The (n-1)EV transforms this one (even your conjecture) to a two variable inequality!
 :)


geometry
geometric transformation
geometry proposed

Let m be a fixed line and A, B are points lying on different halfplane divided by m such that AB is not perpendicular to m. C and D moves in m such that the angle CBD = CAD. Call H the orthocenter of CDB. (S) is a circle passing A and H and intersects HC, HD respectively at M,N. Through H, we draw a line paralel to m and cut (S) at I. Let K be the intersection of AI and MN.
     a) Prove that K lies on m.
     b) Now A and B are fixed. Let J, L be the orthocenter of KCM and KDN. Prove that J, L lies on a fixed line.

Wow! I found this on the $10$'th page! :D I guess people didn't want to read it. It is a bit long, isn't it? :)

It's very rich in properties, so I'll state many of them without proof, because it's all just a big angle chase :).

[b]a)[/b] $ACHD$ is a cyclic quadrilateral. $AMN$ is obtained from $ACD$ by a spiral similarity (I'm not sure we'll use this :)). Let $K'=AI\cap CD,\ K''=MN\cap CD$. We have $\angle ADK'=\angle AHM,\angle AKD=\angle AMH$, so $\angle HAD=\angle MAK'$. We have thus found $AHD,AMK'$ to be similar $(*)$.

We can now use $(\#)$ to find $\angle MK''C=\angle MAC$, so $MK''AC$ is cyclic. From here we find $\angle AK''N=\angle ACH,\ \angle ANK''=\angle AHC$, so, again, $\angle MAK''=\angle HAD$, getting $AHD,AMK''$ similar $(**)$.

$(*),(**)\Rightarrow K'=K''=K$.

[b]b)[/b] If we assume for now that $C,D$ are also fixed, the locus of $J$ (let's just talk about $J$) is easily seen to be a line passing through $B$. We need now to show that these lines when $C,D$ move on $m$ are, in fact, one and the same. Its not hard to prove that the symmetric of $A$ in $m$ also belongs to this locus no matter how $C,D$ move, so when $A,B$ are fixed, the locus of $J$ is the line passing through $B$ and the symmetric of $A$ wrt $m$ (which are different, according to the hypothesis).

[quote="grobber"]Wow! I found this on the $10$'th page![/quote]
What do you mean?

He probably means the 10'th page of the "Geometry Proposed & Own Problems" topics list.

  Darij

That is indeed what I mean. I was just surprised that we ignored it for so long :).

There are NUMEROUS problems, which we ignore for a long time. :D


$ a,b,c>0 $ show that this inequality holds:
\[ \frac{2a}{a^2+bc} +\frac{2b}{b^2+ac} + \frac{2c}{c^2+bc} \leq \frac{a}{bc}+\frac{b}{ac} +\frac{c}{ba} \]

$2a^2 + b^2 + c^2 \ge 4a\sqrt{bc}$; add up the similar inequalities to get $a^2 + b^2 + c^2 \ge a\sqrt{bc} + b\sqrt{ac} + c\sqrt{ab}$. Thus $\frac{a}{bc} + \frac{b}{ac} + \frac{c}{ab} = \frac{a^2 + b^2 + c^2}{abc} \ge \frac{a\sqrt{bc} + b\sqrt{ac} + c\sqrt{ab}}{abc} = \frac{a}{a\sqrt{bc}} + \frac{b}{b\sqrt{ac}} + \frac{c}{c\sqrt{ab}} \ge \frac{2a}{a^2 + bc} + \frac{2b}{b^2 + ac} + \frac{2c}{c^2 + ab}$

WLOG, assume abc=1. Observe that \[\frac{2a}{a^2+bc}=\frac{2a^2}{a^3+1}\le\frac{1}{2}(a+1)\]The result follows immediately from $a^2+b^2+c^2\ge a+b+c$ and $a^2+b^2+c^2\ge 3$.

[quote="jensen"]Thank you, both of you but mecrazywong can you explain you solution ? :lol:[/quote]
What do you want him to explain? :?


inequalities
quadratics
inequalities proposed

Prove that for all reals a,b,c we have $ a^4+b^4+c^4+a^3b+b^3c+c^3a\geq 2ab^3+2bc^3+2ca^3 $. A very nice one.

assuming that a is the smallest and writing $b=a+x,c=a+y$ doing some algebra, the inequality is immediately clear(the hardest part is showing that $x^4+x^3-2x+1\geq 0$ for all positive x(actually i'm not completely sure about it, but setting $c=0$ in the original equation that one is implied and inequalities in one variable aren't that hard)).

Peter

I have also a proof
If a.b>0 we can suppose c>=0 also because else we get a rather easy inequality
|Now use ab^3+bc^3+ca^3-a^3b-b^3c-c^3a|=|(a-b)(b-c)(c-a)|(*)
So we must prove sum{a^4+3b^4-4ab^3}/4>=*
Or sum{(a-b)^2(a^2+2ab+3b^2}/4>=(*)
Now if we increase each of a,b,c by x we can see the RHS increases less than LHS.
So we can suppose one of a,b,c is zero and then it's easy

i just had problems whether my solution actually works for all real a(i thought they were positive), but actually i get a quadratic in a with discriminant $\Delta=-3(x^3-3x^2y+y^3)^2\leq 0$, so the solution works.

Peter

Following Peter Scholze I have also solved it but, after realizing a

deep similarity with INEQ3, I am now going to think of this problem :

" Which inequality between INEQ3 and this one is stronger ? "

Has anyone an answer?

Thank you very much.

These inequalities are
$\#3(a^4+b^4+c^4)+3(a^3b+b^3c+c^3a)\geq 6(ab^3+bc^3+ca^3)$
and
$2(a^4+b^4+c^4)+4(a^2b^2+b^2c^2+c^2a^2)\geq 6(ab^3+bc^3+ca^3)$.

For a, b, c positive, the second is stronger than the first, because
$LHS1\geq LHS2$.

For a, b, c negative we cannot say one is stronger than the other.


trigonometry
inequalities unsolved
inequalities

$ 0<x,y,z<\pi/2$  and $ x\plus{}y\plus{}z\equal{}\pi$ so find the minimum T
$ T\equal{}1*\tan\ x\plus{}2*\tan\ y\plus{}3*\tan\ z$

Because $ tan x\plus{}tan y\plus{}tan z \equal{} (tan x)(tany)(tan z)$
$ \therefore tan x\plus{}tan y\plus{}tan z  \geq 3\sqrt{3}$
$ by am\minus{}gm;T \geq 3\sqrt[3]{6(tanx)(tany)(tanz)} \geq 3\sqrt[3]{18\sqrt{3}}$

is there any equility.please post it to me
sorry for my bad english


Ok, I am taking an independant study course for Algebra II and I haven't had a problem until now, I suppose I have never had this in even the simpliest form before.

The book doesn't give sufficient information, nor my independent study course intruction. I cannot understand any of it. It is the first lesson on factoring and I can understand everything up until the problems. The problems I have are:

2x[squared] + 6y + 8b

3a[squared] + 6a + 9y

I don't just want my problems answered, I would like to know how to do this. I guess a better detailed lesson or description of factoring.

Thanks!

First you find any common factors in the expression. For example, $2$ is a factor of every term in the first problem. So it becomes:
$2({x^2}+3y + 4b)$
In the second one, all three terms have $3$ as a factor, and two have $a$. This becomes either:
${3({a^2} + 2a + 3y})$
or
${3(a(a+2) + 3y})$ (This one looks kind of weird)

Now, say you wanted to factor something like:
${x^2} -2x - 1$
You can split this up into two terms. You can either use intuition or the quadratic formula. This is an easy factorization, so we use intuition. We make it:
$(x+1)(x-1)$
If you multiply out the terms, you see that it is the same as the previous expression.

But let's say you have an ugly polynomial, like:
${x^2} + x + 1$
You can try to guess a factorization, but you will soon find that it's pretty hard to do. We then turn to the quadratic formula. You take a polynomial in the form of:
$a{x^2} + bx + c$ and you plug the coefficients into this formula:

$r = \frac{-b \pm \sqrt{{b^2}-4ac}}{2a}$
plugging in for $a$, $b$ and $c$ (they are all $1$ in our example) we get:
$r = \frac{-1 \pm \sqrt{{1^2}-4(1)(1)}}{2(1)}$
simplifying, we have:
$r = \frac{-1 \pm \sqrt{5}}{2}$
So our factorization of ${x^2} + x + 1$ becomes:
$(x -\frac{-1 + \sqrt{5}}{2})(x - \frac{-1 \pm \sqrt{5}}{2})$
As our roots are in the form
$(x - {r_1})(x - {r_2})$

Hope this helps  ;)

[quote="LordoftheMorons"]Now, say you wanted to factor something like:
${x^2} -2x - 1$
You can split this up into two terms. You can either use intuition or the quadratic formula. This is an easy factorization, so we use intuition. We make it:
$(x+1)(x-1)$
If you multiply out the terms, you see that it is the same as the previous expression.[/quote]
That's wrong... ($(x+1)(x-1)=x^2-1\ne x^2-2x-1$) :|

Basically, if it is in the form $x^2+ax+b$, you might be able to factor it into $(x-r_1)(x-r_2)$.
Now, if you expand $(x-r_1)(x-r_2)$, you get $x^2-(r_1+r_2)+(r_1)(r_2)$.
That implies that you should look for a pair of numbers such that the sum is $-a$ and the product is $b$.
For instance, when you try to factor $x^2+5x+4$, think about two numbers whose sum is $-5$ and product is $4$.
A quick guess and check gives $-1$ and $-4$.
So $x^2+5x+4=(x-(-1))(x-(-4))=(x+1)(x+4)$.

Well, here are some practice problems:
1. $x^2+6x+5$
2. $x^2+8x+16$
3. $x^2+3x-10$

[quote="frt"][That implies that you should look for a pair of numbers such that the sum is $-a$ and the product is $b$.
For instance, when you try to factor $x^2+5x+4$, think about two numbers whose sum is $-5$ and product is $4$.
[/quote]

I've never thought of it that way. When I factor a quadratic ($x^2+bx+c$), I just think about two numbers that add up to b and multiply to be c. Let those two numbers be $p$ and $q$. The factors are then $(x+p)(x+q)$.

and clearly when we are not having our $x^2$ coefficient as unity, the same should apply:

Example $2x^2 +3x +1$

We need two numbers which multiply to give $2$, and add to give $3$, which are $2,1$

So it equals $2x^2 +2x +x +1$ by splitting the $x$s
Factorising yields $2x(x+1) + (x+1)$

So factorised it is $(2x+1)(x+1)$
Or of course just use recognition  :roll:


What is the probability that a card chosen randomly from a standard deck of 52 cards is not a heart or an Ace?

[hide]So there are 13 hearts and 4 aces. But one of the aces is a heart so 13+4-1=16. So it would be $\frac{9}{13}$.[/hide]


function
integration
real analysis
real analysis unsolved

For a Riemann integrable positive function $f$ defined on $R$ and with values in $R$ such that $\int_{-\infty}^{\infty}{f(t)dt}$ converges, define $M(x)=\sup_{r>0}{\frac{1}{2r}{\int_{x-r}^{x+r}{f(t)dt}}}$. Prove that
   a) if $f$ is continuous, then $M$ is too.
   b) The set of all real numbers $x$ such that $M(x)>1$ can be covered with a family of intervals the sum of lengths of which is at most $5\int_{-\infty}^{\infty}{f(t)dt}$. Thus, the set of $x$ for which $M(x)=\infty$ is of measure 0.

Harazi, could you please tell what was the year when this problem was proposed in the ENS exam.

I found it very interesting.

I'll try to cover the second part. My idea for the first one seems a bit unclear.

This average functiion is called Hardy-Littlewood Maximal function. The second point of this problem can be easily done using a version of Hardy-Littlewood Maximal theorem:
$|\{ Mf \geq \epsilon \}| \leq \frac{2}{\epsilon} ||f||_1$ (*) . Here is the reference http://planetmath.org/encyclopedia/HardyLittlewoodMaximalTheorem.html

You can also find a proof here http://www.math.ntnu.no/~eugenia/TMA4225/lecturenotes/lecture15.pdf.

*. From (*) if we tend $\epsilon \to \infty$ we get the result that $| \{ Mf = \infty \} | = 0$.

**.To get the result $\{ x : M(x)>1 \}$ can be covered with intervals of summary length $\leq 5||f||_1$, i think that in order to use (*) we can notice that $M(x)$ is measurable and even more $\{ x: M(x) > a \}$ is open.

The measurability is clear: $M(x)$ being the supremum of average functions(which are in our case continious) is lower semicontinious and the set $\{ x: M(x) > a \}$ is measurable.

To get a stronger version that $\{ x: M(x) > a \}$ is open, we can define $A_r(x) = \frac{1}{2r} \int_{x-r}^{x+r} f(t)dt$, and $M(x)=\sup_{r>0} A_r(x)$, so for any fixed $x$ and $n$ there exists $r_n$ such that $A_{r_n}(x)>M(x)-\frac{1}{n}$. Thus

$\{ x: M(x)>a\}= \bigcup_{n} \bigcap_{r \in \mathbb{Q}} \{ x: A_r(x)> a-\frac{1}{n} \}$ which is open as well as each set $\{ x: A_r(x)> a-\frac{1}{n} \}$ is open because $A_r(x)$ is continuous.

So, the set $\{ x : M(x)>1\}$ is open and can be represented as a union of intervals which summary leghts is due to (*) $\leq 5||f||_1$.

Sorry if it is unclear.

Suppose we replace $\mathbb{R}$ by $\mathbb{R}^n$ and the average over the interval $(x-r,x+r)$ by the average over the ball of radius $r$ centered at $x.$ Then replace the $5$ in part (b) by $5^n.$

The result is still true, basically in the way that Eugene outlines.


logarithms
real analysis
real analysis theorems

It's OK ! Yes , for z=0 is the so-called BBP-formula which enable us to find the n-$ {}^{th}$  hexa-decimal of $ \pi $ , without calculation the first $  n-1 .$ Regards,Alex


calculus
derivative
algebra
binomial theorem

Factorise $(a+b)^7-a^7-b^7$ without expanding $(a+b)^7$.

Method is the thing that counts,answer is fine :)

Yeah, how do you go about doing this?  It's obviously the exapansion of $(a+b)^7$ without the $a^7$ and $b^7$ that are on either end, but how do you find what that is without just using the binomial theorem?

[quote="Pocket Sand"]Can you post a solution? I'm confused.[/quote]

I'm in a hurry,all i can say is see symetry or something.the given thing is a symmetrical expression,so the factors have to be something like $a+b(Aa^2+ab+Bb^2)$(put a=0,b=0....) which are both symmetrical in their respective degrees.....will xplain later :)

Gyan, that's a very interesting method, I've never thought of factoring that way before.

Yeah. Plugging in values is a very good way to begin a factorization problem.

Also, you can tell immdiately by the $-a^7-b^7$ that $ab$ is a factor.

[quote="Gyan"]FWIW:

Quick way to see:
[hide]$a=0$ makes the expression $0$ so ====>  $a$ is a factor
$b=0$ makes the expression $0$ so ===> $b$ is a factor
$a=-b$  makes the expression $0$ so ===> $(a+b)$  is a factor
$a=b\omega$ or $a=b\omega^2$ makes the expression $0$  so ===>  $(a-b\omega)(a-b\omega^2) = (a^2+ab+b^2)$ is a factor
[hide]Here $\omega$ is cuberoot of unity and notice $1+\omega+\omega^2 = 0$
You notice that the whole expression is $b^7((1+\omega)^7 - 1 - \omega^7)$   comes out to be zero because $1+\omega = -\omega^2$
and you get  $-\omega^{14} - \omega^7 -1 = -\omega^2-\omega -1 = 0$[/hide]
You can see that its double root.. too so  last expression should be  a saure term.

[hide](Easiest to see this requires calculus , the derivative would be zero too when $a=b\omega$ etc..[/hide]

[hide](If you don't see its a double roots right away, you can also find the remaing term as it will only of a second degree ... etc..) [/hide]

So you use $(a+b)^7-a^7-b^7 = k a b (a+b) (a^2+ab+b^2)^2$
Since both sides are now of equal degrees , you can find the value of $k$  by using, say $a=b=1$  which will give $k=7$ [/hide]

Hope this helps.[/quote]

That's a better explaination than i can give. :D   It is a handy technique.

Though gyan,wouldn't $a^2$ and $b^2$ have some coefficients so we write it as $Aa^2+ab+Bb^2$ and eventually A and B come out to be 1?


trigonometry
inequalities
inequalities unsolved

There is something wrong with your inequality. a=b=c=d gives a contradiction.
Please correct your fault because I have a rather nice idea what to do with the left side and I would like to post it here.

Misha

[quote="manlio"]For $a,b,c,d$ positive reals prove the ineq

$\sqrt{(a^2-ab+b^2)(c^2-cd+d^2)}+\sqrt{(a^2+ad+d^2)(c^2+cb+b^2)} \geq(a+c)(b+d) $[/quote]

That's t$\infty$ c$\infty$l!!

Let ABP be a triangle with AP = a, BP = b and < APB = 60. Let C be a point on the extension of the side AP beyound the point P such that CP = c, and let D be a point on the extension of the side BP beyound the point P such that DP = d. Then, < BPC = 120, < CPD = 60 and < DPA = 120.

In the triangle APB, the Cosine Law yields

$AB=\sqrt{a^{2}+b^{2}-2ab\cos \measuredangle APB}=\sqrt{a^{2}+b^{2}-2ab\cos 60^{\circ }}$
$=\sqrt{a^{2}+b^{2}-2ab\cdot \frac{1}{2}}=\sqrt{a^{2}-ab+b^{2}}$.

In the triangle BPC, the Cosine Law yields

$BC=\sqrt{b^{2}+c^{2}-2bc\cos \measuredangle BPC}=\sqrt{b^{2}+c^{2}-2bc\cos 120^{\circ }}$
$=\sqrt{b^{2}+c^{2}-2bc\cdot \left( -\frac{1}{2}\right) }=\sqrt{b^{2}+bc+c^{2}}$.

Similarly,

$CD=\sqrt{c^{2}-cd+d^{2}}$

and

$DA=\sqrt{d^{2}+da+a^{2}}$.

Also, AC = AP + CP = a + c and BD = BP + DP = b + d.

By the Ptolemy inequality, applied to the quadrilateral ABCD, we have

$AB\cdot CD+BC\cdot DA\geq AC\cdot BD$;

thus,

$\sqrt{a^{2}-ab+b^{2}}\cdot \sqrt{c^{2}-cd+d^{2}}+\sqrt{b^{2}+bc+c^{2}}\cdot \sqrt{d^{2}+da+a^{2}}\geq \left( a+c\right) \cdot \left( b+d\right) $,

completing the proof.

Sorry, Misha, for posting this - I guess your idea was the same...

  Darij

It's exactly what I wanted to do! I love inequalities with geometric interpretations.

Misha


calculus
integration
abstract algebra
function
Ring Theory
algebra
domain

Hi!

I really don't quite see, why the statement below is correct, or is it just a "typo" ?.
Let X be an integral noetherian scheme. Also, suppose X is Dedekind, that is, an irreducible, locally noetherian, normal scheme of dimension 1 or 0.

[b]Then for every open subset of X, its corresponding ring of functions is a Dedekind ring.[/b]

This is clear for affine open subsets, but what about more general open sets?. Any thoughts?. Anyone out there who could prove this or give a counterexample?.

Note: The author of the book allows Dedekind rings to be 1 or 0-dimensional.

If X is 0-dimensional, then X is a singleton and the claim is trivial. Now let X be 1-dimensional. Note that for every open subset $ U$ it holds $ \Gamma(U,\mathcal O_X)\equal{}\cap_{x\in U}\mathcal O_{X,x}$ where the intersection takes place in the function field of $ X$, the stalk of $ \mathcal O_X$ at the generic point of $ X$. Then use the fact that the intersection of normal subrings of a field $ K$ is normal again (w.r.t. $ K$).

[quote="Third Edition"]If X is 0-dimensional, then X is a singleton and the claim is trivial. Now let X be 1-dimensional. Note that for every open subset $ U$ it holds $ \Gamma(U,\mathcal O_X) \equal{} \cap_{x\in U}\mathcal O_{X,x}$ where the intersection takes place in the function field of $ X$, the stalk of $ \mathcal O_X$ at the generic point of $ X$. Then use the fact that the intersection of normal subrings of a field $ K$ is normal again (w.r.t. $ K$).[/quote]

Well, yes, i understand from the argument in the textbook that $ \Gamma(U,\mathcal O_X)$ is an integral domain, and integrally closed and of dimension 0 or 1. [b]Why is it a Noetherian ring?.[/b]

Of course, $ U$ has a finite open affine cover by spectra of Noetherian rings, but then what?.

I must confess i haven't thought too much about this, and it may well be quite trivial  :oops: .

Would you enlighten me?.  :)

Thanks anyway!.

Are you working over a base field $ k$? If so then take a projective normal model $ \bar{X}$ of (the function field) 0f $ X$. Note that every (non-void) subset $ U$ of $ X$ is the complement of finitely many points of $ \bar{X}$ which implies that $ U$ is affine.


inequalities
trigonometry
symmetry
triangle inequality
trig identities
Law of Cosines
algebra unsolved

let $a$ and $b$ be two positive reel number
solve in the set of reel number the equoition
$\sqrt{x^{2}+a^{2}-ax\sqrt{3}}+\sqrt{y^{2}+b^{2}-by\sqrt{3}}+\sqrt{x^{2}+y^{2}-xy\sqrt{3}}=\sqrt{a^{2}+b^{2}}$
$x$ et $y$ sont les nombres inconu
good luck :roll:

Hint:
Use the inequality to estimate left hand:
$\sqrt{a^{2}+b^{2}}+\sqrt{c^{2}+d^{2}}+\sqrt{e^{2}+f^{2}}\geq\sqrt{(a+c+e)^{2}+(b+d+f)^{2}}$

Kunny's hint gives it away.
Consider a right triangle with leg lengths $b$ and $a$ with $a$ in the horizontal direction and $b$ vertical. Construct the lines which make an angle of $30^{\circ}$ with the $x$ and $y$ (or $a$ and $b$) axes, respectively. (In a clockwise order going around the origin, one encounters first line $a$ then two lines then line $b$)
Place points $X$, $Y$, at a distance $x$,$y$ from the origin on the first and second lines (going counterclockwise) respectively, all in the first quadrant. We claim that the given problem is equivalent to being given in the diagram that $XA+XY+YB=AB$; this is clear by considering the law of cosines condition since $XA^{2}=a^{2}+x^{2}-2ax\cos 30^{\circ}$. Thus, by the triangle inequality, the points $X,Y$ lie on $AB$.

THe rest is basic stuff... we realize setup similar right triangles and painlessly find that $a=m(a+b)$ for some scale factor $m$ then $x=2mb$ so $x=\frac{2ab}{a+b}$ and by symmetry, the same goes for $y$, which is interesting since these two both are the harmonic mean of $a$ and $b$.

A really nice but basic problem (well at least after kunny's tip)

[quote="me@home"]so $x=\frac{2ab}{a+b}$ and by symmetry, the same goes for $y$[/quote]
The correct solution is $(x,y)=(\frac{2ab}{a+\sqrt{3}b},\frac{2ab}{\sqrt{3}a+b})$.
If you look at the figure you see $a>b\Rightarrow x>y$.

Take $L$ the perpendicular point of $X$ on $OA$ ($O$ be the origin with $\overline{OA}=a,\overline{OB}=b,\overline{OX}=x,\overline{OY}=y$),
then $BOXLA$ is an intercept theorem figure with ${\overline{XL}\over\overline{BO}}={\overline{LA}\over\overline{OA}}$.
$\angle XOL=30^\circ$ gives $\overline{XL}={x\over 2}$ and $\overline{OL}={\sqrt{3}x\over 2}$.
So ${{x\over 2}\over b}={a-{\sqrt{3}x\over 2}\over a}$ and $x=\frac{2ab}{a+\sqrt{3}b}$, and symmetric for $y$.


function
algebra
domain
algebra unsolved

Find $ f: R\minus{}>R\plus{}$ such that:
$ f(x)^2\equal{}1\plus{}xf(x\plus{}1)$

Hmm...if I got it correctly,the problems you proposed is:
find all $ f: \mathbb{R}\rightarrow\mathbb{R}^{ \plus{} }$,such that...
substitution $ x \equal{} 0$ implies that $ f(0) \equal{} 1$, and $ x \equal{} \minus{} 1$ implies that $ f( \minus{} 1)^2 \equal{} 0$,therefore $ f( \minus{} 1) \equal{} 0$,contradiction.
[i]Added after reading Allnames reply below:[/i]
Well...then show us the full version.
I suspect you misunderstood what I meant.In fact,I gave a correct solution for the given problem,I had emphasized the condition before I posted a solution.I guess misconceptions appeared due to incompleteness of the statement,furthermore,I am starting to doubt about the domain of $ f$,whereas image seems to be correct.I would rather think of a case,where domain is all negative integers.

It is not enough to solve it Erken
Although it is from a book ,it is not full version of this problem,I have seen it in a notebook but with other condition (I remember $ f$ is continious function)
At all I think we cant kill it without any condition :wink:

It is enough to solve,don't have any other condition all names!


This is related to one of the problems on the forum, which has been unsolved for a long time, but forget about that. Here's what I was wondering:

Given $r$ irrational, is it true that $\lim\inf_{n\rightarrow \infty}2^n\{nr\}=\infty$?

In other words: given $k>0$, is it true that there are only finitely many $n$ s.t. $2^n\{nr\}<k$?

As stated, I think the answer has to be no, for $r$ with sufficiently good rational approximations.  Define a sequence $x_m$ by $x_0 = 1$, $x_{m+1} = 1+x_m + 2^{x_m}$ and let $r = \sum_m 2^{-x_m}$.  Then for $n = 2^{x_m}$ we have $\{nr\} < 2.2^{-x_{m+1}}$.  But $x_{m+1} = n + x_m+1$, so $2^n \{nr\} < 2^{-x_m}$.  Hence there is a subsequence of $\{nr\}$ converging to zero.


logarithms
inequalities
limit
algebra unsolved
algebra

Find $ x\in R$ of this equation $ \boxed{\log_{x}{(x \plus{} 1)} \equal{} \log_{2008}{\left( \frac {3}{2} \right )}}$

hello, let $ f(x)=\frac{\ln(x+1)}{\ln(x)}-\frac{\ln(\frac{3}{2})}{\ln(2008)}$,it is clear that for $ 0<x<1$ the inequality $ f(x)<0$ holds.For $ x>1$ we have $ f^'(x)=\frac{1}{(x+1)\ln(x)}-\frac{\ln(x+1)}{x\ln(x)^2}<0 \Leftrightarrow x\ln(x)<(x+1)\ln(x+1)$ because $ h(x)=x\ln(x)$ ist for $ x>1$ strictly monotonously increasing, further we have $ \lim_{x\to 1+0}f(x)=+\infty$ and $ \lim_{x\to \infty}f(x)=1$, so our equation has no real solution.
Sonnhard.

[color=darkblue]Yes! Thank your post.  :lol:

This problem can be done this way:

+) If $ 0 < x < 1 \Longrightarrow \log_{x}{(x \plus{} 1)} < \log_{x}{1} \equal{} 0$, but $ \log_{2008}{\frac {3}{2}} > \log_{2008}{1} \equal{} 0 \Longrightarrow$ equation have no root.

+) If $ x > 1 \Longrightarrow x \plus{} 1 > x$ $ \Longrightarrow \log_{x}{(x \plus{} 1)} > \log_{x}{x} \equal{} 1$ $ \equal{} \log_{2008}{2008} > \log_{2008}{\frac {3}{2}} \Longrightarrow$. equation have no root.

Hence equation have no root.

Thanhnam2902!  :P [/color]


Bob and Ed are standing inside a simple figure at $(-3,0)$ and $(3,0)$ respectively. Whenever Bob throws a ball at the walls of the figure, it bounces back and hits Ed (assume that gravity, air resistence, etc. are not in play), traveling a distance of 10 units. Find the area enclosed by this figure.

Umm.. isn't there infinite solutions? As long as he keeps throwing it at (0,x) you can make as big as a rectangle you want.

he's assuming that you know the rule of the foci for an ellipse. now he's saying find the area of the elipse with foci at those points, wich is trivial...

Darn you.

This is an interesting property of ellipses, which is that bouncing a "ball" off the sides from one focus will just hit the other focus, and the distance traveled will be constant. However I don't think I gave enough information in the original problem. edited now.

[hide]The walls are in the shape of an elipse with the foci at $(-3,0)$ and $(3,0)$. The center is at $(0,0)$
Since the ball travels $10$ units, the length of the major axis is $10$. In an elipse, the square of major axis equals the sum of the squares of the minor axis and the distance between the foci. Using that fact, the length of the minor axis is $\sqrt{10^{2}-6^{2}}=8$.
Therefore, the area of the elipse is $\pi\cdot\frac{10}{2}\cdot\frac{8}{2}=\boxed{20\pi}$[/hide]

very intresting problem.... did you make it?

It's a pretty classic problem. I first saw it on the HMMT.


inequalities
rotation
triangle inequality
superior algebra
superior algebra unsolved

Let x,y,z,t belongs to R^n where d(x,y)=||x-y||.

Show that(Ptolemy's inequality):
d(x,y)d(z,t)<=d(x,z)d(y,t)+d(x,t)d(y,z)Thanks anyone in advance.

Not the best possible topic match:

1. For $ n\equal{}2$, you have the classical inequality.
2. For $ n\equal{}3$, you have a tetrahedron. Take one side. It bounds 2 triangles. Rotate one of them until it lies in the same plane as the other one and such that the common side is a diagonal in the newly formed quadrilateral. Draw the other diagonal, consider the intersection point and join it to the other vertex that is not already joined to. Now apply the inequality in case $ n\equal{}2$ and the triangle inequality in some triangle with a vertex in the intersection of the diagonals.
3. 4 points span a 3-dimensional space, so case $ n>3$ reduces to $ n\equal{}3$.


Prove or disprove:
$ 1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}...\plus{}\frac{1}{n}<4$

Depends on n doesnt it. If n tends to inf , sum is also inf.

Can you tell me for which $ n$ the expression is $ \ge 4$?

We can use $ \log n \approx 4\minus{} \gamma$, where $ \gamma$ is the [url=http://en.wikipedia.org/wiki/Euler-Mascheroni_constant]Euler-Mascheroni constant[/url]. Then we get $ n \approx e^{4 \minus{} \gamma} \approx 30.65 \ldots$

$ n \equal{} 30 \implies \sum \equal{} \frac {9304682830147}{2329089562800} < 4$

$ \boxed{n \equal{} 31} \implies \sum \equal{} \frac {290774257297357}{72201776446800} > 4$


B is a point on the circle (S)S. A is a point (distinct from B) on the tangent to (S) at B. C is a point not on (S) such that the line segment AC meets the circle (S) at two distinct points. (S') is a circle which touches AC at C and (S) at D, where B and D are on opposite sides of the line AC. Show that the circumcenter of (BCD) lies on the circumcircle (ABC).

Denote s,s' are center of circle S and S'
Let SB S'C meet at G, then G A C B are on one circle,
∠COB=∠CS'O+∠OSB,∠S'GS+∠CS'O+∠OSB=∠CAB+∠CS'O+∠OSB=∠CAB+∠COB=180
So OCAB lie on one circle,where O is circumcenter of  triangle BCD


[quote="SM4RT"]I'm sure everyone knows of the dollar question which is

[u]How many ways can you make a dollar with change?[/u]

So the way to solve it using generating functions, and finding the coefficient of $x^{100}$.

[b]How do you find the coefficient of $x^{100}$ without multiplying it all out?[/b]

This link [url]http://www.cut-the-knot.org/ctk/GeneratingFunctions.shtml#Graham[/url] shows it but it does not really explain it (or I'm missing something).

Please do tell.[/quote]

umm unless im not mistaken the coefficiant of $x^{100}$ is $1$? unless im missing something which im sure i am  :D

He means the coefficient of $x^{100}$ in the generating function of

$P(x) = \frac{1}{(1-x)(1-x^{5})(1-x^{10})(1-x^{25})(1-x^{50})}$

Yes but how would you get it, like they did

$\binom{6}{4}+45\binom{5}{4}+52\binom{4}{4}+\binom{3}{4}=15+225+52+0=\boxed{292}$

How do they derive that?

Yes, I'm also curios about where the 4's in the binomial coefficients come from

I've wondered this too.  If someone knows, it could help a lot in future problems.

They give a reference to pages 344-346 of Concrete Mathematics. I had a quick look through, it looks like they set up a recursive equation which turns out to have a quadratic solution. C[z] = in-terms-of-z. Then they used binomial expansion to expand the $\sqrt{..z..}$ part and it goes on from there.

So yeah, do have a look through the book, it's a very good one :)

Sorry to bring up an old topic, but I'm curious about this.  Could you perhaps elaborate some on what he did?

You could multiply the denominator out and then use the extended Binomial Theorem.


It was n problem of the week :roll:

BTW where did you get these states?

I got them from my coach who got them each year at competition.

Do you think you could scan me a copy of a couple? XD

[quote="biffanddoc"]If you give me them, ill give you 5 years of chapters, 7 years of states and 9 years of nationals :hopeful:[/quote]

no need to give me any because I am not gonna compete next year.  My coach has all the competitions from 1995 (i think) to 2005.  I'll get them from her and scan them.  I am just gonna do states right now.  I'll just post them here when i get them into my computer.

[quote="biffanddoc"]Klebian, you should probably talk to mr. boyd about getting other states. Or request them on the coaches forum of mathcounts.org[/quote]
I have no real way of contacting Mr. Boyd other than through AoPS, which probably wouldn't work, or by asking a student of his. As of now I have asked a student, but he's never on on AIM. XD

Please tell me which years' competitions you want.  I'll do them them weekend. :)

I'll PM you 01, 02 nationals.
Can you get me 1999 states Biff?
And the rest of the states/nats would apply to me also.   :)

Sure. I FOUND IT LAST NIGHT :wow: :wow:

I had taken it to NY the week before and it was in my suitcase (and I made sure that I didn't take along any compeition sets :roll:)

I have 1989 -1991 Nationals and States, but sadly, I have no scanner so I can't really bargain for tests :'(

Sorry for the double post, but can I get a copy of all those State Tests as well?

oooh more mathcounts tests? I have only done back to 2001 on the state level. my coach has more but she's out of town and state for me is on saturday, so i'd love to have a copy!!! and national level tests? I've never had a copy! if that's okay with you guys thanks.
sorry but I probably cannot contribute any unique competitions myself.

can I get a copy of those tests too? sorry, i can't contribute any competitions either.

[quote="piano1321"]Wow! i didnt know that so many people wanted these tests.  sorry to disappoint you all, but my scanner just broke, so i can't really post them here. sorry again. :([/quote]

Must remember to buy you a scanner then.
Oh please please let me make nats. ;)

Can people send me everything they have pre 1999?
My email is mcpreet@gmail.com.
If anyone wants I can give them a gmail invite.

I would post everything I have here, but it's against AoPS policy.  Last time I did it, it got deleted immediately.  So here's what I'm going to do:  Make a website, and post everything on it.  Besides MathCounts,  I also have a GIG of other math competitions' PDF's, most unavailable online right now.

Below, is a list of everything I need.  Please email anything on that list to mysmartmouth@gmail.com.
If you need anything that I have, email me and I will be glad to send you anything you need for free.

[u]I Need:[/u]
All Nationals (I have 2003, but low quality)
Pre-2001 Chapter & State
Pre-2001 Handbooks
[u]Also...[/u]
2006 Duke Math meet (I have a low quality copy)
ALL AMC 8's
ALL AMC 10's (except for 2000 & 2001)


real analysis
function
limit
real analysis unsolved

Suppose $ f$ is Lebesgue measurable on $ (0,1)$ and not essentially bounded. Is it true that for every positive function $ g$ on $ (0,\infty)$  such that $ \lim_{p\to\infty}g(p)\equal{}\infty$ one can find an $ f$ such that $ \lim_{p\to\infty}\|f\|_p\equal{}\infty$  but $ \|f\|_p\leq g(p)$ for all sufficiently large $ p$?

Yes. The key point is that you can increase $ \|f\|_n$ without changing $ \|f\|_k$, $ k\equal{}1,\dots,n\minus{}1$ by much, by adding a very thin and tall peak to $ f$. This can be repeated indefinitely.

Yes, it works, thanks. Any ideas about the convexity for Lebesgue measure in the other problem?

[quote="mlok"]Yes. The key point is that you can increase $ \|f\|_n$ without changing $ \|f\|_k$, $ k \equal{} 1,\dots,n \minus{} 1$ by much, by adding a very thin and tall peak to $ f$. This can be repeated indefinitely.[/quote]

I don't know how.Can somebody tell me how to use this "key point" to consruct $ f$?Thank you in advance.


Its been so long since I posted here. :)
Determine all five-digit numbers with the property that, when the first digit and the last digit are both deleted, the resulting number is $\frac{2}{215}$ of the original number.

[hide="Solution"]If the first digit is $a$, the last $b$ and the middle part $x$, then

$10000a+b+10x={215\over 2}x\iff 10000a+b={195\over 2}x\iff x=\frac{2(10000a+b)}{3\cdot 5\cdot 13}$

Hence $10000a+b$ is divisible by $3$ and $5$, giving $30000,60000,90000,10005,40005,70005$ as candidates. Checking shows that only $70005$ is divisible by $13$, hence $x={2\cdot 70005\over 195}=718$. The only such number is $77185$.[/hide]


Someone broke a calculator ( :oops_sign: ) so that the only keys functioning are the $\sin(x)$, $\cos(x)$, $\tan(x)$, $\sin^{-1}(x)$, $\cos^{-1}(x)$, and $\tan^{-1}(x)$. Given any rational number $q$, show that you can hit a finite number of buttons so that you get the number $q$ on the panel. Assume the calculator is EXACT no matter what, everything is in radians, and the initial display is $0$.

Example.
$q=1$
$\cos^{-1}(0) = 1$

This is a USAMO-problem and has been posted and solved[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=136808#p136808]here[/url].

Since it's been already discussed, it's been locked.  :)


geometry
MATHCOUNTS
percent
AMC 8
number theory
AMC 10

What do you think you'll get?

I'll probably get a 23,24, or 25.

...and why is everyone saying they have to practice so much?, I mean, they're REALLY easy!
-jorian

25 PLZZZ!  :D

I certainly hope so. I haven't prepared at all, but it seems really easy. I have been practicing geometry since mid-october till now, devoted some time to geometry each day. And I have asked lots of people, and on AOL many problems from geometry. so yeah...

i duno. i want a 25 really bad , and I think it is really do-able

I gota watch my silly mistakes, 1st time taking it

Its really not that hard to get a perfect...its just that some kids mess up on the last few
 :(
-jorian

I know. I am in geometry, so I will be utterly disappointed with anything but a 25.

It's my 1st and only time taking it, so I won't ever get another chance. Unless I fail 8th grade, which I don't plan on doing.  :lol:

The only thing I am worried about is if luck will be on my side. Time on my side, questions on my side, luck...

Yeah... I don't feel the pressure, but this score.. will define me for the rest of my life... well... at least this year. till vestavia that is

Yeah, points off usually come from stupid mistakes or trick questions. You have plenty of time, so you really just want to take the test slowly and carefully...

its gonna be fun this year gazing around after i finish while others are doing problems like mad dogs :maybe:
-jorian :ninja:

so usually there is nothing that is not easy on amc 8? just trick questions and careless mistakes that usually prevent people from getting a 25?

I hope I won't make any mistakes or get tricked. *crosses fingers*

as long as you know how/when to apply methods like pyth., ull be fine
-jorian

this website is cool. I found it today, and I will work on it as much as possible tomorrow when I get home from school to get ready for the day ahead. http://www.unl.edu/amc/mathclub/04,1-problems.html

haha i thought the hardest one was 2004- *25. It was worded totally confusing.

[quote="jhredsox"]Ya. when....

[hide]
Loki, Moe, Nick and Ott are good friends. Ott had
no money, but the others did. Moe gave Ott one-¯fth
of his money, Loki gave Ott one-fourth of his money
and Nick gave Ott one-third of his money. Each gave
Ott the same amount of money. What fractional part
of the group's money does Ott now have?
(A) 1/10 (B) 1/4 (C) 1/3 (D) 2/5 (E) 1/2
[/hide]

is the hardest problem on the test (last question, '02), u know its easy!
-jorian[/quote]
Number 25 isn't always the hardest.

I'm pretty sure I will ace the AMC 8. On the practice tests, I got 25, 25, 25, 23, 19. On the 23 and 19, I screwed up on reading. I hope I get 25!

[quote="jhredsox"]i know, i was just trying to make a point :blush:

-jorian

@bpms: when do u go to school if ur on at this time?[/quote]

It depends when you take it. I take around 9 in my math class.

Cool. Thanks for the site. Anyway, I won't ace the AMC 8. :(

i have a chance. I hope luck is on my side. Because I get 24's on practice very frequently  :(

So what? Practice isn't what you normally get on the real test. Unfortunately... :(

You really don't need to get worried about getting a 25 on AMC 8. Shoot for higher goals, like a good score on AMC 10. AMC 8 is definitely nothing to worry about.

The practice tests basically shows what score you get -3. Not all the time but most of the time.

is that really something reasonable anirudh?

i dont think one or two points makes a difference

wait so if you get like a 5 on a practice test, you will get a 2, or an 8 on next test approximately?

ive never heard of that be4

i think what they're pointing out is not to trust your practice tests :D

and no, u guys are being way too optimistic considering 6 of u guys ALREADY think ur getting a perfect score

well guess what, at best, 1 of u will :wink:
-jorian

I recently did two AMC 8's, and on both I got 25. Hopefully, this time I'll get the same. :)

[quote="jhredsox"]how is that possible

is it because you don't have the motivation with no time pressure?
-jorian[/quote]
Everyone is different.

[quote="jhredsox"]yes, and the image was amazingly discrete!
-jorian[/quote]

Do you know what discrete means?

[quote="now a ranger"]discrete- kinda apart

anyway I really wish I can get a 25. and i work well under pressure also[/quote]

pfft I don't

National practice scores last year (approximate)

40
40
40
38
46
40
40
35
31 (2005)
40
40
40
42
39

National Score:25

25, but i won't be shocked or upset if I don't. I know that if I miss something, it'll be due to reading or some small mistake. I *could* spend a great deal of time checking it and whatnot, but why bother? It doesn't mean anything.

to me it sorta does , but then again it doesnt , cuz i havent studied, if i thot i twas important i would have studied a long time ago. Instead I spend time on geometry and other math stuff and number theory and aops

you should focus more on amc 10, which is more of a mathcounts level. AMC 8 doesn't really require studying for. It will just naturally come to you as you study for other competitions.

25 I hope. More likely towards 24 but with a stroke of luck, the 1/4 percent or the 4 leaf clover or magic, or god, or other extraneous things which you probably don't care about so I will continue on, I CAN DO IT!
"Emphasize added"
 :D

http://www.unl.edu/amc/mathclub/04,1-problems.html

is very nice

now a ranger
thank you

Ikh, I make so many stupid mistakes on AMC's.  On math club practices, I ended up finishing in 12 minutes and I would make 1 or 2 mistakes.  If I tried taking about the whole time, I would just end up making the same mistake when checking.

[quote="13375P34K43V312"][quote="bpms"]Ug. I feel sick right now. This my first competition while sick. This sucks so bad. I'll probably know how to do all the problems, or maybe all but 1 in case something special occurs, but I might have a ton of errors.[/quote]

Being sick is an advantage. I was sick at chapters and got a 44 last year, without any practices. Being sick makes you delusionally paranoid about how many you got right, so you will check and check and check without getting bored.[/quote]
MAybe for you, but focusing is impossible right now. Also, um a 44 would be bad for me at chapters, just because I set high goals for myself.

[quote="13375P34K43V312"][quote="bpms"]Ug. I feel sick right now. This my first competition while sick. This sucks so bad. I'll probably know how to do all the problems, or maybe all but 1 in case something special occurs, but I might have a ton of errors.[/quote]

Being sick is an advantage. I was sick at chapters and got a 44 last year, without any practices. Being sick makes you delusionally paranoid about how many you got right, so you will check and check and check without getting bored.[/quote]

Being sick is a big disavantage for me in math but a big advantage for me during sports. :D

I was sick at states(it came on right before the Target) so I was still able to do well on sprint. I only got 39, but still was 2nd. I'd like to know how that happened. Just so oyu know, I found it to be a HUGELY bad thing.

[color=blue]kyyuanmathcount: took out this post.  see the no discussion of AMC 8 thread everyone.  [/color]


Given that $a_1, a_2, ..., a_m$ are nonzero integers.
For every $k = 0,1,...,m$ ($n<m-1$),
$a_1+a_22^k+a_33^k+...+a_mm^k=0$.
Prove that the sequence $a_1, a_2, ..., a_m$ has (at least) $n+1$ pairs of consecutive terms with opposite signs

i think something is missing. what is $n$ here?just any integer $<m-1$?


geometry
3D geometry
tetrahedron
similar triangles
AMC

An insect lives on the surface of a regular tetrahedron with edges of length 1. It wishes to travel on the surface of the tetrahedron from the midpoint of one edge to the midpoint of the opposite edge. What is the length of the shortest such trip? (Note: Two edges of a tetrahedron are opposite if they have no common endpoint.)

$ \displaystyle \textbf{(A)} \ \frac {1}{2} \sqrt {3} \qquad \textbf{(B)} \ 1 \qquad \textbf{(C)} \ \sqrt {2} \qquad \textbf{(D)} \ \frac {3}{2} \qquad \textbf{(E)} \ 2$

[hide]"unfold" the tetrahedron into an equilateral triangle of length 2 composed of the 4 equilateral triangles of length 1.  select the first midpoint from any side of the inner triangle.  the other midpoint lies on either of the two edges on the side parallel to it.  the distance is $1 (B)$[/hide]

LOL

[hide="Answer"]Unfold the tetrahedron and choose any two adjacent triangles. Choose either set of two parallel lines, and the distance between their midpoints is $1\Rightarrow \boxed{B}$.[/hide]

Hmm.. I simply thought about the distance on the tetrahedron without unfolding it. That is, I drew a perpendicular line and basically used similar triangles to get the length as 2*1/2 which is 1 (I have to multiply by 2 since there are two of them). It's basically same as unfolding except I wasn't sure whether it was minimum or not.

[quote=SenorIncongito]Wiki solution: [url]https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_13[/url]
Why is the length the same as the length of the edge of the tetrahedron's side?[/quote]

[quote=Aritra12][quote=SenorIncongito]Wiki solution: [url]https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_15[/url]
Why is the length the same as the length of the edge of the tetrahedron's side?[/quote][/quote]

its problem 15 not 13
I agree with the unfolding method... it makes this problem easy

The official solution shown below also references unfolding. I'm not convinced by it's last line though. My worry is that passing to the net of the tetrahedron doesn't preserve enough structure. In particular, the boundary of the net is not really a boundary, as some sides on the boundary become identified with each other when folding it back into a tetrahedron. I don't get why I should be convinced we couldn't make better use of the new paths this creates and make something smaller than 1. Altogether, the solution convinces me we can achieve 1, but not that it is minimal. My idea is to preserve more structure by mapping to a tiling. So I came up with the proof without words (Tiling.pdf) attached below. Note it really stretches infinitely in all directions.

[quote=probability1.01]Try to prove it's the minimum! (without unfolding)[/quote]
Also, does anyone have any idea of what Alex Zhai had in mind here?


function
calculus
derivative
quadratics
inequalities
algebra
quadratic formula

Is there a restriction on x?  If not, you can make x approach -:inf:, and the value of the expression will approach :inf:.

Or perhaps you mean find a local maximum?

Do you mean to find a [i]relative[/i] maximum of the function because as x -> -:inf:, then f(x) -> :inf:.

yeh i meant a relative maximum, i.e. the highest possible positive value in this case. I know its somewhere b/w f(10.5) and f(11.5), but i would like an exact value (and how to do it).

You'd break out the calculus. It'd look a little like this:

f(x) =  (16-x)(x^2)
f(x) = 16x^2 - x^3
f'(x) = 32x - 3x^2    (took the derivative)
Find the 0s of the derivative with the quadratic formula
-32:pm:32 / -6 = 0 or 32/3

It's a local max iff the sign of the derivative changes from + to -.

f'(-0.1) = -3.23
f'(0.1) = 3.17
So f(0) is a min, not a max

f'(10) = 20
f'(11) = -11
so f(32/3) is a max.

Of course, this post is meaningless if you don't know calculus... and you probably wouldn't have asked if you did.

Expand the expression to get 16x2 - x3. Take the derivative to get 32x - 3x2. Set this equal to 0 and solve for x. You should get x = 0 or x = 32/3. 0 is the local minimum, and 32/3 the local maximum.

Confuted beat me to it. ;)

If you don't know calculus, you can also use the arithmetic-geometric mean inequality.  You said that you wanted to restrict to x :ge: 0.  We can also restrict to x :le: 16, since otherwise the expression is negative.  Now apply the AM-GM inequality to the three numbers 32 - 2x, x, and x.  Their arithmetic mean is 32/3.  So, by the AM-GM inequality, (32 - 2x) x2 :le: (32/3)3.  Now divide by 2 to get an upper bound on your expression (16 - x) x2.  Equality occurs in AM-GM when the numbers are equal.  Solving 32 - 2x = x gives x = 32/3 as previously observed.

[quote="jelyman"]doh.... i dont know calculus, but i do know some of derivatives from my prechal book. So i definitely should have gotten that. Hmm....[/quote]
Naw, that's in chapter 8 of my calculus book. Of course, intuitively it makes sense, but it doesn't surprise me that you didn't think to do it that way.

actually, this kind of thing is handled in my precalc book.

Yeah, my precalculus text has an introduction to calculus at the end.


linear algebra
matrix
algebra
polynomial
quadratics
linear algebra unsolved

I am looking for a problem of positive definiteness of a 3 by 3 matrix $A=A^{T}$ for the form $Q(x,y,z)=[x,y,z]^{T}Q \left[\matrix{x\cr y\cr z}\right]$

on the set of $\{x,y,z~\vert~xz=y^{2}\}$

Are there any standrad result for this?

I guess one possible solution is by Lagrangian Multiplier Method
We form the Lagrangian
$L(x,\lambda)=Q(x,y,z)+\lambda (xz-y^{2})$
and follow the standard procedure  :)

If we follow the bchui's idea:
We seek the extrema of $X^{T}AX$ on the compact
$K=\{X\mid{X^{T}}UX=0,{X^{T}}X=1\}$ where \[U=\begin{pmatrix}0&0&-1\\0&2&0\\-1&0&0\end{pmatrix}\].
$\partial{K}$ is empty therefore it's sufficient to use lagrange's formula.
There exist $\lambda,\mu\in\mathbb{R}$ such that
for all $H$: $X^{T}AH+\lambda{X^{T}}H+\mu{X^{T}UH}=0$ or
(1): $AX+\lambda{X}+\mu{UX}=0$.
Thus we have 5 equations (3 in (1) and the 2 equations which define $K$) in 5 unknowns ${x,y,z,\lambda,\mu}$.
The condition for positive definiteness is: if $X$ is a solution of the system then $X^{T}AX$>0. (1) implies that $X^{T}AX=-\lambda$. Thus for each solution $\lambda$ must be <0.
Remark: necessarily $det(A+\lambda{id}+\mu{U})=0$.
Example: Let \[A=\begin{pmatrix}126&102&33\\102&-16&-50\\33&-50&-124\end{pmatrix}\].
By a Maple calculation we find 2 solutions (max and min if there don't exist any other solution):$\{\lambda\approx-173.9,\mu=\approx-29.3\},\{\lambda\approx134.6,\mu\approx62\}$.
Thus $A$ is not >0 over $K$.

There is also another approach. We need to check that the expression $ax^{2}+bx^{z}+cz^{2}+(dx+fz)y$ is non-negative for all $x,y,z$ such that $xz=y^{2}$. Since $y$ can be replaced by $-y$, we see that $ax^{2}+bxz+cz^{2}$ must be non-negative (i.e., we must have $a,c\ge 0$, $b^{2}\le 4ac$ and the fourth degree expression $(ax^{2}+bxz+cz^{2})^{2}-(dx+fz)^{2}xz$ should be non-negative for all $x,z$ with $xz\ge 0$. Introducing a new variable $t=x/z$, we see that the latter requirement is equivalent to non-negativity of a forth degree polynomial on $[0,+\infty)$, which can be checked by Sturm test. The advantage of this approach is that it can be done by hand and doesn't require any floating point computations if all coefficients are integer.

[quote="loup blanc"]If we follow the bchui's idea:
We seek the extrema of $X^{T}AX$ on the compact
$K=\{X\mid{X^{T}}UX=0,{X^{T}}X=1\}$ where .......[/quote]

That's what I have obtained already. The problem is:
For a quadratic optimization problem with quadratic constraint, such as

Minimize  $Q(x)=x^{T}Ax$
with constraint   $x^{T}G_{1}x=1,~x^{T}G_{2}x=0$

Do we have a general solution or any general solution for certain particular cases?

Note also that $AX+\lambda X+\mu UX=0$ involves a quadratic term. Can we possibly solve such equations explicitly?

We seek the extrema of $x^{T}Ax$ when $x$ goes through $K=\{x\mid{x^{T}Ux=0}$ and $x^{T}Vx=1\}$. I want to be sure that $K$ is a non empty compact; so I choose $V$ symmetric >0 and $U$ symmetric $\not>0$ and $\not<0$; here I assume that $V=id$.Again we have $Ax+\lambda{x}+\mu{Ux}=0$. We solve this 5-5 system with the symbolic calculation of Maple $via$ the Gröbner basis.Then the $\lambda$ are the real roots of a polynomial of degree 6. Let $\lambda_{1}$ and $\lambda_{2}$ the min and the inf of these real roots.The request max is $-\lambda_{1}$ and the request min is $-\lambda_{2}$.
Example: \[A=\begin{pmatrix}-88&31&-16\\31&-100&-34\\-16&-34&51\end{pmatrix},U=\begin{pmatrix}126&102&33\\102&-16&-50\\33&-50&-124\end{pmatrix}\]
The polynomial in $\lambda$'s is:2473506906946$\lambda^{6}$-758144102940511$\lambda^{5}$+98688563506650912$\lambda^{4}$-6373227166814110795$\lambda^{3}$+33017456342608831768$\lambda^{2}$+14937125574310179954480$\lambda$-221426066491786238260800.
The real roots are:$\approx$ -40.3,15.6,98.3,118.4
Thus the request maxima are $\approx$ 40.3 and -118.4


function
calculus
algebra
functional equation
calculus computations

If $ f(x+y) = f(x) + f(y) -xy -1 $ $ \forall x,y \in \mathbb{R} $, and $ f(1) = 1 $, find the number of solutions of $ f(x) = n$ where $ n \in \mathbb{N} $. Here $x \in \mathbb{R} $[hide](Probably this part os wrong -- not sure!!)[/hide]

Let $f(x)=g(x)+1-\frac{x^2}{2}$. Then $g(x+y)=g(x)+g(y)$. The solutions of this functional equation are far too varied to give a definite answer to your question. We can show that $f(x)=1+\frac x2-\frac{x^2}2$ for rational $x$, but not necessarily for all $x$.

It is possible for $f$ to take a particular value on an uncountable dense set, or to avoid it completely.

So the question is wrong?? I though so?? What if $x$ in $f(x)=n $ is an integer???

As was said by jmerry, you can show that, for every $q\in\mathbb{Q}$ the following holds:
\[f(q)=1+\frac{q}{2}-\frac{q^2}{2}\]
If you allow $x$ to vary only among the integers, you are searching $x\in\mathbb{Z}$ such that
\[1+\frac{x}{2}-\frac{x^2}{2}\in\mathbb{N}\]
Now, it's easy to see that the only integer solutions are $-1,0,1,2$.

The impossibility to give an answer when you let $x$ vary among the reals is related to the non continuous solutions of Cauchy's equation (I think this is how it's called) : $f(x+y)=f(x)+f(y)$.

Btw, if you ask your $f$ to be continuous, the problem can be solved (in the same way, with the same solution) with $x\in\mathbb{R}$.

Ehr, now that I think of it...this isn't properly a Calculus subject, is it?


LaTeX
inequalities unsolved
inequalities

Find the minimum of $E= \frac{xy}{z}+ \frac{yz}{x}+ \frac{zx}{y}$, if $x,y,z>0 $ and $x^2+y^2+z^2=1$.

made by Kurliandcik L.D.

cheers! :D :D :cool:

I think you mean $x^2+y^2+z^2=1$. So
\[ \left(\frac {xy}z+\frac{yz}x+ \frac{xz}y\right)^2=\sum \left (\frac {xy}z\right )^2+2(x^2+y^2+z^2). \]
But $ \displaystyle \sum \left (\frac {xy}z\right )^2\geq x^2+y^2+z^2.$, so    \[ \left(\frac {xy}z+\frac {yz}x+\frac {xz}y\right )^2 \geq 3(x^2+y^2+z^2)=3. \]
\[\Leftrightarrow  \frac {xy}z+\frac {yz}x+\frac {xz}y \geq \sqrt 3. \]
The equal occurs when $ x=y=z= \displaystyle \frac {\sqrt 3}3.$

[Admin Edit: Made Latex compliant]

Yes I meant x^2+y^2+z^2=1.

Sorry about that!  :blush:


[quote="ragnarok23"][quote="bpms"][quote="ragnarok23"]A palindrome is a number that is the same forwards and backwards. For example, 212 and 21466412 are palindromes. Consider a palindrome with $n$ digits where $n$ is even. Find a formula, in terms of $n$, of how many $n$-digit palindromes exist.[/quote]
[hide]For the first and last digits they have to be even. So there are four choices for them since they can't be 0. There are n-2 remaining digits. For even values of n there are because there are ten values for each pair of digits and each pair is coordinated. $(10^{\frac{n-2}{2}})(4)$. For odds, there are $(10^{(\frac{n-3}{2})})(40)$ which is equal to $(10^{\frac{n-1}{2}})(4)$ numbers.
If you want a generalization that's too bad :P although there probably is one using the floor function.[/hide][/quote]
I didn't say the actualy number was even, I simply said the number had an even number of digits.  :wink:[/quote]
Heh, I misread that. This should be easier by a lot. Using the same logic...
[hide]Each digit pairs up with one other so there are $(9)(10^{\frac{n-2}{2}})$ since the first digit can't be 0.[/hide]


I'm currently taking an AP US History course. I've read that if you take an AP class you should undoubtedly take the national exam; the reason being that it shows admissions people that you really know the course material. If you're getting an A in the course, is this really necessary? It seems like a lot of money considering all you would be doing is reinforcing your teachers opinion of your performance, unless of course admissions people really do want to see that you take the test.

Take the test.

There are several things at stake here. The first is that it's actual college credit. I know that different colleges have different policies, but at an awful lot of places, it counts. For us, we have an general education requirement for U.S. History, and that AP exam satisfies the requirement. The course title and grade on your high school transcript would not satisfy the requirement.

The college credit frees you up to move forward. If you don't take that U.S. History course during your freshman year, what do you do? College campuses are full of choices, only your own time is limited and you can't take everything. Perhaps the history of a different part of the world? Or an interesting literature course? Or accelerate your math and science courses?

The second is that you telling me you got an A in the course is mostly worth a "so what?" from me. I'm in advising, not admissions, but look at it from the advisor's side, and bring it back to my field. If you tell me you took a calculus course in high school but not the AP exam, then I don't know the standards of your course, and we won't recognize that course. After all, many - maybe most - of the students who took a high school precalculus course aren't really ready to start with our Calculus I. I do know the standards of the AP exam.

Giving "weighted GPA" credit for AP and honors courses is quite widespread. Admissions people know that you have an incentive to take the course just for that reason - but, again, what is the level of that course and what are its standards? The AP exam may only sample your accomplishments, but it is a national standard to weigh that course against.

Because so many AP exams are taken in the senior year, with the results known long after admissions decisions are made, it's hard for those doing admissions to find a way to account for them. But I'm guessing that you're not taking that U.S. History course in your senior year. More likely, you're a junior or a sophomore.

Oh, and you said, "It seems like a lot of money." It's a lot less money than one course's portion of a semester worth of college tuition and/or fees, even at a low-cost state university like mine.

Kent Merryfield has said it very well. I will simply reemphasize the point that the test fee (and one day out of your life) is a small price to pay to take the test and get recognition of your work during the course. Knowing ahead of time that you will take the test will also help you concentrate during the course and very likely learn more than you would if you know that you will avoid the test. And after all, isn't learning the most important goal here? Having the test score in hand, as the first reply pointed out, makes cooler learning goals more achievable, and that's worth a lot more than what the test costs.

Thanks Kent & Tokenadult. I definitely apreciate it, and I'll most likely end up taking the exam.

By the way, yes Kent, I am a sophomore.

the really shows how well u know the subject.

and at my school, u get 5 pts added to your final average if u take the test


I'm thinking Pascal's triangle.  (of course, there are other equivalent or similar ways of looking at it)

I advise Moderators to move it to the Number Theory Section.

[quote="jen7"]How do you solve this problem?
$N(m)={4m\choose 0}+{4m\choose 4}+{4m\choose 8}+\cdots+{4m\choose 4m}$, where m is a positive integer.  Find the smallest m where 81 evenly divides N(m).[/quote] Note that $4N(m)=(1+1)^{4m}+(1+i)^{4m}+(1-1)^{4m}+(1-i)^{4m}=2^{2m+1}(2^{2m-1}+(-1)^m)$.
Therefore, you are to find minimal $m$ s.t. $81\mid 2^{2m-1}+(-1)^m$.
More hints?


geometry
circumcircle
angle bisector
geometry unsolved

(F.Nilov, A.Zaslavsky) Let $ CC_0$ be a median of triangle $ ABC$; the perpendicular bisectors to $ AC$ and $ BC$ intersect $ CC_0$ in points $ A_c$, $ B_c$; $ C_1$ is the common point of $ AA_c$ and $ BB_c$. Points $ A_1$, $ B_1$ are defined similarly. Prove that circle $ A_1B_1C_1$ passes through the circumcenter of triangle $ ABC$.

see http://www.mathlinks.ro/Forum/viewtopic.php?t=223753
sin∠ABB1/ sin∠CBB1 * sin∠CAA1/ sin∠BAA1 * sin∠BCC1/ sin∠ACC1=
sin∠B0BC/ sin∠B0BA * sin∠BAA0/ sin∠CAA0 * sin∠ACC0/sin∠BCC0 =
B0A/B0C * BA0/CA0 * BC0/AC0 =1
We have AA1,BB1,CC1 are concurrent,denote the intersection is D
We can acquire O,D,C1,B1 are concyclic and OP is diameter
O,A1,B1,D are concyclic and OP is diameter
So A1,B1,C1,D,O are concyclic with OP its diameter

As it can be seen from above proof, CC1 is symmedian in triangle ABC and angle bisector of angle AC1B. Knowing that the symmedian passes through the common point of the tangents to (ABC) at A and B, we get OC1 and CC1 perpendicular, hence OC1 is angle bisector as well. Inverting about O so that the circle O transforms to it self, the circle (OAB) is transformed into AB, hence the image of C1 is AB∩OC1=C2, similarly getting A2 and B2.
Taking into account the above statements, we get C_2A/C_2B = (b/a)^2 and other 2 similar relations which, multiplied, show that A2, B2 and C2 are collinear, the conclusion follows (a,b and c being the side lengths of triangle ABC).

Best regards,
sunken rock

Can someone solve this problem using complex numbers ?
I tried to, but I failed.
At least give a hint or something.

By USAMO 2008/2, we know that $C_1$ is the midpoint of the chord intercepted by the $C$-symmedian in the circumcircle of triangle $ABC$. Now, inverting about $O$ maps $C_1$ to the intersection point of the tangent to the circle at $C$ and line $AB$. Thus, inversion maps the circumcircle of $A_1B_1C_1$ to the trilinear polar of the symmedian point $K$ of triangle $ABC$ and so, points $A_1,B_1,C_1,O$ are con cyclic.


geometry
geometric transformation
rotation
factorial

In how many ways can 5 people be seated around a round table?  (Two seatings are considered the same if one is a rotation of the other.)

What does 5!/5=4!=24 mean? What does 5! mean?

We have $ 5\cdot4\cdot3\cdot2\cdot1$ choices but there are $ 5$ rotations. Thus, we have $ 4\cdot3\cdot2\cdot1 \equal{} 24$.

chubbyg-- $ n!\equal{}(n)(n\minus{}1)(n\minus{}2)(n\minus{}3)...(2)(1)$. Basically, it is the multiplication of all integers less than or equal to itself until 1 is reached. It is called a [url=http://www.artofproblemsolving.com/Wiki/index.php/Factorial]factorial[/url]. Try google if you need more information.

[quote="slayer42"]cant you just do (n-1)! instead of (n!)/n? It is much simpler. so for this problem it would be (5-1)! or 4![/quote]

yes
$ (n\minus{}1)!$ is the same as $ \frac{n!}{n}$ <--can you figure out why?

so, since these two are the same, you can do the problem either way

 $\dfrac{5!}{5} = 4! = \boxed{24}$
5! ways to place everyone but divide by 5 because of rotational symmetry.


does anyone know how to simplify:
$ \frac {(3^{2008})^2 \minus{} (3^{2006})^2}{(3^{2007})^2 \minus{} (3^{2005})^2}$
Thanks in advance  :D
EDIT: ALL LATEX IS CORRECTED

[quote="neurosurgeon1118"]does anyone know how to simplify:
$ \frac {(3^{2008})^2 \minus{} (3^{2006})^2}{(3^{2007})^2 \minus{} (3^{2005})^2}$
Thanks in advance  :D
______
BOGTRO: HA HA fixed in this quote.[/quote]

Rewriting, we get

$ \frac{3^{4010}(3^6\minus{}3^2)}{3^{4010}(3^4\minus{}1)}\equal{}\frac{720}{80}\equal{}\boxed{9}$.

yeah sorry but for some reason i don't understand why the Latex considers the 3^2008 as 3^2 008

You realize that this is an AMC 10A 2008 problem, right?

Well sorry but i was wondering if anyone had any idea on how to number 25 if u have the test then can you do it and post a solution.

What's #25? On the 10A?

If so that's totally off topic.

Anyway, solutions would be up. I got that by realizing that there

is a $ \sqrt{7}$, so... Multi choice helped there

Of course, I still failed.

First of all sorry for my ignorance of not looking it up
anyways i scanned the contest pamphlet so i hope this helps:
Wait for some reason i cant attach images on aops anymore so im just going to link it to my website
http://neurosurgeon1118.web.officelive.com/Documents/AMC%2025.JPG


[quote="GameBot"]The diagram consists of three nested squares. Find the ratio of the area of the smallest square to the area of the largest square. Express your answer as a common fraction.
[asy]draw((0,0)--(7,0)--(7,7)--(0,7)--cycle);
draw((3,0)--(7,3)--(4,7)--(0,4)--cycle);
draw((1.2,2.4)--(4.6,1.2)--(5.8,4.6)--(2.4,5.8)--cycle);
label("4",(2,7),N);
label("3",(0,5.5),W);
label("2",(.6,3.2),.5ENE);[/asy][/quote]

The largest square has side lenght 7, the second square has side length 5, the smallest square has side lenght $ \sqrt{11}$.

$ \frac{11}{49}$.

Wait, are you sure of that?  Because by the Pythagorean Theorem, we get $ \sqrt{3^2 \plus{} 2^2}$ or $ \sqrt{13}$, giving an area of $ 13$ not $ 11$...

[quote="ernie"]Wait, are you sure of that?  Because by the Pythagorean Theorem, we get $ \sqrt {3^2 \plus{} 2^2}$ or $ \sqrt {13}$, giving an area of $ 13$ not $ 11$...[/quote]

Typo. Whatever.


$ P$ and $ Q$ are whole numbers such that $ 0 < P < 10$ and $ 0 < Q < 10$. How many common
fractions $ \frac{P}{Q}$ exist if $ \frac{1}{2} < \frac{P}{Q} < 1$?

First you find the possible values for Q: 1, 2... 8, 9.
For these values:
Q is equal to
1 - P is not possible
2- P is not possible
3- 2/3
4- 3/4
5- 3/5, 4/5
6- 4/6, 5/6
7- 4/7, 5/7, 6/7
8- 5/8, 6/8, 7/8
9- 5/9, 6/9, 7/9, 8/9
There are 16.
BUT, the question asks for COMMON fractions. So we list only the simplified here:
2/3, 3/4, 3/5, 4/5, 5/6, 4/7, 5/7, 6/7, 5/8, 7/8, 5/9, 7/9, 8/9
So there are 13.
Answer: 13


geometry
3D geometry
sphere
topology
superior algebra
superior algebra unsolved

Is it possible to define a multiplication so that the following is a group?
(1) open unit interval (-1, 1)
(2) open unit disc
(3) open unit sphere in $R^n$

(1) Yes
(2) Yes
(3) Yes
Just take some bijection $b$ of the desired set to $\mathbb{R}$ (or $\mathbb{C}$ or somewhat else), then define $x*y : = b^{-1}\left( b(x)+b(y) \right)$.

It's more interesting to ask whether these sets are [i]topological[/i] groups as subspaces of the according euclidean space. (3) works exactly for $n=1,2,4$. But I think a lot of algebraic topology is necessary to proove that.

For (1): I don't think normal multiplikation works (or what is normal¿).

PS: hi Martin, welcome here :)

[quote="-oo-"]@Daniel: Ach, ich hab die inversen Elemente vergessen! Wie hast du mich eigentlich erkannt?[/quote]
@Martin: only english outside of the the national forums please  :)
Well I could break that rule now since I doubt that anyone else will understand it even in English, but well: You used that name in the "Schwätz" and sometimes also on the MP ;) (and mods can see the E-Mail addresses).

@all: he just said that he forgot the inverses and asked how I identified him.


group theory
abstract algebra
vector
superior algebra
superior algebra unsolved

[size=150][color=red][b]Problem.[/b][/color] If each of two [color=blue]groups[/color] is isomorphic to a subgroup of the other, are the [color=blue]groups[/color] necessarily isomorphic?
[/size]
I look forward to reading your responses one of these days. Happy solving to you [b]all[/b]!  :wink:

P.S. Last month's problem can be found [url=http://www.mathlinks.ro/viewtopic.php?t=256626][color=orange]here.[/color][/url]

I remembered doing this way back in the day, and luckily I still have Artin sitting on my shelf to fill in the details. :)
[hide="Solution"]The answer is "no": if $ F$ is the free group generated by $ x, y$ then the elements $ u \equal{} x^2, v \equal{} xy$ and $ w \equal{} y^2$ generate a subgroup of $ F$ isomorphic to the free group on 3 generators.  Thus the free group on 3 generators is isomorphic to a subgroup of the free group on 2 generators and vice-versa, but the two groups are not isomorphic.[/hide]

[hide="Another solution"]
Set $ G : \equal{} S_{\mathbb{N}}, \ H: \equal{} S_{\mathbb{N}}\times \mathbb{Z}_{2}.$
Then $ G < H$ and $ H\cong \langle S_{\mathbb{N}\setminus\{1,2\}}, (1\ 2)\rangle < G,$ but $ G\not \cong H,$ because $ \# Z(G) \equal{} 1\neq 2 \equal{} \# Z(H).$[/hide]

Another example (I won't bother tex-ing it - it's not really necessary for this) - take G=RxZ, H=R. Using Zorn's lemma one proves that RxR and R are isomorphic as they are both continuum cardinality Q vector spaces (the existence of a basis is where Zorn is needed). It follows that RxZ is a subgroup of R, and it is obvious that R is a subgroup of RxZ. To see they are not isomorphic, note that one is divisible (R) whilst the other isn't (as Z certainly isn't).
The problem can be generalized to other categories in an obvious manner - it's fun to see where it holds. A particularly fun one (in my opinion) is for fields, enjoy!

Well done, [b]guys[/b]!

I didn't expect anything less from you...  :wink:


Uh, looking at past AIMEs and AMCs (and Altheman's Mock AIME), I see questions like "Bob goes to starbuck from 2-4 and stays for 30 minutes. Jon goes to starbuck from 1-3 and stays for 30 minutes." How do you do these types of questions?

You have to use a geometric approach, since there are infinitely many possibilities.

Try reading this excerpt from Intro to Counting & Probability.  It has some problems like the one you mentioned.

[url=http://www.artofproblemsolving.com/Books/IntroCount/exc2.pdf]Click Here[/url]

Don't know if this is one of the problems that you said you saw already...
but here's an example of an AIME problem that uses the geometric method:

[size=150]AIME 1998 #9[/size]
Two people arrive at a cafe independently at random times between $9 AM$ and $10 AM$ and each stay for $m$ minutes. What is $m$ if there is a $40$ percent chance that they are in the cafe together at some moment?

if you understand the process, try my problem now, it is a step above the normal ones  :)

[quote="Zellex"]AHHHH....Adunakhor....you almost confused me...I didn't get an integer answer and I started checking my answer...then I went to the contests section and found out that it's in the form $a-b\sqrt{c}$...:P...Altheman, what is your problem? you mean the question in your mock AIME?[/quote]

yes!

[quote]AHHHH....Adunakhor....you almost confused me...I didn't get an integer answer and I started checking my answer...[/quote]

Heh, sorry :?. I took this problem from Kalva which didn't include that part.

so how would u do that problem...i am not really sure.
i know that you have to draw part of a circle on a graph, but dont really understand how you lable the axies.


3 circles, touching one another externally, share a common tangent. If the radii of the two larger circles are 36 and 9 respectively, find the radius of the smallest circle.

[hide="Answer"]$4$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=64305[/hide]


floor function
search
number theory
number theory proposed

Prove that the exponent of number 2 in the expansion of $n!$ is $\sum_{j=1}^{\infty}\left\lfloor{n \over 2^{j}}\right\rfloor$.

Similary the exponent of number $p$ in the expansion of $n!$ is $\sum_{k=1}^{\infty}[\frac{n}{p^{k}}]$. You can find in some text book on number theory  :wink:

you can search on lookforbook.com
write:number theory.

How many numbers in the set $\{1,2,\cdots,n\}$ is divisible by $2$? Of course $\left\lfloor{n \over 2^{1}}\right\rfloor$.
How many numbers in the set $\{1,2,\cdots,n\}$ is divisible by $2^{2}$? Of course $\left\lfloor{n \over 2^{2}}\right\rfloor$.

[quote="Lovasz"]Can you post a solution? I don't have many number theory books.[/quote]
[url=http://modular.math.washington.edu/scans/papers/hardy/]Theory of Numbers by Hardy and Wright[/url]


function
analytic geometry
geometry
conics
parabola
calculus
integration

Given the function $ f(x) \equal{} x^2 \minus{} C$ for some constant $ C$, find the vertices of a square (coordinates in terms of $ C$) centered at the origin with sides parallel to the axes such that the area above the curve and in the square is equal to the area below the curve and in the square.

For example, when $ C \equal{} \frac13$, we have the square with vertices $ (\pm1,\pm1);(\pm1,\mp1)$.

[asy]import graph;
unitsize(x=1cm);
real f(real x){return x^2-0.333333333;}
xlimits( -1.5, 1.5);
ylimits( -3, 1);
draw(graph(f,-3,3));
xaxis(Ticks("%",extend=true));
yaxis(Ticks("%",extend=true));
dot((0,0));
draw((1,1)--(1,-1)--(-1,-1)--(-1,1)--cycle);[/asy]

Well won't you have two cases? One when the parabola is enitrely contained within the square, and the one when it is not? Am I being unclear?

Hm, slightly. Tell me what you would do as a first step.

Well we have two cases (I think?) one described in your first post and the other where its range exceeds the square's mximum y value. To eases integration we could move it up by a perhaps?

No, $ a$ makes the algebra harder. I assume $ a$ is a half sidelength? Try a different transformation.

[hide="Hint"]
Shift up by $ C$. Now divide into two cases and solve.[/hide]

Hm do we have to adjust when it hits [b]exactly[/b] on the squares vertices?

Gah we will probably have to use a bunch of calc for this. So we get a shift up by C. So that makes the square (a+c,a+c) and so on?

Yes. Now, see if you can simplify the integration.

Wait a minute! if we shoft like that the x-coordinate remians a right? gah that screws up a bit. and how u use latex for integration?

I think it's
[code]\int_a^bf(x)\,dx[/code] to produce $ \int_a^bf(x)\,dx$.

Hm if we flip the graph over, then we get a nice upside-down parabola which would be ez to integrate. This will probably be a nasty cubic relating a and c then.

Well, the parabola is a quadratic $ \minus{} x^2 \plus{} a\plus{}c$. All that remains is to find the bounds of the integral.

well in the case described in the first post it is 0-a

Well if the top point is a+c, and the graph is x^2 now it goes from 0-sqrt(a+c). Too lazy to integrate.

Bah. Practicing $ \text{\LaTeX}$:

$ \int^{\sqrt {a + c}}_{ - \sqrt {a + c}}x^2\ dx = \left(\frac {x^3}3\right)\bigg|^{\sqrt {a + c}}_{ - \sqrt {a + c}}$

$ = \frac23(a + c)\sqrt {a + c} = 2a^2$.

Thus, $ (a + c)^3 = 3a^2$, and the rest follows. Blah.


How do I make a function in terms of x and y when I have a parametric functions involving trigs and logs? :?:

It depends on how complex it is, but just use the inverse functions.

Usually when you have the parametric functions $ x(t),y(t)$ in terms of trigs and logs, use their inverse functions (and paying attention to their domain/range <---this is very important) and do the algebra to get the relation $ f(x,y)\equal{}0$. But, the point is... why would you want to do that? Parametric functions allow you to see the behavior of a curve in terms of a controlled parameter $ t$. If you are trying to graph it, a graphing calculator with parametric graphing capabilities should be sufficient.


limit
function
parameterization
complex analysis
algebra
functional equation
complex analysis unsolved

Translating into the language of power series:
$f(x)=\sum a_{n}x^{n}$ and $g(x)=\sum b_{n}x^{n}$ have infinite radius of convergence.
$f(x)^{2}g(x)=f(x)g(x)+1$

All right, that makes $f$ and $g$ entire functions. Since $f(z)(f(z)-1)g(z)=1$, $f(z)$ is never zero or 1. By the Picard theorem, $f$ is constant. Then $g$ is also constant. The only solutions have $a_{n}=b_{n}=0$ for $n\ge1$.

Let $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ and $g(z)=\sum_{n=0}^{\infty}b_{n}z^{n}.$

The limit involving roots is the statement that these series have infinite radii of convergence - in other words, $f$ and $g$ are entire.

Now, let's take the chain of identities and make them the coefficients of power series:

$\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}b_{n-k}\sum_{j=0}^{k}a_{k-j}a_{j}\right)z^{n}=1+\sum_{n=0}^{\infty}\left(\sum_{k =0}^{n}a_{n-k}b_{k}\right)z^{n}$

Now let's translate this into something we can see:

$(f(z))^{2}g(z)=1+f(z)g(z)$

So that's what the question comes to: can we find entire functions that satisfy that identity?

The identity can be rewritten as:

$f(z)(f(z)-1)g(z)=1.$

In order for that to be true for all $z\in\mathbb{C},$ we can never have $f(z)=0$ or $f(z)=1.$ That flies in the face of the "little" Picard theorem: the range of a nonconstant entire function can omit at most one value. Therefore, the identity cannot be satisfied, except by constant functions.

[color=green][Ah - took too long to type it. jmerry beat me to it.][/color]

Oh,thank you two :) .This is problem is just come out when I want to prove the Picard's Theorem.And in fact it can be show is equicvalent to Picard's theorem.So I want to know a proof without Picard's theorem,or more elemtrary way.Then there will be an alternative way to prove the  Picard's theorem.

If you try that, be aware that radius of convergence matters- there are many local solutions to the functional equation, and you'd need infinitely many parameters to make it work.

This is not the way to an easier proof of the Picard theorem.

[quote="Kent Merryfield"]What happens if you just run the chain of identities? Can we find formulas for $a_{1},a_{2},\dots,b_{1},b_{2},\dots$ in terms of $a_{0}$ and $b_{0}?$[/quote]
Sorry,I forget the constant entire functions.

In fact $b_{n}$ can expressed by $a_{n}$ but also seems tredious. :blush:


We may say a process is adiabatic if it occurs fast enough such that no heat is exchanged from the system.But we also say a process is quasi-static when it occurs very slow.

Then how can PV^gamma = c hold for a fast occuring adiabatic process because a polytropic process should be quasistatic as wiki says http://en.wikipedia.org/wiki/Polytropic_process .

Also why is it true that only quasistatic process can be a polytropic process.?

I wanna say just briefly:it is slow enough to be considered that pressure and other parameters are the same everywhere in the thermal system and it is ,simultaneously, fast enough to be considered that no heat exchange could be occured.Inside the system heat exchange is very fast,because every kind of heat transfer takes place,but the heat transfer with outside mostly by conduction through some material.


How do you guys learn how to write pre-olympiad problems? Many of you say source: myself.  I'd like to come up with some new problems.  Does anyone have any tips?

Inspiration for my problems (though they may or may not be difficult enough for this forum) comes from randomly examining an interesting subject.  Try to think of a potential trick to solving a problem first, and go from there.  Also, you can use ideas you have seen in other places.

I don't know about Pre-Olympiad but I think just making the problems is same for all levels.

Even though Pre-Olympiad problems are harder to create, if you have solved many problems in that level, it shouldn't be too hard for you to create one. It doesn't mean that it's easy but not totally impossible neither.

I can't make Pre-Olympiad problems but I'm pretty sure I'll be able to do them in future when I have solved many harder problems. There isn't really technique way on how to make a problem. Just like problem solving, you create more problems and you become better problem maker.

And speaking of my own experience, it's true and sometimes it's even more fun than problem solving :lol:!

Sincerely,

Silverfalcon

Well... depends what you expect. Just a good problem is not something you seek for. Well, actually it is.

I sougth for days on my very first problem. Which was incredibly bad. In the beginning it was incredible work for no results. But that way you open your mind, and after weeks I got my first good problem, and since then I hardly ever put time in seeking... I just had the idea at once. And sometimes I have some quite nice ones :)

There are 2 important things in problem writing:
1) finding a problem
2) finding an elegant way to solve it

How to make a problem harder? often for pre-olympiad level it is sufficient to think of a good solution and write a problem around it. I don't know for olympiad level, I'll wait until some more problems of mine get olympiad-approved :)


analytic geometry
AMC
AIME
LaTeX
floor function
modular arithmetic

[quote="miyomiyo"]Bob lives at $(0, 0)$ and works at $(6, 6)$.  Every day, to get to work, he walks down $12$ blocks, each block either increasing his $x$-coordinate or his $y$-coordinate.  How many ways are there for him to get to work if the intersections $(2, 1)$ and $(2, 4)$ are closed off?

What number do you think this would be on an AIME?[/quote]
Unless I'm missing something, probably number 3-5.
Maybe I'm trivializing it though.

[hide="solution"]
The total number of ways from $(0,0)$ to $(6,6)$ is $\binom{12}{6}$
or
$\frac{12!}{6!6!}={11*7*3*2^{2}}$

The total number of ways that pass through $(2,1)$ is $\binom{3}{2}* \binom{9}{4}$
or
$\frac{3!}{2!1!}* \frac{9!}{4!5!}={7*3^{3}*2}$

Likewise, the total number of ways that pass through $(2,4)$ is
$\binom{6}{2}*\binom{6}{4}$

$\frac{6!}{2!4!}* \frac{6!}{2!4!}={3^{2}*5^{2}}$

The number of paths that pass through BOTH $(2,1)$ and $(2,4)$ is $\binom{3}{2}* \binom{6}{2}$
$\frac{3!}{2!1!}* \frac{6!}{2!4!}={3^{2}*5}$

by the inclusion exclusion principle, the paths that don't pass through those intersections are

${11*7*3*2^{2}}-{7*3^{3}*2}-{3^{2}*5^{2}}+{3^{2}*5}=$
$(7*3*2)(22-9)-(3^{2}*5)(5-1) =$
$(2*3)[ 7*13-2*3*5 ] =$
$(6)(91-30) = (6)(61) = 366$

please check the arithmetic. I could have made a mistake.
[/hide]

oh and how do you use combinatoric notation in latex?

EDIT:
thank you Hokkage

I did the same as chickendude's solution. Didn't do the arithmetic out though. Combinatorial form is \binom{n}{k} in LaTeX. For AIME difficulty, I'd say around a 4.

my solution was the same as chickendude's but i got 360 instead. Could someone check his arithmetic please?

[quote="Phelpedo"]Another way to do it: the number of ways to get to a certain intersection is the the sum of the number of ways to get to the intersections below it and to the left.[/quote]

This method is called "block walking", and is useful for smaller situations in which a lot of annoying casework would get in the way of a quick combinatorical solution.

Like:

Find the number of lattice paths of minimal distance from $(0,0)$ to $(8,8)$ if the squares with coordinates $( 2 \pm 1, 3 \pm 1)$ and $(6 \pm 1, 5 \pm 1)$ are blocked off due to a parade.

I'd put this as a 2 or a 3 because it is so easy to brute force by block-walking. If you want to make it later to force combinatorics use, make the numbers larger or something.

Yeah, but then it's not an AIME answer.  The biggest you can go that's under 1000 is (6, 7) which is easy to block-walk.

You can always do stuff like find $\lfloor x/1000 \rfloor$.

Yeah, I know....

How's this.  He works at (12, 12), (3, 3) and (5, 6) are closed.  Find the remainder when the number of ways is divided by 1000.

how could he work a (3,3) and go to work through (5,6)?

[i]Hmm.. what if:[/i]

A man works at the location $(7,7)$ and lives at the origin. He can only travel along the lattice grid. Each change by $\pm 1$ of either of the x- or y-coordinate takes the man $1$ minute to complete. If the number of ways in which he can reach his work, if he has at most $20$ minutes to get there, is $N$, find the remainder when $N$ is divided by $1000$.

[quote="The Zuton Force"][hide="Casewise"]
Classify directions as U, D, L, and R for up, down, left, and right, respectively.

He can either go directly (7 ups and 7 rights), or take 1, 2, or 3 detours, in the form of an additional LR pair or UD pair.

If he were to go:

1. Directly, $\frac{14!}{7! 7!}$

2. Add one pair (either LR or UD), $2 \frac{16!}{7! 8! 1!}$

3. Add two of the same pair (either LR or UD), $2 \frac{18!}{7! 9! 2!}$

4. Add one of each pair, $\frac{18!}{8! 8! 1! 1!}$

5. Add three of one pair (either LR or UD), $2 \frac{20!}{7! 10! 3!}$

6. Add two of one pair and one of the other (either two LR's or two UD's), $2 \frac{20!}{8! 9! 2! 1!}$

Total:

$\frac{14!}{7! 7!}+2 \frac{16!}{7! 8! 1!}+2 \frac{18!}{7! 9! 2!}+\frac{18!}{8! 8! 1! 1!}+2 \frac{20!}{7! 10! 3!}+2 \frac{20!}{8! 9! 2! 1!}$

$\equiv 432+920+640+220+440+400 \equiv \boxed{052 \; \pmod{1000}}$
[/hide][/quote]

Is there a way to solve the general problem without a load of casework? Since it would get pretty ugly if he had to travel $14$ units and had, say, two hours in which to get there.

Well, I mean, it can clearly be generalized into a summation form, but taking it modulo 1000 will be, if I'm not missing something blatantly obvious, the standing problem.

Like, say he had $120$ minutes to go. Even the case in which he takes the full 2 hours with 53 UD routes has $42906253386216936243966766644196440$ possibilities... and that's just one of the $\frac{(53)(54)}{2}=1431$ cases.

Hmm...  It seems like a good problem, but to be an AIME problem it would need a non-brute force solution.

[quote="The Zuton Force"][hide="Casewise"]
Classify directions as U, D, L, and R for up, down, left, and right, respectively.

He can either go directly (7 ups and 7 rights), or take 1, 2, or 3 detours, in the form of an additional LR pair or UD pair.

If he were to go:

1. Directly, $\frac{14!}{7! 7!}$

2. Add one pair (either LR or UD), $2 \frac{16!}{7! 8! 1!}$

3. Add two of the same pair (either LR or UD), $2 \frac{18!}{7! 9! 2!}$

4. Add one of each pair, $\frac{18!}{8! 8! 1! 1!}$

5. Add three of one pair (either LR or UD), $2 \frac{20!}{7! 10! 3!}$

6. Add two of one pair and one of the other (either two LR's or two UD's), $2 \frac{20!}{8! 9! 2! 1!}$

Total:

$\frac{14!}{7! 7!}+2 \frac{16!}{7! 8! 1!}+2 \frac{18!}{7! 9! 2!}+\frac{18!}{8! 8! 1! 1!}+2 \frac{20!}{7! 10! 3!}+2 \frac{20!}{8! 9! 2! 1!}$

$\equiv 432+920+640+220+440+400 \equiv \boxed{052 \; \pmod{1000}}$
[/hide][/quote]

I do not think that is correct becaue you have not accounted for the fact that you have repeated ALOT of paths (for example in some 16 minute paths, they would have already reached the office by the 14th minute so that path cannot be achieved but you hvae counted that into your answer) (for example for a 14 minute path you can have NNNNNNNEEEEEEE but for a 16 minute path you can have NNNNNNNEEEEEEEWE, well that path cannot be done because you would have already reached the office by the 14th minute but you still incorperated that into your 16 minute path counting).


Putnam inequalities function calculus derivative AMC

Prove for non-negative real numbers a1, a2, ..., an and b1, b2, ..., bn,

(a1a2...an)1/n + (b1b2...bn)1/n :le: ((a1 + b1)(a2 + b2)...(an + bn))1/n

The inequality clearly holds if one of a1 .... an or b1....bn  = 0 so suppose not.

Then we can divide everything by GM(a1 ... an) and set x_n = b_n/a_n and find that the original inequality is equivelent to

1 < = [(1 + x_1)(1 + x_2) ... (1 + x_n)]^(1/n) - (x_1*x_2 ... x_n)^(1/n).

This isn't rigorous but suppose x = x_2 = x_2 ... = x_n. Then considering RHS as a function of x_1 by differentiation we find that absolute minimum occurs at x_1 = x. Similarly for other variables. I haven't taken multivariable, so I don't know how to convert this into a tight argument.

Another way is to replace x_1, x_2 with x_1' and x_2' such that GM(x_k's) remains constant while GM(1 + x_k's) decreases. With a little bit of argument this implies that inequality must hold[hide][/hide]


function
real analysis
real analysis unsolved

Not enough information: $ f(x)\equal{}\frac{x^3}{2}\plus{}Cx$ is a solution for any constant $ C$.

how did you get this?please give the solution sir...

Let $ g(t)\equal{}f(t)/t$. In terms of $ g$ the equation simplifies to $ g(x\plus{}y)\minus{}g(x\minus{}y)\equal{}2xy$. Dividing by $ y$ and letting $ y\to 0$, get $ g'(x)\equal{}x$. Hence $ g(x)\equal{}x^2/2\plus{}C$.

[quote="mlok"]Let $ g(t) \equal{} f(t)/t$. In terms of $ g$ the equation simplifies to $ g(x \plus{} y) \minus{} g(x \minus{} y) \equal{} 2xy$. Dividing by $ y$ and letting $ y\to 0$, get $ g'(x) \equal{} x$. Hence $ g(x) \equal{} x^2/2 \plus{} C$.[/quote]
I is not continuous . But with this you can find out it .Please try again.

Oh I am sorry. This function is not  continuous.

What are you talking about?  mlok gave a (valid) family of continuous solutions to the given equation.  There's no claim that these are exhaustive, just that these are enough to demonstrate that all solutions need not pass through a given point $ (2, f(2))$.

This solution to find out :$ f(x)\equal{}\frac{x^3}{2}\plus{}Cx$ is not true.
Because from problem we don't know do there exist $ g'(x)$ or not.

Did you not read what I wrote?  The function $ f(x) \equal{} x^3/2 \plus{} cx$ [i]is[/i] a solution of the given equation for every constant $ c$.  (Plug it in and check.)  mlok made no claim that this family was exhaustive, because for the question mlok was answering (to compute the value $ f(2)$) it was sufficient to demonstrate this family.  (It shows that $ f(2)$ is not uniquely determined by the given equation.)

Telling someone that they didn't solve a question that wasn't asked and that they weren't trying to solve is not helpful.  If you actually were going to provide a complete solution, that would be interesting.

[quote="mlok"]Not enough information: $ f(x) \equal{} \frac {x^3}{2} \plus{} Cx$ is a solution for any constant $ C$.[/quote]
Oh sorry i think he wrote solution.
Here is my solution.
$ h(x) \equal{} f(x) \minus{} \frac {x^3}{2} \minus{} Cx$
Where $ C \equal{} f(1) \minus{} \frac {1}{2}$
So $ h(1) \equal{} 0$
Apply this to the equation:
$ (x \minus{} y)h(x \plus{} y) \equal{} (x \minus{} y)h(x \plus{} y)$
$ \Longrightarrow xh(y) \equal{} yh(x)$
Let $ x \equal{} 1$ then $ h(y)\equiv 0$
Imply that $ f(x) \equal{} \frac {x^3}{2} \plus{} Cx$ where $ C \equal{} f(1) \minus{} \frac {1}{2}$
The solution  only use property of this function not necessary exist $ g'(x)$.


function
calculus
derivative
limit
calculus computations

[quote="epkid08"]Let's say I have two functions, g(x) and f(x). How can I find and prove that g(x)~f(x) at a certain x=c? Would I find the derivative of both functions, set them equal to each other, then solve for x, or something like that?[/quote]

$ f(x)\underbrace{\sim}_{x\to{c}}g(x)\implies\lim_{x\to{c}}\frac{f(x)}{g(x)}\equal{}1$

[quote="Mathstud28"][quote="epkid08"]Let's say I have two functions, g(x) and f(x). How can I find and prove that g(x)~f(x) at a certain x=c? Would I find the derivative of both functions, set them equal to each other, then solve for x, or something like that?[/quote]

$ f(x)\underbrace{\sim}_{x\to{c}}g(x)\implies\lim_{x\to{c}}\frac {f(x)}{g(x)} \equal{} 1$[/quote]

Yes, but I would also like to prove that x=c is the only x that would make g(x)~f(x) true. What would I do to prove this?


calculus
integration
calculus computations

How about $ \int \frac{1}{x^{2008}\plus{}x}dx$


What is the last digit of $2005^{2005}$ ?

I think that it's $3$ but my prove is to comlicated to post it here, if you are interested i can e-mail you :) :) :) :D :D :D

YES E -MAIL ME

 :lol: [color=red]mathematicien@naseej.com[/color]

The last digit of $2005^{2005}$ is $3$??? Am I missing something? Or did I see the problem wrongly?

Last digit seen from which side¿
From left to right: it's obviously  $3$ and the first digit is $7$ then.

[quote="dblues"]:?:  :?:  :?:
The last digit IS 5!! and last digit by convention refers to the units digit, so there's no issue of reading from the left or right...[/quote]
I'm agree with you :?  ;)

I must agree with M@re and Rodolphe: after much tedious calculation and a little advanced numbe theory, I conclude the last digit is a 3. The fact that the last digit of 2005^k is a 5 for small values of k is just a fluke.

No, no.... Its certanly 3, i was solving that problem for two days, and then, when i was so close to give up it just came out to me..... 3

And i have to agree with MysticTerminator about small values for k in $2005^k$, it is only a fluke

hey, M@re(or MysticTerminator) , can you  post your solution plz? ;)

It is too difficult. It would take long time to type

Hey guys, I know that it's very interesting research subject here, but I was browsing the forum and I saw some other problems- it is true, a little bit easier- that wait for solutions, so why don't you leave a little bit this magnificient problem and try to concentrate  on those problems.

dear friends
explain your solutions  :mad: [size=150]ok ok ok [/size]

Solution is so complicated and long ... Are you sure you want it ?

It also contains a clever leap of logic which I initially failed to make.  :)

Unfortunately, the proof does not fit my margin; I cannot post it here.

Oh, sorry for my stupid mistake, its obviously $3$  :blush:  :roll:
But I was unable to solve the problem for $2006^{2006}$, any ideas¿

Ok, guys, sorry, but the topic is closed. There are too many interesting ideas on 1 m^2 of posts.


this is the problem, i couldn't solve:
given prime p, find all solutions iz Z: xyz=p(x+y+z).
thanx

we have to find all possible solutions $(x, y, z)$ to this equation, right ?

Please also consider this problem: http://www.mathlinks.ro/viewtopic.php?p=2406#2406  :)

Yes you have to find all triples (x, y, z).
Please, help me about this problem...


geometry
perimeter
inradius
incenter
Pythagorean Theorem
AMC

How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to 3 times their perimeters?

(A) 6
(B) 7
(C) 8
(D) 10
(E) 12

I posted this solution somewhere else, but since this is my favorite problem of the year, I am going to repost

$Area=semi perimeter*inradius$
which means r=6

Assume the side lengths of the legs are x and y, then the hypotenuse will be $(x-6)+(y-6)=x+y-12$

$x^{2}+y^{2}=(x+y-12)^{2}$
$x^{2}+y^{2}=x^{2}+y^{2}+144+2(xy-12x-12y)$
$0=144+2(xy-12x-12y)$
$0=72+xy-12x-12y$
$72=xy-12x-12y+144$
$72=(x-12)(y-12)$

let $x>y$
(x-8,y-8)=(72,1),(36,2),(24,3),(18,4),(12,6),(9,8)
which gives 6 possible triangles

Good one  :P . I didn't know the formula $Area = Semiperimeter \times Radius$.

[quote="Allan Z"]
Assume the side lengths of the legs are x and y, then the hypotenuse will be $(x-6)+(y-6)=x+y-12$
[/quote]

Can you explain that part? I get $c=xy/6-x-y$

The key is that if $r$ is the radius then there exist positive reals $p, q$ such that

$x = p+r$
$y = q+r$
$z = p+q$

These are the three side lengths that the altitudes from the incenter divide the sides into.  We conclude that

$z = x+y-2r = x+y-12$

Combining this result with the Pythagorean Theorem gets us a nice solution.

I put a,b, and c as sides of the right triangle with c as the hypotenuse

After squaring the two equations I had
I got
[hide]
a+b-2=ab/4
[/hide]
but I realized I wrote down the wrong equation
so I'm wrong
and
I don't feel like doing it again
so I think the others are right

I did if the area=perimeter
instead of 3area=perimeter

I was trying the test earlier today, and I have a semi-new solution.

[hide] At the start, we know that $ ab\equal{}6(a\plus{}b\plus{}c)$ and $ a^{2}\plus{}b^{2}\equal{}c^{2}$. Double the first and add it to the second equation to get \[ (a\plus{}b)^{2}\equal{}c^{2}\plus{}12(a\plus{}b\plus{}c)\] \[ (a\plus{}b)^{2}\minus{}c^{2}\equal{}12(a\plus{}b\plus{}c)\] \[ (a\plus{}b\minus{}c)(a\plus{}b\plus{}c)\equal{}12(a\plus{}b\plus{}c)\] Because a, b, and c are positive, we can cross out $ (a\plus{}b\plus{}c)$ from both sides getting $ a\plus{}b\minus{}12\equal{}c$. This can now be put directly into what Allen Z said to finish the problem.[/hide]


Albatross has an income of 15.51 dollars for every 7 hours he works. If so how much money does he make in a week.

P.S. Don't ask about the name Albatross it's a cool name  :P .

P.P.S If you are clearly above getting started I'm going to have to ask you NOT to answer unless people are cleary stumped and it has been taking a long time to solve. Because geez do you get self-satisyfaction in knowing you can solve problems intended for 12 year olds?  ;)

Does he have an 8 hour work day or a 24 hour work day?

[hide]lets see....
there is 7*24=168 hours in a week
168/7=24
15.51*24=372.24
so...... "Albatross" gets paid a measily $372.24 for his hard toiled 168 hours of work[/hide]

Does he still get paid if he falls asleep on the job :P ? I don't think anybody can stay awake for 168 hours :P .

The problem depends on where albatross lives. Assuming he lives in the gopher state and is employed by a large ($500,000 annual receipts or more) company, he gets higher wages after the first 48 hours.

Assuming he is employed by a small company, then Albatross' best move would be to sue the compnay for having an hourly wage less than 4.90. Same for the large company.

Or maybe Albatross isn't employed by a company, in which none of the above apply.


What positive integer value of $ x$ will satisfy the equation $ (x^2 \minus{} 6x \plus{} 9)^4 \equal{} 5^{16}$?

This equation seems quite scary with the huge exponents, so let's not solve right away yet and instead try to simplify a little.

In the left side,
$ (x^2\minus{}6x\plus{}9)^4$
$ \equal{}((x\minus{}3)^2)^4$ (since $ a^2\minus{}ab\plus{}b^2\equal{}(a\minus{}b)^2$)
$ \equal{}(x\minus{}3)^8$. (multiplying exponents according to power of power property)

Our exponent is $ 8$. Notice that that is half of the exponent on the right side, $ 16$!

We can rewrite the right side as $ 5^{16}\equal{}(5^2)^8$ due to the power of power property, and now we take the eighth root of both sides of the equation, eliminating the large exponent!

Now, it's just a case of simple solving-for-$ x$:
$ x\minus{}3\equal{}5^2$
$ x\minus{}3\equal{}25$
$ x\equal{}\fbox{28}$

[hide=Solution]We can start by taking the fourth root from $ (x^2 - 6x + 9)^4 = 5^{16}$, yielding with $ (x^2 - 6x + 9)= 5^{4}$. So, we get that $x^2 - 6x = 616$. So, $x(x-6)=616$. Prime factorizing $616$, we get $ x=\fbox{28}$ [/hide]

[quote=PenguinMoosey]you can say $(x^2-6x+9)^4 = (x-3)^8 = 5^{16}$, and this gives x-3 = 25, so x = 28[/quote]

Nice!, didn't think about it that way!

[quote=MWZ][quote=PenguinMoosey]you can say $(x^2-6x+9)^4 = (x-3)^8 = 5^{16}$, and this gives x-3 = 25, so x = 28[/quote]

that's really straightfoward! :D[/quote]

No, I didn't think of factoring $x^2-6x+9$ as a quadratic. It occured to me to do prime factorization.

[quote=firebolt360][quote=naren_pr][quote=dragoon]Wait can’t u take 4th root?[/quote]

Probably better to factor $(x^2-6x+9)^4$ as $(x-3)^8$.[/quote]

I mean:
$(x^2-6x+9)^4=5^{16} \implies (x^2-6x+9)=5^4 \implies (x-3)=\pm 5^2 \implies x\in \{28, -22\}$

Why is no one noticing the solution $-22$? Like I get positive int and all but...[/quote]

Everyone notices it but no one states it for the reason you gave :)

[quote=firebolt360][quote=naren_pr][quote=dragoon]Wait can’t u take 4th root?[/quote]

Probably better to factor $(x^2-6x+9)^4$ as $(x-3)^8$.[/quote]

I mean:
$(x^2-6x+9)^4=5^{16} \implies (x^2-6x+9)=5^4 \implies (x-3)=\pm 5^2 \implies x\in \{28, -22\}$

Why is no one noticing the solution $-22$? Like I get positive int and all but...[/quote]

That is how the question is worded. They said positive integer, so we have to go with the flow.


Find all polynomials $ P(x)$ that satisfy equation: $ P(x\plus{}1)\equal{}P(x)\plus{}2x\plus{}1$ for all reals $ x$.

we take $ Q(x)=P(x)-x^2$
we have $ Q(n+1)=P(n+1)-n^2-2n-1=P(x)-n^2=Q(n)$ so $ Q(x)=c$ for some $ c\in\mathbb{R}$
then $ \boxed{P(x)=x^2+c,\ c\in\mathbb{R}}$

[quote="ABRORBEK"][quote="aviateurpilot"]we take $ Q(x) = P(x) - x^2$
we have $ Q(n + 1) = P(n + 1) - n^2 - 2n - 1 = P(x) - n^2 = Q(n)$ so $ Q(x) = c$ for some $ c\in\mathbb{R}$
then $ \boxed{P(x) = x^2 + c,\ c\in\mathbb{R}}$[/quote]
what is Q(x+1)=Q(x) -> Q(x)=c[/quote]

$ Q(n)=Q(n-1)=Q(n-2)=...=Q(1)=Q(0)$
we take $ c=Q(0)$
then equation $ Q(x)=c$ have an infinity of solution then $ \forall x: \ Q(x)=c$ because $ Q$ is polynom

[quote="aviateurpilot"][quote="ABRORBEK"][quote="aviateurpilot"]we take $ Q(x) = P(x) - x^2$
we have $ Q(n + 1) = P(n + 1) - n^2 - 2n - 1 = P(x) - n^2 = Q(n)$ so $ Q(x) = c$ for some $ c\in\mathbb{R}$
then $ \boxed{P(x) = x^2 + c,\ c\in\mathbb{R}}$[/quote]
what is Q(x+1)=Q(x) -> Q(x)=c[/quote]

$ Q(n) = Q(n - 1) = Q(n - 2) = ... = Q(1) = Q(0)$
we take $ c = Q(0)$
then equation $ Q(x) = c$ have an infinity of solution then $ \forall x: \ Q(x) = c$ because $ Q$ is polynom[/quote]IT IS TRY $ n$ not integer  $ n$  real number

A polynomial equation can't have infinitely many solutions unless the polynomial is a constant. That is because an $ n$th degree polynomial has no more than $ n$ roots, and $ n$ is finite.

[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=211560]The method of finite differences[/url] settles the issue.  $ \Delta P(0) \equal{} 1, \Delta^2 P(0) \equal{} 2$ immediately gives $ P(x) \equal{} P(0) {x \choose 0} \plus{} {x \choose 1} \plus{} 2{x \choose 2} \equal{} P(0) \plus{} x^2$ for every integer $ x$, hence everywhere.