quadratics
Gauss
number theory unsolved
number theory

Find the smallest positive prime that divides $n^2 + 5n + 23$ for some integer $n$.

Let $p$ be any prime dividing $P(n) : = n^2 + 5n +23$, for some $n \in \mathbb{Z}$. Trivially $p > 2$, since $P(n) \equiv n(n+1) + 1 \equiv 1 \bmod 2$. Hence $P(n) \equiv 0 \bmod p$ only if $u^2 \equiv - 67 \bmod p$ has a solution in integers (just put $u = 2n+5$). If $p = 3$, this leads to $u^2 \equiv -1 \bmod 3$, which is impossible since $\left(\frac{-1}{3}\right) = (-1)^{\frac{3-1}{2}} = -1$ and subsequently $-1$ is not a quadratic residue $\bmod 3$. If $p = 5$, you get $P(n) \equiv 0 \bmod 5$ iff $n^2 \equiv 2\bmod 5$, which is also impossible as $\left(\frac{2}{5}\right) = (-1)^{\frac{5^2-1}{8}} = -1$, by a well-known lemma of Gauss on the quadratic character of $2$. By similar arguments, letting $p = 7$ or $p = 11$, you find $u^2 \equiv -4 \bmod 7$ and $u^2 \equiv -1 \bmod 11$ respectively, which are both impossible because $\left(\frac{-4}{7}\right) = \left(\frac{-1}{7}\right) = (-1)^{\frac{7-1}{2}} = -1$ and $\left(\frac{-1}{11}\right) = (-1)^{\frac{11-1}{2}} = -1$. Now suppose $p = 13$. Then $u^2 \equiv -67 \bmod 13$ only if $u^2 \equiv -2 \bmod 13$, which is once more impossible, as $\left(\frac{-2}{13}\right) = \left(\frac{-1}{13}\right) \cdot \left(\frac{2}{13}\right) = (-1)^{\frac{13-1}{2}} \cdot (-1)^{\frac{13^2 - 1}{8}} = -1$. Finally let $p = 17$. You get $17 \mid P(n)$ only if $u^2 \equiv 1 \bmod 17$, which is surely fulfilled. And effectively you get $P(-2) = 2^2 - 10 + 23 = 17$. Now write down your own conclusions...

There is another idea:

n=m-2
$n^{2}+5n+23=m^{2}-m+17$
Then, the problem is almost obvious.

or, you can see that n = 23 works, so $23 >= m$, and also $m > 2$. Then just do easy casework for the primes from 3 to 23 until you get a solution.

Very easy. We can easily find that p≠2, 3, 5, 7, 11, 13 and that the solution is 17 for every n which is congruent to 14 or 15 mod 17.

[quote=hitlEuler] If $p = 3$, this leads to $u^2 \equiv -1 \bmod 3$, which is impossible since $\left(\frac{-1}{3}\right) = (-1)^{\frac{3-1}{2}} = -1$ and subsequently $-1$ is not a quadratic residue $\bmod 3$.[/quote]
Can you explain why this works, I am confused

Here is an extension problem
[quote=AIME]Find the least positive integer $m$ such that $m^2 - m + 11$ is a product of at least four not necessarily distinct primes[/quote]

Obviously the given expression is odd for all integers. So, we cannot get the Prime number 2. We also find that it cannot take 3. Now for prime numbers greater than 3 we can write 6k+1 or 6k-1. The number 23 suggests to take 6k-1.
Now, n^2 + 5n + 23 = 6k - 1 is a quadratic equation which gives integer solution if discriminant is a perfect square.  So, 24k - 71 is a perfect square.  And that occurs for k= 3 the least value. So, the least prime number required = 6×3 - 1 = 17.


geometry
parallelogram
trigonometry
circumcircle
trig identities
Law of Sines
Law of Cosines

Let $ ABCD$ a parallelogram with $ \angle A$ acute and $ P$ point outside $ ABCD$ such that $ \angle PDC \equal{} \angle CBP \equal{} a$ ($ a$ acute). Prove that $ \angle DPA \equal{}\angle CPB$.
(Note: please do not use poles or transformations or harmonic division. Just law of sines, law of cosines, cevians... basic theorems only).

This seems false. By the given, P will lie on the circumcircle of $ \triangle BCD$. But then $ \angle DPA$ is less than $ \angle CPB$ in many situations.

Well, I forgot to mention that $ P$ lies inside (convex) $ \angle DPB$. I have added now in an attachment the scheme of the exercise (there's a false line, don't pay attention to it: it's the one without letters).

Let $ \angle PDC \equal{} \angle PBC \equal{} \alpha$.Let $ \angle DPA \equal{} \delta,\angle APC \equal{} \omega,\angle CPB \equal{} \gamma,\angle CDA \equal{} \beta$.
we have:
$ \frac {\sin (\delta \plus{} \omega)}{\sin (\gamma)} \equal{}$

$ \frac {\sin (\delta \plus{} \omega).\sin(\alpha)}{\sin(\alpha).\sin(\gamma)} \equal{}$

$ \frac {CD}{BC} \equal{} \frac {AB}{AD} \equal{}$

$ \frac {AB}{PA}.\frac {PA}{AD} \equal{}$

$ \frac {\sin(\gamma \plus{} \omega).\sin(\alpha \plus{} \beta)}{\sin( \alpha \plus{} \beta).\sin(\delta)} \equal{}$

$ \frac {\sin(\gamma \plus{} \omega)}{\sin(\delta)}$

$ \implies \sin(\gamma)\sin(\gamma \plus{} \omega) \equal{} \sin(\delta)\sin(\delta \plus{} \omega)$

$ \implies \cos(\omega) \minus{} \cos(2\gamma \plus{} \omega) \equal{} \cos(\omega) \minus{} \cos(2\delta \plus{} \omega) \implies \gamma \equal{} \delta$
DONE!

Let be the points $ E\equiv PD\cap BC$ and $ F\equiv PB\cap CD$ and so, the quadrilateral $ BDEF$ is cyclic from $ \angle EBF \equal{} \angle EDF$.

It is easy to show that the triangles $ \bigtriangleup ABD,\ \bigtriangleup CEF$ are similar from $ \angle ADB \equal{} \angle DBE \equal{} \angle CFE$ and $ \angle ABD \equal{} \angle BDE \equal{} \angle CEF$ and so, we have that $ \frac {AD}{CF} \equal{} \frac {BD}{EF}$ $ ,(1)$

But, from the similar triangles $ \bigtriangleup PBD,\ \bigtriangleup PEF$ we have also $ \frac {BD}{EF} \equal{} \frac {PD}{PF}$ $ ,(2)$

From $ (1),\ (2)$ $ \Longrightarrow$ $ \frac {AD}{CF} \equal{} \frac {PD}{PF}$ $ \Longrightarrow$ $ \frac {AD}{PD} \equal{} \frac {CF}{PF}$ $ ,(3)$

From $ (3)$ and because of $ \angle PDA \equal{} \angle PBA \equal{} \angle PFC$, we conclude that the triangles $ \bigtriangleup PDA,\ \bigtriangleup PFC$ are similar.

Hence, we have that $ \angle DPA \equal{} \angle CPF\equiv CPB$ and the proof is completed.

Kostas Vittas.


What is the remainder when $9^{1995}$ is divided by $7$?

[hide]
Remainder when divided by 7 of..
$9^0 = 1$
$9^1 = 2$
$9^2 = 4$
$9^3 = 1$
$9^4 = 2$
$9^5 = 4$
$9^6 = 1$
Start to see a pattern? It's a repeating set of 3 remainders starting with $9^0$. Add $1$ to $1995$ for $9^0$ and then find the remainder of $1996$ when it is divided by $3$ which is $1$ making the answer $2$.
[/hide]

[hide="MODULAR ARITHMETIC"]
$9\equiv2\pmod{7}$
Multiply by 9 ($2\cdot9=18\equiv4\pmod{7}$)
$9^2\equiv4\pmod{7}$
Multiply by 9 again ($4\cdot9=36\equiv1\pmod{7}$)
$9^3\equiv1\pmod{7}$
Again ($1\cdot9=9\equiv2\pmod{7}$)
$9^4\equiv2\pmod{7}$
You dont need to go any farther, because you are back to the first step, becuase you are at $2\pmod{7}$.
So every multiple of 3 power of 9 is $1\pmod{7}$
$1995$ is a multiple of 3, so it is $1\pmod{7}$
The remainder is 1[/hide]

[hide="Doing what biffanddoc did only not"]$9^{1995} \equiv 2^{1995} \equiv 8^{665} \equiv 1^{665} =1 \pmod 7$.

Thus, the remainder is $\boxed{1}$

If you don't understand mods, search the forums  :) [/hide]


Suppose a,b,c,d are positive integers with ab + cd = 1. Then,  For i = 1,2,3,4, let (x_i)^2 + (y_i)^2 = 1, where x_i and y_i are real numbers.  Show that

(ay_1 + by_2 + cy_3 + dy_4)^2 + (ax_4 + bx_3 + cx_2 + dx_1)^2
<= 2( (a/b) + (b/a) + (c/d) + (d/c) ).

I'll post my solution for this problem  :D

Hint: Use Cauchy-Schwarz Inequality  :D

The heart of mathematics is its problems

It's China 2003 problem 6.
I find this an impossible problem mainly because it doesn't seem to have other solution except the official one.
Also you must use a trick that is hard to observe( in my opinion)

[quote]Also you must use a trick that is hard to observe( in my opinion)[/quote]
And just what kind of trick might that be?

Use Cauchy-Schwartz , we have
$\displaystyle(ay_1+by_2+cy_3+dy_4)^2\leq (ab+cd)(\frac{(ay_1+by_2)^2}{ab}+\frac{(cy_3+dy_4)^2}{cd})=\frac{(ay_1+by_2)^2}{ab}+\frac{(cy_3+dy_4)^2}{cd}$
Similar: $\displaystyle(ax_4+bx_3+cx_2+dx_1)^2\leq (ab+cd)(\frac{(ax_4+bx_3)^2}{ab}+\frac{(cx_2+dx_1)^2}{cd})=\frac{(ax_4+bx_3)^2}{ab}+\frac{(cx_2+dx_1)^2}{cd}$
But:  $\displaystyle(ay_1+by_2)^2\leq(ay_1+by_2)^2+(ax_1-bx_2)^2=a^2+b^2+2ab(y_1y_2-x_1x_2)$
Similar. $\displaystyle(cx_2+dx_1)^2\leq c^2+d^2+2cd(x_1x_2-y_1y_2)$, then we get:
$\displaystyle\frac{(ay_1+by_2)^2}{ab}+\frac{(cx_2+dx_1)^2}{cd}\leq \frac {a}{b}+\frac {b}{a}+\frac {c}{d}+\frac {d}{c}$ (1)
The same argument show that: $\displaystyle\frac{(cy_3+dy_4)^2}{cd}+\frac{(ax_4+bx_3)^2}{ab}\leq \frac {a}{b}+\frac {b}{a}+\frac {c}{d}+\frac {d}{c}$ (2)
Combining (1);(2) we get .... :P


calculus
integration
logarithms
function
limit
real analysis
real analysis unsolved

Let's start by writing $x^{x}= e^{x\ln x}$, and now use the power series expansion of the exponential function:

$x^{x}= e^{x\ln x}= 1+x\ln x+\frac{(x\ln x)^{2}}{2}+... = \sum_{n=0}^{\infty}\frac{(x\ln x)^{n}}{n!}$

and so

$\int^{1}_{0}x^{x}\,dx = \int^{1}_{0}\left[\sum_{n=0}^{\infty}\frac{(x\ln x)^{n}}{n!}\right]\,dx = \sum_{n=0}^{\infty}\left[\frac{1}{n!}\int^{1}_{0}(x\ln x)^{n}\,dx\right]$.

If you use integration by parts, you will see that

$\int^{1}_{0}(x\ln x)^{n}\,dx = \frac{(-1)^{n}n!}{(n+1)^{n+1}}$, and so

$\int^{1}_{0}x^{x}\,dx = \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)^{n+1}}= \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(n)^{n}}$.

Note: if I am in no mistake, this integral was evaluated for the first time in 1697 by John Bernoulli.

If you don't mind, can you go through the integration by parts to show that $\int_{0}^{1}{(x\ln x)^{n}dx = \frac{{(-1)^{n}n!}}{{(n+1)^{n+1}}}}$ it would be nice to see how the details are worked out. Are there any tricks involved?


From warm-up 2:
A set of five non-negative integers has a mean of 5, a median of 6 and a mode of 7. What is the smallest possible member of this set?

[hide="Solution"]There cannot be three $ 7$'s. Assume there are, the median would therefore be at least $ 7$.

Therefore, we can assume there are two $ 7$'s. Furthermore, there must exactly one and only one $ 6$, and no other numbers greater than $ 6$, for then the median would not be $ 6$.

So we now know the three largest integers, $ 6$, $ 7$, and $ 7$. We have that $ a\plus{}b\plus{}6\plus{}7\plus{}7 \equal{} 25$, so $ a\plus{}b \equal{} 5$. The smallest possible member of the set is therefore $ \boxed{0}$, where the integers are $ 0$, $ 5$, $ 6$, $ 7$, and $ 7$. No smaller members can exist because no member can be negative.[/hide]


linear algebra
matrix
algebra
polynomial
vector
linear algebra unsolved

Let $A,B\in\mathcal{M}_{2}(\mathbb{R})$ (real $2\times 2$ matrices), that satisfy $A^{2}+B^{2}=AB$. Prove that $(AB-BA)^{2}=O_{2}$.

I may have misunderstood the problem, to be the set of matrices that operate under that relation (even though matrix multiplication isn't commutative.)  Upon second thought, that might not be a viable proof and you might have to manipulate identities a bit more.

We know that tr(AB-BA)=0.
If we consider the Cailey-Hamilton equation(or the characteristic polynomial), it remains to show that det(AB-BA)=0

My solution(a little more complicated that the official one but this is what came to my mind during the contest)

of course we have to prove that $det(AB-BA)=0$;
we see that $detA=0 < = > det B=0$,so

 Case I: $det A=detB=0$;
we have $A^{2}=TrA A$ and $B^{2}=TrB B \Rightarrow TrA A+TrB B=AB \Rightarrow  for TrA and TrB$ different from 0 we easily obtain that $AB=BA$ and we are done. For TrA=0( it is the same if TrB=0) we have that  $A^{2}=0 \Rightarrow  B^{2}=AB \Rightarrow  B^{2}-BA=AB-BA \Rightarrow  det(AB-BA)=0$ and we are done.(this case, TrA=0, I forgot to mention during the contest and I lost 2 points... :( )

Case II: A abd B invertable =>$det(AB-BA)=detA detB det(I-A^{-1}BAB^{-1})$ and let's prove that the last determinant is 0.
denote by $C=A^{-1}B$  and $D=AB^{-1}$. from the initial relation we have $C+D=I$ and we can easily observ that the eigenvalues of $D^{-1}$ are the same as those of $C$. Denothe by $a$ and $b$ the eigenvalues of C and from $C+D=I$ we have that the eigenvalues of $D^{-1}$ are $\frac{1}{1-a}$ and $\frac{1}{1-b}$ => {$\frac{1}{1-a},\frac{1}{1-b}$}={$a$,$b$} and after some easy computations, we obtain that $1-a+a^{2}=0$ sau $1-b+b^{2}=0$ and thus we have $det(I-C+C^{2})=0$ , which is exactly $det(I-CD)= det(I-A^{-1}BAB^{-1})=det(AB-BA)$.

Fact: $AB=BA$.
Proof: $rank(AB-BA)\leq1$ then $A$ and $B$ are simultaneously triangularizable over $\mathbb{C}$.
Now I follow the mon's post.
case 1: $detA=detB=0$. a) $tr(A)tr(B)\not=0$: then $AB=BA$.
b) $tr(A)=0$ then $A^{2}=0$ and we may assume that \[A=\begin{pmatrix}0&a\\0&0\end{pmatrix},B=\begin{pmatrix}b&c\\0&d\end{pmatrix}.\] $B^{2}=AB$ implies $b=d=0$ and $A,B$ commutate.
case 2: $A,B$ are invertible. Let $a,\bar{a}$ be the roots of $1-x+x^{2}=0$. $C$ is a real matrix then $spectrum(C)=\{a,\bar{a}\}$ and $C$ is diagonalizable over $\mathbb{C}$ and we may assume that \[C=\begin{pmatrix}0&-1\\1&1\end{pmatrix}.\]
Moreover $C^{2}=C-I,C^{-1}=I-C=D$ and $XC=CX$ implies $X\in[I,C]$ (the vector space which is generated by $I,C$). $B=AC$ and $A=DB$ imply $CA=AC,CB=BC$ and $A,B\in[I,C]$. $QED$.

Here is an acute solution I got,
Let $C=AB-BA$,we should only show $Tr(C)=0,Tr(C^{2})=0$
The first one is so obvious,for the last one,it is equivalent to $Tr(ABAB)=Tr(A^{2}B^{2})$
But use the equation in condition,we have:
$Tr(ABAB)=Tr(A^{2}B^{2})=Tr(A^{3}B)+Tr(AB^{3})$

for the advantage we will traces because it is communicative. :)

i don`t think that $\mbox{Tr}\,(AB-BA)^{2}=0$ holds in the general case.
loup blanc, you are right. it is true that $AB=BA$ (thus the problem is weak), but the approach is not a very "olympiad-like" (special forms and computations). also, as a note, the observation that $AB=BA$ does not hold for $n \geq 3$, but the statement of the problem holds for all $n$ not divisble by $3$.
the official solution, which is the same one i found, is a simple technique, but not easy to discover all by yourself.
[hide="solution"]
let $\varepsilon$ be a third root of unity. let`s see that
$(A+\varepsilon B)(A+\overline{\varepsilon}B)=A^{2}+B^{2}+\varepsilon BA+\overline{\varepsilon}AB=\varepsilon (BA-AB)$.
now, note that $\det (A+\varepsilon B)(A+\overline{\varepsilon}B) = |\det(A+\varepsilon B)|^{2}$, thus  $\det \big(\varepsilon (BA-AB)\big)$ is a real number. since $\varepsilon\notin\mathbb{R}$, it follows that $\det (BA-AB)=0$.
from here it`s obvious.
[/hide]

[quote="maky"]i don`t think that $\mbox{Tr}\,(AB-BA)^{2}=0$
[/quote]
$\mbox{Tr}\,(AB-BA)^{2}=2\mbox{Tr}\,(ABAB-A^{2}B^{2})$

oh, now i understand. my mistake :blush: very nice and short solution, congratulations

Hi Zhaobin,
how do you prove that $tr(ABAB)=tr(A^{2}B^{2})$ ???
In general case it's absolutely false.

if i understood correctly, then
$\mbox{Tr}(ABAB) = \mbox{Tr}\big((A^{2}+B^{2})AB\big)=\mbox{Tr}(A^{3}B)+\mbox{Tr}(AB^{3})$,
and
$\mbox{Tr}(A^{2}B^{2})=\mbox{Tr}\big(A(AB)B\big)=\mbox{Tr}\big(A(A^{2}+B^{2})B\big)=\mbox{Tr}(A^{3}B)+\mbox{Tr}(AB^{3})$.
(using the fact that $\mbox{Tr}(XY)=\mbox{Tr}(YX)$.)

Sorry, I couldn't understand. Why is  tr(AB-BA)=0?
Also in mon's solution, why det(A)=0 <=> det(B)=0?
And how does the last line of mon's solution follow from the previous one?
Thanks

$tr(AB-BA) =0$ obviously.
So since for $2 \times 2$ matrices, if $det(AB-BA) = 0$, we are done.

Let $w$ be cube root of unity.
But $|A+wB||A+w^2B| = |(A+wB)(A+w^2B)| =  w^2|BA-AB|$
Obviously the LHS is real.
Thus $|BA - AB| = 0$

my proof it is for $C$
1.Very easy :$det(A+xB)=detA+(TrATrB-Tr(AB))x+x^2detB$
2.We have $det(A^2+B^2)=det(AB)$.Result $A^2+B^2-(AB-BA)=BA$, so $det[A^2+B^2-(AB-BA)]=det(BA)$ and from (1) we obtain that $det(A^2+B^2)+det(AB-BA)+Tr[A^2+B^2]Tr(AB-BA)-Tr[(A^2+B^2)(AB-BA)]=detAB$.Because $Tr(AB-BA)=Tr[(A^2+B^2)(AB-BA)]=0$, we have $det(AB)+det(AB-BA)=detAB$, so $det(AB-BA)=0$ and from Hamilton-Cayley, $(AB-BA)^2-Tr(AB-BA)(AB-BA)+det(AB-BA)I_2=O_2$, we obtain  $(AB-BA)^2=O_2$

Let $A,B\in M_n(\mathbb{C})$ be s.t. $A^2+B^2=AB$. We assume that $n=2$ or $3$.
Let $j=\exp(2i\pi/3), U=A+jB,V=A+j^2B$. Then $UV=j(BA-AB)=\dfrac{1}{1-j}(UV-VU)$, that is equivalent to $jUV=VU$.
Note that $\det(VU)=j^n\det(UV)$ implies that $\det(UV)=0$ or $j^n=1$.
$\textbf{Case 1.}$  $n=2$. Then $\det(UV)=0$, $rank(AB-BA)\leq 1$ and $A,B$ are simultaneously triangularizable.
In the same way, $U,V$ are simult. triang. and, moreover, $UV$ is nilpotent.
$\textbf{Case 2.}$  $n=3$. cf.
[url=https://artofproblemsolving.com/community/c7h2812265_a_matrix_equation_in_dimension_3] see here [/url]

@#17 The LHS is real. Please explain how.


I am reading my algebra-book, which says the following.

if $ n$ is a $ a$-pseudoprime, but the strong $ a$-pseudoprime sequence for $ n$ looks like $ \{...,b,1,...,1\}$ with $ b\equiv 1$ or $ \minus{} 1 \pmod{n}$ being FALSE, then $ n$ fails the strong $ a$-pseudoprime test, so $ n$ must be composite.

It then goes on to say the following cryptic statement:

[color=red]The number $ n$ cannot be prime because $ 1$ has at least three square roots, $ 1$, $ \minus{} 1$, and $ b\pmod{n}$[/color]

What is this link between n being prime and the roots it has?

Thanks in advance.

[b]Legendre's Theorem:[/b]  A polynomial of degree $ n$ with coefficients $ \bmod p$ (for $ p$ prime) has at most $ n$ roots.

[b]Special case:[/b]  The polynomial $ x^2 \equiv 1 \bmod p$ has at most $ 2$ roots.  When $ p > 2$, these roots are $ 1, \minus{} 1$.

If another square root of $ 1$ exists $ \bmod n$, then $ n$ cannot be prime.

[hide="Direct proof of the cryptic statement"] Suppose there exists $ b \neq 1, \minus{}1$ that is a square root of $ 1 \bmod n$.  Then $ n | b^2 \minus{} 1$.  Suppose further that $ n$ is prime.  Then $ n | (b \minus{} 1)(b \plus{} 1)$, so either $ n | b\minus{}1$ or $ n | b\plus{}1$, but we assumed that $ b \neq 1, \minus{}1 \bmod n$ - contradiction. [/hide]

oh t0rajir0uu, you know every proof so it comes directly from the hand. That makes me heartened (is heartened a word? disheartened is..) :0)


geometry
circumcircle
symmetry
angle bisector
perpendicular bisector
geometry proposed

Let $ABC$ be an isosceles triangle with $AC=BC$, let $M$ be the midpoint of its side $AC$, and let $Z$ be the line through $C$ perpendicular to $AB$. The circle through the points $B$, $C$, and $M$ intersects the line $Z$ at the points $C$ and $Q$. Find the radius of the circumcircle of the triangle $ABC$ in terms of $m = CQ$.

I think the key to this problem is to prove that $Q$ is on $AM$, where $M$ is the midpoint of $BC$.

A full solution:

Indeed the radius of the circumcircle of triangle ABC is $\frac{2}{3}m$.

[i]Proof.[/i] The line Z is a symmetry axis of the isosceles triangle ABC; hence, it is an angle bisector, a median, an altitude and a diameter of the circumcircle simultaneously. Therefore, both the centroid G and the circumcenter O of triangle ABC lie on this line Z. Since the line Z is an angle bisector, actually the angle bisector of the angle BCA, we have < BCQ = < GCM. Also, < BQC = < GMC (since < BQC = < BMC because the point Q lies on the cirle through B, C and M). Therefore, the triangles BCQ and GCM are similar, and $\frac{CQ}{CM}=\frac{BC}{GC}$. Since CQ = m, we have $\frac{m}{CM}=\frac{BC}{GC}$.

But the midpoint N of the side AB also lies on the line Z, and the centroid G of triangle ABC satisfies $GC=\frac{2}{3}\cdot NC$, so we obtain

$\frac{m}{CM}=\frac{BC}{\frac{2}{3}\cdot NC}=\frac{3}{2}\cdot \frac{BC}{NC}$.

Since the point O is the circumcenter of triangle ABC, it lies on the perpendicular bisector of the side AC, i. e. on the perpendicular to AC at the point M. Therefore, < CMO = 90°. Together with < CNB = 90°, we get < CNB = < CMO. Also, since the line Z bisects the angle BCA, we have < NCB = < MCO. Thus, the triangles CNB and CMO are similar, and $\frac{BC}{NC}=\frac{OC}{MC}$, so that

$\frac{m}{CM}=\frac{3}{2}\cdot \frac{BC}{NC}=\frac{3}{2}\cdot \frac{OC}{MC}=\frac{3}{2}\cdot \frac{OC}{CM}$,

and $m=\frac{3}{2}\cdot OC$, so $OC=\frac{2}{3}\cdot m$. But OC is just the radius of the circumcircle of triangle ABC. Hence we are done.

  Darij

Shortlists solution of Problem 2

Let P be the center of the circle $k_1$ through B, C and M, O be the circumcenter of  ABC and K be the midpoint of MC. Since AC=BC, the center O lies on CH. Let KP intersect CH in point L. Since KP and OM are perpendicular to AC, then KP || OM. From MK = KC it follows that OL = CL. On the other hand OP is perpendicular to BC, hence angle LOP = angle COP = 90º - angle BCH. Also we have angle OLP = angle CLK = 90º - angle ACH. Since ABC isosceles and angle BCH = angle ACH, then angle LOP = angle OLP and LP = OP. Since CP = PQ we obtain that angle CLP = angle QOP and CL = OQ. Thus we have CL = LO = OQ, so CO = 2/3CQ. Finally for the radius R of the circumcircle of ABC we obtain R = 2/3m.

this is my solution for problem 2:
the points M,Q,A,C are concycle so if we denote by ${P}=CQ\cap AM$ we get that $PM\cdot PA=PQ\cdot PC$ (*).
We know that $PM=\displaystyle\frac 13 MA$, $PA=\displaystyle\frac23 MA$, $\displaystyle CP=\frac23 CC'$ where ${C'}=CQ\cap AB$.
We know that $MA^2=m_a^2=\displaystyle\frac{2(b^2+c^2)-a^2}4$ (1)  and
$CC'^2=m_c^2=\displaystyle\frac{2(b^2+a^2)-c^2}4$ (2)

If we denote by $\alpha=\angle ACQ$ and knowing that $\displaystyle R=\frac{abc}{4S}$ and using (*),(1),(2) and $CP+PQ=m$ with some easy computations we get the desired result.

This is a solution of this problem posted by [b]vedran6[/b] in a different thread:

It is obvious that the BCMQ is cyclic and because of that is and we have that angle <CQM=<CBM (1) let $CZ\cap BM=\left(T\right)$ ; and $CZ\cap AB=\left(D\right)$ it is obvious that is $CT=\frac23\cdot CD$ and because of AC=BC we have <ACQ=<BCT (2) , from (1) and (2) we have that trinagle CQM is similar to triangle CBT. From similarity we have that is $\frac{CQ}{CB}=\frac{CM}{CT}$ which is equal to $CQ=\frac{3\cdot CB\cdot CB}{4\cdot CD}$ (3).
  It is well known that in every triangle is $R=\frac{a\cdot b\cdot c}{4P}$ so we use it at this triangle ABC and got $R=\frac{BC\cdot BC}{2\cdot CD}$ (4) . After dividing (3) and (4) we get $CQ=1,5\cdot R$ and the problem is finished :)

Here is [u]my solution[/u] (I think that it differs from the other preceding solutions):

Let $w=C(O,R)$ be a circumcircle of the triangle $ABC$

and let $U$ be the middlepoint of the segment $[AM]$. From

$\widehat {QCM}\equiv \widehat {QCB}\Longrightarrow QM=QB=QA$

(the point $Q$ is the circumcenter for the triangle $ABM$). From

$QU\perp AC,\ MO\perp AC\Longrightarrow QU\parallel MO\Longrightarrow$

$\frac 23 =\frac{CM}{CU}=\frac{CO}{CQ}=\frac{R}{m}\Longrightarrow R=\frac 23 m$.

My Solution
Let $MC=AM=x$ Then  $BC=2x$ From sinus we have $\longrightarrow$ $\frac{2x}{cos\alpha}=2R$ Where $R$ be the circumradius of $\triangle{ABC}$ and $\angle{ACQ=BCQ=\alpha}$. Then from $\triangle{BCQ}$ we have by sinus $\frac{m}{sin(\alpha+\beta)}=\frac{2Rcos\alpha}{sin(2\alpha+\beta)}$ Where $\angle{MBC}=\beta$. Then we have $m=\frac{2Rcos\alpha\cdot  sin(\alpha+\beta)}{sin(2\alpha+\beta)}$ By some trigonometric identies we have $m=\frac{Rsin(2\alpha+\beta)+Rsin\beta}{sin(2\alpha+\beta)}$ $\longrightarrow$ $m=R+\frac{R\cdot sin\beta}{sin(2\alpha+\beta)}$ Loking at $\triangle{MBC}$ by sinus we have $\longrightarrow$ $\frac{x}{sin\beta}=\frac{2x}{sin(2\alpha+\beta)}$ $\longrightarrow$ $\frac{sin\beta}{sin(2\alpha+\beta)}=\frac{1}{2}$ so $m=R+\frac{R\cdot sin\beta}{sin(2\alpha+\beta)}=\frac{3R}{2}$ $\longrightarrow$ $R=\frac{2m}{3}$

Anar Abbas Istek Lyceum

Let $K$ $\in [CA$ , $|MA|=|AK|$ , $M \neq K$ and $O$ is $\triangle ABC$'s circumcircle center

$|CM|=|MA|=|AK|$ $\Rightarrow$ $|MK|=|AC|=|BC|$ $.$ $\dotsm$ $(i)$

$C,M,Q,B$ is circularly $\Rightarrow$ $\angle QMA = \angle QBC$ $\dotsm$ $(ii)$

$ABC$ is isoscles triangle and $D$ is $BA$'s midpoint $\Rightarrow$ $\angle ACD = \angle BAD$ $.$ Also $B,Q,M,C$ is circularly $\Rightarrow$ $|MQ|=|BQ|$ $\dotsm$ $(iii)$

Combine $(i),(ii),(iii)$ $\Rightarrow$ $\triangle QMK \equiv \triangle QBC$

İn here $\angle QKM=\angle QCB=\angle OCA=\angle OAC$ $\Rightarrow$ $OA \parallel QK$ $\Rightarrow$ $\frac{AC}{KC} = \frac{OC}{QC}$ $.$ Also $|AC|=2|MC|$ , $|KC|=3|MC|$ $\Rightarrow$ $\frac{OC}{QC} = \frac{2}{3}$ $\Rightarrow$ $|OC|=\frac{2m}{3}$   $Q.E.D$

Let length of side $CB$ = $x$ and length of $QM = a$. We shall first prove that $QM = QB$.

Let $O$ be the circumcenter of $\triangle ACB$ which must lie on line $Z$ as $Z$ is a perpendicular bisector of isosceles $\triangle ACB$.

So, we have $\angle ACO = \angle BCO = \angle C/2$.

Now $MQBC$ is a cyclic quadrilateral by definition, so we have:
$\angle QMB = \angle QCB = \angle C/2$ and,
$\angle QBM = \angle QCM = \angle C/2$, thus $\angle QMB = \angle QBM$, so $QM = QB = a$.

Therefore in isosceles $\triangle QMB$ we have that $MB = 2 QB \cos C/2 = 2 a \cos C/2$.

Let $R$ be the circumradius of $\triangle ACB$.
So we have $CM = x/2 = R \cos C/2$ or $x = 2R \cos C/2$

Now applying Ptolemy's theorem in cyclic quadrilateral $MQBC$, we get:

$m . MB = x . QM + (x/2) . QB$ or,

$m . (2 a \cos C/2) = (3/2 x) . a = (3/2).(2Ra) \cos C/2 = 3Ra \cos C/2$ or,

$ R = (2/3)m$


rotation percent function probability Support

This thread may offend some. I'm sorry. Deal with it. The Round Table is for [i]serious[/i] topics and this is a [i]serious [/i]topic
Here is my argument. The evolution stance is based primarily on the assupmtion that God does not exist
Refutation:
Many examples showing God's design could be given, possibly with no end. But here are a few:

The Earth...its size is perfect. The Earth's size and corresponding gravity holds a thin layer of mostly nitrogen and oxygen gases, only extending about 50 miles above the Earth's surface. If Earth were smaller, an atmosphere would be impossible, like the planet Mercury. If Earth were larger, its atmosphere would contain free hydrogen, like Jupiter. Earth is the only known planet equipped with an atmosphere of the right mixture of gases to sustain plant, animal and human life.

The Earth is located the right distance from the sun. Consider the temperature swings we encounter, roughly -30 degrees to +120 degrees. If the Earth were any further away from the sun, we would all freeze. Any closer and we would burn up. Even a fractional variance in the Earth's position to the sun would make life on Earth impossible. The Earth remains this perfect distance from the sun while it rotates around the sun at a speed of nearly 67,000 mph. It is also rotating on its axis, allowing the entire surface of the Earth to be properly warmed and cooled every day.

And our moon is the perfect size and distance from the Earth for its gravitational pull. The moon creates important ocean tides and movement so ocean waters do not stagnate, and yet our massive oceans are restrained from spilling over across the continents.

Water...colorless, odorless and without taste, and yet no living thing can survive without it. Plants, animals and human beings consist mostly of water (about two-thirds of the human body is water). You'll see why the characteristics of water are uniquely suited to life:

It has an unusually high boiling point and freezing point. Water allows us to live in an environment of fluctuating temperature changes, while keeping our bodies a steady 98.6 degrees.

Water is a universal solvent. This property of water means that thousands of chemicals, minerals and nutrients can be carried throughout our bodies and into the smallest blood vessels.

Water is also chemically neutral. Without affecting the makeup of the substances it carries, water enables food, medicines and minerals to be absorbed and used by the body.

Water has a unique surface tension. Water in plants can therefore flow upward against gravity, bringing life-giving water and nutrients to the top of even the tallest trees.

Water freezes from the top down and floats, so fish can live in the winter.

Ninety-seven percent of the Earth's water is in the oceans. But on our Earth, there is a system designed which removes salt from the water and then distributes that water throughout the globe. Evaporation takes the ocean waters, leaving the salt, and forms clouds which are easily moved by the wind to disperse water over the land, for vegetation, animals and people. It is a system of purification and supply that sustains life on this planet, a system of recycled and reused water.

The human brain...simultaneously processes an amazing amount of information. Your brain takes in all the colors and objects you see, the temperature around you, the pressure of your feet against the floor, the sounds around you, the dryness of your mouth, even the texture of your keyboard. Your brain holds and processes all your emotions, thoughts and memories. At the same time your brain keeps track of the ongoing functions of your body like your breathing pattern, eyelid movement, hunger and movement of the muscles in your hands.

The human brain processes more than a million messages a second. Your brain weighs the importance of all this data, filtering out the relatively unimportant. This screening function is what allows you to focus and operate effectively in your world. The brain functions differently than other organs. There is an intelligence to it, the ability to reason, to produce feelings, to dream and plan, to take action, and relate to other people.

The eye...can distinguish among seven million colors. It has automatic focusing and handles an astounding 1.5 million messages -- simultaneously.Evolution focuses on mutations and changes from and within existing organisms. Yet evolution alone does not fully explain the initial source of the eye or the brain -- the start of living organisms from nonliving matter.

Scientists are convinced that our universe began with one enormous explosion of energy and light, which we now call the Big Bang. This was the singular start to everything that exists: the beginning of the universe, the start of space, and even the initial start of time itself.

Astrophysicist Robert Jastrow, a self-described agnostic, stated, "The seed of everything that has happened in the Universe was planted in that first instant; every star, every planet and every living creature in the Universe came into being as a result of events that were set in motion in the moment of the cosmic explosion...The Universe flashed into being, and we cannot find out what caused that to happen."

Steven Weinberg, a Nobel laureate in Physics, said at the moment of this explosion, "the universe was about a hundred thousands million degrees Centigrade...and the universe was filled with light."

The universe has not always existed. It had a start...what caused that? Scientists have no explanation for the sudden explosion of light and matter.

Much of life may seem uncertain, but look at what we can count on day after day: gravity remains consistent, a hot cup of coffee left on a counter will get cold, the earth rotates in the same 24 hours, and the speed of light doesn't change -- on earth or in galaxies far from us.

How is it that we can identify laws of nature that never change? Why is the universe so orderly, so reliable?

"The greatest scientists have been struck by how strange this is. There is no logical necessity for a universe that obeys rules, let alone one that abides by the rules of mathematics. This astonishment springs from the recognition that the universe doesn't have to behave this way. It is easy to imagine a universe in which conditions change unpredictably from instant to instant, or even a universe in which things pop in and out of existence."

Richard Feynman, a Nobel Prize winner for quantum electrodynamics, said, "Why nature is mathematical is a mystery...The fact that there are rules at all is a kind of miracle."

All instruction, all teaching, all training comes with intent. Someone who writes an instruction manual does so with purpose. Every cell of our bodies there exists a very detailed instruction code, much like a miniature computer program? As you may know, a computer program is made up of ones and zeros, like this: 110010101011000. The way they are arranged tell the computer program what to do. The DNA code in each of our cells is very similar. There are three DNA bases and it is estimated that there are 100 trillion cells in our bodies. $ (4*(3*10^9))*(10^14)$ different possible outcomes. If life had been by chance 99.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999% of these outcomes would be a failure, and scientists have not found any fossils of such failures as of yet

Natural, biological causes are completely lacking as an explanation when programmed information is involved. You cannot find instruction, precise information like this, without someone intentionally constructing it.

Proof of the existence of God
$ QED$

First of all, edit out that the stretching, it's annoying.

Ironically, though this is titled as an evolution thread, you addressed evolution very little. Most of it was cosmology and related.

//The evolution stance is based primarily on the assupmtion that God does not exist//

Which doesn't explain why a significant proportion of biologists accept both evolution and their own religious convictions. Like Kenneth Miller.

On the next three paragraphs, the probability of any single planet being Earth-like is astronomical. However, the size of the universe is astronomical. Amongst all the universe, the probability of there being at least a handful of Earth like planets is high.

Then again, why does a planet with life need to be Earth like? In the past half century, our discoveries have pushed the boundaries of the conditions that life on Earth can thrive in. We have found lithovores, halophiles, cryophiles, thermophiles and other extremophilic species. Hydrothermal vent communities are fueled not by the sun, but by geothermal energy, and support all life from bacteria to animals. Current opinions hold that the earliest life were not photosynthetic, but these chemosynthetic extremophiles, and molecular evidence supports this. And that's just considering life on Earth. No one knows the full extent of conditions life could exist in in the universe. Their biochemistry could be radically different from ours. Life evolves to fit the conditions, not the other way around.

Life itself regulates many of the processes on Earth. Global temperatures are regulated by the carbon cycle, carbon dioxide being a greenhouse gas and central to photosynthesis. Atmospheric composition is also influenced by life--the oxygen in the atmosphere is almost entirely due to the first photosynthetic organisms.

Consider the Daisyworld simulation. The presence of flowers that can respond to changing global temperatures yields a world with much more stable global temperatures than the control: http://en.wikipedia.org/wiki/File:Daisyworld-gbooth-swingdaisyball.gif

//If Earth were smaller, an atmosphere would be impossible, like the planet Mercury. If Earth were larger, its atmosphere would contain free hydrogen, like Jupiter.//

Uh, why are you citing the most extreme of the planets in our solar system? By larger, apparently you mean a few thousands of times more massive.

//Any closer and we would burn up. Even a fractional variance in the Earth's position to the sun would make life on Earth impossible.//

Again, the universe is pretty big. And again, life evolves to fit its conditions.

How much variance are you talking about? The Earth already varies a good few percent over the course of a year and more so due to Milankovitch cycles. Despite that annual difference in distance from the sun, other factors such as geography play a much larger role in global temperature, as Southern and Northern summers are not significantly different.

//The Earth remains this perfect distance from the sun while it rotates around the sun at a speed of nearly 67,000 mph.//

Not relevant.

//It is also rotating on its axis, allowing the entire surface of the Earth to be properly warmed and cooled every day.//

Most planets rotate about an axis. Not surprising.

//And our moon is the perfect size and distance from the Earth for its gravitational pull. The moon creates important ocean tides and movement so ocean waters do not stagnate, and yet our massive oceans are restrained from spilling over across the continents.//

I'll give you that the moon is somewhat of an anomaly. However, the ocean thing is, I assume, hyperbole.

//Water...colorless, odorless and without taste, and yet no living thing can survive without it.//

Well of course it's odorless and tasteless. Do you now how much it would suck to smell water 24/7? And it's not really true, it's just our sense of smell is not very good. I think dogs can smell water, and their sense of smell is much better than ours. We don't *need* to smell water, it would be more of an annoyance if we did. Smelling water provides no evolutionary advantage.

The properties of water are not independent, they are all due to the chemical structure of water, which is due to their atoms.

Yeah, evaporation. What about it?

Yeah, good description of the human brain.

//The eye...can distinguish among seven million colors.//

Based on about three color receptors (tri-color vision). Computer displays are also based on three colors of LED.

And I seriously doubt humans can distinguish between seven million colors. Considering visible light encompasses about 400-700 nm light, that would be distinguishing light of 4.29x10^-5 nm of difference in wavelength.

//Evolution focuses on mutations and changes from and within existing organisms.//

And selection. Don't forget natural selection. Oh hey, you finally mentioned evolution.

//Scientists are convinced that our universe began with one enormous explosion of energy and light, which we now call the Big Bang.//

Yes, and it's well supported, from CMBR, relative isotope frequencies, universal expansion, and other stuff. Darn, back to cosmology.

//Scientists have no explanation for the sudden explosion of light and matter. //

No we don't. We simply don't know. It's not a bad thing to acknowledge what you don't know. However, appealing to a god of the gaps is something else.

//The way they are arranged tell the computer program what to do. The DNA code in each of our cells is very similar. There are three DNA bases and it is estimated that there are 100 trillion cells in our bodies. different possible outcomes. If life had been by chance 99.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999% of these outcomes would be a failure, and scientists have not found any fossils of such failures as of yet//

That's silly, no one thinks DNA came about by chance, but in a series of stages, driven by chemical interactions. And you should clarify what these "outcomes" are. Is that universes? planets? over what time scale? You're not specific. So obviously, I don't know what these "fossils" you talk about are supposed to be. Are you expecting them to find fossils of the earliest microscopic protocells? Why would scientists expect to?

Of course, none of these points to any specific god, if it were valid.

That topic was discussed often enough and all that completely stupid nonsense was refuted already. Some examples of why these discussions are 99% about people making unproven statements:

"The evolution stance is based primarily on the assupmtion that God does not exist"
That's a claim you gave no proof for. In fact it is plainly wrong, there's enough possibility for science (you are talking about science, not just evolution) and god (you without any reason think of a christian god) to coexist.

"... God's design ... The Earth...its size is perfect ... is located the right distance from the sun ... [etc. etc. etc.]"
You won the lottery, so obviously god wanted it to be¿

"our universe began with one enormous explosion of energy and light, which we now call the Big Bang"
That's an extreme oversimplification and just not true. By the way, there are theories that go before the big bang.

"All instruction, all teaching, all training comes with intent. Someone who writes an instruction manual does so with purpose. Every cell of our bodies there exists a very detailed instruction code, much like a miniature computer program?"
You compare the program with it's manual¿ Well, nonetheless you just deny that evolution could have created this without any more reason than "I think (!) it's improbable". But in fact assuming the current laws of nature given, it was highly plausible.

"Natural, biological causes are completely lacking as an explanation when programmed information is involved."
Unproven claim. In fact, you didn't even say what such an information is supposed to be.

All your points were already mentioned in those threads, so you should have read them.

And to the very last: The whole discussion was not about personal beliefs, but if they may force their oppinion on biology classes.

The promotion of nonsensical religious claptrap as a substitute for scientific study of the world is not serious, it's delusional.  Beginning your post by being rude and obnoxious and announcing you're reviving a previously-locked thread is an efficient way to get your thread locked.  Go learn something instead of making a fool of yourself.

And, just for those looking for proper terminology: the opening post on this thread is an example of the [i]Gish gallop[/i]:
http://rationalwiki.com/wiki/index.php?title=Gish_gallop
http://en.wikipedia.org/wiki/Duane_Gish
It's a favorite tool of promoters of pseudo-science of all kinds.

... ah, and you're a middle school student.  That explains some things.


inequalities
trigonometry
inequalities proposed

Prove that :
  $ m_{a}\cos\frac{B\minus{}C}{2}\plus{}m_{b}\cos\frac{C\minus{}A}{2}\plus{}m_{c}\cos\frac{A\minus{}B}{2}\le \frac{\sqrt{3}}{2}(a\plus{}b\plus{}c)$
  I hope gemath (my idol  :blush: ) and evarist can give me nice solution and of course welkum all mathlinkers  :D !

[color=purple]Hi gemath is your idol ?  :rotfl: But a question for you ; Who did give you this problem ?  :) Because it's a problem in Mathnfriend a mathematics forum in Viet Nam . Please tell me because i've not solve this problem yet, too [/color]

[color=purple]Ok if you have :D [/color]


Breite oles tis synartiseis $ f: R\minus{}\minus{}>R$ pou ikanopoioun tin synthiki:

            $ f(x\plus{}y)\plus{}f(y\plus{}z)\plus{}f(z\plus{}x)\geq 3f(x\plus{}2y\plus{}3z)$
gia kathe pragmatiko $ x,y,z$

Για  $ y \equal{} z \equal{} 0$ εχουμε $ f(0)\geq f(x)$

  Για  $ y \equal{} x,z \equal{} \minus{} x$  εχουμε  $ f(2x)\geq f(0)$ για καθε πραγματικο $ x$.Αρα και $ f(x)\geq f(0)$

  Τελικα απο τις δυο σχεσεις  $ f(x) \equal{} f(0)$ για καθε πραγματικο $ x$.


conics
ellipse
geometry
geometric transformation
reflection

Find an equation for each ellipse given the information below.

(1) Center at (1, 2); focus at (1, 4); contains the point (2, 2)

(2) Center at (1, 2); vertex at (1, 4); contains the point (2, 2)

for some reason i remember that an ellipse has two foci not one.  and i dont think that it has a center because there is not one point that is equadistant from all of the points on the ellipse!

sharkman, can you show your work instead of just asking people your homework questions?

An ellipse has a center; it is the point that is equidistant between the semi-major and semi-minor axes.  True it is not equidistant from every point, but it is the midpoint of the segment connecting any two points on opposite sides.

Oh thank you but i do remember that ellipses have two foci (correct???), unless i just wasnt paying attention when i learned conics!

Given the center and a focus, the other focus is simply the reflection of the first focus across the center. :wink:

[quote="SnipedYou"]Yes all ellipses do have two foci, except the ellipse with eccentricity $ 0$, but that's just usually referred to a circle.[/quote]

In this case, the center is the focus, which is why the reflection of the focus about the center is the same point.


function
integration
inequalities
induction
real analysis
real analysis unsolved

Let $f$ and $g$ be two continuous, distinct functions from $[0,1] \rightarrow (0,+\infty)$ such that

$\int_{0}^{1}f(x)dx = \int_{0}^{1}g(x)dx$

Let

$y_n=\int_{0}^{1}{\frac{f^{n+1}(x)}{g^{n}(x)}dx}$, for $n\geq 0$, natural.

Prove that $(y_n)$ is an increasing and divergent sequence.

Oi marcelo.. lembra de mim?? :)
essa questão se não me engano você faz pensando no espaço de Cauchy(pelo menos foi o que me disseram).. e ai, como vc foi na prova?
nao fui mto bem nao..
abraços
Igor

*This post is in Portuguese

Where did you get this problem from? ;)
I have a nice solution for this one. ;)

It´s from a math olympiad  for university students in brazil...
can you post your solution??  :)
tkz

[quote="Arne"][quote="IgorCastro"]

*This post is in Portuguese[/quote]

Do NOT post in other languages than English in the international forums! :mad:[/quote]

Ok.. sorry.. It´s just because i don´t know how to say all that i said in english..

At least try - you'll see how quickly your English improves :)

Ok, here it goes. First of all, it is easy to see that there is $\alpha\in[0,1]$ such that $f(\alpha)>g(\alpha) .$
 Let us consider the function $\psi : [0,1]\to\mathbb{R_{+}}^{*}$, with $\psi (x)=\frac{f(x)}{g(x)}.$ Since $\psi(\alpha) >1$, from $\psi$ continuous we obtain the existence of an interval $[a,b]$ in $[0,1]$ such that $\psi(x) >1$ for any $x\in [a,b].$ So there is $\lambda >1$ such that we have the following result:

\[ x_{n}=\int_{0}^{a} f(x) \psi ^{n} (x) dx +\int_{a}^{b} f(x) \psi ^{n} (x) dx+\int_{1}^{b} f(x) \psi ^{n} (x) dx\geq \lambda^{n}\int_{a}^{b} f(x)dx.  \]

But, from the mean value theorem we have $c\in (a,b)$ such that $\int_{a}^{b} f(x)dx=(b-a)f(c) >0.$ so the sequence given diverges. ;)

Really a nice solution!!  :lol:
i have think about  $\frac{f(x)}{g(x)}$ and the existence of  $\alpha$
but not link them..

Almost forgot! From this inequality $a^{n+1}+b^{n+1}\geq ab^{n}+ba^{n}$ for any $a, b>0$ it folloows that our sequence is increasing. ;)  You know how to proceed. ;)

Oi Igor. Eu deixe uma mensagem particular pra você, mas você não leu...então, tomara que vc leia esta aqui:

Po kra, espaço de Cauchy? Acho que isso deve ser uma forma pomposa de dizer outra coisa. Porque o que eu acho que pode ser usado é uma extensão da definição de produto interno. Porque se vc considerar as funções como vetores [onde cada entrada é o valo de f num ponto], então vc possui um espaço vetorial no qual continua válido o produto interno e a desigualdade de cauchy...ela resolve o problema =]
abração

**sorry posting again in portuguese, but igor is my friend and i don't see him for a long time, as i know he doesn't speak english, and maybe for that reason he probably does not know that i sent him a PM  :P , i decided to post like that, as we are not trading ideas to solve problems =]

We can prove by induction that $(y_n)$is increasing:
$y_{n+1}=\int_{0}^{1}{\frac{f^{n+2}(x)}{g^{n+1}(x)}dx}$
$y_{n+1}\cdot \int_{0}^{1}{\frac{f^{n}(x)}{g^{n-1}(x)}dx}\geq (\int_{0}^{1}{\frac{f^{n+1}(x)}{g^{n}(x)}dx})^2\geq y_n \cdot\int_{0}^{1}{\frac{f^{n}(x)}{g^{n-1}(x)}dx}$(by hypothesis of induction)and monotony follows.

[quote=Cezar Lupu]

\[ x_{n}=\int_{0}^{a} f(x) \psi ^{n} (x) dx +\int_{a}^{b} f(x) \psi ^{n} (x) dx+\int_{1}^{b} f(x) \psi ^{n} (x) dx\geq \lambda^{n}\int_{a}^{b} f(x)dx.  \]

But, from the mean value theorem we have $c\in (a,b)$ such that $\int_{a}^{b} f(x)dx=(b-a)f(c) >0.$ so the sequence given diverges. ;)[/quote]

How did you obtain the Inequality? Could someone please explain?

$f,g$ are stricly  postive functions , so $\psi$ continuous positive , and $\psi([a,b])=[m,M]$ with $\psi(t)=m>1$ for some $t\in [a,b]$

That's alright, but shouldn't the last integral on the LHS be with limits from b to 1 instead of 1 to b?


Numbers from 1 up to 1,000,000 is written on a paper. How many 2s were written?

[quote="Allan Z"]I got 600000, if this is the correct answer, I will post how I got the answer.[/quote]

Yup. That's correct.

There are 1,000,000 integers from 1 to 1,000,000

Let's look at the unit digits: it can be only 0,1,2, to 9.. so every 10   consecutive integers, there is one number with 2 as the unit digit. 1 million integers have 100,000 groups, so 100,000 numbers have "2" as the unit digit.

Lets look at the tens place: every 100 consecutive integers, there are 10 integers with "2" as the tens place. Therefore, there are 100,000 2's in tens place.

Same thing with hundreds, thousands, ten-thousands, hundred-thousands place, each with 100,000 2's.

Therefore, the answer is $100,000*6=\boxed{600,000}$


[quote="Farenhajt"]It's possible that I'm missing something very elementary, but I don't seem to be able to get the Latex code in a pop-up window, ready to be copied elsewhere, and that was a very useful option in the "old" forum. Can that be done now as well?

Note: I'm using Opera browser - maybe there's the catch?[/quote]

Can you try it in another browser?  I would guess this is a browser problem.

Yup, tried IE and Maxthon (which is IE-based) - it's still the same, hence I guess it's not browser-related. The only thing I could get in all three browsers was a "tooltip" (yellow floating rectangle which appears when you put the mouse cursor over something) with the code, and that ain't very useful.

[quote="SplashD"]You could quote it and then copy and paste the code.[/quote]

Sure, I know that, but that's a bit more troublesome - the pop-up window option was/is much simpler.

You can click on a picture with the right mpuse button and choose [i]Properties[/i]. In some browsers, for example Mozilla Firefox, you can see and copy the code as [i]Image title[/i] or [i]Alternative text[/i].


search
inequalities unsolved
inequalities

Let $ ABC$ is a triangle. Prove that: $ \boxed{p^2\ge 2R^2 \plus{} 10Rr \minus{} r^2 \minus{} 2(R \minus{} 2r)\sqrt {R(R \minus{} 2r)}}$


probability
search
calculus
conditional probability
real analysis
real analysis solved

HII I CAN'T SOLVE THIS PROBLEM, PLZ HELP ME :?

The problem says:
A plane is missing and it is persumed that it is equally likely to have gone down in any of three possible regions. Let (1-Ai) denote the probability that the plane will be found upon a search of the ith region when the plane is in fact in that region,
i=1,2,3.The constants Ai are called overlooks probabilities because they represent the probability of overlooking the plane;they are generally attributable to the geographical and enviromental conditions of the regions. What is the conditional probability that the plane is in the ith region, given that a search of region 1 is unsuccessful.

HOPE THAT YOU HELP ME. ;)

This actually belongs to Calculus Computations and Tutorials. After the search of the first region turned out insuccessful, your probability space shrinks to the following options: the plane is in the first region but was overlooked (probability $\frac {A_1}3$) or it is in the second region or it is in the third region (probability $\frac13$ each). The total probability of all three events is $\frac{A_1+2}3$. Now you should just figure out what portion of this total probability corresponds to each event by simple division. The answer is $\frac {A_1}{A_1+2}$ for the first region and $\frac 1{A_1+2}$ for the second and the third region. Hope this helps  :)

Thank u that would help in my assignment :lol:


limit
logarithms
real analysis
real analysis unsolved

Answer will be $ \plus{}oo$ because your expression is more than $ (\frac{2^{n}}{e^{\sqrt{n}}})^{\frac{\log2}{2\ln n}}\to\plus{}oo$
last step is obviusly  because you get something like this
$ e^{\log 2\frac{n\ln 2\minus{}\sqrt{n}}{2\ln n}}$


trigonometry
real analysis
real analysis unsolved

let $ X\equal{}C'([0,1])$ and $ Y\equal{}C([0,1])$ both with sup norm define $ F: X \to Y$ by $ F(x)\equal{}x\plus{}x'$ and $ f(x)\equal{}x(1)\plus{}x'(1)$ for $ x \epsilon X$
then what can we say about the continuity of $ F$ and $ f$

let $ x_n(x)\equal{}\sin (2\pi n x)$. Then $ \|x_n\|\equal{}1$, and $ |f(x_n)|\equal{}\sin 2\pi n \plus{} 2\pi n \cos 2\pi n \equal{} 2\pi n$. So $ f$ is unbounded.
The same example for the first question. $ \|F(x_n)\|\equal{}\sup |\sin 2\pi n x\plus{} 2\pi n \cos 2\pi n x| \equal{} \sqrt{1\plus{}(2\pi n )^2}$. So $ F$ is unbounded also.

IT means $ F$ and $ f$ are not continuous bcoz continuity implies boundedness
what can we say about closedness of maps


http://www.artofproblemsolving.com/Forum/viewtopic.php?t=110283

[hide]anyway
i don't like writing in latex a lot so i just wrote the answer...
$a=-\frac{3y+6}{3-y}$[/hide]


How to prove this theorem:
If an ordered set $ A$ has the least upper bound property, then it has the greatest lower bound property.

Thanks

The greatest lower bound of a subset $ B$ which has a nonempty set $ C$ of lower bounds is precisely the least upper bound of the set $ C$ - just apply the definitions.

if $ A \equal{} [0,2]$ , $ B \equal{} [0.1),C \equal{} [1,2]$.
Then according definition, the least upper bound of B is 1.
But C does not  have the greatest lower bound.

Oh yes, it does. The greatest lower bound of $ C$ is $ 1$ (a least upper bound, or a greatest lower bound, may belong to the set itself).


function
superior algebra
superior algebra unsolved

Let $ A$ be the set of the functions $ f: E \rightarrow E$

And define the law o in  by :$ fog(x)\equal{}f(g(x))$

Prove that:
if a function $ f$ satisfies $ foh\equal{}fog \implies h\equal{}g$ for all $ h,g \in A$ then f is injective.

if a function $ f$ satisfies $ hof\equal{}gof \implies h\equal{}g$ for all $ h,g \in A$ then f is surjective.

sorry I don't understand German :P

Could you please translate some text from this link(give me a hint).

Proof by contradiction is a pretty good hint.  Letting $ h$ be the identity is another.

I don't think these 2 ideas will work.
Could you write your proof?

You need to choose a suitable function for $ g$.  For example, suppose that $ f$ is not injective, then there exist $ a$,$ b$ such that $ f(a)\equal{}f(b)$, so pick any $ g$ that swaps $ a$ and $ b$.

how do you know that there exists $ g$ that swaps $ a$ or $ b$ ?

There exist functions that do anything you want.  Unless you have left out some conditions on $ A$ for the problem.

the functions on A are not necessary SURJECTIVE. so we don't have that every element $ \in E$ is g(..).

Look at the conditions of your problem very carefully.  It says that $ f$ must satisfy certain conditions for [i]all[/i] $ g$.  That means that if you are using proof by contradiction, you get to pick the $ g$.

You only need to find [i]one[/i] function $ g$, not show that all have some property.  If $ f$ is not injective, there exist $ a,b$ such that $ f(a)\equal{}f(b)$, so defining $ g(a)\equal{}b$, $ g(b)\equal{}a$, and $ g(x)\equal{}x$ for $ x\neq a, b$ will do the trick.  Then $ f\circ g\equal{}f\circ id$, but $ g \neq id$, contradiction.

The second part of the problem is fairly similar.  Once you understand the above proof, I don't think you'll find it too hard.

nice proof!
and the other part is similarely, we set $ h\equal{}Id$ and $ gof\equal{}f$ for $ t \neq x$ and $ g(x)\equal{}f(x)$ and we imediatly get the contradiction
Where $ x$ is a real wich satisfies $ x \neq f(y)$  $ \forall y \in E$

Thanks.

I'm not sure who you're writing these proofs for, but you should try to be more careful.  What is $ t$?  Why are you talking about real numbers?  How are you planning on finding an $ x\in E$ that is not equal to any element of $ E$?

I am writing it for s.o to check it

x is an element in E
since f is surjective there exists x wich # f(y) for all y in E

and gof(t)=f(t) for all t in E

Sorry you keep hounding you about this, but that proof is no clearer than the last, since you never define $ g$. (and as I said, defining $ g$ is the entire problem!)

Don't write your proofs for mind-readers.  Details need to be written.


topology
real analysis
Harvard
college
calculus
integration
geometry

This is to compare the quality of different math programs.  I'll start.  I go to dartmouth.  To find course content you can peruse most course webpages by going to math.dartmouth.edu.  I'm comparing the courses I know to courses I've seen at other universities.

UGrad courses

Math 54 (topology with munkres) probably at the level of most intro topology courses.  Ended on urysohn metrization theorem.

Math 63 (honors real analysis with baby rudin) is probably at the level of most introductory analysis courses

Math 71 (honors abstract algebra with gallian) is probably closer to the non honors version of algebra at harvard

Grad Courses
Math 101 (Abstract Algebra with Dummit and Foote) probably at the level of math 55, though the content changes.

Math 103 (Measure Theory and Lebesgue integral) is probably at the level of Harvard math 214a and the chicago sophomore program.

Overall, I'd say Dartmouth has great teaching (lectures are very clear), but courses lack as much depth, breadth, and creative problem solving as those at tier 1 schools (based on problem sets).  A lot of the time, you don't really get to see the bigger picture of the material, since classes simply don't cover enough material or problems.  Of course, this is only based on the more advanced courses after the introductory ones in algebra, analysis, and topology.

I go to Hopkins.  I'll leave out numbers because I don't think that's particularly relevant:

Undergraduate:
Complex Analysis: We use the book that MIT's complex analysis course uses (I forgot the number), and I know Terry Tao has taught out of it in his course at UCLA.  Standard stuff of complex analysis.

Algebra I & II: Prior to this year, this course used Dummit and Foote but the professor changed the book to Nicholson's book (it's easier, but I think it's good).  Depending on which book is used, might be closer to a graduate course.

Honors Analysis I & II: We use Carothers' book, which is the book used in Caltech's Classical Analysis sequence.  I think the course is modeled after UChicago's Honors Analysis sequence, and the difficulty is probably in between Math 25 and 55.

Topology: Probably similar to every other undergrad topology course that uses Munkres.

Differential Geometry, Fourier Analysis, PDE, Other Stuff: I don't really know any basis for comparison.

Graduate Courses:
Algebra: Uses Lang with D&F as a reference.
Real Variables: Measure Theory with Wheeden & Zygmund's book.
I'm not really familiar with too many others...yet.

I'm happy with the teaching but I wish there were opportunities for research, which there aren't.  Hopkins is obviously really big on biology/medicine so a lot of the research money goes there. :(

I'm just starting at Princeton, so more experienced others can add to this but I'll write about the sense I've gotten so far.

The math department is of course one of the best, although it's fairly small, so many standard courses aren't offered all the time. We do have intro analysis/linear algebra every semester, and intro algebra, topology, differential geometry, analysis courses happen once every year. For more advanced courses, however, we only have advanced topics in algebra/analysis/geometry, so you can only take whatever advanced topic they're offering that semester. E.g., there's no standard undergraduate algebraic number theory or algebraic geometry course; you only get such a course every 2-3 years when it's taught in one of the topics classes. You can also take a graduate (500-level) course, though they tend to be quite difficult (and the graduate courses here are more intended as introductions to research-level math than to graduate-level math, since the graduate students here are expected to learn the basics on their own). On the other hand, many professors are willing to hold small independent studies, so if you can find 4-5 who want to study a particular book, you can probably find a professor to run that as an independent study. For example, I heard that a group of students studied Miranda's Algebraic Curves and Riemann Surfaces with Professor Gunning, and another group learned about modular forms through Apostol's Modular Functions and Dirichlet Series in Number Theory.

As for professors, I get the sense that the professors are generally good teachers and nice people, but of course this can vary (and this is where others can chime in, since I've only seen three professors so far). Many people tell me that professors are very willing to talk to you during office hours. The grading tends not to be too harsh in the more advanced classes. However, I've heard that in the lower level classes (e.g. intro calculus and multivariable calculus), the professors aren't as good, and the grading can be extremely harsh.

Another very nice thing about the math department here is that they pretty much let you take whatever you want to take. I had no trouble getting into algebraic number theory (MAT 453) and complex analysis as a freshman.

Now I'll do a quick run-through of the specific course options. First, I start with the beginning courses that most math majors take.

MAT 214 (Numbers, Equations, and Proofs): An introduction to number theory, which few people take, but is probably a really good idea if you haven't seen number theory and/or haven't seen much beauty in mathematics.
MAT 215 (Analysis in a Single Variable): This is the introductory analysis course, which almost all math majors take as freshman. It covers the first 8 chapters or so of Rudin.
MAT 217 (Linear Algebra): This is the theoretical introduction to linear algebra, which loosely follows Hoffman and Kunze.
MAT 218 (Analysis in Several Variables): This is the rigorous introduction to multivariable analysis, which culminates in proving Generalized Stokes's Theorem and deriving the differential form version of Maxwell's equations. Unlike some multivariable analysis classes, it's not required to already know some multivariable calculus, and there are problem sessions designed to give the student an intuitive feel for the computational aspects of multivariable calculus.

Next come the 300-level courses.

MAT 322 (Algebra with Galois Theory): This is Princeton's introduction to algebra course, which, unlike most other places, is only one semester (and does Galois theory up from no abstract algebra background within that semester!). I've heard that this class is quite hard, and the professor may discuss advanced topics such as Galois cohomology or infinite Galois theory at the end. After this, if you want to learn more algebra, you can take MAT 424, which gives advanced topics varying from year to year. If you want to learn some of the algebra taught in most universities but skipped in MAT 323, you could learn it on your own (which many do), or you could take:
MAT 323 (Algebra): This is an introduction to abstract algebra eventually covering Sylow theory, modules over a PID, and applications of group theory to coding.

MAT 325 (Topology): This is a standard topology course, covering all the chapters of Munkres's topology book.
MAT 326 (Algebraic Topology): This is the undergraduate introduction to algebraic topology, which (possibly surprisingly) doesn't require 325, but does require knowledge of abstract algebra and MAT 218. In recent years, this course has mainly used the book From Calculus to Cohomology.

MAT 327+328 are the introductory courses in differential geometry, the first talking intuitively about curves and surfaces, and the second developing the machinery of manifolds and vector bundles.

MAT 330-332: This is the analysis sequence famous at Princeton, from which three textbooks have been written. The courses are Fourier analysis, complex analysis, and Hilbert spaces/measure theory, though neither is a prerequisite for any other. These courses are somewhat difficult and teach students to think analytically (in the sense of analysis), with lots of difficult problems requiring all sorts of bounding and other analytic arguments. The classes also touch on relations to other areas of math, especially number theory. You'll have seen proofs of Dirichlet's Theorem and the Prime Number Theorem by the end of 330 and 331. I'm in the complex analysis course right now, and we proved Cauchy-Goursat in the third lecture and have been moving on quickly from there; it's quite intense.

There are 400-level topics classes in analysis, algebra, and geometry, respectively (usually 1-2 of each a semester) numbered MAT 451-456. The algebra one usually covers algebraic or analytic number theory, or algebraic geometry. There's also MAT 424, which is a another topics class in algebra that happens in the spring, and this may cover topics such as advanced Galois theory, algebraic number theory, algebraic geometry, representation theory, or Wedderburn theory. There's also an extra topics in analysis class which happens in the fall and covers distribution theory (MAT 433).

That's the Princeton math department. Looking over this, I feel like the department here has a somewhat unconventional and creative tone, which I especially feel about classes like MAT 322 and 326, the willingness of professors to supervise independent studies, and the lack of a set curriculum. In some sense, the department doesn't teach you a specific set of mathematical facts that an undergraduate should know, but it teaches you how to think mathematically.

Wow mat 325 sounds ridiculous...  All of Munkres?!  that's insane, simply because Munkres' text covers so much material.  He goes into algebraic topology and such later on.

I don't know if it covers every little detail in Munkres, but it does deal with basic topology, function spaces/paracompactness, and the fundamental group.

I would think that's about right for most topology courses.  Chapter 1 is a lot of set theory stuff so if you omit that, you could probably do a lot more.  We did the first 4 chapters completely and chapter 9 completely-but we probably could have done more if we skipped the set theory stuff.

I go to Cambridge and am quite interested in how the tripos's level compares to that of top US programs.  (I'm Canadian.)  The question is nontrivial since most Cambridge courses do not follow a textbook.  So when I name some below it means they are drawn from a list of 5 or 6 texts suggested for the corresponding class -- hopefully a meaningful approximation.  Also, due to the exam system (arcane to most North Americans), nth-year students very usually do nth-year courses.  Exceptions typically see a student just sitting in on higher-year lectures.  The keen can locate current, very detailed syllabuses or past example sheets and tripos exams online to get a fuller picture of the course's quality.

Here's IA, the first year.

Numbers & Sets: once called discrete maths.  Basic number theory, set theory/logic, intro to series and sequences, countability.
Differential Equations: standard ODE stuff through to a bit on PDEs.
Vectors & Matrices: an applied/practical approach to linear algebra and vector algebra.  Easy.
Groups: usual theorems (Lagrange, isomorphism thm, Orbit-Stabilizer).  Mobius maps and the group.  Matrix groups. S_n and A_n.
Analysis I: in a single variable.  Up thru Riemann integration.
Vector Calculus: the various integrals, differential operators and integral thms.  Applications to Maxwell's eqns and fluids.  Laplace's and Poisson's eqn.  Then an applied treatment of tensors.
Probability: discrete and continuous random variables.  Inequalities, CLT (with heuristic mgf proof).  Loosely Grimmett and Welsh.  Lectured by Timothy Gowers last year.
Dynamics & Special Relativity: note maths equals math union theoretical physics.  Newtonian stuff, orbits, rotating frames, rigid bodies, SR.

Now a second year, I'm presently taking

Analysis II: all things uniform, normed spaces, metric spaces, contraction mapping, differentiation from R^m to R^n.  Rudin or Korner's Companion appropriate.
Markov Chains: discrete time ones; quite like Norris's first chapter.
Linear Algebra: theoretical treatment.  Dual spaces, JNF, mainly (not completely) restricted to finite-dimensional vector spaces.  Halmos.
Methods: principally for PDEs, though started off with Fourier Series and Sturm-Liouville.  'Applied'.
Quantum Mechanics:  Gasiorowicz's book at a similar level.  Ummm... apparently?

Next term I'll do Complex Analysis (I imagine standard), Geometry (based on Wilson's Curved Spaces), Groups, Rings & Modules (affectionately read GR(i)M; Artin or Herstein), Metric & Topological Spaces (read Sutherland) and Variational Principles.  Possibly, I'll turn up to the non-examinable Algorithms lectures too.  If I wanted, I could be doing Fluid Dynamics, EM, Statistics or Numerical Analysis.  Throughout both terms there are computing projects.

The third year concludes the undergraduate degree and leads into Part III.  There are several courses at an advanced level.  My class list might be: Algebraic Geometry, Riemann Surfaces, Galois Theory, Graph Theory, PDEs, Algebraic Topology, Linear Analysis, Differential Geometry and Representation Theory.   Part III is an [i]intense[/i] 1-year Masters equivalent course which looks awfully fun.  Around ninety courses are offered; pick six.

Overall, I'm really happy with Cambridge.  There are no general requirements and, to my knowledge, the supervision system has no North American analog.  calc rulz, eg, might have had issues diving into higher-level stuff but commonly (in the pure subjects at least) much harder and more interesting problems are included at the end of assignments.  Lecturers are concise, necessarily since a "long" class consists of only 24 lectures.  The once-per-annum evaluation setup demands you to continually synthesize material from different areas.  Unfortunately (correctly?), there seems to exist here the attitude that undergraduate maths research is too trivial to warrant support.  Finally, while there are 250+ math students per year university-wide and loads of choice especially in parts II and III, Cambridge's collegiate system gives rise to a nice "small department" feel.

It seems that Cambridge places a lot more emphasis on applied math than most programs in the US, which makes sense.  It also seems like Cambridge gives more of a survey of many different areas than a detailed look at anything particular.  For example, I didn't see anything in your post that seems like an algebra II course (Galois theory) or anything to do with measure theory, which I think many programs here offer.  Correct me if I'm getting the wrong impression.

Notice I said I will be taking Galois Theory next year. Measure theory is also taught, in Probability & Measure.  (Logic & Set Theory is too... so is Number Fields and Number Theory.)  However my tentative list of 9 courses for next year is lengthy at present.  But Measures is something I'll consider or least try to pick up on my own, especially if it's standard at most US places.

So I think your impression is half-correct.  You definitely can take a detailed look at something; maybe my personal tastes obscured this.  I would argue, and most people who have actually done the tripos would, that the strength is precisely depth.  Different classes by design blend into each and the expectation is, in IB for example, everything from IA you've more or less mastered and can prove on the spot.  An older student from my college took in her fourth year strictly algebra-related classes.  The breadth of options makes such depth feasible.  Also bear in mind we just do math.  It is interesting to note 3rd-year MIT students on exchange tend to complete their math requirements (so that's two more years) here in just one year.  Most also do our second year.  (To be fair, some students at MIT go well-beyond the requirements, diving into grad classes off the bat.)

You're right though Cambridge emphasizes the applied.  In the first two years, you are "forced" into some applied courses and if you in fact prefer this madness, then you can get a math degree having done GR, QM, Fluids, EM etc and not much, beyond the first 1.5 years, in pure math.

A lot of people here double in math and physics or some engineering field so they see a lot of applied stuff (I'm actually in economics as well, which is not quite physics  :wink: ), but Cambridge seems to "force" it on people more (that's not meant to be a bad thing).

Most universities here will teach parts of measure theory as part of an advanced probability course, but that will usually be in an applied math department.  Typically, there is a separate course devoted to measure theory (perhaps called Measure Theory, Real Variables, Integration,...) which is taken by upper level undergrads and grad students.  They typically use a book like Stein's "Real Analysis," Wheeden/Zygmund's "Measure and Integral," "Papa Rudin," or Folland's book, or Royden's book, books which are theory-based instead of probability-based.

You said you were taking Galois theory but I didn't see anything.  I saw "Groups, Rings, & Modules" but I didn't think that Galois theory would be taught in that class.  Most US universities have a two-part algebra curriculum-groups, rings, and fields in the first half and Galois theory in the second.

[quote] Algebraic Geometry, Riemann Surfaces, [b]GALOIS THEORY[/b], Graph Theory, PDEs, Algebraic Topology, Linear Analysis, Differential Geometry and Representation Theory. [/quote]

That's a year 3 list.  Okay, so looking back at it, that paragraph was really really terribly written.  My bad.

As I can't say much more till the third year, I point to http://www.maths.cam.ac.uk/undergrad/documentation/schedules/currentschedules/master.pdf.
(Says, eg, Papa Rudin is appropriate for Linear Analysis.)

As for the measure stuff, the stats here is grouped with the pure in the department DPMMS.  So the Prob & Measure course is probably theoretical: its sole prereq is Analysis II, it's self-described as critical for grad work in analysis and the department lecturing is DPMMS.  If you want, use the link to judge for yourself how much of a measure theory class it is.  Perhaps there's indeed a gap.

Looking through that list (briefly), I noticed a few things.  First, indeed it looks like "Probability and Measure" is the closest thing to what I am referring to, although "Linear Analysis" certainly has elements.  However, I noticed an absence of Differentiation, Maximal Functions, Approximations to the Identity and so forth (or maybe I just didn't read closely enough), which is stuff you're guaranteed to see in a Measure Theory/Real Variables course.  The other thing is the analysis sequence.  Your Analysis I class is what would typically be called "Advanced Calculus" here, or "Honors Calculus" or something.  Spivak's book is just calculus.  Your "Analysis II" is what would typically be done in the first semester of analysis here (say, from baby Rudin or some other book).  The second semester for us would be integration, although how far depends on the course.  For example, my analysis II course was devoted mostly to measure theory, but I don't think that's necessarily true of every analysis II course.

Your course catalog is quite interesting.

Yes, Cambridge seems surprisingly applied, which is not at all a bad thing!  My  measure theory course is far too theoretical sometimes, and, especially as we delve into functional analysis and banach spaces and especially dual spaces (I basically blanked out when my professor talked about isomorphisms from X to X**), I have no sense of the motivations behind these theorems and definitions.  Here, there's some amount of overlap between undergrad studies and graduate studies, i.e. Graduate Algebra goes over undergrad Algebra, but with more depth and breadth such as group actions etc.  I know that MIT is also much more "applied" in that my friend's Analysis II course has psets requiring linear algebra, differentials, measures, etc, in what must be a really fun experience.

I'll also admit that the problems I do are typically pretty easy.  They're typically just quick problems that require only an understanding of the theorems.  I typically finish a week of homework in 1 or 2 hours?

Honestly though, as someone who wants to do more applied stuff, I'm quite lost about what to do in the future and what courses to take.  I'm quite sick of theory and would like to see some applications of functional analysis soon!

[quote="oab729"]My measure theory course is far too theoretical sometimes, and, especially as we delve into functional analysis and banach spaces and especially dual spaces (I basically blanked out when my professor talked about isomorphisms from X to X**), I have no sense of the motivations behind these theorems and definitions.[/quote]
It's measure theory, what do you expect?  I suppose if you want applied stuff you could take a course in PDE or Fourier Analysis (I'll be taking one in Microlocal Analysis next semester).  1-2 hours for a problem set seems really short.  Is this a graduate course or undergrad one?


Starting in exactly one hour from this posting, the [url=http://www.freewebs.com/aopsmath/]math chat room[/url] will be holding a marathon, similar to all the other marathons, execpt in real time. All participants will be expected to bring about 5 problems.
Hope to see you there! ;)
The problems should be anywhere from Mathcounts to AIME level.
Please sign up below!

I'll be there. goes off to find problems...

ok, everythings open, come on over!
[url=http://www.freewebs.com/aopsmath/]HERE![/url]

Im in the chatroom...but Im lagging heavily...


"Most users ever online was 93 on Yesterday, at 4:47 pm"
i do, however, distinctly remember it saying that the most ever online was 424 on january 11
what happened?

At some point, search engine bots ran the number up.  Richard reset it.  There have never really been over 400 people online at once, though the highest number is certainly over 100.  It will eventually be an accurate number now that bots aren't able to run that number up.

But now it says

 Most users ever online was 635 on Tue Jul 05, 2005 12:50 pm

So it is more than 400

Unfortunately we hadn't blocked the counting of all bots.  Once again the number is artificial.  Hopefully we'll eventually master that stat, but it's not near the top of our priority list.

[quote="236factorial"]So the 635 isn't real? Whew! I didn't think it would ever be that much![/quote]I think something around 200 is more close to reality, but we can never be sure :)

i think I remember about 125 people a few days ago... I know days when a lot of people are online, however.

yeah, 625 is definitly not realistic :lol: the most people that were on while i was on was 162 :D


Find all natural numbers $ m,n,k$ such that
$ 5^m \equal{} 2^n \plus{} 3^k$

First suppose $ n\equal{}1$.  Then $ 5^m\equal{}2\plus{}3^k$.  Working mod 4, we see that $ k$ must be odd.  If $ k\equal{}1$, then $ m\equal{}1$ is a solution.  Otherwise, $ 5^m\equiv 2\mod 9$, giving $ m\equiv 5\mod 6$.  But since $ 7|5^6\minus{}1$, we have $ 2\plus{}3^k\equal{}5^m\equiv 5^5\equiv 3\mod 7$, which only happens when $ 6|k$, contradicting $ k$ odd.

Now suppose $ n>1$.  Then mod 4, we see that $ k\equal{}2r$ must be even, so mod 5, $ 2^n \equiv \minus{}3^k\equiv \minus{}4^r$ implies $ n\equal{}2s$ is even.  Now mod 9, $ 5^m\equiv 4^s$ is a square, which implies $ m\equal{}2t$ is even.  Therefore

$ 2^n \equal{} 5^m\minus{}3^k \equal{} (5^t\minus{}3^r)(5^t\plus{}3^r) \equal{} 2^a\cdot 2^b$,

where $ a<b$.  Now $ 2\cdot 3^r \equal{} 2^b\minus{}2^a$ implies $ a\equal{}1$, so $ 5^t\equal{}2\plus{}3^r$.  We just saw this implies $ r\equal{}t\equal{}1$, giving $ (m,n,k)\equal{}(2,4,2)$.

So the only solutions are $ (1,1,1)$ and $ (2,4,2)$.


I'm having trouble understanding the concept of locus.

Can someone give me some locus problems for practice?

Thanks.

Locus is the place or line that a point must be in accordance with a law or variation of other point.Example:the angle bissector is the locus of points whose distance from two concurrent lines is the same.The perpendicular bissector of line AB is the lcus of points whose distance from A and B is the same.I hope you understand.


I am in middle school, and I would like to do something math-related as a Masters degree.  With AP classes these days, it is already time for me to start thinking about a college major.  Any suggestions?  Please don't say "Follow your heart," or anything like that, as I am looking for fresh (or old) ideas.  Also, what is an actuary?  Is it something I should look into?

Questions:
What should I major in?
Should I do a double major?
I like law, but I am better at math.  Should I be a lawyer?  Can I combine the two?

well ...first off i think it is a bit too early for u to think about majors

but here is my say

what college do u want to go to, and then look at what major they are better at math or law...

u can't really combine those to majors so choice the one u like to do everyday of ur life

wow i am getting deep... :D

[quote="Zratonx"]choice the one u like to do everyday of ur life[/quote]

Thanks, but that line is exactly what I am NOT looking for.  I need pure information only, please.

[quote="DPatrick"][quote="mysmartmouth"]what is an actuary?[/quote]
[url=http://www.beanactuary.org/about/whatis.cfm]This site[/url] has a nice description.[/quote]

Cool! That was the phrase of the poster I saw in my math class.  I think it was the same organization, "Be An Actuary."

Edit:  It looks pretty cool, now that I saw what it is.  That is something for me to look in to.

Dude, slow down. You're not even in high school yet. You've still got four years of partying with friends and proms and hot dates and not-so-hot dates and a ton of other dumb stuff left. Don't rush through those into college and life.

To follow the trend here, if you choose too early, then you may close your doors to new fields.  I know that in 2 years of high school, my ideas of a major have changed a lot.  Its good that you're thinking about it, but don't make it a huge part of your life right now.

Have fun.

I agree that middle school is way too early to decide on careers, but: if you love law and math, then go ahead and do both.

One of my students right now is a math major and is pre-law.  She is doing mathematics research this summer in voting theory.  Her plan is to attend law school after college.  (Come to think of it, a friend of mine who is a public defender was a math major as well... hmmm...)

First than all, that "dont tell me [i]follow your heart[/i]" is kind of unpolite, first because it is exactly what most of the people can tell you. I could tell you: Study math, and Electric Engineering or Computer Sciences, becasue those are the fields where I know I am good at, who knows if you would really like such fields (specially because the real math and the math you see in middle school are WAY TOO DIFFERENT). There are things on which people shouldn't tell you what to do, and what to study is one of them.

On the other hand, I agree, it is too early to decide what to study. Most of the people I know decided their future carrier in the Senior year of High School, and the one who did it early did it at the end of the Junior year, there is more time than life, don't run that much, because sometimes the journey is more important than the destination.

If you are interested on doing law, well, take your time, look at the programs of most of the law schools and see what are their course all about... may be you discover that you really like it... or you discover that you would be totally unhappy studing that. Give time to the time. That does not mean that you should not care about that until your senior year, but that shouldn't be on your forehead all the time.

[quote="smbelcas"]One of my students right now is a math major and is pre-law.  She is doing mathematics research this summer in voting theory.  Her plan is to attend law school after college.  (Come to think of it, a friend of mine who is a public defender was a math major as well... hmmm...)[/quote]

I went to a law school open house. Someone asked about what was considered the appropriate major(s) for an undergrad if he/she intended to go to law school. One law professor in that school said: "music major, math minor".

The fact is that the undergrad's majors do not really matter too much for professional schools in law, med or business. However, if you want to go to graduate school, the undergrad's major usually has to match.

[quote="djimenez"]First than all, that "dont tell me follow your heart" is kind of unpolite, first because it is exactly what most of the people can tell you.[/quote]

Sorry! I know everyone has good intentions! But what I was trying to say was that this was a post for gathering ideas about different career options out there.  There may be a job perfect for me, but I have never even heard of it.  And about it being too early too decide on anything, I agree, let me put it this way.

My school has a program where I can finish all "normal" college math up to statistics, and I need to plan out my schedule up to tenth grade by the end of this school year, so says my guidance counselor.  So all of the sudden, I already have to choose what I want to look in to.  What ever happened to the good old days, when you took college level classes in.... college? ;)

well, what classes you take in high school do not necessarily have anything to do with what you would study in college.
as you said, you can finish all the ""normal" college math" so why not just do it, and also take other classes... explore your choices!


superior algebra
superior algebra unsolved

[quote="countdownmath"](i) In the ring $ (\mathbb{Z}/10\mathbb{Z})[x],$ is $ x$ an irreducible element?

(ii) In the ring $ (\mathbb{Z}/10\mathbb{Z})[x],$ show that the ideal generated by $ 2$ and $ x$ is a principal ideal.[/quote]

It seems obviouos that $ x$ has to be a prime element..

But "seeming obvious" isn't enough. $ x \equal{} (5 \plus{} 2x)(8 \plus{} 5x)$.

For part ii, also show that $ 8 \plus{} 5x$ divides 2 so $ (x,2)\leq (8 \plus{} 5x)$ and the reverse inclusion really is obvious.


trigonometry
inequalities
logarithms
real analysis
real analysis unsolved

Let $ x_c(n)=(sinn)^{n^c}$. Prove that this sequence has a limit for all sufficiently large $c$, but does not have a limit for $c=2$.

Fix $\varepsilon\in(0,1)$. The necessary and sufficient condition for $|\sin n|>\varepsilon^{n^{-c}},\ n\to\infty$ is $\left|n-q\cdot\frac\pi 2\right|<\left|\frac\pi 2-\arcsin\varepsilon^{n^{-c}}\right|$ for some natural $q$ which depends on $n$.

The first part of the problem is a consequence of the fact that $\pi$ has finite irrationality measure:

Let $t$ be such that $\left|n-q\cdot\frac\pi 2\right|>n^{-t}$ for all large enough $q,n$ s.t. $\left|n-q\cdot\frac\pi 2\right|<1$, and take $c>2t$. On the one hand, we have $\left|\frac\pi 2-\arcsin\varepsilon^{n^{-c}}\right|^2>n^{-2t}$ for sufficiently large $n$, and, on the other hand, since $\frac{\left|\frac\pi 2-\arcsin\varepsilon^{n^{-c}}\right|^2}{1-\varepsilon^{n^{-c}}}$ tends to a finite limit as $n\to\infty$, while $n\to\infty\Rightarrow\frac{1-\varepsilon^{n^{-c}}}{n^{-2t}}\to 0$, we get the opposite inequality for large enough $n$, thus getting a contradiction unless for all large $n$ we have $|\sin n|\le\varepsilon^{n^{-c}}$, and since $\varepsilon$ was arbitrary, this means that $|\sin n|^{n^c}\to 0$ for $c>2t$.

For the second part, we notice first that there is a constant $C$ and infinitely many pairs $n,q$ s.t. $\left|n-q\cdot\frac\pi 2\right|<\frac Cn$. It would thus suffice to find an $\varepsilon\in(0,1)$ s.t. for infinitely many $n,\ \left|\frac\pi 2-\arcsin\varepsilon^{n^{-2}}\right|>\frac Cn$.

For all $\varepsilon,\ \frac{\left|\frac\pi 2-\arcsin\varepsilon^{n^{-2}}\right|^2}{1-\varepsilon^{n^{-2}}}$ tends to the same finite non-zero limit, while $\frac{1-\varepsilon^{n^{-2}}}{n^{-2}}$ tends to $-\ln \varepsilon$ which $\to\infty$ as $\varepsilon\to 0$, so if $\varepsilon$ is close enough to $0$, then the $n$'s mentioned in the paragraph above which are large enough will satisfy $\left|\frac\pi 2-\arcsin\varepsilon^{n^{-2}}\right|^2>C^2n^{-2}$, or, in other words, $\left|\frac\pi 2-\arcsin\varepsilon^{n^{-2}}\right|>\frac Cn$, as desired.


analytic geometry
combinatorics proposed
combinatorics

The vertices of triangle $ ABC$ in the $ x\minus{}y$ plane have integer coordinates, and its sides do not contain any other points having integer coordinates. The interior of triangle $ ABC$  contains only one point , $ G$, that has integer coordinates. Prove that $ G$ is the centroid of triangle $ ABC$.

Pick's Theorem tells us $ [ABC]\equal{}\frac{3}{2}$ and $ [ABG]\equal{}[BCG]\equal{}[CAG]\equal{}\frac{1}{2}$.  It follows that $ G$ lies on each of the medians of $ ABC$, and thus it is the centroid.

Yes, correct and it is too easy using Pick's Theorem. But why can't we try an alternate proof?
Like for example, suppose the centroid is not a lattice point, can we prove that atleast one side must contain another lattice point? I haven't yet fully proved this though.

It's very certain that you can prove it that way, but I would expect a lot of case work in such a solution.


همونطوری که گفته بودم یسری سوال میذارم که البته اینا رو قبلا توی یه مسابقه آنلاین طرح کرده بودم.
 چون لتکس بلد نیستم همین رو ببینین.از وقتی دیگه المپیاد رو بوسیدم گذاشتم کنار سوالهایی که حتی خودم طرح کردم هم به نظرم سنگین میاد!!!!!!!!

[quote="sepehr ahmadzadeh"]e benladan jan man ba inke jabram fogholade zaeife ehsas mikonam ziadi in namosavi badihieeeee :wink:[/quote]
man in nabarabari ro kheili ghavitar tarh karde boodam ke albatte esbatesh kheili rahat bood baraye hamin fasele ro ziad kardam va meghdari zaeeftar shod vali be har hal ye rahe sakht kardane nabarabari zaeef kardaneshe.
shomaha ham be jaye harf zadan rahe halletoon ro benevisin. :D  :rotfl:


Supposing that an integer n is the sum of two triangular numbers, n=(a^2+a)/2+ (b^2+b)/2, write 4n+1 as the sum of two squares, 4n+1=x^2+y^2 where x and y are expressed in terms of a and b. Conversely, show that if 4n+1= x^2+y^2, then x^2+y^2hen n is the sum of two triangular numbers.

[hide]
Starting from triangular numbers: $k=\frac{a(a+1)}{2}+\frac{b(b+1)}{2}\Rightarrow 4k+1=(a+b+1)^{2}+(a-b)^{2}$.

Starting from squares: Let $4k+1=m^{2}+n^{2}$. Then $m,n$ are of different parity. We get $k=\frac{a(a+1)}{2}+\frac{b(b+1)}{2}$ where
\[a=\frac{m+n-1}{2},\ b=\frac{m-n-1}{2}\]
[/hide]

[quote="scorpius119"]
Starting from triangular numbers: $k=\frac{a(a+1)}{2}+\frac{b(b+1)}{2}\Rightarrow 4k+1=(a+b+1)^{2}+(a-b)^{2}$.[/quote]

are you sure? i think $4n+1=2a^{2}+2b^{2}+2a+2b+1=(a+b+1)^{2}+(a^{2}+b^{2}-2)$.

EDIT: sorry :blush: . me and my stupid mistakes. :blush:

[quote="nayel"][quote="scorpius119"]
Starting from triangular numbers: $k=\frac{a(a+1)}{2}+\frac{b(b+1)}{2}\Rightarrow 4k+1=(a+b+1)^{2}+(a-b)^{2}$.[/quote]

are you sure? i think $4n+1=2a^{2}+2b^{2}+2a+2b+1=(a+b+1)^{2}+(a^{2}+b^{2}-2)$.[/quote]

yes, check the constant term and the $ab$ term in your second equality. :wink:


Points A and B are separated by two rivers. One bridge is to be built across each river so as to minimize the length of the shortest path from A to B. Where should they be placed? (Each river is an infinite rectangular strip, and each bridge must be a straight segment perpendicular to the sides of the river. You may assume that A and B are separated from the intersection of the rivers by a strip wider than the two rivers combined.)

Translate $A$ toward to the first river to a vector $a$, where $a$ is perpendicular to the first river and $|a|$ is a width of the river, and translate $B$ towards to the second river to a vector $b$, where $b$ is perpendicular to the second river and $|b|$ is a width of the river.
We obtained to points $A'$ and $B'$. Then shortest path from $A$ to $B$ is at least $|A'B'|+|a|+|b|$.
And it is clear how to construct optimal path using $A'$ and $B'$ :)

I hope I correctly understood this problem  :blush:


Our team has a very good chance of making NEAML for the first time, in like, 20 years. I was just wondering if I could get some info on it.

I'd suggest PMing rjyanco - he tends to know these things.


For what value of $ k$ does the line represented by the equation $ 1\minus{}kx \equal{} \minus{}3y$ contain the point $ (4,\minus{}3)$?

If we plug in the point, we see that $ 1\minus{}4k\equal{}9 \implies k\equal{}\boxed{\minus{}2}$.


im from louisiana and im quite sure no one here will be going. i wonder if any of you guys are in similar situation and may want to form a team together...
def want to go.
for admins: can i just register by myself and the people at pton just put me in a team?

Unfortunately, you can't register by yourself.  You need to have either exactly 4 people (for a half-team) or exactly 8 people (for a full team).  That way, we can group half-teams together so that they have full teams.  The only reason we're doing this is to make the grouping easier.  This will save on the number of classrooms we have to reserve per contestant (we need to get one classroom for every team), so that we'll be able to allow more teams to register.

If you can't find enough people for a team or half team, we allow partial teams of any size to register for the Power Test alone. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=163891

As WindingFunction mentioned, the team size restriction is so that we can have as many students as possible at the contest with our limited number of rooms; if you are not coming to the on-site portion of the contest, it doesn't matter how many people are on your team (as long as it's no more than 8).

However, I encourage you to ask around and try to find a team. Last year, the Florida, Carolina, and Maryland teams all had several AoPSers and were from rather wide regions. You could try to contact some of the people on those teams to see if they have spaces. (Actually, I think Florida's teams are already fixed, but you could offer to take a spot in case one of their people dropped out.)


Is it possible to have a random variable for which

the probability of it being a rational number between 0 and 1 is 1.
all outcomes (all rational numbers) are equally likely

I'm a bit confused about this. It seems like you should be able to, however, then the probability of any individual outcome is 0, and since there are countable many, the total probability should be 0.

This is where infinities get confusing

We have $ n \rightarrow \infty$ options, each with a $ \frac{1}{n \rightarrow \infty}$ So there are $ n\frac{1}{n}$ options

Does that help?

[quote="0714446459923"]but what happens when we 'reach' infinity?

The chance of each individual outcome is 0, but the sum of all those 0s is still 0, not 1.[/quote]

The probability of one single outcome is indeed 0. However, for any interval you can determine the probability that the variable is in that interval.

but since there are only countably many outcomes, wouldn't that mean the sum is 0?

that's like say $ \sum_{k\equal{}1}^{\infty}0\neq0$

Not a calculus expert, but I don't think this is the same as the equation you wrote. You need to put a limit in. In calculus you'll learn that adding an infinite number of terms each arbitrarily close to $ 0$ is indeterminate; it can give almost any value.

The probability of each term isn't zero. Its arbitrarily close to zero.

[quote="0714446459923"]but even the limit is still 0

$ \lim_{x \to \infty}\sum_{k \equal{} 1}^{x}0 \equal{} 0$[/quote]

You took the limit the wrong way.
\[ \lim_{n\to\infty} \sum_{i\equal{}1}^{n} \frac{1}{n} \equal{} 1
\]Generally, once infinity gets involved normal computation rules need not apply. And the probability of each outcome is indeed 0; there is no equidistribution over an infinite set.

The answer is yes. You can do so in several ways, but one way is to define a probability measure on your random variable $ X \subseteq \mathbb{Q} \subset [0,1]$ by the "minimum possible" total length of open intervals covering X.

This makes the probability of any individual rational number or finite collection of rational numbers equal to 0; but the probability of say, all rational numbers between 1/4 and 2/5 equal to 3/20.

Actually, I think you are right. I forgot that a probability measure also has to be countably additive. Meaning that $ P\Big(\bigcup_{i \equal{} 1}^\infty X_i\Big) \equal{} \sum_{i \equal{} 1}^\infty P(X_i)$ for a countable collection of disjoint events $ X_i$.

Therefore no uniform probability measure could exist on the rational numbers in [0,1]. If it did and the probability of each rational number was $ \epsilon > 0$, by additivity, the total probability of all rational numbers in [0,1] would be infinite. And if the probability of each rational number was 0, by countable additivity, the total probability of all rational numbers in [0,1] would be 0. Both contradicting the requirement that the total probability be 1.

*strange number*, your main mistake is trying to add up infinitely small quantities. For instance, for the expression $ 0\cdot\infty$ one could write $ \lim_{n\to\infty} \frac{a}{n}n\equal{}a$ for any real number $ a.$ So those expressions don't really make sense, and the correct outcome is depending on the problem.

As for the problem at hand, [url=http://en.wikipedia.org/wiki/Probability_Distribution]here you go.[/url]

[quote="0714446459923"]suppose you started adding up the probabilties of all the outcomes

at time 0,
you add the probability of outcome 1, which is 0.
so the total so far is 0
at time 1/2,
you add the probability of outcome 2, which is 0.
so the total so far is 0
at time 3/4,
you add the probability of outcome 3, which is 0.
so the total so far is 0
etc.
at time 1,
you have added together the probabilties of the outcomes of every possible event,
and yet the total is still 0[/quote]

The probability is just ridiculously close to zero
when you add the probabiltiy of every number with in a range all up, it of course will become 1
but there are just too many numbers
infinity is just a concept, not a number

Probability distributions need to be countably additive.  As you've noted, this implies that there is no uniform distribution on a countable set.  This problem would have been much better-suited to the Probabilities & Statistics forum (and putting it there would have avoided some of the nonsense responses on this thread).


The cube volume is 4^3=64. The cylinder volume is 4*2*2*pi=16pi.

Answer: 64-16pi

Area of cube = $ 4^3 \equal{} 64$

Area of cylinder = $ 4\times4\pi \equal{} 16\pi$

Solution = $ \boxed{64 \minus{} 16\pi}$

EDIT: Darn, beat.


Let O is the circumcenter of ABC.P is a point on AB,Q is a point on AC.Prove that
O,P,Q is collinear equiv to $ \sin 2A \equal{} \frac{{PB}}{{PC}}\sin 2B \plus{} \frac{{QC}}{{QA}}\sin 2C$ :)  :welcomeani:

Thank you TRAN THAI HUNG!!

I believe that you mean $ \sin 2A \equal{} \frac{{PB}}{{PA}}\sin 2B \plus{} \frac{{QC}}{{QA}}\sin 2C$,
which is equivalent to $ \sin 2A \plus{} \sin 2B \plus{} \sin2C \equal{} \frac{AB}{AP} \sin 2B \plus{} \frac{AC}{AQ} \sin 2C$,
which is theorem 2 [url=http://www.mathlinks.ro/viewtopic.php?p=570413#570413]here[/url].


[quote="SuperMath"]Let $\large a,b,c>0$. Prove that:
$\large \sum \frac{a}{\sqrt{b+c}} \geq\ \frac{1}{2} \sum \sqrt{b+c}$[/quote]

  Put $\sqrt{a+b}=x$ , $\sqrt{b+c}=y$, $\sqrt{c+a}=z$
$\sum \frac{a}{\sqrt{b+c}}= \sum \frac{x^2+z^2-y^2}{2y}$
So, we have
$\sum \frac{x^2+z^2-y^2}{y} \geq \sum x$
So, it isn't difficult from this step.

Another solution: Chebyshev's plus Cauchy-Schwarz equal proof.