trigonometry ratio algebra polynomial geometry rectangle probability

Those of you that are as paranoid as I am, you can discuss round 1 (and 2) here (once it is ok to do so, I think it isn't yet).

Since all answer sheets must be in the mail by Friday, I would think that it's ok to talk about it after school on Friday.

Tom

ok, I'll start it off

I got a 134 (21 right, 4 blank) barring stupid bubbling mistakes.

Are answers up yet?
The site doesn't have the answers yet.

Walton put together an answer key that we mostly agree with. I think we might have a disagreement with one of them, #23 maybe? Mr. Fulton probably has it lying around somewhere.

Supposed scores for top few people at Walton:

Claus: 144
Edward: 144
Allen: 138
Henry: 138
Slok: 138
Abubakar: 128

Here is a key that we generated from various answers (mostly Claus's) I seem to remember that lots of the rest of my team disagreed with either 21,23, or 24. Let us know if you have a different answer to any questions. And I'd love to see an answer to 21.

1. B
2. D
3. D
4. E
5. C
6. C
7. C
8. A
9. B
10. E
11. E
12. D
13. D
14. B
15. A
16. A
17. D
18. B
19. B
20. A
21. E
22. C
23. A
24. C
25. D

In that case, 138 or 144 depending on my answer for #15. i circled A, but wrote B to the side of the number.

This is sad, since I had an answer to every problem with 20 min to check stuff.

[hide="#24"]$ (a,b,c) \equal{} (3,2,1)$ works :D. $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 14$ [/hide]

Anyway, [hide="#21:"]

Let $ \sin A \equal{} 4K$, $ \sin B \equal{} 5K$, and $ \sin C \equal{} 6K$.

By the law of sines:

$ \dfrac{a}{\sin A} \equal{} \dfrac{b}{\sin B} \equal{} \dfrac{c}{\sin C} \equal{} D$

$ \dfrac{a}{4K} \equal{} \dfrac{b}{5K} \equal{} \dfrac{c}{6K} \equal{} D$

So, $ a \equal{} 4KD$, $ b \equal{} 5KD$, and $ c \equal{} 6KD$

WLOG, let $ KD \equal{} 1$.

Then use the law of cosines to compute:

$ \cos A \equal{} \dfrac{5^2 \plus{} 6^2 \minus{} 4^2}{2*5*6} \equal{} \frac {3}{4}$

$ \cos B \equal{} \dfrac{6^2 \plus{} 4^2 \minus{} 5^2}{2*6*4} \equal{} \frac {9}{16}$

$ \cos C \equal{} \dfrac{4^2 \plus{} 5^2 \minus{} 6^2}{2*4*5} \equal{} \frac {1}{8}$

Multiply all by a factor of $ 16$ to get $ \cos A : \cos B : \cos C: \equal{} 12: 9: 2$ and $ (x,y) \equal{} (12,9)$ [/hide]

#23 is E
solution to #21: using law of sines we can assume we have a triangle abc with bc=4, ac=5, ab=6. Then using law of cosines three times (once for each angle) we find angle A=3/4, B=9/16, and C=1/8 which in ratio form gives us 12:9:2

I also got #23 E (-6024), but I could easily be wrong. Too lazy to do now.

solution to #23: the list forms a six number pattern of x, 2x, x, -x, -2x, -x, and then repeats. the remainder when 36 is divide by 6 is 0 which means 2008 is the last number of the pattern:case1:(-2008, -4016, -2008, 2008, 4016, 2008) or case 2:(4016, 2008, -2008, -4016, -2008, 2008) or case3:(1004, -1004, -2008, -1004, 1004, 2008).  Since the sum of 6 consecutive terms is always 0, the sum of n terms is the sum of the first k terms where k is the remainder when n is divided by 6.  When 100 is divided by 6 the remainder is 4. When 200 is divided by 6 the remainder is 2. This means the sum of the first four terms equals the sum of the first two. this only occurs in case 1 so the answer is -2008+-4016+-2008+2008 or just -2008+-4016 = -6024

If #23 is E, then Edward got a perfect score!!! Don't know about the other people... mine stays the same.

Grrz, I got 144.  I'm pretty sure #23 is E, but I put C cuz for some reason I assumed that 2=-1 (dunno how that happened).

I think I tied Heng.  That makes me want to slit myself.

The other thing that sad-ifies me is that I finished it in the first 45 minutes and decided to check really slowly for the first 22 questions.  And by the time 50 seconds were left, I had only begun to check #23, and I actually noticed that I made a mistake, but I didn't have time to fix it.

And the final thing that depress-ifies me is that this test was, in my opinion, easier than last year's AMC 10A, which I thought was unusually easy. :(  Last year, I totally phailed the first round and missed like over 9000 problems.

[quote="Math Geek"]
I think I tied Heng.  That makes me want to slit myself.
[/quote]

That's good for the self esteem.  :rotfl:, oh wait, um  :(

I liked the test.  It was not easier than an AMC 10.

I think that the only question whose difficulty was a little out of place was number 25.  Had the question been on a test requiring justification, it would have been more difficult.  But I think on a multiple choice test, many students will guess the optimal configuration (maybe unconsciously).

Re bongo :  How do you know that the second term is twice the first?  The sequence could be something like 1,3,2,-1,-3,-2 . . . .

[hide="Problem statement for 23"]In a list of 200 numbers, every one (except the end ones) is equal to the sum of the first two adjacent numbers in the list.  The sum of all 200 numbers is equal to the sum of the first 100 of them.  Find that sum if the thirty-sixth number in the list is 2008.[/hide]

You are right, however, that the sequence must repeat with period 6.  This can be deduced non-constructively from general linear recurrences.  You know that the sequence's recursion is
\[ a_{n+1} + a_{n-1}  =a_n , \]
so we must have
\[ a_n = \alpha r_1 + \beta r_2 , \]
for some $ \alpha, \beta$, where $ r_1, r_2$ are the roots of the polynomial $ x^2 - x +1$.  Since these are (primitive) sixth roots of unity, this not only tells you that the sequence repeats every six terms, but it also tells you that the sum of any six consecutive terms is zero.

[hide="An actual solution for 24"][hide="Problem statement"]If $ a$, $ b$, and $ c$ are three distinct numbers such that $ a^2 - bc = 7$, $ b^2+ac=7$, and $ c^2 + ab = 7$, then compute $ a^2+b^2+c^2$.[/hide]

I will replace $ a$ with $ -a$; the system then becomes
\begin{align*}
a^2 - bc &= 7 \\
b^2 - ca &= 7 \\
c^2 - ab &= 7,
\end{align*}
or
\begin{align*}
a^2 - 7 &= bc \\
b^2 - 7 &= ca \\
c^2 - 7 &= ab .
\end{align*}
Then
\[ a^3 - 7a = b^3 - 7b = c^3 - 7c = abc. \]
So $ a,b,c$ are the roots of the polynomial
\[ x^3 - 7x + k = 0, \]
for some constant $ k$.  Therefore
\[ a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+bc+ca) = 0^2 - 2(-7) = 14.  \qquad \blacksquare \]
[/hide]

Number 25 was the one about the crossing of the river, right? For the justification, could you use say something like this?

The river doesn't really matter, since the crossing length will always be the same. Then you can join the two sides together, and since the shortest distance between any two points is the straight line between them, that line is the minimum. I don't really remember what the question asked for, but you can find the answer directly from that.

[quote="Boy Soprano II"]I liked the test.  It was not easier than an AMC 10.[/quote]

Well, maybe I smartified, but regardless, it was far from overly difficult. :)

#24 didnt come out on my test (white rectangles) but i got everything else right
meh...

oh. i guess i miscounted on #16. 144 for me.

As the attached python program proves, the answer to number 16 is A, 4/13.

[hide="Problem statement"]Consider all sets of nickels, dimes, and quarters having at least one coin of each type and which add up to one dollar.  What is the probability that a [uniformly randomly selected] set of this kind will consist of a prime number of coins?[/hide]

I will now post a solution for problem 22.  It's not as easy as it seems at first, since a finite geometric sequence of integers need not have a ratio that is integral.

[hide="Problem statement"]Quadrilateral $ ABCD$ is inscribed in a circle.  The degree measures of angles $ A$, $ B$, and $ C$, in order, are integers that form an increasing geometric sequence.  Compute the sum of all possible values for the measure of angle $ D$.[/hide]

[hide]Let $ t$ be the measure of angle $ C$ (in degrees); let $ r > 1$ be the ratio of the sequence.  Then
\[ t \left( r\plus{} \frac{1}{r} \right) \equal{} 180 . \]
Let $ r \equal{} \alpha/\beta$, for relatively prime integers $ \alpha >\beta$.  Then this equation translates as
\[ t(\alpha^2 \plus{} \beta^2) \equal{} 180 \alpha \beta. \]
Since $ \alpha$ and $ \beta$ are relatively prime, it follows that
\[ \alpha^2 \plus{} \beta^2 \mid 180 \equal{} 2^2 \cdot 3^2 \cdot 5. \]
Since 3 is a Gaussian prime, it follows that
\[ \alpha^2 \plus{} \beta^2 \mid 2^2 \cdot 5. \]
And since $ \alpha$ and $ \beta$ are relatively prime, they cannot be even, so $ \alpha^2 \plus{} \beta^2$ is congruent to 2 or 1 mod 4.  Thus
\[ \alpha^2 \plus{} \beta^2 \mid 10 . \]
Simple checking reveals that (3,1) and (2,1) are the only possible values of $ (\alpha,\beta)$.  These yield values of $ t$ of 54 and 72, respectively.  So the sum of all possible measures of $ D$ is 126+108 = 234. $ \blacksquare$ [/hide]

Hi everyone
I think i have a solution for problem number 22.
Let angle C be  x. Then angle A is 180-x. Let angle B be y. Then we have x(180 - x) = y^2.
Now looking at the discriminant of the quadratic ( of variable x) we note that the discriminant is precisely (180)^2 - (2y)^2 . This reminds us of pythagorean triples .
This implies 180 = a^2 + b^2 for some positive integers a and b in this case. We see that the only solution for this in positive integers is a = 12 and b = 6. Now 2y is either a^2 - b^2 or 2ab. Thus we see that y is either 54 or 72. This implies angle D is either 126 or 108. Implies the sum of the angles is 234.

Sorry i don't know how to post in latex.

I got a 138 or a 144 depending on whether #23 is A or E.
I made a dumb mistake on #11 for some reason.
For problem 22, I decided to write the equation a(1+r^2) = 180, where a is the measure of <A and r is the ratio m<A/m<B.
Then I used brute force (NOT a smart move :( ).

Solutions are up:

[url]http://math.kennesaw.edu/~ckoppelm/[/url]

Since #23 was E, I got either 144 or 150 depending on my answer to #15.

[quote="Boy Soprano II"]
If $ a$, $ b$, and $ c$ are three distinct numbers such that $ a^2 - bc = 7$, $ b^2 + ac = 7$, and $ c^2 + ab = 7$, then compute $ a^2 + b^2 + c^2$.

I will replace $ a$ with $ - a$; the system then becomes
\begin{align*} a^2 - bc & = 7 \\
b^2 - ca & = 7 \\
c^2 - ab & = 7, \end{align*}
or
\begin{align*} a^2 - 7 & = bc \\
b^2 - 7 & = ca \\
c^2 - 7 & = ab . \end{align*}
Then
\[ a^3 - 7a = b^3 - 7b = c^3 - 7c = abc.
\]
So $ a,b,c$ are the roots of the polynomial
\[ x^3 - 7x + k = 0,
\]
for some constant $ k$.  Therefore
\[ a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = 0^2 - 2( - 7) = 14. \qquad \blacksquare
\]
[/quote]
I think you can solve it in a much shorter way. Here's my solution:
Subtracting the 2nd to the 3rd equation we get: (b-c)(b+c-a)=0.Since a,b,c are distinct numbers, we easily deduce to get b+c=a. Now adding these 2 equation we get: b^2+c^2+a(b+c)=14, which means: a^2+b^2+c^2=14.So the answer must be 14.
Notice that we even don't need the 1st equation. Therefore, I think this problem is kinda badly written.
I'm not really satisfied with the #25's solution. I thought we have to find minimum, not to find the constant.Anyway, I have another solution for it, which is more technical:
Applying Pythagorean theorem we have: x^2+y^2=sqrt(9+l^2)+sqrt(81/4+t^2)(x,y are those 2 dotted segments).Now considering that t+l=10, applying the Minxcopski we have:
sqrt(9+l^2)+sqrt(81/4+t^2)>=sqrt[(3+9/2)^2+(t+l)^2]=sqrt(225/4+100)=[b]12.5[/b](to prove it, we jut simply square both sides two times to get the Cauchy-Schwartz inequality).
The equality occurs when 3t=9/2.l, which means t=6 and l=4(Q.E.D) :)
We can easily solve the generalized problem by using the same method as above(since Minxcopski is still true for n pairs of 2 variables):
Assume that we have n canals. Each canal has 1 mile wide. There is a man running from the point a_1 to a_n(of course, he has to swims directly across each canal). If we know that the distance of each point a_1,a_2,...,a_n to each corresponding one end of canals is: 1,2,3,...,n(well, I just choose the numbers randomly), then what is the least number of miles he may run(don't count the path he swims across the wide of each canal).

Nobody wants to enjoy my generalized problem by solving it? :(

[quote="Criticalline1859"]Since the man has to swim directly across each canal (i. e. swim the shortest distance), we can "compress" the canals so that the starting point falls on the end point, and it is as if he never swam at all (the swimming segment becomes one point). However, since the shortest distance between two points is along a straight line, we know the path he takes.

This problem from the test resembles an problem I did once (it may have been from some easy olympiad).[/quote]
Indeed, your solution is out of my intention. Congratulation though! :oops: . I think I can actually derive this problem into an algebra problem, which makes it more difficult and doesn't seem too obvious as your solution shows it was.

It is almost identical to an exercise in Yaglom's [i]Geometric Tranformations[/i], vol. I.  (I highly recommend all three volumes!)

Once you have the geometric solution, you can apply exactly the same principles in more "algebraic"
 context.  It's just perhaps a bit easier to come up with the solution geometrically.

By the way, the conventional spelling is "Minkowski". :wink:

So, how'd everyone do based on their official scores?

I missed some other stuff too, for 120.  Bleheehhh. I'm not winning this year either :).


1.) Show that $ \binom{a}{b-1}+\binom{a}{b}= \binom{a+1}{b}$

2.) Show that $ \binom{n}{1}+\binom{n}{2}+\binom{n}{3}+...+\binom{n}{n}= 2^{n}-1$

Can you give some sites which discuss combinatorial proofs?
Thanks

[hide="1."] Compute $ \binom{a+1}{b}$ in two cases: (1) the $ (a+1)$th item is chosen, and (2) the $ (a+1)$th item is not chosen. [/hide]
[hide="2."] Consider the nonempty subsets of $ \{1, 2, \ldots, n\}$. [/hide]

2)
click on it to see more dateils
[hide="$ 2^{n}=(1+1)^{n}$"]
\[ 2^{n}=(1+1)^{n}=\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+...+\binom{n}{n}\]
and the result follows
[/hide]

[quote="paladin8"][hide="1."] Compute $ \binom{a+1}{b}$ in two cases: (1) the $ (a+1)$th item is chosen, and (2) the $ (a+1)$th item is not chosen. [/hide]

Let X = {1, 2, 3, ..., a, a +1} be a set of $ a+1$ elements.

Let us choose $ b$ elements from the set

(1) if $ a+1$ is chosen on the first subset, there are $ a$ elements left. Hence, we choose $ b$ elements from  $ a$ elements, that is, $ \binom{a}{b}$.

(2) IM STUCK

for $ 2$ as paladin said:
the non empty subsets of $ A=\{1,2,...,n\}$ are the followings:
$ i_{1})$subsets with $ 1$ element.
$ i_{2})$subsets with $ 2$ elements.
$ i_{3})$subsets with $ 3$ elements.
.
.
.
$ i_{n})$subsets with $ n$ elements.
now we have that:
$ \text{number of nonempty subsets of A}=i_{1}+...+i_{n}=\binom{n}{1}+\binom{n}{2}+...+\binom{n}{n}$
in the other hand for set $ A=\{1,2,...,n\}$ for every subset $ B$ from $ A$,for every element in $ A$,it has two ways,appear in $ B$,or not appear in $ B$,so there is $ 2^{n}$ ways to choose a subset and there are $ 2^{n}-1$ non empty subsets.
so we get that:
$ \text{number of nonempty subsets of A}=\binom{n}{1}+\binom{n}{2}+...+\binom{n}{n}=2^{n}-1$as wanted...

What about the proof by binomial theorem, sorry, im quite new to this topic and still learning how to prove. My book is complicated and i dont understand it. Can we use Binomial Theorem?

yes we can:
we know that:

$ (x+y)^{n}=\binom{n}{0}x^{n}y^{0}+\binom{n}{1}x^{n-1}y^{1}+...+\binom{n}{n}x^{0}y^{n}=\sum_{i=0}^{n}\binom{n}{i}x^{i}y^{n-i}$

now put $ x=y=1$ we get that:

$ 2^{n}=\binom{n}{0}+\binom{n}{1}+...+\binom{n}{n}$

$ 2^{n}-\binom{n}{0}=\binom{n}{1}+...+\binom{n}{n}$

but we have $ \binom{n}{0}=1$

so if we put this in the last equality,we get that:

$ 2^{n}-1=\binom{n}{1}+...+\binom{n}{n}$

[quote="guile"]Let X = {1, 2, 3, ..., a, a +1} be a set of $ a+1$ elements.

Let us choose $ b$ elements from the set

(1) if $ a+1$ is chosen on the first subset, there are $ a$ elements left. Hence, we choose $ b$ elements from  $ a$ elements, that is, $ \binom{a}{b}$.

(2) IM STUCK[/quote]

If the $ (a+1)$th element is chosen, you only have $ b-1$ remaining elements to choose, so it's $ \binom{a}{b-1}$. If it is not, you have all of the $ b$ elements to choose, so it's $ \binom{a}{b}$.


limit
logarithms
real analysis
real analysis unsolved

Let $\zeta ( k ) = \sum_{n = 1}^{\infty} \frac{1}{n^k}_{}$, prove that
\[ \lim_{n \rightarrow \infty} \left( n - \sum_{j = 2}^n \frac{1}{\zeta ( j )} \right) = 1 + \sum_{p \in \mathbb{P}} \frac{1}{p \cdot ( p - 1 )}  \]
where $\mathbb{P}$ is the set of prime numbers.

Do you have a proof for this one? I'm really interrested in it. ;)

Try adding dollar signs next time, cimabue.

Let $\alpha_j = \sum_{p \in P} \frac{1}{p^j}$ for an integer $j \geqslant 2$.

First notice that
\[ \alpha_2 = \sum_{p \in P} \frac{1}{p^2} \leqslant \sum^{\infty}_{j = 1} \frac{1}{j^2} - 1 = \frac{\pi^2}{6} - 1 < 1  \]
Obviously $0 < \ldots < \alpha_n < \alpha_{n - 1} < \ldots < \alpha_2 < 1$. Now remark that
\[ 1 + \alpha_j + \alpha_j^2 + \ldots = \sum_{k = 1}^{\infty} \frac{1}{k^j} = \zeta ( j )  \]
And since $0 < a_j < 1$ we can write $1 + \alpha_j + \alpha_j^2 + \ldots = \frac{1}{1 - \alpha_j}$ so $\frac{1}{1 - \alpha_j} = \zeta ( j )$, which means that
\[ \alpha_j = 1 - \frac{1}{\zeta ( j )}  \]
Thus
\[ \lim_{n \rightarrow \infty} ( \sum_{j = 2}^n \sum_{p \in P} \frac{1}{p^j} ) = \lim_{n \rightarrow \infty} ( \sum_{j = 2}^n \alpha_j ) = \lim_{n \rightarrow \infty} ( n - 1 - \sum_{j = 2}^n \frac{1}{\zeta ( j )} ) = \sum_{k \geq 2} \sum_{p \in P} \frac{1}{p^k}  \]
But
\[ \sum_{k \geq 2} \sum_{p^k \leq x} \frac{1}{p^k} = \sum_{p \leq \sqrt{x}} \sum_{2 \leq k \leq \log x / \log p} \frac{1}{p^k} = \sum_{p \leq x} \frac{1}{p^2 - p} + O ( \sum_{p \leq x} \sum_{k > x} \frac{\log x}{\log p} \cdot \frac{1}{p^k} ) = C + O ( \frac{1}{\sqrt{x}} )  \]
where $C = \sum_{p \in P} \frac{1}{p^2 - p} \text{}$ and $p$ runs over primes. So
\[ \sum_{k \geq 2} \sum_{p \in P} \frac{1}{p^k} = C = \sum_{p \in P} \frac{1}{p^2 - p}  \]
Thus
\[ \lim_{n \rightarrow \infty} ( n - 1 - \sum_{j = 2}^n \frac{1}{\zeta ( j )} ) = \sum_{p \in P} \frac{1}{p^2 - p}  \]
Finally
\[ \lim_{n \rightarrow \infty} ( n - \sum_{j = 2}^n \frac{1}{\zeta ( j )} ) = 1 + \sum_{p \in P} \frac{1}{p^2 - p}  \]
That gives us an asymptotics
\[ \sum_{j = 2}^n \frac{1}{\zeta ( j )} = n + C_1 + o ( 1 )  \]
where $C_1 = 1 + \sum_{p \in P} \frac{1}{p^2 - p}$. This asymptotic is actually close to ''reality'' in the sense that $\frac{1}{\zeta ( j )} \simeq1$.

Unfortunately, I will be away for the next week, so I do not promise an immediate response. I'll be glad to hear some remarks from you (but gosh ! I'll hate to realize I made a mistake ;))

The sum of the reciprocals of primes definitely DOES diverge.

cimabue : Can you be more precise ? Why does it imply the sum of reciprocal of primes converge ?
(I was doubtful about the identity for a_j, so i'll be grateful for pointing the mistake)

Edit: BTW: Note that we must have j>=2, for the a_j identity to hold ;) If that's the "mistake" you want to point... If not , then sorry for the noise.


三题难度中高
做出三题的同学好像还不少

Can you post version English of these problems?  :(  Thank you very much  :wink:

4. Let O and I be the circumcenter and incenter of triange ABC.  The incircle of ABC is tangent to BC, CA, AB at D,E,F respectively.   FD and CA meet at P, DE and AB meet at Q.  Let M and N be PE and QP's midpoints repsectively. Prove: OI is perpendicular to MN.

6. Find the smallest positive integer $n>9$ such that: Given any n integers $a_{1},...a_{n}$ (they can be the same), there exists 9 of them $a{}_{i}{}_{1},...a{}_{i}{}_{9}$ such that if $b_{i}$ is in {4,7} (i=1,2,...9) then $b_{1}a{}_{i}{}_{1}+...+b_{9}a{}_{i}{}_{9}$ is a multiple of 9.

Thank you very much. I know problem 5 you posted :)

Hsin Chu? Fourierseries, do you live in Taiwan?

[quote="fourierseries"]You're welcome.  :D

Does anyone know what the approximate minimum is for 3rd prize?
Also, how many people total take the CMO?[/quote]

Bronze cutting: 6
Silver cutting: 57
Gold cutting: 87

Hello ifai and anthony19891019! I do live in Hsin Chu, Taiwan.

How did you guys like the CMO this year? I think i got 2,4,5,6!

Can you post Day 2 in National Olympiad box? Thank you  :)

I was wondering if the results for the 2007 CMO came out yet.  If they have, could someone post them?


Let $ ABC$ be an acute  triangle and denote the midpoints  $ M$ , $ N$ of the sides $ [AB]$ , $ [AC]$ respectively.

For a point $ D\in BC$ denote $ \|\begin{array}{c} m(\widehat {BDM}) = x \\
\ m(\widehat {CDN}) = y\end{array}$ . Prove that $ \cot B - \cot C = \cot x - \cot y$ $ \Longleftrightarrow$ $ AD\perp BC$ .

Does $ m(\widehat {BDM})$ refer to the measure of angle BDM?  Thanks.

Edit: Oh, is CDM supposed to be CDN? Since otherwise $ N$ is irrelevant.

[quote="Virgil Nicula"][color=green]Let $ ABC$ be an acute  triangle and denote the midpoints  $ M$ , $ N$ of the sides $ [AB]$ , $ [AC]$ respectively.

For a point $ D\in BC$ denote $ \|\begin{array}{c} m(\widehat {BDM}) = x \\
\ m(\widehat {CDN}) = y\end{array}$ . Prove that $ \cot B - \cot C = \cot x - \cot y$ $ \Longleftrightarrow$ $ AD\perp BC$ [/color].[/quote]

[hide]
\[ "\Rightarrow"\]
Drop the projections $ ME$ and $ NF$ on $ BC$.
$ \cot x - \cot y = \cot B - \cot C$ $ \implies$ $ DE - DF = BE - CF$ $ \implies$ $ DE + CF = BE + DF$
$ \{\begin{array}{cccc} MN = \frac {BC}{2} \\
 \\
BC = BE + DE + DF + FC = 2(DE + FC) \\
 \\
MN = EF\end{array}$ $ \implies$ $ MN = DF + FC$.In the same time $ MN = DE + DF$ therefore $ FC = DF$ from where the conclusion.

For the second part $ "\Leftarrow"$ it's the same[/hide]


[quote="April"][color=darkred]Let $AA_{1},\,BB_{1}$ and $CC_{1}$ be the altitudes of the acute triangle $ABC.$ Let $AA_{2},\,BB_{2},\,CC_{2}$ be its medians which intersect $B_{1}C_{1},\,A_{1}C_{1}$ and $A_{1}B_{1}$ in the points $A_{3},\,B_{3},\,C_{3}$ correspondingly.
Prove that the straight lines $A_{1}A_{3},\,B_{1}B_{3}$ and $C_{1}C_{3}$ intersect in a common point.[/color][/quote]

This problem is also true when the triangle $\bigtriangleup A_{1}B_{1}C_{1},$ is an arbitrary cevian triangle with respect to a given triangle $\bigtriangleup ABC.$

So, for the points $A_{3},$ $B_{3},$ $C_{3},$ on $B_{1}C_{1},$ $A_{1}C_{1},$ $A_{1}B_{1}$ rtespectively, we have the follows results:

[b]1)[/b] - If the segment lines $AA_{3},$ $BB_{3},$ $CC_{3},$ are concurrent at one point, then the segment lines $A_{1}A_{3},$ $B_{1}B_{3},$ $C_{1}C_{3},$ are also concurrent at one point.

[b]2)[/b] - If the segment lines $A_{1}A_{3},$ $B_{1}B_{3},$ $C_{1}C_{3},$ are concurrent at one point, then the segment lines $AA_{3},$ $BB_{3},$ $CC_{3},$ are also concurrent at one point.

Kostas Vittas.

PS. As I have been promised, I will post here soon, the proof I have in mind for the general problem.

$($ because of there is not any reference about it, after the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=131476 $)$


Please check if I am correct:

The following are Central force:

a.Action and reaction forces between two bodies
b.Van der Waal's force between two molecules

c.dipolar force between two small magnets
d.Force between two current carrying conductors
e.Lorentz force on a charged particle

(c,d,e) are central as long as B field is time indipendent

the following are not:

a.Viscous force on a body falling through a liquid
b.Coriolis force on a body

I do not know QM...Can you please say whether force between two electron spin will be central...

In a classical two body problem,when we superpose our origin of co-ordinates on any one of the two particles,the other particle(now we see it to have reduced mass [mM/(m+M)],moves w.r.t. me...

I saw in a book #Central Force Motion, that the particle is treated by 2nd law of motion...
i.e. [mM/(m+M)]*a=F where a and F are accelen. and force term.

My question is how do we know that this is indeed an inertial frame?After all the two particles are interacting among themselves and there may well be a radial accelen.(Like in the case of Binary stars).

Another point is that we invoke the concept of centrifugal energy...
(=L^2/(2*mu*r^2)) where mu denotes the reduced mass.

Atam P Arya says that here we are working from a rotating frame as a penalty of eliminating theta from the equations...i.e. we are being bound to put this centrifugal force term(and hence a centrifugal energy) as we are working from a rotating frame...

I am having the fragrance...But cannot see the reality...
Please highlight on this topic


hi all
$n$ is a natural and $p$is a prime number such that $np+1$ is a perfect squar prove that we can show $n+1$ like $a_1^2 + a_2^2+ \dots +a_p^2$ and $a_i$ in natural numbers .

Is $0$ considered as natural¿Please define natural everytime you use that term.
I consider $0$ as natural because  otherwise it makes no sense to me.
Since every integer is sum of four squares, we need just to consider $n=1,2,3$.
For $n=1$ take $p=3$ to get a counterexample.
For $n=2$ we get that $np$ is divisible by $4$ everytime, thus there is no such $p$.
For $n=3$ it's clearly true.

What I think is that what I solved here is not the correct problem.
(what I did is: show that $n+1$ can be represented as sum of $n$ squares)
What's the correct problem¿ Is it: $p+1$ can be represented as sum of $n$ squares¿ (then also not counting $0$ as natural would make sense)

i think u have a mistak zetaX. ;)
$0$ is not a natural number.and u should prove it for every $n$ and$p$ when $np+1$is a perfect squar.
natural numbers:$1,2,3,4\dots$
for exampel let $p=3$and $n=5$ and$np+1=16$ is a perfect squar.we can show $n+1$
in this form(a sum of $3(p)$perfect squar):$n+1=6=1^2 +4^2+1^2$.
i solved it  but i want to have anouther beter solution.

So $n+1$ shall be written as sum of $p$ squares then¿ This is not what is stated in the first post, there it is $n$ squares instead.
But again we could reduce it to case checking, but there will be more cases this time. I will look for something nicer.

yes u r right zetaX
i should sleep and rest for many hours. :rotfl:

Let's suppose $np+1=a^2$. then we have that $p|(a+1)(a-1)$

1) $p|a-1$. then we have $a_1=a_2=...=a_{p-1}=\frac{a-1}p$ and $a_p=\frac{a+p-1}p$ that solve the problem

2) $p|a+1$. then we have $a_1=a_2=...=a_{p-1}=\frac{a+1}p$ and $a_p=\frac{a-p+1}p$ that solve the problem

Bye!


Two fair, six-sided dice are rolled.  What is the probability that the sum of the two numbers showing is less than 11?

The sum of the two numbers is 11 or more when they are 5-6, 6-5, and 6-6. Since there are 3 possibilities that the sum will be 11 or more, and there is a total of 36 possibilities, the answer is 33/36, which is simplified to 11/12.

basically, this problem requires complementary counting


For $a,b,c \in Z^+$, prove that if $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} < 1$, then there is also:
\[ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} < 1 - \frac{1}{abc} \]

[quote="jin"]I think he means $\leq$ and that's a very easy problem.[/quote]

Yeha, I think it would be  :)


algebra
polynomial
induction
number theory proposed
number theory

let $ n\ge 2$ integer and $ x_1,\dots,x_n$ reals.
consider $ N\equal{}\frac{n(n\minus{}1)}{2}$ sums $ x_i\plus{}x_j$ where $ 1\le i<j\le n$ and denote them $ y_1,\dots,y_N$ in arbitrary order.
for which $ n$ numbers $ x_1,\dots,x_n$ are uniquely determined by $ y_1,\dots,y_N$ ?

This problem is similar to the one in my teacher's article printed the Mathematics and Youth, a Mathematics magazines for students in Viet Nam.

[b]Answer.[/b] $ n\in\mathbb{N}^*$ such that $ n \neq 2^k$.

Let $ A = \{x_1,x_2,...,x_n\}$ and $ A^{(2)} = \{x_i + x_j | 1\leq i < j \leq n\}$ where $ A^{(2)}$ is a multiset. We will prove the following statement.

[b]Statement.[/b]
There exists $ A = \{x_1,x_2,...,x_n\}$ and $ B = \{y_1,y_2,...,y_n\}$ such that $ A\neq B$ but $ A^{(2)} = B^{(2)}$ iff $ n = 2^k$.

[b]Proof.[/b]
First we prove that if there exists such sets $ A,B$ then $ n = 2^k$. Indeed, consider the two polynomials $ T(x): = \sum_{i = 1}^{n} x^{x_i}$ and $ H(x): = \sum_{i = 1}^{n^} x^{y_i}$, we have $ A^{(2)} = B^{(2)}$ then $ T^2(x) - T(x^2) = H^2(x) - H(x^2)$. This means $ T(x^2) - H(x^2) = (T(x) - H(x))(T(x) + H(x))$. We have $ T(1) = H(1) = n$ so $ T(x) - H(x)$ is divisible by $ (x - 1)$. Let $ T(x) - H(x) = (x - 1)^k f(x)$ where $ f(1) \neq 0$ then $ (x^2 - 1)^k f(x^2) = (x - 1)^k f(x)(T(x) + H(x))$. This yields $ (x + 1)^k f(x^2) = f(x) (T(x) + H(x))$. Now put $ x = 1$ and remember that we put $ f(1) \neq 0$ then $ T(1) + H(1) = 2^k$ or $ 2n = 2^k$, meanwhile $ n = 2^{k - 1}$, QED.

Now we prove with an induction that if $ n = 2^k$ then there exists such $ A,B$. With $ k = 1$ we can point out $ A = \{1,3\}$ and $ B = \{2,4\}$. Suppose that there exists $ A,B$ with some positive integer $ k$, we will construct $ A^*, B^*$ with $ k + 1$ like this: $ A^* : = A \cup \{b + s | b \in B\}$ and $ B^* : = B \cup \{a + s | a \in A\}$ then it is easy to see that $ A^*$ and $ B^*$ satisfies $ A^*^{(2)} = B^*^{(2)}$ and with $ s$ large enough then $ A^* \neq B^*$, QED.


advanced fields
advanced fields unsolved

let $T: (\Omega, \mathcal{A}) \to (\Omega, \mathcal{A})$ be measure preserving, and $X_{0}: = X: (\Omega, \mathcal{A}) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$. define $X_{n+1}= X_{n}\circ T$, and suppose the sequence $(X_{n})$ is indipendent.
is $T$ ergodic?

Are you sure the problem is exactly as you stated it? I'm asking because there is a trivial example where $X(\omega)=0$ for all $\omega$ and $T$ is just an arbitrary map.

uhm...
it seems like you can show that the map $T: (\Omega, \sigma(X)) \to (\Omega, \sigma(X))$ is ergodic, sorry.


What do you think?  What have you tried?

I reckon you must have the elements $ 5$ and $ 9$ with any subset of the elements, $ 1, 2, 4, 6, 7$, including the empty set, giving

$ \binom{5}{0} \plus{} \binom{5}{1} \plus{} \binom{5}{2}... \plus{} \binom{5}{5} \equal{} (1 \plus{} 1)^{5} \equal{} 2^{5}$.

@ Jbl

this was question of an entrance examination for an indian instiute

where I have marked 2^7 because according to requirement 3,8 should be elements of A and not be B

but the in solution released it is 2^5 :(

And what do you think about the membership of 5 in A?

Which implies ....

(I see AndrewTom has corrected his previously incorrect post above.)

Yes, I was rather hasty:I only had a few minutes and also needed to go back to find out the $ \text{\LaTeX}$ code for $ \binom {n} {r}$.  I have been pipped to the post a number of times because I am slow at $ \text{\LaTeX}$ and should really have thought about it and posted later. (Fools rush in and all that.)


So in pressure-volume work of a gas at isothermal conditions, why is it that we always use the value of the external pressure in calculating the the magnitude of work?

Using a piston as an example, I can understand the reasoning behind using $-P\delta V$ to calculate work if the volume of the gas decreases, but why also use the external pressure when the volume increases? In this latter case, isn't the external pressure different from the pressure that's actually doing the work (gas pushing piston up)?

The work done by the pressure of the expanding gas is equal to the work done by the external gas, but has the opposite sign. So it all comes down to the question of simplicity...

Assume that the external pressure is constant at all times (and is equal to $p_{0}$); then the infinitesimal work of the expanding gas is $p\text dV$, while the work of the external gas is $p_{0}\text dV$. The total work, obtained by integrating these infinitesimal works, is the same in both cases; but since $p_{0}$ is constant this integration is far easier.

(Note that the infinitesimal works are [i]not[/i] equal unless the expansion process is very slow, ie. quasistatic.)

I sort of understand it a bit better now, but I have not taken calculus yet, so I did not fully comprehend your explanation.

Ok, if you don't know calculus than the [i]only[/i] way to calculate the work is by the external pressure (as long as the external pressure is constant) When you learn calculus everything will become much clearer ;)


using limit laws find the lim as x goes to -2- of (x+2)/(x+2)

Lim               tan5x/x
x goes to 0

$\lim_{x\rightarrow-2}\left(\frac{x+2}{x+2}\right)=1$:
Pick $\epsilon>0$ and let $\delta=1$. Then $0<|x+2|<1$ implies \[\left|\frac{x+2}{x+2}-1\right|<\epsilon,\] and the conclusion follows.

$\lim_{x\rightarrow 0}\left(\frac{\tan(5x)}{x}\right)=5$:
Plugging in $x=0$ for the above expression yields the form $\frac{0}{0}$, so we may apply L'Hopital's Rule. \[\lim_{x\rightarrow 0}\left(\frac{\tan(5x)}{x}\right)=\lim_{x\rightarrow 0}\left(\frac{5\sec^{2}(5x)}{1}\right)=5.\]

[quote="towersfreak2006"]$\lim_{x\rightarrow-2}\left(\frac{x+2}{x+2}\right)=1$:
Pick $\epsilon>0$ and let $\delta=1$. Then $0<|x+2|<1$ implies
\[\left|\frac{x+2}{x+2}-1\right|<\epsilon, \]
and the conclusion follows.

$\lim_{x\rightarrow 0}\left(\frac{\tan(5x)}{x}\right)=5$:
Plugging in $x=0$ for the above expression yields the form $\frac{0}{0}$, so we may apply L'Hopital's Rule.
\[\lim_{x\rightarrow 0}\left(\frac{\tan(5x)}{x}\right)=\lim_{x\rightarrow 0}\left(\frac{5\sec^{2}(5x)}{1}\right)=5. \]
[/quote]

Thanks
with the first limit x approaches -2 from the from the left (-2-)

I believe the same proof can be used, since \[\lim_{x\rightarrow a}f(x)\] exists implies \[\lim_{x\rightarrow a^{+}}f(x)=\lim_{x\rightarrow a^{-}}f(x)=\lim_{x\rightarrow a}f(x).\]


Hello every1,

How can u find in the fastest possible way using a c-program/anything  :lol:   the number of ones the binary repesentation of a decimal number contains..

Have a Good day... :)

Hm, the fastest way I know is in O(number of ones). Check out this article by misof, the "bitwise operations can be useful" part:

http://www.topcoder.com/tc?module=Static&d1=tutorials&d2=integersReals


The mean of a set of three numbers is 36. The number 40 is removed from the set. By how much is the mean reduced?

The current sum is 3x36=108. After you remove 40 the remaining two numbers sum to 108-40=68, so their mean is 68/2=34.

However, we want to know by how much the mean is reduced, which is 36-34=$ \boxed{2}$.


superior algebra
superior algebra unsolved

if K is a field extention of F and if a $ %Error. "display" is a bad command.
\in$ k such that  [F(a):F] is odd.show that F(a)=F($ %Error. "display" is a bad command.
\ a^2$).


[quote="hien"]Let $x,y,z>0$ and $x+y+z=1$. Prove that $xy+yz+zx>\frac{18}{2+xyz}$[/quote]sorry, but i think there is a mistake.. try $ \left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right) \Longrightarrow \left(\frac{1}{3}\right)\left(\frac{1}{3}\right)+\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)+\left(\frac{1}{3}\right)\left(\frac{1}{3}\right) \not> \frac{18}{2+\frac{1}{27}}$
maybe you meant to say $ \frac{1}{xy+yz+zx}>\frac{18}{2+xyz}$?

Thanks for show me a mistake of problem. But even the problem you had proposed isn't true with the case as you had showed.

[quote="hien"]Four days ago, I posted the wrong problem. Now i edit it as:
Let $x,y,z>0$ and $x+y+z=1$. Prove that $xy+yz+zx>\frac{18xyz}{2+xyz}$[/quote]ok, $(x+y+z)(yz+zx+xy)\geq (\sqrt{xyz}+\sqrt{xyz}+\sqrt{xyz})^{2}=9xyz> \frac{18xyz}{2+xyz}$ (since $xyz > 0$)

Very short solution.
This is my solution:
$2(xy+yz+zx)=(x+y+z+x+y+z)(xy+yz+zx)\ge 6\sqrt[6]{x^{2}y^{2}z^{2}}.3\sqrt[3]{x^{2}y^{2}z^{2}}=18xyz$
$\Rightarrow 2(xy+yz+zx)+xyz(xy+yz+zx)=(xy+yz+zx)(2+xyz)>18xyz$


A triangle has sides measuring 6, 7, 8 cm. An angle bisector is drawn to the 6 cm side. What is the length, in cm, of the smaller of the two segments into which the 6 cm side is divided?

[hide="hint"]Triangle angle bisector theorem[/hide]

[hide="solution"]
$ \frac{7}{8} \equal{} \frac{x}{6\minus{}x}$

$ 42\minus{}7x \equal{} 8x$

$ 42 \equal{} 15x$

$ x \equal{} \frac{14}{5} \equal{} 2.8$
[/hide]


inequalities
trigonometry
calculus
derivative
function
rearrangement inequality

Suppose WLOG $x < y$.  For the rearrangement inequality we have that the left side is less than
\[
\frac{x}{\sqrt{x^2 + 4}} + \frac{y}{\sqrt{y^2 + 4}}
\]
We have also:
\[
\frac{x}{\sqrt{x^2 + 4}} < \frac{\sqrt{2}}{2}
\]
which is true because it's equivalent to $x^2 < 4$. Applying the inequality to $y$ and summing up we have finished.

Regards,
Spider

Thanks for your reply, [b]spider[/b]. I think your solution is the shortest one. But here is another one : Denote $x=2\tan{\alpha}, y=2\tan{\beta}$ where $\alpha, \beta\in\mathbb(0;\frac{\pi}{4})$.Putting this into inequality we obtain: $\frac{\sin{\alpha}\cdot\cos{\beta}}{\cos{\alpha}}+\frac{\sin{\beta}\cdot\cos{\alpha}}{\cos{\beta}} <\sqrt{2}$.WLOG assume that $\alpha\geq\beta$, so  $\frac{\sin{\alpha}\cdot\cos{\beta}}{\cos{\alpha}}+\frac{\sin{\beta}\cdot\cos{\alpha}}{\cos{\beta}}\leq\frac{\sin(\alpha+\beta)}{\cos\alpha}\leq\frac{\sin2\alpha}{\cos\alpha}=2\sin\alpha<\sqrt{2}.$
QED

[quote="Spider"]Suppose WLOG $x < y$.  For the rearrangement inequality we have that the left side is less than
\[
\frac{x}{\sqrt{x^2 + 4}} + \frac{y}{\sqrt{y^2 + 4}}
[/quote]
I'm sorry for I don't think Spider's solution is right. because  the inequality above is less than the one in the question :)

zhaobin is right -- from rearrangement, we have $\frac{x}{\sqrt{y^{2}+4}}+\frac{y}{\sqrt{x^{2}+4}} \geq \frac{x}{\sqrt{x^{2}+4}}+\frac{y}{\sqrt{y^{2}+4}}$, not the other direction as Spider claimed.

Oh yeah, you are right. I haven't read his solution carefully.  :?

WLOG $x\geq y$.

Then taking the derivative of $f(x)=\sum x\sqrt {x^2+4}-2\prod \sqrt {x^2+4}$ gives

$f'(x)=2\cdot\frac {x^2+2-\sqrt {y^2+4}}{\sqrt {x^2+4}}$.  This is a discriminant with positive leading coefficient and $\triangle = 4(y^2+4)-32<0$.  Thus $f(x)$ is everywhere increasing, so we get:

$\sum \frac {x}{\sqrt{y^2+4}}<\frac {2}{\sqrt{y^2+4}}+\frac {y}{8}$

Now, taking the derivative with respect to $y$ gives $g'(y)=\frac {1}{8}-\frac {4y}{(y^2+4)^2}$.  This function (I'll let you guys do the math, is minimized on its boundary and is otherwise $>0$.  Thus $g(y)$ is also increasing in $y$.  This means that

$\sum \frac {x}{\sqrt{y^2+4}}<\frac {2}{\sqrt{y^2+4}}+\frac {y}{8}<2\cdot\frac {\sqrt {2}}{2}=\sqrt {2}$, which is the desired result.

[quote="Spider"]Ops, you're right, excuse me  :blush:

Spider

PS: a very good first post!...  :lol:[/quote]
It doesn't matter.Mistakes make progress,isn't it?
here is my solution to this question
I proved that x 2 /(y 2 +4)+y 2 /(x 2 +4) <1
becaue  x 2(x 2 +4)+y 2(y 2 +4)-(x 2 +4)(y 2 +4)=x 4 +y 4 -x 2 y 2 -16=(x 2 +4)(x 2 -4)+y 2 (y 2 -x 2 )<0


quadratics
algebra
polynomial
algebra unsolved

Prove that the interval $\lbrack 0,1 \rbrack$ can be split into black and white intervals for any quadratic polynomial $P(x)$, such that the sum of weights of the black intervals is equal to the sum of weights of the white intervals.  (Define the weight of the interval $\lbrack a,b \rbrack$ as $P(b) - P(a)$.)

Does the same result hold with a degree 3 or degree 5 polynomial?

Beautiful problem.  I think that points $(1+cos(\frac{\pi i}{2n}))/2$ define such a partition for polynomials of degree less than $n$.

Can't we just split [0,1] into two intervals?

Consider $A = \frac{P(0) + P(1)}{2}$.  Consider some horizontal line $y = A$.  The arc from $(0, P(0))$ to $(1, P(1))$ must cross this line somewhere, say at ordinate x.  It is clear $0 \leq x \leq 1$ and thus [0,x] and [x,1] are intervals with equal weight.

?? how am I reading the question wrong??

Apparently the order of quantifiers can be different in Chinese and English, this kind of confusion has appeared in several of the China TST problems that were translated here.

I think what is required is to find one partition that works for all polynomials of the given degree; "there Exists a splitting for All $p(x)$").  The other interpretation, "for All $p(x)$ there Exists a splitting", trivializes the problem.


inequalities
geometry
3D geometry
sphere
tetrahedron
symmetry
rotation

Suppose that $n$ points ($n\ge2$) are scattered on the surface of the sphere of radius 1. Prove that the sum of the squares of all distances bewteen pairs of points is $\le n^2$.

Under what circumstances is there equality in this inequality? Also, does it matter whether this is the sphere in 3-dimensional space, or would it also work in $\Re ^m$ for arbitrary $m$?

Kent Merryfield wrote:Under what circumstances is there equality in this inequality?

Easy to show that [hide] n points equally spaced around any great circle satisfy the equality.  Easy to show that 8 points arranged in a cube satisfy the equality.  Likewise for 6 points arranged in a regular octahedron.  And 4 points arranged in a regular tetrahedron also comply.  Perhaps the other two platonic solids?

The equality with respect to the regular tetrahedron provides a cool identity, which is the (well-known?) relationship between the edge length of the tetrahedron and the radius of the circumscribed sphere.   [tex]\frac{s}{R}=\sqrt{\frac{4^2}{6}}[/tex]. [/hide]

rcv, those are all situations in which the inequality is an equality, but that does not come close to a listing of all of the possible cases. The full characterization should be evident from a proof of the inequality.

Thinking a little more about this subject, [hide] I use a center of mass argument.  If the n equally-weighted points have a center of mass at the center of the sphere, then equality holds.  That is true with all of the cases I previously provided, including the other two platonic solids.  If the "center of mass" argument is correct, then these equality cases should extend to higher dimensions.  But I'll further explore the case of 3 dimensions.

By way of example, consider the sphere to have lines of latitude and longitude.  I place one point (typically on the north pole).  And I consider a few cases of n=2, 3, and 4.  If n=2, the other point must lie at the south pole for the centroid to be in the center of the sphere.  The condition of the problem is satisfied ([tex]2^2=4=n^2[/tex]).

If n=3, the three points necessarily define a plane.  The plane must pass through the centroid, which is the center of the sphere.  A symmetry argument requires that the resulting figure be an equilateral triangle, passing through the center of the sphere ([tex]3(\sqrt{3})^2=9=n^3[/tex]).

If n=4, the situation gets more complicated.  The square lying on the great circle and the tetrahedron are two examples.  We can also duplicate the point at the north pole, which then requires two points at the south pole ([tex]4\times 2^2=16=n^2[/tex]).  If we consider a great circle, we can choose any two arbitrary points on the circle, and require the other points lie on the diametric opposite of the circle.  (These points form a rectangle.  The two diagonals are equal to 2, and by the pythagorean theorem the sums of the squares of adjacent pairs of sides are equal to 4.)  The general case of 4 points is as follows:  Without loss of generality, we consider any pair of points.  We rotate the sphere (or the points) such that the points lie on the same latitude, :phi:, with logitudes, :theta:, differing by 180 degrees.  In order for the centroid to fall on the N/S axis of the sphere, the other two points must have longitutes differing by 180 degrees.  By symmetry, they must be equal distance on the other side of the equator.  So, following is the complete set of 4 points that have a centroid in the center of the sphere:  [tex]\{(\phi, \theta_1), (\phi, \theta_1+\pi), (-\phi, \theta_2), (-\phi, \theta_2+\pi)\}[/tex].  [Rotations of the above points are also acceptable.]

I'll give one more specific example for the case of n=5.  Let the first point lie at the north pole.  Let the remaining 4 points have a common latitude and be spaced at intervals of 90 degrees in longitude.  [We are inscribing a square-based pyramid in our sphere, however any rectangular base will yield equivalent results.]  Using the center-of-mass and cartesian coordinates, [tex]\sum_k{z_k}=0[/tex], if the apex of the pyramid has z=1, then 4 points of the base must have z=-0.25.  The diagonal of the square base of this pyramid would have length = [tex]\frac{\sqrt{15}}{2}[/tex].  The side-length of the square base would have length = [tex]\frac{\sqrt{30}}{4}[/tex]  The length of the other edges of the pyramid would have length = [tex]\sqrt{\Big(\frac{\sqrt{15}}{4}\Big)^2+\Big(\frac{5}{4}\Big)^2}=\frac{\sqrt{10}}{2}.[/tex]  The sum of the squares of distances between points is [tex]4\times\Big(\frac{\sqrt{10}}{2}\Big)^2+4\times\Big(\frac{\sqrt{30}}{4}\Big)^2+2\times\Big(\frac{\sqrt{15}}{2}\Big)^2=25=n^2[/tex].

A trivial remark about higher dimensions...  A sphere has an infinite number of "great circles".  Any case where all n points lie in the same plane will apply to every great circle on a sphere in 3 dimensions.  Similarly, a hypersphere has an infinite number of "great spheres".  Every case in which equality holds in a sphere would hold in a "great sphere" on a hypersphere.  By induction, this argument extends to higher dimensions.  However, this argument says nothing about the situation when 5 or more points extend outside 3-dimensional space into 4- or more dimensions.

I have postulated a general result for equality.  "Equality will result in every case where the center of mass of equally weighted points is the center of the sphere in m-space."

I have demonstrated equality for the general cases where n<=4.  I have demonstrated equality for a specific case where m=3, n=5.

I have not proven that having the centroid in the center of the sphere guarantees equality for all n>=5.

I have not proven that there are no other cases of equality.

I have not proven the inequality.[/hide]

rcv saidQuote:[hide]I have postulated a general result for equality. "Equality will result in every case where the center of mass of equally weighted points is the center of the sphere in m-space."[/hide]

This is the correct hypothesis, and it is an if and only if condition. However, you have only proved it for a small number of special cases involving small n, and you have not proved the inequality at all. Can you step away from the special cases and focus on proving the inequality?

Hint: [hide]Use vector notation.[/hide]

Nice problem!

Let ${\bf v_1},\ldots, {\bf v_n}$ be $n$ unit vectors (points on the unit sphere) in $\mathbb{R}^m$, then the sum of the squares of all distances bewteen pairs is $\displaystyle \sum_{1\leq i < j \leq n} \left\| {\bf v_i}- {\bf v_j} \right\|^2$.

$\displaystyle \sum_{1\leq i < j \leq n} \left\| {\bf v_i}- {\bf v_j} \right\|^2
	+ \left\|\sum_{i=1}^{n} {\bf v_i}\right\|^2 $

$\displaystyle = \sum_{1\leq i < j \leq n} ({\bf v_i}- {\bf v_j}, {\bf v_i}- {\bf v_j})
	+ \left(\sum_{i=1}^{n} {\bf v_i}, \sum_{i=1}^{n} {\bf v_i}\right) $

$\displaystyle = \sum_{1\leq i < j \leq n} [({\bf v_i}, {\bf v_i}) - 2({\bf v_i}, {\bf v_j}) + ({\bf v_j}, {\bf v_j})]
	+ \left[\sum_{i=1}^{n}({\bf v_i}, {\bf v_i})  +  \sum_{1\leq i < j \leq n} 2({\bf v_i}, {\bf v_j})\right] $

$\displaystyle  = 2\cdot {n\choose 2} + n = n^2$.

Therefore $\displaystyle \sum_{1\leq i < j \leq n} \left\| {\bf v_i}- {\bf v_j} \right\|^2 \leq n^2$ and the equlity holds when $\displaystyle \sum_{i=1}^{n} {\bf v_i} = {\bf 0}$.

Looks like fuzzy beat me to the solution, but I'll post my version, anyway...  I'll sum the distances between every pair of points in both directions and including the distance from a point to itself:

Define the points in cartesian coordinates:  [tex]P_k=(x_k, y_k, z_k)[/tex], [tex]1\le k\le n[/tex]

[tex]\displaystyle 2S=\sum_{i=1}^n{\sum_{j=1}^n{|P_i-P_j|^2}}[/tex]

[tex]\displaystyle 2S=\sum_{i=1}^n{\sum_{j=1}^n{(x_i-x_j)^2+(y_i-y_j)^2+(z_i-z_j)^2}}[/tex]

[tex]\displaystyle 2S=\sum_{i=1}^n{\sum_{j=1}^n{(x_i^2-2x_ix_j+x_j^2)+(y_i^2-2y_iy_j+y_j^2)+(y_i^2-2y_iy_j+y_j^2)}}[/tex]

[tex]\displaystyle 2S=\sum_{i=1}^n{\sum_{j=1}^n{(x_i^2+y_i^2+z_i^2)+(x_j^2+y_j^2+z_j^2)}}-2\sum_{i=1}^n{\sum_{j=1}^n{x_ix_j+y_iy_j+z_iz_j}}[/tex]

Given all points on the surface of the unit sphere, the leftmost terms are unity.

Given [tex]\displaystyle\sum_{j=1}^n{x_j}=0[/tex], [tex]\displaystyle\sum_{j=1}^n{y_j}=0[/tex], [tex]\displaystyle\sum_{j=1}^n{z_j}=0[/tex], the rightmost inner summation vanishes.

[tex]\displaystyle 2S=\sum_{i=1}^n{\sum_{j=1}^n{(1)+(1)}}-2\sum_{i=1}^n{0}[/tex]

[tex]\displaystyle  2S=2n^2-0[/tex]

The total distance between each distinct pair of points is [tex]S=n^2[/tex] when the centroid of the points is the center of the sphere.

If [tex]\displaystyle\sum_{j=1}^n{x_j}\ne 0[/tex], or [tex]\displaystyle\sum_{j=1}^n{y_j}\ne 0[/tex], or [tex]\displaystyle\sum_{j=1}^n{z_j}\ne 0[/tex], the rightmost summation would be a finite positive value.  So this demonstrates that the inequality holds.

Yes, this was a nice problem.  For me, the key was postulating (and confirming) the general rule for equality.  Then working bottom-up towards the proof.  Extension to higher dimensions is now trivial.

Appendix:  Dotting the last i of "inequality"...

[tex]\displaystyle \sum_{i=1}^n{\sum_{j=1}^n{x_ix_j+y_iy_j+z_iz_j}}[/tex]

[tex]\displaystyle =\sum_{i=1}^n{\sum_{j=1}^n{x_ix_j}}+\sum_{i=1}^n{\sum_{j=1}^n{y_iy_j}}+\sum_{i=1}^n{\sum_{j=1}^n{z_iz_j}}[/tex]

By the distributive property of multiplication over addition:

[tex]\displaystyle \sum_{i=1}^n{\sum_{j=1}^n{x_ix_j}}=\sum_{i=1}^n{x_i\sum_{j=1}^n{x_j}}=\Bigg(\sum_{i=1}^n{x_i}\Bigg)\Bigg(\sum_{j=1}^n{x_j}\Bigg)=\Bigg(\sum_{j=1}^n{x_j}\Bigg)^2\ge 0[/tex].

So, [tex]\displaystyle \sum_{i=1}^n{\sum_{j=1}^n{x_ix_j+y_iy_j+z_iz_j}}\ge 0[/tex], with equality where the centroid is at (0,0,0).

That's it - you both got it. My own favorite version of the proof uses vector notation (as in fuzzylogic's proof) but, like rcv, I arrange to sum over the "whole square" $1\le i\le n,1\le j\le n$ rather than just the "upper triangle" $1\le i<j\le n$. Note that the "diagonal" ($i=j$) terms are zero. You both claim it's true in $\Re^m$ for any $m$, but there's no need to stop there. The same inequality with the same proof and the same equality condition would be true in any real or complex inner product space.


Why didn't YOU hide YOUR message about hiding, then?

I wanted to be surprised about its content.

It's a multiple-way street, see?

[quote="bowei"]Please, everyone who posts here:

When you want to post the answer to a puzzle or a brainteaser, please hide your answer so others will not have their fun spoiled.

use a [hide] followed by some stuff
and then a [(backslash)hide] only actually use a backslash.

That way, you won't spoil the fun![/quote]

in fact you don't even need a backslash anymore. pressingsubmit would automatically do it for you


Prove that if a and b are positive integers and  then a=b.

I have a solution but am wondering if anyone has another way.

My solution is in spoiler.

[hide]Translating those into inequalities, you get a<=b^2, b^2<=a^3, ... or, for odd n>0, a^n<=b^(n+1) and also for even m b^m<=a^(m+1).  So as n and m approach infinity, you get the equivalent of a<=b and b<=a, so a=b.[/hide]

I have a similiar solution but is a bit more mathematically precise. Let p be a prime with alpha  be the largest number so that p^alpha is a divisor of a.  beta is defined similiarly for b. From a^(2k-1)|b^(2k) it follows

alpha(2k-1) <= 2kbeta.

Similiarly you get

2kbeta<=(2k+1)alpha.

From the both inequalities you get

(2k-1)/(2k)<=(alpha)/(beta)<=(2k+1)/(2k) for all k.

Since both sequences (2k-1)/(2k) and (2k+1)/(2k) converge to 1 the constant sequence alpha/beta also converges to 1.  Since the proof is the same for all primes it follows a=b.

Misha

The solution you have is similar to mine in that it relies on stating that as k approaches infinity, the 2 numbers become equal.  I'm thinking (hoping) there would be a way to do it without invoking the infinite, because when things deal with infinity the usual rules break down.

You have to approach infinity because otherwise it is obviously possible to choose two very large integers that are only very slightly different, which fulfil the divisibility conditions for maybe hundreds of powers.

let $gcd(a,b)=k$, then $a=k*m$ and $b=k*n$ and that means that $gcd(m,n)=1$

our sequence of statements gives us

$\frac{b^{2k+2}}{a^{2k+1}}=\frac{n^{2k+1}*b}{m^{2k+1}} \in \mathbb{Z}$

but $gcd(m,n)=1$, so $m^{2k+1}|b$ for $k\in\mathbb{N}$

using the other part of the sequence, we have

$\frac{ a^{2k+1} }{ b^{2k} }=\frac{m^{2k} *a}{n^{2k}} \in\mathbb{Z}$

similarly, $n^{2k}|a$

as we are taking a and b to be finite, then $m=n=1$, so they are equal


probability
function
probability and stats

Suppose that we have a random experiment with outcomes in an arbitrary set S. I want to define a probability measure P on some sigma--algebra F=M(G) generated from a collection G of subsets of S. In order to do it, I know it is enough to define function P on the elements of G.

Is there any condition on G to ensure that the generated sigma--algebra F is rich enough so that we can distinguish the probabilities of different elements and subsets of S?

A possible answer that I don't know whether if it is correct or not is this one: 'given two arbitrary elements s_1 and s_2 in S, G must contain at least one subset of S that contains only one of those elements'.

Does exist something like this?

I'm not sure there's any simple condition you can state just using the $ \sigma$-algebra, but what you have clearly allows bad things.
For example, consider the countable/cocountable $ \sigma$-algebra on $ \mathbb{R}$, and the measure $ \mu(X)\equal{}\begin{cases}0&X\text{ is countable}\\1&\mathbb{R}\setminus X\text{ is countable}\end{cases}$.
That can come from any continuous probability distribution on $ \mathbb{R}$, which isn't good.


An integer sequence is related with the formula $(n-1)a_{n+1}=(n+1)a_n-2(n-1)$ for $n \geq 1$
If $2000|a_{1999}$,  determine the least value of $n$, such that $n \geq 2$ and $2000|a_n$

$(2-1)a_2=(1+1)a_1-2(1-1)\Longleftrightarrow a_2=2a_1$

Deviding the both sides of the given recursive formula by $(n-1)n(n+1)\ (n=2,3,\cdots)$, we have

$\frac{a_{n+1}}{n(n+1)}=\frac{a_n}{(n-1)n}-\frac{2}{n(n+1)}\Longleftrightarrow \frac{a_{n+1}}{n(n+1)}=\frac{a_n}{(n-1)n}-2\left(\frac{1}{n}-\frac{1}{n+1}\right)$

$\Longleftrightarrow \frac{a_{n+1}}{n(n+1)}-\frac{2}{n+1}=\frac{a_n}{(n-1)n}-\frac{2}{n}=\cdots =\frac{a_2}{(2-1)\cdot 2}-\frac{2}{2}=\frac{1}{2}a_2-1=a_1-1$

$\Longleftrightarrow a_n=(n-1)\{(a_1-1)n+2\}$

nice approach, kunny. However, I am not sure if you answered the actual problem. I would use the divisibility to find the $n$. Would that be a good way to approach?

Thanks,xxreddevilzxx.I can't solve this problem, because i'm poor at the problem in relation to number theory.

kunny, I think the very first line of your proof is wrong.  You substituted n = 2 into the very first term but n = 1 everywhere else.  Doing something similar to what you did, however, I get
[hide]$a_n = \frac{n(n-1)}{2}a_2 - (n-1)(n-2)$
From there, we can [hide]plug in $n = 1999$ to get $a_{1999} = \frac{1999\cdot1998}{2}a_2-1998\cdot1997\equiv0\mod 2000$ which we can solve to find $a_2 \equiv 6 \mod 2000$.  From there, we just have to find the smallest solution to $6\frac{n(n-1)}{2}-(n-1)(n-2)\equiv 0 \mod 2000$, which is the same as solving $(n-1)(3n-(n-2))\equiv0$ or $2(n^2-1) \equiv 0\mod2000$.  So we want the smallest square root of 1 greater than 1 mod 1000.  I leave it as an excercize to show that the smallest such number is 249[/hide][/hide]


Έστω $f: R \to R$ παραγωγισημη στο $R$ με $f'(x)+f(x)=3$
    $\forall x \in R$ και $f(0)=5$
Να βρεθεί ο τύπος της $f$

Πρέπει να είναι μια εύκολη άσκηση αλλά έχω κολλήσει και ζητάω την βοήθειά σας.

Αυτό που σκέφτηκα είναι

$\frac{f'(x)}{f(x)}+1=3$

$\frac{f'(x)}{f(x)}=2$

$\left( lnf(x) \right)' = (2x)'$

$lnf(x)=2x+C$

...

όμως αυτό θα μπορούσαμε να κάνουμε μόνο όταν η $f(x)>0$  $\forall x \in R$ άρα  άκυρο αυτό που έγραψα...

Μια παρατήρηση θα σου κάνω για να το παλέψεις μόνος σου..
Χρησιμοποιήσε το συμβολισμό του Leibniz $f'(x)=\frac{df}{dx}$ και θα την βγάλεις..

Ναι, ομως δεν με βοηθαει αυτο  :blush:

Για δοκίμασε να πολλαπλασιάσεις με $e^{x}$

alexandros_plus, οντως αυτο βοηθαει  :trampoline:

$e^{x}f'(x)+e^{x}f(x)=3e^{x}$

$( e^{x}f(x))'=(3e^{x})'$

$e^{x}f(x)=3e^{x}+C$

Για $x=0$ εχουμε $C=2$ αρα

$f(x)=\frac{3e^{x}+2}{e^{x}}$

bouzoukman, θα ηθελα πολυ να δω την λυση με την χρηση του συμβολισμου του Leibniz!

Mailo,nomizo oti to provlima sti lisi zu dn ine toso to oti 8a prepi na ine f(x)>0
 ayto dior8onete...gt oloklironontas +ta dio meli eki pu exis $\frac{f'(x)}{f(x)}=2$ 8a paris $In|f(x)|=2x+c$ +ara mporis na apofigis ayto to simio...
To la8os ine apo tin proti kiolas grami pu dieris me tin f(x).... :o gt dn dini i ekfonisi kamia sin8iki oti  i sinartsi f(x) dn exi rizes..... :maybe: kai ara ine diafori tu midenos...

 :D

RAKAR, εχεις δικιο  :blush:

Να ρωτήσω κάτι σχετικό, γιατί δεν μπορούμε να έχουμε μηδέν στον παρανομαστή(ένταξη το ξέρουμε γιατί μας το είπαν οι καθηγητές στο σχολείο όμως το γιατί δεν μας εξήγησαν ποτέ) ? Τι είναι ακριβώς το μηδέν?

http://en.wikipedia.org/wiki/Division_By_Zero

Σχετικά με τη διαίρεση με το 0

Εγώ πάντως που ήθελα να εξηγήσω το ίδιο θέμα σε μια φίλη μου δεν ασχολείται και πολύ με τα μαθηματικά της είπα τα εξής:

"Σκέψου ότι το $\frac{1}{3}$ έχει τη έννοια ότι έχεις μία πίτα κομμένη σε τρία κομμάτια και παίρνεις το $1$.Με την ίδια λογική το  $\frac{1}{0}$ έχει την έννοια ότι παίρνεις ένα κομμάτι ενώ δεν υπάρχει καθόλου πίτα.... :wink: "

Ελπίζω αυτό να σε βοήθησε να καταλάβει ποιοτικά το θέμα....

Ξαναδές στη Wikipedia στο άρθρο http://en.wikipedia.org/wiki/Power_%28mathematics%29

Έχει αφιερωμένο ένα ολόκληρο τμήμα (Zero to the zero power)

Συγγνώμη αλλά έχω μπερδευτεί :blush:
Τόσο καιρό νόμιζα ότι θεωρούσαμε το $0^{0}=0$...Είναι λά8ος αυτό γιατί το αντίστοιχο τμήμα στο Wikipedia ούτε που αναφέρει αυτή την εκδοχή :huh:
Θα μπορούσε κανείς να μου πει ως τί να το θεωρώ στους μαθηματικούς διαγωνισμούς (συγκεκριμένα της ΕΜΕ) :?:

[quote="silouan"]Κοίταξε το $0^{0}$  αν δεν κάνω λάθος δεν ορίζεται όπως μαθαίνουμε στο σχολείο .
Όμως μπορούμε να πάρουμε το εξής
$\lim_{x\rightarrow 0}x^{x}=1$[/quote]
αυτο ειναι μια εκδοχη.
Φιλε bilot4 αποκλειεται να χρειαστεις σε διαγωνισμο τετοια ποσοτητα ,ειναι θεμα αναλυσης και μπερδεμενο

Σωστό αλλά θα μπορούσε να έχει σημασία στην περίπτωση που $x^{y}=1$ όπυ συμπαιρένουμε ότι το $x$ δεν είναι $0$

$x^{n}=1\Rightarrow $ ριζες της μοναδας $\Rightarrow $μιγαδικους ,(αν ενδιαφερεσαι να μην το δεις σε διαγωνισμο της ΕΜΕ πραγματι δεν θα το δεις αφου στην γ λυκειου ειναι εκτος υλης λογικα).


trigonometry
function
algebra unsolved
algebra

Find the minimum value of $ y \equal{} {3^{\left| {{\mathop{\rm s}\nolimits} {\rm{inx}}} \right|}} \plus{} {3^{\left| {\cos x} \right|}}$

hello, the searched minimium is $ 4$ and will be attained for $ x\equal{}0$.
Sonnhard.

Take $ x\in[0,\pi/2]$. Prove that $ y''<0$. Thus $ \minus{}y$ is a convex function and its minimum is reached for $ x\equal{}0$ or $ x\equal{}\pi/2$.

[quote="loup blanc"]Take $ x\in[0,\pi/2]$. Prove that $ y'' < 0$. Thus $ \minus{} y$ is a convex function and its minimum is reached for $ x \equal{} 0$ or $ x \equal{} \pi/2$.[/quote]

Nice thinking.y is a periodic function of x having period $ \frac{\pi}{2}$.Therefore,minimum and maximum value of f must be present in this interval.

Anyone to prove that $ y'' < 0$ ?
I've calculated the

$ y'' \equal{} \ln(3) \left (3^{\sin x}\ln(3){\cos^2x} \plus{} 3^{\cos x}\ln(3){\sin^2x} \minus{} 3^{\sin x}\sin x \minus{} 3^{\cos x}\cos x \right)$

where $ x \in (0,\pi/2)$ but I cannot prove that $ y'' < 0$


I know for a fact that the result of this is x>-2. What i want is the step by step to get there.

I understand what you mean but for every real number this works, I think there is no exact step by step way to get it.
Same as $x+2=x+\frac{4}{2}$
Can you find $x$ by a step-by-step solution???? no.

[quote="236factorial"]It seems like solving the inequality does no good. we know that if x is positive, then
$\frac{x}{x+2}$ must be less than 1. What if x is negative?

$\frac{-1}{1}$ = -1. This is good.

$\frac{-2}{0}$ = undefined. No good.

$\frac{-3}{-1}$ = 3. It's over 1!!

$\frac{-4}{-2}$ = 2. It's also over 1.

From then on, the any number x less than that won't work, because the absolute value of the numerator is always greater than the absolute value of the denominator. That makes the absolute value of the quotient over 1.
When x<-2, x+2<0, so they're both negative. When they are divided, the result is positive.

Combining those two pieces of information, we conclude that the inequality holds when $x>-2$.[/quote]

oh oh, silly me, I should've plugged in more numbers  :blush:

[quote="Iversonfan2005"]$x$ is equal to all real numbers because $x<x+2$.[/quote]

This is false :)
In fact, the solution is : x > -2 because when you multiply the RHS "x+2" has to be positive, it's only the case when x > -2. ;)

Take a look here. [url]http://www.purplemath.com/modules/index.htm[/url] And you'll get a clearer picture of the generic inequality solution method. (Wow is that a word?! :)

I have another answer:

$\frac{x}{x+2}<1$

$0<1-\frac{x}{x+2}$

$0<\frac{x+2-x}{x+2}$

$0<\frac{2}{x+2}$

You want this positive. 2 is positive. When is fraction positive? When numerator and denominator both positive or both negative.

2 is positive and you must do $x+2$ positive too.

$x+2>0$

$x>-2$

OK now with thes steps?  ;)

Above works.  Also you could look at it like this:

If $x+2 > 0$ then $x < x+2$, which is true for all real x.

If $x+2 < 0$ then $x > x+2$, which is never true.

So it must be $x+2 > 0$, therefore $x > -2$

Thanks again, this new answers are more simple, thanks! If these is not high school where should i post? Please excuse me for posting here.

[quote="sanchez45"]If these is not high school where should i post? Please excuse me for posting here.[/quote]

I think this question is fine here. MAybe the basics forum in the middle school forum is better, but i'm not sure.

No, it doesn't belong in the basics forum. This forum is fine, since that's the kind of material that's tought in Algebra 2.

shinwoo, you even got the answer wrong!

The answer is not "all integers"

It's "$x>-2$" and this does NOT belong in basics, because the fact that you have to consider negatives brings you into a harder territory.

Anyone can say.

"$\frac{x}{x+2}>1 = x>x+2$ See all real numbers?!?!?!"

But the fact is, is that when $x+2$ is negative, you have to switch the signs when you multiply! (key to solving equation!)
So when $x+2$ is negative or $0$ (we see that $x=-2$ gives undefined) the inequatlity does not work, therefore we set $x+2>0$ is $x>-2$


Find all functions $f: R \rightarrow R$ such that :

$f(f(x-y)=f(x)f(y) - f(x) +f(y)-xy$ for all $x,y$

Put x=y, $f(f(0))=f(x)f(x)-x^2$. Put x=0, f(f(0))=f(0)f(0). Thus $f(x)^2-x^2=f(0)^2$(*). This implies |f(x)|=|f(-x)|.
Now put y=0 in the initial equation, f(f(x))=f(x)f(0)-f(x)+f(0).
       put x=0 in the initial equation and replace y by x, f(f(-x))=f(x)f(0)+f(x)-f(0)
Note that |f(f(x))|=|f(|f(x)|)|=|f(f(-x))|, or equivalently (f(f(x))+f(f(-x)))(f(f(x))-f(f(-x)))=0, f(x)f(0)(f(x)-f(0))=0.
If f(a)=0 for $a\neq0$, then from (*), $-a^2=f(0)^2$, contradiction.
If f(a)=f(0), by (*), a=0.
Therefore f(x)=0 iff x=0. By (*), f(x)=x or -x. If f(x)=x, then f(f(x))=f(x)f(0)-f(x)+f(0)=-f(x)=-x, contradiction.
Hence f(x)=-x, it indeed satisfies the initial condition.


probability AMC USA(J)MO USAMO algebra polynomial function \/closed

The Art of Problem Solving Spring Schedule is now available.  We are offering four courses this spring.  Details about the courses are listed below.  More information, including how to enroll, can be found by [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]clicking here[/url].

[b]Introduction to Counting & Probability[/b] (Feb 21 - May 9, 7:30 PM - 9 PM ET/4:30 PM - 6 PM ET).  This course covers basic and intermediate counting concepts, including casework, multiplication, permutations, combinations, Pascal's triangle, probability, combinatorial identities, and the Binomial Theorem. Taught by AHSME perfect scorer and USAMO winner David Patrick.

[b]Intermediate Algebra[/b] (Mar 6 - May 22, 7:30 PM - 9 PM ET/4:30 PM - 6 PM ET). Algebraic subjects covered include polynomials, functions, logarithmic equations, clever substitutions of variables, systems of equations, symmetric sums, advanced factoring methods, and binomial expansion.  Taught by IMO Medalist Valentin Vornicu.

[b]Intermediate Number Theory Seminar[/b] (Feb 28 - Apr 18, 7:30 PM - 9 PM ET/4:30 PM - 6 PM ET). An 8 week problem solving seminar (no exams) with number theory using algebraic and counting approaches. Topics also include Diophantine equations (Pell equations), Fermat's Little Theorem, the Phi Function, and Euler's Theorem.  Taught by 2-time IMO medalist Naoki Sato.

[b]Special AIME Problem Seminar[/b] (Mar 3 & Mar 4, 3:30 PM - 6:30 PM ET/12:30 PM - 3:30 PM PT).  This 6-hour course will consist of two 3-hour sessions during which we will tackle AIME-style problems and discuss test-taking strategies. A 30 minute break during each session means 5 hours of total instruction. Transcripts for the classes will be available to enrolled students as well as a message board where students can discuss class problems. Note: this seminar includes material from the AIME Problem Series A.   Taught by AIME perfect scorer and USAMO winner Richard Rusczyk.

Do the Intro to Counting & Probability as well as the Intro to Number Theory go along with the Intro books?

-jorian

Is it required to have the book(s)? (I already have them so it shouldn't be a problem)

-jorian

Is Intermediate Algebra not appropriate for someone about to start AP Calculus?

Thanks,
lyra

[quote="rrusczyk"]The Art of Problem Solving Spring Schedule is now available.  We are offering four courses this spring.  Details about the courses are listed below.  More information, including how to enroll, can be found by [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]clicking here[/url].
[/quote]

Well, I know reading and following the link is a very challenging task. So you will see they will cover advanced university courses in statistics.  :D And yes, it is free. AoPS people live from love and air.  :wink:

[quote="lyra"][quote="rrusczyk"][quote="lyra"]Is Intermediate Algebra not appropriate for someone about to start AP Calculus?
[/quote]

That's hard to say - your AMC scores would be a better proxy.  I'd say that if you are already in the 130s or higher on the AMC with ease, and expect 5+ on the AIME, then the Intermediate Algebra class may be on the easy side.  Being in AP Calculus means that you probably have the basic concepts down, but AMC performance (and on other contests) will be a better indicator of how suitable the course is for you.[/quote]

Okay, thanks!

~lyra[/quote]

I think that means you're supposed to post your AMC scores...

Nah, she doesn't have to - she can ask me in class if she needs more guidance.

[quote="math92"][quote="bubka"]Are these courses free, and are they suitable for statistics major freshmen? :roll:[/quote]
If you click the link, most of the classes are over 100 dollars.[/quote]

Even if I didn't click the link, most of the classes would be over 100 dollars, as somebody else had already clicked the link (even if that wasn't true, somebody had written it) so the probability function had collapsed for him, and once a probability function collapses, it collapses identically each subsequent time.

Coming back to the topic, I am now beginning to wonder that the AoPS people make a living out of this site. Is that true, gentlemen, or do you do something else as well? (This time I am not villifying this as a money-making racket, I am only curious.)

[quote="bubka"]Coming back to the topic, I am now beginning to wonder that the AoPS people make a living out of this site. Is that true, gentlemen, or do you do something else as well? (This time I am not villifying this as a money-making racket, I am only curious.)[/quote]We are trying to "make a living". All of us working with AoPS could have pursued very high paid jobs in Finance firms (Richard did actually work for those in the past) but we choose to do what we like even if it doesn't give us big financial satisfactions. We mostly do it for you students, even if some haven't yet realized it ;)

[quote="bubka"]That's generous of you. I take back some of my comments in the WOOT thread, though I am frightfully sorry I am not in a postion to join any of your courses. So why don't you open institutes of your own, and expand by getting hold of more people? Surely there are many more interested mathematicians, especially old ones (like Kent Marryfield). I can think of at least a dozen from our institute and that's just one institute out of hundreds. After all, you can only reach so many people online.[/quote]

Time, energy, money, are all constraints to do something like that.  At this point, we are instead focusing on developing tools that will help enable others develop institutes like that.  Specifically, we are creating texts modeled on the approach of our courses.  These texts are then used by learning centers for classes just as you suggest.  And we are trying to build other resources for people who would like to build schools, such as this site.

(FYI - Kent isn't that old!)

He wrote somewhere he was 53! That's pretty old, eh?

Anyway I am willing to talk to some teachers in our institute about this. I have one particular gentleman in mind, our analysis teacher. He can suggest others. As anyone teaching in a university is pursuing something other than money only, it should work.

Now there are many boys on this site from many countries who could do just the same. So I don't see why it shouldn't spread without much difficulty.

[quote="towersfreak2006"]Will there be a mock AIME provided with the AIME Problem Seminar?[/quote]
i thought you were in college  :maybe:

I don't know my AIME score yet (that's a month later i can't wait!!) but I know that my AMC 10 score is 138.

Should I take it?

if I don't have the specific books, can I follow along in the Art of Problem solving: the Basics ?

We are not planning to have a mock AIME with the Special AIME Seminar this year.

If you have a 138 on the AMC10, you are probably a good candidate for trying out the Special AIME Seminar.

MichaelFaraday: For the Intro Counting class, it would be far better to have the actual Intro Counting textbook than to follow along in AoPS Volume 1.  (For the other classes, there is not currently a textbook that goes with the class).

[quote="emilgouliev"]will the transcript be available after the special AIME seminar? if so, for how long? (I might miss the first one or both)[/quote]

The transcript will be up for 3-4 weeks after the seminar.

How useful do you think it will be if you miss one day? (If I take it, I have to miss Sunday because of a math competition) Also, how useful will it be if I've already taken the AIME B Problem Series?

[quote="vibgyor"][quote="rrusczyk"]There won't be problems that are also in the AIME B Problem Series.  There will probably be more discussion of strategy, plus laundry lists of strategies to know about.  We often have students only attend one day (weekends are tough for many students to make both days!)[/quote]
Is there anything we do other than the time we spend in class for the special AIME thingie (i.e. a problem set)
Also, how long are the classes with Number Theory and Algebra[/quote]
those two classes are an hour and a half long

[quote="rrusczyk"]There won't be problems that are also in the AIME B Problem Series.  There will probably be more discussion of strategy, plus laundry lists of strategies to know about.  We often have students only attend one day (weekends are tough for many students to make both days!)[/quote]\
How likely is it that one would suffer from burnout so shortly before the AIME?

That already took place on Feb. 26.  Here's the transcript:
[url]http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=195[/url]

[quote="kstan013"]Is there no math jam for Intermediate Algebra class this time?[/quote]

[quote="vibgyor"][quote="rrusczyk"]There won't be problems that are also in the AIME B Problem Series.  There will probably be more discussion of strategy, plus laundry lists of strategies to know about.  We often have students only attend one day (weekends are tough for many students to make both days!)[/quote]
Is there anything we do other than the time we spend in class for the special AIME thingie (i.e. a problem set)
Also, how long are the classes with Number Theory and Algebra[/quote]

There is no problem set to turn in for the AIME class.  We will post the Mock AIME we gave in 2005 as part of the class (but this AIME is way harder than a regular AIME, so it should be used as practice hard problems!)

Each Number Theory class and each algebra class is 90 minutes long.

[quote="Ubemaya"]What was the average score for the mock AIME?[/quote]
Here's a mention of it:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=435489&highlight=mock#435489


Mods delete this topic. it is useless now.

there is?

oh.

i searched for one but i coudnt find it

I made an even dumber topic in GETTING STARTED and it didn't get deleted.....................WOOHOOOOOOO  ...so i have a bath with water...soap sometimes...u?

yup,some mods we have........so had a bath yet?

-.- I want to see how many people actually view this.

[quote="nutz_for2.718281828"][quote="bubka"]
Just for the record i am viewing this regularly.[/quote]

OMG :rotfl:  [hide=" :rotfl: "]You HAVE to be kidding![/hide]You HAVE to be kidding![/quote]

this is no joke...

wow this is an interesting topic, no wonder it's got 147 views.

Sounds like Elvis Presleys "Love me Tender", but it could also be Westlife's "Bap Bap Baby".

I have sebaceous cyst in my scalp just above my neck. Any idea how to survive it?

rip it off

arnav don't get cancer otherwise i lose $50 and a pizza

guess what song this is:

la la la la la la la la la la la

Some Hey jude song of beatles?

[quote="Treething"]rip it off

arnav don't get cancer otherwise i lose $\$$50 and a pizza[/quote]

It is all very easy for you to say rip it off, but it is rock hard and won't even budge when I press it. They'll probably need a pickaxe to make any headway into it!

guess what song this is:

la la la la la la la la la la la

there was some rap song that came out a few years ago with all these girls singing that[/code]

[quote="kheg_k"]no, the point of this topic was to offer a chess tournament, but that s already been invented, so i edited the post telling the mods to delete this topic. they didn t so...[/quote]

HAVE U TAKEN UR MEDS TODAY???????????????????????????????????????? OR HAVE YOU FALLEN AND CAN'T GET UP?

[quote="Treething"]rip it off

arnav don't get cancer otherwise i lose 50 and a pizza[/quote]

oh dear did you make some sort of bet that I won't get cancer

:rotfl:  :rotfl:  :rotfl:

what's so funny?

[quote="MysticTerminator"]IMO IS SO CROWDED[/quote]
So... now that the IMO test is over... how did you do on it?  :|

bumpity bumpity bump bump bump

 :rotfl: You can only post 3 smileys in one message, but I wanted 7...  :rotfl:
 :ddr:

:roll:  :roll:  :roll:  :?:  :?:  :idea:

i like smileys

Too many random topics in this forum nowadays.
I won't delete this, but I'll lock it...


search
group theory
superior algebra
superior algebra unsolved

[quote="Rust"][quote="cckek"]Find all morphism between $ (\mathbb{Q},+)$ and $ (S_{n},\circ)$[/quote]
What is $ S_{n}$? And morfism in what category?[/quote]

$ (S_{n},\circ)$ is the permutation group, and we search for the groupmorphisms.

Because $ S_{n}$ is not commutative for n>2 we get only trivial morfisms for these case and for n=1.
For n=2 we get $ f(x+x)=s*s=id, \ s=(1,2) \ or \ s=id$, because for all y exist $ x=y/2$ we get $ f(y)=id$ only trivial morfism.

The reason is not, that S_n isn't commutative. This argument can't work, because there are homomorphisms $ \mathbb{Z}\to S_{n}$ for example.

The reason is that $ \mathbb{Q}$ is divisble, while $ S_{n}$ is periodic: There exists a $ m\in\mathbb{N}$ s.t. $ \forall\sigma\in S_{n}: \sigma^{m}=1$. If f is a grouphomomorphism $ \mathbb{Q}\to S_{n}$ then we get $ \forall q\in\mathbb{Q}: f(q)=f(m\frac{q}{m})=f(\frac{q}{m})^{m}=1$ so there is only the trivial homomorphism f=1.


Can somebody help me with these 3 problems dealing with decidable and recursively enumerable languages?

1. Suppose L is the following language: {w}, where

w = 0 if God does not exist
    = 1 if God does exist

Is L decidable? The answer to this question does not have anything to do
with religous belief.

2. Show that the decidable languages are closed under union and comple-
mentation.

3.Show that the recursively enumerable languages are closed under concate-
nation and intersection.

Once again, it seems that I'm a little behind :). I'm not sure what RE languages are - are those the languages accepted by turing machines? And I don't know about the decidable languages either... but it sounds domewhat intuitive that if A is decidable and B is decidable than also (A & B), (not A) are decidable...

Hi,

I will prove that decidable languages are closed under complementation.

Do u recollect the proof of the fact that regular languages are also closed under complementation.?

Well the proof is really simple:- just make a new dfa (dfa 2) with its final states as non-final states of dfa1 (the dfa u intended to complement) and non-final states as the final states of dfa1. Clearly, dfa-2 accepts L-complement if L was accepted by dfa1.

The proof for the decidable language case is also similar. Try it out....

OK, now I know what decidable laguages are. Akash is right, constructing a turing machine that acceptes complement of L is not too hard. Now costructing a turing machine that accepts A or B is a little harder but it's possible - you need to run on your input the first turing machine and then the second turing machine, the only hard part is figuring out how to preserve the input and clear the memory for the second machine but there's a number of ways to do it, I'm sure you'll be able to figure out at least one such way...


AMC
USA(J)MO
USAMO
AIME
ARML
AMC 10
AIME I

Congratulations to the 9 people from Virginia who qualified for USAMO!!!

lol, 6 of those people are from TJ.

[hide="opinion"]
I thought the floor was way too low...  For such an easy AIME I, I thought the floor should have been at least a 7.  The statistics AMCDirector put up in the AMC forum seem to suggest that the AIME II was [i]easier[/i] than the AIME I, which I find hard to believe.
[/hide]

TJ sent alot more last year...

Your quote is right, infinity4ever. I shot to qualify for USAMO even though I had never went to mc districts and had previously never broke 96 on the AMC 10... I didn't make it but I missed breaking 140 on the AMC 10 by a question and answered 1-7 on the AIME II only to miss #1.

All I can think of now is that I wish I was as smart as you...
Also, I didn't happen to see you at TJ ARML practice did, I?

Without looking, my guess of the other 3 qualifiers not from TJ are Seung In, Yuqing, and you or someone I don't know.

~Not Kevin

yeah I was surprised by the low AIME cutoff as well. I suppose they made up for it with the secondary index, but it (AIME) seemed ALOT easier than last years. Well i guess i should just be glad that i made it, but now aime scores are useless :( Time to work on USAMO

@ aufha:  Wow nice improvement on the AMC10!!! A 40-point improvement is phenomenal!!! (lol I had a 40-point improvement too from last year to this year  :P  103.5==>144) but no, I did not go to ARML.  There were no qualifiers from Longfellow this year.   :(  I think that all three of us missed USAMO by one problem due to a silly mistake.  I got a 144/5 and misread # 3 on the AIME II.  :wallbash:  Kongster and some others from Governors made it.  :omighty:

@ Kongster:  nice job on making USAMO!!! AIME scores aren't completely pointless, aren't they still calculated into a USAMO index?

hmm im rly not sure, this is my first year too ;P
i think dpatrick said it wasnt, and mellowmelon, (other gov school who made it) said that they arent calculated anymore. Oh well, only 69? freshmen this year, sorta decent chance of making it :D gl next year


number theory
relatively prime
number theory proposed

Let n be a positive integer such that 2 + 2sqrt(28n^2 +1) is an integer. Show that 2 + 2sqrt(28n^2 + 1) is the square of an integer. My solution seems correct, but I don't like it ;) looking for some other ways.

If 2 + 2sqrt(28n^2 +1) is an integer, it is even and greater than 2, we can write it 2a+2.

Then 28n = (a+1)(a-1) then a is odd, a = 2b+1.
Then 7n=b(b+1)

b and b+1 are relatively prime, so b = 7c or b = c

. If b = c, b+1=7d and c=-1 mod 7 impossible

. If b = 7c, b+1=d then 2a+2 = (2d)


In a forest each of 9 animals lives in its own cave, and there is exactly one separate path between any two of these caves.  Before the election for Forest Gump, King of the Forest, some of the animals make an election campaign.  Each campaign-making animal ----- FGC (Forest Gump Candidate) ------ visits each of the other caves exactly once, uses only the paths for moving from cave to cave, never turns from one path to another between the caves, and returns to its own cave at the end of the campaign. It is also known that no path between two caves is used by more than one FGC. Find the maximum possible number of FGC's.

Okay, so we can have 3 FGCs.  Have them all start from their respective caves.  FGC#1 goes to every cave in order.  FGC#2 goes by twos, and FGC#3 goes by fours.  But then we have three disconnected K3's left.  However, since each vertex is of degree 8 we can have at most 4 FGCs.  So now to decide between 3 or 4.

I believe it's 4.


limit
function
calculus
calculus computations

$ \minus{} 1 \leq a_1 \leq 2$,  $ a_{n \plus{} 1} \equal{} a_n(a_n \minus{} 1)$, compute   $ \lim_{n\rightarrow\infty}a_n$

it suffices to study the function $ g(x)\equal{}x^2\minus{}\frac{3}{4}$ for $ |x|\le \frac{3}{2}$

with $ x_{n\plus{}1}\equal{}g(x_n)$ for different choices of $ x_1$
theres only two fixed points for $ g(x)$, $ x\equal{}\frac{3}{2}, \minus{}\frac{1}{2}$

but $ |x\minus{}\frac{3}{2}|<|g(x)\minus{}\frac{3}{2}|$ when $ x>\minus{}\frac{1}{2}$

and something similar happens near the other possible limit.. which means that the closer you get to each point.. the further you get pushed away..?

leads me to believe that there is no limit for a.e. choice of $ a_1$

of course we can solve for a countable number of solutions for each limit, but they only have measure zero..

would love to see this proved though.. i keep going in circles

Let $ f(x) \equal{} x(x \minus{} 1)$. Set $ f_0(x) \equal{} x, f_1(x) \equal{} f(x), f_2(x) \equal{} f(f(x)), ...$ Your problem is this: Given $ x \in [ \minus{} 1, 2]$, what is $ \lim_{n\rightarrow\infty} f_n(x)$?

Claim: if $ \minus{} 1 < x < 2$, then $ f_n(x) \rightarrow 0$; if $ x \equal{} \minus{} 1$ or $ 2$, then $ f_n(x) \rightarrow 2$.

This is obvious for $ x \equal{} \minus{} 1, 0, 1, 2$. For all other $ x$, $ f_n(x)$ will eventually land in $ [ \minus{} 1/4, 0]$. (Draw the graph of  $ f$ to see why this is plausible. Note that for $ x\in[0,1]$, this is immediate.) So if we prove the limit is $ 0$ for all $ x\in [ \minus{} 1/4, 0]$, we're done. Here it is actually easier to think about $ f_2(x) \equal{} x^4 \minus{} 2x^3 \plus{} x$.  The crux of the matter is this: if $ \minus{} 1/4 \leq x \leq 0$, then $ x \leq f_2(x) \leq 0$, which implies $ x \leq f_2(x) \leq f_4(x) \leq ... 0$. Thus  $ f_{2n}(x)$ converges, hence must converge to a fixed point of $ f_2$, which must be $ 0$. But then $ f_{2n \plus{} 1}(x) \equal{} f(f_{2n}(x)) \rightarrow f(0) \equal{} 0$. Thus the full sequence $ f_n(x)$ converges to $ 0$.

Obviously I've been brief, so ask questions if you like. There may also be a simpler way to do this.


Strange, setting $a=b \rightarrow 3/2^-$ and $c \rightarrow 0^+$ seems to disprove inequality


How many numbers can be written as the sum of two or more distinct members of the set $ \{0,1,2,3,4,5\}$?

$ 1\plus{}0\equal{}1$
$ 1\plus{}2\equal{}3$
$ 1\plus{}3\equal{}4$
$ 1\plus{}4\equal{}5$
$ 1\plus{}5\equal{}6$
$ 2\plus{}0\equal{}2$
$ 2\plus{}2\equal{}4$
After a while, you get the idea that the answer is simply $ 5\plus{}4\plus{}3\plus{}2\plus{}1\equal{}15$.

The largest sum possible is $ 0\plus{}1\plus{}2\plus{}3\plus{}4\plus{}5\equal{}15$, and the smallest is $ 0\plus{}1\equal{}1$. Thus there are 15 sums.

[hide=Alternate SOlution, similar to BOGTRO's]For numbers less than or equal to $5$ and greater than $0$, we can use said number and $0$. For numbers greater than $5$ and less than or equal to $9$, we can use $5$ and some number to make them. FOr numbers greater than $10$, we remember that we can use sums of more than two numbers. We have \begin{align*}
10&=5+4+1\\
11&=5+4+2\\
12&=5+4+3\\
13&=5+4+3+1\\
14&=5+4+3+2\\
15&=5+4+3+2+1
\end{align*} Thus, $15$ numbers can be made.[/hide]


Can anyone explain, using THERMODYNAMIC FUNCTIONS, why the solid/liquid equilibrium line in a phase diagram is nearly vertical while the liquid/gas line has a significant curve. (For example, see phase diagram for water)

Also can you give me an example of a chemical reaction that has :
$\Delta{H}$ (enthalpy) negative, $\Delta{S}$ (change in entropy) positve, and $\Delta{G}$ (change in free energy) negative

[hide]
As for your first question, the heat of fusion of water (energy per unit mass required to turn water from solid to liquid) is much smaller than the heat of vaporization of water (energy per unit mass required to turn it from liquid to gas). This is because in the liquid state of water, the closeness of the moleculues resembles the solid state far more than the gaseous state. Assuming that the heat of fusion is h, the heat of vaporization is H, the total energy required to melt water q, the total energy required to boil water Q, and the mass of the sample m, then
q = hm
Q = Hm
Since H>h and the mass constant, Q>q.

As for the second question, a large number of reactions would fit the criteria given. The one that comes to mind now is a combustion reaction:
C3H8(l) + 502 (g) goes to 3CO2(g) + 4H2O(g)
In this reaction, dH is highly negative, as a large amount of heat is given off overall, mainly due to the formation of the strong double bonds between C and O in CO2. The change in entropy, dS, is positive, since the number of molecules increases (by a 7 to 6 ratio) and moreover because the mixture changes from liquid-gas to pure gas, increasing the molecular disorder and the number of possible configurations for the molecules. Plugging in the negative value of dH and the positive value of dS into the equation dG = dH - TdS (in which T is always positive), we get a negative dG, favoring spontaneity. [/hide]

Think about what is being graphed (the axes).

If a change in pressure doesn't affect the transistion, the line would be vertical.

If you read the back of a cake box, there are high temperature cooking instructions.  When the pressure goes way down (such as for those in Denver) the boiling point is lowered so much that the affects of cooking are different.


function
limit
inequalities
calculus
calculus computations

Suppose that $f: (0,+\infty)\to\mathbb R$ is continuous with $\lim_{n\to\infty}f(nx)=0$ for all $x>0$. Prove that $\lim_{x\to+\infty}f(x)=0.$

why did you put it in this forum?  :maybe:
whats the relation between this problem and inequalities?  :huh:

Oh, no! I havn't post it, there must be someone got my password and post it!!! :mad:


function
real analysis
real analysis unsolved

try this one:

let P(x) a real polynomial. Consider the folowing sequence:
u_0=a, u_(n+1)=P(u_n)
Find an equivalent of u_n i.e. a function f(n,a) so that lim f(n,a)/u_n=1

In the original pb P(x)=x^2+x and a=1.. it was quite easy but what about this generalisation  :D

I remember that I have solved after a long struggle a particular case of this problem, I think $ x_{n+1}=\frac{1+x_n^2}{2} $, but I found it very hard. Do you really consider it easy? I would like to see your solution.

try with P(x)=x^2+x .. i remember it was easy because u0=1 but in your pb u0 was arbitrary .. isn t it ?


[quote="anthony19891019"]I just discovered that Mathlinks has two websites address..[/quote]

They used to be 2 different sites, but now they are the same.... MathLinks is like the "community" part of AoPS, which is the general site

Er, actually that's not how it is. They used to be different forums but then they were merged into one. Now they're just the same site with different skins.


trigonometry
ratio
geometry unsolved
geometry

Let a right angle ABC with $\angle C=90$, let $D$ the midpoint of $AC$, and $E,F$ points of the sides $AB, BC$ respectively such that $\angle DEA=\angle FEB$
$\angle EFB=\angle AFC$, and $BC=3,AC=4$
find: $\frac{DE+EF}{AF}$

I got $\frac{43}{50}$, but my solution is too ugly.

See the attached diagram.

Construct points $D'=\mathcal{S}_{AB}(D)$ and $A'=\mathcal{S}_{BC}(A)$. Then obviously $D', E, F, A'$ are collinear (opposite angles) and $DE+EF=D'F, FA=FA'$. Draw perpendicular $D'M$ to $AA'$. Because of the collinearity, we have ${D'F\over FA'}={MC\over CA'}$

Now $AM=AD'\cos 2\alpha=AD(2\cos^{2}\alpha-1)={b\over 2}\left({2b^{2}\over c^{2}}-1\right)=\frac{b(2b^{2}-c^{2})}{2c^{2}}$

Then ${MC\over CA'}=\frac{b-\frac{b(2b^{2}-c^{2})}{2c^{2}}}{b}=1-\frac{2b^{2}-c^{2}}{2c^{2}}=\frac{3c^{2}-2b^{2}}{2c^{2}}={3\over 2}-{b^{2}\over c^{2}}=\boxed{{1\over 2}+\sin^{2}\alpha}$

For $a=3, b=4$ we have $\sin\alpha={3\over 5}$, thus the desired ratio is ${1\over 2}+{9\over 25}={43\over 50}$


Duke
college
Stanford
MIT
linear algebra

I want to know what everyone thinks of these schools: Duke, Michigan, UVA, and Texas A&M.  Tell me what you think the pros and cons of each school is.

Thanks.

On Texas A&M:
It's located in a fairly small and quiet town
Pretty large school though
Its sciences/engineering are pretty good.  Not elite, but good for a public university...I don't know anything about its mathematics programs, though.

Some people here in Austin say they "brainwash" you (from what I know I think Aggies are a pretty close group)

Overall a pretty good school but as far as public schools in Texas go, I would probably give UT Austin a slight edge.  (I live in Austin, after all)

In Texas UT Austin/Texas A&M are the flagship public universities...fairly respectable but not "great"

I'll probably be able to tell you a bit more about the actual environment of the school in a few weeks, as I am visiting A&M for a day during Thanksgiving holidays.

Michigan is one of the best colleges in the nation, in essentially every respect. You're going to get top-ranked departments in pretty much every discipline, and there are more study options that you can possibly imagine. It's also an institution that is perhaps underrated in the public consciousness, but among profs, it's very respected.

It's also in Ann Arbor, which is a great college town. The downside, of course, is that this is Michigan - you can expect that the winters will get cold. Still, it's not as bad as most people make it, and you get the benefit of a lot of variety in the weather.

Speaking of Michigan, I didn't like the weather today. :D

But seriously, Michigan is underrated. Its selectivity is relatively low only because of its tendency to admit weak in-staters. But its departments are strong across the board.... my sister placed U of M as her first choice school because she liked its biology department. And from what I've gathered from others, a great Honors Math program.

As someone who went to Michigan for grad school, I wouldn't recommend it for undergrads unless they're extremely strong students.   The reason is that while the classes will be beneficial, challenging, etc. for good students, there is trouble with getting good advising or any attention from faculty unless the student is strong within the Honors program (in which case faculty notice the person).  The advising aspect of this for mathematics is reasonably significant at Michigan in particular because for many undergraduate classes, there are at least three different levels/flavors to choose from... and picking the wrong one can take you out of a grad-school track or even out of a regular mathematics track.

As eryaman said, Texas A&M is located in a small town, conveniently located one to three hours away from Houston, San Antonio, Dallas-Fort Worth, and Austin.  Most people in College Station are very friendly and generally everyone loves the Aggies. From what I've been told, the fraternities revolve around university sports. The majority of students seem to be Caucasian, conservative and religious. I think the dormitories can be improved, but there are some apartments around. I also think the campus is pretty, but A&M obviously does not have the same landscaping funds as some of the more "elite" schools do. You may not be challenged by the math curriculum though, since you'll be used to more rigorous work. However, there is a large course offering and the professors I've had were easily accessible. There's not much to do on the weekends or holidays. The biggest attraction around is probably either the shopping mall or the movie theater. The weather is very pleasant, usually clear skies and rarely raining. Since Texas A&M is a public school, a large portion of students will have been accepted by virtue of automatic admission via high school class top 10% rule. Accordingly, the range of motivation is very great, unlike that of study schools. Obviously, these can be pros or cons depending on your perspective.

Hope that helped...

[quote="smbelcas"]there are at least three different levels/flavors to choose from... [/quote]
This seems to be common in many state schools.

[quote="smbelcas"]...and picking the wrong one can take you out of a grad-school track or even out of a regular mathematics track.[/quote]
Could you elaborate a bit more on that, please?

[quote="mel"]
[quote="smbelcas"]...and picking the wrong one can take you out of a grad-school track or even out of a regular mathematics track.[/quote]
Could you elaborate a bit more on that, please?[/quote]

Sure.  (Caveat: I don't know how far beyond UM this generalizes.)  Let's say there are three linear algebra courses.  Their course numbers might be 411, 455, and 493 and the descriptions sound reasonably similar.  Because of the ordering of the numbers, you'd suspect that the 411 course is the easiest and the 493 the most difficult, so that an uncertain student might begin with 455.  As it turns out, the 411 course is a standard undergraduate linear algebra course, the 455 course is computational in nature and intended for engineering majors, and 493 is for the grad-school-preparation track.  If you choose 411, then you won't have the prerequisites for the grad-school-prep abstract algebra course and if you later think you'd like to go to grad school, you'll have to take an additional semester of linear algebra...which might not count for credit because you've already taken a linear algebra course.  If you choose 455, then you'll be underprepared for the regular math-major abstract algebra course (never mind the grad-school-prep version) and may need to go to the one intended for physicists instead (or something).

This is why advising is crucial.

[quote="LordoftheMorons"]Michigan: Ohio State sucks
Now, that might be a little biased, as both of my parents went to U of Michigan, but its true.  :D[/quote]

Ah yes, that is much like my

Berkeley: Stanford sucks

anyways, Stanford has like a "rich culture" going on, also the buildings look like churches : |
(i used to live in Palo Alto, whose high schools were like, feeder schools for stanford)

I was wondering whether anyone could give an informed opinion on U of North Carolina's math department.  Because it's a state university, many math majors are probably not continuing on to graduate school and are instead going to teach at the high school level or at least are not going to pursue their math education further.  For a student who intends (hopes) to go to graduate school, would that student be at a significant disadvantage in applying to graduate school programs coming out of that department as opposed to getting an undergraduate degree from Stanford or MIT?  Or would it really depend on how the student did at any of those places?  As you might imagine, the choice between one's state universty and a first rate private university is partly a financial one, but the choice of the less expensive one is probably not worth it if the prospects beyond college are significantly lessened.

As people have commented in other threads, an undergraduate who

1) asks his advisers for the courses that best prepare for graduate school, and

2) does well in those courses, and

3) attempts to gain some research experience as an undergraduate (perhaps through the [url=http://www.nsf.gov/funding/browse_all_funding.jsp]REU[/url] program), and

4) makes sure to get to know some professors on a one-to-one basis

should have a chance to get into a good graduate school from a state university. The U of North Carolina is good enough not to deny someone that chance. See

[url=http://www.amazon.com/gp/product/082183455X]A Mathematician's Survival Guide: Graduate School and Early Career Development[/url] for more information.

As an undergraduate, I attended my state college (University of Maryland), partly for financial reasons and partly because I was young.  For grad school, I attended MIT, where I received a masters and PhD in computer science.  So it's definitely possible to move up.

Thank you both for your helpful replies.    I guess my next question would be how the graduate school opportunities might compare for a top math major from UNC versus a middle of the pack (at best) math major from Stanford or MIT.  I would expect that one would have to be the absolute best coming out of a state university, but I am curious what the opportunites might be for anyone bright enough to even complete a math major at one of the very top universities, even if that student did not stand out exceptionally among the other majors.

[quote="mathscout"]Because it's a state university, many math majors are probably not continuing on to graduate school and are instead going to teach at the high school level or at least are not going to pursue their math education further.[/quote]

That's actually true of the math majors at almost all schools (private, public, prestigious, not-so-prestigious...).  If you want a community of peers who intend to go to grad school, then you probably want to go to a large school so that the raw numbers of grad-school intenders are large enough.

[quote="mathscout"] I guess my next question would be how the graduate school opportunities might compare for a top math major from UNC versus a middle of the pack (at best) math major from Stanford or MIT. [/quote]

I doubt that there's much of a difference.  That is, if the credentials of a top UNC major are the same as a generic Stanford/MIT major, the chances of getting into a given grad school should be roughly equal.