Can I force the brackets to be as large as the first quotient in this expression (makes it easier to read).  I'm using the \left[ construct but that doesn't seem to work:

\[ M_1(\rho)<\frac{8Rr}{(R\minus{}r)(r\minus{}\rho)}\left[Log\{2M_2(R)\plus{}|g_2(0)|\plus{}1\}\plus{}|Log|g_2(0)\parallel{}\plus{}|g_1(0)|\plus{}\pi\right]\]

If I use \Huge, the brackets are not uniform with the line of code and are bold (not a clean expression):

\[ M_1(\rho)<\frac{8Rr}{(R\minus{}r)(r\minus{}\rho)}\Huge\text{[}Log\{2M_2(R)\plus{}|g_2(0)|\plus{}1\}\plus{}|Log|g_2(0)\parallel{}\plus{}|g_1(0)|\plus{}\pi\Huge\text{]}\]

Any suggestions?
Thanks,
Shaw

thats strange :maybe: I generally use the \left( and it works fine for me :)

You can use
$ M_1(\rho) < \frac {8Rr}{(R \minus{} r)(r \minus{} \rho)}\big[\text{Log}\{2M_2(R) \plus{} |g_2(0)| \plus{} 1\} \plus{} |\text{Log}|g_2(0)\parallel{} \plus{} |g_1(0)| \plus{} \pi\big]$
or
$ M_1(\rho) < \frac {8Rr}{(R \minus{} r)(r \minus{} \rho)}\Big[\text{Log}\{2M_2(R) \plus{} |g_2(0)| \plus{} 1\} \plus{} |\text{Log}|g_2(0)\parallel{} \plus{} |g_1(0)| \plus{} \pi\Big]$
or
$ M_1(\rho) < \frac {8Rr}{(R \minus{} r)(r \minus{} \rho)}\bigg[\text{Log}\{2M_2(R) \plus{} |g_2(0)| \plus{} 1\} \plus{} |\text{Log}|g_2(0)\parallel{} \plus{} |g_1(0)| \plus{} \pi\bigg]$
or
$ M_1(\rho) < \frac {8Rr}{(R \minus{} r)(r \minus{} \rho)}\Bigg[\text{Log}\{2M_2(R) \plus{} |g_2(0)| \plus{} 1\} \plus{} |\text{Log}|g_2(0)\parallel{} \plus{} |g_1(0)| \plus{} \pi\Bigg]$

\big gives $ \big[\ \big]$ \Big gives $ \Big[\ \Big]$ \bigg gives $ \bigg[\ \bigg]$ \Bigg gives $ \Bigg[\ \Bigg]$ and the same applies to lots of other brackets and vertical arrows etc.

By the way, you should use either \log to get $ \log$ or \text{Log} to get $ \text{Log}$ instead of $ Log$. If you use Log a lot then define \Log with \newcommand{\Log}{\text{Log}}

[quote="stevem"]
By the way, you should use either \log to get $ \log$ or \text{Log} to get $ \text{Log}$ instead of $ Log$. If you use Log a lot then define \Log with \newcommand{\Log}{\text{Log}}[/quote]
\operatorname{Log} is even better: compare 2x\text{Log}h(x), which gives $ 2x\text{Log}h(x)$ and 2x\operatorname{Log}h(x), which gives $ 2x\operatorname{Log}h(x)$. You see that spacing is a bit nicer in the second case.

Thanks a bunch guys!  Second to last code using \bigg is perfect for me.

Also, just noticed we enabled fly-by on math  to get code.  That way, I just clicked, cut, paste into TeXnicCenter and run :)
Neat.

Oh yea, see what you guys mean with the Log.  Thanks!

[quote="fedja"]\operatorname{Log} is even better: compare 2x\text{Log}h(x), which gives $ 2x\text{Log}h(x)$ and 2x\operatorname{Log}h(x), which gives $ 2x\operatorname{Log}h(x)$. You see that spacing is a bit nicer in the second case.[/quote]I agree and if you put \DeclareMathOperator{\Log}{Log} in the preamble then you can use \Log as a command whenever you want.


We noticed that the fax number on the Entry form is different from the fax number on the form you send in to give permission to release scores and solutions.  Should they be different?  If not, which one is right?

You have an older version of the entry form (which don't worry, it's perfectly OK to use the older version: as long as it says Year 16 and/or 2004-5 at the top, we'll take it), which accounts for the discrepancy.

Please fax all USAMTS-related material to (619) 445-2379.

Sorry about the confusion.


geometry
3D geometry
tetrahedron
Support
rhombus
graph theory

An ant living on a solid tetraheron of side = $1$ unit wants to build a road system to go from any corner to any other on the surface of its planet.

What's minimum road length it must have?

You want the shortest path that visits each vertex once?
Then this is a Hamiltonian cycle on the complete planar graph with three outer vertices in the form of an equilateral triangle and one at the centroid?

EDIT:
Wouldn't you need one more of those on the opposite, perpendicular edge?

[quote="paladin8"]I don't think that's what the question asks for.

I believe the problem wants to place paths with minimum total length on the surface of the tetrahedron such that, by only traveling on these paths, the ant may reach any vertex. This does not seem like a simple problem, but maybe there's a simple solution.

[hide="first idea"] Place a "center" on the midpoint of one of the edges and draw paths from that point to each of the vertices. This has total path length $1+\sqrt{3}$.[/hide][/quote]
I support that idea, and I think I have an informal proof that it's the best placement.

[hide]Look at the edge of the tetrahedron horizontally, with the midpoint being the chosen point.

If the point is moved either way on a face of the tetrahedron, then while one of the distances remains the same (the distance between the two "far" vertices), the others get longer (hypotenuse > leg).

The same happens when the point is moved along the edge of the tetrahedron. While the combined distance between the two vertices on the edge stays the same, the other distances increase.

Any combination of these two movements will result in a lot more added distance, so the midpoint of an edge would be the closest to all 4 vertices.[/hide]
[hide="Another possible solution"]If two of the faces of the tetrahedron are folded up into a 2-dimensional rhombus, the point closest to all 4 vertices is quite obviously (an easily proven to be) the center of the rhombus, which is the midpoint of the edge on the tetrahedron.

Since the center is the intersection of the shortest distances from opposite vertices, it is a "compromise," and gives the shortest distance to all 4 vertices.[/hide]


Find all integral solutions to the linear diophantine equation $ 3x\plus{}4y\plus{}5z\equal{}6$.

[hide]$ x\equal{}{6\minus{}4y\minus{}5z\over 3}\equal{}2\minus{}y\minus{}z\minus{}{y\plus{}2z\over 3}$

So $ y\plus{}2z\equal{}3a$ where $ a\in\mathbb{Z}$. Putting $ z\equal{}b$ where $ b\in\mathbb{Z}$, we get $ y\equal{}3a\minus{}2b$ and then $ x\equal{}2\minus{}4a\plus{}b$. Therefore all solutions are $ (2\minus{}4a\plus{}b,3a\minus{}2b,b)$ where $ a,b$ are arbitrary integers.[/hide]


Prove that for any $a,b,c>0$ the following inequality holds:

                        $(b+c-a)(c+a-b)(a+b-c)\leq\frac{9a^2b^2c^2}{(a+b+c)(a^2+b^2+c^2)}$ ;)

See http://www.mathlinks.ro/Forum/viewtopic.php?t=22367 post #8, where I show it in the form

$\left(a^2+b^2+c^2\right)\left(a+b+c\right)\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)\leq 9a^2b^2c^2$.

  Darij


Does anyone know what the ARML dates for 2007 are?

Probably June 1-2.  As far as I know, it's always the weekend after Memorial Day.

I have confirmed the date of ARML this year.

http://arml.com/about.htm

I still have the D.E. Shaw calendar card thing they gave us last year :roll: It has the dates of ARML 2007 highlighted in red, June 1-2.


The altitudes of a triangle are $12$, $15$, and $20$.  What is the area of this triangle?

[hide]the sides are $\frac{2A}{12}$, etc, plug these into heron's formula, to get the answer[/hide]

[hide]Or in this particular case we can notice $5\cdot 12=4\cdot 15=3\cdot 20$, hence the triangle is $3: 4: 5$, hence $15$ and $20$ are the legs, hence the area is $150$.[/hide]

so what if a triangle has altitudes of 10, 15, and 25?

Then use the general approach given by Altheman.

oh okay. good idea. so the sides would be 2A/10, 2A/15, 2A/25

then use heron's formula.  :D

haha that formula's pretty long. heron's is probably easier to do

 :lol:

That [i]is[/i] the simplified result of Heron.

oh.. oops  :P

And the H stands for.. oh yeah

See here also: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&t=112464

Much simpler:
Let sides be $a$, $b$, $c$. Equating areas we get $a = 5b/4$, $c=3b/4$.

Now apply cosine rule to $\angle A$. By substituting $a$ and $c$ as above, we get $Cos\angle A = 0$.

This means triangle is right angled at $A$! This implies $c = 15$ and $b = 20$. Thus area = $1/2*15*20$ = $150$.

The problem itself is probably msm level but the later variations definitely aren’t.

This problem was also on Baltic Way 2006.
Here is simple solution.
We know that a $ah_a=bh_b=ch_c$ and now get $b=\frac{ah_a}{h_b}=\frac{12a}{15}=\frac{4a}{5}$ and $c=\frac{ah_a}{h_c}=\frac{12a}{20}=\frac{3a}{5}$. We will now use Heron formula to find the area. $P=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{a+b+c}{2}=\frac{6a}{5}$
Finally, $P=\sqrt{\frac{36a^4}{625}}=\frac{6a^2}{25}$, and area is also $P=\frac{ah_a}{2}=6a$
Now, $\frac{6a^2}{25}=6a$ or $a=25 \implies b=20, c=15$, so $P=150$


algebra
polynomial
integration
calculus
search
function
calculus computations

We can easily verify that $ P(x) \plus{} P(1 \minus{} x) \equal{} 1$, [1], let $ x \minus{}> P(x)$, so $ P(P(x)) \plus{} P(1 \minus{} P(x)) \equal{} 1$ and using again [1],
$ P(P(x)) \plus{} P(P(1 \minus{} x)) \equal{} 1$. Inductive, $ P^{[n]}(x) \plus{} P^{[n]}(1 \minus{} x) \equal{} 1$.
Integrate here from 0 to 1, use substitution x -> 1 - x for the second integral and finally get $ \int_0^1 P^{[n]}(x) dx \equal{} \frac{1}{2}, \forall n \geq 1$

Slick.. how on earth do no notice things like $ P(x) \plus{} P(1\minus{}x) \equal{} 1$. Thanks!  :lol:

if you study a bit properties of P, you can notice that $ P^{'}(x) \equal{} 3x^2 \minus{} 3x \plus{} 1 > 0, \forall x$, so P is  strictly increasing, and for x between 0 and 1 P is between $ \frac {1}{4}$ and $ \frac {3}{4}$

if you need to compute $ P^{[n]}$ you may consider to find out if P has fixed points. Equation $ P(x) \equal{} x$ has in [0, 1] solution $ \frac {1}{2}$

these arguments point to search if P has a simmetry point $ A(a, b) \equal{} A(\frac {1}{2}, \frac {1}{2})$ which is true after computations:

$ \frac {x \plus{} y}{2} \equal{} a \equal{} \frac {1}{2}$
$ \frac {P(x) \plus{} P(y)}{2} \equal{} b \equal{} \frac {1}{2}$

 :)

A note on the title: it's not a degree-6012 polynomial. The degree of the composition is $ 3^{2004}$.

We can rewrite $ P(x)\equal{}\frac12\plus{}\frac14\left(x^4\minus{}(1\minus{}x)^4\right)$ Composing it with itself on $ [0,1]$ shrinks the interval, and $ P^{[2004]}$ is very close to the constant function $ \frac12$.

[quote]A note on the title: it's not a degree-6012 polynomial.[/quote]

An oversight there  :blush:  Thanks


Show that there exists a convex polygon in the  plane, whose every vertex is equidistant from at least three other vertices.

hmm, like 6 months without a solution...
no one?


integration
algebra
linear equation
absolute value
calculus
calculus computations

Given the differential equation:
$ \frac{dy}{dx} \equal{} \frac{y\minus{}1}{x^2}$ where $ x \neq 0$

Find the particular solution $ y \equal{} f(x)$ to the differential equation with the initial condition $ f(2) \equal{} 0$.

This equation is both linear and separable; hence two different methods are available to us. We can use separation of variables, or we can find an integrating factor for this as a linear equation. When that choice is available, I recommend to students that they follow the linear equation pathway and find the integrating factor. This isn't so much because it's less tedious - in this case, separation of variables might be shorter - but because it affords fewer opportunities for human error. And while separation of variables, properly managed, should work, I can see at least one pitfall for the unwary student.

[hide="The pitfall I see."] At one point, you're looking at $ \int\frac{dy}{y\minus{}1}.$ Force of habit leads us to write that as $ \ln|y\minus{}1|.$ But that formula, with the absolute value, is just a convenient summary of two separate formulas, one valid for $ y>1$ and one for $ y<1.$ Check the initial condition: $ f(2)\equal{}0,$ meaning that at that point, $ y\equal{}0<1.$ If $ y$ is less than $ 1$ there, it's always less than $ 1,$ so $ y\minus{}1$ is negative. With that insight, I then write

$ \int\frac{dy}{y\minus{}1}\equal{}\ln|1\minus{}y|.$[/hide]

Just for checking purposes, [hide="a short answer."]$ y\equal{}1\minus{}e^{\frac12\minus{}\frac1x}.$ This solution only valid for $ x>0.$[/hide]


limit
logarithms
integration
real analysis
real analysis unsolved

$\lim_{n \to \infty} {1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}}=$
$\lim_{n \to \infty} {1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}-ln(n)+ln(n)}=$
$c+\lim_{n \to \infty} {ln(n)}=+ \infty$

and

$\lim_{n \to \infty} {\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{2n}}=$
$\lim_{n \to \infty} {\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{2n}} -$
$\lim_{n \to \infty} {\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{n}} =$
$\lim_{n \to \infty}{c+ln(2n)-c-ln(n)}$=
$\lim_{n \to \infty}{ln(2n)-ln(n)}=\lim_{n \to \infty}{ln(2)}$

are these correct because i was thinking about them and
need to know.if they're not i'll have to work more

[quote]$\lim_{n \to \infty} {\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{2n}} - \lim_{n \to \infty} {\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{n}} = \lim_{n \to \infty}{c+ln(2n)-c-ln(n)}$[/quote]
This is nonsense- you can't take the difference of two infinities like that. You want to say that $\sum_{k=1}^n\frac1k=\ln k+c+o(1)$ for some constant $c$, and then it is appropriate to take the difference of two finite sums.

An easier way to get at the limit:
$\sum_{k=n+1}^{2n}\frac1k=\sum_{j=1}^n\frac1n\cdot\frac1{1+\frac jn}$
$\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac1k=\int_0^1\frac1{1+x}\,dx$


What is the sum of all the numbers less than 2009 with exactly 5 factors?

[hide="Solution"]The number of divisors of a number is $ (a \plus{} 1)(b \plus{} 1)(c \plus{} 1)(d \plus{} 1)...$ where $ a,b,c,d...$ are the exponents of its prime factors. The only way a number can have 5 divisors is if exactly one of $ a \plus{} 1,b \plus{} 1,c \plus{} 1...$ is 5, and the others are $ 1$. This means that the only numbers with exactly $ 5$ divisors are fourth powers, but $ 1$ doesn't count.

There are $ 5$ fourth powers (not including $ 1$) less than 2009, so this is our answer.[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Find all normal subgroups of S_n for all n>=5.

$\forall \sigma \in S_{n}, \sigma [a_{1},\dots,a_{k}] \sigma^{-1}=[\sigma(a_{1}),\dots,\sigma(a_{k})]$, where $n \geq 5$ and $k \geq 2$. From here we deduce that if a normal subgroup $H$ of $S_{n}$ contains a $p$-cycle for some prime $p$, then it contains all the $p$-cycles. In particular, if $H$ contains a transposition then $H=S_{n}$, and if $H$ contains a $3$-cycle then $H=A_{n}$ (which is normal in $S_{n}$ since it has index $2$). Otherwise, by Bertrand's postulate, $H$ will contain some element of order $p$ for some prime $p \in [\frac{n}{2}+1,n]$. Such an element is necessarily a cycle (we can assume $p \geq 5$).
Now by what we said previously $H$ must contain all $p$-cycles. But $[a_{1},a_{2},a_{3},\dots,a_{p}][a_{1},a_{2},a_{p},\dots,a_{3}]=[a_{1},a_{3},a_{2}]$, whence $H$ contains all $3$-cycles, and contains $A_{n}$ as a result. Whence $H=A_{n}$ or $H=S_{n}$.

Any proof without Bertrand's postulate ?

If we can use that $A_{n}$ is simple (it has a proof without Bertrand):
If $N$ is normal in $S_{n}$, then $H \cap A_{n}$ is normal in $A_{n}$. Thus $H \cap A_{n}= \{ e \}$ or $=A_{n}$. In the last case, we have $H=A_{n}$ or $H=S_{n}$ by Lagranges theorem.
For the other case, $H$ can have at most $2$ if we consider the projection onto $S_{n}/H$ (we have a subgroup isomorphic to $A_{n}$ in it). So either $H= \{e\}$ and the other case is easy to exclude since we know all elements of order $2$.

Edit: umm yes, simple...

Oh, you mean $A_{n}$ simple, not normal.

This raises a question. In my copy of Algebra (Lang), the non-sovability of $S_{n}$ ($n \geq 5$) is shown just before the simplicity of $A_{n}$. The proof for the former is not soooo easy, but if we prove the latter first, then the lemma you've just proved allows one to show directly that $S_{n}$ is not solvable, right ?

In the proof that $A_{n}$ is simple, nowhere we use the fact that $S_{n}$ is not solvable. I'm wondering why Lang would make it so difficult then. Maybe I'm missing something.

[quote="julien_santini"]by Bertrand's postulate, $H$ will contain some element of order $p$ for some prime $p \in [\frac{n}{2}+1,n]$.[/quote]

may i know why?

Because by Bertrand's postulate, there a prime $p$ in this interval, and since the order of $S_{n}$ is $n!$, $p$ will divide $n!$. Therefore, by Cauchy's theorem, $S_{n}$ will have an element of order $p$.

[quote="Soarer"]I'm not too sure, but why do you need Bertrand's postulate?
If $2| |H|$ or $3| |H|$, we are done. Otherwise there is a $p$, $p \ge 5$ and $p | |H|$, and we can proceed as you did (all $p$ cycles are inside $H$)[/quote]

Do you really think every element of $S_{n}$ with prime order is a cycle ? This is true, however, if the order of this element is $>\frac{n}{2}$.

I see your point. Thanks.

What if the prime factors of H are all smaller than n/2?

Well, I re-read my proof for the first time and I really can't see why I asserted the existence of such a $p$. As you said, all the prime factors of $|H|$ could perfectly be less (or equal to) than $\frac{n}{2}$. I don't know if the mistake can be patched, but I prefer to retain ZetaX's proof, much more trivial. The simplicity of $A_{n}$ is a known fact anyhow.


calculus
integration
ratio
geometric sequence
calculus computations

How can i compute the integral :
$\int_0^a \frac{(a-x)^{n-1}}{(a+x)^{n+1}} dx$, where a>0, n belongs to $N$*

Let $I_n=\int_0^a \frac{(a-x)^{n-1}}{(a+x)^{n+1}}dx.$ Use Integral by parts we have

$I_n=\left[-\frac{(a-x)^n}{n}\cdot \frac{1}{(a+x)^{n+1}}\right]_0^a -\frac{n+1}{n}\int_0^a \frac{(a-x)^n}{(a+x)^{n+2}}dx.$

$=\frac{1}{an}-\frac{n+1}{n}I_{n+1},$ yielding $nI_n=\frac{1}{a}-(n+1)I_{n+1}.$ i.e.

$(n+1)I_{n+1}=-nI_n+\frac{1}{a}\Longleftrightarrow (n+1)I_{n+1}-\frac{1}{2a}=-\left(nI_n-\frac{1}{2a}\right).$

Since the sequence $\left\{nI_n-\frac{1}{2a}\right\}$ is a geometric sequence with common ratio $-1,$ we have

$nI_n-\frac{1}{2a}=(-1)^{n-1}\left(1\cdot I_1-\frac{1}{2a}\right)\ \cdots [*]$ here $I_1=\int_0^a \frac{1}{(a+x)^2}dx=\left[-\frac{1}{a+x}\right]_0^a =\frac{1}{2a}.$

Plugg this into the equation $[*],$ we have $nI_n=\frac{1}{2a}.$

$\therefore\ I_n=\frac{1}{2an}\ \ (n=1,2,\cdots).$

Using the substitution Y=(a-X)/(a+X) you get dy/dx =-2a/(a+X)^2

 The problem is now to integrate Y^(n-1)/2a between upper and lower limits of integration 1 and 0 respectively.


Does anyone know how to manually find an inverse matrix 3x3 or above?

Ah, I usually find inverses by hand anyway, its much more fun.
Just write out your matrix with the identity matrix next to it, and row reduce until you have the identity on the left. Then the answers on the right.

Lets find the inverse of
2 1 0
1 1 3
1 1 2

2 1 0 1 0 0
1 1 3 0 1 0
1 1 2 0 0 1

subtract 3rd row from 2nd:
2 1 0 1 0 0
1 1 3 0 1 0
0 0 1 0 1 -1

subtract 3 times 3rd row from 2nd:
2 1 0 1  0  0
1 1 0 0 -2 3
0 0 1 0 1 -1

first row - second row:
2 1 0 1 0 0
1 0 0 1 2 -3
0 0 1 0 1 -1

and first - twice second
0 1 0 -1 -4 6
1 0 0 1   2 -3
0 0 1 0  1  -1

And then rearranging those rows, the inverse is
1   2   -3
-1 -4   6
0   1  -1

(assuming I haven't made any slips).

[quote="3X.lich"][quote="blahblahblah"][quote="Elemennop"]Another way to do it is:

Find the determinant.

Now, replace each term in the matrix with it's cofactor (the determinant of  the matrix formed by all terms not in the same row or column as the term). When you replace with cofactors, don't forget to apply the alternating +/- depending on the place of the term.

Now transpose this new matrix.

The last step is to multiply each term in the transposed matrix by the reciprocal of the determinant of the original matrix. This new matrix should now be the inverse of hte original one.[/quote]

This method is useless in practice and only has theoretical applications.[/quote]
Show him a 'useful' method :D[/quote]

Exactly what TripleM did:

Transform the adjoined coefficient matrix $[A\vert I_n]$ into $[I_n\vert B]$ by use of row reduction.  This method is easily justified by use of elementary matrices.

This method has time-complexity of $O(n^3)$, as opposed to $O(n!)$ when using cofactor expansion.

Any matrix $A \in M_{3}(R)$ verify Cayley-Hamilton theorem

$A^3-tr(A)A^2+tr(ComA)A-det(A)I_3=0_3$

If $A$ is invertible we have

$A^{-1} = \frac{1}{det(A)}(A^2-tr(A)A+tr(ComA)I_3)$


Given $ a, b, c \geq\ 0$. Prove that:
$ a^6 \plus{} b^6 \plus{} c^6 \plus{} 2abc(a^2b \plus{} b^2c \plus{} c^2a) \geq\ 2(a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3) \plus{} abc(ab^2 \plus{} bc^2 \plus{} ca^2)$

PS: Equality occurs when $ a \equal{} b \equal{} c$ or $ a \equal{} b; c \equal{} 0$  :)

[quote="DangChienbn"]This inequalitiy can be proved by SOS
We must prove:
$ \sum (b \minus{} c)^2 [(b^2 \plus{} bc \plus{} c^2)^2 \minus{} 2a^3b] \geq 0$:|
:blush:  :blush:[/quote]

Prove it  :P

OK
$ S_b; S_c \geq 0$
We must prove $ S_ab^2 \plus{} a^2S_b$
If $ a \geq b \geq c$
$ b^2(b^4 \plus{} 3b^2c^2 \plus{} c^4 \plus{} 2b^3c \plus{} 2c^3a \minus{} 2a^3b) \plus{} a^2(a^4 \plus{} 3a^2c^2 \plus{} c^4 \plus{} 2a^3c \plus{} 2c^3a \minus{} 2b^3c) \geq 0$
$ b^6 \plus{} a^6 \geq 2a^3b^3$
$ 2a^5c \geq 2b^3ca^2 : D$

[quote="DangChienbn"]OK
$ S_b; S_c \geq 0$
We must prove $ S_ab^2 \plus{} a^2S_b$
GS $ a \geq b \geq c$
$ b^2(b^4 \plus{} 3b^2c^2 \plus{} c^4 \plus{} 2b^3c \plus{} 2c^3a \minus{} 2a^3b) \plus{} a^2(a^4 \plus{} 3a^2c^2 \plus{} c^4 \plus{} 2a^3c \plus{} 2c^3a \minus{} 2b^3c) \geq 0$
$ b^6 \plus{} a^6 \geq 2a^3b^3$
$ 2a^5c \geq 2b^3ca^2 : D$[/quote]
If your analyse is trues, you'll have to prove with the case $ a \leq\ b \leq\ c$ (because this ineq is cyclic, not symetric)  :)

My soulution:
$ \sum a^6 \plus{} abc(a^2b \plus{} b^2c \plus{} c^2a) \ge \sum a^6 \plus{} 3a^2b^2c^2 \ge \sum a^2b^2(a^2 \plus{} b^2)$ Shur
We have prove that:
$ \sum a^2b^2(a^2 \plus{} b^2) \plus{} abc(a^2b \plus{} b^2c \plus{} c^2) \ge 2 \sum a^3b^3 \plus{} abc(ab^2 \plus{} bc^2 \plus{} ca^2)$
$ < \equal{} > \sum a^2b^2(a \minus{} b)^2 \minus{} abc(b \minus{} a)(b \minus{} c)(a \minus{} c) \ge 0$
Case 1:
$ b \ge c \ge a$ so $ abc(a\minus{}b)(b\minus{}c)(c\minus{}a) \le 0$ Done!
Case 2:$ b \ge a \ge c$
$ < \equal{} > (a^2b^2 \plus{} b^2c^2)(a \minus{} b)^2 \plus{} (b^2c^2 \plus{} c^2a^2)(a \minus{} c)^2 \plus{} 2b^2c^2(b \minus{} a)(a \minus{} c) \minus{} abc(b \minus{} a)(b \minus{} c)(a \minus{} c) \ge 0$
By Am-Gm
We have:
$ (a^2b^2 \plus{} b^2c^2)(a \minus{} b)^2 \plus{} (b^2c^2 \plus{} c^2a^2)(a \minus{} c)^2 \ge 2bc \sqrt {(a^2 \plus{} c^2)(b^2 \plus{} a^2)}(b \minus{} a)(a \minus{} c) \ge abc(b \minus{} a)(b \minus{} c)(a \minus{} c)$
Done!  :P

And the stronger is also trues and easy
$ a^{6}\plus{}b^{6}\plus{}c^{6}\plus{}2abc(a^{2}b\plus{}b^{2}c\plus{}c^{2}a)\geq\ ab(a^2\plus{}b^2)\plus{}bc(b^2\plus{}c^2)\plus{}ca(c^2\plus{}a^2)\plus{}abc(ab^{2}\plus{}bc^{2}\plus{}ca^{2})$
We can solve it by "only Am-Gm"  :)

[quote="nguoivn"]And the stronger is also trues and easy
$ a^{6} \plus{} b^{6} \plus{} c^{6} \plus{} 2abc(a^{2}b \plus{} b^{2}c \plus{} c^{2}a)\geq\ a^2b^2(a^2 \plus{} b^2) \plus{} b^2c^2(b^2 \plus{} c^2) \plus{} c^2a^2(c^2 \plus{} a^2) \plus{} abc(ab^{2} \plus{} bc^{2} \plus{} ca^{2})$
We can solve it by "only Am-Gm"  :)[/quote]
The problem must be this one, nguoivn. Then, we can use the inequality
$ 2\sum a^2b \minus{}\sum ab^2 \ge \sum a^2b\plus{}\sum ab^2 \minus{}\sum a^3$
to reduce the inequality to
$ \sum a^6\plus{} abc(\sum a^2b\plus{}\sum ab^2\minus{}\sum a^3) \ge \sum a^2b^2(a^2\plus{}b^2),$
which is equivalent to
$ (a\plus{}b\plus{}c)\sum a^3(a\minus{}b)(a\minus{}c) \ge 0.$
And by Schur's Inequality, we see that this inequality is valid. :)

Nice solution, can_hang  :lol:
PS: [hide]Bai` nay` em che' tu` 1 ket' qua? hoan' vi. manh. hon nhung hoi xau' nen danh` lam` yeu' di cho dep : P  [/hide]


I am interested in studying mathematics and computer science, and maybe some of you can maybe help me in my decision by responding to the following questions:

1.)  I have heard that MIT requires more work than either Stanford, or Harvey Mudd in order to obtain a BS degree in CS. Is this true?

2.) Also can anyone fulfull me on the mathematics departments of each schools, and tell me is one better than another?

3.) When I see the rankings of the william lowel puntam competition, Harvard holds more first places than any other school. Does this mean that it has the best mathematics department in the country?

4.) So far my list of colleges is as follows:

1.MIT
2.Harvard
3.Stanford
4. Harvey Mudd
5. Washington State University (Fall back school)

Should I modify this list to suit my interests?

[quote="The Sigma"]When I see the rankings of the william lowel puntam competition, Harvard holds more first places than any other school. Does this mean that it has the best mathematics department in the country?[/quote]

Not necessarily. Harvard has some very brilliant math students, and quite a few math students who enjoy competition math like the Putnam, but it's not necessarily true that the top five schools in the Putnam competition in any one year are ever the top five undergraduate math schools. Some great math teachers simply don't emphasize competition math very much, and some great math students don't devote much time or effort to the Putnam-style examinations.

That said, Harvard is a great math school, and if you like its other characteristics you might as well apply there.

Don't choose among Harvard, MIT and Stanford based on rankings.  Certainly not based on as arbitrary a ranking as a team Putnam score.

For what it's worth, University of Washington is very good for math, although we have a regular poster here who is a grad student there (jmerry), so he probably could tell you more.

Best math department for undergrads means different things to different people.  A lot of things that people care about (e.g. quality of teaching) are very difficult to compare.  But based on the fact that you cite Putnam scores, you might be interested in the question of which schools have the strongest undergraduate math majors.  The Putnam is indeed a coarse indicator of this, but the team rankings are kind of silly because three people (the size of a Putnam team) do not make a math department (well, unless you plan to be one those three people and not fraternize with your intellectual "inferiors" outside of those three).  It is better to take an overall view of the top scorers list.

For many years, there was no question that Harvard attracted a plurality of the best math students (and shockingly rejected many of them), although it is possible that their dominance has waned a bit.  One could reasonably argue that MIT is "stronger" because, being a tech school, it has *more* students who are good at math (many of whom are probably engineers).

On the other hand, Harvard, MIT, and Stanford all have world-class faculty.  This may or may not be important to you, depending on your particular circumstances.  Harvey Mudd's faculty is far less renowned, BUT Harvey Mudd recently won a presitigious competition in which it was named the Best Math Department in the country.  In particular, one can assume that it's a great place for undergraduate education.  Basically, you can't go wrong!

Also, U Washington might make a better safety school than Washington State since it has a much better math department.


LaTeX
calculus
derivative
calculus computations

Can anyone of you prove this? I have tried using some methods used here: $(\sqrt{x})'=\frac{1}{2\sqrt{x}}$, but it didn't work for me.

Here is what i need to prove:

$(n\sqrt{x})'=\frac{n\sqrt{x}}{n\cdot x}$

The $n$ before the squared is to show it is the n'th degree of a square. Don't know if that is what you call it, and maybe you could show me how to write it in $\LaTeX$.

Do you know the power rule
\[ \frac{d}{dx}(x^t)=tx^{t-1}  \]

If so, then $\sqrt[n]{x}=x^{\frac 1n}$, so that
\begin{eqnarray*} \frac{d}{dx}(\sqrt[n]{x})&=&\frac{1}{n}x^{\frac{1}{n} - 1} \\ &=&\frac{1}{n}\cdot \frac{x^{\frac 1n}}{x} \\ \end{eqnarray*}
which is pretty much what you want.

I think his question is basically, how do you [i]prove[/i] the power rule for rational exponents of 1/n.

The answer:
[hide]Use implicit differentiation (if you've learned that already).

Let $y = x^{1/n}$. Then $y^n = x$, and differentiating both sides with respect to x leads to $ny^{n-1}\frac{dy}{dx} = 1$.

Solving for $\frac{dy}{dx}$ gives us
$\displaystyle \frac{dy}{dx} = \frac{1}{ny^{n-1}}$
$\displaystyle = \frac{1}{n(x^\frac{1}{n})^{n-1}}$
$\displaystyle = \frac{x^\frac{1}{n}}{n \cdot x}$[/hide]

Ah, it didn't get to my mind that $(\sqrt[n]{x})'=x^{1/n}$ though i knew it. Think i will go with Blahblahblah's way of proving it. Seems more obvious though i can agree Gauss202's is right too. Have a question though: How do you prove you can go from $x^{\frac{1}{n}-1}$ to $\frac{x^{\frac{1}{n}}}{x}$? This part doesn't seem to obvious though i can see it is true when i put in some numbers...

[quote="TheBB"]General rule:

$a^{b-c}=\frac{a^b}{a^c}$

Or to be more general,

$a^{b+c}=a^ba^c$ and $a^{-c}=\frac{1}{a^c}$

So

$x^{\frac{1}{n}-1}=x^{\frac{1}{n}}x^{-1}=\frac{x^{\frac{1}{n}}}{x^1}=\frac{x^{\frac{1}{n}}}{x}$[/quote]

Fixed your $\LaTeX$, you were missing some curly brackets.

:) Ups, first off i didn't see what you added Bictor717, so i tried to figure it myself from what TheBB said. But nice, see your point now. Thx everyone...

[quote="Bictor717"]Fixed your $\LaTeX$, you were missing some curly brackets.[/quote]

Right, thanks. Fixed my own post now.

And just to make everything perfect i just changed my first post so it says $\LaTeX$ instead of $LaTeX$ :P Had forgot how it was you made that nice commercial $LaTeX$ sign so i made it without the backslash in the first place :)


I have posted this problem before, But somenoe has told me that we shouldn't discuss is before 29 march. Here it is :
$f:Q^{+}\rightarrow R$ such that $f\left( 1997\right) =1$ and
$\left( \forall x,y\in Q^{+}\right) :f\left( x+y\right) \leq f\left( x\right) +f\left( y\right)$  and $f\left( xy\right) =f\left( x\right) f\left( y\right)$
Prove that : $f\left( 2005\right) \leq 1$
Myht has already solved it. Can you rewrite your solution again Myth, please ?  :lol:

Are you kidding? Rewrite it again???

http://www.mathlinks.ro/Forum/viewtopic.php?t=23869

Dear Erdös,

As I told you in another post, many of the exercises that we used in our correspondace program are taken from here.
Thus, before asking for one of them, you would rather search here for it....

Pierre.


In the  laboratory, H2 (g) can be produced by  adding which of the following to 1 M HCl (aq)?

I. 1 M NH3 (aq)
II. Zn (s)
III. NaHCO3 (s)

a. I only
b. II only
c. III only
d. I and II only
e. I, II and III

thanks

The lab method is I and mostly that method is used. I don't know about the other 2  :D

:o wht did you say ? nh3+hcl-> h2 :mad:  the answer is zn only. it is the lab method

(II) is the only method by which you can prepare hydrogen.
(I) yields an ammonium chloride as it is a neutralisation reaction and (III) yields $NaCl$ and $CO_{2}$

We did III in the lab today...definitely not 3! Only 1...


same techniques as all the other problems we are seeing...add $k_{1}\cdot n+k_{2}\cdot n^{2}$ to each sequence and eliminate $n$, $n^{2}$, add like before to isolate either a_n, or b_n (see original recursion 2), then use characteristic equations


Посмотрите моё решение уравнения мадам Зарангешь. Если этого мало будет и если пока это интересно для Вас, тогда скажите мне.


Let $ABCD$ be a quadrilateral. $M$ is the midpoint of $AB$, $N$ the midpoint of $CD$. $AC,MN,BD$ are concurrent. Show that $ABCD$ is a trapezoid.

[hide]Let $P$ be the intersection of $AD$ and $MN$. Menelaos' on $\Delta DAC$ and $\Delta DAB$ gives $\frac{DP}{PA}\cdot\frac{AO}{OC}\cdot\frac{CN}{ND}=-1$ and $\frac{DP}{PA}\cdot\frac{AM}{MB}\cdot\frac{BO}{OD}=-1$. Hence $\frac{AO}{OC}= \frac{BO}{OD}$. The result follows.[/hide]


What is the sum of all prime factors of 2002?

A.33	B.43	C.103	D.152	E.  1003

[hide=" click to reveal hidden content"]

$ 2002 \equal{} 2 \times 7 \times 11 \times 13$

$ 2\plus{}7\plus{}11\plus{}13\equal{} 33$

[/hide]

[hide]So you first prime Factorize 2002.
So and you get, 2*7*11*13
And you add them, and get 33....
SO A[/hide]


A triangle is given about sides to the length $ 13$, $ 14$, $ 15$. To calculate the sum of the distance of the point of intersection of the height from vertexes of this triangle.

Let $ a \equal{} 13, b \equal{} 14, c \equal{} 15$.
So $ h_a \equal{} \frac {2\sqrt {p(p \minus{} a)(p \minus{} b)(p \minus{} c)}}{a} \equal{} \frac {2\cdot 84}{13} \equal{} \frac {168}{13}$. Likewise we obtain that $ h_b \equal{} 12, h_c \equal{} \frac {56}{5}$.
By Pythagoras theorem we can find segments $ BA_1 \equal{} \sqrt {c^2 \minus{} h_a^2} \equal{} \frac {99}{13}, B_1C \equal{} \sqrt {a^2 \minus{} h_b^2} \equal{} 5, BC_1 \equal{} \sqrt {a^2 \minus{} h_c^2} \equal{} \frac {33}{5}$.
$ tg\angle{BCC_1} \equal{} \frac {BC_1}{h_c} \equal{} \frac {33}{56}$,
$ tg\angle{B_1BA} \equal{} \frac {B_1A}{h_b} \equal{} \frac {3}{4}$,
$ A_1H \equal{} A_1Ctg\angle{BCC_1} \equal{} \frac {5}{13}\cdot{33}{4} \equal{} \frac {165}{52}$
$ C_1H \equal{} BC_1tg\angle{ABB_1} \equal{} \frac {99}{20}$
$ B_1H \equal{} CB_1tg\angle{ACC_1} \equal{} CB_1tg\angle{B_1BA} \equal{} \frac {15}{4}$
$ AH \plus{} BH \plus{} CH \equal{} (h_a \minus{} A_1H) \plus{} (h_b \minus{} HB_1) \plus{} (h_c \minus{} HC_1) \equal{} \frac {168}{13} \minus{} \frac {165}{13\cdot4} \plus{} \ \ \plus{} 12 \minus{} \frac {15}{4} \plus{} \frac {56}{5} \minus{} \frac {99}{20} \equal{} 24,25$.
I think there is better solution, but this is also correct. I used notations: $ AA_1,BB_1,CC_1$ - altitudes, $ H$ - ortocentre.


Last year I participated in this competition. I know that I had a prize winning score (according to the FAQ chart), yet I never received any notice. What does this mean?

We did not administer the USAMTS last year.  I recommend sending a letter to this effect to the address to which you sent solutions last year.


calculus
integration
limit
function
real analysis
real analysis unsolved

I've just read about some riemann integration and tried these problems... i'd like to make sure my approaches are correct and also see if there are any nicer solutions.

1.  If $f$ is continuous on $[0,+\infty)$ and $f(x) \rightarrow L$ as $x \rightarrow \infty$, show that for $a > 0$,
$\lim_{n \rightarrow \infty}\int_{0}^{a}f(nx) \, dx = aL$

my attempt:  Let $I(n) = \int_{0}^{a}f(nx) \, dx$.
We can write $I(n) = \int_{0}^{1/\sqrt n}f(nx) \, dx+\int_{1/ \sqrt n}^{a}f(nx) \, dx$, and estimate each of these integrals.  From the given conditions there is a constant $C$ such that $|f(nx)| < C$ for $x \in [0,\infty)$.  There is also some $M$ such that for $x > M$, $|f(x)-L| < \epsilon$ for any $\epsilon > 0$.  If $\sqrt n > M$

$I(n) < \int_{0}^{1/\sqrt n}C \, dx+\int_{1/ \sqrt n}^{a}L+\epsilon \, dx = aL+a\epsilon+\frac{C-L-\epsilon}{\sqrt n}$
and
$I(n) > aL-a\epsilon+\frac{-C-L+\epsilon}{\sqrt n}$

so $\lim I(n) = aL$.

2.  Let $\phi_{n}$ be nonnegative functions integrable over $[-1,+1]$, satisfying the conditions
(a) $\int_{-1}^{+1}\phi_{n}= 1$ for every $n \in \mathbb N$; and
(b) for every $\delta > 0$, $\phi_{n}\rightarrow 0$ uniformly on $[-1,-\delta] \cup [\delta,1]$.
Show that for every $f$ which is integrable over $[-1,1]$ and continuous at $0$, we have $\int_{-1}^{1}f \cdot \phi_{n}\rightarrow f(0)$ as $n \rightarrow \infty$.

my attempt: Write $I = \int_{-1}^{1}f \cdot \phi_{n}= \int_{-1}^{-\delta}f \cdot \phi_{n}+\int_{-\delta}^{\delta}f \cdot \phi_{n}+\int_{\delta}^{1}f \cdot \phi_{n}$ and we will estimate these.  We can pick $\delta$ small enough so that $|f(t)-f(0)| < \epsilon$ for all $t \in (-\delta, \delta)$, and pick $n_{0}$ such that for $n \geq n_{0}$, $0 \leq \phi_{n}(t) < \epsilon$ for all $t \in [-1,-\delta] \cup [\delta,1]$.  Note that since $f$ is integrable there is some $L$ such that $|f| < L$ on $[-1,1]$.  Also note that $\int_{-\delta}^{\delta}\phi_{n}= 1-\int_{-1}^{-\delta}\phi_{n}-\int_{\delta}^{1}\phi_{n}$  Thus

$( f(0)-\epsilon)(1-2(1-\delta)\epsilon) < I < 2L(1-\delta)\epsilon+(f(0)+\epsilon)$

so $I = f(0)$.

sorry about any typos, of which there are probably at least a few

Those are exactly the standard approaches. I don't see any real problems, although I haven't looked closely at the details of the estimate in #2.

great, thanks.  This doesn't have anything to do with the approach itself, but more the aesthetic of the proof: is it considered bad form to not end up with a nice thing like "so $f(0)-\epsilon < I < f(0)+\epsilon$" and instead have $f(0)$ and a bunch of junk that you can make as small as you want.  I usually make it end up the nice way but sometimes it seems like more trouble than it's worth


linear algebra
matrix
calculus
algebra
polynomial
linear algebra solved

Having $A,B \in M_n (C)$, and $A \cdot B = B \cdot A$ ,and $\exists p \in N$, so $A^p  = O_n$.

Prove  that
\[
\det (A^2  + AB + B^2 ) = (\det B)^2
\]

Please post in the right section next time, all Calculus and Matrix / Vectorial Space problems should be posted in the College Playground forums, which is where I just relocated this problem.

[quote="Moubinool"]If $M$ is nilpotent and commute with $B^2$
then
$det(M+B^2)=(det(B)^2)$[/quote]

where does this come from?

The following holds:

If $X$ is nilpotent and commutes with $A$, then $\det(X+A)=\det A$:

If $A$ is invertible, then this is equivalent to $\det(XA^{-1}+I_n)=1$, which is easily seen to be true since $XA^{-1}$ is nilpotent, so $\det(XA^{-1}+xI_n)=x^n$ (the characteristic polynomial of $XA^{-1}$ is $x^n$). We can find infinitely many $\lambda$ s.t. $A+\lambda I_n$ is invertible, so, by the previous sentence, $\det(X+A+\lambda I_n)=\det(A+\lambda I_n)$. Now, since that identity holds for infinitely many $\lambda$ and $\det(X+A+\lambda I_n)-\det(A+\lambda I_n)$ is a polynomial in $\lambda$, it means that it holds for all $\lambda$, and, in particular, for $\lambda=0$, in which case we get $\det(X+A)=\det A$, Q.E.D.

[quote="grobber"]which is easily seen to be true since $XA^{-1}$ is nilpotent[/quote]

I can't prove that $XA^{-1}$ is nilpotent :(

[quote="spix"]X is nilpotent and commutes with A so ........ what`s the problem?[/quote]
I'm dumb... sorry

i believe that $X$ is nilpotent means that $X^n=O_n$ so if $X$ is nilpotent and $AX=XA$ then we have that $A^{-1}X=XA^{-1}$ so i believe that it`s obvious that $(XA^{-1})^n=O_n$ cause $(XA^{-1})^n=X^n(A^{-1})^n$


Can someone help me debug this Java applet?

[code]
import java.awt.*;

public class applet extends javax.swing.JApplet
{
	public void init()
	{
		setBackground(Color.green);
		int c[] = new int[4];
	}
	public void paint(Graphics screen)
	{
		for (int a = 0; a < 10; a++)
		{
			{
				for (int b = 0; b < 4; b++)
				{
					c[b] = (Math.random() * 500);
				}
			}
			screen.drawLine(c[0], c[1], c[2], c[3]);
		}
	}
}
[/code]

Error
[code]
ming@localhost ~/Desktop/Java $ javac applet.java
applet.java:17: cannot resolve symbol
symbol  : variable c
location: class applet
                                        c[b] = (Math.random() * 500);
                                        ^
applet.java:20: cannot resolve symbol
symbol  : variable c
location: class applet
                        screen.drawLine(c[0], c[1], c[2], c[3]);
                                        ^
applet.java:20: cannot resolve symbol
symbol  : variable c
location: class applet
                        screen.drawLine(c[0], c[1], c[2], c[3]);
                                              ^
applet.java:20: cannot resolve symbol
symbol  : variable c
location: class applet
                        screen.drawLine(c[0], c[1], c[2], c[3]);
                                                    ^
applet.java:20: cannot resolve symbol
symbol  : variable c
location: class applet
                        screen.drawLine(c[0], c[1], c[2], c[3]);
                                                          ^
5 errors
[/code].

This is a very simple applet, which should not have errors. So can someone help me find the bugs

As you can see, the problem is variable c

You defined it as a local variable inside init() method

Move the definition just before init()  :)

That was one of the problems with this code, I will post the revised code to anyone who cares.

[code]
import java.awt.*;

public class randomLines extends javax.swing.JApplet
{
	public void init()
	{}
	public void paint(Graphics screen)
	{
		{
			int[] c = new int[4];
			for (int a = 0; a < 10; a++)
			{
				{
					for (int b = 0; b < 4; b++)
					{
						c[b] = (int)(Math.random() * 500);
					}
				}
				screen.drawLine(c[0], c[1], c[2], c[3]);
			}
		}
	}
}
[/code]

Apparently, I had to cast Math.random() * 500 into int, as it is originally double.


For positive real numbers $ a,b,c$ prove that \[ \frac{a^2+b^2+c^2}{ab+bc+ca}\ge \frac 23\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)\]
[hide="Remarks"]
1. It is easy to prove that if $ a\ge b\ge c$ then $ \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\ge \frac 32$. Then from the above we deduce another proof of $ a^2+b^2+c^2\ge ab+bc+ca$.

2. Switching variables the above inequality becomes
\begin{eqnarray*}2\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)&\ge& \frac 23\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} +\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}\right)\\
&=&2\end{eqnarray*}
deriving yet another proof of $ a^2+b^2+c^2\ge ab+bc+ca$.[/hide]

My solution, using the remarks:
[hide]WLOG let $ a \ge b \ge c$. By Nesbit's, $ \sum \frac{a}{b\plus{}c} \ge \frac{3}{2}$, thus the original equation is rephrased to

$ \frac{\sum a^2}{\sum ab} \ge \frac{2}{3} \frac{3}{2}$
$ \Leftrightarrow \frac{\sum a^2}{\sum ab} \ge 1$
$ \Leftrightarrow \sum a^2 \ge \sum ab \Rightarrow a^2\plus{}b^2\plus{}c^2 \ge ab\plus{}bc\plus{}ac$.

By Cauchy's, $ (ab\plus{}bc\plus{}ac)^2 \le (a^2\plus{}c^2\plus{}c^2)(a^2\plus{}b^2\plus{}c^2)$. Square-rooting this inequality gives $ a^2\plus{}b^2\plus{}c^2 \ge ab\plus{}bc\plus{}ac$, which is what we want.[/hide]

I don't think that this is correct. You proved that $ a^2\plus{}b^2\plus{}c^2>\equal{}ab\plus{}bc\plus{}ca$ but not what the inequality asks you to prove!

I dont quite understand... I thought I rephrased the original inequality as $ a^2\plus{}b^2\plus{}c^2 \ge ab\plus{}bc\plus{}ac$? So if this rephrased inequailty is proven true, then the original inequality is likewise true?

The thing is that you didn't rephrase the original inequality. You see, what you said is like this:
a>=b,b>=c => a>=c which is true so a>=b is true.
But this is not a correct method: See the following counter example: 2>3 3>1 so 2>1 so 2>3 holds
Do you get what I mean?

Yes, I see my fallacy now. Thank you for pointing it out, salonikios.
I tried redoing the inequality again by...
[hide]expanding $ \sum \frac{a}{a\plus{}b}$ to get $ \frac{2\sum (a^2b)\plus{}3abc\plus{}\sum (ab^2)}{\sum(a^2b)\plus{}2abc\plus{}\sum(ab^2)}$, which is equivalent to $ \frac{3}{2}\plus{}\frac{0.5\sum(a^2b)\minus{}0.5\sum(ab^2)}{\sum(a^2b)\plus{}2abc\plus{}\sum(ab^2)}$, so the original equality is rephrased to
$ \frac{\sum a^2}{\sum ab} \ge 1\plus{}\frac{\sum(a^2b)\minus{}\sum(ab^2)}{3(\sum(a^2b)\plus{}2abc\plus{}\sum(ab^2))}$, but I am stumped from here...[/hide]

[quote="ghjk"]@nayel:Where did u find this problem?[/quote]
According to the title, it's his.

I've waited long enough for a nice solution of this inequality and I cannot wait anymore! Here's a nice and very useful hint. :wink:

[hide="Useful hint!"]The following stronger inequality is also true for $ a,b,c>0$.
\[ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\ge\frac 23\left(\frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\right)\][/hide]

Ok, that hint is really nice :)

[hide="Solution"]
Rearrangement yields:
$ \frac 23 \left(\frac a{b + c} + \frac b{c + a} + \frac c{a + b}\right)\ge\frac 23\left(\frac a{a + b} + \frac b{b + c} + \frac c{c + a}\right)$
Now let's just add 2 to both sides of the inequality.
\begin{align*} \frac {a^2 + b^2 + c^2}{ab + bc + ca} + 2 & \ge \frac 23 \left(\frac a{b + c} + 1 + \frac b{c + a} + 1 + \frac c{a + b} + 1\right) \\
\frac {a^2 + b^2 + c^2 + 2ab + 2bc + 2ca}{ab + bc + ca} & \ge \frac 23 \left(\frac {a + b + c}{b + c} + \frac {a + b + c}{c + a} + \frac {a + b + c}{a + b}\right) \\
\frac {(a + b + c)^2}{ab + bc + ca} & \ge\frac {2(a + b + c)}3 \left(\frac 1{b + c} + \frac 1{c + a} + \frac 1{a + b}\right) \\
\frac {a + b + c}{ab + bc + ca} & \ge\frac 23\left(\frac 1{b + c} + \frac 1{c + a} + \frac 1{a + b}\right) \\
3(a + b + c)(a + b)(b + c)(c + a) & \ge 2(ab + bc + ca)((a + b)(a + c) + (b + a)(b + c) + (c + b)(c + a)) \\
3\sum_{sym}a^3b + 3\sum_{sym}a^2b^2 + 6\sum_{sym}a^2bc & \ge 2\sum_{sym}a^3b + 3\sum_{sym}a^2b^2 + 7\sum_{sym}a^2bc \\
\sum_{sym}a^3b & \ge\sum_{sym}a^2bc \end{align*}
The last lign is true by Muirhead or AM-GM (since $ \frac {a^3b + a^3c + ab^3 + ac^3}4\ge\sqrt [4]{a^3ba^3cab^3ac^3} = \sqrt [4]{a^8b^4c^4} = a^2bc$).
[/hide]

nice solution due to  bertram :
just I note that :
$ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}  \ge \frac{2}{3}(\frac 1{b \plus{} c} \plus{} \frac 1{c \plus{} a} \plus{} \frac 1{a \plus{} b})$
$ \Longleftrightarrow  a\plus{}b\plus{}c \ge 2(\frac{ab}{a\plus{}b}\plus{}\frac{bc}{b\plus{}c}\plus{}\frac{ca}{c\plus{}a})$
which is obvious by the fact that $ x\plus{}y \ge \frac{4xy}{x\plus{}y}$.

[quote="tchebytchev"]nice solution due to  bertram :
just I note that :
$ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \ge \frac {2}{3}(\frac 1{b \plus{} c} \plus{} \frac 1{c \plus{} a} \plus{} \frac 1{a \plus{} b})$
$ \Longleftrightarrow a \plus{} b \plus{} c \ge 2(\frac {ab}{a \plus{} b} \plus{} \frac {bc}{b \plus{} c} \plus{} \frac {ca}{c \plus{} a})$
which is obvious by the fact that $ x \plus{} y \ge \frac {4xy}{x \plus{} y}$.[/quote]

Is that rearrangement? Anyway, your proof is quite a lot more elegant than mine :D but I always wanted to try a "dumbass" solution (clearing denominators and using muirhead/schur over and over) on a problem ;)

EDIT: Hm, I can't help but think that your inequality looks a bit fishy...
$ 3\left(\frac{ab}{a\plus{}b}\plus{}\frac{bc}{b\plus{}c}\plus{}\frac{cy}{c\plus{}a}\right)\le(ab\plus{}bc\plus{}ca)\left(\frac 1{a\plus{}b}\plus{}\frac 1{b\plus{}c}\plus{}\frac 1{c\plus{}a}\right)$
by rearrangement, but then the inequality sign is pointing in the wrong direction. It's the same logical fallacy that p4f2nw fell for at the begin of the thread.

Oh, and thank you very much for this inequality nayel, now whenever someone asks me to prove $ a^2\plus{}b^2\plus{}c^2\ge ab\plus{}bc\plus{}ca$ I can piss him off :D

[quote="tchebytchev"]nice solution due to  bertram :
just I note that :
$ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \ge \frac {2}{3}(\frac 1{b \plus{} c} \plus{} \frac 1{c \plus{} a} \plus{} \frac 1{a \plus{} b})$
$ \Longleftrightarrow a \plus{} b \plus{} c \ge 2(\frac {ab}{a \plus{} b} \plus{} \frac {bc}{b \plus{} c} \plus{} \frac {ca}{c \plus{} a})$
which is obvious by the fact that $ x \plus{} y \ge \frac {4xy}{x \plus{} y}$.[/quote]
Actually, there's $ a^2\plus{}b^2\plus{}c^2$ in the numerator, but the same method reduces to
\[ a^2\plus{}b^2\plus{}c^2\geq \sum_\text{cyc} a\cdot \frac{bc}{b\plus{}c}\]
which still works (although it uses $ a^2\plus{}b^2\plus{}c^2\geq ab\plus{}bc\plus{}ca$ in the end, so don't use this problem to prove $ a^2\plus{}b^2\plus{}c^2\geq ab\plus{}bc\plus{}ca$ :P )

[quote="scorpius119"][quote="tchebytchev"]nice solution due to  bertram :
just I note that :
$ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \ge \frac {2}{3}(\frac 1{b \plus{} c} \plus{} \frac 1{c \plus{} a} \plus{} \frac 1{a \plus{} b})$
$ \Longleftrightarrow a \plus{} b \plus{} c \ge 2(\frac {ab}{a \plus{} b} \plus{} \frac {bc}{b \plus{} c} \plus{} \frac {ca}{c \plus{} a})$
which is obvious by the fact that $ x \plus{} y \ge \frac {4xy}{x \plus{} y}$.[/quote]
Actually, there's $ a^2 \plus{} b^2 \plus{} c^2$ in the numerator, but the same method reduces to
\[ a^2 \plus{} b^2 \plus{} c^2\geq \sum_\text{cyc} a\cdot \frac {bc}{b \plus{} c}
\]
which still works (although it uses $ a^2 \plus{} b^2 \plus{} c^2\geq ab \plus{} bc \plus{} ca$ in the end, so don't use this problem to prove $ a^2 \plus{} b^2 \plus{} c^2\geq ab \plus{} bc \plus{} ca$ :P )[/quote]

I have exactly no idea where you pulled that a²+b²+c² from. And also, could you maybe double-check whether the relation in chebychev's approach is correct, because I think it's pointing in the wrong way.But his approach then reduces to straightforward AM-HM, so I really don't see how your solution works :(

[quote="tchebytchev"]nice solution due to  bertram :
just I note that :
$ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \ge \frac {2}{3}(\frac 1{b \plus{} c} \plus{} \frac 1{c \plus{} a} \plus{} \frac 1{a \plus{} b})$
$ \Longleftrightarrow a \plus{} b \plus{} c \ge 2(\frac {ab}{a \plus{} b} \plus{} \frac {bc}{b \plus{} c} \plus{} \frac {ca}{c \plus{} a})$
which is obvious by the fact that $ x \plus{} y \ge \frac {4xy}{x \plus{} y}$.[/quote]

[quote="bertram"]
Is that rearrangement?... Hm, I can't help but think that your inequality looks a bit fishy...
$ 3\left(\frac {ab}{a \plus{} b} \plus{} \frac {bc}{b \plus{} c} \plus{} \frac {cy}{c \plus{} a}\right)\le(ab \plus{} bc \plus{} ca)\left(\frac 1{a \plus{} b} \plus{} \frac 1{b \plus{} c} \plus{} \frac 1{c \plus{} a}\right)$
by rearrangement, but then the inequality sign is pointing in the wrong direction. It's the same logical fallacy that p4f2nw fell for at the begin of the thread.[/quote]

I think what tchebytchev did was

$ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \ge \frac {2}{3}\left(\frac 1{b \plus{} c} \plus{} \frac 1{c \plus{} a} \plus{} \frac 1{a \plus{} b}\right)$

$ \iff 3(a \plus{} b \plus{} c)\ge 2(ab \plus{} bc \plus{} ca)\left(\frac 1{b \plus{} c} \plus{} \frac 1{c \plus{} a} \plus{} \frac 1{a \plus{} b}\right)$

$ \iff 3(a \plus{} b \plus{} c)\ge 2\left(\frac {ab}{a \plus{} b} \plus{} c \plus{} \frac {bc}{b \plus{} c} \plus{} a \plus{} \frac {ca}{c \plus{} a} \plus{} b\right)$

$ \iff a \plus{} b \plus{} c \ge 2\left(\frac {ab}{a \plus{} b} \plus{} \frac {bc}{b \plus{} c} \plus{} \frac {ca}{c \plus{} a}\right)$

Well done, tchebytchev! :wink:

[quote="scorpius119"][quote="tchebytchev"]nice solution due to  bertram :
just I note that :
$ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \ge \frac {2}{3}(\frac 1{b \plus{} c} \plus{} \frac 1{c \plus{} a} \plus{} \frac 1{a \plus{} b})$
$ \Longleftrightarrow a \plus{} b \plus{} c \ge 2(\frac {ab}{a \plus{} b} \plus{} \frac {bc}{b \plus{} c} \plus{} \frac {ca}{c \plus{} a})$
which is obvious by the fact that $ x \plus{} y \ge \frac {4xy}{x \plus{} y}$.[/quote]
Actually, there's $ a^2 \plus{} b^2 \plus{} c^2$ in the numerator, but the same method reduces to
\[ a^2 \plus{} b^2 \plus{} c^2\geq \sum_\text{cyc} a\cdot \frac {bc}{b \plus{} c}
\]
[/quote]

I have no idea what you mean, my friend. Have you actually read bertram's solution?

[quote="bertram"]Oh, and thank you very much for this inequality nayel, now whenever someone asks me to prove $ a^2 \plus{} b^2 \plus{} c^2\ge ab \plus{} bc \plus{} ca$ I can piss him off :D[/quote]

Thank you, bertram! :) Your solution is really nice too. I will post my solution later. :wink:

[quote="nayel"]
I have no idea what you mean, my friend. Have you actually read bertram's solution?
[/quote]

Oops. I thought the method shown started from the beginning...

[quote="nayel"]Thank you, bertram! :) Your solution is really nice too. I will post my solution later. :wink:[/quote]

I want to see your solution nayel :)

[quote="bertram"]I want to see your solution nayel :)[/quote]

Sorry this is a bit late, but here's my solution:

WLOG assume that $ a\ge b\ge c$. This implies

$ a\plus{}b\ge c\plus{}a\ge b\plus{}c$   (1)

(1)=> $ \frac{a}{b\plus{}c}\ge\frac{b}{c\plus{}a}\ge \frac{c}{a\plus{}b}$   (2)

On the other hand, $ ab\ge ca\ge bc$   (3)

(3)=> $ a(b\plus{}c)\ge b(c\plus{}a)\ge c(a\plus{}b)$   (4)

Applying Chebyshev's inequality on (2) and (4) we get

$ 3(a^2\plus{}b^2\plus{}c^2)\ge \left(\sum\frac{a}{b\plus{}c}\right)\left(\sum a(b\plus{}c)\right)$

Or $ 3(a^2\plus{}b^2\plus{}c^2)\ge 2(ab\plus{}bc\plus{}ca)\sum\frac{a}{b\plus{}c}$

Which was what we wanted.

[quote="bibobeo"]Anyone think of applying SOS method? I think it would be very nice.[/quote]

$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \minus{} \frac {2}{3}(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}) \equal{} \frac {\sum{[(a \minus{} b)^2(ac \plus{} bc)(a \plus{} b \plus{} c)]}}{3(ab \plus{} bc \plus{} ca)(a \plus{} b)(b \plus{} c)(c \plus{} a)}$

 :)

[quote="Erken"][quote="ghjk"]Here's my solution:
Assume that c=min(a,b,c)
After substituting, i just need to prove: 3(a-b)^2/2(ab+bc+ac)+(c-a)(c-b)(3/2(ab+bc+ac)-(a-b)/(a+b)(b+c)(c+a))>=0
But it's obviously right since 3(a+b)(b+c)(c+a)-2(ab+bc+ac)(a-b)>0
This equality happens when a=b=c :)
@nayel:Where did u find this problem?[/quote]
Could you rewrite it with LaTeX?
And why is the last one true?
Thank you.[/quote]
Sorry for this late replying, Erken. I haven't passed by this topic for a long time.Anyway, the last inequality is true since 3(a+b)(bc+ab+ac+c^2)-2(ab+bc+ac)(a-b)=3(a+b)(ab+bc+ac)+3c^2(a+b)-2(ab+bc+ac)(a-b)=(ab+bc+ac)(a+5b)+3c^2(a+b), which is obviously greater than 0. :)
By the way, I don't know how to use LATEX though.Sorry! :oops:


AwesomeMath
summer program
Mathcamp
search
HCSSiM
Ross Mathematics Program
PROMYS

I've gotten brochures for several summer math camps (AwesomeMath, MathCamp, etc...). Are any of them more "prestigious" than others? Basically, which one should I go to, given the choice?

Whichever one sounds most interesting.  You might go look around the Contests and Programs section of the site (or do a search!) to find  reviews by past participants.

In case you or others haven't found it, there's a nice list here:

http://www.ams.org/employment/mathcamps.html

I've never had the impression that these camps are particularly prestigious.  You should choose your camp based on things like choice of topics.  Some good programs I know of are Hampshire, Ross, Promys, and Sumac.

The answers depend on what region you're from. Some, like MathCamp, draw from across the whole country, while others are more local/regional. Do you want to fly halfway across the country and pay a lot for the "best" experience, or just choose a good one that's not too far away?

do you know if a student coming form europe for example would be able to do summer school in usa?

Ross Mathematics Program definitely accepts students from all over the world. I'm pretty sure MathPath (for younger students) and MathCamp (about the same age range as Ross) do too.

To confirm, Mathcamp had some kids from UK, Romania, and Turkey (possibly more) last year and in years past.

but there's a strong selection to be accpeted to this camps?
which do you think is the easiest to be accepted to?

[quote="tokenadult"]If you don't have a strong preference among the camps, you may as well apply to more than one, as they have different admission and financial aid policies, and the application tests are interesting.[/quote]
i've done all the problems for the Ross Program, do you think this could give some opportunity to be accepted?
do you know if there is one among them which is easiest to be acceppted in?

Yes, some of the programs are more prestigious than others: these are the national-level programs, namely HCSSiM, MathCamp, Ross, PROMYS, SUMaC, and RSI.

I know that HCSSiM and MathCamp take international students.

Each of the programs uses different criteria in choosing students, and all of them have more applicants than students they can take, so there isn't any one of them that's easier to get into than the others. (I know of situations where Student A got into Program B but not Program C, whereas Student D got into Program C but not Program B.)  I would advise applying to more than one program; pick your favorite three and see what happens as you go through the application process.

RSI also takes international students; there are set quotas for different participating countries, as I recall.  China, Singapore, Australia, Poland, Sweden, Germany, Bulgaria (?), Lebanon, and Saudi Arabia all send students (I hope I haven't forgotten any).  I'd like to mention that RSI isn't exactly a "math camp," although we do get MOPpers and HCSSiM/PROMYS alumni.

do you know if Ross program accept international students? beacuse i've done all the problems and  i wouldn't have wasted time.

The Ross Program definitely accepts international students. (My son had classmates from several countries during Ross 2006.) It's not a waste of time to do the application test because the test makes you think about math.

Do you know how Mathcamp give financial aid?I know some of other programms are ready to give full financial aid but I need to know the basical things required to get financial aid besides Admission tests?Will you get full financial aid if you solve every problems correctly?

I think AwesomeMath is better.
I believe you can get a scholarship if you do well on USAMTS.

[quote="ghjk"]Why do you think Awesome is better? At first, I don't care too much about this programm but maybe I will consider it later when the application opens again! I also hear that Awesome gives full financial aid if you do very well on the Admission test! But it does have a good program as well as instructor(I just know Ivan Borsenco is one of instructors there)[/quote]
i'm looking for ross program because the test has the most interesting problems,and i've done all  :lol:
do you know if there are phisics summer camps?

That's not the main problem for choosing one instead of others, alexba913! The main purpose of these Admission test is that they want to see how you think of those specific problems and what's your original idea? I also did a few problems of Ross this year but it's not time for me to do that because of AIME! I think those problems are pretty good because they are discreet problems, which is not my right-hand sided :lol: .
@tokenadult: Do you know what benefits I will get besides experience, the best environment for learning Math and making new friends? I see it as an extra curriculum which will help me a lot when applying for US university as well as a kind of academic activity. But is there anything else? :roll:

This would be a good time for me to join in...

I am a 3-time (soon to be 4-time) Mathcamper, and that being said I would [i][b]strongly[/b][/i] encourage you to consider [url=http://mathcamp.org/]Mathcamp[/url].

On the issue of [url=http://mathcamp.org/fin.php]financial aid[/url], I would say that they're pretty generous. First off they give you an automatic tuition reduction based on your income, which can be for as much as $ \$$1200. In addition to that, they give out a bunch of other scholarship money. If that still isn't enough they'll help you with a local fundraising program. Bottom line: if you're accepted they'll do everything in their power to make sure you can attend.

At camp you have a wide variety of classes and you can attend whatever looks most interesting to you. There is complete academic freedom (something I don't think you'll find at many other camps). There are no "required" classes or any sort of rules about what classes you should or should not take. At the start of each week you are given a schedule of all the classes that week along with descriptions of the classes (and if you want more details on any of the classes, you just need to talk to the mentor teaching the class). After that, to attend a class you just show up. There's no sort of registration or anything like that. Nobody is even going to keep a permanent record of what classes you take (although you will fill out a "log table" each evening where you'll say what classes you took, how much you liked them, whether you tough they were too easy/too hard etc. these will only been seen by your academic adviser, so that you'll be able to remember that stuff when you meet with him/her at the end of the week. The log tables are destroyed at the end of camp (end of the week?)). In each class period, you can choose from about 5 different classes, or you can choose not to take any classes in that period (most people skip at least one class period most weeks). It's technically possible to not go to any classes for the entire camp (although that's kind of stupid). There's a lot of variety in the subjects and difficulties of the classes taught, so everyone should be able to find classes that are right for them. In addition to that, there no such thing as "passing" or "failing" a class and no such thing as grades or mandatory homework (in the sense that you'll get in trouble for not doing the homework [although some classes do hove "required" homework, which really means "you need to do today's homework if you want to understand what's going on in class tomorrow"]).

The classes are taught by grad students at various colleges (the "mentors") college professors (the "faculty") and visiting mathematicians who typically visit camp and teach some classes for one week (of these the most notable  is [url=http://en.wikipedia.org/wiki/John_Horton_Conway]John Conway[/url]). They teach classes on whatever they find most interesting, and they are all excellent teachers. The staff are also all friendly and assessable. Feel free to talk to them outside of class about anything you want to learn about.

The academic freedom I mentioned extends to the rest of camp too. You won't run into any of those strict, arbitrary rules you see at most other camps. Mathcamp really does have only four rules:

1) Be excellent to each other.
2) Don't do anything stupid.
3) No fire.
4) Don't divide by 0 (unless under staff supervision).

The point of Mathcamp is not just to learn math, it's also to have fun. There's a lot of free time outside of class. You're free to do what ever you want (as long as you don't break any of the four rules). There's always a lot of cool things happening. Throughout camp there are a lot of staff organized activities and field trips (all optional, of course!). Campers are also free to organize their own activities (they just need to put up a sign on the schedule board). Mathcamp has a lot of really cool traditions (more than most of the other camps, I expect).

I'm immensely happy that I chose to go to Mathcamp three years. I think it's had a big impact on my life. So yeah, go to Mathcamp.

Finally, a little note on the whole Mathcamp v. AwesomeMath thing. I would say that Mathcamp is a much better program than AwesomeMath. Not to say that AwesomeMath doesn't do it's job well, it is a very good program for what it does, but I don't think I has the right focus. AwesomeMath is heavily geared towards "contest math," the sort of math that's used for high school math competitions. While this sort of stuff might seem important now, it really isn't. Your math competition career will end when you enter college. Yes I know there's still a few things like the Putnam while you're an undergrad, but it's nothing like high school. In 4 or 5 years, do you think it's really going to matter what you got on the AMC or the AIME (or even the USAMO)? Being sucked into this mindset of "you must do whatever it takes to succeed on these competitions" is a real mistake. Contrary to what you might think, most of the math you're learning for these competitions is not "real" math. It's not the sort of thing you'd really be using for any sort of higher math. All these tricks and formulas and quick calculation methods is not what a real mathematician would use. (How often do you think you're average mathematician uses something like Stewart's theorem?)

That being said, I'm certainly not against competitions. I think they can be a lot of fun. They can get people interested in math, and can expose them to some things they wouldn't see otherwise. And they can develop your problem solving skills, which can be useful throughout life. But remember, they are a means to an end, not an end in themselves. If you're giving up other, better opportunities to train for these competitions, I would say that you are making a mistake, and one which you will likely regret in a few years.

While you may get some short term benefits from going to a place like AwesomeMath (like higher scores on some competition or making the USAMO, or even getting into MOP) it will really be kind of a shallow experience.  I can almost guarantee you that in 10 years, you won't be any better off for having gone to AwesomeMath. If you go to some place like Mathcamp (or even somewhere like Ross or PROMYS) you will get a much richer experience that will benefit you later in college or in the rest of your life.

Not to go all tribal on you, but since [b]JAM[/b] has posted such a long endorsement of Mathcamp, I think I can say something about HCSSiM.  HCSSiM is perhaps the highlight of my life so far.

The faculty/staff will make sure you go to class, learn things, and do work in problem session, because they care about your education.  It's hard-core, mathematically.  The program is small, tight-knit, and wacky.  Students think of lots of interesting ways to entertain each other (staff join in, too).

I will not paraphrase all the information on the [url=http://www.hcssim.org/]HCSSiM website[/url], which speaks for itself.  Do read it.

[quote="JAM"]If you go to some place like Mathcamp (or even somewhere like Ross or PROMYS) you will get a much richer experience that will benefit you later in college or in the rest of your life.[/quote]

A note on Ross and PROMYS:  both programs have a fairly heavy emphasis on number theory and on rigor (the PROMYS curriculum is based off of the Ross curriculum, so they are pretty similar).  The problem sets consist of proofs, initially directly from (ring) axioms, so one of the skills you'll develop heavily is proper proof-writing.  Some describe PROMYS as essentially training to be a future mathematician (as opposed to, say, Olympiad training), so if that's the direction you want to go, both programs are great.  They are, however, a lot of work.

[quote="JAM"]Finally, a little note on the whole Mathcamp v. AwesomeMath thing. I would say that Mathcamp is a much better program than AwesomeMath. Not to say that AwesomeMath doesn't do it's job well, it is a very good program for what it does, but I don't think I has the right focus. AwesomeMath is heavily geared towards "contest math," the sort of math that's used for high school math competitions...[/quote]

[quote]While you may get some short term benefits from going to a place like AwesomeMath (like higher scores on some competition or making the USAMO, or even getting into MOP) it will really be kind of a [b]shallow[/b] experience.  I can almost guarantee you that in 10 years, you won't be any better off for having gone to AwesomeMath. If you go to some place like Mathcamp (or even somewhere like Ross or PROMYS) you will get a much richer experience that will benefit you later in college or in the rest of your life.[/quote]

Let me start off by saying that I went to AwesomeMath.  And I think that there are a few benefits that JAM is missing...

Why did I go to AMSP?  Last spring, when I was deciding on which math camp to apply to, I was trying to think of which would be best for me.  And not necessarily for my future or anything, but which would be more FUN.  Now, not that MathCamp isn't fun, it's just that I still have no clue what area of math I want to go into.  And I feel like MathCamp is more geared towards you coming in with an area of interest (I haven't actually gone, but that's the feeling I get).  And I'd prefer not to think about it right now.  I'm having a great time just doing math for fun, and I think that fun should be the major inspiration instead of future.  Also, I don't necessarily agree with this whole "usefulness" argument.  I guess it's true, but only works if you are focusing on an area of math which you are certain you want to go into.

That being said, AMSP is geared more towards contest math, as JAM said.  And I very much see why that seems like a "shallow" experience.  But, I can tell you now, I have improved exponentially in ALL areas of math since I've gotten back, even in math outside of contest problems.  The goal is not necessarily contest math, but rather, the right MINDSET for doing math.

I guess what I'm trying to say is that math is fun.  It's supposed to be.  Do what you love to do more.  If it looks like your passion is more closely tied to something you see in a MathCamp brochure, then pursue it.  If you see it in an AMSP, then do the same.  The same goes for any other camp that I haven't mentioned.

[quote="banned UBER"]MathZoom looks pretty good as its goal is to prepare for all the prestigious contests (AMC, AIME, USAMO, GPML, etc.).[/quote]

I'm guessing you didn't read JAM's post.

[quote="krsattack"]
Let me start off by saying that I went to AwesomeMath.  And I think that there are a few benefits that JAM is missing...

Why did I go to AMSP?  Last spring, when I was deciding on which math camp to apply to, I was trying to think of which would be best for me.  And not necessarily for my future or anything, but which would be more FUN.  Now, not that MathCamp isn't fun, it's just that I still have no clue what area of math I want to go into.  And I feel like MathCamp is more geared towards you coming in with an area of interest (I haven't actually gone, but that's the feeling I get).  And I'd prefer not to think about it right now.  I'm having a great time just doing math for fun, and I think that fun should be the major inspiration instead of future.  Also, I don't necessarily agree with this whole "usefulness" argument.  I guess it's true, but only works if you are focusing on an area of math which you are certain you want to go into.
[/quote]

I haven't gone to MathCamp but this looks like the wrong interpretation to me. If you just look at the list of courses, you can see topics from all different areas. I think many people find what they enjoy by taking new classes, instead of taking them because they already know that will be their main field in the future.

[quote="krsattack"]
That being said, AMSP is geared more towards contest math, as JAM said.  And I very much see why that seems like a "shallow" experience.  But, I can tell you now, I have improved exponentially in ALL areas of math since I've gotten back, even in math outside of contest problems.  The goal is not necessarily contest math, but rather, the right MINDSET for doing math.

I guess what I'm trying to say is that math is fun.  It's supposed to be.  Do what you love to do more.  If it looks like your passion is more closely tied to something you see in a MathCamp brochure, then pursue it.  If you see it in an AMSP, then do the same.  The same goes for any other camp that I haven't mentioned.[/quote]

It's great that you think math is fun. Keep in mind, though, that other people might be looking for camps that will be the most mathematically nourishing and not necessarily the most fun. I went to AMP and I found it pretty fun and pretty instructive. But I also sensed that it was steeped too much in contests. There can be a difference between improving your problem-solving skills and improving your contest performance. In many instances, the Awesomemath lectures would be, "Here's an obscure theorem and here's a bunch of olympiad problems it can be applied to."

[quote="krsattack"]
Why did I go to AMSP?  Last spring, when I was deciding on which math camp to apply to, I was trying to think of which would be best for me.  And not necessarily for my future or anything, but which would be more FUN.  Now, not that MathCamp isn't fun, it's just that I still have no clue what area of math I want to go into.  And I feel like MathCamp is more geared towards you coming in with an area of interest (I haven't actually gone, but that's the feeling I get).  And I'd prefer not to think about it right now.  I'm having a great time just doing math for fun, and I think that fun should be the major inspiration instead of future.  Also, I don't necessarily agree with this whole "usefulness" argument.  I guess it's true, but only works if you are focusing on an area of math which you are certain you want to go into.
[/quote]

I'm going to agree with Phelpedo here, this is [b]completely[/b] the wrong interpretation. I would say that few, if any, mathcampers know what they're going to be doing when they start camp. I certainly had no idea what I was going to do when I started (and that's not much different now).

Basically, what's going to happen when you get to camp:
When you show up, you'll be given a folder. Among other things, it's going to have the course schedule for week one, and descriptions of the week one classes. (To see what this has been like in the past, look [url=http://mathcamp.org/academic.php#classes]here[/url].)

At some point (i.e. by Tuesday [you will arrive on Saturday]) you should go through and decide which classes you want to take. Now do you need to do this completely on your own? Of course not! First off, if the class descriptions don't tell you everything you want to know about the classes, you can (and should!) talk to the staff member teaching the class. They'll tell you what ever you want to know. In addition to this, you will have an academic advisor who will help you make decisions.

Ultimately, the way to choose classes is just to find out as much as you can about them, and then take the ones that sound the most interesting.

Most likely you won't have any clue what classes you're going be taking before you get to camp (you won't actually get the class schedule before you show up anyway). I'd say that my first year, the majority of my classes were on topics I didn't even know [i]existed[/i] before I came to camp.

As for the issue of fun, I'm going to say Mathcamp is much more fun than AwesomeMath can ever be. Admittedly, I've never been to AwesomeMath, so I can't be sure (you'll have to ask some one who's been to both). However, I have been to MOP, and I assume that there'd be at least some similarities between them (they have the same focus, they're run by pretty much the same people etc.), and I have have to say I greatly prefered Mathcamp. Not to say I didn't like MOP, I had a great time there, but I liked Mathcamp much more. To me, MOP just felt a little too structured and strict, whereas Mathcamp is kind built on both academic and non-academic freedom. I felt the staff at Mathcamp was much more approachable, they can't really boss you around and you really just talk to them like your equals. Mathcamp doesn't have any strict, annoying rules, they realize that you can trust people to have some common sense. (ex. "From now on the game of Mafia is banned at MOP. No exceptions"  :|  *sigh* I have a feeling Mathcamp would have handled that [i]very[/i] differently...)

I also felt that MOP was way too competitive, yeah I know there's a lot of collaboration, but it's still ultimately a competitive environment. Your goal there is to tangibly improve in math. That puts a lot of focus on how well you're doing. It puts pressure on you do well. If you find yourself lost in a class, or if you fail a test, you can start to feel like you're failing, your living up to expectations. And all the time you're sort of competing with everyone else. I find that that can take some of the fun out of it. I also find MOP can sometime get a little [i]too[/i] focused on doing math, at the expense of other things. Like, they canceled the laser tag trip we had planned in '06, and they had to keep resceduling the '06 USEMO and it almost didn't happen (because it's useless...)

It is completely different at Mathcamp. Your goal there is just to learn a bunch interesting math (stuff that [i]you[/i] find interesting) and really get a feel for the beauty of math, to hang out with a bunch of cool people, and above all to have fun. It is completely non-competitive. Your just trying to learn as much as you want to learn. If you're completely lost in class, it's no problem as long as you're still having fun (trust me, I've been in this situation a lot). You don't need to worry about failing, or about what other people will think. There's no pressure on you (which is not to say that people slack off, I find that mathcampers are, on the whole, pretty dedicated to learning). I find that this stress-free environment, with a lot of friendly support from your peers, is one the best environments for learning math.

[quote="krsattack"]That being said, AMSP is geared more towards contest math, as JAM said. And I very much see why that seems like a "shallow" experience. But, I can tell you now, I have improved exponentially in ALL areas of math since I've gotten back, even in math outside of contest problems. The goal is not necessarily contest math, but rather, the right MINDSET for doing math. [/quote]

That's actually similar to the experience I had with mathcamp. I'd say my math skills, both in competitions and other math, improved significantly between 9th and 10th grades (after the summer I first went to Mathcamp). I'm not sure how much of that I can really attribute to Mathcamp, and how much is just natural improvement which would have happened anyway, but I do think that I got quite a lot out of Mathcamp.

[quote="krsattack"]Why did I go to AMSP?  Last spring, when I was deciding on which math camp to apply to, I was trying to think of which would be best for me.  And not necessarily for my future or anything, but which would be more FUN.  Now, not that MathCamp isn't fun, it's just that I still have no clue what area of math I want to go into.  And I feel like MathCamp is more geared towards you coming in with an area of interest (I haven't actually gone, but that's the feeling I get).  And I'd prefer not to think about it right now.  I'm having a great time just doing math for fun, and I think that fun should be the major inspiration instead of future.  Also, I don't necessarily agree with this whole "usefulness" argument.  I guess it's true, but only works if you are focusing on an area of math which you are certain you want to go into.[/quote]

Just to note, I'm a Mathcamper ('07) and MathPathian ('06), so I can't say anything about AMSP, but I do want to say a few things about MC (which have already mostly been said by Phelpedo and JAM, so I guess I'm repeating).

I find it rather funny that the reason you choose AMSP--you wanted to go to a camp that was fun besides just good for learning math--is the exact reason I went to MC. I enjoy math contests, but they're hardly my favorite part of math. AMSP looked like it was really meant for preparing for contests (if I'm wrong here, sorry, this is just my impression). The fact that you had to chose the track to take during ASMP before camp didn't appeal to me either. I fortunately got into MC, and choose it because it sounded way more fun: less contest based and a multitude of class topics. Just as an example that you don't need to have a field of interest before going to MC:

Classes I took in '07:
Intro to Problem Solving (5 weeks)
Linear Algebra (2 weeks)
Combinatorial Game Theory (2 weeks)
Intro to Graph Theory (1 week)
Enumeration Celebration (1 week)
Math of Election (.5 weeks)
Math of Juggling (.5 weeks)
Probabilistic Method (1 week)
Diophantine Approximation (1 week)
Colloquia (every weekday at camp, which is Tuesday through Saturday for MC)

Going into camp, I'd say that my favorite field in math was geometry. I somehow managed to take no geometry at MC, and now I love combinatorics (it's tied with geometry), because Enumeration Celebration and Probabilistic Method were such awesome classes. Even though some classes lost me by the end (Diophantine Approximation and some of the Colloquia), they were all interesting.

I truly don't know what I want to do in math for the future. I'm pretty sure I'll major in math in college, though. Just, for me, a less contest based camp is more enjoyable, and I think I get more from it. Not that MC doesn't help at all in contest math--my USAMTS score more than doubled from last year, even though MC classes were so relaxed and exciting that it seems impossible that I was improving my math contest skills at the same time.

Just out of curiosity, what is there in the MC brochure that makes it appear you need a subject of interest before you go?

On a completely different note, MC is worth going to for just the culture, besides the math. I've never encountered such a welcoming community elsewhere (MP comes pretty close, though). With a schedule board where campers themselves can put up activities, there's always something to do, and almost always something you'll enjoy. During MC, I baked pies, played capture the flag, learned many obscure board games, learned beading (I made a bracelet and necklace!), was a werewolf (in Interactive Literature), solved Rubik's cubes, etc. In addition, there are all kinds of math activities to do outside of classes: team problem solving (TPS) every wednesday night, relays on the weekend (once involved swimming across a pool to get new problems), TAU (a time every weekday that stuff can't be scheduled on to do homework, projects, or the standard random conversations you'll find at MC.)

Basically, I thought MC was amazing. I don't know what AMSP is like, but there's no way to say that MC isn't fun. (if anyone is interested in MP, I can ramble on about that, too :P )

[quote="JAM"]

While you may get some short term benefits from going to a place like AwesomeMath (like higher scores on some competition or making the USAMO, or even getting into MOP) it will really be kind of a shallow experience.  I can almost guarantee you that in 10 years, you won't be any better off for having gone to AwesomeMath. If you go to some place like Mathcamp (or even somewhere like Ross or PROMYS) you will get a much richer experience that will benefit you later in college or in the rest of your life.

[/quote]

I would just like to point out the existence of Gabriel Dospinescu's Number Theory 3 class, that is far from the shallow competition-oriented training you describe.


geometry
geometric transformation
reflection
inequalities
inequalities unsolved

$ a,b,c,d>0$ prove that

\[ a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}abcd\plus{}1\ge ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd\]

are you sur , because we have

$ a^2\plus{}b^2\plus{}c^2\plus{}d^2 \geq \frac{8}{3} (ab\plus{}bc\plus{}cd\plus{}da\plus{}bd\plus{}ca)$

this is from mathematical reflections, current issue  :wink:

[quote="M.A"]$ a,b,c,d > 0$ prove that
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} abcd \plus{} 1\ge ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd
\]
[/quote]
$ abcd \plus{} 1\geq2\sqrt{abcd}.$ :wink:

[b]Marian Dinca(teacher,  [i]Bucuresti-Romania[/i]), wrote:[/b]
              $ a,b,c,d > 0\Rightarrow a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} 2.\sqrt {abcd}\ge ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd$.
[b]Remark: [/b]The MA(Easy) problem [b]is the problem S88[/b](of [b]"[i]Mathematical Reflections[/i]", n.3/2008[/b]), [b]and Solutions should arrive by July 19 !!![/b]
view: http://reflections.awesomemath.org/2008_3/MR_3_2008_problems.pdf

[quote="mihai miculita"]
view: http://reflections.awesomemath.org/2008_3/MR_3_2008_problems.pdf[/quote]
A problem O85 by Arkady Alt is a very easy problem. :wink:

There is a nice solution using $ SMV$ theorem :wink:

[quote="arqady"][quote="M.A"]$ a,b,c,d > 0$ prove that
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} abcd \plus{} 1\ge ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd
\]
[/quote]
$ abcd \plus{} 1\geq2\sqrt {abcd}.$ :wink:[/quote]
Without loss generality, we can assume that: $ a \geq b \geq c \geq d$.
We have:
$ f(a,b,c,d) \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} abcd \plus{} 1 \minus{} (a \plus{} c)(b \plus{} d) \minus{} ac \minus{} bd$
Hence :$ f(a,b,c,d) \minus{} f(\sqrt {ac},b,\sqrt {ac},d) \equal{} (a \minus{} c)^2 \minus{} (\sqrt {a} \minus{} \sqrt {c})^2(b \plus{} d) \geq 0.$
By SMV theorem, we  only need prove the inequality with:$ a \equal{} b \equal{} c \equal{} t$and$ d$
The inequality becomes :  $ 3t^2 \plus{} t^3d \plus{} d^2 \plus{} 1 \geq 3t^2 \plus{} 3td.$
euqavilent to :$ t^3d \plus{} d^2 \plus{} 1 \geq 3\sqrt [3]{t^3d^3}.$
Equality Holds if and  only if $ a \equal{} b \equal{} c \equal{} d \equal{} 1$


x,y,z>0 , 1/3 \leq xy+yz+zx \leq 3 . Determine  the range  values  of  xyz  and  x+y+z
thanks

From AM/GM, we have xyz  \leq  ( (xy+yz+zx)/3 ) 3/2   \leq  1, and equalities occurs iff x = y = z.
In another hand let x = y = 1, and let z - > 0.
Note that 1/3 /leq xy + yz +zx = 1 + 2z /leq 3
and we have xyz - > 0.
Thus 0 < xyz  \leq  1, and this is best possible.

It is well_known that (x+y+z) 2   \geq  3(xy+yz+zx).
Thus : x+y+z  \geq  1 and equality occurs for x = y = z = 1/3.
In another hand, let y = z = 1/x where x  \geq  1.
Then 1/3 /leq xy+yz+zx = 2 + 1/x 2 /leq 3,
and x+y+z = x + 2/x - > +oo as x - > +oo.
Thus x+y+z can take all values in [1; +oo[.

Pierre.


Is the number
\[\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}\]
rational or irrational ?

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Let $\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}=x \Rightarrow \sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}-x=0$. But it is very-well known that if $a+b+c=0$, then $a^3+b^3+c^3=3abc$, so we have that $2\sqrt5-x^3=-3x \Rightarrow (x-\sqrt5)(x^2-x\sqrt5+2)=0$, but it is easy to verify that the equation $x^2-x\sqrt5+2=0$, has no real solutions, then we must have $x=\sqrt5$, which is irrational.


geometry
incenter
trigonometry
geometric transformation
reflection
geometry unsolved

∠A=60°,let I be the incenter of ABC,IE is parallel to AC and meets AB at E,D lies in BC such:CD=2BD.prove:
∠B=2∠DEB
OF COURSE, ANY IDEA OR SOLUTION WILL BE APPRECIATED.

is it too hard or too simple?
anyway,using sine laws simply,we can solve it though the solution looks very ugly.
what i wonder is how can we use the point D?
does it have a direct link with other points?

[quote="yetti"]Reflect $ I$ in $ AB, AC$ into $ J, K.$ $ \triangle AIJ, \triangle AIK$ are both equilateral. $ E$ is center of equilateral $ \triangle AIJ$ $ \Longrightarrow$ $ J, E, K$ are collinear, $ JK \perp AI$ and $ EK \equal{} 2 JE.$ $ JK$ cuts $ AC$ at the center $ F$ of equilateral $ \triangle AIK.$ $ EJBI, FKCI$ are both kites,

$ \angle EJB \plus{} \angle CKF \equal{} \angle BIE \plus{} \angle FIC \equal{} 360^\circ \minus{} (\angle EIF \plus{} \angle CIB) \equal{} 180^\circ$

$ \Longrightarrow$ $ BJ \parallel CK.$ Together with $ DC \equal{} 2 BD, EK \equal{} 2 JE$ $ \Longrightarrow$ $ DE \parallel BJ \parallel CK.$ Then
$ \angle DEB \equal{} \angle JBE \equal{} \angle EBI \equal{} \angle ABI \equal{} \frac {_1}{^2} \angle ABC.$[/quote]
wow!what a fantastic solution!!thank you so much.yetti.you are excellent! :lol:


inequalities
graph theory
advanced fields
advanced fields theorems

We know that the following inequality holds for every plannar graph:
\[E\leq 3V-6 \]
But considering the graph $K_{2}$ constructed by two vertices and an edge connecting them, we have $1=E>3V-6=0$. Why's that?

[quote="JBL"]Well, clearly then we [i]don't[/i] know that. :-)  In fact, the theorem is that in any simple, planar graph with at least 3 vertices, $E \leq 3V-6$.  I believe that a proof is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126918]here[/url].  (Also, $\textrm{graph theory}\subset \textrm{combinatorics}$, so this would have been more at home in the combinatorics fora.)[/quote]

Thanks...now I [i]know[/i] that. :oops:


4. How many positive integers less than 10,000 have at most two different digits?

5. In order to complete a large job, 1000 workers were hired, just enough to complete the job on schedule. All the workers stayed on the job while the first quarter of the work was done, so the first quarter of the work was completed on schedule. Then 100 workers were laid off, so the second quarter of the work was completed behind schedule. Then an additional 100 workers were laid off, so the third quarter of the work was completed still further behind schedule. Given that all workers work at the same rate, what is the minimum number of additional workers, beyond the 800 workers still on the job at the end of the third quarter, that must be hired after three-quarters of the work has been completed so that the entire project can be completed on schedule or before?

8 How many positive integer divisors of 2004^2004 are divisible by exactly 2004 positive integers?

12. Let ABCD be an isosceles trapezoid, whose dimensions are AB = 6, BC = 5 = DA, and CD = 4. Draw circles of radius 3 centered at A and B, and circles of radius 2 centered at C and D. A circle contained within the trapezoid is tangent to all four of these circles. Its radius is [-k + m*sqrt(n)]/p, where k, m, n, and p are positive integers, n is not divisible by the square of any prime, and k and p are relatively prime. Find k + m + n + p.

Solutions would be greatly appreciated  :)

my attempt at #8
[hide]to get exactly 2004 divisors, the products of the powers of the prime factors + 1 must equal 2004
we can divide it into 3 cases

case 1: there is only 1 prime factor
there are 1*3P1 = 3 possibilities, namely
22003, 32003 and 1672003

case 2: there are 2 prime factors
now we look for factor pairs of 2004 (without 1 which is covered in case 1)
they are: 2*1002, 3*668, 4*501, 6*334 and 12*167
thus there are 5 pairs, and 3P2 possible base arrangements
5*6 = 30 possibilities

case 3: there are 3 prime factors
now we look for factor triplets of 2004 (omitting 1 again which are covered in previous cases)
they are 2*2*501, 2*3*334, 2*6*167, and 3*4*167
2*2*501 only has 3P1 base arrangements, and the other 3 triplets have 3P3
3+3*6 = 21

3+21+30 = 54

#4
[hide]
$9\cdot\left[9\cdot{3\choose 1}+9\cdot{3\choose 2}+9\cdot{3\choose 3}+1\right]$
$+9\cdot\left[9\cdot{2\choose 1}+9\cdot{2\choose 2}+1\right]$
$+9\cdot\left[9+1\right]$
$+9$
$=927$

First line is the ways to choose numbers with at most 2 digits in the 1000s.
Second line is the ways to choose numbers with at most 2 digits in the 100s.
Third line is the ways to choose numbers with at most 2 digits in the 10s.
Fourth line is the ways to choose numbers with at most 2 digits in the 1s.[/hide]

More explanation... Sorry, I was a little rushed when I posted it.
[hide]  Basically, I broke it into 4 cases 1000s, 100s, 10s, 1s.  For the first case, we can fix the first digit and we have the remaining 3 slots to insert numbers into.  C(3,1) is where we select 1 of the 3 slots and put 1 of the 9 remaining digits in it e.g. 9199.... C(3,2) is where we select 2 of the 3 slots and put 1 of the 9 remaining digits in it e.g. 9119... C(3,3) is where we select 3 of the 3 slots and put 1 of the 9 remaining digits in it e.g. 9111.  Finally, we have the case where all 4 digits are the same 9999.  Multiply this by 9 to change the 1st digit.[/hide]

alan, I got 57 also, can anyone else confirm this answer?

ragingg, could you explain your solution more? I don't see how you get that equation.

#5: [hide="Hint"]The hardest part of this problem is organizing all of the information.  Can you devise a simple system for doing so?  If it makes it easier for you, for example, you might assume that 1/4 of the time = 1 day.[/hide]

[hide="Solution"]Let's suppose that the job requires 4000 units of work, so that it takes 1000 people 4 units of time to accomplish the job.  Then after one unit of time (1/4 of the total job), 1000 units of work have been done; after the second unit of time, 900 more units have been done, and after the third, 800 more have.  So 1300 units of work must still be done in 1 unit of time, i.e. 1300 people are needed.  Thus 500.[/hide]

#12: [hide="Hint"]The key to this problem is drawing the segments connecting the centers of the circles, especially for the inscribed one.  Once you do this, it's a pretty routine problem.[/hide]

[hide="Solution"]Following the suggestion of the hint, we let the radius of the inscribed circle be $r$, and we'll call the center of that circle E.  Then $AE = BE = 3+r$, $CE=DE=2+r$.  Drawing perpendiculars from E to $AB$ and $CD$ yields two different right triangles: one with base 3 and hypotenuse $3+r$; and another with base 2 and hypotenuse $2+r$.  So the other legs are $\sqrt{r^2+6r}$ and $\sqrt{r^2+4r}$, respectively.  But the sum of the two distances is the height of trapezoid ABCD, which is $2\sqrt{6}$ from some quick geometry (drop perpendiculars from D and C to $AB$).  Thus $\sqrt{r^2+6r} + \sqrt{r^2+4r} = 2\sqrt{6}$.

From here, we're pretty much done.  Move one of the square roots to the RHS, square, cancel, isolate the other square root, square, cancel, and solve with the quadratic equation.  You get $r=\frac{-60+48\sqrt{3}}{23}$, so the answer is 60+48+3+23 = 134.[/hide]

8. How many positive integer divisors of 2004^2004 are divisible by exactly 2004 positive integers?

The MathJam has a really good explanation of this:

http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=27#8

[quote="alan"]
[hide]
case 3: there are 3 prime factors
now we look for factor triplets of 2004 (omitting 1 again which are covered in previous cases)
they are 2*2*501, 2*3*334, 2*6*167, and 3*4*167
thus there are 4 triplets and 3P3 base arrangements
4*6 = 24

3+24+30 = 57
[/hide][/quote]

How many 'base arrangements' are there for the first set of exponents?

[quote="mathfanatic"][quote="alan"][hide]case 3: there are 3 prime factors
now we look for factor triplets of 2004 (omitting 1 again which are covered in previous cases)
they are 2*2*501, 2*3*334, 2*6*167, and 3*4*167
thus there are 4 triplets and 3P3 base arrangements
4*6 = 24[/hide][/quote]

[hide]Be careful here.  Are there really 6 possibilities resulting from the 2*2*501 case?  (I think there are only three - pick which of 2, 3, and 167 is the 501.)  Thus there are only 21 for this case, and the final answer comes out to be 54, as expected.[/hide][/quote]

ah thanks, i see it now...silly me :blush:


Tommy has trouble remembering telephone numbers, so when Jenny gives him her telephone number, he wants to write it down.  But Tommy wants to make sure that if he loses the paper with the number on it nobody picks it up and calls Jenny, so instead of writing the number, he breaks the number abc-defg into a 3 digit number abc and a 4 digit number defg, and writes the fraction abc/defg = .16330759088341 on the paper.
Sure enough, Tommy forgets the number, but finds the paper, and takes out a 4-function calculator, writes some numbers on scratch paper, and in a few seconds comes up with the number.

Question: What is the number, and how did he do it so quickly?

Hmm.. this looks like a very interesting problem, and I haven't worked out how to do it yet (though I did notice that the decimal [hide]wasn't exact[/hide], thats easy to check by hand). Used a computer to solve it . I'm very interested. will keep trying.

I'm glad you liked my question, it came to me while i was listening to music and studying number theory.

Ravi B wrote:My guess is to use [hide]continued fractions[/hide].

Could you please be more specific? I still don't quite get it. How does the continued fraction expansion of .16330759088341 help us?

Heres a hint:

[hide]

That decimal is not exact. Its last digit has been rounded.

[/hide]

And if you're still stuck:

[hide]

Each step of the continued fraction will cause it to be more and more precise.

[/hide]

Ok....but abc/defg is equal to the value of that continued fraction, and there is no easy way to somehow set the continued fractions of abc/defg and the decimal equal and then set each term in the continued fractions equal to solve for a,b,c,d,e,f,g.

The problem is basically asking us to approximate the given decimal by a fraction with denominator at most 9999.  Continued fractions are known to be good approximations.  Namely, if you stop the continued fraction of a number at any point, and then simplify the continued fraction to a pure fraction p/q (without bothering to reduce terms), then p/q is known to be the closest approximation to the original number with denominator at most q.

So here's the procedure.  Convert the given decimal to a continued fraction.  After each step, simplify the continued fraction to a pure fraction, and see how good an approximation your pure fraction is.  Eventually you will get a pure fraction that is right on.  It took me about 7 steps to get the right fraction.  (You can short-cut the procedure a bit if you see a decimal you recognize.  For example, when I saw a decimal that started like .727272..., I should have guessed that it was 8/11.)

Rule of thumb: When a problem asks you to approximate a decimal by a fraction (with small denominator), consider continued fractions.

Oh, and now I see why zscool mentioned that he was listening to music at the time.  Do you remember the pop song from 1982 that had a phone number in its title?  Oh wait, most of you weren't born yet...

lol, I'm glad that somebody got the reference, the song is called 8675309 by the band Tommy Tutone, and the girl's name in the song is Jenny.

Hey, I've heard that song -- not that I would remember it, but I have heard it.  Even if I was -2 at the time.

yea, i wasn't born either, but it was on a commercial for verizon where the girl wants to keep her number because its famous

I listen to a lot of somewhat old music (somewhat meaning anywhere from the 60s to the 80s) and I've heard that song.

Jenny, Jenny, who can I turn to?
...........
Jennnnnnnyyyy, I've got your number
I need to make you mine
Jennnnnnnnyyy, I'll call your number
8675309

Fun.


In the circle O, chords CD and AB intersect at right angles at point E.  IF CE = 8, DE = 6 and AE=24, what is the radius of circle O?

(BTW, is the chords intersect at right angles does that mean that on of the chords is the diameter?)

[quote="DaReaper"](BTW, is the chords intersect at right angles does that mean that on of the chords is the diameter?)[/quote]
Not necessarily; they can be perpendicular without one of them being a diameter. One of them [i]must[/i] be a diameter only if it also bisects the other.

[hide]Drawing a diagram, we can see a right triangle with legs $ 6 \plus{} 1 \equal{} 7$ and $ \frac {24}{2} \minus{} 2 \equal{} 11$ and hypotenuse $ r$.

Using the Pythagorean Theorem, we find that $ r \equal{} \boxed{\sqrt{170}}$.[/hide]