Find all pairs of nonnegative integer numbers $(a,b)$ such that
$(b-a)^2(a^2-b)=4ab^2$

Here is my solution,hoping it's correct(because i solved it in a short time!!):
If a=0 then b=0 and if b=0 then a=0 so (0,0) satisfying
Supposing both a,b are  different from 0
Put gcd(a,b)=d and a=md,b=nd,(m,n)=1 so :$(m-n)^2(m^2d-n)=4mn^2$
Because (m-n,m)=(m-n,n),hence $(m-n)^2|4$ so |m-n|=1 or 2
Case 1: |m-n|=1 then $m^2d-n=4mn^2$-> so m|1->m=1-> n=0(exclude) or 2-> d= 18 so (18,36) works!
Case 2:  |m-n|=2 then $m^2d-n=mn^2$->m|2 so m=1 or 2
If m=1 then n=3 or -1(exclude) so d=12->(12,36) works.
If m=2 then n=0(exclude!) or 4 so d= 9->(18,36) works!
Plz, help me check it!


hello, from $ 2^x\equal{}8$ we get $ x\equal{}3$ as the solution, therefore we have $ 3^3\equal{}27$ as the answer.
Sonnhard.

so, x must equal 3. 3^3 equals [b]27[/b]


If $k^{a}-1$ divides $k^{b}-1$ ($k,a,b\in N$), is it necessary that $a|b$?

The fact $k^{a}-1 \mid k^{b}-1\iff a|b$ becomes trivial if you work in base $k$. The number $k^{a}-1$ can be written as $\underbrace{1\ldots 1}_{a\text{ ones}}$ and $k^{b}-1$ can be written as $\underbrace{1\ldots 1}_{b\text{ ones}}$. Then it is obvious that the remainer of $k^{b}-1$ modulo $k^{a}-1$ is $\underbrace{1\ldots 1}_{b\mod a\text{ ones}}$.

[quote="mathgeniuse^ln(x)"]Yes, it must be necessary...

Since, if it weren't...then we would get fractional exponents...[/quote]

Not a valid argument.  Why can't there be some $P(k)$ such that $k^{b}-1 = (k^{a}-1) P(k)$?

Anyway, this problem is very easy with Euclid's Algorithm.

[hide="The first step"] If $k^{a}-1 | k^{b}-1$ then $k^{a}-1 | \left( (k^{b}-1)-(k^{a+(b-a)}-k^{b-a}) \right) = k^{b-a}-1$ [/hide]
Edit:

[hide="Remark"] This is a special case of the general theorem that $gcd(k^{a}-1, k^{b}-1) = k^{gcd(a, b)}-1$.  For the case $a | b$ we have $gcd(a, b) = a$. [/hide]


Prove that a square of side 2.1 units can be completely covered by seven squares of side 1 unit.

Extra: Try to prove that 7 is the minimal amount.

we have square ABCD with AB=2.1
first put a 2x2 square(=4 1x1's) with one vertex at A and completely within ABCD.
second put a 1x1 with one vertex at C and completely within ABCD.
now we have two 0.1x1.1 rectangles, which we can easily cover with a 1x1 square by taking the square with the same center as one of these rectangles and rotating it 45°.


[color=blue]Buenos dias,

Navegando por la web, encontre unos lugares donde ranckean a las Universidades de cada region y a nivel mundial en su conjunto...

Aki algunos links de interes:

- Universidades Latinoamericanas.
http://www.webometrics.info/top100_continent.asp?cont=latin_america

- Universidades Europeas:
http://www.webometrics.info/top100_continent.asp?cont=europe

- Ranking Mundial:
http://www.webometrics.info/top3000.asp

La metodologia utilizada es en base a las visitas de sus websites dentro de un mismo periodo del 2006.
Más info:
http://www.webometrics.info/comparative_methodology.html

Aunque aqui existe otro ranking que he encontrado y parece ser mas mesurado en sus resultados...
revisenlo:

http://www.indiaeducation.info/studyabroad/Best-Universities/World-Universities.asp

Saludos,

Carlos Bravo :)
Lima  -  PERU[/color]

Pues no me parece muy bonito que mi universidad esté en el lugar 2953... pero el método me parece en particular muy poco efectivo. Además, no es muy util calificar a las universidades en general, pues en particular, la mía tiene de las mejores facultades de física, química y matemáticas en México... aunque pues al menos en matemáticas (no estoy seguro pero probablemente es lo mismo en las otras dos) es de un tipo un poco más formativo que productivo

Es verdad lo que dices del ranking general. Mi universidad está en el lugar 1718, cuántas universidades hay en el mundo?.


is the distance from z= a+bi to 0+0i on a complex plane just

$ \sqrt{a^{2}+b^{2}}$ which can also be denoted by $ |z|$

Sorry I just want to make sure

[quote="the future"]is the distance from z= a+bi to 0+0i on a complex plane just

$ \sqrt {a^{2} + b^{2}}$ which can also be denoted by $ |z|$

Sorry I just want to make sure[/quote]

Yes.

So like the 0+0i, that i technically goes with the point so when you apply the distance formula you dont need to do $ \sqrt{a^{2}\plus{}{(bi)}^{2}}$ right? its just a b.

[quote="mountain.dew"][quote="the future"]So like the 0+0i, that i technically goes with the point so when you apply the distance formula you dont need to do $ \sqrt {a^{2} \plus{} {(bi)}^{2}}$ right? its just a b.[/quote]

Yep (as in don't include $ i$ in calculating distance). Another way to see it is that the complex plane is pretty much the Cartesian plane with $ y$-values representing the complex part of complex numbers.[/quote]

thanks

The reason why $ i$ is not in the formula is because the distance would instead be $ \sqrt {a^2 \minus{} b^2}$, which is [b]not[/b] the distance formula.

What's discussed is better referred to as the [b]absolute value[/b] of a complex number (it's a good idea to remember this term).
Basically, for  $ z \equal{} x \plus{} iy, |z| \equal{} \sqrt {z\bar{z}} \equal{} \sqrt {x^2 \plus{} y^2}$ where $ \bar{z}$ is the conjugate of $ z$.

[edit] Thank you azjps.

There is no code, its just the pipe: "|" key above the "Enter" button.


Let $ a$ be an odd number and let $ b$ be an even number, such that $ a^2\plus{}b^2\equal{}p$ is a prime number. Prove that the equation $ x^2\minus{}py^2\equal{}a$ has an integer solution $ (x,y)$.

We will need a Lemma: http://www.mathlinks.ro/viewtopic.php?t=25234 . As $ p \equiv 1 \mod 4$ (or $ p\equal{}2$, which is trivial), we can use it to get $ u,v$ with $ u^2\minus{}pv^2\equal{}\minus{}1$.

We have $ \minus{}a^2 \equal{} b^2\minus{}p \cdot 1^2$ by assumption and note that $ \gcd(a,p)\equal{}1\equal{}\gcd(a,b)$.
Consider $ c,d$ given by $ c\plus{}d \sqrt p \equal{} (b\plus{} \sqrt p)(u\plus{}v \sqrt p)$: they fulfill $ c^2\minus{}pd^2\equal{}a^2$ and furthermore $ \gcd(c,a)\equal{}1$.
To see the latter, assume that there is a prime $ q|c,a$, so $ q|pd^2$. As $ \gcd(a,p)\equal{}1$ we get $ q|d$. Now noting that we have $ b \plus{} \sqrt p \equal{} \minus{}(c\plus{}d \sqrt p)(u\minus{}v \sqrt p)$, we end up with $ q|1$, contradiction.

Now again using $ p \equiv 1 \mod 4$ and that $ a$ is odd, we get $ c^2\minus{}d^2 \equiv a^2 \equiv 1 \mod 4$, thus $ c$ is odd and $ d\equal{}2e$ even. From $ c^2\minus{}pd^2\equal{}a^2$ we reform to $ \frac{c\minus{}a}{2} \cdot \frac{c\plus{}a}{2} \equal{} pe^2$, so by $ \gcd \left(\frac{c\minus{}a}{2} , \frac{c\plus{}a}{2} \right) \equal{} \gcd \left(a,\frac{c\plus{}a}{2}\right) \equal{}1$ we have a product of coprime number equal to $ pe^2$.
So as in the linked thread, we get $ \frac{c \pm a}{2} \equal{} s^2$, $ \frac{c \mp a}{2} \equal{}pt^2$, finally giving $ \pm a \equal{} s^2\minus{}pt^2$.
If we have $ \plus{}a\equal{}s^2\minus{}pt^2$, we are finished, and otherwise we just "multiply" $ \minus{}a\equal{}s^2\minus{}pt^2$ with $ u^2\minus{}pv^2\equal{}\minus{}1$.


Prove that the numbers $ 5^{2009}$ and $ 2^{2010}$ starts with the same digit. (We know that $ \lg 2\equal{}0,301$).

[quote="almonkez"]i think u mean  2^2013 = 8 . 2^2010   isnt it ??[/quote]
Sorry.
I'm simply typo.
(Because copy & paste)


How many integral palindromes are there from 1 to 1000?

A palindrome is a number that is the same forward as backward. Examples are 11, 242, 909, etc...

Possibilities for the 1st digit: 9
Possibilities for the 2nd digit: 10
Possibilities for the 3rd digit: 1

$9 \cdot 10 \cdot 1 = 90$

[quote="-Terminator-"][hide="Answer"]$108$[/hide][/quote]
how did you get that answer

Oops, terminator is correct. I thought it was only 3-digit palindromes  :blush:


$ a,b,c>0$

$ \sum{\frac{a^2}{\sqrt{b^2\plus{}bc\plus{}c^2}}}\ge \frac{a\plus{}b\plus{}c}{\sqrt{3}}$

[quote="dima ukraine"]$ a,b,c > 0$

$ \sum{\frac {a^2}{\sqrt {b^2 \plus{} bc \plus{} c^2}}}\ge \frac {a \plus{} b \plus{} c}{\sqrt {3}}$[/quote]
$ \sum_{cyc}{\frac {a^2}{\sqrt {b^2 \plus{} bc \plus{} c^2}}}\ge\sqrt{a^2\plus{}b^2\plus{}c^2}$ is true too.

[quote="arqady"][quote="dima ukraine"]$ a,b,c > 0$

$ \sum{\frac {a^2}{\sqrt {b^2 \plus{} bc \plus{} c^2}}}\ge \frac {a \plus{} b \plus{} c}{\sqrt {3}}$[/quote]
$ \sum_{cyc}{\frac {a^2}{\sqrt {b^2 \plus{} bc \plus{} c^2}}}\ge\sqrt {a^2 \plus{} b^2 \plus{} c^2}$ is true too.[/quote]
I prove 2 this ineq by Jensen :)

dima ukraine, I think for the following inequality Jensen does not help already.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that:
\[ \frac {a^2}{\sqrt {b^2 \plus{} bc \plus{} c^2}} \plus{} \frac {b^2}{\sqrt {a^2 \plus{} ac \plus{} c^2}} \plus{} \frac {c^2}{\sqrt {a^2 \plus{} ab \plus{} b^2}}\geq\sqrt [4]{3(a^4 \plus{} b^4 \plus{} c^4)}
\]

For first inequality let use this lemma:

$ \sqrt{\frac{a^2\plus{}ab\plus{}b^2}{3}} \leq \frac{a^2\plus{}b^2}{a\plus{}b}$ and inequality equivalents to

$ a^6b\plus{}ab^6\plus{}b^6c\plus{}bc^6\plus{}a^6c\plus{}ac^6 \geq a^5b^2\plus{}a^2b^5\plus{}b^5c^2\plus{}b^2c^5\plus{}a^5c^2\plus{}a^2c^5$ which is obviously true by Muirhead. :wink:

$ \sum_{cyc}\frac {a^2}{\sqrt {b^2 \plus{} bc \plus{} c^2}}\geq\sqrt3 \sqrt[6]{\frac{a^6 \plus{} b^6 \plus{} c^6}{3}}$ is wrong already ( $ a\equal{}1.1,$ $ b\equal{}c\equal{}1.$ )
but $ \sum_{cyc}\frac {a^2}{\sqrt {b^2 \plus{} bc \plus{} c^2}}\geq\sqrt3 \sqrt[5]{\frac{a^5 \plus{} b^5 \plus{} c^5}{3}}$ seems true.


$\lim_{x\rightarrow\infty}\frac{cosx}{7-sinx}=?$

Saman Gharib :huh:

It's periodic and not constant, thus the limit doesn't exist.


i can't quite figure out what i should use to solve problem 167 of chapter 11 of AOps v1, any help?

IN triangle ADC, segment DM is drawn such that angle ADM=angle ACD prove that AD^2=(AM)(AC). M is on AC.

edited.

$5 says this is in the section on similar triangles ...

its not in the section, but in the chpater.. yes, ok i will try something with that.

Chapter, section, close enough -- I'm keeping my money ;).

Look for the triangles ...

Thanks JBL, i can't belive i didn't realize this one.

Oh I would throw it in the power of a point section actually...

its a proof of POP, so in a sence we prove it before we know about it, if you read the book from 1 to end...


Take a look at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=287956[/url].

[quote="stvs_f"]$ \forall n \in N : f(n) \equal{} n \plus{} [\sqrt {n}]$
$ \forall m \in N$: prove that :$ m,f(m),f(f(m)),...$ at least has One square number.[/quote]
My solution. Of course, wlog $ m$ is not a square number.
Let $ m \equal{} n^2 \plus{} r$, where $ 1 \le r \le 2n$. If $ r \equal{} n\plus{}1$ then $ f(m) \equal{} n^2\plus{}n\plus{}1\plus{}n \equal{} (n\plus{}1)^2$, and we are done, so assume the opposite. If $ n\plus{}2 \le r \le 2n, r \equal{} n\plus{}1\plus{}k, k \in \{1,2,...,n\minus{}1\}$ then $ f(m) \equal{} n^2\plus{}2n\plus{}1\plus{}k \equal{} (n\plus{}1)^2\plus{}k$. Letting $ p \equal{} n\plus{}1$. We have a situation, where $ m \equal{} p^2 \plus{} q$ where $ q \in \{1,2,...,p\minus{}1\}$. If $ 1 \le r \le n$, we also have that situation. (Let $ p \equal{} n, q \equal{} r$...)
It's obvious that if $ m \equal{} p^2\plus{}q, q \in \{1,2,..,p\minus{}1\}$ then $ f(f(m)) \equal{} p^2 \plus{} 2p \plus{} q \equal{} (p\plus{}1)^2 \plus{} q\minus{}1$. And we still have $ (q\minus{}1) \in \{1,2,...,(p\plus{}1)\minus{}1\}$ of course. Using this a lot of times:
$ 2q$ $ f$'s:
$ f(f(...f(f(m))...)) \equal{} (p\plus{}q)^2\plus{}q\minus{}q \equal{} (p\plus{}q)^2$, which is a square.. And we are done =)


Cindy was asked by her teacher to subtract $3$ from a certain number and then divide the result by $9$. Instead, she subtracted $9$ and then divided the result by $3$, giving an answer of 43. What would her answer have been had she worked the problem correctly?
(A) 15
(B) 34
(C) 43
(D) 51
(E) 138

[hide="solution"]
multiplying 43 by 3 and adding 9 yields the original number, 138. subtracting 3 and dividing by 9 gives the answer - A) 15
[/hide]
[hide="quick and easy based on the answers"]
knowing that the subtraction shouldnt make a huge deal, we can see that our target is approximately 3/9 of 43. the only possible answer within a decent range of 1/3 of 43 is A) 15[/hide]

[hide]Setting up equations and solving the variable, we get $15$

(A)$15$[/hide]

The other answers do seem unreasonable. :(

What we want to compute is
$\frac{n-3}{9}$

We are given is

$\frac{n-9}{3}=43$
Thus,
$\frac{n-3}{3}=45$
$\frac{n-3}{9}=15$
THE ANSWER IS A!!! (and I didn't even find n) :P


greatest common divisor
number theory
least common multiple
prime factorization

Prove that the product of the gcf of two numbers and the lcm of the same two numbers is equal to the product of the two numbers.

Let the two numbers be $m$ and $n$. In the prime factorization of each number, consider any of the primes $p$, which will be in the form $p^{x}$ for $m$ and $p^{y}$ for $n$. WLOG $x \leq y$. The GCF of $m$ and $n$ considering the prime $p$ will then be $x$. The LCM of the two numbers, again considering just that one prime, will be $y$. Therefore, when you multiply the GCF and LCM together, the prime will be of the form $p^{x+y}$ in the factorization of the product. Obviously, in $m\cdot n$, the prime factorization considering $p$ will also be $p^{x+y}$. Applying this to all prime factors of $m$ and $n$, we arrive at the conclusion.

There was a topic about this recently.  I give two methods in it.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=121310


geometry
3D geometry
tetrahedron
ratio
algebra
polynomial
geometric transformation

Consider an acute triangle $ABC$. Draw concurrent cevians $AD$,$BE$, and $CF$, and let their intersection be $P$. Now let $r=\frac{AP}{PD}$, $s=\frac{BP}{PE}$, and $t=\frac{CP}{PF}$.

Prove that $rst=r+s+t+2$.

Then consider the more general problem where you have $4$ concurrent cevians in a tetrahedron, giving ratios $r,s,t,u$. Prove that $rstu=r+s+t+u+69$. The second formula is known as the "69 formula" hahaha

[quote="drunner2007"][color=darkred]Consider an acute triangle $ABC$. Draw concurrent cevians $AD$,$BE$, and $CF$, and let their intersection be $P$. Now let $r=\frac{AP}{PD},\,s=\frac{BP}{PE},$ and $t=\frac{CP}{PF}.$ Prove that $rst=r+s+t+2$.[/color]
[/quote]
[color=indigo][b]Proof.[/b] Let $x=[BPC],\,y=[CPA],\,z=[APB]$
We have: $r=\frac{AP}{PD}=\frac{y+z}{x},$ so $rst=\prod\frac{y+z}{x}=\frac{\sum yz(y+z)+2xyz}{xyz}=\sum\frac{y+z}{x}+2=r+s+t+2$[/color]

Ok now try to figure out the other one. I'll add two results.
The first is $r+s+t \geq 6$ (with equality when the triangle is equilateral). The second (which is less useful) is $Xr+Ys+Zt=2(X+Y+Z)$, where $X,Y,$ and $Z$ represent the triangles used in the above proof.

[color=indigo]We have:
$r+s+t+3=\sum\left(\frac{y+z}{x}+1\right)=\sum\frac{x+y+z}{x}\\ =\left(\sum x\right)\left(\sum\frac{1}{x}\right)\ge 9\implies r+s+t\ge 6$
From this, we have: $3\max\{r,\,s,\,t\}\ge r+s+t\ge 6\implies \max\{r,\,s,\,t\}\ge 2$

Since $r=\frac{y+z}{x},$ implies $xr=y+z,$ so $xr+ys+zt=(y+z)+(z+x)+(x+y)=2(x+y+z)$ [/color]

[quote="nsato"][quote="drunner2007"]Then consider the more general problem where you have $4$ concurrent cevians in a tetrahedron, giving ratios $r,s,t,u$. Prove that $rstu=r+s+t+u+69$.[/quote]

In general, this relationship does not hold.[/quote]
There is the relationship
\[\frac{1}{1+r}+\frac{1}{1+s}+\frac{1}{1+t}+\frac{1}{1+u}=1\]
but because there are four congruent cevians instead of three, it might be possible to obtain another equation, if not the proposed one.

When $u \to \infty$ (when the point of concurrency approaches one of the faces of the tetrahedron), any restriction on $r, s, t, u$ turns into a restriction on $r, s, t$ in a triangle.  The "69 formula," for example, would imply that we  have $rst = 1$, which is obviously false.  I don't think you're going to find any other nice polynomial relationship between $r, s, t, u$.  (Of course, there is the relation $rstu = rs+\ldots+tu+2(r+s+t+u)+3$, but this is just multiplying out the equation scorpius has given; it is the natural generalization of the result $rst = r+s+t+2$, which becomes clearer when you re-write it as $\frac{1}{r+1}+\frac{1}{s+1}+\frac{1}{t+1}= 1$.)

An intuitive reason to think there should be no other relation: we have 3 degrees of freedom in choosing $r, s, t, u$ by choosing the point of concurrency, which lives in a 3-dimensional space.

Unrelatedly, this is my 6789th post -- let's hear it for base 10!

Interesting. Ok when a friend of mine and I came up with it, we looked at the special case where the cevians of the tetrahedron go to the centroid of the opposite face. Then we thought that the relationship would generalize because of "homothecy" (i think).


calculus
integration
calculus computations

can you solve   :D

$ \int_2^3\; \frac {\left[x^4\cdot (5 \minus{} x) \right]^{1/6}}{7 \minus{} x}\; \;\textbf dx$

I tried, but couldnt really succeed. Can u give a hint?


If one were to choose three random numbers in $ [0,1]$ and convert this vector into spherical coordinates, would the probability of choosing a random direction be the same as choosing two random numbers in $ [0,2\pi]$ for spherical coordinates?

Yes and no. Yes because a probability of getting any exact direction is $ 0$. No because they would not have equal values if you consider the probability distribution.

First, if you choose two arbitrary angles, you can get all possible directions, whereas choosing a random point from $ [0,1]^3$ will only give you a direction in the first octant  :P

So now, say you limit yourself to one octant. Let's look at the projection of the directions onto the xy plane. Then, we are comparing choosing one angle from $ \left[ 0, \; \frac{\pi}{2} \right]$ to choosing a random point in $ [0,1]^2$.

In the first case, each direction is equally likely. However, in the second case, the probability distribution for a particular angle is equal to the distance from the origin to the boundary of the square traveling at that angle. Visually, imagine that to go at 45 degrees, we can take any point on y=x, and so a measure of the number of points we can chose to get 45 degrees is the distance from the origin to (1,1), or $ \sqrt{2}$. However, to go at 0 degrees, an analogous measure would be the distance from the origin to (1,0), or $ 1$. So, its $ \sqrt{2}$ times "more likely" that your angle will be 45 degrees than 0 degrees.

From here, we can make an analogous argument for 3D, or we can say that since the probability distribution of the component in the lateral plane is different, the probability distribution for the directions of the original vectors is also different.


ratio
geometry
trigonometry
trig identities
Law of Cosines
angle bisector

Let $ABC$ be equilateral, and $D, E,$ and $F$ be the midpoints of $\overline{BC}, \overline{CA},$ and $\overline{AB},$ respectively.  There exist points $P, Q,$ and $R$ on $\overline{DE}, \overline{EF},$ and $\overline{FD},$ respectively, with the property that $P$ is on $\overline{CQ}, Q$ is on $\overline{AR},$ and $R$ is on $\overline{BP}.$  The ratio of the area of triangle $ABC$ to the area of triangle $PQR$ is $a+b\sqrt{c},$ where $a, b$ and $c$ are integers, and $c$ is not divisible by the square of any prime.  What is $a^{2}+b^{2}+c^{2}$?

hmm what year is this from?
[hide][img]6598[/img]
Assign variables:
$EP = FQ = x$
$EQ = y$
$PQ = k$
Since $AE = \frac{1}{2}AB$ and $AD = \frac{1}{2}AC$, $\triangle AED \sim \triangle ABC$ and $ED \parallel BC$
Alternate Interior Angles Theorem says $\angle PEQ = \angle BFQ$ and $\angle EPQ = \angle FBQ$
Vertical Angles Theorem says $\angle EQP = \angle FQB$
So $\triangle EQP \sim \triangle FQB$ and by CPCTC $\frac{EP}{EQ}=\frac{FB}{FQ}\Longrightarrow\frac{x}{y}=\frac{1}{x}\Longrightarrow x^{2}=y$
Since $\triangle EDF$ is equilateral, $EQ+FQ=EF=BF=1\Longrightarrow x+y=1$
Solving for $x$ and $y$ using $x^{2}=y$ and $x+y=1$ gives $x=\frac{\sqrt{5}-1}{2}$ and $y=\frac{3-\sqrt{5}}{2}$
Using the Law of Cosines, we get
\begin{eqnarray*}k^{2}&=&x^{2}+y^{2}-2xy\cos{\frac{\pi}{3}}\\ &=&\left(\frac{\sqrt{5}-1}{2}\right)^{2}+\left(\frac{3-\sqrt{5}}{2}\right)^{2}-2\left(\frac{\sqrt{5}-1}{2}\right)\left(\frac{3-\sqrt{5}}{2}\right)\cos{\frac{\pi}{3}}\\ &=&7-3\sqrt{5}\end{eqnarray*}
We want the ratio of the squares of the sides, so $\frac{(2)^{2}}{k^{2}}=\frac{4}{7-3\sqrt{5}}=7+3\sqrt{5}$ so $a^{2}+b^{2}+c^{2}=7^{2}+3^{2}+5^{2}=\boxed{83}$
[/hide]

uugh...this is so computationally intensive...

Denote $AP\cap BC=T$, $AD=1$, $PD=x$, $DR=1-x$, $AP=a$

angle bisector theorem implies $PR=a(1-x)$, $\triangle APD\sim\triangle ATC$ so $AP=PT$, so $RT=ax$, $TC=2x$, then $FT=1-2x$

$\triangle PDR\sim\triangle FTR$, so $FT=\frac{x^{2}}{1-x}$, solving, $x=\frac{3-\sqrt{5}}{2}$

$\frac{PR}{AT}=\frac{1-x}{2}$, $AT^{2}=4+4x^{2}-(2)(2x)$, so $PR^{2}=(1-x)^{2}(x^{2}-x+1)=7-3\sqrt{5}$

the area ratio is side length ratio squared, giving the ratio $\frac{4}{7-3\sqrt{5}}=3\sqrt{5}+7\implies a^{2}+b^{2}+c^{2}=\boxed{083}$....

[size=75]you've written about angle bisector theorem.
"...angle bisector theorem implies ...."
Could you explain about it? i've never seen it before.
(plus proof, if you don't mind.)
PLEASE...[/size]

This is what I found from 15 seconds of searching  :wink: :

[url=http://www.artofproblemsolving.com/Wiki/index.php/Angle_bisector_theorem]AoPS Wiki Article[/url]
[url=http://en.wikipedia.org/wiki/Angle_bisector_theorem]Wikipedia Article[/url]
[url=http://planetmath.org/?op=getobj&from=objects&id=6488]Proof of General Form on Planet Math[/url]

which should be comprehensive.


[quote="Ligouras"]Let $ a, b, c$ positive real numbers such that $ ab \plus{} bc \plus{} ca \equal{} abc \equal{} 1$. Prove or disprove that

$ \sum(\frac {a^2 \plus{} 2a \minus{} 2}{(a \minus{} 1)^2}) \ge 4$ .

_____
 :?:[/quote]

I think you have some typo:
we have
$ 1\equal{}ab \plus{} bc \plus{} ca \geq 3(abc)^{\frac{2}{3}} \equal{} 3$
which is not true and so we can't have such condition.


trigonometry
calculus
derivative
function
calculus computations

I need help on following three problems. I solved them but I'm not sure they're right. Thanks. They all require finding the derivatives.

1. $y = \frac{e^{3x}}{1+e^x}$

[hide="Solution"]

$y' = \frac{(1+e^x)(3e^{3x}) - (e^{3x})(e^x)}{(1+e^x)^2} = \frac{e^{3x} (3+3e^x - e^x)}{(1+e^x)^2} = \frac{e^{3x} (3+2e^x)}{(1+e^x)^2} \blacksquare$
[/hide]

2. $y = \frac{e^x + e^{-x}}{e^x - e^{-x}}$

[hide="Solution"]

$y' = \frac{(e^x - e^{-x})(e^x - e^{-x}) - (e^x + e^{-x})(e^x + e^{-x})}{(e^x - e^{-x})^2}= \frac{4}{(e^x - e^{-x})^2} \blacksquare$ [/hide]

3. $y = \sec(e^{\tan x^2})$

[hide="Solution"]

$y' = \sec(e^{\tan x^2}) \cdot \tan(e^{\tan x^2}) \cdot e^{\tan x^2} \cdot \sec^2 (x^2) \cdot 2x \blacksquare$ [/hide]

[quote="pleurestique"]obviously you know how to take derivatives, so it's kinda funny that you're asking for help. your only mistake was in the second one, where that should be -4.[/quote]

Yes. $(A+B)^2-(A-B)^2=4AB.$

If you have a calculators that can build functions you can check your results. You do this by clicking the button called $dy/dx$ and then set a number for $x$. Then you can check and see if you get the same results when using the same $x$ and doing the calculations manually. But of course, if you haven't got such a calculator this post doesn't help you a lot.

PS. I believe this $dy/dx$ is found on newer calculators via:

$2nd \to  Trace \to  dy/dx$


Without sketching a diagram for a given compound, how would I know if it was ionic or covalent?  For instance, $ MgO$, $ SO_{2}$, and $ CCl_{4}$.

Also, what does the Lewis structure for the hydrogen peroxide molecule $ H_{2}O_{2}$ look like?

Thanks in advance!

[quote="feynman_wannabe"]1. Look at electronegativity differences: Mg and O have a large difference (roughly approximated by position on periodic table), thus magnesium oxide will prolly be ionic. S and O are really close, so its probably covalent, same with CCl4

2. H2O2 looks like HOOH.... hehe... too lazy to draw it in 2 dimensions...  if you can imagine what a water molecule looks like, just consider the OH to go in the spot of one H in H2O. Its like a zig zag[/quote]

1. What would you consider a large and small electronegativity difference?

Specifically, what would $ H_{2}S$ be?  Ionic or covalent?

2. What is the difference between the following two diagrams, assuming that the correct electron dots are shown?

[code]
    H
O - O
H
[/code]

[code]
H   H
O - O
[/code]

1. What would you consider a large and small electronegativity difference?

one easiest way to recognize ionic or covalent is that ionic bond involves metal and a nonmetal. covalent bond involves 2 nonmetals.
So, H2S would be covalent bond.

2. What is the difference between the following two diagrams, assuming that the correct electron dots are shown?

Code:

    H
O - O
H

Code:

H   H
O - O

i don't think there is any difference.
if there is a carbon chain and any substituent attached to it. if the substituents are on the same side, they are called cis; if the substituents are on the opposite side, they are called trans.
in this case, i don't think it matters.

Due to repulsion effects the $ \ce{H_{2}O_{2}}$ molecule prefers the trans kind of arrangement.

In general (and except hydrogen), compounds made up of elements from the s and p blocks are ionic; compounds made up of non-metals are covalent.

Would the following molecule be considered symmetric?

[img]http://www.chanceandchoice.com/ChanceandChoice/waterstructure.jpg[/img]

It looks like there is a vertical line of symmetry, but that's not what my book says...

Thanks!

$ \text{H}_2\text{O}$ has two planes of symmetry (the plane perpendicular to the line connecting the hydrogen atoms, and the plane of the atoms themselves).  Whether one calls such a molecule "symmetric" is probably a judgment call; I would probably consider a molecule like $ \text{CCl}_4$ more symmetric, even though it doesn't have a plane of symmetry.

[quote="elcric"][quote="i_thirst_for_knowledge"]H2O is asymmetric because of the two lone pairs of electrons in the central oxygen atom.[/quote]

My book also says asymmetric, but I still don't get it.  Could you please explain a bit more?[/quote]

In order for a molecule to be nonpolar, or symmetrical, it has to be symmetrical to both vertical line of symmetry and horizontal.  If you look at methane, it has no lone electron pairs, and has a symmetry at both vertical and horizontal.  Water at the other hand has a symmetry at only vertical, but not horizontal, and therefore it's considered polar.

one easiest way to explain why water is asymmetrical or a polar molecule is that:  if you draw a vertical line through oxygen:
                                           O
                                      H        H
because of the lone pairs electrons on the oxygen atom, it will be more negative on the oxygen end. so on the vertical line, you have an end which is more negative than the other end, so it's not equal and polar.[/img]

Also, you can draw the lewis structure in H20 and count the number of bonding and non-bonding electrons. In order for the molecule to be symmetrical, it must be one of the following molecular geometries:
 linear (2 bonding pairs, 0 nonbonding)
 trigonal planar (3 bonding pairs, 0 nonbonding)
 tetrahedral (4 bonding, 0 nonbonding)
 square planar (4 bonding, 2 nonbonding
 trigonal bypyramidal (5 bonding 0 nonbonding)
 and octahedral (6 bonding, 0 nonbonding)

If not, it is asymmetrical.


If a class (e.g. Algebra 3) uses a textbook, and you buy the textbook with the class, approximately when will the books arrive? Will they arrive when the class starts or when you enroll in the class?

Thanks!

When you enroll in the class.

EDIT: AIME - Well, yeah, that's understood. I hope.

They will [b]be sent[/b] when they are ordered, but will get there a bit later.  They're sent from South California.

the warehouse is in south carolina, but the headquarters are in California?

He said south [b]California[/b], not [i]Carolina[/i].


I havent seen this in many places but now, i can see the pattern.
whenever mc starts or is about to begin, alabnama forum starts to get posts, and alot of them.
but when mc ends... the forum just dies down and now, it is DEAD.

please people help me revive, we need to perform cpr on the alabama forum using the stayn alive beat

that's because there's not enough people...

[b]17 more days of school and exams...[/b]

yeah it's that way

well.... 3 of us are online and posting, so it can't be that dead...

omg :o  :o  :o

chestnut actually came on!!!
hey chestnut do u watch chinese tv?

I can post practice problems if you want, but no one asks for hints if they can't solve them. I think there is still an unanswered problem or 2 in the mc marathon thread.

actually there's more like 8 but because i'm so stupid i don't know any of them...


geometry
circumcircle
power of a point
radical axis
\/closed

Today, April 25, 2005, around noon, I posted the solution of a problem from Round 1 of the 5th Mathlinks Contest, as a reply in a thread in the Olympiad section, Geometry, Unsolved Problems. Before positing this solution, I checked that the original poster of the thread has been given a [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31524]permission[/url] to discuss the problems from Round 1 on the forum. I also checked that the thread [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34000]Circumcircle on Radical Axis[/url] in the has been unlocked. The entire thread with all replies, including my solution, has been deleted in five minutes flat - I cannot give the URL. Did I violate any contest rules or what was the reason ? Should I post the same solution elsewhere, would it meet with the same fate?

Yetti

You did not violate any rules. Were you by any change looking for [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=216159#p216159]this post[/url]? :)

Thanks for the reply.

The thread in question was marked as [Moved] for a while, before it completely disappeared, but when I tried to access it, a message "The topic or post you requested does not exist." popped up. The same thing happens (on more that one computer, all running Windows 98), when I click the above link in your reply. In addition, I tried to check the latest posts of the members, who posted in the missing thread, but the posts do not show up in their lists.  :?

Yetti

Oh ok, I know what's wrong. Problem fixed now :)


AMC USA(J)MO USAMO AIME geometry inequalities probability

I was wondering if, like the AIME, past USAMO problems were significantly easier than current ones. Since they were 5 problems in 3.5 hours also, it seems quite likely. But how about the more recent ones that were 3 questions, 3 hours for 2 days? How do they compare in difficulty with recent USAMO problems?

The 5 question tests were indeed much easier than the current ones, I get at least 4s on them in practice. The problems on the 3 hour/day USAMOs (1996-2001) are also a bit easier than the current ones, but they're close enough that my scores for them probably (I've only taken 1 from those years) have a lower mean.

-Adam Hesterberg

the past "easy" USAMO that Adam was talking about is similar in difficulty to some of the easier, current MO's, such as Canadian, British, or Irish.

THOSE past ones are REALLY REALLY easy, they are almost 13s and 14s on the AIME. you should try the past CMO's first. at least i am doing so.

Thanks, guys. Wow, this is a completely WA dominated thread. Yes!

[quote="DPopov"]Should you do past tests or read more on theory? I noticed that most of the time there are 2 geometry questions, 1 or 2 inequality, 1 number theory, and 2 combinatorics questions on the past USAMOs. Also I was wondering if analytical geometry ever comes up?[/quote]

Any question concerning "ever comes up" and the USAMO is difficult to answer, especially if the topic in question has to do with any math whatsoever. (I daresay the probability of a question pertaining to the laws of domestic cabbage sales in the United States is not likely to appear on the USAMO.) I don't think anyone will tell you that there's a 0 percent chance of meeting analytical geometry on the USAMO, but it isn't the most popular topic. Keep in mind though, that some problems not intended to be solved analytically may have an analytical solution, and no one will deduct points because you decided to take that approach.

At this point, work problems and read solutions. That's my advice at least. Be thankful that you even have the time to do so.


inequalities
trigonometry
inequalities proposed

Let $ a,\ b,\ c$ be the side lengths of a triangle. Prove the following inequality.

$ \sqrt{1\minus{}\left(\frac{a}{b\plus{}c}\right)^{2}}\plus{}\sqrt{1\minus{}\left(\frac{b}{c\plus{}a}\right)^{2}}\plus{}\sqrt{1\minus{}\left(\frac{c}{a\plus{}b}\right)^{2}}\leq 3\sqrt{2}\minus{}\frac{3}{2}$

$ \sum{\sqrt{1\minus{}(\frac{a}{b\plus{}c})^{2}}}\equal{}\sum{\frac{2\sqrt{p(p\minus{}a)}}{b\plus{}c}}\equal{}\sum{\frac{2\sqrt{bc}}{b\plus{}c}\cos(\frac{A}{2})}\leq\sum\cos\frac{A}{2}\leq3\cos(\frac{A\plus{}B\plus{}C}{6})\equal{}\frac{3\sqrt{3}}{2}$
$ \frac{3\sqrt{3}}{2}\leq3\sqrt{2}\minus{}\frac{3}{2}$ $ \leftrightarrow$ $ 3\leq4$, which is true.


Isosceles triangle ABC with AB=AC. K,L lies on AB,AC respectively, and CK intersect BL at O.Prove that if AK>AL ,then AK+KO<AL+LO.

:help:
I have been waiting for 3 months but with no replies :o
[b]Can anyone help me?[/b]


(1) Arrange $ A \equal{} \frac {\sqrt {5}}{3},\ B \equal{} \frac {\pi}{4},\ C \equal{} 0.8$ in decreasing order. Justify your answer.

(2) Arrange $ A \equal{} \frac {2}{3} \plus{} \log_8 7,\ B \equal{} 1.5,\ C \equal{} \log_2 3$ in decreasing order. Justify your answer.

[quote="kunny"](1) Arrange $ A \equal{} \frac {\sqrt {5}}{3},\ B \equal{} \frac {\pi}{4},\ C \equal{} 0.8$ in decreasing order. Justify your answer.

[/quote]

$ \sqrt5$ is roughly $ 2.2$, and $ 2.2$ over 3 is $ \frac{22}{30}$, which is $ 0.73$.  $ \pi$ is also roughly $ 3$, for $ b\equal{}.75$.  IN decreasing order, it would then be $ C, B, A$

[quote="kunny"]
(2) Arrange $ A \equal{} \frac {2}{3} \plus{} \log_8 7,\ B \equal{} 1.5,\ C \equal{} \log_2 3$ in decreasing order. Justify your answer.[/quote]

[hide="this might be wrong"]
$ \log_2 3$ is about $ 1.6$ but is a little bit less.
$ \log_8 7$ is about $ .94$, plus $ \frac {2}{3}$ is over $ 1.6$.
So the answer is $ A,C,B$.
[/hide]


real analysis
limit
integration
function
trigonometry
real analysis unsolved

Let $ f$ be Lebesgue integrable on $ [a,b]$. Then $ \lim_{n\to\infty}\int_{a}^{b}f(t)\cos(nt)dt \equal{} 0$

All proofs of Riemann-Lebesgue have this general structure.

Show that for all $ f$ in a particular set of "nice" functions, this limit is zero.

Observe that $ \left|\int_a^bf(x)\cos(nx)\,dx\right|\le \|f\|_1$ for all $ f.$

Show that the set of "nice" functions is dense in $ L^1.$

Put the pieces together: given $ \epsilon>0,$ write $ f\equal{}g\plus{}b$ where $ g$ is "nice" and $ \|b\|_1<\epsilon.$ Take the limsup of the sequence. The part that depends on $ g$ goes to zero, so the overall limsup is $ <\epsilon.$ That finishes it.

Where various proofs differ is in the choice of the set of "nice" functions. Any one of the following would work, and the list is not exhaustive:

1. Step functions.
2. $ C^1$.
3. $ L^2.$ For this, we need Bessel's Inequality.

Note that $ \cos$ can be replaced by any periodic function with average value zero. The first two versions of the argument are the same, while the L^2 version doesn't work without orthogonality.

For previous explanations, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=91119]here[/url] for convergence on nice functions and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=131611]here[/url] for density.

Thanks for reply.
For more information of it, study in this book that i have just seen it


[quote="Peter"]The sequence $\{y_{n}\}_{n \ge 1}$ is defined by \[y_{1}=y_{2}=1,\;\; y_{n+2}= (4k-5)y_{n+1}-y_{n}+4-2k.\] Determine all integers $k$ such that each term of this sequence is a perfect square.[/quote]

 Let us assume that $k\geq 4$ is an integer such that all the members of the sequence $y_{n}$ are squares. We have that $y_{3}=2k-2$ then $k=2t^{2}+1$ for some $t\geq 2$. From the recurrence we obtain that $y_{5}=256t^{6}-96t^{4}+8t^{2}+1$. On the other hand from $t\geq2$ follows that \[(16t^{3}-3t-1)^{2}<y_{5}=256t^{6}-96t^{4}+8t^{2}+1<(16t^{3}-3t)^{2}\]
therefore $y_{5}$ can't be a square $\Rightarrow k\leq 3$.

When $k=1$ then the sequence only consist of $0$s and $1$s.
The case $k=2$ is impossible because $y_{3}=2k-2$ should be a perfect square.
For $k=3$ we get that $y_{n+2}=7y_{n+1}-y_{n}-2$ and we will prove that for all $n\geq 1$ $y_{n}$ is a square.
Denote $a_{n}=y_{n}-\frac{2}{5}$ then $a_{n+2}=7a_{n+1}-a_{n}$ and $a_{1}=a_{2}=\frac{3}{5}$. Solving the equation ${\lambda}^{2}-7{\lambda}+1=0$ we find that \[{\lambda}_{1}=\frac{7+3\sqrt{5}}{2},{\lambda}_{2}=\frac{7-3\sqrt{5}}{2}.\]

Consequently, $a_{n}=c_{1}{{\lambda}_{1}}^{n}+c_{2}{{\lambda}_{2}}^{n}$ where $c_{1}$ and $c_{2}$ are constants. From $a_{1}=a_{2}=\frac{3}{5}$ we can show that $c_{1}=\frac{9-4\sqrt{5}}{5}=(\frac{\sqrt{5}-2}{\sqrt{5}})^{2}$ and $c_{2}=\frac{9+4\sqrt{5}}{5}=(\frac{\sqrt{5}+2}{\sqrt{5}})^{2}$. Thus, \[a_{n}=((\frac{\sqrt{5}-2}{\sqrt{5}})(\frac{3+\sqrt{5}}{2})^{n})^{2}+((\frac{\sqrt{5}+2}{\sqrt{5}})(\frac{3-\sqrt{5}}{2})^{n})^{2}\] and
\[y_{n}=((\frac{\sqrt{5}-2}{\sqrt{5}})(\frac{3+\sqrt{5}}{2})^{n}+(\frac{\sqrt{5}+2}{\sqrt{5}})(\frac{3-\sqrt{5}}{2})^{n})^{2}=x_{n}^{2}\]

Now it is easy to show that $x_{n+2}=3x_{n+1}-x_{n}$ for all $n\geq 1$ and $x_{1}=x_{2}=1$. Therefore $x_{n}\in{N}$ for all $n\geq{1}$ and $y_{n}=x_{n}^{2}$.

Answer: $k=1$ and $k=3$.


find all functions $f: \mathbb{R\to \mathbb{R}}$ such that

$f'(\frac{x+y}{2})=\frac{f(x)-f(y)}{x-y}$ holds for every $x\not=y$

[b](*)[/b] $f'(\frac{x+y}{2})=\frac{f(x)-f(y)}{x-y}$ for all $x\not=y$

First we can replace $f(x)$ with $f(x)-ax^{2}-bx-c$ for any $a,b,c\in\mathbb{R}$ and assume $0=f(0)=f'(0)=f'(1)$.
I will then show $f\equiv 0$, so that the solutions of (*) are $f(x)=ax^{2}+bx+c\in\mathbb{R}[x]$.

$y=-x\not =0$ in (*) gives $0=f'(0)=\frac{f(x)-f(-x)}{2x}$ and $f(x)=f(-x)$ for all $x\in\mathbb{R}$. (also for $x=0$)
Replacing $y$ with $-y$ in (*) then gives
$f'(\frac{x-y}{2})=\frac{f(x)-f(y)}{x+y}=\frac{x-y}{x+y}f'(\frac{x+y}{2})$ for all $y\not =-x$. (also for $x=y\not =0$)
And now $y=2-x\not =-x$ gives $f'(x-1)=(x-1)f'(1)=0$ for all $x\in\mathbb{R}$.
So $f'\equiv 0$ and $f$ is constant $f\equiv f(0)=0$.

but why does the solutions of $(*)$ are only on  the form
$ax^{2}+bx+c$.

Could you explain that? What's wrong with my proof?


Maximize $ \sqrt{49 \plus{} a^2 \minus{} 7\sqrt2a} \plus{} \sqrt{a^2 \plus{} b^2 \minus{} \sqrt2ab} \plus{} \sqrt{50 \plus{} b^2 \minus{} 10b}.$

There is no maximum as the expression can get infinitely large as $ a$ and $ b$ approach $ \infty$.

However, there is a minimum: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=218854[/url]


[quote="m bahador"]let ABCD be a square whit AB=1 and points M,N,P are mid points of AB,BC,AD respectively.let E be a mid point of MP.let Q be the intersection of the segments of BD & NE then BQ=?[/quote]
[hide="Solution"]
Let $ AC$ and $ BD$ meet at $ O$ and allow $ AC$ to intersect $ MP$ at $ E'$.  Notice that $ MP\parallel BD$ since $ MP$ is the midline of $ \triangle ABD$.  Hence, $ \triangle AOB\sim \triangle AE'M$ and $ \triangle AOD\sim \triangle AE'P$.  Thus, $ \frac {E'M}{BO} \equal{} \frac {AE'}{AO} \equal{} \frac {E'P}{OD}\implies 1 \equal{} \frac {OB}{DO} \equal{} \frac {E'M}{E'P}\implies E'P \equal{} E'M$.  So, $ E'$ is the midpoint of $ PM$ and therefore, $ E \equal{} E'$.  Now, notice that $ NP$, $ AC$, and $ BD$ are concurrent at $ O$ and since $ QO\parallel EP$, we have that $ \triangle NQO\sim \triangle NEP$.  $ O$ is the midpoint of $ NP$, so $ \frac {1}{2} \equal{} \frac {NO}{PN} \equal{} \frac {QO}{PE}$.  Also, $ \triangle AMP\sim \triangle ABD\implies \frac {1}{2} \equal{} \frac {MP}{BD} \equal{} 2\cdot \frac {PE}{BD}\implies PE \equal{} \frac {BD}{4}$.  Hence, $ \frac {1}{2} \equal{} \frac {QO}{PE} \equal{} \frac {QO}{\frac {BD}{4}}\implies QO \equal{} \frac {BD}{8}$.  Notice that $ BD \equal{} \sqrt {2}$ and thus $ BO \equal{} \frac {BD}{2} \equal{} \frac {\sqrt {2}}{2}$.  Hence, $ BQ \equal{}BO\minus{}QO\equal{} BO \minus{} \frac {BD}{8} \equal{} \frac {\sqrt {2}}{2} \minus{} \frac {\sqrt {2}}{8} \equal{} \boxed{\frac {3\sqrt {2}}{8}}$ [/hide]

Let $ O$ be the intersection point of diagonals $ AC$ and $ BD.$

Menelaos' theorem gives $ \frac {BN}{NC}\cdot \frac {CE}{EO}\cdot \frac {OQ}{QB} \equal{} 1\Longleftrightarrow \frac {1}{1}\cdot \frac {3}{1}\cdot \frac {OQ}{QB} \equal{} 1$, yielding $ OQ: QB \equal{} 1: 3. \therefore BQ \equal{} \frac {3}{4}BO \equal{} \frac {3}{4}\cdot \frac {1}{\sqrt {2}} \equal{} \boxed{\frac {3}{4\sqrt {2}}}$


SuperMath, How did you come up with that inequality? Did you try to solve it? and hint?

[quote="SuperMath"]Let $\large x;y;z > 0$. Prove that:
$\large \sum \frac{x(x+y+z)}{x^{2}+2 \sqrt{xyz(x+y+z)}} \geq\ 2$[/quote]
 let $x+y+z=3$
we have :
 $\large \sum \frac{3x}{x^{2}+2 \sqrt{3xyz}} \geq\ 2$


HMMT calculus Harvard college MIT geometry ARML

Alright folks. If you're going to HMMT, do say so. Also, let us know what team/school you're representing. :-)

I'll be there, barring unforseen events.

I will be there (TJ A). I think I am taking Algebra and Calculus tests.

TJ.  Grrrr.  I don't go to a cool school like you.  That's why I'm on Redundant Redundancy (I didn't come up with name).  Hopefully, I'll do well.  I'm doing calc and geometry.

Edit:  Which team does MithsApprentice represent?

Wah...My teacher does not want to go to HMMT, despite the fact that we went, did well, and had a great time 2 years ago.

I am going to HMMT!  :D oh yeah, I am like one of 96 people going to HMMT from AAST. Hopefully, we will do well....

Jeese, so many acronyms.  What does HMMT stand for?

Bah, 8 people are going from my school (Choate) and since they didn't have enough room for 10, another kid and I got kicked from the team. (I think it's because we're freshmen :D )

I will be going as part of one of the West Windsor-Plainsboro teams from NJ. (I forget which one.) I will be taking algebra and combinatorics. Hopefully, I'll get to meet some of you there, as I do not know many of you in person.  :lol:

I'll be going to HMMT with Exeter.  We should all try to meet up at some point.

Miths, how are you on the Albany Area MC if you live in Texas?

Well, I happen to know one of Albany's coaches - Mary O'Keeffe.

So I got a spot on their team. I'm not actually in their MC though.

[quote="MithsApprentice"]Well, I happen to know one of Albany's coaches - Mary O'Keeffe.[/quote]

Now I understand why you are so good at math.   :)   Say hello to your coach for me--I'm pretty sure she knows the match between "tokenadult" and my real-world name.

We are [b]not[/b] going to HMMT this year, and probably not next year either.  :(   Maybe someday.

Im going to HMMT. I managed to create Florida's first team and we raised enough funds to afford the costs (consisting largely of the airfare).

[quote="mathfanatic"]...right, but you still have to travel from Texas to Pennsylvania, which is usually the biggest impediment.  Are you flying?[/quote]

Yea...flying would be the only option, no? Lol. I would certainly not be driving there.

And I was under the impression that this was being held in Massachusetts. :P

Oh, haha.  I was thinking of ARML for some reason...

I'm going... our team is called "Evil Geniuses for a Better Tomorrow" and we're coming in 5th place overall!
And I'm not even kidding.

Are we planning on having another AoPSer meeting?  It'd be interesting to meet some of you.

If so, I can bring a sign that says AoPS or something similar.

I would be interested in a meeting.  I'm rather obscure on AoPS, though.

We're allowed to bring computers for Guts, right?

Apparently not...I regret not having been able to meet many people, but I did end up running in to Joel (JBL), who was handing out clipboards before the Guts round.  (It was fairly accidental; I didn't know who he was, but I happened to be wearing my nametag from the "weakest link" mini-event, so he recognized my name.)

But I guess a lot of AoPSers were there.  Mildorf and JGeneson definitely went up on stage, and I saw Cinjon, Yannik, and the florida team collecting frisbees.

I had actually seen you somewhat earlier and suspected it might be you from your sweatshirt, so I was about to start talking to you when I happened to notice your name tag.  Anyhow, good to meet you!


Prove that the average of $n\sin n$ for $n=2,4,6...180$ is $\cot 1$.

Okay...here goes...
I'm not using degree sign, too lazy, just assume in degrees
Simplify $\cot{1}$ to $\frac{1}{\tan{1}}$, and then turn into all sines and cosines, $\frac{\cos{1}}{\sin{1}}$
Now simplify the thing itself.
$\begin{displaymath}\frac{\sum\limits_{k=2}^{180}{k\sin{k}}}{90}\end{displaymath}$

$=\begin{displaymath}\frac{\sum\limits_{k=1}^{90}{k\sin{2k}}}{45}\end{displaymath}$

Now multiplying through, by $\sin{1}$, (note that this means that we must prove that this is equal to $\cos{1}$)

$=\begin{displaymath}\frac{\sum\limits_{k=1}^{90}{k\sin{2k}\sin{1}}}{45}\end{displaymath}$
Using the fact that $\sin{1}\sin{k}=\frac{\cos{k-1}-\cos{k+1}}{2}$ thing
$=\begin{displaymath}\frac{\cos{1}-\cos{3}+2\cos{3}-2\cos{5}+3\cos{5}-3\cos{7}...+90\cos{179}-90\cos{181}}{90}\end{displaymath}$
$=\frac{\cos{1}+\cos{3}+...+\cos{179}-90\cos{181}}{90}$
Then because $\cos{x}=-\cos{180-x}$, we're just left with
$=-\cos{181}$
$=\cos{1}$
which is what we were trying to prove.  QED, i guess...
PS: i tried doing it without multiplying in sin 1: that didn't get me much, because cot is hard to work with...then i tried multiplying it in later, didn't work...so i got this.

[quote="tetrahedr0n"]You could just do it using complex numbers. Then it becomes almost trivial, just a bunch of algebra that you have to be careful in manipulating.[/quote]
How would you do that?

[quote="tetrahedr0n"]Write

$\sin n^\circ = (e^{i\pi n/180} - e^{-e\pi n/180})/(2i)$.

You'll have to evaluate two sums, both in the form

$S=x+2x^2  + 3x^3 + ... + nx^n$.

To derive a formula for this expression, you evaluate $S(x-1)$.

The rest is up to you.[/quote]

That was my technique too!  But then I gave up, because I didn't want to do the algebra.  I figured it would all work itself out in the end.


superior algebra
superior algebra unsolved

can a finite group have exactly 2 elements of order 2?generalise. :D

That doesn't work. The elements of the quaternion group which have order [i]at most[/i] two are $\pm 1$. $1$, however, does not have order two.

The answer is "no". The number of elements of order dividing $p$ in a finite group is a multiple of $p$  when $p$ is a prime (I think jmerry even proved a similar assertion for any positive integer $n>1$, not just for primes).

The quaternion group has only one of order two, you possibly counted the solutions of $x^{2}=1$, but that is something else.

It was proven before that the number of solutions of $x^{n}=e$ is divisible by $n$ if $n$ divides the number of elements in the group. If $n$ is prime it gives that there are $\equiv-1 \mod n$ solutions of $x^{n}=e, x \neq e$. But this solutions are exactly the elements of oder $n$. If $n>1$ doesn't divide the group's order, there is no element of order $n$ by Lagrange's theorem.

Thus we could generalize to:
Let $G$ be a finite group, $p$ any prime. Then the number of elements of order $p$ in $G$ is $0$ or $\equiv-1 \mod p$; especially, it is never a multiple of $p$ greater than $0$.

And what about infinite groups? $\langle a,b | a^{2}= b^{2}= 1 \rangle$?

no, $(aba)^{2}= 1$

can we find all n for which ans. is positive?

[quote="grobber"]I think jmerry even proved a similar assertion for any positive integer n>1, not just for primes).[/quote]
A mistake was found. The fixed version  works for squarefree $n$, but not general $n$.

I don't actually know whether it's true in general now.


inequalities
inequalities solved
3-variable inequality
Schur

Let $a, b, c$ be positive real numbers with $abc = 1$. Prove that
\[\sum_{cyc} (a+ bc) \leq 3 + \frac{a}c+\frac{b}a+\frac{c}b\]

Let's put $a=\frac{x}y, b=\frac{y}z, c=\frac{z}x$, and the inequality becomes:
\[\sum_{cyc} \left(\frac{x}y+\frac{y}x\right) \leq 3+ \sum_{cyc} \frac{x^2}{yz}\]
\[ \iff \sum_{cyc} (x^2y+xy^2) \leq 3xyz+ \sum_{cyc} x^3\]
which is exactly Schur's inequality.

Of course ! I forgot to apply the trick a = x/y, b = y/z, c = z/x.

So IMO2000/2 and Schur are related ! (I hope you get the point)
I know the elegant ways to prove IMO2000/2, but I wanted to try the longer way also :D

shur is related to any inequality ... my two laws for inequalities are:

1) 90% of the inequalities are solved using Cauchy
2) the other 10% can be solved using Schur :D

But the proof of IMO 2000/2 is actually a proof of Schur's inequality !
So IMO 2000/2 is not only related to Schur, it s (a sort of) Schur :)
Since : (a - 1 + 1/b)(b - 1 + 1/c)(c - 1 + 1/a) \leq 1
<=> a + b + c + ab + bc + ca \leq ab + bc + ca + 3
<=> a + b + c + 3abc \geq ab(a + b) + bc(b + c) + ca(c + a)
So IMO2000/2 and Schur are equivalent !


function
number theory
least common multiple
number theory unsolved

Is it possible (how?!) to characterize all integers n such that $\phi(n)|n$? (where $\phi$ is Euler's function)

??!!

bye,Uber

I believe the condition is only true when $n=2^k$ for some non negative integer $k$...This can be shown from that fact that $n$ and $\phi{n}$ are multiplicative properties.

No there are other solutions, e.g. $6$.
It already appeared: http://www.mathlinks.ro/Forum/viewtopic.php?t=48636

All solutions given by: $1,2^k,2^k3^m$, were k and m natural.
Interesting question $c(n)|n?$, were c(n) is minimal number with property $n|a^{c(n)}-1$ for all (a,n)=1.
Function  c(n) for $n=p_1^{m_1}p_2^{m_2}\dots p_k^{m_k}$ calculate so divisor minimal number N, with
$p_i^{m_i-1}(p_i-1)|N$.

Your given $c(n)$ is not totally perfectly computed:
$c(p^n)=p^{n-1}(p-1)$ for odd primes $p$
$c(2)=1$
$c(4)=2$
$c(2^n)=2^{n-2}$ for all $n\geq 3$.
And additionally $c(ab) = \mathrm{lcm}(c(a) , c(b))$ for all $a,b$.

It's clear that $c(n) | \phi(n)$ so all earlier solutions work.
The question is: what are examples of new ones¿

Well, when $2^n+1$ is prime, then $c(2^{n} ( 2^n + 1)^k) = 2^n (2^n+1)^{k-1}$, so this gives us a solution.
So all numbers $n$ whose regular $n$-gon is constructible, or more generally that have only Fermat primes as factors (not necessarily only once), are after multiplication with a big enough power of two, solutions.
It's possible that these are already all solutions (haven't thought about it), but already those are very hard to classify (note that till now it's unknown how many Fermat primes $2^n+1$ there are).

c(n) better than \phi(n), because it is minimal number m satisfyed property:
if  (a,n)=1, then $n|(a^m-1)$.
Let $(Z/nZ)^*=Z_{d_1}+Z_{d_2}+\dots+Z_{d_k}$, were $d_1|d_2|\dots |d_k$.
 Then $\phi(n)=d_1*d_2*\dots*d_k,c(n)=d_k$.
There are more solutions
(1) c(n)|n.
If $n=p_1^{m_1}*\dots *p_k^{m_k}$ is solution (1) then $n_1=p_1^{r_1}*\dots *p_k^{r_k}$ is solution (1) if $r_i\ge m_i$.
Let p is prime. Minimal solution n, with p|n construct $n_1=p,n_2=p(p-1), ..n_{k+1}=s_k*n_k,...$.
$s_k$ is minimal number with property $(r_k-1)|n_{k+1}$, r_k -maximal prime divisor $n_k$, suth that $(r_k-1)\not|n_k$.
This proces stopped and give minimal n with property $p|n,c(n)|n$.

Isn't possible to determinate $d_1,...d_k$ "a priori"? Is it an unsolved problem of Algebra?

You say: $d_k=lcd(r_1,...,r_k)$, where $r_i=(p_i-1)p_i^{m_i-1}$. Are you sure? If we consider $n=42=2*3*7$, then  $Z_{42}^*=Z_2 \oplus Z_6$; thus $d_k$=6, but $lcd(r_1,r_2,r_3)=1$. maybe would you write $d_k=lcm(r_1,..r_k)$? Moreover, I don't understand how you define $d_i,i<k$; maybe $d_i=lcm(r_1,..,r_i)$?

I thank you!
Bye, Uber

ok Rust; Thank you. Another question:

Let p be an odd prime and n a natural such that $\gcd(p,n)=1$
I denote with $o_p(n)$ the order of p in $Z_n^*$.

There exist a way to know $o_n(p)$ "a priori"?

Oh Rust, I'm sorry, but I mistook to write! I wrote $o_p(n)$. I would write $o_n(p)$!. Thus my question become more difficult...  because I want to know $o_n(p)$ a priori, in terms of p and n. Is it possible?


$ 3abc \ge 4(ab \plus{} bc \plus{} ca) \minus{} 9$, for $ a \plus{} b \plus{} c \equal{} 3$ and $ a, b, c > 0$

Yes, this is a consequence of Schur.
$ a^3\plus{}b^3\plus{}c^3\plus{}3abc \ge a^{2}b\plus{}ab^2\plus{}a^{2}c\plus{}ac^2\plus{}b^{2}c\plus{}bc^2 \Longleftrightarrow 6abc\plus{}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}ac\minus{}bc) \ge ab(a\plus{}b)\plus{}bc(b\plus{}c)\plus{}ac(a\plus{}c)$.
Changing $ a\plus{}b\plus{}c$ by $ 3$, we'll have $ 3abc \ge 2(ab\plus{}ac\plus{}bc)\minus{}a^2\minus{}b^2\minus{}c^2 \equal{} 4(ab\plus{}ac\plus{}bc)\minus{}(a\plus{}b\plus{}c)^2 \equal{} 4(ab\plus{}ac\plus{}bc)\minus{}9$.

By Schur,
$ ab\plus{}bc\plus{}ca \leq \frac{(a\plus{}b\plus{}c)^3\plus{}9abc}{4(a\plus{}b\plus{}c)}\equal{}\frac{27\plus{}9abc}{12}\equal{}\frac{9\plus{}3abc}{4}$
$ \Leftrightarrow 3abc\geq 4(ab\plus{}bc\plus{}ca)\minus{}9$


My Geometry teacher is all about thinking outside the box. So, for our first assignment, she asked us this, and told us to give her something by Monday: Prove you are who you say you are. Most of us in the class probably thought, well, birth certificate, but, i thought, well, that's too obviously, and it's not "thinking outside the box". So, my next thought was, well, like in movies, if someone were like pretending to be you, a close family member or friend would ask a personal question that only you would know, so, i guess i could say that. But, it has nothing to do with math at all. Im sure there is some way, but i dont know it yet. If anyone knows, i would GREATLY appreciate it. I hope that someone checks this soon, only like 14 hours, lol. Well, please reply if you knwo anything. Thanks Y'all!

This is something you should think about yourself...

maybe you should create a math test and you determine who each person is by their score. break ties with more tests

have you people ever played "guess who" when you were little?

Use conics to draw things that represent you. :P  I'm not very creative.  :oops:

The guess who thing is a good option...This is more problem solving with logic rather than abstract mathematics, so I may be a tad late but try this,

Assume a person says he/she is NOT who he/she is, ask him/her to prove he/her isn't the person in question, if he cannot prove it without reasonable doubt, he/she is the person he/she proclaimed they were not.

QED

(It probably won't work but if judged with the greatest scruitiny this proof by contradiction might just do it!) :D


Prove that there are no prime numbers in the infinite sequence of integers

10001, 100010001, 1000100010001, ...

The factorization of 10001 is 73*137, so I don't think so...

O man, i misread the question, thought it was 1001.

Well, the divisibility by 3 and (I think) the divisibility by 7 are still valid.  As long as the divisibility by 7 for every 3rd term starting with the second is still valid, then all we have left to prove is that every third term starting with the first is not prime.

So...anyone figure it out yet?

Cause I haven't.

[quote]
Will expressing the numbers as products/sums help?
[/quote]

I tried that, but my expression for the kth term wasn't exactly...elegant.

Is the fifth one divisible by 73?

Also, does this problem require changing bases at all?  I tried some stuff pretending those were base 2 numbers and then converting to base 10, and some more stuff with converting them as-is to base 2, but I didn't really have any clue what I was doing any farther than knowing how to convert between bases... So, I don't really have any ideas on how to finish solving this problem.

hmm, i don't think listing the prime factors of the numbers help. I tried that, and didn't seem to come up with a concrete pattern,

As for changing bases, is it possible to convert not from base 10 to base 2, but from base 2 to base 10? LIke instead of the numbers in the question being in base 10, can we prove that in base 2, those numbres are never prime, which infers in base 10, they are never prime?

[quote]As for changing bases, is it possible to convert not from base 10 to base 2, but from base 2 to base 10? LIke instead of the numbers in the question being in base 10, can we prove that in base 2, those numbres are never prime, which infers in base 10, they are never prime?[/quote]

100010001 is base 2 is 256+16+1=273, which is clearly divisible by 3.

yea, but, if we prove that 10001 in base 2 is not prime, then does it follow that 10001 in base 10 is not prime?

Is there such a theorem??? conjecture???? axiom????? postulate?????? educated guess???????

O darn.

Ok, back to the drawing boards....

Hmm.. nobody replying much to this one anymore :)
There is a nice formula for a general term. See if you can work out what it is.. then some things factor..

whoops I forgot about this. (10^4n-1)/(10^4-1) should be the general formula. diff. squares a couple of times gives (10^n-1)(10^n+1)(10^2n+1)/9*11*101.

Well, that's not prime. :)

Pretty much :) You might wanna explain that a bit more  :wink:

What can you say about the three factors on the top. Can say, two of them be divisible by 3?

mod 9, we replace all the 10's w/ ones, so we get (0)(2)(0) which is divisible by 3, to say the least..

Naw, I don't want to explain, i'm too lazy.

(0)(2)(2) actually. What about in general? Can any of those top factors ever share a factor?


[color=darkblue]Prove that the natural number $9+499\cdot 501\cdot 502\cdot 504$ is a perfect square.

[hide="See at this place in the last analysis."]
[b][u]Generalization.[/u][/b]

$x\in\mathbb R\Longrightarrow 9+(x-1)(x+1)(x+2)(x+4)=9+\left(x^{2}+3x+2\right)\left(x^{2}+3x-4\right)=$

$9+\left[\left(x^{2}+3x-1\right)+3\right]\cdot\left[\left(x^{2}+3x-1\right)-3\right]=\left(x^{2}+3x-1\right)^{2}\ .$

For $x: =500$ obtain the proposed problem.[/hide][/color]

Thank you for your help, Mr. Nicula! :)

Teacher Asha

[color=darkblue]It is my pleasure. [hide="Only for Asha."]Good night, my dear mate ![/hide][/color]


algebra
polynomial
functional equation
roots
IMO Shortlist

Let $ f(x)$ be a monic polynomial of degree $ 1991$ with integer coefficients. Define $ g(x) \equal{} f^2(x) \minus{} 9.$ Show that the number of distinct integer solutions of $ g(x) \equal{} 0$ cannot exceed $ 1995.$

if  $ f^2(x)$   express  $ (f(x))^2$,
then the number of distinct integer solutions of  g(x)=0 cannot exceed 1991.
\[ f(x) \minus{} 3 \equal{} (x \minus{} x_1)^{r_1}(x \minus{} x_2)^{r_2} \cdots (x \minus{} x_n)^{r_n}R_1(x) \\
f(x) \plus{} 3 \equal{} (x \minus{} y_1)^{s_1}(x \minus{} y_2)^{s_2} \cdots (x \minus{} y_n)^{s_n}R_2(x)\]
so
\[ (x_i \minus{} y_1)^{s_1}(x_i \minus{} y_2)^{s_2} \cdots (x_i \minus{} y_n)^{s_n}R_2(x_i) \equal{} 6\]
factors on left side must be ±1,  ±2,  ±3, ±6 , but each one shows only once.
if n>0, m >0, then $ m \le 4, n \le 4, m \plus{} n \le 8$
if n*m=0 then $ n \plus{} m \le 1991$


Those are images.  Not sure how to copy paste them.  We'll have to take a look.

Have you tried a different word processor?  Works fine for me with Word.


[quote="CHAVS"]Please can anyone provide me with a link to Chen's Theorem (1960's?)with proof connected to the golbach conjecture.

Thanks[/quote]
[color=blue]
[b]Chen's theorem[/b]
[i]Any sufficiently large even number can be written in form $p+q_{1}\cdot q_{2}$, where $p$ is prime and $q_{1},q_{2}$ are primes or ones.
[/i][/color]
Proof:
J. R. Chen, On the representation of a large even integer as the sum of a prime and the product of at most two primes, Kexue Tongbao 17 (1966), 385-386
or
J. R. Chen, On the representation of a larger even integer as the sum of a prime and the product of at most two primes, Sci. Sinica 16 (1973), 157-176.

I hope it will help.  :)

[quote="Rzeszut"]
[color=blue]
[b]Chen's theorem[/b]
[i]Any sufficiently large even number can be written in form $p+q_{1}\cdot q_{2}$, where $p$ is prime and $q_{1},q_{2}$ are primes or ones.
[/i][/color]
Proof:
J. R. Chen, On the representation of a large even integer as the sum of a prime and the product of at most two primes, Kexue Tongbao 17 (1966), 385-386
or
J. R. Chen, On the representation of a larger even integer as the sum of a prime and the product of at most two primes, Sci. Sinica 16 (1973), 157-176.

I hope it will help.  :)[/quote]

No use whatsoever, useless . . .

Still open.


geometry
inequalities
triangle inequality
AMC

The sides of a triangle with positive area have lengths $ 4$, $ 6$, and $ x$. The sides of a second triangle with positive area have lengths $ 4$, $ 6$, and $ y$. What is the smallest positive number that is [b]not[/b] a possible value of $ |x \minus{} y|$?

$ \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 10$

[hide]x or y can be anywhere from 3-9. From this we see that |x-y| can be 2, 4, and 6, but not 8. [/hide]

[hide]By Triangle Inequality, the length of the third side must be > 2  but <10. This makes the smallest non-possible value of x-y 8, answer D.[/hide]

[hide="solution"]
triangle inequality gives us $2<x<10$ and $2<y<10$ so the max value we can get from $\abs{x-y}$ is $8$

therefore the answer is $\boxed{D}$
[/hide]

[quote=GameMaster402]x=9.5, y=2.5. 9.5-2.5=7, it works.[/quote]

Thanks

[hide=Solution]Note that we have $2<x<10$ and $2<y<10$ from the Triangle Inequality. Then clearly the minimum impossible value is $10-2 \implies 8 \implies \boxed{\text{(D)}}$.[/hide]

[quote=Critical]Why can't x or y be 2? Triangle inequality is
$$z \le x+y$$
so $4+2=6$.[/quote]

It’s actually strictly x+y>z

@2Bove: Remember equality holds when the three points are collinear, which means the triangle is degenerate. The problem states that this triangle has positive area.
@Bove: No, in general, it is not strictly $x+y>z;$ rather, $x+y>z$ in this specific case just because the problem said so.

[hide='Solution'] Well we know that by triangle inequality that to find 3rd side for triangle with x is in range of $6-4<x<6+4$ and y is in range $6-4<y<6+4$ so x is $2<x<10$ and $2<y<10$. The smallest value that works is $3$ for $x$ and maximum that works for $y$ is $9$ and even if it were vice versa we see max absolute difference is $6$ meaning both $8$ and $10$ work, but key word is smallest value out of the options where $8$ is smaller than $10$. Therefore the answer is $D$.[/hide]

[quote=Critical]Why can't x or y be 2? Triangle inequality is
$$z \le x+y$$
so $4+2=6$.[/quote]

Note that the triangles have positive area. If $x$ or $y$ was $2$, the triangles would be degenerate, and they would not have positive area (the triangles would be lines)

@peace it's $x+y>z$ for nondegenerate and $x+y \geq z$ for degenerate, honestly I use the first for most questions as they don't really want degenerate triangles often but yeah you're technically right.

[quote=OlympusHero][quote=Critical]Huh. So that means $x+y>z$ doesn't make traditional triangles, rather $x+y\ge z$? Ok![/quote]

Yeah basically, if you had $x+y=z$ that would be a degenerate triangle aka line.

[quote=peace09]@Bove Yes; maybe I should've re - phrased what I said. People usually refer to the strict inequality when dealing with triangles, but the nonstrict inequality when referring to points in general on the plane. Also, please learn subjunctive before making posts :) JK[/quote]

Wait subjunctive means tone of post did you think I was talking in a mean tone :( Sorry :([/quote]

No... It's "If $x$ or $y$ were 2" not "If $x$ or $y$ was 2". Even if we're only talking about $x,$ it will still be were :)

[quote=GameMaster402]x=9.5, y=2.5. 9.5-2.5=7, it works.[/quote]

[quote=Darrenthegoat]Sorry for reviving an old thread but wouldn't the smallest |x-y| could be, be 7?[/quote]

No, they don't work due to the fact that the questions say they must be integers.


quadratics
number theory unsolved
number theory

find all natural numbers n such that there exists a natural number m such that \[ 2^n-1 \mid  m^2  +9. \]

IMO 1998 Shortlist.

Prove that n must be a power of 2, and use the Chinese remainder theorem.

supose p divides 2^n-1, then if p is different from 3, p must be congruent with 1 modulo 4 since -1 is a square mod p. conversely, given any p=1(mod4) one can find an m, such that p divides m^2+9. Now supose n is 2^ka with a odd a>1.
2^a-1 is =-1 mod4 so it has at least one prime divisor =-1(mod4) but it cannot be 3 because a is odd, so by the first observation this prime doesnt divides  m^2+9.
So n must be a power of 2, now n=2^k, if k=1 the result is oviously true chosing m=3k. so supose k>1. since 2^(2^k)=1(mod p) then the order of 2 is a power of 2, if it is 1 or 2 then p=3 and for this prime we can find a m. if the order is 2^t with t>1 then p=1 mod 4 and for this prime we can find an m, and of course by hensel lemma also for powers of this primes. also 9 doesnt divides 2^(2^k)-1. so for every power of prime dividing it, we can find a particular m that satisfies p^n divides m^2+9, and using chinese remainder theorem one can find an M such that
2^(2^k)-1divides M^2+9

[quote="hossein11652"]I solved it using quadratic residues :)[/quote]
Can you post your solution?  :maybe:

Just see the recent repost of this problem.


[hide="Read This Before Going Any Further"]
The reason I am asking this question is so that I can present my solution to a contest currently underway

Since this has nothing to do with the problem and only the ability to reproduce a diagram, I feel I am entitled to ask this question, if anyone feels otherwise then please don't post.
[/hide]

So I need to know how to draw circles that are externally tangent on Geogebra, I see the tangent tool, but it only lets me draw tangent lines.

[geogebra]90f141a4a22f5f22b34d73d12db62510aec6959c[/geogebra]

Create two perpendicular lines and create circles with centers on one of the lines that pass through the intersection point.

Thanks, I took your advice and added on to it with my own fixes for my problem.

Anyways, I come to you with a minute question, how do you mark right angles?

I don't mark right angles when i draw diagrams in geogebra. I don't think i know of a way.

Use the angle option; it automatically marks right angles using a square instead of an arc.

Although you can use GeoGebra, I find it easier to solve the problem first, then draw it with Asymptote, which will be easy since I have the coordinates and the radii.


Prove that $c(c+2a) + d(d-2b)$ and $(c+a)(c-a)+(d+b)(d-b)$ cannot both be 0 if a, b, c, and d are rational non-zero numbers.

edit: ii) prove that all real a, b, c, and d such that both expressions are zero satisfy $ab=cd+(d-b)(a+c)$

uhh some progress?

[hide]

we assume that both equal $0$ and try to reach a contradiction, so

 $c(c+2a) + d(d-2b)= 0$ and $(c+a)(c-a) + (d+b)(d-b) = 0$

the latter gives $c^2 + d^2 = a^2 + b^2$

the former gives $c^2 + 2ac + d^2 - 2bd = 0$

adding $a^2 + b^2$ to the left and $c^2 + d^2$ to the right, we get:

$(c+a)^2 + (d-b)^2 = c^2 + d^2$

now, if $(c+a)^2 = c^2$, then $a=0$, but the problem states $a$ is nonzero so we wlog $d > c$, and we have:

$d-b = c$ and $c+a = d$ which implies $a=b$

also, substituting $c+a$ for $d$ into the second equation gives:

$(c+a)(c-a) + c(c+2a) = 0$ and substituting into the first equation gives you the same thing.... from there im pretty much stuck though

[/hide]

[quote="vutang"]I think it is true if a, b,c, d are real numbers[/quote]

its not

[quote="gotztahbeazn"]
now, if $(c+a)^2 = c^2$, then $a=0$...
[/quote]
not quite; a can be -2c
on the right track though

[hide="Solution"]Suppose the contrary.  I'd like to change variables, not because it matters much but just for greater symmetry.  So let $x = c + \frac a2$ and let $y = d - \frac b 2$.  Then the two given equations are $(x - \frac a 2)(x + \frac{3a}2) + (y + \frac b 2)(y - \frac{3b} 2) = 0$ and $(x + \frac a 2)(x - \frac{3a}2) + (y - \frac b 2)(y + \frac{3b} 2) = 0$.  Expanding these and taking the difference of the two equations, we get $ax = by$.  Since $b \neq 0$, we can substitute in $\frac{ax}b$ for $y$ into the original equations.  Then they become $x^2 - \frac{3a^2}4 + \frac{a^2x^2}{b^2} -\frac{3b^2}4 = 0$.  Re-arranging a little, $\frac{(a^2+b^2)x^2}{b^2} = \frac{3(a^2+b^2)}4$, so $\frac{x^2}{b^2} = \frac 34$.  However, this equation cannot have rational solution, so our initial assumption was wrong.[/hide]


vector
linear algebra
linear algebra unsolved

What are those matrices $M \in M_{n}(R)$ that are orthogonal and doubly stochastic ?

These have to be the permutation matrices and no others.

Stochastic includes that the entries are all nonnegative. The only way two vectors with nonnegative entries can be orthogonal would be for the indices of the positive elements to be disjoint. (Else the inner product would be positive.) The only way you can have $n$ pairwise disjoint nonempty subsets of $n$ indices is for each set to have exactly one element. Because the column sums are 1, each of those nonzero elements must be 1.


1) In triangle $ABC$, $D$ divides $AB$ in the ratio 1:2 and $E$ divides $BC$ in the ratio 3:4. If $BDE =6$, find the area of $ABC$

2) $ABCD$ is a paralleogram. A stright line through $A$ cuts $BD$ at $E$ and $BC$ at $F$, and meets $DC$ extended at $G$. prove $AE/EF=AG/AF$

What exactly is $BDE$? and you have to specify exactly which segments are in the ratios, for instance is $AD: DB=1: 2$ or $AD: DB=2: 1$.

Please don't hide your questions  :D .

To Inversonfan: I think $BDE=6$ means the area of triangle $BDE$

Yeah Iversonfan, usually BDE denotes a triangle, and area is the most common single value that can be attributed to a triangle.  The ratios I inferred are just in the order he said, ie. $AD: DB = 1: 2$ and $BE: EC=3: 4$

[hide="answer for 1"]
Drawing the diagram, making the ratios $x$ and $2x$ for AB and $3y$ and $4y$ for BC, and the area we are given gives us $3xy\sin B =6\rightarrow xy \sin B = 2$.  Then the area of the whole triangle is $\frac{21}2 xy \sin B = 21$
[/hide]

$\frac{AE}{EF} = \frac{AD}{BF}$

$\frac{AF}{FG} = \frac{BF}{FC}$ $\Rightarrow$ $BF = \frac{FC \cdot AF}{FG}$

$\frac{AD}{FC} = \frac{AG}{FG}$ $\Rightarrow$ $AD = \frac{AG \cdot FC}{FG}$

$\frac{AE}{EF} = \frac{AG \cdot FC}{FG} \cdot \frac{FG}{FC \cdot AF} = \frac{AG}{AF}$

I have tried to do an attachment but zir extension is not allowed

Allright then after figuring out what's what with the ratios, you can use the formula:
$A=\frac{1}{2}absin(c)$, from there, you can say $6=\frac{1}{2}2x3ysinx$ and area of $ABC=\frac{1}{2}7y3xsin(x)$, and by substitution (and since they share same angle, you see that $ABC=21$


A friend told me this simple, fun problem. See if you can solve it :)!

It was nighttime when four adventurers came to a bridge. One was a champion sprinter, one was an athletic person, one was kind of slow, and one was crippled. They only had one lantern among them, and figured it wouldn't be safe to cross the bridge without it (However, they can leave their comrades on either sides of the bridge without the lantern safely.) They also figured that it would be dangerous if more than two people were on the bridge at the same time. They must try to get all their people across with the lantern, safely. The sprinter can go across in 1 min, the athletic guy takes 2 mins, the slow person takes 5 mins, and the crippled person takes 10 mins. When two people go across together, their speed is the speed of the slower person, because the faster person has to slow down for the slower person (because they both have to have the lantern).

The athletic person decided that he was going to go back and forth carrying the lantern, because he was the fastest. So, they did some mental calculations and found that it would take 2 mins for him to get across with the athlete, 1 min to come back with the lantern, 5 mins for him to take the slow guy, 1 min to come back, and 10 mins to finally take the crippled person over, taking 19 minutes in total. But the crippled person, the genius of the group, said that there's one way to get it under 19 minutes. What is it?

No cheating! :)

Here's where you can find a slightly modified version of this.
[b]DON'T LOOK IF YOU HAVEN'T SOLVED IT, IT HAS THE ANSWER.[/b]
[hide="Online version"]http://www.magicofmath.org/figure_bridge.html[/hide]

send first and second, first comes back, third and 4th go, 2nd comes back, first and second go.

hehehehehe thats an old problem


function
calculus
derivative
LaTeX
complex analysis
real analysis
real analysis unsolved

Let $ F: [0,1]\to\mathbb{R}$ be a continuous function with the properties that there exists $ N\subset [0,1]$ such that $ \lambda(N) = 0$ and $ \lambda(F(N)) = 0$, $ F$ is derivable on $ [0,1]\N$ and $ F'\in L^1([0,1])$. Then the function $ F$ is absolutely continuous.

[edit] $ F$ is derivable on $ [0,1]-N$
thank you guys for pointing the error...

Makes no sense. Something is wrong with the condition

That $ N$ does look suspicious. The result holds for all differentibale functions with integrable derivative (it's a theorem in Rudin's "Real and Complex Analysis", in the chapter on differentiation). So the statement is correct, you just have to leave out all that stuff about $ N$. It is, however, crucial that $ F$ be differentiable everywhere. Maybe Kuba meant to say that $ F$ is differentiable only outside $ N$? In that case, it might still hold - I don't know.

I think the $ N$ part is just trying to say that $ F$ maps sets of measure 0 to sets of measure 0. Maybe it should read: For any $ N\subset [0,1]$ such that $ \lambda (N)\equal{}0$ it is the case that $ \lambda (F(N))\equal{}0$.

If this is the what Kuba meant, then it is simply a consequence of absolute continuity and doesn't need to be a hypothesis.

ohhh... very sorryy... grobber is right - I meant to say that $ F$ is derivable only outside $ N$. Sorry...

[EDIT] If you roll over your mouse over the image of [0,1] from the sentence "F is derivable on [0,1]" you'll see that I wrote [0,1]\N...
(it still remains a mistake)

$ \text{\LaTeX}$ comment: use \setminus for $ [0,1]\setminus N$


function
calculus
derivative
induction
real analysis
real analysis unsolved

Show that $ (e^x \plus{} 1/P)^{ \minus{} 1}$ has all derivatives non-zero at $ 0$, where $ 1 < P \in \mathbb{N}$, and the function is $ C^{\infty}$.

mli: If you have a problem, post it [b]once[/b] and in just one place. You posted this one four times in four places. That's rude and inappropriate. This is probably the right location; I'll delete as many of the others as I can.

Let $ f$ be our function. Then it's readily seen that $ f'(x) \equal{} \minus{}e^x f(x)^2$. Since $ e^x \equal{} 1/f(x) \minus{}1/P$, we have
\[ Pf' \equal{} f(f\minus{}P).
\]
We claim that $ \mathrm{sgn}\,D^k f(0) \equal{} (\minus{}1)^k$. The statement is clear for $ k\equal{}0$ and $ k\equal{}1$ (because $ P>1$). Assuming it is true for $ k$, let us show it is true for $ k\plus{}1$. We have
\[ P D^{k\plus{}1}f(0) \equal{} D^k [f(f\minus{}P)](0) \equal{} \minus{}\sum_{j\equal{}0}^k \binom{k}{j} D^j f(0) D^{k\minus{}j} f(0).
\]
Since, by the assumption of induction, $ (\minus{}1)^j D^j f(0)>0$ and $ (\minus{}1)^{k\minus{}j} D^{k\minus{}j} f(0)>0$, we conclude that
\[ P D^{k\plus{}1} f(0) \equal{} (\minus{}1)^{k\plus{}1} \sum_{j\equal{}0}^k \binom{k}{j} |D^j f(0) D^{k\minus{}j} f(0)|.
\]
Therefore $ (\minus{}1)^{k\plus{}1} D^{k\plus{}1} f(0)>0$. We are done.


calculate the following
99.......9 (n repeats of 9) x 99........9 ( n repeats of 9 ) + 199.......9 (1 followed by n repeats of 9 )


Hello, i live in brazil and i am in college 2nd year right now doing physics. I have a little sister, she will make 7 years old  in february, and passing my vacations with my family, i noticed that she was very interested in reading, science and mainly math. I am still teaching her the algorithm of sum and subtraction and just talked to her about the first degree equations and she got the idea really very fast and i think that she can do a great progress in a small period of time, since she likes math and do a really hard job on it (but believe, its a play to her).
As the math olympiads is the most interesting and challenging stuff to a kid interested in math, i decided to teach her based on the math program for 11-12 years old kids. Off course i am not going straight to this point, as i said i am still teaching the basic things without hurry, my clock is her understanding and interest in learning more things. As i am not any educator and have no experience about it, mainly about math teaching, i would like to know some tips about the teaching, what should i teach her (in math) and so on without forcing her too much, or neglect some important stuffs.
I will be glad with replies, that´s all!

Wow! i'm impressed by your sister's ability to grasp concepts really quickly! However, don't be too complacent; just because she's understanding really quickly doesn't mean that she's ready to quit at a certain point. Keep challenging her.. Go to Mathcounts.org, click on STUDENTS, then the Go-Figure math challenge. If she is understanding subjects extremely quickly, i recommend that you teach her math that is atleast a year or 2 ahead of her. if she understands THAT, then she can go onto studying perhaps pre-algebra, algebra/ geometry as a bonus. Not recommended unless she understands the material really quickly....what type of math is she studying right now? as in what grade level??

ok, thank you for this site, this sims to be really useful, but now, i will need to spend some time teaching her the subjects on this website, and let´s see where she can goes!

Well AoPS can be used for all grades, except the math stuff mostly focuses on what you learn in your middle school through high school years. Of course kindergarteners may visit this site except that they will probably not understand the material. Likewise, a fourth grader is limited to either the MOEMS or Classroom math section.

[quote="tridecalogism_lover"][b]7h3.D3m0n.117 said:[/b][quote]Wow, i'm impressed too. The main thing is that children can get extremely mad at not getting something. Eventually if this happens people can extremely dislike (in your case math) the subject. You have to compliment her enough so that she knows she's not failing. Going here, on AoPS, is a great way to excel at math, problems posted by people here can challenge if not benefit your sister.
[/quote]

just wondering.....what grades are AOPS for? I looked on different sites for the grade levels, but couldn't find it!...It's compatible 4 mathcounts, right? :roll:[/quote]

Actually on the back of AoPS 1 it says that the book is compatible for people from seventh grade to tenth grade. AoPS 2 as far as i can tell is just if you grasp the concepts in volume one quickly...

So to answer your main question, yes, AoPS would definitely help.


i want to read some mathematicial books, written by russian. however i have not dictionary russian-english.
please help me!

[url=http://translation2.paralink.com/]This[/url] may help a bit though it translates mathematical texts rather poorly. Still, you can look up unknown words, short expressions, etc. though I would not recommend trying long sentences: they come out rather garbled as far as I can tell. Of course, there are many other free online translators. Just try a few and see which one does the best job (or go to the nearest bookstore and buy a couple of dictionaries: one general and one mathematical :P).

[quote="fedja"][url=http://translation2.paralink.com/]This[/url] or go to the nearest bookstore and buy a couple of dictionaries: one general and one mathematical :P).[/quote]
i already did this and reply the answer "it is impossible". please help me. i'm necessary to read mathematical books, written by russian.


linear algebra
matrix
ratio
logarithms
arithmetic sequence
geometric sequence

For the arithmetic progression with nth term u(n) = a + (n-1)d,
if A is the 3x3 matrix with first row  1  1  0 , second row  0  1  0 and third row  1  0  1,
then A^n multiplied by the column  matrix with first row a + nd, second row d and third row the sum of (a + (k-1)d) as k goes from 1 to n is equal to the column matrix with first row u(n+1), second row d and third row S(n), the sum to n terms.

For a geometric progression with u(n) = ar^(n-1),
 if G is the 2x2 matrix with first row r  1 and second row r  1, then G^n multiplied by the column matrix with first row a and second row a is the column matrix with first row u(n+1) and second row S(n+1), the sum to n+1 terms.

Can you improve on these? For example, can you find a 2x2 matrix which does the same job for APs or a matrix that doesn't need you to use the value of the common ratio for GPs?

For the 2x2 matrix for APs, using simultaneous equations,

I get a first row with elements

1+ d/a and -d^2/a^2

and a second row with elements

1 and 1.

I have not got a matrix without r for GPs yet.

First prove inductively that

$ \left[\begin{matrix}1 & 1 & 0 \\
0 & 1 & 0 \\
1 & 0 & 1\end{matrix}\right]^n \equal{} \left[\begin{matrix}1 & n & 0 \\
0 & 1 & 0 \\
n & {n(n \minus{} 1)\over 2} & 1\end{matrix}\right]$

Then just multiply $ \left[\begin{matrix}1 & n & 0 \\
0 & 1 & 0 \\
n & {n(n \minus{} 1)\over 2} & 1\end{matrix}\right]\times\left[\begin{matrix}a \\
d \\
0\end{matrix}\right]$

Similar ideas for the second part. Though I doubt you'll find a non-r matrix there, since multiplying matrices is essentially a linear transformation, and geometric progression involves exponents (though you COULD work with logarithms, reducing a geometric progression to an arithmetic one).

Thanks Farenhajt. (Now I'm just left with "sum of tans" from the diagram - I still can't see it.)

I don't think my 2x2 matrix for APs works: it's OK for the first couple of times then fails. Why?

Is the answer to "Can you improve on these?" "No" in both cases?

I now think that the fact that the only possible 2x2 matrix only works for the first two cases and then fails shows that the answer to "can we improve this" is "no" for APs, and that Farenhajt has explained why the answer must also be "no" for GPs. Am I right?


calculus
real analysis
real analysis unsolved

Hi, I have a question in my Calculus course that I can't seem to figure out. The question is basically this: a object falling to earth accelerates at rate [i]g[/i] reduced by an amount of air resistance [i]kv[/i] so that [i]dv/dt=g-kv[/i]. (They call [i]k[/i] the drag constant). Find [i]v[/i]([i]t[/i]) if [i]v[/i](0)=0.

I don't have much knowledge of differential equations yet, and all the questions I've previously dealt with have only one term (no addition signs) so that [i]v[/i] could be easily multiplied or divided to the other side and then integrated. Any help on how to do this would be much appreciated.

Substitute $u = g-kv$.  Then $du =-k \, dv$.


A triangle $ ABC$ is given. Construct an excircle with a centre $ I_{a}$. Draw a line connecting $ I_{a}$ with $ A$ and you get a point $ D\equal{}I_{a}A \cap BC$.
Determine the ratio $ I_{a}D : I_{a}A$ !

[hide="Solution."]For simplicity I will call $ I_a$ as $ I$.
Now we shall find the ratio $ \frac {AD}{ID}$. Construct altitudes from $ AH$ and $ IN$ to $ BC$. It is obvious that
\[ \frac {AD}{ID} \equal{} \frac {AH}{IN}
\]
Besides, from triangle area formulas
$ (1) S \equal{} \frac {1}{2} ah$
$ (2) S \equal{} pr$
we obtain that
\[ AH \equal{} \frac {2[ABC]}{BC}
\]

\[ IN \equal{} \frac {2[ABC]}{AB \plus{} BC \plus{} CA}
\]
Inserting these values, we get
\[ \frac {AD}{ID} \equal{} \frac {AH}{IN} \equal{} \frac {2[ABC]}{BC} : \frac {2[ABC]}{AB \plus{} BC \plus{} CA} \equal{} \frac {AB \plus{} BC \plus{} CA}{BC}
\]
Now
\[ \frac {AI \plus{} ID}{ID} \equal{} \frac {AI}{ID} \plus{} 1
\]
and
\[ \frac {AB \plus{} BC \plus{} CA}{BC} \equal{} \frac {AB \plus{} CA}{BC} \plus{} 1
\]
so
\[ \frac {AD}{ID} \equal{} \frac {AI}{ID} \plus{} 1 \equal{} \frac {AB \plus{} CA}{BC} \plus{} 1
\]
and
\[ \frac {AI}{ID} \equal{} \frac {AB \plus{} CA}{BC}
\]
or
\[ \boxed{\frac {ID}{AI} \equal{} \frac {BC}{AC \plus{} AB}}
\]
[img]http://img.flashtux.org/upload/img1328eea207eex9374ab4d.png[/img][/hide]

You solved it for the [b]IN[/b]center $ I$, but the problem says $ I_a$ is the [b]EX[/b]center. Although the result is the same, I'm not sure the solution could be accepted.

[hide]If $ P$ is the area and $ s$ the semiperimeter of $ \triangle ABC$, then

\begin{eqnarray*}{I_aD\over I_aA}&=&{\varrho_a\over \varrho_a+h_a}\\ &=& \frac{{P\over s-a}}{{P\over s-a}+{2P\over a}}\\ &=& {a\over a+2(s-a)}\\ &=& {a\over 2s-a}\\ &=& {a\over b+c}\end{eqnarray*}[/hide]

You are absolutely right, I really read the statement badly.  :)
However, I came up with idea of how to apply my solution to the original one. [hide]Just extend $ AB$ to $ AB'$ and $ AC$ to $ AC'$. We get the similar triangles $ \triangle ABC\sim \triangle AB'C'$. If we apply my solution to $ \triangle AB'C'$, we get ratio $ \frac {a}{b \plus{} c}$. But (see the image) the dotted segment is equal to the red straight segment. So the needed ratio remains $ \frac {a}{b \plus{} c}$.  :)

[img]http://img.flashtux.org/upload/img1328eea25d84x144ee09e.png[/img][/hide]


Suppose $x_0, x_1, \ldots, x_n$ and $x_0 > x_1 > \ldots > x_n.$ Prove that at least one of the numbers $|F(x_0)|, |F(x_1)|, |F(x_2)|, \ldots, |F(x_n),|$ where \[F(x) =  x^n + \sum^n_{k=1} a_kx^{n-k}, \quad a_k \in \mathbb{R}, \] is greater than $ \dfrac{n!}{2^n}.$

Are there any other restrictions on the $x_i$; for example, do they have to be integers?

Is there any condition on $a_0$?

It was in Crux proposed by Mohammed Aassila, with more  conditions
 $a_0=1$ and $x_0,x_1,...,x_n$ are integers
Here a combination of solutions by M.Bataille, Kee-Wai Lau
$P(x)=(x-x_0)(x-x_1)...(x-x_n)$ by considering the decomposition
$\frac{F}{P}$ into partial fractions we get
$F(x) = \sum_{k=0}^{n}F(x_k)\prod_{j\neq k}\frac{x-x_j}{x_k-x_j}$
The leading of coefficient of F is $a_0=1$ gives
$\sum b_k = \sum_{k=0}^{n}\frac{F(x_k)}{\prod_{j \neq k}(x_k-x_j)}=1$
$|\prod_{j \neq k}(x_k-x_j)| \geq \frac{n!}{C^k_n}$
$|b_k|\leq \frac{|F(x_k)|C^k_n}{n!}$
$1= \leq %Error. "sumb" is a bad command.
_k \leq \sum|b_k| \leq (max|F(x_k)|)/n!.(\sum C^k_n) =
 2^n/n! (max|F(x_k)|)$

I  only did a photocopy of this problem, I did not write the years, volume of
Crux.

[quote="harazi"]I knew this problem and this solution from about 3 years. When did it appear in Crux?[/quote]

The problem appeared much earlier. At least regarding my source.  :)


function
complex analysis
complex analysis unsolved

$f$ an analytic function in the unit complex disk, bounded by a constant $M$, and such that $|f(0)|=|a| > 0$. Let $Z(r)$ be is the number of zeroes of $f$ inside the open ball $B(0,r)$. Find a majoration of $Z(r)$ that depends only on $M,a,r$ for $0<r<\frac{1}{2}$.

[url=http://en.wikipedia.org/wiki/Jensen's_formula]Jensen's formula[/url] can be helpful.

Either Wikipedia authors should stop using punctuation in URL, or AoPS should start accepting it.  :(

[quote="mlok"][url=http://en.wikipedia.org/wiki/Jensen's_formula]Jensen's formula[/url] can be helpful.
[/quote]

[b]Using chess64's suggestion[/b]: [url=http://en.wikipedia.org/wiki/Jensen%27s_formula]Jensen's formula[/url] can be helpful.


[color=blue]Hello everyone. I plan to do a physics major. I'have asked this question to my chemsitry and physics highschool teachers. But I want a third opnion. What is ththe difference between nuclear chemsitry and nuclear physics? I understand they both deal with radioactivity. But is nuclear physics more theorectical?

Thanks[/color]

Hello,
    Nuclear Physics deals with the physical and mathematical aspects of radioactivity and other nuclear phenomena.Nuclear Chemistry largely encompasses only the 'chemistry part' of radioactivity like 'nuclear reactions'. Broadly, nuclear physics analyzes on mathematical backgrounds 'why' certain processes occur, while nuclear chemistry deals more with the 'processes' themselves. And finally, as always, yes nuclear physics is more theoretical a subject compared to Nuclear Chemistry, though the former has quite a few avenues for an experimental physicist also like Particle Physics.
Sathej


trigonometry
inequalities
Olimpiada de matematicas

Desde que $cos2x=2cos^{2}x-1$, entonces $cos^{2}x=\frac{cos2x+1}{2}$ luego  $cos^{2}(\frac{A}{2})=\frac{cosA+1}{2}$, $cos^{2}(\frac{B}{2})=\frac{cosB+1}{2}$, $cos^{2}(\frac{C}{2})=\frac{cosC+1}{2}$, con lo que, sean $p=\frac{cosA+1}{2}$, $q=\frac{cosB+1}{2}$, $r=\frac{cosC+1}{2}$ con lo que la expresion queda:

\[2S=\frac{qr}{p}+\frac{rp}{q}+\frac{pq}{r}=pqr(\frac{1}{p^{2}}+\frac{1}{q^{2}}+\frac{1}{r^{2}})\geq pqr(\frac{1}{qr}+\frac{1}{rp}+\frac{1}{pq})=p+q+r\] lo que sigue en: $S\geq 1+cosA+1+cosB+1+CosC=3+cosA+cosB+cosC$. Es muy facil de probar que

Tengo sueño aora lo termino

I don't think that you can find minimum value, which will be accessible, on this way...

I know that my solution isn't beautiful, but in any case it's right. Sorry, for latexless and my english:lol:

Let's prove that minimum is 9/4 (iff A=B=C=60):

a,b,c - side of this triangle, p - semiperimeter. then it's easy to show that
cos(A/2)^2*cos(B/2)^2/cos(C/2)^2 = (1+cosA)*(1+cosA)/(2*(1+cosA)) =
= (b+c-a)*(a+c-b)*(b+c+a)/(4*c^2*(a+b-c)) = (p-a)*(p-b)*p/(c^2*(p-c))

Let follow x=p-a, y=p-b, z=p-c. Then we have:

N/D=(x+y+z)xyz*( 1/((x+y)^2*z^2) + 1/((y+z)^2*x^2) +  1/((x+z)^2*y^2) )>=9/4

After some simple computation we get numerator of this fraction at the left
N=SUM(x^6*y^5*z^1) + 2*SUM(x^6*y^4*z^2) + 3*SUMc(x^6*y^3*z^3) + 5*SUMc(x^5*y^5*z^2) + 13*SUM(x^5*y^4*z^3) + 24*x^4*y^4*z^4

denominator of this fraction at the left is equal to
D=SUM(x^6*y^4*z^2) + 2*SUMc(x^6*y^3*z^3) + 2*SUMc(x^5*y^5*z^2) + 6*SUM(x^5*y^4*z^3) + 10*x^4*y^4*z^4

[color=red](SUMc - cyclic sum)[/color]

After adding similar terms we get this inequality
4*SUM(x^6*y^5*z^1) + 2*SUMc(x^5*y^5*z^2) + 6*x^4*y^4*z^4 >=
SUM(x^6*y^4*z^2) + 6*SUMc(x^6*y^3*z^3) + 2*SUM(x^5*y^4*z^3)

Let's divide  this inequality by xyz ( we may do it for a little time before :huh:)

[size=167][color=red]4*SUM(x^5*y^4*z^0) + 2*SUMc(x^4*y^4*z^1) + 6*x^3*y^3*z^3 >=
SUM(x^5*y^3*z^1) + 6*SUMc(x^5*y^2*z^2) + 2*SUM(x^4*y^3*z^2)[/color][/size]

As is easy to see that it sum of this 3 well-known inequalities:

1) [color=blue]SUM(x^5*y^4*z^0)>=SUM(x^5*y^3*z^1)[/color]
   (Murhed {5,4,0}>{5,3,1})

2) SUM(x^5*y^4*z^0)>=SUM(x^5*y^2*z^2)
   (Murhed {5,4,0}>{5,2,2})
    SUM(x^5*y^4*z^0)>=2*SUMc(x^5*y^2*z^2)
    [color=blue]3*SUM(x^5*y^4*z^0)>=6*SUMc(x^5*y^2*z^2)[/color]

3) SUMc(x^3*y^3) + 3*x^2*y^2*z^2 >= SUM(x^3*y^2*z^1)
   (Schur inequality for a=xy, b=yz, c=xz)
   Multiply by xyz
    SUMc(x^4*y^4*z^1) + 3*x^3*y^3*z^3 >= SUM(x^4*y^3*z^2)
  [color=blue]  2*SUMc(x^4*y^4*z^1) + 6*x^3*y^3*z^3 >= 2*SUM(x^4*y^3*z^2)
[/color]
end :)


calculus
integration
real analysis
real analysis unsolved

4. Let $f: [a,b] \to R$. Prove that if $\int^{b}_{a}f(x)\,dx$ exists then so does $\int^{b+c}_{a+c}f(x-c)\,dx$ and these two integrals are equal.

I really need some help with this problem.

Naturally, a solution will depend on the definition of integral that you are supposed to use. I assume that the integrals are defined as limits of Riemann sums.

Consider a Riemann sum for $f(x-c)$ on $[a+c,b+c]$. They are of the form $\sum_{i=1}^{n}f(x_{i}^{*}-c)(x_{i}-x_{i-1})$, where $a+c=x_{0}<x_{1}<\dots<x_{n}=b+c$ is a partition of the interval $[a+c,b+c]$, and $x_{i}^{*}$ is a point on the interval $[x_{i-1},x_{i}]$. Let $t_{i}=x_{i}-c$ for $i=0,\dots,n$ and observe that $a=t_{0}<t_{1}<\dots,t_{n}=b$. Also let $t_{i}^{*}=x_{i}^{*}-c$ and observe that $t_{i}^{*}$ is a point on the interval $[t_{i-1},t_{i}]$.

Thus the sum $\sum_{i=1}^{n}f(x_{i}^{*}-c)(x_{i}-x_{i-1})$ is equal to $\sum_{i=1}^{n}f(t_{i}^{*})(t_{i}-t_{i-1})$. The latter is a Riemann sum for $f$ on the interval $[a,b]$, and such sums converge to $\int_{a}^{b}f(x)dx$ as the partitions become finer.

mlok, you proved that the two integrals are equal. But how can one first prove that the second integral "exists"?

Thanks for your reply though. THe proof was rather simple but elegant.

The main point here: each Riemann sum for the second integral is equal to a Riemann sum for the first integral, with partition shifted $c$ units to the left. We know that the Riemann sums for the first integral have a limit (i.e. the integral exists). So, the Riemann sums for the second integral also have a limit.


topology
vector
invariant
advanced fields
advanced fields unsolved

Let $ A$ be  two linked circles in the 3-space, $ B$ - just the disjoint union of two circles.

How can one prove that $ A$ and $ B$ are not homotopic.

The same question with $ A$ - $ 3$ pairwise linked circles in the 3-space , $ B$ - two linked circle + 1 circle, disjoint from these two linked circles.

I know that one can compute the fundamental groups  $ \pi_1(S^1 \vee S^1) \equal{} \mathbb{Z} \ast  \mathbb{Z}$, $ \pi_1 (S^1 \times S^1) \equal{}  \mathbb{Z} \times  \mathbb{Z}$.
But i'm interested in more elementary solution without even knowing the notion of fundamental groups.

uhm.. actually, they [i]are[/i] homotopic, their complements are not ;)
probably you meant that they're not [i]isotopic[/i]: the other way i know to see it is just the intuitive idea that "there's no plane (surface) separating them" in the first case, while in the second there's one...

one way to see it, is introducing the (i think) linking number: for sake of simplicity, suppose one of the two discs is the standard $ S^1$ in the $ xy$-plane, and call $ D$ the disc which it's border of: you can choose an orientation of $ D$, and suppose (for density reasons) that the intersections of $ D$ with the other circle $ C$ is transversal, hence finite.
give an orientation to $ C$, and for any point of intersection $ p$ call $ \epsilon_p$ the "sign of intersection": $ \epsilon_p\equal{}1$ if the base of the space given by the tangent vector to $ C$ in $ p$ and any oriented base of $ D$ is positive with respect to the standard orientation of the space, and $ \minus{}1$ otherwise.
the sum over all the intersection is verified to be an isotopy invariant (and i think it's called linking number), and in the first case it's odd (odd number of intersections), in the second it's zero.

by the way, this distinguish some more things, such as two "doubly linked" circles, just making two loops around one of them, and others too..
one more thing: the thing about "orientation" isn't actually necessary, one can just think of determinants and families of bases, so it's pretty elementary, after all.

What does "linked" mean? It looks from notation that it is $ S^1\times S^1$, but I interpreted it as two circles sharing a point (the "figure eight"). Maybe what I'm thinking of is two connected circles? So, two linked circles are homeomorphic to the $ 2\minus{}$torus? The fundamental group of what I'm thinking is the free group on two generators.

No: two circles in $ {\bf R}^3$ such that each passes through the interior of the other.  For example, the circle in the $ x\minus{}y$ plane centered at $ (0, 0, 0)$ with radius 1 and the circle in the $ x\minus{}z$ plane centered at $ (1, 0, 0)$ of radius 1.


Anyone going to ARML at Iowa U. say your AoPS thing on this thread, and if you'll be at ARML.

I'll be there!
user: aryth (as if you couldn't tell)
name: Kevin Hu (also as if you couldn't tell)
team: Chicago

user: pianoforte (as you probably already know)
name: Stephen Rong
team: Missouri

ya, good luck with that Jay... friday night = food festival.

well good food always tastes better with ice cream afterwards...

[quote="korammath"]user: korammath
name: Russell Houston
team: Texas

YAY KEVVIE HU IS GOING!

JAY IS A WHITEY![/quote]

sry bout that.... my friend wrote it.

[size=75][color=darkred]Nice Russell.... -.-

user (<--- no offense, but that's sort of dumb):  Angela
team:  Texas

=P  Whoever chose me isn't that bright... Oh, crap!  Shouldn't have said that.[/color][/size]

[quote="junggi"]im not going to iowa, but i wanted to point out 2 things

[quote="RedJumpsuitApparatusRules"]asdlg[/quote]
your username fails so much that i feel like punching a kitten

how come LV arml never gets these kinds of topic?
some lv dude needs to start one up right now[/quote]

sry but it rly does fail.

why dont u start a lv thing

[quote="FightTheTide"]AMP REUNION @ IOWA ARML:

Meet at Whitey's (the ice cream place) Friday night.
Further details to follow.[/quote]

Calling all 2007 AMPers:

check the AMP forum for the final details.

user: coolguy00001
name: Shi-Ke Xue
team: Iowa

I hope I'm on the Iowa A-team this year...

user: it's kinda obvious
name:  matt staib
team:  iowa

user: Altheman
name: Alex Anderson
team: Chicago (hopefully I don't get kicked out for missing two practices...)

User: Fanatic
Name: Richard Ni
Team: Indiana (hopefully gold =P)

User: ionave

Name: Sara Akgul

Team: Chicago Allstars D team

user: cognos599
name: Anirudh Prabhu
team: Indiana most likely silver.


Predict the products [b]with stereochemistry[/b] :

[url=http://en.wikipedia.org/wiki/Cholesterol]Here[/url] is the structure of cholesterol:
(totally worth 7 or 8 marks, don't remember..)
(1) When cholesterol is treated with $ \ce {B_2H_6}$ in THF followed by oxidation using $ \ce {H_2O_2}$ in an alkaline medium.
(looks easy but shouldn't there be a pair of diastereomers formed?, the answer expects only one product...)

(2) With $ \ce {H_3O^ + }$ at room temperature. (same doubt)

(3) With $ \ce {CF_3COOOH}$, (same)

(4) With $ \ce {Br_2}$ in carbon tetrachloride.

(5) Draw skeleton structure of compound formed in (1)

I think we've got to give the major product.

steric effect plays a part here no ? the compound formed expects to have minimal steric hindrance . assuming this I've posted as above.

lol i wrote down all the wrong ones, then...except for the Br2, which i wrote just to be safe. Guess it paid off... :oops:

I dunno .
just suggested .
carcul should help here . Is this correct ?


:help:

10. Let a and b be two integers for which 7a + 12b = 1. What is the largest possible positive value of a + b which is less than 2005?

How do you solve this?

Thanks

[hide]Okay, notice that the first solution with a>b, is (7,-4).

The next one is (19,-11)

Notice that the sum is always 3 more than a multiple of 5...

So the smallest one is 2003[/hide]