[hide="Solution"]If $ x \plus{} y \plus{} z < 10$, and $ x, y, z$ are all nonnegative integers, it implies that $ 0\le x \plus{} y \plus{} z\le 9$.

There are $ \binom{0 \plus{} 3 \minus{} 1}{2} \equal{} \binom{2}{2}$ solutions when $ x \plus{} y \plus{} z \equal{} 0$.
There are $ \binom{1 \plus{} 3 \minus{} 1}{2} \equal{} \binom{3}{2}$ solutions when $ x \plus{} y \plus{} z \equal{} 1$.

So on, there are $ \binom{2}{2} \plus{} \binom{3}{2} \plus{} \cdots \plus{} \binom{11}{2} \equal{} \binom{12}{3} \equal{} 220$ solutions when $ 0\le x \plus{} y \plus{} z\le 9$.

Answer: 220[/hide]

Thank you FantasyLover

Btw, that was a lightning-quick solution, did you have this problem on control c? :P

Set w so w+x+y+z=10, then this is balls and urns, shouldn't the answer be $ {13 \choose 3} \equal{} 286$?
This trick is in aops, by the way.

...which is why we set $ w\plus{}x\plus{}y\plus{}z\plus{}1\equal{}10$ and use balls and urns to get $ \binom{12}{3}\equal{}\boxed{220}$.

Bam.

If they're non-negative how can $ x\plus{}y\plus{}z<0$

Look at the title of the topic (the equation in the topic has a typo, indeed  :lol: )

Sorry, fixed the typo.

I don't understand what you did there trtle, could you explain why is the problem equivalent to

$ w\plus{}x\plus{}y\plus{}z\plus{}1\equal{}10$ ?

[quote="triplebig"]How many non-negative numbers satisfy the inequation

$ x \plus{} y \plus{} z < 10$[/quote]

[quote="FantasyLover"]$ x \plus{} y \plus{} z < 10$, and $ x, y, z$ are all nonnegative integers, it implies that $ 0\le x \plus{} y \plus{} z\le 9$.[/quote]

Let $ 9 \minus{} (x \plus{} y \plus{} z) \equal{} a\Longrightarrow x \plus{} y \plus{} z \plus{} a \equal{} 9\ (x,\ y,\ z,\ a\geq 0)$.

By considering the arrangement of 9 balls and 3 urns, as grn_trtle showed, the answer is $ _{9 \plus{} 3}C_3 \equal{} 220$.


I have MikTeX and TeXnic Center installed. However, the \therefore command (the three dots to represent a conclusion) is not built into TeXnic Center, so I needed to download a package. I downloaded "amsfonts-all-but-pk-files.zip" from [url]http://www.ams.org/tex/distrib.html#ibmpc[/url] and unzipped it, but I don't know how to install it into TeXnic Center. Can anyone help? Thanks.[/code]

Try

[code]\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
[/code]

at the top of your file


algebra
function
domain
quadratics
Ring Theory
number theory solved
number theory

find all prime number such that:$p=x^{2}+2y^{2}$

Kind of like [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=41677]this one[/url], but easier :).

That's because $\mathbb Z[i\sqrt 2]$ is, once again, a euclidean domain, and we can do almost the exact same thing, except that this time it won't split in two cases. The answer is: exactly those primes for which $-2$ is a quadratic residue.

[quote="grobber"]...for which $-2$ is a quadratic residue[/quote]
if -2 wants to be a quadratic residue then $p\equiv \pm 1$  $(\mod 8)$
but itsn't true for $p=7$

$-2$ is a quadratic residue iff $p \equiv 1 ; 3 \mod 8$ !


Hi Guys,

New around here and in desperate need of some assistance...

I am slightly confused about the theory behind logic gates, and in particular the use and practice of adder modules.

Can anyone offer any assistance in helping me get an understanding these?

Would appreciate any assisatance anyone can offer...

Fusion.

That sounds too general. What exactly you don't understand about logic gates / adder modules? They're just a piece of equipment that do logical operations / add binary numbers. Depending on the voltage they have on their input(s) they produce some voltage on their output. The voltage can be interpreted as logical / binary ones and zeroes. The basic logic gates are implemented using semiconductors - diodes and transistors - and the more complex ones are built from the basic logic gates.


function
real analysis
real analysis unsolved

let f be a continu function in the intervalle [a,b] of R,prove that there existe an intervalle [x_i,x_j] in [a,b] where f is monoton

False. Any nowhere differentiable differentiable continuous function will have no intervals of monotonicity - after all, a monotone function is differentiable almost everywhere.

$f(x)=\sum_{n=1}^{\infty}\left(\frac34\right)^{n}\cos(4^{n}x),$ to name one classic example.

[color=green][In answer to drapt's complaint below, the multiple typos have now been fixed.][/color]

but where is n in the formule f(x)?????  :oops:


Show that every even number $2n$ can be written in the form

$2n = (x+y)^{2}+3x+y$

With $x, y$ non-negative integers.

[hide="Remark"] Engel does far more manipulation in his solution than is necessary.  The problem is quite simple.  A simple substitution solves it. [/hide]

[hide]We proceed by induction.

$(0,0)$ obviously works, returning 0.

As we increase $y$ by 1 while $x=0$, the number increases by $2y_{0}+2$, where $y_{0}$ is the original $y$.
$\left((y_{0}+1)^{2}+y_{0}+1\right)-\left(y_{0}^{2}+y_{0}\right)=y_{0}^{2}+3y_{0}+2-\left(y_{0}^{2}+y_{0}\right)=2y_{0}+2=2(y_{0}+1)$

From the above, it is clear that we skip $y_{0}$ even numbers in between $(0,y_{0})$ and $(0,y_{0}+1)$.  However, these are filled in:

As we decrease $y$ by 1 and increase $x$ by 1, then the number increases by 2.
$\left(\left((x+1)+(y-1)\right)^{2}+3(x+1)+y-1\right)-\left((x+y)^{2}+3x+y\right)=(x+y)^{2}+3x+3+y-1-\left((x+y)^{2}+3x+y\right)=2$

Now, as we start from $(0,y_{0})$ and proceed diagonally to $(y_{0},0)$, we get the $y_{0}$ even numbers that we would've skipped by going from $(0,y_{0})$ to $(0,y_{0}+1)$.

And thus the induction is complete.[/hide]

Induction is nice, but somewhat unnecessary.

[hide="Since we could just..."] Let $k = x+y$.  Then

$2n = k^{2}+k+2x$

Since $k^{2}+k$ is always even, just pick the largest such number less than $2n$, and $2x$ is simply the difference.  It remains only to prove that $k \ge x$, which is not hard either; $(k-1)^{2}+(k-1) = k^{2}-k$.  [/hide]

Hey, it's still a valid solution.

BTW, what did the book say?


geometry
rectangle
conics
parabola
analytic geometry
Pythagorean Theorem

Let ABCD be a rectangle and let P be a point inside the rectangle, if $ PA\equal{}8$, $ PB\equal{}4$
$ PD\equal{}7$, then find $ PC$

is there a formula for this? I think i've seen some formula for questions like this, or may be i remembered wrong :(

good luck solving it :D

EDIT: here's another question i need help on

Find all values of a for which the parabolas $ y\equal{}1\plus{}2x\minus{}x^2$ and $ y\equal{}x^2\plus{}a$

[hide="hint"]Draw the perpendiculars, use variables, and use the Pythagorean Theorem A LOT of times.[/hide]

yeah, just got it, is it 1?

EDIT: thx xpmath

The theorem that you want is: $ AP^2\plus{}CP^2\equal{}BP^2\plus{}DP^2$.

We can easily prove this using coordinates...let $ A(a,b)$, $ B\equal{}(a,\minus{}b)$ ,etc, and $ P(x,y)$. Then use the distance formula.

I dont understand your other question. What does a need to satisfy?

[quote="i_like_pie"]There are two simple proofs of the [[British Flag Theorem]]: the one Alex mentioned (using coordinates), and one that uses the Pythagorean Theorem (written out in the Wiki article).[/quote]

Those two proofs are exactly the same. The distance formula is just using the pythagorean theorem. The diagram might look a little different, but nothing new is going on.


function
complex analysis
algebra
domain
complex numbers
complex analysis unsolved

Prove that the exponential function $z \to e^{z}$ has infinitely many fixed points in the complex plane

This follows from Picard's Theorem: The theorem says that a function $f(z)$ assumes all possible values except for at most one value infinitely often in any neighborhood of an essential singularity. Apply now this theorem to $f(z) = e^{z}/ z$. Note that $f(z) \neq 0$. Therefore, $f(z) / z = 1$ infinitely often in every neighborhood of $\infty$.

i'm a beginner in complex analysis and i'm in awe of the picard's theorem. This is the first time i'm seeing this result and i went back and searched my book, sadly it doesnt give a proof. anyway i have a doubt, where is the essential singularity for $f(z)=\frac{e^{z}}{z}$? i was thinking if you expand the numerator as a power series and then divide by $z$ you will have a $\frac{1}{z}$ term. i dont see how there is an essential singularity. please can you explain? sorry if i asked a stupid question.

The essential singularity is at $\infty$ (consider it as a map of the Riemann-sphere). Or in other words: consider $f\left( \frac{1}{z}\right)$ instead, then you have an essential one at $0$.

There is another (more direct) proof based on the the Argument Principle.
[b]Argument Principle:[/b] Let $f(z)$ be a holomorphic function, $\gamma$ be a closed contour. Assume that $f(z)$ does not have zeros on $\gamma$. Then the number of zeros of $f(z)$ inside the contour $\gamma$ equals the number of times the curve $f\circ \gamma$ winds around $0$.

To solve your problem, we can apply the Argument Principle to the function
\[f(z) = e^{z}-z,\]
and the contour
\[\gamma: M-2\pi i k \longrightarrow M+2\pi i k \longrightarrow-M+2\pi i k \longrightarrow-M-2\pi i k \longrightarrow M-2\pi i k,\]
where $k$ is integer, and $M$ is large enough.
It is easy to check (draw a picture!) that $f\circ \gamma$ winds $2k$ times around the origin. Therefore, $f(z)$ has $2k$ zeros inside $\gamma$.

[quote="kouboy"]i'm a beginner in complex analysis and i'm in awe of the picard's theorem.[/quote]

Me too!  For me, the best way of visualizing Picard's Theorem is to write $f(z)=e^{1/z}$ as:

$e^{1/z}=Exp[\frac{x}{x^{2}+y^{2}}] Cos[\frac{y}{x^{2}+y^{2}}]-iExp[\frac{x}{x^{2}+y^{2}}] Sin[\frac{y}{x^{2}+y^{2}}]$

Now take a moment and look at what's happening in a neighborhood of the origin.  What is the real part doing?  Well, it's an exponentially increasing function operating on a oscillating function that is increasing in frequency.  So the closer I get to the origin, the larger the amplitude and the higher the frequency.  That's it:  Unrestrained increase in amplitude and  frequency in a bounded domain!  What other geometric phenomenon could possibly be at the root of Picard's Theorem?  And since the imaginary part does the same, then we can begin to understand why in terms of a complex number, f(z) attains every value (except 1) infinitely often in some neighborhood of an essential singulariy.

I've looked for a proof.  Seems it's not an elementary one.  Would be a good project to figure out one that us beginners could follow.   Here's a start (as a beginner):  Anytime a complex function exhibits the phenomenon above (Unrestrained increase in amplitude and  frequency in a bounded domain), Picard's Theorem is probably applicable.


Hey, you must be one of my classmates  :D
I posted this problem in Pre-Olympiad
Still have no idea, heard that we have to handle a lot of expanding


$3$ dayere dobedo motekharej va momas moshtarakhaye dakhelie anha dade shode ast.farz konid $(L_{1},L_{2})$ va $(L_{3},L_{4})$ va $(L_{5},L_{6})$ se joft momashaye moshtarake dakhel bashand va $L_{1},L_{2},L_{3},L_{4},L_{5},L_{6}$ azlae sheshzeleai $ABCDEF$ bashand sabet konid ghotrhaye aslie in sheshzeeii hamresand..

masaleye sakht va pishrafteie.....in darso bekhoon age halo nafahmidi begoo benevisam:

 http://mathworld.wolfram.com/FourConicsTheorem.html

kheli vahshatnake benazaret bara entekhab team sangin nist????? albate man to kafe inam chetori moshaveremoon mige asoone :( khode davood vakili goft roosh fekr nakardam yani 15 min fekr karde kam ovordeee :-??

in soale entekhabe teame...
in ye hadsi boode ke esbat nashode,yeki hads zade ke hamresand vali natooneste esbat kone,vase haminam dade be talaiya shayad yeki oonja esbat kone ke oonjam HISHKI natooneste esbat koneo ye joorai OPENe,dar natije fek konam gozashtane in soal too in forum va bahse rooye oon ye khorde gheire manteghi bashe...  :wink:

are shaiad hagh ba u bashe vali vaghti kesi hal nakarde dalil nemishe roosh fekr nakonim shaiad ye rahehalli be zehnemoon beresee hum??

fekr kardam halesho fahmidid...!!
oon sote ke gozoshtam nesfe halesh tooshe Dge
hal :
be raveshi pishrafte(hendeseye tasvir) natije mishe ke baraye 3 ta beyzi ham oon 6 zelie mishe toosh 1 beyzi mohat kard...
hala age 1 6 zeli dashte bashid ke toosh 1 beyzi mohat bashe...mishe ba tabeye tasvir oono be 1 6 zelie mohiti tabdil kard.....
hala tebghe ghazieye BERIANSHEN(aghtare har 6 zelie mohiti hamresan)
natije mishe ke oon 3 khat hamresan.

[quote="Behsan"]fekr kardam halesho fahmidid...!!
be raveshi pishrafte(hendeseye tasvir) natije mishe ke[/quote]
mishe lotf konido halle kamelesho begid,shak daram ba tasvire sade hal beshe,vase hamin kheili doos daram rahe halletoono kamel tozih bedid ke ma ham befahmim...  :maybe:
 :wink:

[quote="Behsan"]faghad behet injuri begam ke mishe ba tasvir har 3 noghteye hamkhat ro sabet va har khat ro saabet negah dasht va beyzi ro be dayere tabdil kard.......
age bekham be soorate gheyre hendesi va elmish barat shohoodi tozih bedam ...injuri begam ke age tasvire 1 dayere ro rooye safhe  kaaghaz ba noor bendazi roo 1 chize Dge va oono tekoon bedi mitooni dayeraro be beyzi tabdil koni ke kharsaaz ham mishe....
age moshkeli bood davood vakili ro DD begoo rahe hale mano barat bege....
injuri behtare chon oon savadetoono midoone va 1 chizi nemige ke hang koni....[/quote]
nagoftam ke hendese tasvirio tozih bede,man hendese tasvirio khoob baladam,vase haminam mikham rahe kameleto bebinam,na SAVADETO...
pas be jaye inke begi ba hendese tasviri chikar mishe kard,begoo to vase in masale chikar mikoni,KAMEL begoo... :?:  :!:

valla ...age tasvir baladi ke nemidoonam chera bazam mikhay bedooni chon ba oon lem e ke too oon site e ke gozoshtam hast va tasviiir vazehe.....!!!!!!!!

man mac user am va latex nadaram.....vali az inja ke tasvir baladi va nafahmidi hallo.......fekr konam ba "Mittenpunkt images" ashna nisti.
be har haal dar avalin mogheiat ke davood ro DD begoo halamo kaamel behet bege

chera nemifahmi man chi migam...?
bebin beza injoori bet begam,in soale emtehane ma boode,sare jalase natoonestim hallesh konim,hala to migi hallesho baladi man migam KAMEL benevis...
yeni fek kon sare jalase emtehan neshasti va dari too bargat javabe in soalo minevisi,chi minevisi?
minivisi berid az davood vakili beporsid hallamo...!!??fek kon gharare sahih beshe,esme ghazie nanevis begoo chetori azash estefade mikoni,kamel na inkebenevisi:"[b]tebghe oon lemme age balad bashi masale badihie...[/b]"
mage tahala too barge rahe hal naneveshti?
pas rahe halleto doros benevis,vase latex ham kafie avalo akhare har ebarati ke mikhai ba latex benevisi yedoone alamate dolar bezari...

[quote="Behsan"]
Ba hendeseye tasviri natije mishe ke baraye 3 ta beyzi ham, On 6 zelie mishe toosh 1 beyzi mohat kard.
.[/quote]
@Behsan: Loftan in edea ro sabet kon.

@Babak Ghalebi: Az ordo che khabar? Natayeje TST moshakhas shode?

akharin khabar emtehane chan hafte pish bood ke soalasham too site hast,too oon emtehane ke too $2$ rooz bargozar shodeo $6$ ta soal dashte,[b]arman fazeli,hadikhanloo[/b] $6$ ta kamel hal kardan,baghie hame zire $6$ta,gharar bood do hafte badesh bazam ye dorooz emtehan begiran...

ba kamale ehteram ke bara shoma daram:
gharar nist ke har ki oon tooe faghat balad bashe hal kone ...
neveshtam ke man mac User am va latex nadaram
maghrooram nasho age dar morede man shak dari boro az in karash bepors man kiam.
man az tasvir harf mizanam be to bar mikhore!!!!migan har chi derakht por bartar,oftade tar
man alan 2 emtehanaye nahaiyam...bad az emtehana minevisam bebinam Dge chi migi.

mamoolan in 2 elm rasme ke har chi bishtar baladi kamtar saro seda raah mindazi......

[quote="lomos_lupin"]
 Az ordo che khabar? Natayeje TST moshakhas shode?[/quote]
team iran panjshanbe moshakhas mishe :wink:

mohammad hossein barat arezooye movafaghiat mikonam :wink:  :wink:  akharesham in soale tabeiiee TsT HAL NASHod age vaght dashty hallesho baramoon bezar tanx;)

2 behsan:aval az hame inke shoma harki ke bashi che khafano che gheire khafan,ma koochike shomaim,in 1,dovoman inke man age ghaboolet nadashtam nemigoftam raheto benevisi...badesham man saro seda nakardam,harfe man in bood ke be jaye inke 4 ta post bedi ke ba ham kalkal konim,adelane tar bood age hamoon aval rahe kameleto mizashti,chon vasam jalebe ke yad begiram na chize digei...
2 sepehr:hamoon moadelat tabeie team ro migi ke sooratesh in bood:
$f(x^{2}+y+f(y))=f(x)^{2}+2y$
hallesh injoorie:
aval az hame age gharar bedim masalan $x=0$ natije mishe ke tabamoon pooshae chera ke $f(u)=c+2y$...
hala age bejaye $x$ too moadele adade $-x$ ro gharar bedim,samte chapesh hich farghi nemikone vali samte rastesh fargh mikoneo natije mishe ke $f(x)^{2}=f(-x)^{2}$ pas darim $f(x)=\pm f(-x)$
khob hala az poosha boodane $f$ natije mishe ke adade $k$ vojood dare ke $f(k)=0$ pas $f(-k)=\pm f(k)=0$
hala biaid too soorate moadele yebar jaye y bezarid $k$,yebar bezarid $-k$ natije mishe ke:
$f(x^{2}+k)=f(x)^{2}+2k$ va hamchenin $f(x^{2}-k)=f(x)^{2}-2k$ hala age tarafeine do rabete akharo az ham kam konim natije mishe: $f(x^{2}+k)-f(x^{2}-k)=4k$ hala age gharar bedim $x=0$ natije mishe ke $f(k)-f(-k)=4k$ pas $k=0$ pas tanha noghte $0$e tabe hamoon $f(0)=0$e...
hala ba tavajoh be $k=0$ natije mishe ke $f(x^{2})=f(x)^{2}$ pas baraye $x=1$ natije mishe ke $f(1)=f(1)^{2}$ pas ya $f(1)=0$ ke tanaghoze,ya inke $f(1)=1$...
khob hala chon $f(x^{2})=f(x)^{2}$ pas baraye har adade mosbate $x$ khahim dasht $f(x)>0$...
hala sabet mikonim baraye har $x$ darim $f(x)=-f(-x)$

[hide]farz konid injoori nabashe pas baraye $x_{0}>0$ darim $f(x_{0})=f(-x_{0})=c$ tavajoh dashte bashid ke chon $f(mosbat)=mosbat$ pas $c>0$ hala too farze masale be jaye $y$ adade $x_{0}$ va yekbaram adade $-x_{0}$ ro gharar bedid pas natije mishe ke:
$f(x^{2}+x_{0}+c)=f(x)^{2}+2x_{0}$ va inke $f(x^{2}-x_{0}+c)=f(x)^{2}-2x_{0}$ hala age too rabete aval bejaye $x$ adade $0$ ra gharar dahim natije mishe $f(x_{0}+c)=2x_{0}$ va age dar rabete dovom be jaye $x$ adade $\sqrt{x_{0}+c}$ ro gharar bedim natije mishe $f(2c)=f(x_{0}+c)-2x_{0}=0$ pas $2c=0$ pas $c=0$ ke inam tanaghoze pas $f(x)=-f(-x)$...[/hide]

hala farz konid $y<0$ gharar bedid $x=\sqrt{-y}$ pas darim $f(f(y))=f(-y)+2y$ yeni $f(f(y))+f(y)=2y$ hala age tarafeine in rabetaro dar ye $-$ zarb konim natije mishe $f(f(-y))+f(-y)=-2y$ yeni vase har adade mosbate $x$ham khahim dasht $f(f(x))+f(x)=2x$ khob pas kafie in moadelaro hal konim:
$\forall x \in \mathbb{R}: f(f(x))+f(x)=2x$...

babak tanx  :wink:  rahet kheli ghashang boood
ey babaaaaaaaa chera pas nataieje in TsT naiOmaddddddddddddddddddddd?


A river flows at speed u. A boat has speed v relative to the water. If its velocity is at an angle $\alpha$ relative the direction of the river, what is its speed relative to the river bank? What $\alpha$ minimises the time taken to cross the river?

[hide="Solution"]
Well if you look at the boat's vertical and horizontal components, it's easy. Assume the river is flowing from left to right, and we're going from bottom to top. (This makes calculations easy, as it puts the boat in standard position for an angle.) Then, the boat's vertical component is $\textbf{v}\cdot \sin \alpha$ and its horizontal component is $\textbf{v}\cdot \cos \alpha$. Then, the total horizontal component of the boat, when it's in the water, is $\textbf{u}+\textbf{v}\cdot \cos \alpha$, and its vertical is still $v \cdot \sin \alpha$ since the river has no vertical component to interfere. Thus the magnitude of the velocity will be
\[\sqrt{ (\textbf{u}+\textbf{v}\cdot \cos \alpha)^{2}+(v \cdot \sin \alpha)^{2}}= \boxed{ \sqrt{ \textbf{u}^{2}+\textbf{v}^{2}+2 \cdot \textbf{u}\cdot \textbf{v}\cdot \cos \alpha }}\]

To take the least time to cross the river, we simply want to maximize the vertical component, i.e. $\boxed{ \alpha = 90^\circ}$.
[/hide]

[i]An interesting question:[/i]
What angle $\alpha$ will get our boat [u]directly[/u] across the river?


Norway has 14 hospital beds for every 1000 people and has a population of 4.5 × 106. How many hospital beds does Norway have? Express your answer in scientific notation.

There are $ 14\cdot\frac{4.5\cdot106}{1000}\equal{}14\cdot.477\equal{}6.678$. This is $ 6.678\cdot10^0$ in scientific notation.

Are you sure you don't mean $ 4.5*10^6$?  In this case the answer would be 6.3*10^5

Hmm, that's right if your numbers are right. I was like Norway doesn't have a lot of people.... Probably a typo on sri's part.


Find all polynomials such that $P(x^3+1) = P((x+1)^3)$.

I think the funtional equation be rather
1.$P(x^3+1)=P^3(x)+1$
or 2.$P(x^3+1)=(P(x)+1)^3$
the first, if we denote $f(x)=x^3+1$ then $P=f^n=fofofof...$? plz do it yourselves
And the second, that is really nice problem.
We can expand to the following:
Let f,g be two monic Polynomials
Find all polynomial P such that
$Pogof=fogoP$ (where o is the associating operation)

You're undoubtedly right, nmtruong1986, it was too easy like that .....

[quote="tµtµ"]I think only $P=0$ works :

If $P = a_n x^n + ...$ then $P((x+1)^3) - P(x^3+1) = 3n a_n x^{3n-1} + ...$ so $a_n = 0$ and finally $P=0$ which works.[/quote]

you are true but $ P(x) = c , c\in\mathbb{R} $ .


group theory
abstract algebra
superior algebra
superior algebra unsolved

Why can't there be a homomorphism from A_4 [b]onto[/b] S_3?

Notes:
A_4 has 1 element of order 1, 3 elements of order 2, and 8 elements of order 3.
S_3 has 1 element of order 1, 3 elements of order 2, and 2 elements of order 3.

I understand what you are saying; you can ignore my previous question....thanks :)

[quote="loriemash"][quote="Christian Hirsch"]By homomorphism theorem, this would imply $A_4/ker\phi=S_3$. But $A_4$ has no normal subgroup of order $2$[/quote]

What is $ker\phi$? I thought only the identity element in $A_4$  maps to the identity in $S_3$.

If the 8 elements of order 3 in $A_4$ map to the 2 elements of order 3 in $S_3$,
the   3 elements of order 2 in $A_4$ map to the 3 elements of order 2 in $S_3$,
That would leave me with 1 element - the identity which maps to the identity in $S_3$.
So |$ker\phi$| = 1

If this is true why is$A_4/ker\phi = S_3$?[/quote]


Every month, I will be making a tourney.  I have no idea what the format will be, but just sign up with your highest rating, along with your lowest one.  Thank you.

Sign-Up

1 ernie 1451/1113

reserved for sign-ups.

If either a highest rating or lowest rating is not confirmed, then highest rating will be presumed 1200 and lowest will be presumed 1000.

Highest: 1340

Lowest: 1130 or something higher

It's cuz I bombed :(

the format will be 1v16, 2v15, 3v14, etc. with [b]triple[/b] elimination.  Only 13 slots available!!!!!!! (I sound like mewto, dont I?)

meh... ill sign up

1422/1109 (i quit out of a 15-ppl game)

I'll join!
     highest-1335
     lowest-1 yes, that really is my lowest score.

I'll join

highest    1457
lowest     1046 =(   (i know, pathetic)

don't laugh at my pathetic low rating :mad:
ik it's terrible but..... :(
o well :)

i guess technically my lowest rating is 1 from glitch day.. :P  :rotfl:  :ninja:

1559/1 TRIPLE ELIMINATION???? EVEN I WAS NOT THAT BOLD!!!!

btw what is seeding? average of the 2?

CMON PPLZ, GET STARTED! (that includes me  :oops:)

BLEH.

The message is too small.  Please make the message longer before submitting.

[quote="isabella2296"][size=200][color=red][u][b]Join the Butterscotch Tourney! It's a CD Tourney, with limited space - only 16 people can participate! At this moment, there are only 3 slots left! JOIN NOW! Sign-Ups close July 31st, and the Tourney starts in August!! Hurry before all the slots are filled up and I have to put you on the waiting list for September's! [/b][/u][/color][/size]
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213378

(That's what you get for advertising in MY tourney!)[/quote]

nice
revenge is always sweet :lol:
BTW ernie, that bracket is extremely complicated :)

[color=red]AIME_is_hard[/color] DEFEATS [color=red]dragon96[/color] 3-0

Game 1:75-25
Game 2:38-23
Game 3:72-53

Jeez i'm sorry but i can never find kookamunga (sorry if spelled wrong) on

THIS TOURNEY IS BEING POSTPONED.

PM her.  And you spelled it right.
_____________________________________________________________________

PRELIMS

mewto55555 DEFEATS isabella2296, 3-0 (PR-1)
ernie VS pinkpiglet (PR-2)
RomanianGenius VS splatyango (PR-3)
lotsofmath VS ash-Pokemon (PR-4)
AIME_is_hard DEFEATS dragon96, 3-0 (PR-5)
1 AIME15 DEFEATS 15 xoangieexo, 3-1 (PR-6)
skaldjfhrulez VS kookamunga (PR-7)
budi713 VS vallon22 (PR-8)

Quarter-Finals

(1-0) mewto55555 VS winner of PR-2 (QF-1)
winner of PR-3 VS winner of PR-4 (QF-2)
(1-0) AIME_is_hard VS (1-0) AIME15 (QF-3) <--- lol.
winner of PR-7 VS winner of PR-8 (QF-4)

Semi-Finals

winner of QF-1 VS winner of QF-2 (SF-1)
winner of QF-3 VS winner of QF-4 (SF-2)

Finals

winner of SF-1 VS winner of SF-2 (FN)

ONE LOSS BRACKET

Round One

(0-1) isabella2296 VS loser of PR-2 (L-A1)
loser of PR-3 VS loser of PR-4 (L-A2)
(0-1) xoangieexo DEFEATS (0-1) dragon96 (L-A3)
loser of PR-7 VS loser of PR-8 (L-A4)

Round Two

winner of L-A1 VS loser of QF-4 (L-B1)
winner of L-A2 VS loser of QF-3 (L-B2)
winner of L-A3 VS loser of QF-2 (L-B3)
winner of L-A4 VS loser of QF-1 (L-B4)

Round Three

winner of L-B1 VS winner of L-B2 (L-C1)
winner of L-B3 VS winner of L-B4 (L-C2)

Round Four

winner of L-C1 VS loser of SF-1 (L-D1)
winner of L-C2 VS loser of SF-2 (L-D2)

Round Five

winner of L-D1 VS winner of L-D2 (L-E)

Round Six

winner of L-E1 VS loser of FN (L-F)

TWO LOSS BRACKET

LOSERS PLACE 13th

loser of L-A1 (need 3 to move on) VS loser of L-B4 (need 2 to move on)- (2L-A1)
loser of L-A2 (need 3 to move on) VS loser of L-B3 (need 2 to move on)- (2L-A2)
(0-2) dragon96 VS loser of L-B2 (need 2 to move on)- (2L-A3)
loser of L-A4 (need 3 to move on) VS loser of L-B1 (need 2 to move on)- (2L-A4)

LOSERS PLACE 11th

winner of 2L-A1 VS winner of 2L-A2 (2L-B1)
winner of 2L-A3 VS winner of 2L-A4 (2L-B2)

LOSERS PLACE 9th

winner of 2L-B1 VS loser of L-C2 (2L-C1)
winner of 2L-B2 VS loser of L-C1 (2L-C2)

LOSERS PLACE 7th

winner of 2L-C1 VS loser of L-D1 (2L-D1)
winner of 2L-C2 VS loser of L-D2 (2L-D2)

LOSER PLACES 6th

winner of 2L-D1 VS winner of 2L-D2 (LP-6)

LOSER PLACES 5th

winner of LP-6 VS loser of L-E (LP-5)

LOSER PLACES 4th

winner of LP-5 VS loser of L-F (LP-4)

LOSER PLACES 3rd :Winner_Third:

* In this game, as the loser bracket winner has only one loss, the first one with 3 losses is the loser.

winner of LP-4 VS winner of L-F (LP-3)

WINNER PLACES 1st :Winner_First:, LOSER PLACES 2nd :Winner_Second:

* In this game, as the winner bracket winner has no losses, the first player with three losses is the loser.

winner of LP-3 VS winner of FN

RomanianGenius [color=red]DEFEATS[/color] splatyango (duh!!!) :noo:

THE TOURNEY IS CANCELED.  THANK YOU FOR PARTICIPATING (or not) IN ERNIE's TOURNEY.  Please lock this thread.

What was lame about it? Ernie canceled the tournament. There was no point in finishing as every new tournament replaces an oldone.

[quote="AIME_is_hard"]wow that was lame... :|[/quote]

Well maybe you're lame...

*sniffle*

i didn't even play a game yet....

JOIN MY TOURNY!!!!
WEBSITE ON MY SIG!!!!

The way this is going, your tourney may not even play a game...


Find the domain, range, vertical asymptote(s), and hole of the graph of:

\[ f(x)\equal{}\frac{x^2\minus{}4}{x\minus{}16}\]

(you might have to graph it on a TI calculator)

Domain is all real numbers except x can't be 16.
Vertical asymptote is at $ x\equal{}16$
There are no holes

[quote="DaReaper"]what about the range? we did this in class using a calculator, but i dont remember how to do it.[/quote]Perform a long division. Then, consider the values you input for $ x$, what possibles can you get from it?

[quote="DaReaper"]and the answer is?

also, how do you do this w/ a calculator?[/quote]I'm sure you can do this in a graphing calculator like TI-83 or something. Or, if you want an online grapher, try one at [url=http://www.wolfram.com]Wolfram Alpha[/url].

So $ f(x)\equal{}\frac{x^2\minus{}4}{x\minus{}16}\equal{}\frac{x^2\minus{}16x\plus{}16x\minus{}16^2\plus{}16^2\minus{}4}{x\minus{}16}\equal{}x\plus{}16\plus{}\frac{16^2\minus{}4}{x\minus{}16}$. Test intervals using critical points. What do you get when $ x$ is near 16 like 15.999999, 16.000001, or when $ x$ is positively large and negatively large?

Another way is to consider $ y\equal{}\frac{x^2\minus{}4}{x\minus{}16}\Rightarrow x^2\minus{}yx\plus{}(16y\minus{}4)\equal{}0$ and look at the discriminant. What values of $ y$ are admissible for possible values of $ x$?


Is there a name for an "n x n" grid with edges added so that it is 4-regular, non-planar.
All variables have an adjacent neighbor above, below and to its left and right. For vertices on the perimeter of degree 3 one edge is added linking it with its symmetrically equivalent perimeter vertex, i.e. if a vertex at the top of the grid has degree 3 then an edge is added linking it to the vertex at the bottom of its column of vertices in the graph, similarly for rows.
So there are 2*n edges added to a normal "n x n" grid to make it so every vertex is of degree 4.
I am trying to find out if there is a specific name for such a "grid"?

I would call it $n\times n$  torus grid, or $n\times n$ torus lattice.

Torus grid is exactly what I'm looking for. Thanks. :)


[u][b]Formal Instructions:[/u][/b]
The aim of the game is to be the last player remaining.
Every round, there is a time limit.
Before the time limit ends, all players must submit via PM a number between 0 and 100 to me.
And the end of the time limit (called a round), I tally up the submissions and scores in this fasion:
- I post all the submissions (a fail to submit counts as 100).
- I post what half the average submission was.
- I post all players who submitted under half the average.  These players are OUT.
- Everybody gains points equal to their distance from half the average.  Note: you do not want points!
- If anybody has over 100 points total, they are declared OUT.
- I post a time limit for the following round (usually 1-3 days).
When there is one player remaining, they win.

[u][b]Previous games[/u][/b]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1734033730&t=149694]Game 1[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=326399784&t=151415]Game 2[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=326399784&t=154425]Game 3[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=326399784&t=168049]Game 4[/url]

[u][b]Sign-up list (5/10)[/u][/b]
AstroPhys
Hamster1800
xpmath
perfect628
shaggy75

I'm in again-hopefully everyone will submit this time...

I can't edit the original post, so [b]MODERATERS, PLEASE EDIT IN ALL NEW SIGN-UPS[/b].

i'll join, and ill try to remember to submit, although it's possible I might be busy some days

[u][b]Sign-up list (9/10)[/u][/b]
AstroPhys
Hamster1800
xpmath
perfect628
shaggy75
Nerd_of_the_Ages
DonkeyKong
Randomdragoon
abacadaea

What happens if we don't have enough people?

I'll start the game if nobody signs up in 24 hours.

[quote="miyomiyo"][u][b]Sign-up list (8/10)[/u][/b]
AstroPhys
Hamster1800
xpmath
perfect628
shaggy75
Nerd_of_the_Ages
DonkeyKong
Randomdragoon[/quote]

im in

AstroPhys
Hamster1800
xpmath
perfect628
shaggy75
Nerd_of_the_Ages
DonkeyKong
Randomdragoon
abacadaea

Here 'tis.
[size=200][b]ROUND 8 RESULTS[/b][/size]
[hide="Submissions"]perfect: 100 (No Submission)
xpmath: 21
shaggy75: 20
abacadaea: 15
BoesFX: 15[/hide][hide="Half the Average"]Half the Average was [b]17[/b].[/hide][hide="Below Half the Average"]abacadaea, BoesFX[/hide][hide="Scores"]perfect: 131
shaggy75: 89
xpmath: 72[/hide][hide="Over 100"]perfect[/hide][hide="Players Remaining"][u]Players (2/10):[/u]
shaggy75
xpmath[/hide][hide="Round 9"]Get them in sometime tomorrow (Tuesday).[/hide]Wow, 9 rounds....

Wow, perfect really ruin his chance.  :o

But why you take me out? :(

Same reason I got out. You were under half the average.

Quick round!
[size=200][b]ROUND 9 RESULTS[/b][/size]
[hide="Submissions"]shaggy75: 100
xpmath: 30[/hide][hide="Half the Average"]Half the Average was about [b]33[/b].[/hide][hide="Below Half the Average"]xpmath[/hide][hide="Scores"]shaggy75: 89[/hide][hide="Over 100"]shaggy75[/hide][hide="Players Remaining"][u]Players (0/10):[/u]
Rematch![/hide][hide="Round 9, Part II"]Get them in sometime tomorrow (Tuesday).[/hide]

Don't you mean: shaggy: 189?

I doubt this will ever end. That's one of the flaws of this game; the ending.

Sorry to interupt you guys, but I have a question for miyomiyo.

It looks like this game is about to end, so I was just wondering...
Are we going to have a HTA 6?
If we do, count me in..

For last round, I forgot to add up shaggy's score.
Also, Temperal, this is the first time I've had to have a rematch.
[size=200][b]ROUND 9 PART II RESULTS[/b][/size]
[hide="Submissions"]shaggy75: 100
xpmath: 50[/hide][hide="Half the Average"]Half the Average was about [b]38[/b].[/hide][hide="Below Half the Average"]Nobody[/hide][hide="Scores"]shaggy75: 151
xpmath: 94[/hide][hide="Over 100"]shaggy75[/hide][hide="Players Remaining"][u]Players (1/10):[/u]
xpmath[/hide][hide="We have a winner!"]xpmath![/hide]

Aargghh! I always am 2nd place. *sigh*

anyway, congrats xpmath(again) :D

Hm.  If need be, I'll do game 6 but if anybody else has enough free time, I'd appreciate it (I've been pretty busy lately.)

2 in a row :lol:
I'll play, but I probably won't have enough time to mod it :(

[quote]Also, Temperal, this is the first time I've had to have a rematch.[/quote]

Meh, that's because shaggy messed up. What he should have done is chosen 0, then cycled between 0 and 100 to throw xpmath off.

As for the other times, probably same thing. People don't know how to be strategical...

I don't care or anything, and I wouldn't ask that anything be redone, but I [i]did[/i] submit Round 8. It's in my sentbox, long before the deadline.

I think I can moderate HTA 6 if you tell me what I need to do.


induction
combinatorics proposed
combinatorics

Let $ G$ be a connected graph with $ n$ vertices and $ m$ edges such that each edge is contained in at least one triangle. Find the minimum value of $ m$.

The desired finimum is $ f(n) \equal{} [\frac {3n\minus{}2}{2}]$  (*)
where $ [...]$ denotes the integer part.
In another words, we have $ f(2n) \equal{} 3n\minus{}1$ and $ f(2n\plus{}1)\equal{}3n$ for all $ n \geq 3$.

A graph which staisfies the statement of the problem will be called "good" (even if it does not have the minimal number of edges).

First, it is easy to see that if $ k\equal{}2n\plus{}1$ then we may consider the good graph whose vertices are $ A,B_1, \cdots, B_{2n}$ and edges are $ AB_i$ for all $ i$ and $ B_{2i\minus{}1}B_{2i}$ for $ i\equal{}1,\cdots , n$.
Thus $ f(2n\plus{}1) \leq 3n$.

For $ k\equal{}2n$, for some $ n \geq 2$, we may start from the above good graph with order $ 2n\minus{}1$, consider any of its edges, say $ AB$, and add the vertex $ X$ and edges $ XA$ and $ XB$, to obtain a good graph with order $ 2n$ and $ 3n\minus{}1$ egdes.
Thus $ f(2n) \leq 3n\minus{}1$.

Clearly, we have $ f(3)\equal{}3$ and $ f(4)\equal{}5$.

Moreover, since a good graph is connected, each vertex has degree at leas $ 1$. And since each edge must belong to some triangle, each vertex has degree at least $ 2$.
Note that a graph with $ n$ vertices and less than $ \frac{3n}2$ edges must have a vertex with degree no more than $ 2$. It follows that a good graph with $ n$ vertices and $ f(n)$ edges must have a vertex with degree $ 2$.
Now, we prove the desired minimum by induction on $ n$.

Assume that $ f(k)$ is given by (*) for all $ k \leq 2n$ for some $ n \geq 2$.
Now consider a good graph $ G$ with $ 2n\plus{}1$ vertices and $ f(2n\plus{}1)$ edges. Let $ X$ be a vertex with degree $ 2$, and $ XAB$ a triangle in $ G$.
Assume that $ f(2n\plus{}1) \leq 3n\minus{}1$.
Delete $ X$ and the edges $ XA$ and $ XB$. This leaves a graph $ G'$ with $ 2n$ vertices and no more than $ 3n\minus{}3$ edges. Since $ f(2n) \equal{} 3n\minus{}1$, it means that $ G'$ is not good. But we only destroy one triangle from $ G$ and preserve the connectivity. Thus, that $ G'$ is not good comes from the fact that the edge $ AB$ does not belong to a triangle anymore. Since $ G'$ is connected and have at least three vertices, it means that $ A$ or $ B$ is connected by an edge to some other vertex, say $ AC$ is an edge.
Then, draw edge $ BC$. The resulting graph $ G''$ is good, has $ 2n$ vertices and no more than $ 3n\minus{}2$ edges, contradicting that $ f(2n) \equal{} 3n\minus{}1$.
Thus, $ f(2n\plus{}1) \geq 3n$, which gives $ f(2n\plus{}1)\equal{}3n$.

Now, consider a good graph $ G$ with $ 2n\plus{}2$ vertices and $ f(2n\plus{}2)$ edges. Let $ X$ be a vertex with degree $ 2$, and $ XAB$ a triangle in $ G$.
Assume that $ f(2n\plus{}2) \leq 3n\plus{}1$.
Delete $ X$ and the edges $ XA$ and $ XB$. This leaves a graph $ G'$ with $ 2n\plus{}1$ vertices and no more than $ 3n\minus{}1$ edges. Since $ f(2n\plus{}1) \equal{} 3n$, it means that $ G'$ is not good. But we only destroy one triangle from $ G$ and preserve the connectivity. Thus, as above, that $ G'$ is not good comes from the fact that the edge $ AB$ does not belong to a triangle anymore.
If $ A$ has degree $ 1$ in $ G'$, then delete $ A$ and the edge $ AB$. This leaves a good graph $ G''$ with $ 2n$ vertices and no more than $ 3n\minus{}2$ edges, contradicting that $ f(2n) \equal{} 3n\minus{}1$. Thus $ A$ has degree at least $ 2$ in $ G'$, and similarly $ B$ has degree at least $ 2$ in $ G'$.
If we delete the edge $ AB$, we obtain a graph $ G'''$ with $ 2n$ vertices and no more than $ 3n\minus{}2$ edges. Using that $ f(2n)\equal{}3n\minus{}1$, we deduce that $ G'''$ cannot be good. But, in $ G'''$, each edge belongs to some triangle. Thus $ G'''$ is not connected. Since we have deleted only one edge from a connected graph, it means that $ G'''$ has exactly two connected components, with respective order $ a$ and $ b$. Note that each of these components is a good graph. Thus $ 3n\minus{}2 \geq f(a)\plus{}f(b)$ with $ a\plus{}b \equal{} 2n\plus{}1$.
It follows that $ a$ and $ b$ have opposite parities, say $ a\equal{}2x$ and $ b\equal{}2y\plus{}1$. Then $ x\plus{}y \equal{} n$ and, using the induction hypothesis,
$ 3n\minus{}2 \geq f(a) \plus{}f(b) \equal{} 3x\minus{}1\plus{}3y \equal{} 3n\minus{}1$, a contradiction.
Therefore $ f(2n\plus{}2) \geq 3n\plus{}2$, and we are done.

Pierre.

Since $G$ is a connected graph,$G$ admits a spanning tree on $n$ vertices,name it $T$.Then $T$ has $n-1$ edges.Now, $\forall e \in T$ must be contained in at least one triangle.So $|E(G)|\ge |E(T)|+|E(G-T)| \ge |E(T)|+[\frac{n}{2}]=(n-1)+[\frac{n}{2}]=  [\frac{3n-2}{2}] $.An example is the windmill graph.

The answer is $n - 1 + \lceil\frac{n-1}{2}\rceil$. For the construction, consider a path of length $n-1$ (so it goes through $n$ vertices), then for each vertex an even distance away from the start, connect it with the vertex a distance $2$ from it. Keep doing this until we get to the end.

Assume $G$ had minimal edges. Let $T$ be a spanning tree of $G$, so $T$ has $n-1$ edges. Now, we add in the rest of the edges of $G$ in a way such that every edge connects the endpoints of a cherry. Let $S$ be the set of edges not in a triangle yet, originally $S$ is size $n-1$. Observe that each time we add in an edge, we remove at most $2$ elements from $S$: If we remove more than $2$, then there must be a cycle with that edge as a chord, however this means one of the edges in that cycle was not in $T$, so it must be the endpoints of a cherry. This creates another cycle, so one of those edges must not be in $T$, so it must be the endpoints of a cherry, which creates another cycle. This process must terminate, which means that there is a cycle in $T$, contradiction.

Furthermore, we can always choose edges that are on the endpoints of a cherry. If there was not, and assume $S$ was nonempty (or else we'll be done), consider an edge $e\in S$. $e$ has to be part of a cherry (otherwise it's not connected), but if over all cherries containing $e$, non of the edges joining the endpoints were part of $G$, then for $e$ to be part of a triangle, we need $e$ to be the endpoints of another cherry (which can't be added in yet otherwise it contradicts our earlier claim). This means an extra $2$ edges, however we can remove those two edges from $G$ and add in the edge joining the endpoints of a cherry containing $e$, for a new minimal, a contradiction. Thus, such an edge must exist.

Now, since every time we add an edge, we remove at most $2$ edges from $S$, we need to add at least $\lceil\frac{n-1}{2}\rceil$ edges, for a total of $m = n-1 + \lceil\frac{n-1}{2}\rceil$

Let $T$ be the spanning tree of $G$. We know that it has $n-1$ vertices. It's clear that the number of triangles in $G$ is at least $\lfloor \frac{n}{2}\rfloor$. However, when we add an edge to the spanning tree, a unique cycle is formed. Since there are at least $\lfloor \frac{n}{2}\rfloor$ unique cycles, the number of edges added needs to be at least $\lfloor \frac{n}{2}\rfloor$. Therefore, $m\ge n-1+\lfloor \frac{n}{2}\rfloor$. If $n$ is odd, take the friendship graph on $n$ vertices. If $n$ is even, take the friendship graph on $n-1$ vertices and add in a vertex at the top connected to two vertices in the same "clover".

We claim the answer is $\left\lceil{\frac{3}{2}(n-1)}\right\rceil$. For odd $n$, take the friendship graph on $n$ vertices. For even $n$, take the friendship graph on $n-1$ vertices, and the remaining vertex connected to endpoints of an edge in the friendship graph.

We proceed by strong induction on $n$ with the base case trivial.

Assume the statement for $n=3,4,\ldots,k$. Let $G$ be a graph on $k+1$ vertices such that each edge is contained in at least one triangle. Note that every vertex is contained in at least one triangle. Clearly there must not exist a vertex of degree $1$. Assume there exists a vertex $v$ of degree $2$ (otherwise every vertex has degree at least $3$ so we are done). Let the neighbors of $v$ be $u_1$ and $u_2$. The key is to look at $G\setminus v$. In $G\setminus v$, the only edge which can be not contained in a triangle is $u_1u_2$. Assume this edge is not contained in a triangle (otherwise $|E(G)\setminus v|\geq\left\lceil\frac{3}{2}(k-1)\right\rceil$ so $|E(G)|\geq \left\lceil\frac{3}{2}(k-1)\right\rceil+2\geq\left\lceil\frac{3}{2}((k+1)-1)\right\rceil$). There are two cases.

[b]Case 1: [/b]$k+1$[b] is odd.[/b] Among $u_1$ and $u_2$, there exists a vertex connected to another vertex distinct from $u_1$ and $u_2$. Assume WLOG $u_2$ is connected to $w\neq u_1$. Adding an edge between $u_1$ and $w$ creates a triangle containing $u_1u_2$. Now all edges in $G\setminus v\cup u_1w$ are contained in at least one triangle, so by the induction hypothesis, $|E(G\setminus v)\cup u_1w|\geq\frac{3}{2}k-1$. Thus,
\begin{align*}
|E(G)|&\geq\frac{3}{2}k-2+2\\
&=\left\lceil\frac{3}{2}((k+1)-1)\right\rceil.
\end{align*}

[b]Case 2: [/b]$k+1$[b] is even.[/b] $G\setminus v\setminus u_1u_2$ has at most $2$ connected components and every edge is contained in at least one triangle. If $G\setminus v\setminus u_1u_2$ is connected, it has at least $\frac{3}{2}k-\frac{3}{2}$ edges so
\begin{align*}
|E(G)|&\geq\frac{3}{2}k-\frac{3}{2}+3\\
&=\frac{3}{2}(k+1)\\
&>\left\lceil\frac{3}{2}((k+1)-1)\right\rceil.
\end{align*}
If $G\setminus v\setminus u_1u_2$ is not connected, it has two connected components $G_1$ and $G_2$. Since $|V(G_1)|+|V(G_2)|=k$ is odd, if follows that $|V(G_1)|$ and $|V(G_2)|$ have different parity so $|E(G\setminus v\setminus u_1u_2)|\geq \frac{3}{2}k-\frac{3}{2}-1=\frac{3}{2}k-\frac{5}{2}$. Thus,
\begin{align*}
|E(G)|&\geq\frac{3}{2}k-\frac{5}{2}+3\\
&=\left\lceil\frac{3}{2}((k+1)-1)\right\rceil.
\end{align*}
Thus the induction step is proved. $\square$

Just take a spanning tree!

Let $E_1$ be the set of all edges which are not in a triangle. We start with a spanning tree $T$ of $G$ with $n-1$ edges and imagine adding edges to the graph and thus deleting edges from $E_1$.

[b][color=#f00]Claim.[/b][/color] Each new edge deletes at most two edges from $E_1$.

[i]Proof.[/i] Clear as $T$ has no cycles. $\blacksquare$

Thus we need at least $\lceil 1/2(n-1) \rceil$ edges to be added, hence the bound.

For construction, take the $n-1$-star and connect outer vertices alternately.


Can anyone tell what an auxiliary random variable is ?

In what context ?
How can you expect someone to answer your question when you don't define anything   :roll:

When I read a paper I came across this term. I have understood it now. Thanks.


calculus
integration
derivative
calculus computations

Compute

$ \sum_{n\equal{}0}^{\infty}\frac{1}{e^{n^2}}$

And on a side note, does anyone know where I can find a list of sums on the internet?

$ \approx 1.3863186$

There is almost certainly no closed form.

[quote="jmerry"]
There is almost certainly no closed form.[/quote]

I have seen a closed form for $ \sum_{n\equal{}0}^{\infty}\frac{1}{e^{\pi{n^2}}}$

Those are two completely different numbers.

Alright, I guess I should have realized that they were different enough that the convergence of one to a nice sum does not imply that of the other. The question of tables on the internet of interesting series sums would still be greatly appreciated. Thanks everyone.

There is a book of nothing but tables (by two Russian guys, I don't remember their names) that you can order (not online as far as I know).  It's hundreds of pages of integrals, derivatives, series, and everything you could want.


I have post in topic Brazil's MO but no one reply so I post it again in this topic for easy discuss.
$ f(x)\equal{}x^2\plus{}2007x\plus{}1$
Find the number solution of the equation:
$ f_{n}(x)\equal{}0$

[quote="TTsphn"]I have post in topic Brazil's MO but no one reply so I post it again in this topic for easy discuss.
$ f(x) \equal{} x^2 \plus{} 2007x \plus{} 1$
Find the number solution of the equation:
$ f_{n}(x) \equal{} 0$[/quote]

Please could someone explain what $ f_{n}$ means in this context?

Yes. $ f_{n}(x)\equal{}f(f..f(x)..)$ n time.

Assuming $ f_{1}(x) \equal{} f(x)$ Wouldn't the number of roots be the number of roots of a normal polynomial, so $ 2^{2^{2^{.^{.^{.}}}}}$  $ n$ times or am I missing something? (Which I probably am)

You're missing something since roots may have multiplicity greater than 1.


How many integers $ x$ satisfy $ x^2 \plus{} 4x \plus{} 4 \equal{} 0$?

We factor into $ (x\plus{}2)^2\equal{}0$. Then, $ x$ obviously equals $ \minus{}2$, and there is 1 solution.

You could also note that the quadratic is in the form $ a^2 \plus{} 2ab \plus{} b^2$, and so is a perfect square quadratic, which automatically means it has just $ \boxed{1}$ solution.

[hide="Perfect Square Quadratics"]
There's a lot more info in my blog, click the link in my signature.

When you have a binomial, $ (a \plus{} b)$, and you square it, you will get $ a^2 \plus{} 2ab \plus{} b^2$. We see that $ x^2 \plus{} 4x \plus{} 4 \equal{} 0$ is in that form, when $ a$ and $ b$ equal $ x$ and $ 2$, respectively.
[/hide]


If a, b and c are nonnegative real numbers and $ a\plus{}b\plus{}c\equal{}1$ prove that $ \frac{3}{2}\ge\frac{2a}{a\plus{}1}\plus{}\frac{2b}{b\plus{}1}\plus{}\frac{2c}{c\plus{}1}$.

[quote="SAPOSTO"]If a, b and c are nonnegative real numbers and $ a \plus{} b \plus{} c \equal{} 1$ prove that $ \frac {3}{2}\ge\frac {2a}{a \plus{} 1} \plus{} \frac {2b}{b \plus{} 1} \plus{} \frac {2c}{c \plus{} 1}$.[/quote]
$ \frac{2a}{a\plus{}1} \leq \frac{9a\plus{}1}{8} \Leftrightarrow (3a\minus{}1)^{2} \geq 0$
We have :
$ \sum\frac{2a}{a\plus{}1} \leq \sum \frac{9a\plus{}1}{8} \equal{}\frac{9\plus{}3}{8}\equal{}\frac{3}{2}$
Done :D

$ \frac {3}{2}\ge\frac {2a}{a \plus{} 1} \plus{} \frac {2b}{b \plus{} 1} \plus{} \frac {2c}{c \plus{} 1}$

$ \frac {3}{2}\ge\frac {2a}{a \plus{} a \plus{} b \plus{} c} \plus{} \frac {2b}{b \plus{} a \plus{} b \plus{} c} \plus{} \frac {2c}{c \plus{} a \plus{} b \plus{} c}$

$ \frac {3}{2}\ge\frac {2a}{2a \plus{} b \plus{} c} \plus{} \frac {2b}{2b \plus{} c \plus{} a} \plus{} \frac {2c}{2c \plus{} a \plus{} b}$

$ \frac {3}{2}\ge\frac {2a \plus{} b \plus{} c \minus{} b \minus{} c}{2a \plus{} b \plus{} c} \plus{} \frac {2b \plus{} c \plus{} a \minus{} c \minus{} a}{2b \plus{} c \plus{} a} \plus{} \frac {2c \plus{} a \plus{} b \minus{} a \minus{} b}{2c \plus{} a \plus{} b}$

$ \frac {3}{2}\ge 1 \minus{} \frac {b \plus{} c}{2a \plus{} b \plus{} c} \plus{} 1 \minus{} \frac {c \plus{} a}{2b \plus{} c \plus{} a} \plus{} 1 \minus{} \frac {a \plus{} b}{2c \plus{} a \plus{} b}$

$ \frac {b \plus{} c}{2a \plus{} b \plus{} c} \plus{} \frac {c \plus{} a}{2b \plus{} c \plus{} a} \plus{} \frac {a \plus{} b}{2c \plus{} a \plus{} b}\geq \frac {3}{2}$

Using Cauchy-Schwarz:

$ \left(\sum_{cyc}\frac {b \plus{} c}{2a \plus{} b \plus{} c}\right)\left(\sum_{cyc}(b \plus{} c)(2a \plus{} b \plus{} c)\right)\geq 4 (a \plus{} b \plus{} c)^2$

$ \sum_{cyc}\frac {b \plus{} c}{2a \plus{} b \plus{} c}\geq \frac {4(a \plus{} b \plus{} c)^2}{2(a \plus{} b \plus{} c)^2 \plus{} 2(ab \plus{} bc \plus{} ca)}$

$ \sum_{cyc}\frac {b \plus{} c}{2a \plus{} b \plus{} c}\geq \frac {4(a \plus{} b \plus{} c)^2}{2(a \plus{} b \plus{} c)^2 \plus{} \frac {2(a \plus{} b \plus{} c)^2}{3}} \equal{} \frac {3}{2}$

Done.

hello, simplifying the inequality we have
$ \frac {3 \minus{} a \minus{} b \minus{} c \minus{} 5(ab \plus{} bc \plus{} ac) \minus{} 9abc}{2(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)}\geq0$, we will show that the denominator is not negative.
We have $ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [3]{abc} \Leftrightarrow 9abc\le \frac {1}{3}$. Since
$ 5\left(\frac {1 \minus{} (a^2 \plus{} b^2 \plus{} c^2)}{2}\right) \equal{} 5(ab \plus{} bc \plus{} ac)$  and $ \frac {a \plus{} b \plus{} c}{3} \equal{} \frac {1}{3}\le \sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{3}}$ we get $ 5(ab \plus{} bc \plus{} ac)\le\frac {5}{3}$ and we have
$ 9abc \plus{} 5(ab \plus{} bc \plus{} ac)\le2$ and we are done.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, simplifying the inequality we have
$ \frac {3 \minus{} a \minus{} b \minus{} c \minus{} 5(ab \plus{} bc \plus{} ac) \minus{} 9abc}{21 \plus{} a)(1 \plus{} b)(1 \plus{} c)}\geq0$, we will show that the denominator is not negative.
We have $ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [3]{abc} \Leftrightarrow 9abc\le \frac {1}{3}$. Since
$ 5\left(\frac {1 \minus{} (a^2 \plus{} b^2 \plus{} c^2)}{2}\right) \equal{} 5(ab \plus{} bc \plus{} ac)$  and $ \frac {a \plus{} b \plus{} c}{3} \equal{} \frac {1}{3}\le \sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{3}}$ we get $ 5(ab \plus{} bc \plus{} ac)\le\frac {5}{3}$ and we have
$ 9abc \plus{} 5(ab \plus{} bc \plus{} ac)\le2$ and we are done.
Sonnhard.[/quote]
In the first line i think you made a typo mystake it is $ 2*(a\plus{}1)....$ not $ 21\plus{}a$ :P

hello, thank you for your hint, i have corrected it.
Sonnhard.


[quote="aias"]we have $a>0,b>0,c >0$ we want to prove that
$-\frac{(a-b)^2+(b-c)^2+(c-a)^2}{4}\leq\frac{a^3-b^3}{a+b}+\frac{b^3-c^3}{b+c}+\frac{c^3-a^3}{c+a}\leq\frac{(a-b)^2+(b-c)^2+(c-a)^2}{4}$ [/quote]
Yes, this is true. I got nasty proof.

We have $\frac{a^3-b^3}{a+b}=\frac{(a-b)[(a^2+2ab+b^2)-ab]}{a+b}\ge\frac{(a-b)[(a+b)^2-\frac{(a+b)^2}{4}]}{a+b}=\frac{3(a-b)(a+b)}{4}=\frac{3(a^2-b^2)}{4} \rightarrow\sum \frac{a^3-b^3}{a+b}\ge 0$
and we have $-\frac{\sum (a-b)^2}{4}\leq 0$
and the problem is solved

[quote="arqady"]
If $a<b$ this is false. ;)[/quote]
Oh, you're right. It's a big mistake  :blush:

[quote="silouan"]Because the ineq is symmetric and cyclic we may assume that $a\geq b\geq c>0$.

Then use the fact that for $a\geq b\geq c$ the following ineqs hold.

$a+\frac{b}{2}\geq\frac{a^2+ab+b^2}{a+b}\geq b+\frac{a}{2}$  and the similar

ones. Use them we have the result :)[/quote]
You're wrong :D .We can't assume$a\geq b\geq c>0$
and If we can do that
from this$a+\frac{b}{2}\geq\frac{a^2+ab+b^2}{a+b}$ we will have $a^2-\frac{b^2}{2}-2c^2+\frac{ab+bc+ca}{2}\leq \sum\frac{a^3-b^3}{a+b}$
 ;)

I'm sorry, not this
$a^2-\frac{b^2}{2}-2c^2+\frac{ab+bc+ca}{2}\leq \sum\frac{a^3-b^3}{a+b}$
but this
$2a^2+\frac{b^2}{2}-c^2\ge\sum\frac{a^3+b^3}{a+b}$
And I think we can assume that a>b>c when f(a,b,c)=f(a,c,b)
but in this ineq
$f(a,b,c)=\sum\frac{a^3-b^3}{a+b}$ is different from
$f(a,c,b)=\sum\frac{a^3-c^3}{a+c}$
Then we can only assume $a=max(a,b,c)$
I can only prove
$-\frac{(a-b)^2+(b-c)^2+(c-a)^2}{4}\leq\frac{a^3-b^3}{a+b}+\frac{b^3-c^3}{b+c}+\frac{c^3-a^3}{c+a}$ when $a\geq c\geq b$ and
$\frac{a^3-b^3}{a+b}+\frac{b^3-c^3}{b+c}+\frac{c^3-a^3}{c+a}\leq\frac{(a-b)^2+(b-c)^2+(c-a)^2}{4}$
when $a\geq b\geq c$

My solution is true ;) We can assume $a\geq b\geq c>0$ beacuse the ineq is symmetric.

I continue the solution

After the ineq I used we have.
$\sum\frac{a^3-b^3}{a+b}=\frac{(a-b)(a^2+ab+b^2)}{a+b}+\frac{(b-c)(b^2+bc+c^2)}{b+c}-\frac{(a-c)(a^2+ac+c^2)}{a+c}\geq (a-b)(b+\frac{a}{2})+ (b-c)(c+\frac{b}{2})-(a-c)(a+\frac{c}{2})=-\frac{\sum (a-b)^2}{4}$      ;)

In the same manner we can prove that $\sum\frac{a^3-b^3}{a+b}\leq\frac{\sum (a-b)^2}{4}$

and we are done

And here is another solution. I will show a stronger inequality:

[color=blue][b]Theorem 1.[/b] If a, b, c are three positive numbers, then

$-\frac{\sum\left(b-c\right)^2}{8}\leq\sum\frac{b^3-c^3}{b+c}\leq\frac{\sum\left(b-c\right)^2}{8}$,

where the $\sum$ sign denotes cyclic sums.[/color]

[i]Proof of Theorem 1.[/i] We will only show the right part of this inequality: $\sum\frac{b^3-c^3}{b+c}\leq\frac{\sum\left(b-c\right)^2}{8}$. In fact, once this right part will be proven, then, applying it to the numbers c, b, a instead of a, b, c, we will obtain $\sum\frac{c^3-b^3}{c+b}\leq\frac{\sum\left(c-b\right)^2}{8}$, what rewrites as $-\sum\frac{b^3-c^3}{b+c}\leq\frac{\sum\left(b-c\right)^2}{8}$, or, equivalently, $-\frac{\sum\left(b-c\right)^2}{8}\leq\sum\frac{b^3-c^3}{b+c}$, and thus, the left part of the inequality must also hold. Hence, it is enough to prove the right part inequality $\sum\frac{b^3-c^3}{b+c}\leq\frac{\sum\left(b-c\right)^2}{8}$ only.

Upon multiplication by (b + c) (c + a) (a + b), this inequality becomes

$\sum\left(b^3-c^3\right)\left(c+a\right)\left(a+b\right)\leq\left(b+c\right)\left(c+a\right)\left(a+b\right)\cdot\frac{\sum\left(b-c\right)^2}{8}$.

Now,

$\sum\left(b^3-c^3\right)\left(c+a\right)\left(a+b\right)=\sum b^3\left(c+a\right)\left(a+b\right)-\sum c^3\left(c+a\right)\left(a+b\right)$
$=\sum b^3\left(c+a\right)\left(a+b\right)-\sum b^3\left(b+c\right)\left(c+a\right)=\sum b^3\left(c+a\right)\left(\left(a+b\right)-\left(b+c\right)\right)$
$=\sum b^3\left(c+a\right)\left(a-c\right)=\sum b^3\left(a^2-c^2\right)$
$=b^3\left(a^2-c^2\right)+c^3\left(b^2-a^2\right)+a^3\left(c^2-b^2\right)$
$=b^3\left(\left(a^2-b^2\right)+\left(b^2-c^2\right)\right)-c^3\left(a^2-b^2\right)-a^3\left(b^2-c^2\right)$
$=\left(b^3-c^3\right)\left(a^2-b^2\right)-\left(a^3-b^3\right)\left(b^2-c^2\right)$
$=\left(b-c\right)\left(b^2+bc+c^2\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(a^2+ab+b^2\right)\left(b-c\right)\left(b+c\right)$
$=\left(b-c\right)\left(a-b\right)\left(\left(b^2+bc+c^2\right)\left(a+b\right)-\left(a^2+ab+b^2\right)\left(b+c\right)\right)$
$=\left(b-c\right)\left(a-b\right)\left(c-a\right)\left(bc+ca+ab\right)=\left(b-c\right)\left(c-a\right)\left(a-b\right)\left(bc+ca+ab\right)$.

Hence, the inequality in question is equivalent to

$\left(b-c\right)\left(c-a\right)\left(a-b\right)\left(bc+ca+ab\right)\leq\left(b+c\right)\left(c+a\right)\left(a+b\right)\cdot\frac{\sum\left(b-c\right)^2}{8}$
$\Longleftrightarrow\ \ \ \ \ \left(b-c\right)\left(c-a\right)\left(a-b\right)\leq\frac{\left(b+c\right)\left(c+a\right)\left(a+b\right)}{bc+ca+ab}\cdot\frac{\sum\left(b-c\right)^2}{8}$.

Now, since the inequality in question is cyclic, we can WLOG assume that a is the smallest among the numbers a, b, c. Then, $a\leq b$ and $a\leq c$, so that $b-a\geq 0$ and $c-a\geq 0$. Denote b - a = x and c - a = y; then, x and y are two nonnegative reals.

Now,

$\left(b+c\right)\left(c+a\right)\left(a+b\right)=\left(b+c\right)\left(bc+ca+ab+a^2\right)\geq\left(b+c\right)\left(bc+ca+ab\right)$,

so that $\frac{\left(b+c\right)\left(c+a\right)\left(a+b\right)}{bc+ca+ab}\geq b+c\geq b+c-2a$. Thus, instead of proving the inequality $\left(b-c\right)\left(c-a\right)\left(a-b\right)\leq\frac{\left(b+c\right)\left(c+a\right)\left(a+b\right)}{bc+ca+ab}\cdot\frac{\sum\left(b-c\right)^2}{8}$, it will be enough to show the stronger inequality

$\left(b-c\right)\left(c-a\right)\left(a-b\right)\leq\left(b+c-2a\right)\cdot\frac{\sum\left(b-c\right)^2}{8}$,

i. e.

$\left(b-c\right)\left(c-a\right)\left(a-b\right)\leq\left(b+c-2a\right)\cdot\frac{\left(b-c\right)^2+\left(c-a\right)^2+\left(a-b\right)^2}{8}$,

Now, b - c = (b - a) - (c - a) = x - y, c - a = y, a - b = - (b - a) = -x and b + c - 2a = (b - a) + (c - a) = x + y. Thus, this inequality becomes

$\left(x-y\right)y\left(-x\right)\leq\left(x+y\right)\cdot\frac{\left(x-y\right)^2+y^2+\left(-x\right)^2}{8}$
$\Longleftrightarrow\ \ \ \ \ -\left(x-y\right)xy\leq\left(x+y\right)\cdot\frac{x^2-xy+y^2}{4}$
$\Longleftrightarrow\ \ \ \ \ \left(x+y\right)\left(x^2-xy+y^2\right)+4\left(x-y\right)xy\geq 0$
$\Longleftrightarrow\ \ \ \ \ x^3+4x^2y-4xy^2+y^3\geq 0$.

But this follows from

$x^3+4x^2y-4xy^2+y^3=\frac{\left(299-112\sqrt7\right)y^3+\left(3x+4\left(\sqrt7+1\right)y\right)\left(3x-2\left(\sqrt7-2\right)y\right)^2}{27}$
$\geq\frac{\left(299-112\sqrt7\right)y^3}{27}\geq 0$.

This proves Theorem 1.

  Darij

ROFL, darij, what did you do?

$x^3+4x^2y-4xy^2+y^3 = x^3 + y(2x-y)^2\geq 0$... :D :D


inequalities function logarithms calculus

if $ {a,b,c}\in\mathbb{N}$
prove that:
$ a^{\frac {a}{a \plus{} b \plus{} c}}\times b^{\frac {b}{a \plus{} b \plus{} c}}\times c^{\frac {c}{a \plus{} b \plus{} c}}\ge\frac {1}{3}(a \plus{} b \plus{} c)$

In general prove or disprove
$ a^{\frac {a}{a \plus{} b \plus{} c\cdots}}\times b^{\frac {b}{a \plus{} b \plus{} c\cdots}}\times c^{\frac {c}{a \plus{} b \plus{} c\cdots}} \cdots\ge\frac {1}{3\cdots}(a \plus{} b \plus{} c \plus{} \cdots)$

where $ 3\cdots$ means the number of base numbers used in the left hand side of the inequality.

Are you sure you want $ a,b,c$ to be in $ \mathbb N$¿

And in the second one, not only the $ 3$ must be changed, the exponents are wrong, too.

yes the exponents are wrong thanks for telling , i forgot to put $ \cdots$
$ a,b,c$ are positive integers. isnt that $ \mathbb{N}$?

The function $ f(x)\equal{}x\ln x$ is convex in $ \mathbb R^\plus{}$, since $ f''(x)\equal{}\frac 1x>0$. Thus from Jensen's inequality
\[ \frac{a\ln a\plus{}b\ln b\plus{}c\ln c}{3}\ge \frac{a\plus{}b\plus{}c}{3}\ln \left(\frac{a\plus{}b\plus{}c}{3}\right)\\ \implies a^ab^bc^c\ge\left(\frac{a\plus{}b\plus{}c}{3}\right)^{a\plus{}b\plus{}c}\]
which was what we wanted. The generalization follows similarly.

[quote="binomial_4eva"]$ a,b,c$ are positive integers. isnt that $ \mathbb{N}$?[/quote]

Some authors use $ \mathbb{N}$ for the nonnegative integers (a usage I personally prefer).  To avoid ambiguity, you can either write out what you mean ("positive integers" or "nonnegative integers") or use symbols like $ \mathbb{Z}_{ > 0}$ and $ \mathbb{Z}_{\geq 0}$.  (I like this last solution the best, since it has no ambiguity whatsoever.)

Nayel can you explain more please, sorry i didnt get it. i dont get things till i undrestand all steps.what is convex?

Let $ f(x) \equal{} x\ln x$. We have $ f'(x) \equal{} x\frac {d}{dx}(\ln x) \plus{} \ln x\frac {d}{dx}(x) \equal{} 1 \plus{} \ln x$. Thus $ f''(x) \equal{} 0 \plus{} \frac 1x > 0$. Hence $ f$ is convex. Therefore by Jensen's inequality we get
\[ \qquad\frac {f(a) \plus{} f(b) \plus{} f(c)}{3}\ge f\left(\frac {a \plus{} b \plus{} c}{3}\right) \\
\implies\frac {a\ln a \plus{} b\ln b \plus{} c\ln c}{3}\ge \frac {a \plus{} b \plus{} c}{3}\ln\left(\frac {a \plus{} b \plus{} c}{3}\right) \\
\implies \ln a^a \plus{} \ln b^b \plus{} \ln c^c\ge (a \plus{} b \plus{} c)\ln\left(\frac {a \plus{} b \plus{} c}{3}\right) \\
\implies \ln a^ab^bc^c\ge \ln\left(\frac {a \plus{} b \plus{} c}{3}\right)^{a \plus{} b \plus{} c} \\
\implies a^ab^bc^c\ge \left(\frac {a \plus{} b \plus{} c}{3}\right)^{a \plus{} b \plus{} c}
\]
[hide="Here the following were used"]
$ 1.$ $ \frac {d}{dx}(uv) \equal{} u\frac {d}{dx}(v) \plus{} v\frac {d}{dx}(u)$

$ 2.$ $ p\ln (k) \equal{} \ln (k^p)$

$ 3.$ $ \ln(m) \plus{} \ln(n) \plus{} \ln(k) \equal{} \ln(mnk)$[/hide]
I tried my best to clarify my solution..

[b]bionomial_4eva:[/b] [i]"What is convex?"[/i]

If a function $ f(x)$ is convex over the interval $ [a,b]$, then for all $ c,d \in [a,b], \; c \neq d$, $ \frac{f(c)\plus{}f(d)}{2} > f \left( \frac{c\plus{}d}{2} \right)$. In other words, the average value of the function is greater than the function at the average value. Examples of convex functions include $ x^2$, $ b^x \; \forall b \in \mathbb{R} ^\plus{}$, $ x^3$ for positive $ x$, and $ \frac{1}{x}$.

If a function $ g(x)$ is concave over the interval $ [a,b]$, then instead, $ \frac{f(c)\plus{}f(d)}{2} < f \left( \frac{c\plus{}d}{2} \right)$. Examples of concave functions include $ \sqrt{x}$, $ \log x$, and $ \minus{}x^2$.

Jensen's inequality is a generalization of this property.

In his solution, nayel used calculus to determine that the relevant functions were convex, and then applied Jensen's inequality.

@binomial_4eva: inequalities normally tend to be true for some set of real numbers (see the prove above, nowhere was used that they are integers,positive reals suffices).

@The Zuton Force: This definition is dangerous. The normal one is: For every $ \lambda \in [0,1]$ we have $ \lambda f(c)\plus{}(1\minus{}\lambda)f(d) \geq f \left( \lambda c\plus{}(1\minus{}\lambda d\right)$. You need to allow equality, and you need that paramter $ \lambda$ to be allowed to be different from $ \frac 12$.
Same for concave.


Aren't ANY of you nerds Biology hardcores?
Everyone's ranting on and on about math and physics and chemistry. Would like to know others' situation in bio.

There used to be one- rituraj, buts he's enjoying himself in AIIMS these days :P
I love bio somewhat, but I'm not a hardcore. Maybe I'll take up bio again in college :)

hi amenda..i guess i kno u???!!
atleast sm of ur school mates r gud frnds of mine :)

do tell hows ur performance in bio olymp this yr..best of luck 4 IBO ...

for ne further guidance u cn certainly contact me  :D

@amendadavis: How did you do the NSEB. Whats ur score?
What do you think will the cut off be?

Hi..I am a new member. I got 160.5/300 in the inbo 2010. last year's cut-off was 165.75 but I really think the paper was tougher this time. Anybody else got the performance cards?

Hello Radha
and good luck for INBO-2010 result
keep ur fingers crossed

@radha
Good luck. All the best for the result.


What is the maximum number of square centimeters in the area of a triangle whose perimeter is 30 cm? Express your answer in simplest radical form.

ummm...
largest area when equilateral

so each side is 10

use formula

A=(s^2)(sqrt3)/4
=25sqrt3


Let $ a,b,c$ be real numbers satisfying $ a \plus{} b \plus{} c \equal{} \frac {9}{4}$. Find the max value $ S \equal{} (a \plus{} \sqrt {a^2 \plus{} 1})^b \plus{} (b \plus{} \sqrt {b^2 \plus{} 1})^c \plus{} (c \plus{} \sqrt {c^2 \plus{} 1})^a$

thaithuan_GC, You should read this.[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=168836[/url]

This should be locked.

Don't post any comments for the problem.

Japanese Communities Moderator

kunny

Augh! Sorry! I didn't know about that. I'll remove my solution.


trigonometry
ratio
MATHCOUNTS
geometry
trig identities
Law of Sines

In triangle ABC, we have angle A=43 degrees, B=93 degrees and a point D on AB such that AD*DB=BC^2.  FInd the number of degrees in angle ADC.

Can u do it by using laws of sines and ratios for all three triangles?

I dont know about that method (never tried it that way), but there is an easier method.

[hide="hint"]
Power of a Point

I might do the problem later...
[/hide]

I don't think power of point would help unless usaha meant DC^2 instead of BC^2, which I doubt.

I can't find a neat way of doing this without trig right now.

I am not sure if I understood the problem correctly. Using the law of sines, I can represent the length of $AB$ as $n*sin44$ for some value $n$ and $BC$ as $n*sin43$. The maximum value of $AD*DB$ is if $D$ is the midpoint of $AB$, which would make $AD*DB$ $[(n*sin44)^{2}]/4$. That is obviously less than $(n*sin43)^{2}$. But that value was the maximum, so doesn't that mean it is impossible for $AD*DB$ to equal $BC^{2}$?

It should be even easier to see this just by making a rough sketch.

I'll bet that he meant DC^{2}.

BTW, bionunquineist, you are not allowed to use trig in the MathCounts Forum.

And yes, I tried this with trig and I found it to be impossible, so something is wrong with the problem.

But it isn't obviously true, but whatever.

[quote="ragnarok23"]I'm pretty sure there was almost an exact problem like this in the GS marathon.[/quote]
What's GS?
EDIT:  :I_Feel_Stupid: Getting Started

usaha, could you please give us a hint, if this problem actually has a solution.

[hide] :?: Draw a circle such that BC is tangent to it at C and BA passes through the circle at D. You have <CAD = 43 and is an inscribed angle, so arc CD = 43*2 or 86 degrees. <BCD is half of arc CD so it is also 43. <ABC = 180-93-43 = 44 so arc <ACD = 1. <ADC = 136.[/hide]

But if the problem is wrong then I don't know... ?

please don't revive very old topics

thx

jorian

There hasn't been a valid answer or a reason why it's impossible, so this topic is still valid.

[quote="jhredsox"]it is?

what about the 3 weeks rule?

jorian[/quote]

Well... Once upon a time I accidentally revived a topic that was about a year old. I was looking at "unanswered posts", and forgot to look at the date of fthe topic. I got yelled at by a couple people, but then someone said that if a problem hasn't been solved, it's still open.

The exact question escapes me, but it had to do with a frustum.

So yeah, I bet 3 weeks only applies to problems that have already been solved.

I'm not a moderator, though.


Στις δημοτικές εκλογές μιας μικρής πόλης υπάρχουν δύο υποψήφιοι, ο Α και ο Β.Γνωρίζουμε ότι ο Α θα πάρει συνολικά α ψήφους και ότι ο Β θα πάρει συνολικά β ψήφους, με [i][b]α>β.[/b][/i]Να βρεθεί η πιθανότητα που έχει ένας τρόπος καταμέτρησης των ψήφων, ώστε καθ'όλη τη διάρκεια της καταμέτρησης και [i][b]κάθε χρονική στιγμή [/b][/i]ο Α να προηγείται του Β.

ΑΠΟ ΠΟΥ ΤΗΝ ΒΡΗΚΕΣ ΑΥΤΗ? :o  ΑΠΟ ΤΟ ΒΙΒΛΙΟ ΤΟΥ ΜΠΟΛΗ ΣΕΛ 152 ΠΡΟΒΛΗΜΑ 3 ?  :rotfl:  :rotfl: ΔΙΟΤΙ ΜΟΝΟ ΑΠΟ ΕΚΕΙ ΘΑ ΜΠΟΡΟΥΣΕ ΝΑ ΜΠΕΙ ΤΕΤΟΙΑ ΑΣΚΗΣΗ ΤΕΤΟΙΑ ΜΟΥ ΜΥΡΙΖΕΙ
ΜΠΑΙ

Όχι Αμβρόσιε δεν είναι από το βιβλίο του Μπόλη, εξάλλου δεν το έχω (αλλά επειδή καλό θα ήταν να το πάρω πες μου τον τίτλο-μην ψάχνω άδικα...)
Βέβαια η άσκηση είναι κλασική και δεν αντιλέγω- μπορεί να υπάρχει σε οποιοδήποτε βιβλίο.
Πως σου φαίνεται το πρόβλημα ?

ΚΑΛΟ ΕΙΝΑΙ ΟΜΩΣ ΤΟ ΒΙΒΛΙΟ ΤΟΥ ΜΠΟΛΗ ΔΕΝ ΤΟ ΕΧΩ ΔΕΝ ΤΟΞΕΡΩ ΑΠΛΩΣ ΤΟ ΕΙΠΑ ΓΙΑ ΠΛΑΚΑ ΟΚ?
ΧΑΙΡΕΤΑΙ

Η άσκηση μου θυμίζει μία άλλη με έναν ταμία σε ένα θέατρο που είχε στη διάθεσή του ν χαρτονομίσματα των 5€ και το εισιτήριο κόστιζε 15€.
Υπάρχουν μ θεατές (σε σύνολο κ θεατών) που έχουν χαρτονόμισμα των 5€.

Και ζητούσε να βρούμε ποια είναι η πιθανότητα να μην υπάρξει πρόβλημα, δηλαδή να μην ξεμείνει ο ταμίας από 5ευρα.

Για να βρεις τη λύση, σχεδίαζες μία γραφική παράσταση που έμοιαζε με τυχαίο περίπατο και αυτό που ζητούσες ήταν να μην πέσει η γραφική παράσταση κάτω από συγκεκριμένη ευθεία y=c.

Αλλά μετά δε θυμάμαι πως προχωράς ..  :(

Πόντιε, οφείλω να ομολογήσω ότι το προσεγγίζεις....(έχεις νομίζω ένα στοιχείο της λύσης).
Για ιστορικούς λόγους αναφέρω ότι το πρόβλημα αυτό τέθηκε σε ένα μεγάλο Μαθηματικό (μου διαφεύγει το όνομά του..) ως open-unsolved problem...Αυτός βρήκε μία καταπληκτική λύση και τους εντυπωσίασε!!!!

Ύστερα από αρκετές πράξεις σε μια όχι και τόσο κομψή προσέγγιση της άσκησης
κατέληξα στον τύπο:
$P(a,b)=\frac{a}{a+b}* \frac{a-1}{a+b-1}* \left[1-\frac{\binom{a+b-2}{a}}{\binom{a+b-2}{b}}\right]==\frac{a-b}{a+b}$

 ,όπου $P(a,b)$ είναι η ζητούμενη πιθανότητα.
Ελπίζω να είναι και ο σωστός για να διευκρινίσω και τις λεπτομέρειες της λύσης!

Στέλιος... :)

Μπράβο ρε Στέλιο!!!Πάρα πολύ καλά-άψογα θα έλεγα!!!Ρε είδες, παρόλο που έκανες πολλές πράξεις κατέληξες σε εντυπωσιακό αποτέλεσμα.....
Για πες λεπτομέρειες...
Θα έρθεις Αθήνα τη μεγάλη εβδομάδα???

Χλωμό το κόβω να έρθω Αθήνα τη μεγάλη βδομάδα  :(   .Ελπίζω να μη χάσω κάτι καλό ….. :!: Ίσως όμως έρθω την επόμενη βδομάδα.

Όσον αφορά την άσκηση τώρα έχουμε ότι για να ικανοποιηθεί το ζητούμενο οι δυο πρώτες ψήφοι πρέπει να είναι του Α. Η πιθανότητα ,η πρώτη ψήφος να είναι του Α είναι $\frac{a}{a+b}$ (1).Η πιθανότητα να είναι και η δεύτερη ψήφος του Α είναι $\frac{a-1}{a+b-1}$ (2). Στη συνέχεια θα πρέπει ο Β να προηγείται συνεχώς του Α κατά το πολύ μια ψήφο και αρκεί να βρούμε την πιθανότητα να συμβεί αυτό. Για το τελευταίο αρκεί να βρούμε την πιθανότητα ο Β να προηγείται τουλάχιστον μια στιγμή κατά τουλάχιστον δυο ψήφους (στις ψήφους που απομένουν) και να την αφαιρέσουμε από την μονάδα. Αυτή όμως η πιθανότητα είναι $\frac{\binom{a+b-2}{a}}{\binom{a+b-2}{b}}$ (3),αφού οι ψήφοι που απομένουν να ελεγχθούν είναι $a+b-2$ και όλες οι δυνατές περιπτώσεις για τη σειρά αυτών των ψήφων είναι $\binom{a+b-2}{b}$. Όλη τώρα η δουλειά βρίσκεται στο να βρούμε τον παρονομαστή της πιθανότητας (3) τον οποίο αν τον ξες(*) αποδεικνύεται εύκολα με επαγωγή στο b:
-Για b=2 η μόνη ευνοϊκή σειρά ψήφων είναι BBAAAAA….. και πράγματι $\binom{a+2-2}{a}=1$ ,
-έστω ότι ισχύει για κάποιο b, αρκεί να αποδείξουμε ότι ισχύει και για b+1,
-για b+1 όμως οι ευνοϊκές περιπτώσεις είναι όσες και για b συν $\binom{a+b-2}{b-1}$ [,αφού η επιπλέον ψήφος μπορεί να μπει στις υπόλοιπες περιπτώσεις που ο Β προηγείται για ακριβώς μια ψήφο(και ποτέ περισσότερο) κάποια στιγμή και να τις κάνει ευνοϊκές],δηλαδή $\binom{a+b-2}{a}+\binom{a+b-2}{b-1}=\binom{a+b-2}{a}+\binom{a+b-2}{a-1}=\binom{a+b-1}{a}=\binom{a+(b+1)-2}{a}$
και το ζητούμενο αποδείχτηκε.

Το μόνο που έχουμε να κάνουμε τώρα είναι να πολλαπλασιάσουμε τις πιθανότητες (1),(2) και την συμπληρωματική της (3) ,οπότε βγαίνει:

$P(a,b)=\frac{a}{a+b}* \frac{a-1}{a+b-1}* \left[1-\frac{\binom{a+b-2}{a}}{\binom{a+b-2}{b}}\right]=\frac{a-b}{a+b}$

,όπου $P(a,b)$ είναι η ζητούμενη πιθανότητα.

(*)Ο τρόπος για να βρεις τον τύπο είναι ο πλέον γνωστός. Κάνεις υποθέσεις βρίσκοντάς τον για μικρές τιμές και στη συνέχεια προσπαθείς να τον αποδείξεις για όλες τις τιμές ;).

Νίκο για πες τώρα και την καταπληκτική λύση του ‘μεγάλου’ μαθηματικού, γιατί τη δικιά μου δεν θα την έλεγα και τόσο κομψή :roll: .

Στέλιος… :)

basika basizetai stin arxi tis anaklasis, pou efarmostike gia proti fora apo to gallo  mathimatiko Andre. H methodos auti epiluei para polla problimata sindiastikis. Basika i idea einai me xrisi kigklidomaton kai ermineia tou problimatos ekei. Tote efarmozoume mia simetria (anaklasi) os pros ti y=x kai amesa exoume to zitoumeno... xoris katholou kopo kai skepsi....

Wraios!Den skeftika na to prossegiso etsi to thema....

.

Νίκο, μπορείς να περιγράψεις λίγο αναλυτικότερα τη λύση που προτείνεις με την ανάκλαση;

Λουκάς

Pontie

 αυτά διάβαζα πριν απο λίγες ημέρες και  μιας και Νίκος είμαι και εγώ  , απαντώ εγώ .
 Δές τη λύση στο συνημμένο αρχείο .

 Νίκο Ράπανε τι λές ?

Γεια σου Ρωμανίδη!!!
Η λύση σου είναι ακριβώς ίδια με τη δικιά μου!!Πιστεύω ότι συμφωνούν όλοι ως προς την κομψότητα-είναι προφανές!!
Επίσης σε ευχαριστώ γιατί με έβγαλες από τον κόπο να κάτσω να γράψω τη λύση - και δεν ξέρω και latex....
Tώρα θα  γίνω ενεργό μέλος του φορουμ από 3 Μαίου και μετά οπότε και θέλω να θέσω ενα πολύ όμορφο πρόβλημα!
Bye


[hide]
445-157=288, he has 288 more pages to read.
288/24=$\framebox{12}$[/hide]

[quote="math92"]Bob is reading a 445 page book. He has already read 157 pages. If he reads 24 pages a day, how long will it take him to finish the book?[/quote]

[hide]Since he has already read 157 pages, he has 445-157=288 more pages to read.

It will take him 228/24=12 days to finish the book[/hide]

[quote="math92"]Bob is reading a 445 page book. He has already read 157 pages. If he reads 24 pages a day, how long will it take him to finish the book?[/quote]
[hide]
445-157=288
288/24=12 days
[/hide]

[hide]he has already read $157$ pages $\rightarrow $ \[445-157=288\] $\rightarrow $ he should read $ 288$ pages to finish the book and he reads $24$ pages a day $\rightarrow$ he will finish the book in
$\frac{288}{24}$ days  $\rightarrow $ he will finish the book in $12$ days. [/hide]


Hi, a question on AIME 1997 number 6:

link to problem:
http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/1997-aa/97aimequest/97aimque.htm

This seems to be a straight forward problem, but Im having trouble. Can someone provide a solution? Thanks.

This AIME is posted on the Resources page: http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1997

If you are in AoPS format, you can get it by clicking "Contests" above.
If you are in Mathlinks format, you can click "Resources".

I should leave this discussion to jmerry.  :)

But, anyway. Let's work in degrees.

$\angle A_nA_1A_2=180-\frac{360}n.$

We add 60 degrees to that to get $\angle A_nA_1B.$

There are now two possibilities: Either $\angle A_nA_1B=240-\frac{360}n$, or

$\angle A_nA_1B=360-\left(240-\frac{360}n\right)=120+\frac{360}n.$

Suppose $A_n, A_1,$ and $B$ are consecutive vertices of a regular $m$-gon. Then $\angle A_nA_1B=180-\frac{360}m.$ We have two cases:

Case 1:

$240-\frac{360}n=180-\frac{360}m.$ Then

$\frac1m-\frac1n=\frac16.$ Multiply this out to get

$nm-6m+6n=0.$

$nm+6m+6n-36=-36.$

$(m+6)(n-6)=-36.$

$m+6$ has to be at least $9.$ Possible factorizations of $-36$ are as $9(-4),\,12(-3),\,18(-2),$ and $36(-1).$ This gives us the possible combinations of

$m=3,\,n=2,$ but that doesn't count; $m$ must be at least $3.$
$m=6,\,n=3$
$m=12,\,n=4$
$m=30,\,n=5.$

Case 2:

$120+\frac{360}n=180-\frac{360}m$

This leads to $\frac16=\frac1n+\frac1m$

$nm-6m-6n=0$

$nm-6m-6n+36=36$

$(m-6)(n-6)=36.$

Now we're looking for positive factors of $36.$

$36=6\cdot6\,\implies\,m=12,\,n=12$
$36=4\cdot9\,\implies\,m=10,\,n=15$
$36=3\cdot12\,\implies\,m=9,\,n=18$
$36=2\cdot18\,\implies\,m=8,\,n=24$
$36=1\cdot36\,\implies\,m=7,\,n=42$

So the answer to the problem is $n=42.$

Work out what the angles of a regular 7-gon and regular 42-gon must be and you'll see that this works.

I see this as a number theory problem.


[size=167]Mock AIME 3 for 2007[/size]

I am composing another mock aime. The test will be held from [b]10/15/06[/b] to [b]10/22/06[/b] (note date edited from before).

If anybody is interested, please post in this topic. Also I would like somebody to help me check over my problems. For anybody who complained about the difficulty of my previous test, this one should be more approachable :wink: . As always, the rules are standard AIME conditions.

um i guess im in....
why was this posted in the w00t board too?

I'll do it. Although I suck at AIME stuff, no need to make it any easier than the last one...

:jump:  :wow: another AIME ... I'll do it

lol it said:
The message is too small. Please make the message longer before submitting.
when i said "ill try"
anyways, im gonna do it.

Hmm, been waiting to try an AIME for a bit, could be a good oppurtunity  :D

I suppose the statistics on scores from the WOOT class and from the AMC forum will be combined?

i believe i posted in the other thread, but i'll post here too for yes...

I'm in need of all the practice I can get
I'll take it

I'll take it, assuming I have a 3 hour chunk of time during the week...

i will try it but it sounds hard my bro didnt make this last year but in mathcounts he made state

I will try if schoolwork doesn't get into the way.  :o

I will do it.  But don't make it TOO easy, I want a good indicator of my real AIME score (assuming I make AIME with this new scoring system...).

Also, AdorableAddict: have I seen you on the Admissionchances website?

say, isnt the test supposed to be posted today?

yea...sorry, i had a tough hw day, i am posting it now...


Given are $\triangle ABC$ with three internal bisector $AA',BB',CC'$ prove that  $\triangle ABC$ is equilateral triangle $\leftrightarrow (b+c)a^{3}\vec{AA'}+(c+a)b^{3}\vec{BB'}(a+b)c^{3}\vec{CC'}=\vec{0}$

Change $\vec{AA'}= \frac{h_{a}}{r}\vec{AI}$. Now it is easy.

ok ! Another problem:
Given $\triangle ABC$ with $AA',BB',CC'$ is three altitude, prove that
$\triangle ABC$ is equilateral $\leftrightarrow a^{4}\vec{AA'}+b^{4}\vec{BB'}+c^{4}\vec{CC'}=\vec{0}$


geometry
circumcircle
incenter
projective geometry
geometry proposed

Incircle of $ \triangle ABC$ tangents sides $ AC,BC,AB$ at $ D,E,F$respectively. Lines $ EF , FD$ intersects $ AC, BC$ at $ P,Q$ respectively. Let $ M,N$ are midpoints of $ DP$and $ EQ$. Prove that if $ O,I$ are circumcenter and incenter of $ \triangle ABC$ then  $ OI$ is perpendicular to $ MN$

See $ 6^{th}$ quote of http://www.mathlinks.ro/viewtopic.php?t=158254

Let us to change a little the notations of the problem, as follows:

[b][size=100][color=DarkBlue]The incircle $ (I),$ of a given triangle $ \bigtriangleup ABC,$ tangents to its sides $ BC,$ $ AC,$ $ AB,$ at points $ D,$ $ E,$ $ F,$ respectively. The line segments $ EF,$ $ DF,$ intersect $ BC,$ $ AC,$ at points $ P,$ $ Q,$ respectively and let $ M,$ $ N$ be, the midpoints of $ PD,$ $ QE,$ respectively. Prove that $ OI\perp MN,$ where $ O,$ $ I,$ are the circumcenter and incenter of $ \bigtriangleup ABC$ [/color][/size][/b].

It is enough to prove that $ (OM)^{2} \minus{} (ON)^{2} \equal{} (IM)^{2} \minus{} (IN)^{2}.$

Because of the line segments $ AD,$ $ BE,$ $ CF,$ are concurrent at one point, the [b][size=100]Gergone point[/size][/b] as well, so be it $ G_{e},$ we have that the triangle $ DEF,$ is the [b][size=100]Cevian triangle[/size][/b] of $ \bigtriangleup ABC,$ with respect to $ G_{e}$ and so, we conclude that the point $ P,$ is the harmonic conjugate of $ D,$ with respect to the points $ B,$ $ C.$

Similarly the point $ Q,$ is the harmonic conjugate of $ E,$ with respect to the points $ A,$ $ C.$

So, based on the [b][size=100]Newton theorem[/size][/b], we have that $ (MD)^{2} \equal{} (MB)\cdot (MC)$

$ \Longrightarrow$ $ (IM)^{2} \minus{} (ID)^{2} \equal{} (MB)\cdot (MC)$ $ ,(1)$

Similarly we have that $ (NE)^{2} \equal{} (NA)\cdot (NC)$ $ \Longrightarrow$ $ (IN)^{2} \minus{} (IE)^{2} \equal{} (NA)\cdot (NC)$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ (IM)^{2} \minus{} (IN)^{2} \equal{} (MB)\cdot (MC) \minus{} (NA)\cdot (NC)$ $ ,(3)$

$ \bullet$ Based now, on the [b][size=100]Second medians theorem[/size][/b], from the triangle $ \bigtriangleup OMC$we have,

$ (OM)^{2} \minus{} (OC)^{2} \equal{} 2(KX)\cdot (MC)$ $ ,(4)$ where $ K,$ $ X,$ are the midpoints of the segments $ BC,$ $ MC,$ respectively.

Similarly, from the triangle $ \bigtriangleup ONC,$ we have $ (ON)^{2} \minus{} (OC)^{2} \equal{} 2(LY)\cdot (NC)$ $ ,(5)$

where $ L,$ $ Y,$ are the midpoints of the segments $ AC,$ $ NC,$ respectively.

From $ (4),$ $ (5)$ $ \Longrightarrow$ $ (OM)^{2} \minus{} (ON)^{2} \equal{} 2(KX)\cdot (MC) \minus{} 2(LY)\cdot (NC)$ $ ,(6)$

From $ (3),$ $ (6)$ and because of $ MB \equal{} 2(KX)$ and $ NA \equal{} 2(LY)$ (easy to prove),

we conclude that $ (OM)^{2} \minus{} (ON)^{2} \equal{} (IM)^{2} \minus{} (IN)^{2}$ and the proof is completed.

Kostas Vittas.

[quote=darij grinberg]Hint to a different solution:
- The circumcircle of triangle $ ABC$ is orthogonal to the circles with diameters $ DP$ and $ EQ$.
- The incircle of triangle $ ABC$ is orthogonal to the circles with diameters $ DP$ and $ EQ$.
(The second assertion is almost obvious. The first follows from the theory of Apollonian circles - note that $ D$ and $ P$ are harmonically conjugate with respect to the segment $ BC$, and so on.)

  darij[/quote]

I'm done the first and second assertion but I don't know how to  do next.

I finish this question.
Thank you


algorithm
invariant
algebra
polynomial
number theory
Galois Theory
superior algebra

Hi all

I was wondering if anyone out there could direct me to a text (either book or electronic) that explains how programs like SAGE / MAGMA / etc. calculate Galois Groups of polynomials. I looked through Cohen's "Course in Computational Number Theory" but all he appeared to have were basic algorithms for the case of degree 7 or less. I know in general you use the fact that the galois groups of $ f \in \mathbb{Z} / p\mathbb{Z}[x]$ embed into the GG over $ \mathbb{Q}$ but I'm looking for something a little more specific, that deals with cases such as when just knowing the size and cycle decomposition of the group is not enough to differentiate between possible Galois Groups.

Thanks

The algorithm in GAP is fairly similar to the one described in:

Soicher, Leonard; McKay, John. "Computing Galois groups over the rationals."
J. Number Theory 20 (1985), no. 3, 273-281.
http://www.ams.org/mathscinet-getitem?mr=797178
http://dx.doi.org/10.1016/0022-314X(85)90022-8

which I suspect is also quit similar to the one in Cohen's text.  The techniques work for larger degrees.  The author of the Galois identification implementation in GAP has a pretty nice introduction to the algorithms used:

Hulpke, Alexander. "Techniques for the computation of Galois groups."
Algorithmic algebra and number theory (Heidelberg, 1997), 65-77, Springer, Berlin, 1999.
http://www.ams.org/mathscinet-getitem?mr=1672101
http://www.math.colostate.edu/~hulpke/paper/gov.pdf

Roughly speaking the main extra information you get is the cycle type information for the action of the Galois group on sets and sequences of roots too.  This is done using rings of invariant polynomials (in several variables).


Do any one know where I can find problems for McNabb Mathematics Competition?

You should post this somewhere else...I think...

i do't think that theres anything about this on the internet:
[url]http://www.google.ca/search?hl=en&q=%22McNabb+Mathematics+Competition%22&meta=[/url]


AMC
AIME
AMC 12
USA(J)MO
USAMO
ARML
articles

Ok, is it to late to begin preparation for the AIME? I spent alot of time practicing for this years AMD, just so that I could qualify (I got a 111.5 (a 30pt. improvement over last year  :)  ) Is one month enough time to even improve ones score.  I think that my schools all time high score is 4-5, so is getting to the point that I could answer 6 in a month unreasonable? BTW, I got my 11.5 answering only the easy questions--just to clarify that, so right now i can only answer like 2-4 q's per AIME) thanks for any replies!

its never too late to prepare... which AMC did you take... I would assume you took 12A.
if you skipped the last five or so problems... then your AIME score would be between 2-6.... because the last five questions on the AMC12 are comparable to the AIME questions

I stand on the borderline for USAMO, I get 7-11 on a few practice AIMEs(I always make at least 1 stupid mistake), how much can I possibly improve in the next month?

I manage 5-7 on mock AIME's at home, how do I improve?

I don't seem to have a good grasp on many of the topics. Any suggested readings?

It's not too late to start preparing. I didn't study at all for AIME last year until about a week until the test, and still got a 7 and qualified for USAMO. (Of course, it WAS very much geared toward my strong areas.)

The best way to prepare for AIME is....practice, practice, practice!! Get old AIMES, split 'em 8-7 or 5-5-5 and give yourself 90 or 60 minutes to do them, or, if you have more time, do an entire old test. If you don't understand a problem, read through the solution at the end and try to follow along with your own work - see if you can guess where it's going.

As for suggested readings (note that I probably have no idea what I'm talking about; I need to study as much as you do), I'd recommend that you read or reread AoPS Vol. 2, and possibly the easier chapters and sections of Engel's [i]Problem Solving Strategies. [/i] But the main element is practice. Also, if you don't have time for an AIME or run out of old ones, the harder ARML individual questions are at least good for staying fresh.

Ah I did the 1996 AIME today and I made 4 careless mistakes, I ended up with a 7 :( , so in the next month I will work on to cut down # of careless mistakes and work on counting/probability(which I'm really bad at).

Good idea, probability1.01.

[quote="zabelman"]If you want to start cutting down your careless errors, [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php]Read this article.[/url][/quote]

Already read it!

For me, I'm just so intensely concentrated on the test that I often do something routine incorrectly.  Still, I have cut down my errors by doing stuff like MathCounts.  Ironically, it was a trivial mistake that sent me back 3 places at nats :D.

hopefully that article will cut down on all my stupid mistakes :(
So, what do you think I need to get on the aime to make it to USAMO if I'm a sophomore.

[quote="probability1.01"]Whitehorseking, beta, fermatprime... all have good AIME skill :( I wish I could do that.  I think a good way to practice is to study with other people of comparable skill (and an instructor if possible, but few people can hire 24/7 private tutors :D).  So you three should study with each other :P.[/quote]

Wait...

No doubt Beta and fermatprime, but me??? You have to be on mars or something...

I can't even get the borderline score. Eh...

In fact, all three of you made or have a 100% chance of making it 2005, while me, the chances are pretty low.

My chance isn't too good either! I usually only get 8's on practices. At best I get 10 or 11, and those are probably on some really easy years like 88, 92, 2002II.

You all have a GREAT chance of making USAMO!!! :)  Just have to stay positive! As for me....heh...chances continue to decline....I feel like there is NO way I can make it  :(

wow these guys.. beta dpopov
all think they cant make it??????????????

ok then i mite as well give up on making it during the next three years....

and beta if you can pull an 8 on the real AIME this year you're probably in since (correct me if im wrong) a freshman?

[quote="phyziksdood"]So the floor value applies only to grades 10 and under tho? a 12th grader with a score above the floor, but with an index lower than the cutoff wouldn't advance?[/quote]

Quite so. The floor is selected after all of the 11th and 12th grade participants have already been chosen.

When do the stats come out? I want to see how AoPSers did compared to everyone else. Although since anyone can become an AoPSer by joining, there really shouldn't be that much difference.

March 28th is when the USAMO qualifiers will be announced.

[quote="DPopov"]Also, if freshmen and sophomores have a high enough index will they qualify through the first selection instead of the floor score selection?[/quote]

Naturally. Given, of course, that their "high enough index" is through the AMC 12.

[quote="MithsApprentice"]I'd say a week after AIME Alt. Anyone care to specify?[/quote]

Why would they wait till after the alt. to post the AIME I stats?

They just do. I think it's good that they do that. No pressure for the AIME Alt takers.

About how many AIME Alt takers are there? Why do they all choose to take the Alt?

Bleh...our poll showed 5 - 15% (yes, I know, that's a very big range!).

Many times they take AIME Alt due to spring break or some other event that prevents them from taking AIME I. For example, I know that this year one school had their AIMEs delivered late (2 PM on the day of) so if it had been a little later and school had been out, they would've done Alt.

Go to Google and search for the Millenium Prize Problems (hit "I'm feeling lucky" :-) )

http://www.unl.edu/amc/whatswhen.html

All the dates are on there.


[hide]In 7 burgers, we have 1 lb of ground beef and 7 other materials.
The pound of ground beef costs 3.50 and the 7 materials cost 1.05

Each 7 burgers cost 4.55
Each burger costs 65 cents
Now since the restaurant must make a profit of 55%, the bugers must cost 55% more.
65*1.55 = 100.75
Rounded is $\$1.00$

I think that's right, was the "3/4 pounds of filler" in there to distract?[/hide]

[hide="The answer is "] 1.45.  I  had trouble getting 1.45 too. That was why I posted this problem.[/hide]

[quote="cj"]A restaurant makes 7 burgers from one pound of ground beef and three-quarter pounds of filler. One pound of ground beef costs $\$$3.5, the buns, filler and other extras cost a combined $\$$0.15 per burger, and the restaurant must make a 55% profit on each sale. How many dollars to the nearest nickel should be charged per burger?[/quote]
[hide="answer"]
um i think im too late, but anyhow, since 1 lb of meat can provide for 7, and since you need $\$$.15 per burger, $\$$3.5 + $\$$.15 * 7 = $\$$4.55
$\$$4.55/7= 65 cents per burger
since 100-45=55,
65/x=45/100
 :arrow: $\ x=144.4444....$, and that divided by 100 equals 1.44....., or $\$1.45$
[/hide]

Ah. I see where I messed up. The proportion.
Instead of multiplying by 1.55, I should have multiplied by $\frac{100}{100-55}$
OK


What is the largest integer k such that $5^{k}$ divides $3(10!)+12(5!)+4(7!)$?


Beginning with $ (x_{1}, y_{1})$ a sequence is defined by

$ (x_{n\plus{}1}, y_{n\plus{}1}) \equal{} (x_{n}^{2} \minus{} y_{n}^{2} \plus{} a, 2x_{n}y_{n} \plus{} b \plus{} 2)$,

where $ a$ and $ b$ are real constants.

(a) For $ a\equal{}1$ and $ b\equal{}\minus{}1$, find $ (x_{1}, y_{1})$ for which the sequence is constant;

(b) For $ (x_{1}, y_{1}) \equal{}(\minus{}1, 1)$, find the values of $ a$ and $ b$ for which the sequence has period $ 2$.

[quote="AndrewTom"]Beginning with $ (x_{1}, y_{1})$ a sequence is defined by

$ (x_{n \plus{} 1}, y_{n \plus{} 1}) \equal{} (x_{n}^{2} \minus{} y_{n}^{2} \plus{} a, 2x_{n}y_{n} \plus{} b \plus{} 2)$,

where $ a$ and $ b$ are real constants.

(a) For $ a \equal{} 1$ and $ b \equal{} \minus{} 1$, find $ (x_{1}, y_{1})$ for which the sequence is constant;[/quote]
So the system $ x\equal{}x^2\minus{}y^2\plus{}1$ and $ y\equal{}2xy\plus{}1$ $ \iff$ $ y^2\equal{}x^2\minus{}x\plus{}1$ and $ y\equal{}\frac 1{1\minus{}2x}$ $ \implies$ $ (x^2\minus{}x\plus{}1)(2x\minus{}1)^2\equal{}1$ $ \iff$ $ x(x\minus{}1)(4x^2\minus{}4x\plus{}5)\equal{}0$

And two solutions (easy to check back they indeed are solutions) : $ (x_1,y_1)\in\{(0,1),(1,\minus{}1)\}$

[quote="AndrewTom"](b) For $ (x_{1}, y_{1}) \equal{} ( \minus{} 1, 1)$, find the values of $ a$ and $ b$ for which the sequence has period $ 2$.[/quote]
$ (x_2,y_2)\equal{}(a,b)$
$ (x_3,y_3)\equal{}(a^2\minus{}b^2\plus{}a,2ab\plus{}b\plus{}2)$

So the system $ a^2\minus{}b^2\plus{}a\equal{}\minus{}1$ and $ 2ab\plus{}b\plus{}2\equal{}1$ $ \iff$ $ b^2\equal{}a^2\plus{}a\plus{}1$ and $ b\equal{}\minus{}\frac 1{2a\plus{}1}$ $ \implies$ $ (a^2\plus{}a\plus{}1)(2a\plus{}1)^2\equal{}1$ $ \iff$ $ a(a\plus{}1)(4a^2\plus{}4a\plus{}5)\equal{}0$

And two solutions (easy to check back they indeed are solutions) : $ (a,b)\in\{(0,\minus{}1),(\minus{}1,1)\}$
Notice that $ (a,b)\equal{}(\minus{}1,1)$ implies a constant sequence, so the smallest period is $ 1$ but $ 2$ is also a period and the poblem statement did not request that the sequence has [b][u]smallest [/u][/b]period $ 2$.

Outline of another solution:
Define $ z_n \equal{} x_n \plus{} y_n i$. Then, by the recurrence relation, $ z_{n \plus{} 1} \equal{} z_n^2 \plus{} a \plus{} (b \plus{} 2) i$.

(a):
$ (2z_1 \minus{} 1)^2 \equal{} (2i \minus{} 1)^2$.
(b):
We have: $ z_1 \equal{} \minus{} 1 \plus{} i$.
By the recurrence relation, $ z_2 \equal{} a \plus{} b i, z_3 \equal{} (a \plus{} b i)^2 \plus{} a \plus{} (b \plus{} 2) i$.
Since $ (a,b) \in \mathbb{R}^2$, WLOG let $ a \plus{} b i \equal{} u$.
We want to solve $ z_3 \equal{} z_1$ or $ \minus{} 1 \equal{} u^2 \plus{} u \plus{} i$.
This rewrites to $ (2u \plus{} 1)^2 \equal{} (2i \minus{} 1)^2$.

One can check back they are indeed solutions.


If $ x$ is a real number such that

$ 9^x\minus{}4^x\equal{}10$

and

$ 3^x\minus{}2^x\equal{}2$,

compute $ 3^x\plus{}2^x$.

let $ 3^x\equal{}a$, $ 2^x\equal{}b$
we get $ a^2\minus{}b^2\equal{}10$
$ a\minus{}b\equal{}2$
factoring we get $ (a\minus{}b)(a\plus{}b)\equal{}10$ so we get $ a\plus{}b\equal{}5$, which is the answer(5)

This problem is flawed. There is no such $x$ satisfying the two constraints.

By the Principle of Explosion, we may assume the answer to be anything.

By the Principle of Principles, we may assume that pi37 is right.


superior algebra
superior algebra unsolved

A Possible Sketch of $Z(S_{n})=\{e\}$: Begin by proving that the $Z(G)$ is the union of all one-element conjugacy classes of a group.  Two elements of $S_{n}$ are conjugate if and only if they consist of the same number of disjoint cycles of the same lengths (this may be somewhat hard to prove, but I recall having seen a rather simple proof of this fact before).  The only one one element conjugacy class of $S_{n}$ is ${e}$: Take a permutation in cycle form (1,2)(3)(4)(5) and switch two elements in disjoint cycles to get a conjugate permutation.  Otherwise, if there is a single cycle, just switch two elements of the cycle.

Another proof is here: http://www.mathlinks.ro/Forum/viewtopic.php?t=101317

$* Z(S_{n})=1$
let $f \in Z(S_{n})$ so $fg=gf \forall g\in S_{n}$.
Let $f=(ij)$ and $g=(ijk)$. So we have:
$fg = (ij)(ijk)=(kj)$ and $gf =(ijk)(ij)=(ki)$. Hence $i=j\Rightarrow f=1$.
Because a permutation in $S_{n}$ is a product of some transitions ($2-chain$). So we have $f=1 \forall f\in Z(S_{n})$.
Therefore $Z(S_{n})=1$
In the case $Z(A_{n})$ is similar. We can take $f=(ijk)$ and $g=(ij)(kl)$, remember that $A_{n}$ is generated by $3-chains$


Let $ a_n$ and $ b_n$ be defined the following way $ a_1\equal{}3$ , $ b_1\equal{}100$ ,$ a_{n\plus{}1}\equal{}3^{a_n}$ , $ b_{n\plus{}1}\equal{}100^{b_n}$.Determine the least integer m for which $ b_m>a_{100}$.

$ a_n \ge 2 \plus{} 5b_{n \minus{} 2}$ for any $ n \geq 3$


Given three-dimensional vectors

$u=i+3k$

$v=2j+bk$

$w=-3i+j-k$

such that the three vectors are coplanar and $b\in R$. Find $b$

($i,j,k$ are, of course, unit vectors in the $x,y,z$ direction)

[hide="Useful to know"] $u, v, w \text{ coplanar }\Leftrightarrow (u \times v) \cdot w = 0$

(The so-called triple scalar product can also be evaluated as the determinant of the matrix whose coefficients are the coefficients of the three vectors.) [/hide]

[quote="t0rajir0u"] $u, v, w \text{ coplanar }\Leftrightarrow (u \times v) \cdot w = 0$
[/quote]

Well, why is this?

Think of it this way, what is produced by $(u\times v)$? A vector which is normal to both $u$ and $v$. If we take this new vector and take the scalar product with $w$: $(u\times v)\cdot w$, it will equal zero only if the angle between the normal and $w$ is $90^\circ$, which would mean that it lies on the same plane as $u$ and $v$.


probability
Alcumus
number theory
prime numbers

Six 6-sided dice are rolled. What is the probability that three of the dice show prime numbers and the rest show composite numbers?

From the numbers 1-6, there are 3 primes and 2 composites. So our answer is

\[ \left(\frac{3}{6}\right)^3 \cdot \left(\frac{2}{6}\right)^3 \cdot \binom{6}{3} \equal{} \boxed{\frac{5}{54}} \]

just to add on to it, the one that is neither prime nor composite is 1.

It still makes me uncomfortable multiplying whole numbers like 20 in probability problems, like Alcumus did in its solution...

So I did it the cold, hard, abiding-by-definition way, [hide="by counting all the successful outcomes and dividing by all the possible outcomes."]
Number of Successful Outcomes: There are $\dbinom{6}{3} = 20$ ways we can choose three dice to be prime and three dice to be composite. Then, for the three "prime" dice, there are $3^3$ ways we can choose which prime number each of them produce. For the three composite dice, there are $2^3$ ways we can choose which composite number each of those two dice produce. Thus, the number of successful outcomes is just $(20)(3^3)(2^3)$.

Number of Possible Outcomes: Each of the six dice can roll six numbers. This is just $6^6$.

Simplifying: \[\dfrac{(20)(3^3)(2^3)}{6^6}\]\[=\dfrac{(20)(6^3)}{6^6}\]\[=\dfrac{20}{6^3}\]\[=\dfrac{20}{216}\]\[=\dfrac{5}{54}\]Thus, our answer is $\boxed{\dfrac{5}{54}}$.[/hide]

[quote=cw357]. . . the moment you realize there are more prime numbers on a die than there are composite . . .  :wacko:[/quote]

LOL so true when i first realized!!

[quote=cw357]. . . the moment you realize there are more prime numbers on a die than there are composite . . .  :wacko:[/quote]

It's because a lot of the first numbers are prime .

[quote=peelybonehead]Easy 120 xp let’s goo[/quote]

Good job helping and improving this conversation. Thank you.