I have now these titles:

mathematical olympiad challenges
mathematical challenges
winning solutions
problem solving stratergies
problem solving through problems
Acops (finished, but cant completely grasp the calc chapter)
USSR problem book
from kiev to caliber

Which one do you think i should read [b]first[/b] in preparation for usamo.

Any help would be appreciated. :)

Hi
The book which you havenot finished yet, have you posted them all in forum.
About the which you want to study if your knowledge in topics is well my opinion Problem Solving Stratgies
  Abdurashid

[hide=" :help: "]Well can you suggest which title is the easiest out of them all.[/hide]

In my opinion, you should read them in teh following order:

1.  Problem Solving Strategies
2.  Problem Solving Through Problems
3.  Then solve all problems on your own from Mathematical Olympiad Challenges within 2 months.

Masoud Zargar

[quote]In my opinion, you should read them in teh following order:

1. Problem Solving Strategies
2. Problem Solving Through Problems
3. Then solve all problems on your own from Mathematical Olympiad Challenges within 2 months.
[/quote]

i thought problem solving strategies is harder than mathemaical olympiad challenges.

Problem Solving Strategies teaches more than Mathematical Olympiad Challenges.  It ranges from Beginner-Advanced.  The only use of Mathematical Olympiad Challenges is to spend solving the problems in order to gain experience.

What about mathematical olympiad treasures and mathematical miniatures. What do they teach you?[/quote]

From my experience with mathematical miniatures (cant speak about math olympiad treasures) is that it shows a few exemplary ideas and then proceeds on to show how powerful these can be in some tough situations. It also is along the lines with igenuity in mathematics by ross honsberger (except at a higher caliber) in that it shows a great amount of creativity in solving different problems, however its development of each topic isnt really complete it is more of a collection of cool asides that would work best used in cahootz with one of titus other books.


I understand, I was absent-mind!Nice proof!

[b]Problem:[/b]
Let $ a_1;a_2;...;a_n$ be positive real numbers satisfying
$ \frac{1}{1\plus{}a_1}\plus{}\frac{1}{1\plus{}a_2}\plus{}...\plus{}\frac{1}{1\plus{}a_n}\equal{}n\minus{}1$
Prove that $ a_1a_2...a_n \geq (n\minus{}1)^{\minus{}n}$

I saw a problem , and it was posted before :

Let $ a_1;a_2;...;a_n$ be positive real numbers satisfying
$ \frac {1}{1 \plus{} a_1} \plus{} \frac {1}{1 \plus{} a_2} \plus{} ... \plus{} \frac {1}{1 \plus{} a_n} \equal{} n$
Prove that $ a_1a_2...a_n \geq (n \minus{} 1)^{ n}$

  :)

[hide="@chienthan"]Co phai em noi den bai nay khong [/hide]


calculus
integration
limit
logarithms
calculus computations

Find $ \lim_{n\to\infty} \left(\frac{n\plus{}1}{n}\right)^{\frac{1}{n^2}}\left(\frac{n\plus{}2}{n}\right)^{\frac{2}{n^2}}\left(\frac{n\plus{}3}{n}\right)^{\frac{3}{n^2}}\cdots \left(\frac{n\plus{}n}{n}\right)^{\frac{n}{n^2}}\ (n\equal{}1,\ 2,\ 3,\ \cdots)$.


Do anyone knows how to prove this?

Let a and n be integers such that a>n and n>3, then

a^n > 1^n+2^n+3^n+...+(a-1)^n

are you sure it's true LT?  try a=10, n=4.  the LHS is $10^4$, RHS is 15,333

yeah, good points, stevy,  the actual problem should be like this

Given two positive integers a and n where a>n>3, prove then

a^n*(a-1)/(n+1) > (1^n+2^n+3^n+...+(a-1)^n)

i though that i could cancel out (a-1) with (n+1), but appearantly it did not work.


--------------------------------------------------------------------------------

you have two coins in a basket, one of them has a heads on both sides and the other has a heads on one side and a tails on the other. Now, you pick one out (at random) and you look at one side of that coin, and it shows heads. what is the probability that the coin you chose on was the double sided coin with heads-heads?

[hide=":arrow: monkey"]
There are two ways to get the double-headed coin - getting side 1 or side 2.  There is one way to get the single-headed coin.

So is the probability $\frac{2}{3}$?
[/hide]

[hide]there are 3 total "heads" sides, for two of them the other side is heads, for the other the other side is tails.  So $\frac{2}{3}$.  You can also use conditional probability, I think.[/hide]

[hide]
After learning Conditional Probability, I have been able to solve more and more probability related problems. For all of you who don't know Conditional Probability, it is:
(The $|$ sign means given that)
$P(A | B) = \frac{P(A and B)} {P(B)}$

In this case, probability of "A" is the probability that we get the double-sided head coin.
Probability of B is the probability of getting a head when you reveal it.

Basically, P(A | B) = Probability of getting a double-sided head coin given that you got a head.

So...$P(A and B) = \frac{1}{2} * 1$.

$P(B) = \frac{1}{2} * 1 + \frac{1}{2} * \frac{1}{2} = \frac{3}{4}$

$So, \frac{\frac{1}{2}}{ \frac{3}{4}} = \frac{2}{3}$[/hide]


probability
modular arithmetic
MIT
college
floor function
Gauss
quadratics

A man stands at the edge of a cliff (facing it). On every "move", he either takes two steps backwards with probability $ \frac {1}{2}$ or one step forward also with probability $ \frac {1}{2}$. What is the probability that after some number of moves, the man dies by falling off the cliff?

[hide="Solution"]The probability that the man will fall off on the first move is $ \frac{1}{2}$.
The probability that the man will fall off on the second or third move is $ 0$ (because the first move would have to be $ \minus{}2$ steps, and the man would need at least $ 3$ more moves to fall off).
The probability that the man will fall off on the fourth move is $ \frac{1}{16}$ (it can only happen if the first four moves are $ \minus{}2,1,1,1$).
The probability of the man falling off on the fifth or sixth move is $ 0$ for a similar reason as the second and third moves. (Note that the man can only fall off the cliff on a move that is $ 1\pmod{3}$.)
The man will fall off on the seventh move iff the first move is $ \minus{}2$ and the next $ 6$ moves are $ \minus{}2,1,1,1,1,1$ as long as the 2nd $ \minus{}2$ occurs in the second or third move (which can happen in $ 2$ ways out of $ 64$).  The probability of this happening is $ \frac{1}{2}(\frac{1}{32})\equal{}\frac{1}{64}$.

So, the answer to the problem becomes the sum of the infinite geometric series $ \frac{1}{2},\frac{1}{16},\frac{1}{64}...$, which is $ \frac{\frac{1}{2}}{1\minus{}\frac{1}{4}}$, or $ \frac{2}{3}$.[/hide]

That's if I didn't make any mistakes.

cf249, how do you know that the next term in the series will be 1/256 if you didn't actually calculate it?

Actually, I was just assuming it was.  (And I'm having problems calculating it.) :oops:

I've got this:  For the man to fall off on the 10th move, there must be a total of 3 $ \minus{}2$ moves and 7 $ 1$ moves.  However, the first move must be $ \minus{}2$, or the man would fall off after the first move.
At least one $ \minus{}2$ move must occur before the fourth move, or the man would fall off after the fourth move.
The other $ \minus{}2$ move must occur before the seventh move, or the man would fall off after the seventh move.

I don't know where to go from there.

What if I assume that the probability of him dying when he first starts is $ p$? After one move, the probability changes to $ \frac {p}{4}$. At the same time, the probability that he doesn't die is $ 1 \minus{} p$. Therefore, $ 1 \minus{} p \equal{} \frac {p}{4}$, giving $ p \equal{} \frac {4}{5}$. Is there any invalid assumption here?

[quote="aidan"]What if I assume that the probability of him dying when he first starts is $ p$? After one move, the probability changes to $ \frac {p}{4}$.[/quote]

How is this so?

If he steps back two steps, he can only go back to the edge of the cliff with probability $ \frac{1}{4}$. So the probability that he will die is $ \frac {p}{4}$ at two steps from the cliff. Or am I missing something?

There are more ways that he can go back to the edge besides going directly there in 2 steps.

Furthermore, the probability that he will (eventually) go from 2 steps from the edge to 1 step from the edge is not necessarily 1/2 (it's actually equal to the probability that we are looking for).

So then the probability that he will go back to the edge of the cliff from 2 steps away from the cliff is $ p^2$?

There should be a purely algebraic probability solution; I'm having trouble finding it.

By the way, this problem is from MIT OpenCourseWare: Algebraic Combinatorics.

Yes, if you call the probability that he (eventually) falls off of the cliff starting from the edge (call that position x=0) p, then the probability that he (eventually) goes from  x=2 to x=0 is $ p^2$ and the probability that he (eventually) falls off from x=2 is $ p^3$. So you can set up an algebraic relationship between the probabilities.

[hide]I think it should be $ \frac {1}{2} \plus{} \frac {1}{2}p^3 \equal{} p$ (the probability that he falls off (eventually) equals the probability that he falls off on the first step plus the probability that he doesn't fall off on the first step (i.e. moves to x=2) but falls off (eventually) from there.[/hide]

After some thought I think that cf249's solution is correct, though his series wasn't justified. If $ a_n$ is the number of ways the man can fall off the cliff after exactly n moves, I get the raw recurrence $ a_n \equal{} n \minus{} \left\lfloor \frac {2}{3}n \right\rfloor \minus{} \sum_{k \equal{} 1}^{n \minus{} 1} a_k$, with $ a_1 \equal{} 1$.

A few calculations indicates that $ a_n \equal{} 1$ for $ n \equiv 1 \bmod {3}$ and 0 otherwise. All we need is a proof to verify this conjecture. I haven't taken the time to try it, but I would think that an inductive proof should work.

I'm not sure what's wrong with the logic of my recurrence in post #10 though. :oops:

(Edit: I see what's wrong with my recurrence in post #10! I'll see if anyone else can spot the mistake.)

Hmm, your recurrence seems okay to me.  This problem appears in Feller's book (Volume I, Chapter 14, Problems 7 and 8).  Feller's answer agrees with the solution to your recurrence.  I solved it the same way as your recurrence.

Actually, for the man to die in $ 3n \plus{} 1$ moves, there are exactly $ \frac {1}{2n\plus{}1} \cdot \binom {3n}{n}$ ways to do so. The integer sequence is shown at http://www.research.att.com/~njas/sequences/A001764.

So what we require is $ \sum^{\infty}_{n\equal{}0} \frac {1}{2^{3n\plus{}1}} \cdot \frac {1}{2n \plus{} 1} \cdot \binom {3n}{n}$, which is not easy to evaluate (at least to me  :D )

By the way, what's the final answer, Ravi B?

Okay, great! So much for second guessing myself. :maybe:

Now I have to figure out what I was thinking in post #11 and if its a coincidence that it happened to generate the same numbers as cf248!

Wonderful problem.

[hide="By the way, I find the answer to be:"]$ p \equal{} \frac {\sqrt {5} \minus{} 1}{2}$[/hide]

Yes, gauss, your answer is correct.  How did you eliminate the case $ p \equal{} 1$?

By the way, I sometimes give a talk to high-school students on solving such problems using "state diagrams".  Here is a list of problems that I hand out:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38284[/url].

The cliff problem is Question 23.  Question 24 is equivalent.  In that thread, I also present a solution to the related Questions 20 and 21.

Excellent problem, in fact  :P . So $ p$ is actually the probability of him advancing one step to the left after any number of moves?

Do we need to prove that $ p$ cannot be 1? Or can we just intuit that he can indefinitely move around without every getting close to the cliff?

Thanks for the problems!

[quote="aidan"]So $ p$ is actually the probability of him advancing one step to the left after any number of moves?

Do we need to prove that $ p$ cannot be 1? Or can we just intuit that he can indefinitely move around without every getting close to the cliff?[/quote]

Yes, that is the correct interpretation of p.

We do need to prove that p is not 1. I had a very weak reason for throwing it out, but here it is.

I justified it loosely on the fact that he is, in essense, drifting back from the edge of the cliff at an average rate 1/2 step on each move. If we ignored the fact that he can fall off the cliff and let him wander around (even to negative x's), the expected value of his position after n moves is $ \minus{} \frac {n}{2}$. His position is roughly normally distributed about this mean for large n, so I figured he would be very unlikely to deviate substantially from that expected position after a large number of moves n. Furthermore, the probability that he stays on the cliff [i]until[/i] the n-th move even smaller. So the sum of all those probabilities would not be enough to add up to 50% for n=2,3,4...

A better justification is probably needed though.

Yes, gauss, your argument can be made rigorous using the Strong Law of Large Numbers.

For another, inductive argument, see my last post in the other thread.

Sorry to bring out this thread again, but we were discussing a variant of this problem in my school's Math Society today, and one member asked what would happen if the man started at the edge of the cliff and walked one step to the right (away from the cliff) with probability $ \frac {2}{5}$ and one step to the left toward the cliff with probability $ \frac {3}{5}$.

Suppose $ p$ is the probability we want to find. Then $ p \equal{} \frac {3}{5} \plus{} \frac {2}{5}p^2$. Solving this equation, we get $ p \equal{} \frac {3}{2}$ (inadmissible) and $ p \equal{} 1$ (looks inadmissible as well, but this has to be the answer?). Does this mean that the man always dies in the long run?

The answer is $ p \equal{} 1$, as your argument correctly shows.  With probability 1, he will fall off the cliff.  That is different from saying that he will always fall off the cliff.  In infinite probability spaces, "probability 1" does not mean "always".  For example, let's say I keep tossing a fair coin.  With probability 1, I will eventually get heads, even though it is possible for me to never get heads.

Where can I read more about infinite probability spaces?  :D  Suppose now that the man walks away from the cliff with probability $ \frac {1}{5}$ and steps toward the cliff with probability $ \frac {4}{5}$ (one step in each case, as before).

Then $ p \equal{} \frac {4}{5} \plus{} \frac {1}{5}p^2$ and $ p^2 \minus{} 5p \plus{} 4 \equal{} 0$ and again one solution is $ p \equal{} 1$ (also the only sensible one). So does this mean that if the man has more than a half chance of walking towards the cliff on each step, then he will always die in the long run with probability 1?

Another question: Suppose the man begins at the edge of the cliff again, and has a $ \frac {1}{4}$ chance of walking towards the cliff, $ \frac {1}{4}$ chance of walking away from the cliff, and $ \frac {1}{2}$ of standing still on each move. In the long run, what is the probability that he dies? Can we just discard the "standing still" condition and treat the walking away and towards the cliff as probabilities of $ \frac {1}{2}$ respectively?

Yes, you could disregard the "standing still" option.  Things would just take longer.

[quote="aidan"]Where can I read more about infinite probability spaces?[/quote]Any undergraduate probability text should cover them.  I learned from Feller's books, but I'm sure there are lots of other (better?) books out now.  Maybe someone else has a recommendation?  Mosteller has a cute problem book: [i]Fifty Challenging Problems in Probability with Solutions[/i].  His Question 35 is the cliff problem.

[quote="aidan"]So does this mean that if the man has more than a half chance of walking towards the cliff on each step, then he will always die in the long run with probability 1?[/quote]Yes, the resulting quadratic has a nonsensical root bigger than 1, so the only realistic root is 1.  You can also prove the result from the Law of Large Numbers: If you imagine the walker to be able to move beyond the cliff (like in cartoons), then after $ n$ steps, he will be approximately $ cn$ steps beyond the cliff with high probability, for some positive constant $ c$.

[quote="aidan"]Suppose the man begins at the edge of the cliff again, and has a $ \frac {1}{4}$ chance of walking towards the cliff, $ \frac {1}{4}$ chance of walking away from the cliff, and $ \frac {1}{2}$ of standing still on each move. In the long run, what is the probability that he dies? Can we just discard the "standing still" condition and treat the walking away and towards the cliff as probabilities of $ \frac {1}{2}$ respectively?[/quote]Yes, once he makes a move, it's equally likely to be left or right.

Ok thanks a lot! I'll probably have to order the books from MAA or something, i haven't seen any undergraduate probability texts in Singapore. Cheers


The product of any four consecutive positive integers if 17160.  What is their sum?

[hide]  prime factorization of 17160 is 2^3 * 3 * 5 * 11 * 13.

  11 and 13 are very close together so we are most likely dealing with numbers in that range.  2 * 5= 10 and 2^2 * 3=12.  10 *11 * 12 * 13=17160.  Their sum is 10 + 11 + 12 + 13= [b] 46[/b][/hide]

I have a poor way that involves

17160 ^ 1/4 is 11.4somehting, so the four numbers are somewhere around this.

10 + 11 + 12 + 13 is 46.

Strangely enough, this method works for me each time.


function
logarithms
algebra
domain
algebra unsolved

problem 1
find all the decrease funtions such that f is continuous and
f: (0,1) → (0,1)
f(f(x))=x ; each x is in (0,1)
problem 2
k>1
find all the increase funtions such that f is continuous and
f:R+ → R+
f(f(x))=kx; each x>0

[quote="kihe_freety5"]problem 1
find all the decreasing continuous functions from $ (0,1)\to(0,1)$ such that $ f(f(x)) \equal{} x$ $ \forall x\in(0,1)$[/quote]

Take any $ a\in(0,1)$ and any continuous decreasing function $ h(x)$ from $ [0,a]\to[a,1]$ such that $ h(a) \equal{} a$ and $ h(0) \equal{} 1$
Clearly $ h(x)$ is a bijection from $ [0,a]\to[a,1]$

Define then $ f(x)$ as :
$ \forall x\in(0,a]$ : $ f(x) \equal{} h(x)$
$ \forall x\in(a,1)$ : $ f(x) \equal{} h^{ \minus{} 1}(x)$

[quote="kihe_freety5"]problem 2
Let $ k > 1$. Find all the increasing continuous funtions from $ \mathbb R^ \plus{} \to\mathbb R^ \plus{}$ such that $ f(f(x)) \equal{} kx$ $ \forall x > 0$ [/quote]

$ f(f(x)) \equal{} kx$ $ \implies$ $ f(kx) \equal{} kf(x)$ and the general solution is :

Let $ a\in(1,k)$. Choose any continuous increasing function $ h(x)$ from $ [1,a]\to[a,k]$ such that $ h(1) \equal{} a$ and $ h(a) \equal{} k$.
Clearly $ h(x)$ is a bijection from $ [1,a]\to[a,k]$

Define then $ g(x)$ from $ [1,k]\to[a,ka]$ as :
$ \forall x\in[1,a]$ : $ g(x) \equal{} h(x)$
$ \forall x\in(a,k]$ : $ g(x) \equal{} kh^{ \minus{} 1}(x)$

Define then $ f(x)$ as $ f(x) \equal{} k^{[\log_k(x)]}g(k^{\{\log_k(x)\}})$ (where $ [u]$ is the integer part of $ u$ and $ \{u\} \equal{} u \minus{} [u]$ is the fractional part of $ u$)

An other "point de vue" : $ (f(f(x))\equal{}x)\&(f\ decreasing) \ \Longrightarrow \ f(x)\equal{}g(1\minus{}g^{\minus{}1}(x))$ where $ g: [0,1]\rightarrow [0,1]$ is a bijective increasing function .  :cool:

[quote="Diogene"]An other "point de vue" : $ (f(f(x)) \equal{} x)\&(f\ decreasing) \ \Longrightarrow \ f(x) \equal{} g(1 \minus{} g^{ \minus{} 1}(x))$ where $ g: [0,1]\rightarrow [0,1]$ is a bijective increasing function .  :cool:[/quote]

Sure.
More generally, let $ h(x)$ any involutive continuous decreasing function and $ g(x)$ any increasing continuous bijection (with "good" domains an codomains), then $ g(h(g^{\minus{}1}(x)))$ is also an involutive continuous decreasing function.


function
combinatorics open
combinatorics

What is the combinatorial interpretation and proof of following identity?

$\prod_{n=0}^{\infty} (1-q^{2n+2} ) (1+q^{2n+1} )^2 = \sum_{n=-{\infty}}^{\infty} q^{n^2}$

You don't need to double post it ;)
http://www.mathlinks.ro/Forum/viewtopic.php?t=397829

I cannot find anything there. Can anybody show me the location of "397829" ?

I think he meant http://www.mathlinks.ro/Forum/viewtopic.php?t=67551 and he's right when saying that you should not double-post.
Btw: the RHS is the $\vartheta$-function after taking logarithm...

Hmm... I thought that this topic is related to both area.

It probably is, but that's no reason to post it twice (post it in the forum you think it's more adequate; should be combinatorics in this case).


Hi

This is not SPAM.

I'd like to tell you that take part in http://www.forogeometras.com

It's a spanish geometry forum.

Don't worry about the language, if you prefer post in english, french, italian, chinese, japanese, german, korean, russian, arabic and portuguese language.

Meet you there!


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

[b]Let A be the matrix
0     1     0
1     1     0
0     1    -1

Is there an invertible 3*3 matrix C with Inverse(C)AC a diagonal matrix.
[/b]

The Inverse(C) is basically C^(-1)... i didn't know how to make it look clearer.

The answer is yes since the characteristic polynomial has 3 distinct roots . (theory)

 :)


superior algebra
superior algebra solved

how can i solve $2(x)^5 \equiv 1 (\mod 91)$

Use that $91=7\cdot 13$. This reduces it to solve it $\mod$ these number by the chinese remainder theorem.
Now the multiplicative group $\mod$ an odd prime is cyclic, here with orders $6$ and $12$. Since both, $6$ and $12$, are coprime with $5$, there will be exactly one solution for $x^5 \equiv a \mod 7;13$ for any $a$, especially for $a=\frac{1}{2}$. To find this solution $\mod 7$ and $\mod 13$ either guess or find a primitive root.


Hello,

I have another latex problem I cannot solve :blush:

$\frac{ \sum_{i=1}^{n}i }{W}$ is the kind of thing I want to write.  But the indices in the summand don't look too good  :mad: , they should be under the sigma, like in this case:

${\sum_{i=1}^{n}i }$

What can I do about this?

Thank you,
fredbel6

Well, on the site I automatically remove \displaystyle and \limits (as a \displaystyle is automatically applied). It's a good thing though I never thought about the problems that might arrise from that :)

On a normal document [code]$\frac{ \displaystyle \sum_{i=1}^{n}i }{W}$[/code] or [code]$\frac{ \sum\limits_{i=1}^{n}i }{W}$[/code] would do the trick.

Neither of them work:

$\frac{ \sum_{i=1}^{n}i }{W}$

$\frac{ \sum_{i=1}^{n}i }{W}$

EDIT: The \displaystyle and \limits commands disappeared after I posted them. :maybe:

Stupid me... I didn't see the "normal document" part. Sorry.

[quote="Valentin Vornicu"]would do the trick.[/quote]
It does, thank you. :)


function
number theory unsolved
number theory

$ (\minus{}1)^n\textrm{LHS }$ is the number of surjective functions $ f : A \to A$, with $ | A | \equal{} n$ - by the Principle of Inclusion and Exclusion; see any Combinatorics textbook about the proof. $ (\minus{}1)^n\textrm{RHS }$ is clearly the number of the bijective functions.
So this identity only states that surjective is equivalent to bijective over finite sets.


Let ABC be any triangle.Suppose that AD and BE are two bisectors issued from A to [BC] and from B to [BE] respectively such that AD=BE.Show that ABC is an isosceles triangle

from B to [BE] ??
If in a triangle you have two congruent bisectors(AD and BE) then from the formula of bisector you got:
$ (b\minus{}a)(\frac{1\plus{}ab(b^2\plus{}ab\plus{}a^2\plus{}2c(a\plus{}b)\plus{}c^2}{(a\plus{}c)^2(b\plus{}c)^2})\equal{}0$ wich meens that $ a\equal{}b$.


[quote="kunny"]For arbiterary real numbers $a_1,\ b_1,\ c_1,$ let $a_n=\frac{b_{n-1}+c_{n-1}}{2},\ b_n=\frac{c_{n-1}+a_{n-1}}{2},\ c_n=\frac{a_{n-1}+b_{n-1}}{2}\ (n=2,3,\ \cdots).$

Prove that $\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=\lim_{n\to\infty}c_n=\frac{a_1+b_1+c_1}{3}.$[/quote]

$a_n+b_n+c_n=a_{n-1}+b_{n-1}+c_{n-1} \rightarrow a_n+b_n+c_n=a_1+b_1+c_1=k ,\forall n \Rightarrow c_{n-1}+b_{n-1}=k-a_{n-1}, \Rightarrow 2a_n=k-a_{n-1} \Rightarrow a_n=(-\frac{1}{2})^na_1+\frac{k}2 \cdot \frac{1-(-\frac12)^n}{1+\frac12} \Rightarrow \lim_{n\to\infty}(-\frac{1}{2})^n=0 ,so \lim_{n\to\infty}a_n=\frac{k}2 \cdot \frac23=\frac{k}{3}=\frac{a_1+b_1+c_1}{3}$.

 Similar to $b_n,c_n$ ;)

[quote="kunny"][quote="csabafarkas"]
$a_n=(-\frac{1}{2})^na_1+\frac{k}2 \cdot \frac{1-(-\frac12)^n}{1+\frac12}$.[/quote]

It seems something wrong. :?[/quote]

$a_n is a linear recursion$

[quote="csabafarkas"][quote="kunny"]For arbiterary real numbers $a_1,\ b_1,\ c_1,$ let $a_n=\frac{b_{n-1}+c_{n-1}}{2},\ b_n=\frac{c_{n-1}+a_{n-1}}{2},\ c_n=\frac{a_{n-1}+b_{n-1}}{2}\ (n=2,3,\ \cdots).$

Prove that $\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=\lim_{n\to\infty}c_n=\frac{a_1+b_1+c_1}{3}.$[/quote]

$a_n+b_n+c_n=a_{n-1}+b_{n-1}+c_{n-1} \rightarrow a_n+b_n+c_n=a_1+b_1+c_1=k ,\forall n \Rightarrow c_{n-1}+b_{n-1}=k-a_{n-1}, \Rightarrow 2a_n=k-a_{n-1} \Rightarrow a_{n+1}=(-\frac{1}{2})^na_1+\frac{k}2 \cdot \frac{1-(-\frac12)^n}{1+\frac12} \Rightarrow \lim_{n\to\infty}(-\frac{1}{2})^n=0 ,so \lim_{n\to\infty}a_n=\frac{k}2 \cdot \frac23=\frac{k}{3}=\frac{a_1+b_1+c_1}{3}$.

 Similar to $b_n,c_n$ ;)[/quote]

sory :blush:


Let a,b,c,d,e be non-negative real numbers and:
a+b+c+d+e=1.
Find the greatest value of S=ab+bc+cd+de+ea

Suppose that $ a\geq b\geq c\geq d\geq e$
$ \Rightarrow S \leq ab\plus{}bc\plus{}cd\plus{}de\plus{}da \plus{}be\equal{}(b\plus{}d)(a\plus{}c\plus{}e)\leq\frac{(a\plus{}b\plus{}c\plus{}d\plus{}e)^2}{4}\equal{}\frac{1}{4}$
$ S\equal{}\frac{1}{4} \Leftrightarrow b\equal{}a\equal{}\frac{1}{2}, c\equal{}d\equal{}e\equal{}0$

How do did you get the upper bound? ive never seen that inequality before.

You're right that you can't assume that in general. In this case I believe you can prove that any other ordering must give a lower $ S$; this step probably should have been included in the solution.

Oh,you're right. I have not thought about that.
So ... is my answer right?

[quote="Brut3Forc3"]AM-GM?
Let $ b \plus{} d \equal{} x, a \plus{} c \plus{} e \equal{} y$.  :wink:

Since it's cyclic and not symmetric, are we able to assume that $ a \geq b \geq c \geq d \geq e$?
I thought you could only assume one variable is the max.[/quote]

But isn't assuming $ a \geq b \geq c \geq d \geq e$ WLOG because it covers every situation?

The inequality is cyclic, not symmetric. $ a \geq c \geq b \geq d \geq e$ has implications that $ a \geq b \geq c \geq d \geq e$ does not. For instance, $ S$ has an $ ab$ term but no $ ac$ term.

What is the set of $ (a,b,c,d,e)$ that give this maximum value?

I know $ (1/3, \; 1/2, \; 1/6, \; 0, \; 0)$ does, in addition to the given one...

yes that is true.
Sorry but what does cyclic mean; I think it mean that f(f(f(...(x) = x?

Cyclic means something that resembles a circle. It means different things in different concepts.

If you are talking about a cyclic sum over an ordered list $ (a,b,c,d,e)$, for example, then:

$ \sum_{cyc} f(a,b,c,d,e)$ $ \equal{} f(a,b,c,d,e) \plus{} f(b,c,d,e,a) \plus{} f(c,d,e,a,b) \plus{} f(d,e,a,b,c) \plus{} f(e,a,b,c,d)$

[b]Examples:[/b]

$ \sum_{cyc} ab \equal{} ab \plus{} bc \plus{} cd \plus{} de \plus{} ea$, here $ f(a,b,c,d,e) \equal{} ab$

$ \sum_{cyc} abd \equal{} abd \plus{} bce \plus{} cda \plus{} deb \plus{} eac$, here $ f(a,b,c,d,e) \equal{} abd$

You cycle the elements of the list by one element, and evaluate it at each such position.

Well, this whole debate over $ a \geq b \geq c \geq d \geq e$ is moot anyway since only $ d \geq e$ is needed for the solution in the first reply to work. To answer TZF's question, the equality case there is $ a \plus{} c \plus{} e \equal{} b \plus{} d \equal{} 1/2$ and $ da \plus{} be \equal{} ea$. Given $ d \geq e$, this forces one of several possibilities which all end up as cyclic permutations of $ d \equal{} e \equal{} 0$. Then the general equality case is $ (a,1/2,1/2 \minus{} a,0,0)$ and cyclic permutations.

[quote="Brut3Forc3"]AM-GM?
Let $ b \plus{} d \equal{} x, a \plus{} c \plus{} e \equal{} y$.  :wink:

Since it's cyclic and not symmetric, are we able to assume that $ a \geq b \geq c \geq d \geq e$?
I thought you could only assume one variable is the max.[/quote]

Can anyone prove/explain why we can assume one variable is the max?

Because you can cycle the variable names until it is. This only works when the variables are subject to the same constraints.

The cyclic sum of $ f(a,b,c)$ is the cyclic sum of $ f(c,a,b)$, and the cyclic sum of $ f(b,c,a)$.

So you can just replace $ (a,b,c)$ with $ (b,c,a)$ or whatever, to ensure that $ a$ is the largest, or whichever variable you choose.


What is the probability of choosing a  black card or a queen out of a deck?

[hide="Solution"]We compute the complement of the desired probability; the probability that the card is red and not a queen. The probability of the card being red is $ \frac{1}{2}$, and the probability of the card not being a queen is $ \frac{12}{13}$, so the probability of the card being red and not a queen is $ \frac{6}{13}$, so the probability of the card being a queen or black is $ 1 \minus{} \frac{6}{13} \equal{} \boxed{\frac{7}{13}}$.

[/hide]

[hide=" click to reveal hidden content"]

Alternatively, there are $ 26$ black cards and two red queens, that is $ 28$ cards out of $ 52$, so the required probability is $ \frac{28}{52}\equal{}\frac{7}{13}$.

[/hide]

1/2 of all cards are black. 4 cards are queens. 2 cards are black queens.

1/2(52) + 4 - 2 = 26 + 2 = 28

[hide]Well since half of the deck is black and half of the deck is red, there are 26
black cards. There are 4 queens but since 2 of them are already black, you only
count the other 2 in the red. You then add 26+2=28. Then 28/52=7/13[/hide]

[quote="cgyao15"]@TheDeity you forgot to put 28/52=7/13 =D[/quote]
Wow. That was the day I was really out of wack.


In the addition problem shown, $ a, b$ and $ c$ represent three different digits. What is the three-digit sum?

$ \begin{tabular}{r}
abc\\
\plus{} acb\\
\hline
cba
\end{tabular}$

In order for the the second column to hold, we must either have $ c\equal{}0$, which is clearly impossible, or have $ c\equal{}9$ and there be a carry from the third column addition. Since we are going to have a carry to the first column, we see that $ a\equal{}4$, from which $ b\equal{}5\implies459\plus{}495\equal{}\boxed{954}$ quickly follows.

First b+c=b so c=9. Then a+a=c so a=4. Then c+b=a so b=5. So we get 954.


math club membership grew from 75 to 99 members at West Junior High.  the increase representa a growth of 25% for girls and 40% for boys.  how many girls were in the math club last year?


Asymptote
function
algebra
rational function

Find any hopes and asymptotes in the function:

$ g(x)\equal{}\frac{5x\minus{}10}{x^2\minus{}5x\plus{}6}$

[hide]The end behavior of this rational function is similar to that of $ \displaystyle\frac{1}{x}$. There is a horizontal asymptote of $ y\equal{}0$.

Vertical asymptotes are present when the denominator is equal to zero. The denominator can be factored into $ (x\minus{}2)(x\minus{}3)$. However, the numerator has a factor of $ (x\minus{}2)$ present so there is a removable discontinuity at $ x\equal{}2$ and a vertical asymptote at $ x\equal{}3$.[/hide]

[hide="Hints"]It should be painfully obvious that the top and bottom factor. What happens when you factor them? What does it mean?

How do you find vertical asymptotes that aren't holes?

The degree of the denominator is bigger than that of the numerator. What does this mean about the horizontal asymptote? (Let $ x\to\infty$).[/hide]

[hide="Solution"]Write it as $ g(x) \equal{} \frac {x \minus{} 2}{x \minus{} 2} \cdot \frac {5}{x \minus{} 3}$

We see that unless $ x \equal{} 2$, we get $ g\equiv \frac {5}{x \minus{} 3}$. So it is easy to see that there'll be a vertical asymptote at $ x \minus{} 3$ and a horizontal asymptote at $ y \equal{} 0$.

At $ x \equal{} 2$, $ g$ is not defined. But, $ g(x) \equiv \frac {5}{x \minus{} 3}$ for $ x$ just above $ 2$ and just below 2, so based on points near $ x \equal{} 2$, $ g$ "looks like" it'll take the natural value of $ \frac {5}{2 \minus{} 3} \equal{} \minus{} 5$, instead of blowing up to infinity. So we know that $ g$ has a hole at $ x \equal{} 2$.[/hide]


Hello everyone. I had a pair of questions regarding the magnetism section in my physics class. I would greatly appreciate it if somebody could aid me in these questions. Thanks!

1. A magnetic compass has its needle, of mass 0.077 kg and length 5.3 cm, aligned with the horizontal component of Earth's magnetic field at a place where that component has the value Bh = 18 µT. After the compass is given a momentary gentle shake, the needle oscillates with angular frequency  = 42 rad/s. Assuming the needle is a uniform thin rod mounted at its center, find the magnitude of its magnetic dipole moment.

2. A magnetic flux of 8.6 mWb is directed outward through the flat bottom face of the closed surface shown in Fig. 1. Along the flat top face (which has a radius of 4.0 cm) there is a 0.21 T magnetic field  directed perpendicular to the face. What is the magnitude of the magnetic flux (in Wb) through the curved portion of the surface?

Thanks again! If you also are having problems solving a question, post your link and I will try to help!

Since no magnetic monopoles exist, Gauss's law for magnetic fields states that Int(B*dA) = 0, so then you get Flux through bottom + Flux through top + Flux through curved sides = 0.  So then 8.6x10^-3 + 0.21*pi*0.04^2 + Phi = 0, where Phi is the flux through the curved sides.

Thank you very much. Can someone help me with the first problem as well please?

Well, the torque on the needle is given by mu x B= mu*B*sin[theta] (where mu is the magnetic dipole moment) which is approximately equal to mu*B*theta for small values of theta.  The moment of inertia of a uniform thin rod mounted at its center = (1/12)*m*L^2.  So, then I*alpha = -mu*B*theta, which is a special case of a 2nd order ODE. (the same as that for simple harmonic motion).  so since alpha = -w^2*theta, then w^2 = mu*B/I = 42^2, so mu = 42^2*I/B = 42^2*(1/12)*m*L^2/B = 42^2*(1/12)*0.077*0.053^2/(18 x 10^-6) which I dont feel like evaluating.

hmm, i seem to be a little confused with what you are saying...

Through some proof, torque on a magnetic dipole, mu, by a magnetic field B can be shown to be mu*B*sin[theta].  (I don't feel like deriving that formula right now, partly because I have research projects for both English and History to do this weekend, and in both classes I barely have an A)  Well, if theta is small, then sin[theta] is approximately equal to theta.  Since angular momentum =I*w (w stands for omega in this case), the rate of change of angular momentum (defined as torque) is I*a (where a stands for alpha, and a = dw/dt).  With that said, I*a = -mu*B*theta.  (The negative sign arises, because it is a restoring torque, always pulling the needle back to the equilibrium point)
Since it is simple harmonic motion, a sine or cosine function is involved.  Let's use a generic cosine function: (A stands for amplitude) theta = A*cos[w0*t].  (w0 is the angular frequency of the system (42), so unlike w, it is [i]not[/i] a variable)  so then w = d(theta)/dt = -w0*A*sin[w0*t], then a = dw/dt = -w0^2*A*cos[w0*t] = -w0^2*theta.  So then we have a = -w0^2*theta.  And from before, (after dividing by I), we get a = -(mu*B/I)*theta = -w0^2*theta.  So we can conclude that mu*B/I = w0^2, and since the needle is supposedly a uniform rod mounted at its center, its moment of inertia is given by (1/12)*m*L^2, where m is the mass and L is the length.  so then mu = w0^2*I/B = (1/12)*m*L^2*w0^2/B, and since all of these values are given to you, you can substitute them in, and the problem is solved.  Is that any clearer?  If not, tell me where specifically you get lost.

Thank you very very much. I understand what you are saying now. I really appreciate the help.

PS: Good luck on your History and English papers!!


In how many ways 5 rings of different types can be worn in 4 fingers such that no finger is vacant???

[hide="solution"]The first finger has five rings that can be on it. The second finger has four rings, the third has three rings, and the fourth has two rings. The fifth ring can be put on four fingers, so the answer is $ 5*4*3*2*4\equal{}480$[/hide]

[hide="Overcount warning"]If I put the 1st, 2nd, 3rd, 4th rings on fingers 1,2,3,4 and then the 5th ring on finger 1, or if I put the 5th, 2nd, 3rd, 4th rings on fingers 1,2,3,4 and then the 1st ring on finger 1...[/hide]

It's a bit ambiguous.  Since the rings look different, reversing the order of two rings on the same finger might produce a different result, so there would be no over-count in the previous solution.

i disagree.
if one finger has rings A and B, or ring B and A, i think it should be the same either way.

[hide="Solutions"]If order matters (if AB isn't the same as BA), there are $ \frac {5!}{(5 \minus{} 4)!}\cdot4 \equal{} 120\cdot4 \equal{} \boxed{480}$ ways.

If order doesn't matter, there are $ \binom{5}{2}\cdot4! \equal{} 10\cdot24 \equal{} \boxed{240}$ ways.[/hide]


modular arithmetic
number theory unsolved
number theory

[quote="filip"]Is it true or false (for prime $p\ge 5$ and $n\in\mathbb{N}$):
\[{np \choose p}\equiv n \pmod p^3\]?[/quote]

Yes, it's true, because
\[
\binom {np} p = \frac{(np-p+1)\cdots(np-1)(np)}{p!} = n \frac{(np-p+1)\cdots (np-1)}{(p-1)!}=
\]
\[
n^3p^2 \sum_{1\le i<j\le n}\frac {1}{ij} +n\ mod \ p^3=n \ mod \ p^3.
\]

Best regards, if you have any dude go to Combinatorics section  :P   ;)   :D .

RX the Crazy Math


Prove that $ \sum_{k \equal{} 1}^{n}{k^2} \equal{} \sum_{k \equal{} 1}^{n}{\frac {n(n \plus{} 1) \minus{} k(k \minus{} 1)}{2}}$. (Source: Myself)

[quote="sunehra"]Prove that $ \sum_{k \equal{} 1}^{n}{k^2} \equal{} \sum_{k \equal{} 1}^{n}{\frac {n(n \plus{} 1) \minus{} k(k \plus{} 1)}{2}}$. (Source: Myself)[/quote]

The RHS is
\[ \frac {n^2(n \plus{} 1)}{2} \minus{} \left[{2\choose 2} \plus{} \cdots{n \plus{} 1\choose 2}\right] \equal{} \frac {n^3 \plus{} n^2}{2} \minus{} {n \plus{} 2\choose 3}
\]
which simplifies to
\[ \frac {n(n \plus{} 1)}{6}(3n \minus{} n \minus{} 2)
\]
but the LHS is
\[ \frac {n(n \plus{} 1)(2n \plus{} 1)}{6}
\]
so its not true.  Try $ n \equal{} 1$.  Adding a $ \frac{n(n\plus{}1)}{2}$ will fix it I think.

My bad, I omitted a negative when factoring so that's what caused it to be false. Check the problem again now.

you can prove the sum of squares by induction if you wand also


I can't help but notice that all the problems being posted in this forum are immediately being answered by high school students or people who frequently wander in the high school forum.

- Please wait at least 1 day before answering these problems so that middle schoolers can have a shot at them first. After all, this is a forum for middle schoolers.

- If you have already posted a solution, it is tempting for somebody to refer to the solution if they can't immediately solve the problem.

- It's great to post problems! Don't hesitate!

Thanks for your cooperation,
4everwise

I'm not sure if this should be an announcement.. I didn't because i can't be 100% certain.

It is good to have high schoolers in this forum who post "challenge problems" to have middle schoolers to solve, but like 4everwise said, if you post in high school, these problems are not for you to answer.

Also, while there really isn't anything too bad, don't post like 30 challenge problems in a row on one day. Group them or save some for later  :ninja:

Yeah leave some problems for people like me  :P  JK  Group your problems like 236! said. ( I know, I know I don't either) I t makes it a lot easier solving a bunch at once rather than going through a bunch of threads!

I'm a high school student as well, so I try to leave the problems to others. I post problems on a weekly basis though.

Ok, this no longer has to be stickied.  ;)

it would be nice if the high schoolers could post some problems.

[quote="Drunken_Math"]Alright then... Since I meet all of the things that 236factorial posted, I'll ban myself from the basics for the rest of September...  :([/quote]

NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO !!!!!!!!!!!!!!!!!

Still contribute, just don't get greedy  :D

[quote="236factorial"]Well, I'm [b]estimating[/b] this:

2. If you've gotten higher than a 26 at Mathcounts Chapters, you're probably higher than the level of this forum.

[/quote]

are you meaning the basic forum or middle school in general? and if I've made higher than 26 on Chapter can I still post problems in the basic forum that I need help with, because for most of them, I don't really think they are mathcounts level

[quote="Drunken_Math"]Alright then... Since I meet all of the things that 236factorial posted, I'll ban myself from the basics for the rest of September... [/quote]

No no no no no! You've been contributing greatly to the basics forum. I'm just asking that you give middle schoolers a day or two to answer problems after they are posted. We appreciate your contributions.

[quote="LuCy4EvA"]are you meaning the basic forum or middle school in general? and if I've made higher than 26 on Chapter can I still post problems in the basic forum that I need help with, because for most of them, I don't really think they are mathcounts level [/quote]

I have absolutely no control over the other middle school forums. If you need help with a problem, it's fine to post it in this forum. Don't worry about it!

[quote="236factorial"]Well, I'm [b]estimating[/b] this:

1. If you're in high school, you shouldn't be the first one to answer a problem. If the problem is easy to you but no one has solved it for a day or two, feel free to leave a hint.

2. If you've gotten higher than a 26 at Mathcounts Chapters, you're probably higher than the level of this forum.

3. On the AMC 8, if you've gotten higher than a 16, this forum is also probably too easy.  IF you've gotten higher than 90 on the AMC 10, take a trip to the high school forums first.

Edit if necessary, moderators, but those are my thoughts.

Unstickied?! It should be an announcement for about a week.[/quote]
Maybe I shouldn't be posting here too much either.... I guess I'll try  posting more in the Mathcounts forum... Question: Moems  is supposed to be harder than the basics and classroom forums right?

I think that it at the same level. Perhaps a little bit harder.


let $ f$ is a function
$ f: R==>R$
$ x==>f(x)$
such that
$ (\forall x\in R) 3f(x)+2f(-x)={-5x+4\text{ if  } x\leq 3}$
$ 4x-1\text{  if } x> 3$

find $ f\circ f(x)$ for all $ x$ in $ R$


I just noticed that they just posted the [url=http://www.cms.math.ca/Competitions/CMO/]CMO solutions[/url]. Wow that took a long time.

The individual results, psh, I'll probably see that at Waterloo before it ever reaches Internet.

I am stunned to death by the simplicity of the solutions and what my mind was thinking when I was doing the problems.


Volume $ A$ equals one fourth of the sum of the volumes $ B$ and $ C$, while volume $ B$ equals one sixth of the sum of the volumes $ C$ and $ A$.
Find the ratio of the volume $ C$ to the sum of the volumes $ A$ and $ B$.

[hide]$ A \equal{} \frac{B\plus{}C}{4} \implies 4A\minus{}B \equal{} C$
$ B \equal{} \frac{C\plus{}A}{6} \implies \minus{}A\plus{}6B\equal{}C$

Let's multiply the first equation by $ m$ and the second by $ n$, and add them:

$ (4m\minus{}n)A \plus{} (6n\minus{}m) b \equal{} (m\plus{}n) C$

Let's pick $ m$ and $ n$ such that $ 4m\minus{}n \equal{} 6n\minus{}m \implies 5m \equal{} 7n$. We can use $ (m,n) \equal{} (7,5)$.

Doing so, we have $ 23A\plus{}23B \equal{} 12C \implies 23(A\plus{}B) \equal{} 12C \implies \frac{C}{A\plus{}B} \equal{} \frac{23}{12}$[/hide]


[quote="nguyenvuthanhha"][i]Prove that if $ f : \mathbb{N} \ \to \ \{1,2,3,..., 2004 \}$ satisfies :

  $ f(m \plus{} n) | f(m) \plus{} f(n) \forall \ \ m;n \ \in \ \mathbb{N}$

 then $ f$ be a periodic function .[/i]

$ \mathbb{N}$ is the set of all positive integers[/quote]

Unfortunately, this is wrong. here is a counterexample :

$ f(1)\equal{}2$
$ f(n)\equal{}1$ $ \forall n>1$

[quote="nguyenvuthanhha"][i]The problem is absolutely true

   But I think should give the demand in more details

  The function is periodic from some indices :lol: [/i][/quote]

So the problem was absolutely wrong. "Periodic from some indices" is different from "periodic'.

Maths is science of precision and rigor.

[quote="nguyenvuthanhha"][i]Prove that if $ f : \mathbb{N} \ \to \ \{1,2,3,..., 2004 \}$ satisfies :

  $ f(m\plus{}n) | f(m) \plus{} f(n) \forall \ \ m;n \ \in \ \mathbb{N}$

 then $ f$ be a periodic function .[/i]

$ \mathbb{N}$ is the set of all positive integers[/quote]


geometry
calculus
integration
analytic geometry
calculus computations

A large open cistern, filled with water has the shape of a hemisphere with radius 25 ft.  The bowl has a circular hole of radius 1 ft. in the bottom.  By Torricelli's Law, water will flow out of the hole with the same speed it would attain in falling freely from the level of the water to the hole.  How long will it take for all the water to flow out of the cistern?

Here's what I have got...   The answer key says[color=blue] (2916 2/3) / (2g)^(1/2) [/color]

Let V be the volume of the water at any time t and B be the area of the small hole with radius 1 ft.  Also, dV = A dh where A is the cross sectional area A of the hemisphere with radius 25 ft. and h is the depth of the cistern.  If k is to be a constant, my working, starting equation is

A dh = - k B  :sqrt: (2gh)  [color=blue]-> did I use the Torricelli's Law correctly?[/color]

A = pi 25^2 = 625 pi.
B = pi

625 (pi) dh = - k (pi)  :sqrt: (2gh) dt

By the way, does my k have a specific value?  The book didn't say so.

Upon integration, I got

625 (pi) :sqrt: (2gh) /g = -k (pi)t + C

From here, I have to find C by letting h = 25 and t = 0, right?

And I got C = 3125 (pi) :sqrt: (2g) / g

Lastly, at t = ?, h = 0.  So upon substitution..

I got t = 3125(2) /  [(0.6):sqrt: (2g)]

Where did I get wrong?

[quote="justmath11"]
A dh = - k B  :sqrt: (2gh)  [color=blue]-> did I use the Torricelli's Law correctly?[/color]

A = pi 25^2 = 625 pi.
B = pi
...

By the way, does my k have a specific value?  The book didn't say so.
[/quote]

Not entirely. The tank is a hemisphere, which means that the surface area $A$ at the current water level $h$ is not constant, but depends on $h$. In addition, the left side of the equation contains a differential and the right side does not (probably a typo).

For an ideal liquid, Torricelli's law is $v = -k \sqrt h$ and the value of k is $k = \sqrt{2g}$. For a real liquid, some energy is lost due to viscosity.(It would take longer for a tank of honey to get empty than for a tank of water.) Hence, Torricelli's law is $v = -k \sqrt h = -\alpha \sqrt{2gh}$, where $0 < \alpha < 1$ is a dimensionless constant depending on the kind of liquid. Let's forget about $\alpha$.

Put the origin of cylindrical coordinates $(r, \phi, h)$ at the bottom of the hemispherical tank. The tank vertical cross-section is a circle with an equation

$(h - R)^2 + r^2 = R^2$
$r^2 = 2Rh - h^2$

where $R = 25\ ft$ is the tank radius. The surface area $A$ at the current water level is

$A = \pi r^2 = \pi (2Rh - h^2)$

and the area of the drain hole with radius $r_0$ is

$A_0 = \pi r_0^2$

Radius of the drain cannot possibly be 1 ft, with a hole like that, the tank would be empty in no time. I assume that $r_0  = 1\ inch = \frac{1}{12}\ ft$. By Torricelli's law,

$dV = A\ dh = -A_0\ v\ dt = -A_0 \sqrt{2gh}\ dt$

$(2Rh - h^2)\ dh = -r_0^2 \sqrt{2gh}\ dt$

$\frac {dh}{dt} =  -{r_0^2}\frac{\sqrt{2gh}}{2Rh - h^2}$

$\frac {dt}{dh} =  -\frac{1}{r_0^2}\frac{2Rh - h^2}{\sqrt{2gh}}$

$t = -\frac{1}{r_0^2} \int{\frac{2Rh - h^2}{\sqrt{2gh}} dh}$

Substituting $h = x^2, dh = 2xdx = 2\sqrt h\ dx$,

$t = -\frac{2}{r_0^2 \sqrt{2g}} \int{(2Rx^2 - x^4)\ dx} = -\frac{2}{r_0^2 \sqrt{2g}} \left( \frac{2Rx^3}{3} - \frac{x^5}{5} \right) + C =$

$= -\frac{2}{r_0^2 \sqrt{2gh}} \left( \frac{2Rh^2}{3} - \frac{h^3}{5} \right) + C$

By the initial condition, the tank is full ($h = R$) at time $t = 0$:

$0 = -\frac{2}{r_0^2 \sqrt{2gR}} \left( \frac{2R^3}{3} - \frac{R^3}{5} \right) + C$, $C = \frac{14R^3}{15r_0^2 \sqrt{2gR}}$

When $h = 0$, $t = C$, i.e.,

$t = \frac{14R^3}{15r_0^2 \sqrt{2gR}} = \frac{14\cdot 125 \cdot 144}{15 \cdot 5 \sqrt{2g}} = \frac{3360}{\sqrt{2g}}$

I did not get the key answer $\frac {8750}{3\sqrt{2g}}$ either, but I do not know the radius of the drain hole for sure and the key may be incorrect.

Hello yetti.

  Thank you for the help...

  [quote]I did not get the key answer  either, but I do not know the radius of the drain hole for sure and the key may be incorrect. [/quote]

  Actually, you're on the right track!  I have observed that you have assumed the radius of the small circular hole to be 1 inch and not 1 foot.  So that's one reason. [:D]

  On the final solution, when h = 0, find t (which is C)..  I was thinking that R must be 25 and not 5.

  If these were modified in the solution... you will get the answer... [:D]

  Again, thank you for helping me.

Hi, justmath11,

[quote="justmath11"]Hello yetti.
  Actually, you're on the right track!  I have observed that you have assumed the radius of the small circular hole to be 1 inch and not 1 foot.  So that's one reason. [:D]

  On the final solution, when h = 0, find t (which is C)..  I was thinking that R must be 25 and not 5.

  If these were modified in the solution... you will get the answer... [:D]

  Again, thank you for helping me.[/quote]

Right. So, the radius $r_0$ of the drain is 1 ft after all. That's a big leak, I must say. And if I substitute correctly $R^3 = 25^3$ into the final formula for time, then I get the key answer:

$t = \frac{14R^3}{15r_0^2 \sqrt{2gR}} = \frac{14 \cdot 25^3}{15 \cdot 5 \sqrt{2g}} = \frac {8750}{3\sqrt{2g}}$

Darn, I make this kind of mistakes all the time.

yetti

Hello yetti...

  It's just fine... :D, I mean... at least you know how to start formulating or doing those expressions... well, I do make mistakes and carelessness like that during the exams.

  Again, thank you for your help


trigonometry
limit
calculus
complex numbers
complex analysis

When, on complex $ z$, does the series converge:

$ \displaystyle\sum_{n\equal{}1}^{\infty}a_n$,  where $ a_n\equal{}\displaystyle\frac{1}{1\plus{}z^n}$.

You must have $ a_n\to 0$ thus $ z^n\to \infty$, hence $ |z| > 1$. For such $ z$ we have $ |\frac {1}{z^n \plus{} 1}| < \frac {2}{|z^n|}$ for all $ n > N_0$ and hence our series is converging

Yes, but $ a_n \to 0$ is not sufficient; consider $ \sum \frac{1}{n}$.

leshik explicitly gave a comparison to a convergent geometric series. It is sufficient.

In trying to better understand this solution, I've hit a snag in my thinking. I'm pretty sure I'm committing the sin of thinking of $ z$ as the sum of two parts instead of its own animal; consequently this may be the least elegant solution imaginable. I'll show my thinking, and my question is: how can I better think of $ z$ in forming this solution?

We want to see when the sum converges. We'll consider the three cases: $ |z| < 1, = 1, > 1$.

Consider $ |z| > 1$. leshik claims that in this case, $ |\frac {1}{z^{n} + 1}| < \frac {2}{|z^{n}|}$ for all $ n > N_0$. This must be proven.

Let $ z = r(\cos\theta + i\sin\theta)$. The usual rule for powers of complex numbers yields $ z^n = r^n(\cos n\theta + i\sin n\theta)$. Since $ |z| > 1$, we have $ |r| > 1$.

So the claim is: $ \exists N_0$ such that $ n > N_0$ implies $ \sqrt {(1 + r^n\cos n\theta)^2 + (r^n\sin\theta)^2} > \frac {\sqrt {2}}{2}r^n$, i.e. $ \sqrt {1 + 2r^n\cos n\theta + r^{2n}} > \frac {\sqrt {2}}{2}r^n$, i.e. $ 1 + 2r^n\cos n\theta + r^{2n} > \frac {1}{2}r^{2n}$.

Proof: ${ 1 + 2r^n\cos n\theta + r^{2n}} \ge 1 - 2r^n + r^{2n} \ge r^n(r^n - 2) > (\frac {r^n}{\sqrt {2}})(\frac {r^n}{\sqrt {2}})$ for sufficiently large $ n$. Hence $ \sum a_n$ converges for $ |z| > 1$.

Consider $ |z| = 1$. Then $ \frac {1}{1 + z^n} = \frac {1}{1 + \cos n\theta + i\sin n\theta}$, and

$ |a_n| = |\frac {1 + \cos n\theta - i\sin n\theta}{(\cos n\theta + 1)^2 + (\sin n\theta)^2}| = \frac {1}{\sqrt {(\cos n\theta + 1)^2 + (\sin n\theta)^2}}$

$ = \frac {1}{\sqrt{2 + 2\cos n\theta}} \ge \frac {1}{2}$

for all $ n$. Hence $ \sum a_n$ diverges for all such $ z$.

Consider $ |z| < 1$. Then $ z^n \to 0$, so $ \sum a_n$ diverges for all such $ z$ (by similar reasoning as the first case.)

Is there a more elegant proof?

Lesson in how to think like an analyst: lose the habit of thinking that you need to solve everything exactly. Relax, and accept cheap estimates.

For $ |z|>1,$ we have $ |z^n|\to\infty,$ hence $ \frac1{1\plus{}|z^2|}\to 0.$ But how fast? that $ 1$ in the denominator is unimportant, so this looks like $ \frac1{|z^n|}$ for large $ n,$ so we can make that comparison.

That already suffices as a quote of the limit comparison test for series convergence - put perhaps you want to be faintly more formal than that. All right (motivated by what we're already thinking),

$ \lim_{n\to\infty}\frac{|z^n|}{1\plus{}|z^n|}\equal{}\lim_{n\to\infty}\frac1{\frac1{|z^n|}\plus{}1}\equal{}\frac1{0\plus{}1}\equal{}1.$

Since the limit of this sequence is $ 1,$ then it eventually becomes less than $ 2.$ (That's just putting $ \epsilon\equal{}1$ into the definition of the limit of a sequence.) That is, there exists (and we will not solve for it explicitly because we don't need to) a number $ N_0$ such that for $ n>N_0,$ we have

$ \frac{|z^n|}{1\plus{}|z^n|}<2,$ which can be rearranged to $ \frac1{1\plus{}|z^n|}<\frac2{|z^n|}$

That last line, of course, is precisely what leshik wrote.

I'm guessing that my whole argument is a pretty good mirror of what was in leshik's head, at least at some informal level. I feel pretty confident about that even though I don't know anything about leshik - because that's how one thinks when doing calculus or analysis.

As for why the series diverges for $ |z|\le 1,$ its because the terms of the series don't go to zero. How small can $ \frac1{1\plus{}|z^n|}$ be? Not small, unless $ |z^n|$ is big. To be specific, if $ |z|\le1,$ then $ |z^n|\le1$ and $ 1\plus{}|z^n|\le 2,$ which implies that $ \frac1{1\plus{}|z^n|}\ge\frac12.$ The terms don't go to zero.

Good, but your whole solution rests on the notion that $ 1\plus{}|z^n|$ always tends the same way as $ |1\plus{}z^n|$. Looking at why this must be true...

If $ |z^n|>1$, then $ r^n\to\infty$, so both components have large magnitude except where each $ \equal{}0$, so at least one of the two components must always have large magnitude always, so the $ 1$ doesn't matter.

If $ |z^n|\equal{}1$ always, then $ |1\plus{}z^n|$ can be bounded on the low end by analyzing components, so $ |z^n|\not\to 0$.

If $ |z^n|<1$, same thing with $ r^n$ shows $ |1\plus{}z^n|\to 1$.

Ok now I believe you, but I still had to examine components, which I was trying to avoid... but I guess complex numbers have to have two components at some point. Otherwise, I'm not sure how it's obvious that $ 1\plus{}|z^n|$ always tends the same way as $ |1\plus{}z^n|$, and (I think?) your/leshik's solution does not follow without that notion, since the original question dealt with $ 1/(1\plus{}z^n)$.

What do you think?

OK, sure. How about this argument: $ \frac{z^n}{1\plus{}z^n}\to 1,$ so eventually it gets within $ 1$ of $ 1,$ and a circle of radius $ 1$ centered at $ 1$ lies inside a circle of radius $ 2$ centered at $ 0.$ So eventually, $ \left|\frac{z^n}{1\plus{}z^n}\right|\le 2,$ which you can rearrange to $ \left|\frac1{1\plus{}z^n}\right|\le\frac2{|z^n|}.$


group theory
abstract algebra
superior algebra
superior algebra solved

G is the group and p is a smallest prime divisor of /G/.H is a subgroup of G have index p.
Prove that:H is a normal subgroup of G.


Two chords in a circle share a common endpoint and they divide the circle into three parts of equal area.  Find the angle between the chords.

[quote="mysmartmouth"][hide="Solution"]
Let the measure of the angle in degrees be $ \theta$.  Let the radius of the circle have length $ r$.  The chords will have equal length, by symmetry.  So each chord is the radius of a bigger circle with radius $ 2r$, centered at the common endpoint of the two chords.  Then:

$ \frac {\theta}{360}(\pi (2r)^2) \equal{} \frac {1}{3}(\pi r^2)$.

Solving gives $ \theta \equal{} 30^\circ$
[/hide][/quote]

I don't see how you are getting $ 2r$ for the length of the chord.
I don't see a way to find the exact value for $ \theta$, but only numerical approximation.  I'll post my solution later.

[hide="My answer"]$ \theta$ satisfies $ \theta \plus{} \sin \theta \minus{} \pi/3 \equal{} 0$[/hide]

I got the same as the above.

[hide]Basically, let the chords be AB and BC, and let the center of the circle be O. Let A'  be the point opposite A. If the measure of ABC is x, then the measure of ABA' is also x. We thus have that $ 2 \left( \frac {1}{2} r^2 \sin x \plus{} \frac {x}{2 \pi} \pi r^2 \right) \equal{} \frac {1}{3} \pi r^2 \iff x \plus{} \sin x \equal{} \frac {\pi}{3}$

The solution to this is about $ x \approx 0.5362669789888905$, or in degrees, $ x \approx 30.725834588294227^{\circ}$
[/hide]

Yeah, you guys are right.  My mistake...

Here's a similar one:

A chord divides a circle into parts of area 1 and area 2. How long is the chord?


inequalities
trigonometry
inequalities unsolved

For an acute triangle \delta ABC,
show that
$ tan^2\frac{A}{2}\plus{}tan^2\frac{B}{2}\plus{}tan^2\frac{C}{2}\le 8\frac{sin^2\frac{A}{2}}{cosA}\frac{sin^2\frac{B}{2}}{cosB}\frac{sin^2\frac{C}{2}}{cosC}$.

[quote="miwalin"]For an acute triangle \delta ABC,
show that
$ tan^2\frac {A}{2} \plus{} tan^2\frac {B}{2} \plus{} tan^2\frac {C}{2}\le 8\frac {sin^2\frac {A}{2}}{cosA}\frac {sin^2\frac {B}{2}}{cosB}\frac {sin^2\frac {C}{2}}{cosC}$.[/quote]

1. $ \tan^2(\frac {\alpha}{2}) \plus{} \tan^2(\frac {\beta}{2}) \plus{} \tan^2(\frac {\gamma}{2}) \equal{} (\frac {4R \plus{} r}{s})^2 \minus{} 2$

2. $ \frac {\sin^2(\frac {\alpha}{2}) \sin^2(\frac {\beta}{2}) \sin^2(\frac {\gamma}{2} ) }{\cos\alpha \cos\beta \cos\gamma} \equal{} \frac {r^2}{4(s^2 \minus{} (2R \plus{} r)^2)}$ , for acute angled triangle : $ s\ge 2R \plus{} r$

$ \Leftrightarrow (\frac {4R \plus{} r}{s})^2 \minus{} 2 \le \frac {2r^2}{s^2 \minus{} (2R \plus{} r)^2}$

$ \Leftrightarrow f(t) \equal{} 2t^2 \minus{} (24R^2 \plus{} 16Rr \plus{} r^2)t \plus{} (4R \plus{} r)^2(2R \plus{} r)^2 \ge 0$ , $ t \equal{} s^2$

$ \boxed{(2R \plus{} r)^2 \le t \le 2R^2 \plus{} 10Rr \minus{} r^2 \plus{} 2(R \minus{} 2r) \sqrt {R(R \minus{} 2r)} \equal{} t_1}$

So we only need to prove :
$ f(t_1) \ge 0 \Leftrightarrow 2(z \minus{} 2)( 16z^3 \minus{} 40z^2 \plus{} 21z \minus{} 1 \minus{} \sqrt {z(z \minus{} 2)}( 16z^2 \minus{} 24z \plus{} 5 ) ) \ge 0$    - which is true for $ (z \equal{} \frac {R}{r} \ge 2)$


If n!! is defined as (n)!(n-1)!(n-2)!(n-3)!(n-4)!............(1)! , then what is the greatest value of x such that $2^x$ is a factor of 11!! ?

(I know I've seen this notation before, maybe even called "super factorial", but not sure if it was a real thing or if it was made up which is what I did for this problem :)  )

Edited later:

This is different from the above definition   but "Double Factorial" in standard terms is defined a little differently. If interested Please read the hidded text.  (For the  above problem,  following is a little bit off topic)

 [hide]Yes, Double Factorial  (!!) is  a fairly standard  term and is defined as:

$n !! =  n(n-2)(n-4) \dots 5 \cdot 3 \cdot 1    $  for $ n >0 $ and for  odd  $n$

       $ n !!= n(n-2)(n-4) \dots 6 \cdot 4 \cdot 2 $ for  $ n>0 $ and for   even $n $
        $n!! = 1  $ for $ n = 0, -1 $

Common practice is  even to  exteded for odd negative values of  $n$

and is defined as  $(-2n -1)!! = \frac {(-1)^n}{(2n-1)!!}$

(For negative even values of n the function is  not defined     )

Of course,  for fun  one  can notice
$(2n)!! = (2^n) n!$
and $(2n+1)!! = \frac {(2n+1)!}{(2^n) n!}$[/hide]

Thanks Gyan,

So not with standing the actual definition of n!!, try the problem as defined by me and ignore my blatant attempt at rewriting definitions :D

[hide="Answer and Solution"]This seems like a sprint-like problem, although it could be a target. Either way, I found how many twos there would be in 11; 8, and counted down from there. It sometimes helps to list the factor of 2 nd how many are in each (ex: there are three in 8) although that really isn't necessary. Either way, that led me to:
$11\Rightarrow 8$
$10\Rightarrow 8$
$9\Rightarrow 7$
$8\Rightarrow 7$
$7\Rightarrow 4$
$6\Rightarrow 4$
$5\Rightarrow 3$
$4\Rightarrow 3$
$3\Rightarrow 1$
$2\Rightarrow 1$
(assuming that each number was a factorial)Also if you were smart, you would quickly notice the pattern of doubling (it makes sense with odd numbers), so you could do $2(8+7+4+3+1)=2(23)=46$ Therefore $46$ was my answer[/hide]

hockey is using my made up definition of n!!

Using the fake definition given in the original problem hockey has it right.  Using the real definition Gyan offered, there is clearly a MUCH different answer.

[quote="Gyan"]hockeyadi23

You may like to check your answer.
 [hide="Hint"]Is 11!! even [hide="Guess What? "][b]even[/b][/hide]?[/hide][/quote]

Based on the problem, my answer is correct, however if the problem asked for the value of 2^n if it was 11!! as you defined it, the answer would be zero... I think there are two interpretations to this problem, and like on SATs, you should use what the problem gives you.

Please ignore my previous message .. (I deleted it)   I did not read the question too carefully..
Sorry for the confusion and thanks for pointing  it out. (hockeyadi23's - and others -   solution is/are  correct)


It looks like it should be saying $\sum \left( \sqrt[3]{\frac{x}{y} +1}\right)\le\frac{(x+y)(y+z)(z+x)}{xyz}$ - he just typed \sqrt as \sqtrt.


limit
logarithms
number theory open
number theory

$ p_n$ is the n-th prime.

does the limis exist?
\[ \lim_{n\to\infty}\frac{p_{n\plus{}1}\minus{}p_n}{\ln p_n}\]
 :?:

how do you know it  / where it can be found ?

it was on the forum?
I searched for it, but I couldn't find :(

[quote="Rust"]Proved, does not.[/quote]
Are you sure? Using the prime number theorem, one has to compute
\[ \lim_{x\to\infty}\frac{(x\plus{}1)/\ln(x\plus{}1)\minus{}x/\ln(x)}{\ln (x/\ln(x))}.\]
Maple outputs that the limit is 0.

$ p_n\equal{}nln(n)\plus{}o(n)$ is not $ \pi (x)$.


[quote="Dojo"]What is the sum of the two smallest prime factors of 250?

$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$[/quote]

[hide]The first smallest prime factor is $ 2$, which we see it is also a prime factor of $ 250$.  The next highest prime factor, $ 3$, we see, is NOT a prime factor of $ 250$.  We see the next highest, $ 5$, is one, so the answer is $ 2 \plus{} 5 \equal{} \boxed{\textbf{(C)}\ 7 }$[/hide]

[hide="Solution"]$ (C)$ The prime factorization of 250 is $ 2\cdot 5\cdot 5\cdot 5$. The sum of 2 and 5 is 7.[/hide]


zhaobin, it's Schur for $a^{2}b,$ $b^{2}c$ and $c^{2}a.$

:blush: Oh,Yes,this solution is nicer...Sorry,I make a foolish inequality. :blush:


Three dice are thrown. probability that one dice shows even faces while the others show prime nos.

Well, the probability of there being an even number on the face of one of them is 1/2, and the probability of it being a prime number for each of the other two is 1/2, so it will multiply to be 1/8. I think...


If im planning on taking the amc 10 and want to score enough for aime, which class will help me more, the amc 10 or 12 class? Also, i want to do well on aime and am definitely signed up for that class too. my weaknesses are geometry and combinatorics but im afraid i wont have enough time to do those classes because i dont want to overload myself. (too many ap classses)

The main differences between the two classes are that (a) the AMC 12 class has slightly harder problems, and (b) the AMC 12 class covers material that you won't find on the AMC 10 (like trig/complex numbers).

The AMC 10 class is already in its 5th week, but the subjects that you mentioned -- counting and geometry -- are both covered in the second half of the course.

My suggestion would be that if you're 100% sure that you're at the point on the AMC 10 that you can certainly qualify for the AIME, then take the AMC 12 class; otherwise the AMC 10 class may be better.


linear algebra
matrix
invariant
Gauss
search

let A is square matric(n*n) .And Trace(A)=0.
PROVE  A similar D .  D is square matrix with  dii =0 ;  i:=1..n

Just consider that the trace is invariant by similitude...  now by gauss method put $ a_{ii}\equal{}0$ for i = 1,..., n-1. Then $ a_{nn}$ should be 0.

You should specify a field- this is false in finite characteristic, where the identity can have trace zero.

- pic: What do you mean by "gauss method"? Gaussian elimination definitely doesn't work.

Well, I meant Gaussian elimination, but now I realize that it doesn't work. :roll:

Then search for it- because it has already been posted in some other thread. Stating the problem accurately helps, too.


[i]A student was asked to calculate the final temperature of a mixture prepared by adding 20.0 g of ice, initially at -10.0 C, to 20.0 g of water, initally at 25 C. She assumed that all heat lost in cooling the water would be used to warm the ice and concluded that the final temperature of the mixture would be 13 C. However, the correct answer is 0 C. What did the student neglect in her considerations?[/i]

Can someone give a method to "solve" this type of problem (meaning, explain how the student got the answer, and how the correct answer is obtained). Does this require solving a linear system for some x degrees of change in temp.? [/i]

[quote="me@home"]She assumed that all heat lost in cooling the water would be used to warm the ice[/quote]

This is her mistake.  For a given mass (or, equivalently, a given number of mols) of ice, there is a corresponding quantity of thermal energy needed to make the phase transition that [b]does not[/b] contribute to increasing the temperature of the ice.  This is known as the "heat of liquefaction," and it is a constant for a given substance.

Problems of this type require you to take away some amount of thermal energy $E$ from the water and add it to the ice so that their resulting temperatures are equal.  I believe you do end up with a linear system with $E$ and $\Delta T$, but it's been awhile since I've done this kind of annoying thermochem  :D

Actually, about 5 min after I posted, I saw the answer. Also, I didn't post this but the main issue of concern was that the solution simply said, "She forgot to include the phase change" (or something along those lines). The solution actually uses basic algebra (that's why I didn't expect it, it was a challenge problem). [hide="Here is the solution"]
[b]Student's Method:[/b]
\[(20.0g)\left(4.18\frac{J}{g\times C^{\circ}}\right)(\Delta T_{water})=q_{water}=-q_{ice}=(20.0g)\left(2.09\frac{J}{g\times C^{\circ}}\right)(\Delta T_{ice})\implies 2\Delta T_{water}=-\Delta T_{ice}\] is simply the assumption that the heat of cooling the water warms the ice.
Now for the algebra: \[T_{ice \ initial}+\Delta T_{ice}=\Delta T_{water}+T_{water \ initial}\\-35.0^{\circ}=3(\Delta T_{water})\implies \Delta T_{water}=-11.7^{\circ}\implies T_{solution}=25.0^{\circ}-11.7^{\circ}=13.3^{\circ}\]

[b]Correct Method:[/b]
First, assume the ice is melted at some point:
then our equation is $q_{water}=q_{ice \ to \ 0^{\circ}}+mol \ H_{2}O\times 6.02 \frac{kJ}{mol}+q_{ice \ 0^{\circ}\ to \ T_{ice}}$ or $(1.1)(6.02)(1000)+10(20)(2.09)+(\Delta T_{ice})(2.08)(20)=-(20)(4.18)(\Delta T_{water})$ solving with $\Delta T_{water}+\Delta T_{ice}=25$ gives $65=\Delta T_{water}$ which is absurd (the energy evolved by cooling the water isn't enough to overcome the great energy of melting the ice). Thus the ice doesn't melt and stays at $\boxed{0^{\circ}}$
[/hide]

[quote="me@home"]Also, I didn't post this but the main issue of concern was that the solution simply said, "She forgot to include the phase change" (or something along those lines). The solution actually uses basic algebra (that's why I didn't expect it, it was a challenge problem). [/quote]

Chemistry isn't supposed to be mathematically difficult  :D

(Although once you get to kinetics, if you know some calculus, you can derive the rate laws yourself.  That's about as hard as it gets.)

[quote="t0rajir0u"][quote="me@home"]
Chemistry isn't supposed to be mathematically difficult  :D

(Although once you get to kinetics, if you know some calculus, you can derive the rate laws yourself.  That's about as hard as it gets.)[/quote][/quote]

Take some p-chem, and then tell me that  :P.  You're right though, general chem usually is not very mathematically challenging.


At his track meet, Bob ran a $ 400\ \text{m}$ race in exactly $ 50$ seconds. What was his average speed, in meters per second, during the race?

His average speed in meters per second is simply $ 400 \div 50 \equal{} \boxed{8}$.


invariant
geometry
geometric transformation
rotation
reflection
group theory
abstract algebra

I'm absolutely not convinced that this is original or correct or..., but I think it's quite nice and deserves some discussion. I hesitated before posting it in the analysis forum, but knowing what is happening on the linear algebra forum (where a virus, worse than Robertux, is trying to invade all the given space...) and that my solution fully uses analysis, I think I'll post it here:
   Let $ K$ be a convex compact set of $ R^3$, with nonempty interior.  Suppose that there are infinitely many invertible linear operators leaving $ K$ invariant. Then there is an inner product on $ R^3$ such that $ K$ has a rotation axis for this inner product.

The group $ G$ which fixes $ K$ (as a set, not pointwise) is closed and bounded, hence compact. After conjugation with some invertible linear transformation, we can assume it's orthogonal. After this modification the problem is reduced to showing that some rotation around a line fixes $ K$. We can even show that $ K$ is fixed by the reflection in a line through the origin (i.e. a rotation of $ 180^\circ$ around a line).

We know $ G$ is infinite, so it must be a compact Lie subgroup of $ O(3,\mathbb R)$ of dimension $ \ge 1$. As such, it contains a torus of dimension $ \ge 1$, so, in particular, its identity component contains an element of order $ 2$. This element must be a reflection with respect to a line through the origin, because it belongs to the identity component of $ G$ and hence to $ SO(3,\mathbb R)$ (the only possibilities are: reflection through a point, a line, or a plane; the first and the last ones have determinant $ \minus{} 1$).

That pretty much what I did, just that I used the easy fact that the Lie algebra is not zero, so one gets all matrices of the form $ e^{ta}$ for some nontrivial $ a$ and so all rotations having the same axis as $ a$.


[quote="mommasally"]Add and simplify: 18/72 + 19/72 + 18/72[/quote]
eh.. just add the numerators and leave the denominator alone.

Yeah, this is just simple fraction concept. The problem has already been solved quite easily.

mods plz lock this


algebra
polynomial
algorithm
superior algebra
superior algebra unsolved

I'm going to prove that if $ m\mid n$ then $ \mathbb{F}_{p^m}$ is a subfield of $ \mathbb{F}_{p^n}$. One need to prove that the splitting field of $ x^{p^n} \minus{} x$ contains the splitting field of $ x^{p^m} \minus{} x$ or equivalently $ x^{p^m} \minus{} x\mid x^{p^n} \minus{} x$. But I'm a little bit lost that in where this polynomial divisibility should hold? In $ \mathbb{F}_p[x]$ or in $ \mathbb{F}_{p^m}[x]$ or where?

edit: relation means polynomial divisibility.

I'm assuming that when you say "this relation," you are referring to the polynomial divisibility question, but it's a little unclear.

Since the division algorithm involves only rational operations, if $ F \subset K$ and $ p(x) | q(x)$ in $ K[x]$, then $ p(x) | q(x)$ in $ F[x]$.

So actually, the statements that the divisibility holds in $ F_p[x]$, or $ F_{p^m}[x]$, or $ F_{p^n}[x]$, are all equivalent.

The divisibility is even stronger, though; it holds in $ \mathbb{Z}[x]$.  But you only need to show it for $ F_p[x]$.

[quote="LydianRain"]if $ F \subset K$ and $ p(x) | q(x)$ in $ K[x]$, then $ p(x) | q(x)$ in $ F[x]$.[/quote]

Sorry, this was slightly sloppy language.  I mean to specify that $ p$ and $ q$ are polynomials with coefficients in $ F$.

[quote="LydianRain"]So actually, the statements that the divisibility holds in $ F_p[x]$, or $ F_{p^m}[x]$, or $ F_{p^n}[x]$, are all equivalent.[/quote]
Okay. I'm a little bit lost in my proof. The professor said that the following is not a valid proof for theorem 1:

Lemma 1: Let $ K$ be a field, $ m,n\in\mathbb{N}$, $ p$ a prime and $ x^{p^m} \minus{} 1,x^{p^n} \minus{} 1\in K[x]$. Then $ x^{p^m} \minus{} 1\mid x^{p^n} \minus{} 1$ iff $ m\mid n$.

Theorem 1: If $ p$ is a prime and $ m,n\in\mathbb{N}$ with $ m\mid n$, then $ \mathbb{F}_{p^m}$ is a subfield of $ \mathbb{F}_{p^n}$.

Proof: As $ m\mid n$, it follows from the lemma 1 that in $ \mathbb{F}_p[x]$ we have $ x^{p^m} \minus{} 1\mid x^{p^n} \minus{} 1$. Thus the splitting field of $ x^{p^n} \minus{} x$ contains the splitting field of $ x^{p^m} \minus{} x$ so $ \mathbb{F}_{p^m}$ is a subfield of $ \mathbb{F}_{p^n}$.

He said that $ \mathbb{F}_p[x]$ is a typo but if it's not, how can I write clearer proof?

Edit. Oops. There is a mistake. $ \mathbb{F}_p[x]$ should be $ \mathbb{F}_{p^n}[x]$. http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=1783.0001 typos fixes.

You considered the polynomial $ x^{p^n}\minus{}1$ and it's splitting field. But the correct polynomial should be $ x^{p^n}\minus{}x$.

Thank you Zetax. In lemma there should be $ \minus{} 1$ but in theorem there should be $ \minus{} x$.


quadratics
number theory proposed
number theory

Prove that equation $17^{x}=y^{2}+5$ hasn't solutions in positive integers.

[quote="Rust"]$y^{2}=-5+17^{x}\Longrightarrow (\frac{-5}{17})=1$. It give contadition, because $(\frac{-5}{17})=(\frac{17}{5})=(\frac{2}{5})=-1$.[/quote]

Please, can you explain your proof step by step? It's a bit messy. Thanks.

I don't get it either, José, but I guess it's the Jacobi symbol?

But here's my solution:

$y^{2}+5 = 17^{x}\equiv 1\mod{8}$, so $y^{2}\equiv 4\mod{8}$, so $y\equiv 2\mod{4}$. Let $y=2l$ with $l$ odd. Then $17^{x}= 4l^{2}+5$, so $4l^{2}\equiv 12\mod{17}$, so $l^{2}\equiv 3\mod{17}$, which is impossible, because $n^{2}\in\{1,2,4,8,9,13,15,16\}$ for all integers $n$.

Legendre symbol, not Jacobi.

His solution is just saying that -5 is not a square mod 17. Which is trivial with the theory of quadratic residues.

He still needed to check the case $x = 0$, but that's trivial as well.


geometry
3D geometry
AMC
AIME
search
geometric transformation
rotation

A right circular cone has base radius $ r$ and height $ h$.  The cone lies on its side on a flat table.  As the cone rolls on the surface of the table without slipping, the point where the cone's base meets the table traces a circular arc centered  at the point where the vertex touches the table.  The cone first returns to its original position on the table after making $ 17$ complete rotations.  The value of $ h/r$ can be written in the form $ m\sqrt {n}$, where $ m$ and $ n$ are positive integers and $ n$ is not divisible by the square of any prime.  Find $ m \plus{} n$.

[hide] The path is a circle with radius equal to the slant height of the cone, which is $ \sqrt {r^{2} \plus{} h^{2}}$. Thus, the length of the path is $ 2\pi\sqrt {r^{2} \plus{} h^{2}}$.

Also, the length of the path is 17 times the circumference of the base, which is $ 34r\pi$. Setting these equal gives $ \sqrt {r^{2} \plus{} h^{2}} \equal{} 17r$, or $ h^{2} \equal{} 288r^{2}$. Thus, $ \dfrac{h^{2}}{r^{2}} \equal{} 288$, and $ \dfrac{h}{r} \equal{} 12\sqrt {2}$, giving an answer of $ 12 \plus{} 2 \equal{} \boxed{014}$.[/hide]

yeah it definately is $ 12\sqrt{2}$, which gives 014 as the answer...

When you post these, can you call them AIME 2008I in the source description?

[quote="E^(pi*i)=-1"]Whoops, sorry about that.  I'll change them.[/quote]
Thanks, it just makes them much easier to search for - AIME 2008I is a better search term than AIME I 2008 (which will also turn up results for the AIME II, I think).

[quote="CatalystOfNostalgia"][hide] The path is a circle with radius equal to the slant height of the cone, which is $ \sqrt {r^{2} \plus{} h^{2}}$. Thus, the length of the path is $ 2\pi\sqrt {r^{2} \plus{} h^{2}}$.

Also, the length of the path is 17 times the circumference of the base, which is $ 34r\pi$. Setting these equal gives $ \sqrt {r^{2} \plus{} h^{2}} \equal{} 17r$, or $ h^{2} \equal{} 288r^{2}$. Thus, $ \dfrac{r^{2}}{h^{2}} \equal{} 288$, and $ \dfrac{r}{h} \equal{} 14\sqrt {2}$, giving an answer of $ 14 \plus{} 2 \equal{} \boxed{016}$.[/hide][/quote]
The cone's path is actually an annulus I think. Could be wrong.

[quote="t0rajir0u"][quote="worthawholebean"]The cone's path is actually an annulus I think. Could be wrong.[/quote]

An annulus isn't a path.  The point where the cone's base meets the table just traces the circumference of a circle, and the actual cone itself sweeps out the entire circle.[/quote]
Well, actually, the point where the cone's base meets the table traces a circle, and the cone traces the region bounded by a circle.  (A circle is a set of points in a plane equidistant from some point in the plane.)  But yeah, I don't see how you get an annulus, unless you mean a degenerate annulus, i.e., a region bounded by a circle.

I seem to have conflated "circle" and "disc."  My apologies  :oops:

Sorry if this is a dumb question, but maybe I just read the problem wrong. How do we know that the cone only made one trip around the table? How do we know those 17 rotations weren't spread out over the course of multiple trips around the table?

The problem says that it [b]first[/b] returns to its original position on the table after making 17 complete rotations. So this is at the end of just one trip around the table.

[quote="CatalystOfNostalgia"]

Also, the length of the path is 17 times the circumference of the base, which is $ 34r\pi$. [/quote]

Sorry for reviving, but how do you get that? I don't really understand this problem, maybe cause I can't visualize it

Imagine a point on the circumference of the base of the cone that touches the table before the cone starts to spin on its slant. This point touches the table 17 times when the cone being rolled around once and comes back to the same point as it started. The total distance the point traveled is 17 times the circumference of the base of the cone, which is $34r\pi$

[hide=Solution]
We want to find $\frac{h}{r}$ so we can let $r=1$ WLOG.

We then want to find the $h$ such that $2\sqrt{1^2+h^2}\pi=34r\pi$.

We rewrite as $\sqrt{1+h^2}=17$.

We solve to get that $h=\sqrt{17^2-1}=\sqrt{288}=12\sqrt2$.

We then calculate our answer to be $12+2=\boxed{014}$.
[/hide]

[hide = Solution]
So we know that the radius of the circle the cone traces out is the slant height of the cone if you visualize it so the slant height follows with the Pythagorean theorem which is $\sqrt{r^2 + h^2}. $
This means that the circumference of the circle is $2\pi \sqrt{r^2 + h^2}$. Since the cone makes $17$ rotations on the circumference. The circumference of the bottom circle of the cone is $2\pi(r)$ so our equation goes as follows....
$17(2\pi(r)) = 2\pi \sqrt{r^2 + h^2} \Rightarrow 289r^2 = r^2 + h^2 \Rightarrow 288r^2 = h^2.$ So it follows that $\frac{h}{r} = 12\sqrt{2}$ so the answer is $\boxed{14}$

[/hide]

[hide=solution]Assume $r=1$. Then we want to find $h$. If the cone rolls around $17$ times, the perimeter of the circle formed is $17\cdot 2\pi(1)=34\pi$. Thus the radius of that circle is $\displaystyle\frac{34\pi}{2\pi}=17,$ and $h=\sqrt{17^{2}-1^{2}}=\sqrt{288}=12\sqrt{2}$. The answer is $12+2=\boxed{014}$.[/hide]


I'm a student in Hong Kong. A school in hong kong has contacted me that they will hold a team in the ARML this year. Is their other foreign teams competing in the ARML before?

Yes. There is an organization from Taiwan that has brought several teams to ARML in nearly all of the recent five years or so. (They also run an ARML-style contest at home called TRML.) And a team from the Philippines has come in most recent years. Taiwan and the Philippines come to the western ARML site (now Las Vegas), and I assume a Hong Kong team would also come to Las Vegas.

I think last year a team from Columbia came to the eastern ARML site.

For the sake of the Taiwanese teams, the questions have been translated into Chinese, and the power round grading room in Las Vegas has people who can read and translate Chinese. (The Filipino team takes their questions in English.)

But one thing: the non-North American. teams are entered in neither Division A nor Division B and are not eligible for team prizes, and individuals on non-North American. teams are not eligible for individual prizes. Last year, we did allow a couple of individuals from the Taiwan teams compete in the individual tie-breaker, and one of them would have placed in the top 10 had he been eligible. I believe someone with close ties to the Taiwan team matched the prize he would have won.

[Edit: changed "non-U.S." to "Non-North American." Canadian teams (for instance, Massey Ontario) are fully eligible in Div. A or Div. B.]

[quote="Kent Merryfield"]I think last year a team from Columbia came to the eastern ARML site.[/quote]
Yes, I saw a team from Columbia at Penn State.

Thanks, then i will take part in the selection test.

...how many people are in a team?...15 :o

i think i can make the team easily.

look forward to seeing all of you in Las Vegas.

btw: i think English questions will be better than Chinese one.


[size=150]

1. find the cross product of  2i+3j and -i+2j

a.4

b.7k

c.-2i+6j

d.5i-4j+k

2.which of the following is the definition of the hyperbolic sine of x?

a.  sinhx=e^sinx

b. sinh x=(e^x - e^-x)/2

c. sinh x= ln |sin x|

d. sinh x=  :sqrt: |sin x|[/size]

Are these your homework problems? Keep in mind that this forum is NOT a homework help forum, and we do not encourage people to use this forum as a homework help forum.

[quote="ComplexZeta"]...and we do not encourage people to use this forum as a homework help forum.[/quote]

One may even go so far as to say we discourage using this as a homework help forum and ostracize those who do.

Well, I was trying to put it nicely, but Magnara is right.


This is a question from INCho 2005 Can someone help me with this?

Cyclooctatetraene reacts readily with potassium metal to form a dianion (B) ($ \ce{C8H8^{2-}}$).
4.3 Draw the structure of the dianion.
4.4 The geometry of this dianion is
4.5 The reaction of cyclooctatetraene with potassium takes place very easily as -
(a) cyclooctatetraene is highly electron deficient
(b) the oxidation potential of cyclooctatetraene is low
(c) the dianion is aromatic
(d) potassium metal is highly electropositive
The dianion reacts with two moles of acetone to form a mixture of two compounds
(C and D).
4.6 Draw the structures of C and D.
Any olefin on reaction with a per acid gives an epoxide
Cyclooctatetraene gives epoxide E, which on treatment with a Bronsted acid gives
compound G ($ \ce{C8H8O}$). G gives 2,4-DNP derivative and also reacts with Tollen’s
reagent.
In an analogous manner, cyclooctatetraene on reaction with HBr gives compound
H (C8H9Br). Identify structure of H

[hide="Answer"]4.3 and 4.4 The dianion is planar.

4.5 (c) the dianion is aromatic

G is phenylacetaldehyde. H is probably a simple addition derivative.[/hide]

h is mewthyl substitued tropylium bromide(wagner merweein rearrangent takes place by ring contraction)

can you attach the structure?  :D  can you draw and show it?  :)

I don't know to draw structures but I told u enough for u to visualize the structure. Probably u can ask others who know to draw organic structures and show u.

Use a paint document and attach it da  :D

[quote="chemrock"]It is a bicyclic compound with each ring consisting of a cyclopentadienyl anion ring and so it is arromatic and planar and also has the ability to complex even the actinides.[/quote]
wat sort of a compound is that, and how do you get an anion? give the mech.
im pretty sure it is methyl sub tropylium bromide as it is aromatic

Excuse me! I am talking about C8H82-. I think it is

[quote="Carcul"]Although that compound is aromatic, it has a double negative charge, while methyl tropylium has a single positive charge. Are you sure that's the answer?[/quote]

I think chemrock was talking abt the structure of $ \ce{C8H8^{2 - }}$ as asked in question 4.3 while varunarasu was referring to structure H  :)

Ah ok. But then that's wrong, because when cyclooctatetraene accepts two electrons the resulting aromatic species is still an eight membered ring, but with 10 electrons. The structure that Chemrock have shown is the one that results when [i]Pentalene[/i] accepts two electrons.

But can u show the exact structure then. PLEASE......  :help:  :help:

Excuse me I want to know which carbon has negative charge(that's atleast ok) but we say that it is arromatic only because the negative charge also involves in resonance how is that possible PLEASE explain(sorry to trouble u so much I can't find these ans on the net).

yes i wwas referring to H. and carcul is rite, C8H82- has to be an annulene. also all carbons have neg charge, (-1/4 on each). i think you will get it if you draw the resonance structures. start with any two adjacent carbons having -1 charge. i dont know how you got that structure for C8H82-(no offense meant). i mean, give the mech. probably you must have taken the tautomer of COT but im still sure you wont get pentalenide

no Carcul is right that's what peter sykes also tells

[quote="Chemrock"]Excuse me I want to know which carbon has negative charge[/quote]

The negative charge is delocalized all over the carbons on the ring.

[quote="Chemrock"]but we say that it is arromatic only because the negative charge also involves in resonance[/quote]

That dianion is aromatic because:

1) It is planar
2) All carbons are sp2
3) The number of delocalized electrons is an Hückel number (4n + 2, with n = 2)
4) Such compound is then classified as aromatic because it can be shown that, relative to an open chain counterpart, it is more stable.

we all know that Arromatic species obey huckel's rule but BENZYNE is arromatic even though it has 8pi electrons because only 6pi are involved in reasonance so that's why I was confused.

Benzyne has only 6 pi electrons: the other 2 are located on sp2 orbitals. On COT dianion, all 10 electrons belong to the pi system.


[quote="jsc"]999. but theoretically, there is no "easiest question in the world".   :D[/quote]
yea but it was really easy

WRONG!!!!

The largest number is $9^{9^9}$

oh, so it's not that easy.  Well it could be $9^{99}$

hamster is right...that was the easiest problem ever.

[quote="Hamster1800"]nope :) The exponents start at the top and $9^9 > 99$ so $9^{9^9} > 9^{99}$[/quote]
well one is 9^99 is about 2e94
and (9^9)^9 is is about 2e77 but 9^9^9 is 9^(9^9) thats really big.

[quote="-Terminator-"][hide="Answer"]$999$[/hide][/quote]

Oops i forgot about exponents. :blush:

Can you use symbols other than digits?  If so, how about $9!^{9!^{9!}}=362880^{362880^{362880}}$?

[quote="mathnerd314"]You can write a number of any length with just three digits (on your hand) given enough time and enough paper, because you only need to use three fingers to hold your pencil.  :D  :D  :D[/quote]
but you can hold a pencil with two fingers (with effort) and you got an extra finger  :D  :P

but how many pencil can you hold if you had 999 fingers  :D

 :roll:  is there a formula i can use to solve that  :P  :rotfl:

[quote="Hamster1800"]WRONG!!!!

The largest number is $ 9^{9^{9}}$[/quote]

well i cant think of any larger three digit number

[quote="math92"]What is the largest number you can write using 3 digits. You may use a digit more than once.[/quote]

Uh 999? Duh?

[quote="abacadaea"]you could turn an 8 sideways and make it infinity, use a 1 as a factorial sign, and take it to the ninth power. so it looks something like this: $ \infty!^{9}$ :)[/quote]

then y not put a sideways 8 for the power? :D

did it not say distinct....oh ok so true then but its cooler when you have distinct numbers

I think it should be 999.  And don't go 'round saying its EASIEST PROBLEM EVER!
$ 9^{9^{9}}$ is $ 9^{81}$ which is a LOT greater than a 3 digit number.

it never said that. it said that largest number you can write with 3 digits. not a 3 digit number. and instead of $ 9^{9^{9}}$, why dont you have $ 9^{99}$ because $ 9^{9^{9}}\equal{}9^{81}$

Is still $ \infty$.

View my blog for reasons why.

You know as well as I do that such an expression is simply not defined.  Operations are not defined on infinity.

This thread is just repeating itself now and should probably be locked.

$ \infty^{\infty}!$ is much much more. although you solution is also infinity. but u would be using an ifinite amount of digits. because !'s count as 1. atleast in my world they do. and stop posting here please. its fun, but getting old. myhaps a mod should lock this topic.


calculus
integration
function
geometry
derivative
limit
calculus computations

What is an integral? I know the symbol, but I don't know what

it means.

Can someone give me a very basic example along with the

explanation? Thanks.

It's basically the opposite of the derivative.

i.e. if the integral of a function $ f(x)$ can be represented as $ F(x)$, then $ F'(x)\equal{}f(x)$. the antiderivative.

Howeverrrr you have to worry about the constants:

$ f(x)\equal{}4x^3\plus{}6x^2\minus{}3$

then:

$ F(x)\equal{}x^4\plus{}2x^3\minus{}3x\plus{}c$

where $ c$ is a constant.

My explanation may be a bit off since I learned this a while ago..

An integral is a operator over functions that transforms a function into a family of functions. The main identity concerning integrals is:

$ \frac{d}{dx}\left(\int f(x)\,dx\right)\equal{}f(x)$

(I believe this is called the fundamental theorem of calculus - correct me if I am wrong).

In general, you can use integrals to determine the surface under a function using the expression:

$ S \equal{} \int^b_a f(x)\,dx \equal{} F(b)\minus{}F(a)$

where $ F'(x)\equal{}f(x)$. This describes the surface closed by functions $ x\equal{}a$, $ x \equal{}b$, $ y \equal{} 0$ and $ y \equal{} f(x)$.

Other than that, we use integrals in solving differential equations (as to do other things as well). Perhaps it would be more appropriate to say that we use integration to solve differential equations, and that is applying the integration operator.

The most common thing that confuses people when using integrals is the differential that is written at the end of the integral. As though it may seem unimportant, it is the mere essence of integrals. Without doing an elaboration on the topic, let's just say that with the differential you define how do the variables appearing in the expression under the integral behave - like a constant or not. In other words, if you have:

$ \int f(x_1, x_2,...,\,x_i, \,..., x_n)\,dx_i \equal{} F(x_1, x_2,...,\,x_i ,\,..., x_n)\plus{}C$

then

$ \frac{\partial}{\partial x_i}F(x_1, x_2,...,\,x_i ,\,..., x_n) \equal{} f(x_1, x_2,...,\,x_i ,\,..., x_n)$

but

$ \frac{\partial}{\partial x_j}F(x_1, x_2,...,\,x_i ,\,..., x_n) \equal{} f(x_1, x_2,...,\,x_i ,\,..., x_n)\qquad j \neq i$

is not always true.

But don't be fooled, this isn't even scratching the surface of integrals! :)

Both posters have described an [b]indefinite[/b] integral.  It may be easier for you to first be introduced to the [b]definite[/b] integral.  The definite integral $ \int_{a}^{b} f(x) \, dx$ describes the "area under the curve" of $ f(x)$ between $ a$ and $ b$.

Some physical intuition:  if $ v(t)$ describes the velocity of a particle with respect to time, $ \int_{a}^{b} v(t) \, dt$ describes the particle's displacement from time $ t \equal{} a$ to time $ t \equal{} b$.  (In other words, the definite integral is a calculation that takes rates of change of some variable to total change.  So it is the "opposite of the derivative" in some sense.)

The relationship between definite and indefinite integrals (which are also known as "antiderivatives"):  if $ F(x)$ is an indefinite integral of $ f(x)$, then $ \int_{a}^{b} f(x) \, dx \equal{} F(b) \minus{} F(a)$.

You should probably read a calculus textbook for more information.

And another kind of integral that's common is [b]improper[/b] integral. An example of a improper integral is:

$ \int^\infty_0f(t)\, dt$

The problem is the boundary at $ \infty$ - this is why it is called improper. These kind of integrals are solved using limits. Denote $ F'(x)\equal{}f(x)$ and then

$ \int^\infty_0f(t)\, dt \equal{} \lim_ {x \rightarrow \infty}\int ^x_0f(t)\,dt \equal{} \lim _{x \rightarrow \infty}F(x) \minus{} F(0)$


geometry
circumcircle
trigonometry
inequalities
angle bisector
geometry unsolved

Let x, y, z be the lengths of angle bisectors of triangle ABC. And a, b, c be the lengths of sides of ABC.
Show that 1/6x + 1/6y + 1/6z >= 3^(1/2)/(a+b+c).

is that $ \frac{1}{6}\cdot x$ or $ \frac{1}{6x}$?

It must be $ \frac {1}{6x}$, otherwise for an equilateral triangle with side $ s$ we would have $ \Sigma \frac {1}{6}x \equal{} \Sigma \frac {1}{6}(\frac {s\sqrt {3}}{2}) \equal{} \frac {s\sqrt {3}}{4} \ge \frac {\sqrt {3}}{3s}$ which is false for small $ s$.

[quote="mathematical intelligence"] [color=darkred]Let $ ABC$ be a triangle with the incenter $ I$ . Denote the points

$ D\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ so that $ I\in AD\cap BE\cap CF$ .
Prove that $ \boxed {\frac {1}{AD} + \frac {1}{BE} + \frac {1}{CF}\ge \frac {3\sqrt 3}{p}}\ \ (*)$[/color] [/quote]

[color=darkred][b][u]Lemma I.[/u][/b] Denote the second intersections $ M$ , $ N$ , $ P$ of the circumcircle with the lines

$ AI$ , $ BI$ , $ CI$ respectively. Then $ \boxed {AD\cdot AM = bc}$ a.s.o.[/color]

[color=darkblue]Indeed, $ \triangle ABM\sim\triangle ADC$ $ \implies$ $ \frac {AB}{AM} = \frac {AD}{AC}$ $ \implies$ $ AD\cdot AM = bc$ a.s.o.[/color]

[color=darkred][b][u]Lemma II.[/u][/b] $ \boxed {2p^2\ge 27Rr}\ .$[/color]

[color=darkblue]Indeed, $ (a + b + c)^3\ge 27abc$ $ \Longleftrightarrow$ $ 8p^3\ge 27\cdot 4Rpr$ $ \Longleftrightarrow$ $ 2p^2\ge 27Rr\ .$[/color]

[color=darkred][b][u]Lemma III.[/u][/b] $ \boxed {\cos\frac {B - C}{2}\ge\sqrt {\frac {2r}{R}}}\ .$[/color]

[color=darkblue]Indeed, $ (IA + IM)^2\ge 4\cdot IA\cdot IM$ $ \Longleftrightarrow$ $ AM^2\ge 4\cdot 2Rr$ $ \Longleftrightarrow$

$ \left(2R\cos\frac {B - C}{2}\right)^2\ge 8Rr$ $ \Longleftrightarrow$ $ \cos\frac {B - C}{2}\ge\sqrt {\frac {2r}{R}}\ .$

[b][u]Proof of the proposed problem.[/u][/b] The required inequality $ \sum\frac {1}{AD}\ge \frac {3\sqrt 3}{p}$ $ \Longleftrightarrow\over \mathrm {lemma\ I}$

${ \sum\frac {AM}{bc}\ge 3\sqrt 3}{p}$ $ \Longleftrightarrow$ $ \sum a\cdot AM\ge 12Rr\sqrt 3$ $ \Longleftrightarrow$

$ 4R^2\cdot\sum\sin A\sin\left(B + \frac A2\right)\ge 12Rr\sqrt 3$ $ \Longleftrightarrow$ $ \boxed {\sum\sin A\cos\frac {B - C}{2}\ge \frac {3r\sqrt 3}{R}}\ .$

From the [b]lemma III[/b] obtain : $ \cos\frac {B - C}{2}\ge\sqrt {\frac {2r}{R}}$ a.s.o. $ \implies$

$ \sum\sin A\cos\frac {B - C}{2}\ge \sqrt {\frac {2r}{R}}\cdot\sum\sin A$ $ \implies$ $ \boxed {\sum\sin A\cos\frac {B - C}{2}\ge \frac pR\cdot\sqrt {\frac {2r}{R}}}\ (1)\ .$

Observe that $ \frac pR\cdot\sqrt {\frac {2r}{R}}\ge \frac {3r\sqrt 3}{R}$ $ \Longleftrightarrow$ $ 2p^2\ge 27Rr$ , what is truly by the [b]lemma II[/b].

[b][u]Remark.[/u][/b] The inequality $ (1)$ is stronger than the $ (*)$ , i.e. $ \boxed {\sum\frac {1}{AD}\ge\frac 1r\sqrt {\frac {2r}{R}}}\ge\frac {3\sqrt 3}{p}\ .$[/color]

very nice!

I have another proof:
If $ l_b$ is the length of the angle bisector of $ \angle B$, and $ s$ is the semiperimeter.

Then, using Stewart to find that:
$ \sum{l_b} = \sum{\frac {2\sqrt {ac}}{a + c}\sqrt {s(s - b)}}\le \sqrt {s}\sum{s - b} \le s\sqrt {3}$
By AM-GM and QM-AM respectively.

Now by C.S.

$ (\sum{l_a})(\sum{\frac {1}{l_a}})\ge 9\Leftrightarrow \sum{\frac {1}{6l_a}\ge \frac {3/2}{\sum{l_a}}\ge \frac {3/2}{s\sqrt {3}} = \frac {\sqrt {3}}{a + b + c}}$

 :)

[color=darkblue][b]ElChapin[/b], maybe you can solve otherwise
the stronger inegality $ \boxed {\sum\frac {1}{AD}\ge\frac 1r\sqrt {\frac {2r}{R}}}\ge\frac {3\sqrt 3}{p}\ ,$
which was mentioned by mine in the my previous message.[/color]


[quote="forlorn"]Can you find an approximation of π using only a square, a triangle and a circle?  :D  :rotfl:[/quote]

I guess I'll try...

[hide][asy]import cse5; size(250); pair O=MP("O",D((0,0))), A=10*dir(90), B=10*dir(210), C=10*dir(330); D(A--B--C--cycle); D(incircle(A,B,C)); D(5*dir(45)--5*dir(135)--5*dir(225)--5*dir(315)--cycle); MP("1",2.5*dir(45),W); D(O--5*dir(45));[/asy]

Note that the area of the circle is $ \pi$, and the area of the triangle and the area of the square bound the area of the circle. It's easy to calculate each area:
$ \text{Area Triangle }=3\sqrt{3}$
$ \text{Area Circle }=\pi$
$ \text{Area Square }=2$

So: $ 2<\pi<3\sqrt{3}$. I guess you could average the bounds to get:
$ \pi \approx 3.59807\ldots$

Not a very good approximation, though (correct to 1 decimal place!).[/hide]


1. (s-12t)(s-4t)

2. (g-5)(g-7)

3. (x+3)(x+5)

I'm so used to algebra 2 now. Nice to have some relaxing probs once in a while.

I like the excercise after solving some strange problems.


topology
calculus
vector
advanced fields
advanced fields unsolved

is an intersection of two orientable hypersurface(of an orientable manifold) orientable?

assuming they intersect transversally

[quote="yenlee"]I think so.  At each point of the intersection, choose an ordered basis $ e_1,\ldots,e_n$ in the orientation class of the ambient manifold such that $ e_1,e_3,\ldots,e_n$ is in the orientation class of the first hypersurface, $ e_2,e_3,\ldots,e_n$ is in the orientation class of the second hypersurface, and $ e_3,\ldots,e_n$ is a basis of the tangent space of the intersection.  Clearly, there exists such a basis, and one can check that the orientation class of $ e_3,\ldots,e_n$ is uniquely determined by the chosen orientations on the two hypersurfaces and the ambient manifold.[/quote]

I actually got the answer for this question yesterday.
Your approach works, but need some careful calculus on determinant of a matrix.
Easier way to see the statement being true is this:

Pick a non-vanishing top form of the ambient manifold : $ \omega$
Pick non-vanishing normal vector fields respectively from given two hypersurface : $ V_1 , V_2$
Then $ V_1 \urcorner V_2 \urcorner \omega$ restricted to the intersection of two hypersurfaces gives you a nonvanishing top form on it.