analytic geometry
pigeonhole principle
number theory proposed
number theory

Let $ T$ be the set of all positive integer divisors of $ 2004^{100}$. What is the largest possible number of elements of a subset $ S$ of $ T$ such that no element in $ S$ divides any other element in $ S$?

The poset of divisors of any positive integer $ n$ is a product of chains and is therefore [url=http://en.wikipedia.org/wiki/Sperner_family]Sperner[/url], unimodal, and rank-symmetric, where the rank of a divisor $ d$ is the total number of prime factors it has.  Since $ 2004^{100} \equal{} 2^{200} \cdot 501^{100}$, we can take $ S$ to be the set of divisors with exactly $ 150$ prime factors, of which there are $ 101$.

Funny to catch t0rajir0u with such a mistake  :) Everything he says is correct, and I can add a reference to the de Bruijn-Tengbergen-Kruyswijk theorem - but $501 = 3\cdot 167$ is not a prime.

Therefore, the maximal antichain is
\[S = \{2^{\alpha}3^{\beta}167^{\gamma} \mid 0\leq \alpha \leq 200, 0\leq \beta \leq 100, 0\leq \gamma \leq 100, \alpha + \beta + \gamma = 200\},\]
and now $|S|$ is a little harder to compute.

Sorry for opening an old post but is |S|=677

[quote="mavropnevma"]No. In fact $|S| = 2(1+2+\cdots + 100) + 101 = 101^2 = 10201$.[/quote]
could you explain please?

The maximum number of elements is $101^2$. We factorise $2004^{100}=2^{200}3^{100}167^{100}$. In a orthogonal coordinate system $(x,y,z)$ map $ 2^{a}3^{b}167^{c}$ to $(a,b,c)$. The divisors of $2004^{100}$ make up a $200\times 100\times 100$ parallelepiped. Suppose that the largest set $ S $ has more than $ 101^2 $ elements, then by pidgeonhole principle two of those will have the same $ (y,z) $ coordinates and one will divide the other. To get an example for $101^2$ elements consider the set of points $ (200-y-z,y,z) $ ,then it is clear that if the $(y,z)$ coordinates of a point $A$ "contain" those of point $B$ ($y_a \geq y_b, z_a \geq z_b$ , at most one of the equalities can hold), then $x_b>x_a$ and $B$ does not divide $A$.


function calculus quadratics derivative conics parabola

[quote="joml88"][hide="Hint"]Expand everything out and collect like terms.  You should get a quadratic.[/hide][/quote]
Yeah, that works, but you don't have to expand everything out.
[hide]
Rearrange the last 4 terms to $(x-9)^2-(x-8)^2+(x-3)^2-(x-4)^2$.
Using $a^2-b^2=(a+b)(a-b)$, we can rewrite it as $(2x-17)(-1)+(2x-7)(1)=10$.

So $f(x)=(x-5)^2+10$ ;) [/hide]

Oh, that's nice :)  Expanding everything out wouldn't take very long either though.

if in doubt use calc, its longer though:

f'(x) = 2(x-5)-2(x-4)-2(x-8)+2(x-3)+2(x-9) =0
       = 2x-10 =0

min occurs at x=5

Well... you shouldn't need to prove that the x=5 solution is the minimum--clearly the function reduces to a quadratic, which means there's only one extrema.  And since the problem implies that there is a minimum, just go with the derivative answer. ;)  Or, you could notice that there are two -x^2 terms and three x^2 terms to add, which means the quadratic coefficient is positive, and therefore the parabola opens upwards and the vertex is a minimum. *shrug*

I would have went with the calculus solution, but the algebraic method is definitely better.  Great thinking! :)

[quote="t0rajir0u"]Ah, but the calculus solution only proves $x = 5$ is an extrema, you need to find $f''(5)$ to prove it's a minimum  :)

Personally, I find the algebraic solution much more elegant.[/quote]
Just for reference, finding $f''$ is always a really crappy way to find out whether something is maximum or minimum. It is much better to simply study the sign of $f'$ around the point.

I agree with Kalle. most of the time $f'$ will not be a pretty function to take ther derivative of. but if it is easy to find $f''$ i say go for it, i personally find this faster.

Ok, that's enough calculus for intermediate (remember: there shouldn't be [i]any[/i] calculus in the intermediate forum).

LOCKED


function
real analysis
calculus
calculus computations

This one's even worse, since you don't have a linear differential equation.
You can't just make up a series and expect it will be a recognizable function. That $\sqrt{n}$ is essentially impossible to deal with.

is there a way to set it up as a nonlinear DE, sorry for posting this

This is a polylogarithm: more generally, $\mathrm{Li}_{\nu}(x)=\sum_{n=1}^{\infty}\frac{x^{n}}{n^{\nu}}$. [url=http://functions.wolfram.com/]Wolfram's functions website[/url] has everything you want to know about polylogarithms (and other special functions) and much, much more.

I looked for a differential equation or something similar, and didn't find one. A transform method and some form of convolution might do something, but I don't know the right one.

When you make up a summation problem, you should expect it to be unsolvable.


$ 3^{4n} \equiv 1 \pmod{3}$
$ 3^{4n\plus{}1} \equiv 3 \pmod{3}$
$ 3^{4n\plus{}2} \equiv 9 \pmod{3}$
$ 3^{4n\plus{}3} \equiv 7 \pmod{3}$

And, $ 100\equal{}4\times25$.

Thus, units digit is $ \boxed{1}$.

A different way of thinking about it:

Obviously, we don't want to compute $ 3^{100}$ and find the units digit. So let's see if we can find a pattern.

$ 3^1$ has a units digit of 3; $ 3^2$ has a units digit of 9; $ 3^3$, 7; $ 3^4$, 1; and so on. Notice that the pattern goes in groups of our: 3, 9, 7, 1, 3, 9, 7, 1, ...

Since $ 100 \equiv 4 \pmod 3$ (that is, it corresponds to the fourth number in the pattern), it has a units digit of $ \boxed{1}$.


[url]http://www.zendurl.com/h/hallows/[/url]
[hide="some chosen spooilers:"]
Ron dies.
Voldemort dies.
Harry survives.
Nagini/Cup/locket/Ravenclaws book are the horcruces.
Harry is nto a horcrux.
[/hide]
I wonder whether this is a fake?

Wow,

So ron gets killed by belletrix?

And then voldemort gets killed by harry and is saved by a door

What a cheesy ending.  Thats the worst ending Ive ever seen.  And there was no significance to previous mentioned plot items in other books.  So its like, wow heres a horucrux, wedestroy it, now we find another.

badly written book

it was bound to be a badly written book. JKR made the thing far too hot for her handling last time.
I rather found it better than I expected (whatever I can get from the spooilers out there).

By the way here you find the epilogue:
[url]http://www.gryffindorgazette.com/2007/07/17/leaked-deathly-hallows-spoilers-yep-the-real-thing-with-pictures-text-and-summary/[/url]

Oops I am sorry Ron does not die apparently. That was a fake spoiler.

Those spoilers are very definitely fake. (Probably) real spoilers are linked to in the Round Table thread. Don't judge the book yet, all of the reviews so far (that is to say, both of them) are highly favorable.

[quote="indianamath"]pm me if you want the pdf version of the book. no kidding.[/quote]
post the link.

I had posted the link in Round Table, but richard has asked me not to post it. I may PM it though.

Yeah, there was a big leak...I think you can find it on the net.


What is the probability that a randomly chosen factor of $ 15^{90}$ is a multiple of $ 15^{65}$? Express your answer as a common fraction.

[quote="mewto55555"]15^90=3^90x5^90

91x91=8281 factors.

WE want factors that are 15^65xsome factor of 15^25. 15^25 has 25x25=625 factors. Thus its 625/8281.[/quote]

Wouldn't 15^25 have 26*26=676 factors? That would make 676/8281.

Oh yes sorry. Other than that is my solution correct?

Alternatively, note that all factors of $ 15^{90}$ can be expressed as $ 3^a\cdot5^b$. In order for it to be a multiple of $ 15^{65}$, both $ a,b$ must be greater than or equal to $ 65$. This gives us a probability of $ \frac{26\cdot26}{91\cdot91}\equal{}\boxed{\frac{676}{8281}}$.

But that can be reduced if you prime factorize the numerator and denominator:
$ \frac{676}{8281}\equal{}\frac{2^2 \cdot 13^2}{7^2 \cdot 13^2}\equal{}\frac{2^2}{7^2}\equal{}\frac{4}{49}$
...assuming math154's answer is right


Prove that for any positive real numbers a,b,c we have the following inequality:

\[(a^{2}+b^{2}+c^{2})(\frac{1}{a^{2}+ab}+\frac{1}{b^{2}+bc} + \frac{1}{c^{2}+ca} )\geq \frac{9}{2}\]

cheers! :D :D

that's quite easy(maybe not the most elegant proof, but is is one anyway): first, let's make the inequality symmetric. this can be done writing
$\frac{1}{a^2+ab}=\frac{c^2}{a^2c^2+abc^2}$ etc. and then using the inequality
$\frac{c^2}{a^2c^2+abc^2}+\frac{a^2}{b^2a^2+bca^2}+\frac{b^2}{c^2a^2+cab^2}\geq \frac{(a+b+c)^2}{a^2b^2+a^2c^2+b^2c^2+a^2bc+ab^2c+abc^2}$
by cauchy-schwarz. thus it suffices to proof
$(a^2+b^2+c^2)(a+b+c)^2\geq \frac{9}{2}(a^2b^2+^2c^2+b^2c^2+a^2bc+ab^2c+abc^2)$
which is equivalent to
$2a^4+2b^4+2c^4+4a^3b+4a^3c+4b^3a+4b^3c+4c^3a+4c^3b\geq 5(a^2b^2+^2c^2+b^2c^2+a^2bc+ab^2c+abc^2)$
(thanks to mathematica :maybe: )
which is trivial using muirhead(or, who doesn't like muirhead, can use AM-GM:
$a^4+b^4\geq 2a^2b^2, a^3b+ab^3\geq 2a^2b^2, a^3b+a^3c+ab^3+ac^3\geq 4a^2bc, a^4+a^4+b^4+c^4\geq 4a^2bc$
should suffice to proof that inequality :) ).
i hope there's no mistake in it this time.

Peter

oops...why didn't i see this??? maybe because of your commentary cheers which made me thought it's very difficult... :( anyway, a nice solution


LaTeX
calculus
derivative
inequalities open
inequalities

[quote="stephen38"]I wonder why my latex didnot work.But It is not 3/8,but 8/3.[/quote]

Maybe you forgot to put your latex between the dollar signs

See my solution here: http://www.mathlinks.ro/Forum/viewtopic.php?t=142520

[quote="boxedexe"][quote="NguyenDungTN"]$ a \geq b \geq c$
Let $ f(a)=\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{4c^{2}}{a}+8(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})-6a-9b-15c$
We have $ f'(a)=\frac{2a}{b}-\frac{4c^{2}}{a^{2}}+4\frac{\sqrt{b+\sqrt{c}}}{\sqrt{a}}-6 \geq 0$
(because by AM-GM: $ \frac{a}{b}+2\frac{\sqrt{b}}{\sqrt{a}}\geq 3 \ and \frac{\sqrt{c}}{\sqrt{a}}\geq \frac{c}{a}(\frac{c}{a}\le 1) )$
It follows that: $ f(a) \geq f(b) = \frac{b^{2}}{c}+\frac{4c^{2}}{b}+16\sqrt{bc}-6b-15c$
 $ f'(b)= \frac{2b}{c}-\frac{4c^{2}}{b^{2}}+\frac{8\sqrt{c}}{\sqrt{b}}-6 \geq 0$
It follows that: $ f(b) \geq f(c)= 0$[/quote][/quote]
I'm very sorry,but i haven't learn about calculus.But i have learned about derrivative a little. :blush:
Can you tell me how to go from line 4 from your sol to line 5(it follows that :....)????
Thanks.


logarithms
real analysis
real analysis unsolved

Where $ D$ is the differential operator

$ y\equal{}\sum^{\infty}_{n\geq1}\frac{1}{D^n(y)}$

if we let $ f_A(x)\equal{}2x\plus{}2\ln A\minus{}\ln (\frac{x}{x\minus{}1})$ then $ f(\frac{n\plus{}1}{n})\equal{}2\frac{(n\plus{}1)}{n}\plus{}2\ln A\minus{} \ln (n\plus{}1)<0$ for large enough n
and $ f(n)\equal{}2n\plus{}2\ln A\minus{}\ln \frac{n}{n\minus{}1}>0$ for large n
so we know by continuity, there exists a point $ \alpha_A\in (1,\infty)$ such that

$ 2\alpha_A\plus{}2\ln A \equal{}\ln (\frac{\alpha_A}{\alpha_A\minus{}1})$ or $ A^2e^{2\alpha_A}\equal{}\frac{\alpha_A}{\alpha_A\minus{}1}$

so we can see that $ y\equal{}Ae^{\alpha_A x} \to D^n(y)\equal{}\alpha_A^n Ae^{\alpha_A x}$

so $ \sum_{n\equal{}1}^\infty \frac{1}{D^n (y)}\equal{} \sum_{n\equal{}1}^\infty \alpha_A^{\minus{}n} \frac{1}{y}\equal{}\frac{1}{y}\frac{\alpha_A}{\alpha_A\minus{}1}$

and we can verify that $ Ae^{\alpha_A x}\equal{}\frac{e^{\minus{}\alpha_A x}}{A}\frac{\alpha_A}{\alpha_A\minus{}1}$

so for each $ A>0$ there is a corresponding $ y_A\equal{}Ae^{\alpha_A x}$ which solves this equation..
but i dunno if this is the only form of solutions..

[quote="kenn4000"]and we can verify that $ Ae^{\alpha_A x} \equal{} \frac {e^{ \minus{} \alpha_A x}}{A}\frac {\alpha_A}{\alpha_A \minus{} 1}$[/quote]  With $ A$ and $ \alpha_A$ constant?  It looks to me like you've confused yourself badly.  $ A^2 e^{2\alpha} \equal{} g(\alpha)$ does not imply $ A^2 e^{\alpha x} \equal{} e^{\minus{}\alpha x} g(\alpha)$.


Prove that for all $ (a,b) \in \mathbb{R}^2$ we have :

$ (a^2\plus{}1)(b^2\plus{}1) \ge a(b^2\plus{}1)\plus{}b(a^2\plus{}1)$

[hide="Solution"]
We have that,
\[ (a^{2}\plus{}1)(b^{2}\plus{}1)\ge a(b^{2}\plus{}1)\plus{}b(a^{2}\plus{}1)\implies 1\ge \dfrac{a}{a^2\plus{}1} \plus{} \dfrac{b}{b^2\plus{}1}\]
Next, we can show that, $ \dfrac{a}{a^2\plus{}1}\le \dfrac{1}{2}$:
\[ \dfrac{a}{a^2\plus{}1}\le \dfrac{1}{2}\implies 2a\le a^2\plus{}1\implies a^2\minus{}2a\plus{}1\ge 0\]
Similarly $ \dfrac{b}{b^2\plus{}1}\le \dfrac{1}{2}$.

Now add them together to get, $ 1\ge \dfrac{a}{a^2\plus{}1} \plus{} \dfrac{b}{b^2\plus{}1}$ and  we are done. :ninja:
[/hide]

$ 2(a^2 \plus{} 1)(b^2 \plus{} 1)\ge 2a(b^2 \plus{} 1) \plus{} 2b(a^2 \plus{} 1) \Longleftrightarrow (a^2 \plus{} 1)(b^2 \minus{} 2b \plus{} 1) \plus{} (b^2 \plus{} 1)(a^2 \minus{} 2a \plus{} 1) \ge 0 \Longleftrightarrow (a^2 \plus{} 1)(b \minus{} 1)^2 \plus{} (b^2 \plus{} 1)(a \minus{} 1)^2 \ge 0$

Equality if and only if $ a\equal{}b\equal{}1$

 which is true


ratio
symmetry
trigonometry
algebra
system of equations
trig identities
Law of Cosines

Hi, I have trouble proving this:

If $ x\equal{}cy\plus{}bz, y\equal{}az\plus{}cx, z\equal{}bx\plus{}ay$

 show that $ \frac{x^2}{1\minus{}a^2} \equal{} \frac{y^2}{1\minus{}b^2} \equal{} \frac{z^2}{1\minus{}c^2}$

This problem is taken from Hall/Knight's "Higher Algebra". It was used at Cambridge in the 1800's and is recommended for Olympiad preparation in India.

This problem is an algebraic extension of the results of triangle geometry.

Let $ x, y, z$ be the sides of a triangle, and let $ a, b, c$ be the cosines of the angles opposite the respective sides.

The given system of equations comes about as a result of dropping an altitudes to each side [url=http://en.wikipedia.org/wiki/Law_of_cosines#Using_trigonometry]as shown here[/url].

The solution to this system of equations results in the Law of Cosines, as in [b]nsato[/b]'s post.

The result that we want to prove is the square of the Law of Sines.


[quote="shobber"]If $ a_1 \geq a_2 \geq \cdots \geq a_n \geq 0$ and $ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \equal{} 1$, then prove:
\[ a_1^2 \plus{} 3a_2^2 \plus{} 5a_3^2 \plus{} \cdots \plus{} (2n \minus{} 1)a_n^2 \leq 1
\]
[/quote]
Only Karamata:)

[quote=Arne]Hint:

[hide]induction works but there are better ways.[/hide][/quote]

Can you please suggest a solution

[quote=blahblahblah]$\left(\sum_i a_i\right)^2\geq \sum_i \left(a_i^2+2\sum_{j>i} a_ia_j\right)\geq \sum_i a_i^2+2\sum_i\sum_{j>i} a_j^2$

But that last sum is just $a_1^2+3a_2^2+\ldots+(2n-1)a_n^2$, so we're done.[/quote]

Please, can someone explain the last inequality?


Let be a given integer m>1.Find the smallest integer n>=m with the property that ,for every partition the set {m,m+1,......,n} in to 2

 subsets ,one of the subset contains three numbers a,b,c (not necessarily distinict) with a=bc.

For $ n \equal{} m^5 \minus{} 1$ we can make $ A \equal{} \{m, m \plus{} 1, ..., m^2 \minus{} 2, m^2 \minus{} 1, m^4, m^4 \plus{} 1, m^4 \plus{} 2, ..., m^5 \minus{} 1\}$ and $ B \equal{} \{m^2, m^2 \plus{} 1, m^2 \plus{} 2, ..., m^4 \minus{} 1\}$.

Useless explanation: clearly $ (A, B)$ is a partition of $ \{m, m \plus{} 1, ..., n\}$. The idea is choose the sets of the numbers in increasing order (first choose the set in which $ m$ is, then $ m \plus{} 1$, ...) and just change the set when you get a problem (for example, $ m$ and $ m^2$ in the same set is a problem, since $ a \equal{} m^2$ and $ b \equal{} c \equal{} m$ give an invalid partition). The pair $ (A, B)$ works by the idea of the construction... Anyway, to see this, note that any two numbers between $ m$ and $ m^2 \minus{} 1$ have product between $ m^2$ and $ (m^2 \minus{} 1)^2 \le m^4 \minus{} 1$, so this product is an element of $ B$. Also, the product of two other elements in $ A$ exceeds $ m^4m$ and therefore is not in $ A$ (nor $ B$). And the product of two elements in $ B$ is at least $ m^4$, then is not in $ B$.

However, for $ n \equal{} m^5$, any partition has a triple $ (a, b, c)$. Else we would have a partition of $ \{m, m^2, m^3, m^4, m^5\}$ satisfying the condition. $ m$ and $ m^2$ must be in different sets, as $ m^2$ and $ m^4$. But $ m^3$ and $ m^5$ can't be in the set of $ m$ and $ m^4$, because $ m^4 \equal{} m^3m$ and $ m^5 \equal{} m^4m$. Then $ m^2$, $ m^3$ and $ m^5$ are in the same set - contradiction.


A positive odd integer less than 150 is randomly selected.  What is the probability that the product of the number selected and the next consecutive integer is less than 10,000?  Express your answer as a common fraction.

[hide="solution"]
There are 75 odd positive numbers under 150.
There are 50 odd positve numbers 99 or below.
99 * 101 = 9999 and 101 * 103 = 10403
50/75 = [b]2/3[/b]
[/hide]

[quote="MCrawford"]next consecutive integers.[/quote]

and the next how many consecutive integers?

to treething: I think its next integer, not next odd integer.

*shrugs*

The answer's still the same, just replace 99 x 101 by 99 x 100 and 101 x 103 by 101 x 102.


inequalities
geometry
3D geometry
sphere
algebra
function
domain

$a+b+c=3,a,b,c \ge 0$
prove:
$\sqrt{3-ab}+\sqrt{3-bc}+\sqrt{3-ca} \ge 3 \sqrt 2$

This problem is easily tackled using Jensen:

$\frac{\sum_{\text{sym}} \sqrt{3 - ab}}{3} \geq \sqrt{\frac{9 - ab - bc - ac}{3}}$

Now note that by Cauchy-Schwartz, we have

$(a^2 + b^2 + c^2)^2 \geq(ab + bc + ac)^2 \Rightarrow a^2 + b^2 + c^2 \geq ab + bc + ac$
$\Rightarrow (a + b + c)^2 - 2(ab + bc + ac) \geq ab + bc + ac \Rightarrow 3 \geq ab + bc + ac$

Utilizing this in our original equation, we have

$\frac{\sum_{\text{cyc}} \sqrt{3 - ab}}{3} \geq \sqrt{\frac{9 - ab - bc - ac}{3}} \geq \sqrt{\frac{9 - 3}{3}} = \sqrt{2}$

giving us

$\sqrt{3-ab}+\sqrt{3-bc}+\sqrt{3-ca} \ge 3 \sqrt 2$

[quote="hungkhtn"][quote="zanttrang"]This problem is easily tackled using Jensen:

$\frac{\sum_{\text{sym}} \sqrt{3 - ab}}{3} \geq \sqrt{\frac{9 - ab - bc - ac}{3}}$

[/quote]

You're wrong because It's reveser.[/quote]
Ergh, yeah, you're right - darn  :P I guess I should be more careful when applying Jensen... it's funny how often applying Jensen the wrong way gives the desired result.

Edit: nevermind, that method doesn't work either....

[quote="hungkhtn"]OK. This's false.! Just take c=0.[/quote]
How does that provide a counter-example?  I've failed miserably so far to prove this inequality, but I think it is true.  I think AM-GM might help, but you have to prove that

$(3 - ab)(3 - ac)(3 - bc) \geq 8$, and I can't figure out how to do that, even after expanding and playing around with the numbers...

[quote="Nameless"][quote="zanttrang"]

$(3 - ab)(3 - ac)(3 - bc) \geq 8$, and I can't figure out how to do that, even after expanding and playing around with the numbers...[/quote]
It's wrong.When $a=b=\frac{3}{2},c=0$ then $(3-ab)(3-bc)(3-ca)=\frac{27}{4}<8$[/quote]
Yeah, you're right - no wonder I couldn't prove it.

[quote="kunny"]I got it! :D
We are to prove
\[ x^2+y^2+z^2\geq 6\ (x\geq 0,\ y\geq 0,\ z\geq 0)\Longrightarrow x+y+z\geq 3\sqrt{2}.  \][/quote]
Try $x=y=0,z=\sqrt6$ ;)

[quote="kunny"]O.K. but, anyway, here is my solution.

Let $x=\sqrt{3-ab},\ y=\sqrt{3-bc}, z=\sqrt{3-ca}$ we have $x^2+y^2+z^2=9-(ab+bc+ca).$
Thus $(a+b+c)^2\geq 3(ab+bc+ca)\Longleftrightarrow 3^2\geq 3(ab+bc+ca),$ since $a\geq 0,\ b\geq 0,\ c\geq 0,$ we obtain $0\leq ab+bc+ca\leq 3.$

Then since $c\geq 0,$ we have $\sqrt{ab}\leq\frac{a+b}{2}\leq \frac{3}{2},$ analogously $\sqrt{bc}\leq \frac{3}{2}, \sqrt{ca}\leq \frac{3}{2},$ thus

$6\leq x^2+y^2+z^2\leq 9 \left(x\geq \frac{\sqrt{3}}{2}, y\geq \frac{\sqrt{3}}{2}, z\geq \frac{\sqrt{3}}{2}\right)\ \cdots (*).$

Consider for $xyz-space,$  two spheres $x^2+y^2+z^2=6\ and\ x^2+y^2+z^2=9\ and\ x\geq \frac{\sqrt{3}}{2}, y\geq \frac{\sqrt{3}}{2}, z\geq \frac{\sqrt{3}}{2}.$

Assumed that $x+y+z$ takes the value of $k$ for $(*).$ the plane $x+y+z=k$ are to have the intersection points with this domain,

Thus $\sqrt{6}\leq\text{the distance between the center of the sphere and the plane}\ x+y+z-k=0 \leq 3.$

$\Longleftrightarrow\sqrt{6}\leq \frac{|0+0+0-k|}{\sqrt{1^2+1^2+1^2}}\leq 3$, yielding $\sqrt{6}\leq \frac{|-k|} {\sqrt{3}}\leq 3.$ Since $k\geq 0,$ yielding $3\sqrt{2}\leq k \leq 3\sqrt{3}.$

Therefore $3\sqrt{2}\leq \sqrt{3-ab}+\sqrt{3-bc}+\sqrt{3-ca}\leq 3\sqrt{3}.$[/quote]
kunny,can I ask why $x\geq \frac{\sqrt{3}}{2}, y\geq \frac{\sqrt{3}}{2}, z\geq \frac{\sqrt{3}}{2}.$
and $\sqrt{6}\leq\text{the distance between the center of the sphere and the plane}\ x+y+z-k=0 \leq 3.$?
thanks

More general, the following inequality holds for $p \geq \frac{81}{32}$:
$\sqrt{p-ab}+\sqrt{p-bc}+\sqrt{p-ca} \ge 3 \sqrt{p-1}$

A possible solution uses EV-Theorem.

Long time ago,and anybody have a solution without n-1 principle theorem?

Can you solve the first ineq:
$\sqrt{3-ab}+\sqrt{3-bc}+\sqrt{3-ca} \ge 3 \sqrt 2$
I can't understand kunny's solution

I can bet kunny's solution is certainly wrong ;)
But I also didn't have a solution.

[quote="zhaobin"]I find the solution today by accident,hope you can like it :D .
Because we have:
$ \sqrt {3 \minus{} ab} \equal{} \frac {(6 \minus{} 2ab)\sqrt 2}{2\sqrt {2(3 \minus{} ab)}}\ge \frac {(6 \minus{} 2ab)\sqrt 2}{2 \plus{} 3 \minus{} ab} \equal{} \frac {(6 \minus{} 2ab)\sqrt 2}{5 \minus{} ab}$
So we only need to prove:
$ \frac {3 \minus{} ab}{5 \minus{} ab} \plus{} \frac {3 \minus{} bc}{5 \minus{} bc} \plus{} \frac {3 \minus{} ca}{5 \minus{} ca} \ge \frac {3}{2}$
Which is also:
$ \frac {1}{5 \minus{} ab} \plus{} \frac {1}{5 \minus{} bc} \plus{} \frac {1}{5 \minus{} ca} \le \frac {3}{4}$
This is the special case of Vasc's inequality,see
[url]http://www.mathlinks.ro/viewtopic.php?t=18627&search_id=1516423496[/url]
(Let d=1)Or,you can also prove it directly.

BTW,Kunny's solution is wrong because:
the plane $ x \plus{} y \plus{} z \equal{} k$ and the ball $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 6$  have the intersection points.
doesn't imply
$ \sqrt {6}\leq\text{the distance between the center of the sphere and the plane}\ x \plus{} y \plus{} z \minus{} k \equal{} 0$[/quote]

The inequality

$ \frac{1}{5\minus{}ab}\plus{} \frac{1}{5\minus{}bc}\plus{} \frac{1}{5\minus{}ca} \le \frac{3}{4}$

is wrong for $ a\equal{}0,b\equal{}c\equal{}\frac{3}{2}$  :wink:

I have just deleted, because my reasoning was wrong, the deleted post, you will see shalex's post 2 posts forward.

kunny

[quote="zhaobin"]$ a \plus{} b \plus{} c \equal{} 3,a,b,c \ge 0$
prove:
$ \sqrt {3 \minus{} ab} \plus{} \sqrt {3 \minus{} bc} \plus{} \sqrt {3 \minus{} ca} \ge 3 \sqrt 2$[/quote]
With the same conditions the following inequality is true too:
\[ \sqrt {81 \minus{} 32ab} \plus{} \sqrt {81 \minus{} 32ac} \plus{} \sqrt {81 \minus{} 32bc}\geq21
\]

[quote="kunny"][quote="zhaobin"]$ a \plus{} b \plus{} c \equal{} 3,a,b,c \ge 0$
prove:
$ \sqrt {3 \minus{} ab} \plus{} \sqrt {3 \minus{} bc} \plus{} \sqrt {3 \minus{} ca} \ge 3 \sqrt 2$[/quote]

Let $ abc$ fix as $ abc \equal{} k$, $ f(x): \equal{} \sqrt {3 \minus{} \frac {k}{x}} \minus{} \frac {k}{2\sqrt {3 \minus{} k}}\ln x$ for $ x > 0$.

$ f'(x) \equal{} \frac {k(x \minus{} 1)(3x \plus{} 3 \minus{} k)}{2x\sqrt {3 \minus{} k}\sqrt {3x^2 \minus{} kx}(\sqrt {3x^2 \minus{} kx} \plus{} \sqrt {{3 \minus{} k}})}$, yielding

$ \therefore f'(x) \equal{} 0\Longrightarrow x \equal{} 1,\ \frac {k}{3} \minus{} 1$, by A.M.-G.M. , $ 0\leq abc\leq \left(\frac {a \plus{} b \plus{} c}{3}\right)^3\Longleftrightarrow 0\leq k\leq 1$, yielding

$ \minus{} 1\leq \frac {k}{3} \minus{} 1\leq \minus{} \frac {2}{3}$. Thus we have $ f(x)\geq f(1) \equal{} \sqrt {3 \minus{} abc}\geq \sqrt {2}$.

$ \therefore f(a) \plus{} f(b) \plus{} f(c)\geq 3\sqrt {2}\Longleftrightarrow \sqrt {3 \minus{} ab} \plus{} \sqrt {3 \minus{} bc} \plus{} \sqrt {3 \minus{} ca} \ge 3 \sqrt 2$. Q.E.D.[/quote]

I don't understand why

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

Anyway, I think

$ f'(x) \equal{} \frac {k(1 \minus{} x)(3x \plus{} 3 \minus{} k)}{2x\sqrt {3 \minus{} k}\sqrt {3x^2 \minus{} kx}(\sqrt {3x^2 \minus{} kx} \plus{} \sqrt {{3 \minus{} k}})}$

And $ f$ reach the maximum at the point $ x \equal{} 1$, but not minimum.

[quote="kunny"]$ 0 < abc \leq 1\Longrightarrow \ln abc < 0$,

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

$ \geq \sqrt {3 \minus{} ab} \plus{} \sqrt {3 \minus{} bc} \plus{} \sqrt {3 \minus{} ca}\geq 3\sqrt {2}$.

[quote="shalex"]
$ f'(x) \equal{} \frac {k(1 \minus{} x)(3x \plus{} 3 \minus{} k)}{2x\sqrt {3 \minus{} k}\sqrt {3x^2 \minus{} kx}(\sqrt {3x^2 \minus{} kx} \plus{} \sqrt {{3 \minus{} k}})}$

And $ f$ reach the maximum at the point $ x \equal{} 1$, but not minimum.[/quote]

Whose solution is this? :wink:[/quote]

Don't you see that

$ f(a)\plus{}f(b)\plus{}f(c) \ge 3\sqrt{2}$

(which your proved) is weaker than

$ \sqrt{3\minus{}ab}\plus{} \sqrt{3\minus{}bc}\plus{} \sqrt{3\minus{}ca} \ge 3\sqrt{2}$?

Because $ f(a)\plus{}f(b)\plus{}f(c) \ge 3\sqrt{2}$ is equivalent to

$ \sqrt{3\minus{}ab}\plus{} \sqrt{3\minus{}bc}\plus{} \sqrt{3\minus{}bc} \ge 3\sqrt{2}\plus{}\frac{abc}{2\sqrt{3\minus{}abc}}\ln abc$

But the right side is less than or equals to $ 3\sqrt{2}$

Anyway I suggest you to check $ f'(x)$ by hand or with Mathematica or something else  :wink:

[b]shalex[/b], I'm very sorry, I would have to appologize to you, I admit that my calculation was wrong.
Please forgive my some critics to you.

kunny

That's alright  :lol:
Now we should concentrate on how to prove it in a good way...
It seems not a easy one.

Thank you, :oops: I have tackled with several solution so far, not solved. Indeed how difficult the problem is.

[quote="shalex"]

The inequality

$ \frac {1}{5 \minus{} ab} \plus{} \frac {1}{5 \minus{} bc} \plus{} \frac {1}{5 \minus{} ca} \le \frac {3}{4}$

is wrong for $ a \equal{} 0,b \equal{} c \equal{} \frac {3}{2}$  :wink:[/quote]
Thank you,I am wrong :blush:

arqady,Do you have a solution?

[quote="zhaobin"]

arqady,Do you have a solution?[/quote]
Of mine inequality? Yes, and very nice!

Can you post it?

BTW,
Do you have some idea of mine?

I'm eager to see your solution.
Could you post it? Thanks a lot!  :lol:

[quote="zhaobin"]Can you post it? [/quote]
I have sent it to Crux a year ago. :wink:
I am sorry, shalex and zhaobin.
[quote="zhaobin"]BTW,
Do you have some idea of mine?[/quote]
I think, other idea and without  Vasc's EV-Theorem.:wink:


DO NOT DISCUSS THIS YEAR'S TEST

COME ON PEOPLE!

Edited by moderator.


I haven't taken an olympiad before, but

[hide]I'm going to assume $ x$, $ y$, and $ z$ are nonnegative. If they can be negative, we can let $ x \equal{} \minus{} 5$, $ y \equal{} 3$ and $ z \equal{} 3$ and we see that:

$ \frac {x^2}{1 \plus{} y} \plus{} \frac {y^2}{1 \plus{} z} \plus{} \frac {z^2}{1 \plus{} x} \equal{} 6.25 > 1$ and $ x \plus{} y \plus{} z \equal{} 1$

If $ x$, $ y$, $ z$ are nonnegative, then we substitute $ x \plus{} y \plus{} z$ for 1:
$ \frac {x^2}{1 \plus{} y} \plus{} \frac {y^2}{1 \plus{} z} \plus{} \frac {z^2}{1 \plus{} x}\leq 1$
$ \frac {x^2}{(x \plus{} y \plus{} z) \plus{} y} \plus{} \frac {y^2}{(x \plus{} y \plus{} z) \plus{} z} \plus{} \frac {z^2}{(x \plus{} y \plus{} z) \plus{} x}\leq (x \plus{} y \plus{} z)$

Now, we finish off by proving $ \frac {x^2}{(x \plus{} y \plus{} z) \plus{} y}\leq x$:
If $ x\equal{}0$, then equality holds. If $ x\neq 0$:
$ \frac {x^2}{(x \plus{} y \plus{} z) \plus{} y}\leq x$
$ \frac {x}{(x \plus{} y \plus{} z) \plus{} y}\leq 1$
$ x\leq x \plus{} 2y \plus{} z$

And since $ y$ and $ z$ are nonnegative, the inequality holds. Similarly, $ \frac {y^2}{(x \plus{} y \plus{} z) \plus{} z}\leq y$ and $ \frac {z^2}{(x \plus{} y \plus{} z) \plus{} x}\leq y$.

Therefore,
$ \frac {x^2}{(x \plus{} y \plus{} z) \plus{} y} \plus{} \frac {y^2}{(x \plus{} y \plus{} z) \plus{} z} \plus{} \frac {z^2}{(x \plus{} y \plus{} z) \plus{} x}\leq (x \plus{} y \plus{} z)$
$ \frac {x^2}{1 \plus{} y} \plus{} \frac {y^2}{1 \plus{} z} \plus{} \frac {z^2}{1 \plus{} x}\leq 1$

Finally, we want to show that equality can hold. We see this is true when $ x \equal{} 1$ and $ y \equal{} z \equal{} 0$. [/hide]

Solution of [b]ehehheehee[/b] very nice. [b]kujta1[/b] can post your problem by latex if you see at Latex help of this forum.

If $ x\plus{}y\plus{}z\equal{}1$, prove:
$ \frac {x^2}{1\plus{}y} \plus{} \frac {y^2}{1\plus{}z} \plus{} \frac {z^2}{1\plus{}x} \ge 1$

[quote="thanhnam2902"]Solution of [b]ehehheehee[/b] very nice. [b]kujta1[/b] can post your problem by latex if you see at Latex help of this forum.

If $ x \plus{} y \plus{} z \equal{} 1$, prove:
$ \frac {x^2}{1 \plus{} y} \plus{} \frac {y^2}{1 \plus{} z} \plus{} \frac {z^2}{1 \plus{} x} \ge 1$[/quote]

I think it should be
$ \frac {x^2}{1 \plus{} y} \plus{} \frac {y^2}{1 \plus{} z} \plus{} \frac {z^2}{1 \plus{} x} \ge \frac{1}{12}$
and that can be proven easily by BCS.

[quote="bibobeo"][quote="thanhnam2902"]Solution of [b]ehehheehee[/b] very nice. [b]kujta1[/b] can post your problem by latex if you see at Latex help of this forum.

If $ x \plus{} y \plus{} z \equal{} 1$, prove:
$ \frac {x^2}{1 \plus{} y} \plus{} \frac {y^2}{1 \plus{} z} \plus{} \frac {z^2}{1 \plus{} x} \ge 1$[/quote]

I think it should be
$ \frac {x^2}{1 \plus{} y} \plus{} \frac {y^2}{1 \plus{} z} \plus{} \frac {z^2}{1 \plus{} x} \ge \frac {1}{12}$
and that can be proven easily by BCS.[/quote]
You mean $ \frac{1}{4}$ right?

Sure :) . I don't think we have any difficulty doing this problem, do we?

[quote="bibobeo"]...and that can be proven easily by BCS.[/quote]What's BCS?

I believe thats just the [url=http://en.wikipedia.org/wiki/Cauchy_Schwarz_Inequality]Cauchy-Schwarz Inequality[/url].

@ ehehheehee, unfortunately you can't just assume that $ x,y,z$ are nonnegative.

Edit: I meant, ehehheehee's approach isn't quite correct, not that the problem statement was wrong (assuming I understand your comment correctly, Temperal).

Edit: Oh wait, ignore these last two comments, I was looking at thanhnam2902's latexed post, which is different.

Then the inequality is false. I'm guessing that's what the original poster meant.


inequalities
integration
algebra unsolved
algebra

let $ a$ and $ b$ be two complexe numbers whose real parts are negative or 0,prove the inequality:
l$ e^{a}\minus{}e^{b}$l$ \leq $ l$ a\minus{}b$l


Let A = 1, B = 2, C = 3, ..., Z = 26. The product value of a word is equal to the product of the values of its letters. For example, CAB has a product value of 3 $ \times$ 1 $ \times$ 2 = 6. What common English word has a product value of 715?

The prime factorization of 715 is $ 5 \cdot 11 \cdot 13$. Thus, our letters are those that correspond to 5, 11, and 13 - that is, EKM. However, no word can be made out of these letters, so we add 1 = A, to give us the letters AEKM, or $ \boxed{\text{MAKE}}$.


In the past time, a fraction as exponent was quite okay

e^\frac{2}{3}

and it is $e^\tfrac{2}{3}$

But now it becomes $e^\dfrac{2}{3}$, which looks too ugly. I don't know why the system changes all "frac" into "dfrac", that is the problem.

And what is worse,

e^\frac{2}{3} \pi

was $e^{\tfrac{2}{3}} \pi$ in the past, but now it becomes $e^\dfrac{2}{3} \pi$. I don't know why the system interpret $\pi$ to be in the exponent.

And it is really bad...when we are coming across an integral

\int_0^\frac{\pi}{2} \sin x dx

and the output was $\int_0^{\tfrac{\pi}{2}} \sin x dx$ in the old days, but now ... it looks like
$\int_0^\dfrac{\pi}{2} \sin x dx$, which is awful...

Please don't post twice the same topic! This one is closed.


[quote]The sides of unit square ABCD have trisection points X, Y, Z, and W.  If AX:XB = BY:YC = CZ:ZD = DW:WA = 2:1, then what is the area of the shaded region?  Express your answer as a common fraction.

[asy]draw((0,0)--(3,0)--(3,3)--(0,3)--cycle);
label("$A$",(0,3),NW);
label("$B$",(3,3),NE);
label("$C$",(3,0),SE);
label("$D$",(0,0),SW);
label("$X$",(2,3),N);
label("$Y$",(3,1),E);
label("$Z$",(1,0),S);
label("$W$",(0,2),W);
draw((0,0)--(3,1));
draw((3,0)--(2,3));
draw((3,3)--(0,2));
draw((0,3)--(1,0));[/asy]
[/quote]

I mean the square in the middle.

How would you solve a problem like this?

The little triangle (BX intersection of BW and XC) in the corner is similar to triangle BAW by AA. So the ratio of the sides is 1 to sqrt 10. Then, you find one of the legs using that ratio and find the area of that little triangle. Since all the little triangles in the corners are congruent to each other and triangle XBC, YCD, ZDA, and WAB are all congruent to each other, you can find the areas of the bigger triangles combined and subtract the area of the smaller triangles. Then you get the region you don't want. Then you subtract that from 1 and you get the area of the square.

Coordinate bashing seems like it would work and wouldn't be too hard.

In other words, you can find the equations for DY and AZ and then find the intersection. Do that again to find the coordinates for either of the points next to that intersection. Then you can use the distance formula.

[quote="Kouichi Nakagawa"]Cf. http://www.mathlinks.ro/viewtopic.php?t=248969  :)[/quote]

That's a beautiful solution :), but they're different problems.

I think I saw this inside the intro to geometry text. their solution was pretty much the same as angie's. :|


[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=235424]Lookie![/url]

I want to prove the $ 5.5m \minus{} 30h$ thing.

hello, do you mean with $ m$ minutes?
Sonnhard.

Yes, $ m$ is for minutes and $ h$ is for hours.

It finds the angle between the minute hand and the hour hand in a standard twelve hour clock.

[hide="One Way"]
Pretend we are currently at $ h: 00$. Then we can see that the hour hand is currently at $ 30h$ degrees. Also, let the angle between the hands of the clock be $ x$. Since the minute hand is always moving at a speed of $ 6 \minus{} 0.5 \equal{} 5.5$ degrees per minute relative to the hour hand (do you see why?), and the minute hand needs to go a distance of $ 30h \minus{} x$ relative to the hour hand (do you see why?) in order to be $ x$ degrees away from the hour hand, this will take $ t \equal{} \frac {30h \minus{} x}{5.5}$ time to complete this. But we already know that $ t \equal{} m$! This means that we can simply solve: $ m \equal{} \frac {30h \minus{} x}{5.5}\implies5.5m \equal{} 30h \minus{} x\implies x \equal{} 30h \minus{} 5.5m$. However, we need to take the absolute value of this, which is $ \boxed{|5.5m \minus{} 30h|}$. Note that we must subtract this from $ 360^\circ$ when the value obtained from the formula is a reflex angle.
[/hide]

[hide="Another Way"]
The minute hand is at $ \frac m{60}\cdot360 \equal{} 6m$ degrees relative to $ 12$ o'clock.

The hour hand is at $ \frac h{12}\cdot360 \equal{} 30h$ degrees relative to $ 12$ o'clock. However, we need to consider the effect of the minute hand on the hour hand. Since every hour, the hour hand moves $ 30$ degrees, the hour hand must move an extra $ \frac m{60}\cdot30 \equal{} 0.5m$ degrees.

Thus the absolute difference, and angle between the two hands, is $ |5.5m \minus{} 30h|$. Again, remember that you may get reflex angles.
[/hide]

EDIT: Fixed.

[quote="math154"]
 Since the minute hand is always moving at a speed of $ 6 \minus{} 0.5 \equal{} 5.5$ degrees per minute (do you see why?)[/quote]

minute hand is moving at speed of 360/60=6 degrees per minute, and the hour hand catches up 30/60=0.5 degrees per minute.

minute hand is not moving at a speed of 5.5 degrees per minute; the DIFFERENCE BETWEEN THE HOUR AND THE MINUTE HAND increases by 5.5 degrees per minute.

Yes, I meant relative motion to the hour hand. Sorry for any confusion! :blush:


limit
algebra
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logarithms
calculus
integration
calculus computations

It diverges to $+\infty$. After togethering two fractions, we see that the degree of numerator (viewed as the polynomial of $k$) is higher than that of denominator's.

In fact,

$\frac{(k+2)^{k+2}}{k^{k}}-\frac{k^{k}}{(k-2)^{k-2}}\;\; \approx \;\; 2k^{2}e^{2}\ln\left( 1+\frac{2}{k}\right)+C$

for some constant $C$. (I failed to determine the exact value of $C$. But by numerical calculation, I found that it is approximately 30.)

------------------------------------------------------------------------------------------------------------------------------

After some manipulation and calculation, I conjectured that

$\frac{(k+2)^{k+2}}{k^{k}}-\frac{k^{k}}{(k-2)^{k-2}}\;\; \approx \;\; 4 k e^{2}$.

Now I'm trying to prove this.

hmmm..... interesting !  :|

well, regarding the simpler

$\lim_{k\to\infty}\;\;\left( \;\frac{(k+1)^{k+1}}{k^{k}}\;-\;\frac{k^{k}}{(k-1)^{k-1}}\;\right)$

i computed it to [u]converge [/u]to $\;\;e\;\;!$

and so, i conjectured perhaps the original limit will converge to something like
$\;\;m_{1}e+m_{2}e^{2},\;\;\;$ where $\;\;\;m_{1},m_{2}\in\textbf N^{+}$

and similarly, without thinking much at all :P , i was led to believe

$\lim_{k\to\infty}\;\;\left( \;\frac{(k+a)^{k+a}}{k^{k}}\;-\;\frac{k^{k}}{(k-a)^{k-a}}\;\right) \;=\;\boxed{\sum_{i=1}^{a}m_{i}e^{i}}$
( where $\;\;\;a,m_{i}\in\textbf N^{+}\;$)

I was able to prove a weaker formula: For $k \geq 0$,

$\lim_{n\to\infty}\frac{1}{n^{k-1}}\left( \frac{(n+k)^{n+k}}{n^{n}}-\frac{n^{n}}{(n-k)^{n-k}}\right) \;\; = \;\; k^{2}e^{k}.$

I calculated this only for integral values of $m$, but I believe it would work for all nonnegative real $m$.

-----------------------------------------------------------------------------------------------------------------

I succeeded in solving what I conjectured. Let $g(x) = (1+ax)^{\frac{1}{x}}$ and $g(0) = e^{a}$. Then $g(x)$ satisfies

$g'(x) = g(x)\left(-\frac{a^{2}}{2}+\frac{2a^{3}}{3}x-\frac{3a^{4}}{4}x^{2}+\cdots \right)$

and hence $g'(x) =-\frac{a^{2}}{2}e^{a}+o(1)$ as $x \to 0$.

Not let $a = 2$. Then

$\frac{(n+2)^{n+2}}{n^{n}}-\frac{n^{n}}{(n-2)^{n-2}}-4 n e^{2}$ $=$ $n^{2}\left[ g \left(\frac{1}{n}\right)-g \left(\frac{1}{n-2}\right) \right]+4n \left[ g \left(\frac{1}{n}\right)-g(0) \right]+4 g \left(\frac{1}{n}\right)$.

and applying MVT immediately gives the desired result.

[quote="The Zuton Force"]Sorry, misan, can you explain how you found that your limit converges to $e$?

I have that it is equivalent to:

$\lim_{x \to \infty}\ln \left( 1+\frac{1}{x}\right) \frac{(x+1)^{x+1}}{x^{x}}$

but I don't know what to do after that, or even if that's the right approach.

Thanks![/quote]
That expression is equal to
\[\left( 1+\frac{1}{x}\right)^{x}\left( \ln \left[\left( 1+\frac{1}{x}\right)^{x}\right]+\ln \left( 1+\frac{1}{x}\right)\right) \]

[quote="sos440"]

$\frac{(k+2)^{k+2}}{k^{k}}-\frac{k^{k}}{(k-2)^{k-2}}\;\; \approx \;\; 4 k e^{2}$.

[/quote]
I agree with you and your aproximatly result, but it's not so exact as my.
my aproximation result:

$\frac{(k+2)^{k+2}}{k^{k}}-\frac{k^{k}}{(k-2)^{k-2}}\;\; \approx \;\; e^{2}k^{2}(\frac{4}{k}-\frac{40}{3k^{3}}+o(\frac{1}{k^{3}}))$
 :P
also :
there is not constant.

[quote="misan"][quote="The Zuton Force"]Sorry, misan, can you explain how you found that your limit converges to $e$?
[/quote]

so, were you able to do it afterwards ? no ? :|
alright then, here is my solution:

$\lim_{k\to\infty}\;\left[ \frac{(k+1)^{k+1}}{k^{k}}-\frac{k^{k}}{(k-1)^{k-1}}\right]$

$=\lim_{k\to\infty}\;\left[ (k+1)\left( 1+\frac{1}{k}\right)^{k}-(k-1)\left( 1-\frac{1}{k}\right)^{-k}\right]$

$=\lim_{k\to\infty}\;\left[ k\left( 1+\frac{1}{k}\right)^{k}+\left( 1+\frac{1}{k}\right)^{k}-k\left( 1-\frac{1}{k}\right)^{-k}+\left( 1-\frac{1}{k}\right)^{-k}\right]$

note that [u]both [/u]the [b]2nd [/b]and [b]4th [/b]expressions converge to $e$ in the limit $k\to\infty$,
so we have

$=2e+\lim_{k\to\infty}\;\left[ k\left( 1+\frac{1}{k}\right)^{k}-k\left( 1+\left(-\frac{1}{k}\right) \right)^{-k}\right]$

$=2e+\lim_{k\to\infty}\;\left[ k\left( 1+k\frac{1}{k}+\frac{k(k-1)}{2!}\frac{1}{k^{2}}+\frac{k(k-1)(k-2)}{3!}\frac{1}{k^{3}}+\cdots+\right)\right.$
$-k \left( 1+(-k)\left(-\frac{1}{k}\right)+\frac{(-k)(-k-1)}{2!}\left(-\frac{1}{k}\right)^{2}+\right.$
$\left. \left. \frac{(-k)(-k-1)(-k-2)}{3!}\left(-\frac{1}{k}\right)^{3}+\cdots+\right)\right]$

$=2e+\lim_{k\to\infty}\;\left[ k\left( 1+1+\frac{1}{2!}\left( 1-\frac{1}{k}\right)+\frac{1}{3!}\left( 1-\frac{1}{k}\right)\left( 1-\frac{2}{k}\right)+\cdots+\right) \right.$
$\left.-k \left( 1+1+\frac{1}{2!}\left( 1+\frac{1}{k}\right)+\frac{1}{3!}\left( 1+\frac{1}{k}\right)\left( 1+\frac{2}{k}\right)+\cdots+\right) \right]$

now, observe how [u]several [/u]terms cancel out! :|
eventually, we will be left with

$=2e+\lim_{k\to\infty}\;\left[ k\left\{ 2\frac{1}{2!}\left(-\frac{1}{k}\right)+2\frac{1}{3!}\left(-\frac{3}{k}\right)+2\frac{1}{4!}\left(-\frac{6}{k}-\frac{6}{k^{3}}\right)+\right. \right.$
$\left. \left. 2\frac{1}{5!}\left(-\frac{10}{k}-\frac{50}{k^{3}}\right)+\cdots+\right\}\right]$

it is probably [u]easier [/u]to see it like this, so i avoided writing out
the [b]general [/b]term using $C^{n}_{k}$ and it's associated k-degree polynomials,
although the solution i posted originally for the class had all of that...

$=2e+\lim_{k\to\infty}\;\left[-\left\{ 1+1+\frac{1}{2!}+\textbf{o}\left( \frac{1}{k^{2}}\right)+\frac{1}{3!}+\textbf{o}\left( \frac{1}{k^{2}}\right)+\frac{1}{4!}+\textbf{o}\left( \frac{1}{k^{2}}\right)+\cdots+\right\}\right]$

well, you should know by now what the terms within the [i]curly [/i]brackets
will add up to in the limit $k\to\infty$ :|

$=2e+(-e)\;=\;\boxed{e}$[/quote]
Hi.

you don't explain this step :
[quote="misan"]
$=2e+\lim_{k\to\infty}\;\left[ k\left\{ 2\frac{1}{2!}\left(-\frac{1}{k}\right)+2\frac{1}{3!}\left(-\frac{3}{k}\right)+2\frac{1}{4!}\left(-\frac{6}{k}-\frac{6}{k^{3}}\right)+\right. \right.$
$\left. \left. 2\frac{1}{5!}\left(-\frac{10}{k}-\frac{50}{k^{3}}\right)+\cdots+\right\}\right]$

it is probably [u]easier [/u]to see it like this, so i avoided writing out
the [b]general [/b]term using $C^{n}_{k}$ and it's associated k-degree polynomials,
although the solution i posted originally for the class had all of that...

$=2e+\lim_{k\to\infty}\;\left[-\left\{ 1+1+\frac{1}{2!}+\textbf{o}\left( \frac{1}{k^{2}}\right)+\frac{1}{3!}+\textbf{o}\left( \frac{1}{k^{2}}\right)+\frac{1}{4!}+\textbf{o}\left( \frac{1}{k^{2}}\right)+\cdots+\right\}\right]$
[/quote].
but it's not so improtant. look:
when you wrote
$\sum o(\frac{1}{k^{2}})+o(\frac{1}{k^{2}})+o(\frac{1}{k^{2}})+o(\frac{1}{k^{2}})+...+o(\frac{1}{k^{2}})=o(\frac{1}{k^{2}})$
where $k\to+oo$
it's not true.

for example
let's take $g(1/k)=\frac{1}{k^{3}}$
we can write it as
$g(1/k)=o(\frac{1}{k^{2}})$
let's consider sum:
$\lim_{k\to+oo}\sum_{n=0}^{+oo}g(1/k)$
in your case it's equal to $o(\frac{1}{k^{2}})$
but note that
 $\sum_{n=0}^{+oo}g(1/k) <\sum_{n=0}^{k^{M}}g(1/k)=k^{M}g(1/k)=k^{M}\frac{1}{k^{3}}=k^{M-3}$
let $M>3$
we consider limit in $k\to+oo$
and it's equal to $+oo$.


geometry
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articles

1. Let  1, 4, ... and 9, 16 ... be two arithmetic progressions. The set S is the union of the first 2004 terms of each sequence. How many distinct numbers are in S?

I started out by finding the 2004th term of each sequence [6010 for the first sequence and 10024 for the second, using the formula for arithmetic sequences a+d(n-1)]. Then I listed out the sequences until I found a few common terms.

1, 4, 7, 10, 13, [b]16[/b], 19, 22, 25, 28, 31, 34, [b]37[/b], 40, 43, 46, 49, 52, 55, [b]58[/b], 61, 64, 67, 70, 73, 76, [b]79[/b] etc etc

9, [b]16[/b], 23, 30, [b]37[/b], 44, 51, [b]58[/b], 65, 72, [b]79[/b] etc etc.

The bolded numbers are common terms. In the first sequence, there are 5 numbers before the first common term, 6 numbers after the first common term but before the second common term, then another 6 until the next common term, and yet another 6 until the next common term, etc etc. The problem is, I don't know how to count all of the numbers...

2. Each face of a cube is painted either red or blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?

Since each face can be painted red or blue, there are a total of $ 2^{6}\equal{}64$ ways to color it. Now I need help finding the numerator.

3. In triangle ABC we have AB = 7, AC = 8, and BC = 9. Point D is on the circumscribed circle of the triangle so that AD bisects angle BAC. What is the value of AD/CD?

I drew out the figure, but I'm terrible at geometry, so I had absolutely no idea what to do.

4. An envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 twenties. Two bills are drawn at random without replacement. What is the probability that their sum of $ \$20$ or more?

I know the denominator is $ \binom {8}{2}$ , but how do I find the numerator?

5. The quadratic equation $ x^{2} \plus{} mx \plus{} n \equal{} 0$ has roots that are twice those of $ x^{2} \plus{} px \plus{} m \equal{} 0$, and none of m, n, and p is zero. What is the value of n/p?

Izzy says I should learn Vietas.[/b]

[hide="Hint for number five"]
Have you heard of vieta's? If not, you should read up on it. Check out the AoPSWiki article, and the chapter in the Intermediate Algebra book.
[/hide]

Show what you tried on these problems; don't keep on posting problems that you don't appear to have tried.

I didn't know I had to do that...but okay.

1) [hide="Outline"]Here's an outline of the solution:
First we have to count the number of elements in each of the two sets. Let the sequence $ 1,4,7,\dots$ be set $ A,$ and let the sequence $ 9,16,23,\dots$ be set $ B.$ We want to count the number of elements in their union. So by PIE, that is just
\[ \#(A\cup B) \equal{} \#(A) \plus{} \#(B) \minus{} \#(A \cap B).\]
So, we just count the number of elements in $ A$ and in $ B.$ Now we need to find the common elements. So let $ n$ and $ m$ be arbitrary elements in sets $ A$ and $ B,$ respectively, and they are the ones that are common to both. The value of the $ n^{\text{th}}$ element in $ A$ is $ 1\plus{}3(n\minus{}1).$ Similarly, the value of the $ m^{\text{th}}$ element in $ B$ is $ 9 \plus{} 7(m \minus{} 1).$ Since they are equal, we must have $ 1 \plus{} 3(n \minus{} 1) \equal{} 9 \plus{} 7(m \minus{} 1) \implies 3n \minus{} 2 \equal{} 2 \plus{} 7m \implies \dfrac{3n \minus{} 4}{7} \equal{} m.$ So, we just see seek integer values that satisfies that equation. (Take into consideration the upper and lower limits of both the sets.) [/hide]

Maybe...there's a better way to do this problem. (I'm referring to #1.)

4) [hide="Solution"]We are just looking for the number of ways to obtain a sum of 20 or more with the two bills. So, we just do some casework.

[i]Case 1: The first bill is a one dollar bill. [/i] The second bill has to be a twenty. There are 2 choices for the 1 dollar bill and 2 choices for the 20 dollar bill, for a total of 2*2=4 choices for this case.

[i]Case 2: The first bill is a five dollar bill. [/i] The second bill has to be a twenty. There are 2 choices for the 5 dollar bill and 2 choices for the 20 dollar bill, for a total of 2*2=4 choices for this case.

[i]Case 3: The first bill is a ten dollar bill. [/i] There are two things that can happen. The next bill is a 20 dollar bill, or the next bill is a ten dollar bill. For the next bill to be 20 dollars, we can have the first one be either of the two 10 dollar bills, and there are 2 choices for each of the 2 choices of the 10 dollar bill, for a total of 2*2=4 choices. And if the second bill is a 10 dollar bill, there is only one way that can happen. Therefore, there are 4+1=5 choices for this case.

[i]Case 4: The first bill is a twenty dollar bill. [/i] We have already accounted for the other bills being paired up with a 20 dollar bill, except for when we have two 20 dollar bills, hence there is only one possibility for this case.

That gives us a total of 4+4+5+1 = 14. As you have said there are $ \binom{8}{2}$ ways to choose any 2 bills [without regard to order. Note that we have kept that in mind while computing the number of successful outcomes]. Therefore, the probability of success is $ \dfrac{14}{\dbinom{8}{2}} \equal{} \dfrac{1}{2}.$[/hide]

5)[hide="More of a hint"] As isabella had said, use Vieta's and model the situation of the problem, that should give you a nice system to work with. And just solve the system. [/hide]

EDIT: Oops, didn't catch that. Thanks modularmarc :)

[quote="limac"]
$ \#(A\cup B) \equal{} \#(A) \plus{} \#(B) \plus{} \#(A \cap B)$
[/quote]

Actually,

$ \#(A \cup B) \equal{} \#(A) \plus{} \#(B) \minus{} \# (A \cap B)$

[hide="#3 Solution"]Connect the points $ B$ to $ D$ and $ D$ to $ C$. Notice that  $ < BAD \equal{} < DAC$ (AD is angle bisector) and since the angles $ < BAD$ and $ < BCD$, and $ < DAC$ and $ < CBD$ share the same arc, they are equal. So $ < BAD \equal{} < DAC \equal{} < BCD \equal{} < CBD$, so $ \triangle BDC$ is isosceles, with $ BD \equal{} CD$. Now apply Ptolomey's theorem:

$ (AB)(CD) \plus{} (BD)(AC) \equal{} (BC)(AD)$

$ 7(CD) \plus{} 8(BD) \equal{} 9(AD)$

$ 7(CD) \plus{} 8(CD) \equal{} 15(CD) \equal{} 9(AD)$

$ \therefore \frac {AD}{CD} \equal{} \frac {15}{9} \equal{} \boxed{\frac {5}{3}}$[/hide]

[hide="#2 Solution"]There are 3 sets of 4 faces for which can be painted all of the same color. Then there are 2 faces left to choose its colors. So there are

$ 2* 1 * 1 * 1 * 2^2 + 2*1*1*1*{}^2{}^2 + 2*1*1*1*2^2 = 24$ ways to paint it.

There are $ 2^6 = 64$ ways to paint it overall, so the probability is $ \frac{24}{64} = \boxed{\frac{3}{8}}$.[/hide]

Limacs outline for #1 looked confusing
[hide="1"]Ok. Basically what we want to find is how many terms are in 2 sets when combined. Thats 2004x2. But obviously we have overcounted, because as you noticed, a lot of terms were repeated in both sets. Just how many terms? Well, you also noticed that of every 7 terms, just one of them was repeated. So 2004/7=286 terms were repeated. Therefore, 4008-286=the answer[/hide]

[hide="Dont open until learning vietas formulas!"][hide="Did you learn them yet?"][hide="Are you sure?"]Let the roots of $ x^2 + px + m = 0$ be $ r_1$ and $ r_2$, so $ x^2 + px + m = (x - r_1)(x - r_2)$. We also know that the roots of $ x^2 + mx + n = 0$ are $ 2r_1$ and $ 2r_2$, so $ (x - 2r_1)(x - 2r_2)$. So by Vieta's formulas, we have the following:

$ p = - (r_1 + r_2)$
 and
$ m = r_1r_2 = - 2(r_1 + r_2) \implies \frac {r_1r_2}{ - (r_1 + r_2)} = 2$
 and
$ n = 4r_1r_2$

So $ \frac {n}{p} = \frac {4r_1r_2}{ - (r_1 + r_2)} = 4(2) = \boxed{8}$[/hide][/hide][/hide]

[quote="Thunder365"][hide="1"]Ok. Basically what we want to find is how many terms are in 2 sets when combined. Thats 2004x2. But obviously we have overcounted, because as you noticed, a lot of terms were repeated in both sets. Just how many terms? Well, you also noticed that of every 7 terms, just one of them was repeated. So 2004/7=286 terms were repeated. Therefore, 4008-286=the answer[/hide][/quote]

Agreeable...I guess. But the point is [at least when you are doing it for practice] you have to prove why it's every seventh term. And also they repeat in blocks of seven starting after the 16 (the 16 is 6 from the start).


Obviously it didn't work out before, and the sites have proven to be just about useless, so I am, for now, going to do all the work right here on the AoPS forums again.

Please note:
[b]This is a registration thread, not a discussion thread.  Please post your comments in the discussion thread.[/b]
Yes, this is for March, not April.
The max. teams is 16, but only 8 will qualify for the tournament. (the Qualifications Round is shorter and not a part of the real tournament)
Teams are seeded by order of posts.  This will change in the future, but rating will never be a factor as it changes often, and can be inaccurate at times.
For all who are concerned, the rules will MOST LIKELY be either verizonwires or similar to it for the real tournament, the qualifications round will be something shorter.
The registration deadline is also MOST LIKELY going to be Friday, February 29 at 11:59 PM.
Edit: Please do not post here if you are unsure of whether or not all of the members of your team will be on it.  Also, if you are wondering the avatar size is 66x66px.

Register by making a post in the following form:
[code][size=16][u][b]Team Name[/b][/u][/size]

[b]Members:[/b]
Member 1
Member 2
Member 3

[b]Avatar:[/b]
[img]URL[/img][/code]

That's it, you're done.

[size=134][i][b]Team WeWillWin[/b][/i][/size]

[b]Members[/b]

lotsofmath
fireemblem13
alexhhmun (if he wants to)

[b]Avatar [/b](though its not really an avatar)

 :first:

DUDE WTF
DONT STEAL STEVENMEOW FROM US FOOL
k thanks
Team Plastic: Returning Team
ra5249
budi713
stevenmeow
(same avatar as last time)

[size=59](just for reference)[/size]

[size=134][u][b]DeadPlanaria[/b][/u][/size]

[b]Members:[/b]
undefined117
cognos599
MustItAlwaysBeMe

[b]Avatar:[/b]
(same as before)

RETURNING TEAM!

Boyz of Darkness

verizonwires
vallon22
emton

[b]You can now register here:[/b]
http://aopsftw.freehostia.com/home/

It is much easier for me to see and manage all of the teams.  If you have already signed up here, you do not need to do so again.

If you wish, you can still register here, and I will update the site from what you enter here.  If you need to make a change, PM me or post in the discussion thread.

i already registered for march tourny b4 the feburary tourny was cancelled and everything got messed up.
should u register again? of am i good?

Register again.
1.  It's easy to sign up on the site.
2.  You don't [i]need[/i] to sign up there but its helpful.
3.  People might want to change their teams.
4.  I already have your avatar.

I'll add a checkbox for returning teams.
And for anyone concerned, multiple team submissions are deleted.

[size=134][u][b]WeCantDecide[/b][/u][/size]

[b]Members:[/b]
CircleSquared
sansuunoousama
Unemployed (if he wants)

[b]Avatar:[/b]
[img]<to%20be%20decided>[/img]

EDIT: sansuu and i decided on a team name

Team AznPride
Lawrence Wu
returnofkiladi
ash-Pokemon

Avatar= "http://gallery.webman.me.uk/d/5997-2/NewBliss.jpg"

[size=150][b][u]PENGUIN INVASION[/u][/b][/size]

[b]Members:[/b]
RollandWu
anniebananie
e714212835
[b]
Avatar:[/b]

(same as before)

Team Indiana

me, mathcrazed, and some other person?

[quote="perfect628"]$ \$600,000$ this one fails too.[/quote]
The only thing that would make it fail is people that bet on it failing.  If the people playing actually care, trust me, it will get under way.

[quote="MustItAlwaysBeMe"]Oh my...my Indiana team versus yours in the first round(both of our teams have at least one person who moved from Indiana and at least one who lives in Indiana)  That is if there are only 8 teams.

mathcrazed was on CarmlSaml before, but they haven't registered.  I'll put xpmath with Indiana.  And I'll go ahead and add WeWillWin, even though alexhhmun was with iin77 before this.  Maybe I should PM them to see if they are going to do it.

Oh yeah, and didn't you decide on Team Beast before?

To see the stats right now: http://aopsftw.freehostia.com/home/[/quote]
We'll get owned XD

do u have our avatar for march team tournament? It is the same one as my individual tournament if you didn't get it.

Ok, I added it.  Thanks.  The registration deadline is Friday, and if there are more people, qualification round is due 2 weeks from Friday.  Every other round will be due one week from the previous.  If you cannot attend anything [b]please let us know[/b].

umm.. okay so how are the matches going to be organized?

It depends on whether or not new teams join.

Team SagginWolley

1: KLEBIAN
2: IGNITE168
3: ZENITH XAN

Please: new registrations should go here: http://aopsftw.freehostia.com/home/

You can also view the stats and teams there.  Thanks.

Deadline for registration is over.
See the Statistics page for information about the Qualifications Round.

Edit:  I reseeded the teams, by when they had all their members.  WeWillWin is 2nd now.  To see all the teams go to the site.

wow exactly do we have 2 put down 3 ppl?


Can someone provide a hint for this problem?

ABC is a triangle. E, F are points on the side BC such that the semicircle diameter EF touches AB at Q and AC at P. Show that the intersection of EP and FQ lies on the altitude from A.

Since $EF$ is a diameter, we conclude $\angle FEQ=\angle EFP=\pi/2$. It follows that $KQLP$ is an inscribed quadrilateral. Thus $\angle QLK=\angle QPK=\angle BQE=\angle LQA$ (since $AQ$ is tangent to circle).
In the same way we obtain $\angle PLK=\angle LPA$.
Let $O$ be circumcenter of $KQLP$. Since $\angle KQA=\pi/2$, $KL$ is a diameter and $O$ lies on line $KL$. And we can write $\angle LQA=\angle QLK=\angle QLO=\angle LQO$ (due to isosceles triangle). Also we obtain $\angle LPA=\angle PLK=\angle LPO$. It means that $O$ and $A$ are the same points, i.e. $A$ lies on $KL$.
In the triangle $EFL$ lines $EP$ and $FQ$ are altitudes, this $K$ is orthocenter $\Rightarrow$ $LK$ is altitude to $EF$. So we conclude $AK\perp BC$.

That's all.

http://www.mathlinks.ro/Forum/viewtopic.php?t=6143

Try searching here on this forum for postings by "Iris Aliaj". They contain lots of JBMO problems.

  Darij

Hmm... Do you force mathlinkers to read your long proofs?

I just wanted to show him that most of the old JBMO problems are already solved here. By the way, the thread I linked to contains not just a long solution of mine, but also a rather short and nice solution of Omid Hatami which is the same as yours.

  Darij

And how could I know that it is an old JBMO?


Hey,
Let's try to see how many people we can get on FTW at once! Just post a link to this page in your sig and be on FTW at 9:00 EST (8:00 Central, 7:00 Mountain, 6:00 pacific) on Friday October 17.
That's it! If theres any day or time that you would think would be more people signing up on  don't hesitate to say.

wow...every topic seems to get more and more pointless...

(^_^) and kl are right.  Stop it.

I bet you can get banned for starting these very spammy topics.

So just stop.

i go back to what i said on the other thread
this is not going to help your productive post count which is necessary for chatting privilages
DO NOT SPAM

and to (^_^), saying "wtf" is a muteable offense in FTW and i suggest you do it no more


[quote="Bob75"]Let ABCD be a quadrilateral with 2 diagonals perpendicular at the point O.
Let E, F, G and H be the symmetric of O respectively with respect of the straight lines (AB), (BC), (CD) and (AD).
Prove that the points E, F, G and H are cocyclic.[/quote]

 I'm very busy this moment , but I will send you a hint.

 Let $K,L,M,N$ the midpoints of $OE,OF , OG, OH$ resp. These points are on the sides of the initial quadrilateral .We have to  prove that $\angle E+\angle G = 180$ or   $\angle NKL+\angle NML = 180$ since in $EFGH , KLMN$ the sides parallel.But $\angle NKO = \angle NAO, \angle LKO = \angle LBO , \angle LMO = \angle LCO, \angle NMO = \angle NDO$ . Add member to member and you will have the desired result.

[u] Babis[/u]

I'm also very busy at the moment, so I will send you a link:

http://www.mathlinks.ro/Forum/viewtopic.php?t=4344

  darij


Find the least common multiple of $ 6!$ and $ (4!)^2$.

Another way:

We prime factorize each:

$ 6!\equal{}6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\equal{}2^4 \cdot 3^2 \cdot 5$

$ (4!)^2\equal{}4 \cdot 3 \cdot 2 \cdot 1 \cdot 4 \cdot 3 \cdot 2 \cdot 1\equal{}2^6 \cdot 3^2$

The LCM will be $ 2^6 \cdot 3^2 \cdot 5\equal{}\boxed{2880}$.

[quote=pinkpig]how is this   Level 8 Counting & Probability  +59 XP[/quote]

its +5 XP


My fax machine isn't working too well and I'm not sure if my entry form went through. Is there any way I can get it confirmed that I sent it in? My ID # is 6303.

Also, how do you write a line segment in LaTeX? (Making a little bar over the letters). I don't just want AB but AB with a bar over it.

\overline...

\overline{AB} = $\overline{AB}$

[quote="qweretyq"]My fax machine isn't working too well and I'm not sure if my entry form went through. Is there any way I can get it confirmed that I sent it in? My ID # is 6303.[/quote]Unfortunately, we have not yet filed the papers (right now it's just a big stack next to the fax machine), so there's no way yet that we can confirm a particular submission.

We will be filing the papers this week.  When we record your submission, a green checkmark will appear on your "My USAMTS" page on the USAMTS website.  [b]It will take several days for us to log all the submissions[/b], so don't panic if you don't see a green checkmark right away.  We'll let you know when you can panic.  :lol:


Since the MAML is in the same format as the 94-99 AHSME's, how do you think they compare to them?

I find them just a little harder, but there are more easy questions.

I think the easier AMC12 questions are easier than the easier MAML questions, but the last few AMC12 questions are much more difficult than the last MAML questions.

The second half of the AMC 12, in general, seems to require more mathematical creativity than anything on the MAML. My main thing with the MAML was finishing as many problems as I could (without a calculator, as my batteries ran out midway through). On the AMC 12, its about figuring out how to do the last few.

Also, TI-89 users have a greater advantage over TI-83/83+/84 users in MAML than in AMC12 IMO

TI-89s are not allowed in the MAML...?

In my opinion, the MAML is much easier, as you can make WLOG assumptions or just plug in numbers and get the right answer.


If $ f(x)$ if it satisfies the equation

$ f(x \plus{} 2) \minus{} 5f(x) \plus{} 6f(x \plus{} 1) \equal{} 0$

Did you mean $ f : \mathbb{R} \rightarrow \mathbb{R}$ ?
If so, the function is like what I'm going to say down here.
We can let the images of $ x \in \left[0,2\right)$ be any real numbers.
And then the other images will strictly come by the condition.  :)


Define $A=\left(\begin{matrix}1&c\\4&-1\\ \end{matrix}\right)$ for the constant number $c.$ Let $Q$ be the point to which a moving point on the line $y=x+1$ is mapped. Assumed that $\triangle{PQR}$ is a triangle with $\angle{RPQ}=90^\circ$ for a fixed point $R$ and any real numbers $t.$ Find the minimum value of the area of the excircle of $\triangle{PQR}.$


search
trigonometry
inequalities
inequalities proposed

If $a,b,c$ are sides of [color=red]acute[/color] triangle show that this ineqality holds:

$\frac{1}{a^{2}+b^{2}-c^{2}}+\frac{1}{b^{2}+c^{2}-a^{2}}+\frac{1}{c^{2}+a^{2}-b^{2}}\ge \frac{27}{(a+b+c)^{2}}$

Maybe it was posted before but I didn't find it by a search function.

[quote="Eros88"]
$\frac{1}{a^{2}+b^{2}-c^{2}}+\frac{1}{b^{2}+c^{2}-a^{2}}+\frac{1}{c^{2}+a^{2}-b^{2}}\ge \frac{27}{(a+b+c)^{2}}$
[/quote]

Hmm... I was reading College Math Journal Nov 2006 issue and I see that this is proposed problem #838.
And the proposer? "Arkady Alt" :)

[quote="GacKiem"]Very nice solution, arqady! Congatulation![/quote]
Thank you,  :lol: , but it's very easy inequality.  :maybe:

I believe that the following refined one is also true for all acute triangles with side lengths $a,b,c$
\[\frac1{a^{2}+b^{2}-c^{2}}+\frac1{b^{2}+c^{2}-a^{2}}+\frac1{a^{2}+c^{2}-b^{2}}\geq\frac9{ab+bc+ca}.\]

And the following is still true:
$\frac1{a^{2}+b^{2}-c^{2}}+\frac1{b^{2}+c^{2}-a^{2}}+\frac1{a^{2}+c^{2}-b^{2}}\geq\frac9{a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}}$
 :D

from AM-GM
$\sum_{cyc}\frac{1}{a^{2}+b^{2}-c^{2}}=\frac{1}{2}\sum_{cyc}\frac{1}{ab\cos C}\geq \frac{1}{2}\frac{3}{\sqrt[3]{(abc)^{2}\cos A \cos B \cos C}}\geq \frac{3}{\sqrt[3]{(abc)^{2}}}$
because $\frac18 \geq \prod \cos A$ ...  :wink:

equivalent to
$\sum_{cyc}\frac{1}{x}\geq \sum_{cyc}\frac{2}{x+y}$
to prove use $\frac{1}{m}+\frac{1}{n}\geq \frac{4}{m+n}$  :D

[quote="gemath"]That's right it very simple but quite strong but starting problem look so hard !
Actuall follow the way of arqady
$\sum\frac{1}{b^{2}+c^{2}-a^{2}}\ge\frac{1}{4r^{2}}\ge\sum\frac{1}{a^{2}}$[/quote]
The right inequality is well-known  :D
We have: $\sum\frac{1}{(y+z)^{2}}\le\frac{\sum x}{4\prod x}\quad\forall\, a,\,y,\,z\ge 0$
Let $x: =p-a,\,y: =p-b,\,z: =p-c\;\text{where}\;p=\frac{a+b+c}{2}$, we have:
\[\sum\frac{1}{a^{2}}\le\frac{p}{4\prod (p-a) }\Longrightarrow \sum\frac{1}{a^{2}}\le\frac{1}{4r^{2}}\]
:D


modular arithmetic
abstract algebra
number theory
least common multiple
function
group theory
number theory proposed

$n$ is a natural number. $d$ is the least natural number that for each $a$ that $gcd(a,n)=1$ we know $a^{d}\equiv1\pmod{n}$. Prove that there exist a natural number that $\mbox{ord}_{n}b=d$

Is that $a^{d}\equiv 1\pmod n$? If it is, then it follows from the following result:

The exponent of a finite abelian group occurs as the order of an element of the group.

This, in turn, is clear from the structure theorem for finite abelian groups: they are finite direct products of cyclic groups.

Let \[n=\prod_{i}p_{i}^{k_{i}}\] , then ring \[Z/nZ=\sum_{i}(Z/p_{i}^{k_{i}}Z.\] Used $(Z/p^{k}Z)^{*}$ (p>2) are ciclic followed your result.

If two elements $u_{1}$ and $u_{2}$ have orders $n_{1}$ and $n_{2}$, then let $(n_{1}, n_{2})=m$, and let $n_{i}=mk_{i}$ for $i=1, 2$.  Then $(k_{1}, k_{2})=1 \implies \exists a, b\in \mathbb{Z}, ak_{1}+bk_{2}=1$, with $(a, k_{2})=(b, k_{1})=1$.  Then consider $u_{1}^{b}u_{2}^{a}=u_{3}$, and some work with orders yields that $n_{1}, n_{2}$ divide the order of $u_{3}$.

$n=2^{\alpha_{1}}p_{2}^{\alpha_{2}}p_{3}^{\alpha_{3}}...p_{m}^{\alpha_{m}}$
We will prove that $LCM(\phi(p_{2}^{\alpha_{2}}), \phi(p_{3}^{\alpha_{3}}), ..., \phi(p_{m}^{\alpha_{m}}))|d$. Let $A$ be the set of residues $\mod{n}$ which are primitive roots $\mod{p_{i}^{\alpha_{i}}}$ for $i=2,3,...m$. By the Chinese Remainder Theorem $|A|>0$ and the claim is obvious. Now, from the set $A$ we can choose such $g$ that $ord_{2}(g)$ is largest and then it's clear that $d=ord_{2}(g) \cdot LCM(\phi(p_{2}^{\alpha_{2}}), \phi(p_{3}^{\alpha_{3}}), ..., \phi(p_{m}^{\alpha_{m}}))|$. Also $ord_{n}(g)=d$

sorry, I don't know what is the meaning of $ord_{n}b$?

$ord_{n}(b)=a$ iff $a$ is the smallest positive integer such that $b^{a}\equiv 1 \mod{n}$ (if such exists).

Hm...isn't this a very well known fact of the carmichael function $\lambda$? (since it is obvious that $d=\lambda(n)$ - or it is even by definition, depending how one defines $\lambda$). A strange problem to propose for a competition  :huh:

Yes, it is a bit strange... It's obvious... It's not even necessary to invoke Carmichael function...


That was a really stupid question, the answer is obvious and it's $ 4^{ln(n)}$.

If anyone wants to know why: Take $ \ln$ of both sides to get $ \ln 4\ln n$ compared to $ \ln n$, and $ 4>e$ so $ \ln 4>1$, which makes the LHS larger.


What is the mean of the following list of numbers: \[ \{1,4,7,10,13,16,19,22,25,28,31\}.\]

[quote=gaussintraining]Since the terms are evenly spaced and there an odd number of terms, the mean is just the middle number, $ \boxed{16}$.[/quote]

How do you prove that?

Say there are $2k+1$ terms with middle term $n$, with common difference $d$. Then the sequence is
$$\{n-kd,n-(k-1)d,n-(k-2)d,\dots,n,\dots,n+(k-2)d,n+(k-1)d,n+kd\}.$$
When summing to take the average, we can create $k$ pairs that sum to $2n$ (note how things cancel), along with the $n$ in the middle. So the average is $\frac{k(2n)+n}{2k+1}=\frac{(2k+1)n}{2k+1}=\boxed{n}$.


inequalities
search
inequalities proposed

Oh sorry I did not know that this one was posted before

Thank you [b]Albanian Eagle[/b]

Salem


There are two 2-by-2 squares, with a total area of $ 4\cdot2\equal{}8$

There are eight trapezoids, with a total area of $ 8\cdot \frac{1}{2}(2)(2\plus{}4)\equal{}8\cdot6\equal{}48$

The total surface area is $ 48\plus{}8\equal{}56$


Now and then, I'll post several mathcounts level questions for the people who are taking MC this year. Of course, I'm not a moderator of this forum but to have myself doing some fun math, I'll just do it.

Let's start out with an easy problem.

This problem appeared in real chapter and I think it's neat to go over this (this is changed from the original. So, the idea is same but the answer is different):

Find the number of lines that can be formed by 4 points.

Answer
[hide]This is a basic combinatorics question. We have 4 points and we only need 2 points to have a line.

So, it's:

\[\binom {4}{2} = \frac {4*3*2!}{2!*2!} = 2*3 = 6\]

To check your work, try to draw it. The answer should be 6.

The combinatorics does not appear often as they do in AMC's. But it IS very important so if you see any questions like this, you should be able to do this.

Regards,[/hide]

Thanks for pointing out. I forgot \[(n-k)!\] part.

yeah, the second 2! should be on the bottom of the fraction.

[quote="RC-7th"][quote="Treething"]Are the four points distinct? If not, the answer could be zero :P .[/quote]

If they were the same points then the answer would be infinite wouldn't it?[/quote]

But don't you need two distinct points to define a line?

And if you allow three points to be on one line, you get another different answer :)

I'm sorry.

It should've said "Four distinct lines which aren't collinear" and same wording.

Thanks. :)

Find the number of lines that can be formed by 4 points.

[hide]from one point, there are 3 lines.  for a total of 4 points there are 12 lines.  to remove the duplications divede by 2 to get [b]6[/b][/hide]

[hide]3 distinct non-collinear points will always form some sort of triangle.  As long as the 4th point follows the above requirements, it will always be 6 lines.[/hide]

It probably would have been simplest to say, "Find the [u]maximum[/u] number of lines that can be formed by 4 points."  Just my two cents. :)


Dude if you do not know how to play categories, post it in a new thread. Signups here. 5 for first game, more as we go on.

................Google does exist, y'know.

Yeah don't cheat, cuz then it isnt as fun? Everyone will cheat anyway. Ok the following posters have been selected to play in this order:

Mewto55555
math154
zzzzzzzzz
pianoforte
xpmath
pythag

My category is:

Star Wars Planets!

[b]Bespin[/b]

[b]I dunno what you're talking about![/b]

Hoth?  Hoth?  Hoth?  Hoth?  Hoth?  Hoth?  Hoth?  Hoth?  Hoth?

:rotfl:  :rotfl:

Wrong universe OUT

Xpmath?

ah...pianoforte, you should have said Kashyyk or Tabioo (something like that)

Sorry thats not a planet in the star wars galaxy. Pythag go

After Round I only 3 remain.

Mew
ZZZ
Pythag

Continuing:

Bespin
Hoth
Tatooine
[b]Naboo[/b]

Dagoba...

When do we get to change the topic?

When someone wins, the next person in line picks the topic.

Btw, its Dagoba[b]h[/b].

[quote="pianoforte"]I love you.[/quote]

Young, same gender, love. [size=59]cough gay cough[/size]

Man everyone I know of has already started doing their driving permit whatever crap studying, even Ubemaya and ccy  :!:  Oh wait junggi has not either...and he's a junior rofl.

People in our school, 8th grade, have already passed all their tests.  Now when they are 16, they just go to the office and the people hand over their license.

[quote]yeah, basically. silly ks driving laws.[/quote]

Your mom is silly...

[quote="yaofan"]can we stay on topic with the ultimate game?[/quote]

No. Read the forum's name.


Hi, I have a query....If a student is studying at a US colege and is able to come up with the Putnam expectations can he participate at it only if he is necessarily studying mathematics at college ?? I am trying to say that if he is studying medicine for example can he represent his college by participating ???
Thank you in advance......

As far as I know, there are no prerequisites for participating in the Putnam besides being an undergraduate.


Putnam
function
calculus
derivative
college contests

Hey!

Let T_0 = 1, T_1 = 3, T_2 = 6, and for n >= 3,
T_n = (n + 4)T_{n-1} - 4n T_{n-2} + (4n - 8)T_{n-3}.
The first few terms are 2, 3, 6, 14, 40, 152, 784, 5168, 40576, 363392.
Find, with proof, a formula for T_n of the form T_n = A_n + B_n, where
(A_n) and (B_n) are well-known sequences.

I have tried plugging in numbers and making a chart, but I can't seem to understand how to find the A_n and B_n.  Any suggestions or hints or tips?

It's not hard to guess the formula. Is there any famous sequence which is close?

Once you have a guess that fits, prove it with induction.

A more pointed hint: the recursive formula involves a lot of multiplication of things near $ n$ by previous terms. The simplest recursion that's anything like that is $ (n\plus{}1)!\equal{}(n\plus{}1)\cdot n!.$ So $ T_n$ should grow more or less like a factorial.

Correction to sabrewulf's original post: His first line should read

"Let $ T_0\equal{}2,T_1\equal{}3,T_2\equal{}6,$ and for $ n\ge 3,$"

I agree that in this case it is easy to guess the formula by looking at the first few values.  If it wasn't so easy to guess, is there a more systematic approach?  I was thinking generating functions, either ordinary or exponential.

The form of the answer suggests that exponential generating functions will be the more useful, [hide="primarily because"] The ordinary generating function of $ n!$ is terrible; its radius of convergence is $ 0$. [/hide]
But even without that knowledge, the form of the problem alone suggests that exponential generating functions will be more useful
[hide="because"] Using the ordinary generating function leads to a differential equation which is most easily solved by a Laplace transform - and that will more or less give you the exponential generating function. [/hide]

Starting to do the work:

We have $ T_n \equal{} (n \plus{} 4)T_{n \minus{} 1} \minus{} 4n T_{n \minus{} 2} \plus{} (4n \minus{} 8)T_{n \minus{} 3}$, with $ T_0 \equal{} 2, T_1 \equal{} 3, T_2 \equal{} 6$.  Let $ T(x) : \equal{} \sum_{n \equal{} 0}^\infty \frac {T_n}{n!} x^n$.  Then $ T''(x) \equal{} \sum_{n \equal{} 2}^\infty \frac {T_n}{(n \minus{} 2)!} x^{n \minus{} 2} \equal{} 6 \plus{} \sum_{n \equal{} 3}^\infty \frac {(n \plus{} 4)T_{n \minus{} 1} \minus{} 4n T_{n \minus{} 2} \plus{} (4n \minus{} 8)T_{n \minus{} 3}}{(n \minus{} 2)!} x^{n \minus{} 2}$.  We can rewrite
\[ \frac {(n \plus{} 4)T_{n \minus{} 1} \minus{} 4n T_{n \minus{} 2} \plus{} (4n \minus{} 8)T_{n \minus{} 3}}{(n \minus{} 2)!} \equal{} \frac {T_{n \minus{} 1}}{(n \minus{} 3)!} \plus{} \frac {6T_{n \minus{} 1}}{(n \minus{} 2)!} \minus{} \frac {4T_{n \minus{} 2}}{(n \minus{} 3)!} \minus{} \frac {8T_{n \minus{} 2}}{(n \minus{} 2)!} \plus{} \frac {4T_{n \minus{} 3}}{(n \minus{} 3)!}.
\]
Multiplying by $ x^{n \minus{} 2}$ and summing these termwise for $ n \geq 3$ gives
\[ T''(x) \equal{} xT''(x) \plus{} 6(T'(x) \minus{} 3) \minus{} 4xT'(x) \minus{} 8(T(x) \minus{} 2) \plus{} 4xT(x).
\]
Except, something has gone wrong ... any suggested fixes?  [This is one reason this is a bad way to solve this question :).]  Then you're going to have to solve that differential equation (or rather, one that looks a lot like it, modulo my arithmetic errors) and from there the answer should be pretty clear.

Still not sure what went wrong with the previous try; starting over and being careful (but not showing the arithmetic), I get the following:
with $T = \sum_{n \geq 3} \frac{T_{n - 3} x^n}{n!}$, I get
\[
8T  - 4x T' + 4(x - 1)T'' + (1-x)T''' = 2 - 7x + 2x^2.
\]
We can try to solve this together with initial conditions $T(0) = T'(0) = T''(0) = 0$, or we can take three derivatives; with $S = T''' = \sum_{n \geq 0} \frac{T_n x^n}{n!}$ this yields the equation
\[
-4S + (12 - 4x)S' + (4x - 7)S'' + (1 - x)S''' = 0
\]
with initial conditions $S(0) = 2$, $S'(0) = 3$, $S''(0) = 6$.  The question is, are standard differential equations techniques enough to solve these equations, or does one still have to guess the answers?


induction
algebra
polynomial
Vieta
number theory open
number theory

Prove that $ p$ divides the numerator of $ 1 \plus{} \frac {1}{2^4} \plus{} \frac {1}{3^4} \plus{} ... \plus{} \frac {1}{(p \minus{} 1)^4}$, $ p\ge 7$ is a prime

Take a look at the proof for exponent $ 2$ and make the necessary adaptions ;)

i don't know if this is what zetax refers to, but here's a proof...

we know that the set of the inverses modulo $ p$ of $ \{1,2,...,p\minus{}1\}$ is $ \{1,2,..,p\minus{}1\}$, so, if we extend the concept of congruences to rational numbers, we would have to prove that $ 1\plus{}2^4\plus{}3^4\plus{}...\plus{}(p\minus{}1)^4$ is divisible by $ p$.

to find a closed form for $ S(n)\equal{}1\plus{}2^4\plus{}...\plus{}n^4$, suppose that $ S(n)\equal{}an^5\plus{}bn^4\plus{}cn^3\plus{}dn^2\plus{}en\plus{}f$, and consider the diference $ S(n\plus{}1)\minus{}S(n)\equal{}(n\plus{}1)^4$...
after comparing coefficients and using that $ S(1)\equal{}1$ (to find that $ f\equal{}0$) it follows that $ S(n)\equal{}\dfrac{6n^5\plus{}15n^4\plus{}10n^3\minus{}n}{30}\equal{}\dfrac{n(n\plus{}1)(6n^3\plus{}9n^2\plus{}n\minus{}1)}{30}$ (which in fact can be proven by induction on $ n$)... from here it's easy to conclude  :D

Far more work than necessary.  The polynomial with roots $ 1^4, 2^4, ... (p\minus{}1)^4 \bmod p$ is just $ x^{ \frac{p\minus{}1}{\gcd(4, p\minus{}1)}} \minus{} 1$, and Vieta's concludes for $ p > 5$.

[quote="t0rajir0u"]Far more work than necessary.  The polynomial with roots $ 1^4, 2^4, ... (p \minus{} 1)^4 \bmod p$ is just $ x^{\frac {p \minus{} 1}{\gcd(4, p \minus{} 1)}} \minus{} 1$, and Vieta's concludes for $ p > 5$.[/quote]
and what if $ gcd(4,p\minus{}1)\equal{}4$, $ x_1\equiv x_2\equiv x_3(mod p)$  $ x_i\in(1^4,2^4,...,(\frac{p\minus{}1}{4})^4)$ and $ x_i\neq x_j$.

Let me rephrase that.  The polynomial with roots the [i]distinct[/i] values in $ \{ 1^4, 2^4, ... (p \minus{} 1)^4 \}$ is $ x^{\frac {p \minus{} 1}{\gcd(4, p \minus{} 1)} } \minus{} 1$ and the sum of the elements of that multiset is a multiple of the sum of its distinct values.

[quote="Flakky"]Ok why then the sum of the elements of that multiset is a multiple of the sum of its distinct values?[/quote]

So you are asking why, for any element $ a$ of $ \mathbb{F}_p\setminus\left\{0\right\}$, the value

$ \left|\left\{x\in\mathbb{F}_p\setminus\left\{0\right\}\mid x^4 \equal{} a^4\right\}\right|$

does not depend on $ a$.

Hint: The set $ \left\{x\in\mathbb{F}_p\setminus\left\{0\right\}\mid x^4 \equal{} a^4\right\}$ is the disjoint union of the set $ \left\{a, \minus{} a\right\}$ (which always has two elements) and the set $ \left\{x\in\mathbb{F}_p\setminus\left\{0\right\}\mid x^2 \equal{} \minus{} a^2\right\}$ (which has two elements if $ \minus{} 1$ is a square mod $ p$, and is empty if $ \minus{} 1$ is not a square mod $ p$).

  darij

Take a look at http://www.mathlinks.ro/viewtopic.php?t=40171 (in fact, that theorem holds for all integer exponents $ k$ such that $ (p\minus{}1) \nmid k$).


calculus
integration
trigonometry
function
calculus computations

Is it true that

if $ x>0$ then $ \int^x_0\frac{\sin t}{1\plus{}t}dt>0$

?

Yes, it is true. The slickest way I know: integrate by parts with $ u\equal{}\frac1{1\plus{}t}$ and $ dv\equal{}\sin t\,dt,$ and (this is the key trick), choose the antiderivative as $ v\equal{}1\minus{}\cos t.$

$ \int_0^x\frac{\sin t}{1\plus{}t}\,dt\equal{}\left.\frac{1\minus{}\cos t}{1\plus{}t}\right|_0^x\plus{}\int_0^x\frac{1\minus{}\cos t}{(1\plus{}t)^2}\,dt$

But the boundary terms evaluate to $ \frac{1\minus{}\cos x}{1\plus{}x},$ which is always nonnegative, and the remaining integral is the integral of a nonnegative function. (And the integrand is continuous and positive at some points in every interval, so the integral must be strictly positive.)

The only property of $ \frac1{1\plus{}t}$ that I really used in this proof was that it is a decreasing positive function on $ [0,\infty)$ (which could be weakened to $ (0,\infty)$ as long as we could retain convergence of the integral at zero.)

I never know that we can use integration by part in such a way that the constant of the antiderivative is freely chosen by us :)  . Learn new things today. Thanks

By the way , can you elaborate about the the decreasing property of $ \frac{1}{1\plus{}t}$ that you use? I can't see where we need it . Thanks  :)

If $ f(t)\equal{}\frac1{1\plus{}t},$ then the second integral was $ \int_0^{\infty}(1\minus{}\cos t)(\minus{}f'(t))\,dt.$

We needed $ \minus{}f'(t)>0$ to argue that the integral was positive.

[quote="shyong"]I never know that we can use integration by part in such a way that the constant of the antiderivative is freely chosen by us :)  . Learn new things today. Thanks.[/quote]
A classic from a beginning class in integration: find the antiderivative of $ \int x\arctan x\,dx.$

You'll integrate by parts, letting $ dv\equal{}x\,dx.$ The algebra works out nicely if you choose $ v\equal{}\frac12(x^2\plus{}1).$ (You can do it otherwise - this isn't about whether it works, but about making it work with the least effort.)

That's a slick proof, Kent.  Let me see if I can solve the original problem with less cleverness.  Let $ f(t)$ equal $ \frac{1}{1 \plus{} t}$ for $ t \le x$ and equal $ 0$ for $ t > x$.  Note that $ f$ is a decreasing function.

Our object is to show that
\[ \int_0^\infty f(t) \sin t \, dt > 0.
\]
Let $ S$ be the set of nonnegative $ x$ such that $ \sin x \ge 0$.  Then we can rewrite the integral as
\[ \int_S [f(t) \minus{} f(t \plus{} \pi)] \sin t \, dt.
\]
Because $ f$ is decreasing, the term being integrated is always nonnegative.  So the integral itself is nonnegative.

Actually, I am supposed to show the integral is positive, but I am feeling lazy...

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31180&search_id=1291315243]You can refer to this.[/url]

OK, so I've said this before in exactly the same way.


Guess which letter won!
Guess which letter got 0 USAMO qualifiers!
Guess which three letters only had 1 USAMO qualifier!

[hide][code]A :52
B :25
C :28
D :45
E :12
F :1
G :11
H :11
I :7
J :70
K :25
L :8
M :33
N :12
O :1
P :19
Q :5
R :24
S :50
T :22
U :0
V :5
W :13
X :1
Y :18
Z :7
[/code][/hide]

Did you count "Brady Zarathustra" as B or a Z? If you're going by first name, he should be a Z.

I'm predicting a few As on the IMO team...

And some Bs.

And a Z maybe.

Oh and Js ftw (in terms of USAMO qualifiers of course, probably not gonna be many on the IMO team).

dude and also a M/R you can't forget the M/R

so uh ABM/RS

[quote="Hamster1800"]These count "Brady Zarathustra" as a B. I'm getting the same numbers as posted using grep (without changing the file)

I feel bad if randomdragoon actually manually counted them...[/quote]
Well, I did write my own program to do it for me (read a text file and count stuff) but after that it was pretty fast.

uh the list is up?

I thought the IMO team will be AAABRZ?

What? Whose the T?

Alex Zhai
Arnav Tripathy
Brian Lawrence
Zarathustra Brady
Bohua Zhan
[b]M[/b]atthew [b]R[/b]ichard McCutchen

seem pretty far ahead of everyone else currently. So it's like

AABB(M/R)Z

dude Sherry Gong what the hell

she could probably own me

I have not actually consulted her on the subject

yeah - but what happens if one of them goes to RSI or something like BL did last year?  then what's the IMO team going to be?  Huh?  Huh?

[quote="magixter"]How do you guys know the USAMO participants?[/quote]

http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2007-ua/2007usamoqual.shtml

well...

Lawrence-senior
McCutchen-senior
Arnav-senior
Hesterberg-senior
Bohua-senior?
Sherry-senior
Zeb-senior

The only junior is Zhai, but weren't people telling him not to go to RSI?

[quote="Kent Merryfield"][quote="tjhance"]Oh yeah go T's. T's own.  :D[/quote]
I could see a T on the IMO team.[/quote]
Dude yeah... I could see an I.. and an M.. O... T..... E A (M?)

uh we're all seniors except for zhai who is very short

yeah and RSI is for juniors, if I'm not mistaken

yeah

on a related note sherry is also very short

[quote="probability1.01"]Dude Alex is the best USAMO qualifying first name. Go Alex![/quote]
Word.

Go Alex!

Just wondering, why has no one mentioned a Y for IMO? There are two returning USAMO winners whose names start with Y.

And Go Alex! indeed.  :rotfl: (even though I sadly did not contribute to the A total this year).

I can name two Y's off the top of my head so there's probably more than 2....

and J's all the way :) we won last year too

bohua is a genius

i saw him numerous times

and i tried conversing with him but he wanted to read

life sucks

To me that seems a bit redundant, tired, even a bit plagiaristic (if that is a word)

[quote]amen to probability1.01... Alex has to be the best first name in all time.

We need to take over the world some day.[/quote]

What I am trying to say is that this task has already been completed (see Alexander the Great).  He took over the world in more ways than one:

1. Literally
2. Instead of opressing his new vassals, he asked for a few concessions and that they name their children after him.  Because of this, his name has taken over the world (and apparently, much of the math community)

[quote="SamE"]lol with the short people arnav

S is on the rise! Our conquering of D proved the only change in the top 6... and now we're only two qualifiers behind A! We'll see about USAMO, but you're right some of the best are A's.[/quote]

 :o   I'm a D.  Whatever, we'll get S back next year  :P .

I hope I can contribute to the "S"
even though my chances are slim
only two Kentucky people... :(

Do you not think a 2005 IMO Silver Medalist and a 2006 IMO Gold Medalist will make the team? Odd.

randomdragoon should probably re-run his programs on the revised list.


which statement is INCORRECT:

a) $O_{2}$ is considered to be a molecule and an element
b) Dalton's statement "All atoms of a given element are identical" is no longer accepted due to the existence of isotopes and ions.
c) one mole of aluminum is equal to the number of aluminum atoms in 26.98 g of aluminum.
d) gases behave most ideally at high temperatures and low pressures

or are all the statements correct?

[hide="hint"]There is no $\text{O}_{2}$ on the periodic table...[/hide]

ok I see

can subscripts in a chemical equation be fractions?

a) no, because equations must always be written in standard form
b) yes, as long as the fractions are halves (like 13.5 or 6.5, etc)
c) yes, as long as there are the same number of each type of atom on both sides of the balanced equation
d) sometimes, because some equations cannot be balanced without fractions for subscripts
e) no, because the subscripts represent the number of atoms in a molecule

Are you sure you mean subscripts and not coefficients?  The answer choices seem to be referring to coefficients.

The answer choices are obviously referring to coefficients, but it seems that the word "subscripts" was used as a synonime.

[quote="Yeppyyep"]could you briefly say how you got that answer?[/quote]

A chemical equation must obbey the law of conservation of mass, because at a microscopic level the atoms are not destroyed or created in a chemical reaction, that are only chemical bonds that are broken and formed. This is translated "macroscopically" by writing the same number of atoms of each element in both sides of a chemical equation.

Well, if subscripts is what is actually meant, then (e) would be a much better answer, and the other answers would be distractors.  I have never heard of subscripts and coefficients being used as synonyms.  (e) is almost right; however, it is not quite correct for network, solid ionic, or metallic substances.

[quote="gregx007"]There is no $\text{O}_{2}$ on the periodic table...[/quote]
Unless I'm mistaken, $\text{O}_{2}$ (dioxygen) is an allotrope of oxygen, so shouldn't it be considered an element? Also, $\text{O}$ naturally exists as $\text{O}_{2}$ in nature...

By your argument, graphite should also be considered an element.  I was going to dispute this, but then I actually looked up the IUPAC definition of a "chemical element":

   1. A species of atoms; all atoms with the same number of protons in the atomic nucleus.
   2. A pure chemical substance composed of atoms with the same number of protons in the atomic nucleus. Sometimes this concept is called the elementary substance as distinct from the chemical element as defined under 1, but mostly the term chemical element is used for both concepts.

Interesting.


ARML Duke college HMMT AMC USA(J)MO USAMO

Let's introduce ourselves! I'll start.

I'm Thomas Mildorf, I've lived in Northern Virginia since 1986, I go to Thomas Jefferson (Affectionately known as the dark side by some competitors at ARML :D) where I have been on the VMT since freshman year; this year I am the captain. I'm a three time MOPper ('02-'04) and I run weekly practices for the USAMO. When I go more than a day without math I am a fish out of water.

Your turn!

(Only people who are in this community and logged in can see the posts in this forum.)

I am Jesse Geneson. I am a junior. I go to Thomas Jefferson with the Mildorf. I am a co-captain of the math team this year. My math record is not too good, but I have done a few good things. Besides math, I like to do Tae Kwon Do and play the viola. I am not a fish like the Mildorf, but fish are tasty.

I am not from Virginia(plz don't kick me out). But I have been to Virginia, know a few people who live in Virginia, have a friend who lived in Virginia, like Virginia, and envy people who go to TJ. So I decided to see if I would be allowed into this community(Thanx Mildorf). Whoa, this is my hundreth post!

Hi

my name is mark hou

i am a sophmore attending Jefferson, a year below J and 2 below T

i have managed to screw up an AMC 12, an ARML, a Duke meet, and a HMMT, but hopefully i'm slightly better this year >_<

i also listen to NOFX

mark

I'm a freshman and TJ, and I am not that good at math team. However, I did go to a crappy middle school (Irving), so I have a mediocre excuse.
I decided to post in this particular forum because of the complete lack of posts over the past year.

hello, I'm In Young Cho. I'm a seventh Grader at Longfellow Middle School. I just made USAMO this year and am psyched! I'm not that good at math, but I like doing it, although I can live without it. Other things I do are violin, piano, reading, and just fooling around on computer. Nice meeting you guys!

Hi, I'm Renjie You, an 8th grader at Longfellow Middle School, and I'm absolutely terrible compared to the above people.

The only thing I have to be remotely happy about is getting into TJ.

[quote="Zenith Xan"]Hi, I'm Renjie You, and 8th grader at Longfellow Middle School, and I'm absolutely terrible compared to the above people.

The only thing I have to be remotely happy about is getting into TJ.[/quote]

Pretty much everyone's terrible compared to Mildorf  :P

Alo

I'm Andrew Kim. I go to Longfellow MS, going on 8th grade. I think I'm pretty much a dumbo...
But, there's always bright sides of everything. I only like math that interests me. :roll:
Other things I do...hmm... very hard question
I do what most of you do. I also swim in a team, play screechy violin, war with my little brother, read ... books, and jump/roll/slide inside and outside my house.

My favorite animal is monkey  :D
My favorite color is blue and yellow.
That was random...

[quote="Andrew Kim"]But, there's always bright sides of everything. I only like math that interests me. :roll:
[/quote]

I highly condone this view of mathematics, though if you are like me you will eventually decide that all mathematics (yes, even geometry  :o ) is beautiful.

Good luck on TJ admissions, I look forward to seeing you in a year if you make it in!

Haha.. I was stalking Renjie and I found this. We should all stop posting because we pale under Thomas' and Jacob's genius... Haha. All branches of math are beautiful, sure, but I'm bad at all of them  :rotfl:
Anyways, basically everyone's applying to TJ, and then I won't get in, and then I'll cry. But that's life.

Woah, us non-Jefferson students are definitely in the minority here.

My name's Natalie, and I'm a junior at Hermitage.  Not much to know about me other than I eat, sleep, and breathe math and science.  :D

My name is Haitao Mao. I was born in 1990. I graduate in 2008. I am a relatively inactive AoPS user. For the random fact, I think it's safe to say that we are all dools.
Oh yea, and I agree with Jesse that fish are tasty.

@alex, by insane i didnt mean like not-cool or anything, lol, she was just awesomely fast.

oh, and did u make top 16 too? :?:

not THAT easy, but a bit easy yes, most of the states cutoff was in mid 30ties but va was like mid 40ties...

lol 2008 is even.

chapter state and nats were easy.

Huh? 2008 was hard!!!, i mean for me(6th grade)

how are odds easy, lol, 2005 is so-far the hardest nat/state year yet for MC...

i've taken all states from 1995-2009, by far 2005 is the highest, not just me, other people too, u can make a poll about this and 2005 will most likely get the most votes.  :oops:

Wait are you confusing with school years, because if you're a seventh grader of 08~09, the tests are 2009 not 2008.

um, no i'm not confusing for school yrs, i've taken the official 2005 state, non-officially, my score was about 5-6 points lower than my average. i got a 37 on it, and i usually get around 43. Although i got 38 this yr.. :(

Wow, did you have a bad day? I thought this years was easy...

[quote="HiDN428"]@gauss, yea, there is this one girl who is insane...and 3 other guys representing, the cutoff for Va was almost as high as CA and TX this year...44 it think.

VA will place top 5 as a team for sure, and atleast 2 of the 4 people will place top 10...my predictions.

BTW, 3 of the 4 people were from the same school..[/quote]

actually i think the cutoff was a 41?

[quote="alex960531"]Not to mention California's cutoff score, as usual in the mid-40s.

Remember: odd years are easy, even years are hard for everything(Chapter, State, National)[/quote]

45 this year, although it's not "usually in mid-40s" - for example, just last year it was 39 (6 way tie for 3rd place :P) and that wasn't a particularly difficult year.

I am Fred Zhang and I go to Longfellow Middle School


I put this in the Classroom forum, but it seems to winding for them until they get to a higher math level. Here it is:

You have the numbers 1,2,3,4,5,6,7,8,9,10.

If 1 and 10 cannot be at the ends, 5 cannot be in the middle, and 8 must be next to 4, how many ways can you arrange these numbers?

There are two hints in the classroom forum, but I don't think you will need them.

ok, so essentially this is a multi-step counting problem.

pick the locations of 2,3,6,7, and 9.

5P5=120

5 can be at either end

2*120=240

the sequence should look something like this:

5,2,3,6,7,9

1 and 10 can be at either end, so you have to "insert" them into the middle.

5_2_3_6_7_9

Since they can go into the same spot:

6C2 =15 240*15 = 3600

The sequence should look like this:

5,1,10,2,3,6,7,9

4 and 8 can't be next to each other, so you have to "insert them into any two different spots

_5_1_10_2_3_6_7_9_

7C2=36

3600*36=129600

The question says 8 and 4 [b]must[/b] be next to each other.


advanced fields
advanced fields unsolved

1=1
2+3+4 = 1 + 8
5 + 6 + 7 + 8 + 9 = 8 + 27
10 + 11 + 12 + 13 + 14 + 15 + 16 = 27 + 64

conjecture the general formula of the 4 equations

not good at this kind of stuff =(

my brain cannot recognize patterns at all


A solid compound $P$ ,when heated with concentrated sulphutic acid , liberates gas which is soluble in water to form a solution $Q$ . When solution $Q$ is added to a solution of silver nitrate,white precipitation formed .The white precipitate dissolves in aqueous ammonia .Compound $P$ is likely to be

A.$MgF_{2}$
B.$NaBr$
C.$Ba(NO_{3})_{2}$
D.$CaI_{2}$
E.$KCl$

I haven't been in touch with chemistry for a long time but I would still like to take a guess. P is KCl and Q is Chlorine. The white precipitate is AgCl and it dissolves in Ammonia.


How many different sums can be thrownwith two dice, the faces of each die being numbered
0, 1, 3, 7, 15, 31?

How many different sums can be thrown with three dice, the faces of each die being numbered
1, 4, 13, 40, 121, 364?

My quick answer would be $ \frac{6^{2}}{2}=18$, however I don't know if there's an efficient way of checking whether you can get the same sum with two different pairs of numbers:
[hide="Combinations"]
$ 0+0 = 0$
$ 0+1 = 1$
$ 0+3 = 3$
$ 0+7 = 7$
$ 0+15 = 15$
$ 0+31 = 31$

$ 1+1 = 2$
$ 1+3 = 4$
$ 1+7 = 8$
$ 1+15 = 16$
$ 1+31 = 32$

$ 3+3 = 6$
$ 3+7 = 10$
$ 3+15 = 18$
$ 3+31 = 34$

$ 7+7 = 14$
$ 7+15 = 22$
$ 7+31 = 38$

$ 15+15 = 30$
$ 15+31 = 46$

$ 31+31 = 62$
[/hide]
They all look different to me... actually it's 21, this is because I forgot about the pairs of the same numbers, which would not be produced twice by $ 6^{2}$, so my answer was 3 short.

$ 21$

I would expect the same is true for 2.3.2, unless there are different pairs which add to the same total.