AMC
USA(J)MO
USAMO
AIME
IMO Shortlist
\/closed

Hi,

I tried to establish some abbreviations for mathematical olympiads. One could write it in the
following syntax:

<olympiad>year_stage_class_problem-number

where the last three items are additional options / extra features, e.g:

IMO1992_3, DeMo2003_4_13_3,...

But I think the syntax is not clear as one does not always know which item was omitted. So I have to
rethink it probably. What's your opinion about giving the problem's reference as abbreviation where
it is known ? You are invited to add further short-cuts as probably there are many more...

IMO: International Mathematical Olympiad
ISL: IMO Shortlist Problems
ILL: IMO Longlist Problems

TOT: Tournament Of The Towns
JTOT: Junior Tournament Of The Towns
ETOT: Easy Tournament Of The Towns
APMO: Asian Pacific Mathematical Olympiad
IAMO: Iberoamerican Mathematical Olympiads
BMO: Balkan Mathematical Olympiads
JBMO: Junior Balkan Mathematical Olympiads
KM: Le Kangourou des Mathmatiques

RuIST: Russian IMO Selection Tests
ASU: All Soviet Union Math Competitions
ARO:  All-Russian Olympiads 1999-2003:
LMO: Sankt Peter(s)burg / Leningrad olympiad
MMO: Moscow Mathematical Olympiad

RMO: Romanian Mathematical Olympiad
RIST: Romanian IMO Selection Tests

VMO: Vietnam Mathematical Olympiad
VIST: Viet Nam IMO Selection Tests

BulIST: Bulgarian IMO Selection Tests
BulMO: Bulgarian Mathematical Olympiad
SBulMO: Spring Bulgarian Mathematical Olympiad
WBulMO: Winter Bulgarian Mathematical Olympiad

ChMO: China Mathematical Olympiad
ChIST: China IMO Selection Tests

HKMO: Hong Kong Mathematical Olympiad

TMO: Taiwan / Chinese Taipeh Mathematical Olympiad

SKMO: South Korea Mathematical Olympiad
SKIST: South Korea IMO Selection Tests

IrMO: Iran Mathematical Olympiad
IrIst: Iran IMO Selection Tests

USAMO: USA Mathematical Olympiad
AIME: American Invitational Mathematics Examination
USAIST: USA IMO Selection Tests

DeMO: Deutsche Mathematik-Olympiade / German Mathematical Olympiad
BWM: Bundeswettbewerb Mathematik / Federal German Mathematical Competition
FIST: Federal Republic of Germany (FRG) IMO Selection Tests

CaMO: Canada Mathematical Olympiad

BrMO: British Mathematical Olympiad

FLMO:   Flandern Mathematical Olympiad

JMO:   Japan Mathematical Olympiad

InMO:   India Mathematical Olympiad

FrMO:   French Mathematical Olympiad
...

orl wrote:
[quote]I tried to establish some abbreviations for mathematical olympiads. One could write it in the
following syntax:

<olympiad>year_stage_class_problem-number

where the last three items are additional options / extra features, e.g:

IMO1992_3, DeMo2003_4_13_3,...

But I think the syntax is not clear as one does not always know which item was omitted. [/quote]

A solution could be to enforce full syntax, with asterisks (*) where information is missing or unknown, so use IMO1992_*_*_3 for <no stage> <no class> <problem 3>.

Is there really TOT/JTOT? I thought it's just TOT (and TT is also an abbreviation)
Also,is it USAIST, or USATST?
RMM - Romanian Master in Mathematics


Given five distinct points $A,M,B,C,D$ in this order on the circumference of the circle $O$ such that $MA=MB.$
Let $P,Q$ be the intersection points of the line $AC$ and $MD$, and that of the line $BD$ and $MC,$ respectively.
If two intersection points of the line $PQ$ and the circumference of the circle $O$ are $X,\ Y,$ then prove that $MX=MY.$

Since all this transformations:
 $A\mapsto P$ (from circle $O$ to line $MB$),
$B \mapsto A$ (involution of circle $O$ )
$Q\mapsto B$ (from line $MC$ to circle )
are projective, so is map $Q\mapsto P$ from line $MC$ to line $MB$.

Now since point $M$ maps to it self this transforamtion is persepective
from some point $S$. Then map $X \mapsto Y$ induced with this perspective
is involution on circle $O$. It is exactly detrmenited with to pair of $(X,Y)$.
Now if $X=A$ then $Y=B$ and if $X=M$ then $Y=M$.
So it is reflecting across line $MS$ where $S$ is center of  circle $O$.

I can't understand how you do your transformations: $A -- P$ ,
can you give source from where you studied omography and involution
or somebody would explain it more in detailes.

Another way to solve it:
1) $PQCD$ is cyclic
2) $PQ||AB$
3) $OM\perp AB$, thus $OM\perp PQ$, hence $\angle XOM =\angle YOM$ and $XM=YM$.

$AB\cap MC = F$
$AB \cap DM= G$

$\angle CAB=\angle CMB=x$
$\angle DMA=\angle DBA=y$
$\angle ABM=\angle BAM=z$
$\angle BQM=\angle MPA= \pi-(x+y+z)$
$\angle BFM= \pi-(x+z)$
$\angle MGA=\pi-(y+z)$

so
$MG=\frac{\sin{z}*AM}{\sin{(y+z)}}$
$MP=\frac{\sin{(x+z)}*AM}{\sin{(x+y+z)}}$
$MF=\frac{\sin{z}*BM}{\sin{(x+z)}}$
$MQ=\frac{\sin{(y+z)}*BM}{\sin{(x+y+z)}}$

$MG*MQ=MP*MF \Rightarrow FG // PQ \Rightarrow MO\bot PQ\Rightarrow MX=MY$

Are you sure you have no typo, Number1?
btw, involution is a projective transformation with period 2, if I still remember correctly what I've read 5 months ago.

Since $AM=BM$,we have $\widehat{PCQ}=\widehat{PDQ}$.So the quadrilateral CDPQ is inscribed in a circle.Thus,$\widehat{CPQ}=\widehat{CDQ}$.Therefore,$MX=MY$ as desired.

[img]http://i43.photobucket.com/albums/e387/phamduyhiep/Mathlinks/Japan2006.jpg[/img]

[quote="phamduyhiep"]Therefore,$MX=MY$ as desired.

[/quote]

Why?

Mmm,

  yes ,you must try a little more with the angles. The arcs [b]BY[/b] and [b]AX[/b] are finally equal.Use the fact that $\angle PQD = \angle PCD$

[u]Babis[/u]

ok
We have $\widehat{CPQ}=\frac{arcCY+arcXA}{2}$
also $\widehat{CDQ}=\frac{arcCY+arcBY}{2}$

But $\widehat{CPQ}=\widehat{CDQ}$ so $\frac{arcCY+arcXA}{2}=\frac{arcCY+arcBY}{2}$

so arcBY=arcXA  (1)

Now AM=BM  so  arcAM=arcBM (2)

So (1)+(2)  =>  arcXM=arcMY  so XM=MY

I cannot write with more details :D
I hope and I am sure you will understand

By Pascal's theorem on degenerate hexagon $ACMMDB$ we get the intersection of the tangent to the circle through $M$ with $BA,$ which is a point at infinity on line $AB$, is collinear with $XY$ which obviously implies that $XY \parallel AB$ and so $MX=MY$.


algebra
polynomial
calculus
derivative
quadratics
algebra proposed

Find all $ P\in\mathbb{R}[x]$ such that \[1+P(x)=\frac{P(x-1)+P(x+1)}{2}\] holds for all $x\in\mathbb R.$

[quote="N.T.TUAN"]Find all $ P\in\mathbb{R}[x]$ such that $ 1+P(x)=\frac{P(x-1)+P(x+1)}{2}\forall x\in\mathbb R.$[/quote]

If $ P(x)$ verifies the requested condition, then all the derivatives of $ P(x)$ verify $ P^{(i)}(x)=\frac{P^{(i)}(x-1)+P^{(i)}(x+1)}{2}$
It is easy to see that no quadratic polynomial $ Q(x)=ax^{2}+bx+c$ with $ a\neq 0$ verify this last equation  $ Q(x)=\frac{Q(x-1)+Q(x+1)}{2}$ since $ Q(0)=c\neq a+c=\frac{Q(-1)+Q(1)}{2}$
So $ deg(P)\leq 2$ and $ P(x)=ax^{2}+bx+c$. Putting back this expression in the original equation, we find $ a=1$

And all the solutions are polynomials $ P(x)=x^{2}+bx+c$

[hide="Is it a hint?"]Transforming the equation to $ P(x+1)=2P(x)+2-P(x-1)$ and using a kind of binet formula like fibonacci sequence[/hide]

[quote="N.T.TUAN"]Find all $ P\in\mathbb{R}[x]$ such that $ 1+P(x)=\frac{P(x-1)+P(x+1)}{2}\forall x\in\mathbb R.$[/quote]
we take $ f(x)=p(x)-x^{2}$
we have $ f(x)=\frac{f(x-1)+f(x+1)}{2}$ then $ f(x)=ax+b$
so $ p(x)=f(x)+x^{2}=x^{2}+ax+b$


Al, Bob, Chris, Dave, Ed, and Frank are all students.  A 4-person subcommittee is to be selected from these boys at random.

What is the probability that Al and Bob will not be on any committes?

[hide]I got 1/15...I'm not sure if that's correct or not..[/hide]

It is right. There are 6!/(2!4!) = 15 total possible combinations.

Only one will not have neither Al nor Bob.

1/15 is the answer.


Dear Mathlinkers,
let ABC be a triangle, (1) the incircle of ABC, I the center of (1), DEF the contact triangle, A' the midpoint of BC, V, W the meetpoints of BI, CI with EF and Fe the Feuerbach's point of ABC.
Prouve: D, A', V, W and Fe are concyclic.
Perhaps a rediscovery?
Sincerely
Jean-Louis

The circle $ (DA'VW)$ is the nine point circle of $ \triangle BCI$ and it's known that this circle and nine point circles of $ \triangle CIA$, $ \triangle AIB$ and $ \triangle ABC$ pass through common point which lies on the pedal circle of $ I$ wrt $ \triangle ABC$ and since this circle is incircle of $ \triangle ABC$ the common point of these nine-point circles is Feueurbach point in $ \triangle ABC$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=109112 and http://www.forumgeom.fau.edu/FG2009volume9/FG200905.ps .

Dear limes,
yes, you are right...
Now, with this figure I am able to prove more simply another interesting result which is well known...
I must finished this further proof.
Sincerely
Jean-Louis


[quote="joejia"]Out of their first 40 games the Williamsburg Wildwings had a record of 18 winds and 22 losses. Assuming no ties, how many of their next 50 games must they win if their record is to show twice as many wins as losses?[/quote]

[hide]  Altogether there will be 90 matches.  To have twice as many wins as losses, our equation becomes.

$90=2x+x$  Where $x$ is the number of losses, and $2x$ is the number of wins.

$90=3x$

$30=x$

  Therefore they need$2x=60$ wins altogether.  They already have 18 wins, meaning they need to win $60-18=\boxed{42}$ more games[/hide]

[hide]
Let $x$ be the number of losses after $90$ games.
$40+50=2x+x$
$90=3x$, $x=30$
Therefore, the number of total wins needed is $2x$ or $60$. Thus, the number of wins needed in the next $50$ games is $60-18$ or $42$[/hide]


1.  $D$ is a point on the side $BC$ of the triangle $ABC$ such that $AD > BC$. $E$ is a point on the side $AC$ such that $\frac{AE}{EC} = \frac{BD}{AD-BC}$. Show that $AD > BE$.

2.  Given $101$ distinct non-negative integers less than $5050$ show that one can choose four $a, b, c, d$ such that $a + b - c - d$ is a multiple of $5050$.

3.  Show that one can find $50$ distinct positive integers such that the sum of each number and its digits is the same.

4.  For which $n$ do the equations have a solution in integers:
 \begin{eqnarray*}x_1 ^2 + x_2 ^2 + 50 = 16x_1 + 12x_2 \\ x_2 ^2 + x_3 ^2 + 50 = 16x_2 + 12x_3 \\ ... \\ x_{n-1} ^2 + x_n ^2 + 50 = 16x_{n-1} + 12x_n \\ x_n ^2 + x_1 ^2 + 50 = 16x_n + 12x_1 \end{eqnarray*}

5.  Prove that for any real numbers we have $\sum_{i<j}{|a_i-a_j|}+\sum_{i<j}{|b_i-b_j|}\leq\sum_{i,j=1,n}{|a_i-b_j|}$.

6.  The convex hexagon $ABCDEF$ satisfies $\angle A + \angle C + \angle E = 360^{\cdot}$ and $AB \cdot CD \cdot EF = BC \cdot DE \cdot FA$. Show that $AB \cdot FD \cdot EC = BF \cdot DE \cdot CA$.

1. http://www.mathlinks.ro/Forum/viewtopic.php?p=346587#p346587
2. http://www.mathlinks.ro/Forum/viewtopic.php?p=346590#p346590
3. http://www.mathlinks.ro/Forum/viewtopic.php?p=346591#p346591
4. http://www.mathlinks.ro/Forum/viewtopic.php?p=346595#p346595
5. http://www.mathlinks.ro/Forum/viewtopic.php?p=220999#p220999
6. http://www.mathlinks.ro/Forum/viewtopic.php?p=3488#p3488


calculus
integration
calculus computations

1. integral of z/(z-5)^3
2. integral of square root of ((square root of x ) + 1)/square root of x


[x] denotes largest integer not exceeding x.

(a)For each integer $ n >= 1$, and each prime number $ p$, prove that the maximal power of $ p$ dividing $ n!$ is

$ \sum^\infty_{m=1} [\frac{n}{p^m}]$.

(b)Prove that, for each real number x, we have ${ [x] - 2[\frac{x}{2}] <= 1}$.

(c) Prove that

$ 2^n <= \frac{(2n)!}{(n!)^2} <= 2^{2n}  , n>=2$

[hide="a)"]$ [\frac{n}{p^m}]$ is equivalent to the number of integers from $ 1$ to $ n$ divisible by $ p^m$. Since each number divisible by $ p^m$ is divisible by $ p,p^2,p^3...p^{m\minus{}1}$, each number divisible by at most $ p^m$ is accounted for $ m$ times. Hence, the sum is equal to the number of $ p$'s in the product $ n!$, and we are done.[/hide]

[hide="b)"]Take two cases:
i) $ 2k \minus{} 1 \le x < 2k$ for $ k \in \mathbb{Z}.$
In this case, we know that $ [x]\equal{}2k\minus{}1$ and $ [\frac{x}{2}]\equal{}k\minus{}1$. Hence $ [x]\minus{}2[\frac{x}{2}] \equal{} 1$.

ii)$ 2k \le x < 2k\plus{}1$ for $ k \in \mathbb{Z}.$
In this case, we know that $ [x]\equal{}2k$ and $ [\frac{x}{2}]\equal{}k$. Hence $ [x]\minus{}2[\frac{x}{2}] \equal{} 0$.[/hide]

[hide="c)"]Equality holds for $ n\equal{}0,1$. It then suffices to show that for $ n\ge1, 2 \le \frac{(2n\plus{}1)(2n\plus{}2)}{(n\plus{}1)^2} \le 4$, which is obvious.[/hide]


algebra
polynomial
real analysis
real analysis unsolved

Find the minimum interval $[a,b] \in [0,1]$ such that : for all polynomials $P(x)$ - degree $P \leq 2$ and $P$ increases on $[a,b]$ then :
    $P(1) \geq P(0)$


Find a finite sequence of 16 numbers such that:
(a) it reads same from left to right as from right to left.
(b) the sum of any 7 consecutive terms is $ \minus{}1$,
(c) the sum of any 11 consecutive terms is $ \plus{}1$.

[quote="Leon"]Find a finite sequence of 16 numbers such that:
(a) it reads same from left to right as from right to left.
(b) the sum of any 7 consecutive terms is $ \minus{} 1$,
(c) the sum of any 11 consecutive terms is $ \plus{} 1$.[/quote]

The only possible sequence is $ (5,5, \minus{} 13,5,5,5, \minus{} 13,5,5, \minus{} 13,5,5,5, \minus{} 13,5,5)$

(and conditions b and c are enough to determine the sequence)

[quote="Bugi"]5 5 -13 5 5 5 -13 5 5 -13 5 5 5 -13 5 5

Edit: beaten :)[/quote]

few seconds  :wink:


Find each quantity if v = 3i - 5j + 2k and w = -2i + 3j - 2k.

(1) 3v - 2w

(2) ||v|| + ||w||

Pretty straightforward.

1) $ 3\textbf{v}\minus{}2\textbf{w}\equal{}3(3,\minus{}5,2)\minus{}2(\minus{}2,3,\minus{}2)\equal{}(9,\minus{}15,6)\minus{}(\minus{}4,6,\minus{}4)\equal{}(13,\minus{}21,10)$.

Definition for $ \vec{ \mathbf{w} } \equal{} x \hat{ \mathbf{i}} \plus{}  y \hat{ \mathbf{j}} \plus{}  z \hat{ \mathbf{k}} \equal{} \left( x , y ,z \right)$ , $ \| \vec{ \mathbf{w} } \| \equal{} \sqrt{ x^{2} \plus{} y^{2} \plus{} z^{2}}$ .


integration
function
advanced fields
advanced fields unsolved

Let $\Omega$ be a bounded open subset of $\mathbf R^{n}$ with smooth boundary. Consider the problem
\[\begin{cases}-\Delta u = 0 & \text{ in }\Omega, \\ \frac{\partial u}{\partial\nu}= \lambda u & \text{ on }\partial\Omega. \end{cases}\]
(a) Show that there exists minimal $\lambda=\lambda_{1}$ for which the problem has nontrivial solution. Also, show that $\lambda_{1}>0$.

(b) Show that
\[\lambda_{1}= \inf_{u \in C^\infty(\Omega)}\frac{\int_\Omega |\nabla u|^{2}}{\int_{\partial\Omega}u^{2}}. \]
(c) Show that there is only one nontrivial solution corresponding to $\lambda_{1}$, moreover, it is strictly positive in $\Omega$.

a), b) It follows from $\int_\Omega |\nabla u|^{2}=\lambda\int_{\partial\Omega}u^{2}$.

c) Let $x_{0}\in\partial\Omega$ is point of strong minimum and $u(x_{0})<0$. $u_\nu(x_{0})>0$ (Gireau's lemma). Then $u_\nu(x_{0})-\lambda u(x_{0})>0$. It's contradiction.

could you elaborate a little bit more about your 'proof' ? thanks  :)

[quote]a), b) It follows from $\int_\Omega |\nabla u|^{2}= \lambda \int_{\partial\Omega}u^{2}$.[/quote]

How did you conclude from here the existence of $\lambda_{1}$? How did you derive that variational identity for $\lambda_{1}$? ...

Could you elaborate, indeed.  :)

I am not sure about something. Consider the set $E \subset H^{1}(\Omega)$ whose elements are the $H^{1}$ functions that satisfies $\partial_{\nu}u = \lambda u$ on the boundary. It is clearly convex, but is it closed in $H^{1}$ ?


who had already made a Perpetual motion machine in Crocodile Physics?!

I'ts possible?

What do you mean by crocodile physics? :maybe:

google it

http://www.crocodile-clips.com/crocodile/physics/

Me!! I love them! I keep playing with that machine all the time


I just thought of this problem. Is it possible to solve for the exact value of x in the following equation:

e^x = x+4

It is definitely possible to use iteration to estimate the value. However, I was wondering if finding the exact value is possible.

Thanks.

If by "exact" you mean "in terms of elementary functions," then no.  In general, arbitrary problems won't tend to have closed-form solutions.

By Lindemann-Weierstrass' theorem, $x$ is transcendental, thus you will have at least no chance with algebraic terms.


Find all integers $ n$ for which there are distinct odd integers $ a,b,c$ such that $ n\plus{}2009\equal{}a\plus{}b\plus{}c$ and $ n\plus{}abc\equal{}ab\plus{}bc\plus{}ca$.

$ a\plus{}b\plus{}c\minus{}2009\plus{}abc\equal{}ab\plus{}bc\plus{}ca$
$ \Leftrightarrow (a\minus{}1)(b\minus{}1)(c\minus{}1)\equal{}\minus{}2008\equal{}\minus{}2^3 \times 251$

As $ a,b,c$ are distinct odd integers, $ a\minus{}1,b\minus{}1,c\minus{}1\equal{}2,\minus{}2,502$,
$ n\equal{}505\minus{}2009\equal{}\minus{}1504$

I got something different:

We have the following equations:

$ n+2009 = a+b+c$ and $ n+abc=ab+bc+ca$

We can manipulate these to eliminate the variable $ n$:
\begin{eqnarray*}
(n+abc)-(n+2009) &=& ab+bc+ca-a-b-c\\
abc-2009&=& ab + bc+ca-a-b-c\\
abc-ab-bc-ca+a+b+c &=& 2009\\
abc-ab-bc-ca+a+b+c-1 &=& 2009 - 1\\
(a-1)(b-1)(c-1) &=& 2008\\
(a-1)(b-1)(c-1) &=& (-2)\cdot 2\cdot (-502)
\end{eqnarray*}

We know that the values of $ a,b,c$ have to be either $ -1,3,-501$ because the question states that $ a,b,c$ are odd distinct numbers, and thus the factors of 2008 that are equal to $ (a-1)(b-1)(c-1)$ have to be even and distinct. The only such triple is $ (2,-2,-502)$. Making the possible triples for the values of $ a,b,c$ the following $ (3,-1,-501),(3,-501,-1),(-501,3,-1)$ but in this case order doesn't matter because it doesn't affect the value of $ n$ if the values of $ a,b,c$ are changed in between themselves. To find $ n$ I will substitute one of the triples into the first equation.

$ n+2009 = 3+(-1)+(-501)$

$ n = -2508$

Now to check the value of $ n$, I substitute it into the second equation to make sure the value works.

$ -2508+(-501)(-1)(3) = (-501)\cdot (-1) + (-1)\cdot 3 + (-501)\cdot 3$

$ -2508 + 501\cdot 3 = 501 - 3 - 1503$

$ 0 = 0$

$ \therefore \text{ Answer.} \left\{\boxed{n = -2508}\right\}$

Yeah, the problem with yours, stephencheng, is that (a-1)(b-1)(c-1)=2008, so (a-1)(b-1)(c-1) can't equal (2)(-2)(502) because those numbers multiplied equal NEGATIVE 2008: they need to be (-2)(2)(-502).


Define an equivalence relation on [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/c7c3319ca051f66d30ae4b291f9189a3.gif[/img] by declaring that x~y iff x-y is rational . Let N be a subset of [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/c7c3319ca051f66d30ae4b291f9189a3.gif[/img]  that contains precisely one member of each equivalence class.
Let E be a Lebesgue measurable set.
Prove
1. If [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/d6acf9d5ec29553562a81b09db7a70a7.gif[/img], then Lebesgue measure of E = 0
2. If Lebesgue measure of E > 0, then E contains a nonmeasurable set.

Order the set of rational numbers to $\{q_{i}\}$ with $q_{0}=0$. Define $E_{i}=E+q_{i}$. (Note: the addition is modulo 1. that is, $a+b=a+b-1$ if $a+b\ge 1$ and $a+b=a+b$ otherwise)

so, $mE_{i}=mE$ and $E_{i}$'s are disjoint. if $mE>0$, then $\sum mE_{i}=\infty$. but $\cup E_{i}\subset [0,1)$. contradiction. This solves 1.

For 2. Define $N_{i}=N+q_{i}$. And define $D_{i}=E\cap N_{i}$. Suppose all $D_{i}$ are measurable, then $mD_{i}=0$ for all $i$ by 1. But $E=\cup D_{i}$ and $mE=\sum mD_{i}=0$. contradiction.


number theory
greatest common divisor
algebra unsolved
algebra

Compute the gcd of the numbers $ 7^{2n}\minus{}48n\minus{}1, n \geq 4$.

[quote="Lousin Garckz"]Compute the gcd of the numbers $ 7^{2n} \minus{} 48n \minus{} 1, n \geq 4$.[/quote]

Let $ x_n\equal{}7^{2n} \minus{} 48n \minus{} 1$ : we get the relation $ x_{n\plus{}1}\equal{}49x_n\plus{}2304n$ and so the required gcd is a divisor of $ 2304$

But $ x_4\equal{}2304\times 2502$ and so $ 2304|x_n$ $ \forall n\geq 4$

So the required number is $ 2304$


Let a, b, c are positive real numbers such that $ a\plus{}b\plus{}c\equal{}1$. Prove that \[ 4\left(ab\plus{}bc\plus{}ac\right)^2\plus{}6abc\plus{}1\ge 5\left(ab\plus{}bc\plus{}ac\right)\]

[quote="dangtrung"]Let a, b, c are positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ 4\left(ab \plus{} bc \plus{} ac\right)^2 \plus{} 6abc \plus{} 1\ge 5\left(ab \plus{} bc \plus{} ac\right)
\]
[/quote]

Let  $ s\equal{} a\plus{}b\plus{}c$
Homogenize this inequality:
\[ 4(ab\plus{}bc\plus{}ca)^2\plus{}6abcs\plus{}s^4 \geq 5(ab\plus{}bc\plus{}ca)s^2
\]
Then
\[ LHS \minus{} RHS \equal{} \sum_{cyc}a^2 (a\minus{}b)(a\minus{}c) \plus{} b^2(b\minus{}c)^2 \geq 0 \]


Solve the equation $ x! * y! \equal{} z!$ where x,y,z are positive integers. (n!=1*2*3*...*n)

Some solutions (possibly not all) are of form $ (1,t,t)$ and $ (t,t! \minus{} 1,t!)$ (of course, x and y are interchangeable). Both are easy to check.

What it boils down to is finding the solutions for the equation $ x! \equal{} (y \plus{} 1)(y \plus{} 2)...(y \plus{} t)$, with $ x$, $ y$ and $ t$ natural, which obviously has no solutions if $ t\geq x$.

There [b]are[/b] other solutions: for example $ y\equal{}7$, $ t\equal{}3$, $ x\equal{}6$, meaning $ (6,7,10)$ is a solution.

This problem gives us many roots,
Example: x=a, y=a!-1, z= a!


Solve without a calculator on the interval [0, 2pi]

$ 8(sin (x/2))^2 (cos (x/2))^2 \equal{} 1 \minus{}sin x$

[hide] Here, we use the identity $ 2\sin\theta\cos\theta \equal{} \sin2\theta$ (specifically we use the identity divided by two).
We have $ 2\sin^2x \equal{} 1 \minus{} \sin{x}$. Letting $ y \equal{} \sin{x}$, we have $ 2y^2 \plus{} y \minus{} 1 \equal{} 0$ so we have $ (2y \minus{} 1)(y \plus{} 1) \equal{} 0$. Hence, $ \sin{x} \equal{} \frac {1}{2}$ or $ \sin{x} \equal{} \minus{} 1$. Hence, x is $ 30, 150, 270$ and whatever other ones I may have missed (in degrees).[/hide]


For first identity.

Using identity $|z|^2=z\overline{z}$ and $\overline{A^k}=A^{-k}$ we obtain $\frac{1}{|1-A^k|^2}=\frac{1}{(1-A^k)(1-A^{-k}}=-\frac{A^k}{(1-A^k)^2}$. Hence required sum is
\[-\sum_{k=1}^{n-1}\frac{A^{k(m+1)}}{(1-A^k)^2}=-R_{m+1}.\]


MIT college Harvard calculus derivative function Mafia

Alright heres a little fun game.

First person inserts an object into slot machine like so:

*inserts can of coke*

2nd person responds with something like:

*recieves 25 cents* it can be funny too. that wasnt.

and then 2nd person adds another object:

*inserts baseball*

3rd person recieves and inserts 'n' so forth.

Alright lets start.

*inserts this smily  :rotfl: *

*receives :huuh: *

Inserts bag of chips.

Receives spihc fo gab.

Inserts a soda machine.

Recieves a concussion.

  Inserts solution to Riemann Hypothesis.

Receives a million trillion bucks.

Inserts a dollar.

Recieves a swordsman.

Inserts a castle.

recieves a princess

inserts the princess

recieves mario asking "where's peach"

inserts bowser

receives bowser junior

Inserts a mafia member.

Receives a hostage.

Inserts President Bush.

Receives absolutely nothing.

Inserts 1$\$$ again.

sees that valentin's dollar has been spit back out

tries to flatten the dollar and tries to put it back in

spits back out

*sigh*....

digs through pocket, finds a new dollar, and puts it in...

*recieves four quarters*

puts in one quarter...

Receives an Aops/ML-slot machine

Inserts it again

Receives a white elephant,

inserts Asia...

recieves infinity

inserts the 11th dimension

Recives a note that says: "We are sorry, for the moment, only three and a half dimentions can be acepted, try it later"

Inserts A green, floor of the roof computer :P

recieves the sixth dimension and uses it.

inserts a wealthy version of myself.

recieves a poor person of nutz_for_e

inserts a stapler

recieves a stapler covered in Jell-O.

Inserts bottled water.

recieves potion

*sees nutz_for_e walking by*
*gives nutz_for_e potion*
*turns into a clump of grass*

inserts a clump of grass

recieves a dollar

*dances around*
*throws it in the air*
*says WOO HOO*
*sees knife*
*throws knife and dollar in air*
*runs*
*kinfe cuts dollar in shreds*
*says BAD KNIFE!*

inserts ripped dollar

receives a report card.

Inserts a tube of toothpaste.

recieves an ad for Colgate

inserts an ad for Crest

if u live in the US, u know what i mean

Receives delicious candy bar

Inserts candy bar wrapper :P

recieves throw-up mixed with ABC chocolate.

inserts my tossed cookies.

recieves a solution to Riemann Hypothesis

*beggar comes by*
*asks for some paper to eat*
*hands over solution to Riemann Hypothesis*
*beggar notices that it is the solution to the Riemann Hypothesis*
*beggar says THANKS MAN, U R THE <stupid> MAN*
*beggar is richer than bill gates the next day*
*sigh*

inserts another *sigh*

receives a huge gush of wind

yells "what the heck was that for?" and kicks over the slot machine


Define the iteration $\underbrace{f(f( \ldots f(}_{N \text{ times}}x) \ldots )$ as $f_{N}(x) \text{ for }x \in \mathbb{R}\text{.}$

Does $\cos_\infty (x)$ converge? For what values of $x$?
Same questions for $\sin_\infty (x)$ and $\tan_\infty (x)$.

(Everything is in radians.)

[hide="sin"] $|\sin x| < |x|$ for all $x$.  Since $\sin x$ is increasing on the interval $-1 \le x \le 1$ it follows that after the second iteration we have

$|\sin_{n+1}(x)| < |\sin_{n}(x)|$

With a pretty obvious limit of $0$. [/hide] [hide="cos"] More difficult to prove, but this converges to the positive constant $c$ such that $\cos c = c$.  Similar argument - the foundation is that the argument of the second $\cos$ function is bounded by the range of the first; i.e. $-1 \le x \le 1$.

Cosine is also positive on this interval.  Hence after the second iteration we have a strictly positive value. [/hide]


Sam and Sally each have a pocketful of U.S. coins.  Sam has 49 cents while Sally has 43 cents.  Neither has a sufficient number of any coin to exchange for a larger value coin.  For example, neither has five pennies to exchange for one nickel.  What is the least number of coins they could have if they combine their coins?


I hope this is not considered as spam.

[quote]France has a public university system with teaching and research of high quality. French universities have always enjoyed independence, freedom and recognition. However, within the last few months, the French government, without any discussion or consultation, has decided to brutally destroy this system and replace it with a market-based model in which arbitrary decisions and instability would prevail.

...[/quote]

http://math.univ-lyon1.fr/appel/ (english: http://math.univ-lyon1.fr/appel/spip.php?article2)

The market-based system works pretty well in the United States.

Meaning "those who a smart enough still make their way up [b]in spite[/b] of the system"? :P No, the current French college education is far superior to what we have in the USA. I would say that real mathematical education in the USA starts in the graduate school and even that is severely hindered by the "market requirements" (like one cannot run a graduate course with less than $ n$ people, ($ n$ varies from 5 to 10), etc.). Also, the prerequisite/credit system instead of the strict curriculum is a disaster. The students come with the required prerequisites, all right, but in many cases they took them like 2-3 years ago and remember next to nothing... The necessity to pay high tuition makes many students who do not have rich parents or do not wish to be in debt for the rest of their lives work night shifts. Now, can you understand mathematics after a sleepless night? I cannot. The calculus textbooks are colorful, printed on the best available paper, terribly expensive and nearly content-free. The results? The Americans are "an underrepresented minority" among the science faculty in American universities and 50%-90% of the students taking our graduate courses are Chinese. I can go like that forever but my point is simple: the USA is [b]not[/b] an example to follow as far as high education is concerned and the root of the problem may lie in the "market approach" to the education.

[quote="ZetaX"]If you mean with "pretty well" that you need fortune to study at a good university...  :roll:[/quote]

American universities offer enormous amounts of financial aid, merit-based scholarships, and need-based scholarships. Maybe 50 years ago, a motivated student would run into financial barriers as he climbed up the academic ladder, but today U.S. universities put so much money into meeting the financial needs of their students that a motivated student should really have no problem.

The advantage of the market-based system is that the parents of rich but academically trailing students can pay full tuition and the university can use this to provide more financial aid or to obtain more educational resources.

What I meant by "pretty well" was just to respond to the rather biased quote that myan gave: the US has a market-based system with "teaching and research of high quality", where the universities enjoy "independence, freedom and recognition", and where "arbitrary decisions and instability" do not prevail.

[quote="FibonacciFan"]...

The advantage of the market-based system is that the parents of rich but academically trailing students can pay full tuition and the university can use this to provide more financial aid or to obtain more educational resources.

...[/quote]

What about parents of poor and academically trailing students?


abstract algebra
group theory
superior algebra
superior algebra unsolved

What are the infinite abelian groups which are isomorphic to each of their non-trivial homomorphic images?

I've got just an example:

Consider $\mathbb{Z}(p^\infty)$, the subgroup of $\mathbb{Q}/ \mathbb{Z}$ which consists of those elements whose order is a power of the prime $p$. It is generated by the $x_{k}= \frac{1}{p^{k}}+\mathbb{Z}$, and has presentation $\mathbb{Z}(p^\infty) = \langle x_{1}, x_{2}, \dotsc ~|~ p x_{1}= 0 , p x_{2}= x_{1}, p x_{3}= x_{2}, \dotsc \rangle$. One can show that every proper subgroup is finite and cyclic, generated by some $x_{k}$. With the presentation, it is easy so see that $\mathbb{Z}(p^\infty) / (x_{k}) \cong \mathbb{Z}(p^\infty)$.

There is no other groups. Only $\mathbb Z(p^{\infty})$, $p$-prime. It follows from the structure theorems for division groups and for groups with a finite exponent ($\mathbb Z_{m}$-modules).

@lofar: Wow, this would be a great result :-)

But I can't see any connections between the question and your sketch ...  :maybe:

[b]Theorem.[/b] [i]If $A$ is an infinite abelian group isomorphic to any of its non-trivial factors, then $A\cong\mathbb Z(p^{\infty})$ for some prime $p$.[/i]

[i]Scetch of the proof.[/i]
$A$ is divisible. Suppose to the contrary, that there exists $n>1$ such that $An\subsetneq A$. Then $A\cong A/An$ and $An=0$. The hypothesis of the theorem implies that $A\cong\mathbb Z_{p}$ for some prime $p$. But $A$ is infinite --- contradiction.

$A$ is divisible, so $A$ is a direct sum of subgroups isomorphic to either $\mathbb Q$ or $\mathbb Z(p^{\infty})$ ($p$-prime). The result follows.

Thank you :)

Of course, 0 and Z/p are the only finite groups satisfying the property.


calculus
integration
function
trigonometry
analytic geometry
calculus computations

need help with this one

$ \iint_{A}^{} \frac{1}{(x^2\plus{}y^2)^2}dxdy$

where

$ A\equal{} \left \{ (x,y) : x^2\plus{}y^2\leq 1 \, \, \, \,and \,\, \,  x\plus{}y\geq 1\right \}$

Use the Change of Variables theorem with the function $ g(r,\theta) \equal{} (r\cos\theta,r\sin\theta)$.  Now, we must have $ (r,\theta)\in [0,1]\times[0,\frac{\pi}{2}]$ (edited typo) with the extra condition that $ r$ must be a function of $ \theta$ so that it ranges between a value a function of $ \theta$ and $ 1$.  To find this range, you must solve the system $ y \equal{} x\tan\theta,x \plus{} y \equal{} 1$ (why?).  Then integrate!  I hope this helps.

[hide]A change of variables to polar coordinates seems like it might be good. Let $ x \equal{} r \cos \theta$ and $ y \equal{} r \sin \theta$. Then your region $ A$ is described by $ A\equal{}\left\{ (r;\theta) : 0 \leq \theta \leq \frac{\pi}{2} \,\,\,\,and\,\,\, (\cos \theta \plus{} \sin \theta)^{\minus{}1} \leq r \leq 1 \right\}$

So the integral becomes,

$ \int_0^\frac{\pi}{2} \int_\frac{1}{\cos \theta \plus{} \sin \theta}^1 \frac{1}{r^4} (r \, dr)(d\theta)$

If you notice that $ \cos \theta \plus{} \sin \theta \equal{} \sqrt{2} \sin \left(\theta \plus{} \frac{\pi}{4}\right)$ it will make the integral easier to compute.[/hide]


trigonometry
search
algebra unsolved
algebra

[quote="epsilon07"][quote="castigioni"]The answer is 4005[/quote]

could you give the solution..[/quote]

Just search a bit before posting. See http://www.mathlinks.ro/Forum/viewtopic.php?t=279606


geometry
circumcircle
trigonometry
angle bisector
geometry theorems

Let the angle bisector of $ \angle A$ in a triangle $ ABC$ intersects the circumcircle of the triangle $ ABC$ at $ R$ what is the length of $ AR$ in the terms of lengths and angles of triangle $ ABC$ (find most closed form)

I found in some post in this forum (without proof) that, $ AR\equal{}\frac{b\plus{}c}{2\cos \frac{A}{2}}$ Can anyone give a complete proof? :maybe:

[quote="math_07"] [color=darkred]Let the angle bisector of $ \angle A$ in a triangle $ ABC$ intersects the circumcircle $ w = C(O,R)$ of the triangle $ ABC$ at $ E$ .

What is the length of $ AE$ in the terms of lengths and angles of triangle $ ABC$ (find most closed form).

I found in some post in this forum (without proof) that $ AE = \frac {b + c}{2\cos \frac {A}{2}}$ . Can anyone give a complete proof  ?[/color] [/quote]

[color=darkblue][b][u]Method 1.[/u][/b] Observe that $ BC=2\cdot BE\cdot\cos \frac A2$ . Apply the [b]Ptolemeus' relation[/b] to the cyclical

quadrilateral $ ABEC\ \ : \ \ AE\cdot BC=EF\cdot (AB+AC)$ . In conclusion, $ \boxed {\ AE=\frac {b+c}{2\cos\frac A2}\ }$ .

[b][u]Method 2.[/u][/b] Suppose w.l.o.g. that $ B>C$ . Denote $ \|\begin{array}{c}
X\in AB\ ,\ EX\perp AB\\\
Y\in AC\ ,\ EY\perp AC\end{array}$ . Observe that $ B\in (AX)$ , $ Y\in (AC)$

and $ \triangle EBX\equiv\triangle ECY$ because they are right and $ EB=EC$ , $ EX=EY$ . Thus, $ BX=CY$ , $ AX=AY$  and

$ AB+AC=(AB+BX)+(AC-CY)=AX+AY=2\cdot AX=2\cdot AE\cdot\cos\frac A2$ . In conclusion, $ \boxed {\ AE=\frac {b+c}{2\cos\frac A2}\ }$ .

[b][u]Method 3.[/u][/b]

[b][u]Method 4.[/u][/b] IF $ D\in AE\cap BC$ , then $ \triangle ABE\sim\triangle ADC$ . Thus, $ \frac {AB}{AD} = \frac {EA}{CA}$ ,

i.e. $ AD\cdot AE = bc$ . But $ AD = \frac {2bc\cdot\cos\frac A2}{b + c}$ . In conclusion, $ \boxed {\ AE = \frac {b + c}{2\cos \frac {A}{2}}\ }$ .

[b][u]Method 5.[/u][/b] $ AE = 2R\sin (B + \frac A2)$ $ \implies$ $ AE\cdot 2\cos \frac A2 =$ $ 2R\cdot 2\sin(B + \frac A2)\cos \frac A2$ .

Thus, $ AE\cdot 2\cos \frac A2 = 2R(\sin C + \sin B) = b + c$ . In conclusion, $ \boxed {\ AE = \frac {b + c}{2\cos \frac {A}{2}}\ }$ .

[b][u]Method 6.[/u][/b] $ AD\cdot DE = DB\cdot DC$ $ \implies$ $ AE = AD + DE = AD + \frac {DB\cdot DC}{AD}$ $ \implies$

$ AE = \frac {2bc\cos\frac A2}{b + c} + \frac {a^2bc}{(b + c)^2}\cdot \frac {b + c}{2bc\cos\frac A2}$ $ \implies$ $ AE = \frac {2bc\cos\frac A2}{b + c} + \frac {a^2}{2(b + c)\cos\frac A2} =$ $ \frac {4bc\cos^2\frac A2 + a^2}{2(b + c)\cos\frac A2}$ $ \implies$

$ AE = \frac {2bc(1 + \cos A) + a^2}{2(b + c)\cos\frac A2} = \frac {2bc + b^2 + c^2 - a^2 + a^2}{2(b + c)\cos\frac A2}$ $ \implies$ $ AE = \frac {(b + c)^2}{2(b + c)\cos\frac A2}$ , i.e. $ \boxed {\ AE = \frac {b + c}{2\cos \frac {A}{2}}\ }$ .

[b][u]Method 7.[/u][/b] Denote $ N\in OE$ for which $ NA\perp AE$ . Thus, $ N\in w$ and $ AE = NE\cdot\cos\frac {B - C}{2}$ , i.e. $ AE = 2R\cos\frac {B - C}{2}$ .

Therefore, $ AE\cdot 2\cos\frac A2 = 2R\cdot 2\cos\frac {B - C}{2}\sin\frac {B + C}{2} = 2R(\sin B + \sin C) = b + c$ . In conclusion, $ \boxed {\ AE = \frac {b + c}{2\cos \frac {A}{2}}\ }$ .

[b]Remark.[/b] You can obtain directly the relation  $ AE = 2R\cdot\cos\frac {B - C}{2}$ from the $ O$ - isosceles triangle $ AOE$ ![/color]

[quote="math_07"] [color=darkred]Let the angle bisector of $ \angle A$ in a triangle $ ABC$ intersects the circumcircle $ w = C(O,R)$ of the triangle $ ABC$ at $ E$ .

What is the length of $ AE$ in the terms of lengths and angles of triangle $ ABC$ (find most closed form).

I found in some post in this forum (without proof) that $ AE = \frac {b + c}{2\cos \frac {A}{2}}$ . Can anyone give a complete proof  ?[/color] [/quote]

[color=darkblue][b][u]Method 1.[/u][/b] Observe that $ BC = 2\cdot BE\cdot\cos \frac A2$ . Apply the [b]Ptolemeus' relation[/b] to the cyclical

quadrilateral $ ABEC\ \ : \ \ AE\cdot BC = EF\cdot (AB + AC)$ . In conclusion, $ \boxed {\ AE = \frac {b + c}{2\cos\frac A2}\ }$ .

[b][u]Method 2.[/u][/b] Suppose w.l.o.g. that $ B > C$ . Denote $ \|\begin{array}{c} X\in AB\ ,\ EX\perp AB \\
\ Y\in AC\ ,\ EY\perp AC\end{array}$ . Observe that $ B\in (AX)$ , $ Y\in (AC)$

and $ \triangle EBX\equiv\triangle ECY$ because they are right and $ EB = EC$ , $ EX = EY$ . Thus, $ BX = CY$ , $ AX = AY$  and

$ AB + AC = (AB + BX) + (AC - CY) = AX + AY = 2\cdot AX = 2\cdot AE\cdot\cos\frac A2$ . In conclusion, $ \boxed {\ AE = \frac {b + c}{2\cos\frac A2}\ }$ .

[b][u]Method 3.[/u][/b] Denote the diameter $ [AS]$ of the circle $ w$ and $ M\in BS\cap AE$ , $ N\in CS\cap AE$ . Observe that $ \widehat {SMN}\equiv\widehat {SNM}$

with common value $ \frac {B + C}{2}$ , i.e. the triangle $ SMN$ is $ S$ - isosceles. Since $ SE\perp MN$ obtain $ EM = EN$ , i.e.

$ 2\cdot AE = AM + AN$ . Since $ c = AM\cdot\cos\frac A2$ , $ b = AN\cdot\cos\frac A2$ obtain $ 2\cdot AE\cdot \cos\frac A2 = b + c$ , i.e. $ \boxed {\ AE = \frac {b + c}{2\cos\frac A2}\ }$ .

[b][u]Method 4.[/u][/b] IF $ D\in AE\cap BC$ , then $ \triangle ABE\sim\triangle ADC$ . Thus, $ \frac {AB}{AD} = \frac {EA}{CA}$ ,

i.e. $ AD\cdot AE = bc$ . But $ AD = \frac {2bc\cdot\cos\frac A2}{b + c}$ . In conclusion, $ \boxed {\ AE = \frac {b + c}{2\cos \frac {A}{2}}\ }$ .

[b][u]Method 5.[/u][/b] $ AE = 2R\sin (B + \frac A2)$ $ \implies$ $ AE\cdot 2\cos \frac A2 =$ $ 2R\cdot 2\sin(B + \frac A2)\cos \frac A2$ .

Thus, $ AE\cdot 2\cos \frac A2 = 2R(\sin C + \sin B) = b + c$ . In conclusion, $ \boxed {\ AE = \frac {b + c}{2\cos \frac {A}{2}}\ }$ .

[b][u]Method 6.[/u][/b] $ AD\cdot DE = DB\cdot DC$ $ \implies$ $ AE = AD + DE = AD + \frac {DB\cdot DC}{AD}$ $ \implies$

$ AE = \frac {2bc\cos\frac A2}{b + c} + \frac {a^2bc}{(b + c)^2}\cdot \frac {b + c}{2bc\cos\frac A2}$ $ \implies$ $ AE = \frac {2bc\cos\frac A2}{b + c} + \frac {a^2}{2(b + c)\cos\frac A2} =$ $ \frac {4bc\cos^2\frac A2 + a^2}{2(b + c)\cos\frac A2}$ $ \implies$

$ AE = \frac {2bc(1 + \cos A) + a^2}{2(b + c)\cos\frac A2} = \frac {2bc + b^2 + c^2 - a^2 + a^2}{2(b + c)\cos\frac A2}$ $ \implies$ $ AE = \frac {(b + c)^2}{2(b + c)\cos\frac A2}$ , i.e. $ \boxed {\ AE = \frac {b + c}{2\cos \frac {A}{2}}\ }$ .

[b][u]Method 7.[/u][/b] Denote $ N\in OE$ for which $ NA\perp AE$ . Thus, $ N\in w$ and $ AE = NE\cdot\cos\frac {B - C}{2}$ , i.e. $ AE = 2R\cos\frac {B - C}{2}$ .

Therefore, $ AE\cdot 2\cos\frac A2 = 2R\cdot 2\cos\frac {B - C}{2}\sin\frac {B + C}{2} = 2R(\sin B + \sin C) = b + c$ . In conclusion, $ \boxed {\ AE = \frac {b + c}{2\cos \frac {A}{2}}\ }$ .

 [b][u]Method 8.[/u][/b] Denote $ U\in (BA$ for which $ AU = AC$ . Observe that $ UB = b + c$ , $ UC = 2b\cos\frac A2$ .

Since $ \triangle AEC\sim\triangle UBC$ obtain $ \frac {AE}{UB} = \frac {AC}{UC}$ $ \implies$ $ \frac {AE}{b + c} = \frac {b}{2b\cos\frac A2}$ , i.e. $ \boxed {\ AE = \frac {b + c}{2\cos \frac {A}{2}}\ }$ .

[b][u]Method 9.[/u][/b] Denote $ K\in (AC)$ for which $ AK = AB$ . Observe that $ EK = EB = EC$ what means the triangle $ ECK$ is $ E$ - isosceles. Thus,

 for the midpoint $ M$ of $ [CK]$ we have $ EM\perp KC$ and $ 2\cdot AM = AC + AK$ $ \implies$ $ AM = \frac {b + c}{2}$ and $ AM = EA\cdot\cos\frac A2$ $ \implies$ $ \boxed {\ AE = \frac {b + c}{2\cos \frac {A}{2}}\ }$ .

[b]Remark.[/b] You can obtain directly the relation  $ AE = 2R\cdot\cos\frac {B - C}{2}$ from the $ O$ - isosceles triangle $ AOE$ ![/color]

[size=117][color=darkred][b]Now you can choose the simplest from these nine proofs ![/b][/color][/size]

Nice, job Virgil Nicula!  I don't think I've ever seen someone come up with nine proofs on this site before!
Mine was similar to proof #4.


Let $ \sum a_i \equal{} nx$, $ x>0$. By Jensen, $ \sum \ln(a_i) \le n \ln(x)$. Hence $ \sum \ln(a_i)\plus{} n\ln(n)\le n( \ln(x)\plus{} \ln(n)) \equal{} n \ln(nx) \equal{} RHS$.

[quote="R.Maths"]If $ a_{1},a_{2},....,a_{n} \ge 0$ then prouve that :

$ ln(a_{1}) \plus{} ln(a_{2}) \plus{} .... \plus{} ln(a_{n}) \plus{} n.ln(n) \leq n.ln(a_{1} \plus{} a_{2} \plus{} .... \plus{} a_{n})$

 :lol:[/quote]

your inequality?  :wink: just equivalent to AM-GM ......

It's AmGm and therefore not yours.  :wink:  :P


Let $ a,b,c,d\geq0$.Prove that:
$ (1\minus{}a\plus{}a^2)(1\minus{}b\plus{}b^2)(1\minus{}c\plus{}c^2)(1\minus{}d\plus{}d^2)\geq (\dfrac{1\plus{}abcd}{2})^2$

[quote="Inequalities Master"]Let $ a,b,c,d\geq0$.Prove that:
$ (1 \minus{} a \plus{} a^2)(1 \minus{} b \plus{} b^2)(1 \minus{} c \plus{} c^2)(1 \minus{} d \plus{} d^2)\geq (\dfrac{1 \plus{} abcd}{2})^2$[/quote]
It's true for all reals $ a,$ $ b,$ $ c$ and $ d.$ :wink:

Use the M. Lasku's inequality:
$ 2(a^2\minus{}ab\plus{}b^2)(c^2\minus{}cd\plus{}d^2)\geq a^2c^2\plus{}b^2d^2.$

We have $ 2(1 \minus{} a \plus{} a^2)^2\geq1 \plus{} a^4$  if and only if $ (1 \minus{} a)^4\geq0$

 Then $ (1 \minus{} a \minus{} a^2)^2(1\minus{}b \plus{} b^2)^2\geq(1 \plus{} a^4)(1 \plus{} b^4)$

 Since $ (1 \plus{} a^4)(1 \plus{} b^4)\geq(1\plus{}a^2b^2)^2$, we get $ 2(1 \minus{} a \plus{} a^2)(1 \minus{} b \plus{} b^2)\geq1 \plus{} a^2b^2$ (*)

 Similarly $ 2(1 \minus{} c \plus{} c^2)(1 \minus{} d \plus{} d^2)\geq1 \plus{} c^2d^2$(**)

Multiplying the inequalities (*)  , (**),and folow's $ (1 \plus{} a^2b^2)(1 \plus{} c^2d^2)\geq(1 \plus{} abcd)^2$ we obtein  conclusion.


analytic geometry
geometry
3D geometry
pyramid
calculus
integration
algebra

2B WAS DEFINITELY THE HARDEST OMG...EVERYTHING ELSE WAS PRETTY EASY THOUGH

2b just needs careful work but 4 took me a while.

2b just took a session of sitting down and working, and 4 wasn't so hard.  I didn't have the patience to solve 5, though.

2b would have taken a lot of effort for me. 4 was not very hard. For 5 I found a pretty quick solution, which was nice.

3 and 4 I found easy (especially 4), 2 took some thought, but once I figured out how to split up the cases, it wasn't hard.  I guess the hardest was 5, simply because it was long.  Once I had the realization that the height of plane P is simply height of the pyramid times a constant and I could find this constant (all while I was fiddling around at work), from there on I always knew where to go.  The problem is I just went with my instinct, and although I got to the right answer eventually, I went through a strange and really long path.

I went from integrals to simplifying an algebreic equation to coordinate geometry to solving a sytem of 3 equations to using the distance formula and finally to finding the constants in a linear equation $y=mx+b$, and finding what value of $x$ I should use.  It was draining.

Hard to choose...I needed about the same time and effort to solve all of them. I chose 2b because I solved it last, but once I gave up on finding something simple and clever and just went through with the casework, it wasn't that bad.

2b wasnt as bad for me... mine was definitely 5... because i'm horrible at geometry... so it took me a ridiculously long time to do 5

5 was the last one I solved, but solving 2b definitely felt considerably more rewarding for me.

#4 was was interesting, because it took one good insight, but from there it was really easy. I got it quickly, so I think the hardest was 5, mainly because 3d geometry is just a pain to visualize and draw. Pretty easy round in general though.

[quote="The Original Pi Guy"]Oddly enough, number 1 was the hardest for me, because I kept on misreading it over and over. I thought it said "find all 4-digit numbers."  Luckily I figured it out before I sent in my solutions.  :lol:[/quote]

I had the same problem.  Even typed up a proof for myself to show that a four digit number was impossible, so that I could be sure I wasn't missing something obvious (I knew the authors would have sent out an edit if there were an issue with the problem itself).  Eventually I went back and read the statement... don't know how I missed it the first n times...

However, #2b was definitely the hardest for me.  The geometry one didn't pose much of a problem (All of last year's geometry problems were harder for me), but I spent about 2 weeks of thinking about 2b on and off, and finally managed to solve it (in a non-brute force/casework method) after 5 consecutive hours of work, 2 20oz Cokes, and quite a bit of panic (why did I put it off for so long?)  Ended up submitting my solutions 5 seconds before the deadline.

2b, mostly because I didn't like my first solution.  My first solution was exactly the one in the math jam (which didn't take long), but I wanted something nicer (I hate brute force/case solutions), so I spent forever trying to make a recursive solution work.  Fortunatlely it did :D

I thought #5 was pretty easy... I couldn't visualize it in my head, but I had a small crappy drawing, and from there it's all just similarities.

i voted for 2b because of the yucky casework involved.  4 and 5 were fun!  To visualize problem 4, i ended up making a large grid of yarn on the floor...

I didn't find any of the problems particularly difficult, because I had a general idea of how to solve each problem upon reading the questions.

I would definitely like to see some more counting questions though; those are always fun.

Wow, I actually got 2b, but somehow added wrong, maybe accidentally skipped a case in my solution (I got 264 by accident)

But I made pretty much 0 progress on #4 and #5, maybe I should have tried harder, then again, I suck at geometry. (#5 round 2 is gonna kill me...)

Or you can just write a program to check your answer for you :D

[quote="lowest_reynolds"]

I would definitely like to see some more counting questions though; those are always fun.[/quote]

NO!!!

[quote="SamE"]Assuming that you solve all of them in the best possible way, the one most likely to mess you up, I'd say, was 2b. Once you get 252, is there any way to know that you're right? No. So you just have to go through each of the cases or your complicated recursion or whatever over and over until you're confident. Whereas, with #1 you knew you had the right answer and with #3 and #4, the answer was obvious but the proof was the only difficult part. #5 was kind of both of these, but really easy to prove if you see the trick.

So assuming you can do them, 2b would take the most time and work. And that's essentially a measure of difficulty, I guess.[/quote]

The way I did #2b (different from the math jam method) involved recursion, and allowed me to check that my answer was correct.  Below is a chart of the number of ways to arrange $n$ numbers such that there are a given number of "errors" in the circle (where an error is a number that is not greater/less than both neighbors).  If the number of ways to get each error adds up to the total number of circular permutations for a given $n,$ then you can be confident that you're right.

$\begin{tabular}[t]{|c|c|c|c|c|c|c|c|c|}\hline n & \multicolumn{8}{c}{\bf Errors}\\\hline &0 &1 &2 &3 &4 &5 &6&Total \\\hline 2 &1 & & & & & & & 1! \\\hline 3 & &2 & & & & & &2! \\\hline 4 &2 & &4 & & & & &3!\\\hline 5 & &16 & &8 & & & &4! \\\hline 6 &16 & &88 & &16 & & &5! \\\hline 7 & &272 & &416 & & 32& &6!\\\hline 8&272 & &2880 & &1824 & &64 &7! \\\hline \end{tabular}\indent $

Using the above chart, To find the number of ways to arrange $n$ numbers with $e$ errors, you can add together the number of ways to add an error to $n-1$ numbers with $e-1$ errors and the number of ways to subtract an error from $n-1$ numbers with $e+1$ errors.  For example, 8 numbers with 2 errors is (5)(272) + (3)(416), since you can add an error in one of 5 places in each of 272 circles, or subtract an error from one of 3 places from 416 circles.

[quote="Iversonfan2007"][quote="lowest_reynolds"]I would definitely like to see some more counting questions though; those are always fun.[/quote]

NO!!![/quote]
I agree...with Iversonfan2007. :D

Cool solution, eyefragment.

[quote="oppenhejo"]4 was the toughest for me, I need something tangible not abstract, 2b took me a while but I knew how I would solve it from teh start and i did 5 in a much harder longer way, without any similar anything and got the right answer using parametric equations and some 3d coordinate busy work[/quote]
I agree, I hate abstract math. Why does everyone hate combinatorics? It's my forte anyways.

I chose 2b because I eventually just wrote a program to solve it.
I struggled with 4 until I came upon the solution; then it was the easiest of all of them.
I was about to do 5 with analytic geometry, but somehow, less than a week beofre the solutions were due, I saw the better way.


the sides of a colorless 5x5x5 cube is colored by one of three colors. Two opposite sides are colored red. Two adjacent sides are colored blue and the remaining two adjacent sides are colored green. The cube is cut to 125 cubes with side length 1. How many of the 125 cubes have exactly 1 out of 3 colors?

:lol:   Remove the edges of the cube. So there are only 3x3 on each side. So you have to do (3x3)x6=54

[quote="orangefronted"]Two opposite sides are colored red. Two adjacent sides are colored blue and the remaining two adjacent sides are colored green. [/quote]

Can someone put up a picture to show how this coloring is done?
Are these "sides" refering to sides of 1x1 cubes or the side of the 5x5 cube?
Does "opposite" mean opposite in the flat plan or in space diagonal?


geometry
geometric transformation
inequalities
algebra
polynomial
function
algebra solved

$a_{n}$ ($n$ is integer) is a sequence from positive reals that \[a_{n}\geq \frac{a_{n+2}+a_{n+1}+a_{n-1}+a_{n-2}}4\] Prove $a_{n}$ is constant.

Consider the quantity b_n = a_(n+2) + 3a_(n+1) + a_n.  It is easily verified that b_(n+2) - b_(n+1) <= b_(n+1) - b_n is a translation of the intial inequality condition.
However, the b_i are positive since the a_i are positive.  If d_n = b_(n+1) - b_n < 0, then b_(n+k) <= b_n +kd_n which is negative for sufficiently large k, a contradiction.  We derive a similar contradiction in the other direction for d_n > 0.  Therefore, d_n = 0 for all n and the b_i form a constant sequence.
Suppose b_n = 5c for some constant c.  Let a_n - c = c_n.  The equation 5c = a_(n+2) + 3a_(n+1) + a_n becomes c_(n+2) + 3c_(n+1) + c_n = 0.  This has characteristic equation t^2+3t+1=0.  It follows that c_n has explicit formula c_n = A x^n + B x^(-n) where x=(-3-5^.5)/2.  As n--> + infinity, x^n dominates x^(-n) and alternates sign implying A=0 (else a_n has negative terms).  Similarly, n-->-infinity implies B=0.  Therefore, c_n = 0 for all n and a_n = c as required.

Note: The consideration of a_(n+2) + 3a_(n+1) + a_n may seem implausible.  However, this comes from considering the equality case of the given inequality (where a_n=c will obviously occur).  The recurrence has characteristic polynomial (t-1)^2 (t^2+3t+1) which leads naturally to the consideration of b_n.

[quote="TomciO"]Really, I can't understand why sequence given by the $a_{n}= \sqrt{n}$ is not a counter example? More generally, why every concave (and positive for natural numbers) function is not a counter example?[/quote]
 What is $a_{-1}$?

Ok, now I can see that it is mentioned that $n$ is an integer, thanks.


${\sum_{k=0}^{\infty }\frac{1}{( \alpha k+\beta )^{n}}=\alpha^{-n}\zeta \left(n,\frac{\beta }{\alpha }\right)}$

use Zeta function. it's a well-know and simple function. it's easy to proove the formula above


function
inequalities
limit
algebra
polynomial
complex numbers
complex analysis

I have found one solution to the following problem, but I'd like to see other solutions. I won't post my solution right away.

Determine all entire functions $f(z)$ satisfying the inequality
$|f(z)| \ge \frac{|z|^{2}}{1+|z|}$
for all complex numbers $z$.

We have $\lim_{|z|\to\infty}|f(z)|=\infty.$ That means that $f$ has a pole at infinity, and any entire function with a pole at infinity is a nonconstant polynomial. A nonconstant polynomial must have as many zeros as its degree (by the Fundamental Theorem of Algebra). But this inequality only allows $f(z)=0$ when $z=0.$ Furthermore, the requirement that it be greater than $C|z|^{2}$ near zero restricts us to having a zero of order 1 or 2 at zero.

The only possibilities are then $f(z)=Cz$ for some nonzero constant $C$ or $f(z)=Cz^{2}.$ Both of these possibilities are seen to work, as long as $|C|\ge 1.$

Thanks Merryfield. Your solution is the same as mine, although I used a more contrived argument to say that $f$ is a polynomial. I argued that if $f$ is not a polynomial and entire, it must have an essential singularity at $\infty$ (since otherwise it would be meromorphic and hence rational, and hence a polynomial). Then by Casorati-Weierstrass there will be points close to $\infty$ where $|f|$ will be close to zero, contradicting the inequality.


calculus
derivative
function
real analysis
real analysis theorems

I wonder if the following theorem is true: given a function $ f: [a,b] \to \mathbb{R}$ which has an antiderivative almost everywhere (or even everywhere), is $ f$ necessarily locally Lebesgue integrable?

My guess is "no" but I couldn't find an example. I took a look at "Counterexamples in analysis" and it only has an example in the case of Riemann integrability.

The example of Kent Merryfield shows that $ f(x)$ is not locally integrable around zero. Try, I you think, to provide an example $ f(x)\equal{}g'(x)$ which is nowhere locally Lebesgue integrable or prove this is impossible.

The set of continuity points of an antidifferenetiable function is a comeagre set, so $ f$ must have (a lot of) continuity points. $ f$ is bounded in a small neighborhood of such a point and in particular it is locally integrable there. I think that we must also have local integrability in a subset of $ [a,b]$ which has positive measure, but I'm not sure.


function
LaTeX
integration
calculus
calculus computations

Suppose an object produces constant power k. Find acceleration function in terms of k if velocity at t=0 is v and position at t=0 is p. I've set (sorry i can't do latex)
k = (x/t)*x"    (x(t) is position in terms of t)
k = ( x / t) * d^2x/(dt)^2 but I don't know if this is right, and I can't solve this differential equation because I haven't done real math in a year and I'm getting slower.

I've ?simplified it to x * (dx/dt) - x + c = (k/2) * (t^2) but i don't think that's right

EDIT: for the function above we have $x(0)=p$ and  $\left. \frac{dx}{dt}\right|_{t=0}=v$ still working...

EDIT2: SORRY! Doesn't work in DE, will look for error (or one of you mathlinker's will find it first).
__________________________________________________________
ORIGINAL POST:

The Diff. Eq. is seperable, so do this:

$k= \frac{x}{t}\frac{d^{2}x}{dt^{2}}\Rightarrow kt\ dt^{2}= x\ d^{2}x$
$\Rightarrow k\left( \int t\ dt\right) \ dt= \left( \int x\ dx\right) \ dx \Rightarrow k\left( \frac{t^{2}}{2}+c_{1}\right) \ dt=\frac{x^{2}}{2}\ dx$

which is the same as

Eqn (1) $\ \ k\left(\frac{t^{2}}{2}+c_{1}\right) =\frac{x^{2}}{2}\frac{dx}{dt}$

now we use the initial conditions to solve for $c_{1}$: we have $v=\left. \frac{dx}{dt}\right|_{t=0}$ and $x(0)=p$ so it follows that for $t=0$

$kc_{1}=\frac{p^{2}}{2}v\Rightarrow c_{1}=\frac{vp^{2}}{2k}$

plug-in this value into Eqn (1) to get

$k\left(\frac{t^{2}}{2}+\frac{vp^{2}}{2k}\right) =\frac{x^{2}}{2}\frac{dx}{dt}$
$\Rightarrow k\int\left( \frac{t^{2}}{2}+\frac{vp^{2}}{2k}\right) \ dt= \int\frac{x^{2}}{2}\ dx$
$\Rightarrow k\left( \frac{t^{3}}{6}+\frac{vp^{2}}{2k}t+c_{2}\right) = \frac{x^{3}}{6}$

call this Eqn (2). Now we use the initial conditions to solve for $c_{2},$: put $t=0$ in Eqn (2) to get

$kc_{2}= \frac{p^{3}}{6}\Rightarrow c_{2}= \frac{p^{3}}{6k}$

substitute this back into Eqn (2) to get

$k\left( \frac{t^{3}}{6}+\frac{vp^{2}}{2k}t+\frac{p^{3}}{6k}\right) = \frac{x^{3}}{6}$

or, equivalently

$x(t)=\left(kt^{3}+3vp^{2}t+p^{3}\right)^{\frac{1}{3}}$

I think that's right,  -Ben


algebra
polynomial
binomial coefficients

Prove that for any prime $p$ in the interval $\left]n, \frac{4n}{3}\right]$, $p$ divides \[\sum^{n}_{j=0}{{n}\choose{j}}^{4}.\]

[b]EDIT:[/b] An edited and extended (Theorem 1 generalized even further, Theorem 5 extended and proved anew) version of the below solution can be found here:
http://projectpen.files.wordpress.com/2009/04/pen-e16.pdf
or here:
http://www.cip.ifi.lmu.de/~grinberg/pene16.pdf

[quote="Peter"]Prove that for any prime $ p$ in the interval $ \left]n, \frac {4n}{3}\right]$, $ p$ divides
\[ \sum^{n}_{j \equal{} 0}{{n}\choose{j}}^{4}.
\]
[/quote]

Nice and instructive problem. Let's generalize:

[color=blue][b]Theorem 1.[/b] If $ n$ and $ k$ are positive integers and $ p$ is a prime such that $ \frac {2k \minus{} 1}{2k}\left(p \minus{} 1\right) < n < p$, then $ p\mid\sum_{j \equal{} 0}^{n}\binom{n}{j}^{2k}$.[/color]

Before we prove this, we first show some basic facts about binomial coefficients and remainders modulo primes.

[color=blue][b]Definition.[/b][/color] The binomial coefficient $ \binom{x}{u}$ is defined for all reals $ x$ and for all integers $ u$ as follows: $ \binom{x}{u} \equal{} \frac {x\cdot\left(x \minus{} 1\right)\cdot ...\cdot\left(x \minus{} u \plus{} 1\right)}{u!}$ if $ u$ is a positive integer, $ \binom{x}{0} \equal{} 1$, and $ \binom{x}{u} \equal{} 0$ if $ u$ is a negative integer.

[color=blue][b]Theorem 2, the upper negation identity.[/b] If $ n$ is a real, and $ r$ is an integer, then $ \binom{ \minus{} n}{r} \equal{} \left( \minus{} 1\right)^{r}\binom{n \plus{} r \minus{} 1}{r}$.[/color]

[i]Proof of Theorem 2.[/i] We will only consider the case of $ r$ being positive (the other cases are trivial). Then, using the definition of binomial coefficients,

$ \binom{ \minus{} n}{r} \equal{} \frac {\left( \minus{} n\right)\cdot\left( \minus{} n \minus{} 1\right)\cdot ...\cdot\left( \minus{} n \minus{} r \plus{} 1\right)}{r!} \equal{} \left( \minus{} 1\right)^{r}\cdot\frac {n\cdot\left(n \plus{} 1\right)\cdot ...\cdot\left(n \plus{} r \minus{} 1\right)}{r!}$
$ \equal{} \left( \minus{} 1\right)^{r}\cdot\frac {\left(n \plus{} r \minus{} 1\right)\cdot ...\cdot\left(n \plus{} 1\right)\cdot n}{r!} \equal{} \left( \minus{} 1\right)^{r}\cdot\binom{n \plus{} r \minus{} 1}{r}$.

This proves Theorem 2.

[color=blue][b]Theorem 3.[/b] If $ p$ is a prime, if $ u$ and $ v$ are two integers such that $ u\equiv v\mod p$, and if $ k$ is an integer such that $ k < p$, then $ \binom{u}{k}\equiv\binom{v}{k}\mod p$.[/color]

[i]Proof of Theorem 3.[/i] If $ k\leq 0$, then $ \binom{u}{k} \equal{} \binom{v}{k}$, so that Theorem 3 is trivial. Thus, it remains to consider the case $ k > 0$ only. In this case, $ k!$ is coprime with $ p$ (since $ k! \equal{} 1\cdot 2\cdot ...\cdot k$, and all numbers $ 1$, $ 2$, ..., $ k$ are coprime with $ p$, since $ p$ is a prime and $ k < p$).

Now, $ u\equiv v\mod p$ yields

$ k!\cdot\binom{u}{k} \equal{} k!\cdot\frac {u\cdot\left(u \minus{} 1\right)\cdot ...\cdot\left(u \minus{} k \plus{} 1\right)}{k!} \equal{} u\cdot\left(u \minus{} 1\right)\cdot ...\cdot\left(u \minus{} k \plus{} 1\right)$
$ \equiv v\cdot\left(v \minus{} 1\right)\cdot ...\cdot\left(v \minus{} k \plus{} 1\right) \equal{} k!\cdot\frac {v\cdot\left(v \minus{} 1\right)\cdot ...\cdot\left(v \minus{} k \plus{} 1\right)}{k!}$
$ \equal{} k!\cdot\binom{v}{k}\mod p$.

Since $ k!$ is coprime with $ p$, we can divide this congruence by $ k!$, and thus we get $ \binom{u}{k}\equiv\binom{v}{k}\mod p$. Hence, Theorem 3 is proven.

Finally, a basic property of binomial coefficients:

[color=blue][b]Theorem 4.[/b] For every nonnegative integer $ n$ and any integer $ k$, we have $ \binom{n}{k} \equal{} \binom{n}{n \minus{} k}$.[/color]

This is known, but it is important not to forget the condition that $ n$ is nonnegative (it is necessary!).

Now to something completely different:

[color=blue][b]Theorem 5.[/b] If $ p$ is a prime, and $ f$ is a polynomial of degree $ < p \minus{} 1$ whose coefficients are all integers, then $ \sum_{j \equal{} 0}^{p \minus{} 1}f\left(j\right)\equiv 0\mod p$.[/color]

[i]Proof of Theorem 5.[/i] Since $ f$ is a polynomial of degree $ < p \minus{} 1$ whose coefficients are all integers, it can be written in the form $ f\left(X\right) \equal{} \sum_{i \equal{} 0}^{p \minus{} 2}f_{i}X^{i}$ with $ f_{i}$ being an integer (not necessarily nonzero) for every $ i\in\left\{0;\;1;\;...;\;p \minus{} 3;\;p \minus{} 2\right\}$.

Hence,

$ \sum_{j \equal{} 0}^{p \minus{} 1}f\left(j\right) \equal{} \sum_{j \equal{} 0}^{p \minus{} 1}\sum_{i \equal{} 0}^{p \minus{} 2}f_{i}j^{i} \equal{} \sum_{i \equal{} 0}^{p \minus{} 2}f_{i}\cdot\sum_{j \equal{} 0}^{p \minus{} 1}j^{i}$.

But for every $ i\in\left\{0;\;1;\;...;\;p \minus{} 3;\;p \minus{} 2\right\}$, we have $ \sum_{j \equal{} 0}^{p \minus{} 1}j^{i}\equiv 0\mod p$ (in fact, for $ i \equal{} 0$, this is clear since $ \sum_{j \equal{} 0}^{p \minus{} 1}j^{0} \equal{} \sum_{j \equal{} 0}^{p \minus{} 1}1 \equal{} p$, and for $ i\geq 1$, this is equivalent to $ \sum_{j \equal{} 1}^{p}j^{i}\equiv 0\mod p$ (since $ \sum_{j \equal{} 0}^{p \minus{} 1}j^{i}\equiv \sum_{j \equal{} 1}^{p}j^{i}\mod p$, because $ 0^{i}\equiv p^{i}\mod p$), what follows from http://www.problem-solving.be/pen/viewtopic.php?t=676 part [b]a)[/b] because $ p \minus{} 1\nmid i$). Thus,

$ \sum_{j \equal{} 0}^{p \minus{} 1}f\left(j\right) \equal{} \sum_{i \equal{} 0}^{p \minus{} 2}f_{i}\cdot\sum_{j \equal{} 0}^{p \minus{} 1}j^{i}\equiv\sum_{i \equal{} 0}^{p \minus{} 2}f_{i}\cdot 0 \equal{} 0\mod p$,

and Theorem 5 is proven.

[i]Proof of Theorem 1.[/i] We have $ p \minus{} n \minus{} 1\geq 0$, since $ n < p$ yields $ n \plus{} 1\leq p$.

Also, $ \frac {2k \minus{} 1}{2k}\left(p \minus{} 1\right) < n$ yields $ \left(2k \minus{} 1\right)\left(p \minus{} 1\right) < 2kn$, so that $ 2k\left(p \minus{} 1\right) \minus{} \left(p \minus{} 1\right) < 2kn$, thus $ 2k\left(p \minus{} 1\right) \minus{} 2kn < p \minus{} 1$, or, equivalently, $ 2k\left(p \minus{} n \minus{} 1\right) < p \minus{} 1$.

We have to prove that $ p\mid\sum_{j \equal{} 0}^{n}\binom{n}{j}^{2k}$. This rewrites as $ \sum_{j \equal{} 0}^{n}\binom{n}{j}^{2k}\equiv 0\mod p$. This is equivalent to $ \sum_{j \equal{} 0}^{p \minus{} 1}\binom{n}{j}^{2k}\equiv 0\mod p$ (in fact, since $ n < p$, the two sums $ \sum_{j \equal{} 0}^{p \minus{} 1}\binom{n}{j}^{2k}$ and $ \sum_{j \equal{} 0}^{n}\binom{n}{j}^{2k}$ differ only in some terms $ \binom{n}{j}^{2k}$ with indices $ j$ in the interval $ \left]n;\;p \minus{} 1\right]$, and these terms are zero, so we have $ \sum_{j \equal{} 0}^{p \minus{} 1}\binom{n}{j}^{2k} \equal{} \sum_{j \equal{} 0}^{n}\binom{n}{j}^{2k}$).

For every integer $ j$ with $ 0\leq j < p$, we have

$ \binom{n}{j} \equal{} \binom{ \minus{} \left( \minus{} n\right)}{j} \equal{} \left( \minus{} 1\right)^{j}\binom{\left( \minus{} n\right) \plus{} j \minus{} 1}{j}$       (after Theorem 2)
$ \equiv\left( \minus{} 1\right)^{j}\binom{p \minus{} n \plus{} j \minus{} 1}{j}$         (by Theorem 3, since $ \left( \minus{} n\right) \plus{} j \minus{} 1\equiv p \minus{} n \plus{} j \minus{} 1\mod p$ and $ j < p$)
$ \equal{} \left( \minus{} 1\right)^{j}\binom{p \minus{} n \plus{} j \minus{} 1}{\left(p \minus{} n \plus{} j \minus{} 1\right) \minus{} j}$      (by Theorem 4, since $ p \minus{} n \plus{} j \minus{} 1$ is nonnegative, since $ p \minus{} n \minus{} 1\geq 0$ and $ j\geq 0$)
$ \equal{} \left( \minus{} 1\right)^{j}\binom{p \minus{} n \plus{} j \minus{} 1}{p \minus{} n \minus{} 1}\mod p$,

so that

$ \binom{n}{j}^{2k}\equiv\left(\left( \minus{} 1\right)^{j}\binom{p \minus{} n \plus{} j \minus{} 1}{p \minus{} n \minus{} 1}\right)^{2k} \equal{} \binom{p \minus{} n \plus{} j \minus{} 1}{p \minus{} n \minus{} 1}^{2k}\mod p$      (since $ 2k$ is even).

Therefore,

[color=green][b](1)[/b][/color] $ \left(p \minus{} n \minus{} 1\right)!^{2k}\cdot\sum_{j \equal{} 0}^{p \minus{} 1}\binom{n}{j}^{2k}\equiv\left(p \minus{} n \minus{} 1\right)!^{2k}\cdot\sum_{j \equal{} 0}^{p \minus{} 1}\binom{p \minus{} n \plus{} j \minus{} 1}{p \minus{} n \minus{} 1}^{2k}$
$ \equal{} \sum_{j \equal{} 0}^{p \minus{} 1}\left(\left(p \minus{} n \minus{} 1\right)!\cdot\binom{p \minus{} n \plus{} j \minus{} 1}{p \minus{} n \minus{} 1}\right)^{2k}$
$ \equal{} \sum_{j \equal{} 0}^{p \minus{} 1}\left(\left(p \minus{} n \minus{} 1\right)!\cdot\frac {\prod_{u \equal{} 0}^{\left(p \minus{} n \minus{} 1\right) \minus{} 1}\left(\left(p \minus{} n \plus{} j \minus{} 1\right) \minus{} u\right)}{\left(p \minus{} n \minus{} 1\right)!}\right)^{2k}$
$ \equal{} \sum_{j \equal{} 0}^{p \minus{} 1}\left(\prod_{u \equal{} 0}^{\left(p \minus{} n \minus{} 1\right) \minus{} 1}\left(\left(p \minus{} n \plus{} j \minus{} 1\right) \minus{} u\right)\right)^{2k}\mod p$.

Now,

$ \prod_{u \equal{} 0}^{\left(p \minus{} n \minus{} 1\right) \minus{} 1}\left(\left(p \minus{} n \plus{} X \minus{} 1\right) \minus{} u\right)$

is a polynomial of the variable $ X$ whose degree is $ p \minus{} n \minus{} 1$. Thus,

$ \left(\prod_{u \equal{} 0}^{\left(p \minus{} n \minus{} 1\right) \minus{} 1}\left(\left(p \minus{} n \plus{} X \minus{} 1\right) \minus{} u\right)\right)^{2k}$

is a polynomial of the variable $ X$ whose degree is $ 2k\left(p \minus{} n \minus{} 1\right)$. Since $ 2k\left(p \minus{} n \minus{} 1\right) < p \minus{} 1$, it is therefore a polynomial of degree $ < p \minus{} 1$. Since the coefficients of this polynomial are all integers, Theorem 5 thus yields

$ \sum_{j \equal{} 0}^{p \minus{} 1}\left(\prod_{u \equal{} 0}^{\left(p \minus{} n \minus{} 1\right) \minus{} 1}\left(\left(p \minus{} n \plus{} j \minus{} 1\right) \minus{} u\right)\right)^{2k}\equiv 0\mod p$.

Using [color=green][b](1)[/b][/color], this becomes

$ \left(p \minus{} n \minus{} 1\right)!^{2k}\cdot\sum_{j \equal{} 0}^{p \minus{} 1}\binom{n}{j}^{2k}\equiv 0\mod p$.

Now, $ \left(p \minus{} n \minus{} 1\right)!$ is coprime with $ p$ (since $ \left(p \minus{} n \minus{} 1\right)! \equal{} 1\cdot 2\cdot ...\cdot\left(p \minus{} n \minus{} 1\right)$, and all numbers $ 1$, $ 2$, ..., $ p \minus{} n \minus{} 1$ are coprime with $ p$ because $ p$ is a prime and $ p \minus{} n \minus{} 1 < p$). Hence, we can divide this congruence by $ \left(p \minus{} n \minus{} 1\right)!^{2k}$, and obtain $ \sum_{j \equal{} 0}^{p \minus{} 1}\binom{n}{j}^{2k}\equiv 0\mod p$. As we said, this proves Theorem 1.

  Darij


the function f(y) satisfies f(2+x)=f(2-x) for all real numbers x. If the equation f(x)=0 has exactly four distinct real roots, then what is the sum of these roots?

[hide="Solution"]Substitute $g(x) = f(x+2)$.  This gives us $g(x) = g(-x)$ for all real x, so $g(x)$ is symmetric about the y-axis.  If $f(x) = 0$ has four roots, then $g(x) = 0$ also has four roots (shifted by two), and since the function is symmetric about the y-axis, all these roots must add up to 0.  But since $g(x) = f(x+2)$, each root is shifted up two, and the sum of the roots is $2 + 2 + 2 + 2 = \boxed{8}$.[/hide]

This problem was discussed before.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=354012#p354012

Locked.


graph theory
combinatorics unsolved
combinatorics

infinite sportsmen want to enroll in sport clubs ,just finitly many sportsmen can be enrolled in any sport club.we know that in  any infinite set of sportsmen there exists two such that they enroll in the same club(any sportsman can enroll in more than one club)prove that if you ignore finitly many of them others enroll in infinte clubs.

This isn't calculus. [Edit- this thread was moved immediately after my post]

We argue by contradiction. Consider an infinite set, each of whose members is in finitely many clubs. Since each club is finite, each member of this set shares a club with only finitely many others.
Consider the complete graph with vertices for each member of this finite-club group, and edges colored white if the two share a club, black if they don't. By Ramsey's theorem, there is an infinite subset with either every edge inside white or every edge inside black. Since there are only finitely many white edges at each vertex, this monochromatic subset can't be white.
Now, an infinite subset with all edges black contradicts our hypothesis that every infinite set contains some two members sharing a club, and we are done.


Given $ 4$ coplanar not concyclic points $ A,B,C,D$ such that no $ 3$ lie on the same line, how many circles are at the same distance from all $ 4$ points? (The distance between a point and a circle is considered to be the lenght of the tangent)

Ambiguous problem, you should reword it somehow.

Denote by $ tan(X,S)$ the lenght of the tangent from point $ X$ to circle $ S$. The problem asks for the number of circles $ S$ such that $ tan(A,S)\equal{}tan(B,S)\equal{}tan(C,S)\equal{}tan(D,S)$

aren't there 2 tangents?

oh yeah, they're equal length...

What if the four points were in a square arrangement, couldn't it be infinite then?
Or maybe I just suck at Geo.

[quote="ZhangPeijin"]What if the four points were in a square arrangement, couldn't it be infinite then?[/quote]
[quote="sunnykim"]not concyclic points[/quote]

[hide="Well.."]
I have a feeling the answer is $ 0$ .. Let $ A,B,C,D$ be points, $ O$ be center of circle, and $ A_1,B_1,C_1,D_1$ be points of tangency; then by Pythagorean Theorem $ OA \equal{} \sqrt{r^2 \plus{} AA_1^2}$. Since $ r$ is constant, that is one-to-one, eg any length of tangent corresponds to a certain distance from a point to the center of the circle, so $ A,B,C,D$ need to be cyclic, contradiction. Unless there is a definition for an internal tangent?
[/hide]


group theory
abstract algebra
inequalities
superior algebra
superior algebra unsolved

Let $ G$ be a group and $ P\leq G$ a $ p$-subgroup such that $ P\in \text{Syl}_p(N_G(P))$. Prove that $ P\in\text{Syl}_p(G)$.

Suppose not. Then there is a $ P_1\in\text{Syl}_p(G)$ such that $ P\leq P_1$ by theorem and $ P\neq P_1$. I think we can get a contradiction through knowing that $ |P|<|N_G(P)|$ as a strict inequality?

you are right:

First note, that the index $ |N_G(P): P|$ is not divisible by $ p$.  Determine $ P_0\in Syl_p(G)$ with $ P\leq P_0$. If $ P<P_0$ holds, then $ p$ will divide $ |N_{P_0}(P): P|$. This yields a contradiction, since $ N_{P_0}(P)\leq N_G(P)$ holds. Therefore, we must have $ P\equal{}P_0$ and $ P\in Syl_G(P)$.

[quote="hjbrasch"]If $ P < P_0$ holds, then $ p$ will divide $ |N_{P_0}(P): P|$.[/quote]

Why? Why not $ P \equal{} N_{P_0}\left(P\right)$ ?

  darij

@darij:
Because finite p-groups are nilpotent. And in every nilpotent group H every proper subgroup $ U < H$ fullfills $ U < N_H(U)$.
(For finite H this is equivalent to being nilpotent.)

Thx, Gockel, for providing the explanation. A slightly more elegant proof (compared to my first proof) would have omitted the division by $ p$ language:

Choose $ P\leq P_0\in Syl_p(G)$. Since the $ p$-group $ N_{P_0}(P)$ contains $ P$ and lies in $ N_G(P)$, we must have $ N_{P_0}(P)\equal{}P$ and for $ p$-groups this is only possible for $ P_0\equal{}P$


I recently ran across a very hard radical equation that I myself couldn't solve. If anyone can solve it please post your answer and how you got it up here.

5(squareroot) s-2 = (squareroot) 19s- 29 (/squareroot) +3

Thanks much.

[quote="darthraver245"]
\[5 \sqrt{s-2}= \sqrt{ 19s-29 }+3 \]
[/quote]
[hide="Solution"]
Apply the transformation $x=s-2$. Then, we have:
\[\left( 5 \sqrt{x}-3 \right)^{2}= \left( \sqrt{19x+9}\right)^{2}\implies 25x-30 \sqrt{x}+9 = 19x+9 \]

\[\implies 6 x = 30 \sqrt{x}\implies x=25 \implies \boxed{ s= 27}\]
Being lazy, I checked with derivatives that this was the only solution.  :P [/hide]

[hide="brute force method, sorta"]Or you can just square it out.
\begin{eqnarray*}25s-50&=&19s-29+9+6\sqrt{19s-29}\\ 6s-30&=&6\sqrt{19s-29}\\ s^{2}-10s+25&=&19s-29\\ s^{2}-29s+54&=&0\\ (s-2)(s-27)&=&0\\ \end{eqnarray*}
Now we need to check solutions because squaring is not one-to-one.
$0\neq\sqrt{9}+3$
$5\sqrt{25}=25=22+3=\sqrt{484}+3$
So the only solution is $s=\boxed{27}$.[/hide]

Where does the +6(squareroot)19s-29(/squareroot) come from?

It comes from squaring:
$(a+b)^{2}=(a^{2}+2ab+b^{2})$. The term in question is the $2ab$.
Y'know, this should be in HSB.


geometry
cyclic quadrilateral
geometry proposed

[color=darkred]The $ A$- exincircle $ w_{a}$ of the triangle $ ABC$ touches it in the points $ D\in BC$ and $ E\in AC$. For a point $ M\in (BC)$ denote the point $ T\in w_{a}$ for which $ M\in TT$
and the intersection $ V\in TM\cap AC$. Show that the lines $ AM$, $ BV$, $ DE$ are concurrently.[/color]

The concurrency is a very nice result, which I proved with notions of poles and polars as follows.

Denote $ K$ the point where $ \omega_{a}$ touches $ AB$.

Since $ KE$ is the polar of $ A$ and $ TD$ the polar of $ M$ w.r.t $ \omega_{a}$, we have that $ AM$ is the polar of $ N \in KE \cap TD$.

Again, since $ KD$ is the polar of $ B$ and $ TE$ is the polar of $ V$ w.r.t $ \omega_{a}$, we have that $ BC$ is the polar of $ L \in KD \cap TE$.

And similarly, $ DE$ is the polar of $ C$.

From the above remarks, $ AM$, $ BV$, $ DE$ concur, if and only if, the points $ N$, $ L$, $ C$ are collinear, but this is true since in the cyclic quadrilateral $ TKED$, $ NL$ is the polar of the point $ KT \cap DE$, and $ C \in NL$, if and only if $ KT \cap DE \in DE$, obvious.

[i]Remark[/i]: If $ Z \in AM \cap BV \cap DE$, then also $ Z \in KT$.

I will change the notations.

[color=blue][b]Problem.[/b]let A-excircle of triangle $ ABC$ touches $ BC,AC,AC$ at $ D,E,F$ respectively and let an arbitrary segment $ B'C'$ which $ B'\in AF\ ,\ C'\in AC$ touches excirlc at $ D'$ and intersect with $ BC$ at $ M$ prove that four lines $ DE,D'F,BC',AM$ are concurrent.[/color]

[i]Prrof[/i].let $ D'F\cap DE\equiv P\ ,\ DD'\cap EF\equiv S$ the polars of $ A$ and $ M$ pass thourough $ S$ and also polar of $ S$ pass thorough $ P$ hence $ A,P,M$ are collinear.
let $ AM\cap B'C\equiv U$ we know in cyclic quadrilateral $ D'DEF$ point $ DF\cap ED'$ lies on polars of $ P$ and $ S$ hence $ DF\cap ED'\equiv U$ .
notice that polar of B,C',P and S  pass through  $ U$ hence $ S,B,P,C'$ are collinear.

[quote="Virgil Nicula"][color=darkred]The $ A$- exincircle $ w_{a}$ of the triangle $ ABC$ touches it in the points $ D\in BC$, $ E\in AC$ and $ F\in AB$.
For a point $ M\in (BC)$ denote the point $ T\in w_{a}$ for which $ M\in TT$ and the intersections $ V\in TM\cap AC$, $ X\in BC\cap AT$.
Show that the lines $ AM$, $ BV$, $ DE$ and $ FT$ are concurrently.[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Apply the [b]Brianchon's theorem[/b] to the quadrilateral $ ABMV$ which touches the circle $ w$ in the points $ F\in AB$, $ D\in BM$, $ T\in MV$ and $ E\in VA$,

i.e. $ AB\equiv FF$ a.s.o. Thus, $ \left\{\begin{array}{c}A\in FF\cap EE\ ,\ M\in DD\cap TT\\\ B\in FF\cap DD\ ,\ V\in EE\cap TT\end{array}\right\|$ $ \implies$ [b]the lines $ FT$, $ DE$ and the "diagonals" $ AM$, $ BV$ are concurrently in a point $ S$.[/b][/color].

[color=red][EDITED by pohoatza][/color]


Let $ \{a_n\}$ be a sequence given by the formulas:
$ a_1 \equal{} 1$;
$ a_{n \plus{} 1} \equal{} a_n \plus{} \frac {1}{a_n}$; $ n \equal{} 1,2,...$
Prove that $ 14 < a_{100} < 18$.

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=316711[/url]


calculus
integration
geometry
trigonometry
algebra
partial fractions
real analysis

I'm not sure how to do this.
Evaluate the following definite integral:
$ \int_{0}^\pi \frac{x sin x}{1+cos^{2}x}dx$

I get the feeling that it is not 0 or 1 (By doing Rieman Sum type area approximation to the integral).

Hint:
$ \int_{0}^\pi x f(\sin x)dx=\frac{\pi}{2}\int_{0}^\pi f(\sin x)dx$

If you'd like to proof this equality, take $ x=\pi-t$

And the second way to evaluate the expression:
Take $ t=x-\frac{\pi}{2}$

Using partial fractions I've got down to:
$ \frac{\pi}{4}\left( \int_{0}^\pi \frac{dx}{\sqrt{2}-sin x}-\int_{0}^\pi \frac{dx}{\sqrt{2}+sin x}\right)$
Is this the right way to go about it? I'm still trying at the moment,  :?:

In this way, we've got:
$ \frac{\pi}{2}\int_{0}^\pi \frac{\sin xdx}{1+\cos^{2}x}=-\frac{\pi}{2}\int_{0}^{\pi}\frac{d\cos x}{1+\cos^{2}x}=-\frac{\pi}{2}\left[\arctan\cos x\right]^{\pi}_{0}=\frac{\pi^{2}}{4}$


If we consider a basis $ \beta$ for (as an example) $ \bb{R}$ over $ \bb{Q}$, is it safe to assume that any vector $ x$ in the vector space can be represented by a finite number of vectors in $ \beta$? Is this implicit in our proof that every vector space has a basis?

Yes, that is the definition of $ \beta$ spanning the vector space. In the proof, being a maximal linearly independent set is equivalent to being a spanning linearly independent set


Find all integer roots of this equation:
$ x^{2}\equal{}y^{3}\plus{}16$

[quote="i_like_pie"][hide]$ x^{2}\equal{} y^{3}\plus{}16$

$ x^{2}\minus{}16 \equal{} y^{3}$

$ (x\plus{}4)(x\minus{}4) \equal{} y^{3}$

$ \boxed{\left(\pm4,0\right)}$[/hide][/quote]

How do you know thats the only solution?

I never said it is. I tried to find others, but failed to do so, and just kept my post as it is.

[hide="hint"]
property of eliptic curves[/hide]

[hide="solution"]We have $ (x\minus{}4)(x\plus{}4)\equal{}y^{3}$.

Thus, either both $ x\minus{}4$ and $ x\plus{}4$ are perfect cubes, or $ x\plus{}4$ and $ x\minus{}4$ have a common factor.  $ GCD(x\plus{}4,x\minus{}4)|8$, so thus, the only possible values of $ x$ that allow $ x\plus{}4$ and $ x\minus{}4$ to have a common factor other than 1 are $ x\in\{\pm2,\pm4,\pm8\}$.

$ x\equal{}2\Rightarrow y^{3}\equal{}\minus{}12\Rightarrow\emptyset$
$ x\equal{}\minus{}2\Rightarrow y^{3}\equal{}12\Rightarrow\emptyset$
$ x\equal{}4\Rightarrow y^{3}\equal{}0\Rightarrow y\equal{}0$
$ x\equal{}\minus{}4\Rightarrow y^{3}\equal{}0\Rightarrow y\equal{}0$
$ x\equal{}8\Rightarrow y^{3}\equal{}48\Rightarrow\emptyset$
$ x\equal{}\minus{}8\Rightarrow y^{3}\equal{}48\Rightarrow\emptyset$

Thus, we now only need to consider the case where $ x\minus{}4$ and $ x\plus{}4$ are both perfect cubes.  Consider $ (n\plus{}1)^{3}\minus{}n^{3}\equal{}3n^{2}\plus{}3n\plus{}1$.  Thus, if $ |n|\geq3$, $ (n\plus{}1)^{3}\minus{}n^{3}>8$, so both $ x\minus{}4$ and $ x\plus{}4$ cannot be perfect cubes.

We only have to consider the cases where $ x\in\{\minus{}2,\minus{}1,0,1,2\}$.
We have already considered $ x\equal{}\pm2$, so we have:

$ x\equal{}1\Rightarrow y^{3}\equal{}\minus{}15\Rightarrow\emptyset$
$ x\equal{}\minus{}1\Rightarrow y^{3}\equal{}\minus{}15\Rightarrow\emptyset$
$ x\equal{}0\Rightarrow y^{3}\equal{}16\Rightarrow\emptyset$

Thus, the only solutions are $ (\pm4,0)$[/hide]

[quote="b-flat"][hide="solution"]We have $ (x\minus{}4)(x\plus{}4) \equal{} y^{3}$.

Thus, either both $ x\minus{}4$ and $ x\plus{}4$ are perfect cubes, or $ x\plus{}4$ and $ x\minus{}4$ have a common factor.  $ GCD(x\plus{}4,x\minus{}4)|8$, so thus, the only possible values of $ x$ that allow $ x\plus{}4$ and $ x\minus{}4$ to have a common factor other than 1 are $ x\in\{\pm2,\pm4,\pm8\}$.[/hide][/quote]

Why cant $ x$ equal, say, 16 for $ x\minus{}4$ and $ x\plus{}4$ to have a common factor?

[hide="Overkill"] By the Nagell-Lutz theorem, an integer point $ (x, y)$ on the elliptic curve $ y^{2}\equal{} x^{3}\plus{}16$ (note the change of coordinates) must satisfy $ y^{2}|\Delta$, where

$ \Delta \equal{}\minus{}27(16)^{2}\equal{}\minus{}2^{8}\cdot 3^{3}$

Hence we can only have $ y \equal{}\pm 1,\pm 2,\pm 4,\pm 8,\pm 16,\pm 3,\pm 6,\pm 12,\pm 24,\pm 48$.  Checking, the only solution is $ (x, y) \equal{}\boxed{ (0,\pm 4) }$. [/hide]


Are you kidding? The left side is only of degree one and the right side is of degree $n$.
Just put $x_1=x_2=\ldots=x_n$ and you see obviously it becomes wrong.

[quote="tchebytchev"]It´s just AM-GM ;)[/quote]
What do you mean by it, I told you take all $x_i$s equal and very large, then the inequality will not hold!!!
Maybe you forgot some $n$th root or something in your inequality.

[quote="tchebytchev"]prove for all psitive numbers:$ x_{1},x_{2},......,x_{n}$
$ \frac{x_{1}^{2}}{2}\plus{}\frac{x_{2}^{4}}{4}\plus{}.......\frac{x_{n}^{2^{n}}}{2^{n}}\plus{}\frac{1}{2^{n}}\geq x_{1}x_{2}....x_{n}$[/quote]
sorry for the mistake, now the inequality is true .


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=13999[/img]

 :(

Note that $ BH \perp M_c M_a$ and $ \angle{M_c B M_a} \equal{} \pi \minus{} \angle{M_c H M_a}$, we could construct the point $ B$. Then construct $ A, C$ is trivial: take the symmetric images of $ B$ about $ M_c$ and $ M_a$ respectively to get $ A, C$.

<McBMa  is [b]acute[/b]
and  pi - McHMa is [b]obtuse[/b]


What is the number of degrees in the exterior angle of the largest angle of a triangle if the interior angles are in the ratio of 2:3:5?

We have that $ (2\plus{}3\plus{}5)x\equal{}180 \implies 10x\equal{}180 \implies x\equal{}18$. Hence, the largest angle of the triangle is $ 5(18)\equal{}90^\circ$. The sum of any interior angle of a triangle and the corresponding exterior angle is $ 180$, so the angle we are looking for is $ \boxed{90^\circ}$.

But wouldn't you use 2x=36, getting an exterior angle of 144?

You're finding the exterior angle to the largest interior angle. The value of the largest interior angle is 18(5) = 90 degrees so the value of its exterior angle is 180-90=90 degrees.

Oh, I didn't read the part about the largest angle. Thanks.


Battleship:

play battleship!. Play [url=http://dasarathy0.tripod.com/old_site/]here[/url] or [url=http://www.4aunlimited.co.nr]here[/url]

me too. That's why i gave a place to play it. In the mornings, you can play it here.

Does anyone have a strategy for the game?

Assuming the enemy ships are always placed completely randomly, just start guessing in a diagonal pattern so that no two spaces are touching, excluding diagonals. Once you find the Destroyer, continue the pattern so that no three spaces are in a row, etc. Obviously, if you find a ship along the way, take it out. Logically, that will be the fastest method, although of course it's possible to do it faster if you get lucky.

I guess I should start around the center?

I don't think it really matters where you start, if, as I said, the ships are placed completely randomly. If you're up against a human who you know is likely to place the ships in certain areas, it's more logical to start around one of those areas.

After taking out about 6 - 7 ships, you can figure out where the person put his ships if you know what type of a person he is.

Actually, assuming the placement of the ships is random, you should start near the center.  Consider the two-unit ship, whatever it's called.  The corner space will only be occupied with that ship if the ship is in either of two positions.  However, any given space near the center will be occupied with the two-unit ship if the ship is in any of four positions.  This concept is even more important if you consider the five-unit ship.

Of course, keep in mind that if you're playing a human opponent, you need to vary your strategy.  Otherwise, your opponent will place his ships in the places he knows you'll look last.

is there any website that lets you play battleship against other ppl online? not with a computer? kinda like what a lot of sites do for chess.


On the sides of an acute-angled triangle ABC,similar isosceles triangles DBC,ECA,FAB are constructed externally,such that  <DBC=<DCB=<EAC=<ECA=<FAB=<FBA=<BAC.
Let M be the mid-point of BC,and let P and Q be the intersections of DE with AC and of DF with AB,respectively.
Prove that MP:MQ=AB:AC

A reply was posted at http://www.mathlinks.ro/Forum/viewtopic.php?p=118331 .

Please [b]stop posting into multiple subfores[/b]. It is only contraproductive to have one problem posted into twenty different threads.

  Darij


$P(x) \in Z_{[x]}$ .

Show that the number  of prime divisor of $P(x)$ when $x \in Z$ is infinite

Even a stronger assertion holds; see the thread http://www.mathlinks.ro/Forum/viewtopic.php?t=20203 (this thread is a bit confusing, since the posts #2 and #3 have to be interchanged).

  Darij


Use image below to do these.

1. The six depicted triangle are all 30-60-90 right triangles sharing a common vertex A and sharing the common sides with neighbors. If the length of side AB is 1 unit, how long is the side AH?

2. From #1, figure out:

$BC^2+CD^2+DE^2+EF^2+FG^2+GH^2$


Hi,

I'm a sophomore in high school, and next year I'll be in AP Calculus BC. Since I'm pretty new to all these science fairs, could anyone guide me as to what I can do to start a mathematics research project which I could still enter into ISEF, but without a mentor? For example, could you give some examples of feasible project ideas? Thanks for your help in advance.

-Joe

Did you enter in science fair before?  I got into ISEF with a project in the math category last year as a freshman.  I created a mathematical method to easily calculate the entropy of an image and applied it to identifying cancer tissues, so it had a bunch of statistics in there too, but nothing past trigonometry level as only now as a sophomore am I also taking AP Calculus BC.  I recommend at least basic Calculus in a math project if you want to compete at international level.

Most people there however created a new theory or solved age-old problems and practically everyone had way past Calculus level math in their projects and almost everyone had a mentor.  I think it's really hard to think of a feasible mathematical topic without the aid of a mentor.  For example the guy that won first place (Michael Viscardi, who's board was almost directly in front of me) solved parts of the Dirichlet problem but did it with a mentor since there was extremely abstract algebra in it.

This year I'm probably entering in computer science since my project is entirely about programming and modelling and most of the math is just applied (like Groebner bases).

I appreciate the reply, chi. Did you have a mentor for your ISEF project? If not, where did you get the idea for your project? Thanks.

Actually I sorta did and sorta did not have a mentor.  See to get the cancer pictures I had to ask this university professor for them.

How did I get the project idea though?  hm, kinda that I always liked math (obviously) and if I had to enter science fair for my class then I would only want to enter in math, since I don't really like science.  Since I was taking biology last year it also had to relate to [i]that[/i] for our teacher.  So thinking for a looooong time and doing tons of research I decided that a real-world problem, like cancer, would be the best topic to pursue.  Now the number one reason that people die of cancer is that it cannot be identified by current methods at an early stage.  So I got stage 1 and 2 cancer pictures from my mentor and the rest I did myself out of reading a lot of math books on entropy, or chaos, and it worked since cancer is "chaotic".

Of course I don't recommend applied mathematics if you plan to enter the math category, and that is the reason I'm entering computer science this year since I cannot think of a pure mathematics topic myself that is feasible without a mentor.


ενα ομορφο θεμα-σχετικα απλο

εστω $ p(x) \equal{} x^2 \plus{} bx \plus{} c$,b,c στο R

ν.δ.ο. $ max\{|p(w)|,|p( \minus{} w)|,|p(0)| \} \geq w$ για καθε $ |w|\geq2$

P.S. ειναι γενικευση ενος παλιου θεματος σε εξετασεις στο πολυτεχνειο
το βαζω στις ολυμπιαδες αλλα πιθανη μεταφορα του στα σχολικα θα μου φανει λογικη.

Ηλίας

Βάζω τη βασική ιδέα!
Από τριγωνική ανισότητα:
$ |p(w)| \plus{} |p( \minus{} w)|\geq 2|w^2 \plus{} c|$. Οπότε υποθέτουμε ότι $ |p(w)| < w, |p( \minus{} w)| < w$ και έχουμε ότι $ w > |w^2 \plus{} c|$.

Από αυτό σε συνδιασμό με το ότι $ |w|\geq 2$ βγαίνει έυκολα ότι $ |c|\geq w\iff |p(0)|\geq w$

ακριβως δημητρη. εχουμε σχετικα ιδια λυση μονο που εγω υπεθεσα στην αρχη οτι ολα ειναι μικροτερα του w και κατεληξα σε ατοπο χωρις καποια ουσιαστικη διαφορα.  :)


calculus function geometry trigonometry ARML USAMTS MATHCOUNTS

Is anyone from here going? It's in hawaii this year. State your division if you are going, and from what school. Maybe we can meet up there.

I am from Thomas Jefferson, Auburn, WA and I will be an Alpha.

Saiem Gilani, Rickards High School, Theta, Tallahassee, Florida

Tell me if this rumor I heard is true... Thomas Jefferson flew in all the way from Washington for a MAO competition...I believe it was Palm Harbor and you all didnt know it was a calculator test.. so you all didnt bring calculators... resulting in rather poor scores and no placing...ouch?

Also from Washington. Bellevue High School. Mu (I hate calc, though, at least on competitions :P; it's pretty useful in other places). Should be some good stuff in Hawaii :gleam:  :surf:

Well Lucky for you, this year won't have much calculus.  The test editor doesn't put alot of calculus on the Mu Individual from what I've heard, and his tests are very well written so the dispute rate is ridiculously low, like .002% of questions had accepted disputes.  Hes the same guy that edited Denver '01 Nationals and Seattle '97 Nationals.

Ah Bellevue High School, I've heard of them.   Correct me if I'm wrong, the only 3 states who really compete heavily in MAO are Washington, Florida and Alabama right?

Do you all have state conventions and regional/invitational competitions?

EDIT: The editor guy's name is Tom Clymer I believe... at first I thought it was Thomas Jefferson's sponsor... I think Tom Norris?

What does [b]MAO[/b] mean?
As I understand it is a contest.
Is there a website concerning this?
Where can someone find the problems?

MAO is Mu Alpha Theta, the O is supposed to be a greek Theta symbol

Mu Alpha Theta in greek phonetically says math, but with a proper greek its more like "moth"

http://www.mualphatheta.org/National_Convention/MAO_Past_Tests.htm

That should give you all the past year's tests for nationals.

If anyone cares to help me with Theta Functions, Theta Sequences and Series, and Theta Matrices and Determinants, I'd appreciate it haha.  I hate how easy the tests are though lol. All of these tests can be made difficult but they rarely ever do.

Yea yea, Bellevue High, Washington, I'm in Alpha

Dammit Young's gonna own me

I'm taking number theory, S&S, and something else  :D

Derek, u are in luck.

I am taking Functions, Complex #, and Geometry. I will surely pwn in complex and geo. :D

[quote="Pakman2012"]
EDIT: The editor guy's name is Tom Clymer I believe... at first I thought it was Thomas Jefferson's sponsor... I think Tom Norris?[/quote]

to answer few questions, yes, Washington do have MAO competition statewide. in the contest, we have 3 topics (NT, S&S, Prob) for everyone (meaning only 1 round), ciphering, ind, class test, and special event (1 per person ---- Jeopardy, Relay, Speed Math, Chalk Talk, and something else)

Tom Clymer was a student at Mt. Rainier, who is also coming to Hawaii. He was our coach's first student, from what i hear, and my coach is Tom Norris. and dont think Norris can edit test; he dont even take tests.... :D  Clymer is a good test writer though. He runs a monthly mail-in contest that you can check out [url=http://www.natassessment.com]here[/url]

FL MAO is pretty big as far as tests go.  We dont have ciphering at the state level but they intend to implement it in 2007.  We do 3 topic tests which are usually seperated by division and change from year to year.  However, open tests can be taken by anyone such as Number Theory which... Oleg won as a freshman a few years back..wow.  Due to the watering down the difficulty of tests due to editors, the scores have become much more inflated and there is less of a degree of seperation which bothers me.  But there is always one test that always has dramatically low scores or someone who makes a ridiculously high score compared to the rest of the field.  All Theta tests were so ridiculously easy this year...and I didnt do so well.  Placing on the individual began at 100 for the top 25, 97 for top 20 in Functions, 110 for top 20 in Logs and Exponents and 93 for top 20 in Circumference/Perimeter/Area/Volume.  I hate that kind of inflation.  Someone who had a very small grasp on any of algebra II could've gotten a 110 on those tests.  It was just a matter of who made the least mistakes.

[quote="Derek"]Lol if it weren't for you Young I would've gotten 3rd place in # theory at state :P[/quote]

hmm you should go tell Meir about that. I am not the one who won.

I'm curious...was 2004 Washington S&S considered difficult? average? or easy?

Cause I got a 37/40 and I skipped this one calculus looking question and a couple that were just boring looking haha

EDIT: Same question, 2004 Washington Individual 37/40...theres a geometry question I need help with too...it seemed easy but then I'm like uh oh.

Well seeing as how the national convention is in about a week and I'm leaving in 3 days for it.  Is anyone else in MAO from other states?  What topics are you all taking? Are there any other thetas not from Alabama and Florida I'm gonna have to worry about?  How are we gonna meet up at nationals?

Pakman2012, please don't post two replies in a row, it is considered spamming.  In the future, just edit your previous post.

Thanks.

Oh it's a MAO interschool individual competition given only in Florida.  Mu and Alpha kids take the Dostal while the Theta kids take the Nunn.

[quote="xxazurewrathxx"][quote="Elemennop"]Alpha
R2 - Sequences and Series (argh logs and exponents is never in the division I'm in ARGH)[/quote]

Be thankful for that, my friend. You would've been pulverized by Wendy.
[/quote]

All I know is that I saw Elemennop dominate about a half dozen USAMOers at the Vestavia tournament this weekend.  It wasn't even close.

Haha, I didn't attend the competition this year but I heard Elemennop really laid it down on Comprehensive. I'm not too surprised though, many Comprehensive students from Vestavia mentioned to me that Seva was dominating, anyway, congratulations.

There were 3 USAMOers there I believe: David Harris, Kevin Hu, and Ben Cunningham (codeblue87). Were there any more?

Anyway, enough of the lying with the dostal/nunn scores.

Nunn: Our highest score was a 120 from Karlanna (cutiepi), the rest were not as spectacular.

Dostal: Our total was 653
Me - 120
Weiping (Masterfuzzy) - 115
Saiem (pakman2012) - 115
Ben (codeblue87) - 110
Suchandan Pal - 98
Vivek Pal - 95

It seems like we lost to Stoneman, we might beat Buchholz but I doubt it. Rickards should be at least 3rd in Dostal.

Auburn High School had a USAMOer (from 2004) there and I don't know about the rest, so I just estimated.

FYI -- I'm a big fan of the Rickards program.  It's really built up over the past few years.

Yeah Albert Lee from Auburn HS?

Any Georgia schools?

P.S.
Dostal:
Me - 120
Nimish Ramanlal - 115

Those were the only high scores from central Florida.

yeah i got a 95 on dostal, not too happy with that.
gaku and sean got 115s and roland got a 111 (hes really ticked about that).
im not too sure how other ppl did thoug, but i know i got 4th at Chiles.

Does anyone know anything about Buchholz and their scores? I hear they have a 120 in nunn and a handful of 115s, I don't know much about their Dostal though.

As for the Vestavia thing, I was actually kinda excited when I heard Ben made a 75 (which was extra buff), I thought he would win, but then when I heard there were two people ahead, I was just sitting there wondering who it could have been. It just goes to show that one should never underestimate Vestavia  :P

I dunno, it seems like theres been a very recent trend of writing easier tests.  2003-2005 have all been pretty manageable.

Yeah, I'm kinda mad that I missed that easy question on Dostal.

And my horrible performance at Vestavia... riddled with many mistakes.

I'm pretty sure I talked to you, although I didn't introduce myself.

I'll talk to you at the first MAO Invitational in January which I think Vestavia is coming to.
Palm Harbor with calculators... should be interesting.  :lol:

Guess that means I have to buy a calculator for the tournament.

isnt palm harbor normally w/ calculator?
or is it that the jan. invitational normally doesnt allow calculators?

or am i missing something. :?

Usually from what I know, Vestavia attends the no calculator competition of the FAMAT regular season.  However, the January Invitational this year has calculators, and thats sort of unusual because Vestavia is attending.

Haha, gonna have to buy a calculator.. classic  :P

ahh.
well seeing as how i was not able to go to the no calculator competiton of last year (b/c of SAT) i dont think i would have noticed this.