limit
inequalities
function
real analysis
real analysis unsolved

For $ n\geq 2$ ,let $ a_n\equal{}\frac{(n^2\plus{}1)(n^2\plus{}2)...(n^2\plus{}n)}{(n^2\minus{}1)(n^2\minus{}2)...(n^2\minus{}n)}$. Show that

a)$ \lim_{n \rightarrow \infty} a_n\equal{}e$.

b)$ \lim_{n \rightarrow \infty}n(a_n\minus{}e)\equal{}e$.

Consider $ \lim_{n \to \infty} ln\left(1 \plus{} \frac {1}{n^2}\right) \plus{} ... \plus{} ln\left(1 \plus{} \frac {n}{n^2}\right) \minus{} ln\left(1 \minus{} \frac {1}{n^2}\right) \minus{} ... \minus{} ln\left(1 \minus{} \frac {n}{n^2}\right)$
We note the inequality $ x \ge \ln(1 \plus{} x) \ge x \minus{} \frac {x^2}{2}$ for $ x \ge 0$, and reverse for $ \minus{} 1 < x < 0$

In this way,

$ \frac {1 \plus{} ... \plus{} n}{n^2} \ge ln\left(1 \plus{} \frac {1}{n^2}\right) \plus{} \cdots \plus{} ln\left(1 \plus{} \frac {n}{n^2}\right) \ge \frac {1 \plus{} ... \plus{} n}{n^2} \minus{} \frac {1^2 \plus{} ... \plus{} n^2}{2n^4}$
By sandwich rule the middle expression goes to $ \frac {1}{2}$. Similarly for the negative part. Thus the whole expression goes to 1. Exponential function gives $ lim_{n \to \infty} a_n \equal{} e$.

The bound obtained here, if we do not take log and use the exponential version of the inequality, yields a bound on both sides for the given expression as
$ e^{\frac{n\plus{}1}{n}} \ge a_n \ge e^{\frac{n\plus{}1}{n} \plus{} o\left(\frac{1}{n}\right)}$
Using these two bounds we can do part b easily.


calculus function logarithms derivative limit inequalities integration

My professor proposed to me a question that states:

[quote]Assume f(x) is a positive function, and f"(x) is continuous.

Show that no f(x) on the interval on (0, $ \infty$) exists such that f(x)f"(x) $ \le$ -1[/quote]

I broke down the problem by analyzing each positive function. The method was correct, my professor said, but I need to show a general proof, since he mentioned that there are infinite amount of positive functions.

So, I suggested that if f(x) is a positive function, then f(x)" is a positive function as well. That didn't suffice either in words. How can that statement be proven in a general context (as in, the statement sounds ok, but it needs a mathematical proof).

Your suggestion can't be proven in a general context because it's wrong.  Consider $ f(x) \equal{} \ln (x \plus{} 1)$.

Edit:  Unless you mean $ f(x)$ is positive on $ ( \minus{} \infty, \infty)$.  In that case, what you actually want to show is that $ f''(x)$ cannot be negative everywhere.

Yeah, $ f(x)$ is positive in the interval (-$ \infty$,$ \infty$)

Shouldn't I show f"(x) is positive everywhere to satisfy the condition that f(x)f"(x) $ >$ -1

The problem is how we can find the value of $ x$ where 0 $ >$ f(x)f"(x) $ >$ -1. What function can satisfy that?

The problem is to show that no positive function can satisfy $ f(x) f''(x) \le \minus{}1$ for [i]every[/i] value of $ x \in (0, \infty)$.  It suffices to show that $ f''(x) > 0$ [i]somewhere[/i] (not everywhere).  Again, it's not true that $ f''(x)$ is positive everywhere; consider $ f(x) \equal{} \frac{1}{x^2 \plus{} 1}$.

So, basically you are suggesting that I post a contradiction to the proof?
The example you have provided made so much sense in regards to a function that concaves down; where $ 0\le x <1$ would lead to the formation of large negative values.

I guess, basically, the statement is false then, because there exists a positive function $ f(x)$ such that $ f(x)f''(x) \le \minus{}1$

Correct me if I'm wrong :P

Thanks for your assistance

If a ($ C^2$) function is positive on the [b]whole[/b] real line, it has a nonnegative second derivative somewhere. At that point $ c$, we have $ f(c)f''(c)\ge 0>\minus{}1$

If you don't require the whole real line, there are counterexamples. $ f(x)\sqrt{\frac43}(x\minus{}a)^{\frac32}$ is a counterexample on $ (a,\infty)$; $ f(x)f''(x)\equal{}\minus{}1$ everywhere on this set.

If $ f''(x)<0$ for all real $ x$ and $ f'(a)\equal{}b>0$, what can you say about $ \lim_{x\to \minus{}\infty} f(x)$?

Doesn't $ (x\minus{}a)^{\frac32}$ have a positive second derivative?

$ (x\minus{}a)^{\frac{3}{2}}$ : concaves upwards --> Positive

jmerry, wouldn't that result to zero if the second derivative is negative (concave down), which is similar to condition presented in $ f(x) \equal{} \frac{1}{x^2\plus{}1}$

Elaborating on that - I've gotten this far but can't close the deal:

Suppose $ f(x)>0$ on $ [0,\infty)$ and $ f(x)f''(x)\le \minus{}1.$

We must have that $ f'(x)>0.$ Else, if the graph is ever decreasing and is always concave down, then the function cannot remain positive.

Now, let $ u(x)\equal{}\frac{d}{dx}\ln(f(x))\equal{}\frac{f'(x)}{f(x)}.$ We have that $ u(x)>0.$

$ u'(x)\equal{}\frac{f(x)f''(x)\minus{}f'(x)^2}{f(x)^2}\le\frac{\minus{}1\minus{}f'(x)^2}{f(x)^2}
\le\frac{\minus{}1}{f(x)^2}\minus{}u(x)^2.$

If I throw out the first term [color=green][in retrospect, a bad idea][/color] and use that $ u'\le\minus{}u^2,$ I can solve the differential inequality to find that $ u(x)\le\frac1{x\plus{}C_1}.$ Integrate that and exponentiate it to get that $ f(x)\le C_2(x\plus{}C_1).$

That much gets us nowhere; we already knew that $ f$ had to be concave down. We must make use of the fact that $ f''(x)$ is not merely negative but that $ f''(x)f(x)\le\minus{}1.$

I have found no counterexamples but don't yet have a proof.

OK, I was confused. No counterexample is quite that simple, and in fact there isn't one.

A proof: (assumptions as in Kent's post)
Let $ f(0)\equal{}a,f'(0)\equal{}b$.
We have $ f(x)\equal{}a\plus{}\int_0^x\left(b\plus{}\int_0^t f''(s)\,ds\right)\,dt\equal{} a\plus{}bx\plus{}\int_0^x (x\minus{}t)f''(t)\,dt$
Using the estimate $ f(x)\le a\plus{}bx, f''(x)\le \frac{\minus{}1}{a\plus{}bx}$, we get $ f(x)\le a\plus{}bx\minus{}\int_0^x \frac{x\minus{}t}{a\plus{}bt}\,dt\equal{}a\plus{}bx\plus{} \frac1b\int_0^x\frac{a\plus{}bt}{a\plus{}bt}\,dt\minus{}\int_0^x \frac{x\plus{}\frac ab}{a\plus{}bt}\,dt$
$ f(x)\le a\plus{}bx\plus{}\frac1bx\minus{}\frac1b(x\plus{}\frac ab)\ln(a\plus{}bx)\plus{}\frac1b(x\plus{}\frac ab)\ln(a)$
That goes to $ \minus{}\infty$ as $ x\to\infty$, so no such function exists.

jmerry, how did the last equation reach $ \minus{}\infty$?

[quote="jmerry"]

Let $ f(0) \equal{} a,f'(0) \equal{} b$.
We have $ f(x) \equal{} a \plus{} \int_0^x\left(b \plus{} \int_0^t f''(s)\,ds\right)\,dt \equal{} a \plus{} bx \plus{} \int_0^x (x \minus{} t)f''(t)\,dt$
[/quote]

I'm still confused on this step.

What property did you use to bring that?

I switched the order of integration- Fubini's theorem. In the opposite order, the inner integral is easy.


Prove that if $a_{1},a_{2}, \ldots ,a_{n}>0$ and $a_{1}+ \ldots +a_{n}=1$ then:
$ \frac{a_{1}^2}{a_{1}+a_{2}}+ \frac{a_{2}^2}{a_{2}+a_{3}}+ \ldots +\frac{a_{n}^2}{a_{n}+a_{1}} \geq \frac{1}{2}$.

made by Teresin D.

Ineq which is obious using the fact that:

$ \frac{x_{1}^p}{y_{1}^{p-1}}+ \ldots + \frac{x_{n}^p}{y_{n}^{p-1}} \geq \frac{ ( \sum x_{1} )^p}{ ( \sum y_{1} )^{p-1}}$
$x_{k}$ and $y_{k} \in R^{+}$, $ k=1,n$ and $p \in N$.
do you have something different?

cheers! :D :D

As no one seems to be replying, I guess I should do it :)

First, the obvious solution with Cauchy: \[\left(\sum_{cyc}\frac{a_{1}^2}{a_{1}+a_{2}}\right)\left(\sum_{cyc}a_1+a_2\right) \ge \left(\sum_{cyc}a_1\right)^2\]\[\left(\sum_{cyc}\frac{a_{1}^2}{a_{1}+a_{2}}\right) 2 \ge 1\]\[\left(\sum_{cyc}\frac{a_{1}^2}{a_{1}+a_{2}}\right) \ge \frac12\]

Then, the less obvious solution (which I really like :))
$f(x) = \left(\frac{1}{1+x}\right)$ is convex for positive $x$.

The LHS rewrites as $\sum_{cyc}\frac{a_1}{1+\frac{a_2}{a_1}}$ so Jensen gives us:

\[\sum_{cyc} a_1 \cdot f\left( \frac{a_2}{a_1} \right) \ge (a_1+...+a_n) \cdot  f\left( \sum_{cyc} a_1 \frac{a_2}{a_1} \right) = 1 \cdot f(1) = \frac12\]

 :D

This is a special case of an inequality that I posted a while ago (http://www.mathlinks.ro/Forum/viewtopic.php?t=49296).

Let $S_1=\sum\frac{a_i^2}{a_i+a_{i+1}}$, $S_2=\sum\frac{a_{i+1}^2}{a_i+a_{i+1}}$.  Then $S_1-S_2=\sum\frac{a_i^2-a_{i+1}^2}{a_i+a_{i+1}}=\sum a_i-a_{i+1}=0$, so $S_1=S_2$.

But now, $S_1+S_2=\sum\frac{a_i^2+a_{i+1}^2}{a_i+a_{i+1}}\ge\sum\frac{a_i+a_{i+1}}{2}=\frac{1}{2}$, so $S_1\ge\frac{1}{2}$ as required.


Two cards are drwan at random from a pack of well shuffled 52 cards
1)what is the probability the cards drawn are both spades

[b]book's solution[/b]
the numebr of ways of choosing two cards $ \binom{52}{2}\equal{}A$
number of  ways of chosing two spades $ \binom{13}{2}\equal{}B$
so probability is $ \frac{B}{A}$

[b]my method[/b]
total possible  cases in the sample space is $ 52 \times 51\equal{}A'$
total cases favouring us from the sample space is $ 13 \times 12\equal{}B'$
so probability is $ \frac{B'}{A'}$

i feel my solution is right though both give the same answer becaus e the book considers 1 spade,2 spades different same as 2spades,1spades though they as a combination are same but are surely two different sample points

who is right?

You're both right- there's more than one way to interpret it. Your interpretation has the order they are drawn mattering, while the book's doesn't.
The important thing is consistency- if you use order in one place, also use it in the others.

yes just realized it's just that our sample spaces are different and that the the sample spaces and number of favourable sample points are 'magnified' by the same factor :)
thanks


Maybe someone could help me with this problem:

In a desert exercise a tank travels 7km on a bearing of 050° from a base [i]O[/i] then 6km on a bearing of 140° to a bunker [i]B[/i]. Calculate the magnitude and direction of the displacement vector [i]OB[/i].

I drew out the diagram, but any trigonometry I try to use requires more information (another side or an angle).

Hi, maws!

Let $A$ the point where the tank changes direction

The triangle $OAB$ is right, since $150-40=90$

(You can easy see that, if you draw a parallel from $A$ to x axis)

Then $OB^2 = OA^2+AB^2 = 49+36 = 85 \Rightarrow$

$OB = \sqrt{85} \approx 9,22$

$tan( \angle AOB) = \dfrac{6}{7} \Rightarrow$

$\angle AOB \approx 40,6^o$

So the direction of  $\overline{OB}$ is $40,6+50=90,6^o$


Find an explicit form for the sum:

$\mathop{\sum}_{n=1}^{\infty}\frac{1}{m^{2}+n^{2}}$

where m is a real number.

This has been done recently.Search the forum...


What is the limit of the sequence

$ 1\plus{}\frac{1}{2^3}\plus{}\frac{1}{3^3}\plus{}\dots\plus{}\frac{1}{n^3}$
and how is it proved?

Ever heard of the [url=http://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant]Apéry[/url] constant?

o i failed i calculated it on mathematica and at first didnt realize why its zeta(3)


function
superior algebra
superior algebra unsolved

Hello,

 :blush:  I have yet another question about algebra (do tell me if this is the wrong place  :maybe: )

Let $Y$ be a quasi-affine variety ,this means: an open subset of a variety in $k^{n}$, where $k$ is an algebraically closed field

I would like to understand why $Y$ and its closure $\bar{Y}$ have the same dimension.

My book is a bit vague :
Let $Z_{0}\subset Z_{1}\ldots Z_{s}$ be a "maximal chain" of distinct (!) irreducible closed subsets of $Y$

Now $\bar{Z_{0}}\subset \bar{Z_{1}}\ldots \bar{Z_{s}}$ would also have to be "maximal chain" of distinct irreducible closed subsets of $\bar{Y}$

There are three things I would like to ask :

1. Why can we be sure that they are distinct
2. Why is this a maximal chain?
3.If we do prove that claim, does this mean the whole proof is finished.

All hints are welcome. :|
Thank you very much,fredbel

[quote="Jacobson"]are you sure the dimension is really the topological dimension and not transcedence degree of some function field

(p.s. don't rely too much on me :) )[/quote]
Yes that is what it is supposed to be? The topological dimension? :maybe:

Ok, the other definition of dimension I was thinking of was:

If $X$ is some variety, $dim X = tr.deg. k(X)$
where $k(X)$ is a function field, but since this are only functions on open dense subsets of $X$, it is clear that $dim U = dim X$

but this is not what you're looking for I suppose. Sorry :)

1. If $\overline{Z_{i}}=\overline{Z_{j}}$ then simply intersecting with $Y$ gives $Z_{i}=Z_{j}$ because the only points you add when you take the closure are points in $\overline Y-Y$ (they are closed subsets of $Y$).

2.If you had a longer chain of irreducible subsets of $\overline Y$, just intersect it with $Y$ to get a longer chain of irreducible subsets of $Y$ which would contradict the maximality of the original chain.

3.The following are equivalent for a quasiprojective variety $Y$:

a) $dim Y= n$ (it's called the Krull dimension of $Y$)
b) The length of the maximal chain of non-empty irreducible proper subvarieties of $Y$ is $n-1$.
c) The trascendance degree of the field of rational functions is $n$.


Support
number theory
relatively prime
prime numbers

I've seen people talk about the Chicken McNugget Theorem, but I've never actually seen a post where the person explains what it is...

can anyone explain to me how to use it?
thanks  :)

Edit::
I've also searched on AoPS wiki but I haven't found it in there either  :(

The Chicken McNugget Theorem says, in the simplest terms, that if you can buy Chicken McNuggets in groups of $a$ or in groups of $b$ (where $a, b$ are some relatively prime positive integers), then the biggest number of nuggets that you [b]cannot[/b] buy is $ab-a-b$.

For example, if $a = 3, b = 5$, then $ab-a-b = 15-3-5 = 7$.  You can't buy $7$ nuggets.  You can, however, buy $8$, or $9$, or $10$, or any other larger number.

is there a way to find out the smallest number of nuggets that you [b]can't[/b] buy?

eh... smallest value that is larger than a or b

dude, think about it

if they're relatively prime, then it has to be 1 higher than the largest of the 2 numbers.

try it with ANY two numbers!

jorian

eh...
let's try with my favorite numbers:

(3, 5).

5+1=6

3+3=6.

I see...?

do you get it or are you still confused?

im confused about whether or not you're confused :P

jorian

his point is that what  you said doesn't work with 3 and 5
the smallest isn't 6 cuz he COULD get 6

Applying the mcnugget theorem to 2 and 3, we see that the largest number we cannot create with 2's and 3's is 1.  So if such a formula were to exist, something screwy would happen if you plugged in 2 and 3.

Also, I am pretty sure it wouldn't be that hard to try numbers above the largest number.  Let the two numbers be $a$ and $b$ with $a$ and $b$ relatively prime and $a<b$.  If $b$, $b+1$, $b+2$,...,$b+a-1$ can all be created, we can create any number above those numbers by adding a multiple of $a$ to one of those numbers.

So we only need to try $a+1$ possibilities.  And since $b<a$, all of those numbers can only be created using $a$'s.  If we use at least one $b$, you see the rest is less than $a$.  So the number, if any, is the smallest of the numbers  $b$, $b+1$, $b+2$,...,$b+a-1$ that is not a multiple of $a$.  Hope I didn't mess up.

[quote="ProtestanT"][quote="jhredsox"]ok, i was doing something stupid. i was thinking that you had to use both :blush:

ok, here's my formula:

x is smaller than y

if y does not equal x-1 (mod x) then you add 1 for your answer

if y does equal x-1 (mod x) then you add 2 for your answer

hope that works

jorian[/quote]
hm...I got 2 and 3... 3 is 1 mod 2, so i add 2 to 3. i got 5. but 5 is 3+2. so it doesn't work.[/quote]

i just realized before you had posted, it works for everything but when the lowest number is 2

jorian

if it is 2, you can obviously find it

[quote="ehehheehee"]Looking at my work from above, I think your formula does work with all numbers but if one is 2.  If one number is 2, the largest number that cannot be created is two less than the other number so there would be no answer.[/quote]
which formula?
the $ab-a-b$ formula works. try any 2 relatively prime numbers. You can always get a number that is greater than $ab-a-b$ from adding those 2 number.
the reason $ab-a-b$ cannot be added as a sum of $a$'s and $b$'s is because if there were $a$'s in the number that makes it a multiple of $b$ away from $ab-a-b$, there would need to be either a negative amount of $b$'s needed, or the number of $a$'s itself is negative.
so that shows that $ab-a-b$ is the highest possible number that cannot be created from adding relatively prime numbers $a$ and $b$.

[quote="ehehheehee"]So the number, if any, is the smallest of the numbers  $b$, $b+1$, $b+2$,...,$b+a-1$ that is not a multiple of $a$.  Hope I didn't mess up.[/quote]

This looks correct.

[quote="ProtestanT"]the reason $ab-a-b$ cannot be added as a sum of $a$'s and $b$'s is because if there were $a$'s in the number that makes it a multiple of $b$ away from $ab-a-b$, there would need to be either a negative amount of $b$'s needed, or the number of $a$'s itself is negative.[/quote]

This is not quite detailed enough to constitute a proof.  You're missing some pretty big and important steps.

I'm not trying to prove it. I'm trying to show how it works.
I don't go as far as to proving it because i don't know the proof. I just know that it works all the time.

Yes, but you didn't support your claim that $ab-a-b$ can't be written as a sum of $a$s and $b$s.  That's weaker than the Chicken McNugget Theorem, so it's not as hard to show.

[hide="A quick demonstration"] If there are integers $x, y$ such that

$ab-a-b = ax+by$

Then

$-b \equiv by \bmod a \Leftrightarrow$
$b(y+1) \equiv 0 \bmod a \Leftrightarrow$
$a | (y+1)$

So the smallest positive value of $y$ is $a-1$.  But then

$ax+by > b(a-1) = ab-b > ab-a-b$ [/hide]
[size=75][color=red]
<Please take proofs to a higher-level Forum.  -- moderator rcv>[/color][/size]


In how many ways can four people sit in a row of five chairs?

24 ways to seat them. 5 ways to choose empty seat

24*5=120

Or, you can just arrange them in $ 5!$ ways, giving you $ \boxed{120}$.

You see that you need to arrange 4 people in 5 seats.
$ 5p4 = 120 $


projective geometry
geometry unsolved
geometry

[quote="mr.danh"][quote="kaloi01"]let angle $ ABD \equal{} 90$, let $ P$ on $ AD$ and line $ k$ is symmetric to the $ PB$ with respect to the line $ AB$. Prove that $ k$ is harmonic conjugate to $ AB$ with respect to $ AB$ and $ BD$.[/quote]
Your post has a type mistake.
[quote="mr.danh"]Let angle ABD = 90, let P on $ AD$ and line k is symmetric to the PB with respect to the line AB. Prove that k is harmonic conjugate to PB with respect to AB and BD.[/quote]
It is obvious because Ba is the bisector of angle between k and BP, AB is perp to BD.[/quote]

which one

is that true if angle$ ABD$ not equal to $ 90$ degree


[quote]Sorry! We could not process your edit due to a loss of session data. Please try again. If it still doesn't work, try logging out and logging back in.[/quote]

This is the second time in a row this has happened when I tried to mod something on the Wiki. Is this a mere coincidence?

No it's sort of a bug. It's because the way the login system (tied into the Forum one) works. Not sure yet though why it happens but posting anything fixes it.

I think this is fixed right now. Has anyone had this happening to them in the last 2 months?


geometry
cyclic quadrilateral
geometry unsolved

circles $(C1),(C2),(C3)$, intersects each other.

$(C1),(C2)$ at points $N,K$

$(C1),(C3)$ at points $M,K$

$(C2),(C3)$ at points $O,K$

$A$ an arbitrary point on $(C2)$.

$AN$ and $AO$ intersects $(C1),(C3),$ at points $B,C$ respectively.

prove, $B,M,C$ are colinear.

ha ha ha ha....

is it so hard?...

what grade are u in...?

ok it doesnt mather...

lets solve the problem...

consider a n inversion with center $K$ and an arbitrary radius.

circles $(C1),(C2),(C3)$, changes to three lines.and will make triangle$M'N'O'$.

in problem, points $A,N,B$, are colinear.then after inversion,quadrilateral $A'B'N'K$ is cyclic.

also points $A,O,C$,are colinear then after inversion ,quadrilateral $A'O'C'K$ is cyclic.

now the problem changes to another problem:

$O'N'M'$ is a triangle.$A',B',C'$ are three points which are on $N'O',N'M',M'O'$ respectively.

we also know that quadrilaterals $A'B'N'K$ and $A'O'C'K$ are cyclic.

prove that:

                $KB'M'C'$ is cyclic.

solution: $\angle M'N'O'$+$\angle N'O'M'$=$180-\angle N'M'O'$.

$\angle M'N'O'=180-\angle A'KB'$

$\angle N'O'M'=180-\angle A'KC'$

$180-A'KB'+180-A'KC'$=$180-\angle N'M'O'$.

$180-(360-\angle B'KC')=-\angle N'M'O'$

$\angle B'KC'+\angle N'M'O'=180$

and it shows us that $KB'M'C'$ is cyclic.

it means that in first shape points $B,M,C$ are colinear.

did u get it?...............................................................

[quote="marko avila"]it  seems ashegh takes a liking  to  bothering little girls  and unnecessarily using  hard tools  to  practically  solve the same  problem  (note that  all  you  did  after your inversion was  change the condition  from proving  a cyclic quadrilateral  knowing  that    B , M , C   are collinear  instead  of the  other  way  around )  when  all  you  couldve said was that  this is just a simple  angle chase  basically similar  to  the last  proof you  wrote ,  but im  pretty  sure you  knew this  very nice ashegh!!  :rotfl:[/quote]

 :rotfl: ...

marko ure right ... :D

i wanted to write six other solutions with inversion,to make sure that britney spears find out what inversion is...

inversion with each of the centers(M,N,K,O,B,M,C),will solve the problem,and this fact made me wonderfull...

to Mr.Ashegh:

i read ure solution ,but i got nothing...

what is inversion.

as marko avilla said ,it might be a powerfull tool.

where can i learn it?

to Mr. marko avilla:

marko can u post u solution? because i really need a good solution which can help me .

the solution of Ashegh is nice but it doesnt help me.

to Ashegh:

i will be glad if u send the other six solutios.because the other users can enjoy them. ;)

[quote="Virgil Nicula"][b]Marko Avila[/b] is all right ! [b]Ashegh[/b], you run over hedge and ditch with a inversion ... !
This problem is very easily (at the most for the Intermediate Topics). [hide="Britney Spears, here is the simplest solution for the your proposed problem."]$\widehat {BMK}\equiv \widehat {ANK}\ ,\ \widehat {CMK}\equiv \widehat {AOK}\ ,\ m(\widehat {ANK})+m(\widehat {AOK})=180^{\circ}\Longrightarrow$
$m(\widehat {BMK})+m(\widehat {CMK})=180^{\circ}\Longrightarrow \boxed {M\in BC}\ .$[/hide] [hide="I will offer to Britney Spears a similar problem and I will wait the his proof only here."]
[quote][color=blue]Let $C_1=C(O_1,R_1),\ C_2=C(O_2,R_2)$ be two disjunct circles and let $C=C(O,R)$ be a circle exterior tangent to the circles $C_1$, $C_2$ in the points $T_1$, $T_2$ respectively. A common exterior tangent $t$ for the circles $C_1$, $C_2$ touches them in the points $A_1$, $A_2$ respectively. Denote the intersection $S\in T_1T_2\cap O_1O_2$. Prove that $S\in A_1A_2$ and the quadrilateral $A_1A_2T_2T_1$ is cyclically.[/color][/quote][/hide][b]Ashegh[/b], please don't begin with the inversion that I will feel giddy !
[b]Britney Spears[/b] , after two days I will present the simplest proof for the added problem.[/quote]

yeah I had the same solution but i did not post before ...  your solution is good Virgil Rspect...-

Thanks to all ! [color=blue]The my greatest pleasure is when see someone who is happily because of mine.[/color]

thanks  Mr. Virgil Nicula.

ure solution was really nice.

and it helped me alot. :)

what a bout similarity dear britney?

dou know what it is?

or even u dont know it.
 :rotfl:

i know it.

can u post u solution?

i wonderd ,because i see so many solutions for this problem.

thanks for pay attention to me. :)

hi dears,

please don't bother yourself for ashegh.;)

Ashegh I think you can't even do simple angle chase.

the only thing you know is inversion(I know U). :rotfl:

ok,,,i will answer u later amir2.first i should help britney spears ... ;)

look at this link. http://www.mathlinks.ro/Forum/viewtopic.php?t=68589

there are two usefull lemma in one of the posts.first read them and then read the solution.

triangles $KO2O1$ and $KAB$ are simmilar then:$\angle AKO2=\angle BKO1$

we have this fact for triangles $KO2O3$ and $KAC$ too:$\angle AKO2=\angle CKO3$.

then we conclude that:$\angle O1KO3=\angle BKC$

suppose $BM$ intersets $(C3)$ at $C'$.then az the lemma said $\angle O1KO3=\angle BKC'$

and from $2$ recent lines we conclud that $C=C'$

and $B,M,C$ are colinear.

i wish u understand my soution.

what is ure idea dear $Levi$.i dont think the solution is hard. ;)

[quote="amir2"][quote="stupid"]ok,,,i will answer u later amir2[/quote]

I think like U,too.

I think I have to speak to U different.

Tomoorow at school, at 8 o'clock,in the yard,we will do a Mortal Combat. :yinyang:  to finish life of silly and foolish person.

 :starwars:[/quote]

we can have a marathon with geometry problems...

each one who couldnt solve, must leave olympiad.

the first round will begin tomorrow.

to virgil nicula:

dont u think the problem u said,is hard for her.

it is one of the facts of homotecy.

and i think she cant do it?

then could u plz post u solution ...

i really want to know how can we solve it easy.

thanks :D

[quote="Davron"][quote="amir2"]$\text{Your fate has been written you will die.}$[/quote]

Hello brothers !!! We the same roots i am from Tajikistan and i had partizipated to the IMO 2005 and Inshallah will participate in the IMO 2006. My name is Davroni Latif and yours Ashegh and Amir ?

If really i enjoyed your chat in this link about the Mortal Combat !!![/quote]

hi davroni...

my name is $Ali Ashegh$

and amir2 is $Amir saeedi$

we are class mate ...

in the mortal combat, amir2 ,died...

sorry :(

[quote="Ashegh"]in the mortal combat, amir2 ,died... [/quote]

Hey boys Ashegh didn't come to school that day,he is a lier,lier lier .... . :mad:

[quote="Virgil Nicula"][b]Marko Avila[/b] is all right ! [b]Ashegh[/b], you run over hedge and ditch with a inversion ... !
This problem is very easily (at the most for the Intermediate Topics). [hide="Britney Spears, here is the simplest solution for the your proposed problem."]$\widehat {BMK}\equiv \widehat {ANK}\ ,\ \widehat {CMK}\equiv \widehat {AOK}\ ,\ m(\widehat {ANK})+m(\widehat {AOK})=180^{\circ}\Longrightarrow$
$m(\widehat {BMK})+m(\widehat {CMK})=180^{\circ}\Longrightarrow \boxed {M\in BC}\ .$[/hide] [hide="I will offer to Britney Spears a similar problem and I will wait the his proof only here."]
[quote][color=blue]Let $C_1=C(O_1,R_1),\ C_2=C(O_2,R_2)$ be two disjunct circles and let $C=C(O,R)$ be a circle exterior tangent to the circles $C_1$, $C_2$ in the points $T_1$, $T_2$ respectively. A common exterior tangent $t$ for the circles $C_1$, $C_2$ touches them in the points $A_1$, $A_2$ respectively. Denote the intersection $S\in T_1T_2\cap O_1O_2$. Prove that $S\in A_1A_2$ and the quadrilateral $A_1A_2T_2T_1$ is cyclically.[/color][/quote][/hide][b]Ashegh[/b], please don't begin with the inversion that I will feel giddy !
[b]Britney Spears[/b] , after two days I will present the simplest proof for the added problem.[/quote]

hi mr Virgil Nicula

i couldnt solve the offerd problem :( .

my teacher said its homotecy.

dont be sad dear britney...

here is a very easy problem for u...

think a bout it.

http://www.mathlinks.ro/Forum/viewtopic.php?p=435294#p435294

[hide="The proof of the present-problem."][quote="the first point"][color=red]$1^{\circ}.\blacktriangleright$ Denote $S'\in O_1O_2\cap A_1A_2$. We observe that $\boxed {\frac{S'O_1}{S'O_2}=\frac{R_1}{R_2}}$.
Aplly the Menelaus' theorem to the transversal $\overline {ST_1T_2}$ and the triangle $OO_1O_2$:
$\frac{SO_1}{SO_2}\cdot \frac{T_2O_2}{T_2O}\cdot \frac{T_1O}{T_1O_1}=1$ $\Longrightarrow$ $\frac{SO_1}{SO_2}\cdot \frac{R_2}{R}\cdot \frac{R}{R_1}=1$ $\Longrightarrow$ $\boxed {\frac{SO_1}{SO_2}=\frac{R_1}{R_2}}$ $\Longrightarrow$
$\frac{S'O_1}{S'O_2}=\frac{SO_1}{SO_2}$. Therefore, $S'\equiv S$ $\Longrightarrow$ $S\in A_1A_2\ .$[/color][/quote]
[quote="the second point"][color=blue]$2^{\circ}.\blacktriangleright$ Denote $x=m(\widehat {OT_1T_2})$, $y=m(\widehat {O_1A_1T_1})$, $z=m(\widehat {O_2A_2T_2})$.
$m(\widehat {T_1A_1A_2})=90^{\circ}-y\ ;\ m(\widehat {T_1T_2A_2})=180^{\circ}-x-z\Longrightarrow$
$\boxed {m(\widehat {T_1A_1A_2})+m(\widehat {T_1T_2A_2})=270^{\circ}-x-y-z}\ ;$
$m(\widehat {A_1T_1T_2})=180^{\circ}-x-y\ ;\ m(\widehat {A_1A_2T_2})=90^{\circ}-z\Longrightarrow$
$\boxed {m(\widehat {A_1T_1T_2})+m(\widehat {A_1A_2T_2})=270^{\circ}-x-y-z}\ .$
Therefore, $m(\widehat {T_1A_1A_2})+m(\widehat {T_1T_2A_2})=m(\widehat {A_1T_1T_2})+m(\widehat {A_1A_2T_2})$,
i.e. the quadrilateral $A_1T_1T_2A_2$ is inscribed.[/color][/quote]
\[ \boxed {\ \mathrm {I\ wish\ you\ every\ success\ !!\ }} \][/hide]

oh thanks mr virgil nicula...

conversation was good... :D


geometry
parallelogram
geometry unsolved

Equilateral triangles $ BCB_1, CDC_1, DAD_1$ are constructed outside the convex quadrilateral $ ABCD$. If $ P$ and $ Q$ are, respectively, the midpoints of $ B_1C_1$ and $ C_1D_1$ and $ R$ the midpoint of $ AB$, prove that the triangle $ PQR$ is equilateral.

1.D1QPH is a parallelgram
2.HID1 ≌ RPJ
done!

[quote="jbmorgan"]What are your $ H,I,J$?

And why is that a parallelogram?[/quote]
midpoint of d1b1,d1c,bc1
so the parallelogram comes


calculus
integration
function
geometry
real analysis
real analysis unsolved

This is very very strange,I've tried a lot of method but no one worked on this problem...But I just feel that this shouldn't be too diffcult,maybe I just overlooked something...
$f$ is a convex function defined on $[a,b]$(or concave?I'm not sure,I mean that for any two positive reals $\lambda_{1}$ and $\lambda_{2}$ such that $\lambda_{1}+\lambda_{2}=1$ and $a \leq x_{1}\leq x_{2}\leq b$,$f(\lambda_{1}x_{1}+\lambda_{2}x_{2}) \geq \lambda_{1}f(x_{1})+\lambda_{2}f(x_{2}))$,
and $f'(x)$ exists for any $x \in [a,b]$
Then,given $f(a)=f(b)=0,f'(a)=\alpha>0,f'(b)=\beta<0$,what we need to prove is that $\int_{a}^{b}f(x)dx \leq \frac{1}{2}\alpha \beta \frac{(b-a)^{2}}{\beta-\alpha}$

Think geometrically in this problem.  Construct tangents at points $(a,0)$ and
$(b,0)$ - due that you know the values of deruvatives at $a,b$ they will be the lines $L_{1}: y = \alpha (x-a)$ and $L_{2}: y = \beta (x-b)$. The integral you interested in is a area under the graph of $f$ which is non greater then the area of the triangle, obtained by the intersection of the lines $L_{1},L_{2}$ and $y=0$, this area is exactly $\frac{1}{2}\alpha \beta \frac{(b-a)^{2}}{\beta-\alpha}$

It is not hard,I think. :)
Just prove:
$f(x) \le \alpha (x-a),x \in [a,a+\frac{-\beta }{\alpha-\beta}(b-a)]$
$f(x) \le-\beta (b-x),x \in [a+\frac{-\beta }{\alpha-\beta}(b-a),b]$
I hope it is right

Thanks for the help.
I've thought it over last night,and I found out that indeed I'd overlooked something,it is really not hard as you've said,Zhaobin.


function
LaTeX
calculus
calculus computations

A function $ f : R \to R$ satisfies the equation $ f(x \plus{} y) \equal{} f(x) \cdot f(y)$ $ \forall x, y$ $ \epsilon$ $ R$ and $ f(x) \neq 0$ for any $ x$ in $ R$. Let the function be differentiable at $ x \equal{} 0$ and $ f'(2) \equal{} 2$, then $ f(x) \equal{} e^{nx}$.
Therfore, $ n \equal{} ?$

Some comments about LaTeX: there is no reason to break a single statement like "$ \forall x, y, \in R$" into three separate pieces.  Also, the correct command for "is an element of" is "\in," not "\epsilon."

As written, that's an equation that can't be solved exactly with elementary functions. It's probably a typo- I'd guess that $ f'(0)\equal{}2$ is intended.

hello
[hide="may be some help"]
I think u r preparing for JEE only .U would be given options.U can approximately find it.
u answer  will lie in b/w $ \frac{1}{2}$ and 1.
I don't think such could be asked in subjective papers .

[/hide]
Thank u.

[quote="kabi"]hello
[hide="may be some help"]
I think u r preparing for JEE only .U would be given options.U can approximately find it.
u answer  will lie in b/w $ \frac {1}{2}$ and 1.
I don't think such could be asked in subjective papers .

[/hide]
Thank u.[/quote]

Yes, agreed. But the JEE does not have a fixed pattern and subjective questions may be asked with students being asked to mark [b]numerical values[/b] of their answers through bubble [hide="matrices"](from $ 0$ to $ 9$ with a "$ \cdot$" decimal too available for marking or something of that sort)[/hide] the way [b]FIITJEE AITS[/b] committee did for [b]Part Test-$ II$[/b] dated $ 21^{st}$ $ December,$ $ 2008$.

Hello
Well my coaches are IITian .I m in contact with most of IIT Prof.
According to them the pattern would be only objective at least for the next two years.
Thank u.

$x \to 0$ $y\to 0$ yields $f(0) = f(0)^2$ thus $f(0) = 1$ as given $f(x) \neq 0$. Clearly $f'(x) = \lim_{h \to 0} \dfrac{ f(x+h) - f(x)}{h} = f(x) \lim_{h\to 0} \dfrac{f(h)-1}{h}$.Given $f'(0)$ exists then we must have that the limit  $\lim_{h\to 0} \dfrac{f(h)-1}{h}$ exists.Now $f'(0) = 2 $ thus $2 = \lim_{h\to 0} \dfrac{f(h)-1}{h}$.$f'(x) = 2f(x)$.

$$\boxed{f(x) = e^{2x}}$$.


hmmm...from personal experience, i always perform better at math if someone in my class (or school) is better than me. if there's nobody who's better, then i never get better. do many of u guys have that same feeling?

I don't know. What difference does it make whether the someone who's better than you is in your class, or living half a country away? As long as you know he exists. (Unless of course you steal his notes...) :) .

[quote="EFuzzy"]I don't know. What difference does it make whether the someone who's better than you is in your class, or living half a country away? As long as you know he exists. [/quote]

Well, as long as you feel some competitiveness around you, then, probably you'll try to become better than that person. But if you are the best in your class, probably you're not going to try that hard. I think that it is easier to feel challenged if the other party is there all the time, you know him/her and everybody around you have an idea of what is happening, that if you are 3000 miles away.

Best,

In my mathematically challenged school, I haven't yet met someone[i] better[/i] than me at math, but a few people come close... I find it helps. On tests, I check my work two, three, even four times to guarantee myself 100. It's really just no fun when you're the only 100 and everyone else gets 80's or below. Motivation goes away--for example, I can sleep through and get a 92 on that test and still be the best in the class, or something like that.

Knowing that people better than me exist on the other side of the world....
Wouldn't that be too [i]much[/i] pressure? I mean, there's always going to be someone who's better than you are.

too much pressure? no such thing :D

All I can say is that I feel motivated knowing that I am not the best.

well, 80-90% of people in this forum are probably better at math than i am. however, unless i know somebody personally, it wouldn't affect me if somebody got a higher AIME score than me.

however, bragging rights are good :D

I know I personally strive to be the best of the best in all things that I'm really good at (IE math), so competition for me is good. However, I tend to get discouraged from constant defeat.

Luckily there's this really good ninth grader at my school.

at my school it's impossible to be the best. there's always some little twerp coming up that's faster and brighter than you, and even THAT kid isn't as smart as the guy that graduated a couple of years ago and got accepted into every top tier he applied to. between the legacy and the underclassmen, you've REALLY got the pressure on.

I got to a bad(compared to most people here) school.  There is only one person better than me in the whole school (and I'm not that good!) whos is a senior, while I am a freshman.  Seeing all these people on this forum keeps me motivated, otherwise I would get lazy.

i've accepted that i'm not going to be the best at my school.  i'm pretty good at math contests (not as good as most of the ppl on this forum tho) but there are 3 people better than me on my team (my team is one of the best in the state). i could probably be the best mathlete at many other schools but being in a winning team is better, imo.
having people better than me motivates me a little.  joml88 is on my team and forces me into studying more than i would usually do cuz he wants the team to be the best that it can be. i just wish there was a pill you could take that makes you better at math contests.

Doesn't anyone do math for the sake of math? I'm more competitive, but I still enjoy math.

yes, but you can get lazy sometimes............ok fine, I get lazy :blush:

I totally agree with plokoon51, when I was in 5th grade, the people around me were all kinda... not so bright... then I went to a different middle school, and found lots of people smarter than me... and then i started doing math team.... and felt the pressure of your friends and competitors... and I think it made me work harder.... but of course, back then, if you ask me what mathcounts is, i would probably guess a stupid class where you learn how to count. I didn't know what AOPs was either... and defintely not all the smart people in the world

if they're too high above you (darkshadow?) or too low (no names stated) you tend to give up, whereas if they are slightly higher or lower than you , you try much harder- true for just about anything

[quote="miraculouspostmaster"][quote]joml88 is on my team and forces me into studying more than i would usually do cuz he wants the team to be the best that it can be.[/quote]
Definitely have to agree, joml (nicknamed "the quintessential man") is undoubtedly the single greatest instigator of the success of the math team at Summerville High School.  Raisin' the standard constantly (thank god he didn't go to governor's school).[/quote]
joe didn't go to the governer's school because he wouldn't be good enough to make their math team.  :D


Express in simplest form: $ \sqrt 3 \times \sqrt 5 \times \sqrt{15}$.

$ \sqrt{3}\cdot \sqrt{5}\cdot \sqrt{15}\equal{}\sqrt{15}\cdot \sqrt{15}\equal{}15$

Note: An important property of square roots states that $ \sqrt {a} \times \sqrt {b} \equal{} \sqrt {ab}$. We note that $ 3 \times 5$ is 15, so we can use another property: $ \sqrt {a} \times \sqrt {a} \equal{} \sqrt {a^2} \equal{} a$. So we have:

$ \sqrt {15} \times \sqrt {15} \equal{} \boxed{15}$.

EDIT: I am not posting solutions - I am giving more detail in case the asker was not aware of certain things, like the properties of square roots.

Why are you posting solutions when they are already posted?


$n=1$ and $n=2$ are clearly solutions.
In another hand, for $n \geq 3$, we have $n! \geq 2n$ so that, from Bertand's postulate, there is at least one prime number, say $p$, greater than $n$ but least than $2n$. This prime $p$ divides the rhs but not the LHS, thus the equality cannot hold...

Pierre.


Let $ a,b,c,d\in\mathbb{R}$. Prove that
$ \max \left\{ \left| ac\right| ,\left| ad\plus{}bc\right| ,\left| bd\right| \right\} \geq \left( \frac{\sqrt{5}\minus{}1}{2}\right) .\max \left\{ \left| a\right| ,\left| b\right| \right\} .\max \left\{ \left| c\right| ,\left| d\right| \right\}$

Hint: we may assume that both maxima on the right side are equal to one.

I don't understand :( So, the ineq will be $ \max \left\{ \left| ac\right| ,\left| ad\plus{}bc\right| ,\left| bd\right| \right\} \geq \frac{\sqrt{5}\minus{}1}{2}$
How can?


The Canadian IMO Team this year is:

[b]Yan (Cynthia) Li[/b], Dr. Norman Bethune Collegiate Institute, Scarborough (Ontario)
[b]Alexander Remorov[/b], William Lyon Mackenzie Collegiate Institute, North York (Ontario)
[b]Jonathan Schneider[/b], University of Toronto Schools, Toronto (Ontario)
[b]XiaoLin (Danny) Shi[/b], Sir Winston Churchill High School, Calgary (Alberta)
[b]Chen Sun[/b], A.B. Lucas Secondary School, London (Ontario)
[b]Chengyue (Jarno) Sun[/b], Western Canada High School, Calgary (Alberta)

The media release can be found here:
http://www.newswire.ca/en/releases/archive/May2008/30/c6261.html

Congratulations to the members of this year's IMO team.

Good Luck to all of you.

Looks like a very strong team - congratulations, and best of luck!


Dear everyone.

please read the following

for a triangle ABC, let I be its incenter.
its incircle (I) touches BC, CA, AB at D, E, F respectively.
DK is perpendicular to EF.
BK, CK meet CA, AB at G, H respectively.

prove that quadrilateral HBCG is bicentric.

I think that it is easy for cyclic.
Is it easy for circumscriptible, too ?

I long for your nice answers.
thanks for reading my shoddy post. :blush:

Let the drawn altitude meet $ (I)$ at $ S$.draw the tangent of $ (I)$ at point $ S$ and let it intersect $ AB$ at $ H'$.u can easily prove that $ \angle SH'B \equal{} 180^\circ \minus{} \angle C$.similarly for the point $ G'$ which is the intersection of the tangent of $ (I)$ at $ S$ with $ AC$.so we have proven that $ H'G'CB$ is concyclic.it remains to prove the collinearity of $ H',K,C$ and the collinearity of $ G',K,B$.
but u can easily prove that (by brianchon's theorem) $ DS,FE,CH',BG'$ are concurrent.so we are done.
we have proven that it is both concyclic and  circumscriptible! :)

[hide]$ \bullet$ Let be the oint $ Z\equiv (I)\cap DK$ and we denote as $ G',\ H',$ the points of intersection of $ AC,\ AB$ respectively, from the line tangent to $ (I),$ at point $ Z.$

In circumscribed quadrilateral $ BCG'H',$ based on the [b][size=100]Newton theorem[/size][/b], we have that the line segments $ BG',\ CH',$ pass through the point $ K.$

So, we conclude that $ BG'\equiv BG$ $ \Longrightarrow$ $ G'\equiv G$ and similarly, $ H'\equiv H.$

$ \bullet$ Because of now $ DZ\perp EF,$ the quadrilateral $ BCGH$ is also cyclic as well ( and easy to prove ).

Hence, $ BCGH$ is a bicentric quadrilateral and we are done.[/hide]
Kostas Vittas.

thank you for nice solutions shoki, vittasko. :)


analytic geometry
graphing lines
slope
geometry
logarithms
angle bisector
AMC

you know the rules
PM me your answers in 75 mins

[hide]MOCK_AMC_C

1) What is the slope of the line perpendicular to :
$y=2x\sqrt{3}+8\sqrt{56}-3\sqrt{3}x$

A)$\sqrt{3}$
B)$\frac{\sqrt{3}}{6}$
C)$\frac{\sqrt{3}}{3}$
D)$\frac{-1}{\sqrt{3}}$
E)$3$

2) In isosceles right triangle $ABC$ with $\angle B=90^{\circ}, AB=\sqrt{6}$. What is the area of $ABC$?

A)$2\sqrt{2}$
B)$2\sqrt{12}$
C)$2\sqrt{3}$
D)$\sqrt{6}$
E)$3$

3) A sequence of positive numbers is defined by $(a_{n+1}+n)a_n=1$, how many rational elements exist in this sequence?

A)$\infty$
B)$0$
C)$1$
D)$2$
E)$3$

4) If $f(2x)=4x^2-8x+3$, find $f(x)$:

A) $x^2-4x+3$
B) $x^2+4x+3$
C) $3x^2-x+3$
D) $3x^2-3x+4$
E) $x^2-3x+4$

5) What is the value of $x$ so that the line passing through $(x,5)$ and $(4,11)$ has a slope of $6$?

A)0
B)1
C)2
D)3
E) none of the above

6) Which of the following could not be a measure of an interior angle of a regular polygon?

A) $171^{\circ}$
B) $173^{\circ}$
C) $175^{\circ}$
D) $177^{\circ}$
E) $179^{\circ}$

7) In a multiple choice test with 20 questions, every correct answer is worth +7 and every incorrect answer is worth -2 points(omitted problems are 0 pts). If Jeffery scores 87 pts, how many problems did he not answer?

A)2
B)5
C)7
D)9
E)13

8) Assume $1+2+\cdots +n$ is a three digit number with all digits equal.  Then $n$ is divisibly by which of the following numbers?

A)15
B)6
C)7
D)8
E)11

9) Evaluate: $2\sqrt{3+2\sqrt{3+2\sqrt{3+\cdots}}}$

A)2
B)4
C)6
D)3
E)5

10) The unit digit of $23^{23}-17^{17}$ is:

A)0
B)2
C)4
D)6
E)8

11) If we know $9x+5y$ is divisible by $11$, then what is the value of $k$ such that $10x+ky$ would be divisibly by $11$?

A)2
B)4
C)6
D)8
E)10

12) If a sequence is defined by $I_1=I_2=I_3=1$ and $I_n=I_{n-2}+I_{n-3}$ ($n>3$); find $I_{25}$

A)616
B)660
C)657
D)578
E)601

13) Find the sum $i^{0!}+i^{1!}+i^{2!}+\cdots +i^{2003!}+i^{2004!}$, where $i^2=-1$:

A)1999+i
B)2004+i
C)1999+2i
D)2004+2i
E)1999+3i

14) For numbers $n,10,3,9,6$ the mean of the collection is equal to its median, find the sum of all possible values of $n$:

A)22
B)23
C)24
D)25
E)26

15) Find $\sum_{n=2}^5 {(\frac{n^2}{n!}-\frac{(n-1)^2}{(n-1)!})}$

A)$\frac{5}{24}$
B)$\frac{19}{24}$
C)$\frac{-19}{24}$
D)$\frac{-5}{24}$
E)$1$

16) Evaluate: $\log_{3}{32}\cdot \log_{2}{27}+ 7^{\log_{49}{3}}-\sqrt{3}$

A)$15$
B)$20\sqrt{3}$
C)$20+\sqrt{3}$
D)$15-\sqrt{3}$
E)$-\sqrt{3}$

17) If $35!=\overline{10333147966386144929*66651337523200000000}$, what digit does the $*$ represent?

A)6
B)7
C)8
D)9
E)1

18) For traveling from city A to city B there are 17 Ground options and 17 air options. 3 passengers travel in the morning and 4 passengers travel in the afternoon. In how many ways can the 3 passengers traveling morning have all traveled with ground OR all 4 passengers traveling in the afternoon have traveled with air?

A)$\binom{17}{4}$
B)$\binom{17}{3}$
C)$\binom{37}{4}$
D)$\binom{18}{4}$
E)$\binom{18}{7}$

19) In isosceles triangle $ABC$ (AB=AC), the angle bisector of $B$ intersects with $AC$ at $D$; we have $BC=BD+AD$ , what is $\angle A$?

A)$100^{\circ}$
B)$108^{\circ}$
C)$110^{\circ}$
D)$115^{\circ}$
E)$120^{\circ}$

20) What is the number of devisors of $1374*1375*1376$?

A)448
B)648
C)24
D)240
E)144

21) If $\frac{n^3-1}{5}$ is prime, in which interval does $n$ lie?

A)$[1,5]$
B)$[5,10]$
C)$[10,15]$
D)$[15,20]$
E) There is no such number!

22) $a,b,c,d$ are positive numbers, with product of one. What is the minimum of $a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd$?

A)8
B)9
C)10
D)12
E)13

23) From point $O$ inside $\triangle ABC$ we draw perpendicular  lines $OM,ON, OP$ on to $AB,AC,BC$ respectively. If $\angle AOB=60^{\circ}, MN=\frac{\sqrt{5}}{3}, OP=3, NP=\sqrt{3}, OM=\sqrt{5}$ what is $\angle BOC$?

A)$135^{\circ}$
B)$150^{\circ}$
C)$138^{\circ}$
D)$120^{\circ}$
E)$170^{\circ}$

24) In $\prod_{1\le i\le j\le 9}{x_i-x_j}$ what is the coeffiecient of $(x_1^4x_2^4\cdots x_9^4)$?

A)$0$
B)$\binom{9}{4}$
C)$-\binom{9}{4}$
D)$9\cdot \binom{2}{9}\binom{3}{9}$
E)$-9\cdot \binom{2}{9}\binom{3}{9}$

25) The number of devisors of $n$ is $d(n)$. We say $n$ is “nice” if $d(d(d(n)))=d(d(n))$ which of the following is false?

A) There are infinite “nice” multiples of $3$
B) There are infinite “nice” even numbers
C) There are infinite “nice” odd numbers
D) if $n$ is nice then $n^3$ is “nice”
E) if $n$ is nice then $d(n)$ is “nice”
[/hide]

bump :)

i will be online the entire time

for example: the  devisors of 6 are:

1,2,3,6

time is now up
please send me your answeres


geometry
3D geometry
algebra
function
domain
logarithms
integration

a map $ f$ given by

$ s_i \equal{} p_i \minus{} \frac {1}{2}\sum_{j\in{}I_{ \minus{} i}}p_ip_j \plus{} \frac {1}{3}\!\sum_{(j,k)\in{}I_{ \minus{} i}}p_ip_jp_k \\
\minus{} \frac {1}{4}\!\sum_{(j,k,l)\in{}I_{ \minus{} i}}p_ip_jp_kp_l \plus{} \cdots \plus{} ( \minus{} 1)^{n \minus{} 1}\frac {1}{n}\prod^n_{i \equal{} 1}p_i\ \forall{} \ i \equal{} 1,\ldots,n$

where $ (x_1,x_2,\ldots,x_k)\in{}I_{ \minus{} i}$ denoting $ x_1 < x_2 < \cdots < x_k$, chosen from the index set $ \{1,2,\ldots,n\}\setminus\{i\}$.

and the domain of $ f$ is a cube given by $ 0\leq{}p_i\leq{}1\ \forall{}i\in\{1,\ldots,n\}$

I have found $ \sum^n_{i \equal{} 1}s_i \equal{} 1 \minus{} \prod^n_{i \equal{} 1}(1 \minus{} p_i)$

so $ 0\leq\sum^n_{i \equal{} 1}s_i\leq{}1$ and $ s_i\geq{}0$

Is  f  a one-to-one and onto map to the simplex $ 0\leq\sum^n_{i \equal{} 1}s_i\leq{}1$ and $ s_i\geq{}0$ ??

for $ n \equal{} 2$, the claim is true by finding a inverse function, but it seem impossible to find the inverse for general $ n$

the inverse function theorem can't be used, since it is used to test whether a function is local invertible not global invertible.

can anyone help me

thanks

Yes, but without much success so far :(. Be patient: most people here have plenty of other things to do and it is not unusual to wait for a solution a couple of weeks or even more ;).

OK, I believe I've got it. Consider the function $ F(p_1,\dots,p_n)\equal{}\sum_i s_i\log p_i\plus{}\int_0^1\Bigl[\prod_j(1\minus{}p_jx)\minus{}1\Bigr]\frac{dx}{x}$. The solution of your system is just the critical point of this function. Let $ p_i\equal{}e^{t_i}$. The key observation is that $ F$ is a concave function of variables $ t_i$ as long as all $ t_i<0$. Indeed, its second differential (it $ t$ variables) can be written as
\[ \int_0^1 \Bigl[\sum_i h_i\psi_i  \Bigr]^2x\Phi \,dx\minus{}\int_0^1\sum_i h_i^2 (\psi_i\plus{}x\psi_i^2)\Phi\,dx\]
where $ \Phi(x)\equal{}\prod_j(1\minus{}p_j x)$ and $ \psi_j(x)\equal{}\frac{p_j}{1\minus{}p_jx}$. Now just notice that
\[ \Bigl[\sum_i h_i\psi_i  \Bigr]^2\le \Bigl[\sum_i h_i^2\psi_i  \Bigr]\cdot \Bigl[\sum_i \psi_i  \Bigr]\]
(Cauchy-Schwarz). Hence, all we need is to show that
\[ \int_0^1 \psi_j \Bigl[\sum_i \psi_i  \Bigr]x\Phi \,dx\le \int_0^1  (\psi_j\plus{}x\psi_j^2)\Phi\,dx\]
or, equivalently,
\[ \int_0^1 \psi_j \Bigl[\sum_{i\ne j} \psi_i  \Bigr]x\Phi \,dx\le \int_0^1  \psi_j \Phi\,dx\]
This can be rewritten as $ \int_0^1 (x\Phi_j'\plus{}\Phi_j)\equal{}\Phi_j(1)\ge 0$ where $ \Phi_j(x)\equal{}\prod_{i\ne j}(1\minus{}p_ix)$, which is obvious.

Now, once we know that our function is strictly concave in $ t_i$ for $ t_i<0$, we can immediately conclude that it has at most one critical point in the unit cube and that point is the point of maximum. This takes care of the one-to-one part. As for the onto part, we just need to show that for $ \sum_i s_i<1$, the maximum is attained inside the cube, not on the boundary. Note that it cannot be attained with any of $ p_i\equal{}0$ (then $ F\equal{}\minus{}\infty$). Hence, $ \frac{\partial F}{\partial p_i}\ge 0$ at the point of maximum whether it is on the boundary or not. Thus, instead of your identities, you get the inequalities $ s_i\ge \dots$. But, as you cleverly noticed, if you add them up, you get $ \sum_i s_i\ge 1\minus{}\prod_i(1\minus{}p_i)$. Since the left hand side is $ <1$, we cannot have any $ p_i\equal{}1$ at the point of maximum and, thereby, we get the desired identities.

Now can you tell me where this question came from? (it doesn't look like a contest problem or an exercise from some textbook...)

i appreciate your help very much, fedja.
there are some places i don't understand.

[quote="fedja"]Let $ p_i \equal{} e^{t_i}$. The key observation is that $ F$ is a concave function of variables $ t_i$ as long as all $ t_i < 0$. Indeed, its second differential (it $ t$ variables) can be written as
\[ \int_0^1 \Bigl[\sum_i h_i\psi_i \Bigr]^2x\Phi \,dx \minus{} \int_0^1\sum_i h_i^2 (\psi_i \plus{} x\psi_i^2)\Phi\,dx
\]
where $ \Phi(x) \equal{} \prod_j(1 \minus{} p_j x)$ and $ \psi_j(x) \equal{} \frac {p_j}{1 \minus{} p_jx}$.

[/quote]

what is $ h_i$ ?
(i can understand your proof of concave)

by the way,
i get second differential $ \frac {\partial^2 F}{\partial t^2_i} \equal{} \minus{} p_j\int^1_0 \Phi_j(x)dx\leq{}0$ (the minus of right-hand side of origin problem, can shown to be $ \leq{}0$)
for $ t_i < 0$ ($ 0 < p_i < 1$)

[quote="fedja"]

As for the onto part, we just need to show that for $ \sum_i s_i < 1$, the maximum is attained inside the cube, not on the boundary. Note that it cannot be attained with any of $ p_i \equal{} 0$ (then $ F \equal{} \minus{} \infty$). Hence, $ \frac {\partial F}{\partial p_i}\ge 0$ at the point of maximum whether it is on the boundary or not. Thus, instead of your identities, you get the inequalities $ s_i\ge \dots$. But, as you cleverly noticed, if you add them up, you get $ \sum_i s_i\ge 1 \minus{} \prod_i(1 \minus{} p_i)$. Since the left hand side is $ < 1$, we cannot have any $ p_i \equal{} 1$ at the point of maximum and, thereby, we get the desired identities.

[/quote]

1) the onto part, i don't understand clearly.  how to get $ \frac {\partial F}{\partial p_i}\ge 0$ at the point of maximum whether it is on the boundary or not ?
    may you explain more

2) Is the one-to-one and onto for the boundary $ \sum_is_i \equal{} 1$ follows from $ s_i$ is continuous in $ p_i$ ?

finally,
the problem is one of the result derived in my work  :)  , i'd like to mention your help in acknowledgment section
And i will send you a private message about this.

1) The second differential is a quadratic form; $ h_i$ are just the variables of that form, so $ D^2F(h)\equal{}\sum_{i,j}\frac{\partial^2F}{\partial t_i\partial t_j}h_ih_j$. I needed to show that it is $ \le 0$ for all $ h_i$.

2) Negative [b]pure[/b] partial derivatives are necessary but not sufficient for concavity. You need to show that cross terms are not able to turn the negative expression given by the diagonal terms into a positive one.

3) Since you freely can go to the left from $ p_j$ at the point of maximum and stay within the domain (here it is crucial that $ p_j\ne 0$), the partial derivative of $ F$ with respect to $ p_j$ should be non-negative (otherwise slightly to the left you'll have a bigger value).

4) Yes, you are right that I formally proved only that the open cube is homeomorphically mapped to the open simplex. The onto claim is easy to extend, of course: the continuous image of a compact set has to be compact. The one-to-oneness on the boundary is a bit trickier and, generally speaking, you can have a continuous one-to-one map inside that glues together some points on the boundary. What is clear is that $ s_i\equal{}0$ is equivalent to $ p_i\equal{}0$, so, removing all zeroes, we just reduce the problem to itself with fewer variables and now we only need to show that we are fine with all $ s_i>0$. The most straightforward approach would be to try to prove something like strict concavity of $ F$ on the boundary (note that in order to lose it, all inequalities have to become identities, which is possible only in very degenerate situations). Frankly speaking, I didn't take a close look at that boundary issue. :blush: See if you can figure it out by yourself now; if not, let us know and either I or somebody else will try to take proper care of it :).

$ \Bigl[\sum_i h_i\psi_i \Bigr]^2 < \Bigl[\sum_i h_i^2\psi_i \Bigr]\cdot \Bigl[\sum_i \psi_i \Bigr]$
if $ h_i$ are not all the same.

hence, to show the strictly concave, only need to consider $ h_i$ are all the same, i.e.
$ \int_0^1 \Bigl[\sum_i \psi_i \Bigr]^2x\Phi \,dx \minus{} \int_0^1\sum_i (\psi_i \plus{} x\psi_i^2)\Phi\,dx < 0$
this can be rewritten as
$ \sum_j p_j\Phi_j(1) > 0$  (correspond to the original proof $ p_j\int_0^1 (x\Phi_j' \plus{} \Phi_j) \equal{} p_j\Phi_j(1)\ge 0$)
this is satisfied except for $ p_1 \equal{} \cdots{} \equal{} p_n \equal{} 1$

and for the point $ p_1 \equal{} \cdots{} \equal{} p_n \equal{} 1$ which is the unique point map to $ s_1 \equal{} \cdots{} \equal{} s_n \equal{} \frac {1}{n}$

Is this ok?

Almost. We may have $ \Phi_j(1)\equal{}0$ for all $ j$ as soon as just two $ p_i\equal{}1$. But, once we have all $ h_i$ the same (this part is perfectly correct), the day is saved because it means that if there are at least two critical points, one of them has to be inside the cube and there we have strict concavity. So now the proof is complete :).

Oh, yes, $ \Phi_j(1) \equal{} 0$ for all $ j$ as soon as just two $ p_i \equal{} 1$

why if all $ h_i$ are the same
[quote="fedja"]
it means that if there are at least two critical points, one of them has to be inside the cube and there we have strict concavity. So now the proof is complete
[/quote]
how does this relate to that the boundary is one-to-one map?

If we have two critical points, to get a contradiction, we need just strict concavity on the interval joining them. If their difference (i.e., the direction of that interval) has all equal coordinates (the only direction in which we may lack strict concavity), both points cannot be on the boundary simultaneously.

The preimages of the same $ s_j$ are still the critical points of $ F$ even when $ \sum s_j\equal{}1$. If one of them is inside the cube, we immediately get $ \sum_j s_j<1$, so we are not on the boundary at all then.

ok, i understand better. thanks. :lol:

but, i am confused again about the onto part that
$ \frac {\partial F}{\partial p_i}\ge 0$ at the point of maximum whether it is on the boundary or not

for example, $ s_i < 0$, then the point of maximum (it exists since unit cube is compact )
is on the boundary,
it follows that $ 0 > \sum_i s_i\ge 1 \minus{} \prod_i(1 \minus{} p_i) \equal{} 1$
what's wrong?

If $ s_i<0$, we can go to $ p_i\equal{}0$ to get $ F\equal{}\plus{}\infty$, so the maximum is not attained! The whole point was that we stay away from the $ 0$ endpoints if all $ s_i>0$.


I was just wondering what is the approximate ranking of national math olympiads when compared to each other ie which ones are easier, which ones are harder. :?

CMO > USAMO > RMO.

CMO is heavy-work. USAMO seems to need creativity.

Thanks, guys.
Do you by any chance know the ranking of the other contests the problems of which are available in Aops:
Australia, Brazil, Canada, France, Greece, India, Italy, Lithuania, Netherlands, Poland, South Africa, Turkey, Belgium, Bulgaria, Germany, Hong Kong, Iran, Japan, Moldova, Peru, Romaina, Korea, Taiwan, UK, Vietnam? :?
Thanks for your help in advance. :)


To cover the distance between halfway AB and point B the truck will need 0.50 hr (assuming its speed is constant and equal to 40 miles/hour). In that amout of time, the car traveled the distance $d = 50 miles/hr \times 0.50 hr = 25 miles$.

[quote="sharkman"]A truck travels 40 miles from point A to point B in exactly 1 hour. When the truck is halfway between point A and point B, a car starts from point A and travels at 50 miles per hour. How many miles has the car traveled when the truck reaches point B?[/quote]

[hide="Solution"]When the truck is halfway between there, it has been $\frac{1}{2}$ hour. The car starts to move now.
In $\frac{1}{2}$ more hours, the truck will reach point $B$, in which the car has moved $50 \textrm{ miles per hour}\cdot \frac{1}{2}\textrm{ hours}= \boxed{25 \textrm{ miles}}$[/hide]

[hide="Solution"]When the truck is half way between points A and B, the car starts driving at 50 mph.  The truck takes $ \frac{1}{2}$ an hour to get to B.  For the car, 50 mph times $ \frac{1}{2}$ is 25.  The car has gone 25 miles.
[b]Answer:[/b] $ \boxed{25 \textrm{miles}}$[/hide]

[hide]The rate for the truck is 40 mph. The rate for the car is 50 mph.

Rate is solved, then the time: It takes the truck 1 hour to get to point B.

The formula for the truck is 40(1)=D, which is 40 miles.

The truck is halfway there and then the car comes. The truck was only 0.5 way there. The rest of the 30 minutes to get to point B, the car will also be hitting the road. 30 min/60 min=1/2, so then the car's speed times the time will get you the distance: 50 times 1/2 is just 50/2, which is 25 miles.

The answer is 25 miles.:wow::wallbash_red::wallbash:[/hide]

[quote="sharkman"]A truck travels 40 miles from point A to point B in exactly 1 hour. When the truck is halfway between point A and point B, a car starts from point A and travels at 50 miles per hour. How many miles has the car traveled when the truck reaches point B?[/quote]

[hide]Assuming the rate is constant, The truck drive' need 1/2h' to B, so ca' 50/2=25 miles[/hide]

easy-the truck has traveled half an hour and has half an hour left the other dude/truck/person has traveled 50*.5 =25

halfway there = 30 minutes left

[u] 50 [/u]miles =[u] 1 [/u]hr.
  2            2

25 = 1/2 hr.

answer : 25 miles

[hide]
25 miles since in half an hour the truck travels 20 miles then the car starts.in half an hour the truck gets to piont b and the car travels 25 miles
[/hide]

I completely agree with your answer but you probably shoulnt post on month old threads

I think this has already been solved. Plus, it's such an old thread.

Lock this.

[quote="sharkman"]A truck travels 40 miles from point A to point B in exactly 1 hour. When the truck is halfway between point A and point B, a car starts from point A and travels at 50 miles per hour. How many miles has the car traveled when the truck reaches point B?[/quote]

Starting with $ D_{\text{truck}} \equal{} r_{\text{truck}}\cdot t$

$ \Rightarrow \frac{1}{2}D_{\text{truck}}\equal{}\frac{1}{2}r_{\text{truck}}\cdot t$

$ \therefore D_{\text{car}} \equal{} r_{\text{car}}\cdot\frac{1}{2} t$.

$ \Rightarrow D_{\text{car}} \equal{} r_{\text{car}}\cdot\left(\frac{\frac{1}{2}D_{\text{truck}}}{r_{\text{truck}}}\right)$

$ \boxed{D_{\text{car}} \equal{}50\text{ mph }\cdot\left(\frac{\frac{1}{2}\cdot 40\text{ miles}}{40\text{ mph}}\right) \equal{} 25\text{ miles }}$

we know $ D\equal{}rt$ so halfway between point $ A$ and point $ B$ the first truck has traveled 20miles is 30minutes. So the first truck has half an hour to go. If the second truck goes 50mph. So 50/2=25

the second truck has traveled 25miles

This problem is quite simple
all you need is that
distance = rate x time

and you can also make
rate = distance / time
or
time=  distance / rate

but i digress, you end up wit 25

Someone plz teach me Latex, i am getting jealous of all you people i can understand every other computer languages like java but not Latex, someone help me!!

i need to learn too

do you need to download anything to get it?

Cough Cough.

http://www.artofproblemsolving.com/Wiki/index.php/LaTeX


abstract algebra
floor function
induction
logarithms
limit
greatest common divisor
number theory unsolved

Find the largest real number $ \alpha$ for which threr exists an infinite sequence $ a_{n}$ of positive integers satisfying the following propertices

$ i.$ For each $ n\in N,a_{n}>1997n$

$ ii.$ For every $ n\ge2,a_{n}^{\alpha}$ does not exceed the greatest common divisor of the set {$ a_{i}+a_{j}|i+j=n$}
 :wink:

If $ n \in \mathbb{N}$ and $ A_{n}= \{a_{1}, a_{2}, \ldots, a_{n}\}$ is any finite set of integers, set $ \gcd(A_{n}) = \gcd(a_{1}, a_{2}, \ldots, a_{n})$. Here as usual, $ \mathbb{N}= \{1, 2, \ldots\}$. Moreover, if $ a_{1}, a_{2}, \ldots$ is a sequence of elements in any abelian additive group $ (G,+)$, then put $ \mathcal{D}_{n}(a_{1}, a_{2}, \ldots) = \{a_{i}+a_{n-i}\}_{i=1}^{n-1}$, for any $ n \in \mathbb{N}$ [btw, note that $ \mathcal{D}_{1}(a_{1}, a_{2}, \ldots) = \emptyset$]. As a notational shortcut, we'll write $ \mathcal{D}_{n}$ instead of $ \mathcal{D}_{n}(a_{1}, a_{2}, \ldots)$, whenever it's clear from the context which is the sequence $ a_{1}, a_{2}, \ldots$ we're dealing with. We're proving the following (which is, in fact, a rewording and a slight generalization of the original problem):

[quote]Given any positive real number $ \alpha$, let
\[ K_\alpha = \{k \in \mathbb{R}: \exists\, \{a_{n}\}_{n=1}^\infty \subseteq \mathbb{Z}\mbox{ s.t. }a_{n}> n\alpha \;\wedge\; a_{n+1}^{k}\le \gcd(\mathcal{D}_{n+1}), \mbox{ as }n \in \mathbb{N}\}\]
Then $ K_\alpha$ has a maximum and $ \max(K_\alpha) = 1$.[/quote]

[i]Proof:[/i] note that $ 1 \in K_\alpha$. In fact, taking $ a_{n}= n \cdot (\lfloor \alpha \rfloor+1)$, for any $ n \in \mathbb{N}$, it’s plain that $ a_{n}> n\alpha$, as $ n = 1, 2, \ldots$; and $ \mathcal{D}_{n}= \{a_{i}+a_{n-i}\}_{i=1}^{n-1}= \{a_{n}\}$, as $ n = 2, 3, \ldots,$ yielding $ a_{n}\le \gcd(\mathcal{D}_{n})$.

Hence $ K_\alpha \ne \emptyset$ and there exists $ \sup(K_\alpha) \ge 1$. By way of contradiction, admit there is any real $ k > 1$ such that $ k \in K_\alpha$. Then it would exist as well a sequence $ \{a_{n}\}_{n=1}^\infty \subseteq \mathbb{Z}$ fulfilling both the conditions i) and ii) with the problem in hand. Then, in particular, it would result $ a_{2n}^{k}\le 2a_{n}$, for any $ n \in \mathbb{N}$, i.e. $ a_{2n}\le (2a_{n})^{1/k}$, and consequently (by an easy induction)
\[ a_{2^{n}}\le \exp\!\left(\ln(2) \cdot \sum_{r=1}^{n}\frac{1}{k^{r}}\right) \cdot a_{1}^{1/k^{n}}\le \exp\!\left(\frac{\ln 2}{k-1}\right). \]
Anyway, this would be absurd, because of implying that $ \{a_{2^{n}}\}_{n=1}^\infty$ is bounded above. The claim follows.

[quote="s.tringali"]Anyway, this would be absurd, because of implying that $ \{a_{2^{n}}\}_{n=1}^\infty$ is bounded above. The claim follows.[/quote]
hallo,
where is the contradiction here?

[quote="aviateurpilot"][quote="s.tringali"]Anyway, this would be absurd, because of implying that $ \{a_{2^{n}}\}_{n=1}^\infty$ is bounded above. The claim follows.[/quote]
[...] where is the contradiction here?[/quote]
By construction, $ a_{n}> n\alpha$, for any $ n \in \mathbb{N}$. Hence $ \lim_{n \to \infty}a_{n}= \lim_{n \to \infty}a_{2^{n}}=+\infty$. :wink:


abstract algebra
induction
Ring Theory
superior algebra
superior algebra unsolved

Let $ R : \equal{} {\mathbb Q}[[X_1,...,X_n]]$ and let $ f: M\to N$ be a morphism of free (not necessarily finitely generated) $ R$-modules. Then $ f$ is an isomorphism if and only if the induced map $ M/{\mathfrak m}M\to N/{\mathfrak m} N$ is an isomorphism. Here, $\mathfrak m$ denotes the unique maximal ideal of $R$.

I have found a way to do this "by hand", but maybe some of you can give a structured proof.

edit. The following proves the claim above only in case where $f$ is an epimorphism. In general the claim is false (see postings below).

Tensoring the exact sequence
\[ 0 \to M\cong N\to 0\]
by $ R/\mathfrak{m},$ and using the right exactness of the tensor product we obtain one implication.
(Note that we don't need any assumptions other than $ f: M\to N$ is an isomorphism of $ R$-modules, and $ \mathfrak{m}$ is an ideal in a commutative unitary ring $ R$).

For converse observe that the exact sequence
\[ 0\to K \to M\stackrel{f}{\to} N\to 0\]
where $ K \equal{} \text{Ker}\,f,$ splits since $ N$ is free. Hence tensoring by $ R/\mathfrak{m}$ one gets the exact sequence
\[ 0\to K\otimes_{R} R/\mathfrak{m} \to M/\mathfrak{m}M\to N/\mathfrak{m}N\to 0\]
and since the arrow $ M/\mathfrak{m}M\to N/\mathfrak{m}N$ is an isomorphism, we have $ K\otimes_{R}R/\mathfrak{m} \equal{} 0.$
Moreover $ M\cong K\oplus N,$ so $ K$ is a projective $ R$-module (as a direct summand of the free $ R$-module $ M$), and from Kaplansky theorem (which says that every projective module over commutative local ring is free) $ K$ is a free module.
So $ K \cong \bigoplus_{i\in I}R,$ and thus
$ 0 \equal{} K\otimes R/\mathfrak{m} \cong \left(\bigoplus_{i\in I}R\right) \otimes R/\mathfrak{m} \cong \bigoplus_{i\in I}(R\otimes R/\mathfrak{m})\cong \bigoplus_{i\in I}R/\mathfrak{m}$
which implies $ I \equal{} \varnothing$ and $ K \equal{} 0.$
To sum up - this time we use assumptions that $ M, N$ are free modules over a commutative local ring $ R$, $ \mathfrak{m}$ is an ideal (not necessarily maximal) in $ R,$ and the map induced by $ f$ is an isomorphism.

Sorry to necromance, but how do we have an exact sequence

\[ 0\to K \to M\stackrel{f}{\to} N\to 0\]

if we don't know that $f$ is surjective? (Also, $\mathfrak m$ shouldn't be the ideal $R$ of $R$, but that's probably understood.)

Unfortunately I have no clue why I assumed the morphism $M\to N$ to be an epimorphism.

What is more the original claim seems to be wrong:
Suppose that the claim is true.
By functoriality of $\mathfrak{m}$-adic completion of $R$-module the induced map of completions $\widehat{f}:\widehat{M}\to \widehat{N}$ is an isomorphism.
Now $M,N$ as free modules over Noetherian local ring embed into their completions, and the following setting yields a counterexample:
\[R = \mathbb{Q}[[t]],\ \mathfrak{m} = t\mathbb{Q}[[t]], \ M = \bigoplus_{i=0}^{\infty}R\cdot e_{i} = N,\ f(e_{i}) = e_{i} + te_{i+1},\]
for the induced map over the residue field is the identity map, but
\[\widehat{f}^{-1}(e_{i}) = e_{i} - te_{i+1} + t^{2}e_{i+2} - t^{3}e_{i+3}+ \ldots = \sum_{j=0}^{\infty}(-1)^{j}t^{j}e_{i+j}\in \widehat{M}\setminus M\]
contradicts $\widehat{f}$ being monomorphic extension of $f$ and $f$ being epimorphism.

One may ask whether $\bar{f}:M/\mathfrak{m}M\to N/\mathfrak{m}N$ being isomorphic implies that $\widehat{f}:\widehat{M}\to \widehat{N}$ is an isomorphism.
This can be proven true using the content of my previous message:
It is not hard to prove by induction, that since induced morphism $M/\mathfrak{m}M\to N/\mathfrak{m}N$ is an epimorphism, then all induced morphisms $M/\mathfrak{m}^{\nu}M\to N/\mathfrak{m}^{\nu}N$ are epimorphisms.
If we view $M/\mathfrak{m}^{\nu}M\to N/\mathfrak{m}^{\nu}N$ as an epimorphism of free modules over a local ring $(R/\mathfrak{m}^{\nu}, \mathfrak{m}/\mathfrak{m}^{\nu})$, we have the situation from my previous post, and reasoning there shows that all the induced maps $M/\mathfrak{m}^{\nu}M\to N/\mathfrak{m}^{\nu}N$ are isomorphisms.
By functoriality of inverse limit, this implies that $\widehat{f}$ is also an isomorphism.


algebra
polynomial
Ring Theory
algebra open

Prove or disprove:  if the locus of an irreducible polynomial $ P(x, y, z, ... ) \equal{} 0$ over some polynomial ring is a subset of the locus of another polynomial $ Q(x, y, z, ... ) \equal{} 0$, then $ P | Q$ over that ring.

Consider $ P(x,y)\equal{}x^2\plus{}y^2\plus{}1$ over the reals.  :P

What if I require that the coefficients are in an algebraically closed field?

I'd be happy just for an answer in the case that $ P$ is linear.  I believe the problem is true in that case (if not any others), but I'm having trouble phrasing an argument in terms of a division algorithm...


Find all functions $f$ such that $f(x-1)+f(x+1)=6x+f(x)$.

Well if it says to find all functions, you can define whatever you like on the interval $[0,2)$ and then this defines the whole function.

But when the question request to find all functions satifying given condition then, we can't restrict $f$ determine on $[0;2)$

Sorry for bring unclear, what i ment was something slse.

For  Altheman's function what were saying is that we can pick the function to have any values we want on that interval. Then using the recursion this will give us the value of the function at every other value on the real line. Comprede?

[quote="boxedexe"][quote="Altheman"]substitute g(x)=f(x)+6x, to get g(x-1)+g(x+1)=g(x), now this is just a linear recursion...[i.e. like the fibonacci sequence]...but you have no initial values, so i don't think you can do anything more than that...[/quote]
Use characteristic equations.  Then you'll have $g(n)=a\lambda_{1}^{n}+b\lambda_{2}^{n}, a,b,\in\mathbb{R}$, where $\lambda_{1,2}$ satisfy the characteristic equation $\lambda^{2}-\lambda+1=0$.[/quote]

that does not completely define the function because the function is not necessarily continuous, and sequences can be disjoint, by this i mean:

g(1), g(2), g(3),...

can be seperate from

g(1.5), g(2.5), g(3.5),...

and two reals, a,b do not define such a function

[quote="Altheman"]substitute g(x)=f(x)+6x, to get g(x-1)+g(x+1)=g(x), now this is just a linear recursion...[i.e. like the fibonacci sequence]...but you have no initial values, so i don't think you can do anything more than that...[/quote]
A minor thing, but I think you mean $g(x) = f(x)-6x$.  Then you have

$g(x-1)+6(x-1)+g(x+1)+6(x+1) = g(x-1)+g(x+1)+12x$

$= 6x+g(x)+6x = g(x)+12x$

$\Rightarrow g(x-1)+g(x+1) = g(x).$

[quote="boxedexe"]Yes, you are absolutely correct.  $a,b$ are dependent upon the initial values.[/quote]

But the $a$ and $b$ used to generate the integer values of $f(x)$ can be different to the $a$ and $b$ used for say the values who have fractional part $\frac{1}{2}$

[quote="arkhammedos"]:?:  i think that it falls but ...
if we take $x-1=X$ and $x+1=Y$ we will get
$f(x)+f(y)=3(x+y)+f(\frac{x+y}{2})$
so we will get $f(x)=6x$ if we take $x=y$
and it gives answer but there is a mistake :?:[/quote]

If you take $X=Y$, then $x-1 = X = Y = x+1$, which is impossible.

[quote="mathgeniuse^ln(x)"][quote="Altheman"]substitute g(x)=f(x)+6x, to get g(x-1)+g(x+1)=g(x), now this is just a linear recursion...[i.e. like the fibonacci sequence]...but you have no initial values, so i don't think you can do anything more than that...[/quote]

I think it should read g(x-1)+g(x+1)-12x=g(x), so it won't be linear.[/quote]

uugh...they should cancel, i just placed the wrong sign...it does become a linear recursion...remember a linear recursion is not a linear function


inequalities
algebra
polynomial
inequalities unsolved

Find the best costant $k$ such that for all positive reals $a,b,c$ we have

$\displaystyle \frac{a^3}{ka^2+b^2}+\frac{b^3}{kb^2+c^2}+\frac{c^3}{kc^2+a^2} \ge \frac{a+b+c}{1+k} $

Magnificient inequality.


limit
Functional Analysis
advanced fields
advanced fields unsolved

a).Let $ T \in B(X,Y), T_{n}\in B(X,Y)$, image of all the operators $ T_{n}$ are finite-dimensional and $ \lim_{n \to \infty}\parallel T-T_{n}\parallel = 0$. Prove that $ T$ is a compact operator. * Here $ X,Y$ are Banach spaces, $ B(X,Y)$ is a Banach space of all continuous operators from $ X$ to $ Y$.
b). Prove that if $ Y$ is a Hilbert space, there the converse of a) is also true.

a) is easy, of course, since compact operators form a closed subspace of bounded operators and if an operator is known to be both bounded and have a finite-dimensional range then it is compact.

For b) suppose that $ Y$ is Hilbert and $ T \in \mathcal{B}(X,Y)$ is compact. Let $ n \in \mathbb{N}$ and $ U \subset X$ denote the open unit ball of $ X$. Then $ T(U)$ is totally bounded and we can find $ x_{1}, \ldots, x_{m_{n}}\in U$ such that balls centered at $ Tx_{1}, \ldots, Tx_{m_{n}}$ and of radius $ 1/n$ cover $ T(U)$. Denote $ Y_{n}= \textrm{span}(T(x_{1}), \ldots, T(x_{m_{n}}))$. We take an ortoprojection $ \varphi: Y \to Y_{n}$ and define $ T_{n}= \varphi_{n}T$. Now $ T_{n}\in \mathcal{B}(X,Y)$ has a finite-dimensional range and $ \parallel T-T_{n}\parallel = \sup\{\parallel Tx-\varphi_{n}Tx\parallel : x \in U\}\le 1/n \to 0$.

Wikipedia has an article on the [url=http://en.wikipedia.org/wiki/Approximation_property]Approximation property[/url]. Most classical spaces have it.
If all closed subspaces of a Banach space $ X$ have the approximation property, then it is said that $ X$ has the Hereditary APproximation PropertY, or that $ X$ is happy. Very few Banach spaces are happy.


I attempting to prove the problem below.  Can anyone help me tighten it?  I am assuming induction is needed, but I'm not including it here.

Let $ k$ be an integer not divisible by 3.  Show that $ k^{2}\minus{}1$ is divisible by 3.

We can rewrite the expression as $ (k\minus{}1)(k\plus{}1)$.  In order for this product to be divisible by three, at least one of the terms must itself be divisible by 3.  If we let $ k\equal{}n\plus{}1$ where $ n$ is an integer divisible by 3, we can plug it in to obtain $ (n)(n\plus{}2)$.  Since we have $ n$ by itself as one of the terms, the product must be divisible by 3.  Next, let $ k\equal{}n\plus{}2$ where $ n$ is an integer divisible by 3.  Plugging it in, we have $ (n\plus{}1)(n\plus{}3)$.  $ n\plus{}3$ is divisible by 3, hence the entire product must be as well.  Since $ n\plus{}1$ and $ n\plus{}2$ represent all integers not divisible by 3, we have shown that $ k^{2}\minus{}1$ is divisible by 3.

Let $ k\equal{}3p\plus{}r\ (p\in\mathbb{Z},\ r\equal{}\pm 1)$, $ k^{2}\minus{}1\equal{}3(3p^{2}\plus{}2pr)\plus{}r^{2}\minus{}1$ since $ r^{2}\minus{}1\equal{}0$, which completes the proof.

The ideas are fine, but it looks like you need some practice in mathematical writing- it's not phrased very well.

Here's how I would put it, using your argument:

First, we note that $ k^{2}\minus{}1\equal{}(k\minus{}1)(k\plus{}1)$ for any integer $ k$. If $ k$ is not a multiple of $ 3$, we have two cases:
Case 1: $ k\equal{}3m\plus{}1$ for some multiple $ 3m$ of $ 3$. Then $ (k\minus{}1)(k\plus{}1)\equal{}(3m)(3m\plus{}2)$, which is divisible by $ 3.$
Case 2: $ k\equal{}3m\plus{}2$ for some multiple $ 3m$ of $ 3$. Then $ (k\minus{}1)(k\plus{}1)\equal{}(3m\plus{}1)(3m\plus{}3)\equal{}3(m\plus{}1)(3m\plus{}1)$, which is divisible by $ 3.$
This covers all cases, and we have that $ k^{2}\minus{}1$ is divisible by $ 3$ whenever $ k$ is an integer not divisible by $ 3$.


ARML
geometry
geometric transformation
reflection
number theory

I'm going to ARML this year but unfortunately, I'm unable to attend all of the practices for my team. The worst part of ARML is the Power Round. I don't really understand how to start the problems.

Any tips?

Have the strongest people on your team do the hardest parts.

What our (Washington) team does beforehand is write up the problem numbers on the board and assign people to each problem on a partially volunteer basis and partially ability (we do this for the team test too).  So Adam Hesterberg always gets the last part  :D

As for starting... do the first few parts first?  I'm not totally sure what you're asking.  The first few tend to be a sort of introduction to the general topic of the Power Round, so they help.  Are you asking for how to tackle the questions, or how to write the solutions up?

yeah, as t0rajir0u mentioned we have devide it into parts for people,
so if you are not that good with those questions, i would suggest you try the first couple problems, cause they are always very doable and easy to write a proof for

Hmm...interestingly that EXACTLY what So. Cal. did...maybe it's a common technique??
Yeah...and look for the stuff that you are the best at, i.e. if your good at number theory, see if there's any number theory ones

Well, the first Power Round question that I faced was one in NYSML 1992. It was something about reflecting point in square and I hate those types of problems.  :blush:

Maybe if it was algebra, I would've know what to do but eh..

[quote="scorpius119"]Remember, the Power Question is for an entire team to work on, and considering that the entire team could be made of USAMOers, the later parts of the question are usually difficult to approach by those with less experience. (Last year, I was lucky to answer ONE piece of the question by myself, let alone three or four.)

And yea, usually different areas are covered, so it is usually best to figure which ones your'e strong at. Although the reflection on last week's question was not that easy to visualize, once the formulas for the length and number of bounces were found, the rest of Part I was all NT.[/quote]

Yeah well the hard thing for me now is that I won't be able to come all the practices. This means that I need to do extra well and prepare by myself for some parts of ARML.

 :|


I was wonderin where i can get practice tests for 1997 - 2000 state tests. i want to get in the top 4. :lol:

Try looking at the stickys, [i]trade for them[/i]...

It's right above the regular topics; specifically, the thread labeled "MATHCOUNTS Resources".


topology
advanced fields
advanced fields unsolved

$A\subset X, A$ is contractable, then $X\sim X/A$.

Maybe (X,A) should be a CW-pair. But even with this assumption, I still cannot solve it.  Any help would be very appreciated, thank you!

By the way, I have a lemma here: given $f: X\rightarrow Y, f|_A\sim const$, then exist $g: X\rightarrow Y$, s.t.$g|_A=const$ where $f\sim g$.  Using Y=X in this problem, one easily deduce that $g\circ\pi\sim id: X\rightarrow X$, where $\pi: X\rightarrow X/A$ is the quotient map, and g is the map in the lemma.  But the reverse seems not as easy as above.

Perhaps the CW condition is too strong since using only homotopy groups can solve this problem.  I guess I should change the restriction to be: A is a deformation retract of some neighbourhood in X.  Under this condition, the lemma above still holds.


number theory
relatively prime
number theory unsolved

Let $ n, m$ be positive integers of different parity, and $ n > m$.  Find all integers $ x$ such that $ \frac {x^{2^n} \minus{} 1}{x^{2^m} \minus{} 1}$ is a perfect square.

The given expression reduces to $ (x^{2^m} \plus{}1)(x^{2^{m\plus{}1}}\plus{}1)...............(x^{2^{n\minus{}1}} \plus{}1)$

We see clearly that there are an odd number of terms in the expansion because m and n have different parity.

So the expression is divisible by some $ 2^l$ where l is odd if x is odd
because each term of the expression can be written as $ 2y$ where  y is odd
So the given expression can never be a perfect square if x is odd.

If x is even observe that each term in the expression is odd and each term is relatively prime to each other.

So then each term should be a perfect square

But $ (x^{2^m}$ is itself a perfect square so if $ (x^{2^m} \plus{}1)$ is a perfect square then x is 0.

I guess this is the only answer. :D


calculus
logarithms
geometry
MATHCOUNTS
AoPS Books
number theory

I am most likely going to buy all the intro books and i want to buy the Aops Vol 2 but i took a trig/precalc course at a local college and i still have textbook and its explains chapters 1,3,5,6(i am not sure if it explains newton's sums though), 8,9,10,11, 13 but it doesn't have that kind of problem solving approach so is it still worth it to buy an Aops Vol 2 or do i just stick with the textbook?

thanks
Dhaivat(uh huh that's my name)

I say this with full honesty:
I believe that AoPS Volume 2 is far better than the other trig/precalc courses out there.  And I have (some) credentials: I'm going to be taking Calculus AP as a freshman this year, and I think that that will be much easier than AoPS Volume 2.  School standards are just not enough for those who are talented with mathematics.  I got a perfect score on an Algebra II/Precalculus final not too long ago, and it was with very minimum studying (probably studying wasn't necessary).  I couldn't even solve half the problems on several chapters of the AoPS book (they are tough!).
I "thought" I knew most of the subjects in Volume 2, seeing as I was almost done with a Precalculus course at a very good school (best statewide testing scores in the state), but I was wrong.

Even from the beginning chapter on logarithms, challenging problems were present.  After you go through Volume 2, you'll (most likely) believe that the stuff you learned at your local college is [i]useless[/i] compared to the way you learned it through AoPS.

But, I would highly suggest that you go through the Intro books first (since you're ordering them and AoPS Vol. 2) before AoPS Vol. 2.  The concepts in Volume 2 seem easy enough, but the problems require extremely good thinking (plenty are USAMO/IMO problems or other problems of equal difficulty).

One thing about the AoPS Vol. 1 and Vol. 2 is that they are very fast-paced. I am constantly having to reread and reread and reread to grasp the concepts. The problems are very challenging for me, even though I am only in the Vol. 1. I thought they would be kind of easy, like the classes I took in school. I was extremely surprised and I am always having to go back and reread the chapter to solve the problems. A great thing about Vol. 1 is that many chapters are devoted to geometry, which is helpful because that's what I'm taking this year, and I always like to learn the subject during the summer in preparation for my next class. The Intros are slower and easier in that they don't zip through the concepts. I don't have to reread so much with those. But I find that the AoPS books that I have covered so far(so far I have finished Intro to C and P, and I'm going through Vol. 1 and Intro to NT - I only found out about AoPS in March, when I stumbled upon Intro to C and P) are all great.

[quote="Poincare"]yeah, the   [size=200] AoPs books are awesome![/size] they're the best...[/quote]
I totally agree. I read all four of the Intro books and both the AoPS Vol. 1 and 2. They're good for Mathcounts prep. :)

[quote="Mind Storm"]You should get the book beacause it will teach you things the textbook does not.[/quote]
That's true :)

aops books are great and all, but i think u should read the textbooks and master the very basics of a subject before moving on to aops with its faster paced and more challenging materials, i dont know about u, but i find aops books dont cover enough about a subject before moving on to the next subject, and i have a hard time starting on a new subject with the aops book (but im not using the subject books) and sometimes the explanations/solutions are hard to follow, especially on the harder problems, but i guess inteligence level varies, so... :P

well, i am really dumb  :(  so i'll follow your advice

[quote="Mind Storm"]You should get the book beacause it will teach you things the textbook does not.[/quote]
If you're taking, say, Algebra I this year, I recommend you get the books accossiated (spelling?) with that grade level.

And if you're taking Algebra I (like me), I reccommend you get: Intro. to Number Theory, Intro. to Algebra, and AoPS Vol. 1.

[quote="isabella2296"]Also take whatever interests you, and things that are your weak points. I'm not so great at geometry, so I'm going to read Intro to Geo. I like Number Theory, and that's what I'm reading now. I recommend reading AoPS Volume 1 AFTER all the Intro's because it's very fast paced, so you'll learn very little. It's better as a review.[/quote]
Well, Vol. 1 can still help you with topics you need =]

[quote="Poincare"]i am doing the intro books. And, i have my birthday on August 15 (indian independence day  :) ) and i was wondering if i should buy Aops vol 2[/quote]
It depends what math class you're going to take this year, other AoPS books you have, etc.

[quote="Poincare"]i am taking pre calc or calc and i have all the intros  :)[/quote]

Have you finished all the intros? Taking calc doesn't mean you can do Volume 2 if you aren't good at problem solving

You can always just buy Volume 2 now, and when you feel you are ready, start reading it.

@Urc: yes i know
@ isabella2296: that's what i plan to do  :)

AoPS goes into a lot more depth than any regular school textbook. There have been plenty of replies already, so I'll just say, even if you know most of it, you'll still learn a lot, and it's a great source of problems by subject (about 600 per volume), and as a reference for math competition.

If you want to, the classes are great. I've noticed that the classes are 'deeper' *cough*and harder*cough* (wow congestion) then the books, in each subject. (Wow I should be hired as an ad person for AoPS)

But the classes are freaking expensive!!!!!!!!!!!!

They're a few hundred dollars (and yes, the price has increased a little in the past few years), but they're cheap compared to other online courses offered by other institutions. And the teaching is great, and you won't find the same kind of material in most places, unless you happen to live by a math circle, or go to a really good school. It's a good value.

I find the AoPS books to be great. They go at a nice pace, (presuming you do every single problem, which tends to be a problem, as I'm usually too lazy to do all the problems of a section  :P .) and they're the first thing I've learned quite literally anything from over the last 2 or so years. I like doing the Vol 2 things despite the fact that I haven't finished the last few chapters of vol 1. I wish the online classes were smaller though, so we could get a little more attention then we already have.


geometry
circumcircle
geometric transformation
homothety
vector
reflection

Let $A',B',C'$ be the feet of the altitudes of a triangle $ABC$, $\{ D \} = A'B' \cap CC'$, $\{ E \} = A'C' \cap BB'$. $O_a,O_b,O_c$ are the points symmetric to the circumcenter $O$, with respect to $BC,CA,AB$. $M_a,M_b,M_c$ are the midpoint of $BC,CA,AB$. $H_a,H_b,H_c$ are the midpoints of the segments determined by the orthocenter $H$ and $A,B,C$.

(a) $OA \perp B'C'$ and $OA \| M_a H_a$

(b) $HO_a \perp B'C'$, $BO_a \perp A'B'$, $CO_a \perp A'C'$

(c) $AO_a \perp DE$

(d) the lines $AO_a,BO_b,CO_c$ are concurrent

Besides these, there are also (e), (f), (g) and (h), but I'll post them after I get some nice solutions for the first four.

For part a, it's more generally true of a point and the projections of its isogonal conjugate, in fact. For this particular case, it's easy to see that the angle OA makes with AC (90 + <B) is 90 more than the angle B'C' makes with AC (<B). The image of $M_aH_a$ under the homothety with center H and factor 2 is the diameter through A on the circumcircle (easy with vectors) which is of course parallel to OA.

um...AO_a is not perpendicular to B'C', sorry

the midpoint of AO_a is N, the ninepoint center (again easy with vectors), so thus AO_a, BO_b, CO_c all concur at N

for part (c), I've proven AO_a is the same as AN and that DE goes through the point of intersection of B'C' and BC...more to come later possibly (history now  :( )

i've made a typo: it was $HO_a \perp B'C'$; the other two $\perp$ are correct.

Now let me prove your generalization (very easy :) )

Let $Q$ be the isogonal conjugate of $P$ wrt $\triangle ABC$. $D,E,F$ are the projections of $Q$ on $BC,CA,AB$.

Since $\measuredangle AFQ=\measuredangle AEQ=\frac{\pi}{2}$, it follows that $AEQF$ is cyclic. Thus, $\measuredangle AEF = \measuredangle AQF$. But we also have $\frac{\pi}{2}-\measuredangle AQF = \measuredangle  FAQ=\measuredangle PAC$, i.e. $\measuredangle PAC + \measuredangle AEF = \frac{\pi}{2}$. Hence, $AP \perp FE$.

yup! and of course we know that H, O are isogonal conjugates

part 2 is now easy - we already know OA is perpendicular to B'C', and OA is the image of HO_a under reflection in N, the ninepoint center

still working on part 3...btw, go ahead and post the other 4 parts

(e) $M_a,M_b,M_c,H_a,H_b,H_c$ are the vertices of a hexagon which has a center of symmetry.

(f) $A,B,C,O_a,O_b,O_c$ are the vertices of a hexagon with all sides equal and which has a center of symmetry.

(g) The sides of the triangle $O_a O_b O_c$ are the perpendicular bissectors of $(AH),(BH),(CH)$.

(h) The triangles $O_a O_b O_c$ and $ABC$ are symmetric with respect to a point.

These are rather easy now, only (c) is harder than the rest, but it was solved in the Olimpiad - Geometry forum.


What's the rest of the division of 16^240 by 287


Problem:

a spacecraft is in the space (ok, i know you're probably enough intelligent to understand it...). Suddendly, a window of the spacecraft gets broken. The area of the window is $S$, the internal pressure is $p_0$, the volume of the spacecraft is $V$, the absolute temperature is $T$, the gas' molar mass is $M$.  The air that's inside starts coming out through the window, so rapidly that its temperature doesn't change.

Find $p = f(t)$.
Discuss possible analogies with RC circuits.

SNS Admission, Physics test

I apologize for my horrible english; if you have doubts about the text, please ask me.

It seems to me that we need one more piece of data - the density, molecular mass or total mass of the gas.

Yes, you need the molar mass of the gas. Sorry

Is it......

$p=p_0\cdot e^{\frac{S\sqrt{\pi RT}}{2V\sqrt{2M}}t}$

?? M is the molar mass.

The exponent once again:

$\frac{S\sqrt{\pi RT}}{2V\sqrt{2M}}t$

Don't you think that the exponent should be negative? The pressure is decreasing...

However i got something quite different, but i'm not sure about my solution.
I got the exponent $-t \sqrt \frac{2RTS^2}{MV^2}$

:| would either mind showing a little work, please? ;)

Yes, I forgot the minus sign.... Uhm, frankly my working is a little lengthy and I'm not sure about the validity of the steps I made at all. I'll try to post it later today (hey, it's 3 am 1st January of the year 2006!)... but your result is not that much different, you only have a constant of $\sqrt{2}$ instead of my $\frac{\sqrt{\pi}}{2\sqrt{2}}$.

OK, my solution goes something like this:

How much air is inside of the spaceship? Let $p(t)$ be the pressure and $m(t)$ the mass of the air that's left inside the spaceship and let's assume that the outside pressure is 0 during the whole decompression. Volume of the spaceship $V$ and inside temperature $T$ are constants, let $M$ be the molar mass of the air:

$pV=nRT$
$pVM=nMRT=mRT$
$m=\frac{pVM}{RT}$

Now let's find $Q=\frac{\text{d}m}{\text{d}t}$. Pressure $p=\frac{\text{d}F}{\text{d}S}$ means that the molecules of air hit the walls of the spaceship and apply force dF on each surface of area dS. Also the applied force $F=\frac{\text{d}P}{\text{d}t}$ is caused by the change of linear momentum (dP - I use capitalized P so that it's not confused with pressure p) of the molecules of air each dt. Had the window of the spaceship not broken, the force applied on the window would have been $p_0S$ or in other words, the molecules of air inside of the spaceship would have gained $p_0S$ momentum per second because of bouncing off of the window. This means that every $\text{d}t$ hit the window molecules with total momentum in the direction perpendicular to the wall equal to $\frac{1}{2}p_0S\text{d}t$ and bounce away with momentum equal to $-\frac{1}{2}p_0S\text{d}t$ (elastic collision) resulting in total change of momentum equal to $p_0S\text{d}t$. If $\overline{|v_x|}$ is the mean speed of molecules in direction perpendicular to the window then the average total mass of molecules that hit the window per second is $\frac{p_0S}{2\overline{|v_x|}}$.
But of course, the situation doesn't change even with the window broken. If the inside air pressure is $p$ then the window is hit... or actually this many molecules (their total mass per second) escape through the window opening: $\frac{pS}{2\overline{|v_x|}}$. Therefore $\frac{\text{d}m}{\text{d}t}=-\frac{pS}{2\overline{|v_x|}}$. (This is where I originally forgot to put the minus sign.)
Now we only need to find $\overline{|v_x|}$. I've managed to calculate (some nasty integration that I'm not sure I've done correctly...) that $\overline{|v_x|}=\frac{1}{2}\overline{|v|}$. And $\overline{|v|}=\sqrt{\frac{8RT}{\pi M}}$ - I found this on the web, that's another thing I'm not very sure about. So finally:

$\frac{\text{d}m}{\text{d}t}=-\frac{pS}{\sqrt{\frac{8RT}{\pi M}}}=-\frac{\sqrt{\pi M}S}{\sqrt{8RT}}p$

But we know what is the mass equal to:

$m=\frac{pVM}{RT}$

so

$-\frac{\sqrt{\pi M}S}{\sqrt{8RT}}p=\frac{\text{d}m}{\text{d}t}=\frac{VM\text{d}p}{RT\text{d}t}$

$-\frac{\sqrt{\pi RT}S}{V\sqrt{8M}}p=\frac{\text{d}p}{\text{d}t}$.

After integration we get the result I've posted.

Oh yes, perhaps I found the matter.

First of all, i post my own solution:

$dp = dm \frac{RT}{MV}$
$p=\frac{1}{2}\rho v^2 \Rightarrow v=\sqrt \frac{2p}{\rho}$ (Bernoulli)

Therefore $\frac{dm}{dt} = \frac{dV}{dt} \cdot \rho = SV\rho = S\sqrt{2p\rho}$

Therefore $\frac{dm}{dt} = -pS\sqrt \frac{2M}{RT} \Rightarrow dp = - dt \cdot p\sqrt \frac{2RTS^2}{MV^2}$

Therefore $\frac{dp}{dt} = -p \sqrt \frac{2RTS^2}{MV^2}$
Then, integrating....

I think our results are different for two reasons:

- The expression you used for $v_x$ is for the average speed; if you use (don't ask me why) the expression for the most probable speed $\sqrt{2RT}{M}$ your solution differs only for a factor 2.

- In my opinion, that factor 2 is the $vena contracta$ effect! Do you know it? When it was explained to me, i was shocked! The proof is obtained in a way that's very very similar to your work about this problem.

Hm, interesting, Bernoulli's equation, how simple... I forgot all about it :). Your solution seems to be flawless, I think you got it right.
Vena contracta effect? I've just googled it up - I don't think it has anything to do with this problem, neither my nor your solution addresses this issue so it shouldn't make the results differ. The reason why my solution gives different result is that I probably made some invalid arguments, it is just way too complicated and I wasn't too sure about it anyway. Well, it was a nice problem Bacco :).


(a) Find the domain and range of the function $x \ge 2y^2-4$
(b) Find the domain and range of the function $x^3+y\leq0$
Functions are from reals to reals

I may be being too picky here but "$x\ge2y^2-4$" and "$x^3+y\le0$" are not functions, so I don't know what you're asking. If you said the functions $x=2y^2-4$ and $y=-x^3$ it would make sense to me.

Hmm... then let's say it's a relation rather.

Since $x \geq 2y^2-4$ is the same as $x \geq 2(y^2-2)$, and the smallest $y^2$ can ever be is $0$, then the smallest $2y^2-4$ can ever be is $-4$. So the domain is $x \geq -4$. And the range is all real numbers since $y$ can be whatever it wants to be.

and wouldn't both the domain and range of $x^3+y \leq 0$ just be all real numbers?

theres not much work i can show...
[hide]
$x\ge2y^2-4$
domain: $x \in \left[-4, +\infty\right)$
range: $y \in \Re$
$x^3+y\le0$
domain: $x \in \Re$
range: $y \in \Re$
[/hide]


[quote="ZetaX"]Why should there always be an inequality¿
We can ask the same for diophantine equations, sequences, combinatorial geometry or functional equations.[/quote]

 Maybe because there are a lot of creative solutions :). If u remember Iura got special prize for solving inequality, solution was beautiful.

For any area you have problems with very nice creative solutions (and it's just a matter of random that the special price was for inequalities).

Unfortunately, in the area of inequalities, there are also tons of absolutely uncreative solutions, and given the time restrictions at the IMO, these are the ones usually produced by the participants. We need $ n$-number inequalities, but unfortunately few people really spend their time with creating such ones and proving them without three-letter theorems...

  darij

Hello,

I see you talking a lot about "Beautiful Solutions".

Does giving a beautiful solution is better than a normal one?? I don't think  :lol:  .

Tricky, creative solutions may show some nice new ideas.
And solutions that are much shorter (without sketching ;) ) are better to read, too (good books for example show "better" solutions sometimes, too).


[quote="dmath328"]Prove that $ \displaystyle(sinA \plus{} sinB \plus{} sinC)(\frac {1}{A} \plus{} \frac {1}{B} \plus{} \frac {1}{C})\ge \frac {27\sqrt3}{2\pi}$

$ (A \plus{} B \plus{} C \equal{} \pi,A,B,C > 0)$

I think it's not easy one for me.[/quote]

hi. Srhiri  $ sinx$     $ 0\leq x \leq\pi$  ---> $ sinx$ -->     convex ---->  $ sinA\plus{}sinB\plus{}sinC\leq3sin(\frac{A\plus{}B\plus{}C    }{3})$

It has an error (It's not easy one;)

And x^2 : convex,  -x^2:concave      maybe;;

So sin(x) is concave, but (sina+sinb+sic)/3=<sin(a+b+c/3)

I am sorry, the answer that I gave is incorrect

I think that only for $ ABC$ nonobtuseangle, i.e. on $ \left(0,\frac {\pi}{2}\right]$ !

You can use Tchebashev's inequality.

Tchebashev's inequality  is same with Chebyshev's inequality?

If it is different one, please give me info about that inequality.

And give me proof, please.

(Also, I think it is true for (0,Pi) )

no one can solve it..??

I want to know solution of the problem very much! but i can't solve it....

 :blush: