < b < c$.  Then by Mean Value Theorem, $ f'(x)$ would have a root between $ a$ and $ b$ and a root between $ b$ and $ c$, a contradiction.  Therefore, there can be only one more solution, which is somewhere between $ 0$ and $ 1$.[/hide]

EDIT: Fixed my mistake (kind of).

[quote="AstroPhys"]How is $ x \equal{} 1$ a root?[/quote]

You're right, somehow I confused 2015 with 4015...

Well basically there's a root at $ x\equal{}0$ and another root between 0 and 1 that I don't know how to find.  :D

This is a very hard problem. Are you sure it can be solved without calculator? Anyone got any ideas?

[quote="AstroPhys"]Can you explain how you know the other root is between $ 0$ and $ 1$?  :)[/quote]

Well it's first of all clear that there's only 2 roots, one at $ x\equal{}0$. We notice that LHS>RHS when $ x\equal{}1$, so it must be between $ 0$ and $ 1$


geometry
3D geometry
sphere
angle bisector

Let S be a great circle with pole P. On any great circle through P, two points A and B are chosen equidistant from P. For any spherical triangle ABC, where C is on S, prove that the great circle arc CP is the angle bisector of angle C.

BTW, what is the point of the phrase "on any great circle through P"? There is only one great circle with the same center as the sphere that passes through P.

[quote="Phelpedo"]
BTW, what is the point of the phrase "on any great circle through P"? There is only one great circle with the same center as the sphere that passes through P.[/quote]

Consider rotating that circle around the line formed by $P$ and the center of the sphere.  There are an infinite number.  The only real requirements of A, B is that they be equidistant from P and be on "opposite sides" (if that helps to visuailze it).

Btw, when you say "the arc $CP$ bisects $\angle{C}$, does that mean the projection of CP onto plane ABC?

For a solution, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=424758][b]here[/b][/url].

Best regards,
sunken rock