MIT college Harvard AMC USA(J)MO USAMO geometry

Hey! I'm a freshman now, and my goal is to make it to MIT or Harvard, etc. I was wondering if the following were good enough to make it to MIT:

4 or 5 AP Courses

IB DIPLOMA (Equivalent to 7 rigorous APs)

300+ Hours of Volunteering (About volunteering, does it matter where you volunteer or if you volunteered in many places, or rather in a couple places but consistently?)

A score of 2200+ on the SAT.

Maybe the President of a few clubs in my school.

If not, what can I do to increase my chances?

Thanks in advance.

--Percy--

It's definitely a good start... unless you're saying that's for overall high school?  :huh: In which case, unless you had amazing essays, I don't think you'd have much of a chance.

People who apply to Ivys are crazy. People with 2300+, national olympiads, science competition national finalists, often get rejected.

I agree with Pascal.

If that's what you're anticipating accomplishing through your four years of high school, you will definitely need something extra.

There's also the question of "Ivy League" as you mention in your title.
MIT is not an Ivy.  While Harvard is an Ivy, there is a wide variety to choose from.

You specifically asked for an opinion concerning MIT.
That's probably not enough, considering MIT probably gets those kinds of applications by the boatload.

I would suggest some research.  If you don't feel ready for research, then you could at least win/do well in some math competitions or enter some other competitions you could potentially do well in.
Volunteering is nice, but there's not much to show for it unless it ends up truly being a big part of your life.

However, it is good to start thinking along the lines of these things. (as a freshman, I was completely oblivious to college matters...)

You should also visit the MIT website if you're interested.
[url]http://www.mitadmissions.org/topics/before/menu.shtml[/url]
This page especially will help you find out what paths you can take to try to boost your chances.

I will repeat what I have said before.  Many applicants have those stats, those volunteer hours, and that kind of rigorous courseload.  However, it is up to you to figure out what else you want to do to help the MIT admissions staff get a feel for who you are.

My advice: Find a few things you like to do. Do them well. Maybe it's math; maybe it's physics; maybe it's writing. Getting into the top schools is tough, yes, but it's not like everyone who gets in is a USAMO winner or a Siemens finalist or whatever. Take the most challenging courses you can handle. There's no way to look at a small snapshot of what you plan to do and say "this will get you in" or not. Don't spend your entire time in high school obsessing over college admissions. Just do the best you can in the toughest classes you can, and you'll maximize your chances.

Hmph. Thanks for being honest.  :wink:  What are some things that can increase one's chances into getting into a good college?

Maybe like...

Publishing a novel?
Winning a math competition?
DECA winner?
Anything else? :)

All those things will increase your chances, but all those things are unreasonable for the vast majority of people, including the majority of those admitted to elite schools. Don't think about it as "how do I increase my chances"; think about it as "how can I challenge myself and what can I be passionate about?"

There are entire books written on how to structure your high school journey to maximize your chances for you dream college.
As worthawholebean said, you don't want to do things just for the sake of college. If you do this, you are merely torturing yourself and all for nothing, because colleges don't look for just accomplishments, but passion. If you haven't done so already, use freshmen year of high school to explore different areas of science, math, liberal arts, extracurriculars, sports, and find what you are interested in. About those suggestions, do whatever interests you. If you love writing, participate in writing competitions, publish some pieces of work. If you are into science/math, join any competitions related to these fields, maybe look into science/math related stuff to do during the summer.

Also, I would say that a definite must is to have one or two nonacademic extracurricular activities, and go really in depth with these. For example, pick up an instrument and join an orchestra, or tryout for varsity sports, or maybe try art. It may be too late to start in freshmen year and get some notable achievements, such as honor orchestras or varsity sports, but it's definitely a good idea to start NOW.(people who have notable achievements have usually started instruments since early elementary school, and sports usually in elementary/middle school). Why do an extracurricular? It shows that you are a person with actual interests, not just a machine who does work all day. Colleges want variety on their campuses.

It is definitely a competitive world out there.

[quote="worthawholebean"][quote="pascal12"]Also, I would say that a definite must is to have one or two nonacademic extracurricular activities, and go really in depth with these. For example, pick up an instrument and join an orchestra, or tryout for varsity sports, or maybe try art.[/quote]I don't think it's good to characterize anything as a "definite must." The only real advice is to find something you like to do and do it. Obviously, don't sit around playing video games all day, but don't spend all your time working, either. You'll burn yourself out in a year. I don't know what grade you guys are in, but having just finished the application process, I can say that you can make it a lot harder on yourself than it has to be, and many of my classmates did. Wherever you end up going, you'll find your own path to being admitted there. When you decide how to spend your time, if you're going to consider at all what colleges want to see, try as hard as you can to think in terms of "This will make me a more interesting and intellectually disciplined person" rather than "I have to do [i]something[/i] to get into XXXX." If you think the latter thought, you will choose wrong.[/quote]

Ok, you're right, perhaps I worded it too strongly. But I still suggest finding a nonacademic hobby or something that you enjoy. Unless you really enjoy studying and working all the time. My extracurriculars offer me a chance to get away from working and all the stress.

In addition...
Have fun.  Life does not begin (or end) with college admissions or even undergraduate studies.  There still grad school, work, life, etc.  High school might be the last time you can have fun, drop an activity, and bring another one in that you're serious about that you can get good at.  (I've heard college is fun, but I haven't witnessed it for myself.)

Having fun and being really enthusiastic about what you do also helps make it easy to write about or tell about what you like to do.

Don't be afraid to dabble a little bit.  Just don't take too long, or else your high school will be gone before you know it.

In addition to what the other people are saying, do what comes naturally do you. You should consider doing an extracurricular; you should consider holding some sort of leadership position - but don't do it for college. You should be doing something because it interests you, not necessarily because you consider it prestigious for college.

Keep your grades up, and try studying for the AMC chain of contests. MIT is incredibly competitive. Do your best to stand out.

[quote="Lazarus"] MIT is incredibly competitive. [/quote]
I'm sure you would know (late congratulations from me  :) ).

More relevantly, college admissions officers are good at distinguishing between what applicants do out of their own interests and what they do simply to get into a good college, and they will prefer the former. As everyone as said on this thread already, it is better to find and engage in your own interests then to worry about how college admission officers may view them.


Let $ n$ be a composite. Prove that there exists positive integer $ m$ satisfying $ m|n, m\le\sqrt {n},$ and $ d(n)\le d^3(m).$ Where $ d(k)$ denotes the number of positive divisors of positive integer $ k.$

we can write $ n \equal{}\prod_{k\equal{}1}^{g}q_k \prod_{i \equal{} 1}^{h}p_i^{a_i}$
i)$ p_i,q_k$ are primes
ii) $ a_i>1$
iii) $ q_1<q_2<...<q_g$

we take $ d \equal{} \prod_{i \equal{} 1}^{h}p_i^{b_i}$ where $ b_i \equal{} [a_i/2]: \ (2b_i\le a_i)$
if $ g\ge 2$ we take $ d'\equal{}\prod_{k\equal{}1}^{[q/2]}q_k$
and $ m\equal{}dd'\le \sqrt{n}$ and $ d^3(m) \equal{}2^{3[g/2]} \prod_{i \equal{} 1}^{h}(b_i \plus{} 1)^3\ge 2^g\prod_{i \equal{} 1}^{h}(a_i \plus{} 1) \equal{} d(n)$


function
algebra
domain
induction
algebra unsolved

Well, i hope this problem wasn´t posted yet, but, whatever!
Determine all functions from Q->Q such that:

$f(x+f(y))=f(x)f(y)$

Letting $x = 0$ we determine that $f(f(y)) = f(0) f(y)$, so that for any $f(y) = r$ in the range of our function we have $f(r) = f(0) r$.

(Edited: the rest is troublesome...)

well, i had also arrived in your solution but in a different way. I wasn´t able to get a different solution, since i think that constant ones aren´t the only solutions of the problem. To avoid the problem with $f(0)$, i just excluded 0 from the domain but got stucked in my induction.
Another thing. $f(x)=1$ is clearly another solution to this equation and is very easy to see that.

[quote="t0rajir0u"]$f(a+f(b)) = f(a+kb) = ka+k^{2}b$[/quote]

Oh.  I was wondering why my solution was so strange.  This step isn't valid because $a+kb$ is not necessarily part of the range.   :oops:

(*)  f(x+f(y))=f(x)f(y)

If f(y)=0 for some y, then f(x)=f(x+f(y))=f(x)f(y)=0 for all x. So we get f=0, which is indeed a solution of (*).

So assume now f(y) is not 0 for any y.
Then f(y)=f(y-f(y)+f(y))=f(y-f(y))f(y) and so f(y-f(y))=1 for any y.
So we get some y with f(y)=1.
And then f(x+1)=f(x+f(y))=f(x)f(y)=f(x) for all x.
And so we get f(x+y)=f(x) for all x,y with y integer.
Now for all x,y we have:
f(x+2f(y))=f(x+f(y))f(y)=f(x)*f(y)^2
f(x+3f(y))=f(x+2f(y))f(y)=f(x)*f(y)^3
...
f(x+nf(y))=f(x)*f(y)^n  for any natural number n.

but nf(y) is an integer for some n, since f(y) is rational,
and then f(x)=f(x+nf(y))=f(x)*f(y)^n.
So f(y)^n=1 and f(y)=1 or -1.
But then we can take above n=1 and get f(y)=f(y)^n=1 for any y.
So f=1, which is also a solution of (*).

i used the same induction that olorin has used and arrived also in the same result. Seems to be hopeless to find some other more elaborated functions as answer  :mad:


prove that a square has the smallest perimeter of all quadrilaterals circumscribed about a given circle of

radius $r$.

The theorem is
Theorem. Let R be a positive number. Among all cyclic quadrilaterals with circumradius R, the square has the greatest perimeter.

I would like to see a proof though, can anybody send his solution please?

Posted before by my ad solved by Darij and Ph-Ah.
Darij solution was
[quote="darij grinberg"]

[color=blue][b]Theorem.[/b] Let R be a positive number. Among all cyclic quadrilaterals with circumradius R, the square has the greatest perimeter.[/color]

In this case, the [i]proof[/i] is easy: Let ABCD be a cyclic quadrilateral with circumradius R. Let O be the circumcenter of the quadrilateral ABCD, and let $\alpha=\frac12\measuredangle AOB$, $\beta=\frac12\measuredangle BOC$, $\gamma=\frac12\measuredangle COD$ and $\delta=\frac12\measuredangle DOA$ be the chordal angles of the sides AB, BC, CD and DA, respectively, in the circumcircle of the quadrilateral ABCD. Then, since the length of a chord in a circle equals $2\cdot\text{ radius of the circle }\cdot\text{ sine of the chordal angle of the chord}$, we have $AB=2R\sin\alpha$, $BC=2R\sin\beta$, $CD=2R\sin\gamma$ and $DA=2R\sin\delta$. On the other hand,

$\alpha+\beta+\gamma+\delta=\frac12\measuredangle AOB+\frac12\measuredangle BOC+\frac12\measuredangle COD+\frac12\measuredangle DOA$
$=\frac12\cdot\left(\measuredangle AOB+\measuredangle BOC+\measuredangle COD+\measuredangle DOA\right)=\frac12\cdot 360^{\circ}=180^{\circ}$.

As the angles $\alpha$, $\beta$, $\gamma$ and $\delta$ all lie in the interval [0°; 180°], and the function $f\left(x\right)=\sin x$ is strictly concave in this interval, the Jensen inequality yields

$\sin\alpha+\sin\beta+\sin\gamma+\sin\delta\leq 4\sin\frac{\alpha+\beta+\gamma+\delta}{4}=4\sin\frac{180^{\circ}}{4}=4\sin 45^{\circ}$,

with equality if and only if $\alpha=\beta=\gamma=\delta$, what is equivalent to saying that the quadrilateral ABCD is a square (in fact, $\alpha$, $\beta$, $\gamma$ and $\delta$ are the chordal angles of the sides AB, BC, CD and DA of the cyclic quadrilateral ABCD, and if these chordal angles are equal, the sides are also equal, but a cyclic quadrilateral with equal sides is a square). Hence, we get the following inequality for the perimeter AB + BC + CD + DA of the quadrilateral ABCD:

$AB+BC+CD+DA=2R\sin\alpha+2R\sin\beta+2R\sin\gamma+2R\sin\delta$
$=2R\cdot\left(\sin\alpha+\sin\beta+\sin\gamma+\sin\delta\right)\leq 2R\cdot4\sin 45^{\circ}$,

with equality if and only if the quadrilateral ABCD is a square. Thus, the perimeter of the quadrilateral ABCD is maximal if the quadrilateral ABCD is a square. This proves Theorem 1.

  Darij[/quote]


vector
calculus
inequalities unsolved
inequalities

Doable with Lagrange Multipliers, but the calculations would be unbelievably tedious.  :(

Hey, man, I wonder if Lagrange Multipliers method is allowed in high school.

That expression looks quite formidable  :|

No, Lagrange Multipliers are a method in vector calculus to find the extrema of some $f( \vec{v} )$ given the condition that $g( \vec{v} ) = k$.

(Edit:  Actually, this wouldn't work because $x_k$ are all positive, and the method would not return an all-positive vector.  I think.)


symmetry function quadratics conics parabola Support inequalities

[hide="Proof"]We have $f(x)-x=a(x-z)(x-\bar{z})$ where $z=u+iv, u,v\in\mathbb{R}, v\neq 0$. Hence $f(x)-x=a((x-u)^2+v^2)$. It follows that $f(f(x))-f(x)=a((f(x)-u)^2+v^2)$, giving $f(f(x))-x=(f(f(x))-f(x))+(f(x)-x)=a((f(x)-u)^2+(x-u)^2+2v^2)$.

Therefore, $v\neq 0\Rightarrow {{f(f(x))-x}\over a}\geq 2v^2>0$, hence $f(f(x))-x$ has no real zeros.

Generalization for $f^n(x)=x, n\in\mathbb{Z}^{+}$ is quite obvious. [/hide]

[hide="another proof"]

Since $f(x)=x$ has no solution it can only be $f(x)>x$ or $f(x)<x$ $\forall x$ . We only need to consider $f(x)>x$ since $f(x)<x$ is just the symmetry of what we prove for $f(x)>x$

Since we have $f(x)>x$ then for $f(x)=ax^2+bx+c$ , $a$ must be positive or else there will be solution . But over $[-\frac{b}{2a},\infty)$ , $f(x)$ is increasing so $f(x)>x$ $\implies f(f(x))>f(x)>x$ . For $(-\infty,\frac{-b}{2a}]$,every point of $f(x)$ can be reflected to those in the interval $[\frac{-b}{2a},\infty)$ and this again shows that $f(f(x))>x$ over $(-\infty,\frac{-b}{2a}]$ . Hence there wont be any solution for $f(f(x))=x$
[/hide]

[quote="peeta"]@shyong:
why cant this be simplyfied to: we consider $\forall x\in \mathbb R : f(x)>x$ then since $f(x)\in\mathbb R \implies f(f(x)) > f(x) \stackrel{f(x)>x}{\implies} f(f(x)) > x$. analog we can consider $\forall x\in \mathbb R : f(x)<x$ and this results the proof ?[/quote]

If $f(x)$ is decreasing , then $f(x)>x$ $\implies f(f(x))>f(x)$ will be false . So I think we must first show that $f(x)$ is increasing over some interval .

[quote="shobber"][quote="shyong"]If $f(x)$ is decreasing , then $f(x)>x$ $\implies f(f(x))>f(x)$ will be false [/quote]
Why?[/quote]

isnt it that if $a>b$ then if $f(x)$ is decreasing then $f(a)<f(b)$ ?

EDITTED :Farenhajt, I didnt see that we post the same thing in same time  :rotfl:

[quote="Farenhajt"][quote="shobber"]Why?[/quote]

Because of the definition of a decreasing function: $a>b\implies f(a)<f(b)$. Now put $f(x)$ instead of $a$ and $x$ instead of $b$.[/quote]

but this is strange i think ...

say for a quadratic function $f$ is true that \[ \forall x\in \mathbb R: f(x)>x \]
and there is such a function $f$ if we consider the graph we can see that (e.g. $f(x)= x^2 +10$ )
a then $\forall x\in\mathbb R: f(x) \in \mathbb R$. therefore if we plug in $f(x)$ for $x$ we get \[ \forall x\in \mathbb R: f(f(x)) > f(x) \]

but on the other hand if we only consider the decreasing part of $f$ so let \[ \noindent f^*: A \to \mathbb R, \text{ with } A: =\{x\in\mathbb R: f'(x) \le0 \text{ (that is decreasing) } \},\\ \forall x\in A: f(x) = f^*(x) \]
then $\forall x\in A: f^*(x)>x$ ... (all that what u wrote, which i omit, since i just got the answer why your guys' thinking is wrong (i think ;) ))

u pretend that $f^*(x) \in A$, but exactly this is not true, if $\forall x \in \mathbb R: f(x)>x$

let $x_0 : = \frac{b}{2a}$, so that $A = (\infty, x_0]$, then $\forall x < x_0: f^*(x) > f^*(x_0) > x_0$ (because $f^*$ is decreasing on $[x,x_0]$). So $f^*(x) \not\in A$, therefore your argument is invalid :P

now that i am pretty sure it is another solution, which i had i'll write it down correctly:
[hide]
let $f$ be quadratic function, so that \[ \forall x\in\mathbb R: f(x) \neq x \]
since $f$ is continues on $\mathbb R$ this is equivalent to \[ \forall x\in\mathbb R: (f(x) > x)\vee (f(x) < x) \]
let us first consider $\forall x\in\mathbb R: f(x) > x$.
since $\forall x\in \mathbb R: f(x) \in \mathbb R$ we get that \[ \forall x\in \mathbb R: f(f(x))> f(x) >x \]
therefore this case doesnt have any real roots.
analog we can proof this for the second case, and were done :)
[/hide]

[quote="peeta"]\[ \forall x\in\mathbb R: (f(x) > x)\vee (f(x) < x)  \]
let us first consider $\forall x\in\mathbb R: f(x) > x$.
[/quote]

No, your solution is wrong, since you erroneously assume that $f(x)$ is either increasing or decreasing on the whole $\mathbb{R}$, but that's not the case (quantifier algebra: $(\forall z)(P\lor Q)\nRightarrow \lbrack(\forall z)P\rbrack\lor\lbrack(\forall z)Q\rbrack$. Shyong's proof is OK, read it carefully - especially these parts:

[quote="shyong"]...But over $[-\frac{b}{2a},\infty)$ , $f(x)$ is increasing...
...For $(-\infty,\frac{-b}{2a}]$,every point of $f(x)$ can be reflected to those in the interval $[\frac{-b}{2a},\infty)$[/quote]

He splits the proof into two parts - where $f(x)$ is increasing and where it's decreasing, and that's the only way if you want to prove this via increasing/decreasing argument.

I think maybe I know what peeta want to say .

For the first case , since $f(x)>x$ ....(i) for all real $x$ . This means that the range of which $x$ can take to satisfy (i) is all real number .

Then for $f(f(x))$ , since $fx)$ have to be U-shape parabola ($a>0$) ,the range for $f(x)$ is $[c-\frac{b^2}{2a},\infty)$ . But $f(x)>x$ holds for all real $x$ . This means that $f(f(x))>f(x)$ should also hold since the range of $f(x)\in\mathbb{R}$ . So we say that $f(f(x))>f(x)>x$ .

Is this what you are trying to say , peeta ?

[quote="Farenhajt"]Oh, I see - he argues that the parabola is either completely above or completely below the line $y=x$. Cool.[/quote]
excactly that my argument :)

did it come across that bad?

[quote="Kalle"]Indeed, peeta's argument can be used for any continuous function $f$ where $R_f \subseteq D_f$ and $f(x) = x$ has no solution.[/quote]
thanks for your support :)

[quote="Kalle"]Indeed, peeta's argument can be used for any continuous function $f$ where $R_f \subseteq D_f$ and $f(x) = x$ has no solution.[/quote]

But isn't $R_f \subseteq D_f$ a necessary condition for $f(f(x))$ to exist? And if it's not fulfilled, the question is pointless, hence we can consider it implicit.

Cool, we finally sorted that line of argument out. But isn't my proof shorter? :)

[quote="Farenhajt"]Cool, we finally sorted that line of argument out. But isn't my proof shorter? :)[/quote]
it might be shorter in lines, but i think mine is shorter in terms of thoughts. actually ours are quite similar, just that i dont use complex numbers ;)

i agree that its cool to lay the discussion down ;) .. i thought i had explained why ur argument was invalid, but i guess i didnt quite get it across ;)

My argument invalid? Which one? (I really haven't seen you ever talked about MY proof)

peeta's kind of argument is often useful. Try using it on the following problem I had on an exam once:
Let $f_1(x) = \ln(1+ x)$ for $x > 0$ and define recursively the functions $f_2, f_3, ...$ by $f_n(x) = \ln(1+f_{n-1}(x))$. Show that
\[ x - \frac{nx^2}{2} < f_n(x) < x \quad \text{for} \quad n=1,2,3,... \quad \text{and} \quad x> 0 \]

first let $f(x) = f_1(x)$ and $f^n(x) = f_n(x)$ (the $n$th itaration of $f$, which is obviously the same )
[hide]
let us do both inequalities by induction:

1. $f^n(x) < x$ (the following proof is actually quite similar to the one above, but i use a diffrent technique to finish ;) )

first we need to prove that $f(x) < x\quad \forall x>0$:

since $\displaystyle \lim_{x\to 0 } f(x) = 0$ (we do it that complicated, because $f$ is unfortuanatly is only defined on $(0,\infty)$ )

now $f'(x) = \frac{1}{x+1} < 1 = \frac{d}{dx}\ x\quad \forall x>0$, so at the beginning (at $0$) $f(x)$ and $x$ are equal, but $f$ increases faster.

now we have to conclude $k\to k+1$ \[ f^{k+1}(x) = \ln ( f^k(x)+1) < f^n(x) < x %Error. "qedsymbol" is a bad command.
\]

2. $f^n(x) > x-\frac{nx^2}{2}$ (here we use the same argument of $f'$ )

first $f(x) > x-\frac{x^2}{2}$: \[ \lim_{x\to 0 } f(x) = 0 \] \[ f'(x) = \frac{1}{x+1} > 1-x = \frac{d}{dx} \left(x-\frac{x^2}{2} \right) \iff 1> 1-x^2 \quad\forall x>0 \]

second $k\to k+1$: \[ f^{n+1}(x) = f^n(f(x)) > f(x) -\frac{nf(x)^2}2 > x- \frac{(n+1)x}2 \iff \frac1{x+1}-\frac{nf(x)}{x+1} > 1- (n+1)x \iff 1-nf(x) > 1-nx \iff f(x) < x \quad \forall x>0 \][/hide]

and were done =)

Yes, I think that is a well-constructed and nice problem, although I guess there is really only one solution. A small remark that isn't of importance: You don't need to take the limit of the function as x approaches 0. If an inequality is true for the "extended function" over $[0, \infty)$ the inequality is of course also true over $(0,\infty)$ since $(0,\infty) \subset [0,\infty)$

[quote="Kalle"]I guess there is really only one solution[/quote]
i acually dont really like the starting proof of that ineq ... but i cant imagen a proof without calculus :( , maybe with argumentation, that substituting calculus, but thats not nice either ...

[quote="shobber"]$f(x)=ax^2+bx+c$. Prove that if $f(x)=x$ has no real roots, then $f \left( f(x) \right) =x$ doesn't have no real roots neither.[/quote]

Here's a more general statement (and in the contrapositive): Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function, and let $f_n$ denote the $n$th iterate of $f$.  Show that if $f_n$ has a fixed point for some positive integer $n$, then $f$ has a fixed point.

(This is a specific case of [url=http://mathworld.wolfram.com/SharkovskysTheorem.html]Sharkovsky's Theorem[/url])


number theory
prime numbers
number theory unsolved

Find all prime numbers $p$ for which there exist numbers $x,y\in \mathbb{Z}$ satysfing symstem of equations:
$p+1=2x^2$, $p^2+1=2y^2$.

this appeared in at least 2 competitions.
Here is my solution
Subtract the first condition from the second one, then
\[
p(p - 1) = 2(y - x)(y + x)
\]
If $p|y-x$, $p \le y-x$ and $p-1 \ge 2(y+x)$. Then we have $y-x>2(y+x)$ or $0>y+3x$ which is not true.
Therefore $p|y+x$
So we have $y+x \ge p$ and $p-1 \ge 2(y-x)$. Therefore $y+x \ge p>p-1 \ge 2(y-x) => 3x>y$
If $p \ge 10$, $(p-10)(p+1) \ge 0$ or $p^2-9p-10 \ge 0$
$p^2+1 \ge 9p+11>9p+9$
$2y^2>9(2x^2)$
$y>3x$, which contradicts with $3x>y$.
Therefore $p<10$ or $p=2,3,5,7$. By simple checking, p=7 is the only solution.

I think it is from Germany 1996 if i am not wrong.


calculus
derivative
analytic geometry
graphing lines
slope
function
real analysis

Find the equations of all lines that pass through the origin and are tangent to the curve of equation $ y\equal{}x^3\minus{}9x^2\minus{}16$

Take the derivative:
$ f'(x)\equal{}3x^{2}\minus{}18x\equal{}m$

Then take the equation of a line and take the origin as a point:
$ y\minus{}y_{0}\equal{}m(x\minus{}x_{0})\Rightarrow y\equal{}mx$

Now plug in the slope.  To avoid confusion, I will substitute $ a$ for $ x$ in the equation for $ m$:

$ y\equal{}(3a^{2}\minus{}18a)x$

Ok. If I take some a and find the derivative at this point I'm gonna find the slope of the line. But I can't assure that a tangent to this point passes through the origin. The line would be paralell to the other that passes through the origin. If you do the graphic of this function you'll notice this. And the problem asks to find all lines that pass through the origin, not paralell.

The value of the tangent at point $ x\equal{}a$ must be the same as the value of the function at the same point. Or:

$ (3a^2\minus{}18a)a\equal{}a^3\minus{}9a^2\minus{}16$

$ 3a^3\minus{}18a^2\equal{}a^3\minus{}9a^2\minus{}16$

$ 2a^3\minus{}9a^2\plus{}16\equal{}0$

The solutions are

$ a_1 \equal{} 4$

$ a_2 \equal{} \frac{1}{4}(1\minus{}\sqrt{33})$

$ a_3 \equal{} \frac{1}{4}(1\plus{}\sqrt{33})$

and these are only good values of $ a$.

It's easier to do the prep-work in the general case: we have a curve described by $ y \equal{} f(x)$.  We want to find all values of $ w$ such that the tangent to the curve at $ (w, f(w))$ passes through $ (a, b)$.  Thus, the slope $ f'(w)$ must be equal to $ \frac {f(w) \minus{} b}{w \minus{} a}$.  In our case, $ (a, b) \equal{} (0, 0)$, so we can write this as $ wf'(w) \equal{} f(w)$.  In our case we have $ f(x) \equal{} x^3 \minus{} 9x^2 \minus{} 16$, and we solve from there just as milin did.

A related question: find all continuously differentiable $ f$ such that [i]every[/i] tangent to the curve $ y \equal{} f(x)$ passes through the origin.

Another related question: find all points on the original curve such that the [i]normal[/i] to the curve passes through the origin.

From the equation of the tangent we have:

$ f(x) \equal{} f'(x) x$

and this gives us

$ f(x) \equal{} C x$

As for the normal:

$ f'(x) f(x) \equal{} \minus{} x$

$ f^2(x) \equal{} C \minus{} x^2$

so

$ f_1(x) \equal{} \sqrt {C \minus{}  x^2}$

$ f_2(x) \equal{} \minus{} \sqrt {C \minus{} x^2}$

And here's a question for you: find all functions such that the surface of the triangle that is enclosed by the tangent to the function in any point and the axes is constant.

So there are three tangents to the curve that pass through the origin. I didn't understand why you match the equations $ y\equal{}(3a^2\minus{}18a)a \equal{} a^3\minus{}9a^2\minus{}16$ at point a. How do you assure that the solutions of this equation would give the only values of a that satisfy the problem?

Because, as you've said, the solution provided by JRav gives lines parallel to your function, but parallel according to the derivation of your function at point $ x \equal{} a$. So, if you put a condition that the line must intersect your function at the same point where you took the derivation - you'll get your solutions. Or, here's another explanation: you need your tangent to be of the form $ y \equal{} k x$ but the general form of the tangent is $ y \equal{} f'(x_0)(x \minus{} x_0) \plus{} f(x_0)$. As you can see, for the general form to become your form you need to have $ \minus{} x_0 f'(x_0) \plus{} f(x_0) \equal{} 0$ and that exactly what I've done! :)


Let $ x,y,z>0$, s.t. $ xy\plus{}yz\plus{}zx\plus{}2xyz\equal{}1$. Suppose that $ z\equal{}\max\{x,y,z\}$. Prove that:

$ \dfrac{1}{x}\plus{}\dfrac{1}{y}\plus{}\dfrac{1}{z}\minus{}4(x\plus{}y\plus{}z)\geq \dfrac{(2z\minus{}1)^2}{z(2z\plus{}1)}$

We can assume
$ x\equal{}\frac{a}{b\plus{}c}$ etc... but I'm not sure if it will help.

$ F(x,y,z) \equal{} \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \minus{} 4(x \plus{} y \plus{} z)$ . WLOG we can assume $ x\leq y\leq z$
$ F(x,y,z)\geq F(\frac {x \plus{} y}{2},\frac {x \plus{} y}{2},z) \Leftrightarrow \frac {1}{x} \plus{} \frac {1}{y}\geq\frac {4}{x \plus{} y}$.

Edit: (it's just a fact. nothing more :oops: ).

i don't think that's a valid solution because $ \left(\frac{x\plus{}y}{2},\frac{x\plus{}y}{2},z\right)$ does not necessarily satisfies the initial condition..

:huh: oh yeah.

however, we should prove
$ \frac {1}{x} \plus{} \frac {1}{y} \minus{} 4(x \plus{} y)\geq 2(2z \plus{} 1) \minus{} 4\frac {2}{2z \plus{} 1} \equal{} \dfrac{(2z \minus{} 1)^2}{z(2z \plus{} 1)} \plus{} 4z \minus{} \frac {1}{z}$

Let $ xy \equal{} k^2$ .

we have  $ xy \plus{} yz \plus{} zx \plus{} 2xyz \equal{} 1 \Leftrightarrow (x \plus{} y)z \plus{} xy(1 \plus{} 2z) \equal{} 1$
Firstly, when we see $ 1\geq 2z\sqrt {xy} \plus{} (1 \plus{} 2z)xy$, we'll find that $ k\leq\frac {1}{2z \plus{} 1}$,
then, if we donate $ x\equal{} kx'$ and $ y \equal{} ky'$, for the new $ x'$ and $ y'$ we will have $ kz(x' \plus{} y') \plus{} (2z \plus{} 1)k^2 \equal{} 1$

$ \Rightarrow x' \plus{} y' \equal{} \frac {1 \minus{} k^2(2z \plus{} 1)}{kz}$ and in addition $ x' \plus{} y' \equal{} \frac {1}{x'} \plus{} \frac {1}{y'}$

$ \Rightarrow \frac {1}{x} \plus{} \frac {1}{y} \minus{} 4(x \plus{} y) \equal{} (\frac {1}{k} \minus{} 4k)(x' \plus{} y') \equal{} (\frac {1}{k} \minus{} 4k)\frac {1 \minus{} k^2(2z \plus{} 1)}{kz}\geq$
$ \geq (2z \plus{} 1 \minus{} 4\frac {1}{2z \plus{} 1})\frac {1 \minus{} \frac {1}{2z \plus{} 1}}{\frac {z}{2z \plus{} 1}} \equal{} 2(2z \plus{} 1) \minus{} 4\frac {2}{2z \plus{} 1}$

[quote="Inequalities Master"]No one?
arqady, can_hang2007, where are you?[/quote]
From the given condition, we have $ z \ge \frac {1}{2}$ and
\[ xy(1 \plus{} 2z) \equal{} 1 \minus{} z(x \plus{} y)
\]
Thus $ z(x \plus{} y) \le 1$ and
\[ (x \plus{} y)^2(1 \plus{} 2z) \ge 4 \minus{} 4z(x \plus{} y)
\]

\[ \Rightarrow x \plus{} y \ge \frac {2}{2z \plus{} 1}
\]
Now let $ t \equal{} x \plus{} y \ge \frac {2}{2z \plus{} 1}$, we have
\[ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \minus{} 4(x \plus{} y \plus{} z) \equal{} \frac {(x \plus{} y)(1 \plus{} 2z)}{1 \minus{} z(x \plus{} y)} \minus{} 4(x \plus{} y) \plus{} \frac {1}{z} \minus{} 4z \equal{} \frac {(1 \plus{} 2z)t}{1 \minus{} tz} \minus{} 4t \plus{} \frac {1}{z} \minus{} 4z \equal{} f(t)
\]
We have
\[ f'(t) \equal{} \frac {1 \plus{} 2z}{(1 \minus{} tz)^2} \minus{} 4 \ge \frac {1 \plus{} 2z}{\left( 1 \minus{} \frac {2z}{2z \plus{} 1}\right)^2} \minus{} 4 \equal{} (1 \plus{} 2z)^3 \minus{} 4 \ge (1 \plus{} 1)^3 \minus{} 4 > 0
\]
Hence $ f(t)$ is increasing for all $ \frac {1}{z} \ge t \ge \frac {2}{2z \plus{} 1}$, and we obtain
\[ f(t) \ge f\left( \frac {2}{2z \plus{} 1}\right) \equal{} \frac {(2z \minus{} 1)^2}{z(2z \plus{} 1)}
\]
We have done! :)
[quote="BG Yoda"]
then, if we donate $ x' \equal{} kx$ and $ y' \equal{} ky$, for the new $ x$ and $ y$ we will have $ kz(x \plus{} y) \plus{} (2z \plus{} 1)k^2 \equal{} 1$

$ \Rightarrow x \plus{} y \equal{} \frac {1 \minus{} k^2(2z \plus{} 1)}{kz}$ and in addition $ x \plus{} y \equal{} \frac {1}{x} \plus{} \frac {1}{y}$

$ \Rightarrow \frac {1}{x'} \plus{} \frac {1}{y'} \minus{} 4(x' \plus{} y') \equal{} (\frac {1}{k} \minus{} 4k)(x \plus{} y) \equal{} (\frac {1}{k} \minus{} 4k)\frac {1 \minus{} k^2(2z \plus{} 1)}{kz}\geq$
$ \geq (2z \plus{} 1 \minus{} 4\frac {1}{2z \plus{} 1})\frac {1 \minus{} \frac {1}{2z \plus{} 1}}{\frac {z}{2z \plus{} 1}} \equal{} 2(2z \plus{} 1) \minus{} 4\frac {2}{2z \plus{} 1}$[/quote]
Sorry, but I dont understand your solution here?

[quote="BG Yoda"]$ x'$ and $ y'$ are the initial $ x$ and $ y$ and the new $ x$ and $ y$ are such that $ xk \equal{} x'$ and $ yk \equal{} y'$ where $ k^2 \equal{} x'y'$ (the initial product of $ x$ and $ y$).
The idea is to get $ xy \equal{} 1$ because then $ x \plus{} y \equal{} \frac {1}{x} \plus{} \frac {1}{y}$.

I've already fixed it :)[/quote]
I checked it and it's okay now, friend. Nice proof :)


function
limit
real analysis
real analysis unsolved

Let $ \alpha \in \left ( 0,\frac{1}{2} \right )$ and the function $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that there exist: $ \lim_{x\to\infty} (f(x\plus{}1)\minus{}\alpha f(x))\equal{}0$. Prove that: $ \lim_{x\to\infty} f(x)\equal{}0$.

A concrete example: $ f(x)\equal{}\frac1{2^n}\cdot\frac1{x\minus{}n}$ on $ (n,n\plus{}1]$.

but then how do we take x$ \to infinity$. cant we say as $ x \to infinity$, $ f(x \plus{} 1) \to f(x)$?

[quote="jmerry"]A concrete example: $ f(x) \equal{} \frac1{2^n}\cdot\frac1{x \minus{} n}$ on $ (n,n \plus{} 1]$.[/quote]
Do you mean $ f(x) \equal{} \alpha ^n\cdot\frac1{x \minus{} n}$ on $ (n,n \plus{} 1]$?

Yes. I just thought $ \alpha$ was $ \frac12$ there.


geometry
trigonometry
ratio
similar triangles
trig identities
Law of Sines
Law of Cosines

[quote="Calculus"]I am working through challenging problems in geometry. I have a habit of spending way to much time on one problem or giving up way to easily. About how much time should I allow for each problem?[/quote]

Spend as long as it takes you to solve it. If you don't solve it, it doesn't matter how long it took. Geometry is probably one of the subjects for which this is very true. Sometimes you'll realize that the problem is really beyond your abilities, but just by playing around with it, experimenting possible strategies, is good practice. Of course my geometry is pretty weak, so maybe someone else can offer some deeper insights specific to geometry.

well...my tips are:

1) try to use related formulas.  If they give you a lot of angles, use law of sines, angle chasing (similar triangles, extending same arc, etc.), If only lengths are given, try to use Stewart's formula, pythagorean, law of cosines, etc.  Try to think of all formulas that might be related to the problem.

2) use your imagination.  If you got a lot of trig expressions, maybe law of sines or cosines  will help.  if it has a lot of squared lengths, try manipulating stewarts or pythagorean or law of cosines.

3)try to draw imaginary lines around.  I often draw 5 or 6 lines before I found the [b][size=150]1[/size][/b] line that's necessary.

If you tried all three and still have NO idea...skip it.  That doesn't mean you can't solve it, but most likely means it will take you 30+ min to solve...(In some miraculous situation the solution will hit you 5 min before the test.  For me it hit 5 min AFTER the test...:mad:  :mad:)

[quote="pkerichang"]Try to draw imaginary lines around.  I often draw 5 or 6 lines before I found the [b][size=150]1[/size][/b] line that's necessary.[/quote]This is very very very true, especially with Challenging Problems in Geometry.

I'll expand on it a bit though.

Try to draw auxilliary lines that give you parallel lines, perpendicular lines, and congruent triangles. (or similar ones) Area ratios are often helpful as well.

One more thing: For CPIG, try not to resort to Stewarts, Cevas, and all those other advanced theorems for the first half of the book. They can all be solved elegantly with easier methods.  :)


[quote="EhsanNamdari"]Given a,b,c >0 and a+b+c = 3; prove that:
 ∑1/(1+2b^2 c)≥1[/quote]

$ \sum{\frac{1}{1\plus{}2b^2c}}\equal{}\sum \frac{a^2}{a^2(1\plus{}2b^2c)}\geq \frac{(a\plus{}b\plus{}c)^2}{\sum a^2(1\plus{}2b^2c)}$

we only prove

$ \frac{(a\plus{}b\plus{}c)^2}{\sum a^2(1\plus{}2b^2c)}\geq 1$

$ \Longleftrightarrow 2(ac\plus{}ab\plus{}bc)(1\minus{}abc)\geq 0$

from $ a\plus{}b\plus{}c\equal{}3$

we can prove

$ 1\minus{}abc\geq 0$


abstract algebra
calculus
integration
superior algebra
superior algebra unsolved

Let $K$ be a finite degree extension of $\mathbb{Q}$

Let $O_{K}$ be the ring of algebraic integers in $K$ over $\mathbb{Z}$

a fractional ideal is a subset $J$ of $K$ such that it is a finitely generated $O_{K}$ module

Show that $J$ is a fractional ideal if and only if there is a $c$ in $K$ such that $c J$ is an ideal of $O_{K}$
One side seems doable, the other side seems to use Noetherian properties?

This is mentionned in a study of class groups without proof, I was wondering whether or not it is doable

What's the problem :? Or did I misunderstand¿

When $J$ is generated by $a_1,a_2,...,a_n \in K$, then write $a_i=\frac{b_i}{d_i}$ with integral $b_i , d_i$ and $c=d_1 d_2 ... d_n$ will do.

For the inverse, use that every ideal of these rings are finetely generated, which just means that the ring is noetherian (otherise by taking one generator after another you would get an infinite chain). So when $cJ$ is an ideal generated by $b_1,b_2,...,b_n$, $J$ will be finetely generated by $\frac{b_i}{c}$.

Okay, I saw this coming, I admit it : why is the ring noetherian?   I'm not good at that, I find it confusing, especially when $O_{K}$ itself is a ring extension of $\mathbb{Z}$

thanks for the reply you already gave me , it was a confirmation


combinatorial geometry
combinatorics unsolved
combinatorics

Let $ A$ be a set of 6 points in a plane such that no 3 are collinear. Show that there exists 3 points in $ A$ which form a triangle having an interior angle not exceeding 30 degrees.

Let $ C$ be the convex hull of the set of the six points. We assume that $ C$ is an $ n$-gon, where $ n\geq 3$. Thus, the sum of the interior angles of $ C$ is $ (n \minus{} 2)\pi$.
It follows that one of these angles is not greater than $ \frac {(n \minus{} 2)\pi}{n}$. Assume that $ A$ is some vertex of $ C$ such that the interior angle at $ A$ has the above property. Now, draw each of the segments from $ A$ to each of the five other points. They divides the angle at $ A$ into four parts. One of these parts is therefore not greater than $ \frac {(n \minus{} 2)\pi}{4n} \leq \frac {\pi}{6}$, as desired.
Note that the equality holds for the set of the vertices of the regular hexagon.

The same reasoning gives the optimal bound $ \frac {\pi}{k}$ for a set of $ k \geq 3$ given points.

Pierre.

We know there can be 20 triangles at all. Now let's assume contrary. So all the triangles have interior angles bigger than 30 degrees. It's obvious that sum of interior angles is 3600 degrees. Look at
x_1 + x_2 + x_3 + ... + x_60 = 3600 such that x_i  >= 30 where i = 1, 2, 3, ..., 60.
Grouping all the three compounds together to fit, we can easily conclude that the wanted result will come.
Thanks.
Hope you understood.

what was your meant ?
 [quote]Grouping all the three compounds together to fit. [/quote]


inequalities
geometry
logarithms
inequalities proposed

The unit square is subdivided in two ways into $n$ subsets of positive area $A_1,A_2,...,A_n$ and $B_1,B_2,...,B_n$. Let $S_{i,j} = A_i \cap B_j$, and denote by $|S|$ the area of a set $S$. Show that:
\[ \sum_{i,j=1}^n |S_{i,j}| \cdot \log \frac{|S_{i,j}|}{|A_i|} \leq \sum_{j=1}^n |B_j| \cdot \log |B_j| \]

After all this inequality is not difficult, it just scares you at first sight. :D

The main idea is to transform it in
$\prod_{i,j=1}^{n}|S_{ij}|^{|S_{ij}|}\le \prod_{i=1}^{n}|A_{i}|^{|A_{i}|}\cdot|B_{i}|^{|B_{i}|}$
And use the inequality:
$x_{1},x_{2},...,x_{k}\in [0,1] \longrightarrow \prod x_{i}^{x_{i}}\le (\sum_{i=1}^{k}x_{i})^{\sum_{i=1}^{k}x_{i}}$


symmetry
geometry
geometric transformation
rotation
superior algebra
superior algebra unsolved

This is general question about structures in abstract algebra. Introductory books generally introduce semigroups, monoids and groups sequentially.Associativity property comes with semi groups itself and the property is considered fundamental. Let us look at symmetry group of a regular polyhedra ( or a regular polygon) we observe commutativity failing while associativity holds. From the point of view of rotations, it looks magical. You do two rotations along different axis and you observe commutativity does not hold, while of you do three operations , suddenly associativity holds.

Is there anything fundamental about associativity ?

At first I would say yes, but I know better. Tons of people (OK probably not tons) are doing stuff in non-associative algebras. Also, look up loop theory (a non-associative group). It seems that both of these structures are starting to appear in physics, so there isn't even an argument that associativity is a fundamental/naturally occurring axiom. Sometimes non-associativity occurs in natural.


[quote="babyloverain"]Prove that:
$ \frac {2(a^2 \plus{} b^2)}{(a \plus{} b)^2} \plus{} \frac {2(b^2 \plus{} c^2)}{(b \plus{} c)^2} \plus{} \frac {2(c^2 \plus{} a^2)}{(c \plus{} a)^2} \ge 3 \plus{} \frac {(a \minus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2}$[/quote]
For $ (a \plus{} b)^2(a \plus{} c)^2(b \plus{} c)^2\neq0$ we obtain:
$ \frac {2(a^2 \plus{} b^2)}{(a \plus{} b)^2} \plus{} \frac {2(b^2 \plus{} c^2)}{(b \plus{} c)^2} \plus{} \frac {2(c^2 \plus{} a^2)}{(c \plus{} a)^2} \ge 3 \plus{} \frac {(a \minus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2}\Leftrightarrow$
$ \Leftrightarrow\frac {(a \minus{} b)^2}{(a \plus{} b)^2} \plus{} \frac {(b \minus{} c)^2}{(b \plus{} c)^2}\geq\frac {(a \minus{} c)^2(2ac \minus{} b^2)}{(a \plus{} c)^2(a^2 \plus{} b^2 \plus{} c^2)}.$
Let $ 2ac \minus{} b^2\geq0.$
By Cauchy we obtain: $ \frac {(a \minus{} b)^2}{(a \plus{} b)^2} \plus{} \frac {(b \minus{} c)^2}{(b \plus{} c)^2}\geq\frac {(a \minus{} c)^2}{2b^2 \plus{} 2(a \plus{} c)b \plus{} a^2 \plus{} c^2}.$
Hence, it remains to prove that
$ (a \plus{} c)^2(a^2 \plus{} b^2 \plus{} c^2)\geq(2ac \minus{} b^2)(2b^2 \plus{} 2(a \plus{} c)b \plus{} a^2 \plus{} c^2),$
which is equivalent to $ 2b^4 \plus{} 2(a^2 \minus{} ac \plus{} c^2)b^2 \plus{} (a^2 \plus{} c^2)^2\geq2(2ac \minus{} b^2)(a \plus{} c)b.$
Hence, we can assume that $ (a \plus{} c)b\geq0.$
But $ 2b^4 \plus{} 2(a^2 \minus{} ac \plus{} c^2)b^2 \plus{} (a^2 \plus{} c^2)^2\geq2b^4 \plus{} \frac {(a \plus{} c)^2b^2}{2} \plus{} \frac {(a \plus{} c)^4}{4}$
and $ (2ac \minus{} b^2)(a \plus{} c)b\leq\left(\frac {(a \plus{} c)^2}{2} \minus{} b^2\right)(a \plus{} c)b.$
Hence, it remains to prove that
$ 2b^4 \plus{} \frac {(a \plus{} c)^2b^2}{2} \plus{} \frac {(a \plus{} c)^4}{4}\geq2\left(\frac {(a \plus{} c)^2}{2} \minus{} b^2\right)(a \plus{} c)b.$
Let $ \frac {2b}{a \plus{} c} \equal{} t.$
Hence, $ 2b^4 \plus{} \frac {(a \plus{} c)^2b^2}{2} \plus{} \frac {(a \plus{} c)^4}{4}\geq2\left(\frac {(a \plus{} c)^2}{2} \minus{} b^2\right)(a \plus{} c)b\Leftrightarrow$
$ \Leftrightarrow t^4 \plus{} 2t^3 \plus{} t^2 \minus{} 4t \plus{} 2\geq0,$ which is obvious.

We can write it into: $ \sum\ \frac {(a \minus{} b)^2}{(a \plus{} b)^2} \geq\ \frac {(a \minus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2}$
Using CS, we have: $ LHS \geq\ \frac {(a \minus{} c)^2}{(a \plus{} b)^2 \plus{} (b \plus{} c)^2} \plus{} \frac {(a \minus{} c)^2}{(a \plus{} c)^2} \geq\ \frac {4(a \minus{} c)^2}{\sum\ (a \plus{} b)^2} \geq\ RHS$  (By AM-GM)  :)
Nice hint to creat, babyloverain  :)


Solve the following, given that y(1) = 1

x(-5y^2 + 9x^2)y' = 2(y^2)(3x-5y)

It's there a trick behind this? I have tried homogenous with u = y/x, but it is so much work and I can't isolate the variable y.


Are the AMC 8 competitions generally for middle school students?

yes...8th graders are the oldest kids allowed to take it.

It was a fun contest. Just don't let any pressure get to you, and make sure you solve what the question is asking for, and you'll do fine.

I took it halfway into geometry as well last year as an 8th grader.

Good luck!

Yeah, the AMC 8 is nice and easy. Don't get too fussed about it, even if you screw up, it doesn't have anything to do with whether or not you can take the AMC 10 anyway.

For the record, one of my friends and I both got 24s (gah, stupid mistakes) in 8th grade after a few months of geometry.

You may find the answers to AMC 8 Frequently Asked Questions 1 and 2 helpful.  See
http://www.unl.edu/amc/f-miscellaneous/faq.shtml#amc8

Steve Dunbar
Director, MAA American Mathematics Competitions

[quote="mihail911"][quote="xpmath"]If you have had a full course in Algebra 1, you should be able to score quite well on it.[/quote]

depends what u mean by "well?"  :lol:
i was half-way through geometry with a complete course in algebra under my belt, in 7th grade, when i scored a 16

(im a lot better now btw )[/quote]
Well, I mean above average-the national honor roll.

Yeah, and you don't get points for leaving blank either (like on 10/12), so guessing is encouraged.
There was a problem on the AMC 8 last year about scoring :roll:

[quote="moogra"]Theres no guessing penalty so feel free to guess[/quote]

realy? so its just the SAT youre penalized for?

oh no, most multiple choice tests either have a penalty for wrong answers, or a small reward for lnot guessing

amc 8 and ACT plan (ACT too?) do not have penalties for guessing

amc 10,12, SAT, and most others do

Also, only school teams can register for AMC 8, right? Or can individual students register themselves? If so, when's the deadline and how much does it cost for admission?

I believe almost all of your questions can be answered by reading the AMC 8 information brochure:
http://www.unl.edu/amc/b-registration/b1-archive/2007-2008/2007-AMC8Bro.pdf
and the AMC 8 registration form:
http://www.unl.edu/amc/b-registration/b1-archive/2007-2008/07-AMC8RegisForms.pdf

Both are linked from the first page of the AMC website.

Steve Dunbar
MAA Director, American Mathematics Competitions

i dont really understand it from that link...

im too confused. I don't see whether or not an individual can register for him/herself instead of having a school team. Are only schools allowed to register an individual student, or can the student register alone?


Applying cauchy and again and again we get the result.

[quote="Lucas Molina"]Why is wrong?[/quote]
I am sorry for my carelessness, as under the square root sign, $a^{2^n}$ looks like $a^{2n}$...  :blush:


$ n$ sets $ S_1$, $ S_2$ $ \cdots$, $ S_n$ consists of non-negative numbers. $ x_i$ is the sum of all elements of $ S_i$, prove that there is a natural number $ k$, $ 1<k<n$, and:

\[ \sum_{i\equal{}1}^n x_i < \frac{1}{k\plus{}1} \left[ k \cdot \frac{n(n\plus{}1)(2n\plus{}1)}{6} \minus{} (k\plus{}1)^2 \cdot \frac{n(n\plus{}1)}{2} \right]\]
and there exists subscripts $ i$, $ j$, $ t$, and $ l$ (at least $ 3$ of them are distinct) such that $ x_i \plus{} x_j \equal{} x_t \plus{} x_l$.

The problem description is not correct, please, fix it.


Let $ f: [a,b] \longrightarrow \mathbb{R}$ e differentiable at each point of $ [a,b]$, and suppose that $ f'(a) \equal{} f'(b)$. Prove that there is at least one point $ c \in (a,b)$ such that

\[ f'(c) \equal{} \frac {f(c) \minus{} f(a)}{c \minus{} a}
\]

Put $ \varphi(x) \equal{} \frac{f(x)\minus{}f(a)}{x\minus{}a}$ and assume without loss of generality that $ \varphi(a) \equal{} f'(a) \equal{} 0$.  If $ \varphi(b) \equal{} 0$ then the result follows from Rolle's theorem, otherwise, assume without loss of generality again that $ \varphi(b) > 0$ so that $ \varphi'(b) < 0$.

Pick $ x_1\in (a,b)$ with $ \varphi(x_1)>\varphi(b)$, so that $ \varphi(a) < \varphi(b) < \varphi(x_1)$, and hence there is $ x_2\in (a,x_1)$ with $ \varphi(x_2) \equal{} \varphi(b)$.  Now we can apply Rolle's theorem again.

I don't see how you can apply rolle's theorem to get the resutl.  :(


Some three non negative real numbers p,q,r satisfy:p+q+r=1.Prove that:
\[7(pq+qr+rp)\le 2+9pqr\]

It's followed easily by the ineq :
 $ (p+q-r)(p+r-q)(r+q-p) \leq pqr $

From Schur's inequality: $1+9xyz\geq{4(xy+yz+zx)}$ for $x+y+z=1$, and it remains to see that:
$1=(x+y+z)^2\geq{3(xy+yz+zx)}$ and we are done.

Well for example lets take that r<=2/3. If we work by the Sturm method setting p+q constant we will easily get that it is enough to solve the inequality for p=q. so it becomes
7(p2 +2pr)<=2 +9p2r and with r=1-2p we get that we have to prove that
-18p3 +30p2 -14p +2>= 0 withch breaks down like this
(p-1)(-18p2+12p-2)>=0(1) For p=1 the inequality holds.
-18p2 +12p-2<=0 (that is easily proved) and with p-1<=0 we have that (1) is true.

[quote="RaduSorici"]Maybe you guys know it with a different name but for example if you have a constant sum like p+q=c and some kind of expression it is sometimes helpful to take p+ :varepsilon:  and q- :varepsilon: instead of p and q and see how your expression changes.[/quote]
It's something similar(even weaker sometimes)to Mixing Variables!

[quote="RaduSorici"]Maybe you guys know it with a different name[/quote]
Sometimes we call it "bludgeon's method" here, in Russia :lol:
It helps to prove many inequalities but often solutions are ugly.


some time when you use lists you really wish to save space...well, there's just nothing on the right and it looked weird.  So is it possible to make the list go two-column??  Like, if there's 5 items, then it puts 3 on the left and 2 on the right, and everyone's happy??

I guess I could use array, but everything I have to do "\text{(a) } ....." and it annoys me...

[quote="pkerichang"]I guess I could use array, but everything I have to do "\text{(a) } ....." and it annoys me...[/quote]Use tabular for text mode tables so no need for \text. It has the same syntax as array but expects text rather than mathematics. You can also use parbox or minipages for columns by putting them next to each other - see [url=http://tug.ctan.org/tex-archive/info/lshort/english/lshort.pdf]Not So Short Introduction[/url] page 121.


Let $a,b,c\geq 1$, prove that
$\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\leq \sqrt{c(ab+1)}$

if $\sqrt{a-1}+\sqrt{b-1}\leq\sqrt{ab}$ then
$\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\leq\sqrt{ab}+\sqrt{c-1}= \sqrt{ab+1-1}+\sqrt{c-1}\leq\sqrt{(ab+1)c}.$
$\sqrt{a-1}+\sqrt{b-1}\leq\sqrt{ab}\Leftrightarrow (\sqrt{(a-1)(b-1)}-1)^2\geq0.$


Let $a,b,c,d\geq 0$ such that $a+b+c+d\geq 4$
Prove that $\frac{a}{b^{2}c+1}+\frac{b}{c^{2}d+1}+\frac{c}{d^{2}a+1}+\frac{d}{a^{2}b+1}\geq 2$
I thinks this inequality very nice . What do you think? :)  :)

It is also included in my book, and created by me a long time ago. Using AM-GM inequality in indirect way,

Yes! but if you not using AM-GM inequality then your solution very well too .Can you solve it ?  :)  :)


I'm kind of new to this site, but I want to get the most help I can from this stuff and do really well on the AMC and AIME next year, and hopefully get to the Olympiad by my senior year (right now I'm a sophomore), and I was just wondering what material you think would be best for me.  I got a 115 on the AMC-12 this year, and I got a 1 on the AIME that should have been a 4 except for stupid arithmetic errors.  So do you think I would get more help here, or in the intermediate forum?  Also, would the AoPS book volume 1 help me much at this level, or should I just go to volume 2?

Thanks!

AoPS volume 1 will definitely help you, I'd strongly recommend going through it (and this means problems too) before moving on to volume 2.

The most important thing, though, is to do problems.  You should probably coast around here and Intermediate and post solutions.  And when people correct your solutions you'll realize you did some arithmetic wrong, like me :)

But this is a great place to learn new concepts and to practice familiar ones - don't be afraid to ask questions if you don't understand someone else's solution!

Sounds good, thanks for the help.  What usually happens for me regarding not understanding is that I get the concept, just not the notation or formula.  I ended up brute-forcing a few competition problems this year because of things like that...


algebra
polynomial
linear algebra
linear algebra unsolved

Let $ P$ and $ Q$ be two polynomials with rael coeficients which have integer values at same points. Prove that
 $ P \minus{} Q\ \text{or}\ P \plus{} Q$  is constant.

P.S. I'm not quite sure I'm post at right place. Please mod to move post, if it's neccessary to right place. Thank you.

WLOG the leading coefficients of $ P$ and $ Q$ are positive.
Let $ x_0$ be such a point that $ P(x)$ and $ Q(x)$ are both
positive and monotonically increasing to infinity in
$ (x_0; \infty )$. Let $ y_1 \in (x_0; \infty )$  be such a point that $ P(y_1) =n \in\mathbb{N}$.
Then $ Q(y_1)=m  \in\mathbb{N}$. Let $ y_2 >y_1$ be such a point that $ P(y_2) =n +1$.
Since our polynomials have integer values at the same points
and they are monotonically increasing  for $ x>x_0$ we conclude that
$ Q(y_2)=m+1$. Analogously we denote by $ y_3 >y_2$  such a point that
$ P(y_3) =n +2$ and conclude that  $ Q(y_3)=m+2$ and so on. Thus we
obtain an infinite sequence of points $ (y_k)_{k=1}^{\infty}$ such that
$ P(y_k)-Q(y_k) =n-m$ for every $ k$.
So $ P-Q$ is constant.


Where can I find a beginner user guide to Texniccenter(how to use it - all its functions) and a complete manula for using Latex??
Please suggest an online, free source!!!

How about our guide?

[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]

Can somebody suggest a more comprehensive package for Texniccenter?? I basically want to learn all its features!!

Use the help manual that comes with it. However I don't see what can be so comprehensive at a text editor ...  :?


geometry
3D geometry
algebra unsolved
algebra

[quote="jackie-chan01"]If n be even; and α - β , α + β are the middle pair of terms of an A.P, then find the sum of the cubes of the terms[/quote]

Let $ n\equal{}2p$. The sequence is then $ a_k\equal{}\alpha\plus{}\beta(2k\minus{}2p\minus{}1)$ and the required sum is $ S\equal{}\sum_{k\equal{}1}^{2p}(\alpha\plus{}\beta(2k\minus{}2p\minus{}1))^3$ and so :

$ S\equal{}\sum_{k\equal{}1}^{p}(2\alpha^3\plus{}6\alpha\beta^2(2k\minus{}2p\minus{}1)^2)$

$ S\equal{}2p\alpha^3\plus{}6\alpha\beta^2\sum_{k\equal{}1}^{p}(2k\minus{}1)^2$

$ S\equal{}2p\alpha^3\plus{}6\alpha\beta^2(\sum_{k\equal{}1}^{2p}k^2\minus{}4\sum_{k\equal{}1}^{p}k^2)$

$ S\equal{}2p\alpha^3\plus{}\alpha\beta^2(2p(2p\plus{}1)(4p\plus{}1)\minus{}4p(p\plus{}1)(2p\plus{}1))$

$ S\equal{}2p\alpha^3\plus{}\alpha\beta^22p(2p\plus{}1)(2p\minus{}1)$

And so $ \boxed{S\equal{}n\alpha^3\plus{}\alpha\beta^2n(n^2\minus{}1)}$


function
floor function
modular arithmetic

Prove that
there is no real solutions to

$ \lfloor x\rfloor \plus{} \lfloor 2x\rfloor \plus{} \lfloor 4x\rfloor \plus{} \lfloor 8x\rfloor \plus{} \lfloor16 x\rfloor \plus{} \lfloor 32x\rfloor \equal{} 12345$

where $ \lfloor x\rfloor$ is the floor functuion of $ x$.

I believe we're supposed to prove that there are no real solutions to the equation. (Read the title).
So a counterexample would be a value of $ x$ that satisfies the equation :wink:.

Clearly, any $ x$ that satisfies the equation must be greater than zero, so suppose that $ x$ can be expressed in the form $ a+b$, where $ a$ is a nonnegative integer and $ 0\leq b<1$.

So therefore,
\begin{align*}
\lfloor x \rfloor &+ \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor \\
 &= a+\lfloor b \rfloor + 2a+\lfloor 2b \rfloor + 4a+\lfloor 4b \rfloor + 8a+ \lfloor 8b \rfloor + 16a+\lfloor 16b \rfloor + 32a+\lfloor 32b \rfloor \\
 &= 63a+\lfloor b \rfloor + \lfloor 2b \rfloor + \lfloor 4b \rfloor + \lfloor 8b \rfloor + \lfloor 16b \rfloor + \lfloor 32b \rfloor\end{align*}

But since $ 0\leq b<1$, we have bounds on the floors of the multiples of $ b$. Namely,
\begin{align*}\lfloor b \rfloor &\leq 0, \\
\lfloor 2b \rfloor &\leq 1, \\
\lfloor 4b \rfloor &\leq 3, \\
\lfloor 8b \rfloor &\leq 7, \\
\lfloor 16b \rfloor &\leq 15, \\
\lfloor 32b \rfloor &\leq 31\end{align*}
So the sum $ \lfloor b \rfloor + \lfloor 2b \rfloor + \lfloor 4b \rfloor + \lfloor 8b \rfloor + \lfloor 16b \rfloor + \lfloor 32b \rfloor$ is less than or equal to 57.

So mod 63, $ \lfloor x \rfloor \&+ \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor \leq 57 \pmod {63}$.

However, a quick check reveals that $ 12345 \equiv 60 \pmod {63}$, therefore we have a contradiction, so there is no value of $ x$ that satisfies the equation.

[quote="Yongyi781"]Clearly, any $ x$ that satisfies the equation must be greater than zero

[/quote]
Actually, I think it can be less than 0

For $ x<0$, $ \lfloor x \rfloor < 0$ as well, and $ \lfloor nx \rfloor < 0$ for any $ n>0$. The sum of any number of integers less than zero is also less than zero, but 12345 is greater than zero, contradiction.


calculus
integration
limit
trigonometry
logarithms
calculus computations

Let $0<t<\frac{\pi}{2}$.

Evaluate

\[\lim_{t\rightarrow \frac{\pi}{2}} \int_0^t \tan \theta \sqrt{\cos \theta}\ln (\cos \theta)d\theta\]

$\int \tan \theta \sqrt{\cos \theta}\ln (\cos \theta)d\theta$

Let $\theta = \cos^{-1} e^u \Rightarrow u= \ln(\cos \theta ) \Rightarrow du = -\tan \theta $

The integral then becomes

$-\int  u \sqrt{e^u} \ du = (4-2u)e^{u/2}$

$=\left(4-2\ln(\cos \theta)\right)\sqrt{\cos \theta}$

$\lim_{\theta\to\frac{\pi}{2}} \left(4-2\ln(\cos \theta)\right)\sqrt{\cos \theta} = -2\lim_{\theta\to\frac{\pi}{2}} \ln(\cos \theta)\sqrt{\cos \theta}$

$=-2\lim_{x\to 0} \ln(x)\sqrt{x} = -2\lim_{x\to 0} \left(\frac{\ln x}{\frac{1}{\sqrt{x}}}\right) = -2\lim_{x\to 0} \left(\frac{\frac{1}{x}}{-\frac{1}{2x^{3/2}}}\right) = 4\lim_{x\to 0} \sqrt{x} = 0$

So finally,

$\lim_{t\rightarrow \frac{\pi}{2}} \int_0^t \tan \theta \sqrt{\cos \theta}\ln (\cos \theta)d\theta = -4$


[quote="Eldest"]Let $ k$ be a positive integer and let $ a,b,c$ be posiive real numbers such that $ abc\le 1$. Prove that the equality
$ \frac {a}{b^k} \plus{} \frac {b}{c^k} \plus{} \frac {c}{a^k}\ge a \plus{} b \plus{} c$[/quote]
[hide]I'll post my proof later.
[/hide]

This is a problem on M&Y in Vietnam. Pleases lock it!


I'm having some problem with this problem, someone help me please.

A child's game consists of a block that attaches to a table with a suction cup, a spring connected to that block, a ball, and a launching ramp. By compressing the spring, the child can launch the ball up the ramp. The spring has a spring constant k, the ball has a mass m, and the ramp raises a height h. The spring is compressed a distance S in order to launch the ball. When the ball leaves the launching ramp its velocity makes an angles *alpha* with respect to the horizontal.

Calculate the vector velocity of the ball when it just leaves the lanching ramp (both magnitude and direction. Be sure to specify your coordinate system).

Thanks in advanced everyone.  :D

The direction of the vector is given in the problem statement. Its magnitude is (using energy):

$\sqrt{\frac{kS^2}{m}-2gh}$


symmetry
geometry
3D geometry
tetrahedron
octahedron
icosahedron
dodecahedron

Prove that

[b]The group of rotational symmetries of a tetrahedron is
A(4).

The full group of symmetries of a tetrahedron is
S(4).

The group of rotational symmetries of a cube or of an octahedron
is S(4).

The group of rotational symmetries of an icosahedron or of a dodecahedron is A(5). [/b]

based on:

A regular solid is a 3-
dimensional polyhedron in which each face is a regular polygon. Any two
faces as well as any two vertices can be matched by an isometry (a rigid
motion) of the 3-dimensional space. It is convenient to use a symbol
{p, q}

for a regular solid whose faces are regular
p-gons with q of them situated
around each vertex. We have proved that there are FIVE regular solids:

•
TETRAHEDRON {3, 3},

•
CUBE {4, 3},

•
OCTAHEDRON {3, 4},

•
DODECAHEDRON {5, 3},

•
ICOSAHEDRON {3, 5}.

Any regular solid may be inscribed in a sphere, and then any symmetry of
any regular solid will leave the center of the sphere fixed and will transform
the surface of the sphere onto itself. We call a rotational symmetry of a
regular solid any rotation of the sphere (with respect to an axis passing
through its center) mapping the regular solid into inself. A symmetry of a
regular solid is either a rotational symmetry or a reflection with respect to
a plane (also passing through the center of the sphere), mapping the regular
solid onto itself.

First get the orders of the groups...
[hide]by the orbit-stabilizer theorem: pick a convenient point (a vertex or center of a face), find the orbit of the point (the number of different points it can end up), the stabilizer subgroup (the number of symmetries fixing that point), and multiply their sizes to get the order of the group.

For example, on rotations of the cube, the orbit of the center of a face is the centers of all faces (there are 6). The stabilizer is the set of rotations that fix the center of a face. The center of the whole cube must also be fixed, so there's a unique line of rotation, and you can find that the stabilizer has 4 rotations. That gives 24 rotations of the cube.[/hide]

Although orbit-stabilizer just gives the order, not much more information is actually needed about the group to show it's isomorphic to $ A_n$ or $ S_n$ for small $ n$.
[hide] For the tetrahedron, both groups are clearly isomorphic to a subgroup of $ S_4$ (they permute the 4 vertices), so this just involves showing the subgroups of $ S_4$ of the right orders (they should be 12 and 24) are unique. (Of course the 24-element subgroup of $ S_4$ is unique  :P  )

For the cube, there are four objects that get permuted such that each arrangement of those objects uniquely determines how the cube was rotated, that gives an isomorphism to a subgroup of $ S_4$.

The icosahedron is more difficult. One way is to find multiple subgroups of order 5 (that is, two 72-degree rotations around different lines), then use the result of my post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=174996]here[/url]. (The proof works on any group of order 60 with multiple subgroups of order 5. I found out that multiple subgroups of order 3 isn't needed, since if $ P$ were a normal subgroup of order 3 and $ Q$ were a subgroup of order 5, $ PQ$ would be a subgroup of order 15.)

With the icosahedron, there may be a more direct way of showing isomorphism to a subgroup of $ S_5$ (maybe the centers of the faces form five tetrahedrons, such that the rotation is uniquely determined by an arrangement of the tetrahedrons).[/hide]

A relatively explicit way of constructing the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=29946]isomorphism[/url].

For the group $ R$ of rotations of a icosahedron/dodecahedron to be $ A_5$: you can subdivide the midpoints of the $ 20$ faces into $ 5$ regular tetrahedrons. Then it's not hard to check that a rotation preserves those tetrahedrons. Especially, $ R$ act's on a $ 5$-element set.
We are left to show that if a rotation leaves all tetrahedrons (seen as sets of four points) fixed, then every single point of the icosahedron remains fixed. To see this, note that two opposite midpoints $ P_1, P_2$ of faces belong to distinct tetrahedrons $ T_1, T_2$, and for every pair $ T_1, T_2$ of distinct tetrahedron, there is exactly one pair of opposite points $ P_1 \in T_1 , P_2 \in T_2$. Thus if the tetrahedrons remain fixed, every point does.
This gives an injection $ \phi : R \to S_5$ as desired.


Given 3 line segment with the length respectively of $ a,b,c$. With ruler and compasses, draw a line segment with the length of $ x$ such that:
$ \frac{1}{x^2}\equal{}\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}$

[quote="Unidranix"]Given 3 line segment with the length respectively of $ a,b,c$. With ruler and compasses, draw a line segment with the length of $ x$ such that:
$ \frac {1}{x^2} \equal{} \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}$[/quote]

At first introduce a basic skill in this kind of construction.
[geogebra]0aae5d907bd8dab56b0254ef2080897cda2eb4c1[/geogebra]

Given Points $ A,D,C$ we can find Point $ B$ (by drawing perpendicular ($ L_1$) to $ AC$ at $ D$ and using mid-point of $ AC$ as center and $ \frac{AC}{2}$ as radius to draw circle $ C_1$ and then the intersection of $ L_1$ and $ C_1$ is $ B$) .

Given Points $ A,B,D$ we can find $ C$ (by drawing perpendicular ($ L_2$) to $ AB$ at $ B$ and then the intersection of $ L_2$ and $ AD$ is $ C$).

And by similar triangle $ pr\equal{}q^2$

[b]Solution:[/b]
WLOG, we can assume that $ a\equal{}1$,

Then we can construct line segments of $ b^2$ (set $ p\equal{}1$ and $ q\equal{}b$) and $ c^2$ (set $ p\equal{}1$ and $ q\equal{}c$) .

So we can construct line segments of $ \frac{1}{b^2}$ (set $ p\equal{}b^2$ and $ q\equal{}1$) and $ \frac{1}{c^2}$ (set $ p\equal{}c^2$ and $ q\equal{}1$) and thus get a line segment of length $ \frac{1}{x^2}$ and thus that of length $ x^2$ (set $ p\equal{}\frac{1}{x^2}$ and $ q\equal{}1$)

Then we can get a line segment of length $ x$ by setting $ p\equal{}1$ and $ r\equal{}x^2$.

Use the the following property of the right-angled triangle:

If $ z$ is the altitude on the hypotenuse of the right-angled triangle with legs $ b,c,$ then
\[{\frac {1}{z^2} = \frac {1}{b^2} + \frac {1}{c^2}}\]
Now,  construct the right triangle with legs $ b,c,$ draw the altitude $ z$ on the hypotenuse and construct the right triangle with legs $ z,a.$ The altitude of this triangle is the required $x.$


Find all positive integers $x, y, z$ such that $2^x + 2^y + 2^z = 2336$.

Note that $2336 = 2^{11} + 2^8 + 2^5.$

Thus, the base $2$ expansion of $2336$ uses $3$ digits $1$. If any two of the numbers $x,y,z$ are equal then the base $2$ expansion of the LHS only uses at most two digits $1$, which gives a contradiction.
And, if $x,y,z$ are pairwise distinct, the unicity of the base $2$ expansion leads to $x=5,y=8,z=11$ or some permutation of it.

Pierre.

the binary form of any number is unique! :ninja:

Yes, but you are not sure that the HS is given on its binary expansion form...

Pierre.

P.S. OOps I confused between Left and right in my first post...

WLOG,  we take x《y《z such that x +a = y & x+a+b =z. So, 2^x (1 + 2^a + 2^(a+b ))= 2336= 32 ×73. Now the expression on Left side within bracket is odd. So, 2^x =32, that is x =5. Hence, 2^a + 2^(a+b) = 72 = 8×9. So, 2^a (1+2^b) = 8×9. Again 1 + 2^b is odd. So, 2^a = 8 that is a =3. Hence b=3. So, x = 5, y = 8 & z = 11 and their permutations are the required solutions.


superior algebra
superior algebra unsolved

Use induction.  What can we say if $ (1\plus{}p)^{p^{n\minus{}2}} \equal{} 1 \plus{} a p^{n\minus{}1}$, where $ a$ is not divisible by $ p$?


calculus
derivative
integration
function
real analysis
real analysis theorems

We have defined $ D^n f \equal{} \frac {d^n f}{dx^n}$ for any $ n\in\mathbb Z$.

[ One can let $ D^0 (f) \equal{} f$,     $ D^{ \minus{} 1}(f) \equal{} \int_a^x f$ ]

So, can we define $ D^r (f)$ for any $ r\in\mathbb Q$ or $ \mathbb R$?

How would one take a derivative a non-whole number of times?

As a matter of fact you can do it. It is called [url=http://mathworld.wolfram.com/FractionalDerivative.html]fractional calculus[/url].

Since the definition relies on the gamma function, you get a derivative for any real number, as long as it is not a negative integer.

Note, that it fits perfectly with our current definition, and the notion of things like "half" make sense, because if you do the half derivative twice you get the derivative.

The other thing you can do involves the Fourier transform.

Define $ \widehat{f}(\xi) \equal{} \int_{ \minus{} \infty}^{\infty}e^{ \minus{} 2\pi i \xi x}f(x)\,dx.$

Then you can reconstruct $ f$ by

$ f(x) \equal{} \int_{ \minus{} \infty}^{\infty}e^{2\pi i \xi x}\widehat{f}(\xi)\,d\xi.$

We have that $ \widehat{f'}(\xi) \equal{} 2\pi i\xi\widehat{f}(\xi).$

Based on that, we can try "defining" the derivative of order $ r$ by

$ \left(\frac {d}{dx}\right)^rf(x) \equal{} \int_{ \minus{} \infty}^{\infty}e^{2\pi i \xi x}(2\pi i\xi)^r\widehat{f}(\xi)\,d\xi.$

One problem is that that isn't really well defined - we must choose a branch of the $ r$th power of $ i.$ But given an appropriate branch choice, it does give us as way to work with fractional derivatives. (I also didn't say what restrictions we should put on $ f;$ in some cases, we are interpreting things as tempered distributions.)


Find the sum in the funcion of n:

$\sum_{k=1}^n k \cdot 2^k = 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \dotsb + n \cdot 2^n$

[hide="Answer"]$2(n - 1)(2^n - 1) + 2n$[/hide]

[hide="hint"]
Multiply by $S$ and substract the two resulting equations.
[/hide]


[quote="hxy09"]$ x_1,x_2,...,x_6\ge0$
Prove that $ \sum_{i \equal{} 1}^{6}\frac {x_i^2}{x_{i \plus{} 1}}\ge \sqrt {6\sum_{i \equal{} 1}^{6}x_i^2}$

[b]Does it hold for replacing $ 6$ by $ n$[/b]?[/quote]

$ x_7\equal{}?$

[quote="hxy09"]$ x_1,x_2,...,x_6\ge0$
Prove that $ \sum_{i \equal{} 1}^{6}\frac {x_i^2}{x_{i \plus{} 1}}\ge \sqrt {6\sum_{i \equal{} 1}^{6}x_i^2}$

[b]Does it hold for replacing $ 6$ by $ n$[/b]?[/quote]

$ x_{i\plus{}1}\equal{}x_1$

$ n\equal{}2$

 a counter-example  $ x_1\equal{}5,x_2\equal{}4$

$ n\equal{}3$

 a counter-example  $ x_1\equal{}14,x_2\equal{}13,x_3\equal{}12$

$ n\equal{}4$

 a counter-example  $ x_1\equal{}27,x_2\equal{}26,x_3\equal{}25,x_4\equal{}24$

$ n\equal{}5$

 a counter-example  $ x_1\equal{}44,x_2\equal{}43,x_3\equal{}42,x_4\equal{}41,x_5\equal{}40$

[quote="kuing"][quote="hxy09"]$ x_1,x_2,...,x_6\ge0$
Prove that $ \sum_{i \equal{} 1}^{6}\frac {x_i^2}{x_{i \plus{} 1}}\ge \sqrt {6\sum_{i \equal{} 1}^{6}x_i^2}$

[b]Does it hold for replacing $ 6$ by $ n$[/b]?[/quote]

$ x_7 \equal{} ?$[/quote]

Oh,by convention $ x_7\equal{}x_1$  :wink:

hedeng123,very sorry,I don't agree with your counter-example,could you check it again?
By the way,I deem that this inequality is vey nice and hard beyond doubt :)

[quote="hxy09"]$ x_1,x_2,...,x_6\ge0$
Prove that $ \sum_{i \equal{} 1}^{6}\frac {x_i^2}{x_{i \plus{} 1}}\ge \sqrt {6\sum_{i \equal{} 1}^{6}x_i^2}$

[b]Does it hold for replacing $ 6$ by $ n$[/b]?[/quote]

I am sorry

$ n \equal{} 2,3,4,5$

The inequality holds

[quote="can_hang2007"][quote="hxy09"]$ x_1,x_2,...,x_6\ge0$
Prove that $ \sum_{i \equal{} 1}^{6}\frac {x_i^2}{x_{i \plus{} 1}}\ge \sqrt {6\sum_{i \equal{} 1}^{6}x_i^2}$

[b]Does it hold for replacing $ 6$ by $ n$[/b]?[/quote]
This is a nice and hard inequality. Can you share us your proof, hxy09? Thanks.[/quote]

Hi friend,I will post the solution next week for I still want to wait for a surprise :D

[quote="hxy09"][quote="can_hang2007"][quote="hxy09"]$ x_1,x_2,...,x_6\ge0$
Prove that $ \sum_{i \equal{} 1}^{6}\frac {x_i^2}{x_{i \plus{} 1}}\ge \sqrt {6\sum_{i \equal{} 1}^{6}x_i^2}$

[b]Does it hold for replacing $ 6$ by $ n$[/b]?[/quote]
This is a nice and hard inequality. Can you share us your proof, hxy09? Thanks.[/quote]

I have to apologize for my carelessness...For a calcution-mistake was discovered in the original proof :blush:
The proof is valid only for the case$ n\le5$ which ,I think,is a piece of cake for you :maybe:
Whether the ineq holds for $ n\ge6$ still remains mythterious for all of us. :([/quote]

[quote="hxy09"][/quote][quote="hxy09"][quote="can_hang2007"][quote="hxy09"]$ x_1,x_2,...,x_6\ge0$
Prove that $ \sum_{i = 1}^{6}\frac {x_i^2}{x_{i + 1}}\ge \sqrt {6\sum_{i = 1}^{6}x_i^2}$

[b]Does it hold for replacing $ 6$ by $ n$[/b]?[/quote]
This is a nice and hard inequality. Can you share us your proof, hxy09? Thanks.[/quote]

I have to apologize for my carelessness...For a calcution-mistake was discovered in the original proof :blush:
The proof is valid only for the case$ n\le5$ which ,I think,is a piece of cake for you :maybe:
Whether the ineq holds for $ n\ge6$ still remains mythterious for all of us. :([/quote][quote="hxy09"][/quote]

福建杨学枝老师在今年的初等数学研究会议上面提出过上面的猜想(杨学枝22道不等式猜想)。

[quote="FelixD"]@hedeng123: please write in english  :)[/quote]

thanks
sorry     :lol:

[quote="hedeng123"][/quote][quote="hxy09"][/quote][quote="hxy09"][quote="can_hang2007"][quote="hxy09"]$ x_1,x_2,...,x_6\ge0$
Prove that $ \sum_{i = 1}^{6}\frac {x_i^2}{x_{i + 1}}\ge \sqrt {6\sum_{i = 1}^{6}x_i^2}$

[b]Does it hold for replacing $ 6$ by $ n$[/b]?[/quote]
This is a nice and hard inequality. Can you share us your proof, hxy09? Thanks.[/quote]

I have to apologize for my carelessness...For a calcution-mistake was discovered in the original proof :blush:
The proof is valid only for the case$ n\le5$ which ,I think,is a piece of cake for you :maybe:
Whether the ineq holds for $ n\ge6$ still remains mythterious for all of us. :([/quote][quote="hxy09"][/quote][quote="hedeng123"]

福建杨学枝老师在今年的初等数学研究会议上面提出过上面的猜想(杨学枝22道不等式猜想)。[/quote]
[size=150]hedeng123 said:[/size]

"This conjecture was proposed by Mr Xuezhi Yang at the Symposium in Elementary Mathematics this year, that is one of the 22 conjectures of Yuezhi Yang's."

[quote="hxy09"]The case of $ n\ge 6$ is indeed a conjecture.Intuitively I privately think that $ \exists N,n\ge N$ the ineq doesn't hold :maybe:[/quote]
I have proved now the case $n=6$ by CS. You can see here for a stronger result: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=366576

For $n=5,$ we have this stronger result: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=366596


Anyone knows how to type the symbol for "$A$ strictly implies $B$" i.e. $\Box (A\Rightarrow B)$ in Latex (which is usually typed in modal logic as "an arrow sign with a hook replacing the $>$ at the right end) :?:

See [url]http://www.tug.org/tex-archive/info/symbols/comprehensive/[/url] for a document which lists virtually every symbol that you can do in LaTeX.

[quote="Rushil"]Are you talking about \[ \succ  \] ???

If yes , this comes from
[code] \succ [/code][/quote]

As a matter of fact, the symbol is the mirrot image of $A\succ B$ with the hooked side on the right. how could this be done?

Any other symbol give more curly in the hook so that the "strictly implication" $A\prec B$ could looks more neat?


Number 10 in base 2 system is equal to the number 2 in base 10 system.

When a fraction (1/4) is changed into a decimal in base 't' system, it becomes (0.56565656...)
Find the 't' that satisfies this.

(1/4)(t) = 0.56565656(t)

I was able to change the decimal into fraction in base 10, but I didn't know what to do with the fraction (1/4).

[quote="910316"]Number 10 in base 2 system is equal to the number 2 in base 10 system.

why?
10=1010..(2),2=2..(10)[/quote]

ok, 10[size=59](2)[/size] [size=100]= 2[/size][size=59](10)[/size]

This is because 10[size=59](2)[/size] = 1*2+0*1, and 2[size=59](10)[/size] = 2*1 is obviously 2.

You have to understand a bit on base systems (binary..)

Does this answer your question?

[quote="Anonymous"]

ok, 10[size=59](2)[/size] [size=100]= 2[/size][size=59](10)[/size]

This is because 10[size=59](2)[/size] = 1*2+0*1, and 2[size=59](10)[/size] = 2*1 is obviously 2.

You have to understand a bit on base systems (binary..)

Does this answer your question?[/quote]

just to start off, consider this case with 0.565656... in base 10 system.

Then the expression for this in fraction is (5*10+6*1)/(10^2-1). (The

denominatory is 10^2-1 because it's in a base 10 system

Therefore, the 0.5656... in base 't' system will become, in fraction,

0.565656...[size=59](7)[/size][/size]=(5*t+6*1)/(t^2-1) now, this is in base 't'

Now if we look at the original problem, 1/4 (t) = 0.5656... (t),

1/4 (t) should equal (5t+6)/(t^2-1)

t^2-1=20t + 24;  t^2-20t-25=0. However, 't' becomes an irrational number which

does not satisfy the conditions of the problem. But I am pretty sure the way I'm

approaching is correct. Any flaws in my argument?


Find f(n) for n>=3
where f(n) is the smallest number if there are f(n) points in the plane(no three collinear), there must exists a convex n-gon.

This is the Erdös-Szekeres "happy-ending" problem, and still an open problem for $ n \geq 7.$ Historically speaking, this question leads to the developpment of Ramsey theory.
It is known that $ f(3)\equal{}3,f(4)\equal{}5,f(5)\equal{}9,f(6)\equal{}17$ and no other exact value of $ f(n)$ is known.

As far as I know, the current bounds are $ 1\plus{}2^{n\minus{}2} \leq f(n) \leq \binom{2n\minus{}5}{n\minus{}3} \plus{} 1$.
The lower bound is conjectured to be the exact value of $ f(n)$.

Pierre.


Let S be a subset of {1, 2, 3, ..., 2046} with |S| = 21.
Show that there exist distinct a, b, c in S such that bc < 2a < 4bc.

Well, I'll give the solution myself since noone seems to be interested ...
Nice problem though :)

Note that if b < a < c \leq 2b then bc < 2a < 4bc.
Now we will prove that such an S must contain a, b, c with b <a < c \leq 2b.
Suppose this is not the case. Let a1 < a2 <  ... < a21 be the elements of S.
Then we have
a21 \geq 2a19 + 1 (by assumption)
\geq 2(2a17 + 1) + 1
\geq 2(2(2a15 + 1) + 1)
\geq ...
= 210a1 + 29 + 28 + ... + 2 + 1
\geq 210 + 29 + ... + 2 + 1
= 2047.
Now we have a contradiction since a21 \leq 2046.
Hence there exist a, b, c with b < a < c \leq 2b. So we're done.


At a party there are only single women and married
men with their wives. The probability that a randomly
selected woman is single is 2 /5 . What fraction of the
people in the room are married men?

[hide="Solution"]
We have that $\frac{2}{5}$ of the women are single, so $\frac{3}{5}$ of the women are married, and the same number are married men.  We combine parts of $2 + 3 + 3 = 8$ and conclude that $\boxed{\frac{3}{8}}$ of the people at the party are men. [/hide]

(This seems a little easy for AMC #22.)


limit
calculus
derivative
induction
logarithms
abstract algebra
trigonometry

Let $ p $ be a positive integer , $ e_n=\left(1+\frac{1}{n}\right)^n \; , \; E_n=\left(1+\frac{1}{n}\right)^{n+1} $ and $ \Delta^p X_n:=\sum\limits_{k=0}^{p}(-1)^{p-k}{p \choose k}X_{n+k} .$
[b]Prove that [/b]  $ \lim\limits_{n\to \infty}n^{p+1}\Delta^pE_n = -\; \lim\limits_{n\to \infty}n^{p+1}\Delta^pe_n = \frac{(-1)^pp!}{2}e . $

Open question : If  $\eta_n=\eta_n(a):= \left(1+\frac{1}{n}\right)^{n+a} $ , what about  $ \lim\limits_{n\to \infty}n^{p+1}\Delta^p \eta_n\; \; \; ,\; \; a\in \left\{ \frac{1}{3},\frac{1}{2}\right\}\; \; ?  $

Hi [b]flip2004[/b].Very interesting problem.At last I got the solution.But I don't think it is the best one.Though I'll post it without some computations.
First of all lets notice that ${\triangle}^{p}\frac{1}{n}{\sim}\frac{(-1)^{p}p!}{n^{p+1}}$.It is not hard to show using the formulas ${\triangle}^{p}x_{n}=\sum_{k=0}^{p}{(-1)^{k}C_{p}^{k}x_{n+k}}$
and $\frac{1}{n+k}=\frac{1}{n}\frac{1}{1+\frac{k}{n}}=\frac{1}{n}{\sum_{i=0}^{\infty}{\frac{(-1)^{i}k^{i}}{n^{i}}}}$(I think here will be more nice proof without infinit series,but using derivatives)
Now we must prove that $\lim_{n{\rightarrow}\infty}{\frac{{\triangle}^{p}E_{n}}{{\triangle}^{p}\frac{1}{n}}}=\frac{e}{2}$
Denote by $f(x)=(1+\frac{1}{x})^{x+1}$ and $g(x)=\frac{1}{x}$.Thus we must prove that
$\lim_{n{\rightarrow}\infty}{\frac{{\triangle}^{p}f(n)}{{\triangle}^{p}g(n)}}=\frac{e}{2}$
Now I will prove that $\lim_{n{\rightarrow}\infty}{\frac{{\triangle}^{p}f(n)}{{\triangle}^{p}g(n)}}=\lim_{x{\rightarrow}\infty}{\frac{f^{(p)}(x)}{g^{(p)}(x)}}$
Indeed for $p=1$ $\frac{f(n+1)-f(n)}{g(n+1)-g(n)}=\frac{f'({\zeta}_{n})}{g'({\zeta}_{n})}$,by Causchy's theorem where $n<{\zeta}_n<n+1$.For $p=2$ $\frac{f(n+2)-2f(n+1)+f(n)}{g(n+2)-2g(n+1)+g(n)}=
\frac{(f(n+2)-f(n+1))-(f(n+1)-f(n))}{(g(n+2)-g(n+1))-(g(n+1)-g(n))}=\frac{f'(c+1)-f'(c)}{g'(c+1)-g'(c)}=\frac{f''(\zeta)}{g''(\zeta)}$,and so on.Using induction we can show that $\frac{{\triangle}^{p}f(n)}{{\triangle}^{p}g(n)}=\frac{f^{(p)}({\zeta}_{n})}{g^{(p)}({\zeta}_{n})}$,where $n<{\zeta}_{n}<n+p$.
It is time to count $\lim_{x{\rightarrow}\infty}{\frac{f^{(p)}(x)}{g^{(p)}(x)}}$
$g^{(p)}(x)=\frac{(-1)^{p}p!}{x^{p+1}}.
f'=f{\cdot}t$,where $t=\ln{(1+\frac{1}{x})}-\frac{1}{x}$.
$f''=f't+ft'=f(t^{2}+t')\\
f'''=f'(t^{2}+t')+f(2tt'+t'')=f(t^{3}+3tt'+t'')\\
\ldots\\
f^{(n)}=f(t^{n}+\ldots+t^{(n-1)})\\$
Simple computations show that
$t^{(n)}={(-1)^{n-1}(n-1)!}\frac{n+\sum_{j=1}^{n-1}{x^{j}(nC_{n}^{k}-C_{n}^{k-1})}}{x^{n+1}(1+x)^{n}}{\sim}{\frac{(-1)^{n-1}(n+1)!}{2x^{n+2}}}$
Using that let's notice that $f^{(p)}{\sim}{ft^{(p-1)}}$(it is just simple but boring computation and induction)${\sim}e{\cdot}\frac{(-1)^{p}p!}{2x^{p+1}}$,from which we deduce that $\lim_{x{\rightarrow}\infty}{\frac{f^{(p)}(x)}{g^{(p)}(x)}}=\frac{e}{2}$,so $\lim_{n{\rightarrow}\infty}{\frac{{\triangle}^{p}f(n)}{{\triangle}^{p}g(n)}}=\frac{e}{2}$ which is the statement we want to prove.
I know guys one need a patience to read the solution above but every solution is solution.[b]flip2004[/b] I'll keep thinking on the open question,but I think we will get a result using the method above.And I want to see your solution.

I have just found another way to show that  ${\triangle}^{p}\frac{1}{n}{\sim}\frac{(-1)^{p}p!}{n^{p+1}}$.Denote $g(x)=\frac{1}{x}$.So ${\triangle}^{p}\frac{1}{n}={\triangle}^{p}g(n)$.
Using the theorem of Lagrange $p$ times we can say that ${\triangle}^{p}g(n)=g^{(p)}({\zeta}_{n})$,where $n<\zeta_{n}<n+p$.
$g^{(p)}({\zeta}_{n})=\frac{(-1)^{p}p!}{{{\zeta}_{n}}^{p+1}}$,so
$\frac{(-1)^{p}p!}{n^{p+1}}<(>)g^{(p)}({\zeta}_{n})<(>)\frac{(-1)^{p}p!}{(n+p)^{p+1}} $,and the conclusion follows.
Any opinions guys? :)

Hi [b]LevonNurbekian[/b], thanks for your interest and nice  solution. In my solution I have used an interesting identity established  by Horst Alzer and Christian Berg. More exactly let us  define    [i] Alzer-Berg kernel[/i]   $ {\mathcal A}: [0,1] \to [0,\infty) $  by $ {\mathcal A}(0)={\mathcal A}(1)=0 $ and
\[ {\mathcal A}(t)= \frac{1}{\pi}t^t(1-t)^{1-t}\sin{ \pi t} \; \; \; ,\; \;  t\in (0,1).
\]
 According to H.Alzer-C. Berg ( see [i] Some classes of completely monotonic functions[/i], Annales   Academiae Scientiarum  Fennicae ,Mathematica, Volumen 27 (2002) 445-460 )  ,   we have   $    \int\limits_{0}^1 \frac{{\mathcal A}(t)}{1-t}\; dt = \frac{e}{2}-1  . $ Moreover
 \[ e- \left(1+\frac{1}{x}\right)^x
=   \frac{1}{x+1}+\int\limits_{0}^{1} \frac{{\mathcal c  A}(t)}{(1-t)(x+t)}\; dt\]
 \[ \left(1+ \frac{1}{x}\right)^{x+1}-e=   \frac{1}{x}+\int\limits_{0}^1 \frac{{\mathcal A}(t)}{(1-t)(x+1-t)}\; dt
\; ,\; (x>0) .
\]
[b] The differences $\nabla^p $. [/b]  Let us recall  that $ \nabla X_n =X_{n}-X_{n+1} $ and for a positive integer $p$  \[\nabla^pX_n=\nabla\left(\nabla^{p-1}\right)X_n=
    \sum\limits_{\nu =0}^p(-1)^{\nu}{p \choose \nu}X_{n+\nu}=(-1)^p\Delta^pX_n\; ,\; n\in {\mathbb N} . \]
[b]Lemma 1.[/b][i]  For $p\in {\mathbb N}  $ and $ t > -1 $ holds the identity
\[(1)\; \; \;
\nabla^p\frac{1}{n+t}=\frac{p!}{(n+t)_{p+1}}\; \]
 where $
(a)_0=1 $ and  $ (a)_m=a(a+1)\cdots (a+m-1) $ when $ m\in {\mathbb
N} .$ [/i]
 [b] Proof.[/b] We have \[
\nabla^p\frac{1}{n+t}=\sum\limits_{\nu =0}^p(-1)^{\nu}{p \choose
\nu}\frac{1}{n+\nu+t} = \sum\limits_{\nu =0}^p(-1)^{\nu}{p \choose
\nu}\int\limits_{0}^1t^{n+\nu+t-1}\; dt =\]
\[=\int\limits_{0}^{1}t^{n+t-1}(1-t)^p\; dt= \frac{\Gamma(n+t)\Gamma(p+1)}{\Gamma(n+t+p+1)}=
\frac{p!}{(n+t)_{p+1}}
\]
which proves (1).
 [b]Theorem.[/b][i]  If $e_n:=\left(1+\frac{1}{n}\right)^n\; ,\; \;
E_n:=\left(1+\frac{1}{n}\right)^{n+1}\; , $ then the following  representations are valid
\[(2)\; \;  \frac{\nabla^p e_n}{p!} =   -\; \frac{1}{(n+1)_{p+1}} - \;
\int\limits_{0}^{1}\frac{{\mathcal A}(t)}{(1-t)(n+t)_{p+1}} \; dt \]
\[   \frac{\nabla^p E_n}{p!}=    \frac{1}{(n)_{p+1}}+\int\limits_{0}^{1}\frac{{\mathcal
A}(t)}{(1-t)(n+1-t)_{p+1}} \; dt \; .\] [/i]
[b] Proof.[/b] Using (1),  for $p\ge 1 $  one finds
\[
   \nabla^p(e-e_n)=   -\nabla^pe_n=   \nabla^p\frac{1}{n+1}+\int\limits_{0}^{1} \frac{{\mathcal A}(t)}{1-t}\nabla^p\frac{1}{n+t}\; dt =
\]
\[=   \frac{p!}{(n+1)_{p+1}}+p!\int\limits_{0}^{1}\frac{{\mathcal A}(t)}{(1-t)(n+t)_{p+1}}\; dt \; .
\]
In other words $ \nabla^p e_n = -\;  \frac{p!}{(n+1)_{p+1}}\left(1+\int\limits_{0}^{1}\frac{{\mathcal A}(t)}{1-t}
\cdot\frac{(n+1)_{p+1}}{(n+t)_{p+1}}\; dt \right) $ which proves the first representation. The second equality from (*) together with (1)  completes the proof.
 [b]Corollary 1.[/b] [i]If $ p,n \in {\mathbb N} $ , then
\[\frac{e}{2(n+1)_{p+1}}
< -\; \frac{\nabla^pe_n}{p!}<
\frac{e}{2(n)_{p+1}}-\frac{p+1}{(n)_{p+2}}
\]
\[
\frac{e}{2(n+1)_{p+1}}+\frac{p+1}{(n)_{p+2}}\le
\frac{\nabla^pE_n}{p!} < \frac{e}{2(n)_{p+1}}.
\] [/i]
[[b] Proof.[/b]  For $ \alpha\in [0,1]$ we have $ 1\le
\frac{(n+1)_{p+1}}{(n+\alpha)_{p+1}}\le
\frac{(n+1)_{p+1}}{(n)_{p+1}}=1+\frac{p+1}{n}, $ and   and (1), (3)
we obtain the desired inequalities.
  For the first positive integral values of $p $ one finds
[b]Corollary 2.[/b] $\;    -\; \frac{e}{2}+ \frac{3-e}{2n}<
(n+1)^2\left(e_{n}-e_{n+1}\right)<-\;    \frac{e}{2}+ \frac{e}{2n} $
\[    \frac{e}{2}-\frac{e-2}{n}<   (n+1)^2\left(E_n-E_{n+1}\right)<
\frac{e}{2}+\frac{e}{2n},\]
    \[  -e+\frac{6}{n+3}<    (n)_3\left(e_{n+2}-2e_{n+1}+e_n\right)<
-   e + \frac{3e}{n+3}\]
\[    e-\frac{3(e-2)}{n+3} < (n)_3\left(E_{n+2}-2E_{n+1}+E_n\right)<
 e\; .\]
 From Corollar  1 one finds
 [b]Corollary 3= SOLUTION pf the proposed problem .[/b] [i]If $   e $   is Napier's constant, then
\[ e=\frac{2}{p!} \lim\limits_{n\to \infty}n^{p+1} \nabla^p E_n =-\;
\frac{2}{p!} \lim\limits_{n\to \infty}n^{p+1}\nabla^p e_n . \] [/i]
Note: perhaps some of above remarks will be published in  [i]Octogon[/i] . Regards,Alex

Look guys I have improved my solution,and it will be possible to read from beginning to end.I don't need $g(x)$ at all.I only need $f(x)$.As I told one can show(just boring,but no so large computations) $f^{(p)}(x){\sim}f{\cdot}t^{(p-1)}{\sim}\frac{(-1)^{p}p!}{x^{p+1}}$,when $x$ goes to $\infty$.That means that $f^{p}$ for fixed $p$ is monotonic from fixed(suppose $x>A$) place.Since
${\triangle}^{p}E_{n}={\triangle}^{p}f(n)=f^{(p)}({\zeta}_{n})$,where $n<\zeta_{n}<n+p$,then $f^{(p)}(n)<(>){\triangle}^{p}E_{n}<(>)f^{(p)}(n+p)$, from $n>A$,so
$\frac{f^{(p)}(n)}{\frac{(-1)^{p}p!}{n^{p+1}}}<(>)\frac{{\triangle}^{p}E_{n}}{\frac{(-1)^{p}p!}{n^{p+1}}}<(>)  \frac{f^{(p)}(n+p)}{\frac{(-1)^{p}p!}{n^{p+1}}}$,from which the conclusion follows(since
 $f^{(p)}(x){\sim}\frac{(-1)^{p}p!}{x^{p+1}}$).Signs $>$ or $<$ depend on the oddness of $p$.
How is it now?
 :)


Note that 2ksin2k +(180-2k)sin(180-2k)=(180)sin2k
 Now let S=sin2+sin4+...+sin90
then (sin1)S=sumsin(2i)(sin1) = 1/2 sum cos(2i-1)-cos(2i+1)
=1/2(cos1-cos91)=1/2(cos1+sin1)
hence: S=1/2 +1/2cotg1
and so your desired sum is: 2(1/2+1/2cotg1)-1=cotg1.

But..., what's the reason for showing the triangles?

Pure algebraic problem lies in Combinatoric Section!
I will move it Algebra Solved Problems.


Hi I am currently a grade 12 student, and wondering if the aops would be useful for the 3 maths in ontario: Geometry and discrete, Calculus, and Data Management(mostly discrete and calculus). Secondly, would these books be helpful for the Euclid contest.

Thanks.

For Gr. 12 Math, you really don't need AoPs books in my opinion. If you lack experience with contest like Euclid/ AMC/ AIME, you might want to try your hand on the AoPs books. They do a really good job to familarize you with some basic technique and way of thinking in solving contest problems.

illmatican, which school are you from?


[quote="sharkman"]I was given this:

2sin^2(x) + sin(x) - 1 = 0

After doing the trig work, I found the answers to be:

x = pi/6 or 30 degrees

x = 5pi/6 or 150 degrees

x = 3pi/2 or 270 degrees

I found the answers for x by looking at a trig chart of degree measures located in the back of the textbook.

My questions:

How do I find the same answers WITHOUT (1) using a calculator and (2) looking at trig charts?

There's got to be a way, right?

Do I need to memorize the unit circle?

Do I need to use a reference triangle?

If it is a reference triangle, how do I use it correctly?

Help!

Thanks![/quote]

Gr... Use $ \text{\LaTeX}$.
[hide="Hint"]Treat it as a quadratic.[/hide]

[hide="Solution"]Let $ y = \sin x$. Then:
$ 2y^{2}+y-1 = 0$
$ (2y-1)(y+1) = 0$
$ y = 1/2 \textbf{ or }-1$

Re-substituting,
$ \sin x = 1/2 \textbf{ or }-1$

Now, you can solve for $ x$ easily:
$ x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3 \pi}{2}$[/hide]

Don't memorize. That's basically the motto of AoPS. Think of a unit circle, don't just cram in the information.

[quote="sharkman"]Yes, but how do I use the UNIT CIRCLE to find the answers?

I don't know how to use the unit circle.

I hate looking at trig tables.[/quote]

Pick a point $ (x,y)$ on the unit circle. Let L be the line formed by this point and the origin. If L makes an angle of $ \theta$ with the x-axis, then
$ \sin \theta = y$
$ \cos \theta = x$

Picking a point on the unit circle creates a right triangle with the point, the x axis, and the origin.

If you know trig of right triangles, this may be easier to visualize for you.

[quote="spell it out 11"]If you memeorize sin 30, no other trig needds to be memorized because they all come fmr sin 30 (except for the multiple of 90 angles)[/quote]

Really, the only memorization needs to be done is with
$ \sin 30^{\circ}$
$ \sin 45^{\circ}$

And, you don't even need to memorize that. Just remember your 30-60-90 and 45-45-90 triangles.

If you just memorize for a test and don't really care about how it is derived, you are going to forget it in just one year.

Tell me, how do I use right triangle trigonometry to find the values of angles (in degrees and radians) without looking at trig charts.  I hate looking at charts. It is more professional being able to find angles through the unit circle or right triangle trigonometry.

Can someone show me here?

Most of those values are transcendental, and furthermore not expressable by radicals. You can always use infinite series, but anything that you need for any reasonable problem can be derived using sin30, sin45, and the unit circle.


function
logarithms
algebra
polynomial
absolute value

http://en.wikipedia.org/wiki/Talk:Lambert_W_function

Can someone please explain the 3rd step of the solution on the above link?

I am an Algebra II student with no knowledge of math beyond precalculus.  I understand logs, but unfortunately my teacher reduced them to simple equations.

I appreciate your help

Potential

12 views, I am sure one of you understands the 3rd step.  Please help.

The only part that troubles me is why they used e.

Recall the properties of logs and exponentials: $e^{\ln|x|}= |x|$,  $e^{-\ln|x|}= \frac{1}{|x|}$  because $-\ln|x| = \ln|x^{-1}|$. Recall that $\ln(x^{2}) = 2\ln|x|$ or in general $\ln|x^{a}| = a\ln|x|$.  Step 3 in the solution to your post uses these properties.  We have $\frac{5t}{2^{t}}$  which can be written as $5te^{-t\ln(2)}$.

EDIT: Fixed the confusing lack of absolute values. Thanks  for pointing that out t0rajir0u.

[quote="illcrowflu"]$e^{\ln|x|}= x$[/quote]

I am not really sure why you use the absolute value signs on the LHS only.  Clearly $e^{\ln |-1|}=-1$ is a false statement.

note: also, the use of $e$ probably is chosen because if you ever are going to try to approximate the value, then $e$ makes the Taylor polynomials much, much prettier.


Let P, Q, R be polynomials with real coefficients satisfying P^2+Q^2=R^2, where deg(P) = deg(R) = 3, deg(Q)=2. Denote by s number of real roots (counted with multiple) of polynomial P.Q.R. Find the smallest value of s.

<i>Source: by Namdung on Russian theme</i>

I hope that the following solution is good, though I doubt:
  The number is in my opinion 6. Let us take P(X)=X(X^2+X), Q(X)=X(X+0.5), R(X)=X(X^2+X+0.5). Then P^2+Q^2=R^2 and PQR has 6 real roots (with multiplicities).
  Now let us prove that it is at least 6. Take a a real root of R. Then P(a)=Q(a)=0. So we can take P(x)=(x-a)(ex^2+fx+g), Q(x)=(x-a)(cx+d) and R(x)=(x-a)(hx^2+ix+j) where e,c,h are non-zero. Changing P with -P and R with -R doesn't affect anything, so we can suppose that e>0 and h>0. Since for all x we have (ex^2+fx+g)^2+(cx+d)^2=(hx^2+ix+j)^2, we must have ( (h-e)x^2+(i-f)x+j-g)((h+e)x^2+(i+f)x+j+g)=(cx+d)^2. Thus, h^2=e^2 (so h=e), (i-f)(h+e)=0, i^2-f^2+(j-g)(h+e)=c^2, (i-f)(j+g)+(i+f)(j-g)=2cd and finally j^2-g^2=d^2. It follows that f=i, h=e, j^2-g^2=d^2, 2e(j-g)=c^2 and  f(j-g)=cd. We prove that at least one of hx^2+ix+j  and  ex^2+fx+d has real roots. Otherwise, f^2<4eg and i^2<4hj. Thus, f^2<4eg and f^2<4ej and so f^2<2e(j+g). But (cd)^2=(j^2-g^2)*2e(j-g)=f^2(j-g)^2, from where f^2=2e(j+g), contradiction.
     So at least one of the two trinomials has real roots and their product has at least 2 real roots. So PQR is divisible by (x-a)^3(cx+d)(x-x_1)(x-x_2), where x_i are the two real roots that the product must have. Therefore, PQR has at least 6 real roots.

May we know why you put this problem here? It is all but superiour algebra!

Let n be a positive integer and {A, B, C} be a
partition of ${1, 2, 3, ..., 3n}$ such that $|A| = |B| = |C| = n.$
Prove that there exist x from A, y from B, z from C such that one of
x, y, z is the sum of the other two.
\end{problem}
\begin{proof}
Statement: Let ${1,2,...,n}$ be 3-colored, such that each color is
used at least $n/4$ times. Then, the equation $x+y = z$ has a
tricolor solution.

Suppose for a contradiction that it is possible to color each of the
integers $1,2,...,n$ by one of the three colors, say red, blue,
green, such that there is no tricolor solution of the equation $x+y = z,$ and such that each of the color is used at least $n/4$ times.

Wlog, we may assume that 1 is red. Thus, there do not exist two
consecutive integers which are blue and green. In the following, the
interval $[a,b]$ denotes ${a,a+1,...,b}.$ Note that $|[a,b]| = b-a+1.$ From above the blue and green integers form some disjoint
monochromatic intervals, with at least a red integer as a separator.

- First, we prove that if there is a green (resp. blue) interval $A$
such that $|A| > 1$ then there is no blue (resp. green) interval $B$
such that $|B| > 1$ : Since there is a finite number of interval, we
may consider the green interval $A = [a_{1}, a_{2}]$ with maximal length
among the green intervals, and a blue interval $B = [b_{1}, b_{2}]$ with
maximal length among the blue intervals. Then, for $a$ in $A$ and
$b$ in $B$, $|a-b|$ is not red, since we have no tricolor solution
to $x+y = z.$ Let $C = \{|a-b| / a \in A, b \in B\}.$ Then, $C$ is
not empty and contained in ${1,2,...,n}.$ Moreover, it contains no
red element thus, from above, it is a monochromatic interval. More
precisely : $C = [b_{1}-a_{2}, b_{2}-a_{1}]$ if $a_{2}< b_{1}C = [a_{1}-b_{2}, a_{2}-b_{1}]$ if $b_{2}< a_{1}.$ Then, in both cases $|C| = |A|+|B|-1.$ Then, if $|A|, |B| \geq 2,$ then $|C| > |A|$ and $|C| > |B|,$ which contradicts the maximality of $A$ and $B.$

Wlog, we then assume that there is no green interval $A$ such that
$|A|> 1,$ that is there do not exist two consecutive green integers.

- Then, we prove that there exist two consecutive blue integers : If
not, since blue and green integers are never consecutive ones, and
since 1 is red, it would follow that for each blue integer $b$, the
number $b-1$ is red. Similarly, for each green integer $g$, we would
have $g-1$ red. Thus, there would be at least $n/2$ red integers,
which contradicts that the are more than $n/2$ non-red integers.

- Now, we prove that there exist $g$ and $g+2$ which both are green
: First, note that, from above, if $g \leq n-1$ is green, then $g \geq 2$ and $g-1$ and $g+1$ are red. Suppose, for a contradiction,
that there is no green integer $g$ such that $g+2$ is green too. Let
$N(R), N(G), N(B)$ be the number of red, green, blue integers
respectively. When $g$ runs over the set of the green integers with
$g \leq n-1,$ the intervals $[g-1, g+1]$ form a family of pairwise
disjoints intervals, each containing 2 red integers. Thus :- If $n$
is not green, then $N(R) \geq 2N(G),$ from which we deduce that $n = N(R)+N(G)+N(B) \geq 3N(G)+N(B) > 3n/4+n/4 = n,$ a
contradiction. - If $n$ is green, then $N(R) \geq 2N(G)-1$ and 2
is green (if not 1 has not been counted and we have the same
contradiction as above). If $B$ is the blue interval with maximal
length, with the above notations, then $b_{2}< n.$ Thus $b_{2}+1$ is
red. Moreover, since $|B| \geq 2,$ it follows that $b_{2}-1$ is
blue. Thus $(b_{2}-1)+2 = b_{2}+1$ is a tricolor solution of $x+y = z.$ A contradiction again, hence result.

Now, let's suppose that $g$ and $g+2$ are green. Then $g+1$ is red
and, if $B$ is the same as above, we either have $b_{1}> g+2$ or
$b_{2}< g.$ - If $b_{1}> g+2$ : Let $E = \{b-g / b \in B\}\cup \{b-g-2 / b \in B\}.$ We will prove that E is a monochromatic
interval : Let $x$ be in $E$. If $x = b-g,$ then $x+g = b,$ where
$g$ is green, $b$ is blue. If $x = b-g-2,$ then $x+(g+2) = b,$
where $g+2$ is green and b is blue. In both cases, we deduce that
$x$ is not red. From $B = [b_{1}, b_{2}],$ we deduce that $E = [b_{1}-g, b_{2}-g] \cup [b_{1}-g-2, b_{2}-g-2].$ And, since $b_{1}< b_{2},$
we have $b_{2}-g-2 \geq b_{1}-g-1,$ thus $E$ is an interval.
Then $E$ is an interval which contains no red integer. From above,
we deduce that E is monochromatic, but not red. Moreover $|E| = b_{2}-g-(b_{1}-g-2)+1 = |B|+2.$ Thus, from the maximality of
$B,$ we deduce that $E$ is not blue. And, since $|B| > 2,$ it
follows that $E$ is not green. A contradiction.

- If $b_{2}< g,$ we use the same reasoning on $E' = \{g-b / b \in B\}\cup \{g+2-b / b \in B\},$ to obtain the same contradiction.

And we are done....

Note that the result is optimal in the sense that for $n= 4k,$ we
may color the odd integers in red, the numbers of the form $4p+2$ in
blue and the numbers of the form $4p$ in green. Then $N(R), N(B), N(G) \geq 4$ and, by parity argument, there is no tricolor solution
of $x+y = z.$
\end{proof}

Surely, I thought he will immediately remove his post  :maybe:


Treething has to spend 63 hours in community service. At the very least, how many weeks will this take, given that he can only go to community service on weekdays and he can only spend 108 minutes a day on community service?

[quote="Treething"]Treething has to spend 63 hours in community service. How many weeks will this take, given that he can only go to community service on weekdays and he can only spend 108 minutes a day on community service?[/quote]

[hide]$63\times 60 \div 108 \div 5=\boxed{7 \text{ weeks}}$[/hide]

Correct. I added a little clause in order to make people bent on poking my problems to pieces unhappy.

[hide]$63\times60=3780$ minutes of community work.
$108\times5=540$ minutes able to be done per week.
$3680/540=7$ weeks.[/hide]


let (a_n)_n be a sequence of real numbers s.t. a_1=a>1 and
a_(n+1)=a_n+\sqrt (a_n)-1
Compute lim{n->oo}(4a_n-n^2)/(nlnn)

This looks like 2*Cesaro -Stolz(CS) to me...

hmmm... a_n is stricltly incresing; and from the recurence we get that lim(a_n)=oo. Using the fact that lim(a_n)=oo we get that lim a_(n+1)/a_n=1 the using CS fo x_n=sqrt(a_n)/n we get that lim(x_n)=1/2 (I think...) using the last result you compute the limit:
lim (4a_n-n^2)/(nlnn)=2 lim (yn)
yn=(2sqrt(a_n)-n)/ln(n).
we again apply CS for y_n and you get the limit, it is something negative, i don't have time to compute everything now, sorry!
hope it's okay, pls check the computations..
cheers! :D :D