The first few terms of a sequence are: $3, 12, 4, 16, \frac{16}{3}, \ldots$. Each even-numbered term is obtained by multiplying the preceding term by 4, whereas each odd-numbered term, after the first, is obtained by dividing the preceding term by 3.  If the 101st term is represented as $\frac{4^{a}}{3^{b}}$, determine the ordered pair $(a,b)$.

This is my first time on the forum, so I don't know how to do Latex so I'll have to post this solution in just normal writing.

We want to put the sequence in terms, of 4^x/3^y, which will let us figure out the 101st term.  3=4^0/3^-1, 12= 4^1/3^-1, 4= 4^1/3^0.  This means that the exponents in front of x and y increase alternately.  Every odd numbered term can be expressed as 4^((x-1)/2)/3^((x-1)/2-1)). When you let x=101, you find a=50, b=49 so the a+b is 99


Imi poate spune si mie cineva despre ecuatia claselor de echivalenta a unui grup G?
Acea relatie intre Z(G) si clasele care il dau pe G.

Uite [url=http://www.mateforum.ro/viewtopic.php?t=377]aici[/url] e un post bun; poti gasi si o demonstratie acolo.


Use four distinct digits to form a four-digit number which does not end in the digit “0”.  Reverse those four digits to write a second four-digit number.  What is the maximum possible positive difference between those two numbers?

A.3087	B.8532	C.8622	D.8712	E.8802

[hide="Solution"]Write the number as $ ABCD$. We want to maximize $ ABCD \minus{} DCBA$, which is the equivalent of maximizing $ 999(A\minus{}D) \plus{} 9(B\minus{}C)$, with $ D \neq 0$. Let $ A \equal{} 9$, $ D\equal{}1$, $ B \equal{} 8$, and $ C \equal{} 0$. We therefore get $ 999 \times 8 \plus{} 90 \times 8 \equal{} \boxed{8712}$. The answer is $ \boxed{D}$.

[/hide]


Let a,b,c be real numbers such that $ |a^3|\leq bc$. Prove that $ b^2 \plus{} c^2 \geq {\frac {1}{3}}$ whenever $ a^6 \plus{} b^6 \plus{} c^6 \geq{\frac {1}{27}}$.

Have $ |a^3|\le bc\Rightarrow a^6 \plus{} b^6 \plus{} c^6 \equal{} b^6 \plus{} c^6 \plus{} (a^3)^2\le b^6 \plus{} c^6 \plus{} b^2c^2$
Now, $ a^6 \plus{} b^6 \plus{} c^6 \geq{\frac {1}{27}}\Rightarrow b^6 \plus{} c^6 \plus{} b^2c^2\ge \frac {1}{27}$
Then, $ b^2c^2\ge \frac {1}{27} \minus{} (b^6 \plus{} c^6)$   (*)
On the other hand, $ (b^2 \plus{} c^2)^3 \equal{} b^6 \plus{} c^6 \plus{} 3b^2c^2(b^2 \plus{} c^2)$
Then, if $ b^2 \plus{} c^2 \equal{} x, b^6 \plus{} c^6 \equal{} y$ have $ b^2c^2 \equal{} \frac {x^3 \minus{} y}{3x}$
Substituting $ b^2c^2 \equal{} \frac {x^3 \minus{} y}{3x}$ into (*) and replacing $ b^6 \plus{} c^6$ by $ y$ get:
$ \frac {x^3 \minus{} y}{3x}\ge \frac {1}{27} \minus{} y$.
As $ 3x \equal{} 3(b^2 \plus{} c^2) > 0$ (we can't have $ 3(b^2 \plus{} c^2) \equal{} 0$ as then $ b \equal{} c \equal{} 0$ so we do not get $ a^6 \plus{} b^6 \plus{} c^6\ge \frac {1}{27}$), we can multiply both sides by $ 3x$:
$ x^3 \minus{} y\ge \frac {1}{9}x \minus{} 3xy$
$ \Leftrightarrow (x^3 \minus{} \frac {1}{9}x) \plus{} (3xy \minus{} y)\ge 0$
$ \Leftrightarrow x(x \plus{} \frac {1}{3}(x \minus{} \frac {1}{3})) \plus{} 3y(x \minus{} \frac {1}{3})\ge 0$
$ \Leftrightarrow (x \minus{} \frac {1}{3})(x(x \plus{} \frac {1}{3}) \plus{} 3y)\ge 0$ (**)
Because $ x,y\ge 0$ then $ x(x \plus{} \frac {1}{3}) \plus{} 3y\ge 0$ so in order for (**) to be true, must have $ x \minus{} \frac {1}{3}\ge 0\Rightarrow b^2 \plus{} c^2\ge \frac {1}{3}$ QED.


geometry
3D geometry
sphere
vector
topology
advanced fields
advanced fields unsolved

which is the definicion of spherical tangent bundle????? :blush:

I am not sure. May be it would be easy if you can tell us the context. But, what I am guessing is that it is a tangent bundle of a sphere. I could be wrong!
Peace Up

I think this like Ajoye23 thought.
Hope it is not wrong.

Do you mean tangent sphere bundle?

A sphere bundle is the set of vectors in a bundle of norm 1. You need to use the fact that you have a Riemannian metric on a manifold to get an inner product (or some other inner product here).

So,

ST(M)|U = { v in TM_U | ||v|| = 1 }

This is a useful object in showing bundles are orientable and also shows up in characteristic class theory.


limit integration trigonometry logarithms

$\alpha$を$0\le \alpha < 1$なる実数とする。

$n$個の実数$\sqrt{1},\sqrt{2},\sqrt{3},\cdots,\sqrt{n}$のうち、その小数部分が$\alpha$以下であるものの個数を$N(n)$とおくとき、
　　　　$\displaystyle\lim_{n\to\infty}\frac{N(n)}{n}=\alpha$
となることを示せ。

高１の時にこれを聞き、数学を好きになったきっかけの問題です。その時は結果がとても綺麗だと思いました :P今見ると特に何でもない問題なんですが… :noo:

$n$が無限大の時を考ると
実数$\sqrt k$　(k=1,2...)が実数$\alpha$を小数部分に持つ確率は等しいので
$\alpha$以下の小数部分をもつ確率は$\alpha$になる。

よって$\lim_{n \to \infty} \frac{N(n)}{n}=\alpha$

そんなおおざっぱに解いてしまう解法もあるのですね！その方がラクですね :P
僕は地道にはさみうちの解法しか知りませんでした :lol:

数学的にこれでOKなのかどうかは微妙なのですが。

nが無限大の時
例えば100000000000000000001と100000000000000000002の平方根の小数部分は僅かの違いしかもたないので
nが無限大の時の確率は同等と見なしてもまぁいいかな？という感じで・・

でも理屈としてはイケてそう？ですよね :P

はさみうちを用いた解法もかいておきます :)

まず、$\sqrt{1},\sqrt{2},\sqrt{3},\cdots$のうち、その整数部分が$l$で、小数部分が$\alpha$以下であるものの個数が$[2\alpha l +\alpha^2]+1\cdots (\ast)$で表されることを示します。
$\alpha$を固定して、$\sqrt{1},\sqrt{2},\sqrt{3},\cdots$の中の小数部分が$\alpha$以下であるもののうち、整数部分が$l$であるものの個数を$a_l$とする。また、正整数$k$に対して、
　　　　$l \le \sqrt{k} \le l + \alpha \Longleftrightarrow l^2 \le k \le l^{2}+2\alpha l +\alpha ^2$
　　　　$\Longleftrightarrow l^2 \le k \le l^{2}+[2\alpha l +\alpha ^2]$
　　　　$\Longleftrightarrow k=l^{2},l^{2}+1,l^{2}+2,\cdots,l^{2}+[2\alpha l +\alpha ^2]$
であるので、
　　　　$a_{l}=(l^{2}+[2\alpha l +\alpha ^2])-l^{2}+1=[2\alpha l +\alpha^2]+1$
よって$(\ast)$は示された。

さて、$p=[\sqrt{n}]$ $(p\le \sqrt{n}< p+1)$　で$p$を定めると、$n$個の実数$\sqrt{1},\sqrt{2},\sqrt{3},\cdots,\sqrt{n}$の整数部分は$1,2,\cdots,p$であるから、$N(n)$ は、
　　　　$\displaystyle\sum_{l=1}^{p-1}a_{l} < N(n) \le \displaystyle\sum_{l=1}^{p}a_{l}$
　　　　$\Longleftrightarrow \displaystyle\sum_{l=1}^{p-1}[2\alpha l +\alpha^2]+1 < N(n) \le \displaystyle\sum_{l=1}^{p}[2\alpha l +\alpha^2]+1$   $\cdots (\star)$
また、$2\alpha l +\alpha^2 - 1 <[2\alpha l +\alpha^2] \le 2\alpha l +\alpha^2$　であるから、
　　　　$(\star)\Longleftrightarrow \displaystyle\sum_{l=1}^{p-1}2\alpha l +\alpha^2 < N(n) \le \displaystyle\sum_{l=1}^{p}2\alpha l +\alpha^2+1$
　　　　$\Longleftrightarrow \alpha (p-1)p+\alpha^2 (p-1) < N(n) \le \alpha p(p+1)+(\alpha^2 +1)p$
　　　　$\Longleftrightarrow \alpha\cdot\displaystyle\frac{(p-1)p}{n} + \alpha^2 \cdot\displaystyle\frac{p-1}{n}<\displaystyle\frac{N(n)}{n}\le \alpha\cdot\displaystyle\frac{p(p+1)}{n}+(\alpha^2 +1)\cdot\displaystyle\frac{p}{n}$ $\cdots(\sharp)$
また、
　　　　$p\le \sqrt{n}< p+1\Longleftrightarrow p^2 \le n< p+1\Longleftrightarrow 1\le \displaystyle\frac{n}{p^2}<\displaystyle\left(1+\frac{1}{p}\right)^2$
であり、$n\to\infty$のとき$p\to\infty$であるから、$\displaystyle\lim_{n\to\infty}\displaystyle\frac{n}{p^2}=1$　よって$\displaystyle\lim_{n\to\infty}\displaystyle\frac{p^2}{n}=1$　となる。よって$(\sharp)$において
　　　　$\displaystyle\lim_{n\to\infty}\displaystyle\frac{p}{n}=\displaystyle\lim_{n\to\infty}\displaystyle\frac{p^2}{n}\cdot\displaystyle\frac{1}{p}=0$
となるから、従って、$(\sharp)$で$n\to\infty$とすることで、はさみうちの原理により
　　　　$\displaystyle\lim_{n\to\infty}\displaystyle\frac{N(n)}{n}=\alpha$
を得る。$(Q.E.D.)$

大阪市立大

$ \boxed {3a}$ 1996後期(改)
$ a_{1} \equal{} 2,a_{n} \equal{} 2^{n \minus{} 1}$$ (n\ge2)$に対して、
　　　　$ \displaystyle\lim_{n\to\infty}\displaystyle\sum_{k \equal{} 1}^{n}\displaystyle\int_{ \minus{} \pi}^{\pi}(\cos a_{1}x)(\cos a_{2}x)\cdots(\cos a_{k}x)dx$
を求値せよ。

$ \boxed {3b}$ 2005前期(改)
$ M$を2以上の自然数とする。$ \mathbb{N}$を自然数全体の集合とし、$ n\in\mathbb{N}$について、集合$ A_{n} \equal{} \displaystyle\left\{m|m\in\mathbb{N},m\le\displaystyle\frac {M}{n}\right\}$の要素の個数を$ a_{n}$とする。$ S(M) \equal{} \displaystyle\sum_{i \equal{} 1}^{M}a_{i}$とおくとき、
　　　　$ \displaystyle\lim_{M\to\infty}\frac {S(M)}{M\log M}$
を求値せよ。

答えだけ… :P
[hide="answer 3a"]$ 2\pi$[/hide]　　[hide="answer 3b"]1[/hide]

$ \boxed {4a}$
$ \displaystyle\lim_{n\to\infty}\displaystyle\sum_{k \equal{} 1}^{n}\log\displaystyle \left(1 \plus{} \displaystyle\frac {k}{n^2}\right)$

[hide="ans"]$ \frac {1}{2}$[/hide]

$ \boxed {4b}$
$ \displaystyle\lim_{n\to\infty}\displaystyle\sum_{k \equal{} 1}^{n}\displaystyle\frac {\sqrt {k}}{n}\sin\displaystyle \left(\displaystyle\frac {\sqrt {k}}{n}\right)$

[hide="ans"]$ \frac {1}{2}$[/hide]

$ \boxed {4c}$
$ \displaystyle\lim_{n\to\infty}\displaystyle\sum_{k \equal{} 1}^{n}\displaystyle\sin\displaystyle \left(\displaystyle\frac {1}{n \plus{} k}\right)$

[hide=ans]$ \log{2}$


The function f(x,y) satisfies the conditions:
(1) f(0,y)= y+1
(2) f(x+1,0)= f(x,1)
(3) f(x+1,y+1)= f(x, f(x+1,y))
for all non-negative integers x,y.
Find f(4,1981)

This is IMO 1981[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=60818[/url]


limit
inequalities
calculus
ratio
real analysis
real analysis solved

if ${\lim }\limits_{n \to \infty } \frac{a_n}{a_{n+1}}=l$,prove  ${\lim }\limits_{n \to \infty } a_n^\frac {1}{n}=l$.
and if ${\lim }\limits_{n \to \infty } a_n^\frac {1}{n}=l$.can we also prove  ${\lim }\limits_{n \to \infty } \frac{a_n}{a_{n+1}}=l$
$(a_i>0)$

Your first excercise is very standart.

Second one is false. We can easily construct example satisfying  $\lim_{n \to \infty } a_n^{1/n}=l$ and $\lim_{n\to\infty}a_{n+1}/a_n$ doesn't exist.

Is nt the 1st one called the Cauchy's 2nd limit theorem.
I might be wrong..

As [b]Myth[/b] told it can be easily constructed a series s.t. exists $\lim_{n{\rightarrow}\infty}{\sqrt[n]{a_{n}}}$,but it does not exists $\lim_{n{\rightarrow}\infty}{\frac{a_{n+1}}{a_n}}$(I'm sure that [b]Myth[/b] had that example).Here is one:$1,\frac{1}{2},1,\frac{1}{3},1,\frac{1}{4},...,...$.
Now more generalised inequality
$\liminf_{n{\rightarrow}\infty}{\frac{a_{n+1}}{a_n}}{\leq}\liminf_{n{\rightarrow}\infty}{\sqrt[n]{a_n}}{\leq}\limsup_{n{\rightarrow}\infty}{\sqrt[n]{a_{n}}}{\leq}\limsup_{n{\rightarrow}\infty}{\frac{a_{n+1}}{a_n}}$.
It is not hard to show.Indeed this inequality proves that the Dalamber's test of converging of positive series is weaker than Causchy's one.

[quote="LevonNurbekian"]Indeed this inequality proves that the Dalamber's test of converging of positive series is weaker than Causchy's one.[/quote]
To translate this into the terminology usually used in American calculus books, "The ratio test for the convergence of series is weaker than the root test."


Is it possible to tile an equilateral triangle with several equilateral triangles who have all different sizes ?  :)

I couldn't figure out how to approach the problem myself, but the answer appears to be

[hide]no[/hide]

as can be seen at

[hide]http://www.math.niu.edu/~rusin/known-math/99/equilateral[/hide]

Thank you Kevinatcausa  :D

I've find another approch more geometric .

Would you please post the geometric approach you found?

Require that we can cut the equilateral triangle in equilateral triangles which are all in  different sizes . We call « piece » a set of two of these equilateral triangles which share a vertex and an edge . The common vextex is call the origin of the piece and the vextex in the  common edge is call the end of the piece . Two pieces $ P$ and $ Q$ are independant  if  the interior of $ P\cap Q$ is empty . Let $ P$ a piece of end $ E$ , the piece $ D$ is a daughter of $ P$ if $ D$ and $ P$ are independant and $ E$ is the origin of $ D$ . $ M$ is a mother of $ P$ if $ P$ is a daughter of $ M$ . The initial triangle is convex so every piece have exactly one daughter and every piece have at most one mother . Now , let a piece build with the smallest triangle and its daughter , the daughter of its daughter … The number of pieces is finite so in the list there are two identical pieces . Every peace have only one mother so the last daugther is the initial piece . It’s not possible because the mother of the first piece will be build with a smaller peace of the smallest : contradiction .

Sorry for my bad english  :(

[img]http://img115.imageshack.us/img115/6320/pieceau8.jpg[/img]

You say

It’s not possible because the mother of the first piece will be build with a smaller peace of the smallest : contradiction .

but why?  there's no guarantee that the pieces will be getting smaller.

Okay , it's not correct ( sorry ) . What I have wanted to say : if the first piece have a mother , we can find a smaller piece of the smallest  :lol:

[img]http://img409.imageshack.us/img409/3892/piecesmotherny5.jpg[/img]


solve equation $ 3^x\cos(x\arccos\frac {1}{3}) \equal{} 3^x \minus{} 16x$    [color=green]its new ))[/color]   :D

Here my solution  ;$ 3\cos(x\arccos\frac{1}{3})\equal{}3^2\minus{}18x\plus{}2x$
$ \cos(x\arccos\frac{1}{3})\equal{}1\minus{}2x\frac{9}{3^x}\plus{}2x\frac{1}{3^x}$
$ \cos(x\arccos\frac{1}{3})\equal{}1\minus{}2x\frac{1}{3^x\minus{}2}\plus{}2x\frac{1}{3^x}$
$ \cos\alpha\equal{}\frac{1}{3}\implies\alpha\equal{}3\arccos\frac{1}{3}$
$ \cos(x\arccos\frac{1}{3})\equal{}\cos2\alpha$
$ \cos2\alpha\equal{}1\minus{}2x\cos^x\minus{}2 \alpha\plus{}2x\cos^x \alpha$
It are known  $ \cos4\alpha\equal{}8\cos^4\alpha\minus{}8\cos^2\alpha\plus{}1$
$ \Longrightarrow  x\equal{}4$

[quote="Pirkuliyev Rovsen"]Here my solution  ;
[...]
$ \Longrightarrow x \equal{} 4$[/quote]

You just forgot infinitely many other solutions. You should better check your solution. Something is surely missing ...  :wink:


Let a and b be integers for which:

1 + 1/2 + 1/3 + ... + 1/2004 = a/b

Suppose that exactly one of a and b is even. Determine which of a and b is even, and find the largest integer r such that 2r divides ab.

(As always, put your solutions in spoiler mode)

well so far, i found out that....

[hide]if a/b is in lowest terms then b = [1,2,3,...,2004] = 2^10*3^6*5^4....

so b has 10 factors of 2 and must be even[/hide]

ehhh but im stuck on the rest

[hide]Sum(i=1,2004)(1/i) = Sum((2004!/i)/2004!).

Say 2004! = 2^k * m, where a is odd.

we see that Sum(2004!/i) | 2^(k-10), since there isn't any i in the denomenator that factors to a power of 2 greater than 10.

However, we see that if we take Sum(2004!/i) mod 2^(k-9), all terms in the series drop out (since they lack a power of 2 greater than 9) EXCPECT when i=1024.  Thus, we have Sum(2004!/i)  ! :equiv: 0 mod 2^(k-9).

Thus, sum(2004!/i) = 2^(k-10) * n, where n is odd.

So, we have Sum((2004!/i)/2004!)= 2^(k-10) * n / (2^k * m) =      n/(2^10 * m).  Where n is odd.  Thus, we have a=n, b=2^10m.

ab=nm*2^10.  Since nm is odd, we have that 2^10 is the largest power of two that divides the product.  r=10.[/hide]

i just thought of an easy way to do this....

[hide]

since a/b is in lowest terms, b is the least common multiple of 1,2,3,...2004 so the largest factor of 2 in b that is less than or equal to 2004 is 2^10=1024

this means b is even and a is odd (or else a/b would reduce even more)

so a has no factors of 2, b has 10, thus ab has 10 factors of 2 and r=10[/hide]

Interesting. [hide]b=the least common multiples of set(1,2,3,4,5....2004)=2^10*3^6...... and the top is an odd integer, therefore, r=10. [/hide]

ThAzN1 wrote:i just thought of an easy way to do this....
[hide]
since a/b is in lowest terms, b is the least common multiple of 1,2,3,...2004[/hide]

How do you know this is true?

errr i didn't mean lowest terms... [hide]i meant that only one of them has any factor of 2

well... if they weren't in lowest terms with one even and one odd, that wouldn't affect the amount of factors of 2 in ab anyway[/hide]

whoops...

ThAzN1 wrote:errr i didn't mean lowest terms... [hide]i meant that only one of them has any factor of 2[/hide]

But how do you know that is true? I think in essense this is what the problem is asking you to prove.

gauss202 wrote:ThAzN1 wrote:errr i didn't mean lowest terms... [hide]i meant that only one of them has any factor of 2

well... if they weren't in lowest terms with one even and one odd, that wouldn't affect the amount of factors of 2 in ab anyway[/hide]

Sure it would. It would depend on how many factors of 2 you had in a. The more there were, the less you would have in a/b when it was reduced.

??? but since... one of a or b is odd and the other is even, it would be impossible to reduce a/b by a factor of 2?

I know. I reread what you wrote and edited my post concerning your assumption that only one of a and b is even for that choice of b.

I agree with ThAzN1 if you read my solution. b=2^10*3^6....... so each term equals b/n/b, where n=1,2,3,...2004. when n=1024, we will have an odd number for b/n, for other ns, we have all even for b/n, so a=b/1+b/2+b/3....b/2004 which is an odd number.

I agree with ThAzN1 if you read my solution. [hide]b=2^10*3^6....... so each term equals b/n/b, where n=1,2,3,...2004. when n=1024, we will have an odd number for b/n, for other ns, we have all even for b/n, so a=b/1+b/2+b/3....b/2004 which is an odd number.[/hide]

[hide]for my choice of b = [1,2,3,...2004]... a = b + b/2 + b/3... + b/2004, thus all of the b/something will have some factor of 2 with the exception of 1024, which is odd since b has 10 factors of 2 and so does 1024... that means b/1024 has no factor of two...

so 2003 even integers + 1 odd means that a is odd when b = [1,2,3,...2004][/hide]

wow... i never thought about all this before...

Nowww you've got it. ;-) It wasn't that what you were saying was untrue - it's just that you hadn't justified it. Your last post does that though. Good job! :)

Now what about in general:  If you write

1/1 + 1/2 + 1/3 + ... + 1/n in lowest terms as a/b,

Is a or b a multiple of 2.  What is the largest integer value of r such that 2^r divides ab?

[hide]The same thing will always happen as when n=2004.

Say k is the largest positive integer such that 2^k <= n.  Then we can write the fraction with a common denominator or 1*3*5*...([(n+1)/2]*2-1) * 2^k.

[(n+1)/2]*2-1 is just a way of saying the greatest odd integer no greater than n.

Then all the terms in the sum will have even numerators except for 1/(2^k).

So again the numerator of the fraction is odd, and the denominator has a factor of 2^k.

[/hide]

What if instead we ask about whether a or b is a multiple of 3, and what is the largest value of r such that 3^r divides ab?

--Dan

well i think that in general...

[hide]if 1+1/2+1/3...+1/n = a/b and n :ge: 2

uhh then a = b+b/2+b/3...+b/n

and b = [1,2,3,...n]

so for every prime p :le: n... the terms b, b/2, b/3..., b/n will all be divisible by p^something except for the fraction b/p^x where x is the power of p in [1,2,3,...n] because all of the p's cancel out

this means that a is not divisible by p so a/b is in lowest terms if the process above is repeated for all primes :le: n

....sooo b would be a multiple of 3 and the largest r that divides ab would be the power of 3 in [1,2,3,...n]

which is 6 when n=2004[/hide]

yea... im not really sure if all this is right tho


[quote="fallinlovewithmaths"]For a polynom p, p(x)^5 - x is divided by (x-1)(x-2)(x-3). Find p(x)[/quote]
Let $ Q(x) \equal{} (p(x))^5 \minus{} x$, then $ Q(1)\equal{}Q(2)\equal{}Q(3)\equal{}0$ or
$ p(1)^5 \equal{} 1$, $ p(2)^5 \equal{} 2$, $ p(3)^5 \equal{} 3$. Or equivalently:
$ p(1) \equal{} 1$, $ p(2) \equal{} \sqrt[5]{2}$ $ p(3) \equal{} \sqrt[5]{3}$. And there are infinitly many polynomials which satisfies that. Should there be another restricition?


calculus
integration
real analysis
real analysis unsolved

I've been figuring this out for the past two days. No luck. :(

Integral of: arctan[1/x] dx

I must use integration by parts.

Any ideas? Thank you very much for your time.

$\arctan(x^{-1})=\text{arccot}(x)$.  Put $du=dx$, $v=\text{arccot}(x)$ and see what you get.

Question, does that mean:

arcsin(1/x) = arccsc(x)
arccos(1/x) = arcsec(x) ? Thanks!

Yes, up to periodicity/uniqueness issues.


The positive integer 182634795 has the property that the digits 1 to 5 are in natural ascending order, but 1 to 6 are not, and each digit from 1 to 9 is used precisely once.

How many such positive integers exist?

Put 12345, then you have 6 spaces to put the remaining four numbers in, but you have to subtract the case where 6 is in the last space.

I think the answer is

[hide]

[/hide]

anyone can post a solution and explain please

First place the digits 1 to 5 in their natural order, as requested by the problem. In the obtained number, their are six "slots" where you may place other digits: $\overline{a1b2c3d4e5f}$ We can see immediately that their are 5 slots, namely $a$ to $e$ that may contain the digit $6$, hence 5 possibilities to place the digit $6$. We then obtain a six-digit number with 7 slots. Now consider three cases:
[list]
[*] The remaining three digits occupy one single slots. As they may be permutated, there are $7\cdot3!$ possibilities.
[*] The remaining three digits occupy different slots. There are $3!\cdot\left(\begin{array}{c}7\\3\end{array}\right)$, as, for each permutation of $\{7;8;9\}$, three slots among seven have to be selected.
[*] Finally, suppose the three digits occupy two slots. There are $\left(\begin{array}{c}7\\2\end{array}\right)$ choices for the slots. Furthermore, the three digits may be permutated and partitionned in $2\cdot 3!$ ways.[/list]
Finally, the total number of such integers is $5\cdot\left[7\cdot3!+3!\cdot\left(\begin{array}{c}7\\3\end{array}\right)+2\cdot 3!\cdot\left(\begin{array}{c}7\\2\end{array}\right)\right]$ ways.

The casework may seem exaggerated, by as I am awfully bad at combinatorics, I prefer to be prudent.

After you find the 5 possibilities to place the digit 6, you can do this.

Let a,b,..,g be the remaining spots to place the other digits.
a+b+c+d+e+f+g = 3
Which has $\binom{9}{3}$ solutions. Finally you can permute those 3 last digits.

$\boxed{5\cdot 3! \binom{9}{3}}$

[quote="Lepuslapis"]yes, that seems to be easier.  :D

By the way, a slightly more interesting problem is the following one:

[i]The positive integer 5016283497 has the property that the digits 0 to 4 are in natural ascending order, but 0 to 5 are not, and each digit from 0 to 9 is used precisely once. How many such positive integers exist? (We suppose that a number does not begin with 0.)[/i][/quote]
The integer looks like this:

[] 0 [] 1 [] 2 [] 3 [] 4 []

The "[]" represents an optional place for any number of the remaining digits, or for none at all.

There are 5 places to assign the digit 5 to (Anywhere except the end of the number, because 1 to 5 don't occur in their natural order).

Then there are 7 places to assign the digit 6 to, because it can go in any of the places, and at either side of the 5.  Afterwards, there are 8 places to place the 7 into, 9 places for the 8 and 10 places for the 9.  This means that there are 5 x 7 x 8 x 9 x 10 = 25 200 integers of this kind, including those which begin with 0.

To exclude those which begin with zero, do the same process again, but now, nothing can go in the first empty position.

So there are 4 x 6 x 7 x 8 x 9 = 12 096 of them which begin with zero.

This leaves 25 200 - 12 096 = 13 104 valid integers.

Is this correct?

Yes, I think so. I had thought of a slightly different approach: (1) The digit 5 is placed in the first place. Then, continue normally. (2) The digit 5 is not placed in the first slot. So we have to place another digit in that slot (there are 4 possibilities remaining). Then, continue normally.

[hide]_1_2_3_4_5
There are 5 places to put 6.
Then there are 7 places, at the front, back, or in between two numbers, to place 7.
Then there are 8 places to place 8.
Then there are 9 places to place 9.

$5\cdot7\cdot8\cdot9=2520$.[/hide]


calculus
integration
logarithms
real analysis
real analysis unsolved

I'm very sorry if it had been posted .
Calculus the integral :
$ \int {x^x}dx$

There is no simple solution to the integral. Still, you could be interested in this:

[url]http://mathworld.wolfram.com/SophomoresDream.html[/url]

I cannot open the link to Mathworld...but ususally they dont have very in-depth reasoning.

But
 $ \int_0^1\frac{dx}{x^x}$

$ =\int_0^1{e^{x\ln\left(\frac{1}{x}\right)}dx}$
$ =\int_0^1\sum_{n=0}^{\infty}\frac{x^n\ln^n\left(\frac{1}{x}\right)}{n!}$

Now if we consider

$ \int_0^1x^n\ln^n\left(\frac{1}{x}\right)dx$

The change of variable $ \ln\left(\frac{1}{x}\right)=z$

Gives

$ \int_0^{\infty}e^{-(n+1)z}z^ndz$

From there if we let $ (n+1)z=m$ we get

$ \frac{1}{(n+1)^{n+1}}\int_0^{\infty}e^{-m}m^ndm$

Which we should realize is

$ \frac{\Gamma(n+1)}{(n+1)^{n+1}}=\frac{n!}{(n+1)^{n+1}}\quad\forall{n}\in\mathbb{N}$

$ \therefore\sum_{n=0}^{\infty}\frac{1}{n!}\int_0^1(-x\ln(x))^ndx$

$ =\sum_{n=0}^{\infty}\frac{n!}{n!(n+1)^{n+1}}$

$ =\sum_{n=1}^{\infty}\frac{1}{n^n}$


I am completely baffled on what to do with fractional exponents like in the following problem.

factor and simplify

(x^2 + 5)^ -1/4 - (x^2 + 5)^ -5/4

Can anyone point me in the right direction?

thanks

[quote="t0rajir0u"]It might help to let $u = (x^{2}+5)^{-\frac{1}{4}}= \frac{1}{ \sqrt[4]{ x^{2}+5}}$.  Then the original expression is

$u-u^{5}= u(1-u^{4})$

Which is

$\frac{1-\frac{1}{x^{2}+5}}{ \sqrt[4]{x^{2}+5}}= \frac{x^{2}+4}{ (x^{2}+5) \sqrt[4]{x^{2}+5}}$

There's... really not much you can do past that point, as far as I can see.[/quote]

thanks... also did you use special board code to make it pretty print?

[quote="t0rajir0u"]It's called $\LaTeX$.  Most of the time, you just need to put dollar signs around your math.  There's a whole section of the site devoted to learning how to use it.[/quote]

ah interesting... I am a going back to school for a Computer Science degree, which requires a lot of math, so I might be posting/reading this forum a lot in the future.  Nice site.  I have not had any math for over a decade and I am jumping into College Algebra.  A lot of it is coming back, but I still have some catching up to do.  I guess it's a matter of re-learning.  I made it through Calculus before, but it's been awhile...

Now I don't fully understand what your are telling me.  This is what I did.

$(x^{2}+5)^{-}1/4$ - $(x^{2}+5)^{-}5/4$

      $1$
_____________
$(x^{2}+5)^{1}/4$ - $(x^{2}+5)^{5}/4$

   $1$
_____________
$(x^{2}+5)^{-}4/4$ =

    $1$
-  _____________
 $(x^{2}+5)$

this probably isn't right, but that was my logic.  What did I do wrong if it isn't right?

thanks

It's that very first step that's problematic.  $a^{-1}+b^{-1}\neq \frac{1}{a+b}$.

Rather, $a^{-1}+b^{-1}= \frac{1}{a}+\frac{1}{b}$, a very different expression, so that first step should result instead in

$\frac{1}{ (x^{2}+5)^{ \frac{1}{4}}}-\frac{1}{(x^{2}+5)^{ \frac{5}{4}}}$


In an triangle $ ABC$, denote by $ l_a,l_b,l_c$ the internal angle bisectors, $ m_a,m_b,m_c$ the medians, and $ h_a, h_b, h_c$ the altitudes to the sides $ a, b, c$ of the triangle. Prove that

$ \frac{m_a}{h_b\plus{}l_b}\plus{}\frac{m_b}{h_c\plus{}l_c}\plus{}\frac{m_c}{h_a\plus{}l_a}\geq\frac{3}{2}$

[quote="arqady"][quote="filomen"]In an triangle $ ABC$, denote by $ l_a,l_b,l_c$ the internal angle bisectors, $ m_a,m_b,m_c$ the medians, and $ h_a, h_b, h_c$ the altitudes to the sides $ a, b, c$ of the triangle. Prove that

$ \frac {m_a}{h_b \plus{} l_b} \plus{} \frac {m_b}{h_c \plus{} l_c} \plus{} \frac {m_c}{h_a \plus{} l_a}\geq\frac {3}{2}$[/quote]
It's obviously true.[/quote]

Arqady, can you detail , please?  :(

filomen, because, $ h_b\leq m_b,$ $ l_b\leq m_b$ and AM-GM. :wink:


This looks like a chinesse hypothesis that I heard about.
Anyway, the proof that you are searching for is widely known, it can be found in Arthur Engel's book, and as I am sure in many many other books.
Here is a link http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_little_theorem

[quote="drunner2007"]Give a combinatorial proof of the assertion that $p|(2^{p}-2)$ for all positive primes $p$.[/quote]
[hide]Just let 2=1+1 and plug into the binomial theorem. The two 1s cancel out since your subtracting 2, and everything else has a factor of p.[/hide]
[hide="or you could"]try induction[/hide]


What to you think the AMC 10A 2008 passing score will be?

The only realistic options are 114, 115.5, 117, and 118.5. Also, why is there no option for 120?

Also, many of the options overlap, if you think it will be 99 do you vote less than 105 or 90-100?

sorry, i screwed up the poll, please ignore!

I don't know... that AMC 10A seemed harder than last year's.  I'm going with 115.5

[quote="cognos599"][quote="the future"]Hmm, im just going to guess:

96 for AMC 12, 118 for AMC 10

it seemed like the amc 12 might have been harder than previous years amc 12s, and amc 10 about even, but ill just take down 2 points because there was no calcs.

but its most likely not right[/quote]

won't the cutoff be a possible score. for example, it is impossible to get a score of 118 as the score 118.5 is possible. but i don't know.[/quote]

Yep, you're right  :lol:


Here is my rather holey proof of 1+1=3

We know that 2+2=4,
but 2=II in roman numerals,
and, since i=sqr(-1),
2+2=-1+-1=-2=-(II)=-(-1)=1
So, 2+2=1.

Subtract 1 from both sides
3=0

Divide by 3
1=0

Add two to both sides
3=2

Since 2=1+1,
3=1+1

Anyone want to improve?

:shock:

Relevant for everyday life today in the modern world.

Duh! <jk>

Isn't math [i]supposed[/i] to be useless to ``everyday life'' and stuff like that?


function
limit
real analysis
real analysis theorems

In an analysis book i found this proof for the uniqness of a limit of a function.

"It is easy to show that when a Limit of a function f(z) exists at a point a,it is unique.To do this ,we suppose that

lim f(z) =l and lim f(z)=m as z----> a (z goes to a).

Then,for any positive number ε,there are positive numbers r,δ such that

           |f(z)-l |< ε   whenever 0< |z-a |<r

                       and

             |f(z)-m |<ε whenever 0< |z-a |<δ

     So if   0< |z-a |<θ ,where θ denotes the smaller of the two Nos r and δ,we find that

 |l-m |= |(f(z)-m)-(f(z)-l) |=< |f(z)-l | + |f(z)-m | < ε +ε=2ε

          But   |l-m | is a nonnegative constant, and ε can be chosen arbitrarily small.
Hence

l-m =0 , or l=m."

is that proof correct??

[quote="dallasboy"]What doubts do you have about it?  Does every step make sense to you?[/quote]

What the Author of that proof really proves here is:

If limf(z) = l and limf(z) = m as z goes to a ,then given ε>0 and also GIVEN

0<lz-al < θ ,where θ denotes the smaller of the two Nos r and δ,

THEN l=m.

...............and not.

If limf(z) = l and limf(z) = m as z goes to a, THEN l=m

Note the Author in the last step of his proof ,he assumes the theorem:

IF, for all ε>o <ε ,then l=m

Many of analysis books give a wrong proof following ,more or less the lines of the above proof.

Do you agree??

You can prove by contradiction that if $ |x| < \epsilon$ for all $ \epsilon > 0$ then $ x \equal{} 0$. Since $ \epsilon$ was arbitrary in the proof this property will hold.

I know the classical proofs of the uniqueness of limit values, but I was wondering if this would hypothetically "cut it". Let f be a real valued function. Let $ \lim_{x\to{c}}f(x)$ exist and equal $ p$. Also let $ \lim_{x\to{c}}f(x)$ equal $ p'$. So we know that $ \lim_{x\to{c}}f(x)\minus{}\lim_{x\to{c}}f(x)\equal{}p\minus{}p'$, but because the existence of both limits we would have that $ \lim_{x\to{c}}f(x)\minus{}\lim_{x\to{c}}f(x)\equal{}\lim_{x\to{c}}\left\{f(x)\minus{}f(x)\right\}\equal{}0$ which implies that $ p\minus{}p'\equal{}0\implies p\equal{}p'$. I know that is not as rigorous as usual, but is that not technically correct?

You're implicitly using the uniqueness of limits to prove the uniqueness of limits.  There's nothing stopping $ \lim_{x \to c} f(x) \minus{} \lim_{x \to c} f(x) \neq 0$ because you can't show additivity of limits until you show limits are unique!

[quote="t0rajir0u"] There's nothing stopping $ \lim_{x \to c} f(x) \minus{} \lim_{x \to c} f(x) \neq 0$ because you can't show additivity of limits until you show limits are unique![/quote]
Dang additivity! Here is another one. This one is shady at best and very specific, proceed to shoot down if need be. Let $ f: \mathbb{R}\mapsto\mathbb{R}$ and $ f\in\mathcal{C\left(\mathbb{R}\right)}$. Since $ f$ is a continuous function we may redefine $ f(x) \equal{} \lim_{t\to{x}}f(t)$, now suppose that $ \lim_{t\to{x}}f(t)$ was multi-valued, this would contradict our definition of $ f$ as a function. I know, that was pretty bad, but you cannot learn without getting your hands dirty.

EDIT: Wait, same thing. We cannot establish that $ f\in\mathcal{C}\implies\lim_{x\to{c}}f(x)\equal{}f(c)$ before we establish uniqueness. Sorry to waste your time...long night  :roll:


function
combinatorics unsolved
combinatorics

There is a connected network with $ 2008$ computers, in which any of the two cycles don't have any common vertex. A hacker and a administrator are playing a game in this network. On the $ 1st$ move hacker selects one computer and hacks it, on the $ 2nd$ move administrator selects another computer and protects it. Then on every $ 2k\plus{}1th$ move hacker hacks one more computer(if he can) which wasn't protected by the administrator and is directly connected (with an edge) to a computer which was hacked by the hacker before and on every $ 2k\plus{}2th$ move administrator protects  one more computer(if he can) which wasn't hacked by the hacker and is directly connected (with an edge) to a computer which was protected by the administrator before for every $ k>0$. If both of them can't make move, the game ends. Determine the maximum number of computers which the hacker can guarantee to hack at the end of the game.

I have a qustion: Can a the hacker hack a computer twice?
Another quesiton: Is the network a undirected graph or directed graph?

A (maybe implicit) hint:[hide]If the network is undirected connnected and both hacker and administrator can hack and protect a page twice(which is a reasonable guess), then the answer of the generalizede question about $ N$ pages , if I am not wrong, is around $ \frac {N}{3}$, though I haven't calculated the detail.The extreme case is that there are two cycles with $ 2k$ and $ k$ vertice and have excalty a common vertex.
And the key method to solve this problem I used is construct a graph $ T(V,E)$, each vertex $ v$ of whic represents a cycle in the network and each edge $ (v_i,v_j)$ of which represent the cycles represented by $ v_i$ and $ v_j$ have a common computer.It follows that $ T$ should be a tree. And we can estabish a function $ f$ on $ V$ such that $ f(v_i)$ is the number of computers of the cycle represented by $ v_i$. Then we can find an edge of $ T$ such that the two sub-tree created by eliminating it has the least difference of the sum of $ f(v)$, and then give some algebric analysis.  [/hide]

[quote="Umut Varolgunes"]It is undirected.
Actually, there was nothing about hacking or protecting a computer twice in the exam paper. Yet, I think it doesn't make sense. Because hacking a computer twice brings you nothing.

[hide]@Hawk Tiger: Cycles have no common vertex as it is given in the question. So your extreme case is wrong, though your answer prediction is correct. (In fact, even if this graph is accepable, it is still wrong.) The extreme case is interesting, one of the main parts of the solution.
Besides, it is true that the solution requires some algebraic work, but the solution I know isn't using the turning the cycles into vertices idea, in spite of the fact that it seems like a good idea to start. [/hide]
I'm going out of the town for more or less a week. When I come back I'll write the solution if nobody writes it.[/quote]
I think I have misread the problem. I thought it is "Cycles have no common edges". And I don't know why if my graph is accepable, it is still wrong(I haven't done problems for long time, and may be stupier now...). And I think my solution is still useful, for my premary condition is weaker.

sorry, it must be "acceptable". (Does accepable exist?) In your solution you are finding an edge instead of a vertice. I'm giving the extreme case and you can see that your idea doesn't work, because you must start with a specific vertice of a cycle, instead of considering a cycle only with its number of vertices.
in this graph, hacker should hack one of the six vertices of hexagon and after, admin. protects the opposite vertice. The admin can guarantee to protect two of the three main vertices.

We claim that the answer is $\boxed{671}$. First, let's show that the hacker cannot do any better. Take a cycle of six vertices, say $v_1 v_2 v_3 v_4 v_5 v_6.$ Then, draw a path of $669$ vertices which starts at $v_1$, a path of $668$ vertices which starts at $v_3$, and a path of $668$ vertices which starts at $v_5.$ Draw these paths in a way so that they share no vertices amongst each other, and each share only one vertex with the cycle. In this case, it can be seen that the hacker's optimal result is getting $671$ computers. Notice that every computer must eventually be hacked or protected because of connectivity, and so a winning strategy for the hacker is equivalent to a strategy which restricts the administrator to at most $1337$ computers.

Let's now show that the hacker can always get $671$ computers. Let's consider the obvious graph $G$ which corresponds to the network of computers. Consider now the graph $G'$ which results when we condense all cycles of $G$ into vertices. This is doable by the condition of the problem. For example, if $G$ is the example exhibited above, $G'$ becomes three paths of $668, 667, 667$ edges glued together at one vertex. Observe that $G'$ is acyclic and connected, and so therefore is a tree.

For a subgraph of $G'$, define its [b] size [/b] to be the number of vertices it has, after "un-condensing" all vertices of it which correspond to cycles of $G'$. For instance, in our example, $G'$ has size $2008$ and not $1 + 668 + 667 + 667 = 2003$.

Call a vertex $v$ of $G'$ [b] tasty [/b] if its removal leaves only connected components of $G'$ which have sizes $< 1338.$ We now consider two cases.

[b] Case 1. [/b] There does not exist a tasty vertex of $G'$.

We will apply the following operation on every vertex $v$ of $G'.$ Since $v$ is not tasty, there exists a connected component of size at least $1338$ resulting from the removal of $v$. Let $w$ be the unique vertex of this connected component which is adjacent to $v.$ Then, draw an arrow from $v$ to $w.$ Because there are only $|G'| - 1$ edges of $G$ and $|G'|$ arrows drawn, by Pigeonhole there must be an edge $ab$ so that there is an edge from $a$ to $b$ and from $b$ to $a.$

Now, consider the effect of removing the edge $ab.$ From the definitions of these arrows, we know that the two resulting connected components must both have size at least $1338.$ However, it's clear that the sum of their sizes must be at most $2008$, and so this case is actually an absurdity.

[b] Case 2. [/b] There exists a tasty vertex of $G'$.

Call this vertex $v.$ If this vertex corresponds to a vertex of $G$ (and not a cycle), then there is literally no way that the hacker can lose. Indeed, the administrator can only protect the computers in one connected component of $G$ resulting from the removal of $v$, and so this is at most $1337$ computers. Hence, the hacker can ensure at least $2008 - 1337 = 671$ hacked computers because the remainder of the graph is connected.

Otherwise, suppose that $v$ corresponds to a cycle $v_1 v_2 \cdots v_k$ of $G.$ In the cases when $k = 2, 3, 4, 5,$ or $6$, it is easy to find a winning strategy for the hacker and so these cases are left as exercises to the reader. Henceforth, assume that $k > 6.$

We'll first exhibit a strategy for the hacker to hack at least half of the vertices of the cycle. Indeed, this is trivial, as the hacker simply keeps on hacking computers in the cycle arbitrarily as long as he/she can.

Consider a set $S$ of contiguous vertices in the cycle. We will define the [b] quality [/b] of $S$ as follows. Consider the subgraph of $G$ which results when we remove the vertices of the cycle and all edges involving any of these vertices. It is a collection of connected components, some of which were attached to vertices of $S$ in the original graph $G.$ Call these connected components [b] branches [/b] of $S.$ The quality of $S$ is then the sum of the numbers of vertices of all its branches plus the number of vertices in $S.$ Notice that, in a sense, quality is additive.

Let's now consider two subcases.

[b] Subcase 1. [/b] There exists a contiguous set of at least half of the vertices of the cycle with quality at most $670.$

WLOG let this contiguous set of vertices be $\{v_1, v_2, \cdots, v_a\}$ for some $a \ge \frac{k}{2}$. Then, it's clear that the quality of $\{v_{a+1}, v_{a+2}, \cdots, v_k\}$ is at least $1338.$ By "discrete continuity," there must exist some $a+1 \le b \le k$ so that the quality of $\{v_{a+1}, \cdots, v_b\}$ and the quality of $\{v_b, v_{b+1}, \cdots, v_{k}\}$ are each at least $669.$ The hackers strategy is now simple. Start by hacking $v_b.$ Now, it's clear that the hacker can ensure that either the entirety of the set $\{v_{a+1}, \cdots, v_b\}$ is hacked or the entirety of the set $\{v_b, v_{b+1}, \cdots, v_{k}\}$ is hacked. Indeed, simply start hacking clockwise around the cycle or start hacking counterclockwise around the cycle, depending on which computer the administrator first protectts. In either case, the hacker will have hacked a contiguous set of vertices of quality at least $669$ eventually. After the hacker ensures that he/she has hacked a contiguous set of at least $669$ vertices as described above, then the hacker will continue to hack vertices of the cycle arbitrarily. Afterwards, we consider two cases. If the administrator has protected a computer in the branches of the set of computers that the hacker has hacked, then the hacker wins because the administrator can protect at most one branch with at most $1337$ vertices (by the tastiness of $v$). Otherwise, the administrator proceeds to infect all of the branches of the set of computers he has hacked so far. This ensures that the hacker has hacked at least $669$ computers.

We will now work on improving this bound. If one of the sets consists of at least half of the vertices of the cycle, then it's clear that the other set consists of only one vertex. The hacker alters his strategy as follows. Start by infecting this vertex, say $\alpha$. From here, the hacker will set out to infect at least half of the vertices of the cycle. The hacker will hack at least two more computers in the cycle, because $k \ge 7.$ Again, if the administrator protects any of the branches of $\{\alpha \}$, then he/she can again protect at most $1337$ computers, and the hacker wins. Otherwise, the hacker hacks all of the branches of $\{\alpha\}$. After this, the hacker will have hacked at least $669$ computers (quality of $\{\alpha\}$) plus the additional two computers, for at least $671$ computers.

Otherwise, both sets consist of less than half of the vertices of the cycle. If both sets have at most $\left \lceil \frac{k}{2} \right \rceil - 2$ vertices, then the hacker can win with a similar strategy as in the above paragraph (infect an entire set and then get some more computers in the cycle). Otherwise, suppose that one of the sets has at least $\left \lceil \frac{k}{2} \right \rceil - 1$ vertices. This means that the other set has at most $2$ vertices. If it has one vertex, then the above argument works. Suppose it has $2$ vertices. There then exists a set of two vertices with quality at least $669$. Call these two computers the [b] legends. [/b] Observe that if the hacker is able to hack both of the legends as his/her first two moves, then he/she wins. Indeed, he/she then simply focuses on infecting additional vertices of the cycle, and then we finish with the same "if the administrator protected a branch of the legends, he/she loses; otherwise, we just infect the branches of the legends" logic. Therefore, if the hacker takes one of the legends as his first move, the administrator must respond by protecting the other legend. This leads to the following solution. Split the cycle into two contiguous sets of vertices differing in size by at most $1$ so that the legends are in different set. The hacker then hacks the legend which in the set of quality at least $1004$, call this set $\Gamma$. After the hacker protects the other legend, it's clear that the hacker can go on to infect all of the other vertices in $\Gamma$, and from there can infect all of the branches of $\Gamma.$ As the quality of $\Gamma$ is at least $1004$, the hacker will have infected at least $1004 \ge 671$ computers, and so he/she wins.

[b] Subcase 2. [/b] There does not exist such a set.

Then, the hacker simply keeps on picking vertices of the cycle arbitrarily until he/she cannot. At this point, the hacker will have infected a set of at least half of the computers of the cycle with quality at least $671$, by the assumption of this case. If the administrator ever protected any computer among the branches of this set of vertices, then since $v$ is tasty he/she will have protected at most $1337$ computers and the hacker can therefore win by infecting all other computers. Otherwise, the hacker can infect all branches of this set as well, for at least $671$ computers.

As we've exhausted all cases, we've shown that the hacker can always infect $\ge 671$ computers, and so we're done.

$\square$


geometry
3D geometry
tetrahedron
circumcircle
sphere
IMO Shortlist
IMO Longlist

In a tetrahedron, all three pairs of opposite (skew) edges are mutually perpendicular. Prove that the midpoints of the six edges of the tetrahedron lie on one sphere.

This can basicly be reduced to a plane geometry problem.

The vertex $D$ is projected onto the orthocenter $H$ of $ABC$ (this follows from the conditions $AB\perp CD,\ AC\perp BD,\ AD\perp BC$). This means that the midpoints $U,V,T$ of $DA,DB,DC$ are projected onto the midpoints $X,Y,Z$ of $HA,HB,HC$. Let $M,N,P$ be the midpoints of $BC,CA,AB$. The points $M,N,P,X,Y,Z$ liue on the nine-point circle of $ABC$, so the points $U,V,T$ lie on a circle of radius equal to $\frac R2$ ($R$ is the circumradius of $ABC$) which lies on a plane parallel to $ABC$ and which has its center directly above the nine-point center of $ABC$. It's is now clear that $M,N,P,U,V,T$ lie on a sphere which has its center in the midpoint of the segment formed by the centers of $(MNP),(UVT)$.


i think simplest form means common fraction, not mixed number

[quote="xscapezaer"][hide]
$1+1+1+1+1+1+\frac{(6(5(4(2+3)+3!)+4!)+5!)+6!}{7!}$

It's really nasty to calculate, but it works...

Btw, the ans is $7 \frac{22}{35}$
[/hide][/quote]
That is not right. And how would you do that calculation without a calculator?

Edit: Sorry, I accidentally put the wrong equation in... look at first post... :blush:

[hide="Hint"]Look for a pattern.[/hide]

[hide="Solution"]If you add the first parentheses, you'll get$\frac{3}{2}$.When you add the next one you get $\frac{4}{3}$. So the denominator is going to be the previous numerator and the numerator is going to be one more than it. And this happens for 6 times. So its just $\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\frac{6}{5}\cdot\frac{7}{6}\cdot\frac{8}{7}=4$.[/hide]
Sorry if it is too hard to follow(if it is).

Oh you were supposed to multiply those? Oh well...

[quote="Quevvy"]Express in simplest form:
$\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{7}\right)$

Sprint round = no calculators =  :( = I need a fast way to do this!

[size=150][b]EDIT:[/b][/size]It is actually$\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{7}\right)$
Sorry :blush:[/quote]

now its vway easier, with the multiplication.
it is simply 3/2 times 4/3 times 5/4 times (denominator + 1)/(denominator) up to 7, so the 3, 4, 5, 6, and 7 cancel out, and that leaves $\frac{8}{2}=4$


I have a general question about learning mathematics.

What type of learning methodology do you feel is more effective: the "top-down" approach, or the "bottom-up" approach?

By top-down I mean learning a subject by first looking at theorems and their proofs, and then running through many examples and applying the theorems directly, and deductively coming up with new results from the theorems.

By bottom-up I mean learning a subject by working on specific problems that involve new concepts, learning the concepts through the problems, and slowly building up, culminating in a theorem or complete generalization.

Why do you think one approach is more effective than the other? Are different approaches more or less suited for particular subjects, or at particular times? How does one's mathematical maturity affect the effectiveness of each of these approaches?

I know this is a long list of questions, but I want to try to get a good understanding of the topic. Thanks

[quote="probability1.01"]This is a very interesting topic. I learned pretty much all of high school math using the bottom-up approach, and I think it really helped me gain a solid understanding. However, top-down is much faster. I think the main difference is that bottom-up builds intuition, whereas top-down just gets the ideas across. You should go bottom-up in very new subjects, but for something like analysis after calculus, top-down is probably more efficient.[/quote]

i agree with most of what you said. but it seems like at a certain point all math classes take the top-down approach (even in subjects that students have never seen before).

[quote="bubka"]ET Bell, speaking of Carl Gustav Jacob Jacobi, says:

"Jacobi seems to have been the first regular mathematical instructor in a university to trin students in research by lecturing on his own latest discoveries and lettign the students see the creation of a new subject taking place before them. He belived in pitching young men into the icy water to learn to swim or drown by themsleves. Many students put off attempting anything on their own account till they have mastered everything relating to their problem that has been done by others. The result is that but a few ever acquire the knack of independent work. Jacobi combated this dilatory erudition. To drive home the point to a gifted but diffident young man who was always putting off doing anything until he had learned soemthing more, Jacobi delivered himself of the following parable: "Your father would never have married, and you wouldn't be here now, if he had insisted on knowing [i]all[/i] the girls in the world before marrying [i]one[/i].""[/quote]

wow that's pretty interesting. i notice that i to feel like attaining master of a subject before i venture out to discover/research new things. with the new book i'm working in though, it forces you to discover important results on your own. so maybe i can change that tendency.

Bottom up requires lots of time, but develops deeper understanding imo. Classes seem to want to get large amounts of work through to students as quickly as possible..

I would say going over and studying a topic at least 3 times during the study time will force you to learn the material thoroughly. So I think the top-down approach is a better strategy for true understanding.

actually now that I think about it, doing problems is where the real understanding comes from. So a bottom up teaches you problem solving skils (i.e. looking at problems first, seeing if you can do them, and then looking at book for help)

Here in India , it's mostly a top-down approach that most people take..... I've never really seen anyone go by a bottom-up approach here....  :|

[quote="drunner2007"]most of the AOPS books are centered around the bottom up approach, right?[/quote]
That has been my impression (from an exhaustive sampling of [hide="..."] one book  :roll: [AoPS Volume 2][/hide]).  Art of Problem Solving is extremely unusual in their efforts to actually teach students all the reasoning behind the mathematics, as opposed to force-feeding them information.

I can also attest to the fact that the above-mentioned book caused my AMC score to jump by nearly twenty points.

When teaching myself from the book I use the "top-down" approach, whereas my math teacher uses the "bottom-up" appproach when teaching my class.

I can work through more material at a faster rate using the "top-down" system, but I think that I understand material better the way my teacher does it (or maybe that is just because I have a teacher, I don't know).  I think that it really depends on the person and the material, what works best and what they can handle.

In the 1960's, my dad did his dissertation on the question of which approach was more effective for the teaching of middle-school mathematics.  His study showed that it's more effective to do bottom-up, but takes longer.

When I study mathematics, I tend to use a combination approach by starting with top-down to narrow the area I want to think about, and then use bottom-up to get a grasp on that area.  This has been effective for me in producing research in subfields outside my specialty.

When I'm teaching, I tend to vary my approach based on the mathematical maturity and interest of the students.  I usually spend class time on bottom-up activities, and have students do any top-down reading outside of class.  But the more mature/interested the students, the less I assign reading and the more I assign investigations.


topology
real analysis
real analysis unsolved

Let $ \Omega \subset \mathbb{R}^n$ open set, and let $ f_1,...,f_r \in \mathcal{ C}(\Omega):$ $ \displaystyle{\sum_{j\equal{}1}^r}|f_j(x)|>0, \forall x \in \Omega$. Find $ g_1,...,g_r \in \mathcal{C}(\Omega)$ so that $ \displaystyle{\sum_{j\equal{}1}^rg_j(x)f_j(x)}\equal{}1, \forall x \in \Omega$

Can you please find titles that say [i]something[/i] about the content of your threads?


So I was on the plane listening to music on random mode, where it just randomly plays songs without regard to whether the song has already been played or not.  So first I hear Song 5, then I hear Song 17, and then I hear Song 5 again, even though I had practically just heard it.  If I have 80 songs on my CD, then on average, what is the number of songs I need to listen to in order to hear every single song at least once?

there must have been a similar problem posted before.
the answer is [tex]n\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)[/tex]

see [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=14034[/url]
the "coupon collector" problem at the near bottom of the page.

Intutive and short reply:  let us take general case with 'n'  songs (n=80)

If you want to hear  one particular song you have to play (on average) n songs.
Suppose there are 'm' songs of  one type (say rock-and-roll), then to hear  one (that is, any one) song  of that type..  you have to  (on average) play    n/m   songs.

Right?
Now let us say the by the type I mean  "fresh" of 'not_heard_before':
So for first  "fresh" song  =>            1    (obviiously  first song is always  new)
For the second "fresh" song ==>    n/(n-1)
For the next fresh song        ==>     n/(n-2)

So the answer = 1+  n/(n-1)+ n/(n-2) + n/(n-3) + ......   n/1
Which is same as given above.. (Detail argument is given in the link shown there) ...


In a class of 28 sutdents, the teacher selects 4 people at random to participate in a geography contest. What is the probability that this group of four students includes at least 2 of the top 3 geography students in the class? Express your answer as a common fraction.

We examine two cases.

[b]Case 1[/b]: 2 of the top 3 students

$ 2 \times 1 \times 26 \times 25 \equal{} 1300$

$ \binom{28}{4} \equal{} 20475$

$ \frac{1300}{20475} \implies \frac{4}{63}$

[b]Case 1[/b]: 3 of the top 3 students

$ 3 \times 2 \times 1 \times 25 \equal{} 150$

$ \binom{28}{4} \equal{} 20475$

$ \frac{150}{20475} \implies \frac{2}{273}$

$ \frac{4}{63} \plus{} \frac{2}{273} \equal{} \boxed{\frac{58}{819}}$

Please correct me if I'm wrong.

[hide]
In total, there are $ {28\choose 4}$ ways to choose them.
To include at least 2 of the top 3 geography students,
1- We only choose 2
$ 3{25\choose 2}$
2- We choose all three.
$ 25$

So $ \dfrac{925}{{28\choose 4}}$, i don't feel like calculating.[/hide]
EDIT: you can also use complementary counting.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ (G,*)$ a group and $ (H_1,*)$, $ (H_2,*)$  two subgroups of $ (G,*)$

It's obvious that the intersection of $ H_1$ and $ H_2$ give a subgroup of G

now find a counterexample wich proves that The union of two subgroups is not always a subgroup.

the union of two subgroups $ (H_1,*) (H_2,*)$ is a subgroup iff $ H_1 \subset H_2$ or$ H_2 \subset H_1$.
proof:
suppose that the union is subgroup and there exist $ a \in H_1 \minus{} H_2$ and $ b \in H_2 \minus{} H_1$.since the union is subgroup then $ ab \in H_1 \cup H_2$.
 if $ ab \in H_1$ and $ a \in H_1$ then $ b$ is also in $ H_1$ which it is contardiction and we the same way we obtain  a contradiction in the other case .

Take $ 2\in 2\mathbb{Z}$ and $ 3\in 3\mathbb{Z}$, then $ 5\not\in 2\mathbb{Z}\cup 3\mathbb{Z}$, so we don't
have closure...


If a+b=n, where n > 0 and where a > 0 and b > 0, obviously n > a and n > b

Prove that max ab= (n^2)/4

This might be too easy for pre-olympiad  :D
It's just equivalent to $4ab \leq n^{2}= (a+b)^{2}$, i.e. $0 \leq (a-b)^{2}$.

Yeah, I know. I should have looked at it for a while on my own. I also meant to say to find the maximum of ab, but it is essentially the same. I now know that since nb-b^(2) is a function and quadratic. It, therefore, can be written as -(b-n/2)^(2) + N^2/4.


The vertices of a right triangle $ABC$ inscribed in a circle divide the circumference into three arcs. The right angle is at $A$, so that the opposite arc $BC$ is a semicircle while arc $BC$ and arc $AC$ are supplementary. To each of three arcs, we draw a tangent such that its point of tangency is the mid point of that portion of the tangent intercepted by the extended lines $AB,AC$. More precisely, the point $D$ on arc $BC$ is the midpoint of the segment joining the points $D'$ and $D''$ where tangent at $D$ intersects the extended lines $AB,AC$. Similarly for $E$ on arc $AC$ and $F$ on arc $AB$. Prove that triangle $DEF$ is equilateral.

[quote="April"]We have
\[ \left\{\begin{array}{c}\widehat{D''DC}=\widehat{DAC}=\widehat{DD''A}\\ \widehat{E''EC}=\widehat{EAC}=\widehat{EE''A}\end{array}\right\| \]

\[ \Rightarrow\widehat{DCE}=\widehat{DCA}+\widehat{ACE}=2\cdot\left(\widehat{DAC}+\widehat{CAE}\right)=2\cdot\widehat{DAE}\]
On the other hand, the quadrilateral $ AECD$ is cyclic, so $ \widehat{DCE}+\widehat{DAE}=180^\circ\Rightarrow\widehat{DAE}=60^\circ$

Hence $ \widehat{DFE}=\widehat{DAE}=60^\circ.$ Similarly, $ \widehat{DEF}=60^\circ.$

It implies that triangle $ DEF$ is equilateral.[/quote]

[b] Nice problem - Nice solution. I add the figure to it.[/b]

 A! Now I see a question : Why

   $ \widehat{ACE}=2\widehat{CAE}$ ?

 I do not deduce it obviously :(

[color=red] [b]EDIT :  The figure in the official solution is different. Please do not look at it. I will change it , if it is wrong.[/b][/color]


calculus
integration
number theory
modular arithmetic

i am a math teacher that is working on her master's.  My teacher gave me this problem to work and other than going one by one to find a pattern, I am not sure how to do this.  This is due tomorrow!   :oops:  the problem is:

 Find every positive integer (n) such that n^3+n^2+n^1+1 is square

thank you!!!

We can factor the given equation to $(n^2+1)(n+1)$ Now we are given this number is square. We note that the given equation is also equal to $\frac{n^2+1}{n+1}(n+1)^2$, thus since the result is also square, $\frac{n^2+1}{n+1}$ is square. Now modulus 4 is fun for squares: We know that every square is 0 or 1 mod 4. We can perform casework upon that fraction to find when it is in the set {0,1} mod 4.

n==0 mod 4: we have the result equal to 1 mod 4, so a possibility
n==1 mod 4: we have 3/2, which is not a perfect square mod 4.
n==2 mod 4: we have 5/3, which isn't a perfect square mod 4.
n==3 mod 4: we have 10/4, which isn' a perfect square mod 4.

Thus n==0 mod 4. Now let n=4a. We have now $\frac{16a^2+1}{4a+1}$ square. And now we can evaluate mod 16, which results in $\frac{1}{4a+1}$. Now we know that the denominator is integral, thus it must be 1 for the fraction to be a square integer thus a=0. Therefore the only possibility is n=0.

Actually I have a minor little hole, but you can use infinite descent to plug it.

N=0,1, and 7 works... but I'm not sure what you did wrong Sodafid farms. Can you explain to me why you put it in mod 16? And why that means that n has to equal 0? I understand the mod 4 part.

that gives me a formula--so do I simply start plugging a number in for (N).  Is there no true pattern?  And why is 1 not working--When I plug 1 into my problem 1^3+1^2+1^1+1 that =4 which is a square.  But it's not working in your formula.  Thanks!!

solafidefarms - you are examining the equation in modular arithmetic, so you cannot conclude that $n = 0$ because you have the top and bottom mod 16 (for example, the top of the fraction can be 17).

[quote="georgia03"]And why is 1 not working--When I plug 1 into my problem 1^3+1^2+1^1+1 that =4 which is a square.  But it's not working in your formula.  Thanks!![/quote]
$n^3+n^2+n+1=\frac{n^4-1}{n-1}$ is not true for $n=1$ because the fraction is undefined if $n=1$.


function
calculus
integration
complex analysis
complex analysis unsolved

I'm looking for help in integrating

int(dz/z*)

Where z* denotes complex conjugate.

Thanks very much.

Since that's not an analytic function, it has no antiderivative. The integral is strongly path-dependent, and should be handled using real-variable methods. Green's theorem may be useful.

Thanks..

Could you explain how I could use real variable methods and Green theorem to go about it??


search
pigeonhole principle
number theory
number theory solved

Looking at the IMOshortlist2002 at number theory I found an interesting problem(well.. all problems are.. but this one caught me :D) because it may be a generalization of a problem given at the first selection test for the OIM 2003 and OBM 2003 12 march 2003 in the Republic Moldavia, if it is true:

N4. Does the equation 1/a + 1/b + 1/c + 1/(abc) = m/(a + b + c) have infinitely many solutions in positive integers a, b, c for any positive integer m?

the problem was like so at the test:

Prove that the equation 1/a + 1/b + 1/c + 1/(abc) = 12/(a + b + c) has infinitely many solutions in natural numbers! so it is actually m=12.

what do you think?  cheers! :D

I think I managed to slve the problem at the test (for m=12). I just observed some solutions and then I guessed how to produce more:

We can see that (1,1,1); (1,1,2); (1,2,3) are solutions. Seeing these I thought we can find a sequence of numbers a1,a2,...,an,... such that any 3 consecutive terms of the sequence form a sol fo our eqn. We have a1=a2=a3=1, a4=2, a5=3. I found the next sol: (2,3,7). Then I observed that an is alternatively (a(n-1)+a(n+1))/2 or (a(n-1)+a(n+1))/3. You can just check and you'll see it works :D . We can also observe that an*a(n+1)+1=a(n-1)*a(n+2). I just thought this might help. Here are some terms:1,1,1,2,3,7,11,26,41,...

As for problem in the shortlist, in that form, it's easy:the answer is NO. Think about what happens if you put m=1. The left part is obviously > than the right part. Actually we can prove without any difficulty that any  natural m for which we have sol must be at least 7. I think it should ask for which m's the eqn has sols, and for which it doesn't. What do you think?

hey
nice solution grobber!!! i wud like to some of the solutions that "grobber" has already given.we search for particular solutions in artihmetic progression,we have
a=x-d,b=x,c=x+d.then after some calculations this reduces to pell's equations x^2-3d^2=1 which ,as is well known,as infinitely many solutions.
cheers!(this is for the case m=12)

How about the general case?

Doesn't this way work for other m's? As long as they are large enough, of course :D

Hey, sorry for asking such a stupid question :D I misunderstood the question from the shortlist. I thought that it was asking if there were an infinity of solutions for ANY m. Boy, that was stupid of me!

How exactly is the question worded? If it's worded this way the solution is easily no since there's no solution for m=1,2,3

No solution when n<=9, very possible to have infinite solution when n=10 or n>=12 (involoves some generalisation of Pell's Equation that I don't know the results), unsure about n=11

There are also infinitely many solutions for m=12 if you set x=a, y=b, z=a+b.  Also we can find infinitely many m for which this gives infinitely many solutions.  For if m = 2k this gives

1/a + 1/b + 1/(a+b) + 1/(ab(a+b)) =k/(a+b) or

(a+b)^2 + ab + 1 = kab, or

a^2  + b^2 - (k-3)ab = 1, so we get a Pellian equation of the form a^2+b^2-rab=1.  But  1 = a^2+b^2-rab = (a+b*(r+sqrt(r^2-4))/2)*(a+b*(r-sqrt(r^2-4)/2) so it suffices for (a+b*(r+sqrt(r^2-4))/2) to be a unit in the ring of integers in [bold]Q[/bold](sqrt(r^2-4)).  Let r^2-4 = a^2*d with d squarefree.  Then as long as d>1 it is a theorem than there are infinitely many units in the ring of integers in [bold]Q[/bold](sqrt(d)).  So if d=r^2-4, we're home free since any such unit can be written in the desired form (a+b*(r+sqrt(r^2-4))/2).   If not, we're still done because infinitely many of the units are congruent to 1 mod r+sqrt(r^2-4)/2 (pigeonhole principle) and those ones will give us solutions for a, b.

--Alison


which problems are harder, AIME or ARML?
would it be good practice for AIME if i did old ARML problems, or are they a step-down in difficulty?

thanks! :)

It probably depends on which round of ARML. I've never done ARML, just practiced from tests posted online. I haven't worked all of the kinds of rounds, but I personally think the Individual round is slightly easier than AIME, and the Relay round is a bit easier than that.

Try HMMT (Harvard-MIT Math Tournament) problems

ARML Team Round problems are probably good practice for the AIME.  Note, though, that ARML is one of the few competitions that makes you pay for its tests (with the exception of the previous year's).  I second the recommendation of HMMT.

[quote="mihail911"][quote="carpo"]Just a suggestion, but it would be good practice for AIME if you did past AIME(!) [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45]problems[/url]. Plus many of the problems are very nice. (I don't know much about ARML.)[/quote]

yeah, i know, but what if i finish all past AIME problems? i mean, im going to have to do a lot of them this year..and what happens when i run out of old aime questions? i just wanted to know of what contests held a similar range of difficulty as the AIME[/quote]
If you actually finish all past AIME problems, you should be moving onto past Olympiads; this is what I did last year, except that I never even got around to the AIMEs from the mid to late 1990s.

ARML problems tend to rely more on tricks.  If you have solutions, learning those tricks can be useful.

Yea, I tend to think of ARML problems as: "one step then don't make a computation error" problems.

The other ARML problems (back to 2004) are up on the website, they just aren't linked to anywhere. Try taking the links and replacing the current year with a previous one.

[quote="mihail911"]yeah, i know, but what if i finish all past AIME problems? i mean, im going to have to do a lot of them this year..and what happens when i run out of old aime questions? i just wanted to know of what contests held a similar range of difficulty as the AIME[/quote]

I echo MellowMelon. If you manage to get through all past AIME problems, then either you have gone through them too fast, in which case you should go over the difficult problems again; or you will be extremely well-prepared for the contest. (By the way, it's a good idea anyway, whenever you find a problem you can't solve and read the solution, to go over it a few days later, since otherwise you may forget how to solve it.) If you diligently go through only half of the problems then you are probably ready to move on to olympiad problems. I speak from experience: I printed of all the AIMEs and planned to do them, but only ever got from 1983 to about 1994.


[quote="TTsphn"]$ p_{\minus{}1}\equal{} q_{\minus{}1}\equal{} 1$
$ p_{0}\equal{} q_{0}\equal{} 1$
$ p_{n}\equal{} 2p_{n\minus{}1}\plus{}(2n\minus{}1)^{2}p_{n\minus{}2}$
$ q_{n}\equal{} 2q_{n\minus{}1}\plus{}(2n\minus{}1)^{2}q_{n\minus{}2}$
Find $ lim\frac{p_{n}}{q_{n}}$[/quote]
Obviosly $ p_{n}\equal{}q_{n}$, therefore $ lim\equal{}1$.

Obviosly $ p_{n}\equal{} (2n\plus{}1)!!,n\ge 0$. Let $ q_{n}\equal{} x_{n}p_{n}$, then
$ x_{0}\equal{} 1;x_{1}\equal{}\frac{2}{3}, x_{n}\equal{} x_{n\minus{}2}\plus{}\frac{2(x_{n\minus{}1}\minus{}x_{n\minus{}2})}{2n\plus{}1}$.
Let $ y_{n}\equal{} x_{n}\minus{}x_{n\minus{}1}$, then $ y_{1}\equal{}\minus{}\frac{1}{3},\ y_{n}\plus{}y_{n\minus{}1}\equal{}\frac{2}{2n\plus{}1}y_{n\minus{}1}$ or
$ y_{n}\equal{}\minus{}\frac{2n\minus{}1}{2n\plus{}1}\equal{} ... \equal{}\frac{3}{2n\plus{}1}(\minus{}1){n\minus{}1}y_{1}\equal{}\frac{(\minus{}1)^{n}}{2n\plus{}1}$, therefore $ x_{n}\equal{} 1\plus{}\sum_{k \equal{} 1}^{n}\frac{(\minus{}1)^{k}}{2k\plus{}1}$. Let $ a \equal{}\sum_{k \equal{} 0}^{\infty}\frac{(\minus{}1)^{k}}{2k\plus{}1}\equal{} arctg(1)$, then $ \lim_{n\to\infty}\frac{p_{n}}{q_{n}}\equal{}\frac{1}{a}\equal{}\frac{4}{\pi}$.

I think, that these problem is best your problem.

No I have some other problems and I will post later
Have fun.


Let $ M$ be an open set in $ \mathbb{R}$ that is unbounded above. Does there exist a point $ x \in M$ such that $ \{nx\}_{1}^{\infty}$ intersects $ M$ in an uncountable set of points? (I can prove it for the infinite, but countable set of points.)

For any $ x$, the set of all $ nx$ is countable. There's no chance.


function
advanced fields
advanced fields unsolved

Let $X$ and $Y$ be Banach spaces and $\Omega\subset X$ be open and $\Phi\in C^{1}(\overline\Omega,Y)$. The space $C^{1}(\overline\Omega,Y)$ is endowed with the norm $\|\Phi\|: =\sup_{x\in\Omega}\|\Phi(x)\|+\sup_{x\in\Omega}\|\Phi'(x)\|_{B(X,Y)}$ with $B(X,Y)$ being the space of all bounded linear operators from $X$ to $Y$.

Now let $y\in Y$ be a regular value of $\Phi$ (i.e. $\Phi'(x)\in B(X,Y)$ is isomorphism for each $x\in\Phi^{-1}(y)$). Suppose that the equation
\[\Phi(x)=y \]
has $n$ solutions in $\Omega$.

Prove that there exists $\epsilon>0$ such that the equation
\[\Phi_\epsilon(x)=y \]
also has $n$ solutions if $\|\Phi_\epsilon-\Phi\|\le\epsilon$.

[b]Comment:[/b] Do not use degree theory.

The Inverse Function Theorem is your friend.

The Inverse Function Theorem is local result. Anyway I don't know how to solve the problem using that theorem. I solved it using Banach contraction principle instead. Can you solve the problem using the Inverse Function Theorem?

The problem is a local problem.  Remember that the IFT is a corollary of the contraction mapping principle, so the IFT solution is probably essentially the same as yours.

The basic idea is to apply the IFT to a new function $\Psi: X\times\mathbb{R}\to Y\times\mathbb{R}$ given by $\Psi(x,t)=(\Phi(x)+{t\over\epsilon}(\Phi_\epsilon(x)-\Phi(x)), t)$.


algebra
polynomial
Vieta
complex numbers

(****)

42. Solve the system where $ a,b,c$ are distinct complex numbers:
\[ a^3 \plus{} 7b \plus{} 7c \plus{} 6 \equal{} 0
\]

\[ b^3 \plus{} 7c \plus{} 7a \plus{} 6 \equal{} 0
\]

\[ c^3 \plus{} 7a \plus{} 7b \plus{} 6 \equal{} 0
\]

Subtract the second equation from the first to get $ (a\minus{}b)(a^2\plus{}ab\plus{}b^2\minus{}7)\equal{}0$, since $ a\not \equal{} b$, $ a^2\plus{}ab\plus{}b^2\equal{}7$. Similarly, $ b^2\plus{}bc\plus{}c^2\equal{}7$. Subtracting, $ (a\minus{}c)(a\plus{}b\plus{}c)\equal{}0$, since $ a\not \equal{} c$, $ a\plus{}b\plus{}c\equal{}0$. The original equation become $ a^3\minus{}7a\plus{}6\equal{}(a\minus{}1)(a\minus{}2)(a\plus{}3)\equal{}0$, so $ (a,b,c)$ can be any permutation of $ (1,2,\minus{}3)$

A question from a lower-level math guy...
How do you recognize what will and won't factor? Do you just know it, see certain patterns, or use trial and error?

That is a well-known technique for solving this type of problem. If you notice that you switch a and b in the first equation, then you get the second equation. If you subtract them, you have a polynomial in a that has a root of b, so (a-b) is a factor.

I was trying to illustrate another technique for this problem, I forgot to check this technique.

The nicest solution method for a symmetric system, in my experience, is a symmetric manipulation of the equations.  In this case, we have a symmetric manipulation (taking the three pairwise differences) that gives us three two-variable equations instead of three three-variable equations.

Setting the LHS's of the equations equal in pairs, we find that $ a^{3}\minus{}7a\equal{}b^{3}\minus{}7b\equal{}c^{3}\minus{}7c$. Since they are distinct, they must be the roots of $ x^{3}\minus{}7x\plus{}C\equal{}0$, for some constant $ C$. Thus, by Vieta, $ a\plus{}b\plus{}c\equal{}0$, and $ \minus{}7a\equal{}7b\plus{}7c$. Substituting this into the first equation gives $ a^{3}\minus{}7a\plus{}6\equal{}0$, etc.


abstract algebra
invariant
inequalities
set theory
superior algebra
superior algebra unsolved

Let F be a free module of infinite rank $ \alpha$ over a ring R that has invariant dimension property. Prove: for each cardinal $ \beta$ such that $ 0 \leq \beta \leq \alpha$, F has infinitely many proper free submodules of rank $ \beta$.

Attempted Solution:

I'm not too good with set theory, so there is probably something wrong with this.

First, I think that inequality $ 0 \leq \beta$ should be strict. There is only one empty basis...

Let F be free on the set X. So X is a basis for F over R and $ |X| \equal{} \alpha$.

If $ \beta$ is finite, then this is pretty clear. There are infinitely many ways to take $ \beta$ elements out of $ X$. Given a subset Y of X of size $ \beta$, consider the submodule it generates. This submodule is free on Y and has invariant dimension property and has dimension $ |Y| \equal{} \beta$ and it must be proper since otherwise X is not linearly independent.

Now for the infinite case. Given $ \beta \geq \aleph_0$, we find a proper subset $ Y$ of $ X$ with cardinality $ \beta$. $ Y$ will generate a submodule with invariant dimension property that has have dimension $ \beta$, that is free on Y, and is proper like in the finite case. Now we can throw out infinitely many elements of $ Y$ to get infinitely many proper subsets of $ Y$ each of which generate submodules of dimension $ \beta$.

you have to keep attention that your submodules are different. I don't think that your easy approach works. translate the problem into matrices, then it's pretty clear.

[quote="-oo-"]you have to keep attention that your submodules are different. I don't think that your easy approach works.[/quote]For $ A\subset B\subset X$ it holds that the modules generated by A and B, resp., are equal iff $ A\equal{}B$, since $ X$ is linearly independent. Assuming that you inductively construct a family of strictly decreasing subsets $ Y_{n\plus{}1}\subsetneqq Y_n\subsetneqq X$ with $ |Y_n|\equal{}\beta$, then I cannot see where the approach fails to work, since the induced modules are strictly decreasing.


let a+b+c+d=0 and a ^3 +b ^3 +c ^3 +d ^3 >0
prove that a^5 + b^5+c^5+d^5>0
i have found the answer for it but it's too long
can you give me a short solution.

Let S_1 = a+b+c+d, S_2 = ab+bc+cd+da+ac+ad, S_3 = abc+acd+abd+bcd and S_4 = abcd.
Let P(x) = (x-a)(x-b)(x-c)(x-d)
= x 4 - S_1*x 3  + S_2*x 2  - S_3*x + S_4.
Since a,b,c,d are the roots of P(x) and from S_1 = 0, we deduce that :
a 4 = -S_2*a 2  + S_3*a - S_4,
thus
a 5  = -S_2*a 3 + S_3*a 2  - S_4,
and we have similar relations with b,c,d. Summing, it leads to :
 \sum a 5  = -S_2* \sum a 3  + S_3* \sum a 2  , (1).

In another hand :
0 = (a+b+c+d) 3 =  \sum a 3 + 3 \sum a 2 (b+c+d) + 6S_3
= \sum  a 3  + 3 \sum a 3 (-a) + 6S_3
= - 2 \sum a 3 + 6S_3
thus 3*S_3 =  \sum a 3 .
Moreover :
0 = (a+b+c+d) 2  then 2*S_2 = - \sum a 2 .

From (1), we deduce that :
 \sum  a 5 /5 =  (\sum a 3 /3)*( \sum a 2  /2), from which the result follows easily.

Pierre.

40 th British  mo , 2004 , round  2.  No  i dont  have  an  easier solution  Pierre. :)


geometry
geometric transformation
homothety
ratio
inequalities
limit
Putnam

Let $I_k =[a_k,b_k]$ such that $0<a_1<b_1<a_2<...<a_k<b_k<...$ and $a_k\to\infty$.
Prove that there exists $x>0$ such that the sequence $\{nx\}$ intersects infinitely many $I_k$.

[i]Edited by Myth[/i]

Let $I'_{k_1}=I_1$. Consider the image of $I'_{k_1}$ through a homothety of ratio $n$ (and center $O$, the origin of the real line). For some $n$, there is a $k$ s.t. $I_k\cap nI'_{k_1}\ne \emptyset$, and let $I'_{k_2}=\frac 1n\cdot I_k\cap nI'_{k_1}$. What we did for $I'_{k_1}$ we do now for $I'_{k_2}$ (we can find an $n$ larger than the previous one, of course), and we find an $I'_{k_3}$, and so on. We can thus find a decreasing sequence of compact intervals ($I'_{k_1}\supseteq I'_{k_2}\supseteq I'_{k_3}\supseteq\ldots$), with non-void intersection (Cantor's Theorem). Take $x$ to be a point belonging to this intersection.

I bet you think it's not clear enough. That's my opinion as well :).

We can WLOG suppose $b_k-a_k\to 0$

Lemma. For each $[u,v]\subset[0,1]$ there exists $n$ and $k$ s.t. $[a_k,b_k]\in [nu,nv]$.
$\square$ Proof. Indeed, for all $n>N$ the following inequality holds: $nu+v<nv$. Let's find $I_k$ s.t. $b_k-a_k<v-u$ and $nu<a_k<(n+1)u$, $n>N$. Then $b_k<a_k+(v-u)<(n+1)u+(v-u)<nu+v<nv$.$\blacksquare$

Let $[u_0,v_0]=[0,1]$. Due to lemma we can find $[a_{k_1},b_{k_1}]\subset [n_0u_0,n_0v_0]$. Define $[u_1,v_1]=[a_{k_0}/n_0,b_{k_0}/n_0]$. And so on. We have obtained infinite sequence of embedded segments. Therefore we conclude there is $x$, which lies in all this segments. It means $\{nx\}$ intersects infinitely many $I_k$.

There is "somehow" related problem :)

 Is is possible that :
- $f$ is continuous
- $\lim_{n\rightarrow \infty}{f(nx)} = \infty$  for all $x$
- We don't have $\lim_{x \rightarrow \infty}{f(x)} = \infty$ for all $x$ in $\mathbb{R}$

 This one from Putnam

Actually, it is direct consequence of initial problem.
Just suppose contrary, then we will able to find sequence of segments $I_k$, s.t. $f$ is bounded on $I_k$ with the same constant $C$.
That's all.
And yes, Levon posted problem with 0 instead of $+\infty$ some time ago.

[quote="grobber"]And I think it's been discussed before, receiving similar solutions. I don't know if it was precisely this one though. I remember something very similar, but with limits $0$ instead of $\infty$.[/quote]
I'm just saying that they are two " somehow related" problems!!  :?
 No intention to ask anyone to solve  :D

And I remember giving basicly the same solution as the one Myth briefly presented here :). The problem has been posted more than once, I think. I seem to remember alekk posting it as well.


Let $ G$ be a graph with $ n$ vertices, and let $ d_1$, $ d_2$, ..., $ d_n$ be the degrees of these $ n$ vertices of $ G$. Let $ e$ be the number of edges of the graph $ G$. Prove that the number of triangles in $ G$ and $ \overline G$ (that is, the sum of the number of triangles in $ G$ with the number of triangles in $ \overline G$) is:
$ {n \choose 3} \minus{} \left(n \minus{} 2\right)e \plus{} \sum_{i \equal{} 1}^n {d_i\choose 2}$.

We calculate the number of triples $(a,b,c)$ of vertices of graph $G$, which do not form a triangle neither in $G$, nor in $\overline{G}$. We must prove that the number of such triples equals $(n-2)e-\sum{d_i\choose 2}$. Note that for each such triple, there exists two vertices in it, which are joined by the edge with exactly 1 other vertice in this triple (say, if $a$ is joined with $b$ and $c$, and $b$ and $c$ are not joined, such vertices are $b$ and $c$). Call such vertices nice vertices of the triple. On the other hand, every vertex $a$ of degree $d$ is nice in exactly $d(n-1-d)$ triples (we may add to the triple arbitrary vertex $b$, joined with $a$, and arbitrary vertex $c$, not joined with $a$). So, our number equals $\frac12(\sum d_i(n-1-d_i))=(n-2)e-\sum{d_i\choose 2}$ .


Prove that in any convex 2n-gon  there is a diagonal not parallel to any side.

[quote="t0rajir0u"]You seem to have assumed the polygon is equiangular.[/quote]

oooh, whoops, I can't read. But wouldn't that be the extreme case?

[quote="13375P34K43V312"][hide]There are $ n$ pairs of parallel sides. Label the vertices, in clockwise (or counterclockwise) order, $ A_{1},A_{2},\ldots,A_{2n}$. Consider the side $ A_{1}A_{2}$, and its corresponding parallel side, $ A_{n+2}A_{n+1}$. There are $ n-2$ diagonals parallel to these, namely $ A_{2n}A_{3},A_{2n-1}A_{4},\ldots,A_{n+3}A_{n}$. Hence, there are $ n(n-2)$ diagonals parallel to a side. There are $ \frac{2n(2n-3)}{2}=n(2n-3)$ diagonals in total, and $ n(n-2)<n(2n-3)$ for all $ n>1$, so by the Pigeonhole Principle there must exist some diagonal not parallel to any side.[/hide][/quote]

yeah i think i came to the same result. the equilateral case should be the extreme case with the largest number of pairs of parallel diagonals.


My brother is 4 x as old as iam right now.  in 6 years he will be 2x as old as i am. how old am i?

Set ur age as X
Ur brother is 4X
4X+6=2(X+6)
X=3

[b] Just Subtitute  and you will get your age as 3[/b]

Guys, please don't post the same solution over and over again if you use the exact same method.
And don't revive a thread that's a month old.

1st set teh variables
                  age now      | age in 6 years
You           |     x           |x+6
your brother|     4x         |2x
2nd make teh equation
4x+6=2(x+6)
solve
2x+3=x+6
x=3
so you r 3 years old

Hi lalagundam! I see you're new to AoPS.

As future reference, try not to post solutions to problems that already have solutions unless you have an alternate method OR you find problems with the previous solutions. Also, "reviving", as AoPSers commonly call it, is frowned upon.

Welcome to AoPS, and....have fun! :)


calculus
derivative
analytic geometry
graphing lines
slope
function
algebra

Hi, How do you do the following question?

Find the equation of all lines that are tangent to both y=x^2+2 and y=-2-x^2. Indicate the points where the tangent line intersect the curve.

Is y^2=4x^2 the solution?

Thanks

Let $f(x) = x^2 + 2$ and $g(x) = -x^2 - 2$.  At any point $(c,f(c))$, the tangent to $f$ is given by:

\[ y = f'(c)(x-c) + f(c) = 2c(x-c) + c^2 + 2 = (2c)x + (2-c^2) \]

Likewise, at any point $(d,g(d))$, the tangent to $g$ is given by:

\[ y = g'(d)(x-d) + g(d) = -2d(x-d) + -d^2 + -2 = (-2d)x + (d^2-2) \]

What the question is asking for is to find the values of $c$ and $d$ for which the tangent lines described above are the same.  So we have a system of equations where both the slopes and the y-intercepts must be the same:

$2c = -2d$

$2-c^2 = d^2 - 2$

Solve this to get $c = \pm\sqrt{2}$

Thus, using tangent equation $y = f'(c)(x-c) +f(c)$, the two desired tangent lines are:

\[ y = 2\sqrt{2}(x-\sqrt{2}) + 4 \]
\[ y = -2\sqrt{2}(x+\sqrt{2}) + 4 \]

Think about the graph of the two functions, and it makes sense that there are only two such tangents.  Hope this helps.

madpenguin


The sum of a certain number of consecutive positive integers is 1000. Find these integers.

This problem looks easy.. but it's a little bit tricky.

[hide]There are actually 4 different sets.  I'll give one of them:  198+199+200+201+202... The rest shouldn't be [i]too[/i] hard to find :P [/hide]

It's not hard...just a little bit of number theory.

[hide]
$n(2a+n-1)=2000$ and $2a+n-1>n$ :)
[/hide]

[hide="Somewhat Elegant Solution"]Case 1: Odd number of integers.

Let $n$ be the average of the integers.  The integers are

$n-k,n-k+1,n-k+2\ldots n-1,n,n+1\ldots n+k-2,n+k-1,n+k$

There are $2k+1$ terms.  The sum is $n(2k+1)$.  Therefore, $n(2k+1)=1000$, and $2k+1$ is an odd factor of $1000$.  We have

$(k,n)=(0,1000),(2,200),(12,40),(62,8)$

The sets of numbers for this case are

$1000$
$198$ to $202$
$28$ to $52$

The fourth ordered pair produces some negative integers, so it doesn't work.

Case 2: Even number of integers.

The integers are

$n-k-\frac{1}{2},n-k-\frac{1}{2}+1\ldots n-\frac{1}{2},n+\frac{1}{2}\ldots n+k+\frac{1}{2}-1,n+k+\frac{1}{2}$

Since these are integers, $n=\frac{2m+1}{2}$ for some integer $m$.  The average of the integers is $n$, and there are $2k+2$ integers.  The sum is $n(2k+2)$.  Therefore,

$\frac{2m+1}{2}(2k+2)=1000$
$(2m+1)(k+1)=1000$

$2m+1$ is an odd factor of $1000$.

$(m,k)=(0,999),(2,199),(12,39),(62,7)$
$(n,k)=(\frac{1}{2},999),(\frac{5}{2},199),(\frac{25}{2},39),(\frac{125}{2},7)$

Only the fourth ordered pair produces only positive integers.  The set of numbers for this case are

$55$ to $70$.

There are four solutions.[/hide]

[hide]
let $1000=n+(n+1)+...+(n+m)$
we then get
$1000=(m+1)n+\frac{m(m+1)}{2}$
so that goes to:
$1000=(m+1)(n+\frac{m}{2})$
now we factor 1000, $1000=2^3*5^3$
so now we look for $(m+1)$ to be equal to one of the divisors of 1000, however, by looking at the $\frac{m}{2}$ in the equation we see that m+1 is some multiple of 5; 5, 25, 125.  Looking at 25 we find $m=24$.  Then by plugging in $n=28$, so, $1000=28+29+...+52$
to find other values you change the m; 4, 24 and 124 are possible values so there are 3 possible nontrivial series.  To find them just pug into the main equation
[/hide]

Nice solution, demji!

[quote="demji"]
$1000=(m+1)(n+\frac{m}{2})$
[/quote]

At this point, I would multiply by $2$ to get $(m+1)(2n+m)=2000$, so that you can forget about the possibility of fractions.

however, by keeping the 2 as a divisor we limit ourselves to powers of 5, i didn't count $5^0=1$ since that is a trivial answer.  Its much faster to keep the 2 since you can automatically eliminate cases where m is odd, meaning that m+1 is an even divisor.  The only odd divisors would be powers of 5, giving u 4 total possibilities, 3 non-trivial.


Suppose I have a uniformly distributed r.v. on [0,1] and the cumulative distribution function is strictly increasing on [0,1]. What steps would I take to come up with a new r.v. on [a,b] with a known cdf F that is also strictly increasing on [a,b]?

I was thinking that if the original r.v were X: S -> R on [0,1]

The new r.v. is, say Y: S -> R on [a,b], and defined over the same S, so Y(s): F^{-1}(X(s))

however, I need this to map to [a,b]. So, say we generate X(s_i) = r_i. Then I need to map each of these r to some number in [a,b]. Does a linear transformation work? How would I define this transformation?

X is uniformly distributed on [0,1].
F-1 (X) will be a random variable with distribution function F in [a,b]. :)


So... How can I apply for it? I would appreciate an answer from an admin. or another student... Thank you in advance

Here's what the MAA site says:
[quote]What if my school doesn’t offer the AMC tests?
A. Urge your principal, math teacher, gifted education coordinator or anyone else you can think of to help you out and register for the contest.  If your school doesn’t offer the AMC tests to you, then you can offer the tests to it.  You could make your local school an offer it would be silly to refuse by signing up for the AMC test of your choice at your own expense, and then letting the local school know you have an opportunity for other students at the school that is already paid for.  Please make arrangements with your school prior to registering as we must send the contest materials to the school directly.  Many colleges and universities also host the contests, particularly the “B” date of the AMC 10/12.  Check our web pages for a list of participating Institutions of Higher Learning. [/quote]
IMO, If your school doesn't offer it, your best bet is probably to contact other schools in your area and see if they'll let you take it there. If not, try the above.

I had just read that, but thanks anyways, mate... Have a good New Years


One cookie and two brownies cost the same as two cookies and one brownie. One cookie and one brownie together cost $ \$$1. What is the cost, in cents, of one cookie?

If one cookie and two brownies cost the same as two cookies and one brownie, then one cookie costs the same as one brownie (Why? Think logic.). Thus, if two of them cost $ \$1$, one of them would cost $ \fbox{50}$ cents.

Here's the "why":

$ c\plus{}2b\equal{}2c\plus{}b$
$ c\equal{}b$.


Solve the equation with x,y are integer  :
$x(x+1)(x+2)(x+3)(x+4)(x+5)+1=y^2$

[quote="pontios"]The obvious one is $x=0,y=1$[/quote]

well .... when y = 1 or -1 , x =0,-1,-2,-3,-4,-5

another one is y=71 , x=2

I think we can use the fact that 6 consecutive integer can be divided by 6!=720 hence

y^2 = 720k + 1    where k is any non-negative integer

but then I have no idea how to continue .... :blush:  :blush:


Let $ S$ be a finite set of natural numbers. Among arbitrary 3 numbers from $ S$ we can choose 2 of them who's sum belongs to $ S$. What is the maximum possible number of elements for $ S$.?

Check that if we have $ a_1 > a_2 > a_3 > a_4 > 0$ as the four largest elements of $ S$ then we can't have the given condition for both the triples $ \{a_1, a_2, a_3\}$ and $ \{a_1, a_2, a_4\}$.  Then conclude.

What if we replace "3" by "4?"


what formula do i need to solve the following question?:

"The grades for a biology pop quiz are normally distributed with a standard deviation of 12.5% and a mean of 58.6%. Give the probability that a student received a mark of 71% or more."

thanx

You first find the Z-Value, which is just the "(sample percentage - mean percentage)/ (Standard Deviation)". After you get the Z- Value, use Table A to find the probability.

Note: If you have no clue what I'm talking about then this isn't the right problem for you.

isn't table A for z-scores that are negative? i got a z-score of .992

No it doesnt have to be negative. There is a positive and negative side.


Anyone know how to tell the number of ligands that a transition metal ion will accept?  Thanks.

It's been a while and I may be wrong on some of these-

First case- the ion has a filled D subshell - the ion will have 4 ligands in a tetrahedral shape (the 1 s- and 3 p- orbitals are used)
Second case- the ion has a D subshell with 8 electrons- the ion will have 4 ligands in a square planar shape- (one of the P orbitals is left empty)
Third case- the ion has a D subshelll that has 6 electrons or fewer such that 2 D orbitals can be opened up and a total of 2 d, 1 s, and 3 p orbitals may be used for the complex to have a octahedral shape.

my chemistry teacher always used to tell me the number of ligands in a coordinated complex is usually twice the charge. so like Fe(OH)[size=75]4[/size] 2-

but there are always exceptions. so you might wanna look that up


In the triangle $ ABC$, $ AC \equal{} 2BC$, $ \angle C \equal{} 90^{\circ}$ and $ D$ is the foot of the altitude from $ C$ onto $ AB$. A circle with diameter $ AD$ intersects the segment $ AC$ at $ E$.
Find $ AE: EC$.

[hide="Solution"]
[asy]unitsize(3cm);

pair C=(0,0);
pair B=(0,1);
pair A=(2,0);
pair D=waypoint(B--A,0.2);
pair O=midpoint(D--A);
pair F=foot(O,A,C);
pair E=reflect(O,F)*A;

label("$B$",B,N);
label("$C$",C,SW);
label("$A$",A,SE);
label("$D$",D,NE);
label("$O$",O,NE);
label("$E$",E,SW);
label("$F$",F,S);

draw(C--B--A--cycle);
draw(C--D);
draw(O--E);
draw(O--F);
draw(Circle(O,arclength(D--O)));[/asy]
Let $ BC \equal{} x$ and $ AC \equal{} 2x$. By similar triangles, $ DA \equal{} \frac {4\sqrt5}{5}x$, so $ OA \equal{} \frac {2\sqrt5}{5}x$ and thus $ FA \equal{} \frac {4}{5}x$. By HL we know that $ \triangle OFA$ and $ \triangle OFE$ are congruent, so $ FE \equal{} FA \equal{} \frac {4}{5}x$. So $ \frac {AE}{EC} \equal{} \frac {\frac {8}{5}x}{2x\minus{}\frac{8}{5}x} \equal{} \boxed{4}$.[/hide]

[b]worthawholebean[/b] —
[hide="Correction"]$ EC \neq 2x$, rather $ EC \equal{} 2x \minus{} \frac{8}{5}x \equal{} \frac{2}{5}x$. Since they ask you for the ratio $ AE : EC$,  the actual answer would be $ \frac{8}{5}x : \frac{2}{5}x \Rightarrow\boxed{4: 1}$, or simply $ 4$.[/hide]

Following relations hold: $AC^2=AD\cdot AB; BC^2=BD\cdot AB$; dividing side by side: $\left(\frac{AC}{BC}\right)^2=\frac{BD}{AD}\ (\ 1\ )$.
$DE\bot AC\iff DE\parallel BC\implies \frac{AE}{CE}=\frac{AD}{BD}=\left( \frac{AC}{BC}\right)^2=\frac{4BC^2}{BC^2}=4$.

Best regards,
sunken rock

:) [quote="sunken rock"] $DE\bot AC\iff DE\parallel BC\implies \frac{EA}{EC}=\frac{DA}{DB}=\left( \frac{CA}{CB}\right)^2\implies$ $\boxed{\frac {EA}{EC}=k^2}$ , where $CA=k\cdot CB$ .[/quote]
Yes, it is the shortest proof !


[quote="ccy"]Three faces of a cube are randomly selected. What is the probability that they have a common vertex? Express your answer as a common fraction.[/quote]
[hide]this problem is familliar... but I still forgot how... 4/5?[/hide]

[hide="Hint"]How many faces of a cube meet at a single vertex?[/hide]

[hide]Ummmm
well by choosing a random face
you have four other faces that can potentially share the same vertex with another face. $\frac{4}{5}$.
Then you have two other faces that share the same vertex with the other two. $\frac{2}{4}$
$\frac{4}{5}\cdot\frac{2}{4}=\boxed{\frac{2}{5}}$[/hide]

[hide="this is probably wrong"]$6*5*4$ choices. 6 vertices, each with one possible combination.
$\frac{6}{6*5*4}=\boxed{\frac{1}{20}}$.
I somehow know that's wrong.[/hide]

[quote="ccy"]Three faces of a cube are randomly selected. What is the probability that they have a common vertex? Express your answer as a common fraction.[/quote]

so what was the answer? post it in hide

[hide]There are $\binom63=20$ ways of choosing 3 faces.

There are only 8 vertices so there are only 8 combinations.

$\frac{8}{20}=\frac25$[/hide]

[hide]
8 vertices, 8 ways for the faces to come together.

$\binom 63=20$ ways to choose the faces, so the answer is
$\frac{8}{20}=\frac{2}{5}$
$\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\frac 25}}}}}}}}}}}}}}}$[/hide]


geometry
geometric transformation
reflection
ratio
trapezoid
perpendicular bisector
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14718[/img]

 :(

Let the perpendicular bisector $ m$ of the segment $ AB$ cut the line $ r$ at $ M.$ Let $ C \in r$ on the same side of $ m$ as $ A$ be arbitrary, so that $ CA < CB,$ and let $ D$ be reflection of $ C$ in $ m.$ Internal and external bisectors of the angle $ \angle BCA$ cut the line $ AB$ at $ Z, Z'.$ Ratio $ \frac {PA}{PB} \equal{} \frac {CA}{CB}$ is constant for $ P$ on the C-Apollonius circle $ (K)$ of the $ \triangle ABC$ with diameter $ ZZ',$ centered at the midpoint $ K$ of the segment $ ZZ'.$ Since $ C \in r,$ $ \frac {CA}{CB}$ is minimum, when the C-Apollonius circle $ (K)$ is tangent to $ r$ at $ C.$ $ A$ is a reflection (inversion) of $ B$ in $ (K)$ and the other way around $ \Longrightarrow$ $ KA \cdot KB \equal{} KC^2,$ the right $ \triangle AKC \sim \triangle CKB$ are oppositely similar and the angles $ \angle CAK \plus{} \angle CBK \equal{} 90^\circ$ are complementary. $ ABCD$ is isosceles trapezoid,

$ \angle CAD \equal{} 180^\circ \minus{} (\angle CAK \plus{} \angle BAD) \equal{} 180^\circ \minus{} (\angle CAK \plus{} \angle CBK) \equal{} 90^\circ.$

Since $ \frac {CA}{CB}$ is minimum, $ \frac {DA}{DB} \equal{} \frac {CB}{CA}$ is maximum. To construct $ C, D,$ draw a circle with center $ M$ and radius $ MA \equal{} MB,$ intersecting $ r$ at $ C, D.$

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14779[/img]

 :thumbup:


Given three circles, construct a fourth one which bisects the circumference of the other three, ie the 4th circle cuts each of the other circles in diametrically opposite points.

Does something like this work... I think that even if the circles were randomly placed, you would first take the opposing endpoints of the diameter of each circle and then find a point equidistant from all 6 points and you should have your circle


calculus
integration
trigonometry
logarithms
derivative
real analysis
real analysis unsolved

Find general expressions for the two following integrals:

1) $ \int_0^{\pi/2} \frac{(\tan x)^p \minus{} 1}{\ln(\tan x)}\,dx$, 0 < p < 1.

2) $ \int_0^{\infty} \sin(x^n)\,dx$, n natural, n > 1.

carcul,  you must be psychic !  :o i asked   (1) just 3 weeks back on  grade 8's midterm quiz!

(1) here is the solution i gave later. i gave 0 to anyone who did not provide the answer like i wanted.
 lots of them did as a matter of fact ...i was   disappointed....  :noo:

$ \textbf I\equal{}\int_0^{\dfrac{\pi}{2}} \;\dfrac {\tan^p\,(x)\minus{}1 }{ \ln\,(\tan\, x)} \;\textbf dx\equal{}\int_0^p\int_0^{\dfrac{\pi}{2}}\tan^p\,(x)\;\textbf dx\;\textbf dp$
$ \left[ we\quad know \quad \int_0^{\dfrac{\pi}{2}}\;\tan^p\,(x)\;\textbf dx\equal{}\dfrac{\pi}{2\;\sin\,\left[\frac{\pi}{2}\,(p\plus{}1)\right]}\qquad (for\quad 0<p<1)\right]$
$ so,\quad \textbf I\equal{}\int_0^p\dfrac{\pi}{2}\text{cosec}\,\left[ \dfrac{\pi}{2}(p\plus{}1)\right]\;\textbf dp$
$ \equal{}\int_0^{\dfrac{\pi}{2}p} \sec\,(z)\;\textbf dz\qquad (where\quad z\equal{}\dfrac{\pi}{2}p)$
$ \equal{}\boxed{\ln\,\left(\tan\,\left[\dfrac{\pi}{4}(p\plus{}1)\right]\right)}$

(2) http://www.artofproblemsolving.com/Forum/viewtopic.php?t=158083

Yes, that's right. For the first one you can also differentiate in order to p.

Carcul can u please post ur solution using the differentiation please :)

I think this is what Carcul might have meant.

$ I(P) \equal{} \int_0^{\pi/2} \frac{(\tan x)^p \minus{} 1}{\ln(\tan x)}\,dx$

$ I'(P) \equal{} \int_0^{\pi/2} (\tan x)^p\,dx$ (By Leibnitz method)
and then continue with misan's steps.

Yes, that's right. Next you are to show that $ \int_0^{\pi/2}(\tan x)^p\,dx \equal{} \frac{\pi}{2} \sec\left(\frac{p \pi}{2}\right)$, 0 < p < 1.

2) Let $ I_n=\int_0^{\infty} \sin(x^n)\,dx$

Substitute $ u=x^n$ to get $ I_n=\frac{1}{n}\int_0^{\infty} u^{\frac{1}{n}-1}\sin u\,du$.

Now put $ J_n = \frac{1}{n}\int_0^{\infty} u^{\frac{1}{n}-1}e^{-iu}\,du$ so that $ I_n = -\mbox{Im}\left[J_n\right]$. Put $ w=iu$ in $ J_n$ to get

$ J_n = -\frac{1}{n}(-i)^{\frac{1}{n}}\int_0^{\infty} w^{\frac{1}{n}-1}\e^{-w}\,dw$
$ = -\frac{1}{n}(-i)^{\frac{1}{n}}\Gamma\left(\frac{1}{n}\right)$
$ = -\frac{1}{n}\left[\cos \left(\frac{\pi}{2n}\right)-i\sin\left(\frac{\pi}{2n}\right)\right] \Gamma\left(\frac{1}{n}\right)$
$ \Rightarrow I_n= -\mbox{Im}\left[J_n\right] = \frac{1}{n}\sin\left(\frac{\pi}{2n}\right)\Gamma\left(\frac{1}{n}\right) = \boxed{\Gamma\left(1+\frac{1}{n}\right)\sin\left(\frac{\pi}{2n}\right)}$