geometry
3D geometry
tetrahedron
octahedron
icosahedron
conics
ellipse

Treething sells eggs by the dozen. Each one of his eggs is a regular tetrahedron with side length x. What is the combined volume of a dozen of Treething's eggs? Express your answer in terms of x.

The altitude of the base is $\frac{x\sqrt{3}}{2}$. Since the height of the tetrahedron hits the base in the centroid. The distance between the centroid is $\frac{x\sqrt{3}}{2}*\frac{2}3=\frac{x\sqrt{3}}{3}$

using pythagorean theorem, you get the height of the tetrahedron is $\frac{x\sqrt{6}}3$

and the base area is ${\frac{x^{2}\sqrt{3}}4}$

so the volume of each egg is ${\frac13*\frac{x\sqrt{6}}3*\frac{x^{2}\sqrt{3}}4}=\frac{x^{3}\sqrt2}{12}$

volume of a dozon is $\boxed{x^{3}\sqrt{2}}$

Extension: Treething sells eggs by the dozen. Each one of his eggs is a regular octahedron with side length $x$. What is the combined volume of a dozen of Treething's eggs? Express your answer in terms of $x$.

And while we're at it, half of Treething's eggs by the dozen turn out to be perfect icosahedrons with side length $x$. The other half turn out to be the three dimensional shape formed by rotating a $2.2$-ellipse about one of its axes of length $x$. What is the combined amount of juicy egg yolk in a dozen of his eggs (assume the egg whites are negligible)

Note: a 2.2 ellipse satisfies the equation $x^{2.2}+y^{2.2}=r^{2.2}$. It is supposed to be one of the most aesthetic shapes (I think I read that somewhere a while ago?)

????

Treething lays eggs??!?!?!? lol

Octahedron:

[hide]
Split it into two pyramids.  The height of each pyramid forms a right triangle with half a diagonal and a side, so \[h^{2}=\frac{1}{2}x^{2}+x^{2}\Rightarrow h=\frac{\sqrt{6}}{2}x\] The volume of a pyramid is $1/3Bh$, so the area of this pyramid is $\frac{\sqrt{6}}{6}x^{3}$, and a dozen of them have volume \[2\sqrt{6}x^{3}\] [/hide]

About the 2.2-ellipses --  :ninja:

[quote="not_trig"]Octahedron:

[hide]
Split it into two pyramids.  The height of each pyramid forms a right triangle with half a diagonal and a side, so \[h^{2}=\frac{1}{2}x^{2}+x^{2}\Rightarrow h=\frac{\sqrt{6}}{2}x\] The volume of a pyramid is $1/3Bh$, so the area of this pyramid is $\frac{\sqrt{6}}{6}x^{3}$, and a dozen of them have volume \[2\sqrt{6}x^{3}\] [/hide]

About the 2.2-ellipses --  :ninja:[/quote]

The height is a leg, not the hypotenuse.  :wink:

A "2.2 ellipse" is defined only when $x,y\ge0$. And even in the first quadrant, it's a circle.

[quote="funcia"][quote="Allan Z"]Now I have to memorize this:
The volume of a regular tetrahedron with side length x is
$\boxed{\frac{x^{3}\sqrt2}{12}}$

 :rotfl:[/quote]

Do you have any other -hedron formulae? they kill me...[/quote]

Octahedron: $\frac{x^{3}\sqrt2}{3}$, but you should know how to derive that. :wink:

can anyone derive the formula for the volume of a regular isocahedron with side length x?

that was the other part of my question...

[quote="mustafa"][hide]
Volume of a regular icosahedron: 5/12(3+sqrt5)x^3

Where x is the edge length

[/hide][/quote]

can you tell us how you got that?

http://mathworld.wolfram.com/Icosahedron.html
Derivation of Volume is at the bottom. Basically split it into 20 pyramids.

And the more natural extension:

Treething sells eggs by the dozen. The eggs are ellipsoids (spelling?), formed by rotating an ellipse about its major axis. If the minor axis has length $a$ and the major axis has length $b$, find the combined volume in terms of a and b.

:D

I used calculus and i get $\frac{4}{3}a^{2}b\pi*12=16a^{2}b\pi$..

I know.. calculus is way too difficult for this forum

What a sec are you sure there's no pi in there? Cuz rotating stuff in calculus usually gives a pi (at least what i remember from the last section of calc AB) and using geometry projections keeps the pi also...

anyway thats how you do it without calculus - stretch the oblong shape until the two axes have equal length and see what it does to volume...

[quote="lotrgreengrapes7926"]A "2.2 ellipse" is defined only when $x,y\ge0$. And even in the first quadrant, it's a circle.[/quote]

It most certainly is neither a circle nor an ellipse, and in the form it is given, it most certainly is defined in all four quadrants.

Allan Z:  Your expression is incorrect.

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

opps.. forgot the $\pi$, ya in rotations, there is a $\pi$
:wallbash: <<< that's me

t0rajir0u: are you sure? the question is asking about rotating and ellipse by it's major axis. That one in mathworld isn't the ellipsoid formed by rotating, it's the general formula of an ellipsoid. I derived it with the volume equation of an object rotating around x-axis

[quote="t0rajir0u"]The ellipsoid formed by rotating an ellipse about its major axis is just an ellipsoid with two of its semiaxes the same.  So, in other words, the semiaxes are $a, a, b$.

If you did the rotation correctly the two answers should coincide  :wink:[/quote]

exactly, so one egg will be $\frac{4}{3}a^{2}b\pi$.. note that $16a^{2}b\pi$ is the volume of a DOZEN

Well, the answer with the $\pi$ is correct [b]now.[/b]  :P

yeah.. i know.. hehe I was rushing and I dropped out the $\pi$ by mistake..  But I corrected for a long time though


modular arithmetic
trigonometry
logarithms

There are exactly $ k$ perfect squares that satisfy $ 1996^{1996}\text{ (mod k)}\equal{}0$ Find k.

[quote="Temperal"]There are exactly $ k$ perfect squares that satisfy $ 1996^{1996}\text{ (mod k)}\equal{} 0$ Find k.[/quote]

Ok, the wording of this problem is very unusual. I assume you mean:

There are exactly $ k$ perfect squares $ p$ that satisfy:

$ 0\equiv 1996^{1996}\pmod{p}$

And instead of using \text{(mod k)}, it's better to use \pmod{k}, and \equiv produces $ \equiv$.

[hide="Solution"]Basically, the problem asks us to find the number of perfect square factors of $ 1996^{1996}$ (every square number factor satisfies the equation). Factoring,

$ 1996^{1996}\equal{} (2^{2}\cdot 499)^{1996}\equal{} (2^{2})^{1996}\cdot (499^{2})^{998}$.

Hence, the number of factors is $ 1997\cdot 999 \equal{}\boxed{1995003}$[/hide]

I hope I haven't made any arithmetic errors...

That's correct. Yes, I know I screwed up the equiv, and I don't really see the need for pmod when text works fine.

It's generally better practice to use the command than text.  I mean, you don't see people using text for $ \sin,\cos,\log$...


q: find a maximum of f(a,b,c) = a^2.b/(a^2 + (b+c)^2) + b^2.c/(b^2 + (c+a)^2) + c^2.a/(c^2 + (a+b)^2) for a , b , c  > 0.
a: max f = + infinity.
we can see that f(n,n,n) = 3n/5 --> infinity as n --> infinity. Thus the maximum does not exist or if you want it is + infinity.

suggestion: if we modify the hypothesis and put: a + b + c = 1 and a , b , c > 0. what is then the maximum value of f ? i bet you can find an answer.


[quote="indybar"]In $ABC$, $2AC^2=AB^2+BC^2$.
Prove that $cot^2B \geq cotAcotC$[/quote]
$\begin{eqnarray*}
(*)\ \ cot^2 B\ge cotA\cdot cotC & \Longleftrightarrow & accos^2 B\ge b^2 cosAcosC\\
 &  \Longleftrightarrow & (a^2+c^2-b^2)^2\ge (b^2+c^2-a^2)(a^2+b^2-c^2)\\
 &  \Longleftrightarrow & b^4-2b^2(a^2+c^2)+(a^2+c^2)^2\ge b^4-(a^2-c^2)^2\\
 &  \Longleftrightarrow & b^2(a^2+c^2)\le a^4+c^4.
\end {eqnarray*}$

\begin{eqnarray*}
(**)\ \ cot^2\frac{B}{2}\le cot\frac{A}{2}\cdot cot\frac{C}{2} & \Longleftrightarrow & (s-a)(s-c)\le (s-b)^2\\
& \Longleftrightarrow & b^2-(a-c)^2\le (a+c-b)^2\\
& \Longleftrightarrow & b(a+c)\le (a^2+c^2).
\end{eqnarray*}

If $2b^2=a^2+c^2$, then:

$(*)\Longleftrightarrow (a^2+c^2)^2\le 2(a^4+c^4)$, true.

$(**)\Longleftrightarrow b(a+c)\le 2b^2\Longleftrightarrow a+c\le 2b\Longleftrightarrow (a+c)^2\le 2(a^2+c^2)$, true.

[b]Remark.[/b] Originally I thought that the relation (*) is $cot^2\frac{B}{2}\ge cot\frac{A}{2}\cdot cot\frac{C}{2}\ !$. Thus appeared the relation (**).

[quote="levi"]
$\begin{eqnarray*}
(*)\ \ cot^2 B\ge cotA\cdot cotC & \Longleftrightarrow & accos^2 B\ge b^2 cosAcosC\\
 &  \Longleftrightarrow & (a^2+c^2-b^2)^2\ge (b^2+c^2-a^2)(a^2+b^2-c^2)\\
\end {eqnarray*}$
[/quote]

How do you get this?

With the relations $\frac {a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=2R.$ and cosA=cotA.sinA a.s,o.

A further problem:
Assuming the given condition on the sides of the triangle, show that  CotACotC <= (CotB)^2  when B>=60

                                                                                CotACotC<=[(CotB)^2]/4 + 1/4  when B<=60

$4cotAcotC\le 1+cot^2B\Longleftrightarrow 4b^2cosAcosC\le ac\Longleftrightarrow (b^2+c^2-a^2)(b^2+a^2-c^2)\le a^2c^2\Longleftrightarrow b^4\le a^2c^2+(a^2-c^2)^2.\ \ (*)$

$B\le 60\Longleftrightarrow 2cosB\ge 1\Longleftrightarrow a^2+c^2-b^2\ge ac\Longleftrightarrow b^2\le a^2+c^2-ac\Longleftrightarrow$
$b^4\le (a^2+c^2-ac)^2=(a^2+c^2)^2-2ac(a^2+c^2)+a^2c^2=a^2c^2+(a^2-c^2)^2+$
$4a^2c^2-2ac(a^2+c^2)=(a^2-c^2)^2+a^2c^2-2ac(a-c)^2\le a^2c^2+(a^2-c^2)^2,$ i.e. $b^4\le a^2c^2+(a^2-c^2)^2.$ Thus, the relation (*) is true.


geometry
inequalities
calculus
integration
logarithms
number theory
complex analysis

What is your favorite topic in math? (Of the following)  Mine is close between number theorey and geometry, but I will pick number theorey.

[quote="jhollenbeck"]Inequalties is algebra, is it not ;)[/quote]

Yeah..
However, in some books (or math sites  ;) ) are considered a seperate topic....

 :)

calculus :D i even find myself using calculus on problems that have an obvious and somewhat easy non-calculus approach.

I used to like number theory but then now I just seem to have developed a serious liking towards calculus -- Especially after reading russian books on integration from first principles -- The concept is simply amazing !!

Surely Number Theory.
But what is Algebra II for you (especially when inequalities shall belong to Algebra)¿
But Complex Analysis is missing :( (and no, there is a huge difference to normal calculus).
Game theory and other themes of combinatorics are also missing...

PS: I'm sure such poll allready occured ;)

I am currently in Algebra and I like it.

I have looked into the geometry and algebra II book, and it lookd nice!  :D

[quote="ZetaX"]PS: I'm sure such poll allready occured ;)[/quote]
Yeah, there was at least one :roll:

Anyway, I definitely like Geometry the most :D

[size=150]Geometry is  :first: [/size]

Funny, I used to like Algebra and stuff best and I hated Geometry...

I love algebra becuase I have MASTERED IT. I understand everything in Algebra. Geometry is difficult for me......good luck to those who like Geo the best. You are sure to get stumped on  some really weird problem.

I like Geometry, I think most of people who love geometry used to hate it. It's weird

 :first:  Geometry :first:

Algebra II, definitely. Yay for logs and complex numbers! :lol:

I have always liked number theory the best (starting... way back)

Recently I have seen some of the beauty of geometry, and also some inequalities... why do they always want me to learn what I'm bad at? WHY?!

Calculus - I just enjoy the fact that it's much different and can do much more amazing things than everything before it. I mean, yeah Algebra can get pretty complex and interesting (Geometry too), but even the simplest idea in Calculus uses some idea of...infinity!

After that, I love complex numbers and logarithms - logarithms are the only type of problem where I can easily do say AMC 12 #25.
Complex numbers are fun too, hmm *wonders what wonders could be created if we combined complex numbers and calculus  :) , hint hint complex analysis*

[quote="dogseatcheese"]OMG Logs and Complex numbers make my head spin in tiny little circles....and I faint....they are so hard and annoying.[/quote]

Logs and complex numbers together are rather straightforward :)

$\ln (z) = \ln (re^{i \theta}) = \ln r + i \theta + 2 \pi k, k \in \mathbb{Z}$.  :)

Personally, my vote is for complex analysis.  Even if I don't understand it, it's very cool :)

Actually mine is Combinatorics...it rules!


function
search
limit
calculus
calculus computations

Does any one have interesting problems on limits?

Which limits -- sequences, functions? You might also try to search forum for "limit".

Suppose $ f(x)$, $ f'(x)$ and $ f''(x)$ exist and are continuous in $ x \in (0, \infty)$, and that
\[ \lim_{x \rightarrow \infty} (x^2 \, f''(x) \plus{} 4x\, f'(x) \plus{} 2f(x)) \equal{} 1.\]Find $ \lim_{x \rightarrow \infty} f(x)$ and $ \lim_{x \rightarrow \infty} x\, f'(x)$.

hello, here you will find some problems
http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/limcondirectory/LimitConstant.html
Sonnhard.

[quote="JBL"]aidan, look through debank2's posts -- there's basically zero worthwhile content, mostly posts of the same useless form as in this thread.  There's really not much point in responding to someone who comes to a forum with thousands of calculus problems (and a search function) and makes posts like this -- just ignore them.[/quote]

Right.. will do. Anyway I'd appreciate some help with that problem I posted above: I've repackaged it into
\[ \lim_{x \rightarrow \infty} [(x^2 \plus{} 2x)(f'(x) \plus{} f(x))]' \equal{} 1\] and can't think of any way to proceed. Does anyone have a hint to point me in the right direction?

Let
 $ \begin{array}{l} g\left( x \right) \equal{} {x^2}f\left( x \right),x \in \left( {0, \plus{} \infty } \right) \Rightarrow \\
\Rightarrow g'\left( x \right) \equal{} {x^2}f'\left( x \right) \plus{} 2xf\left( x \right) \Rightarrow \\
\Rightarrow g''\left( x \right) \equal{} {x^2}f''\left( x \right) \plus{} 4xf'\left( x \right) \plus{} 2f\left( x \right),\mathop {\lim }\limits_{x \to \plus{} \infty } g''\left( x \right) \equal{} 1 \\
\end{array}$
Also
$ \mathop {\lim }\limits_{x \to \plus{} \infty } f\left( x \right) \equal{} \mathop {\lim }\limits_{x \to \plus{} \infty } \frac {{g\left( x \right)}}{{{x^2}}}\mathop \equal{} \limits^{DLH} \mathop {\lim }\limits_{x \to \plus{} \infty } \frac {{g'\left( x \right)}}{{2x}}\mathop \equal{} \limits^{DLH} \mathop {\lim }\limits_{x \to \plus{} \infty } \frac {{g''\left( x \right)}}{2} \equal{} \frac {1}{2}$

and
$ \begin{array}{l} \mathop {\lim }\limits_{x \to \plus{} \infty } \frac {{g'\left( x \right)}}{{2x}} \equal{} \mathop {\lim }\limits_{x \to \plus{} \infty } \frac {{{x^2}f'\left( x \right) \plus{} 2xf\left( x \right)}}{{2x}} \equal{} \frac {1}{2} \Rightarrow \\
\Rightarrow \mathop {\lim }\limits_{x \to \plus{} \infty } \left[ {\frac {{xf'\left( x \right)}}{2} \plus{} f\left( x \right)} \right] \equal{} \frac {1}{2} \\
\end{array}$
so
$ \mathop {\lim }\limits_{x \to \plus{} \infty } xf'\left( x \right) \equal{} 2\mathop {\lim }\limits_{x \to \plus{} \infty } \left[ {\frac {{xf'\left( x \right)}}{2} \plus{} f\left( x \right) \minus{} f\left( x \right)} \right] \equal{} 2 \cdot \frac {1}{2} \minus{} 2 \cdot \frac {1}{2} \equal{} 0$

Thanks I think I get it now, but I have a question about this part:

$ \mathop{\lim }\limits_{x\to\plus{}\infty }f\left( x\right) \equal{}\mathop{\lim }\limits_{x\to\plus{}\infty }\frac{{g\left( x\right)}}{{{x^{2}}}}\mathop \equal{}\limits^{DLH}\mathop{\lim }\limits_{x\to\plus{}\infty }\frac{{g'\left( x\right)}}{{2x}}\mathop \equal{}\limits^{DLH}\mathop{\lim }\limits_{x\to\plus{}\infty }\frac{{g''\left( x\right)}}{2}\equal{}\frac{1}{2}$

How do I justify using l'Hopital's rule here? Do we know that $ \frac{g(x)}{x^2}$ or $ \frac{g'(x)}{2x}$ is indeterminate as $ x \rightarrow \infty$?

Since $ g'' \to 1$, there exists some $ N$ such that $ x > N$ implies $ g''(x) > \frac{1}{2}$.  Then $ g'(x) > \frac{1}{2}(x \minus{} N) \plus{} g(N)$, a linear function of positive slope.  Then do the same thing again.

How does $ g'(x) > \frac{1}{2} (x \minus{} N) \plus{} g(N)$ allow l'Hopital's?  :|

It implies $ g' \to \infty$ as $ x \to \infty$.


inequalities
algebra
polynomial
function
rational function
superior algebra
superior algebra open

I think almost all the inequalities I have ever seen were derived from the fact that $x^2\geq 0$ for all real numbers $x$. Is every inequality a consequence of this fundamental inequality? One way to make the question precise is the following.

Is there a polynomial $f$ with integer coefficient, which is positive indefinite in $\mathbb{R}$, and an ordered ring $R$ in which $x^2\geq 0$ such $f$ is [i]not[/i] positive indefinite in $R$?

If $R$ is an $\mathbb{R}$-algebra, then since every positive indefinite polynomial in $\mathbb{R}[X]$ is a sum of squares of polynomials, we see that there is no one-variable counterexample. It is not clear to me that there are no one-variable counterexamples if $R$ is not  an $\mathbb{R}$-algebra. Of course, it is possible that there are multivariable counterexamples even if $R$ is an $\mathbb{R}$-algebra.

If the answer some of the above question is negative, then it is interesting to ask the same questions when $R$ is a field.

Boris

i think it has been proven that every positive definite polynomial is the sum of squares of quotients of polynomials. this immediately proves your claim.

Can you refer me to a proof of that for multivariable polynomials?

Also, what if $R$ is not $\mathbb{R}$-algebra? For example, $2x^2=(\sqrt{2}x)^2$, but $R$ does not necessarily have a $\sqrt{2]}$. Of course, in this case $2x^2=x^2+x^2$. However, the polynomial $3x^2-3x+1$ is positive definite, but there is no way of writing it as a sum of squares of polynomials with integer coefficients. Also, what if $R$ is not a field?

Boris

I think there are several ways this can go wrong. I'm pretty sure there are multivariable counterexamples, and even $\mathbb{R}[X,Y]$ allows them.

We can also cause trouble with a choice of field. Consider $\mathbb{Q}(\sqrt{2})$, with the usual order. It is impossible to determine algebraically which square root of 2 is positive because there is a field automorphism mapping it to a negative number, so $\sqrt{2}$ cannot be written as a sum of squares.

This example could be made larger; consider the smallest field $F$ containing $\mathbb{Q}$ such that for any $a,b\in F$, $a^2+b^2=c^2$ for some $c\in F$. Since it can be embedded in $\mathbb{R}$, it can be ordered, but it has infinitely many automorphisms which scramble the order, so there must be many elements which cannot be written as sums of squares, and $\sqrt{2}$ is still one of them.

i'm sorry - indeed we can get some problems. but it least it answers the question whether one can prove every algebraic inequality with the inequality $x^2\geq 0$ in the real numbers.

Hilbert's 17th problem asked something very similar to what you're asking. He asked whether every rational function (i.e. quotient of polynomials) with real coefficients, which is everywhere greater or equal to 0, is the quotient of two polynomials, each of which is a sum of squares. This was answered affirmatively by Emil Artin in the 1920s. Note that the analogous questions for polynomials is false. There exist polynomials that are everywhere greater or equal to 0, but cannot be written as the sum of squares of polynomials. (It can however be written as a sum of squares of rational functions.) See the Wikipedia page on Hilbert's 17th Problem.


inequalities
search
inequalities proposed

Let $ a,b,c\geq 0$ such that:$ a^2\plus{}b^2\plus{}c^2\equal{}1$.Prove that$ : \sum\frac{a}{a^3\plus{}bc}\geq 3$

it is a russian inequality if my memory is good , a hint is to use the substitution : $ x\equal{}\frac{ab}{c} ...$ and there is another nice simple sol but hard to find ( at least for me) , search for it in the forum :wink:


MATHCOUNTS
algebra
system of equations
AMC

If you want to get technical, this is actually based on a mathcounts national problem - but it's hard enough to count as an AMC-level problem.

Express the following as a difference of two square roots:
:sqrt:(5 - 2:sqrt:6)
Or in exponents, this is (5 - 2*6^1/2)^1/2.

Show how you arrived at your answer.

P.S. If you already have done this, don't say it.  There's a little "trick" to finding it.

Is it sqrt(5-2*sqrt(6)) or sqrt(5-sqrt(6))? You wrote each one in some place. If it's the first, it's sqrt(3)-sqrt(2). I don't know how to do it nicely and properly. I just solved it ``by inspection.'' It's possible to consider the two cases a*sqrt(2)+b*sqrt(3) and a+b*sqrt(6) and solve a system of equations, but that takes a fairly large amount of time. A friend of mine from school has a set of Chinese training books that he allowed me to borrow, and there were a lot of problems having to do with nested radicals, so after going through a lot of their exercises, I can now do most simple ones in my head very quickly.

can someone please give the solution to this problem. i still dont know how to do it. thanks in advance

It has to be a+b*sqrt(6) or a*sqrt(2)+b*sqrt(3). Just square each one with the original number and then solve the systems of equations.

Sorry, I was on vacation.  Cut off from computers for a whole week!  Sadness!

But back to the point.  ComplexZeta, it was the former.  That is, sqrt(5-2*sqrt(6))

And the answer [i]is[/i] sqrt(3)-sqrt(2).  Here's my solution:

sqrt(5-2*sqrt(6))
Ignore the first square root for a sec.  We have

5-2*sqrt(6)
= (3+2) - 2*sqrt(6)
= 3 - 2*sqrt(6) + 2
= (:rt3:):^2:  - 2* :rt3: * :rt2: + (:rt2:) :^2:
This is of the form a^2 - 2ab + b^2 = (a-b)^2, with a = :rt3: , b =  :rt2:
So 5-2*sqrt(6) = ( :rt3: - :rt2:) :^2:

Put that back under the radical and you get:
sqrt(( :rt3: - :rt2:) :^2: )
= :rt3: - :rt2:

In other words, sqrt(3) - sqrt(2), as Simon said.

thanks, i finally understand. CELEBRATE!

The more problems the better. I finished 2 math books this week, and I don't know what to do now other than IMO problems.

4 sheezy. i may not be able to solve it though. oh well. make it quick. i leave tomorrow

ComplexZeta - see my <unknown author> challenge #1 under the Challenge section.  If you haven't seen it already or have any other reason for not wanting to try it.

Seems like it's just you and me on the forum right now.  We should PM.  I'll send you a message.


logarithms
geometry
3D geometry
algebra
polynomial
parallelogram
ratio

[color=blue][size=150]2005 mock AMC D[/size][/color]
by amirhtlusa(Amir Talebi)

you guys know the rules,
take 75 mins to answer the questions, when done PM your answers to me.
you can use a SAT approved calculator and other aids.
MANY THANKS TO CHESS64 FOR PDF FILE
i'll try to be online Most of the time
[hide]
1.If $x=1+2^p$ and $y=1+2^{-p}$ then  $y$  in terms of $x$ is:

A)$\frac{x+1}{x-1}$
B)$\frac{x+2}{x-1}$
C)$\frac{x}{x-1}$
D)$2-x$
E)$\frac{x-1}{x}$

2.The smallest value of $x^2+8x$ for all real values of $x$ is:

A)-4
B)4
C)-8
D)-16
E)NOTA

3.The real value of $x$ such that $64^{x-1}$ divided by $4^{x-1}$ equals $256^{2x}$

A)$\frac{-2}{3}$
B)$\frac{-1}{3}$
C)$0$
D)$\frac{1}{4}$
E)$\frac{2}{3}$

4. Define the operation $\star$ for numbers $\in\mathbb{R^+}$  as $a\star b=\frac{ab}{a+b}$ then $4\star(4\star 4)=$

A)$\frac{3}{4}$
B)$1$
C)$\frac{4}{3}$
D)$2$
E)$\frac{16}{3}$

5. $(\sqrt{1+\sqrt{1+\sqrt{1}}})^4$ equals:

A)$\sqrt{2}+\sqrt{3}$
B)$\frac{1}{2}(7+3\sqrt{5})$
C)$1+2\sqrt{5}$
D)$3$
E)$3+2\sqrt{2}$

6. If $x=(\log_{8}{2})^{(\log_{2}{8})}$ then $\log_{3}{x}$ equals:

A)$-3$
B)$-\frac{1}{3}$
C)$\frac{1}{3}$
D)$3$
E)$9$

7. If the sum of two numbers is $1$ and their product is $1$ then the sum of their cubes is:
A)$2$
B)$-2-\frac{3\sqrt{3}}{4}$
C)$0$
D)$-\frac{3\sqrt{3}}{4}$
E)$-2$

8.a sequence of natural numbers is made as following:  the first term in this sequence is equal to $1$, then every term afterwards is equal to the sum of all the previous terms before it plus one.  The $n$ th term of this sequence is
A)$2n-1$
B)$n$
C)$2^n-1$
D)$2^{n-1}$
E)$n(n+1)(2n+1)$

9. If $x,y$ are positive integers, then the number of ordered pairs $(x,y)$ which satisfy: $3x+5y=501$ is :

A)$33$
B)$34$
C)$35$
D)$100$
E)NOTA

10. we have an equilateral triangle.  We construct a new equilateral triangle by connecting the midpoints of the sides of the bigger triangle. After performing this operation $n$ times we see that there is a TOTAL number of $45$ triangles in the picture.  $n$ is :

A)9
B)10
C)11
D)12
E)22

11. if the polynomial $ax^3+bx+c$ is divisible by $x^2+tx+1$ then which of the following is correct?
A)$a^2-2c\geq b$
B)$a+c>3$
C)$a^2-c\geq ab$
D)$a^2+c^2=ab$
E)$a^2-c^2=ab$

12.How many distinct values of $n$ will make $1!+2!+3!+4!+\cdots+n!$ a perfect square?
A) Infinite
B)2
C)3
D)4
E)5

13. in $\triangle ABC$ point $M$ is on the side $BC$.  From $M$ we draw lines parallel to $AB$ and $AC$.  if the area of the created parallelogram is $\frac{5}{18}$ of the area of $\triangle ABC$;  What ratio does $M$ divide the side $BC$ into?
A)1 to 5
B)1 to 6
C)1 to 4
D)2 to 9
E)it can not be determined

14. What is the sum of the coefficients in the expansion $(2x+y-3z)^{2006}$
A)$2006$
B)$2^{2006}$
C)0
D)1
E) $3^{2006}$

15. let $d(n)$ denote the number of (positive) divisors of the number $n$. find $d(24898)$
A)4
B)6
C)8
D)16
E)10

16. in the game of chess, the king can attack any of the 8 possible squares around itself.  In how many ways can we place three kings in a $3\times 3$ square such that none of the kings could attack each other?
A)5
B)7
C)9
D)8
E)10

17. what is the maximum number of terms that can be picked from the following set of numbers such that the mean of the picked numbers would be $15$?
$\{21,14,13,17,15,16,23,12,8,11,12,9,5,4,16,2,14,15,18,8,3,16\}$

A)13
B)14
C)15
D)16
E)17

18. if $f(x)=\log{(\frac{1+x}{1-x})}$ for $-1<x<1$ then $f(\frac{3x+x^2}{1+3x})$ in terms of $f(x)$ is:
A)$-f(x)$
B)$2f(x)$
C)$3f(x)$
D)$(f(x))^2$
E) $(f(x))^3-6f(x)$

19. 7 people numbered 1 through 7 are standing in a line.  We know the following:
between the numbers 1 and 5, there is one person
between the numbers 1 and 7, there are 3 people
between the numbers 1 and 3 there is one person
between the numbers 2 and 4 there is one person
and between the numbers 6 and 4 there is one person.

In how many ways could these 7 people be standing in the line?
A)4
B)8
C)12
D)14
E)16

20. for natural numbers $n$, the sum of digits of $n$ in base 10 is represented by $f(n)$(for example $f(1376)=17$). If $n_0=13^{76}$ then which one of the following numbers is in the sequence: $n_0, f(n_0),f(f(n_0)),\cdots,f(f(\cdots(f(n_0))\cdots)),\cdots$

A)1
B)2
C)3
D)4
E)NOTA

21. let $P$ denote the product of $3659893456789325678$ and $342973489379256$.  The number of digits in $P$ is:
A)36
B)35
C)34
D)33
E)32

22. how many natural numbers in a form of $\overline{abcd}$ can be found such that: $\overline{abcd}=\overline{ad}\cdot\overline{ada}$.
A)5
B)6
C)7
D)infinite
E)none

23. in the isoceles triangle $ABC$, $AB=AC$ and $\angle A=100$.  Along the side $AB$ (from $B$), we pick the point $M$ such that $AM=BC$. $\angle BCM$ equals:
A)10
B)15
C)12
D)8
E) 18

24. $f: R\rightarrow R$ is defined by: $f^2(x+y)=f^2(x)+f^2(y)$. Find $f(2006)$
A)$1$
B)$2006^2-1$
C)$0$
D)$1401$
E)$1405$

25. how many numbers, $n$, between $1$ and $100$ are there such that $n$ is sum of some positive integers, where every digit appears exactly once.(for example: $90=0+1+52+3+4+6+7+8+9$)
A)7
B)11
C)15
D)19
E)23[/hide]

I'll be here to take the test!

Could you post in a PDF too... that be so nice... :)

[quote="Boy Soprano II"]Question :  Why A.M.C. D?  Wouldn't this one be L?  Or, if we were starting over this year, A?[/quote]
There have already been three Mock AMCs this year.

bump :)
pm your answers to me after 75 mins.

[quote="zanttrang"]There have already been three Mock AMCs this year.[/quote] Pardon me.  Just goes to show what happens if one doesn't pay attention.

75 minutes after 6:00, or 75 minutes after we start taking it?  Only I'm about to have dinner right now...

Anyone that's interested in a PDF, here... yes it's ugly, but I need to get started :!:

you don't have to take it right now.
the questions will be here later. if you want your name to be in the results. then you have to send your answers to me by 7:30

thank you Chess64 for the PDF.
you should tell me how to do it later. :D  :blush:

For #10, so after 1 construction, there are 4 triangles or 5 triangles (counting the largest one that you originally started)?

[quote="henry23"]total :) (so 5)[/quote]
yes i meant 5
[b]also to clarify[/b] ::
on #24 $f^2(x)$ means $(f(x))^2$.

I think 24 is a good problem.

[quote="bubala"][quote="probability1.01"]Are you sure about 24? (No, you're not)[/quote]

Don't be so rude probability.  But amir, you ought to redo 24.  :)[/quote]

he's probably sure and no, he doesn't need to redo it

i guess it is harder than its supposed to be :D  :D(or maybe not?!)

Hmmm maybe it actually is possible if it is R. I thought it was R+ for some reason...

only use the Equilateral triangles to make more triangles.

[quote="chess64"]#10: are you supposed to construct triangles in the middle triangle? like after 2 steps, there are 17, right??[/quote]

huh I thought after 2 steps there were BLAH BLAH BLAH

dont tell how many there are :D (it gives off the answer).
but yeah there are only 9 after the second time.

oh, so i'm doing it wrong...

after 1 step: 5 triangles
2 steps: 4 triangles inside of each of those (not the outer one) is 16, plus the entire one is 17

that's wrong?

i'll just guess then, i don't get it :(

[quote="amirhtlusa"]dont tell how many there are :D (it gives off the answer).
but yeah there are only BLAH BLAH BLAH after the second time.[/quote]

all right edited accordingly

to clarify problem whatever-it-is (#10?)

just form new triangles within the *innermost* equilateral triangle, I believe

I had to eat so I didn't get to do problems for about 30 min.. I'll just help getting answers.

BTW, MysticTerminator, are you just looking at problems or actually doing it? I'm asking this because you didn't even know what the number of problems were.  :D

[quote="MysticTerminator"]
to clarify problem [b]whatever-it-is[/b] (#10?) [/quote]

 :lol:

[b]time is not UP, please send in your answers[/b]

you guys can start the discussion at 7:30

Argh... I didn't see that we had 75 minutes from when we started. I totally forgot about this and didn't see it until 3:22, but just sent in my answers right now after 55 minutes. >_<

[b]EDIT:[/b] Err, wait, now I'm confused. Is time up now? What did you mean when you said we had until 7:30 Eastern? :huh:

[quote="amirhtlusa"][quote="chess64"]hm i don't get one of the answer choices given for #18 :([/quote]

umm.. do you mean you dont see a choice?
that might be because in the PDF file the choice A is right infront of the question
 ;)[/quote]

i know, but i didn't get one of those answers :( oh well

[quote="amirhtlusa"][b]time is not UP, please send in your answers[/b][/quote]

?????????????? :?

not probably a typo for now

yea for 18 I also didn't get one of those answers, but I wasn't really trying that one, so I might have overlooked some clever factorization - didn't look too interesting, though.

if some of the powers on 18 were switched (the bottom would be a 3x^2 instead of 3x and the top would be an x^3 instead of an x^2 I think), it might work out better... but eh, maybe it works out this way

[quote="amirhtlusa"]ooops. :P
typo for 18 the bottom was supposed to $1+3x^2$
sorry about that[/quote]
This problem is familiar - could it be from the qualifying exam for the UCSB C of CC qualifying exam?  :lol: I'm sad that I was too busy last weekend to take the test - I'll sit down eventually and actually do it under exam conditions, though I've sort of skimmed the problems and some of the solutions thread, so the value will be slightly diminished.

i think igot it from a old mao book, maybe around 1960's


maybe this would help
[hide="hint"]
the inequality is equivalant to:
$\frac{a^n-1}{a^{n+1}-1}>\frac{b^n-1}{b^{n+1}-1}$[/hide]

And to ruin it!

$f(x)=\displaystyle \frac{x^n-1}{x^{n+1}-1}$

Now how would we prove that this is always decreasing??? Hmmm, let me think....

Unless I made a mistake, the expression is equal to $\frac{1}{1+\frac{1}{\frac{1}{x}+\frac{1}{x^2}+....\frac{1}{x^n}}}$, which clearly decreases as x increases


Let $ a,b,c$ be positive numbers such that:$ a\plus{}b\plus{}c\equal{}3$.Prove that:

$ 6(4\minus{}\sqrt{3})(\sqrt{3a\plus{}b}\plus{}\sqrt{3b\plus{}c}\plus{}\sqrt{3c\plus{}a}) \ge 3(27\minus{}5\sqrt{3})\plus{}7(3\minus{}\sqrt{3})(ab\plus{}bc\plus{}ca)$

Since this inequality,we have an nice inequality of nguoivn:

$ 10(\sqrt{3a\plus{}b}\plus{}\sqrt{3b\plus{}c}\plus{}\sqrt{3c\plus{}a}) \ge 39\plus{}7(ab\plus{}bc\plus{}ca)$
 :)

[quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that:$ a \plus{} b \plus{} c \equal{} 3$.Prove that:

$ 6(4 \minus{} \sqrt {3})(\sqrt {3a \plus{} b} \plus{} \sqrt {3b \plus{} c} \plus{} \sqrt {3c \plus{} a}) \ge 3(27 \minus{} 5\sqrt {3}) \plus{} 7(3 \minus{} \sqrt {3})(ab \plus{} bc \plus{} ca)$

Since this inequality,we have an nice inequality of nguoivn:

$ 10(\sqrt {3a \plus{} b} \plus{} \sqrt {3b \plus{} c} \plus{} \sqrt {3c \plus{} a}) \ge 39 \plus{} 7(ab \plus{} bc \plus{} ca)$
 :)[/quote]
We can solve this inequality by Schur inequality :)

They are also results from this problem of can_hang:
http://mathvn.org/forum/viewthread.php?thread_id=1180


Four whole numbers, when added three at a time, give the sums 180,197,208 and 222. What is the largest of the four numbers?

$\text{(A)} \ 77 \qquad \text{(B)} \ 83 \qquad \text{(C)} \ 89 \qquad \text{(D)} \ 95 \qquad \text{(E)} \ \text{cannot be determined from the given information}$

[hide="How to solve"]Add all four of the numbers together, divide by three, then subtract the smallest triple.

Easy.  :) [/hide]

[hide]
Let the numbers be $a$, $b$, $c$, and $d$.  Add all of the weights together and you get $3(a+b+c+d)=807$, so $a+b+c+d=269$.  Now notice that if you subtract 269 from the weight, you get negative the non-added number.  It's clear that the highest number you can find is $|180-269|=89$, which is $\boxed{C}$.
[/hide]


hi
question:assume that "n" is anteger ,n>1 and "p" is a prime number
and we have    n|p-1    and     p|(n^3)-1
prove that  4p-3 is a perfect squar.
tnx

$ P \minus{} 1\geq(n)$ so $ (P,n \minus{} 1) \equal{} 1$. so $ P|n^2 \plus{} n \plus{} 1$ and so $ (nm \plus{} 1)k \equal{} n^2 \plus{} n \plus{} 1$ ($ p \equal{} nm \plus{} 1$) so we must have $ n(n \plus{} 1 \minus{} km) \equal{} k \minus{} 1$ so $ k\geq(n \plus{} 1)$ so $ n^2 \plus{} n \plus{} 1\geq(nm \plus{} 1)(n \plus{} 1)$ so $ k \minus{} 1 \equal{} 0$ so $ nm \plus{} 1 \equal{} n^2 \plus{} n \plus{} 1 \equal{} p$ so $ 4p \minus{} 3 \equal{} 4n^2 \plus{} 4n \plus{} 1$


search
combinatorics unsolved
combinatorics

There are two tennis teams,each team has 1000 players.Knowing that any two competitors of the two teams copete with each other once only and there is no game ending in a tie
   Prove that :Existing 10 competitor of one team so that there is one player of the other losing at least one among that 10 competitors .

You've posted almost exactly the same thing before, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1344065672&t=158664]here[/url].  This time, the statement is trivial.  Probably what you mean is,

"There are two tennis teams with 1000 players.  Each player plays one game against every player on the other team, and no game ends in a tie.  Show that we can find 10 players on one team such that every player on the other team lost to at least one of those players."

Also, give your posts better titles.


Assume : $ a+b =p-1 $ , where $ a,b,p $ are natural numbers and p is a prime number . Prove that : $ a!b! +(-1)^a \equiv 0 [p] $

By Wilson's Theorem, taking $\mod p$, we have $(p-1)! = b! (b+1) (b+2) \ldots (p-1) \equiv b! (-a) (-a + 1) \ldots (-1) \equiv b! a! (-1)^a \equiv -1$, hence result follows. :)

[quote="coolamit"]Err, this may be stupid...
how is $a!b!(-1)^a \equiv -1 \mod p$?[/quote]
  I don't understand you ! :roll:  This expression is exactly what we need to prove . ;)


For #1, try factoring it.

I just looked at these for about 5 seconds.

Can you elaborate more on the given hints?

I am not sure what the questions are asking for, you see?

For each question, find all values of $x$ for which $f(x)=0$.  So for the first one, it's to solve for $x$ when $x^{4}-1=0$.

[quote="Interval"]For each polynomial function below, find the complex zeros.

[1] f(x) = x^4 - 1

[2] f(x) = 2x^4 + x^3 - 35x^2 - 113x + 65[/quote]
[hide="Numero Uno (1)"]
$x^{4}-1=0$
$(x^{2}+1)(x+1)(x-1)=0$
One of them must be zero so $x=1,-1$ or $x^{2}+1=0\rightarrow x^{2}=-1$ so $x=\pm i$[/hide][/hide]

[quote="Quevvy"][quote="#H34N1"]Yes. And its really easy because you know that:

[hide]
$x^{4}-1=(x^{2}+1)(x+1)(x-1)=0$

So, just figure out values of $x$!

$x^{2}=0$
$x=0$

$x+1=0$
$x=-1$

$x-1=0$
$x=1$

$x\in\{-1,0,1\}$

So, that should be an even bigger (hint)
[/hide][/quote]
By the way $x\ne0$.

I think you mean $x^{2}+1=0\rightarrow x=\pm i$[/quote]

Yes; i meant that :blush:


algebra
system of equations
algebra proposed

Solve the system of equations:

$ y^2\equal{}(x\plus{}8)(x^2\plus{}2),$
$ y^2\minus{}(8\plus{}4x)y\plus{}(16\plus{}16x\minus{}5x^2)\equal{}0.$

[quote="moldovan"]Solve the system of equations:

$ y^2 \equal{} (x \plus{} 8)(x^2 \plus{} 2),$
$ y^2 \minus{} (8 \plus{} 4x)y \plus{} (16 \plus{} 16x \minus{} 5x^2) \equal{} 0.$[/quote]

$ y^2 \minus{} (8 \plus{} 4x)y \plus{} (16 \plus{} 16x \minus{} 5x^2)\equal{}[y\minus{}(4\plus{}5x)] \cdot [y\minus{}(4\minus{}x)]\equal{}0$

[b]Case 1:[/b] $ y\equal{}4\plus{}5x$, then $ y^2\equal{}(x\plus{}8)(x^2\plus{}2)\equal{}x^3\plus{}8x^2\plus{}2x\plus{}16$
$ \Leftrightarrow x(x\minus{}19)(x\plus{}2)\equal{}0$
$ \Leftrightarrow (x,y)\equal{}(0,4),(19,99),(\minus{}2,\minus{}6)$ as solutions.

[b]Case 2:[/b] $ y\equal{}4\minus{}x$, then $ (4\minus{}x)^2\equal{}y^2\equal{}x^3\plus{}8x^2\plus{}2x\plus{}16$
$ \Leftrightarrow x(x\plus{}2)(x\plus{}5)\equal{}0$
$ \Leftrightarrow (x,y)\equal{}(0,4),(\minus{}2,6),(\minus{}5,9)$ as solutions.

So $ (x,y)\equal{}(0,4),(19,99),(\minus{}2,\minus{}6),(\minus{}2,6),(\minus{}5,9)$ are the only solutions for the system of equations.

hello,eliminating $ y$ we get $ y\equal{}\minus{}\frac{(x\plus{}2)(x^2\plus{}x\plus{}16)}{(\minus{}8\minus{}4x)}$ inserting this in our equationsystem
and factoring we get $ x(x\plus{}5)(x\minus{}19)(x\plus{}2)\equal{}0$ from here we get all solutions as
$ x\equal{}0,y\equal{}4$
$ x\equal{}\minus{}5,y\equal{}9$
$ x\equal{}\minus{}2,y\equal{}6$
$ x\equal{}\minus{}2,y\equal{}\minus{}6$
$ x\equal{}19,y\equal{}99$
Sonnhard.


Given a mxn grid .We color its vertex with red,blue and green.Firstly,color the vertex of boundary with red.A cell is called properly if it has exactly 2 adjacent vertex which have same colour.Show that: the number of properly cell is always even.

To each side of an unit square, give the number $2$ if its endpoints have the same color, and $1$ otherwise.
Then define the weight of an unit square as the sum of the numbers attached to its four sides.
It is easy to verify that an unit square is a properly cell iff its weight is odd.
Thus, the total sum $S$ of all cells has the same parity than the number of properly colored cells.
But $S$ is even since in this sum each side is counted twice except those which form the boundary of the whole grid, and these ones are attached to $2$.

Pierre.

[quote="thealchemist"]Wasthis asked in the Indian Namtional Olympiad 2000?[/quote]
No, it was not. Next time just check the olympiad resources section. :wink:


topology
advanced fields
advanced fields unsolved

$(X_i,d_i)$ are metric spaces, where the topology on each space is induced by each $d_i$, respectively. Prove that
\[ d(x,y)=\sum_{i=1}^\infty \dfrac{1}{2^i}\dfrac{d_i(x_i,y_i)}{1+d_i(x_i,y_i)}  \]
defines a metric in $\prod_{i=1}^\infty X_i$, and the topology which it induces is exactly the product topology.

I think it must be easy, but I don't know how to verify the topology that $d(x,y)$ induces.

To verify the topology, show that an epsilon-ball lies between two set of the basis for the product topology, which have the form (open in first k coordinates)X(unrestricted in all other coordinates). In addition, show that as the ball shrinks to 0, the other sets do as well.

[quote="jmerry"]In addition, show that as the ball shrinks to 0, the other sets do as well.[/quote]
What are 'the other sets'? And what does it verify the topology? It looks so different from the definition of an induced topology that I know.

That was badly phrased. Here's what I should have said the first time:

Given an open set (an $\epsilon$-ball) $U$ in the metric topology and a point $x$ in that set, find $V$ an open set (in the basis) in the product topology with $x\in V\subset U$.
Given an open set (in the basis) $U$ in the product topology and a point $x$ in that set, find $V$ an open set (an $\epsilon$-ball) in the metric topology with $x\in V\subset U$.

The first shows that every open set in the metric topology is open in the product topology, and the second shows that every open set in the product topology is open in the metric topology.
We have to work with the open sets, because we don't know in advance that the product topology is determined by sequences.

[quote="jmerry"]Given an open set (an $\epsilon$-ball) $U$ in the metric topology and a point $x$ in that set, find $V$ an open set (in the basis) in the product topology with $x\in V\subset U$.
Given an open set (in the basis) $U$ in the product topology and a point $x$ in that set, find $V$ an open set (an $\epsilon$-ball) in the metric topology with $x\in V\subset U$.
[/quote]
I get it now, thank you.


function
number theory unsolved
number theory

Are you sre about the $3^{a_i}$? In that case, I think the problem follows easily from the greedy algorithm.
But according to your other recent post, I think the problem deals with $a_i^3$, and in that case, it is another one from ISL 2002 :
http://www.kalva.demon.co.uk/short/soln/sh02n1.html

Pierre.

I had written something for this one but it didnt posted!

If it is the problem Pierre is talking about by locking mod9 one finds that
a^3=0,-1,1 so we get the minimal number is 4 because 2002^2002=4mod9, and we can achieve this with 1000*2002^2001+1000*2002^2001+2002^2001+2002^2001.

But the problem as it is is not easy. Obviously the best representation is the base 3 one. but how can we count the number of powers used on it?

I thought about a function f(1)=1 f(2)=2 f(3)=1 and f(3n)=f(n) f(3n+1)=f(n)+1 and f(3n+2)=f(n)+2 and our number is f(2002^2002) but we should find propertys to make easy the calculations.


AMC
AIME
number theory
AMC 10
AMC 8
\/closed

I'm currently debating weather to take Algebra 3 or Intermediate C+P.  I think my Algebra is OK but...it's still just Algebra 2.3 level.  On the other hand, counting made me FAIL in the AIME.  I don't even want to talk about it.

The only class I've taken is Intermediate Number theory seminar.

138 AMC 10, 24 AMC 8, never mind the AIME, or the MC, but I think I improved a lot since then.

They are both pretty long classes I think, and I can probably only take 1.

Let the post-test be your guide.  Do the Do You Need This test for Algebra 3.  If you can't get more than half of them, then you should take Algebra 3.  If you can get 1/2 - 3/4 of them without too much difficulty, then you can probably feel OK getting the book, focusing on the parts you don't know (and the Challenge Problems in the parts you do), and heading into C&P.  If you're 3/4+, then skip the class and go to C&P.

So I'm assuming C+P is harder?

OK guess I'll take Algebra III.


On the 5 by 5 square grid below, each dot is 1 cm from its
nearest horizontal and vertical neighbors. What is the product of
the value of the area of square $ ABCD$ (in
cm$ ^2$) and the value of the perimeter of
square $ ABCD$ (in cm)? Express your
answer in simplest radical form.

[asy]unitsize(1inch);
dot((0,0));
dot((0,1));
dot((0,2));
dot((0,3));
dot((0,4));
dot((1,0));
dot((1,1));
dot((1,2));
dot((1,3));
dot((1,4));
dot((2,0));
dot((2,1));
dot((2,2));
dot((2,3));
dot((2,4));
dot((3,0));
dot((3,1));
dot((3,2));
dot((3,3));
dot((3,4));
dot((4,0));
dot((4,1));
dot((4,2));
dot((4,3));
dot((4,4));
draw((0,3)--(3,4)--(4,1)--(1,0)--cycle);
label("$A$",(3,4),N);
label("$B$",(4,1),E);
label("$C$",(1,0),S);
label("$D$",(0,3),W);[/asy]

the length of one side of the square is $ \sqrt{10}$.

so the area of the square is 10, and the perimeter of the square is $ 4\sqrt{10}$.

answer : $ 40\sqrt10$

[hide = Solution]The length of the square's side is $\sqrt{10}$ because it is the hypotenuse of a right triangle with legs $3$ and $1:$ $\sqrt{3^2 + 1^2} = \sqrt{10}.$ So the area of square $ABCD$ is $(\sqrt{10})^2 = 10.$ The perimeter of the square is $4\sqrt{10}.$ The question asks for the product of the area and perimeter: $10 \cdot 4\sqrt{10} = \boxed{40\sqrt{10}}.$[/hide]

[hide = Alternative Way to Find Area]We could have found the area of square $ABCD$ a different way, though it is best to do it the first way, since the side length of $ABCD$ needs to be found anyway for perimeter. The area of the entire 4 by 4 square is $16,$ and the area of one 1 by 3 triangle is $\frac{3}{2}.$ There are four of these, giving the left-out portions a total area of $6.$ $16 - 6 = 10,$ square $ABCD$'s area.[/hide]


spaceguy, I'll pretend you posted 13.

$ 12$

[b]47[/b]
boyz pwn
boyz rule :D

girls are lazy
girls are dumb

lol the girlz  all eating lunch  :rotfl:  :rotfl:
 :D

all the boyz , rejoice
coz we won the game
and put the girls to shame
showed them who the man is

Come on, stop insulting girls... And end this topic

ooh

I never meant to insult them ...
just serves them right though...

and next time you should vote for girls :rotfl:

aww.. i leave for four hours and this is wut hapenns?? sigh

Sigh... good job boys... @westiepaw: go to round seven!!!

Wow, don't check the forum for 24 hours and...   :(   FF changes really fast!  These Boy v. Girl threads are getting kinda annoying.  Shouldn't we just start a Boy vs. Girl thread that resets after it reaches 0 or 50?


analytic geometry
combinatorics unsolved
combinatorics

How many squares are on chessboard n * n :)

Do you mean squares formed by the unit squares of the board or squares whose vertices are vertices of the unit squares of the board? In the second case, are you allowing only squares whose sides are parallel to the sides of the board or not?

Pierre.

I mean all squares, not the small one. All squares that are formed by lines parallel lines to the sides of board.

Thus, the answer is $\frac {n(n1)(2n+1)} 6.$
Let $U_n$ be the desired number.
Clearly $U_1 = 1.$

Now, draw the $n \times n$ board, and consider it as the grid of integer points $(i,j)$ where $0 \leq i,j \leq n$. These integer points are the vertices of the uit squares forming the board, and they are the vertices of the squares we are looking for.
There are $U_{n-1}$ squares whose vertices are not points with at least one coordinate equal to $n$ (that is whose vertices belong to the $(n-1) \times (n-1)$ board).

Let $A$ be the point with coordinates $(n,n).$
There are $C_n^2$ squares with two vertices having their $x$-coordinate equal to $n$ but not being $A$ (just have to choose two points among these $n$ points, which determine uniquely the whole square).

The same for the $y$-coordinate, and all these squares are pairwise distinct, because we did not choose $A$.

Then, there are $n$ squares with one vertex being $A$ (just choose one point among the $n$ with $x$-coordinate equals to $n$, which determine uniquely the square).

Therefore, $U_n = U_{n-1} + 2C_n^2 + n = U_n + n^2.$
It follows that $U_n = 1^2 + 2^2 +\cdots + n^2 = \frac {n(n+1)(2n+1)}6$, as claimed.

Pierre.


Find all primes $p > 0$ and all integers $q \geq 0$ such that $p^{2}\geq q \geq p$ and $\binom{p^{2}}{q}-\binom{q}{p}= 1$.

[i]Costel Chiteş, Adrian Stoica[/i]

[i]Note.[/i] I slightly modified the problem. The original version didn't make sense.

[hide]I know by inspection that $(p,q)=(2,3)$ works.[/hide]

Does anyone have a solution to this nice problem?  :dry:


I'm confused with how to solve this problem, so help would be great!

The volume of any spherical balloon can be found by using the formula V= 4/3 pi r^3. Write an equation for r in terms of V and pi

[hide]
It is just asking you to solve the equation $V=\frac{4}{3}\pi r^{3}$ for $r$.

$V=\frac{4}{3}\pi r^{3}\implies\frac{3}{4}\cdot V=\pi r^{3}\implies r^{3}=\frac{3V}{4\pi}\implies r=\sqrt[3]{\frac{3V}{4\pi}}$.[/hide]

thanks i kept multiplying it by 3 and wondered why i was wrong


A man parked his car in a carpark at 0830 and retrieved it at 1545 on the same day.
a) How long did he park his car in the carpark?
b) If the parking charges are $1.50 for the first hour and 80 cents for each subsequent half hour or part thereof, how much must he pay on parking his car there?

[quote="risinglegend"]A man parked his car in a carpark at 0830 and retrieved it at 1545 on the same day.
a) How long did he park his car in the carpark?
b) If the parking charges are 1.50 for the first hour and 80 cents for each subsequent half hour or part thereof, how much must he pay on parking his car there?[/quote]

[hide]If he started at 8:30 AM and ended at 3:45 PM, then he stayed for 7 hours 15 minutes.

For the first hour, he pays 1.50 on the first hour, leaving 6 hours and 15 minutes for the 80 cent charge.

That charge would be $80*(6*2+1)=80*13=1040$ cents,  or 10.40

So the total would be $1.5+10.4=11.9$ so it would be 11.90[/hide]


Let $ d: X\times X \rightarrow \mathbb{R}$ determine $ \displaystyle{(x,y)\equal{}\sum_{k\equal{}1}^{\plus{}\infty}\frac{\min(|x_k\minus{}y_k|,1)}{k^2}}$
with $ x\equal{}\{x_n\}_n,y\equal{}\{y_n\}_n\subset X$. Prove $ (X,d)$ is a metric
space

you might want to make your notation more precise...

[quote="sondhtn"]Let $ d: X\times X \rightarrow \mathbb{R}$ determine $ \displaystyle{(x,y) \equal{} \sum_{k \equal{} 1}^{ \plus{} \infty}\frac {\min(|x_k \minus{} y_k|,1)}{k^2}}$
with $ x \equal{} \{x_n\}_n,y \equal{} \{y_n\}_n\subset X$. Prove $ (X,d)$ is a metric
space[/quote]
Sorry. Where $ \displaystyle{d(x,y) \equal{} \sum_{k \equal{} 1}^{ \plus{} \infty}\frac {\min(|x_k \minus{} y_k|,1)}{k^2}}$

Proving that $ d(x,x) \equal{} 0\quad \forall x$ and $ d(x,y) \equal{} d(y,x)\quad \forall x,y \in X$ is trivial. Further,
\[ \min \left( |x_i \minus{} y_i|,1\right) \plus{} \min \left( |y_i \minus{} z_i|,1\right) \ge \min \left( |x_i \minus{} z_i|,1\right)\]
is true for every $ i$.

Since,  the distance is bounded ($ d(x,y) < k \plus{} \frac {\pi^2}{6}$ with some appropriate constant), it follows that its a metric space.


Prove that for any positive numbers $ x$ , $ y$ , $ z$ exists the relation

$ 3\cdot\sum_{\mathrm{sym}}\frac {y\plus{}z}{x(2x\plus{}y\plus{}z)}\ge4\cdot \frac {(x\plus{}y\plus{}z)^2}{3(y\plus{}z)(z\plus{}x)(x\plus{}y)}\ .$

[quote="Virgil Nicula"]Prove that for any positive numbers $ x$ , $ y$ , $ z$ exists the relation

$ \cdot\sum_{\mathrm{sym}}\frac {y \plus{} z}{x(2x \plus{} y \plus{} z)}\ge4\cdot \frac {(x \plus{} y \plus{} z)^2}{3(y \plus{} z)(z \plus{} x)(x \plus{} y)}\ .$[/quote]

We have
\[ \sum {\frac {{y \plus{} z}}{{x\left( {2x \plus{} y \plus{} z} \right)}}} \equal{} \sum {\frac {1}{x} \cdot \left( {1 \minus{} \frac {{2x}}{{2x \plus{} y \plus{} z}}} \right)} \equal{} \frac {{xy \plus{} yz \plus{} zx}}{{xyz}} \minus{} 2\sum {\frac {x}{{2x \plus{} y \plus{} z}}} \ge \frac {{xy \plus{} yz \plus{} zx}}{{xyz}} \minus{} \sum {\frac {1}{{x \plus{} y}}}
\]
It's enough to show that
\[ \frac {{xy \plus{} yz \plus{} zx}}{{xyz}} \minus{} \frac {1}{{x \plus{} y}} \minus{} \frac {1}{{y \plus{} z}} \minus{} \frac {1}{{z \plus{} x}} \ge \frac {{4\left( {x \plus{} y \plus{} z} \right)^2 }}{{3\left( {x \plus{} y} \right)\left( {y \plus{} z} \right)\left( {z \plus{} x} \right)}}
\]

\[ \Leftrightarrow x^2 \plus{} y^2 \plus{} z^2 \ge xy \plus{} yz \plus{} zx
\]
Done.

Using Cauchy Schwarts, we have:
LHS $ \geq\ \frac {4(x \plus{} y\plus{}z)^2}{3(x \plus{} y)(y \plus{} z)(z \plus{} x)} \equal{}$ RHS
 :)


prove that if $n | (p-1)^{n}+1$ then $p|n$

Can be n=1. Exactly if n natural >1 and $n|(p-1)^{n}+1$, then $p|n.$
Let $q$ is minimal prime divisor n. If q=2, then p=2, n=2. If q>2 n is odd p>2 and $(q,p-1)=1$. Let $2k=ord_{q}(p-1)$, then $k|n$ and $k<q$. Therefore k=1. It give $q|p-1+1$. It mean q=p.


A set of $ n$ dice is rolled repeatedly. For each die the probability of showing a six is $ p$. Show that the probability that the first of the dice to show a six does so on the rth roll is \[ q^{nr}(q^{\minus{}n} \minus{} 1)\] where $ q \equal{} 1 \minus{} p$.

Determine, and simplify, an expression for the probability generating function for this distribution, in terms of $ q$ and $ n$. The first of the dice to show a six does so on the $ R$th roll. Find the expected value of $ R$ and show that, in the case $ n \equal{} 2$, $ p \equal{} \frac{1}{6}$, this value is $ \frac {36}{11}$.

Show that the probability that the last of the dice to show a six does so on the $ r$th roll is \[ (1 \minus{} q^r)^n \minus{} (1 \minus{} q^{r\minus{}1})^n\].
Find, for the case $ n \equal{} 2$, the probability generating function. The last of the dice to show a six does so on the $ S$th roll. Find the expected value of $ S$ and evaluate this when $ p \equal{} \frac{1}{6}$.

[hide="Part 1"]
On the first roll, we must have each die roll something other than six, which happens with probability $ q^n$.
On the second, we again have $ q^n$, and so on for $ r - 1$ rolls, giving $ q^{n(r - 1)}$.

The probability that none of the $ n$ dice shows a 6 on a roll is $ q^n$, so the probability that this doesn't happen (i.e. probability that we get at least one 6 on a roll) is $ 1 - q^n$.

The product is $ q^{n(r - 1)} - q^{n(r - 1) + n} = q^{nr - n} - q^{nr} = q^{nr}(q^{ - n} + 1)$. $ \Box$
[/hide]

Edit: Can anyone tell me what's wrong in the following, that's giving me $ \frac {36}{25}$ for the expected value?

[hide="Part 2"]
Well, if we encode the probabilities for various $ r$ as the coefficients of a power series, the generating function is
\begin{eqnarray*} g(x) & = & \sum_{r = 0}^{\infty}q^{nr}(q^{ - n} - 1)x^r \\
& = & (q^{ - n} - 1)\sum_{r = 0}^{\infty}(q^nx)^r \\
& = & (q^{ - n} - 1)\left(\frac {1}{1 - q^nx}\right) \\
& = & \frac {q^{ - n} - 1}{1 - q^nx} \\
\end{eqnarray*}
[/hide]

[hide="Part 3"]
If a 6 occurs on the $ R^{\text{th}}$ roll, this happens with probability $ q^{nR}(q^{ - n} - 1)$. To find the expected value, we sum over all the possible $ R$.
\begin{eqnarray*} E(R) & = & \sum_{R = 0}^{\infty}q^{nR}(q^{ - n} - 1) \\
& = & g(1) \\
& = & \frac {q^{ - n} - 1}{1 - q^n} \end{eqnarray*}
If $ p = \frac {1}{6}$, $ q = \frac {5}{6}$.  In this case, with $ n = 2$, we find:
\[{ E(R) = \frac {\left(\frac {5}{6}\right)^{ - 2} - 1}{1 - \left(\frac {5}{6}\right)^2}} = \frac {36}{25}
\]
[/hide]

I'll edit and post the rest of it if/when I solve it ;)


integration
logarithms
calculus
calculus computations

How can I show that this thing diverges?

$\int_{2}^\infty \frac{ dx }{xln|x|}$

Thanks.

That thing has an elementary antiderivative. Try to find it. (Hint: use a substitution.)


Given 3 different real numbers a,b,c such as 1/(a+b)(c-a) + 1/(b+c)(a-b) + 1/(c+a)(b-c)=0 what can be said aabout them? same question if 1/(a+b)(c-a) + 1/(b+c)(c-a) + 1/(c+a)(b-c)=0
sorry for the text,im writing from a net cafe and they dont have latex

[quote="ahmet"]
sorry for the text,im writing from a net cafe and they dont have latex[/quote]

The forum always has LaTeX. Just write all your math in between two dollar signs and you'll see that the forum automatically turns it into LaTeX.


conics
ellipse
geometry
analytic geometry
trigonometry
graphing lines
slope

We haven an ellipse shaped conductive wire. There is [b]I[/b] current flowing in it. The area of the ellipse is [b]A[/b] the  length of it is [b]L[/b]. Find the magnitude of the magnetic field in the middle of the ellipse.

Let's start with some geometry: the equation of an ellipse is
\[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1, \]
where $a = L/2$ and $b = A/a\pi = 2A/L\pi$ are the semi-axes of the ellipse. I'll use these instead of $A$ and $L$ for the sake of clarity.

Furthermore, differentiating the ellipse equation we get
\[\frac{x\text dx}{a^{2}}+\frac{y\text dx}{b^{2}}= 0, \]
which gives
\[\frac{a^{2}}{b^{2}}\frac{y}{x}\frac{\text dy}{\text dx}=-1. \]
We'll need this for applying Biot-Savart's law later on.

Now we can shift to polar coordinates $(r, \phi)$, where the equation of the ellipse is
\[r(\phi) = \frac{b}{1+\varepsilon\cos\phi}, \]
where $\varepsilon \equiv \sqrt{1-4A/L^{2}\pi}$ is the eccentricity of the ellipse. The above differentiated equation is now
\[\frac{a^{2}}{b^{2}}\tan\phi \tan\theta =-1, \]
where $\theta$ is the slope of the tangent on the ellipse at the polar angle $\phi$. Thus, the angle between the radius vector and the tangent on the ellipse at point $\phi$ is $\theta-\phi$, which can be derived from the above equation.

Now we can get to the physics of the problem. The magnetic field (or rather, its induction) $B$ is given by Biot-Savart's law
\[B = \frac{\mu_{0}I}{4\pi}\oint \frac{\text d\vec l\times \vec r}{r^{3}}, \]
where $\text dl$ is the element of length of the ellipse. Using the above geometry results, we can transform the above equation into
\[B = \frac{\mu_{0}I}{4\pi}\oint \frac{\sin(\theta-\phi)\text dl}{r(\phi)^{2}}= \frac{\mu_{0}I}{4b\pi}\int_{0}^{2\pi}\sin(\theta-\phi)(1+\varepsilon \cos\phi) \text d\phi. \]
Now, as impossible as it may look, this is solvable in closed form, but it looks awfull (at least in Mathematica), so I'm not writing it here... Do you have an elegant solution?

The parametric form of an ellipse in polar coordinates:
[b]x=a*cos(phi),    y=b*sin(phi)[/b]
after differentiation:
[b]dx=-a*sin(phi)d(phi),   dy=b*cos(phi)d(phi)[/b]
according to the Biot-Savart law, the magnetic field: H=B/mu0

[b]H= (I/(4*pi))*ab*integral((0 to 2*pi)(d(phi)/(a^2*(cos(phi))^2+b^2*(sin(phi))^2)^1,5) =
(I*b/(a^2*pi))*integral((0 to 2*pi)(d(phi)/(1-k^2*(sin(phi))^2)^1,5)[/b]
where:
[b]k^2=(a^2-b^2)/a^2=c^2/a^2[/b]
this integral can be given with a whole second general elliptic integral [b]E(k)[/b].
So we get:
[b]H=(I*b)/(a^2*pi)*(E(K)/(1-k^2))[/b]
the length of the ellipse:
[b]L=4*integral(0 to pi/2)(sqrt(a^2*sin^2(phi)+b^2*cos^2(phi))d(phi)= 4*a*E(k)[/b]
so we get:
[b]E(k)=L/(4a)[/b]
so the magnetic field in the middle of the ellipse:
[b]H=(I*L)/(4*A)[/b]

I dont know how to put in formulas. I hope though you can understand it.[/code]

It was tedious to get through this :) You should try to use LaTeX, just put the TeX code in dollar signs...

Anyway, I'm sorry that I misunderstood your post... So, $L$ is the perimeter of the ellipse? That changes things... But there's something I don't think I understand -- how did you get that $(...)^{3/2}$ term in your integrand?

Thaakisfox! This is unreadable. Use latex pls. See http://www.artofproblemsolving.com/LaTeX/AoPS_L_TeXer.php and https://www.komal.hu/munkafuzet/munkafuzet.cgi?a=tk&p=32. Ty.

Ok here it is nicely:
The parametric form of an ellipse in polar coordinates:
$x=a\cos\phi , y=b\sin\phi$
after differentiation:
$dx=-a\sin%Error. "phid" is a bad command.
\phi , dy=b\cos%Error. "phid" is a bad command.
\phi$
according to the Biot-Savart law the magnetic field: $H=\frac{B}{\mu_{0}}$
so:
$H=\frac{Iab}{4\pi}\int_{0}^{2\pi}\frac{d\phi}{{(a^{2}\cos^{2}\phi+b^{2}\sin^{2}\phi})^\frac32}$$=\frac{Ib}{a^{2}\pi}\int_{0}^{2\pi}\frac{d\phi}{{(1-k^{2}\sin^{2}\phi)}^\frac32}$
Where :
$k^{2}=\frac{a^{2}-b^{2}}{a^{2}}=\frac{c^{2}}{a^{2}}$
This integral can be given with a Whole second genre elliptic integral:$E(k)$
so we get:
$H=\frac{Ib}{a^{2}\pi}\frac{E(k)}{1-k^{2}}$
the perimeter of the ellipse:
$L=4\int_{0}^{\frac{\pi}{2}}\sqrt{a^{2}\sin^{2}\phi+b^{2}\cos^{2}\phi}d\phi$
So we get:
$E(k)=\frac{L}{4a}$
So the magnetic field in the middle of the ellipse:
$H=\frac{IL}{4a}$
it looks much better now.

The biot-savart law is :
$B=\frac{{\mu_{0}}{I}}{4\pi}\oint\frac{d\vec{l}\times{\vec{r}}}{r^{3}}$
now:
$d\vec{s}\times\vec{r}=\vec{k}abd\phi$
and according to the parametric form of the ellipse:
$r=\sqrt{a^{2}\cos^{2}\phi+b^{2}\sin^{2}\phi}$
So if we right these in the law:
${{B=\frac{{\mu_{0}}{I}}{4\pi}ab\int_{0}^{2\pi}}\frac{d\phi}{r^{3}}=\frac{{\mu_{0}}{I}}{4\pi}ab\int_{0}^{2\pi}}\frac{d\phi}{{\sqrt{a^{2}\cos^{2}\phi+b^{2}sin^{2}\phi}}^{3}}$
and from here we get that $(...)^\frac32$ term in the integrand

that's rich... and what's the magnitude of the magnetic field in arbitrary point in this ellipse?:P

Very nice :) I'll look through and try to follow your solution now that it's in LaTeX. Please do post more problems!

The book Classical Electrodynamics by John David Jackson contains more nice problems with solutions. :) This is one of the best books in the world for learning physics.

Unfortunately, I don't have the book :( Never found it, and didn't want to buy it over the net...

This is very nice! Actually, I was troubled by that vector product. I didn't realize it was that simple until I took another look at Kepler's laws. And I liked the idea of eliminating the elliptic integral. Once again, great job!


calculus
integration
real analysis
real analysis unsolved

Evaluate $ \int^5_3 \sqrt {x \plus{} 2\sqrt {2x \minus{} 4}} \plus{} \sqrt {x \minus{} 2\sqrt {2x \minus{} 4}}\ dx$

The options given are:
A) $ 2(\sqrt 5 \plus{} 1/5)$
B) $ 4 (\sqrt3 \minus{} 1/3)$
C) $ 3(\sqrt2 \minus{} 1/2)$
D) $ 5(\sqrt 6 \plus{} 1/6)$
E) None of these

Using a $ u\equal{}2x\minus{}4$ substitution, I get the answer as B).

I'd appreciate it if someone could verify this.

[quote="Dr Sonnhard Graubner"]hello, are you sure that you write down your problem correctly?
Sonnhard.[/quote]

I sure am. I'm also pretty sure that my answer is right.

I did a quick plug-in to Mathematica, and I got $ 4 \sqrt{3}\minus{}\frac{2}{3}\sqrt{2}$.

hello, i have got the same result.
Sonnhard.

So the two of you believe that the answer is E) ie None of these?

[quote="Dr Sonnhard Graubner"]hello, by hand i have got result B.
Sonnhard.[/quote]

Same here!

Here's the problem: you must have used the substitution:

$ x \equal{} \frac{1}{2}(z\minus{}2)^2\plus{}2$

This is not one-to-one on $ [3,5]$ so this is what's been doing all the trouble.

hello, i have used the substitution $ t\equal{}2\sqrt{2x\minus{}4}$
Sonnhard.

hello, no, i don't need nor more substitutions. I have $ dx\equal{}\frac{t}{2}dt$ and so our integral is
$ \int_{2\sqrt{2}}^{2\sqrt{6}}\left(\sqrt{\frac{t^2}{8}\plus{}2\plus{}t}\plus{}\sqrt{\frac{t^2}{8}\plus{}2\minus{}t}\right)\frac{t}{2}dt$.
Sonnhard.

Well, all I can say is that it's something ugly.. :)

Hi,
Note that $ \sqrt {x \plus{} 2\sqrt {2x \minus{} 4}} \equal{} \sqrt{x\minus{}2} \plus{} \sqrt{2}$ and $ \sqrt {x \minus{} 2\sqrt {2x \minus{} 4}} \equal{} |{\sqrt{x\minus{}2} \minus{} \sqrt{2}}|$.


function
trigonometry
algebra
domain
real analysis
real analysis unsolved

Suppose that there is sequence $f_{n}$ converges to $f$ pointwise almost everywhere as $n \rightarrow \infty$, and for each fixed $n$, there is a sequence $f_{n,k}$ which converges to $f_{n}$ pointwise almost everywhere as $k \rightarrow \infty$. (All functions are measurable and defined on [a,b])

Prove that there is a sequence $k(n) \rightarrow \infty$ as $n \rightarrow \infty$ such that $f_{n,k(n)}$ converges to $f$ pointwise almost everywhere as $n \rightarrow \infty$.

Thanks a lot! And if someone could give a counterexample for the above proposition doesn't work for everywhere convergence, it will be greatly appreciated!

A counterexample to the "everywhere" version: $f_{n,k}(x)=(\cos(\pi xn!))^{2k}$ converges (as $k\to \infty$) to $f_{n}$, where $f_{n}(x)=1$ if $\pi xn!$ is an integer and $f_{n}(x)=0$ otherwise. The sequence $f_{n}$ converges to the characteristic function of $\mathbb Q$. However, there is no subsequence of $f_{n,k}$ that converges to $\chi_{\mathbb Q}$ pointwise. In fact, there is no sequence of continuous functions that converges to $\chi_{\mathbb Q}$ pointwise. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=136356]A related thread.[/url]

For each n, choose $k(n) > n$ with
\[m\{|f_{n,k(n)}-f_{n}| > \frac{1}{n}\}< 2^{-n}\]
This is possible because the measure of the domain is finite.

By [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=143796]strong convergence in measure implies pointwise convergence[/url], we have $|f_{n,k(n)}-f_{n}| \to 0$ a.e. Therefore $f_{n,k(n)}$ and $f_{n}$ have the same limit a.e.


[quote="Raja Oktovin"]Let $ ABC$ be a triangle. A circle $ P$ is internally tangent to the circumcircle of triangle $ ABC$ at $ A$ and tangent to $ BC$ at $ D$. Let $ AD$ meets the circumcircle of $ ABC$ agin at $ Q$. Let $ O$ be the circumcenter of triangle $ ABC$. If the line $ AO$ bisects $ \angle DAC$, prove that the circle centered at $ Q$ passing through $ B$, circle $ P$, and the perpendicular line of $ AD$ from $ B$, are all concurrent. [/quote]
Let E be the intersection of (Q) and AC, O is the circumcenter of triangle ABC
We have $ \frac{AP}{AO}\equal{}\frac{AD}{AQ}$ then $ PD//OQ$ or $ OQ\perp BC$
So Q is the midpoint of arc BC
Since PO bisects $ \angle DAC$, it's easy to see triangle AQC and ADE are isosceles then DECQ is a isosceles trapezium.
We get $ \angle QED\equal{}\angle EDC\equal{}\angle DAE$
So QE is a tangent of (P)
Thus $ QE^2\equal{}QD.QA\equal{}QB^2$
We obtain $ QB\equal{}QE$ or E lies on (Q)
Moreover, $ \angle DQE\equal{}\angle DCE\equal{}\angle AQB$ then  $ BQE$ is an isosceles triangle
Therefore $ AD\perp BE$
Hence the circle centered at $ Q$ passing through $ B$, circle $ P$, and the perpendicular line of $ AD$ from $ B$, are all concurrent at E


inequalities
calculus
integration
search
inequalities proposed

(1) Suppose that real numbers $ a_1,\ a_2,\ x_1,\ x_2,\ y_1,\ y_2$ satisfy :

$ 0<a_1\leq a_2$

$ a_1x_1\leq a_1y_1$

$ a_1x_1+a_2x_2\leq a_1y_1+a_2y_2$.

Prove : $ x_1+x_2\leq y_1+y_2$.

(2) Let $ n\geq 2$ be integers.

If $ 3n's$ real numbers $ a_1,\ a_2,\ \cdots ,\ a_n\ ,\ x_1,\ x_2,\ \cdots ,\ x_n,\ y_1,\ y_2,\ \cdots ,\ y_n$ satisfy $ 0<a_1,\ a_2\leq
\cdots \leq a_n$ and $ n's$ inequalities $ \sum_{i=1}^j a_ix_i\leq \sum_{i=1}^j a_iy_i\ (j=1,\ 2,\ \cdots n)$.

Sorry for that, I forgot to write the last part.

(2) Let $ n\geq 2$ be integers.

If $ 3n's$ real numbers $ a_1,\ a_2,\ \cdots ,\ a_n\ ,\ x_1,\ x_2,\ \cdots ,\ x_n,\ y_1,\ y_2,\ \cdots ,\ y_n$ satisfy $ 0 < a_1,\ a_2\leq \cdots \leq a_n$ and $ n's$ inequalities $ \sum_{i \equal{} 1}^j a_ix_i\leq \sum_{i \equal{} 1}^j a_iy_i\ (j \equal{} 1,\ 2,\ \cdots n)$,

 prove : $ \sum_{i\equal{}1}^n x_i\leq \sum_{i\equal{}1}^n y_i$.

@Moderator, please edit the context of question (2). Thanks in advance.

kunny

Hi,

The theme of this problem is Abel Summation, or Integration by Parts in discrete form:
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=135850088&t=192043

we have
$ \sum_{i\equal{}1}^n x_i\equal{}\sum_{i\equal{}1}^n (\frac {1}{a_i})(a_ix_i) \equal{} \sum_{i\equal{}1}^m (\sum_{j\equal{}1}^i a_jx_j)(\frac {1}{a_i}\minus{}\frac {1}{a_i\plus{}1}) \plus{} \frac {1}{a_n} (\sum_{i\equal{}1}^n a_ix_i) \leq \sum_{i\equal{}1}^m (\sum_{j\equal{}1}^i a_jy_j)(\frac {1}{a_i}\minus{}\frac {1}{a_i\plus{}1}) \plus{} \frac {1}{a_n} (\sum_{i\equal{}1}^n a_iy_i)\equal{} \sum_{i\equal{}1}^n (\frac {1}{a_i})(a_iy_i) \equal{} \sum_{i\equal{}1}^n y_i$
where $ m \equal{} n\minus{}1$ and $ \frac {1}{a_i} \minus{} \frac {1}{a_i\plus{}1} \geq 0$ for $ i \equal{} 1,2 \cdots m$

Regards,
Lei Lei


algebra
polynomial
number theory proposed
number theory

$ p(x)$ is a polynomial in $ \mathbb{Z}[x]$ such that for each $ m,n\in \mathbb{N}$ there is an integer $ a$ such that $ n\mid p(a^m)$. Prove that $0$ or $1$ is a root of $ p(x)$.

[hide="Hint"]For some prime $ q$, let $ n \equal{} q$, $ m \equal{} q \minus{} 1$.[/hide]

Let $P(x)=\sum_{k=0}^n a_k x^k$. Note that if $a_0=0$, the condition trivially holds. Now let $a_0\neq 0$. Fix a prime $p$, such that $p>|a_0|$ and $p>\sum_{k=0}^n |a_k|$. Observe that, for $m=p-1$ and $n=p$, we have $p\mid P(a^{p-1})$. If $p\mid a$, then we earn a contradiction since  $p>|a_0|$. Thus, $p\nmid a$, hence,  $a\not\equiv 0\pmod{p}$, and by Fermat, $a^{p-1}\equiv 1\pmod{p}$. This yields, $P(a^{p-1})\equiv P(1)\pmod{p}$. Since $p>\sum_{k=0}^n |a_k|\geqslant |P(1)|$ (by triangle inequality), it follows that $P(1)=0$.


The area of square $ A$ is $ \frac14$ of the area of square $ B$. The diagonal of square $ A$ is $ n$. Find the perimeter of square $ B$ in terms of $ n$. Express your answer in simplest radical form.

Let x be the side length of A, and 2x be the side length of B. Thus $ n \equal{} x\sqrt {2}$. We know that the diagonal of B is $ 2x\sqrt{2}$, which is $ 2n$. Thus each side length is (in terms of n) $ \frac{2n}{\sqrt{2}}\equal{}\frac{2n\sqrt{2}}{2}\equal{}n\sqrt{2}$. We want 4 times this, or $ 4n\sqrt{2}$.

We can just do everything with $n$. Knowing that the diagonal of square $A$ is $n$, we know each side of square $A$ must be $\frac{n}{\sqrt{2}}$. Knowing that the area of square $B$ is four times larger, we can write the area of square $B$ as $\frac{4n^{2}}{2}$. Taking the square root of that and multiplying it by four gives us the the perimeter of square $B$. Hence $\boxed{4n\sqrt{2}}$.

Okay, that wasn't fair. In the FTW game I put 4sqrt2n, but then the answer is 4nsqrt2.

Well that is because $4\sqrt{2}n$ can be interpreted as $4\sqrt{2n}$, which is wrong.


induction
limit
integration
function
real analysis
algebra proposed
algebra

Suppse that $k$ be a positive integer. Prove that (without induction) \[ \lim_{n \rightarrow \infty}\frac{1^k+2^k+...+n^k}{n^{k+1}}=\frac{1}{k+1} \]

Just use that for large enough $n$:  $\sum_{i=1} ^n i^k$ tends to $\int_0 ^n x^k dx = \frac{n^{k+1}}{k+1}$

Use Stolz-Cesaro Theorem (see http://planetmath.org/encyclopedia/StolzCesaroTheorem.html ),
then binomial expansion of $(n+1)^k$ and $(n+1)^{k+1}$.

Another solution:
$\frac{1^k+2^k+...+n^k}{n^{k+1}}$ is a Riemann sum associated to function $f: [0,1]\rightarrow \mathbb {R}, f(x)=x^k$ (and k may be an arbitrary positive number), so the limit in the statement is equal to $\int_0^1x^kdx=\frac 1{k+1}$.

Thanx  ;)  but who can solve it by the Sqeezing Lemma ????


geometry
incenter
circumcircle
trigonometry
geometry proposed

Let $ I$ be incenter of triangle $ ABC$, $ M$ be midpoint of side $ BC$, and $ T$ be the intersection point of $ IM$ with incircle, in such a way that $ I$ is between $ M$ and $ T$. Prove that $ \angle BIM\minus{}\angle CIM\equal{}\frac{3}2(\angle B\minus{}\angle C)$, if and only if $ AT\perp BC$.

Beautiful problem. [hide="A hint"] when $ AT\perp BC$ then $ T$ is on the nine-point circle.[/hide]
[hide="Solution"] Assume $ AT\perp BC$ and incircle meets $ BC$ at $ E$. Now note that $ \angle BIM\minus{}\angle BIC\equal{} 2\angle EIM\minus{}\frac{1}{2}(\angle C\minus{}\angle B)$ So we should prove $ \angle EIM\equal{}\angle B\minus{}\angle C$.
Now note that if we invert to $ M$ with respect to the incircle and since the invertion preserves the angles, nine-point circle changes to the tangent line that is perpendicular to $ IM$ at it's other intersection with incircle beside of $ T$ which means that $ T$ was initially on the
nine-point circle.
We have $ \angle EIM\equal{}\angle H_{a}TM$ where $ H_{a}$ is the feet of the altitude from $ A$. So it comes down to showing that arc $ H_{a}M$ is twice the $ \angle B\minus{}\angle C$. Let $ N$ be the mid point of $ AC$ then note that $ \angle H_{a}NC\equal{}180\minus{}2\angle C$ since $ H_{a}N$ is median in the right triangle $ AH_{a}C$ and $ \angle MNC\equal{}A$. hence $ \angle H_{a}NM\equal{}\angle B\minus{}\angle C$. Since $ N$ is on the nine-point circle $ \angle H_{a}NM\equal{}\angle H_{a}TM$ which finishes the proof. [/hide]

[color=darkred][b][u]Prpblem.[/u][/b] $ w = C(I,r)$ is the incircle of $ \triangle ABC$. Denote the points $ \{\begin{array}{c}M\in (BC)\ ,\ MB = MC\\ \ D\in BC\ ,\ AD\perp BC\\ \ T\in MI\cap w\end{array}$.
Prove that $ T\in AD\Longleftrightarrow$ $ m(\widehat{BIM}-m(\widehat{CIM}) =\frac{3}{2}\cdot (B-C)$.[/color]

[color=darkblue][b]Proof.[/b] Suppose w.l.o.g. that $ b > c$. Calculate easily $ ME =\frac{b-c}{2}$ and $ MD =\frac{b^{2}-c^{2}}{2a}$. Denote the points $ \{\begin{array}{c}E\in BC\cap w\\ \ S\in MI\cap AD\end{array}$ and $ AS = x$.

Observe that $ IE\parallel AD$ in the $ D$- right triangle $ MDA$. Thus, $ IE\parallel AD\Longleftrightarrow$ $ \frac{IE}{SD}=\frac{ME}{MD}$ $ \Longleftrightarrow$ $ \frac{r}{h_{a}-x}=\frac{a}{b+c}$ $ \Longleftrightarrow$

$ ax = ah_{a}-(b+c)r$ $ \Longleftrightarrow$ $ ax = 2pr-(b+c)r$ $ \Longleftrightarrow$ $ x = r$, i.e. $ \boxed{\ AS = r\ }$. Therefore, $ \boxed{\ T\in AD\Longleftrightarrow T\equiv S\ }$.

==============================================================================================================

Since $ \triangle ATI$ is $ A$- isosceles and $ m(\widehat{TAI}) =\frac{B-C}{2}$, obtain $ m(\widehat{DTM}) = B-C$. Since $ IE\parallel TD$, obtain $ T\in AD$ $ \Longleftrightarrow$ $ \phi\equiv\boxed{m(\widehat{EIM}) = B-C}$.

Since $ \{\begin{array}{c}m(\widehat{BIM}) =(90^{\circ}-\frac{B}{2})+\phi\\ \ m(\widehat{CIM}) =(90^{\circ}-\frac{C}{2})-\phi\end{array}\|$ $ \implies$ $ m(\widehat{BIM})-m(\widehat{CIM}) =$ $ 2\phi-\frac{B-C}{2}=\frac{3}{2}\cdot (B-C)$, obtain  $ m(\widehat{BIM})-m(\widehat{CIM}) =\frac{3}{2}\cdot (B-C)$.

[b]Remark.[/b] $ T\in AD\Longleftrightarrow$ the circumcenter $ O$ of $ \triangle ABC$ belongs to the $ A$- Nagel's line, i.e. $ O\in AF$, where $ M\in (EF)$ and $ MF = ME$ $ \Longleftrightarrow$

$ OM = r$ $ \Longleftrightarrow$ $ r = R\cos A$ (the our triangle is acute !) $ \Longleftrightarrow$ $ AT = OM$ $ \Longleftrightarrow$ the point $ T$ is middle of the segment $ AH$,

where $ H$ is the orthocenter of $ \triangle ABC$ $ \Longleftrightarrow$ the point $ T$ belongs to the Euler's circle of $ \triangle ABC$ $ \Longleftrightarrow$ $ \cos B+\cos C = 1$ $ \Longleftrightarrow$ $ \frac{r}{R}+\frac{a}{b+c}= 1$.

[b]Commentary.[/b] The remarkable relation $ \boxed{\ AS = r\ }\ \ (*)$ is well-known. From this moment the our nice problem became  more easily !

By the way, I don't like the right hand of the equivalence from the conclusion, is even nastily. In the my opinion, $ m(\widehat{EIM}) = B-C$ is more nicely.

In fact, this problem  is the property $ (*)$ somehow "painted".
[/color]

I think this is closely related to the solution given.

Denote by D the intersection of AI and the circumcircle; it is not difficult to see that D is the center of the circle passing through B, I, C, and the A-excenter. In particular, we will use that DM is perpendicular to BC. The angle condition is easily seen to be equivalent to DIM = IDM, so we wish to show that AT is perpendicular to BC if and only if MID is isosceles. This follows easily once we note that IT / IA = DM / DI, so both conditions are in fact equivalent to the similarity of the triangles AIT and DIM.

Let $D_{1}$ be the touch point if the $A$-excircle,let $AD_{1}\cup \odot I={Z}$,$\odot I$ touches $BC$ in $D$.

$IM$  is the $I$  midline in $\triangle DZD_{1}$ and hence $IM||D_{1}Z$.$ID\perp BC$,$AT\perp BC$ and hence $AT||IZ$ $\implies$
$AZIT$ is a parallelogram but as $IT=IZ$ its a rhombus  and hence $AD_{1}$ is the isogonal to $AT$ $\implies$  $AD_{1}$ passes thru
the center of $\odot ABC$.
$$\angle CIM=\angle \frac{\pi}{2}-\frac{\gamma}{2}+ \angle DIM=\angle \frac{\pi}{2}-\frac{\gamma}{2}+\angle HAO$$
$$=\angle \frac{\pi}{2}-\frac{\gamma}{2}+\frac{\pi}{2}-\beta-\frac{\pi}{2}+\gamma=\frac{\pi}{2}+\frac{\gamma}{2}-\beta$$
Analogously we have $\angle BIM=\frac{\pi}{2}-\gamma+\frac{\beta}{2}$ so we are done.$\clubsuit$


function
calculus
derivative
calculus computations

Let $ z(x,y)$ be a two-variable $ C^1$ class function on $ \{(x,y): x,\ y > 0\}$. Show that $ z(x,y)$ can be expressed in the form
\[ z \equal{} yf\left(\frac {y}{x}\right),\]
where $ f(s)$ is a one-variable $ C^1$ class function on $ (0,\infty)$, if and only if
\[ x\frac {\partial z}{\partial x} \plus{} y\frac {\partial z}{\partial y} \equal{} z.\]

$ \frac{\partial z}{x} \equal{} \minus{}\frac{y^2}{x^2} f '\left(\frac{y}{x}\right)$
$ \frac{\partial z}{y} \equal{} f\left(\frac{y}{z}\right) \plus{} \frac{y}{x} f '\left(\frac{y}{x}\right)$

Thus, $ x\frac{\partial z}{x}\plus{}y\frac{\partial z}{y} \equal{} \minus{}\frac{y^2}{x} f '\left(\frac{y}{x}\right) \plus{} y f\left(\frac{y}{z}\right) \plus{} \frac{y^2}{x} f '\left(\frac{y}{x}\right) \equal{} y f\left(\frac{y}{z}\right) \equal{} z$

Indeed that's true, but observe it's a double implication. I certainly had no problem showing that if the function satisfies $ z\equal{}yf(y/x)$ then the relation of the partial derivatives is satisfied, but I find the converse somewhat difficult to prove in a formal way. Can you help me finish the proof please?

Define a new function $ Z(t,y)\equal{}z(ty,y)$; we want to show that $ Z(t,y)\equal{}yg(t)$, where $ g(t)\equal{}f(\frac1t)$. This is effectively a substitution.

Differentiating, $ \frac{\partial Z}{\partial t}\equal{}y\frac{\partial z}{\partial x}$ and $ \frac{\partial Z}{\partial y}\equal{}t\frac{\partial z}{\partial x}\plus{}\frac{\partial z}{\partial y}$. Multiply the latter equality by $ y$ to get $ y\frac{\partial Z}{\partial y}\equal{}ty\frac{\partial z}{\partial x}\plus{}y\frac{\partial z}{\partial y}\equal{}z(ty,y)\equal{}Z(t,y)$.

That's a simple differential equation $ h'(y)\equal{}\frac{h(y)}{y}$ for the one-variable function $ h(y)\equal{}Z(t,y)$, and its solutions are $ h(y)\equal{}cy$ for constants $ c$, which extend to the whole region $ y>0$. This $ c$ is independent of $ y$, but may depend on $ t$; call it $ g(t)$, and we have written $ Z$ in the desired form.

For all such problem,You can solver like this:
Let
$ u\equal{} z\minus{}yf(\frac{y}{x})$
$ \frac{\partial u}{\partial x}\equal{} \frac{y}{x^2} f'(\frac{y}{x})$
$ \frac{\partial u}{\partial y}\equal{} \minus{}f(\frac{y}{x} )\minus{}\frac{y}{x}f'(\frac{y}{x})$
$ \frac{\partial u}{\partial z}\equal{} 1$
So
$ \frac{\partial z}{\partial x}\equal{} \frac{y}{x^2} f'(\frac{y}{x})$
$ \frac{\partial z}{\partial y}\equal{} \minus{}f(\frac{y}{x} )\minus{}\frac{y}{x}f'(\frac{y}{x})$


Let $ g: [1;\plus{}\infty)\to[1;\plus{}\infty)$ a funtion with the property $ g(g(x))\equal{}x^2\minus{}2x\plus{}2$ . Prove that $ g$ is injective.

Isn't it quite easy?
If $ g(a) \equal{} g(b)$ then $ a^2\minus{}2a\plus{}2 \equal{} b^2\minus{}2b\plus{}2$
so that $ (a\minus{}b)(a\plus{}b\minus{}2)\equal{}0$.
Since $ a,b \geq 1$ the conclusion holds easily.... :wink:

Pierre.


Hi there,

I am a new user to Asymptote.

and I need some help in drawing the following:

[asy]size(200);
pair A=(0,5), B=(-5,0), C=(5,0);
draw(circle(A,1));
draw(circle(B,1));
draw(circle(C,1));
label("$1$",A);
label("$4$",B);
label("$5$",C);

pair D=(-5,-5), F=(5,-5);
draw(circle(D,1));
draw(circle(F,1));
label("$2$",D);
label("$3$",F);[/asy]

That's what I can do so far.

The original picture is in the PDF file.

Thank you for help  :lol:


A set of 1990 persons is divided into non-intersecting subsets in such a way that

1. No one in a subset knows all the others in the subset,

2. Among any three persons in a subset, there are always at least two who do not know each other, and

3. For any two persons in a subset who do not know each other, there is exactly one person in the same subset knowing both of them.

(a) Prove that within each subset, every person has the same number of acquaintances.

(b) Determine the maximum possible number of subsets.

Note: It is understood that if a person $A$ knows person $B$, then person $B$ will know person $A$; an acquaintance is someone who is known.  Every person is assumed to know one's self.

Let $A,B$ be two people who do not know each other. They have exactly one common acquaintance $C$, and for each acquaintance $D_a\ne C$ of $A$ there is exactly one common acquaintance $D_b\ne C$ of $D_a,B$. This way, we can pair up the acquaintances of $A$ and those of $B$ which are different from $C$, and the conclusion is that every two people who do not know each other have the same number of acquaintances.

In the graph $G$ which we associate to this situation in an obvious manner, two vertices can have different degrees only if they lie in different connected components of the complementary graph $\overline G$. Now, if $\overline G$ has at least two components each of which has $\ge 2$ vertices, there will be $4$ vertices $a_1,a_2,b_1,b_2$ such that $a_i,b_j$ are connected for $i,j=1,2$, which contradicts the hypotheses, and if it has at least two components, one of which has exactly one vertex, then that vertex is connected to all the others, so we get a contradiction again. This means that $\overline G$ is connected, and it follows from this and the first sentence in the paragraph that all vertices of $G$ have the same degree.

It's easy to see that there are no graphs $G$ with the required properties and with fewer that $5$ vertices, and that a $5$-cycle satisfies all the conditions; it follows from this that the largest possible number of subsets is achieved when we divide the set $1990$ vertices (well, people :)) into $398=\frac{1990}5$ subsets.

[quote="grobber"]Now, if $ \overline G$ has at least two components each of which has $ \ge 2$ vertices, there will be $ 4$ vertices $ a_1,a_2,b_1,b_2$ such that $ a_i,b_j$ are connected for $ i,j \equal{} 1,2$, which contradicts the hypotheses,[/quote]Sorry to revive this old thread. I don't understand this part of the solution. Can you explain why is it a contradiction?

A square in this graph either requires an additional edge (so a triangle) or has two disconnected vertices with two common neighbors (contradicting the "exactly 1" in condition 3).

I'm fairly sure such a graph must be a 5 cycle or 2 attached 5 cycles. if any vertex has degree less than 2, since it can't be zero it has degree one but then the one it is attached to is itself attached to all others since that first one is not attached to any =><=

Now since each vertex has degree 2 we have a cycle. We clearly do not have a 4 or 3 cycle but if the length is 6 or greater take 2 verticies 2 apart if they are connected we have a smaller cycle if not then they are connected to another which if put together with the part of the cycle that vertex is not part of gives us a smaller cycle so we have a 5 cycle.

Every vertex not connected to the 5 cycle that we will call ABCDE is connected to at most  of them. We consider pairs X, Y with X not in ABCDE and Y in ABCDE. And note that X,Y have a common neighbor outside ABCDE if and only if they are not connected X is not connected to a neighbor of Y in ABCDE. If any vertex is not attached to any of ABCDE it contributes 5 pairs while those attached to one contribute 2 since it is not possible for a vertex to be common neighbors for three such pairs ( if two are connected to one vertex then those 2 have two common neighbors one outside ABCDE and one of ABCDE they are both connected to) thus we only consider the case where each vertex outside is attached to exactly one of ABCDE and each is a common neighbor of exactly 2 pairs X,Y. So each vertex outside ABCDE has degree 3. Now if a vertex is connected to A then we need another vertex to connect it to C going around like this we find there is a vertex connected to each of ABCDE. Now consider a neighbor of A and one of C if they are not connected then their neighbor T is needs to be connected to a vertex of ABCDE which is not a neighbor of A or C which is impossible thus each neighbor of A is connected to each neighbor of C. Thus there is at most 1 neighbor of C so C has degree 2+1=3 so we are done.

If the 5 cycle is the entire graph we are also done. :starwars:


Let ABC be a right triangle with hypotenuse c.

Prove the inequality

$a+b+2(c+\sqrt{2}) \geq 4\sqrt{a+b+c}$

But the inequality is not homogenous...

So in fact it should be true even if the $\sqrt{2}$ is not there... :huh:

If it's not homogenous and there's no $\sqrt{2}$, we can easily derive a contradiction.

That's why there's $\sqrt{2}$.

Because it's not homogenous.

The trick is pure evil.

Hm, I see, I sort of misread the question. I'll think about it :)

I don't know what your evil trick is, but here's my solution.
If $c \ge 4-2\sqrt{2}$,
$a+b+c+c+2\sqrt{2} \ge 2\sqrt{(a+b+c)(c+2\sqrt{2})} \ge 4 \sqrt{a+b+c}$ because $c \ge 4-2\sqrt{2}$ implies $\sqrt{c+2\sqrt{2}} \ge 2$
If $c \le 4-2\sqrt{2}$,
$8(\sqrt{2}-1)^2 \ge c^2=a^2+b^2 \ge 2(a+b)^2$ which gives $2\sqrt{2} - 2 \ge a+b$, then
$a+b+2c+2\sqrt{2} \ge 2(a+b+c+1) \ge 4\sqrt{a+b+c}$

You beat me there, Soarer. It's not straightforward, but only a little tricky :)

[quote="Soarer"]
If $c \le 4-2\sqrt{2}$,
$8(\sqrt{2}-1)^2 \ge c^2=a^2+b^2 \ge 2(a+b)^2$ which gives $2\sqrt{2} - 2 \ge a+b$, then
$a+b+2c+2\sqrt{2} \ge 2(a+b+c+1) \ge 4\sqrt{a+b+c}$[/quote]

Wrong!!! since when do we have $a^2+b^2 \ge 2(a+b)^2$???

If it's that easy, i won't post it here.

No one?
Yeah the trick is evil...
but you can manage to find it...

Still no one?

Okay it seems too hard for you guys...

I'll give you a little hint. (For those who want to solve it on their own, don't look.)

[hide]

Here's the hint

observe that

$a+b+c= \sqrt{a^2+b^2+c^2+2ab+2ac+2bc}$

that's it.

Now you are much closer to the solution.[/hide]

I think I have found a really nice and simple solution to this:

$a+b+2c+2\sqrt{2} \ge \sqrt{2}(a+b+c) + 2\sqrt2 \geq 4\sqrt{a+b+c}$

The first step follow from AM-QM in the form $(2-\sqrt{2})\sqrt{a^2+b^2}\ge(\sqrt{2}-1)(a+b)$, the second step is just AM-GM in the classical way.

Detective, what was your solution? It seems you had a much bigger one in mind? (seeing your comments and hint in the end)

ah okay i admit that your solution is far better than mine

here's mine by AM-GM four variables

$a+b+2(c+\sqrt{2}) = (a+c) + (b+c) + \sqrt{2} + \sqrt{2} \geq 4\sqrt{\sqrt{2(a+c)(b+c)}}$

Notice that by pythagoras theorem we have

$2(a+c)(b+c) = 2ab + 2ac + 2bc + 2c^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc$

Since

$\sqrt{a^2 + b^2 + c^2 + 2ab + 2ac + 2bc} = a+b+c$

we have the desired result.

it's pretty tricky right?


[quote="t0rajir0u"][hide="Solution"] The quadratic equation in $a$ given by

$a^{3}-\frac{ b^{3}(a-c)}{b-c}-\frac{c^{3}(a-b)}{(c-b)}= (a+b+c)(a-b)(a-c)$

Has roots $a = b, a = c, a = 0$ (yes, it is quadratic because the cubic terms cancel), hence it is an identity.  Dividing by $(a-b)(a-c)$ we have the original expression is equal to $\boxed{ a+b+c }$. [/hide][/quote]

Thanks for the solution.....


Let $ ABC$ be an acute triangle.Let $ S_{BC}$ has segment $ BC$  as its diameter.Circle $ W_A$ is  to lines $ AB$ and $ AC$ and is tangent externally to $ S_{BC}$ at $ A_1$ .Points $ B_1,C_1$ are difined analogously.Prove that $ AA_1,BB_1,CC_1$ are concurent.

This solution belongs to a friend of mine
Denote M=A1   N=B1  P=C1   W is the circumcenter of WA
M1,M2 is the projection of M to AB,AC
WM/MD=2r/a
MM1=(r+2ra⋅a2⋅sinB)/(1+2ra)，MM2=(r+2ra⋅a2⋅sinC)/(1+2ra)
So MM1/MM2=(1+sinB)/(1+sinC)
sin∠MAB/sin∠MAC=(1+sinB)/(1+sinC)
the same way sin∠NBC/sin∠NBA=(1+sinC)/(1+sinA) sin∠PCAsin∠PCB=(1+sinA)/(1+sinB)
done!


group theory
abstract algebra
superior algebra
superior algebra theorems

No.

Construct the nonabelian group with a semidirect product.

Yes it IS!

In fact we can say something stronger. Assume [G] = pq with p and q relatively prime.

Then by sylows theorem there exists sylow subgroups say P and Q with [P] = p and [Q] = q. Now the number of p-sylow subgroups is congruent to 1modq. Moreover the number of p-sylow subgroups must divide [G]/p. So in our case it must divide q. This forces the number of p-sylow subgroups to be one. Since there is only one p-sylow this p-sylow must be normal in G.

A similar argument shows there is only one q-sylow  and it is therefore normal in G.

So P and Q are the unique p and q sylows, respectively. Since P and Q have prime order they must be cyclic. Suppose <x> = P and <y> = Q.

We will show [<xy>] = pq so that <xy> generates all of G making G a cyclic group.

First we will show xy = yx. It suffices to show xyx~y~ = 1. [by x~ i mean x inverse]

Since <y> is normal in G xyx~ must also lie in <y>. But [xyx~] = q. Thus we must have xyx~ = y. Hence, xyx~y~ = yy~ = 1.

Now look at (xy)^(pq) = x^(pq)*y^(pq) = 1*1 = 1. Hence [xy] = pq so G = <xy>.

No it ISN'T!
You have $ 13 \equiv 1 \mod 3$ and $ 13$ divides $ \frac {39}{3}$. Thus there can be $ 13$ $ 3$-Sylows.

oops my mistake  :blush:  i was thinking of when p<q and q is not congruent to 1modp

How could you construct a group of order 39 that consists of 13 Sylow-3 groups and a group of order 13?
Maybe the representation is:
{$ a^3 \equal{} b^{13} \equal{} 1,b^{ \minus{} 1}ab \equal{} a^j$},where j is not valued 1,
P.S.
The 13-ordered sub-group is a normal one,then j cannot be valued all values that is not equal to 1.

I like the semidirect product suggestion. Let $ N$ be the normal 13-subgroup. Then it trivially intersects the 3-group, $ K$. Thus $ NK\equal{}G$. Since $ N$ is normal we have that $ G\cong N\rtimes_\phi K$ for some $ \phi$. Since we assume non-abelian, the mapping $ \phi: K\to Aut(N)\cong \mathbb{Z}/12$ is not the trivial homomorphism. Suppose $ K\equal{}\langle x\rangle$. Then let $ \phi(x)\equal{}4$. This automorphism produces a non-abelian semidirect product and we are done.

[quote="JackSchmidt"]Consider the group of 2 x 2 matrices over the field with 13 elements.  It has a subgroup generated by a=[ 3, 0 ; 0, 1 ] and b=[ 1, 1 ; 0, 1 ].  This subgroup has order 39, 13 Sylow 3-subgroups, and a normal Sylow 13-subgroup.  It has a presentation $ \langle a,b : a^3 \equal{} b^{13} \equal{} 1, ba \equal{} ab^8 \rangle$.[/quote]

Bingo,it is right.And the limitation that $ ba\equal{}ab^j$,j could have more values.
I wonder all nonabel-39-ordered group's presentation.


Let $M$ be the set of solution of the following ecuation $3^{x+3}+9=\sqrt{81^{2x-1}}+3^{4x}$. If $k$ is the number of elements from set $M$ then find $k$.

[hide]
$k=1$.
My solution:
The equation can be simplified to:
\[3^{4x}-3^{x+3}+9^{2x-1}-9=0\]
Let $x=1$, we have: $3^4-3^4+9-9=0$. Hence $x=1$ is a solution of the equation.

Next I shall show that $x=1$ is the only solution.
Use the monotonicity of $f(x)=a^x$. If $a>1$ then $f(x)$ is an [b]increasing function[/b].
If $x>1$, then $3x-3>0$.
Hence $4x - x+3 > 0$.
Which means $4^x - 4^{x+3} > 0$
And since $2x-1>1$, so $9^{2x-1}-9>0$.
Hence the $3^{4x}-3^{x+3}+9^{2x-1}-9 > 0$. Impossible.

The proof of $x<1$ is the same.

Hence we see $x=1$ is the only solution so $k=1$.

P.S. We can set $x \neq 1$ and the proof can become shorter.[/hide]

Well i did it this way: Let $3^x=a>0$

\begin{eqnarray*}
3^{x+3}+9&=&\sqrt{81^{2x-1}}+3^{4x}\\
27\cdot 3^x+9&=&3^{2(2x-1)}+3^{4x}\\
27\cdot 3^x+9&=&3^{4x}\cdot\frac{1}{9}+3^{4x}\\
27a+9&=&a^4\cdot\frac{1}{9}+a^{4}\\
10a^4-243a-81&=&0 \end{eqnarray*}

The solution of the last equation is not so obviously ($a=3$). Can anybody post a method of solving this ?:)

I believe there is formula for the roots of 4 degree equations.

[quote="Slizzel"]$27a+9=a^4\cdot\frac{1}{9}+a^{4}$
$10a^4-234a-81=0$[/quote] you must have mistyped or something it's not 234 thou
$10a^4-243a-81=0$

shobber: the quartic formula is WAY too ugly and i dont recommend it unless that is absolutely the last option to go for

Sorry , fixed. I must have hurried :). I think there is a way to find the roots. Using [i][b]Horner's Method[/b][/i] we obtain the root $a=3$. Then we know that  $10a^4-243a-81$ is divisible by $a-3$ so we just divie it. After that we get: $(a-3)(10a^3+30a^2+90a+27)=0$. But we also know that $a>0$ which means that $(10a^3+30a^2+90a+27)$ is always different by $0$. So the only root is $a=3\Rightarrow x=1$ :)

Whats Horner Method ?  :?  :?

Anyway I make a conclusion after seeing those solution ....

After simplying , we can get

1+3^(x+1) = 3^(4x-4) . 10

If x is rational , then notice that LHS doesnt have factor of 3 , so the only possible rational solution for x is when 3^(4x-4) = 1 which is x=1

If x is irrational , just subsitute 3^x = a  like what Slizzel had done and you will see there is no solution .


linear algebra
matrix
vector
function
linear algebra solved

$A=(a_{ij}) \in M_n(R)$ where $a_{ij}$ is the number of common divisors of $i$ and $j$

Prove that $A$ is invertible

I don't know if this would help much, but I think this comes more or less from the same applications than the Hilbert Matrix. Basically the idea is to look for a combinatorical explanation of the entries of the matrix, and then, somehow it becomes self-evident that the matrix is invertible.

Sorry, but I can't find much about the Hilbert Matrix right now in the web, but if you need a solution, probably find something about it may help in your problem.

[quote="jmerry"]Following djimenez's idea, we show that $a_{ij}=\langle v_i,v_j\rangle$ for some linearly independent vectors $v_i$ in an inner product space. Let the space be the set of functions from $\{1,2,\dots,n\}$ to $\mathbb{R}$ with the standard inner product $\langle a,b\rangle=\sum_k a(k)b(k).$ Let $v_i(k)=1$ if $k|i$ and $v_i(k)=0$ otherwise. This clearly gives the appropriate result (interpreting the question as asking for positive divisors). These vectors are clearly linearly independent; the change of basis matrix from the standard basis is triangular, with each diagonal entry equal to 1.
This shows that $A$ is positive definite, hence invertible.[/quote]

jmerry, brilliant, just a detail that may make easier to understand the problem to the reader. The fact that $a_{ij}=\langle v_i,v_j\rangle$ means that, if you consider $V=[v_1,\dots,v_n]$ (looking the $v_i$'s as colum vectors), then $A=VV^T$. The fact that $V$ is triangular with diagonal entries $1$ makes it invertible (what makes $A$ automatically invertible), but also, $x^TAx=x^TVV^Tx=\langle V^Tx,V^Tx\rangle=\|V^Tx\|^2>0$ if $x\neq 0$, therefore $A$ is possitive definite.

Best,


I think you write that juste for fun  :rotfl: ?


algebra
polynomial
LaTeX
number theory unsolved
number theory

Given a polynomial $P(x)$ with integer coefficients. For each natural number $n$ the number $P(n)$ is greater than $n$. Consider the sequence $x_1=1 , x_2=P(x_1),  ...  ,x_n=P(x_{n-1}),....$ .Given that for each netural number $N$ there exists a member of the sequence which is divisible by $N$. Prove that $P(x)=x+1$ ;)

nice problem.

We first show that $x_2 = 2$, which implies that $P(1)=2$.
Suppose $x_2 = P(1) = \sum _{i=0}^{m} a_i = k$, where $k$ is a positive integer. (Note that $P(x)$ has only integer coefficients and $P(x) \geq x+1$ for all positive integers $x$.) Now $x_3 = P(2) = \sum _{i=0}^m a_i * k^i = k (mod (k-1))$. Similarly, $x_n = 1 (mod (k-1))$ for all positive integers $n$. But if $k>2, k-1$ has a prime divisor, which will divide no term in the sequence ${x_n}$. Contradiction. Hence $P(1)=1$, as desired.
By similar reasoning (start from $x_3$), we get $P(2)=3$, and $P(n)=n+1$ for all natural $n$. QED :)

[color=red]Edited by Megus: al.M.V -  it isn't hard to use LaTeX - you can learn it to make your posts more readable [/color]

It's been posted before at least twice :).

Check http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6448 and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10285 (problem 6).

alright Megus, I will try to learn Latex... :P I'm just super lazy to bother most of the times... ;)


Let $f(x)$ be a continous function for all real values of $x$ such that $f(a) \le f(b)$ whenever $a \le b$.  Prove that, for ever real number $r$, the equation $x+f(x)=r$ has exactly one solution.

Thanks

Both x and f(x) are strictly increasing, meaning x + f(x) is also strictly increasing, meaning that each distinct input will result in a distinct output.

This means that x + f(x) = r must have exactly one solution for each value of r.

Is it common knowledge that this is true because how do we know that no real value will be "skipped".

I meant how can we show that x+f(x) is continuous...oh...I see...thanks very much...I forgot that x is really a constant in this case at any point...

Thanks Treething...


Oh,I am sorry
I have work and I could`nt replay
This is the Link :

[url=http://www.journals.cms.math.ca/CRUX/]www.journals.cms.math.ca/CRUX/[/url]

Another thing
only 96...99 is public and you can download it

Sometimes ago, I can take volume 9 of the year 1991, v10 is unavaillable, but now it's out. What's the matter?

[quote=Pham_Quoc_Sang][quote=Ali_aomt]Crux is a good problem solving journal by Canadian Mathematical Society
and some issue of this journal is public and you can use free[/quote]

I have some articles I want to recommend to Crux. What should I do?[/quote]

One can submit their articles on the website freely although read [url=https://cms.math.ca/publications/crux/information-for-contributors/]this[/url] first.


Can anyone explain balls and urns to me? i believe that's what it's called. I see AoPSers use very often.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1904611533&t=145047[/url]

This should help  :wink: .


[quote="alex2008"]Let be $ a,b,c > 0$ . Prove that :

$ \frac {2a^3}{a^6 \plus{} bc} \plus{} \frac {2b^3}{b^6 \plus{} ca} \plus{} \frac {2c^3}{c^6 \plus{} ab}\le \frac {a}{bc} \plus{} \frac {b}{ca} \plus{} \frac {c}{ab}$[/quote]
$ a^6\plus{}bc \ge 2\sqrt{a^6bc}\equal{}2a^3\sqrt{bc}$
So we have to prove that:
$ \sum\sqrt{\frac{1}{bc}} \le \frac{a^2\plus{}b^2\plus{}c^2}{abc}$
Setting $ \sqrt{a}\equal{}x,\sqrt{b}\equal{}y,\sqrt{c}\equal{}z$,
And ginally the inequality is:
$ a^4\plus{}b^4\plus{}c^4 \ge a^2bc\plus{}ab^2c\plus{}abc^2$
Which is true by Muirhead for the triples
$ (4,0,0) \succ (2,1,1)$


given angle xOy , one arbitrary circle with center I inside angle xOy . circle (I) has constant radius R . M on the circle , MH drawn perpendicular to Ox at H and MK drawn perpendicular to Oy at K ,
 find M where MH+MK min

Isn't the circle with centre I touching the arms of $ \angle XOY$??
If so, let $ M$ be the point such that $ MH\plus{}MK$ is minimum. That is,
$ (\sin \angle HOM\plus{}\sin \angle MOK)OM$ is minimum.
$ \Leftrightarrow (2\sin \frac {XOY}{2} \cos \frac {(\angle HOM \minus{} \angle MOK)}{2})OM$ is minimum. But $ 2\sin \frac {XOY}{2}$ is constant.So, we have to consider when $ \cos \frac {(\angle HOM \minus{} \angle MOK)}{2}OM$ is minimum.
Let $ OM_1M_2$ be the angle bisector of $ \angle XOY$ with $ M_1$ and $ M_2$ on the circle such that $ OM_2>OM_1$.
$ M$ lies in the $ arc AM_1B$ where $ A,B$ are the points of contact of the arms of $ \angle XOY$.
$ \cos \frac {(\angle HOM \minus{} \angle MOK)}{2}\equal{}\cos \frac {2\angle MOM_1}{2}\equal{}\cos \angle MOM_1$
But, $ (\cos \angle MOM_1)OM > OM_1\cos 0$ since $ \angle OM_1M > 90$.
But this is contradictory Since $ (\cos \angle MOM_1)OM$ is minimum.
It follows that point $ M$ coincides with point $ M_1$.

the circle I does not touch the arms of xOy
here you can download the figure of this problem
http://www.mediafire.com/?d1bjdnj1j4g