inequalities
calculus
derivative
inequalities proposed

It's my conjecture. I've proved it for the inequalities between the first side and last side , the second side and last side  but the first ineq 's really hard!
$\sum\frac{a^{3}}{a^{2}+2b^{2}}\ge \sum\frac{a^{3}}{2a^{2}+b^{2}}\ge \frac{a+b+c}{3}$

See here the post number 6:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22937

[quote="silouan"][quote="arqady"]  $\Leftrightarrow \Leftrightarrow\sum_{cyc}\left(\frac{a^{3}(a+b)(a-b)}{a^{2}+2b^{2})(2a^{2}+b^{2})}-\frac{2}{9}\cdot(a-b)\right)$[/quote]

Arquady extremely nice solution.... :D
I think that in the above line you used first derivative technique to find the constant $\frac{2}{9}$ .
Could you share this method with us ??Thank you[/quote]
I think that it's not very difficult to explain why he do that.
First of all, I guess that he intend to use SOS method to solve this problem.
Than he factor the fractions. And we have : the number $\frac{2}{9}$ just because he let a=b=c=1 so that you can see which number is necessary.

@: Sorry, my english not very good.

But do you have any other solution without S.O.S?


inequalities
geometry
vector
rectangle
Pythagorean Theorem

How do you prove you guys the Pythagorian Teorem ($ a^{2}\plus{}b^{2}\equal{}c^{2}$)

[quote="Elio (n)"]How do you prove you guys the Pythagorian Teorem ($ a^{2} \plus{} b^{2} \equal{} c^{2}$)[/quote]

Drop the perpendicular $ CD$ to $ AB$; then $ \Delta ABC, \Delta ADC, \Delta BCD$ are all similar since they have equal angles. This implies $ \frac{\overline{AD}}{\overline{AC}}\equal{}\frac{\overline{AC}}{\overline{AB}}\implies \overline{AC}^2\equal{}\overline{AB}\cdot\overline{AD}$ and in the same way $ \overline{BC}^2 \equal{} \overline{AB}\cdot\overline{BD}.$ Add these two inequalities to get the result.

:P

http://en.wikipedia.org/wiki/Pythagorean_theorem
http://mathworld.wolfram.com/PythagoreanTheorem.html

Proof using Differential Equations sounds funny.

The uber-cheap way is to arrange four "copies" of an arbitrary right triangle into a square, and then use the area of the squares formed for a really easy proof.

The "standard" way (given in most geometry textbooks) involves drawing the altitude and using the geometric mean theorem and the similarity of the resulting triangles.

How about this proof for the Pythagorean Theorem for any number of dimensions:
Let $ \textbf{a}$, $ \textbf{b}$, and $ \textbf{c}$ be three vectors such that $ \left|\textbf{a}\plus{}\textbf{b}\right|\equal{}\textbf{c}$ and $ \textbf{a}\cdot \textbf{b}\equal{}0$.
Then we have
$ \left(\left|\textbf{a}\plus{}\textbf{b}\right|\right)^{2}\equal{}\textbf{c}^{2}$.
This implies $ \left(\textbf{a}\plus{}\textbf{b}\right)\cdot\left(\textbf{a}\plus{}\textbf{b}\right)\equal{}\textbf{c}\cdot\textbf{c}$, which is equal to $ \textbf{a}\cdot\textbf{a}\plus{}2\textbf{a}\cdot\textbf{b}\plus{}\textbf{b}\cdot\textbf{b}\equal{}\textbf{c}\cdot\textbf{c}$.
We can re-write this as
$ \left|\textbf{a}\right|^{2}\plus{}2\textbf{a}\cdot\textbf{b}\plus{}\left|\textbf{b}\right|^{2}\equal{}\textbf{c}^{2}$
But recall that the dot product is 0, so this reduces to
$ \boxed{\textbf{a}^{2}\plus{}\textbf{b}^{2}\equal{}\textbf{c}^{2}}$

You mean $ |a|^2 \plus{} |b|^2 \equal{} |c|^2$.  I don't consider that a valid proof; it's something of a tautology, since concepts like the dot product and length are defined so that the Pythagorean theorem is true.

In any case, there are over 400 published proofs of the Pythagorean theorem.

This isn't really related to the proof but anyway..........

I have a friend at University. One of his Mathematics Course (I think) is about the History of Mathematics. He got a test and it only has one question: 'Find as many proofs of Pythagoras' Theorem as possible' (3 hours).

I think the top guy wrote 40-50 proofs (ranging from elementary Mathematics to some weird Topological stuff, I think).

With regard to BanishedTraitor, that is impressive.

I tried to prove that in the square . He did not sodisfate me   :maybe:

Take a look:
$ a$ is the leg of the square
$ c$ is the diagonal of the sqare
$ h$ is the high sur $ c$

$ S \equal{} a^{2}$
$ S \equal{} c*h$ if we look we can find that $ c \equal{} 2h$

$ a*a \equal{} c*\frac {c}{2}$
$ 2(a^{2}) \equal{} c^{2}$
$ a^{2} \plus{} a^{2} \equal{} c^{2}$

What do you think for this? Is this a kind of proveing?  :wink:  :huh:

Unfortunately I think it only proves the isosceles right triangle case.

It can be generalized to rectangles with some additions, I think.


advanced fields
advanced fields theorems

I have a course using the book [i]Logic in Computer Science: Modelling and Reasoning about Systems[/i] (Michael Huth), and there is a review about it saying that "[i]This peculiarity of the order of exposition, partly repeated in Chapter 2, is in my view the most controversial feature of the book, as it can create in the reader with little or no prior experience in logic (to whom the book is clearly addressed) the wrong impression that logical semantics is mainly used to give meaning to deductive systems, rather than that deductive systems are designed in order to capture logical validity and consequence[/i]" and "[i]Further in Chapter 1 the authors explain and prove the soundness and completeness of the earlier introduced system of natural deduction (not of the propositional logic itself, as the subsection titles suggest)[/i]". Unfortunately, I don't understand this review :(, so I hope someone will help me explain it more clearly :).
By the way, could you please suggest some good books for my "[i]Logic in Computer Science[/i]" course? Thanks very much for your help :).

I took a philosophy course a few years ago called "A Historical Approach to Mathematical Logic".  Among other books (Aristotle, Kant, Frege), we used "Introduction to Logic" by Patrick Suppes (Dover, 1957).  It's old, yes, but logic hasn't changed, and this is a nice, crap-free book on propositional logic.

It takes you gently into interpreting sentences as logical forms, slowly adding more and more logic notation, building up to the axioms and theorems and rules of inference.  It also goes over some ideas about what "definition" means, and it has a couple chapters at the end on set theory.

As for that review, I think it's suggesting that a reader without much experience with logic would interpret the book as telling you how to USE logic to understand typical speech (in particular, prose as used in mathematics, sciences, and philosophy to get from hypotheses to conclusions).  On the other hand, you should understand that typical speech is used to EXPRESS logical thought.

For example, we might say that Bob is human, and that humans are mortal, and therefore Michael is mortal.  After having read the book, we then re-express this as logical semantics:
B -> H
H -> M,
therefore  B -> M.

We want to avoid thinking about logic as a tool to understand speech.  We really want to think about logic as what we find after we strip away the additional bows and ribbons and shiny wrapping paper that were added by our minds in speaking.  This whole B->H stuff is what was in our minds subconsciously, and we used our other languages (English) to assign words to these things in order to say "Bob is Human".  We designed the system for deduction (A is B and B is C means A is C) as an easier way for us to reach our logical conclusion without stripping away too much of the extra trappings added by English language.    The important thing to remember is that logic is there all along underlying everything, and everything we do is done with that in mind, at least subconsciously.


ratio HMMT calculus LaTeX geometry perimeter trigonometry

You are in the middle of a circular swimming pool (radius $r$ say). You dad is angry with you  and is on the outside of the pool, and runs around trying to get to you. Say his running speed is $m$. Whats the minimum speed you have to be able to swim at to get out of the pool before he catches you? (fill in the units yourself)

[hide] Well obviously you're going to go for the point that's furthest away from him... [/hide]

[quote="Chocoboom"][hide]well you gotta swim a distance of $r$ and he has to run a distance of $r\pi$ to get around the pool.. you you gotta swim at a rate of at lease $\frac{m}{\pi}$[/hide]

ytayayaya?? too easy? (im eating popcorn and typing at the same rtime[/quote]

what if you turn, then he has to run extra distance

oh damn.. then you gotta locus it huh... my bad

[quote="seamusoboyle"]Well if you swim outwards and sidewards in a direction such that dad is on the opposite side of the center for as long as possible, then you dont have to swim as far to get to the edge and dad is still on the opposite side of you...

From there it gets tricky...[/quote]
Nice thought, seamusoboyle!

[hide="informal solution"]By the way, can the dad swim? If not, then stay at the center  :D [/hide]

Well, suppose that you are at a distance $R$ from the center, and the dad is on the opposite side of the center. Your speed is $u=a\cdot m$

If you move sidewards, on the circle with radius $R$, then your angular speed is $\omega_{1}= \frac{u}{R}$

The dad's angular speed is $\omega_{2}= \frac{m}{r}$

$\omega_{1}\geq \omega_{2}\Leftrightarrow$

$\frac{u}{R}\geq \frac{m}{r}\Leftrightarrow$

$R\leq r\frac{u}{m}\Leftrightarrow R\leq ra$

So let $R_{o}=ra$.

As $R<R_{o}$ you can achieve greater angular velocity than dad's.
So you can (at the same time) move slightly outwards, such that the dad is always at the opposite side.
In other words, you can move from the point $(R,\theta)$ to the point $(R+dR, \theta+\omega_{2}dt)$

When you reach the circle $R=R_{o}$, then you have to go [u]only[/u] outwards.

The distance you have to swim is $r-R_{o}$.
The dad has to run at a distance $\pi \cdot r$

So it's necessary that $\frac{r-R_{o}}{\pi \cdot r}<\frac{u}{m}=a \Leftrightarrow$

$r-R_{o}<\pi ar \Leftrightarrow$

$r-ar<\pi a r \Leftrightarrow$

$1-a<\pi a \Leftrightarrow$

$1<(\pi+1) a \Leftrightarrow$

$a>\frac{1}{\pi+1}\approx 0.24145$

[quote="Altheman"]what if you turn, then he has to run extra distance[/quote]
Well, maybe there is a better way to escape.

Look at the scetch below. The outer circle is the pool.

The inner circle is the circle with radius $R_{o}$

In my last post I said that you will swim on the segment $AC$.

Suppose that the dad goes anti-clockwise. Then you can move on the curve $AB$.

If the ratio $\frac{S_{AB}}{S_{DC}+S_{CB}}$ is less than $\frac{AC}{S_{DC}}$ then it's better to move on $AB$.

So we have to find a point $B$ and a curve $AB$ such that the above condition holds.

The curve $AB$ should be such that the dad won't change direction (at any time he realizes the value of the [u]directed[/u] angle $DOA$ and decides wich direction will he follow).
So you shouldn't swim on the [u]segment[/u] $AB$, because then your dad will go clockwise and he will catch you  :o

With $S_{x}$ above I denote the length of the curve $x$
I wrote $S_{DC}+S_{CB}$ instead of $S_{DB}$ to show that it is the large arc $DB$, which contains the point $C$

And of course since the quickest path is a straight line we can replace that curve with a line methinks, though only after a point wherby we can be sure dad wont turn around again.

But now that i think about it that wont actually happen will it? because your outside the $R_{0}$ circle so he'll always be best off staying in the same direction.

In conclusion, im pretty sure we can replace that curve with a straight line...

wow - almost sounds like the opposite of a recent HMMT calculus problem (the one with the 3 stooges).

Basically, the optimal strategy would be like this (I think): call dad mobile particle $D$, call yourself $Y$ and the center of the pool $O$. Let $D'$ be another mobile particle diametrically opposite $D$, and let $Y'$ be the point on circle $O$ that lies on the radius containing $Y$, at all times, even when dad runs and you swim. Let's assume we have a smart Dad. Since your optimal distance from $Y$ to the edge of the pool would be the straight line $\overline{YY'}$, dad always runs on minor arc $DY'$ (I don't know the arc sign in LaTeX), so the objective is to constantly keep $Y'$ and $D'$ at the same point. I am Certain (notice the capital "c") that this problem involves calculus, so let's move it to that forum.

When I first read this problem (some yrs ago) I thought, like the user Chocoboom, that the ratio was $\pi$, so I conclude that this problem is not interesting. Now I see that this problem is [color=orange][b]excellent![/b][/color]  :thumbup:

[quote="seamusoboyle"]In conclusion, im pretty sure we can replace that curve with a straight line...[/quote]

I agree!  :lol: There is no need to go on a curve since the segment is shorter.

I tried the problem again. At the beginning, I believed that you should move slowly on the segment, but then I saw that you can move at your maximum speed and the dad won't change direction.

Well, summarizing.. The problem has two stages:

[color=green]At the 1st stage you should move such that the dad is at the opposide side of the center. With these conditions you can succeed a maximum distance from the center, call it $R_{o}$. So at the end of the first stage you are on the circle $(O,R_{o})$[/color]

[color=brown]At the 2d stage, you start from the circle with radius $R_{o}$, having the dad at the opposite side. Say that you are at the point $A(0,R_{o})$ and the dad is at $E(0,-r$). Your aim at this stage is to move to a point $B$ which is on the perimeter of the pool, before the dad reach this point.[/color]

[color=blue][b]Recall the notations:[/b]

$r:$ the radius of the pool
$\omega_{2}:$ dad's angular speed
$m=\omega_{2}\cdot r:$ dad's speed
$u=a \cdot m:$ your speed
$R_{o}= ar$[/color]

Suppose that the dad is moving anti-clockwise, starting from the point $E$.

Let $F \in DO \cap [AB]$.
If the dad see that you are between $B$ and $F$ then he will change direction of moving. [color=blue]To simplify the expressions, we suppose that the dad doesn't changes direction, even if you are [b][u]on[/u][/b] the point $F$[/color]. So, at any time you must be on the segment $[AF]$, and, if possible, [b][u]on[/u][/b] the point $F$.
The dad has to cover all the semicircle, so he will run at his maximum speed $m$.

In the triangle $OAF$ denote:
$\angle FOA = \theta$
$\angle OAF = A$
$\angle AFO = \pi-A-\theta$

[hide="The angle A is not acute"]
If $A=90^{o}$ then the ray $AB$ is tangent to the circle $(O,R_{o})$, so it has one common point with it
If $A>90^{o}$ then the ray $AB$ has one common point with $(O,R_{o})$
If $A<90^{o}$ then the ray $AB$ has two common point with $(O,R_{o})$

At the last case, the ray $AB$ intersects the circle again at a point $G$. There is no need to go there, we are already on the circle

So we have $A\geq \frac{\pi}{2}$
--------------------------------------------------------------
[/hide]

Using the fact that $\hat{A}$ is not acute I proved that you can always swim with your maximum speed $u$, and the dad won't change direction (if someone ask it, I'll post the proof)

Using cosines law in $\triangle OAB$ we get $AB = r\sqrt{a^{2}+1-2a\cos\phi}$, where $\phi = \angle AOB$

The dad must run at a distance $(\pi+\phi)r$

you can escape if $\frac{\pi+\phi}{\omega_{2}}> \frac{r\sqrt{a^{2}+1-2a\cos\phi}}{u}\Leftrightarrow$

$\frac{\pi+\phi}{\omega_{2}}> \frac{r\sqrt{a^{2}+1-2a\cos\phi}}{a\omega_{2}r}\Leftrightarrow$

$\pi+\phi > \frac{\sqrt{a^{2}+1-2a\cos\phi}}{a}\ \ (1)$

The condition $A\geq \frac{\pi}{2}$ is translated as $\cos\phi \geq a$.

With trial end error I found that the smaller $a$ such that $(1)$ holds is about $a \approx 0.217233628211221$ and $\angle OAB=90^{o}$

[hide="Take a practice"][color=olive]By the way, I wrote a flash program to play this game. Click on the smurf to begin. Move the mouse to show the direction of the move. If you don't move quickly the gargamel will catch you. If you loose, press Start.
At any time you can change the values on the boxes:

[b][u]m[/u][/b]: The speed of the gargamel

[u][b]ratio[/b][/u]: The value of $a$ is calculated by the formula $a = \frac{ratio}{\pi+1}$. If you put ratio=1, then $a$ will be $\frac{1}{\pi+1}$, according to the first approach. The default value of the ratio is 1.2

[u][b]dt[/b][/u]: Flash player calls the drawing every dt milliseconds. I don't know exactly how this changes the appearing. I put it because when you change the value of m it runs so quickly. Just test some values

You can find this flash game [url=http://www.rapiduploader.com/890878]here[/url][/color][/hide]

[quote="seamusoboyle"]This particular problem dosnt require calculus, its an olympiad problem i got from somewhere...[/quote]

Don't olympiad problems go in the Olympiad forum?
geez I slaved over that problem for at least an hour trying to find the perfect curve, and I got to approximately 1/4.12, just less that 1/(pi+1)


how do you find the inradius of a triangle given lengths of sides and area?

use the area formula $rs$
we know area (many ways of doing so-herons, etc) and we know semiperimeter so we can do it

if we plug directly into herons, we get $r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$

this isn't hard enough for interm but whatev

[quote="breez"]Also, a more computationally friendly way:

$A = \frac{abc}{4R}$

$R = \frac{abc}{4A}$[/quote]

That's for the circumradius, for the inradius we've got:

$A = s \cdot r$, where $s$ is the semiperimeter of circumscribed triangle.


What is the remainder when the sum of the first 102 counting
numbers is divided by 5250?

First we need to find the sum of the first 102 counting numbers. We can plug this into our formula for the first $ n$ numbers:

$ \frac{n(n\plus{}1)}{2}$

$ \frac{102(102\plus{}1)}{2}$

$ \frac{10506}{2}$

$ 5253$

And obviously, the remainder is $ \boxed{3}$.


limit function logarithms calculus derivative integration algebra

For $ m$ a positive integer and $ n$ a non-negative integer, prove (without recourse to l'Hopital's rule) that

$ \lim_{t \to \infty} e^{\minus{}mt}t^{n} \equal{} 0$.

[quote="AndrewTom"]For $ m$ a positive integer and $ n$ a non-negative integer, prove (without recourse to l'Hopital's rule) that

$ \lim_{t \to \infty} e^{ \minus{} mt}t^{n} \equal{} 0$.[/quote]

make a Taylor expansion toward $ e^{mt}$

Consider the function $ f(x)\equal{}xe^{\minus{}ax}$ for some $ a>0.$ We have $ f'(x)\equal{}(1\minus{}ax)e^{\minus{}ax},$ which means that $ f(x)$ is both positive and decreasing on $ [a^{\minus{}1},\infty),$ which in turn implies that it has a finite limit as $ n\to\infty,$ since it's the limit of a monotone bounded function.

Now let $ b>a>0.$

$ \lim_{x\to\infty}xe^{\minus{}bx}\equal{}\left(\lim_{x\to\infty}xe^{\minus{}ax}\right)\left(\lim_{x\to\infty}e^{\minus{}(b\minus{}a)x}\right)$

provided both limits exists. But we've already argued that the first limit exists and is finite, and the second limit is clearly zero. So we now have

$ \lim_{x\to\infty}xe^{\minus{}bx}\equal{}0$ for all $ b>0.$

Then the original problem is

$ \lim_{t\to\infty}e^{\minus{}mt}t^n\equal{}\left[\lim_{t\to\infty}e^{\minus{}(m/n)t}t\right]^n\equal{}0^n\equal{}0.$

--------

A numerical aside, which I've used as a class example: according to this, $ \lim_{x\to\infty}x^{1000}e^{\minus{}.001x}\equal{}0.$ But that function rises to a maximum, which is fairly large, before coming back down and tending to zero.

The actual numerical question: express, as a floating-point number with at least four significant figure accuracy, the maximum value of $ f(x)\equal{}x^{1000}e^{\minus{}.001x}.$

(To stay in the spirit of the problem, use only an "ordinary" calculator or spreadsheet, not a system that allows arbitrary precision arithmetic. This does require you to be slightly smarter than your calculator.)

Another way:

 A [b] Lemma :[/b] If $ a_k > 0$ and $ \lim_{n\to \plus{} \infty}{a_{n \plus{} 1}\over a_n} \equal{} l < 1,$ then $ \lim_{n\to \plus{} \infty}a_n \equal{} 0.$

(easy to prove).  Easy  is also the application  to $ a_n \equal{} {n^k\over e^{nq}}$ ($ k,$ $ q$ integers).  The same conclusion holds if you have a subsequence $ \{n_k\}$ in place of $ \{n\}$ and then can say that your limit is zero.

$ \ln (e^{\minus{}mt}t^{n} ) \equal{} n \ln t \minus{} mt$

The derivative of that is $ \frac {n}{t} \minus{} m \to \minus{}m$, so $ n \ln t \minus{} mt \to \minus{} \infty$ and $ e^{n \ln t \minus{} mt} \to e^{\minus{}\infty} \equal{} 0$

The methods in the last three posts all work, of course, and the possibilities haven't been exhausted yet.

For the purposes of my teaching here's what I'll say about all of those methods: we won't repeat them every time we have this problem. We do this once and we remember the result, with some phrasing like "an exponential grows faster than any power of the variable." And when we wind up with something like, say, $ \left.xe^{\minus{}x}\right|_0^{\infty}$ as part of an integration problem, we don't spend any time figuring what that limit at infinity is - we know it immediately.

Thanks to all three of you.

It would be nice to see proofs of Arrigo-Sacchi's remarks (although they seem fairly obvious) and to see other possibibilites as alluded to by Kent.

Proof of the Lemma.

Hypotheses. $ \forall\ \varepsilon\ \exists\ n_\varepsilon\colon n > n_\varepsilon\ \Longrightarrow\ 0 < {a_{n \plus{} 1}\over a_n} < l \plus{} \varepsilon.$

Choosing $ \varepsilon < 1 \minus{} l$ we have $ l \plus{} \varepsilon < 1.$ This implies that $ a_n$ decreases from some
$ n$ on and then converges to some nonnegative value $ A.$ If $ A > 0$ we would have $ \lim_{n\to\infty} {a_{n \plus{} 1}\over a_n} \equal{} {\lim_{n\to\infty} a_{n \plus{} 1}\over \lim_{n\to\infty} a_n} \equal{} 1$ but this is in contrast with
$ {a_{n \plus{} 1}\over a_n} < l \plus{} \varepsilon < 1$ definitively. Only $ A \equal{} 0$ is possible. End of the proof of the lemma

The application to the sequence $ {n^k\over e^{nq}}$ is immediate: $ \lim_{n\to\infty} {(n \plus{} 1)^k\over e^{p(n \plus{} 1)}} { e^{pn} \over n^k} \equal{} \lim_{n\to\infty} {(n \plus{} 1)^k\over n^k } { e^{pn} \over e^{p(n \plus{} 1)}} \equal{} 1/e$

Therefore we have $ \lim_{n\to\infty} {n^k\over e^{nq}} \equal{} 0.$ The last step is to show that, given a
subsequence of the naturals strictly increasing, $ \{n_s\}_{s \equal{} 1}^\infty,$ $ n_s\in {\bf N}$ and $ n_{s \plus{} 1} > n_s,$
$ \lim_{s\to\infty} {n_s^k\over e^{n_sq}} \equal{} 0$ holds. But this follows immediately from the definition of limit
and the observation
that the strict monotonicity of $ n_s$ implies $ n_s\ge s.$

[quote="AndrewTom"]Thanks to all......[/quote]

OK, here's another argument:

For $ t>0,$ $ e^{mt}\equal{}\sum_{k\equal{}0}^{\infty}\frac{m^kt^k}{k!}>\frac{m^{n\plus{}1}t^{n\plus{}1}}{(n\plus{}1)!}\equal{}Ct^{n\plus{}1}.$

That is, the sum is greater than one term of the sum, and we only care that the coefficient of that term is positive.

Hence, $ \frac{t^n}{e^{mt}}<\frac{t^n}{Ct^{n\plus{}1}},$ which tends to $ 0$ as $ t\to\infty.$

From this argument, we see that it's not just the exponential.

If $ f(x)\equal{}\sum_{k\equal{}0}^{\infty}a_kx^k$ is a power series with an infinite radius of convergence, and $ a_k\ge 0$ for all $ k$, and infinitely many of the $ a_k$ are nonzero (that is, it's not a polynomial), then $ \lim_{x\to\infty}\frac{x^n}{f(x)}\equal{}0$ for any $ n.$

(Any nonpolynomial entire function has some direction in the complex plane in which it grows faster than any polynomial - but if I start saying things like that, I'm starting to invoke complex analysis.)

$ \lim_{t\to\infty}e^{\minus{}mt}t^{n}\equal{}\lim_{t\to\infty}\frac{t^{n}}{e^{mt}}\equal{}\frac{1}{\infty}\equal{}0$

since : $ lim_{x\to\infty}\frac{e^{x}}{x^{n}}\equal{}\infty$ $ \forall n\in{\mathbb{N^{*}}}$.


вычислите предел : $ \lim_{x \minus{} > 0} \frac {cos(\pi/2cosx)}{sin(sinx^2)}$
 english version :
 find lim  $ \lim_{x \minus{} > 0} \frac {cos(\pi/2cosx)}{sin(sinx^2)}$

Разложение в ряд Тэйлора приводит к ответу $ \pi/4$.  :) English version: expand into Taylor series to find the limit $ \pi/4$.

Have other way to solve .Use L'Hospital rule.


Hi there, I would appreciate some help with this one:

find the length of the shortest path from the point (3,5) to the point (8,2) that touches the x-axis and the y-axis.

Thanks! :-)

If you want a hint.. think of a pool/snooker table.

Or better yet, think about a beam of light in a room of mirrors. (Remember that light always follows the shortest path between two points).

[hide]Verbose to explain without a drawing.

A=(3,5), B=(8,2), A'=(-3,5), B'=(8,-2)

Let C and D be the points on the y-axis and on the x-axis respectively.

.First, |AC| = |A'C|, |DB| = |DB'|

.Second the shortest path between two points (in an Euclidian space at least ) is the straight line.

So draw line A'B', it cuts the axes on the two wanted points  :twisted:

Not sure but I think this works.[/hide]

Nothing to do with this problem but isn't the forum slower with this new design ??

In deed you are right.  :)

And yes, the forum does seem slower with the new format. I think it is those illustrated flash player animations on the left that are slowing it down.

If I were you frznfire, I would have posted this in either the "Getting Started" or "MATHCOUNTS" section.  The "Intermediate Topics" section is for beyond MATHCOUNTS and AMC problems.

Umm... Actually this is definetly worthwile to put it in the Intermediate section Frodo.  I think placement should be in the eyes of the person who is asking the question.

Yeah I think this would be a bit harder for mathcounts.. intermediate seems fine. But anyway.

First find how far apart they are horizontally or by their x-values, then how far apart they are vertically or by their y-values.  Then use Pythagorean Theorem.

Frodo, you might look at the restrictions, it must touch both the x and y axis...

It's not [i]that[/i] simple...

Ugh!  So blind!  I need some better glasses! :shock:  Forget everything I said previously on this section.


salam , in javabaye man vase marhale yek sal dovvom sevvomast :
1-zavayaye bozorgtar az $ 59$ : hame kamtar az $ 62$
2-adade tabii khali az mokaab : $ 729$ doroste (man $ 2187$ zadam  :( )
3-moadele ghadr motlagh dare : 1 javab
4-majmoe argham ba haselzarbe 1000 : $ 23$
5-nesbate $ BK$ be $ BC$ : $ {1 \over 2}$
6-haselzarb payapeye argham ke mirese be 2 : $ 8$
7-$ 32451$ : $ 58$
8-mosallase $ 20$daraje $ 60$daraje : $ 80$ araje
9-adade mosallasi : $ 2$ javab
10-bozorgtarin zarib : nazadam
11-besyar avval : $ 1$ adad
12-majles 8om : $ 10153$
13-dastgahe moadelat  : $ 3$ javab
14-5 mohre sefid va 5 mohre siah : $ 120$ halat
15-bishtarin tedad amele 2 : $ 277$
16-noghteye $ A$ sabet va $ M$ va $ N$ moteghayyer : andaze shoaae dayereye mohati sabete
17-$ a \plus{} b \plus{} c \plus{} d \equal{} 1$ : nazadam
18-olaghe : $ 2\sqrt 3 \plus{} {\pi \over 3}$
19-10 zelyie montazam : $ 7$
20-mosallase kaghazi : nazadam
21-$ 6$ ta jabe : $ 120$
22-$ f(x)$ : $ 36$
23-mosallase motesaviolazla mohat dar morabba : nazadam
24-mostatile $ 1 \times 2$ : $ \sqrt {{{84} \over 5}}$
25-modeleye $ 2^x \plus{} 2x^2 \equal{} y^2$ : $ 4$ javab
26-$ M,M_1,M_2,M_3$ : nazadam
27-raghame dahgane $ a^n$ : ehtemalan eshteb zadam (bejaye dahgan yekan gereftam) vali bazam fek konam baraye bazi motanaveb va baraye bazi na motanaveb bashe  :D
28-takhshise adad be mokaab : $ 21$
29-karte sookht  :D  : $ C$
30-majmooe ha : nazadam

dar kol 24 ta zadam yekish sad dar sad ghalate yeki digash maloom nist .. baghie ghalatam ham badan maloom mishe  :huh:

Toxic jan oonayi ke nazadio va oonayi ke fekr mikonam ghalataro javabasho mizaram :
tedad arghame bozorgtarin zarib : 25 ta 35
bishtarin tedad amele 2 : 280 ( mesalesh jalebe :D)
noghteye $ M$ va $ N$ : fekr konam hamishe bar yek dayereye sabet momase
$ a \plus{} b \plus{} c \plus{} d \equal{} 1$ : 1/4
10 zelyie montazam : 8
mosalase kaghazi : 11
$ M_{1} , M_{2} , ...$ : 40 daraje
majmooe ha : $ 6^{5}$ (sorry yadam nist chande :D)
fekr konam oon raghame dahgane $ a^{n}$ ham hamooni ke zadie yani bara bazi $ a$ ha motenavebe va baraye bazi az $ a$ haye dige namotenavebe :)

[quote="khashi70"]Toxic jan oonayi ke nazadio va oonayi ke fekr mikonam ghalataro javabasho mizaram :
tedad arghame bozorgtarin zarib : 25 ta 35
bishtarin tedad amele 2 : 280 ( mesalesh jalebe :D)
noghteye $ M$ va $ N$ : fekr konam hamishe bar yek dayereye sabet momase
$ a \plus{} b \plus{} c \plus{} d \equal{} 1$ : 1/4
10 zelyie montazam : 8
mosalase kaghazi : 11
$ M_{1} , M_{2} , ...$ : 40 daraje
majmooe ha : $ 6^{5}$ (sorry yadam nist chande :D)[/quote]

10 zelii ro mitoni begi chera  :(  ? mesaletam begoo vase 280 :D  :(  :
alan fahmidam che bad joobi zadam vase $ M$ va $ N$

10 zelyie ke kheyli toolanie vali ye rahesh ine ke ba koochiktarin zel kar konio peyda koni :) , man tahala az kheylia porsidam taghriban nesfeshoon $ 8$ zadan
baraye oon $ 280$ e ham inaro dar nazar begir :
$ 2^{276} , 2^{276} , 2^{277} , 2^{278} , 2^{279}$ , majmooeshoon daghighan mishe $ 2^{280}$
be nazaretoon kaf chande emsal ?:D

[quote="m bahador"]dar morede sole edad mosalashay....
man ye hadsi zadam valy nemidoonam doroste ya na
ine ke aval goftam ke dar 5 zely 2 mosalas hast ,dar 6 zely 3 mosalas ,pas dar n zely n-3 mosalas :D[/quote]
manam hamin karo kardam ... kashke ke nemikardam !

agha man kilido kamel daram vali soalaro nadaram yadam nist shomarehashoono vali javaba yadame har ki mikhad be sepehr_rpn pm bede migam  :D  :wink:  dar zemn khodamam 25 ta dorost 2 ta ghalat..!! dar zemn mosallasaaa 8 dorostee

bacheha soalaro mizanin man soalaro mikham bebinammm.dar che sathi boode emsal soala eyne parsal asun boodan ya naaa????

AGHa Man 23 Zadam 4 tash GHalate ( Ya hossien ) Fekr MIkonid SaGhfe Soalat Chande ?

saghf na avallan kaf :wink:  badesham fek konam 45% hatman ghaboole ...!!!

man ke be nazaram kaf too bazeye (10,12) taghri mikone!

man fekr konam kaf 12 menhaye epsilon bashe :)


Which book is good to understand the concept of NMR spectroscopy?

Any standard organic chemistry textbook has one (or more) chapter about spectroscopy, which give an introductoty treatment to the subject. You should start from there. Then, when you have mastered the basic concepts, then if you want you can go into a more detailed book. I have posted some time ago a post here with a collection of problems suitable for those initiating spectroscopy.


AMC AIME ARML USA(J)MO USAMO geometry LaTeX

I have been following the competition for years and I think this year’s is one of the worst, in terms the quality of questions and the resulted score distribution.

I disagree. As you can see there were less perfect scores than usual and the questions overall were not bad. Althought the overall difficulty might have decreased slightly, the test still required the use of some creative ideas.

I thought the AIME I was really bad.  The questions were still good by themselves, but they were horrible compared to all the other AIMEs I have seen.  The first 12 were ridiculously easy, so stupid mistakes was a huge factor.  #13 I did not like, since it was just brute force (at least so says the math jam).  #14 and #15 weren't too bad.  The AIME II looks a lot better, but still a little disappointing.

I agree that the AIME problems were too easy,  but it wasn't that terrible.  You still needed problem solving skills to solve the problems even though careless mistakes were a big factor and a few could be brute forced.  Plus, don't you enjoy looking at #12 and being like, "Oh, I can get that!"?  BUt yes, it would have been better if the problems were slightly more challenging.

GREAT YEAR i think, cuz its my first year of AMC and i have no comparisons to make.

I think some problems are AIME I are bad, problems on AIME II are better except #9, 10(that does [i]not[/i] mean they are much harder, it's just that problems on AIME II have more elegant/less calculation solutions)

I know that this is all meant as a harmless discussion / comparison of this year to others, but I should give a contest writer's take on the situation.

The people who write AIME problems contribute these ideas for free.  It's amazing when you consider that some of the best contests in America (AMC, AIME, ARML) take problems from people who contribute them for free, while there are a lot of other contests that pay their writers and still can't attain the same level of quality.  And yet the AMC series of competitions is consistently the best in America by just about any measure - the number of students the contests reach, the elegance of the questions, the prestige of the contest and the fact that good AMC scores are a great help when applying to college.  I guess my point is that complaining about a "worse than usual" AIME is like letting someone feed you a 12-ounce steak every year and then complaining one year that it isn't as tender and juicy as usual.  Or that your steak weighed only 11.8 ounces while the guy next to you got a 12.1 ouncer.

In recent years, the AMC panel has been stretched very thin.  Old-school math competitors like me remember when there used to be one AHSME (now we have two AMC-12s and two AMC-10s annually) and one AIME (we now have two of those also).  Really old-school people will remember when the AHSME was the only qualifying test for the USAMO - there was no AIME.  The quantity of tests AMC offers now each year makes it much harder to consistently put out tests with the kind of quality people have come to expect of this fine competition.

Of course, even a "bad" AMC test is a great test by any reasonable standard.  Scores were up this year because (at least from what I hear) the AIME 1 involved some ideas that have appeared on competitions in the past.  This is where the "quantity vs. quality" issue becomes very important - a finite collection of problem posers can only come up with a certain number of "brand-new" problem ideas in a given year.  The remaining problems have to come from old ideas or variants of old problems.  This can make a huge difference when the current testing schedule requires about 25 "very good/hard" problems (last five for each AMC-10 and 12, allowing for some overlap, and last five for each AIME).  We used to need about 10 to 12 (5 to 7 for AHSME and 5 for AIME).

Does this mean that the AIME becomes less a test of which students are most clever, and more a test of which students are most consistent?  Most likely the answer is yes.  But these things happen in, say, the NFL all the time.  If a game is played on a windy day, it favors the running game over the passing game.  If a game is played indoors, it favors the passing game and the kicking game.  If a game is played outside in rain or snow or mud, it favors the defense.  If you want to be the ultimate NFL football team, you have to strengthen all aspects of your team.  If you want to be the ultimate math competitor, you have to hone all aspects of your game - vision, intuition, cleverness, accuracy, speed, calculation, logic, and courage.  Neglect any one of these areas, and you'll eventually find a math contest that exposes your weakness.

Anyway, I couldn't pass up the chance to give a lecture.  Sorry  :P

Sorry if I sounded harsh.  I did say they were good problems compared to most contests.  I'm sure we are all thankful for AMC's wonderful contests every year, even with tight funding.  But my point is that it still wasn't as good as past contests, and hopefully it won't be a trend.

Quite alright!  My intent certainly wasn't to come down on anyone in particular.  But as my students who ran a math meet at our school this year found out, writing problems (esp. original ones) is a MUCH tougher proposition than it sounds at first.  As someone who sympathizes with the AMC problem contributors, I feel it's my responsibility to share what they go through.  My impression from talking to people is that the people who decided that AMC should have about 6 contests a year rather than two are people who are quite distant from the problem-creation process.  This has created some frustration for the people who have to bear the brunt of that decision.

Two points to keep in mind:
(1) The readers of this forum are probably much different than the  general population students taking the AMC contests.  Remember, even for the AIME, there are 16,000 who are invited, about 12,000-13,000 participate and there are about 5,000 registered users of this forum.  What you may find routine, even pedestrian, may seem exotic or even incomprehensible to a student encountering such problems for the first time, maybe without a mentor.  We aim to make some problems accessible to even that student.  Not easy to achieve, and in past years, we have had severe criticism from teachers and participants who said the contests were too hard.

This is a little like presenting a movie to a theater full of full-time movie critics.   We want to impress the critics, but we must also please the movie-going public; we need to please the movie-going public, without turning off the critics.

(2) If you have problems you would like to submit for consideration on the 2007 AMC contests, or the 2007 AIME, send them to me now (preparation begins about 2  years in advance).  I will enter them into the problem submission pool to be considered by our reviewing panel.

Steve Dunbar
AMC Director

[quote="AMCDirector"]
(2) If you have problems you would like to submit for consideration on the 2007 AMC contests, or the 2007 AIME, send them to me now (preparation begins about 2  years in advance).  I will enter them into the problem submission pool to be considered by our reviewing panel.

Steve Dunbar
AMC Director[/quote]

Will a private message on AoPS do?

[quote="eryaman"]Shouldn't students be banned from submitting problems?  After all, if the problem does end up on a test, that person will have an unfair advantage.[/quote]

well, they would only have an advantage on that one particular problem. And that's only if they are still in high school 2 years later when the problem gets used.

the computing olympiad allows anyone to submit problems. It's less of an advantage than you might think.

The problems go through an enormous amount of review and editing.  They usually get changed a lot.  I barely recognize the problems I submit when they finally appear, and I'm involved every step of the way.  By the way, submitting a problem does not guarrantee it will appear, it only gets considered.  We use about 1 from every 3 submitted.  And you don't know if it will appear on AMC 10 A, 10 B, 12A, 12 B, AIME, or even AMC 8 since we often take the germ of an idea and either make it harder or easier.

and yes, sending a message to me at AoPS will be fine.  I'll probably just acknowledge that I got it.  If you send it in LaTeX code, that would be preferred.

Steve Dunbar
AMC Director

Mr. Dunbar --

I have a decent trig problem that I wrote, but I don't know if you want to use trig on the AIME.  What's your take?

JB

There've definitely been trig on the AIME in the past (and even this year!) so go ahead and submit it. I doubt Mr. Dunbar would be able to actually answer how much trig is planned for the tests in future years.

Yeah, and there has been trig on the AMC 12 too.

It depends on how you define 'good'.  The problems weren't particularly good or fun in comparison to other AIMEs, which you could say is not 'good', but it was easy compared to the other AIMEs!


hello, and wellcome :)

Yes it is.

Here's a demonstration:

Let $[ABC]$ be a triangle

Let $p$ be the height of the triangle relatively to $C$ and $q$ the height relatively to $A$.

We have: $\frac{p}{b}= \sin A \implies p = b \sin A$ and $\frac{p}{a}= \sin B \implies p = a \sin B$

So

$a \sin B = b \sin A \implies \frac{a}{\sin A}= \frac{b}{\sin B}$

On equal way:

$\frac{q}{b}= \sin C \implies q = b \sin C$ and $\frac{q}{c}= \sin B \implies q = c \sin B$

So

$b \sin C = c \sin B \implies \frac{b}{\sin B}= \frac{c}{ \sin C}$

So we conclude: $\frac{a}{\sin A}= \frac{b}{\sin B}= \frac{c}{ \sin C}$

[quote="Farenhajt"]And the Extended Sine Law is ${a\over\sin A}={b\over\sin B}={c\over\sin C}=2R$, where $R$ is the circumradius.[/quote]
Proof: see these two pictures. The rest is clear.  :D

yeah the proof is like the area of a triangle? i forgot.

[quote="Johnbaggins"][color=darkred]Sorry to bother you all but is the sine rule: a/sinA = b/sinB = c/sinC Thanks in advance[/color][/quote]
[color=darkblue]Let $w=C(O,R)$ be the circumcircle of the triangle $ABC$.

Denote the point $D\in w$ for which $[BD]$ is a diameter of the circle $w$ and $m(\widehat{BDC})=\phi\ .$

Apply the definition of Sinus in the right triangle $BDC\ :$

$BC=BD\cdot\sin \phi$, i.e. $a=2R\cdot\sin A$ because $BC=a\ ,\ BD=2R\ ,\ \phi =A$ or $\phi=180^{\circ}-A\ .$[/color]

[quote="Virgil Nicula"]
... Apply the definition of SINUS ...[/quote]

 :rotfl:

@Johnbaggins: Yes, it is the Law of Sines.


Suppose there are $l,m,n\in \mathbb{R}$ satisfying the equations

$m-n=8$ ; $mn+l^2+16=0$ . Find

$m^{2005}+n^{2005}+l^{2005}$  :D

[quote="socrates"]We have $m=n+8$
so substituing this in the second relation we get
$l^2+(n+8)n+16=0$ or
$l^2+(n+4)^2=0$
consequently $l=0$ and $n=-4$
From the relation $m=n+8$ we find easily $m=4$

The value of $\displaystyle m^{2005}+n^{2005}+l^{2005}$ is
obviously zero.

This is a very easy problem
that could propably be in the intermediate section.[/quote]

Well done  :D  . I have another solution which is by treating it as a quadratic equation with roots $m$ and $-n$ . Then we will get

$x^2-8x+l^2+16=0$ . Since it has real roots , the discirminant is

$64-4(l^2+16)\geq 0$ which gives us $-l^2\geq 0$ and hence $l=0$ . The rest is easy . :P


If $a\ge b\ge c\ge d\ge 0$ and $a+b+c+d=2$, then

$ab(b+c)+bc(c+d)+cd(d+a)+da(a+b) \le 1$.

My solution to my recent one holds for this one.  :lol:

[quote="e.lopes"]Nice one, Vasc!

I solved using that $LHS = (ab+cd)(a+b+c+d)+(d^{2}-b^{2})(a-c)$  :)[/quote]
But
$(ab+cd)(a+b+c+d)\le 1$
is not true. :!:
See $a=b=1, c=d=0$.

First,let us prove a lemma:
Lemma:
For any $a+b+c+d=2$ and $a\geq b \geq c \geq d \geq 0$
$a^{2}b+b^{2}c+c^{2}d+d^{2}a \geq ab^{2}+bc^{2}+cd^{2}+da^{2}$

Proof of lemma:
Let $F(a)=(b-d)a^{2}+(d^{2}-b^{2})a+b^{2}c+c^{2}d-bc^{2}-cd^{2}$
$F^{'}(a)=(b-d)(2a-b-d) \geq 0$ $\rightarrow F(a) \geq F(b)=(c-d)b^{2}+(d^{2}-c^{2})b+cd(c-d)$
$F^{'}(b)=(c-d)(2b-c-d)\geq 0$ $\rightarrow F(b) \geq F(c)=0$

Now,let us turn back the solution of the problem. From lemma we have:
$a^{2}b+b^{2}c+c^{2}d+d^{2}a+ab^{2}+bc^{2}+cd^{2}+da^{2}+2(abc+bcd+cda+dab) \geq 2(ab^{2}+bc^{2}+cd^{2}+da^{2}+abc+bcd+cda+dab)$
It follows that;
$(a+b+c+d)(a+c)(b+d) \geq 2(ab(b+c)+bc(c+d)+cd(d+a)+da(a+b))$
But, by AM-GM inequality: $(a+c)(b+d) =(a+c)(b+d) \leq \frac{(a+b+c+d)^{2}}{4}=1$

Nice solution, NguyenDung.
But, more elementary,
$F(a)=(a-b)(b-c)(a-c)+(c-d)(a-d)(a-c) \ge 0$.


[quote="computer_nuke"][quote="nebula42"][quote="computer_nuke"]So basically, you are trying to prove that, if Xn's are the roots of some polynomial, then the sum of the 19th powers times the sum of the 93rd powers is always greater than the sum of the 20th powers multiplied by the sum of the 92rd powers.[/quote]
Yes, except the $x_{i}$ are not roots of a polynomial; they are just  any positive reals.[/quote]

Well, yes, but what I mean is that they create a polynomial with those roots. But I'm just trying to now figure out how to prove the inequality.[/quote]
What do you mean by "they create a polynomial with these roots"?  :huh:  That would only be true if the polynomial were
$(x-x_{1})(x-x_{2})\dots(x-x_{n})$
but it's not.

[quote="nebula42"][/quote][quote="computer_nuke"][quote="nebula42"][quote="computer_nuke"]So basically, you are trying to prove that, if Xn's are the roots of some polynomial, then the sum of the 19th powers times the sum of the 93rd powers is always greater than the sum of the 20th powers multiplied by the sum of the 92rd powers.[/quote]
Yes, except the $x_{i}$ are not roots of a polynomial; they are just  any positive reals.[/quote]

Well, yes, but what I mean is that they create a polynomial with those roots. But I'm just trying to now figure out how to prove the inequality.[/quote][quote="nebula42"]
What do you mean by "they create a polynomial with these roots"?  :huh:  That would only be true if the polynomial were
$(x-x_{1})(x-x_{2})\dots(x-x_{n})$
but it's not.[/quote]
you guys are making it very complicated. :)
i will post a hint soon.

[quote="computer_nuke"]Xn's are the roots of some polynomial[/quote]

Entirely irrelevant to the problem.

sorry it took me so long, i was busy doing something else.
[hide="hint"]Let $S_{k}=x^{k}_{1}+x^{k}_{2}+...+x^{k}_{n}$. Apply the cauchy's inequality to this and see what you get. Notice the similarity between what you get, and $(x_{1}{}^{1}{}^{9}+x_{2}{}^{1}{}^{9}+...+x_{n}{}^{1}{}^{9})(x_{1}{}^{9}{}^{3}+x_{2}{}^{9}{}^{3}+...+x_{n}{}^{9}{}^{3})\geq(x_{1}{}^{2}{}^{0}+x_{2}{}^{2}{}^{0}+...+x_{n}{}^{2}{}^{0})(x_{1}{}^{9}{}^{2}+x_{2}{}^{9}{}^{2}+...+x_{n}{}^{9}{}^{2})$.
This is a very lengthy hint, but now I think you should be able to solve it.[/hide]
Hope this helps a bit. :)
P.S. Forget about the polynomial, it [b]is[/b] irrelevant, as t0rajir0u stated.


SASE is most likely Self-Addressed Stamped Envelope; 8.5 by 11 should be the size (in what units, I'm not sure).

You mean those big yellow envelopes?

Because, the mail envelopes aren't 8 1/2 by 11

If the phrase was used in the USA (which is the country that likes such crazy abbreviations most), then the units are certainly inches and then it, indeed, means a big yellow envelope used for sending documents printed on regular paper that one doesn't want to fold for some reason.

oh
ok
ty 4 all ur help
but how do i send a envelope to someone else with my address on it?

You just send it in another envelope with their address on it :roll:

good point =)
ty for all the help
unless anyone else has anything 2 say, plz lock this thread


Okay basically this is what happens. There are a certain number of citizens. One is a mafia, one is an angel, one is a sheriff and the rest are citizens. In each round, a mafia is chosen. He pms the mod saying i want to kill _____. The angel says I want to save _____ and the sheriff says I accuse _____. Finally, the citizens collabrate and choose someone to be executed. That person, whether or not they are mafia or not, are killed. This happens each round until either the mafia or the citizens are left.

Can someone mod this? Thanks.

There's a reason there is a mafia forum; make big games so we can have lots of replacements. Maybe you can replace there.

Wait this is the same type of mafia?

:fool:


solved at:

http://www.mathlinks.ro/Forum/viewtopic.php?t=57683
http://www.mathlinks.ro/Forum/viewtopic.php?t=57552

  darij

PS. ## to the moderators: please delete the following topics:

http://www.mathlinks.ro/Forum/viewtopic.php?t=121359
http://www.mathlinks.ro/Forum/viewtopic.php?t=87005
http://www.mathlinks.ro/Forum/viewtopic.php?t=63759

but  open this inequaliti for n number is true or fales


The point with coordinates $ (6,\minus{}10)$ is the midpoint of the segment with one endpoint at $ (8,0)$. Find the sum of the coordinates of the other endpoint.

[quote="$LaTeX$"]The coordinates of the midpoint is the average of the coordinates of the endpoints. The other endpoint must then be $ (4,\minus{}20)$.[/quote]
To complete the solution...
Therefore, the sum of the coordinates is $-20+4=\boxed{-16}$

To clarify the solution:

Let the other endpoint be $(x,y)$
We have $\frac{8+x}{2}=6$ and $\frac{0+y}{2}=-10$

Therefore $8+x=12$ and $y=-20$ so $(x,y)=(4,-20)$

We need $x+y;x+y=4+(-20)=\boxed{-16}$

Of course, you'll go through the process much faster when solving it.


Calculus isn't a high school topic.  Try

http://www.artofproblemsolving.com/Forum/index.php?f=296

Anyway, this problem is very easy if we let $y = 5x^{4}+ax^{3}+bx^{2}+cx+d$.  (We would then need to find the homogeneous solution, which is just $K e^{-x}$.)


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

I am reading Rosenberg's book on Algebraic K-Theory and it says that the product in the $ K_0$ of a commutative ring is well defined because

$ \left( R' \otimes_R P \right) \otimes_{R'} \left(R' \otimes Q\right) \cong R' \otimes_R \left( P \otimes_R Q \right)$

where R, R' are commutative rings and P,Q are R-modules, and there is a ring homomorphism from R to R' which gives R' an R-module structure.

I've tried to understand that isomorphism but I'm not yet comfortable working with tensor products. Can someone give me a hint and/or tell me a nice book to learn a bit about tensor product of modules?

Thanks!

Basically you use the associativity of the tensor product, along with the $ R'$-$ R$ bimodule structure of $ R'$.

$ R'\otimes_{R}\left( P\otimes_{R}Q\right)$ is the same as $ \left(R'\otimes_{R} P\right)\otimes_{R}Q$.  This is also the same as $ \left(\left(R'\otimes_{R'}R'\right)\otimes_{R} P\right)\otimes_{R}Q$, which is $ \left(R'\otimes_{R'}\left(R'\otimes_{R} P\right)\right)\otimes_{R}Q$, or $ \left(\left(R'\otimes_{R} P\right)\otimes_{R'} R'\right)\otimes_{R} Q$; this is the same as $ \left( R'\otimes_{R}P\right)\otimes_{R'}\left(R'\otimes Q\right)$.

Steve

Forgive me if this sounds rude, but if you aren't comfortable with tensor products then why are you trying to learn K-theory?

Thanks Steve =)

t0rajir0u: It was a bit rude. Among math students and professors I've met a lot of people who think they're better than everybody, who think that everything is obvious and trivial, and who seem to have been born with all the knowledge they have. I love math. I know I'm not a genious but I'm not that bad either. I sometimes have to work quite hard to understand some things but I like the process. I dont pretend to become Grothendieck, dont worry. I will try not to post such silly questions in the future.

I apologize for my tone.  I wasn't trying to comment on your intelligence; it just seems to me that a natural way to learn mathematics is to learn the easy stuff first and the hard stuff later.  But maybe there's something to be said for diving into the hard stuff; again, I apologize.


Let $S$ be a set with six elements.  In how many different ways can one select two not necessarily distinct subsets of $S$ so that the union of the two subsets is $S$?  The order of selection does not matter; for example, the pair of subsets $\{a, c\}$, $\{b, c, d, e, f\}$ represents the same selection as the pair $\{b, c, d, e, f\}$, $\{a, c\}$.

[hide] Call the two subsets $m$ and $n$. For each of the elements in $S$, we can assign it to either $m$, $n$, or both. This gives us $3^6$ possible methods of selection. However, because the order of the subsets does not matter, each possible selection is double counted, except the case where both $m$ and $n$ contain all $6$ elements of $s$. So our final answer is then $\frac{3^6-1}{2}+1=\boxed{365}$
[/hide]

On the same page, we have [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126236]this[/url] topic.

[quote="Pineappleperson"]Why'd u subract one from 3^6?[/quote]

Don't necro threads. It isn't cool.

Don't post in old threads.

See, necromancers summon the dead.

[quote=fishy15]oops this solution does not work

[hide=fake solution]
We "fix" $a$ to be in the first term so we don't overcount the possibilities. After that, $b$ has 3 options: with $a$, in the other set, or in both the sets. Continuing down the terms, we get an answer of $1(3)^5 = 243$. However, this undercounts some of the possibilities. I believe this can be found when we fix $a$, but I cannot see how to improve the statement to fix the problem.[/hide][/quote]

a can also be in both sets

more combo: Pick a subset. We'll do cases on the number of elements. If it has $0$ elements, we have $\binom{6}{6}2^0$ ways. If it has $1$ element, we have $\binom{6}{5}2^1$ ways. So on, for a total of $(2+1)^6=729$. However, swapping the order of the subsets should not give a new pair, but we've counted it as one, so every pair except for the one with two subsets having all elements is overcounted with factor $2$. The answer is hence $\frac{728}{2}+1 = \boxed{365}$.


algebra
decimal representation
Number theoretic functions
China
TST
Digits

Given a positive integer $A$ written in decimal expansion: $(a_{n},a_{n-1}, \ldots, a_{0})$ and let $f(A)$ denote $\sum^{n}_{k=0} 2^{n-k}\cdot a_k$. Define $A_1=f(A), A_2=f(A_1)$. Prove that:

[b]I.[/b] There exists positive integer $k$ for which $A_{k+1}=A_k$.
[b]II.[/b] Find such $A_k$ for $19^{86}.$


Certain positive integers have these properties:
       I. The sum of the squares of the digit is 50;
       II. each digit is larger than the one to its left.
What is the product of the digits of the largest integer with both properties?

Edit:
Oops, fixed.

[hide]
I couldn't think of any fast way to do this, does anyone have a way?
I basically just picked the largest square that the difference when subtracted from 50 would be able to be broken down into the sum of as many squares as possible. I ended up getting 36 and 14.  14 breaks down into $3^2 + 2^2 + 1^2$ so the answer is $36$.
Anyone have a faster way?
[/hide]

Are you sure that problem is correct?

From the information given, that's impossible.

You have the sum of the digits is 50.

Now lets say the number is 123456789 - the sum is 45, which is the maximum.

I think he meant for i: The sum of the squares of the digit is 50.

If so, then think of number 1236 which gives you 36.

I believe that's right answer.


The official answers to the 2007 AIME Alternate administered on Wed, March 28, 2007 will be posted on the AMC website on Monday, April 2.

Another country is giving the AIME (will not count toward USAMO qualification of course) a couple of days later than the US and specifically asked that we not post the 3-character answer codes until after they are done.

Yes, I understand that unofficial answers and solutions will be appearing here.  You may wish to hide the solutions and answer codes, simply so as not to accidentally spoil the interest, even for students taking the AIME Alternate for practice, or unofficially, here in the US.

Thanks.

Steve Dunbar
Director, American Mathematics Competitions

You might want to edit the topic...should it not read Alternate AIME?

what's up with having two screennames?  i thought you werent supposed to do that

You mean user SDunbar? I can't be 100% certain, but I think it was created for the April Fools' joke about the USAMO.

However, if they are the same person, I retract this statement.

That post was certainly an April Fool's joke, and has been deleted, and the SDunbar account deactivated.

The AIME II answers are now up on the AMC web site.


A ladder which is 3.4 meters long is leaning against a wall. Its base is 1.6 meters from the wall. How many meters up the wall does the ladder reach?

The ladder forms the hypotenuse of a right triangle with the legs being 1.6 and the answer. Using the pythagorean formula, if the answer is x, then $ x^2\plus{}1.6^2\equal{}3.4^2$. This simplifies to $ x^2\equal{}9$, and since $ x\ne \minus{}3$, $ x\equal{}\boxed{3}$.


ratio
limit
calculus
calculus computations

Suppose that $ a_n\geq 0$ and there is

$ \lim_{n\rightarrow\infty}\frac {a_{n \plus{} 1}}{a_n} \equal{} c$
If c>1,series diverges.
if c<1 series converges.

For $ a_n \equal{} \frac {n!}{n^n}$

$ \lim_{n\rightarrow\infty} \frac {(n \plus{} 1)!/(n \plus{} 1)^{n \plus{} 1}}{n!/n^n}$

$ \lim_{n\rightarrow\infty} \frac {n^n}{(n \plus{} 1)^n}$

Then I used I'Hopital Rule and got answer 1.

First of all, your use of notation is awful.  Second of all, this is a discrete limit ($ n$ is running through the integers), so applying L'hopital's rule is not obviously going to be valid (although sometimes it will give you the right answer anyhow).  Third of all, you almost certainly have applied it incorrectly (I'll be dollars to donuts that you differentiated treating some copies of $ n$ as variables and other copies as constants, when in fact they are all the same $ n$).  What you should do instead is algebraically manipulate the quantity $ \frac{n^n}{(n \plus{} 1)^n}$ that you're trying to find the limit of ... I personally recommend using the fact that $ \frac{x^a}{y^a} \equal{} \frac{1}{(y/x)^a}$ for all $ x, y \neq 0$ and all $ a$ and hoping that you get something you might recognize.

Oh,thanks

I get it

$ \lim_{n\rightarrow\infty}\frac{n^n}{(n\plus{}1)^n}\equal{}\lim_{n\rightarrow\infty}\frac{1}{(1\plus{}\frac{1}{n})^n}\equal{}\frac{1}{e}$


I'm not quite sure what this problem's asking.

Let $p$ and $q$ be the radii of two circles through $A,$ touching $BC$ at $B$ and $C,$ respectively. Show $pq=R^2.$

I'm assuming $ABC$ is a triangle and $R$ is the circumcircle's radius, but the position of the circles is unclear.  :?

[quote="Elemennop"]Two unique circles, that are both tangent to line BC (at the endpoints of line segment BC), but also contain A.[/quote]Oh, right.

Ok, got it.  :D

[quote="pkerichang"]is this problem from geometry revisited??? looks very similar to me...[/quote]Yes you are right. That's from Geometry Revisited, Coxeter & Greitzer, pg 3 Ex 4 NML#19, MAA


Prima zi, la capatul unei calatorii destul de lungi cu autocarul/autobuzul am ajuns in sfarsit la destinatie. Se pare ca aceasta olimpiada este in top, cel putin din punctul de vedere al conditiilor de cazare, atat pentru elevi / profesori insotitori cat si pentru comisia de evaluare.

Deocamdata urmeaza deschiderea la 16,00. Sunt curios totusi cam cati participanti intra pe net (si mai exact pe site :P) in timpul olimpiadei.

Prima proba (de fapt singura, ca celelalte vor fi baraje) va fi maine (nr. 29 Martie 2005) :)

Tare m-as bucura sa vad si eu subiectele. Avem vreo sansa sa le vedem in viitorul apropiat?  ;)

De indata ce vom avea ceva timp liber draga Gabi (si subiectele toate la indemana) le voi posta. In cel mai rau caz Sambata seara o sa fie toate online :)

PS Te saluta Claude si cu Andrei (Chites) :)

Multumesc. Ii salut si eu pe ei. Sper sa va distrati bine zilele astea... ;)

Could someone translate in english and post in college playground/algebra/analysis /proposed problem
the problems

Clasa A XI-A
subiectul 1; 2 ;3 ; 4

Clasa A XII-A
subiectul 2 ;  4

Thank you

How many students participate this olympiad in

grade 10
grade 11
grade 12

Frumoasa nationala..de fapt a fost prima mea nationala.Eram in clasa a7a atunci.
Sper ca voi ati nimerit mai bine ca noi nimerisem cam prost la marginea orasului in niste conditii ceva mai proaste.


Determine all pairs $(n,p)$ of positive integers such that [list][*] $p$ is a prime, $n>1$, [*]  $(p-1)^{n} + 1$ is divisible by $n^{p-1}$. [/list]

1. Case: $p=2$, then we want $n|1^n+1=2$, thus $n=2$.

2. Case: $p$ odd.
Let $q$ be smallest prime divisor of $n$ and let $o$ be the order of $p-1 \mod q$. Since $(p-1)^{2n} \equiv 1 \mod q$, we get $o|2n$, and by Fermat's little theorem, we get $o|q-1$.
But if $o$ would have a factor in common with $n$, this factor would be $\leq q-1 <q$, impossible. But then $o|2n$ reduces to $o|2$, yielding $(p-1)^2 \equiv 1 \mod q$ or equivalent $q|p(p-2)$.

Case $q|p-2$;
We get $(p-1)^n+1 \equiv 1^n+1=2 \mod q$. But we need $q$ to divide this number, thus $q=2$. But then $p$, is even, too, so $p=2$, which we excluded.

Case $q|p$;
We write $n=p^k s, p \nmid s , k>0$. It's well known [I can give a proof if desired] that for odd primes $p$ if $v_p(a-b)>0, p \nmid a,b$, then $v_p(a^t-b^t)=v_p(a-b)+v_p(t)$ (here $v_p(m)$ is as always the $p$-adic valuation of $m$, thus the number of times $m$ is divisible by $p$). With $a=p-1, b=-1, t=n$ this gives us $v_p((p-1)^{n}+1)=k+1$ and has to be $\geq k(p-1)$ to enable $p^{k(p-1)}|n^{p-1}|(p-1)^{p^ks}+1$. Now $k+1 \geq k(p-1) \iff 2k+1 \geq kp \iff 2+\frac 1k \geq p$. Since we assumed $p\geq 3$, we have $k=1$ and $p=3$.
If $s=1$, we have $n=3$ giving a solution. So let's assume $s>1$, let $q$ be a smallest prime divisor of $s$, and by $n=3$ we want that $s|9s^2=n^2|2^n+1=8^s+1$. Following the same idea than before, we take $o$ as the order of $8 \mod q$ and get again $o|2$ by the same argument. In both cases $q|8^2-1=63$, thus $q=7$. But $8^s +1 \equiv 1^s+1 =2 \mod 7$, contradiction.

Conclusion: all solutions are given by $(n,p)=(2,2),(3,3)$.

[quote="ZetaX"] [I can give a proof if desired] that for odd primes $ p$ if $ v_p(a \minus{} b) > 0, p \nmid a,b$, then $ v_p(a^t \minus{} b^t) \equal{} v_p(a \minus{} b) \plus{} v_p(t)$[/quote] Proof would be welcome :)

[size=9]Note to self: checked up to there.[/size]

[quote=shinichiman]What about $n=1$ ?? $p$ could be any prime.[/quote]

N>1 ..


Τα θεματα ηταν πραγματικα πολυ ευκολα για διαγωνισμο επιλογης

1) Να βρεθει το: $ \lim_{n \rightarrow \plus{}\infty}{\frac{1^1 \plus{} 2^2 \plus{} ... \plus{} n^n}{n^n}}$

2) Εστω $ f: \{1,2,3, ...\} \rightarrow \{1,2,3, ...\}$ μια 1-1 και επι συναρτηση. Να δειχθει οτι η σειρα $ \sum_{n\equal{}1}^{\plus{}\infty}{\frac{f(n)}{n^2}}$ απειριζεται θετικα.

3) Να βρεθουν τα ορια:

$ \lim_{n \rightarrow \plus{}\infty}{n!det{\begin{pmatrix} 1 & 1 & ... & 1 \\ 1 & \frac{3}{2} & ... & 1 \\ ... & ... & ... & ... \\ 1 & 1 & ... & \frac{n\plus{}1}{n} \end{pmatrix}}}$

$ \lim_{n \rightarrow \plus{}\infty}{(n \minus{} 2)!det{\begin{pmatrix} 2 & 1 & ... & 1 \\ 1 & \frac{3}{2} & ... & 1 \\ ... & ... & ... & ... \\ 1 & 1 & ... & \frac{n\plus{}1}{n} \end{pmatrix}}}$

4) Εστω $ A \equal{} \{a_1, a_2, ..., a_n\}$ αναδιαταξη του συνολου $ \{1, 2, 3, ..., n\}$. Τοτε καθοδος του $ A$ ονομαζεται καθε δεικτης $ i \in \{1, 2, ..., n\minus{}1\}$ για τον οποιο ισχυει $ a_i > a_{i\plus{}1}$. Να βρεθει το πληθος των συνολων $ A$ τα οποια εχουν ακριβως μια καθοδο.

5) Εστω $ n, a$ θετικοι ακεραιοι. Αν ο $ n$ διαιρει τον $ (a\minus{}1)^{2009}$ να αποδειχτει οτι ο $ n$ διαιρει και τον $ \frac{a^n \minus{} 1}{a \minus{} 1}$.

[quote="NickNafplio"]Τα θεματα ηταν πραγματικα πολυ ευκολα για διαγωνισμο επιλογης

1) Να βρεθει το: $ \lim_{n \rightarrow \plus{} \infty}{\frac {1^1 \plus{} 2^2 \plus{} ... \plus{} n^n}{n^n}}$

2) Εστω $ f: \{1,2,3, ...\} \rightarrow \{1,2,3, ...\}$ μια 1-1 και επι συναρτηση. Να δειχθει οτι η σειρα $ \sum_{n \equal{} 1}^{ \plus{} \infty}{\frac {f(n)}{n^2}}$ απειριζεται θετικα.

3) Να βρεθουν τα ορια:

$ \lim_{n \rightarrow \plus{} \infty}{n!det{\begin{pmatrix} 1 & 1 & ... & 1 \\
1 & \frac {3}{2} & ... & 1 \\
... & ... & ... & ... \\
1 & 1 & ... & \frac {n \plus{} 1}{n} \end{pmatrix}}}$

$ \lim_{n \rightarrow \plus{} \infty}{(n \minus{} 2)!det{\begin{pmatrix} 2 & 1 & ... & 1 \\
1 & \frac {3}{2} & ... & 1 \\
... & ... & ... & ... \\
1 & 1 & ... & \frac {n \plus{} 1}{n} \end{pmatrix}}}$

4) Εστω $ A \equal{} \{a_1, a_2, ..., a_n\}$ αναδιαταξη του συνολου $ \{1, 2, 3, ..., n\}$. Τοτε καθοδος του $ A$ ονομαζεται καθε δεικτης $ i \in \{1, 2, ..., n \minus{} 1\}$ για τον οποιο ισχυει $ a_i > a_{i \plus{} 1}$. Να βρεθει το πληθος των συνολων $ A$ τα οποια εχουν ακριβως μια καθοδο.

5) Εστω $ n, a$ θετικοι ακεραιοι. Αν ο $ n$ διαιρει τον $ (a \minus{} 1)^{2009}$ να αποδειχτει οτι ο $ n$ διαιρει και τον $ \frac {a^n \minus{} 1}{a \minus{} 1}$.[/quote]
Να ρωτησω κατι ρε σεις.Στον Seemοus δεν μπορει να παει οποιος θελει????Γιατι γινεται διαγωνισμος επιλογης??

Παει μια 6μελης ομαδα. Περισυ πηγαν οσοι ηθελαν επηδη ο διαγωνισμος εγινε στην Ελλαδα νομιζω. Αλλα οπως και να εχει η βαθμολογια καθε χωρας στο διαγωνισμο εξαρταται μονο απο την Εθνικη ομαδα της, τα υπολοιπα δεν υπολογιζονται νομιζω

Μια λυση για το πρωτο

Θετουμε $ a_n\equal{}1\plus{}2^2\plus{}...\plus{}n^n$

Τοτε ψαχνουμε το $ \frac{a_n}{a_n\minus{}a_{n\minus{}1}}$

η το αντιστροφο αυτου $ 1\minus{}\frac{a_{n\minus{}1}}{a_n}$

Ομως αν υπαρχουν τα ορια ισχυει οτι το οριο $ \frac{a_n}{a_{n\minus{}1}}$ ισουται με το οριο της n-οστης ριζας του $ a_n$ το οποιο εδω απειριζεται θετικα οποτε
το ζητουμενο οριο ειναι $ 1$.

Το δευτερο θεμα ειναι απο τον IMC 1999.

@NickNafplio
[hide]Καλα αποτελεσματα![/hide]

[quote="AlexandrosG"]Μια λυση για το πρωτο

Θετουμε $ a_n \equal{} 1 \plus{} 2^2 \plus{} ... \plus{} n^n$

Τοτε ψαχνουμε το $ \frac {a_n}{a_n \minus{} a_{n \minus{} 1}}$

η το αντιστροφο αυτου $ 1 \minus{} \frac {a_{n \minus{} 1}}{a_n}$

Ομως αν υπαρχουν τα ορια ισχυει οτι το οριο $ \frac {a_n}{a_{n \minus{} 1}}$ ισουται με το οριο της n-οστης ριζας του $ a_n$ το οποιο εδω απειριζεται θετικα οποτε
το ζητουμενο οριο ειναι $ 1$.

Το δευτερο θεμα ειναι απο τον IMC 1999.

@NickNafplio
[hide]Καλα αποτελεσματα![/hide][/quote]

Thanx αν και το βλεπω δυσκολο με μονο 2/5 + κατι ψιλα απο το 5ο
Επρεπε να ειχα ασχοληθει και με το 1ο το οποιο ειναι πολυ ευκολο αφου λυνεται και πολυ πιο απλα αν πεις οτι η καθε δυναμη ειναι μικροτερη η ιση απο το ν υψωμενο στον ιδιο εκθετη.

Το 2ο βγαινει με ανισοτητα αναδιαταξης, ενω ο Σιλουανος το ελυσε με Cauchy-Schwartz. Υπαρχει μηπως κανενας αλλος τροπος γι αυτο?

[quote="AlexandrosG"]Μια λυση για το πρωτο

Θετουμε $ a_n \equal{} 1 \plus{} 2^2 \plus{} ... \plus{} n^n$

Τοτε ψαχνουμε το $ \frac {a_n}{a_n \minus{} a_{n \minus{} 1}}$

η το αντιστροφο αυτου $ 1 \minus{} \frac {a_{n \minus{} 1}}{a_n}$

Ομως αν υπαρχουν τα ορια ισχυει οτι το οριο $ \frac {a_n}{a_{n \minus{} 1}}$ ισουται με το οριο της n-οστης ριζας του $ a_n$ το οποιο εδω απειριζεται θετικα οποτε
το ζητουμενο οριο ειναι $ 1$.
[/quote]

Μια διαφορετική λύση: $ 1 \leq \frac{1\plus{}2^2 \plus{} \ldots \plus{} n^n}{n^n} \leq \frac{n \plus{} n^2 \plus{} \ldots \plus{} n^n }{n^n} \leq \frac{n^{n\minus{}1} \plus{} n^{n\minus{}1} \plus{} n^n}{n^n} \leq 1 \plus{} 2/n.$

[quote="Dimitris X"][quote="silouan"]Έχουν βγει και τα αποτελέσματα : Δυστυχώς ξέρω μόνο τους 5 από τους 6 . τους γραφω με τυχαία σειρά
Λαμπρόπουλος Λεωνίδας
Παναγιωτάκος Νίκος
Γιώργος Σακελλάρης
Κώστας Καραπατάνης
and me  :lol:[/quote]
Μπραβο Sil  :clap2:
Καλη επιτυχια σου ευχομαι οπως και στα αλλα τα παιδια[/quote]

bravo paidia..toulaxiston me tous treis apo esas eimaste palioi gnorimoi (me to leonida sti bmo sti veroia,me to sil kai ton niko stin teleutaia imo) kai eimai polu xaroumenos gi auto!kali epitixia kai sti sunexeia se ola ta melh ths omadas...

Μήπως ξέρει κανείς τα ονόματα όλων των παιδιών που πέρασαν?

Πριν λιγο εμαθα απο τον κ. Φελλουρη οτι ειμαι στην 6μελη ομαδα, δεν εχει ανακοινωθει επισημα η ολυμπιακη ομαδα αλλα αν μου τα ειπε σωστα ο κ. Φελλουρης τοτε η ομαδα εχει ως εξης:

1) Μπραζιτικος Σιλουανος, Μαθηματικο ΕΚΠΑ, 1ο ετος
2) Παναγιωτακος Νικολαος, Μαθηματικο ΕΚΠΑ, 1ο ετος
3) Γιωργος Σακελλάρης , Μαθηματικο ΕΚΠΑ, 2ο ετος
4) Κωστας Καραταπανης, Μαθηματικο ΕΚΠΑ, 2ο ετος
5) Κολλιοπουλος Νικολαος, ΣΕΜΦΕ - ΕΜΠ, 1ο ετος
6) Λαμπροπουλος Λεωνιδας, ΣΗΜΜΥ - ΕΜΠ, 2ο ετος

Edit: Μολις ανακοινωθηκαν και επισημα τα αποτελεσματα, σε λιγη ωρα λογικα θα αναρτηθουν και στην ιστοσελιδα της ΕΜΕ

Παιδια συγχαρητηρια και καλη επιτυχια στην ολυμπιαδα. :first:


limit
logarithms
calculus
derivative
function
calculus computations

how do you find:
${\lim_{x\to\{10}}\frac{x-10}{1-\log{x}}?$

I've never worked with limits with logarithms, so please explain it in simple terms.

is there a way to do it without L'Hospital and without getting too complex?

The first thing I'll do is make a substitution - not to solve the problem, but to reshape it a little. Let $u=\frac{x}{10}$ so that $x=10u.$ Then we have

$\lim_{x\to10}\frac{x-10}{1-\log_{10}x}= \lim_{u\to1}\frac{10u-10}{1-(1+\log_{10}u)}$

$=-10\cdot\lim_{u\to 1}\frac{u-1}{\log_{10}u}$

Flip that over: what is $\lim_{u\to 1}\frac{\log_{10}u}{u-1}?$ That's the derivative of the function $f(x)=\log_{10}x$ at $x=1.$ You really can't escape calculus in some form here. We need to know two things: first, that $\log_{10}x=\frac{\ln x}{\ln 10},$ and second, that the reason we make such a big fuss in calculus about the natural logarithm is that its derivative at the special point $1$ is as simple as possiblem, namely $1.$

So $\lim_{u\to 1}\frac{\log_{10}u}{u-1}=\frac1{\ln 10}$ and the answer to the original question is $-10\ln 10\approx-23.03.$


geometry
AMC
AIME
calculus
analytic geometry
3D geometry
tetrahedron

I am looking for a resource (book (or section of one), online text, etc...) on the basic 3D geometry seen on short-answer exams like AMC, AIME, Mandelbrot, etc... Anyone have any suggestions?

Fierytycoon

[color=cyan]I believe that analytic 3D geometry is taught in precalculus textbooks.  That is of some use.  Certainly, any book that does multivariable calculus should have a chapter introducing you to 3-D coordinate systems, and equations of planes and so on.  That will deal with some of the questions on the AIME, but not most.  In general, the techniques you need for the rest of the problems will be 2-D techniques, but they will have to come with either good diagrams or fabulous powers of visualization.  For example, the problem I gave earlier on the challenges forum can be solved by finding the volume of a regular tetrahedron, which is messy but doable by simply dropping an altitude and finding its length by taking a vertical plane cut of the tetrahedron and then doing normal triangle stuff, and then applying a dilation to it, which requires visualization or a really really good diagram and knowing how dilation affects a volume, which I guess is something that might be nice to know in advance, but it's easy to figure out by similar figures:  if lengths increase by a factor of r, volumes increase by a factor of r^3.  So, my usual, not-very-helpful advice is to find as many problems as you can and do them.  And then find other problems.  There was a 3-D geometry problem posted on the Mandelbrot Competition Problem Corner, which can be found somewhere on http://www.mandelbrot.org, maybe 2 years ago.  There may also be others there.  You might check old USAMTS.  There is one every so often on the AIME.  Probably on ARML and NYSML, too.[/color]

Quite often on 3D problems, it is useful to look at a 2D cross-section (for example, one face of a tetrahedron). Also keep in mind that some properties of 2D figures are very closely related to their 3D analogies. On AIME 2001 there was a question asking about the radius of the insphere of a tetrahedron. Well, we know the inradius of a triangle ABC is the area divided by the semiperimeter, or K=r*p/2. We can show this by constructing triangles ABI, BCI, and CAI, if I is the incenter. Similarly, if ABCD is a tetrahedron and I is the incenter, then V=r*S/2, where V is volume and S is surface area. We prove this by constructing tetrahedra ABCI, BCDI, CDAI, and DABI.

I recall a problem from Zeitz's [u]The Art and Craft of Problem Solving[/u] that is based on this idea. Given nine lattice points in space, prove that at least one of the segments connecting two of them contains a lattice point between the two endpoints. (Ok this problem has also appeared in many other places, including Discovery Quest 2003, which is a SF Bay Area contest.) Zeitz said that the 3D version is too hard to do from scratch, and so what you do is consider the 2D version and prove that on a plane with 5 points such a property is guaranteed. I won't tell you how to do it. But the 3D version has a proof that is practically identical.

I know there's some solid geometry in AoPS Volume II in the Analytic Geometry chapter.  There's probably more in Volume I, but my sister stole my copy for the summer and it's been a long time since I've gone through it carefully.

I just looked through 40 or so books that are designed for (or happen to be very helpful with) preparing for high school math competitions, and I can't really find anything that matches what you want.  The Hungarian Problem Book III has a [i]very[/i] short discussion of solid geometry and I definitely would [i]not[/i] recommend buying the book for this section alone.  Lots of my books have a few solid geometry problems, but no real section.  For instance, many of the Honsberger books (the Mathematical Gems series, probably Mathematical Chestnuts, and others) discuss a few problems in depth.  The Contest Problem Books (five books, each with ten years of old AHSMEs with solutions) have their problems classified at the end, which can be helpful.   (Contest Problem Book I lists 10 problems under Solid Geometry, III lists 2 under Volumes and there are likely others hidden under things like Proportion and Coordinate Geometry, IV lists 5 under Cubes, 1 under Pyramids, 2 under Spheres, 3 under Tetrahedrons, and 1 under Volumes... I can't find my copies of II or V right now.  V includes AIME problems too, in case you're interested.) The IMO 1978-1985 book has four solid geometry problems with solutions, though I guess you could get those online, and the USAMO 1972-1986 book has eight solid geometry problems with solutions.

That probably wasn't very helpful.  I'm sure someone else will have better suggestions.

All that said, it's a lot easier to find good plane geometry resources.  Many solid geometry problems can be solved by extending (or even simply applying) things you know from plane geometry.  If you have a solid (haha) understanding of plane geometry, the solid stuff will fall into place with just a few more tools that you probably already know.

On a [i]totally[/i] unrelated note, [i]Winning Solutions[/i] by Edward Lozansky and Cecil Rousseau is a great book.  I haven't looked at it since two summers ago and I just found it while I was looking through my shelves.  I learned so much from it.  It doesn't have a geometry section, but it's still fantastic.  Maybe I'll start a new topic about it at some point.

Now I feel that my math book collection (or library or whatever you want to call it) is so limited. I have about 120 math books I think, but I don't think I have anywhere near 40 problem-solving books. Maybe I have 15 or 20, but that's all.

120?  You win.

I doubt I have more than 25 math books that aren't related to problem solving... not counting nonsense my school made me buy before the AP Calc test.

Okay, looking now, maybe it's more like 35.  You still win.  That's also not counting stuff like [i]Beautiful Mind, Man Who Loved Only Numbers, In Code, I Want to Be a Mathematican[/i], and that sort of thing.  However, it IS counting about half a dozen knot theory books (which I don't plan to read... except for the two or three chapters I read through when I got them) that my aunt bought for me after we talked for five minutes about some knot theory-related thing.

I also counted five volumes of Arbelos among my 40-something problem solving books.  That's probably cheating.

I bought [i]Proofs from THE BOOK[/i] tonight at Borders, and it's still in a shopping bag downstairs.  I forgot about that.  Math books++

We're so ridiculous.  :D

I hope you enjoy [i]Proofs from THE BOOK[/i]. There are some really great things in there. There are also a whole lot of things that I don't understand, but they're probably also great.

Good to hear that you enjoyed it.  It was wrapped in plastic at the store, so I couldn't flip through it... but it was on sale, so how could I resist?  Looking at it now, it's not exactly what I was expecting, but I'm sure I'll still like it.

[quote="Mindspa"]Now you two only need to review every book you own. :) Who starts? ;)[/quote]

Haha.  Simon first!

[quote="Mindspa"]Now you two only need to review every book you own. :) Who starts? ;)[/quote]

It's something of a problem that I'm not in possession of every book I own. I often lend books to people, and they don't necessarily (or even usually) give them back. Then again, I have been guilty of doing that too from time to time. I'm not a very good reviewer though. I always enjoy math books as long as they aren't too hard for me to understand or far too easy. (I don't have many that are far too easy for me since I only started buying them really recklessly when I was in high school, and before that I would buy ``popular math'' books, such as somewhat non-technical descriptions of Fermat's Last Theorem.)

[quote="ComplexZeta"] and before that I would buy ``popular math'' books, such as somewhat non-technical descriptions of Fermat's Last Theorem.)[/quote]

I haven't outgrown that yet.  Last night, I seriously considered buying [i]Fermat's Enigma[/i] and a couple months back, I got Keith Devlin's [i]The Millennium Problems[/i] book. =)

[quote="Eve"]I haven't outgrown that yet.  Last night, I seriously considered buying [i]Fermat's Enigma[/i] and a couple months back, I got Keith Devlin's [i]The Millennium Problems[/i] book. =)[/quote]

Neither have I. But now I don't buy only popular math books.

[quote="Mindspa"]And what about this one: http://www.amazon.com/exec/obidos/tg/stores/detail/-/books/0817641556/102-9224201-6549707
It costs 700 kr ($70) in the store where I saw it.[/quote]

That's a good book.  I've got that one.  The topics covered are a bit unusual though.  More on this later... =)

[quote="Mindspa"]I want to know more about Riemann and his life. Is this book good? http://www.amazon.com/exec/obidos/ASIN/0309085497/ref=pd_sxp_elt_l1/102-9224201-6549707[/quote]

I just started reading that book and I think it's pretty good. But I found this book to be much better - The Music of the Primes by Marcus du Sautoy. http://www.amazon.com/exec/obidos/tg/detail/-/0066210704/ref=pd_bxgy_text_1/002-2695708-1660037

[quote]But I found this book to be much better - The Music of the Primes by Marcus du Sautoy. [/quote]

But you know, personally, I like the cover of "Prime Obsession" more than that of "The Music of the Primes"  :D

Fierytycoon

[color=cyan]I am impressed by the quantity of books you own . . . my own count is more like 20 or less.  However, I am recently greatly depleted -- I had on the order of 30 problem-solving or problem books, maybe more like 40, but they all belonged to my high school.  That's very good when you're on the high school team and they lend you everything you could ever need, but not so good when you graduate and have to give them all back.  :sigh:.[/color]

You guys mind posting your list of problems solving books?

I happened to notice on amazon.com that Simon Rubinstein-Salzedo had a top 10 math book list. Im guessing that with the lack of Rubinstein-Salzedo's out there, that that was ComplexZeta.

Fierytycoon

Yes, it was I. I should try to update that list with some more of my favorite books now if I can figure out how to use Amazon properly.

[quote]I happened to notice on amazon.com that Simon Rubinstein-Salzedo had a top 10 math book list. Im guessing that with the lack of Rubinstein-Salzedo's out there, that that was ComplexZeta.
[/quote]

Yup.  Besides, if you click on "About Me" or something, it mentions ComplexZeta.

Where is your list complexzeta? I'd like to see if I should get any of the books.

-interesting_move

in my school library there're over one HUNDRED math problem solving books.
what's more, over TWENTY of them have many solid geometry problems.

The two IMO books, and I think the USAMO book has a section on 3D geometry:

International Mathematical Olympiads, 1955-1977
[url]http://www.amazon.com/exec/obidos/tg/detail/-/0883856271/qid=1104929680/sr=1-3/ref=sr_1_3/103-1325159-2301452?v=glance&s=books[/url]
International Mathematical Olympiads; and Forty Supplementary Problems, 1978-1985
[url]http://www.amazon.com/exec/obidos/tg/detail/-/088385631X/qid=1104929603/sr=8-3/ref=sr_8_xs_ap_i3_xgl14/103-1325159-2301452?v=glance&s=books&n=507846[/url]
USA Mathematical Olympiads 1972-1986 Problems and Solutions
[url]http://www.amazon.com/exec/obidos/tg/detail/-/0883856344/qid=1104929725/sr=1-8/ref=sr_1_8/103-1325159-2301452?v=glance&s=books[/url]

I find that most 3D geometry problems reduce to using results in plane (plain?) geometry.

And finally, I wouldn't get too worked up about the number of books one has.  I would go for quality, not quantity.

[quote="Danbert"][quote="Mindspa"]I want to know more about Riemann and his life. Is this book good? [/quote]

Yes.  I really liked this book.[/quote]

I second that :) . It's the wonderfully clear way he explains the math that made me hooked though, not the history details, though those are excellent too.


rotation
trigonometry
combinatorics proposed
combinatorics

Consider a point $O$ in the plane. Find all sets $S$ of at least two points in the plane such that if $A\in S$ ad $A\neq O$, then the circle with diameter $OA$ is in $S$.

[i]Nikolai Nikolov, Slavomir Dinev[/i]

[i]Lemma:[/i]  If  $A\in S$ and $O\neq A$ then a closed disc with radius $OA$ belongs to $S$.

[i]Proof:[/i] By the conditions of the problem we have that the circle $w$ with diameter $OA$ lies in $S$. Then take any point $B$ in the interior of this circle and draw a line $l$ perpendicular to $OB$ at $B$. Take $C\in l\cap w$ (note that the latter in non-empty) and draw the circle with diameter $OC$ ($C\in S$). Note that $B$ lies on this circle. Hence $B\in S$. Q.E.D.

Now comes the proof of the problem. First of all let's consider the case when there is a ray $z$ passing through $O$ such that $\forall x\in \mathbb R^+\ \ \exists X\in S\colon X\in z,\ OX>x$. Draw a line $y\bot z : y\cap z=O$ and denote the half plane whose boundary is $y$ and in which $z$ lies by $U$. Note that every point that lies in $U$ but does not lie on $y$ is in $S$ (using the choice of $z$ we prove it in the same fasion as we have proven lemma). Now note that every point that lies on $y$ is in $S$ (take a point in $U$ sufficiently close to $y$ and sufficiently far from $O$, draw a circle with diameter whose one endpoint is the chosen point and the other is $O$ and use lemma). Now in the same manner (taking $y$ instead $z$) we prove that every point in each of the two half-planes in which $z$ and $-z$ divides the plane belongs to $S$. And finaly we prove that $-z$ lies in $S$ (again analogously). Hence $S$ coincides with the plane.

That was the nasty part of the solution. Now comes the other one with a briliant idea.  In this case that is left such ray $z$ as aforementioned does not exist and consequently $\exists m\in \mathbb{R^+}\colon \forall A\in S\ \ OA\le m$. And let $m$ be as minimal as it can be. In other words $m=\sup \{ |OA| \colon A\in S \}$. Now perform inversion centered at $O$ with radius $m$. Denote $X'$ an inversive image of $X$ and $S'$ an inversive image of $S$. Note that the conditions of the problem imply that for every $A'\in S'$ the line $l'\bot OA'\colon A'\in l'$ lies in $S'$ and our lemma implies that all half-plane which has $l'$ as the boundary and does not contain $O$ lies in $S'$. Now choose $l'$ sufficiently close to an open disc centered at $O$ with radius $m$. Fix $\epsilon >0$ and set $l_0=l'$. By $A_i$ we will denote a point on $l_i$ such that $OA_i\bot l_i$. Now choose $B_0\in l_0 \colon B_0A_0=\epsilon$ and let $l_1$ pass through $B_0$ perpendicularly to $OB_0$. $l_{n+1}$ is obtained from $l_n$ in the same fashion. We just have to choose such $B_n$ that $\angle A_{n-1}B_nA_{n+1}$ is obtuse. Note that in such a way we step by step rotate around the circle of inversion by a fixed a angle. And note that every $l_i$ and a half plane which has the boundary $l_i$ and does not contain $O$ lies in $S'$ (we get this using lemma and the aforementioned equivalence of the conditions of the problem). In such a way we rotate around the circle and get that all plane without the open disc centered at $O$ with radius $m$ lies in $S'$. (Indeed it is easy but nasty to make the above argument rigorous: one just have to choose $l'$ sufficiently close to the circle and $\epsilon$ sufficiently small. However intuitively it's clear what is going on). Hence an open disc centered at $O$ with radius $m$ lies in $S$. By the definition of $m$ no other points different from those lying on the boundary of the closed disc centered at $O$ with radius $m$ can lie in $S$. Let $T$ be any subset of this boundary and note that we can have $T\subset S$ and $S$ will satisfy the conditions of the problem.

So $S$ is either all plane or a union $T\cup U$ where $U$ is any disc centered at $O$ without its boundary and $U$ is any subset of points of this boundary. Clearly all such $S$ satisfy the conditions of the problem.

hey this looks complicated!

Any easier ways?

Firstly how does the lemma work? Surely you've only shown that the interior of the circle with [b]diameter[/b] OA belongs to S?

I think I have a nicer way of tackling the unbounded case.

Suppose there exists a point P in the plane that is not in S. Then the line through P perpendicular to OP is also not in S (if it were, it would imply that P were in S, which it isn't). Also, for any point R of the plane on the other side of this line, one can choose a point Q on the line such that $\angle OQR = \pi/2$, so if R were in S, Q would be in S, again a contradiction, so the whole of the plane on the other side of this line would be excluded from S. We can also choose $Q\not =P$ on the line, draw the perpendicular to OQ through Q, and draw similar conclusions. If we do this a few times (so we next choose a point A such that AQ is perpendicular to OQ, etc.), we can form a closed polygon around O, which is a bound for S.

Therefore, if S is unbounded, S must coincide with the plane. I didn't manage to find a solution for what S looks like when it's bounded though :(.

I beg your pardon, indeed the word radius in the statement of the lemma must be changed to diameter. But it's just a typo since afterwards I'm using the lemma properly.

I hope I haven't done any silly mistakes,because I spent quite some time writing down this proof(just when I was about to finish it my PC crashed and I had to start all over again :O )

Lemma 1: Suppose $A\in S$ then the closed disk of diameter $OA$ is contained in $S$.
Proof: Take point $X$ in the disk, bring perpendicular to $OX$ at $X$ and let the section with the circle of diameter $OA$ be $Y$,then $X$ belongs to the circle of diameter $OY$(obviously $Y\in S$).

Lemma 2: Suppose $A\in S$,then the open disk $(O,OA)$ is contained in $S$.

First,we prove the following "sublemma":
Given point $Z$ on circle $(O,OA)$ there is a point $X$ in the interior of the disk such that $\left | ZX  \right |<\epsilon$ for every $\epsilon >0$
Proof: Split the angle $\measuredangle AOZ$ into $n$ equal parts by the lines $OA=\lambda_0,\lambda_{1}, \lambda_{2},...,\lambda_n=OZ$ and define the sequence of points $X_0=A$ and $X_{k+1}$ the projection of $X_{k}$ onto $OA_{k+1}$. All such $X_{k}$ belong to $S$ because $X_{k+1}$ belongs to the circle of diameter $OX_{k}$. Then I have $\left | OX_{k+1}  \right |=\left | OX_{k}  \right |\cos( \frac{\measuredangle AOZ}{n})$. Therefore $\left | OX_{n}  \right |=\left | OA  \right |(\cos( \frac{\measuredangle AOZ}{n}))^n$ and $\left | X_{n}Z  \right |=\left | OA\right |(1-(\cos( \frac{\measuredangle AOZ}{n}))^n)$ which tends to $0$ as $n$ tends to infinity.

Using the sublemma, I can have a point $X\in S$ at any radius $r<OA$ and by taking the union of the closed disks with diameter $OX$,where $X$ varies over all points at radius $r$, I get that the closed disk $(O,r)$ belongs to $S$ and since $OA$ is the supremum of such $r$ the open disk $(O,OA)$ belongs to $S$.

Now, we take cases.
If $S$ is unbounded, then it is the plane. Suppose there exists point $Z$ in the plane not contained in $S$, then there is a point with $OX>OZ$ and by lemma 2 $X\in S$,contradiction.
If $S$ is bounded, then consider the set of points (not necessarily finite) $A$ that $OA=R$ is maximum distance(or the supremum of such distances). Then,by Lemma 2 $S$ contains the open disk $(0,R)$ and a set of points on the circle.


Can you make it possible to disable automatic log-in because I want to log-in as a different screen name, but I can't on the computers that I have enabled automatic log-in.

1. You can change your username without registering a new account at "PROFILES" (at the top bar, near the centre).

2. You can use the "LOGOUT" button to log out and then log in as a different user (also at the top bar)

Delete your browser cookies. For information on how to do that, try your help menu.

Hey Osiris. I've tried logging out but it automatically logs me in again. It may work for you but not for me.

Confuted, I'll know how to delete cookies so I will do that.


On a circle, select 6 points consequently, A, B, C, D, E, and F. Mark the intersection of the extensions of AB and ED as L, that of BC and FE as M, and that of CD and AF as N. Prove L, M, and N are on one straight line.


inequalities
algebra
polynomial
induction
superior algebra
superior algebra unsolved

Let $P(X) = \sum_{i=0}^n a_i X^i \in \mathbb{R}[X]$, $c(P)$ the number of sign's changes in the list of $a_i$ and $z_+(P)$ the order of $P^{-1}( \{0 \} ) \cap \mathbb{R}_+^*$.
Prove that: $z_+(P) \leq c(P)$.

What is the order of that set? What does it mean? :?

It says that the number of positive roots of a polynomial is at most the number of changes of signes in the sequence of its coefficients. Order of a set means cardinal, I think.

Oh, ok, thanks. I've never heard it called "order" before :).

Thanks, grobber, at least someone seems to be interested in this beautiful problem.  :)

[quote="harazi"]Too bad extremely beautiful problems are forgotten, while spams are respected... :([/quote] This is a very classical one. I agree with harazi, it is really nice. Here is a [hide="hint"] Substitute $x=e^t$, multiply by $e^{\lambda t}$ with appropriately chosen $\lambda$ and differentiate. [/hide] As to respected spams, I haven't seen many of them on this forum (though, maybe, I was looking in the wrong places  :?  :P ).

If $f'$ has at least one multiple positive root, then the number of (distinct) positive roots of $f'$ as at least equal to the number $k$ of positive roots of $f$, so we can use induction (clearly, the sign changes go from $f'$ to $f$). We may thus assume that $f$ has $k$ distinct simple positive roots, and $f'$ has $k-1$ distinct positive roots.

There are at least $k-1$ sign changes in the sequence $a_n,a_{n+1},\ldots,a_1$. With the above assumptions, we reach the conclusion that $f(0),f'(0)$ have different signs (a drawing should show why, otherwise, $f'$ would have the same number of positive roots as $f$, and we assumed it had one less). This means that there's one extra sign change, from $a_1$ to $a_0$, which, together with the $k-1$ found before, gives $k$.
I'm probably missing something, but I need to get some sleep :).

Of course, I've used this as a well-known fact for years. This is Descartes' Rule of Signs.
If multiplicity is counted, then the difference is an even integer.


If $ E$ and $ F$ are (Lebesgue)measurable, and $ m(E)>0, m(F)>0$, prove that
$ E\plus{}F$={$ x\plus{}y : x \in E, y \in F$}
contains an interval.


There is a circle with center O.  Outside of O there is a point P.  R and S are on O such that PR and PS are tangent to the circle.  PR=1.   :ang: RPS=120 :^o:.  Find the area enclosed by RP, PS, and arc RS.

It's doable.  There's little information, but it's doable.

[hide] :ang: RPS = 1/2 (big arc RS - little arc RS), so little arc RS = 60. Then :ang: ROS = 60. Thus :tri: ROP and :tri: POS are 30-60-90 triangles, and PR = PS = 1. Area of PROS = :rt3: , area of sector ROS = :pi: / 6, so the area we are looking for is :rt3: - :pi: / 6.[/hide]