hi again . ye soale func dige shabihe ireland 1995 amma male tst singapore 2008 $ f: R \longrightarrow R$ $ (x\plus{}y)(f(x)\minus{}f(y))\equal{}(x\minus{}y)f(x\plus{}y)$

1.rahat sabet mishe age yek noghte $ f$ karandar nabashe $ f$ hameja unlimitede.
2. age $ f$ karandar bashad az surate masale dar miad ke $ f$ peyvastast.
3.rue $ Q$ rahat dar miad  $ f\equal{}c_{1}x\plus{}c_{2}x^2$ pas $ f\equal{}c_{1}x\plus{}c_{2}x^2$

manam ye rahe hale dige daram ke marahelesho minevisam :
$ 1$-sabet konid $ f(0)\equal{}0$ .
$ 2$-ye rabeteyi beyne $ f(x\plus{}1)$ va $ f(x)$ ba farze $ f(1)\equal{}c$ bedast biarin . darvaghe gharar bedin $ y\equal{}1$ va ye rabete bedast miad . hala rabeteye moshabehe $ f(x\plus{}2)$ va $ f(x\plus{}1)$ ro ham bar hasbe $ c$ be dast biarin . va az in $ 2$ ta rabete $ f(x\plus{}1)$ ro hazf konin ta be ye rabete bar hasbe $ c$ va $ x$ baraye $ f(x\plus{}2)$ va $ f(x)$ be dast biarin .
$ 3$-gharar bedin $ y\equal{}2$ va farz konin $ f(2)\equal{}d$ dar in soorat ye rabete beyne $ f(x\plus{}2)$ va $ f(x)$ be dast miad .
$ 4$-hala dastgahe bedast oomade ro hal konin va natije begirin ke baraye har $ x$ :
$ f(x)\equal{}(\frac{d}{2}\minus{}c)x^{2}\plus{}(2c\minus{}\frac{d}{2})x$ yani $ f(x)\equal{}mx^{2}\plus{}nx$

tanx rahatoono kamel khundam. really nice age shoma do nafar ham nabudin forume iran rasman empty empty bud. :lol:


pigeonhole principle
trigonometry
geometry
trig identities
Law of Sines

I was browsing through

Advanced Mathematics: Precalculus with Discrete Mathematics and Data Analysis by Richard G. Brown,

which is an ordinary high school textbook. Yet, it contains quite a few "nonstandard" problems and is actually a pretty challenging text as it offers only very brief explanations. It even contains many allusions to contest-level concepts, such as the pigeonhole principle, the extended law of sines, and ptolemy's theorem, in the  harder end-of-section problems.

If anyone has used this textbook before, would you consider it good to go through as a supplement to the traditional AoPS texts?

I've used it before and I think using AoPS alone is fine but if you want more practice problems or to learn something in a different way this book would be helpful.


Buenas!!

Bien, tengo un problema que hace ya un tiempo q lo estoy pensando y no logro sacarlo.
Ademas ya se lo di a varias personas y nada... :S

Si me pueden dar una mano con el les agradeceria... es para un torneo de geometria y se me acaba el tiempo :S

"Construir un triángulo ABC conociendo los puntos medios de los lados AB y AC, y además conociendo el punto medio del arco menor BC de la circunferencia circunscripta al triángulo ABC."

Muchas gracias!

gabi.-

Supongo que te refieres a construcción con regla y compas.
Si ese fuera el caso, entonces no es posible hacer tal construcción (al menos, no en general).
Este es un contraejemplo:

Considera el triángulo $ ABC$ tal que:
La altura $ AH$ mide $ x$ con $ 3 < x < 4$ raíz del polinomio $ F = X^3-20X+20 \in Q[X]$,
$ CH$ mide $ 2$,
$ HB$ mide $ 4$.

Sean $ M$, $ N$ y $ P$ los puntos medios de $ CA$, $ AB$ y $ BC$, respectivamente.
Sea $ Q$ el punto medio del arco menor $ BC$ de la circunferencia circunscrita al triángulo $ ABC$.
Sea $ D$ el punto de intersección de $ MN$ y $ PQ$ y $ c$ la medida de $ DQ$
Sean $ C\^BQ=\alpha$ (y entonces $ C\^AB = 2\alpha$), $ C\^AH=\beta$ y $ H\^AB = \theta$.
Es fácil ver que $ MD=2$ y $ DN = 1$.
También,se puede ver que $ tg\alpha = \frac{c-x}{3}$, $ tg\beta = \frac{1}{x}$ y $ tg\theta = \frac{2}{x}$.
De esto y de $ tg(2\alpha)=tg(\beta+\theta)$, se obtiene que
$ x^3+(5-c^2)x+4c = 0$
Como $ x$ es raíz de $ F$, se concluye que $ c=5$.

Supongamos ahora que dados $ M$, $ N$ y $ Q$ sea posible construir el triángulo $ ABC$.
Esto implica que dados $ 1 (=MD)$, $ 2 (=DN)$ y $ c=5 (=DQ)$ (los cuales son todos costructibles) es posible construir el número $ x$.
Lo cual contradice el siguiente:
[b]TEOREMA[/b]
Si $ x$ es raíz (real) de un polinomio cúbico $ F$ con coeficientes racionales, y $ F$ no posee raíz racional entonces $ x$ no es constructible.

Disculpa, me equivoque al poner uno de los datos, debe ser:

La altura  $ AH$ mide $ 2x$ ( y no $ x$).


Point $ A$ is fixed on circle $ O$. $ B$ is arbitrarily chosen such that $ B\neq A$, $ B\neq O$, and $ \angle OBA\equal{}90^{\circ}$. In triangle $ OBA$, $ BM$ is a median. Point $ C$ is on circle $ O$ such that $ BM\parallel{}OC$ and $ C$ is on the same side of $ OA$ as $ B$. Prove that $ A$, $ B$, and $ C$ are collinear.

Nice problem!!

let P the second intersection of AP with the circle (O). See that M is the center of the semicircle with diameter AO hence
<BAO=1\2<BMO but in the triangle CAP we also get <CAP=1/2<COP , since <COP=<BMO 'cause CO//MB , we get
<BAO=<CAP ----> the rays AB and CA coincide, therefore A,B,C are colliner

So $ P$ is the point that is diametrically opposite $ A$ in circle $ O$?

This is just homothety.

[hide]
Let $ \mathcal H$ centered at $ A$ map $ M$ to $ O$. Therefore $ \mathcal H$ maps circle $ M$ to circle $ O$. Also, $ \mathcal H$ maps $ MB$ to $ OC$, and since $ B$ is on circle $ M$ and $ C$ is on circle $ O$, $ \mathcal H$ maps $ B$ to $ C$,  therefore $ A,B,C$ are collinear, QED.[/hide]

Edit: Here's a natural extension of this problem.

Circle $ O$ is given. Point $ A$ is any fixed point on $ O$. Let $ X$ be any point on the line through $ OA$ and in the interior of circle $ O$, and $ M$ be the midpoint of $ XA$. Find the locus of points $ B$ in the interior of circle $ O$ such that, if the parallel to $ BM$ through $ X$ intersects circle $ O$ at $ C$, the points $ A,B,C$ are collinear.

Does this locus depend on the location of $ X$?


Let a,b,c be positive reals. Prove that\[\frac{a+b}{b+c}\cdot\frac{a}{2a+b+c}+\frac{b+c}{c+a}\cdot\frac{b}{2b+c+a}+\frac{c+a}{a+b}\cdot\frac{c}{2c+a+b}\ge\frac{45}{64}\frac{a^2+b^2+c^2}{(a+b+c)^2}+\frac{33}{64}\]

It is proposed and unsolved for months. Could you post your solution mecrazywong?Thanks

[quote="mecrazywong"]Let a,b,c be positive reals. Prove that\[ \frac{a+b}{b+c}\cdot\frac{a}{2a+b+c}+\frac{b+c}{c+a}\cdot\frac{b}{2b+c+a}+\frac{c+a}{a+b}\cdot\frac{c}{2c+a+b}\ge\frac{45}{64}\frac{a^2+b^2+c^2}{(a+b+c)^2}+\frac{33}{64} \][/quote]
I just prove that $\frac{x}{(z+y)(2x+y+z)}\geq \frac{42x-3y-3z}{32(x+y+z)^2}$ for all $x,y,z>0$,and the problem is done :P  :P  :D


geometry
3D geometry
prism
real analysis
real analysis unsolved

Find the minimum possible sum of lengths of edges of a prism all of whose edges are tangent of a unit sphere.  [Muller-Pfeiffer].

The answer is $ 9\sqrt{3}$.
Hint: It isn't very hard to characterize completely all such prisms in question...


Find all real numbers $\theta$, with $0 \le \theta < \pi$, such that:

$(\sin 3\theta)(\sin 3\theta-\cos \theta) = (\sin \theta)(\sin \theta-\cos 3\theta)$

[hide="Solution"]Multiplying the equation out, we get

$\sin^{2}3\theta-\sin3\theta\cos\theta=\sin^{2}\theta-\sin\theta\cos3\theta$,
which can be transformed into
$\sin^{2}3\theta-\sin^{2}\theta=\sin(3\theta-\theta)=\sin2\theta$.
The LHS can then be turned into
$\sin(4\theta)\sin(2\theta)$ after trig manipulation. Then, we have
$\sin2\theta(\sin4\theta-1)=0$
$\Longrightarrow\theta=0, \frac{\pi}{8}, \frac{5\pi}{8}$ if $\theta\in[0, \pi)$.

$\mathbb{QED}$[/hide]

[hide]Expanding and rearranging gives \[\sin^{2}3\theta-\sin^{2}\theta = \sin3\theta\cos\theta-\sin\theta\cos3\theta,\] and the RHS is the expansion of $\sin (3\theta-\theta) = \sin 2\theta$, so \[\sin^{2}3\theta-\sin^{2}\theta = \sin2\theta.\] Let $\phi = 2\theta$, so that the LHS becomes \[\left[\sin (\phi+\theta)+\sin (\phi-\theta)\right]\left[\sin (\phi+\theta)-\sin (\phi-\theta)\right],\] or \[\left(2\sin\phi\cos\theta\right)\left(2\sin\theta\cos\phi\right).\] Rearranging the products on the LHS and setting it equal to the RHS gives \[\sin (4\theta) \cdot \sin (2\theta) = \sin (2\theta)\iff \sin (2\theta)\left(\sin (4\theta)-1\right) = 0.\] Then we may proceed as [b]Xaero[/b] has done above.[/hide]

method is right but you left out $\frac{\pi}{2}$

Yeah, since $\sin 2\theta=0$, also allows $2\theta=\pi$.


I dont know how to do this without calculus, but

Consider $ f(x) \equal{} x^{\frac{1}{x}}$
$ \Rightarrow f'(x) \equal{} x^{\frac{1}{x}}\left(\frac{1}{x^2} \minus{} \frac{\log_e(x)}{x^2} \right)$

set the derivative to zero and we see that the max (you'll have to trust it's the max because the double derivative test will take forever) is $ x\equal{}e$

$ \Rightarrow e^{\frac{1}{e}}>\pi^{\frac{1}{\pi}}$
$ \Rightarrow e^{\pi} > \pi^e$

[hide="Hint"]Let $ f(x) \equal{} \frac {lnx}{x}$ and find the extremum of $ f$ and conclude the answer[/hide]

Thanks. This is how I did it. I was hoping for a non-calculus solution.


conics
parabola
geometry
circumcircle
analytic geometry
Ross Mathematics Program
geometry solved

Prove that the circumcircle of the triangle formed by three tangents of a parabola passes through the focus. Hence prove that the orthocenter of such a triangle lies on the directrix of the parabola.

easy (and not quite interesting) if u use coordinate geometry. anyway, SL Loney gives a solution (and analytic geometry has no place in imo geometry)

Yup I do agree....that the problem becomes a routine Analytical Geometry Problem....and I forgot to mention that your are supposed to solve it in plane geometry( all the computation of AG is not encouraging..right...:D)...so you can be confident that my solution will find a place in the IMO if asked....:D

Please give a complete reference of that problem....Not only 'book'...

Pierre.

[quote="kalyanram_a"]Prove that the circumcircle of the triangle formed by three tangents of a parabola passes through the focus.[/quote]

This is proven on p. 47-48 of

Ross Honsberger, [i]Episodes in Nineteenth and Twentieth Century Euclidean Geometry[/i], Washington 1995

using Simson lines.

  Sincerely,
  Darij Grinberg

I think this can be proved by an angle chase, by using the optical property of the parabola: the tangent to the parabola in a point is the bisector of the angle formed by the segment from that point to the focus and the distance from the point to the directrix.


y is 1 not a prime?

and don't give me the answer "because the definition of a prime....."

there's gotta be another reason for this..

a prime is a number that is divisible by only itself and one.

For instance, 3 is only divisible by 3 and 1.

however, 1 is not prime because it is divisible by 1 and 1, which is just... 1.

So, as indianamath said, there is just one factor of one, even though it is simply itself and 1.

Also, if $ 1$ was prime, it would make a bunch of cool theorems inconsistent, and we'd have to say
For all primes $ p$ not equal to 1, ...

[quote="youdonotknowme"]and don't give me the answer "because the definition of a prime....."[/quote]

Definitions are only kept when they produce useful results.  So the reason we do not let $ 1$ be a prime is that the definition of a prime is more useful when we exclude $ 1$.

If that's still not good enough for you (and it should be; in higher mathematics there is no good reason to pick one definition over another other than to simplify the statements of theorems and so forth as much as possible), consider:

Number-theoretically, $ 1$ is unique because it is the only positive integer whose inverse is also a positive integer: $ 1^{-1}= 1$.  When you study number systems more general than the integers, numbers like these are called [b]units[/b], rather than being called either prime or composite.

One last consideration:  the statement of the Fundamental Theorem of Arithmetic makes no sense if $ 1$ is allowed to be a prime, since $ 1 = 1^{n}$ for any $ n$.

[quote="Nexmus"]Plus, mathmeticians said so.[/quote]

really? i've never thought of that before.

Well, it's not kind of a reason, its just a "What i say is law thing.." Like if Ceaser said that 1 was prime, it would probably be known like that for atleast 100 years... :lol:

[quote="Nexmus"]Well, it's not kind of a reason, its just a "What i say is law thing.." [/quote]

While it is true that certain definitions in mathematics are more by fashion than by anything else (such as the two ways to define the Gamma function), the definition of primality is a deep subject in number theory, and there are [b]very good reasons[/b] (some of which I have already given) for $ 1$ not to be considered a prime.

[hide="Among them (do not click if you are a Mathcounts-level stickler)"] - In a general ring, a number $ p$ is prime if $ p | ab \implies p | a$ or $ p | b$.  But this is vacuously true when $ p = 1$ because $ 1$ divides everything.
- The concept of primality can be extended to [i]prime ideals[/i] of a ring.  The ideal corresponding to $ 1$ is not a proper ideal (since it is the entire ring).[/hide]

[quote="firecricket91"][quote="andrewliu"]a prime number has 2 factors.
1 has one factor.... 1[/quote]

first dont revive old topics especially what you already said has been covered..but thats not the true mathematical reason why its not prime, and you also said some things that were actually techically untrue..you said,
"a prime number has 2 factors", its got 2 [b]or more[/b].
And besides you messed up your definiition a bit because -1 has two factors, 1 and -1, but I dont think -1 is considered prime.  Its also gotta be positive.[/quote]

Dude give me one prime number that has more than 2 factors. I think you mean composite....

[quote="firecricket91"][quote="andrewliu"]a prime number has 2 factors.
1 has one factor.... 1[/quote]

first dont revive old topics especially what you already said has been covered..but thats not the true mathematical reason why its not prime, and you also said some things that were actually techically untrue..you said,
"a prime number has 2 factors", its got 2 [b]or more[/b].
And besides you messed up your definiition a bit because -1 has two factors, 1 and -1, but I dont think -1 is considered prime.  Its also gotta be positive.[/quote]

Make sure that your spear tip isn't cracked before you throw it.

andrewliu is actually right.


AMC USA(J)MO USAMO Putnam calculus \/closed

I noticed in the classes section of this site that Dave Patrick is supposed to be the teacher of the Intermediate Counting class.  Could you give a brief summary of who he is?  I am just wondering because for one I am most likely going to take that class and I would like to know my teacher at least a little bit.  Also, will he be teaching much in the future like Mr. Rusczyk and Mr. Crawford do?

Dave Patrick will be joining our company full-time in July.  Dave is a 1988 USAMO winner (and notched the only perfect score on the AHSME that year).  He attended RSI in 1987 and was top 10 on the Putnam in 1991.  He has a PhD in mathematics from MIT.

[quote="jelyman"]Thats cool. On a sort of related note, have you guys ever thought of having a putnam class? I guess it might go against your whole calculus too early ideal, but Im sure it would be a very good addition. Also, how good do you think putnam problems (non-calculus ones) are for studying towards a goal of making mop?[/quote]

IMHO, with my lack of worldly wisdom, I don't think that as many students seriously study for the Putnam compared to the number of students who study for high school contests. I say this because students of mathematics in college have many other opportunities, such as research, and do not have as much time to devote to competitions. Most of the people who perform well on the Putnam were already very good at high school mathematics contests.

Is Dave the one with the username DPatrick?

[quote="rrusczyk"]We've had a few college people ask about a Putnam class.  It's unlikely we'll offer a Putnam class any time in the near future, but our Olympiad Problem Solving class would certainly be useful for the Putnam.[/quote]

Mabye you could try a Putnam MathJam?

I think it would be pretty reasonable to assume that DPatrick= Dave Patrick and since the sn is in red(meaning he is an administrator) it has to be him.

[quote="nr1337"]Mabye you could try a Putnam MathJam?[/quote]
I'll second the motion.

Given that DPatrick is the Putnam top-10er among us, we'll try to talk him into it.


what is the difference between the AMC A and the AMC B?

There's no difference.  They're just on different dates.

is one harder then the other? (i wouldn't think that they were)

Nope, they are comparable in difficulty.

The difficulty varies in different years, but in the same year, to be fair, they probably try hard to keep the A and B at the same difficulty. Same with the AIME and alternate AIME.


vector
geometry
geometric transformation
rotation
linear algebra
matrix
function

I just have a general question on transformations, and I was wondering if someone could please clarify to me.  Thank You!

I understand that in the study of Linear Algebra that multiple geometric transformations can be thought of as a composition of linear mappings.
For example, if we wish to first shear a vector $\vec{x}=(x_{1},x_{2})$ in the $x_{2}$ direction by $0.1$ followed by a rotation through an angle of $\frac{\pi}{6}$.  The overall transformation would be $R\circ S$, with a matrix of $[R][S]$ where $R$ and $S$ represents the rotation and shear transformation.  Altogether the initial vector $\vec{x}$ would be transformed to the new vector $R(S(\vec{x}))$ or $[R][S][\vec{x}]$.

With this in mind, I was wondering if in general any combination of transformations could be regarded as a composition of mappings/functions of individual transformations.  In Linear Algebra, this works for transforming vectors, but can we take this transformation ("composition") approach to transforming things like functions.

Let's say we wanted to know the function $f$ after we transform $f(x)$ first with a horizontal compression by $\frac{1}{a}$, second a reflection in the y-axis, and third a shift to the right by $d$ units.

Since these are transformations are being applied horizontally, we just need to transform the argument $x$ in the function $f$
Let $C(x)=ax$ be the a function which compresses horizontally by $\frac{1}{a}$
Let $R(x)=-x$ be the function which reflects vertically in the y-axis
Let $T(x)=x-d$ be the function which translates $d$ units to the right.
Altogether the transformations to the argument $x$ would be $T(R(C(x)))=-ax-d$ and the new transformed function would be $f[T(R(C(x)))]=f(-ax-d)$  Is this right?

This probably isn't the right forum for this question, but...

yes, you can take as many compositions as you want and it's the same as applying the transformations right after the other.

Hi calculuslover800!  Thank you for replying.  I'm sorry if this is not the correct forum, so moderators feel free to redirect this post to another forum.  I was just wondering if this was applicable for transforming functions, because someone told me that the transformation for the function should be $f(-a(x-d))$ contrary to what my method predicted.  I was just wondering, was he correct?


geometry
MATHCOUNTS
Pythagorean Theorem
AMC

AB is a chord of the larger of two concentric circles and is tangent to the smaller circle at T. Given AB = 8, find the area of the region bounded by the two circles.

[hide]The area bounded by the circles is pi(R^2-r^2) where R is the radius of the larger circle and r is the radius of the smaller. Let the point of where chord AB meets the smaller circle be C, and let the centers of both circles = O. Notice that triangle OCB is a right triangle with legs r and 4, and hypotenuse R. So by the Pythagorean theorem R^2-r^2= 16, so the area is 16(pi). [/hide]

This question has been on a ton of math contests, possibly with a different length for AB. I doubt MathCounts was the first to use it.

In case you don't know why the leg is 4:

Handy Theorem:
If a diameter is perpendicular to a chord, it also bisects it.

Rahter easy to prove...


logarithms
AMC
AIME
Vieta
algebra
polynomial
number theory

The equation \[2^{333x-2}+2^{111x+2}=2^{222x+1}+1\] has three real roots. Given that their sum is $m/n$ where $m$ and $n$ are relatively prime positive integers, find $m+n$.

[hide="Ingenius solution that was 3x shorter than mine"]Let $y=2^{111x}$.
The equation then becomes $\frac{y^3}{4}+4y= 2y^2+1$. After rearranging, we get $\frac{y^3}{4}-2y^2+4y-1=0$ and the product of the roots, which we will call $y_1y_2y_3$ equals $-\frac{-1}{\frac{1}{4}}=4$

However, we need to find the sum of the roots of the previous equation. Let $\{x_1,x_2,x_3\}$ be the roots of the original equation. Since $y_1y_2y_3=4$ and $\log_2 y_1y_2y_3=2$,

$\log_2 y_1y_2y_3 = \log_2 2^{111x_1}+\log_2 2^{111x_2}+\log_2 2^{111x_3}=111(x_1+x_2+x_3)= 2$.

Thus, $\frac{m}{n}= \frac{2}{111}$ and $m+n=113$[/hide]

IS there even another solution? :? I mean besides the obvious guess-and-test, which is what I did last year... :roll:

[hide]Multiplying by 4 and moving the equation to the left hand side gives $\left( 2^{111x}\right) ^3 -8\left( 2^{111x}\right) ^2 +16\left( 2^{111x}\right)-4=0$. If the roots are denoted $r,\ s,\ t$, then $2^{111(r+s+t)}=\left(2^{111r}\right) \left(2^{111s}\right) \left(2^{111t}\right)=4$. This gives $111(r+s+t)=2$, so $r+s+t=\frac{2}{111}$ making the answer 113.[/hide]

[quote="qweretyq"]How exactly do you "guess" this Jesusfreak?[/quote]

I don't remember exactly. The answers are obviously all very close to $0$, so I think I guessed and tested all values of $x$ with $1,2,3$ in the numerator and $111,222,333,666$ in the denominator, including negative of those, as well. Some get canceled out by simplification, so it isn't TOO many to check. :P

This was also the first time I had ever seen an AIME, so I was rather clueless.

[quote="jli"][quote="314.math"][quote="4everwise"]and the product of the roots, which we will call $ y_1y_2y_3$ equals $ \minus{} \frac { \minus{} 1}{\frac {1}{4}} \equal{} 4$

However, we need to find the sum of the roots of the previous equation. Let $ \{x_1,x_2,x_3\}$ be the roots of the original equation. Since $ y_1y_2y_3 \equal{} 4$ and $ \log_2 y_1y_2y_3 \equal{} 2$,

$ \log_2 y_1y_2y_3 \equal{} \log_2 2^{111x_1} \plus{} \log_2 2^{111x_2} \plus{} \log_2 2^{111x_3} \equal{} 111(x_1 \plus{} x_2 \plus{} x_3) \equal{} 2$[/quote]

I don't get this part of the solution.  Could someone please explain it to me?  Thanks![/quote]

which part.  the first part is because of vieta's theorem and the res is straightforward.[/quote]

How was Viete's Theorem used?

The product of the roots is the constant term of a polynomial $ \left(\frac{y^3}{4}\minus{}2y^2\plus{}4y\minus{}1\right)$ over the first term's coefficient.  Add a negative sign because the polynomial is of an odd degree.

After that, each root $ y_i\equal{}2^{111x_i}$ for $ i\equal{}1,2,3$.  4everwise used logarithyms to get the $ x$  by itself, then used properties of logs to express $ x_1\plus{}x_2\plus{}x_3$ in terms of a product of $ y$s.

[hide=Solution]Let $y\overset{def}{=}2^{111x}$. Expanding and rearranging gives $y^3-8y^2+16y-4$. We want $x_1+x_2+x_3=\log_{2^{111}}(y_1y_2y_3)=\log_{2^{111}}4=\frac{2}{111}$ by vieta's. Thus the wanted answer is $\boxed{113}$.[/hide]

Denote $2^{111x}=y$. Then we have $\frac{y^3}{4}+4y=2y^2+1$, or $y^3-8y^2+16y-4=0$. If the roots of this polynomial are $p_1, p_2, p_3$, then $x = \frac{\log_2{p_1}}{111}, \frac{\log_2{p_2}}{111}, \frac{\log_2{p_3}}{111}$. Summing and using $\log_2{p_1} + \log_2{p_2} + \log_2{p_3} = \log_2{p_1p_2p_3}$ gives $\frac{\log_2{p_1p_2p_3}}{111}$. But $p_1p_2p_3=4$ from Vieta, so our fraction is equivalent to $\frac{2}{111} \implies \boxed{113}$.

Let $u=2^{111x}$. Our equation is now $\frac{1}{4}u^3+4u=2u^2+1\implies u^3-8u^2+16u-4=0$. By Vieta's Formulas, the product of the roots is $4$ which implies that
\begin{align*}
2^{111r}\cdot2^{111s}\cdot2^{111t}&=2^{111(r+s+t)}=4 \\
111(r+s+t)&=2 \\
r+s+t&=\boxed{\frac{2}{111}} \\
\end{align*}


geometry
circumcircle
perpendicular bisector
geometry unsolved

[quote="HTA"]given triangle ABC and its circumcenter O . The perpendicular bisector of AO intersects BC at A1 . B1 and C1 are constructed similarly . Prove that A1 B1 C1 are collinear[/quote]
[img]http://i209.photobucket.com/albums/bb172/Ineq/a-1.jpg[/img]
Construc $ BN,CP \perp OA$
we have
$ \frac {\overline{A_1B}}{\overline{A_1C}} \equal{} \frac {\overline{MN}}{\overline{MP}} \equal{} \frac {\overline{MN}.\overline{MO}}{\overline{MP}.\overline{MP}} \equal{} \frac {(\overline{MO} \plus{} \overline{ON}).\overline{MO}}{(\overline{MO} \plus{} \overline{OP}).\overline{MO}} \equal{} \frac {\frac {1}{4}R^2 \minus{} \frac {1}{2}\overline{OA}.\overline{ON}}{\frac {1}{4}R^2 \minus{} \frac {1}{2}\overline{OP}.\overline{OA}} \equal{} \frac {\frac {1}{2}R^2 \minus{} \vec{OA}.\vec{OB}}{\frac {1}{2}R^2 \minus{} \vec{OA}.\vec{OC}}$
similar and Apply Me-ne-la-uyt we have disired!
:D


function
limit
real analysis
real analysis unsolved

Let $ m \ge 2$ be a positive integer. Define the sequence $ \left( x_n^{(m)} \right)_{n \ge 1}$ by
$ x_n^{(m)} \equal{} \displaystyle\sum_{k\equal{}1}^n k^{\minus{}m}$, $ \forall n \ge 1$.
a) Prove that $ \left( x^{(m)} \right)_{n \ge 1}$ converges.
b) Denote by $ \ell_m$ the limit of $ \left( x_n^{(m)} \right)_{n \ge 1}$. Determine the positive integers $ k$ for which there exists the nonzero and finite limit
$ \displaystyle\lim_{n \rightarrow \infty} n^k\left( \ell_m \minus{} x_n^{(m)} \right)$.

[quote="PhilAndrew"]Let $ m \ge 2$ be a positive integer. Define the sequence $ \left( x_n^{(m)} \right)_{n \ge 1}$ by
$ x_n^{(m)} \equal{} \displaystyle\sum_{k \equal{} 1}^n k^{ \minus{} m}$, $ \forall n \ge 1$.
a) Prove that $ \left( x^{(m)} \right)_{n \ge 1}$ converges.
b) Denote by $ \ell_m$ the limit of $ \left( x_n^{(m)} \right)_{n \ge 1}$. Determine the positive integers $ k$ for which there exists the nonzero and finite limit
$ \displaystyle\lim_{n \rightarrow \infty} n^k\left( \ell_m \minus{} x_n^{(m)} \right)$.[/quote]
it's obviusly for them who know something about zeta function. but it's easly problem also without knowing about zeta function

[quote="PhilAndrew"]Let $ m \ge 2$ be a positive integer. Define the sequence $ \left( x_n^{(m)} \right)_{n \ge 1}$ by
$ x_n^{(m)} \equal{} \displaystyle\sum_{k \equal{} 1}^n k^{ \minus{} m}$, $ \forall n \ge 1$.
a) Prove that $ \left( x^{(m)} \right)_{n \ge 1}$ converges.
[/quote]

\[ \left( x^{(m)} \right)_{n\ge 1} \le \left (x^{(2)} \right)_{n\ge 1} < \zeta(2) \equal{} \frac{\pi^2}{6}\]

and the sequence is monotonically increasing, so it converges. But we are kind of cheating by knowing that $ \zeta(2)$ converges... blah.

a) it's classic result (it's obvioulsy increasing , and we show that it's bounded)

b)
\[ \lim_{n\to\infty} \frac {l \minus{} x_n}{\frac {1}{n^k}} \equal{}
\]

\[ \equal{} ^{S.C.} \lim_{n\to\infty} \frac { \minus{} x_{n \plus{} 1} \plus{} x_n}{\frac {1}{(n \plus{} 1)^k} \minus{} \frac {1}{n^k}} \equal{}
\]

\[ \equal{} \lim_{n\to\infty} \frac {\frac {1}{(n \plus{} 1)^m}}{\frac {(n \plus{} 1)^k \minus{} n^k}{n^k(n \plus{} 1)^k}} \equal{}
\]

\[ \lim_{n\to\infty} \frac {n^{2k} \plus{} o(n^{2k})}{n^{m \plus{} k \minus{} 1} \plus{} o(n^{m \plus{} k \minus{} 1})}
\]
which has to be finite $ \Leftrightarrow$ $ 2k \equal{} m \plus{} k \minus{} 1$ so

$ \boxed{k \equal{} m \minus{} 1}$

[quote="not_trig"][quote="PhilAndrew"]Let $ m \ge 2$ be a positive integer. Define the sequence $ \left( x_n^{(m)} \right)_{n \ge 1}$ by
$ x_n^{(m)} \equal{} \displaystyle\sum_{k \equal{} 1}^n k^{ \minus{} m}$, $ \forall n \ge 1$.
a) Prove that $ \left( x^{(m)} \right)_{n \ge 1}$ converges.
[/quote]
\[ \left( x^{(m)} \right)_{n\ge 1} \le \left (x^{(2)} \right)_{n\ge 1} < \zeta(2) \equal{} \frac {\pi^2}{6}
\]
and the sequence is monotonically increasing, so it converges. But we are kind of cheating by knowing that $ \zeta(2)$ converges... blah.[/quote]

$ \frac{1}{k^2}<\frac{1}{k(k\minus{}1)}\equal{}\frac{1}{k\minus{}1}\minus{}\frac{1}{k}$


induction
number theory unsolved
number theory

Prove that $ [(3\plus{}\sqrt{5})^n]$ is odd for every positive integer $ n$.

[quote="moldovan"]Prove that $ [(3 \plus{} \sqrt {5})^n]$ is odd for every positive integer $ n$.[/quote]

Let $ a_n\equal{}[(3\plus{}\sqrt{5})^n\plus{}(3\minus{}\sqrt{5})^n]$.

Then $ a_0\equal{}2, a_1\equal{}6, a_{k\plus{}2}\minus{}6a_{k\plus{}1}\plus{}4a_k\equal{}0$

$ 0<(3\minus{}\sqrt{5})n \le 1$, so $ [(3 \plus{} \sqrt {5})^n]\equal{}a_n\minus{}1$

Clearly we have $ a_0, a_1$ are even.
$ a_k, a_{k\plus{}1}$ even $ \Rightarrow a_{k\plus{}2}$ even.
So by induction $ a_n$ is even for all $ n$
$ \Rightarrow [(3 \plus{} \sqrt {5})^n]$ is odd for all $ n$.


sorry, wrong section.  mods please move this post to inequalities

This is not sharp.
Rehomogeneise the equation by writing it as :
4(x 3  + y 3  + z 3 ) + 24xyz - (x + y + z) 3   \geq  0.
Exapanding, this is equivalent to :
 \sum x 3  - 2 \sum x 2 y + 2 \sum xyz  \geq  0, where the sums are symmetric.
But, this last inequality is a direct consequence of Schur's inequality.

Pierre.


Need inequalities for this problem ASAP!!

Bonham Garden Fertilizers produce Regular and Super-Gro formulations. There are 10 employees, or 400 worker-hour of production time each week. It takes 1/4 hour to produce and package Regular and 1/4 hour to produce Super-Gro .Bonham has  $\$$7000 to spend on the raw materials necessary to produce the fertilizers. Raw materials for Regular cost them  $\$$ 2, and materials for Super- Gro cost them  $\$$5 per package. Bonham makes a profit of   $\$$1 on each regular and  $\$$2 on each Super- Gro. How many packages should Bonham produce to maximize their profit?

Thank you very much!!

the dreadful word "simplex" springs to mind


Locked. This one is currently in a "problem of the month" contest.


geometry LaTeX trigonometry articles AoPSwiki MATHCOUNTS ratio

Given that ABCDEF is a regular hexagon with side length 6, find the area of triangle BCE.

[size=59][hide="answer"][/size]
$18\sqrt{3}$[/hide]

[quote="SplashD"][hide]$\frac{(6)(6\sqrt{3})}{2}=\boxed{18\sqrt{3}}$

[color=blue][size=75]kyyuanmathcount: I think you meant $18\sqrt{3}$[/size][/color]
[/hide][/quote]

[hide]I also think he meant $18\sqrt{3}$.  We see that the base of the triangle is 6 (BC) and the height of the triangle is two times the altitude of an equilateral triangle with side length of 6, which is $6\sqrt{3}$.  Hence the area is $18\sqrt{3}$.[/hide]

Yeah, I fixed it now. I accidently typed \srqt instead of \sqrt when I was doing $LaTeX$

[quote="mathgeniuse^ln(x)"][hide]Is it true that the area of the triangle is 1/3 the area of the hexagon?[/hide][/quote]

Yeah, I think it is.

[hide]BCE is 1/3 of ABCDEF. BCE=$s^2\sqrt{3}/2=18\sqrt{3}$.[/hide]

[quote="mathgeniuse^ln(x)"]I have an extension.

In regular pentagon, ABCDE, with a side length of 6, what is the area of triangle BCE?

I just want to find a general relationship for these problems.[/quote]

Do you want me to use trig on this?

To get a generalization for something like this, you will have to use trig (at least, I think so).

It would definately take trig to make a generalization for this problem due to the angles.

Okay, don't do the question if it involves trig.

Almost all generalizations in geometry (especially when finding area) use trig because the angles aren't always nice.

If you want an example, look at the article I wrote in AoPSWiki about finding the area of any regular geometric figure.  It is a lot more complicated than it looks, but with things like $\sin45^\circ$, $\sin30^\circ$, $\sin60^\circ$, they cancel out easily.

Okay thanks.  It is cool how people were able to find that $\sin 60^\circ=\frac{\sqrt{3}}{2}$ and other angles.

[quote="mathgeniuse^ln(x)"]I have an extension.

In regular pentagon, ABCDE, with a side length of 6, what is the area of triangle BCE?

I just want to find a general relationship for these problems.[/quote]

[hide="Trig Solution"]Let the side length be $s$. We can calculate that $\angle CDE=108^o$, and so $\angle DCE=36^o$. This means that $\angle BCE = \angle BCD - \angle DCE = 72^o$. Let the midpoint of $\overline{BC}$ be $M$. Then $EM=CM \tan{\angle BCE}=\frac{s \tan{72^o}}{2}$, so $[BCE]=\frac{BC \cdot EM}{2}=\boxed{\frac{s^2 \tan{72^o}}{4}}$.[/hide]
These kinds of problems don't really belong in the MathCounts forum though...

I know that is why I said don't do it if it involves Trig.

Hey, it's Neal Wu.

[quote="lotrgreengrapes7926"][quote="SplashD"]Alg I and Geometry are the typical 7th and 8th grade math classes for my school.[/quote] Those are typical 9th and 10th grade classes at my school. And it's a year faster for "advanced" people.[/quote]

Ditto. Not at the private school I'm going to next year though. If you take Algebra I in grades 9-12, you won't get credit for it.

[quote="Ignite168"][quote="mathgeniuse^ln(x)"]Yep, by the way, have there ever been any problems that people could use trigonometry but it turns out to be much simpler?[/quote]

That question on jclarke's AMC round.[/quote]

If you're talking about the question with the quadrilateral, I disagree. The formula needed to solve it was derived from trig.

[quote="mathgeniuse^ln(x)"]Yep, by the way, have there ever been any problems that people could use trigonometry but it turns out to be much simpler?[/quote]
I think #25 on the AMC 8 this year was one of those (I actually tried using trigonometry at first to solve it  :P ).

[quote="mathgeniuse^ln(x)"]Yep, by the way, have there ever been any problems that people could use trigonometry but it turns out to be much simpler?[/quote] Problem 8 on the 1998 state sprint.
It comes with a picture but oh well.
In triangle ABC AB=20 m<A=30 m<C=45. find the length of BC in simplest radical form.
ALthough in this case, trig is simpler :D.
However on problem 28 of the 03 sprint I tried using trig to find BC and CD  :blush: .

[quote="mathgeniuse^ln(x)"][quote="SplashD"][quote="mathgeniuse^ln(x)"]Yep, by the way, have there ever been any problems that people could use trigonometry but it turns out to be much simpler?[/quote]

Probably not, but you would still get the same answer.[/quote]

Since we are talking about hexagons, does anyone know any Mathcounts Questions relating to hexagons?[/quote]  THere are plenty. Here's one.
Hexagon ABCDEF has side length 10. What is the length of AE in simplest radical form?
This meets both of your questions :D. I did this with the law of cosines when I first saw it.

[quote="pgpatel"][/quote][quote="mathgeniuse^ln(x)"][quote="SplashD"][quote="mathgeniuse^ln(x)"]Yep, by the way, have there ever been any problems that people could use trigonometry but it turns out to be much simpler?[/quote]

Probably not, but you would still get the same answer.[/quote]

Since we are talking about hexagons, does anyone know any Mathcounts Questions relating to hexagons?[/quote][quote="pgpatel"]  THere are plenty. Here's one.
Hexagon ABCDEF has side length 10. What is the length of AE in simplest radical form?
This meets both of your questions :D. I did this with the law of cosines when I first saw it.[/quote]

It's obviously [hide=" :P "]$10\sqrt{3}$[/hide]

[quote="pgpatel"]  THere are plenty. Here's one.
Hexagon ABCDEF has side length 10. What is the length of AE in simplest radical form?
This meets both of your questions :D. I did this with the law of cosines when I first saw it.[/quote]

[hide]Is the answer $10\sqrt{3}$?[/hide]

Yes. It's really simple, but I solved it for the first time back when I got 3 right at last years state target

[quote="mathgeniuse^ln(x)"]Here is another question.

What is the ratio of the area of the circumscribed circle to the inscribed circle of a regular hexagon?[/quote]

[hide]Let $C$ be the center of the hexagon, $A$ and $B$ be adjacent vertices of the hexagon, and $M$ be the midpoint of $\overline{AB}$. Then since $\triangle AMC$ is a $30-60-90$ triangle with $\angle M$ its right angle, $\frac{MC}{AC}=\frac{\sqrt{3}}{2}$. The ratio we are looking for is $\frac{\pi AC^2}{\pi MC^2}=\frac{AC^2}{MC^2}=\left(\frac{2}{\sqrt{3}}\right)^2=\boxed{\frac{4}{3}}$.[/hide]


[quote="binomial_4eva"]Drug A: on its first test was eefective on 100 out of 1000
patients=10%
Drug A:on its second test was effective on 4000 out of 10000 patients=40%
Drug B on its first test was effective on 2000 out of 10000
patients=20%
Drug B: on its second test was effective on 600 out of 1000
patients=60%

which drug is more effective?[/quote]

[hide]Well $\frac{100+4000}{11000}> \frac{2000+600}{11000}$, so drug $A$ is more effective.[/hide]

Simpson's Paradox.

[hide]Out of the total 11000, 4100 were affected by drug A. Similarly, 2100 out of 11000 were affected by drug B.  So we have:

$\frac{41}{110}$ and $\frac{21}{110}$

It becomes obvious that [b]drug A was more effective[/b].[/hide]

[quote="binomial_4eva"]Drug A: on its first test was eefective on 100 out of 1000
patients=10%
Drug A: on its second test was effective on 4000 out of 10000 patients=40%
Drug B on its first test was effective on 2000 out of 10000
patients=20%
Drug B: on its second test was effective on 600 out of 1000
patients=60%

which drug is more effective?[/quote]
 [hide="answer"]
Not counting resilience and repeated factors, Drug A was effective on 4100/11000 while the other was 2600/11000. therefore Drug A is the answer. [/hide]


geometry
parallelogram
trigonometry
geometry solved

Let $ABC$ be a triangle of sides $a,b,c$ and $R$ the radius of the circumscribed circle.
Find the distance between $O$(the center of the circle) and the vertex $D$ of the paralelogram $ACBD$.

Let $M$ be the midpoint of $AB$. Of course $C,M,D$ are collinear and $CM=MD$. Let $CM=d$ and let $\angle CMR=\alpha$ Then from cosine theorem for triangle $CMO$: \[ R^2=d^2+MO^2-2d*MO*cos \alpha \]
 and for triangle $OMD$: \[ OD^2=d^2+MO^2+2d*MO*cos \alpha=R^2+4d*MO*cos \alpha \] But $OM^2=R^2-BM^2=R^2-\frac{c^2}{4}$ thus we obtain  $4d*MO*cos\alpha=2(d^2+MO^2-R^2)=2(d^2-\frac{c^2}{4})$ which means that \[ OD^2=R^2+2d^2-\frac{c^2}{2} \] But as we know $d^2=\frac{1}{4}(2(a^2+b^2)-c^2)$ which gives together \[ OD^2=R^2+a^2+b^2-c^2 \]


:roll:

Problem: you drive a car 45 miles at an average speed r (in miles per hour) to reach your destination.
Due to traffic, your average speed on the returnh trip is 3/4r. The round trip took a total of 1 hour and 45 minutes.
Find the average speed for each leg of the trip.
Solve the linear system us elimination

Answer Quickly!!!!!!!!!!!!!!!!!!!!!![/i]

eviloflame, it may help if you tell us how you attempted to solve this problem.  What did you try, and where did you get stuck?

Using logic...
You would find that 60 minutes could be 1r and 45 minutes would be 3r/4.
Yeah.


geometry
inradius
incenter
geometry theorems

[quote]Given a triangle $\triangle ABC$ with $AB\neq BC$. Let the incircle of triangle $ABC$ have center $I$ and radius $r$. Let $H_{b}$ be the foot of the $B$-altitude of triangle $ABC$, and $M_{b}$ the midpoint of the side $CA$. Let $T=IM_{b}\cap BH_{b}$. Prove that $TB=r$.[/quote]

I changed the name of points. :wink:
[b]Problem[/b]:[color=orange] In triangle [b]ABC[/b] , [b]I[/b] is incenter and [b]M[/b] and [b]H[/b] are midponit and feet of [b]A[/b]-altitude respectively.let $T=(IM\cap AH)$ prove that [b]AT=inradius[/b][/color]

[i]proof[/i]. let segment $XY$ be parrallel to $BC$ and tangent to incircle at $Z$ and $X\in AC\ ,\ X\not =B$  so triangles $AXY$ and $ABC$ are homothec wrt $A$

hence $AZ$ pass through $Z'$ ($Z'$ is tangency point of $A$-excircle with $BC$) and we know that $M$ is midponit of $D$ and $Z'$($D$ isn tnagecy point of incircle with $BC$)  and because $IZ=ID$ hence $IM\parallel ZZ'$ and we have $AT\parallel ID$ therefore quadrilateral $ATIZ$ is parallelgoram hence $AT=IZ=\text{inradius}$


Chung ta - nhung nguoi Vietnam chan chinh hay ra suc xay dung nen toan hoc Vietnam vung manh ,gop phan vapo viec dua Vietnam len thanh cuong quoc Xa Hoi chu nghia , cuong quoc toan hoc so mot the gioi ...Phai bai tru nhung thanh vien qua khich nhu Analyse...vi chinh chung chi lam cho nguoi Viet ta mat doan ket ma thoi...tien len VIETNAM

nói hay quá! tiến lên VN, tiến lên Xã Hội Chủ Nghĩa! tiến lên ĐCSVN! tiến lên chủ nghĩa *****!  :)

LGC :noi dang hoang dum chut :d

Các bác post những thứ gì mà kỳ thế này? Thật là không hiểu nổi.

"duy vật" ở chỗ nào nhỉ, mà sao tự nhiên lại "duy vậy" ở đây  :?:
à, hiểu rồi, "tiến lên chủ nghĩa...."
nói như vậy thật là  :wink:
mà analysis ko nên xúc phạm anh Linkgreencold như vậy  :wink:

[quote="analysis"]huh? sao tên linkgreencold xóa mất chữ "duy vật" của tôi viết rồi? tên mod này có vấn đề?  :)[/quote]
ha ha khong co van de gi,tai vi bua do anh analysis buon ngu hay sao nen go nham no thoi,tui dinh sua lai,nhung ma cu boi den no de danh gia tinh hinh.an  h analysis khong phien gi chu?


complex numbers
geometry solved
geometry

If A_0A_1...A_(n-1) is a regular n-gon inscribed in a unit circle, n \geq 3, then A_0A_1*A_0A_2*A_0A_3*...*A_0A_(n-1)=n.

Complex numbers ... Consider roots of unity. [This is one of the only problems I ever managed to solve with complex numbers :)] I don't have the time now to write out the solution.

Hi Arne,

you are right. Applying complex numbers just works fine. For another problem involving complex numbers see: http://www.kalva.demon.co.uk/imo/isoln/isoln906.html
Actually the case for the pentagon, n = 5, was asked. But it is harder to prove. And it is not a single case that proving a theorem for all values might be easier than proving a special case.


Mafia Game 19 has now begun!

Players:

Peter VDD
Ignite168
b-flat
Captain Sugar
Sunny
samath
bubka
ragnarok23
jli
S2
slamchess
Treething

Roles:

Mafia (3): Standard Mafia role.  May communicate among themselves during the first day and all nights.  The roles of those killed by the Mafia will be known only to the Mafia and the Oracle.

Citizens (5): Standard Citizen role.  No powers except the ability to post in this thread and vote on lynches.  None of the Citizens is a Miller.

Inspector (1): Standard Inspector role.  May inspect one person per night; is told that it was a successful inspection if the target was a Mafia, unsuccessful otherwise.

Angel (1): Standard Angel role.  May protect one person (possibly the Angel himself/herself) per night; during that night, that person cannot be killed by the Mafia or Vigilante.  When someone has successfully been saved by the Angel, I will say so in this thread.

Oracle (1): Every night, after the Mafia have (hopefully not) killed someone, the Oracle is told the person's role.  The Oracle is innocent, and wins if the Citizens win.

Vigilante (1): Standard Vigilante role.  The Vigilante can kill one person per night (or choose not to excercise the power).  The Vigilante is innocent, and wins if the Citizens win.

To summarize:

Mafia: 3
Citizens: 5
Inspector: 1
Angel: 1
Oracle: 1
Vigilante: 1

My rules on replacement bear repeating, in case you deleted the PM in question, so I repeat them here.

I take it as given that you already know the rules of Mafia. If you don't, go read the rules thread NOW! If you should, at any point in the game itself, post something like "Okay, what happens at night?" that would indicate that you haven't, you will be mercilessly replaced.

Also, I expect you to maintain a moderate level of activity in the game. I'm not going to have any set-in-stone rules about posting frequently, but I do ask that you at least look at the thread (and, and, during night phase, check your PM's) every time you log on. If, for two consecutive days, I get on the site and see both times that you haven't logged on in the past 24 hours (i.e., you haven't signed in on AoPS for at least 48 hours straight), I shall PM you to ask if you are still interested in playing. From that point, you have 24 hours to either respond to the PM or post on the thread; if you do neither, you will be replaced at my discretion. You will also be replaced if you show an extended amount of inactivity in the game (a 24-hour warning PM will be sent in this case also).

During the course of this game, editing posts is forbidden! Okay, now that I have your attention... you may edit your posts, provided that either a) nobody has posted after the post in question, or b) less than 15 minutes has passed since the post was originally submitted, or c) I request or give you permission to do so. If you violate all of those in editing a post, you will immediately be dismissed from the game and replaced.

Lynches require a majority to go through; on the first day, 7 AYE votes will be required.  Acquittals, on the other hand, simply need at least half of the votes (6 NAY votes are enough to get someone off the block).  I reserve the right to change this rule as I see fit.

Votes may be changed at any time you want.  Keep in mind, however, that switching your vote back and forth may make you look suspicious.

Accusations and secondings do not count as AYE votes.

I think those are all the things I need to go over.  I'd say have fun, but there are Mafia running around, so... be worried.  Be very worried; it might save your life.

Let the game begin!

Despite you link "Thread posted" pointing to game [b]18[/b], I got here and am ready to play.

Okay I'm here, what do I do now.

(just kidding)

[quote="Treething"]Okay I'm here, what do I do now.[/quote]You hop around, sing like a bird, squeal like a pig, take a gun out of your pants, shoot in the air and cry it out: "I'm mafia, and I'm proud of it!". ;)

here

on the pm it said 6 citizens and now there are 5...

I added a Mafia.

Also, my link pointing to Game 18 was... intentional.  It was... to get into the deceptive spirit of the game... yeah.  It most certainly was [i]not[/i] a stupid mistake on my part...

[color=white]...and anyone who believes this probably should not be playing Mafia.[/color]

Here I am, about 2 and 1/2 hours after the start of the game.

[quote="Ignite168"]I am also here, I was hoping for at least 1 mason, but oh well.[/quote] :rotfl:

[quote="Ignite168"]I am also here, I was hoping for at least 1 mason, but oh well.[/quote]

 :P  I didn't get it for a second, but then i realized. there's no point to only one mason. :rotfl:

With the power of induction of course.

I will now use my power to edit since it has not been 15 minutes to say, before someone even suggests a random accusation, no that is not a good idea as you could be mafia making up the random accusation and you will probably fall under suspicion if you want a "random" accusation.

Just getting that out the way as it happens at the beginning of every game

^

You could have edited it anyway, since nobody had posted after you.

[quote="ragnarok23"]I'm here.
I got 1337 ninja skillz, so no mafia is going to kill me.
Unless I am one.
 :o
But let's leave that a secret for now.  :roll:[/quote]

that hurts my brain. :?

Hmm..
is [i]Klebian[/i] mafia then?
edit:Oh, it's night, sorry

I am a somniloquist.

Sorry, Sunny, but the new entrant has somewhat spoilt it all. If I knew this, I would have retracted my vote. I thought S2's vote would decide it all ... alas!

Feeling sleepy. Yawn.

*Drops off on a wayside bench*
(Yes I am a somnambulist too)

[b]NIGHT OVER![/b]

kheg_k went home late that night, feeling somewhat pleased with himself.  He, the Inspector, had successfully found a Mafia!  And what's more, he didn't even get caught!

Or hadn't he...

He started to polish his magnifying glass.  Odd... it looked perhaps a bit different than the one he knew.  He didn't remember the handle being quite so thick.  It also didn't usually make ticking noises, except for the time he was hired as a private eye by old Loony Jack, but that's another story.  In particular, there weren't these green, digital numbers, counting down...

Meanwhile...

ragnarok23, the Oracle, was sitting in his living room, watching some old [i]Baby Bob[/i] reruns-how he loved him some [i]Baby Bob[/i]!-when suddenly the screen went fuzzy and an image of kheg_k appeared.

"I've been killed!" said kheg_k.  "They Mafia must have caught up to me last night!"

"What was your role?" inquired ragnarok23?

"Does it matter?" said kheg_k.  "The Mafia outnumber us now!  You'd better flee the town while you're still alive!  Errr... oh.  Look around..."

ragnarok23 did so, and saw his own body slumped on the couch, a bullet hole through his head.  He raised his hand in front of his eyes.  It was transparent.

"Now what happens?" said ragnarok23.

"You'll see," said kheg_k.  "But... shouldn't you know already?  You've been killed once before, I recall.  By the way, did anyone ever tell you you have terrible taste in TV?"

Meanwhile...

life=tennis+math was wandering down a dark alley around midnight.  When you're the Angel, and are protecting yourself, he thought, you could afford to do things like that, and he was always one to do stupid things just because he can.  Heck, he'd scaled Everest only the year before!  Of course, some might consider it cheating that he flew rather than climbed, but come on, scaling Everest is [i]scaling Everest[/i].  Let them say what they want...

In fact, he ran into some "they", just then.

The two Mafia approached him, one from each side of the alley, their faces obscured in the darkness.  They drew their guns.

"You can't kill me," said life=tennis+math.  "I've got divine protection!"

"Maybe that's true.  It didn't save kheg_k and ragnarok23, though," said one of the figures, and life=tennis+math recognized the voice with a shock as belonging to Klebian.

"You killed [i]both[/i] of them tonight?" said life=tennis+math, shocked.

"Yep.  Don't feel lonely, though.  You'll have them for company soon enough."  said the other one, who life=tennis+math realized was bubka.

This has to be said for life=tennis+math.  He regained his composure quite quickly.  In fact, his guard was down for only about a second.

However, it was the one second the Mafia needed...

[b]GAME OVER.  [color=red]MAFIA WIN![/color][/b]

Roles:

Peter VDD/nutz_for2.718281828: [color=blue]Citizen[/color]

Ignite168: [color=blue]Citizen[/color]

b-flat: [color=blue]Citizen[/color]

Captain Sugar/kheg_k: [color=blue]Inspector[/color]

Sunny: [color=blue]Vigilante[/color]

samath/Klebian: [color=red]Mafia[/color]

bubka: [color=red]Mafia[/color]

ragnarok23 (originally): [color=blue]Citizen[/color]

jli: [color=blue]Citizen[/color]

S2/life=tennis+math: [color=blue]Angel[/color]

slamchess/ragnarok23 (replacement): [color=blue]Oracle[/color]

Treething: [color=red]Mafia[/color]

Night actions:

Night 1
Inspector -> Ignite168
Vigilante -> Nobody
Angel -> S2/life=tennis+math
Mafia -> jli

Night 2
Inspector -> bubka
Vigilante -> Nobody
Angel -> S2/life=tennis+math
Mafia -> Ignite168

Night 3
Inspector -> b-flat
Vigilante -> b-flat
Angel -> S2/life=tennis+math
Mafia -> S2/life=tennis+math

Night 4
Inspector -> samath/Klebian
Vigilante (dead)
Angel -> S2/life=tennis+math
Mafia -> Captain Sugar/kheg_k

Fun Facts:

Every night, the Angel chose to save him/herself.

Five of the first six people killed were Citizens.

The story after the end of Night 2 was inspired by a PM sent to me by nutz_for2.718281828

Of the 12 people participating in the game originally, 5 were eventually replaced.

Two of those people requested replacements (Peter VDD and samath).

bubka wins the Perfect Attendance Award: of the 12 people who started in the game, he was the only one who, as of the start of Night 4, was both still alive and hadn't been replaced.

I might think of some more fun facts later.  For now, the Mafia should just celebrate a game well won.  The advantage shifted quite a lot over the last two days.  For everyone else, well, gunshot wounds don't hurt [i]that[/i] much, do they?  And you get a chance to get some rest... no?  Well, there are always future games.

Thanks to everybody who participated in this game!  Without you, the victims *cough* contestants, there couldn't have been a Mafia Game 19.

Would've been easier without treething sacrificing himself...
We actually didn't think s2 was the angel, which is why we didn't kill ragnarok on night 3...

And of course, I got lucky with life=tennis+math missing the post where I roleclaimed angel.  :ninja:  :winner_first:

[quote="life=tennis+math"]*Groan* sorry guys, I thought kheg_k was the Mafia, not bubka. Oh well, that was a lot of fun even though I didn't play for too long. I think I might sign up for next round.
In the future I know how to rectify my mistakes (namely, not skipping important pages like the one where Klebian roleclaimed angel). Well anyway, congratulations to everyone especially bubka and Klebian. Again, non-Mafia, sorry for not using my brains more. I promise in future games that I will read all the pages.[/quote]

Well I'll give you credit that my role claim was kinda insignificant looking:
[quote]
Well, there's nothing either of us could do, eh?
One of us is mafia, and considering the fact that I'm angel...[/quote]

Okay - here is how the game should have worked out.

Firstly, Treething could have done without the suicide act. He was still among the unsuspected, and there was a chance of me getting kheg_k lynched, which would give us a quicker win.

Secondly, after kheg_k's roleclaim, I thought "This is the end". I even imagined posting "uh oh ..." after that post. But I regained my composure and made a counterroleclaim that was quite believable.

Thirdly, night 3 decisions were a shock. I had taken S2 to be a sure kill, for no angel in her right mind would save S2. After the result, I instantly worked out what had gone wrong. The angel has been saving herself all along! Meanwhile, the vigilante did us a favour by popping out of nowhere and gettign rid of the last citizen.

Now for the last and most interesting part. Klebian made a hurried roleclaim and got himself into trouble. I thought we could only win if we could get Sunny lynched before S2 came. But ragnarok23 refused. So I worked out the final part of the Plan Omega as follows:
* Klebian gets lynched without me supporting him much.
* I take out Sunny during night.
* I get kheg_k lynched during the day.
* I kill ragnarok23 at night to win.
I was sure of the kills as the angel would protect herself! However, what really happened after that was history. Really life-=tennis+math, you should seriously consider taking the first train down to Wimbledon and never returning again. (No offense - just joking!)

A lesson to kheg_k: If you look suspicious, people don't believe you even when you are truthful.

P.S.: I felt somewhat uneasy last night because of the new angel, who might not save himself, but it seems angels are all alike. Once you have seen one, you have seen 'em all!

So much for naying bubka. We should have looked somewhere in there.

 :rotfl: Baby bob...

Well, this was fun, eh? I was suspicous about Klebian, I didn't think anything was up with Sunny, but, oh well.
OMFG BABY BOB IS SOOOOOOO FLIPPIN AWESOME YAAAAAAAAAAAAAAAAaaaa

You know, I actually thought you'd go for ragnarok23 during Night 4.  My reasoning was that if you successfully make a kill, you'd win anyway, so consider the cases where you try to kill someone and it becomes known that you fail:

If you fail to kill life=tennis+math, Klebian would probably be lynched, because those two would be opposing people who both roleclaim Angel, and life=tennis+math can't be guilty.

If you fail to kill kheg_k, the same would happen to bubka.

On the other hand, if you failed to kill ragnarok23, you might still be able to convince everyone (and by everyone, I mean ragnarok23; the Inspector and Angel would know what's up) that life=tennis+math was actually Mafia, even though the Mafia had targeted him once.

the turning point in it all was when life=tennis+math missed klebian's roleclaim.

Bubka played a really good game.

Hehehe agreed, definitely a good game all around.


function
floor function
ceiling function
symmetry
geometry
geometric transformation
reflection

If $ f(x)\equal{}\lfloor x \rfloor$ and $ g(x)\equal{}\lceil x \rceil$, express $ g(x)$ in terms of $ f(x)$ and $ x$.

(Source: AoPS Volume 1)

I posted this because there's a (in my opinion) more elegant solution than $ g(x)\equal{}f(x)\plus{}1\minus{}f(1\minus{}x\plus{}f(x))$, the answer given in the solutions manual.

If x is an integer, $ g(x)\equal{}f(x)$ If x is not an integer, $ g(x)\equal{}f(x)\plus{}1$

True, but at this point in the text, piece-wise defined functions had not yet been presented.

The answer I have in mind is even better, though.  :)

[hide="Hint."]
-----
Take a look at the graphs of the [url=http://mathworld.wolfram.com/FloorFunction.html]floor function[/url] and the [url=http://mathworld.wolfram.com/CeilingFunction.html]ceiling function[/url]. Note the [i]symmetry[/i] between the graphs and find a way to exploit it. In other words, find an expression that [i]reflects[/i] the symmetry in the graphs.  :wink:
-----
[/hide]

No, but I'll give you a big hint:

[hide="Giveaway."]
-----
If you reflect the graph of the floor function ($ f(x)$) over the y-axis and then over the x-axis, you get the ceiling function ($ g(x)$).
-----
[/hide]

[quote="math154"]I think that's what he did, but I don't think that's in terms of f(x) and x.[/quote]

It is to the former and why not to the latter?

$ \minus{} x$ is still an expression in terms of $ x$. The solution given in the solutions manual contains $ \minus{} x$ as well.

[hide="Answer."]
$ g(x)\equal{}\minus{}f(\minus{}x)$
[/hide]


Hello Everyone,

This is a try to make a nice inequalities marathon in the [color=red]Pre-Olympiad level[/color]

Please if you write any problem don't forget to [color=red]indicate its number[/color] and if you write a solution please indicate for what problem also to prevent the confusion that happens in some marathons.

Please [color=red]show your solution[/color] don't just write by AM-GM then Cauchy-Schwarz and we are done.

OK finishing the talk now we go:

[b]Problem 1:[/b] For any positive real numbers $ a,b,c$ show that the following inequality holds \[ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \geq \frac{c\plus{}a}{c\plus{}b}\plus{}\frac{a\plus{}b}{a\plus{}c}\plus{}\frac{b\plus{}c}{b\plus{}a}\]

[hide="Solution to problem 1"]Ok. After not so many computations i got that:
\[ \frac {a}{b} + \frac {b}{c} + \frac {c}{a} - \frac {a + b}{c + a} - \frac {b + c}{a + b} - \frac {c + a}{b + c}\]

\[ = \frac {abc}{(a + b)(b + c)(c + a)}\left(\frac {a^2}{b^2} + \frac {b^2}{c^2} + \frac {c^2}{a^2} - \frac {a}{b} - \frac {b}{c} - \frac {c}{a}\right)\]

\[ + \frac {abc}{(a + b)(b + c)(c + a)}\left(\frac {ab}{c^2} + \frac {bc}{a^2} + \frac {ca}{b^2} - 3\right)\]
So in order to prove the above inequality we need to prove $ \frac {a^2}{b^2} + \frac {b^2}{c^2} + \frac {c^2}{a^2}\ge \frac {a}{b} + \frac {b}{c} + \frac {c}{a}$ and $ \frac {ab}{c^2} + \frac {bc}{a^2} + \frac {ca}{b^2}\ge 3$

The second inequality is obvious by AM-GM , and the for the first we have:
\[ \left(\frac {a^2}{b^2} + \frac {b^2}{c^2} + \frac {c^2}{a^2}\right)^2\ge 3\left(\frac {a^2}{b^2} + \frac {b^2}{c^2} + \frac {c^2}{a^2}\right)\ge \left(\frac {a}{b} + \frac {b}{c} + \frac {c}{a}\right)^2\]
where i used AM-GM and the inequality $ 3(x^2 + y^2 + z^2)\ge (x + y + z)^2$ for $ x = \frac {a}{b}\ ,\ y = \frac {b}{c}\ ,\ z = \frac {c}{a}$

So the inequality is proved.[/hide]

[b]Problem 2:[/b]Let $ a,b,c,d > 0$ such that $ a\le b\le c\le d$ and $ abcd = 1$ . Then show that:
\[ (a + 1)(d + 1)\ge 3 + \ds\frac {3}{4d^3}\]

[quote="Maths Mechanic"][hide="Another solution to 1"]
Substitute $ \frac {a}{b} \equal{} x,\frac {b}{c} \equal{} y,\frac {c}{a} \equal{} z$.So $ xyz \equal{} 1$.The inequality after substitution becomes
$ x^{2}z \plus{} y^{2}x \plus{} z^{2}x \plus{} x^{2} \plus{} y^{2} \plus{} z^{2}\ge x \plus{} y \plus{} z \plus{} 3$

$ x^{2}z \plus{} y^{2}x \plus{} z^{2}x\ge3$.So now it is left to prove that $ x^{2} \plus{} y^{2} \plus{} z^{2}\ge x \plus{} y \plus{} z$ which is easy.
[/hide]

EDIT:alex2008 where is $ b$ and $ c$ in your inequality.[/quote]

There is no $ b,c$ in my inequality . The inequality is correct . You can try to find a counterexample but you'll only lose your time . Anyway the inequality is easy . $ b,c$ are not so important , they are more for confusing .

@Maths Mechanic: Are you Raghav Grover?

[hide="Solution to problem 2"]

From the condition $ a\leq b\leq c\leq d$ we get that $ a\geq\frac {1}{d^3}.$

$ \implies (a \plus{} 1)(d \plus{} 1)\geq \left (\frac {1}{d^3} \plus{} 1\right )(d \plus{} 1)$

Now let's prove that $ \left (1\plus{}\frac{1}{d^3}\right )(d\plus{}1)\geq 3 \plus{} \frac {3}{4d^3}$ .

This is equivalent with: $ (d^3 \plus{} 1)(d \plus{} 1)\geq 3d^3\plus{}\frac 34$

$ \Longleftrightarrow \left [d(d \minus{} 1)\right ]^2 \minus{}  [d(d \minus{} 1)]\plus{}1 \ge \frac 34 \Longleftrightarrow \left [d(d \minus{} 1)\minus{}\frac {1}{2}\right ]^2\ge 0.$

Equality holds for $ a \equal{} \frac {1}{d^3}$ and $ d(d \minus{} 1) \minus{} \frac {1}{2} \equal{} 0\Longleftrightarrow d \equal{} \frac {1 \plus{} \sqrt {3}}{2}$

[/hide]

[quote="alex2008"]
@Maths Mechanic: Are you Raghav Grover?[/quote].

Yes :lol:

[b]Problem 3:[/b] If $ a,b,c$ are three positive real numbers, then
\[ \frac {a}{\left(b \plus{} c\right)^2} \plus{} \frac {b}{\left(c \plus{} a\right)^2} \plus{} \frac {c}{\left(a \plus{} b\right)^2} \geq \frac {9}{4\left(a \plus{} b \plus{} c\right)}\]

[hide="solution on problem 3"]
$ \sum\frac {a^2}{ab^2 \plus{} ac^2 \plus{} 2abc} \ge \frac {(a \plus{} b \plus{} c)^2}{\sum _{sym}a^2b \plus{} 6abc}$
So we only have to prove that:
$ 4(a \plus{} b \plus{} c)^3 \ge 9 \sum_{sym}a^2b \plus{} 54abc \Longleftrightarrow 4(a^3 \plus{} b^3 \plus{} c^3) \plus{} 12\sum_{sym}a^2b \plus{} 24abc \ge 9\sum_{sym} a^2b \plus{} 54abc \Longleftrightarrow 4(a^3 \plus{} b^3 \plus{} c^3) \plus{} 3\sum_{sym}a^2b \ge 30abc$.

But $ \sum_{sym} a^2b \ge 6abc$.And $ a^3 \plus{} b^3 \plus{} c^3 \ge 3abc$.
So $ 4(a^3 \plus{} b^3 \plus{} c^3) \plus{} 3\sum_{sym}a^2b \ge 30 abc$[/hide]

[b]PROBLEM 4[/b]
For $ a,b,c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 1$ prove that $ 7(ab \plus{} bc \plus{} ca) \le 2 \plus{} 9abc$

[hide="Solution to problem 4"]Homogenize to
\[ 2(a \plus{} b \plus{} c)^3 \plus{} 9abc\geq 7(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)\]
Expanding it becomes :
\[ \sum_{sym} a^3 \plus{} 6\sum_{sym} a^2b \plus{} 21abc\geq 7\sum_{sym} a^2b \plus{} 21abc\]
So we just need to show:
\[ \sum_{sym} a^3\geq \sum_{sym} a^2b\]
which is obvious by

$ a^3 \plus{} a^3 \plus{} b^3\geq 3a^2b$ and similars.[/hide]

[b]Problem 5:[/b]Let $ x,y,z\in \mathbb{R}_ \plus{}$ . Pove that :
\[ \sqrt {x(y \plus{} 1)} \plus{} \sqrt {y(z \plus{} 1)} \plus{} \sqrt {z(x \plus{} 1)}\le \frac {3}{2}\sqrt {(x \plus{} 1)(y \plus{} 1)(z \plus{} 1)}\]

Sorry to interrup the flow of the marathon guys, but where did you learn about inequalities at from this level? Cuz I finished the Inequalities chapters in Intermediate Algebra and AoPS Vol. 2 but I guess this is the next level  :lol:

I guess these would be old IMO or MO's or TST's. Because they do not give any inequalities in IMO because have become so good at them!
x+1=a/b,y+1=b/c,z+1=c/a.
sqrt(a-b/c)+sqrt(b-c/a)+sqrt(c-a/b)<=3/2
sqrt(a-b/c)<=1/2
a-b/c<=1/4
4(a-b)<=c
4(b-c)<=a
4(c-a)<=b
Adding all equations
0<=a+b+c

Your solution assumes $ (x \plus{} 1)(y \plus{} 1)(z \plus{} 1) \equal{} 1$, a very strange assumption to make, and cannot be done since the inequality is not homogenous.

[hide="solution to problem 5"]
Dividing with the square root on the RHS we have :

$ \sqrt{\frac{x}{(x\plus{}1)(z\plus{}1)}} \plus{}\sqrt{\frac{y}{(x\plus{}1)(y\plus{}1)}} \plus{}\sqrt{\frac{z}{(y\plus{}1)(z\plus{}1)}} \leq \frac{3}{2}$

By AM-GM

$ \sqrt{\frac{x}{(x\plus{}1)(z\plus{}1)}} \leq \frac{1}{2} \left( \frac{x}{x\plus{}1} \plus{}\frac{1}{y\plus{}1} \right)$

$ \sqrt{\frac{y}{(x\plus{}1)(y\plus{}1)}} \leq \frac{1}{2} \left(\frac{y}{y\plus{}1} \plus{}\frac{1}{x\plus{}1} \right)$

$ \sqrt{\frac{z}{(y\plus{}1)(z\plus{}1)}} \leq \frac{1}{2}\left(\frac{z}{z\plus{}1} \plus{}\frac{1}{y\plus{}1} \right)$

Summing we obtain

$ LHS \leq \frac{1}{2} \left( \left(\frac{x}{x\plus{}1} \plus{}\frac{1}{x\plus{}1} \right) \plus{}\left(\frac{y}{y\plus{}1} \plus{}\frac{1}{y\plus{}1} \right) \plus{}\left(\frac{z}{z\plus{}1} \plus{}\frac{1}{z\plus{}1} \right) \right) \equal{}\frac{3}{2}$

[/hide]

[b] Problem 6 [/b]. Let $ a,b,c$ be positive numbers  , then prove that
\[ \frac{1}{a} \plus{}\frac{1}{b} \plus{}\frac{1}{c} \geq \frac{4a}{2a^2 \plus{}b^2 \plus{}c^2} \plus{}\frac{4b}{a^2 \plus{}2b^2 \plus{}c^2} \plus{}\frac{4c}{a^2 \plus{}b^2 \plus{}2c^2}\]

[hide="Solution to problem 6"]

By $ AM\minus{}GM$ we have $ 2a^2\plus{}b^2\plus{}c^2\ge 4a\sqrt{bc}$ .

$ \implies \frac{4a}{2a^2\plus{}b^2\plus{}c^2}\le \frac{4a}{4a\sqrt{bc}}\equal{}\frac{1}{\sqrt{bc}}$ .

Addind the similar inequalities $ \implies RHS\le \frac{1}{\sqrt{ab}}\plus{}\frac{1}{\sqrt{bc}}\plus{}\frac{1}{\sqrt{ca}}\ (1)$ .

Using Cauchy-Schwarz we have $ \left(\frac{1}{\sqrt{ab}}\plus{}\frac{1}{\sqrt{bc}}\plus{}\frac{1}{\sqrt{ca}}\right)^2\le \left(\frac 1a\plus{}\frac 1b\plus{}\frac 1c\right)^2$

so $ \frac{1}{\sqrt{ab}}\plus{}\frac{1}{\sqrt{bc}}\plus{}\frac{1}{\sqrt{ca}}\le\frac 1a\plus{}\frac 1b\plus{}\frac 1c\ (2)$ .

From $ (1),(2)$ we obtain the desired result .

[/hide]

[b]Problem 7[/b]

Let $ a,b,c,d,e$ be non-negative real numbers such that $ a\plus{}b\plus{}c\plus{}d\plus{}e\equal{}5$ . Prove that:

$ abc\plus{}bcd\plus{}cde\plus{}dea\plus{}eab\le 5$ .

[hide]for problem 7:
apply AM GM  3 at a time to (a,b,c,d,e).on adding we get it[/hide]

[quote=sqing][quote=Wellingson]
[hide=Problem 752]
Let $n \geq m \geq 1$ and $x \geq y \geq 0$, such that:

$x ^ {n + 1} + y ^ {n + 1} \leq x ^ m-y ^ m.$ Prove that $x ^ n + y ^ n \leq 1.$[/hide][/quote]
https://artofproblemsolving.com/community/c6h1944168p13395590[/quote]

Solution?

[quote=Pain rinnegan][quote="Tourish"][b]Problem524[/b]  $a,b,c>0$ such that $\frac{a}{b+c}+\frac{b}{c+a}\leq\frac{3}{2}$, find the minimum of $\frac{c}{a+b}$.[/quote]

[hide="Problem 524"]Expanding the condition we have:

\[2(a^2+b^2)\le 3ab+3c^2+c(a+b)\]

Then

\[3c^2+c(a+b)\ge 2(a^2+b^2)-3ab=\frac{1}{4}(a+b)^2+\frac{7}{4}(a-b)^2\ge \frac{1}{4}(a+b)^2\]

Let $x=\frac{c}{a+b}$, dividing by $(a+b)^2$, we have $3x^2+x\ge \frac{1}{4}\Leftrightarrow \left(x-\frac{1}{6}\right)\left(x+\frac{1}{2}\right)\ge 0$, thus $x\ge \frac{1}{6}$.
https://artofproblemsolving.com/community/c6h299899p2359174[/hide][/quote]

[hide=Problem 729'][quote=sqing][quote="sqing"][b]Problem 729[/b]
Let $a,b$ be positive numbers satisfying $\sqrt{a^2+3}+\sqrt{b^2+8}=5$. Prove that\[\frac{3}{a}+\frac{2}{b}\ge 5.\][/quote]
[hide=*]https://artofproblemsolving.com/community/c6h299899p3242135
[/hide]
By C-S,
$$30=\sqrt{(a^2+3)(9+27)}+\sqrt{(b^2+8)(4+32)}\geq3a+9+2b+16\implies 3a+2b\leq 5,$$
$$\frac{3}{a}+\frac{2}{b}\ge \frac{(3+2)^2}{3a+2b}\geq 5.$$[/quote][/hide]
Let $a,b$ be positive numbers satisfying $\frac{3}{a}+\frac{2}{b}=5$. Prove that\[\sqrt{a^2+3}+\sqrt{b^2+8}\ge 5.\]

[quote=lvcat]Let $a,b,c >0$ such as $a+b+1=c$ . Find the minimum value of
\[P=\dfrac{a^3}{a+bc}+\dfrac{b^3}{b+ac}+\dfrac{c^3}{c+ba}+\dfrac{14}{(c+1)\sqrt{c+ab}}\][/quote]
[url=https://artofproblemsolving.com/community/c6h299899p2951261]here[/url]

[quote=Mathsy123][hide="Problem 755"]
Let $a,b,c$ positive reals. Prove that:
$$\frac{a+b}{\sqrt{a+2c}}+\frac{b+c}{\sqrt{b+2a}}+\frac{c+a}{\sqrt{c+2b}} \geq 2\sqrt{a+b+c}$$
[/hide][/quote]

very nice!
Holder inequality work:

$(\frac{a+b}{\sqrt{a+2c}}+\frac{b+c}{\sqrt{b+2a}}+\frac{c+a}{\sqrt{c+2b}})^\frac{2}{3}[(a+b)(a+2c)+(b+c)(b+2a)+(c+a)(c+2b)]^\frac{1}{3}\geq{2(a+b+c)}$ . It is enough to show that

$\frac{[2(a+b+c)]^3}{a^2+b^2+c^2+5(ab+bc+ca)}\geq{4(a+b+c)}$  or

$2(a+b+c)^2\geq {a^2+b^2+c^2+5(ab+bc+ca)}$  or

$(a-b)^2+(b-c)^2+(c-a)^2\geq0$. Equality holds for  $a=b=c$.

Dear sqing here :$a,b,c \geq ,$  watt is?

[quote=sqing][b]Problem 756[/b]
Let $a,b,c \geq , ab+bc+ca >0$  and $a+2b+2c=1.$ Find the maximum of $\frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}.$[/quote]

Holds and that$a,b,c \geq 0$ and $a+b+c=1$ then

$\frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}<\sqrt{\frac{3}{2}}$

[quote=MariusStanean][b]Problem 757[/b]
Let $a,b,c$ be nonnegative real numbers. Prove that
$$\frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}\le 1-\frac{27abc}{(a+b+c)^3}.$$[/quote]
Very beautiful.

[quote=teomihai][quote=sqing][b]Problem 756[/b]
Let $a,b,c \geq 0, ab+bc+ca >0$  and $a+2b+2c=1.$ Find the maximum of $\frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}.$[/quote]

Solution ?[/quote]

This marathon was died. :)


algorithm
superior algebra
superior algebra unsolved

Identify the following groups for every $n$
\[
H_{2n}:=\langle a,b\mid a^n=b^2=1,\,ab=ba^2\rangle
\]

and also this one

\[
G:=\langle a, b\mid a^4=b^3=1,\,ab=b^2a^2\rangle
\]

[quote="mchis68"]\[
H_{2n}=\left\{\begin{array}{ll}
S_3 & ,\,\mbox{ if }3|n\,,\\
Z_2 & ,\,\mbox{ if }3\not|n\,.
\end{array}\right.
\][/quote]

 :yup:  :thumbup:   :clap2:  Sketch of proof??

\[b^{-1}ab=a^2\Longrightarrow a^4=a\Longrightarrow a^3=1\Longrightarrow\dots\]
and $G=1$: start with $H=<a>$(or $<b>$) to find, using Todd-Coxeter's algorithm(standard tool for group presentations), that $G=H$. Then you get $a=b=1$, so $G=1$.

Excuse me,mchis68,but I'm really interrested in that algorithm due to Todd-Coxeter.Can you enlighten me? :?

What did google ever do to you?

http://en.wikipedia.org/wiki/Todd-Coxeter_algorithm


It's probably a little early, but I was just wondering who was going to the UGA tournament?

i'll be there.

by the way, did you do GA ARML last year captcha?

i haven't for the past two years, as i was on the exeter ARML team, but i was for the three years before that

[quote="captcha000"]ha! let's see if vestavia has improved since the last time i won uga =P[/quote]

huh

what

two years ago?

when we sent eight sophomores, who didn't even have an algebra course yet in high school?

yea

I will be going either as an individual, an individual-team, or I shall convince 3 more people at Wheeler to come. Most likely one of the first 2.

I'll be there, hopefully with a team. And I expect the Augusta homeschoolers will be there too, as one of them was in the top 40 last year too.


Hi, sorry i've got another question. If someone could help me out please. The question is:

Find the coefficient of x^9  in the series expansion of (1-2(x)^3))^-4 (if that confuses you, then i've got the link below for the proper question). Thanks a million.

http://members.optusnet.com.au/gujju_style/maths6.JPG

I really don't see what could be the problem in solving this problem, unless of course you don't know what a binomial expansion is...
Just substitute the neccesary quantities.
$(1+x)^{m}=\cdots$
$m=-4$ and replace $x$ by $-2x^{3}$

To be honest, i don't really know what it is. Which is the reason why i'm so stumped. I've tried to look at a few examples on the net, but don't really seem to make sense to me. Any help please?

Dunno if this's right:
Let $x^{3}= t$. We're looking for the coefficient of $t^{3}$ in \[(1-2t)^{-4}= \left(\frac{1}{1-2t}\right)^{4}= (1+2t+4t^{2}+8t^{3}+16t^{4}+...)^{4}\] The terms $16t^{4}, 32t^{5}, 64t^{6},...$ aren't interesting. We clearly want to sum the coefficients of $(1)^{3}(8t^{3})$, $(1)(2t)^{3}$ and $(1)^{2}(2t)(4t^{2})$ in the expansion of $(1+2t+4t^{2}+8t^{3})^{4}$. Now use [url=http://mathworld.wolfram.com/MultinomialSeries.html]multinomial series[/url] to get that $\frac{4!}{3!1!}+\frac{4!}{1!3!}+\frac{4!}{2!1!1!}= 18$ is the coefficient we're looking for.

[quote="Kurt Gödel"]Now use [url=http://mathworld.wolfram.com/MultinomialSeries.html]multinomial series[/url] to get that $\frac{4!}{3!1!}+\frac{4!}{1!3!}+\frac{4!}{2!1!1!}= 18$ is the coefficient we're looking for.[/quote]
should be
$\frac{4!}{3!1!}\cdot8+\frac{4!}{1!3!}\cdot2^{3}+\frac{4!}{2!1!1!}\cdot2\cdot4= 160$

cheers


Each row of the Misty Moon Amphitheater has $ 33$ seats. Rows $ 12$ through $ 22$ are reserved for a youth club. How many seats are reserved for this club?

$ \textbf{(A)}\ 297\qquad
\textbf{(B)}\ 330\qquad
\textbf{(C)}\ 363\qquad
\textbf{(D)}\ 396\qquad
\textbf{(E)}\ 726$

[hide="Solution"]
There are $ (22\minus{}12\plus{}1)33\equal{}\boxed{\textbf{(C)}\ 363}$ seats reserved.
[/hide]


I don't know the role of b in this problem?

I think that there is an error in your text.
Here is the problem,with an another faces.(see the attachment)


$ a,b,c\ge 0$ Prove $ \frac{a^2\plus{}b^2\plus{}c^2}{5}\ge min((a\minus{}b)^2,(b\minus{}c)^2,(c\minus{}a)^2)$
EMV work here

[quote="dima ukraine"]$ a,b,c\ge 0$ Prove $ \frac {a^2 \plus{} b^2 \plus{} c^2}{5}\ge min((a \minus{} b)^2,(b \minus{} c)^2,(c \minus{} a)^2)$
EMV work here[/quote]
Assume that $ a \ge b \ge c$, then
\[ \min\left\{ (a\minus{}b)^2,(b\minus{}c)^2,(c\minus{}a)^2\right\} \equal{}\min\{ (a\minus{}b)^2,(b\minus{}c)^2\}\]
If $ a\plus{}c \ge 2b$, then $ (b\minus{}c)^2 \equal{}\min \{ (a\minus{}b)^2,(b\minus{}c)^2\}$, we have to prove
\[ a^2\plus{}b^2\plus{}c^2 \ge 5(b\minus{}c)^2\]
which is true because
\[ a^2\plus{}b^2\plus{}c^2\minus{}5(b\minus{}c)^2 \ge (2b\minus{}c)^2\plus{}b^2\plus{}c^2\minus{}5(b\minus{}c)^2\equal{}3c(2b\minus{}c) \ge 0\]
If $ a\plus{}c \le 2b$, then $ (a\minus{}b)^2\equal{}\min\{ (a\minus{}b)^2,(b\minus{}c)^2\}$, we have to prove
\[ a^2\plus{}b^2\plus{}c^2 \ge 5(a\minus{}b)^2\]
which is true because
\[ a^2\plus{}b^2\minus{}5(a\minus{}b)^2\equal{}2(2a\minus{}b)(2b\minus{}a) \ge 0\]
This ends the proof. :)


Solve the exponential equation $ \sqrt {3^{x}} \plus{} 1 \equal{} 2^{x}$  where  $ x$ is in $ \mathbb{R}$

hello, the only solution is $ x\equal{}2$,defining the function $ f(x)\equal{}\sqrt{3^x}\plus{}1\minus{}2^x$ then we get $ \lim_{x \to \minus{}\infty}f(x)\equal{}1$ and $ \lim_{x \to \infty}f(x)\equal{}\minus{}\infty$ further has $ f(x)$ at the point $ x\equal{}x_{Max}\equal{}\frac{\ln\left(\frac{4\ln(2)^2}{\ln(3)^2}\right)}{\ln(\frac{3}{4})}<0$ a local maximum, with $ f(x_{Max})\approx 1.09$.
Sonnhard.

Hi Dr.Sonnhard Graubner
I'm sorry but you're solution is not correct.There's an elementary solution.

hello, but why is my solution not correct?
Sonnhard.

sorry,my bad.You're solution is OK(as always).

Sonhard's solution is not okay: the fact that $ f \to 1$ as $ x \to \minus{}\infty$ surely does not show that there $ f$ never takes the value 0 for some negative $ x$.  In fact, nothing Sonhard has said rules out solutions other than $ x \equal{} 2$.

$ \sqrt{3^x}\plus{}1\equal{}2^x\Longleftrightarrow 1\plus{}\frac{1}{\sqrt{3^x}}\equal{}\left(\frac{2}{\sqrt{3}}\right)^x$,

LHS is strictly decreasing on the other hand RHS is strictly increasing,hence there is just a solution $ x\equal{}2$.

Is there a way to solve this with logarithms?

hello JBL,where should $ f(x)$ have another intersection point with the $ x$ axes?Please remember that $ f(0)\equal{}1>0$ and $ f(x)$ has only one local extremum $ x_{Max}$ with $ f(x_{Max})>1$, further $ f(x)$ is a continously function and for $ x>0$ strictly monotonously decreasing? I don't understand your argumentations.
Sonnhard.

This is elementary solution.
$ \sqrt {3^x } \plus{}1 \equal{}2^x \Rightarrow (\frac{\sqrt 3}{2})^{x} \plus{}(\frac{1}{2})^{x}\equal{}1 \Rightarrow (\sin 30)^{x} \plus{} (\cos 30)^{x}\equal{}1 \Longrightarrow x\equal{}2$

Hi Sonnhard (and sorry for misspelling your name above), I agree that what you have now written is a proof -- but the words "only one local extremum" are very important and are missing from your first post, where instead you wrote "a local maximum," which is not enough to conclude.  (Perhaps I should have realized your intended meaning.)

[quote="enndb0x"]This is elementary solution.
$ \sqrt {3^x } \plus{} 1 \equal{} 2^x \Rightarrow (\frac {\sqrt 3}{2})^{x} \plus{} (\frac {1}{2})^{x} \equal{} 1 \Rightarrow (\sin 30)^{x} \plus{} (\cos 30)^{x} \equal{} 1 \Longrightarrow x \equal{} 2$[/quote]

Just to comment, we should justify why:
If $ a^2\plus{}b^2\equal{}1$ and $ a^x\plus{}b^x\equal{}1$ with $ a,b>0$, then why does it follow that $ x\equal{}2$?

If $ x>2$, then, $ 1>a>0$ so $ a^x<a^2$, similar for $ x<2$.


Show that from any five integers, not necessarily distinct, one can always choose three of these integers whose sum is divisible by 3.

All integers can be grouped into three 'classes': 0, 1, and 2 mod 3. If we have the sums of one of each class it will be divisible by three: $0 + 1 + 2 \equiv 3 \equiv 0 \bmod{3}$. If we have any three of one class, it will also be divisible by three: $0 + 0 + 0 \equiv 0 \bmod{3}$, $1 + 1 + 1 \equiv 3 \equiv 0 \bmod{3}$, and $2 + 2 + 2 \equiv 6 \equiv 0 \bmod{3}$.

Case 1: We have three integers $a, b, c$ from a set of five integers such that $a \equiv 0 \bmod{3}$, $b \equiv 1 \bmod{3}$, and $c \equiv 2 \bmod{3}$. We have shown that the sum of one integer from each class will be divisible by three.

Case 2: There are no integers $a, b, c$ such that $a \equiv 0 \bmod{3}$, $b \equiv 1 \bmod{3}$, and $c \equiv 2 \bmod{3}$. In other words, there are five integers that belong to only two of the three classes. Thus, there must be one class that has at least 3 integers belonging to it. As shown, any three integers in this one class will be divisible by three when summed.

A couple of thoughts about this proof.

First, you don't need to put 'quotes' around classes.  0, 1 and 2 are known as residue classes or equivalence classes, so the word is simply the right word without analogy.

Second, while this may seem trivial, the better way to phrase Case 2 is to invoke the pigeon hole principle.  Given nk + 1 pigeons in n holes, there must be at least one hole that contains at least k + 1 pigeons.  Phrasing your proof as

"If none of the 5 integers belongs to one of the classes, then by the pigeon hole principle, one of the two remaining classes contains at least three of the integers."

Your first paragraph could also be subsumed by your casework.  There is no real need for lemmas here and the proof will flow better if you simply say

"When three of the integers fall into one of the residue classes, then

\[ n_1 + n_2 + n_3 = 3(n_1) \equiv 0 \pmod{3}.\" \]


Yeah, just use the fact that n! = n*(n-1)! as Arne said for questions of that type and cancel out the 8! everywhere.

As for the second one, I've never found anything that works better than graphing.  Graphing is nice and safe.


probability
limit
expected value
conditional probability

In every second, the probability that a certain bacterium will die is p and the probability that it will divide into two identical bacteria is 1-p. (The offsprings will die or divide independently of each other.) What is the probability of extinction?

Edit: It already passes the deadline, so feel free to post it if you can. I still couldn't solve it though. But I'm pretty sure that it has something to do with recursion

Let $ a$ be the probability of extinction given a start of one cell. If the cell dies, we are done. If the cell does divide, then both have to die, so that has a probability of $ a^2$ since both bacteria must die.

So $ a\equal{}(1\minus{}p)\plus{}p( a^2)$

So $ a\equal{}1$ or $ a\equal{}1\minus{}\frac{1}{p}$. Clearly, the second option is nonsense since it is negative so $ a\equal{}1$...?

First, you need to flip $ p$ and $ 1\minus{}p$. ;)  And the second root is actually (once flipped) $ \frac{p}{1\minus{}p}$, which is between $ 0$ and $ 1$ when $ p < 1/2$. When $ p > 1/2$, the cell dies too often, so the probability of extinction is $ 1$.

Hmm... What are you guys doing here? I think we have some "different" interpretation on this problem. My interpretation is that the problem asks for the probability when the time is infinite, which means it can't be as simple as both solutions above.Here's my little effort that I had made:
Let P(i) be the probability of the extinction at i-th second. We have:
$ P(1) \equal{} p$
$ P(2) \equal{} (1 \minus{} p)p^2$
$ P(3) \equal{} 2(1 \minus{} p)^2.p^3$(since we have 2 choices which bacteria will die) + $ (1 \minus{} p)^3.p^4$
$ P(4) \equal{} 4(1 \minus{} p)^3.p^4 \plus{} 6(1 \minus{} p)^4p^5 \plus{} 6(1 \minus{} p)^5.p^6 \plus{} 4(1 \minus{} p)^6.p^7 \plus{} (1 \minus{} p)^7.p^8$
Because the probability now becomes much unpredictable in the case when bacteria don't die before i-th second, I have to find a formula to describe this probability at any time. But I failed to find out such an expression like that. Btw, we can see that p must be less than 1 and greater than 0, so Altheman's solution has some flaws.
@MellowMelon: What do you meant when $ p > 1/2$, the cell dies too often? I never see such an argument like that before though. Can you elaborate more?

Yes, we are doing it for infinite time too. I think you misunderstood what Altheman was doing (I did too at first). Remember $ a$ is the probability of extinction. If the cell dies on the first one, great, that's it. Happens with probability $ p$. If it doesn't, we have two cells. We can consider them as separate entities now, both which must become extinct. By definition it happens for each one with probability $ a$, with total probability $ (1\minus{}p)a^2$.

My comment that $ p > 1/2$ meant the cell dies too often was in reference to the fact that the other root, $ \frac{1\minus{}p}{p}$, was greater than $ 1$ when $ p > 1/2$. Was just an observation.

Another way to explain the case $ p > \frac{1}{2}$ is to note that the expected number of cells at time $ t$ goes to 0 as $ t\to \infty$ in this case, so the probability that any cells are left must also go to 0.  (Note that this [i]doesn't[/i] explain the fact that extinction is a certainty when $ p \equal{} \frac{1}{2}$.)

$ p \equal{}$probability of death.

Just so we are clear, I did switch the $ p$ and $ 1 \minus{} p$ in my first post but otherwise it is correct.

In[6]:= Reduce[{a == (p) + (1 - p) a^2, p >= 0, 1 >= p, a >= 0,
  1 >= a}, a]

Out[6]= (0 <= p < 1/2 && (a == p/(1 - p) || a == 1)) || (1/2 <= p <=
     1 && a == 1)

The case remains to tell when $ p\in [0,0.5)$ whether $ a \equal{} 1$ or $ a \equal{} \frac {p}{1 \minus{} p}$.

Here is another way to look at it that I believe settles the problem.

I think we can look at this from an expected value perspective. Let $ E_n$ be the expected number of bacteria after $ n$ divisions. $ E_0 \equal{} 1$ of course. Now $ E_n \equal{} (1 \minus{} p)(2E_{n \minus{} 1})$ since for each bacteria, $ (1 \minus{} p)$ chance of living and that would double the number of bacteria. Otherwise, death would take that bacteria to zero. I suppose that I cannot quote the theorem, but this is valid to sum up in this way for expected values.

Now $ E_n \equal{} (2 \minus{} 2p)^n$. if $ 2 \minus{} 2p\ge 1\iff p\le\frac {1}{2}$, then as $ n\rightarrow\infty$, $ E_n > 0$ so we cannot have total extinction so $ a\ne 1$ in that case and we are done.

I think I now understood the insight in Altheman's solution. The reason he gives the probability of extinction after the first time is because this probability represents an infinite time. So if we let a be the total probability of extinction at ANY TIME, then it still equals to the probability of extinction of one cell AT ANY TIMES AFTER the 1st one multiply the conditional probability to make sure that this extinction can happen after the 1st one. I think that's why he has the equation: $ a\equal{}(1\minus{}p)a^2\plus{}p$, which represents the total probability of extinction at any time=probability of extinction at the 1st one+the probability at anytime after the 1st one.
Btw, I'm curious whether we actually have a closed-form formula to describe the probability of extinction at any time(as I did mention in my post above, that's the only way I had thought of). Now I realize that the coefficients in each of the probability of extinction at i-th second, or $ P(i)$, are actually numbers in Pascal's triangle. Therefore, I'm pretty sure there exists the closed-form formula. Should we find out an alternative way to solve this problem, which is: "Find the closed form formula for this Probability". :)

[quote="Altheman"]Now $ E_n \equal{} (2 \minus{} 2p)^n$. if $ 2 \minus{} 2p\ge 1\iff p\le\frac {1}{2}$, then as $ n\rightarrow\infty$, $ E_n > 0$ so we cannot have total extinction so $ a\ne 1$ in that case and we are done.[/quote]  This is wrong: if we have a sequence of nonnegative integer random variables $ (X_i)$ with $ \mathbb{E}(X_i) \geq c > 0$, this does not imply $ \lim_{i \to \infty}\Pr(X_i > 0) > 0$.  (This is the content of the parenthetical in my previous post.  Note that the converse of this statement is true.)  For example, consider the "double or nothing game": you start out with one dollar, and repeatedly flip a coin.  Every time it's heads, the casino doubles your money, but if it's ever tails, the casino wipes you out.  Your expected value at time $ t$ is always one dollar, but you have probability 0 of playing forever (i.e., probability 1 of extinction).

By the way, this problem is very similar to [i]percolation on a binary tree[/i].  I stumbled across the following article by David Austin, which presents a nice discussion: http://www.ams.org/featurecolumn/archive/percolation.html

ghjk: We could ask for even more, viz., the probability $ P(k; t)$ that at time $ t > 0$ there are exactly $ 2k$ cells alive.  Then we have $ P(0; 1) \equal{} p$, $ P(1; 1) \equal{} 1 \minus{} p$ and $ P(n; t \plus{} 1) \equal{} \sum_{m \geq 0} \binom{2m}{n} P(m; t) p^{2m \minus{} n} (1\minus{} p)^n$.  Anyone want to give solving this recursion a try?

[quote="JBL"]
ghjk: We could ask for even more, viz., the probability $ P(k; t)$ that at time $ t > 0$ there are exactly $ 2k$ cells alive.  Then we have $ P(0; 1) \equal{} p$, $ P(1; 1) \equal{} 1 \minus{} p$ and $ P(n; t \plus{} 1) \equal{} \sum_{m \geq 0} \binom{2m}{n} P(m; t) p^{2m \minus{} n} (1 \minus{} p)^n$.  Anyone want to give solving this recursion a try?[/quote]
What is m here? Did you check it in some small cases like at $ P(2,3)$ or $ P(3,4)$? This result can solve this problem completely, of course, if it's true, which I suspect. :(

That recursion lets us compute any given value $ P(n; t)$ that we like, but it's a far cry from a closed form for those values, which would be much nicer.  I haven't tried to solve it yet (and I don't know if it has a nice solution) -- generating-functional methods are probably called for, though.

Now, about where the recursion comes from: suppose we know that there were $ 2m$ cells at time $ t$.  Then the probability that there are $ 2n$ cells at time $ t \plus{} 1$ is $ \binom{2m}{n} p^{2m \minus{} n} (1 \minus{} p)^n$.  Thus, the probability that there were $ 2m$ cells at time $ t$ followed by $ 2n$ cells at time $ t \plus{} 1$ is $ \binom{2m}{n} P(m; t) p^{2m \minus{} n} (1 \minus{} p)^n$.  Thus the probability that there are $ 2n$ cells at time $ t \plus{} 1$ is equal to the sum of this expression over all values of $ m$, which is what I've written.


[quote="pluricomplex"]Find the infimum of $\alpha$ and supremum of $\beta$ so that for all $a,b,c$ be the sides of a triangle we have
\[ \beta<\Big|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\Big|<\alpha \][/quote]
For a lower bound, what if $a=b=c$ ?

See also here
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=1893755

[quote="darij grinberg"][quote="pluricomplex"]Find the infimum of $\alpha$ and supremum of $\beta$ so that for all $a,b,c$ be the sides of a triangle we have
\[ \beta<\Big|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\Big|<\alpha  \][/quote]

As for $\alpha$: this is problem 58 (c) from

Arthur Engel, [i]Problem-Solving Strategies[/i], New York 1998,

chapter 7. In fact, the smallest upper bound $\alpha$ is $\alpha=\frac{8\sqrt2-5\sqrt5}{3}<0.04446$, and equality holds for a degenerate triangle with sidelengths c = 1, $b=\frac{\sqrt{10}+\sqrt5+\sqrt2+1}{2}$ and a = b + c.

As for $\beta$: Nice one ;) .

  darij[/quote]
The constant is kind of ugly. Here is the one with nicer number.
for all $a,b,c$ be the sides of a triangle we have
\[ \Big|\frac{a-b}{a+2b}+\frac{b-c}{b+2c}+\frac{c-a}{c+2a}\Big|< \frac12  \]


How many distinct 6-letter arrangements are there of the letters in the word STREET?


function
inequalities
real analysis
real analysis unsolved

Show that exists a global solution for the differential equation

$ y'\equal{}x^{\alpha} \minus{} y^{\minus{}\beta}$ with $ x > 0,\; y>0,\; \alpha>0,\; \beta >0$

My problem is how to found a lower bound "$ v$"to use the Comparision Theorem that says "Let $ v$ and $ w$ be functions that are differentiable in $ [0,\infty]$ and satistfy the inequalities $ v\leq w$ and $ Pw\leq 0 \leq Pv$ in $ [0,\infty]$, with $ P\varphi \equal{} \varphi ' \minus{} f(x,\varphi)$. Then the differential equation $ y'\equal{}f(x,y)$ has a global solution $ \phi$ with $ v\leq \phi \leq w$ for $ x\geq 0$.)

Thanks

The key is to use the right initial conditions. Try $ y(1)\equal{}1$. Then $ 1$ is a lower bound going both forward and backward: $ x^{\alpha}\minus{}y^{\minus{}\beta}\ge 0$ for $ x,y\ge 1$ and $ x^{\alpha}\minus{}y^{\minus{}\beta}\le 0$ for $ x,y\le 1$

[quote]
satistfy the inequalities $ v\leq w$ and $ Pw\leq 0 \leq Pv$ in $ [0,\infty]$, with $ P\varphi \equal{} \varphi ' \minus{} f(x,\varphi)$.
[/quote]
To start with, I'm pretty sure it should be $ P{\color{red}v}\leq 0 \leq P{\color{red}w}$ in this particular case ;)


topology
function
trigonometry
complex analysis
complex numbers
complex analysis unsolved

Show that if $f$ is entire and $f(z) \in \mathbb{R} \Leftrightarrow z \in \mathbb{R}$ then $f$ has at most one zero.

I know it's classical, but I'm just beginning ...

An intuitive sketch:

Pick a an $R>0$ such that $f$ has no zeros on boundary of the disk $\mathcal D_R$ of radius $R$ centered in the origin, and let $\gamma: [0,2\pi]\to\mathbb C,\ \gamma_R(t)=Re^{it}$. The number of zeros of $f$ inside $\mathcal D_R$ is the winding number of the closed curve $\Gamma_R=f\circ\gamma_r$ around $0$. Since this curve cuts the real axis only in two points, its index around $0$ cannot be larger than $1$.

To make things rigorous, let $\omega_R=\frac{f(R)+f(-R)}2,\tilde R=\frac{|f(R)-f(-R)|}2$, and define the homotopy $H: [0,2\pi]\times[0,1]\to\mathbb C,\ H(t,s)=(1-s)\Gamma_R(t)+(\omega_R+\tilde Re^{it})s$. This is a homotopy from $\Gamma_R(t)$ to $\tilde\Gamma_R(t)=\omega_R+\tilde Re^{it}$ carried out in $\mathbb C\setminus\{0\}$, so the winding numbers of $\Gamma_R$ and $\tilde\Gamma_R$ around $0$ are equal. The latter, however, is clearly either $0$ or $1$, depending on whether $0$ lies inside or outside the circle of radius $\tilde R$ centered at $\omega_R$.

Edit:

Sorry, I forgot to mention this: in the paragraph above, I asumed WLOG that $f$ takes the open upper half-plane into itself, and the open lower half-plane into itself, since there were only two possibilities (the other one being that $f$ takes the open upper half-plane into the lower one and vice-versa, and in this situation we could work with $-f$ instead of $f$).

Sorry for the double post, but the message would have become ridiculously long :).

I saw a problem in the AMM (problem AMM E3242, proposed by J. B. Miles and L. A. Rubel), asking for a proof of the fact that an entire function with this property ($f(z)\in\mathbb R\iff z\in\mathbb R$) must be of the form $f(z)=az+b,\ a\in\mathbb R^*,\ b\in\mathbb R$. I think this should work:

The proof above basically says that for every real $a,\ f^{-1}(a)$ has at most one element, and if $f(w)=a$, then $w$ is a simple zero of $f(z)-f(a)$. Let's assume for brevity that $f(0)=0$ (we simply subtract a real constant from our initial $f$ and work with this new function). Then $f$ is of the form $f(z)=ze^{g(z)}$ for some entire function $g$, which we assume to be non-constant. Let's see what $f(z)\in\mathbb R (z\ne 0)$ means: we write it as $\bar z\overline{e^{g(z)}}=ze^{g(z)}\ (1)$. Since $\overline{e^t}=e^{\bar t},\ \forall t\in\mathbb C,\ (1)$ says that $\frac{\bar z}z=e^{g(z)-\bar g(z)}\ (2)$.

Let $\arg z$ be the unique real in $[0,2\pi)$ such that $\frac z{|z|}=\cos(\arg z)+i\sin(\arg z)$, and let $\Im g$ be the imaginary part of $g$. Then $(2)$ can be written in the form $\cos(2\arg z)-i\sin(2\arg z)=e^{2i\Im g(z)}=\cos(2\Im g(z))+i\sin(2\Im g(z))$, which, in turn, is equivalent to $\Im g(z)+\arg z=k\pi\ (3)$ for some integer $k$. Denote the open upper half-plane by $\Pi_+$, and the open lower half-plane by $\Pi_-$. $\Im g$ cannot be bounded in the entire plane, so its unbounded on either $\Pi_+$ or $\Pi_-$. Suppose it's the former. Let $u$ be the restriction of the LHS of $(3)$ to $\Pi_+$. $u$ is then a continuous real function, and $u(\Pi_+)$ is unbounded, because $\Im g$ is unbounded in $\Pi_+$ by assumption, and $\arg$ is bounded. $u(\Pi_+)$ is also connected, so it must contain a real of the form $k\pi,\ k\in\mathbb Z$. This value, however, is achieved for some $z\in\Pi_+$, which is not real, and this contradicts our hypothesis.


function
combinatorics theorems
combinatorics

There are two kinds of Stirling numbers, called the first and second kinds.

The Stirling number of the first kind are the number of ways to arrange n objects into k cycles. We will be using the notation [n.k] to denote Stirling numbers of the first kind. (This notation isn't really standard.) To give an example of what that means, [4,2]=11 because we can divide {1,2,3,4} into [1,2,3][4], [1,3,2][4], [1,2][3,4], [1,2,4][3], [1,4,2][3], [1,3][2,4], [1,3,4][2], [1,4,3][2], [1,4][2,3], [2,3,4][1], and [2,4,3][1]. [1,2,3] is the same as [2,3,1], for example, because it's a cycle. (You can think of it as stringing  beads on a necklace, so you can move the beads around on the necklace, but you aren't allowed to flip it over.

A recursive formula for [n,k] is given by

[n,k] = (n-1) [n-1,k] + [n-1,k-1]

(Can you see why?)

Stirling numbers of the second kind are the number of ways to partition a set of n things into k nonempty subsets. We shall write {n,k} for Stirling numbers of the second kind with the slight provision that the same notation is used for sets, so it should be obvious. For example, {4,2}=7 because we have {1,2,3}U{4}, {1,2}U{3,4}, {1,2,4}U{3}, {1,3}U{2,4}, {1,3,4}U{2}, {1,4}U{2,3}, and {2,3,4}U{1}. Note that the sets are not ordered, so {1}U{2,3,4} is the same as {2,3,4}U{1}.

A recursive formula for {n,k} is given by

{n,k} = k {n-1,k} + {n-1, k-1}.

(Can you see why?)

Ok that's all very well, but why are these numbers at all useful, and why are they famous? Well, they satisfy many interesting identities. See if you can prove this one combinatorially:

:sum: (k=0 to n) [n,k] = n!

Next, try expanding x(x+1)(x+2)...(x+n-1). What do you notice about the coefficients you're getting? Well, if you have calculated a small ``Pascal'' triangle for [n,k], you may notice that

x(x+1)(x+2)...(x+n-1) = :sum: [n,k] x^k.

What if we go backwards? Can you find a way to write x^n as the sum of x(x-1)(x-2)...(x-n+1)? Well, again, if you have a ``Pascal'' triangle for {n,k}, you may notice that

x^n = :sum: {n,k} x(x-1)(x-2)...(x-k+1).

If you have any questions, feel free to post them here.

I saw a very nice proof of the fact that sum from k=0 to n of [n,k]P(t,k)=t^n. (I think that's the identity, at least) May I post?

ok. we know that [n,k] means the number of ways to partition a set of n integers into k partitions. So, we write the k partitions in a row. Ok, now we know that P(t,k) means the number of ways to choose k integers from t integers in some order. So, we now have k integers in some specific order. We write those in a row under the k partitions that we already have. This creates a correspondence between the k partitions and the k numbers. This is actually a function, since there is never more than one output for a given input. Therefore, [n,k]*P(t,k) represents the number of ways to form a function from a set of n integers into another set of t integers, that in this case happens to have image size k. If we sum this over all k from 0 to n, we get that the sum of k from 0 to n of [n,k]*P(t,k) will be the number of functions from a set of n integers to a set of t integers with any image size possible. This will of course be t^n, giving us our identity.


R/r>(b/a)+(a/b), where a,b are different sides of a triangle.

i know it is difficult, but i can't believe no one solve it... :(

Just wondering, does $R$ stand for circum-radius and $r$ for in-radius?

hmm...

we will use the fact that : $l_{a}\geq h_{a}$, so : $l_{a}^{2}=\frac{4pbc(p-a)}{(b+c)^{2}}$ but $(b+c)^{2}\geq 4bc$, therefore: $l_{a}^{2}\leq p(p-a)$

but from $l_{a}\geq h_{a}$ we have : $_{b}^{2}+h_{c}^{2}\leq l_{b}^{2}+l_{c}^{2}\leq ap$, but $h_{b}=\frac{2S}{b}$ and $h_{c}=\frac{2S}{c}$

$\Rightarrow 4S^{2}(\frac{1}{b^{2}}+\frac{1}{c^{2}})\leq ap$, and multypling by $\frac{bc}{4S}\Longleftrightarrow \frac{b}{c}+\frac{c}{b}\leq \frac{R}{r}$


Prove that if $\forall k \in \mathbb{N*}, a \in P$ and $a>5$ then $a^{4k}-1 \vdots 240$


give the criterion for which

$n^{^}n^{^}n^{^}{.}^{^}{.}^{^}{.}$

converges, where $n\;\in\;\mathbb{R}^{+}\;\;$?
(took me well over 1 hr to solve this!) :rotfl:

FYI: answer below...

[hide]
$\;\;\;\frac{1}{e^{e}}\;\;\leq\;\;n\;\;\leq\;\;\sqrt[e]{e}$
[/hide]

[quote="misan"]
(took me well over 1 hr to solve this!) :rotfl:
[/quote]
It would take you much less time to find it on the forum if you bothered to search :rotfl: The last time it was posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=146808]here[/url] Seriously, folks, I can understand when something posted by harazi and solved by grobber at the dawn of time pops up again, but to post problems that are currently [b]on the front page[/b] (in Calculus Unsolved and Proposed subforum) is a bit too much :(


Any ideas?

             Let $f : [0, 1] \to\mathbb{R}$ be a continuous function such that $f (0) = f (1) = 0$. Prove
that the set
                $A : =\{ h \in [0, 1] : f (x+h) = f (x) \text{ for some }x \in [0, 1]\}$

is Lebesgue measureable and has Lebesgue measure at least $\frac{1}{2}$ .

Let $B = \{h: 1-h \in A\}$. Of course $\mu A = \mu B$.

For $h \in [0,1]$ let $g(x) = f(x+h)-f(x)$ for $x+h \leq 1$, $g(x) = f(x+h-1)-f(x)$ for $x+h > 1$.

$g$ is continous and has zero in $[0,1]$, hence $h \in A$ or $h \in B$ so $A \cup B = [0,1]$.

Nice! How do you conclude that A is measurable?

Oh, I forgot about it. $f$ is continous, hence $A$ is closed.

Sorry for late reply Fedor.

$ f$ takes its minimum and maximum value at, say, $ a, b$. Then $ g(a) \geq 0$ and $ g(b) \leq 0$.