I know that i posted on the wrong topic, but i couldn't find any that say anything about curriculum.  :(  Anyway I have a question. I am currently in Algebra 1 right now (9th grade) and would be going to geometry next year. I want to skip geometry and go to algebra 2. Will this affect me? Any help?

well since ur gonna skip geometry, u obviously need to be good at it... if u don't think u are, then u shouldn't skip geometry. it was a hard subject 4 me too...

I know that I am going to study over the summers to get better at it. I mean will the administrater or the college look that i skipped geometry and don't accept me? I was kind of getting worried about that

Well first of all, I don't think the administration will let you skip if don't take it... you can't just skip math classes if you've never taken them.

Second, geometry is a very important subject on math competitions, you'll learn the basics in high school geometry. I personally did it in half a year, and I really sucked at contest geometry, so I've been reading the geometry chapters in AoPS volume 1. If you have that, it will be an excellent way to learn geometry. You'll learn more than you'll even learn in school.

[quote="aznxenocide"]i doubt colleges will look at that and think "wow this kid skipped a math level. he must a retard." i mean, seriously. skipping a math level shows them your smarter and probably more motivated than other kids. so i say, if you're confident you don't need it, go for it.[/quote]

No and yes.

Taking high math class doesn't necessarily mean that you're very smart. What matters is how you get on each class. But it will show that because you got into higher level, you're a motivated student.

I wouldn't recommend skipping geometry, however. Whether you're going to compete in math competition or not, geometry is one of the requirement math class for colleges. If I remember correctly, most colleges [b]do[/b] look for one year of planar/solid geometry and two years of algebra.

[quote="aznxenocide"]i doubt colleges will look at that and think "wow this kid skipped a math level. he must a retard." i mean, seriously. skipping a math level shows them your smarter and probably more motivated than other kids. so i say, if you're confident you don't need it, go for it.[/quote]

you look like somebody i know ._.

x3no?


Let $a,b,c$ be positive reals such that $a+b+c<1$. Prove that if $x\geqq a,y\geqq b,z\geqq c$, then $yz+zx+xy>a(b+c)x+b(c+a)y+c(a+b)z+3abc$.

[quote="Valiowk"]Observe that if we fix $y$ and $z$ and reduce $x$, the LHS decreases more than the RHS.  Hence it suffices to prove $ab + bc + ca > a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc$.  Multiplying the LHS by $a+b+c$, we obtain the RHS, hence the inequality is proven.[/quote]

Thank you for your reply, Valiowk.

I can't understand your solution well.Please explain  more in detail.

kunny


algebra
polynomial
linear algebra
matrix

Let $A,B \in \mathcal M _2 \left( \left( 0,\infty \right) \right)$. Prove that $(AB)^2=(BA)^2$ iff $AB=BA$.

Well, if $AB=BA$ then $(AB)^{2}=AB\cdot AB=BA\cdot AB=(BA)^{2}.$  :)

Edit: Maybe you want to prove the converse. If $(AB)^{2}=(BA)^{2}$ then $AB=BA.$ ;)

 [hide="Hint"]

  Use Hamilton-Cayley for $AB$ and $BA$. ;)

 [/hide]

"iff" means both ways.

And of course, I was interested in proving the harder "direction". How to use Hamilton? I tried but I got nothing  :(

Ok, here it goes: if $(AB)^{2}=(BA)^{2}$ .By applying Hamilton-Cayley to$AB$ and $BA$ we will get
 $tr(AB)(AB-BA)=O_{2}$ and since $tr(AB)>0$ hence $AB=BA.$ I hope it is clear. ;)

Here's how I'd phrase the argument. Let $C=AB$ and $D=BA$.

$\text{Tr}(C)=\text{Tr}(D)=a$ and $\det(C)=\det(D)=b.$

So, $C$ and $D$ have the same characteristic polynomial, $x^2-ax+b$

By Cayley-Hamilton

$(AB)^2-aAB+bI=0$
$(BA)^2-aBA+bI=0.$

Subtract these and use that $(AB)^2=(BA)^2$ to get

$a(AB-BC)=0$ or $\text{Tr}(AB)(AB-BA)=0.$

So far, this is exactly what cezar said, only wordier.

If $\text{Tr}(AB)\ne0,$ then we are done. But why should we have that? Becuase the matrices have positive elements, and hence positive traces.

If you didn't have that positivity condition, the theorem would be false.

Try this example:

$A=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ and $B=\begin{bmatrix}0&1\\1&0\end{bmatrix}$

Then $AB=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ and $BA=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$

while $(AB)^2=(BA)^2=-I.$

[quote="Kent Merryfield"]So, $C$ and $D$ have the same characteristic polynomial, $x^2-ax+b$[/quote]

Why this precisely?

It follows from the definition of the characteristic polynomial, $P_A = \det(A-xI_n)$. In the case $n=2$, $P_A=x^2 - \left( \textrm{tr} A \right) x + \det A$.

Can we make this result work for bigger matrices? (maybe with some stronger conditions)

I have an extension of this problem. If AB=BA and p is a nonnegative integer, prove that (AB)^p = (A^p)(B^p). Sorry I don't know how to use the LaTex.


[quote="Niichavo"]Prove that for any natural number $ a,b,c$
$ [a,b][b,c][c,a]\geq [a,b,c]^2$[/quote]
You mean $ [a,b]\equal{}lcm(a,b)$

[quote="Allnames"][quote="Niichavo"]Prove that for any natural number $ a,b,c$
$ [a,b][b,c][c,a]\geq [a,b,c]^2$[/quote]
You mean $ [a,b] \equal{} lcm(a,b)$[/quote]
If $ [a,b] \equal{} lcm(a,b)$ we go like this.
Set $ a \equal{} \prod_{i\equal{}1}^r p_i^{a_i}$, $ b \equal{} \prod_{i\equal{}1}^r p_i^{b_i}$, $ c \equal{} \prod_{i\equal{}1}^r p_i^{c_i}$ ($ a_i, b_i, c_i \in \mathbb{N}_0$)

Thus we have $ [a,b][b,c][c,a] \equal{} \prod_{i\equal{}1}^r p_i^{max(a_i,b_i)\plus{}max(b_i,c_i)\plus{}max(c_i,a_i)}$
And $ [a,b,c]^2 \equal{} \prod_{i\equal{}1}^r p_i^{max(a_i,b_i,c_i) \cdot 2}$
So we just have to prove $ max(a_i,b_i)\plus{}max(b_i,c_i)\plus{}max(c_i,a_i) \ge  2 \cdot max(a_i,b_i,c_i)$
To prove that $ max(x,y)\plus{}max(y,z)\plus{}max(z,x) \ge  2 \cdot max(x,y,z)$ we can assume w.l.o.g that $ x \ge y \ge z$
So $ x \plus{} y \plus{} x \ge 2x$ which is obviously true, since $ x,y,z \in \mathbb{N}_0$

[quote="Mathias_DK"][quote="Allnames"][quote="Niichavo"]Prove that for any natural number $ a,b,c$
$ [a,b][b,c][c,a]\geq [a,b,c]^2$[/quote]
You mean $ [a,b] \equal{} lcm(a,b)$[/quote]
If $ [a,b] \equal{} lcm(a,b)$ we go like this.
Set $ a \equal{} \prod_{i \equal{} 1}^r p_i^{a_i}$, $ b \equal{} \prod_{i \equal{} 1}^r p_i^{b_i}$, $ c \equal{} \prod_{i \equal{} 1}^r p_i^{c_i}$ ($ a_i, b_i, c_i \in \mathbb{N}_0$)

Thus we have $ [a,b][b,c][c,a] \equal{} \prod_{i \equal{} 1}^r p_i^{max(a_i,b_i) \plus{} max(b_i,c_i) \plus{} max(c_i,a_i)}$
And $ [a,b,c]^2 \equal{} \prod_{i \equal{} 1}^r p_i^{max(a_i,b_i,c_i) \cdot 2}$
So we just have to prove $ max(a_i,b_i) \plus{} max(b_i,c_i) \plus{} max(c_i,a_i) \ge 2 \cdot max(a_i,b_i,c_i)$
To prove that $ max(x,y) \plus{} max(y,z) \plus{} max(z,x) \ge 2 \cdot max(x,y,z)$ we can assume w.l.o.g that $ x \ge y \ge z$
So $ x \plus{} y \plus{} x \ge 2x$ which is obviously true, since $ x,y,z \in \mathbb{N}_0$[/quote]
You are faster than me
But futher more we will have $ [a,b][b,c][c,a]/[a,b,c]^2$ ,your proof is its one


superior algebra
superior algebra unsolved

Show that primes ideals of $ R_1\times...\times R_n$ have the forms $ R_1\times...\times P_i\times...\times R_n$ with $ P_i$ is prime in $ R_i$.
What are prime ideals of $ \mathbb{Z}/p^t\mathbb{Z}$ with $ p$ is prime, $ t\in\mathbb{N}$

1) determine first the ideals of a finite product of rings (simply by manipulating the definitions) and calculate their quotient rings. the integer quotient rings yield the prime ideals.

2) the prime ideals of R/I correspond to prime ideals in R, which contain I. and $ \mathbb{Z}$ is a PID.


inequalities
trigonometry
algebra
polynomial
inequalities proposed

Prove that : In the triangle ABC :
  $ p \leq 2R + (3\sqrt3-4)r $

Here is a beautiful solution (found after seeing that in fact this problem is the equivalent of an algebraic inequality from Old and New Inequalities):
   By homogeneity we can take $r=1$. if we put the sides $ a=y+z, b=z+x, c=x+y$ then this translates into $ xyz=x+y+z$. It is easy to see that $ R=(xy+yz+zx-1)/4 $ and thus the inequality becomes the mentioned problem from the book:
     If $x+y+z=xyz$ then $ (x-1)(y-1)(z-1)$ is at most $ 6\sqrt{3}-10$ . We gave two solution in the book, but the second solution, found by me is ugly and I won't post here. I will post the first solution: trivial considerations show that we may assume $ x,y,z$ greater than 1. If we take $ x=a+1, y=b+1, z=c+1$ the condition becomes $ ab+bc+ca+abc=2$. Now, suppose that $ abc>6\sqrt{3}-10$. Then by AM_GM we have $ ab+bc+ca> 12-6\sqrt{3}$ and we contradict $ ab+bc+ca+abc=2$. This great solution was found by Marian Tetiva.

Your solution use the method "mixing variables" ??  I have an ugly solution by using trigonometry .

No, it uses derivatives. Anyway, what should one expect after seeing the above solution? I find it so easy and beautiful.

I define it a "natural" solution for its elegance and simplicity.

i know that this is named Blundon inequality:

$2R^2+10Rr-r^2-2(R-2r)^{2 \over 3}\leq p^2 \leq 2R^2+10Rr-r^2+2(R-2r)^{2 \over 3}$

can you prove it?
is it the same ineq before couple of transformation?

Well, I have no intention to think at such a messy thing. Probably it is very strong, but it is more than ugly.


[quote="maokid7"]how would you prove that $\lim_{x\to{1}}x^{2}+5x+1=7$ and not some other value, such as $-3$, using the $\delta$,$\epsilon$ definition of a limit?[/quote]
Draft copy:
$|{x^{2}+5x+1-7}| = |{(x-1)(x+6)}| \le \left7|{x-1}| < 7\delta = \varepsilon$

So, for every given $\varepsilon > 0$ consider $\delta =\varepsilon/7$ so for all $|x-1|<\delta$ we'll get $|x^{2}+5x+1-7| <\varepsilon$ i.e. $\lim_{x\to{1}}x^{2}+5x+1=7$

PS. Sorry, i was hurry and made some stupid mistapes. Now it's correct.

what i did was $|x-1|<\delta$ and $|(x-1)|*|(x+6)|<\epsilon$

$|x+6|=|(x-1)+7|\leq |x-1|+7=\delta+7$ so u have

$(\delta)(\delta+7)<\epsilon$  then we choose $\delta \leq{1}$ since if we find one $\delta$ that works any $\delta$ smaller that it will work as well so

$\delta<\frac{\epsilon}{8}$

but i dont know how this fails if we say the limit is -3

If you choose another value $a \not= 7$, you won't get $|x^{2}-5x+1-a| = |x-1| \cdot |x+?|$. Hence you cannot "proof" that $a$ is the limit analogously.

There's also the theorem that if \[\lim_{x\rightarrow n}f(x)=a\] and also \[\lim_{x\rightarrow n}f(x)=b,\] then $a=b$.


[quote="shobber"][quote="sherlocktan"]a,b,c,d是正数，且a+b+c+d＝１，求证
a/(a+b)^0.5+b/(b+c)^0.5+c/(c+d)^0.5+d/(d+a)^0.5<3/2[/quote]
I'll tex it first:
$a,b,c,d \in R^+$ and $a+b+c+d=1$. Prove:
\[\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+d}}+\frac{d}{\sqrt{d+a}} < \frac{3}{2}\][/quote]
Thank you very much!
 :D

这道烂题李胜宏给我们讲过，他就是用的这种方法，who knows!

好象是浙大的那个猛B吧，他是中国奥林匹克高级教练，这几年的冬令营和集训队的命题组成员，还带队去过IMO。


linear algebra
matrix
geometry
perimeter
search
combinatorics proposed
combinatorics

At a table having form a square $10x10$ sit $100$ workers, each having distinct salaries. In every unit square sits exactly one worker. Two workers are neighbours if the unit squares where they sit have a common side. The workers are very envious and a worker is happy only if at most one of his neighbour has larger salary.
What is the largest possible number of happy people?

I dont know if this can help, but, if we paint the good squares in black, there cant be a cycle, it is easy to prove anyway, so based on this I think the answer will be $59$ However, I havent proved that if we select 60 squares we always get a cycle. Obviously the best we can get is a single tree,how can I show its lenght is at most 59?

i think we can do better than 59 .infact for any induced tree if we label the vertices sequentially first by labelling along a maximal path $P$ with the labelling decreasing along $P$ and then doing the same on each component of $V \setminus P$.i think i have an instance where we get 69.consider the induced subgraph on the following vertices:(i shall put it as a matrix;this is not the adjacency matrix.the entry 0 means that we don't take that cell and 1 means we take that cell)
    $\star tarray 1\&1\&1\&1\&1\&1\&1\&1\&1\&1\\
    1\&0\&1\&0\&1\&0\&1\&0\&0\&1 \\
    1\&1\&0\&1\&0\&1\&0\&1\&1\&1 \\
    1\&0\&1\&1\&0\&1\&0\&0\&0\&1\\
    1\&1\&0\&1\&1\&1\&1\&1\&1\&1\\
    1\&0\&1\&0\&0\&0\&0\&1\&0\&1\\
    1\&1\&1\&1\&1\&1\&1\&0\&1\&1\\
    1\&0\&1\&0\&1\&0\&0\&1\&0\&1\\
    1\&1\&0\&1\&0\&1\&0\&1\&0\&1\\
    1\&0\&1\&1\&1\&1\&1\&1\&1\&1\\
  %Error. "endarray" is a bad command.
$
as for the non-existence of a cycle it is trivial-in any cycle take the cell with the least entry among the cells of the cycle.this cell has 2 neighbors larger than it.contradiction.

A good idea for bounds here will be seing how many edges our trees contain.

actually it is also possible to get 70 vertices with its induced graph acyclic.
On the upper side, it is not difficult to show that in a $5 \times 5$ grid,the max is 18;hence in a $10 \times 10$ grid the maximum is 72.so that leaves a deficit of 2.
it seems that 70 is best possible,so we'll have to thin this down a little further.

I proved $72$ is the maximun in a very elegant way. Every tree, of size $k$, covers and is bounded by $3k+1$ edges, and every edge belong to exactly one  tree. Now, if alll the edges in the perimeter are covered, we must have all the perimeter with good squares, and we get a cycle, so there are at most 219 edges covered. So we have $3N+K<220$ where N is the number of good squares, and K the number of trees they form, this implies $N<73$.

My answer is also $72$. I think you're not interested in seeing my example, all I can say is that the example was constructed basing on that ideas: A $2\times 2$ cell can contain at most $3$ selected cells. If it contains $3$ selected cells, call it 'full'. Take the cell with greatest number and call it the 'good' one. Then if we partition the cells into $25$ $2\times 2$ squares we must ensure that among pairs of adjacent selected cells from different 'full' $2 \times 2$ squares one of the must be good. Now a laborious search will construct the example.
To prove the maximum is $72$ we note the following.
Let $k$ be the number of selected cells. between any $2$ adjacent cells draw an arrow from the smallest to largest.
Then we get $180$ arrows.
From the other side, this number cannot exceed $k+4(100-k)$. But from the maximum cell no arrow exits, so it doesn't exceed finally $k+4(100-k)-1$. And to the end, the smallest number on the boundary has at least two arrows pointing out from it, but at most three since it is at the boundary (not $4$ as we counted). So we must substract one more $1$ to get $k+4(100-k)-2 \geq 180$ or $3k \leq 218$ so $k \leq 72$.
QED


[quote="Techsan"]Does anyone know of any good free software that allows you to find patterns in sequences of numbers?[/quote]

Well, I do not, and in general I don't think it is something easy to find. Nevertheless, something may be useful is to look for the studies and software developed for the groups trying to "break the Bible Code", that, honestly I consider a stupid project that is finding just stadistical anomalies, but some of those guys are taking it seriusly, and spending years of their research to it. They might have something that may help you (just a guess) :?

Have you tried: [url=http://www.research.att.com/~njas/sequences/]The On-Line Encyclopedia of Integer Sequences[/url], this has a large database of sequences, perhaps the one you need is in there.

I would say Hidden Markov Models... just because.


This is obvious, since for every number $ a \in \{1,2,3,\ldots,n\}$ there exist a number $ b \in \{n\plus{}1,n\plus{}2,\ldots,2n\}$ such that $ a \mid b$. The claim easily follows.


I think that

[hide]since he had $15 to start with and then sold the C.D for $20 dollars he then had a $5 profit however when he buys the C.D again he pays $25 which is a $10 defecit on his original sum making his defecit -$5 he then sells it for $28 giving him a $3 profit and making his end profit -$2 [/hide]
is this solution right or is the solution just
[hide]$8[/hide]

[hide="CLICK HERE!!!!!"]-15+20-25+28=8 dollars. He spent 15 dollars, earned 20, spent 25, earned 28. He made 5 from his first deal of buying for 15 and selling for 20. He made 3 from his second deal of buying the cd for 25 and selling it for 28. 3 + 5 is also 8.[/hide]

Sudonym, your second solution is correct.

I think I've seen this problem from the book called

"Lady and the Tiger" by Simon Smullyan, one of my favorite authors :roll:

Yeah... he clearly made an eight-dollar profit

[quote="MCrawford"]Jacob bought a CD for $\$15 and sold it for $\$20.  He then bought it back for $\$25 and sold it again for $\$28.  How many dollars profit did he make?[/quote]

[hide]he made 5 the first time and 3 the second so 8[/hide]

i don't think so, but what do you mean by another way?

i mean another way besides combining like terms


calculus
integration
function
analytic geometry
topology
real analysis
real analysis unsolved

so the question is to show that there exists a unique spectral integral $ J$ on the bounded borel functions on $ K$ such that $ J(\pi_j) \equal{} T_j$ with $ \pi_j$ the usual coordinate projection in $ \mathbb{C}^n$ where $ K$ is the joint spectrum of $ T_1,...T_n$ bounded normal commuting operators.  (i.e. if $ L: \sigma (A) \to K\subset \mathbb{C}^n$ is the homeomorphism from the $ C^*$ algebra $ A$ generated by $ T_1,..,T_n$ by $ L(\phi ) \equal{} (\phi(T_1),...,\phi(T_n))$

so i guess maybe i need to show that (i) $ J$ is a $ C^*$ homomorphism, (ii) $ J(f_n)\to J(f)$ strongly for $ f_n\to f$ pointwise boundedly in the bounded borel functions on $ K$, and (iii) $ \parallel{}J(f)\parallel{}\le \sup_{y\in K} |f(y)|$

but im having problems really understanding what $ J$ does to a general $ f$ of this type, so its hard to even begin to show (i)

am i approaching this wrong?

all around, i would appreciate any help

ah, i solved it, i just had to use a previous unique existence theorem and prove that

$ J(f(x))\equal{}J'(L^{\minus{}1}(f(L(x))))$ obeyed the above. thanks anyway


Could you prove that 10^(8n) + 1 is not prime , for all integer n ?
Thanking in advance.

17 does half of the job, can you find the other number? :D

10^8 + 1 = 17.5882353.
10^8 = -1 [17], if k is odd, k = 2k’ + 1 and 10^8k = -1 [17], then all the numbers 10^(16k + 8) + 1, divided by 17, are not primes.
But the others numbers 10^(16k) + 1 ? primes or not primes ? I don’t know.

For odd values of $n$, we find that $17 | 10^{16n} - 1$ but that $17 \not | 10^{8n} - 1$, so we can conclude $17 | 10^{8n} + 1$ for odd $n$.

For even $n$, calculation gives $353, 449, 641, 1409, 69857 | 10^{16} + 1, 10^{48} + 1$, but neither of them share common factors with $10^{32} + 1$... perplexing  :|

Um? Let's reduce the problem some more...
Let $n = 2^k\cdot \ell$ with $\ell$ odd.  Let $m = 10^{2^{k+3}}$ so $10^{8n}+1 = m^\ell+1$ which is obviously composite of $\ell > 1$.  Now suppose $n$ is a power of $2$...  :)

$m^l + 1$ is composite only if $m + 1$ is, and that remains to be proven...

Though at least the problem has been reduced to $10^{2^{k+3}} + 1$, which is more specific, but I am still unable to find a general proof that this is composite  :|

(My original idea involved finding an odd number $p, (10, p) = 1$ such that $\phi(p) = 2^{k+4}$, in which case $p | m^2 - 1$ but $p \not | m - 1$, however, I cannot find such a $p$ even for $k = 1$ and I suspect none exists :| )

[edit] oops wow I must've been half asleep when I posted this  :oops:

Well, in fact "d'où" has to be translated as "hence" ;)
(but : "où" = "where", yes :))

same thing! :mad: (well, not really)

p.s. thanks...  :blush:

You're welcome ;)

But, to get back to the topic, I think :
$a(k) = 10^{16k} + 1$


$ 0<a\le b\le c\le 2a$ with $ a\plus{}b\plus{}c\equal{}4s$, show that $ abc\ge 2s^3$.

From $ a \leq  b \leq c \leq 2a$ we have that $ a, b, c$ are sides of a triangle.  I think that you can use that then exist $ x, y, z \geq 0$ such that $ a\equal{}y\plus{}z,$ $ b\equal{}z\plus{}x,$ and  $ c\equal{}x\plus{}y$.  :)

I suppose that the minimum of $ abc$ exists.
Since $ a\le b\le c\le 2a$, make $ a$ fixed.
$ 4bc\equal{}(b\plus{}c)^2\minus{}(b\minus{}c)^2$, so if $ |b\minus{}c|$ increases then $ bc$ decreased.
Thus $ abc$ reaches the minimum only if $ a\equal{}b$ or $ c\equal{}2a$.
Let P: $ c\equal{}2a$ or $ a\equal{}b$, similarly we make $ b,c$ fixed and get
     Q: $ c\equal{}b$ or $ a\equal{}b$
     R: $ c\equal{}b$ or $ c\equal{}2a$
Since all of P,Q,R must be true if we want the minimum, it's easy to see that
al least two of the three equals $ c\equal{}2a,a\equal{}b,c\equal{}b$ must be true.
So we have only 3 cases:
A: $ c\equal{}2a,a\equal{}b$
B: $ a\equal{}b,c\equal{}b$
C: $ c\equal{}2a,c\equal{}b$
easy to find that Case A gives the minimum of $ abc$, so $ abc\ge 2s^3$.
But the problem is how to prove that the minimum exists, who can help me?

I think that this inequality is wrong. We can use $ a\equal{}y\plus{}z,$ $ b\equal{}z\plus{}x,$ $ c\equal{}x\plus{}y$  $ (x, y, z \geq 0)$ and find example.  :idea:

We must show that $ (p\plus{}q\plus{}1)^3\leq 32pq$ where $ p,q\in [1,2]$ (get this from $ p\equal{}\frac{b}{a},q\equal{}\frac{c}{a}$).

To do this, first we have
\[ (p\plus{}q\plus{}1)^3\leq (p\minus{}1)(q\plus{}3)^3\plus{}(2\minus{}p)(q\plus{}2)^3\iff (p\minus{}1)(2\minus{}p)(p\plus{}3q\plus{}6)\geq 0\]
Second, we have
\[ (p\minus{}1)(q\plus{}3)^3\plus{}(2\minus{}p)(q\plus{}2)^3\]
\[ \leq (q\minus{}1)(125(p\minus{}1)\plus{}64(2\minus{}p))\plus{}(2\minus{}q)(64(p\minus{}1)\plus{}27(2\minus{}p))\]
\[ \equal{}(q\minus{}1)(61p\plus{}3)\plus{}(2\minus{}q)(37p\minus{}10)\iff (q\minus{}1)(2\minus{}q)(3p\plus{}q\plus{}6)\geq 0\]
Finally, the above expression is $ 24pq\plus{}13p\plus{}13q\minus{}23$. Showing that this is at most $ 32pq$ reduces to
\[ 8pq\plus{}23\geq 13p\plus{}13q\iff (8p\minus{}13)(8q\minus{}13)\plus{}15\geq 0\]
But this is evident: if $ 8p\minus{}13,8q\minus{}13$ have the same sign, it's clearly true. Otherwise, $ \minus{}5\leq 8p\minus{}13\leq 0\leq 8q\minus{}13\leq 3$ implying the result.


I really doubt that this is true. Observe that (n!!)^1/2000 is bigger than n^k for any k if n is large enough.

Well a friend of mine told me that this is a problem which was given in a Romanian olypiad in Galati, 9th grade and the solution in that book goes like this:

[b]Solution[/b]

Well you can see that:

((2p)!!)^(1/2000)=(2*4*..*(2p)*1*..*1)^(1/2000)<1+p^2/2000 (GM<=AM)

give values to p from 1 to n and get the required result!

but the solution seems to be false as it is not true for any n large enough as [b]harazi[/b] also indicated that:

[quote]"to apply AM-GM we the number of variables must be the same with the inverse of the exponent of their product.Thus, if n is very big, we cannot apply AM-GM to establish that ((2p)!!)^1/2000<1+p^2/2000." [/quote]

cheers! :D

AS I had already told, you can apply AM-GM if and only if p<=2000. But if n>200000000000000 we cannot be sure that p<=2000 for all p from 1 to n.

I guess that this ineq is true only for n<=1000.


limit
logarithms
integration
inequalities
real analysis
real analysis unsolved

Find the limit
$ \lim_{n\to\infty}{(1\plus{}\frac{1}{n^2})(1\plus{}\frac{2}{n^2})\cdots(1\plus{}\frac{n}{n^2})}$

$ \prod^{n}_{i \equal{} 1}\left(1 \plus{} \frac {i}{n^2}\right) \equal{} \frac {(n^2 \plus{} n)!}{(n^2)! n^{2n}} \approx \frac {\sqrt {2 \pi(n^2 \plus{} n)}\frac {(n^2 \plus{} n)^{n^2 \plus{} n}}{e^{n^2 \plus{} n}}}{n \sqrt {2 \pi}\frac {n^{2n^2}}{e^{n^2}} n^{2n}}\approx$

$ \approx \frac {(n^2 \plus{} n)^{n^2 \plus{} n}e^{ \minus{} n}}{n^{2n^2 \plus{} 2n}} \equal{} \left(1 \plus{} \frac {1}{n}\right)^{n^2 \plus{} n}e^{ \minus{} n}$

$ \lim_{n \rightarrow \infty}\left(1 \plus{} \frac {1}{n}\right)^{n^2 \plus{} n}e^{ \minus{} n} \equal{} e^{\lim_{n \rightarrow \infty} \left((n^2\plus{}n)\ln\left(1\plus{}\frac{1}{n}\right)\minus{}n\right)}$

$ \lim_{n \rightarrow \infty} \left((n^2\plus{}n)\ln\left(1\plus{}\frac{1}{n}\right)\minus{}n\right)\equal{}$

$ \equal{}\lim_{n \rightarrow 0} \frac{1}{n}\left(\left(1\plus{}\frac{1}{n}\right)\ln(1\plus{}n)\minus{}1\right)\equal{}^{l'Hopital}$

$ \equal{}\lim_{n \rightarrow 0} \frac{n\minus{}\ln(n\plus{}1)}{n^2}\equal{}^{l'Hopital}$

$ \equal{}\lim_{n \rightarrow 0} \frac{1}{2\plus{}2n}\equal{}\frac{1}{2}$

Therfore:

$ \lim_{n \rightarrow \infty} \prod^{n}_{i \equal{} 1}\left(1 \plus{} \frac {i}{n^2}\right) \equal{} \sqrt {e}$

An alternate computation. This may be a little simpler.

Let $ L$ be the limit to be computed, and take the logarithm.

$ \ln L\equal{}\lim_{n\to\infty}\sum_{k\equal{}1}^n\ln\left(1\plus{}\frac{k}{n^2}\right)\equal{}
\lim_{n\to\infty}\sum_{k\equal{}1}^b\left(\frac{k}{n^2}\minus{}r(k,n)\right)$

where $ 0<r(k,n)<\frac{k^2}{2n^4}.$

On the one hand,

$ \lim_{n\to\infty}\sum_{k\equal{}1}^nr(k,n)\le\lim_{n\to\infty}\sum_{k\equal{}1}^{n}
\frac{k^2}{2n^4}\le\lim_{n\to\infty}\sum_{k\equal{}1}^{n}\frac{n^2}{2n^4}
\equal{}\lim_{n\to\infty}\frac{1}{2n}\equal{}0.$

On the other hand,

$ \lim_{n\to\infty}\sum_{k\equal{}1}^n\frac{k}{n^2}\equal{}\lim_{n\to\infty}\frac{1}{n}
\sum_{k\equal{}1}^n\frac{k}{n}\equal{}\int_0^1x\,dx\equal{}\frac12.$

Hence, $ \ln L\equal{}\frac12$ from which $ L\equal{}\sqrt{e}.$

We can use the inequality $ x\minus{}\frac{x^2}{2}<\ln (1\plus{}x)<x\ (x>0).$ :lol:

kunny, when I said that $ 0<r(k,n)<\frac{k^2}{2n^4},$ I was using precisely that inequality.

Look here for in romanian forum for another solution at this problem!
http://www.mateforum.ro/viewtopic.php?t=808

On that link Cezar Lupu submitted two solutions; I assume that soruz is referring to the second one. The first one is essentially the same as mine with these small differences: he used the explicit formulas for $ \sum_{k\equal{}1}^nk$ and $ \sum_{k\equal{}1}^nk^2,$ while I used a crude estimate for the second and turned the first into a Riemann sum.


Is it true that ;
$ {(2a\plus{}1)}^{n}\equiv {r}^{n} mod({2}^{n})\Rightarrow 2a\plus{}1\equiv rmod({2}^{n})$

No? Take $ a\equal{}r\equal{}1$ and $ n\equal{}2$.


a) Construct a continuous function f : R to R which takes each of its values exactly three times.

b) Construct a function f : R to R which is continuous on R/Q and discontinuous
on Q.

b) $ f(x)= \begin{cases}1/q ~\text{for}~x=p/q \\ 0 ~\text{for}~x\in \mathbb{R}\setminus \mathbb{Q}\cup \{0\}\\ \end{cases}$

Define $ g(x)=\begin{cases}-x,&x\in[0,1]\\2x-3,&x\in[1,2]\end{cases}.$

Then for $ 2n\le x<2n+2,$ let $ f(x)=g(x-2n)+n.$

The result should be a piecewise linear continuous function with nodes at

$ \dots,(0,0),(1,-1),(2,1),(3,0),(4,2),(5,1),(6,3),(7,4),\dots$

I'd suggest sketching the graph of $ f$ to understand why it satisfies the conditions for (a).
If the requirement is for a function to take on each value a specific odd number of times, you should be able to satisfy it with a similar construction.


How would you find the number of 0's at the end of (100!)(200!)(300!)

Example:

Suppose I wanted to find the number of zeroes at the end of 25!. Remember that to produce a 0, i.e. a 10, we need a 5 and a 2. There are plenty of 2's in the factorization of 25!, more than 5's, so we just need to count the number of 5's that occur.

We have $ 25 \div 5 \equal{} 5$ single 5's and $ 25 \div 5^2 \equal{} 1$ double 5, giving us a total of $ 6$ 5's or 0's.

See if you can do the same for 100!, 200!, and 300!, but remember that you'll also have to find the number of triple 5's by dividing by 125 as well in 200! and 300!.

Helpful book to learn this
Intro the Number Theory :)

Just because nobody has posted the solution. I think I'll do so.
I think the easiest way to count the 0's is to do the factorials separately.
100! has 24 zeros at the end.
200! has 49 zeros at the end.
300! has 74 zeros at the end.
Add 24,49, and 74 up, and you get 147 zeros at the end of (100!)(200!)(300!).

Actually, izzy already gave him/her hints for which he can solve this easily.

[quote="isabella2296"]Example:
We have $ 25 \div 5 \equal{} 5$ single 5's and $ 25 \div 5^2 \equal{} 1$ double 5
[/quote]

How 25! has a double 5. Double 5 is 55 which is bigger than 25.

I also don't see how 100! has 24 zeros.

When I count 100!
I see 11 zeros from: 10, 20, ...., 90, 100
plus 10 zeros from 10 fives: 5, 15, ... , 95
so total is 21. not 24.
Did I miss something?

55 should count as one five, right?
51, 52, ..., 59 should not be counted, right?

By double 5 she means $ 5^2$. You have to count that twice since it has 2 5's in it's prime factorization.
That is why you only got 21 because:
$ \frac{100}{5^1}\equal{} 20$

$ \frac{100}{5^2} \equal{} 4$

hence $ 20\plus{}4 \equal{} 24$


Find the largest positive integer $ n$ with the following property: we can choose $ 2007$ distinct positive integers from $ [2.10^{n \minus{} 1},10^{n})$ such that for any two $ 1\leq i < j\leq n$, there exists a positive integer $ \overline{a_{1}a_{2}...a_{n}}$ from the chosen for which $ a_{j}\geq a_{i} \plus{} 2.$


So...um...I misplaced my information sheets when I was cleaning my room... :maybe:
Could someone tell me how we're getting from the airport to camp? (and anything else important I need to know)

Thanks

What information sheets?

Anyways...I think there will be people meeting you at the airport and sending you to the campus

The packet that they sent us with all the info and forms and stuff.

So we don't need money for a taxi/shuttle or anything?

"Students arriving by air will be greeted at the airport at the time indicated on the Flight Information Form by AMSP staff who will arrange their transportation to the UTD campus."

Last year, we got off the plane and met the AMSP staff. I believe that it was at the baggage claim. They weren't hard to find. Once a group of kids got together, they sent us off on a shuttle to the campus (we didn't have to pay for the shuttle of course).

When I lost my packing list, I got a friend (1=2) to send it to type it out on IM (why he didn't just e-mail it, I'll never know).  If you contact someone (else because I'm not sure I have those papers either), I'm sure that they could email you a copy.  I'm good at losing things, so I hope that helps.

I emailed Dr. Andreescu... and he replied the following day (but I emailed at like 11 the night before so that might be why xD)


Hi all,

How does the whole "category block" thing work. It seems really cool, but I don't get how people get it. Thanks!

In your blog options, you need to activate it.

$ \text{Profile} \implies \text{Blog Options} \implies \text{Blog Layout} \implies \text{Show Categories}$: click on the bullet for $ \text{Yes}$. Then, on a post, you can edit from the options below where you write what you put in the post to put the post in a certain category.

Thanks a lot :) .
Now what about chnaging your style of blog so it doesn't look super boring?

[quote="pinkmuskrat"]Now what about changing your style of blog so it doesn't look super boring?[/quote]I'm not sure if you are addressing this to the previous poster or to AoPS as a whole. If to AoPS as a whole, feel free to provide suggestions on what you think the style should look like.

Go to:

Profile -> Blog Options -> Blog Customization

Scroll down to Custom CSS and input your own code.  :)

Okay, I did not mean "boring" in that sense. It actually, in my opinion, is pretty elegant. I was just talking about making it different from the mainstream AoPS blogs.

Right, and [url=http://www.w3schools.com/css/]CSS[/url] can make it like that.


given triangle $ABC$ and let $D$ and $E$ be any points on $BC$ with $E$ closer to $C$. A circle  through A cuts the line segments $AB, \ AC,\ AD,\ AE$ at points$P,\ Q,\ R,\ S$ respectively.prove that :
\[\frac{AP.AB-AR.AD}{AS.AE-AQ.AC}=\frac{BD}{CE}\]
i have only a trignometrical proof.

[quote="anuj kumar"][color=darkred]Given triangle $ABC$ and let $D$ and $E$ be any points on $[BC]$ with the point $E$ closer to the point $C$. A circle $w\ ,\ A\in w$ cuts the second time the lines $AB, \ AC,\ AD,\ AE$ in the points $P,\ Q,\ R,\ S$ respectively. Prove that exists the relation $\left|\frac{AP.AB-AR.AD}{AS.AE-AQ.AC}\right|=\frac{BD}{CE}$. [size=117][b][u]Nice problem ![/u][/b][/size][/color][/quote]
[color=darkblue][b]Lemma.[/b] For the triangle $ABC$ and a circle $w=C(O,R)$ , $A\in w$ define

the point $T\in AA\cap BC$ and the second intersections $X$, $Y$ of the lines $AB$, $AC$

respectively with the circle $w$. Then exists the relation $\boxed{\frac{TB}{TC}=\frac{AB}{AC}\cdot\frac{AX}{AY}}\ \ (*)$.
[u]Particular case.[/u] If $X: =B$ and $Y: =C$ obtain the well-known relation $\frac{TB}{TC}=\left(\frac{AB}{AC}\right)^{2}$.

[b]Proof.[/b] $\left\{\begin{array}{c}\mathrm{Sinus'\ theorem\ : \ }AX=2R\cdot\sin\widehat{TAB}\ ,\ AY=2R\cdot\sin\widehat{TAC}\\\ \mathrm{useful\ well-known\ property\ : \ }\frac{TB}{TC}=\frac{AB}{AC}\cdot\frac{\sin\widehat{TAB}}{\sin\widehat{TAC}}\end{array}\right\|$  $\Longrightarrow (*)$.

[b]Proof of the proposed problem.[/b]
Apply the lemma $(*)$ to the point $T\in AA\cap BC$ and the triangles $\left\{\begin{array}{c}ABD\\\ ACE\\\ ADE\end{array}\right\|$ $\Longrightarrow$
$\left\{\begin{array}{cccc}(*) & \frac{TB}{TD}=\frac{AB}{AD}\cdot \frac{AP}{AR}& \Longrightarrow & \frac{TD}{AD\cdot AR}=\frac{BD}{|AB\cdot AP-AD\cdot AR|}\\\\ (*) & \frac{TE}{TC}=\frac{AE}{AC}\cdot\frac{AS}{AQ}& \Longrightarrow & \frac{TE}{AE\cdot AS}=\frac{CE}{|AC\cdot AQ-AE\cdot AS|}\end{array}\right\|\ \ (\ : \ \downarrow\ )$ $\Longrightarrow$

$\left\{\begin{array}{c}\frac{TD}{TE}\cdot\frac{AE}{AD}\cdot \frac{AS}{AR}=\frac{BD}{CE}\cdot\left|\frac{AC\cdot AQ-AE\cdot AS}{AB\cdot AP-AD\cdot AR}\right|\\\\ (*)\ \ \ \frac{TD}{TE}=\frac{AD}{AE}\cdot \frac{AR}{AS}\end{array}\right\|$ $\Longrightarrow$

$\boxed{\frac{BD}{CE}=\left|\frac{AB\cdot AP-AD\cdot AR}{AC\cdot AQ-AE\cdot AS}\right|}$.[/color]

thank you very much ,sir virgil nicula,you gave me the proof i would be interested in , a nice and short one now i present my proof for this problem:
the proof relies on the following lemma:
lemma:in a triangle$ABD$ let any line$l$ cuts the sides $AB$ and $AD$ at $P$ and $R$ respectively.also let angle$APR=\alpha$ and angle$ARP=\beta$.
      then $|AB.AP-AR.AD=\frac{PR.BD.sin(\alpha-D)}{sinA}|=|\frac{PR.BD.sin(\beta-B)}{sinA}|$.
proof:we have
\[AP=\frac{PR.sin\beta}{sinA}\ and\ AB=\frac{BD.sinD}{sinA}\]
$\implies\ AP.AB=\frac{sin\beta.sinD.PR.BD}{sin^{2}A}$
       similarily we have $AR=\frac{PR.sin\alpha}{sinA}$ and $AD=\frac{BD.sinB}{sinA}$
       $\implies AR.AD=\frac{BD.PR.sin\alpha.sinB}{sin^{2}A}$
       now observe that $\alpha =\pi-(\beta+A) \ and\ B=\pi-(D+A)$
\[AP.AB-AR.AD=\frac{PR.BD}{sin^{2}A}(sin\beta.sinD-sin(D+A).sin(\beta+A)) \]

\[=\frac{PR.BD}{sin^{2}A}(sin\beta.sinD-(sinD.cosA+cosDsinA)(sin\beta.cosA+sinA.cos\beta)) \]

\[=\frac{PR.BD}{sin^{2}A}(sin\beta.sinD-(cos^{2}A.sinD.sin\beta+sinAcosA(sin\beta.cosD+cos\beta.sinD)+sin^{2}A.cos\beta.cosD) \]
)
\[=\frac{PR.BD}{sin^{2}A}(sin\beta.sinD-cos^{2}A.sinD.sin\beta-sinA.cosA.sin(\beta+D)-sin^{2}A.cos\beta.cosD) \]

\[=-\frac{PR.BD}{sin^{2}A}sinA.sin(A+D+\beta) \]

\[=|\frac{PR.BD.sin(\beta-B)}{sinA}| \]
now for the prsent problem with notations same as in given question,we apply it like this:
    firstly join PR,RS,SQ,RQ,PS and PQ.
    applying above result :
\[|AP.AB-AR.AD|=|\frac{PR.BD.sin(ADB-APR)}{sinBAD}| \]
similarily $|AQ.AC-AS.AE|=|\frac{SQ.CE.sin(AQS-AEC)}{sinCAE}|$
    now
\[angleCAE=angleQRS\]  and \[angleBAD=anglePSR \]
(by given cyclicity)
    in  triangle  $PSR$, we  have
\[sinPSR/PR=sinRPS/RS........1 \]
and in triangle$QRS$,we have
\[sinQRS/QS=sinRQS/RS.........2 \]
since       angle$RPS$=angle$RQS$        therefore we have
\[sinPSR/PR=sinQRS/QS...........3 \]
combining 1,2,3 we have:
\[|\frac{AB.AP-AR.AD}{AS.AE-AQ.AC}|=|\frac{PR.BD.sinQRS.sin(ADB-APR)}{QS.CE.sinPSR.sin(AEC-AQS)}|=|\frac{BD.sin(ADB-APR)}{CE.sin(AEC-AQS)}| \]
now we have to prove $|sin(ADB-APR)|=|sin(AEC-AQS)|$................4
    but angle$ADB$=$\pi$-angle$ADE$ and angle$APR$=$\pi$-angle$ASR$=angle$RSE$
    so we have |angle$ADB$-angle$APR$|=angle$RSE$+angle$ADE$........5
    similarily |angle$AEC$-angle$AQS$|=angle$SRD$+angle$SED$........6
    using 5 and 6 and observing that
\[angleRSE+angleADE=2\pi-(angleSRD+angleSED) \]
,4 becomes evident.
   thank you.

[quote="anuj kumar"][color=darkred]Given triangle $ABC$ and let $D$ and $E$ be any points on $[BC]$ with the point $E$ closer to the point $C$. A circle $w\ ,\ A\in w$ cuts the second time the lines $AB, \ AC,\ AD,\ AE$ in the points $B',\ C',\ D',\ E'$ respectively. Prove that exists the relation $\left|\frac{AB'.AB-AD'.AD}{AE'.AE-AC'.AC}\right|=\frac{BD}{CE}$. [size=117][b][u]Nice problem ![/u][/b][/size][/color][/quote]

[color=indigo][b]Proof.[/b] Denote: $B_{1},\,C_{1},\,D_{1},\,E_{1}$ be the projections of $O$ on $AB,\,AC,\,AD,\,AE$ respectively.
We have: $AB\cdot AB' =\overline{AB}\cdot\overline{AB'}=2\overline{AB_{1}}\cdot\overline{AB}=2\overrightarrow{AO}\cdot\overrightarrow{AB}$

Similar: $AD\cdot AD'=2\overrightarrow{AO}\cdot\overrightarrow{AD}$
$\Longrightarrow \left| AB\cdot AB'-AD\cdot AD'\right|=2\left|\overrightarrow{AO}\cdot\overrightarrow{DB}\right|$

Similar: $\left| AE\cdot AE'-AC\cdot AC'\right|=2\left|\overrightarrow{AO}\cdot\overrightarrow{CE}\right|$

Hence: $\left|\frac{AB\cdot AB'-AD\cdot AD'}{AE\cdot AE'-AC\cdot AE'}\right|=\frac{AO\cdot DB\cdot\cos\left(\overrightarrow{AO},\,\overrightarrow{DB}\right)}{AO\cdot CE\cdot\cos\left(\overrightarrow{AO},\,\overrightarrow{CE}\right)}=\frac{BD}{CE}$[/color]


geometry
inequalities
trigonometry
trig identities
Law of Sines
inequalities proposed

let $a,b,c$ be the sides of a given triangle.  let $P=a+b+c$, and $A$ be the area of the triangle.
prove the strict inequality:

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}<\frac{P}{2A}$

this is the same that proves that $sinA+sinB+sinC > sinA.sinB+sinA.sinC+sinC.sinB$
(by the law of sines and the formula $A = abc/4R$).

but this is obvious, because $sinx \leq 1$.


If the operation $ Z$ is defined as $ a Z b \equal{} b \plus{} 10a \minus{} a^2$, what is the value of $ 2Z6$?

[hide=solution]$They$ $literally$ $tell$ $you$ $the$ $steps$, $so$ $you$ $just$ $need$ $to$ $do$ $them.$ $When$ $plugged$ $in$, $the$ $equation$ $is$ $simplified$ $to$ $6+20-4$ $which$ $turns$ $into$ $26-4$ $which$ $is$ $22$.[/hide]


A white solid $ X$ was found by chemical tests to be an aromatic carboxylic acid of molecular of molecular formula $ \ce{C8H8O2}$. To determine its structural formula, a $ 0.7311g$ sample was ignited in excess oxygen in a bomb calorimeter. The reaction compartment was surrounded by $ 611.8g$ of water initially at $ 20.519 ^{\circ}C$, and the heat capacity of the calorimeter (excluding water) was $ 52.00JK^{\minus{}1}$. After the combustion was complete, the temperature of the water was measured to be $ 28.454^{\circ}C$.

Given:

$ \Delta_{f}H^{\circ}( \ce{H2O_{(l)}}) \equal{} \minus{}285.83 kJmol^{\minus{}1}$

$ \Delta_{f}H^{\circ}(\ce{CO2_{(g)}}) \equal{} \minus{}393.51 kJmol^{\minus{}1}$

$ \Delta_{f}H^{\circ}(ortho) \equal{} \minus{}417 kJmol^{\minus{}1}$

$ \Delta_{f}H^{\circ}(meta) \equal{} \minus{}426 kJmol^{\minus{}1}$

$ \Delta_{f}H^{\circ}(para) \equal{} \minus{}429kJmol^{\minus{}1}$

Is the original chemical ortho, meta or para?

X can also be a monosubstituted benzene derivative. Why is that option excluded? And don't we also need the specific heat of water?

Specific Heat of Water is quite obvious though. Anyway, that's the question, don't ask me why :D .


An integer is reduced to 1/9 of its original value when a certain one of its digits is delected,
and the resulting number is again divisible by 9.
1)prove that the division of this resulting number by 9 results in deleting an additional digit.
2) find all the integers having the above mentioned property.

Suppose n has this property. Since n/9 is divisible by 9 we get that the digit we have deleted is 9 or 0.
case 1) n=A9B where A and B are some numbers so A9B = 9.AB But it is not hard to show that A9B>10.AB>9.AB. So in this case there are no solutions.
case 2) n=A0B. A0B = 9.AB which simplifies to $A.10^{k}=8B$ where k is number of digits in B. Since $B<10^{k}$ the solutions are
$A=1$, $B=\frac{10^{k}}{8}$
$A=2, B=\frac{10^{k}}{4}$
$A=4, B=\frac{10^{k}}{2}$
so these are all solutions. Notice that removing the digit A leaves exactly$\frac{n}{9}$


Let

             $ a \equal{} \frac {kx}{x \plus{} y \plus{} z},b \equal{} \frac {ky}{x \plus{} y \plus{} z},c \equal{} \frac {kz}{x \plus{} y \plus{} z},k > 0$,

then

  $ (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \plus{} 3(2\sqrt 3 \minus{} 3)abc \plus{} 3\sqrt 3 \minus{} 3\sqrt 3(ab \plus{} bc \plus{} ca)$

  $ \equal{} \frac{\sum[3\sqrt {3}x \plus{} (k \plus{} \sqrt {3})(k \minus{} 2\sqrt3)^2(y \plus{} z)](x \minus{} y)(x \minus{} z)\plus{} 3\sqrt 3(2k \plus{} 3)(k \minus{} 3)^2xyz}{(x\plus{}y\plus{}z)^3}$

$ \ge0$.

Very nice inquality, Vasc. I like this one so much  :D Here is my ugly solution for this one.
The following expression was called $ f(a,b,c)$ which:
\[ f(a,b,c) \equal{} LHS \minus{} RHS \equal{} (a \plus{} b \plus{} c \minus{} 3\sqrt {3})(ab \plus{} bc \plus{} ca) \plus{} 3(2\sqrt {3} \minus{} 3)abc \plus{} 3\sqrt {3}
\]
The first we will show that:
\[ f(a,b,c)\ge \min \left \{ f\left(a,\frac {b \plus{} c}{2},\frac {b \plus{} c}{2}\right),\ f\left(a,b \plus{} c,0\right) \right\}
\]
In fact, it is very easy to see that:
\[ f(a,b,c) \minus{} f\left(a,\frac {b \plus{} c}{2},\frac {b \plus{} c}{2}\right) \equal{} (b \minus{} c)^2[(8 \minus{} 6\sqrt {3})a \minus{} b \minus{} c \plus{} 3\sqrt {3}]
\]
Factoring we can get:
\[ f(a,b,c) \minus{} f\left(a,b \plus{} c,0\right) \equal{} \minus{} bc[(8 \minus{} 6\sqrt {3})a \minus{} b \minus{} c \plus{} 3\sqrt {3}]
\]
It means that there are at least one of them is correct.
So we only need prove two cases. Let $ t \equal{} \frac {b \plus{} c}{2}$ we have:

$ \bigstar$ The first case: If $ 0\le t\le \frac {2\sqrt {3}}{9 \minus{} 4\sqrt {3}}$ we have:
\[ f(a,t,t) \equal{} 2ta^2 \minus{} (6\sqrt {3}t \minus{} 6\sqrt {3}t^2 \plus{} 4t^2)a \plus{} 2t^3 \minus{} 3\sqrt {3}t^2 \plus{} 3\sqrt {3} \equal{} f(a)
\]
Moreover:
\[ \Delta \equal{} 12t(t \minus{} 1)^2(9t \minus{} 4\sqrt {3}t \minus{} 2\sqrt {3}) < 0
\]
So $ f(a,t,t)\ge 0$. The inequality was proved.

$ \bigstar$ The second case:  If $ t\ge \frac {2\sqrt {3}}{9 \minus{} 4\sqrt {3}}$ we have:
\[ f(a,2t,0) \equal{} 4t^2a \minus{} (6\sqrt {3}a \minus{} 2a^2)t \plus{} 3\sqrt {3} \equal{} f(t)
\]
On the other hand:
\[ f'(t) \equal{} 2a(a \plus{} 4t \minus{} 3\sqrt {3})\ge 0
\]
So we have:
\[ f(t)\ge f\left(\frac {2\sqrt {3}}{9 \minus{} 4\sqrt {3}}\right) \equal{} (12\sqrt {3} \minus{} 16)x^2 \minus{} (92 \minus{} 48\sqrt {3})x \plus{} 129\sqrt {3} \minus{} 216 > 0
\]
Because we also have:
\[ \Delta \equal{} 88128\sqrt {3} \minus{} 153216 < 0
\]
We are done. Oh what is ugly solution  :P By the way, It is only the consequence of CID the theorem which was proved by hungkhtn, I think like that. Althought it is not nice proof but we can get a simple solution before some minutes to think and solve. I am also interested in  fjwxcsl's proof, not easy to think a solution like that  :wink:

[quote="fjwxcsl"]Let

             $ a \equal{} \frac {kx}{x \plus{} y \plus{} z},b \equal{} \frac {ky}{x \plus{} y \plus{} z},c \equal{} \frac {kz}{x \plus{} y \plus{} z},k > 0$,

then

  $ (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \plus{} 3(2\sqrt 3 \minus{} 3)abc \plus{} 3\sqrt 3 \minus{} 3\sqrt 3(ab \plus{} bc \plus{} ca)$

  $ \equal{} \frac {\sum[3\sqrt {3}x \plus{} (k \plus{} \sqrt {3})(k \minus{} 2\sqrt3)^2(y \plus{} z)](x \minus{} y)(x \minus{} z) \plus{} 3\sqrt 3(2k \plus{} 3)(k \minus{} 3)^2xyz}{(x \plus{} y \plus{} z)^3}$

$ \ge0$.[/quote]
Very nice your solution, fjwxcsl. :lol:
Indeed, for $ a\plus{}b\plus{}c\equal{}k$, we may write the inequality in the homogeneous form

$ (1\minus{}\frac {3\sqrt 3}{k})(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \plus{} 3(2\sqrt 3 \minus{} 3)abc \plus{} \frac{3\sqrt 3}{k^3}(a\plus{}b\plus{}c)^3 \ge 0$.

Then we use Schur's inequality $ (a\plus{}b\plus{}c)^3\ge \minus{}9abc\plus{}4(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)$ and, finally, $ (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca) \ge 9abc$.

Zaizai-hoang, why the first case is $ t\le \frac {2\sqrt 3}{9\minus{}4\sqrt 3}$ ?

[quote="zaizai-hoang"]Because of
\[ 0\le t\le \frac {2\sqrt {3}}{9 \minus{} 4\sqrt {3}} \to (9t \minus{} 4\sqrt {3}t \minus{} 2\sqrt {3}) < 0
\]
that all why I devided problem into 2 cases  :)[/quote]
I don't think so. :)
But, by AC-Theorem (from Algebraic Inequalities), it suffices to prove the original inequality for $ a\equal{}b\equal{}c$, for $ a\equal{}0$ and $ b\equal{}c$, and for $ a\equal{}b\equal{}0$.

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\plus{}3(2\sqrt 3\minus{}3)abc \plus{}3\sqrt 3 \ge 3\sqrt 3(ab\plus{}bc\plus{}ca)$.[/quote]
This inequality is linear of $a+b+c$.
Thus, by $uvw$ for the proof it enough to check only one case: $b=c$, which gives
$2ba^2+2((3\sqrt3-2)b^2-3\sqrt3b)a+2b^3-3\sqrt3b^2+3\sqrt3\geq0$.
Since $2b^3-3\sqrt3b^2+3\sqrt3=b^3+b^3+3\sqrt3-3\sqrt3b^2\geq3\sqrt[3]{b^6\cdot3\sqrt3}-3\sqrt3b^2=0$,
so for $(3\sqrt3-2)b-3\sqrt3\geq0$ our inequality is true.
Let $0\leq b<\frac{3\sqrt3}{3\sqrt3-2}$.
Hence, we need to prove that
$b^2\left((3\sqrt3-2)b-3\sqrt3\right)^2-2b\left(2b^3-3\sqrt3b^2+3\sqrt3\right)\leq0$, which is
$(b-1)^2\left((3\sqrt3-4)b-2\right)b\leq0$, which is true for $0\leq b<\frac{3\sqrt3}{3\sqrt3-2}$.


This problem is very easy.Used CAUchy:$a=b=c$.

Can you explain more clearly ??

I don't think it so easy

[quote="tkhtn"]Used $MINKOWSKY$.Yes.[/quote]
follow you.
but How to find the minimum of:$\sum_{cyc} \sqrt[4] {\frac{b^2}a}$??

i think you can prove this by cauchy(it's very simple)if you know"chon diem roi"


Anybody have any reading list recommendations to refresh everything for me to take the GRE Math subject exam?  The thing is I've been out of school getting rusty for about 6 yrs, and I'd prefer to not go over ALL my old textbooks.  I'd basically like to find a single Cliffsnotes for my entire undergrad studies, but I'll settle for individual subject refresher books that will make everything come flooding back.  I've ordered the "Cracking the math GRE" book, haven't gotten it yet, but suspect it won't be enough to bring everything back well enough for me to nail the test.  I took the practice test they have online, and managed my way through most of it, but I definitely need to bone up before going in for a real score.
So, who's got the shortest-but-100%-good-enough reading list for me? :)
Thanks all.

I think there is a physics Web site with tips about the math subject GRE that comes up pretty easily by a Google search and that has some study suggestions. Good luck.


I hope you can help me on this one.

Denote by $P$ the class of all halfopen intervals $\left[a,b\right)$ on the real line. Let $E$ be the $\sigma$-ring generated by $P$ and let $B$ be the Borel sets on the real line. Show that $E=B$.

I have managed to prove that $E \subset B$. If I then succeed to prove that the open subsets of the real line is contained in $E$, then it's clear that $B \subset E$ but I'm having trouble with this. I'm guessing that we need to prove that every open subset of the real line is the countable union of some intervals, but how do you do this in the easiest way?

Let $U$ be open. For each $x\in U$ let $I_{x}$ be the largest interval containing $x$ and contained in $U$. Specifically, $I_{x}$ is the union of all open intervals containing $x$ and contained in $U$. If $x\neq y$, either $I_{x}=I_{y}$ or $I_{x}\cap I_{y}=\emptyset$. Then the set $I=\{I_{x}\colon x\in X\}$ is countable, being a collection of disjoint open intervals. It is clear that $U=\bigcup_{J\in I}J$.

In topological terms, you are looking at the connected components here: the connected components of a topological space form a partition of that space. If $U\subset\mathbb{R}$ is open, its connected components are open intervals. And it is a fact that any collection of disjoint open intervals in the real line is countable.


A circle centered at $ A$ with a radius of 1 and a circle centered at $ B$ with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. The radius of the third circle is
[asy]
size(220);
real r1 = 1;
real r2 = 3;
real r = (r1*r2)/((sqrt(r1)+sqrt(r2))**2);
pair A=(0,r1), B=(2*sqrt(r1*r2),r2);
dot(A); dot(B);
draw( circle(A,r1) );
draw( circle(B,r2) );
draw( (-1.5,0)--(7.5,0) );
draw( A -- (A+dir(210)*r1) );
label("$1$", A -- (A+dir(210)*r1), N );
draw( B -- (B+r2*dir(330)) );
label("$4$", B -- (B+r2*dir(330)), N );
label("$A$",A,dir(330));
label("$B$",B, dir(140));

draw( circle( (2*sqrt(r1*r),r), r ));
[/asy]
$ \displaystyle \textbf{(A)} \ \frac {1}{3} \qquad \textbf{(B)} \ \frac {2}{5} \qquad \textbf{(C)} \ \frac {5}{12} \qquad \textbf{(D)} \ \frac {4}{9} \qquad \textbf{(E)} \ \frac {1}{2}$

Using the Pythagorean theorem and having r being the radius of the mutually tangent circle (with center P) to the two other circles and the line, we find the horizontal distance between the centers of circles A and B is 4.  Another way the horizontal distance can be found is by horizontal distances from A to P and from P to B.  This comes out to be sqrt[(1+r)^2 - (1-r)^2] + sqrt[(4+r)^2 - (4-r)^2].

sqrt[(1+r)^2 - (1-r)^2] + sqrt[(4+r)^2 - (4-r)^2] = 4
2(sqrt(r)) + 4(sqrt(r)) = 6(sqrt(r)) = 4
sqrt(r) = 2/3
r =[b] 4/9[/b].

Nice solution!
I'll rewrite it using variables $r_1, r_2$ for the radii of the circles

We have the circles:
$(A,r_1)$
$(B,r_2)$
$(P,r)$

Also, let $A',B',C'$ the points of contact between the three circles and their common external tangent

So, the horizontal distance between $A,B$ is $A'B'$,
while their vertical distance is $|r_1-r_2|$

$(A'B')^2 = AB^2 - |r_1-r_2|^2 = (r_1+r_2)^2 - (r_1-r_2)^2 = 4r_1r_2\Rightarrow$

$A'B' = 2\sqrt{r_1 r_2}$

Similarly

$A'P' = 2\sqrt{r r_1}$
$P'B' = 2\sqrt{r r_2}$

$A'B' = A'P'+P'B' \Rightarrow$

$2\sqrt{r_1 r_2} = 2\sqrt{r r_1} + 2\sqrt{r r_2} \Rightarrow$

$\sqrt{r_1 r_2} = \sqrt{r r_1} + \sqrt{r r_2} \Rightarrow$

$\sqrt{r_1 r_2} = \sqrt{r} (\sqrt{r_1} + \sqrt{r_2}) \Rightarrow$

$\sqrt{r} = \frac{\sqrt{r_1 r_2}}{\sqrt{r_1} + \sqrt{r_2}}$

or else,

$\frac{1}{\sqrt{r}} = \frac{1}{\sqrt{r_1}} + \frac{1}{\sqrt{r_2}}$

[img]http://www.artofproblemsolving.com/Forum/download/file.php?id=2996&t=1&[/img]
[code][img]http://www.artofproblemsolving.com/Forum/download/file.php?id=2996&t=1&[/img][/code]
Can a mod try putting that in the original post, please? That would help for the Contests Page, until we have an Asy diagram for that!

I just gave a talk on this two weeks ago o_O

This is a specific case of the Descartes Four Circles Theorem; see here:

http://en.wikipedia.org/wiki/Descartes'_theorem
http://www.math.siu.edu/kocik/symmetry/papers/wilker-4proofs.pdf
http://www.math.siu.edu/kocik/papers/Descartes-short.pdf
http://www.ams.org/samplings/feature-column/fcarc-kissing
http://mathworld.wolfram.com/DescartesCircleTheorem.html
http://www.artofproblemsolving.com/blog/34543

etc.

Although this problem already has an attached Geogebra diagram, an embedded Asymptote diagram is needed for the diagram to actually show up in the contest collections. Here it is:
[asy]
size(250);
real r1 = 1;
real r2 = 3;
real r = (r1*r2)/((sqrt(r1)+sqrt(r2))**2);
pair A=(0,r1), B=(2*sqrt(r1*r2),r2);
dot(A); dot(B);
draw( circle(A,r1) );
draw( circle(B,r2) );
draw( (-1.5,0)--(7.5,0) );
draw( A -- (A+dir(210)*r1) );
label("$1$", A -- (A+dir(210)*r1), N );
draw( B -- (B+r2*dir(330)) );
label("$4$", B -- (B+r2*dir(330)), N );
label("$A$",A,dir(330));
label("$B$",B, dir(140));

draw( circle( (2*sqrt(r1*r),r), r ));
[/asy]


Let $A_1A_2A_3$ be any plane triangle.
Prove that exists one and only one point $P$ such that $PA_i^2+PA_j^2+A_iA_j^2$ is constant.

I hope this is a well-known point  :D.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=5381 . Yes, the de Longchamps point is known, at least to me ;) .

  Darij

If I remember well, it's also IMOSL1995/G2!


[quote="MLT1421"]i want to know the members if there is any chance you know...[/quote]

No problem. Go to http://www.imo2007.edu.vn/index.php?module=ViewInforTableForm.php&act=viewresult&ct=GRE


geometry
geometric transformation
reflection
circumcircle
trigonometry
incenter
Pythagorean Theorem

[quote="stergiu"]To this problem I'd like all possible geometric(synthetic) solutions :) :

A triangle $ ABC$ has $ AB \equal{} AC$ and $ \angle A \equal{} 120$. If points $ D,E$ are on side $ BC$ and $ BD \equal{} DE \equal{} EC$, prove that the triangle $ ADE$ is equilateral.

 Babis[/quote]
[hide="Solution"]
Consider the midpoint of $ BC$ and call it $ M$.  Also, let the reflection of $ A$ across $ M$ be $ A'$.  Since $ \triangle ACB$ is isoceles, we have that $ AM\perp BC$ and $ \angle MAB \equal{} \angle MAC \equal{} \frac {\angle BAC}{2} \equal{} 60$.  Now, $ BM \equal{} \frac {BC}{2}$ and it is given that $ BD \equal{} \frac {BC}{3}$, so $ \frac {MB}{BD} \equal{} \frac {3}{2}$.  Also notice that $ AM \equal{} A'M$ because of the reflection, so $ D$ is the centroid of $ \triangle ABA'$.  Now, since $ A$ is on the perpindicular bisector of $ BC$, we see that the reflection of $ A$ across $ M$ is the same as the reflection of $ A$ across $ BC$, so $ \angle BA'M \equal{} \angle BAM \equal{} 60$, so $ \angle BA'A \equal{} 60 \equal{} \angle BAA'$, which implies that $ \triangle BA'A$ is equilateral.  Furthermore, $ D$ is the centroid of equilateral $ \triangle A'BA$, so $ D$ is the circumcenter of $ \triangle ABA'$.  Thus, $ \angle ADE \equal{} \angle ADM \equal{} \frac {\angle ADA'}{2}$, the last because of the fact that $ A$ and $ A'$ are symmetric about $ BC$.  Also, $ \frac {\angle ADA'}{2} \equal{} \angle ABA' \equal{} 60$, the first because $ D$ is the circumcenter of $ \triangle ABA'$ and the last because $ \triangle ABA'$ is equilateral.  It follows that $ \angle ADE\equal{}60$, and similarly, $ \angle AED \equal{} 60$, so $ \triangle AED$ is equilateral. [/hide]

[hide]
[asy]pair A,B,C,D,E,M;
A=(0,1);
B=(-sqrt(3),0);
C=-B;
D=(2*B+C)/3;
E=-D;
M=(0,0);
draw(D--A--B--C--A--E);
draw(A--M);
label("$A$",A,N);
label("$B$",B,dir(210));
label("$C$",C,dir(330));
label("$D$",D,S);
label("$E$",E,S);
label("$M$",M,S);
label("6",(A+B)/2,dir(120));
label("6",(A+C)/2,dir(60));[/asy]
WLOG let $ AB\equal{}AC\equal{}6$.  Thus $ BC\equal{}6\sqrt{3}\Rightarrow BD\equal{}DE\equal{}EC\equal{}2\sqrt{3}$.  Let $ M$ be a point on $ \overline{BC}$ such that $ \overline{AM}$ is a median.  Thus $ BM\equal{}3\sqrt{3}\Rightarrow DM\equal{}\sqrt{3}$.  Also, $ AM\equal{}3$.  By the Pythagorean Theorem, $ AD\equal{}\sqrt{AM^2\plus{}DM^2}\equal{}\sqrt{12}\equal{}2\sqrt{3}$.  Similarly, $ AE\equal{}2\sqrt{3}$.  Thus all three sides of $ \triangle ADE$ are of equal length, hence it is equilateral.
[/hide]

Let $ ABC$ be an equilateral triangle and $ E$, $ F$ be on $ BC$ so $ EB \equal{} BC \equal{} CF$. Then $ \angle EBA \equal{} 120$, $ EB \equal{} EA$, so $ \angle EAB \equal{} 30$ so $ \angle EAF \equal{} 120$. Now clearly there is only one nonsimilar diagram so yay we're done.

OR

Clearly, for $ BC$ fixed, $ D$, $ E$ are uniquely defined. So let the perpendicular bisectors of $ AB$, $ AC$ hit $ BC$ at $ D'$, $ E'$. Then $ BD'\equal{}D'A\equal{}CE'\equal{}E'A$ by symmetry. $ \angle D'AE'\equal{}120\minus{}30\minus{}30\equal{}60$ so $ E'AF'$ is equilateral, thus $ BE'\equal{}E'F'\equal{}F'C$ so $ E'\equiv E$ and $ F'\equiv F$

I tried to produce lines $ AD,AE$  to meet the circumcircle of $ \triangle ABC$ at $ F,H$ , but I failed to find a new solution. $ DEHF$ is inscribed , but no more. Do you think that we can continue ?.
  At the end , I will send the shortest solution , made by a friend , teacher of physics !
  Before I click , I made a metric solution , using the extended Pyth. theorem and the median theorem on triangles ABC(for side $ BC$ ) and $ ADE$ for median $ AE$. We take very easy $ AE \equal{} EC$ ect.

 babis

Let AD=AE=n, BD=DE=x, so AB=$ x\sqrt {3}$. By Stewart's on ABE with cevian AD, we get $ 3x^3 \plus{} n^2x \equal{} 2n^2x \plus{} 2x^3$ $ \implies$ n=x. Stewart's isn't the most elegant, but it forces the solution easily.

Or, by Law of cosines on ABD, we have $ n^2\equal{}3x^2\plus{}x^2\minus{}2\sqrt{3}x^2cos30 \implies$ n=x.

Elegant Solution:
Draw point F so FD=FE=DE and it's on the other side of DE. By angle chase, triangles ABF and ACF are equilateral and D is the incenter and hence the circumcenter of ABF. So AD=BD=DF.

This is similar to Altheman's solution's, but I don't like showing the converse is true and then forcefully try to prove it must work the other way. This solution avoids that.


Given that $abc=1$ for positive real numbers $a,b,c$ show that

$1+\frac{3}{a+b+c}\geq \frac{6}{ab+bc+ca}$

The following inequality is stronger:
$1+\frac{8}{a+b+c}\geq \frac{11}{ab+bc+ca}$

Would someone mind providing a proof for both then?

Thanks.

[quote="zanttrang"]Using Lagrange multipliers I believe you can show that the following inequality is true for positive reals $m \leq n$:

$1+\frac{m}{a+b+c}\geq \frac{n}{ab+bc+ac}+\frac{1}{3}(3+m-n)$

[hide="partial proof"]Proceed as follows:

$f(a,b,c) = 1+\frac{m}{a+b+c}-\frac{n}{ab+bc+ac}-\frac{1}{3}(3+m-n).$

$g(a,b,c) = abc = 1$.

$\bigtriangledown f = \left<-\frac{m}{(a+b+c)^{2}}+\frac{n(b+c)}{(ab+bc+ac)^{2}},-\frac{m}{(a+b+c)^{2}}+\frac{n(a+c)}{(ab+bc+ac)^{2}},-\frac{m}{(a+b+c)^{2}}+\frac{n(a+b)}{(ab+bc+ac)^{2}}\right>$

$\bigtriangledown g = <bc, ac, ab>$

Now we apply the Lagrange equation:

$\bigtriangledown f = \lambda \bigtriangledown g \Rightarrow$

$-\frac{m}{(a+b+c)^{2}}+\frac{n(b+c)}{(ab+bc+ac)^{2}}= bc \lambda$

$-\frac{m}{(a+b+c)^{2}}+\frac{n(a+c)}{(ab+bc+ac)^{2}}= ac \lambda$

$-\frac{m}{(a+b+c)^{2}}+\frac{n(a+b)}{(ab+bc+ac)^{2}}= ab \lambda$

$abc = 1.$

Subtracting the first equation from the second and simplifying gives:

$a = b \text{ or }(ab+bc+ac)^{2}= \frac{n}{c \lambda}.$

Similarly, subtracting the other two pairs of equations gives

$b = c \text{ or }(ab+bc+ac)^{2}= \frac{n}{a \lambda}$

$a = c \text{ or }(ab+bc+ac)^{2}= \frac{n}{b \lambda}.$

So we either have $a = b = c$ or a permutation of $a = b, (ab+bc+ac)^{2}= \frac{n}{a \lambda}= \frac{n}{b \lambda}.$

In the first case, we use $abc = 1$ to see that $a = b = c = 1$, which gives $f(1,1,1) = 0$.

I'm not sure how to deal with the second case, but I think this is a path to proving that inequality. [/hide][/quote]

I don't think the solution to my original post requires Lagrange Mutiplier. Still, I could be wrong.

[quote="zanttrang"][quote="BanishedTraitor"]
I don't think the solution to my original post requires Lagrange Mutiplier. Still, I could be wrong.[/quote]
I'm sure it doesn't, but I couldn't figure out any simpler way, and started playing around with Lagrange multipliers and came up with something cool  :lol:[/quote]

Well it is pretty cool lol.


If (4 × n) – 22 = 46, what is the value of n?

[hide="Answer"]$n=17$ Do it backwards. $(46+22) / 4=n$[/hide]


[hide]$a=5^3\cdot 3^5=30375$

since $3|a^3, 27|b^3$ so $9\cdot 27^m|a$ and for 5a it's the same. If a had other prime divisors, they must have an exponent $\alpha_i$ which is $6|\alpha_i$, which make the hypotesis false, so 3 and 5 are the only divisors.[/hide]

[hide]
We have $3 | b^3$ so $3^2 | a$ and similarly $5 | c^2$ implies $5 | a$.  If $p \neq 3,5$ divides $a$, then $p^6 | a$ which contradicts the $d = 1$ thingy.  Now, $a = 3^2\cdot 5^3 = 1125$.[/hide]

[quote="me@home"]I don't understand, is $d$ just another variable, with its value defined??[/quote]

the point of d is to have that there is no factor to the power of 6 or more, because we have that

$a=3^{p}*5^{q}$, where to be an integer, $p \equiv 2 \mod 3$ and $p\equiv 0 \mod 2$, or $p\equiv 2 \mod 6$, a similar argument follows for $q$, so having the d^6 does not divide $a$ limits us to p=2, similarly, we cannot have any other excess factors, because they would be to be to the power of 6n to be factors of a, keeping b and c integers, but that is a contradiction

[quote="kunny"]Let $a,b,c$ be positive numbers such that $3a=b^3,\ 5a=c^2.$ Assume that a positive integer is limited to $d=1$ such that $a$ is divisible by $d^6.$

(1) Prove that $a$ is divisible by $3$ and $5.$
(2) Prove that the prime factor of $a$ are limited to $3$ and $5.$
(3) Find $a.$[/quote]

I meant, he states in the problem that d is limited to 1. My question was, why does he define d and later use it as a variable?

The question could be better stated.  Replace the last sentence with, "The only positive integer $d$ such that $d^6|a$ is $d = 1$."  Does that clear it up for you?


If x-y=-6 and xy=6, then what is x^3-y^3?

Got to go but a start:

$ y\equal{}\frac{6}{x}$

$ x^{2} \plus{}6x\minus{}6\equal{}0$

$ x\equal{} \minus{}3 \pm \sqrt{15}$

$ y\equal{} 3 \pm \sqrt{15}$

[hide="Solution"]We know that $ (x \minus{} y)^3 \equal{} ( \minus{} 6)^3 \equal{} \minus{} 216$.  We can expand $ (x \minus{} y)^3$ and regroup it to get
\[ x^3 \minus{} 3x^2y \plus{} 3xy^2 \minus{} y^3 \equal{} (x^3 \minus{} y^3) \minus{} 3(x \minus{} y)(xy)\]
We are already given $ x \minus{} y$ and $ xy$, so we can substitute to get
\[ (x^3 \minus{} y^3) \minus{} 3(x \minus{} y)(xy) \equal{} (x^3 \minus{} y^3) \minus{} 3( \minus{} 6)(6) \equal{} (x^3 \minus{} y^3) \plus{} 108\]
Remember that this is all equal to $ \minus{} 216$.  Setting $ (x^3 \minus{} y^3) \plus{} 108$ equal to $ \minus{} 216$, we get that $ x^3 \minus{} y^3 \equal{} \minus{} 216 \minus{} 108 \equal{} \boxed{\minus{}324}$[/hide]

hello, solving the equation system
$ x\minus{}y\equal{}\minus{}6$
$ xy\equal{}6$ we get
$ x\equal{}\minus{}3\plus{}\sqrt{15},y\equal{}3\plus{}\sqrt{15}$
$ x\equal{}\minus{}3\minus{}\sqrt{15},y\equal{}3\minus{}\sqrt{15}$
now you can plugg these solutions in the term $ x^3\minus{}y^3$.
Sonnhard.


Given a convex quadrilateral $ ABCD$ such that $ OA\equal{}\frac{OB\cdot OD}{OC\plus{}OD}$, where $ O$ is the intersection of the diagonals. The circumcircle of $ ABC$ intersects line $ BD$ at the point $ Q$.  Prove that $ CQ$ is the bisector of $ \angle DCA$.

anyone??

This is as far as I got:

$ \Delta ACD \text{ is iscoceles }\Rightarrow CQ\text{ is the bisector of }\angle DCA$


Observe that the fraction $ \frac{1}{7}\equal{}0,142857$ is a pure periodical decimal with period $ 6\equal{}7\minus{}1$,and in one period one has $ 142\plus{}857\equal{}999$.For $ n\equal{}1,2,\dots$ find a sufficient and necessary
condition that the fraction $ \frac{1}{2n\plus{}1}$ has the same properties as above and find two such fractions other than $ \frac{1}{7}$.

Period for $ \frac{1}{m}, (m,10)\equal{}1)$ is T if and only if $ m|10^{T}\minus{}1$ and T is minimal number with these property.
If T=2n is even, then $ m|10^{n}\plus{}1$ for many m (it is true for odd prime m), therefore $ 10^{T}\minus{}1\equal{}A*10^{n}\plus{}B,A\plus{}B\equal{}10^{n}\minus{}1$.

 From above we can easily get,
A+1=(10^n+1)÷(2*n+1),then we have,
10^(n-1)+1≤A+1=(10^n+1)÷(2*n+1) then by subtraction we get, [10^(n-1)]*(9-2*n)≥2*n, i.e
9>2*n,then we must have n≤4.Thus,we have only n=3 satisfied such  conditions.


geometry
parallelogram
geometric transformation
reflection
ratio
analytic geometry
angle bisector

Let $ABCD$ be a parallelogram and let $K$ and $L$ be points on the segments $BC$ and $CD$, respectively, such that $BK\cdot AD=DL\cdot AB$. Let the lines $DK$ and $BL$ intersect at $P$. Show that $\measuredangle DAP=\measuredangle BAC$.

I'd say this is also a perfect setting for simple application of Ceva  :)

Let the lines BL and DK meet the lines AD and AB at U and V, respectively. Let the line AP meet the line BD at R. Since the lines AR, BU and DV concur at P, the Ceva theorem, applied to triangle ABD, yields $\frac{BR}{RD}\cdot\frac{DU}{UA}\cdot\frac{AV}{VB}=1$, where we use directed segments. Since $\frac{AV}{VB}=-\frac{AV}{BV}$ and $\frac{DU}{UA}=-\frac{DU}{AU}$, we can rewrite this as $\frac{BR}{RD}\cdot\frac{DU}{AU}\cdot\frac{AV}{BV}=1$.

Now, since CD || AB, we have $\frac{DU}{AU}=\frac{DL}{AB}$. Since BC || DA, we have $\frac{AV}{BV}=\frac{AD}{BK}$. Thus, we get $\frac{BR}{RD}\cdot\frac{DL}{AB}\cdot\frac{AD}{BK}=1$. In other words,

$\frac{BR}{RD}=\frac{AB}{DL}\cdot\frac{BK}{AD}=\frac{BK\cdot AB}{DL\cdot AD} = \frac{AB^2}{AD^2}\cdot\frac{BK\cdot AD}{DL\cdot AB}$.

Since $BK\cdot AD = DL\cdot AB$, this becomes

$\frac{BR}{RD}=\frac{AB^2}{AD^2}$.

Thus, the line AR is a symmedian in triangle ABD. On the other hand, since the diagonals of a parallelogram bisect each other, the line AC is a median in triangle ABD. Since the symmedian in a triangle is the reflection of the corresponding median in the corresponding angle bisector, we have < DAP = < BAC. Proof complete.

It's very well possible that there is a much simpler solution.

  Darij

Where Ceva works, usually, Menelaos do it too.

By Menelaos, DL/DC*KC/KB*PB/PL = 1. If Q=AP^CD, as ABP~QLP, PB/PL = AB/QL.  From hypothesis DL/BK = AD/AB. Combining all these relations we get QL/DL = KC/BK from which follows that QD/DL = BC/BK or QD/DA = DL/BK = AD/DB=CB/BA which implies that QDA~CBA and from this the thesis.

[quote="grobber"]Perfect setting for a homography :D.

The one we're interested in is the map $DK\mapsto BL$. Since this takes $DB$ to $BD$, it means that it's a perspectivity, so the locus of $BL\cap DK$ is a line. Now look at some particular situations to find this line (for example, consider the cases when $P$ is on $BC,DC$; notice that we don't need the points $K,L$ to lie on the segments as long as we choose directions on $BC,DC$ in a convenient way).[/quote]

Grobber I don't understand the idea underliing this proof. Could you, please, make it explicit and give more details, if you have time to do it, about why, for instance,  "since this takes $DB$ to $BD$, it means that it's a perspectivity, so the locus of $BL\cap DK$ is a line".

There are some theorems in projective geometry which I used but didn't prove. For example:

Assume you have a homography between two lines (a bijection preserving cross-ratios taking the points of a line $\ell_1$ to the points of the line $\ell_2$, including infinity points). If the intersection point of the two lines is mapped to itself, then there is a theorem stating that the lines formed by the pairs $A,f(A)$ pass through a fixed point.

In the solution I used the dual of the above: instead of a homography between two lines we have a homography between two pencils through two points $D,B$, and the line connecting these two points is mapped to itself. Then the dual of that theorem states that this is a perspectivity, i.e. the intersection points of the pairs $a,f(a)$ ($a$ is a line through $D$ and $f(a)$ is its image, passing through $B$) lies on a line. This intersection point was called $P$.

After we have determined this, all we need to do in order to determine the line is look at two particular positions of this intersection point, and I indicated two such positions in which it's easier to see what the line on which $P$ lies actually is.

I hope it sheds some light on what I wrote up there. :)

[quote="grobber"]
...

I hope it sheds some light on what I wrote up there. :)[/quote]

I have to study carefully this post: I'm not an expert so I don't have many theoretical knolodge, I'm only an enthusiast.

Anyway, to help me in undertstanding this case, could explain where and how you use the hypothesis that BK.AD=DL.AB.

Thank you very much.

Beleive me, I'm no expert either :).

I'll try to explain that part whih you asked me about.

Take $D,B$ as origin points on the lines $DC,BC$, and consider the positive directions on these lines to be $D\to C,\ B\to C$ respectively. We want to show that $K\mapsto L$ is a homography between the two lines, so that it will unduce a homography $DK\mapsto BL$ between the pencils of lines through $D,B$. In order to do this, all we need to do is show that the map is of the form $f(x)=\frac{ax+b}{cx+d}$, where $x$ is the coordinate of $L$ on the line $DC$ (remember that we have fixed an origin and a direction, so we can regard points on this line as coordinates), and $f(x)$ is the coordinate of $K$ on $BC$. From the condition given we get $f(x)=\frac{AB}{AD}\cdot x$, which is precisely a homography.

This problem is a nice application of the [b]Menelaus'[/b] theorem.

Let $AP \cap DC=Q$.From hypothesis we have $\frac{AB}{BK}=\frac{AD}{DL}$ and $\angle{ABK}=\angle{ADL} \Rightarrow \triangle{ABK} \sim \triangle{ADL} \Rightarrow \angle{BAK}=\angle{DAL}$.

Now applying the Menelaus' theorem wrt $\triangle{BLC}$ taking $DK$ as the transversal we have $\frac{DL}{LC} \times \frac{CK}{BK} \times \frac{BP}{PL}=1$.From the similar triangles $APB$ and $QPL$ we obtain $\frac{BP}{PL}=\frac{AB}{QL}$.Again from hypothesis we have $\frac{DL}{BK}=\frac{AD}{AB}$.Plugging in these relations we obtain $\frac{KC}{QL}=\frac{CD}{AD}=\frac{AB}{AD}=\frac{AK}{AL}$(the last step is from the fact that $\triangle{ADL}\sim \triangle{ABK}$).These together with $\angle{AKC}=\angle{ALQ}$ implies that $\triangle{ALQ} \sim \triangle{AKC}$.Thus $\angle{LAQ}=\angle{KAC}$ and the result follows.

Dear Mathlinkers,
another way is to known how to construct simply K and L wrt the relation, and then perhaps the problem is solved.
Sincerely
Jean-Louis

Dear Mathlinkers,
I confirm that the construction of the K and L, leads to an nice proof of this problem without calculation...I have to write my proof...
Sincerely
Jean-Louis

Dear Mathlinkers,
you can se my proof on
http://perso.orange.fr/jl.ayme  vol. 18 Regard 1, p. 3-5
Sincerely
Jean-Louis


algebra
polynomial
group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

Hi. I'm a newcomer to the world of Galois theory so I would like to have some guidance with my solution to the problem below. :)

I'm required to find the Galois group of the polynomial $ x^4-2$ over the field $ \mathbb Q.$

Since $ x^4-2=(x-\omega)(x+\omega)(x-i\omega)(x+i\omega),$ where ${ \omega=\sqrt{\sqrt2}}$ and $ i^2=-1,$ we see that a splitting field extension is $ \mathbb Q(\omega,i): \mathbb Q.$ The degree of the extension is $ 8$ as $ \{1,\omega,\omega^2,\omega^3,i,i\omega,i\omega^2,i\omega^3\}$ is a basis for $ \mathbb Q(\omega,i)$ over $ \mathbb Q.$ As $ x^4-2$ is irreducible in $ \mathbb Q,$ the polynomial is separable in $ \mathbb Q(\omega,i);$ hence its Galois group $ \Gamma$ has order $ 8$ and is isomorphic to a subgroup of $ S_4.$ Now $ |S_4|=2^3\cdot3$ and so any subgroup of order $ 8$ is a Sylow $ 2$-subgroup. Hence all subgroups of order $ 8$ of $ S_4$ are conjugate – and so $ \Gamma\cong D_4.$

Is that correct? :maybe:


ARML
probability
expected value
geometric sequence

This "paradox" was written one of the HCSSM's advertisements posted/handed out at the Penn State ARML Meet. Find the flawed step, and fix it  :)

[b]Problem:[/b] Suppose that one of two envelopes contains twice as much money as the other. You pick an envelope without knowing the value of its contents. If given the opportunity to switch, should you?

[b]Solution:[/b] If the envelope you've chosen has $ D$ dollars in it, the other is equally likely to have $ 2D$ or $ D/2$ dollars, so it has an expected value of $ \frac {(2D) \plus{} (D/2)}{2} \equal{} \frac {5}{4} D$, which is 25% more than the value of your current envelope.

So the best strategy is to switch (and keep switching)

[i]PS. If you know the answer right away, please don't post immediately[/i]  :wink:

"Solution"

[hide]There are only two envelopes.  Thus, we have D and the expected value of the other envelope, 5/4D.  That's it.  Switching more than once does not create an infinite spiral or geometric sequence or whatever of the values.  It's just D and 5/4D.  Thus, eliminate the words in parentheses, and we have the answer:  Yes, you should switch, but only once.[/hide]

But logically that doesnt make sense. The chance you picked the one with more money is 1/2. Switching doesn't change that. And although the expected value is 5/4d it still doesnt change if you were right or not.

[hide="Solution."]It's not equally likely to have both $ 2D$ and $ \frac {D}{2}$ dollars; in fact, the probability of one of those events is $ 1$ while that of the other is $ 0$.  On the other hand, if first you pick an envelope and [i]then[/i] a second one is added with either twice or half the amount of money in yours, it [i]is[/i] in your best interest to switch.[/hide]

[hide="My solution"]
You can't treat $ D$ as a constant in your equation because you're comparing different situations. You have to assume that $ D$ is the smaller amount in all cases and analyze switching and staying as seperate situations.

If $ D$ is the smaller amount, then the expected value when staying is $ \frac 12 (D \plus{} 2D) \equal{} \frac 32 D$.
Likewise, the expected value when switching is also $ \frac 12 (D \plus{} 2D) \equal{} \frac 32 D$. We have shown that they are the same, therefore, it doesn't matter whether you stay or switch.[/hide]

[hide="Solution"]Basically, the fault here is treating the amount of money in the first envelope as a constant.  It is variable, and instead the only constant is the total amount of money in both envelopes, say $ S$.  The expected value of switching is $ \frac{\frac{2S}{3}\plus{}\frac{S}{3}}{2}\equal{}\frac{S}{2}$, the same as staying.[/hide]

See http://en.wikipedia.org/wiki/Two_envelopes_problem .  The resolution of this paradox is open.


I'm kind of nervous about part I.  I always bomb it.  I remember looking through the scores thread last year tetrahedr0n.  What was your eventual placing?  I ended up barely making it into the top 50, and I'm doubtful I will make it there again this year.

Xuan, post in the Michigan forum. All sorts of MMPC stuff there.

I guess I'm not part of it cuz I don't see any threads in the Michigan forum.  Can someone add me?

I'm not a member of the Michigan forums yet, but I need to get this off my chest.  I GOT OWNED!!!!!  owie.  I got like maybe 15 or so for sure.  I am going to go cry in my corner now.

Altough we agreed not to talk about it for a day, it should theoretically be over by now. Still, let's not talk about the problems. I agree with xuan's comment - it was somewhat challenging.

PS - I ended with rank 2 last year.

I thought the test was somewhat challenging also. Do you guys have an idea of the cutoff will be this year? Also, is it safe now to start discussing the questions?

WOW!!

im bored.. tetrahedr0n, things are going to change this year :P :P  :ninja:

See http://www.delta.edu/math/mmpc or
cbecb,abcae,a(e)acda,cdada,aebae,edace,bbddc

i took it, and got my score last night, i got 29

I also took the MMPC and I checked my answers and I got a score of 34 with question 11 included.

I am not Dawson, since Dawson did not take the MMPC because he is living in China.

You might not be Dawson, but Dawson did take it.

[quote="sunil"]im bored.. tetrahedr0n, things are going to change this year   [/quote]

One thing isn't; I'm going to improve my final ranking.

Yeah, he's the one listed as being homeschooled: code of 999001.

Oh, O.K.

I was also wondering whether or not MMPC will post a score distribution on the test.

[quote="dasherm"]All I want to say is:

WHAT THE HECK HAPPENED WITH PROBLEM 11 ANSWERS:

    [b](e)[/b] @!#%  (b) #&%^ (c) ##*! (d) #^&@ [b](a)[/b] #*&*

??????!!!!!![/quote]

I noticed that too...


number theory
greatest common divisor
modular arithmetic

[quote="mathemonster"]$ (5^{25a} \plus{} 26, 5^{25a} \plus{} 5^{24a}) \implies (5^{24a} \minus{} 26, 5^{a} \plus{} 1)......(5^{2a} \plus{} 26, 5^{a} \plus{} 1) \equal{} (5^{2a} \plus{} 26, 5^{2a} \plus{} 5^{a}) \equal{} (x \minus{} 26, x \plus{} 1), \minus{} 26 \equiv 1\pmod{1,3,9,27}$ so if $ 3^{k\leq 3}|a \implies gcd \equal{} 3^k$ [quote="feuxfollets"][hide]
Subtract $ 5^a \plus{} 1$ from $ 5^{25a} \plus{} 26$ 26 times

We get $ 5^{25a} \minus{} 26 * 5^a$

So the GCD is $ 5^a$
[/hide][/quote]

can't subtract $ 5^a$, only can subtract $ 5^{a} \plus{} 1$ so wrong[/quote]

How did you get 5^25a+26,5^25a+5^24a?

[quote="mathemonster"]$ (5^{25a} \plus{} 26, 5^{25a} \plus{} 5^{24a}) \implies (5^{24a} \minus{} 26, 5^{a} \plus{} 1)......(5^{2a} \plus{} 26, 5^{a} \plus{} 1) \equal{} (5^{2a} \plus{} 26, 5^{2a} \plus{} 5^{a}) \equal{} (x \minus{} 26, x \plus{} 1), \minus{} 26 \equiv 1\pmod{1,3,9,27}$ so if $ 3^{k\leq 3}|a \implies gcd \equal{} 3^k$ [quote="feuxfollets"][hide]
Subtract $ 5^a \plus{} 1$ from $ 5^{25a} \plus{} 26$ 26 times

We get $ 5^{25a} \minus{} 26 * 5^a$

So the GCD is $ 5^a$
[/hide][/quote]

can't subtract $ 5^a$, only can subtract $ 5^{a} \plus{} 1$ so wrong[/quote]

\textbf{?!?!?}

$ (5^{25a} \plus{} 26) \minus{} 26(5^a \plus{} 1) \equal{} 5^{25a} \plus{} 26 \minus{} 26*5^a \minus{} 26\equal{} 5^{25a} \minus{} 26*5^a$

no one is subtracting $ 5^a$............

[quote]How did you get 5^25a+26,5^25a+5^24a?[/quote]
He's just multiplying the second expression by $ 5^{24a}$


Prove that, for any positive real numbers $a,b$ and $c$,
\[ \frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{12(a+b)(b+c)(c+a)}{(a+b+c)^3}\ge \frac{41}{9} \]

Nice inequality.

Here is just my draft in a few minutes:

We can normalize$a+b+c=3$. Set $w=ab+bc+ca$, $r=abc$, then we have $a^2+b^2+c^2=9-2w$, and
\[ (a+b)(b+c)(c+a)=(3-a)(3-b)(3-c)=(9-3(a+b)+ab)(3-c)=27-9(a+b+c)-abc+3(ab+bc+ca)=3w-r. \]
The desired inequality reads
\[ \frac{9-2w}{w}+\frac{12(3w-r)}{27}\geq\frac{41}9. \]
Equivalently,
\[ 27(9-2w)+12w(3w-r)\geq 123w. \]

[quote="ductrung"]Prove that, for any positive real numbers $ a,b$ and $ c$,
\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {12(a \plus{} b)(b \plus{} c)(c \plus{} a)}{(a \plus{} b \plus{} c)^3}\ge \frac {41}{9}
\]
[/quote]
Suppose that $ c \equal{} min\{a;b;c\}$
We can rewrite this inequality:

$ (a \minus{} b)^2[\frac {1}{ab \plus{} bc \plus{} ca} \minus{} \frac {32a \plus{} 32b \plus{} 20c}{9(a \plus{} b \plus{} c)^3}] \plus{} (a \minus{} c)(b \minus{} c)[\frac {1}{ab \plus{} bc \plus{} ca} \minus{} \frac {20a \plus{} 20b \plus{} 32c}{9(a \plus{} b \plus{} c)^3}] \geq 0$ :oops:

[quote="ductrung"]Prove that, for any positive real numbers $ a,b$ and $ c$,
\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {12(a \plus{} b)(b \plus{} c)(c \plus{} a)}{(a \plus{} b \plus{} c)^3}\ge \frac {41}{9}
\]
[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3.$
We obtain $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {12(a \plus{} b)(b \plus{} c)(c \plus{} a)}{(a \plus{} b \plus{} c)^3}\ge \frac {41}{9}\Leftrightarrow$
$ \Leftrightarrow\frac {3u^2 \minus{} 2v^2}{v^2} \plus{} \frac {4(9uv^2 \minus{} w^3)}{9u^3}\geq\frac {41}{9}\Leftrightarrow27u^5 \minus{} 59u^3v^2 \plus{} 36uv^4 \minus{} 4v^2w^3\geq0.$
But $ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2 \equal{}$
$ \equal{} \sum_{cyc}(a^4b^2 \plus{} a^4c^2 \minus{} 2a^3b^3 \minus{} 2a^4bc \plus{} 2a^3b^2c \plus{} 2a^3c^2b \minus{} 2a^2b^2c^2) \equal{}$
$ \equal{} (9u^2 \minus{} 6v^2)(9v^4 \minus{} 6uw^3) \minus{} 3w^6 \minus{} 2(27v^6 \minus{} 27uv^2w^3 \plus{} 3w^6) \minus{}$
$ \minus{} 2w^3(27u^3 \minus{} 27uv^2 \plus{} 3w^3) \plus{} 2w^3(9uv^2 \minus{} 3w^3) \minus{} 6w^6 \equal{}$
$ \equal{} 81u^2v^4 \minus{} 108u^3w^3 \plus{} 162uv^2w^3 \minus{} 108v^6 \minus{} 27w^6\geq0.$
Id est, $ w^6 \plus{} 2(2u^3 \minus{} 3uv^2)w^3 \minus{} 3u^2v^4 \plus{} 4v^6\leq0.$
Hence, $ w^3\leq3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}.$
Thus, it remains to prove that
$ 27u^5 \minus{} 59u^3v^2 \plus{} 36uv^4 \minus{} 4v^2\left(3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\right)\geq0.$
But $ 27u^5 \minus{} 59u^3v^2 \plus{} 36uv^4 \minus{} 4v^2\left(3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow(u^2 \minus{} v^2)\left(27u^3 \minus{} 24uv^2 \minus{} 8v^2\sqrt {u^2 \minus{} v^2}\right)\geq0,$ which easy to check out.
By the way, this reasoning gives  that the following inequality
\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {3k(a \plus{} b)(b \plus{} c)(c \plus{} a)}{(a \plus{} b \plus{} c)^3}\ge \frac {9 \plus{} 8k}{9}
\]
holds for all positive $ a,$ $ b$ and $ c$ and for all $ k\leq\frac {81\sqrt3}{32} \equal{} 4.384...$
and it's the best value for $ k$ because for $ k>\frac {81\sqrt3}{32}$ there is a problem around $ b\equal{}c.$ :wink:

How is your solution, Ductrung?

Thank you very much. :)

I have spent a lot of time on this inequality trying to find a solution without full expansion, but I failed.
 :(
Has anyone think it that possible?

[quote="pvthuan"]Nice inequality.

Here is just my draft in a few minutes:

We can normalize$ a \plus{} b \plus{} c \equal{} 3$. Set $ w \equal{} ab \plus{} bc \plus{} ca$, $ r \equal{} abc$, then we have $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 9 \minus{} 2w$, and
\[ (a \plus{} b)(b \plus{} c)(c \plus{} a) \equal{} (3 \minus{} a)(3 \minus{} b)(3 \minus{} c) \equal{} (9 \minus{} 3(a \plus{} b) \plus{} ab)(3 \minus{} c) \equal{} 27 \minus{} 9(a \plus{} b \plus{} c) \minus{} abc \plus{} 3(ab \plus{} bc \plus{} ca) \equal{} 3w \minus{} r.
\]
The desired inequality reads
\[ \frac {9 \minus{} 2w}{w} \plus{} \frac {12(3w \minus{} r)}{27}\geq\frac {41}9.
\]
Equivalently,
\[ 27(9 \minus{} 2w) \plus{} 12w(3w \minus{} r)\geq 123w.
\]
[/quote]

I cannot prove that $ 27(9 \minus{} 2w) \plus{} 12w(3w \minus{} r)\geq 123w$

Can anyone please help me?

Thank you very much. :oops:


inequalities
pigeonhole principle
inequalities unsolved

Prove that for all real numbers $a , b , c$  we have

  $\min \{(a-b)^2 , (b-c)^2 , ( c-a)^2 \}$  $\leq \frac{1}{2} (a^2 + b^2 + c^2 )$

Babis

[quote="stergiu"]Prove that for all real numbers $a , b , c$  we have

  $\min \{(a-b)^2 , (b-c)^2 , ( c-a)^2 \}$  $\leq \frac{1}{2} (a^2 + b^2 + c^2 )$[/quote]

Nice one!

By the pigeonhole principle, at least two of the three numbers b + c - 2a, c + a - 2b and a + b - 2c must have the same sign. Assume WLOG that the numbers c + a - 2b and a + b - 2c have the same sign. Then, $\left(c+a-2b\right)\left(a+b-2c\right)\geq 0$. Thus,

$\frac12\left(a^2+b^2+c^2\right)$
$=\left(b-c\right)^2+\frac16\underbrace{\left(a+b+c\right)^2}_{\text{trivially }\geq 0}+\frac13\underbrace{\left(c+a-2b\right)\left(a+b-2c\right)}_{\geq 0}\geq\left(b-c\right)^2$.

But trivially, $\left(b-c\right)^2\geq\min\left\{\left(b-c\right)^2;\;\left(c-a\right)^2;\;\left(a-b\right)^2\right\}$. Thus, $\frac12\left(a^2+b^2+c^2\right)\geq\min\left\{\left(b-c\right)^2;\;\left(c-a\right)^2;\;\left(a-b\right)^2\right\}$, and we are done.

  Darij


linear algebra
matrix
algebra
polynomial
algorithm
linear algebra unsolved

Hi guys,

Does anyone know if there's a straightforward necessary-and-sufficient condition for a real, square matrix to have purely real eigenvalues?  Or a necessary-and-sufficient condition for a univariate polynomial with real coefficients to have purely real roots?  Maybe this condition could be formulated in terms of resultants?

There is an [url=http://mathworld.wolfram.com/SturmFunction.html]algorithm[/url] that allows you to find the number of real roots of any polynomial with real coefficients. Other than that one can hardly say anything.

If your matrix is symmetric, I believe the spectral theorem states that it can be diagonalized to a real matrix.

Yes, being symmetric is a sufficient condition for real eigenvalues. But, as fedja indicated, it's a long way from being necessary.


Let $\ A=2004^{2005}-2005^{2004}-2052^{1991}.$ Prove that $ A>0 $ and  is  not a prime. Futhermore $ S(A) $  is  divisible by $ 24 $ where $ S(n) $ is the sum of all the divisors of $ n $.

Not like the past years, in 2005 the VN TST  has 2 rounds, the second round was hold to pick only 1 person +5 ones chosen in the first round=6 : VN team.There were 3 problems. Time allowed:240 mins

It's divisible by $7$:
$2004^{2005}-2005^{2004}-2052^{1991} \equiv 2^1-3^0-1^{\mathrm{blah}} \equiv 0 \mod 7$ by Fermat and $7|2002 , 6|2004$.

For the second part: $A$ is $\equiv -1 \mod 8$ and  $\equiv -1 \mod 3$, thus $A+1$ is divisible by $24$, so it reduces to http://www.mathlinks.ro/Forum/viewtopic.php?t=17815 .


calculus
integration
logarithms
real analysis
real analysis unsolved

[quote="borislav_mirchev"]1. $ \int (2x^{2} \plus{} 1)e^{x^{2}}\,dx$[/quote]

The answer is $ xe^{x^2}$. Very easy.

[quote="borislav_mirchev"]2. $ \int 2^{x}\sqrt {4^{x} \minus{} 1}\,dx$[/quote]

Observe that
\[ \left( {2^x } \right)^\prime \equal{} 2^x \ln 2
\]
and
\[ \displaystyle\int {\sqrt {x^2 \pm a } dx} \equal{} \frac {x}{2}\sqrt {x^2 \pm a } \pm \frac {a }{2}\ln \left| {x \plus{} \sqrt {x^2 \pm a } } \right| \plus{} C
\]
where $ a > 0$.

[quote="borislav_mirchev"]3. $ \int (1 \plus{} lnx)\sqrt {1 \plus{} (xlnx)^{2}}\,dx$[/quote]
\[ \int {\left( {1 \plus{} \ln x} \right)\sqrt {1 \plus{} \left( {x\ln x} \right)^2 } dx} \equal{} \int {\sqrt {1 \plus{} \left( {x\ln x} \right)^2 } d\left( {x\ln x} \right)} .
\]

1. - I know it is very easy. How did you solve it?
But I posted this - I believe it is not difficult, but no one didn't solve it.
http://www.mathlinks.ro/Forum/viewtopic.php?t=169021
the situation is same with this:
http://www.mathlinks.ro/Forum/viewtopic.php?t=138970
The second one is not calculus but vietnamese math olympiad problem.

[quote="borislav_mirchev"]... but no one didn't solve it....
.[/quote]
maybe no one didn't want to solve your standart integral which is as for homework

[quote="borislav_mirchev"]You may be right or may not be. But until the problem is not solved - no one may say he can solve it, except he is so good and solve it in mind. I'll try next time to invent an integral - no one can solve. But what is the sence of such problems?![/quote]

let's see  :rotfl:


Fie $ABC$ un triunghi isoscel astfel incat $m(BAC)=20^{0}$. Fie $M$ proiectia lui $C$ pe $AB$ si $N$ un punct pe $AC$ astfel incat $2CN=BC$. Aflati masurile unghiurilor tringhiului $AMN$.

pardon, am uitat sa zic ca este si isoscel.  :oops:
am corectat enuntul.

Si mie mi-au dat unghiurile  20, 60, 100 :) O sa postez solutia dar o mai verific putin ca era initial gresita :rotfl: - totusi dadea raspunsul bun :)

[quote="maky"]heh...problema asta am avut`o in concurs, intr`a 7`a (2002), la barajele pentru juniori
bineinteles ca atunci nu am facut`o, insa mai tin minte si acum ideea :)
fie $P$ mijlocul lui $BC$. se construieste $K$ astfel incat $MKP$ sa fie echilateral ($K$ in interiorul triunghiului).
de aici iese destul de usor
[hide="mi se pare ca raspunsul era"]
$\angle AMN=60^{\circ}$.
[/hide][/quote]

da , a fost in 2002 la juniori si multumesc pt solutie.

Alta idee: exprimam laturile triunghiului AMN in functie de BC si functii trigonometrice (pt a usura calculele consideram K mijlocul lui BC).

[quote="mychrom"]Fie $ABC$ un triunghi isoscel astfel incat $m(BAC)=20^{0}$. Fie $M$ proiectia lui $C$ pe $AB$ si $N$ un punct pe $AC$ astfel incat $2\cdot CN=BC$. Aflati masurile unghiurilor tringhiului $AMN$.[/quote][hide="Metoda 1."] [color=darkblue] Construim punctele :

- simetricul $P$ al punctului $B$ in raport cu punctul $M$ ;

- mijlocul $R$ al segmentului $CP$.

Se arata usor ca : $\{\begin{array}{c}RC=RM=RP=RN=CN\\\\ m(\widehat{CPN})=60^{\circ}\end{array}\Longrightarrow CMPN$- ciclic

(inscris in cercul cu centrul $R$ si raza $\frac{a}{2}$) $\Longrightarrow$  $\widehat{NMA}\equiv\widehat{NMP}\equiv\widehat{NCP}$.

In concluzie, $m(\widehat{NCP})=60^{\circ}\Longrightarrow \{\begin{array}{c}m(\widehat{NMA})=60^{\circ}\\\ m(\widehat{MNA})=100^{\circ}\end{array}$.[/color][/hide]


I am posting problems I made up b/c i think it fits level of some AMPers.

1. $ (x \plus{} y)^2(x\plus{}y\plus{}2)(x\plus{}y\minus{}2) \equal{} 1 \plus{} 16xy$ Find integer solutions for ordered pair (x,y).

2. Find the remainder when $ {kp \minus{} 1 \choose p}$ is divided by p for some positive integer $ k$ $ \geq$ $ 2$ and p is prime

3. If a[size=75]0[/size]=1, a[size=75]1[/size]=1, and a[size=75]n+2[/size]=a[size=75]n+1[/size]+2a[size=75]n[/size] for $ n$ $ \geq$ $ 0$, find explicit formula for a[size=75]n[/size].

4. If a[size=75]n+2[/size]=a[size=75]n+1[/size]+a[size=75]n[/size],
 find a[size=75]1[/size]+a[size=75]2[/size]+......+a[size=75]n[/size] in terms of 'a's.

Nice Problems

2.[hide] I assume that you mean $ p$ as a prime, otherwise I don't think there is a nice solution.

If $ p$ is prime, note that $ p|\binom{p}{k}$ where $ 1\le k\le p \minus{} 1$ (write it out in factorials...). Anyway,

$ (x \plus{} 1)^{kp \minus{} 1} \equal{} ((x \plus{} 1)^{p})^{k \minus{} 1}(x \plus{} 1)^{p \minus{} 1}\equiv (x^p \plus{} 1)^{k \minus{} 1}(x \plus{} 1)^{p \minus{} 1}\mod p$

Now look at the $ x^p$ terms on both sides.

On the LHS, it is $ \binom{kp \minus{} 1}{p}x^p$ and on the right it is $ \binom{k \minus{} 1}{1}x^p\cdot \binom{p \minus{} 1}{0}\cdot x^0$

Thus $ \binom{kp \minus{} 1}{p}\equiv k \minus{} 1\mod p$[/hide]


There's nothing to beat a good book for learning LaTeX (or mathematics for that matter) and I have recommended a couple at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=117360]LaTeX How-Tos[/url]. Now there's a new version of the book [url=http://tinyurl.com/2ljcv7]More Math into LaTeX[/url] due out later this year and written by an expert in the field. I mention this because a shorter free version is now available online.

This [url=http://www.cam.ctan.org/tex-archive/info/Math_into_LaTeX-4/Short_Course.pdf]Short Course[/url] is "[i]a short introduction for the beginner, dramatically reducing the steep learning curve of a few weeks to a few hours"[/i]. and will "[i]help you get started quickly with LaTeX, including detailed instructions on how to install LaTeX on a PC or a Mac;[/i]"

It may be called [i]Short[/i] Course but there are 180 pages so lots of information. Feedback here from those learning LaTeX would be interesting.

I should add that the online version can be freely distributed but is copyright. Another [url=http://www.cam.ctan.org/tex-archive/info/Math_into_LaTeX-4/Short_Course.pdf]link to the book[/url].

Wow, thank you very much, stevem  :wink:


hi
i had a math exam today and a question that seemed really easy actually lost me. i hope you can help.

QUESTION
hadija is paid £240 a week for her wages.
this is 20% more than last year.
Bill says ' this means that her weekly wage last year was £192'
Bill is wrong
explain why.

Work out what hadija's weekly wage was last year.

Please help its driving me mad and i've been thinking about it for the best part of today!
could you please lay out the way to solve a question like this so i can learn for future reference.
Thankyou

[quote="manutdfan1"]hi
QUESTION
hadija is paid £240 a week for her wages.
this is 20% more than last year.
Bill says ' this means that her weekly wage last year was £192'
Bill is wrong
explain why.

Work out what hadija's weekly wage was last year.[/quote]
Since it is 20% [i]more[/i], then...
[hide]
If x is her wage last year, then $1.20x=240$
$12x=2400$
$x=200$[/hide]
[hide="What Bill did"]Bill just took 80% of hadija's wage now and got 192 [/hide]


quadratics
number theory
prime numbers
algebra
Diophantine equation
number theory unsolved

1) Solve in primes $p+q=(p-q)^3$. (R. Jenodarov)

2) Find all pairs $(p,q)$ of prime numbers, for which $x^4+(q-2)x=p-4$ has at least one integer root. (E. Voronovich, Minsk)

3) Solve in primes $p^2+q=5q^2+p$

4) Prove that for every natural number $a>2$ there exist a prime number $p$, such that
$a^3+1$ is devisible by $p$, but $a+1$ is not. (Sh. Leman)

5) Find all primes $p$, for which the numbers $p+1$ and $p^2+1$ are of the form $2n^2$, where n is natural number.

6) Sovle in primes $p^2-p+1=q^3$. (A. Golovanov)

7) $a,b,c$ are natural numbers and $3c^2=c(a+b)+ab$. Prove that the number $8c+1$ is a square if $(a-b)$ is a prime number. (S. Berlov)

8) $a,b$ are natural numbers and $a^2+b^2$ is devisible by $ab-1$. Prove that $\frac{b^2+a^2}{ab-1}$ is a prime number. Show that $x^2-5xy+y^2+5=0$ has infinite many solutions in natural numbers.

9) $p$ is a prime number. Solve in integers $p(x+y)=xy$. (Spanish 1995)

10) $p$ is a prime number. Find all integers $k$, for which $\sqrt{k^2-pk}$ is natural number. (Spanish 1997)

11) $p>3$ is a prime number. Prove that $3^p-2^p-1$ is devisible by $42p$.

12) $p$ is an odd prime which can be written as a difference of two fifth powers.Show that $\sqrt{\frac{4p+1}{5}}=\frac{n^2+1}{2}$ for some odd integer $n$.
(Irish 2001)

[i]13) Do there exist a polinom $Q(x)$ with degree not smaller than 2 with natural coeficients, for which for every prime number $p$, $Q(p)$ is prime. (A. Kanel)
(This problem is most difficult and interesting (and it`s my favourite). If someone is interested in the solution i can post it.)[/i]

Some of the problems are very easy, but there are very hard problems here indeed. I believe that all of them are interesting. I didn`t solve all of them and that is why a post them in unsolved problems. If you have a problems of this type in your mind, please post them. Thank you.

1) let $p - q = r$. we have $p+q = r^3$, so, $2q = r(r+1)(r-1)$.

the rest is easy.

3) we get $q|p-1$. so, $p = qk+1$. now, we find $k(qk+1) = 5q-1$.

so, $k \leq 2$, and the rest is easy.

4) $gcd(a+1, a^2 - a + 1) = gcd(a+1, 3a) = 1$ or $3$. so, if the prime factors of $a+1$ and $a^3+1$ are the sames,

we have that $a+1 = 3$ (because $a+1$ and $a^2-a+1$ are powers of three, and $a+1$ isn't divisible by $9$)

the rest is easy.

5) suppose $p$ odd. let $p+1 = 2q^2$ and $p^2+1 = 2r^2$. we have $p|r^2-q^2 = (r+q)(r-q)$.

we know that $q < r < n$. so, $r+q = p$. the rest is easy. (we get one quadratic equation and find $p = 7$).

6) http://www.mathlinks.ro/Forum/viewtopic.php?p=224998#p224998

8) in http://www.mathlinks.ro/Forum/viewtopic.php?t=26232 is proved that $\frac{a^2+b^2}{ab-1} = 5$.

to prove the second part, use that $\Delta = 21y^2 - 20 = x^2$ and the pell equation $x^2 - 21y^2 = -20$ have infinite solutions (because have the first $(1,1)$)

9) suppose, WLOG, that $p|x$. let $x = px_1$.

from this, we get $(x_1 - 1)(y - p) = p$, and the rest is easy.

11) by fermat little theorem, we get easily that $p|3^p - 2^p - 1$.

we know that all primes $p>3$ are in the form $6k+1$ or $6k+5$

and that $ord_7(2) =3$, $ord_7(3) = 6$. the rest is easy.  :)

to be continue..  :D

[quote="Beat"]
13) Do there exist a polinom $Q(x)$ with degree not smaller than 2 with natural coeficients, for which for every prime number $p$, $Q(p)$ is prime. (A. Kanel)
[/quote]

This isn't that hard if we use Dirichlet's Theorem on primes in arithmetic progressions, for example. Maybe I'll try to find a simpler proof later :).

Suppose this is possible. Notice that we certainly have $Q(0)\ne 0$, in this case. Let $p$ be a prime which does not divide $Q(0)$, and let $q=Q(p)$ be another prime. Since $(p,Q(0))=1$, we have $(p,q)=1$, so we can find infinitely many primes of the form $p+tq,\ t\in\mathbb Z$. Now, if $p+tq$ is large enough, we'll have $q|Q(p+tq),\ q<|Q(p+tq)|$, i.e. the value of $Q$ when computed in the prime $p+tq$ is not a prime. We thus have a contradiction.

e.lopes i appreciate your solutions, but please use more details.
I am really interested in the second problem. It looks nice.

7) Solution.
Let $p=a-b$ is a prime and $q=a+b$. Then we have $12c^2=4cq+q^2-p^2$, or
$p^2=q^2-12c^2=(q-2c)(q+6c)$. But $p$ is prime, so $q+6c=p^2$ and $q-2c=1$, then
$8c+1=p^2$ is perfect square.

14) $p,q$ are different prime numbers, such that for an integer number $k>2$,
$p^2+kpq+q^2$ a a perfect square. Prove that $(p-2)(q-2)\leq k+2$.
(I. Bogdanov, S. Berpov)

15) $n$ is a natural number and $5^n+3^n+1$ is a prime number. Prove that $n$ is devisible by 12.

[quote="Beat"]15) $ n$ is a natural number and $ 5^n \plus{} 3^n \plus{} 1$ is a prime number. Prove that $ n$ is devisible by 12.[/quote]

In $ \mathbb{F}_3$ we have $ 2 \mid n$. In $ \mathbb{F}_5$ we have $ 4 \mid n$. So in $ \mathbb{F}_7$ we have $ 5^{4N} \plus{} 3^{4N} \plus{} 1 \equal{} 4^N \plus{} 2^N \plus{} 1$. But if $ 3 \nmid N$ then $ 0 \equal{} 2^2 \plus{} 2 \plus{} 1 \equal{} 2^{2N} \plus{} 2^N \plus{} 1$.

[quote="Beat"]

10) $ p$ is a prime number. Find all integers $ k$, for which $ \sqrt {k^2 \minus{} pk}$ is natural number. (Spanish 1997)

[/quote]

Let $ k^2 \minus{} pk \equal{} x^2$, then it's easy to see that $ p \nmid x$ as $ x \neq 0$,
Also , Discriminant $ \equal{} p^2 \plus{} (2x)^2 \equal{} y^2$ is a square.

So $ p \equal{} m^2 \minus{} n^2 \equal{} (m \plus{} n)(m \minus{} n) \equal{} p \times 1$ gives $ m \equal{} \frac {p \plus{} 1}{2}, n \equal{} \frac {p \minus{} 1}{2}$,

Now $ k \equal{} \frac {p \pm y}{2} \equal{} \frac {(m^2 \minus{} n^2) \pm (m^2 \plus{} n^2)}{2} \equal{} (\frac {p \plus{} 1}{2})^2$ or $ \minus{} (\frac {p \minus{} 1}{2})^2$ are the solutions for primes $ p > 2$

[quote="Beat"]

12) $ p$ is an odd prime which can be written as a difference of two fifth powers.Show that $ \sqrt {\frac {4p \plus{} 1}{5}} \equal{} \frac {n^2 \plus{} 1}{2}$ for some odd integer $ n$.
(Irish 2001)
[/quote]

$ x\minus{}y|x^5\minus{}y^5>x\minus{}y$, so we have $ p\equal{}(x\plus{}1)^5\minus{}x^5$ for some $ x$.

Now $ \sqrt{\frac{4p\plus{}1}{5}}\equal{}\sqrt{4x^4\plus{}8x^3\plus{}8x^2\plus{}4x\plus{}1}$
$ \equal{}\sqrt{(2x^2\plus{}2x\plus{}1)^2}\equal{}\frac{(4x^2\plus{}4x\plus{}1)\plus{}1}{2}\equal{}\frac{(2x\plus{}1)^2\plus{}1}{2}$, done.