Divisibility
polynomial
IMO
number theory
IMO 1998
david monk

Determine all pairs $(x,y)$ of positive integers such that $x^{2}y+x+y$ is divisible by $xy^{2}+y+7$.

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

[b]Unchecked solution by nhat:[/b]

find all a,b in positive integer such that:
         $\frac {a^{2}*b+a+b}{a*b^2+b+7}$ in positive integer
I want the another solution which is different to the solution of IMO
here my solution
for this case a=b we easy to have the solution
if a<b then $a*b*(a-b)+a-7$<0 (contradition)
hence a>b
let a in $(m*b,(m+1)*b)$ (m i  positive integer)
then we easy to prove that :  m<$\frac {a^{2}*b+a+b}{a*b^2+b+7}$<(m+1) (contradition)
so a divisible to b
give $a=k*b$  (k in positive integer)
then
      $\frac{k^2*b^3+k*b+b}{k*b^3+b+7}$=$\frac{b*{k^2*b^2+k+1}}{k*b^3+b+7}$
if $(a,7)=1$ then $(b,k*b^3+b+7)=1$
so that $\frac {k^2*b^2+k+1}{k*b^3+b+7}$ in positive integer
and we esay to prove that k>b
so easy to prove that \[k\equiv 0 (mod b)\] so $k=q*b$ (b  >= 3and q i positive integer)
then $\frac {k^2*b^2+k+1}{k*b^3+b+7}$=${\{q^2*b^4+q*b+1}{q*b^4+b+1}$=$q$+$\frac {1-q}{q*b^4+b+1}$  (contradition)
if b divisible to 7 so that $b=7*u$ (u in positive integer)
and easy to prove that u=k (it's similar for this case two)
so we have the solution $a=7*k^2$,$b=7*k$
 my solution hasor hasn't correct :D  :D  :D

The solution given isn't quite correct... there are at least two missing solutions: $(11,1)$ and $(49,1)$.

$11^2+11+1 = 133$, which is divisible by $11+1+7 = 19$.
$49^2+49+1 = 2451$, which is divisible by $49+1+7 = 57$.

My full solution...

If $(xy^2+y+7)|(x^2y+x+y)$, then there exists a positive integer $k$  such that $k(xy^2+y+7) = x^2y+x+y$ , or

$7k-y = (xy+1)(x-ky)$ .

First we claim that all solutions have both sides nonnegative.

Suppose $7k-y < 0$ . But then we have

$|7k-y| < |y| < |xy+1| < |xy+1||x-ky|$ ,

which is a contradiction. So both sides are nonnegative. Now consider the two cases: (1) both sides are zero, and (2) both sides are positive.

Case 1: Both sides are zero.

Then we have $7k-y = 0 \Rightarrow y = 7k$ . And also $(xy+1)(x-ky) = 0$, but since $xy+1 > 0$ we know $x-ky = 0 \Rightarrow x = ky = 7k^2$ . So we have all solutions of the form $(x,y) = (7k^2, 7k)$ .

Case 2: Both sides are positive.

Then $x > ky$ so the RHS is positive. So we have $7k > 7k-y = (xy+1)(x-ky) > (xy+1) > y^2k+1$. Hence $y < 3$ . So we check $y = 1, 2$ .

For $y = 1$ , we have $(x+8)|(x^2+x+1)$. Since

$\frac{x^2+x+1}{x+8} = x-7+\frac{57}{x+8}$ ,

we have $(x+8)|57 \Rightarrow x = 11, 49$ , giving the solutions $(11,1)$ and $(49,1)$ .

For $y = 2$ , we have $(4x+9)|(2x^2+x+2) \Rightarrow (4x+9)|(4x^2+2x+4)$ . Since

$\frac{4x^2+2x+4}{4x+9} = x-\frac{7x-4}{4x+9}$ ,

so $(4x+9)|(7x-4)$ . But since $2(4x+9) > 7x-4$ , we must have $4x+9 = 7x-4$ , which does not have an integer solution.

Hence our only solutions are $(x,y) = (11,1); (49,1); (7k^2, 7k)$ for positive integers $k$ .

Here my solution I hope there are no mistakes
 Let $ab^2+b+7=n$ then $ab^2\cong -b-7 (mod n) \rightarrow a^2b^2\cong -ab-7a (mod n)$
                              and  $a^2b\cong -a-b (mod n) \rightarrow a^2b^2\cong -ab-b^2(mod n)$ from here
$\rightarrow 7a \cong b^2(mod n)$ so $n$ divides both $(ab^2+b+7)$ and $a(b^2-7a)$ so
$(ab^2+b+7)-a(b^2-7a)=b+7+7a^2$ then n divides $b+7+7a^2$ , too.
so $b+7+7a^2\geq ab^2+b+7 \rightarrow 7a \geq b^2$. Lets share it in two parts
1) $7a-b^2=0 \rightarrow (a,b)=(7k^2,7k)$
2) $7a-b^2>0$
Since we have found that $n|(7a-b^2)$ then $7a-b^2\geq ab^2+b+7$ from here we will have
$(7-b^2)(a+1)\geq b+14$.  $b+14>0$ then $(7-b^2)(a+1)>0$ also $(a+1)>0$ then we have to have $(7-b^2)>0$ here $b$ is positive integer then $b= 1$ or $2$ otherwise  inequality will lead to contradiction
If $b=1$ then the problem will become as $\frac{a^2+a+1}{a+8}$
Let $k=a+8$ then $a^2\cong -a-1 (mod k) (i)$
$a\cong -8(mod k)\rightarrow a^2\cong 64 (mod k) (ii)$
from $(i)$ and $(ii)$ we will have $-a-1\cong 64 (mod k)$ and $a+65$ is divisible by $k$ then

$\frac{a+65}{a+8}=\frac{a+8}{a+8} + \frac{57}{a+8}$  so $\frac{57}{a+8} (iii)$ is an integer, too.
so there are just two solutions for $(iii)$ those are $a=11$ or $49$ ( think didisors of $57$)
If $b=2$ by simliar of previous we will show that there is no solution for $b=2$
 So the solutions are $(a,b) = (11,1), (49,1), (7k^2,7k)$
We are done

If $ x < y$ then it is easy to see there are no solutions.
By the division algorithm, we can let $ x \equal{} ay \plus{} b$, where $ 0 \le b < y$.
Then we have
$ a^2y^3 \plus{} 2aby^2 \plus{} (b^2 \plus{} a \plus{} 1)y \plus{} b$ is divisible by $ ay^3 \plus{} by^2 \plus{} y \plus{} 7$
Let the ratio between these two numbers be $ a \plus{} k$, then
$ a^2y^3 \plus{} 2aby^2 \plus{} (b^2 \plus{} a \plus{} 1)y \plus{} b \equal{} (a \plus{} k)(ay^3 \plus{} by^2 \plus{} y \plus{} 7)$
$ aby^2 \plus{} b^2y \plus{} y \plus{} b \equal{} aky^3 \plus{} kby^2 \plus{} ky \plus{} 7a \plus{} 7k$
If $ k \equal{} 0$, then $ aby^2 \plus{} b^2y \plus{} y \plus{} b \equal{} 7a$
Case 1:$ b \equal{} 0$
Then $ y \equal{} 7a$ and we have the solution $ (7a^2,7a)$
Case 2: $ b \ge 0$
Then either $ y \equal{} 1$ or $ 2$. If $ y \equal{} 1$, then $ b$ can't be more than $ 7$, we try out all values from $ 1$ to $ 7$ for $ b$. $ b \equal{} 4$ and $ b \equal{} 6$ works, this gives the solutions $ (11,1)$ and $ (49,1)$. If $ y \equal{} 2$ then $ b \equal{} 1$ and we have no solution.
If $ k > 0$, then since $ b < y$, we must have $ k \equal{} 1$ since if $ k > 1$ then the left side is less than the right side (we would have $ aky^3 > aby^2 \plus{} b$, $ kby^2 > b^2y$, $ ky > y$). But if $ k \equal{} 1$, then $ b \equal{} 7a \plus{} 7 \plus{} (y \minus{} b)(ay^2 \plus{} by)$. But $ y \minus{} b \ge 1$ since $ b < y$, a contradiction.
So $ k < 0$. Then we must have $ y \equal{} 1$ or the left side would be greater than the right. Since $ 0 \le b < y$, we must have $ b \equal{} 0$. Thus, $ 1 \equal{} ak \plus{} 7a \plus{} 8k$, or $ (a \plus{} 8)(k \plus{} 7) \equal{} 57$. We find $ (a,k) \equal{} (11, \minus{} 3)$ and $ (49, \minus{} 6)$ work, giving the solutions $ (11,1)$ and $ (49,1)$.
So the three solutions are $ (7a^2,7a)$, $ (11,1)$, and $ (49,1)$

I thought to myself this morning - "I haven't really tried any IMO problems". I decided to remedy that by solving this one.

First, suppose that $ gcd(y,7)\equal{}1$. Then the problem can be restated as follows.

$ xy^2\plus{}y\plus{}7 \, | \, x^2y\plus{}x\plus{}y$

$ \Leftrightarrow xy^2\plus{}y\plus{}7 \, | \, y(x^2y\plus{}x\plus{}y)\minus{}x(xy^2\plus{}y\plus{}7)$

$ \Leftrightarrow xy^2\plus{}y\plus{}7 \, | \, y^2\minus{}7x$

It is easy to see that since $ x,y$ are positive, then $ xy^2\plus{}y\plus{}7 > y^2\minus{}7x$. Then we must have either $ y^2\equal{}7x$ or $ y^2\minus{}7x < 0$. The first case cannot arise since $ gcd(7,y)\equal{}1$ so we must have $ y^2\minus{}7x < 0$.

Then our condition is equivalent to:

$ xy^2\plus{}y\plus{}7 | 7x \minus{} y^2$ and we must have $ 7x \minus{} y^2 \ge xy^2 \plus{} y \plus{} 7$

The second condition implies that $ y < 3$ since otherwise we have:

$ (7x \minus{} y^2) \minus{} (xy^2\plus{}x\plus{}y) \le 7x \minus{} 9 \minus{}9x \minus{}x\minus{}3 \equal{} \minus{}3x \minus{} 12 < 0$.

Hence $ y\equal{}1$ or $ y\equal{}2$.

[b]Case 1:[/b] $ y\equal{}1$.

Our problem restates as $ x\plus{}8 | 7x\minus{}1$ and so we write:

$ \dfrac{7x\minus{}1}{x\plus{}8} \equal{} k$.

It is clear that $ k$ must be positive. We rewrite this as:

$ x\equal{}\dfrac{8k\plus{}1}{7\minus{}k}$ and since $ x$ is positive, the denominator must be positive and so $ k < 7$. Substituting the values $ k\equal{}1,2,3,4,5,6$ into this expression in turn shows that only $ k\equal{}4$ and $ k\equal{}6$ give integer values for $ x$. These are $ x\equal{}11$ and $ x\equal{}49$.

So our two solutions are $ (x,y)\equal{}(49,1), (11,1)$. Checking in the original problem reveals that they work.

[b]Case 2:[/b] $ y\equal{}2$.

Starting in the same way as above we get $ x\equal{} \dfrac{9k\plus{}4}{7\minus{}4k}$. In a similar way to above we must have $ k < 2$, but substituting $ k\equal{}1$ does not give integer $ x$. Hence, no solutions in this case.

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

Now we must consider when $ gcd(7,y)\equal{}7$. This case is easy. Let $ y\equal{}7t$. Any pair $ (x,y)$ which satisfies the problem must also satisfy:

$ xy^2\plus{}y\plus{}7 | y(x^2y\plus{}x\plus{}y)\minus{}x(xy^2\plus{}y\plus{}7)$

$ xy^2\plus{}y\plus{}7 | y^2\minus{}7x$

We cannot have $ y\equal{}1$ or $ y\equal{}2$ as above since $ gcd(y,7)\equal{}7$. Thus we must have the other possibility, $ y^2\equal{}7x$.

Then we have $ 7x\equal{}y^2\equal{}49t^2$

So $ x\equal{}7t^2$

Then our solution is $ (x,y)\equal{}(7t^2,7t)$.

Checking this in the original problem:

$ \dfrac{x^2y\plus{}x\plus{}y}{xy^2\plus{}y\plus{}7}$

$ \equal{}\dfrac{343t^5\plus{}7t^2\plus{}7t}{343t^4\plus{}7t\plus{}7}$

$ \equal{}t$ which is an integer.

Hence all solutions are given by:

$ (x,y) \equal{} (11,1), (49,1), (7t^2,7t)$

[quote="KrazyFK"]
First, suppose that $ gcd(y,7)\equal{}1$. Then the problem can be restated as follows.

$ xy^2\plus{}y\plus{}7 \, | \, x^2y\plus{}x\plus{}y$

$ \Leftrightarrow xy^2\plus{}y\plus{}7 \, | \, y(x^2y\plus{}x\plus{}y)\minus{}x(xy^2\plus{}y\plus{}7)$

$ \Leftrightarrow xy^2\plus{}y\plus{}7 \, | \, y^2\minus{}7x$
[/quote]
What is the importance of the statement $\gcd(y,7)=1$?  Can you not perform the same manipulations even if $\gcd(y,7)=7$?

Also, my solution involved using the division algorithm to get $\frac{a^2b+a+b}{ab^2+b+7}=\frac{a}{b}+\frac{b^2-7a}{ab^3+b^2+7b}$, can this only be done if $a\geq b$?  From there I get $b^2-7a=0\implies b^2=7a\implies a=7k,b=7k^2$.  Why can't I use the division algorithm to get the same result if $b<a$?

The statement of $\gcd(7, y)=1$ is not actually used for the algebraic manipulations, but to set aside a case that forces $y^2-7x$ to be positive (in other words, it's used right after your quote).

Oh, so from $ \frac{a^{2}b+a+b}{ab^{2}+b+7}=\frac{a}{b}+\frac{b^{2}-7a}{ab^{3}+b^{2}+7b} $, we must split it into cases.  So I forgot the case where $b^2-7a<0$, so $7a-b^2>\frac{ab^{3}+b^{2}+7b}{b}$.

Yep.

The $b^2-7a<0$ case leads to the two extra solutions, $(49, 1); (11, 1)$.

Thanks, but just from the division algorithm, what is the concrete rational that we must have $ 7a-b^{2}>\frac{ab^{3}+b^{2}+7b}{b} $ (division by $b$ on the RHS)?  Is it just that if we factor out the $\frac1b$ on the RHS of $ \frac{a^{2}b+a+b}{ab^{2}+b+7}=\frac{a}{b}+\frac{b^{2}-7a}{ab^{3}+b^{2}+7b} $, we just want $a+\frac{b^2-7a}{ab^2+b+7}$ to be an integer?

That's right. From your division algorithm, if we want the LHS to be an integer, then $b$ has to divide $a+\frac{b^2-7a}{ab^2+b+7}$, so it must be an integer, and so $ab^2+b+7$ must divide $7a-b^2$.

Define $S=\left \{(a,b)\in \mathbb{Z}^{+}|\displaystyle \frac{a^2b+a+b}{ab^2+b+7}\in \mathbb{Z}^{+}\right \}$

We note that (i)

$ab^2+b+7|a^2b+a+b\Rightarrow ab^2+b+7|a(b(a^2b+a+b)-a(ab^2+b+7))$

then $ab^2+b+7|ab^2-7a^2\Rightarrow ab^2+b+7|ab^2-7a^2-(ab^2+b+7)$

now we have that

$ab^2+b+7|-7a^2-b-7\Rightarrow ab^2+b+7|7a^2+b+7$

note that $ab^2+b+7>0$ and $7a^2+b+7>0$

therefore $ab^2+b+7\leq 7a^2+b+7\Rightarrow b^2\leq 7a$

if $b^2=7a$ then $a=7t^2$ and $b=7t$ for all $t\in \mathbb{Z}^{+}$

therefore $(7t^2,7t)\in S$

now $b^2<7a$

By (i) we have that

$ab^2+b+7|b^2-7a\Rightarrow ab^2+b+7|7a-b^2$ because $b^2<7a$

now because $b^2<7a$ we have $7a-b^2>0$ and note that $ab^2+b+7>0$

therefore $ab^2+b+7\leq 7a-b^2$ that is impossible for $b\ge 3$ therefore $b<3$

We have two cases:

Case 1: $b=1$

we have $a+8|a^2+a+1\Rightarrow a+8|a^2+a+1-a(a+8)+7(a+8)$

therefore $a+8|57=1\cdot 3\cdot 19$ if we prove all the divisors we have that $(49,1)$ and $(11,1)$ are solutions.

therefore $(11,1),(49,1)\in S$

Case 2: $b=2$

we have $4a+9|2a^2+a+2\Rightarrow 4a+9|4(2(2a^2+a+2)-a(4a+9))+7(4a+9)=79$

therefore $4a+9|79$ no solutions here

Finally we have that $S=\{(11,1),(49,1),(7t^2,7t)\}$

Thanks

Suppose $S=\tfrac{x^2y+x+y}{xy^2+y+7}$ then \[y\left(S-\frac{x}{y}\right)=\frac{y^2-7x}{xy^2+y+7}\] must also be an integer. If $y^2-7x=0$ then $y^2=7x$ which implies $7\mid y$ so we can let $y=7a$ and so $x=7a^2.$ Note that $(7a^2,7a)$ is a solution for all positive integers $a.$
\\$~$\\
If $y^2-7x>0$ then $y^2-7x\ge xy^2+y+7.$ Also, $y^2\le xy^2$ and $-7x<0<y+7$ which is a contradiction. Thus, $y^2-7x<0$ which implies $7x-y^2\ge xy^2+y+7$ so $(x+1)y^2+y+7\le 7x.$ If $y\ge 3$ then $(x+1)y^2\ge 9(x+1) > 7x$, contradiction. Thus, $y=1$ or $2.$
\\$~$\\
If $y=1$ then $x+8 \mid x^2+x+1$ and we see that the solutions are $(49,1)$ and $(11,1).$ If $y=2$ then $4x+9\mid 2x^2+x+2$ and we see that there are no solutions. Thus, our solutions are $(49,1),(11,1),(7a^2,7a)$ for some $a.$

We claim the solutions are $(x,y)=(7k^2,7k),(49,1),(11,1)$ where $k \in \mathbb{N}$. These can easily be seen to work. Note that $$xy^2+y+7|y(x^2y+x+y)-x(xy^2+y+7) = y^2 - 7x.$$ If $y^2-7x=0$ we get the solutions $(x,y)=(7k^2,7k)$. If $y^2-7x>0$, then $y^2 < xy^2 + y +7 \le y^2 -7x$, a contradiction. If $y^2-7x<0$, then $xy^2+y+7 < 7x - y^2 \implies y^2 < 7 \implies y=1,2$. It is easy to see that these give the solutions $(x,y)=(11,1),(49,1)$, and so we have checked all cases so we are done.

[i]Solution:[/i] We will show that $(a,b) = (7k^2,7k)$ for all $k \in \mathbb{Z}_{> 0}$ and $(a,b) = (11,1), (49,1)$ are the only possible solutions. It is easy to check that they work. We now show they are the only solutions. Call the problem condition $(1)$.

The basic idea for the whole solution is to eliminate $ab^2$ from the other side of the divisibility.
\begin{align*}
        ab^2+b+7 \mid a^2+a+b \implies ab^2+b+7 &\mid b^2-7a \tag{2}
\end{align*}

Now we will just consider 3 cases and get bounds to finish the problem.

[b]Case 1:[/b] $b^2 > 7a$.

The divisibility condition would now imply that
    \[ab^2+b+7 \le b^2-7a \implies 0< b^2(a-1)+b+7 \le -7a^2 <0\]
which is a contradiction.
---------
[b]Case 2:[/b] $b^2 = 7a$.

As we are working over integers the case condition already gives us $(a,b) = (7k^2,7k)$ for any positive integer $k$. We already know it works.
-----
[b]Case 3:[/b] $b^2 < 7a$.

We will show $b \in \{1,2\}$ in this case. From $(2)$ we get that
    \[ab^2+b+7 \le 7a-b^2 \iff 0<a \le \frac{-(b^2+b+7)}{b^2-7}\]
which is only possible when $b = 1,2$. Putting $b=1$ in $(1)$ we get $a=11, 49$. $b = 2$ just fails to give any solution. We already know these solutions pairs work so we are done. $\blacksquare$

[hide=Casing]Given, $\left(xy^2+y+7\right)\mid\left(x^2y+x+y\right)\Rightarrow\left(xy^2+y+7\right)\mid\left(x^2y^2+xy+y^2\right)$ (Multiplying RHS by $y$). So, $\left(xy^2+y+7\right)\mid\left(\left(x^2y^2+xy+y^2\right)-x\left(xy^2+y+7\right)\right)$ which finally gives us $\left(xy^2+y+7\right)\mid\left(y^2-7x\right)$. Since $x>0$, clearly $xy^2+y+7<y^2<y^2-7x$. Hence, $y^2-7x$ cannot be positive, as if it would, then $0<y^2-7x<xy^2+y+7$ would result in a contradiction. So, we consider the following cases:
Case 1) $y^2-7x<0\Rightarrow7x-y^2>0\Rightarrow7x-y^2>xy^2+y+7\Rightarrow x\left(7-y^2\right)>y^2+y+7>0$. This gives $7<y^2\Rightarrow y=1,2$.
Subcase 1) $y=1\Rightarrow\left(x+8\right)\mid\left(x^2+x+1\right)\Rightarrow\left(x+8\right)\mid\left(x^2+x+1-x^2-8x\right)\Rightarrow\left(x+8\right)\mid\left(1-7x\right)\Rightarrow\left(x+8\right)\mid\left(1-7x+7x+56\right)\Rightarrow\left(x+8\right)\mid57$. This leaves us with $x+8=19,57$ and subsequently $x=11,49$.
Case 2) $y^2-7x=0\Rightarrow y^2=7x\Rightarrow7\mid y^2\Rightarrow7\mid y\Rightarrow y=7k\Rightarrow x=7k^2$ where $k$ is a positive integral parameter. This solution always satisfies.

So, the solution set is $\{\left(11,1\right),\left(49,1\right),\left(7k^{2},7k\right)\}$ ($k$ is a positive integral parameter). [/hide]

We have
\[xy^2+y+7\mid x^2y+x+y\mid x^2y^2+xy+y^2\rightarrow xy^2+y+7\mid x(-y-7)+xy+y^2=y^2-7x.\]
We will now casework on the sign of $y^2-7x$.

[b]Case 1,[/b] $y^2-7x>0$. Then $xy^2+y+7\le y^2-7x$, absurd.
[b]Case 2,[/b] $y^2-7x<0$. Then $xy^2+y+7\le 7x-y^2$. Thus $y\le 2$.
[b]Subcase 1,[/b] $y=1$. Then $x+8\mid 7x-1$. So $x+8\mid 57$, so $x+8=\in\{19,57\}$ so $x=11$ or $x=49$, both of which work.
[b]Subcase 2,[/b] $y=2$. Then $4x+9\mid 7x-4$. But $7x-4$ is supposed to be positive, so $4x+9=7x-4$, absurd.
[b]Case 3,[/b] $y^2-7x=0$. Then $y$ is a multiple of $7$, say $7k$, then $x=7k^2$. So
\[7k^2(7k)^2+7k+7=343k^4+7k+7\mid (7k^2)^2(7k)+7k^2+7k=343k^5+7k^2+7k,\] which always works.

So our solution set is $(11,1)$, $(49,1)$, and $(7k^2,7k)$ for any positive integer $k$.

We want
		\[xy^2+y+7\mid y(x^2y+x+y)-x(xy^2+y+7)=y^2-7x\]
		If $y^2>7x$, then we must have
		\[xy^2+y+7\le y^2-7x\]
		which never happens (easy to see uppon dividing both sides by $xy^2$). If $y^2=7x$, we get $y=7k, x=7k^2$, which works. If $7x>y^2$, then
		\[xy^2+y+7\le 7x-y^2\implies xy^2\le 7x\implies y=1,2\]
		and routine calculations deliver the pairs $(11, 1)$, and $(49, 1)$.

We get that $\frac{x^2y+x+y}{xy^2+y+7} = k.$ Clearing the denominator and rearranging, we get $y=xy(ky-x) + (ky-x)+7k$ and factoring we get $y=(xy+1)(ky-x)+7k$. Clearly, if $ky-x>0$, then the right hand side is larger than the left hand side. If $ky-x=0$, then $y=7k$ meaning $x=7k^2$. So all pairs of the form $(7k^2, 7k)$ work for positive integers $k$. Now if $ky-x<0$, then $x>ky$. Let $x=ky+m$. Then $7k-mky^2-m^2y-m=y$. Since $y$ is positive, we get $7k>m(ky^2+my+1).$ If $y\geq 3$, then we have $9k$ added up which won't work. So $y=1,2$. If $y=1$, $7k-mk-m^2-m=1$. This rearranges to $k = \frac{m^2+m+1}{7-m}$. This means $1\leq m \leq 6$. Testing each one, the ordered pairs we get are $(x,y)=(49, 1), (11,1).$ Now if $y=2$, we get $7k-4mk-4m^2-m=2$. The only possible value is $m=1$ which gives $3k=7$ which is not an integer. So our final answer is the ordered pairs $\boxed{(7k^2, 7k), (49,1), (11,1)}.$

I claim the solutions are $(49,1)$, $(11,1)$ and $(7n^2,7n)$ for any positive integer $n$.

First rewrite the divisibility as $$xy^2+y+7 \mid y(x^2y+x+y)-x(xy^2+y+7) = y^2-7x$$

Now we have two cases.

Case 1: $y^2-7x \geq 0$.

Note that since $x,y \geq 1$ we clearly have $xy^2+y+7+7x> y^2$ so we must have $y^2- 7x = 0$ or $y^2 = 7x$. If we let $x = 7n^2$ where $n$ is any positive integer then $y = 7n$ and we have our first pair $(7n^2,7n)$.

Case 2: $y^2 - 7x < 0$.

In this case, we are essentially looking at $$xy^2+y+7 \mid 7x-y^2$$ Thus we must have $xy^2+y+7 \leq 7x-y^2$ which rearranges to $xy^2+y^2+y+7 \leq 7x$. Note that in order for this to be true, we must have $y \leq 2$ or else $xy^2 > 7x$. Now we simply just check the cases where $y=1$ and $y=2$.

If $y = 1$, then $x+8 \mid 7x-1$ which is equivalent to $x+8 \mid 7(x+8)-(7x-1) = 57$ from which we extract $x = 11$ or $x = 49$. This gives us the pairs $(49,1)$ and $(11,1)$.

If $y = 2$, then $4x+9 \mid 7x-4$ which is equivalent to $4x+9 \mid 7(4x+9)-4(7x-4) = 79$ from which we see there are no positive integer solutions for $x$.

We've exhausted all cases and we're done.

We claim the answer is $(7n^2,7n),(49,1),(11,1).$ These clearly all work. Note that $$x^2y+x+y\equiv 0\pmod{xy^2+y+7}$$ $$x^2y^2+xy+y^2\equiv 0\pmod{xy^2+y+7}$$ $$y^2-7x\equiv0\pmod{xy^2+y+7}.$$ (at the second step we subtracted $x$ times the modulus.) The point of this is that now everything is linear in $x$.\\

Case 1: $y^2-7x>0$. Then, we have $$y^2-7x<y^2\leq xy^2<xy^2+y+7,$$ hence there are no solutions as the expression is positive and smaller than the modulus.\\f

Case 2: $y^2-7x=0$. Then, clearly $(x,y)=(7n^2,7n)$, which clearly works.\\

Case 3: $y^2-7x<0$. Then, we have $$7x-y^2=c(xy^2+y+7).$$ We can rearrange this to $$x=\frac{y^2+cy+7c}{7-cy^2}.$$

Since $cy^2\leq 6$, there are not that many cases. If $y=1$, we have $$x=\frac{8c+1}{7-c}.$$ For $c=4$ and $c=6$ this gives $x=11$ and $x=49$, and other values of $c$ up to 6 do not give integer $x$. If $y=2$, we have $$x=\frac{9c+4}{7-4c},$$ here only $c=1$ is allowed which does not give an integer anyway. Finally, $y\geq3$ is not possible, hence done.

We have $$0\equiv x^2y+x+y\equiv x^2y^2+xy+y^2-(x(xy^2+y+7))\equiv y^2-7x\pmod{xy^2+y+7}.$$ Due to size reasons $y^2-7x\le0$, if it equals 0 then the solution set is $(7n^2,7n)$. If it's <0 we must have $xy^2+y+7\le7x-y^2<7x\implies y\in\{1,2\}$; easy calculation reduces it into $\boxed{(x,y)\in\{(7n^2,7n),(11,1),(49,1)\}.}\blacksquare$

wait im not sure why but why are my sols so short lol on overleaf typing they take up like 10 lines

Note that this implies that $xy^2 + y + 7$ divides $y(x^2y + x + y) - x(xy^2 + y + 7) = y^2 - 7x$.
This implies that either $y^2 = 7x$, or that $x \ge 2, y^2 \ge 16$ can not hold.
In the first case, we get the solution set $(x, y) = \left(7k^2, 7k\right)$ which can be seen to hold.
We now bash out the remaining cases. It can be seen that if $x = 1$, then there are no solutions.
Then, if $y = 1$, it follows that $x + 8 \mid 1 - 7x$ so $x \in \{11, 49\}$. We can check that $(x, y) = (49, 1)$ and $(x, y) = (11, 1)$ works.
If $y = 2$, then $4x + 9 \mid 4 - 7x$ which implies $4x + 9 \mid 3x - 13$ which has no solutions.
If $y = 3$, then $9x + 10 \mid 9 - 7x$ which has no solutions.

Rewrite the condition as $$xy^2+y+7 \mid (x^2y^2+xy+7x) - (x^2y^2+xy+y^2) = 7x-y^2.$$ Now we have a few cases:

If $y^2 > 7x$, then $y^2 - 7x < y^2 < xy^2+y+7$, which is obviously impossible. If $y^2 < 7x$, then $$xy^2+(y+y^2) + 7 \leq 7x,$$ which implies $y^2 \leq 7$ and $y \in \{1, 2\}$. This yields the solutions $(49, 1)$ and $(11, 1)$.

If $y^2 = 7x$, then the entire curve of solutions $(7n^2, 7n)$ can be checked to work.

Bruh, the first time I solved the problem, I solved it for $x^y + x + y \mid xy^2 + y + 7$ ;-; :stretcher: . Also, these edge cases are so hard for me to find without making sillies.

I claim that the answers are $(x,y)=(11,1)$, $(49,1)$ and $(7n^2,7n)$ for any $n\in\mathbb N$.

Firstly, we have $xy^2 + y + 7 \mid x^2y + x + y \mid x^2y^2+xy+y^2\equiv (x^2y^2+xy+y^2)-x(xy^2 + y + 7) = y^2 - 7x$.

Now if $y^2 > 7x$, then we get that $xy^2 + y + 7 \le y^2 - 7x$ which gives us $y^2 -y(1+x^2) -(7+7x) \ge 0$. But then the discriminant must be $\le 0$, that is $(1+x^2)^2 + 4(7+7x) \le 0$ which clearly has no solution.

Now if $y^2 < 7x$, then we get that $xy^2 + y + 7 \ge y^2 - 7x$. Then we get that $x^2y -7x +(y^2+y+7) \le 0$. This means that the discriminant must be $\ge 0$, that is $7^2 -4y(y^2+y+7)\ge 0 \implies 4y(y^2+y+7)\le 49$. This gives only solution $y=1$ since the left side is strictly increasing and it exceeds $49$ for $y=2$. For $y=1$, from the problem statement we get that $x+1+7 \mid x^2 + x + 1 \equiv (x^2+x+1)-x(x+8) = -7x+1 \equiv (-7x+1)+7(x+8) = 57$. This gives us that $x\in \left\{11,49\right\}$ each of which are solutions.

Now for the other case when $y^2 = 7x$, then clearly $7\mid y$ and so $49n^2 = 7x \implies x = 7n^2$ which is another solution.

Notice the LHs also divides
\[y(x^2y+x+y)-x(xy^2+y+7) = y^2-7x.\]

If $y^2-7x=0$, we have the solutions $\boxed{(7k^2,7k)}$. Otherwise, we notice
\[|xy^2+y+7| > |y^2-7x|,\]
implying there are no solutions, unless $y=1,2$ where we get the pairs $\boxed{(11,1),(49,1)}$. $\blacksquare$


Prove that $\exists k\in R$ such that:
$|\sum_{i=1}^n \frac{\sin i}{i}|<k$ for all natural numbers $n$

are you sure there really exist such a "k"?  :?
in fact,we can prove for any positive interger " t" ,max{sin t,sin(t+1),sin(t+2)}>=0.5
so |sin1/1|+|sin2/2|+……>=0.5(1/3+1/6+1/9+……)=1/6(1+1/2+1/3+……）
it seems to have no limit....... :huh:

Hey,$\sum_{i=1}^n|\frac{\sin i}{i}|$ is different from $|\sum_{i=1}^n\frac{\sin i}{i}|$


inequalities
trigonometry
inequalities unsolved

[b][size=167] ;) Solve the inequality (2sinx - √3)(2sinx - 1)>0. Where x is greater than or equal to zero but less than or equal to 2∏. [/b][/size] :?:

That's obviously false....
Take $x=\frac{\pi}{2}$

Feemind, you are right. It is a inequation and not a inequality ![hide="Here is the answer of the inequation (another problem)."]$\left(2\sin x-\sqrt 3\right)(2\sin x-1)\le 0\ ,\ x\in \left[0,2\pi\right]\Longleftrightarrow x\in \left[\frac{\pi}{6},\frac{\pi}{3}\right]\cup\left[\frac{2\pi}{3},\frac{5\pi}{6}\right]\Longleftrightarrow$
$\frac{\pi}{6}\le \left|x-\frac{\pi}{2}\right|\le \frac{2\pi}{3}\Longleftrightarrow \left|\left|x-\frac{\pi}{2}\right|-\frac{5\pi}{12}\right|\le \frac{\pi}{4}\ .$[/hide]


logarithms
calculus
calculus computations

hello folks.  :D

can you show

$ \prod_{k\equal{}1}^\infty e^{\minus{}2\plus{}\dfrac{2}{k}}\cdot\left (1\plus{}\frac{2}{k} \right )^k\;\equal{}\;\boxed{
\frac{e^{2\gamma\plus{}3}}{2\pi}}$

where $ \gamma$ is euler's constant

Now we will check your knowledge  of asymptotic in analysis  :lol: :
I will use next symptotical expansion :
$ \sum _{1\leq k \leq n}k \ln k = (n^{2} + n + \frac {1}{6})\frac {\ln n}{2} - \frac {1}{4n^{2}} + C_{1} + 0(1)$
take $ ln$ of your product and consider :
$ \sum_{k = 1}^{N} - 2 + \frac {2}{k} + k(\ln(k + 2) - \ln k)$
$ 1)\sum_{k = 1}^{N} - 2 = - 2N$
$ 2)\sum_{k = 1}^{N} = 2(\ln N + \gamma + o(1))$
$ 3)\sum_{k = 1}^{N} - k\ln k = - (N^{2} + N + \frac {1}{6})\frac {\ln N}{2} + \frac {1}{4N^{2}} - C_{1} + 0(1)$
$ 4)\sum_{k = 1}^{N}k\ln(2 + k) = \sum_{k = 1}^{N}(k + 2)\ln(2 + k) - 2\ln(2 + k) = \sum_{k = 1}^{N + 2}k\lnk - 2\ln(N + 2)!$
$ \sum_{k = 1}^{N + 2}k \ln k = ((N + 2)^{2} + N + 2 + \frac {1}{6})\frac {\ln (N + 2)}{2} - \frac {1}{4(N + 2)^{2}} + C_{1} + 0(1)$

$ 5) N! \approx \sqrt {2\pi N} \left( \frac {N}{e} \right) ^{N}$
just susbtitue and $ n \to \infty$ and  (if your problem is true) then you are done.  :)

(mine and important is that when you calculate :
$ \sum_{1}^{N} k(\ln(k + 2) - \ln k) =$ here you will not have $ C_{1}$ :) .)


Hi everyone!
Can you give me a link to where i can find the piano and violin score of Haliluah??
Thanks in advance.

While I wouldn't know the answer even if the details were there, there's no way your post is specific enough for anyone to know what piece of music you're asking for. That's a very common title, and the piece is almost certainly choral and not originally for the instruments you're asking for.

Not sure if such a thing exists. [url=http://www.mutopiaproject.org/cgibin/piece-info.cgi?id=459]Here[/url] is the full thing if you want to arrange it yourself.

Does anybody know where to get a copy of the violin part, or even orchestra, part of the Brahms Violin Concerto in D Major, Op. 77???

I want to play the piece real bad...

Thank you beforehand! :thumbup:

[quote="shobber"]This is a really famous one, so famous that you can buy it in the big book stores.[/quote]

Actually, my nearby bookstores have a section as big as a toenail for sheet music. My nearby university library has midget sheets that are so cramped and unreadable it isn't even funny.

I don't find this piece very famous, either.

[quote="Micheal Niland"]This is for hwenterprise

Concerning score of Brahms Violin Concerto.You can download a free piano and violin version at http://www.sheetmusicarchive.net.

A moderately priced edition of the full scores of all the Brahms concertos has been published by Dover in a single volume.It should be available in any of the large bookstores.Also you could try some of the internet sites that sell second hand books.
There are many full scores online at http://www.dib.indiana.edu/variations/scores/symphonic.html ,but unfortunately the Brahms violin concerto is not amongst them.
.

I hope this information helps you.Best wishes!

Michael[/quote]

Thanks, Micheal, that info was helpful...though now I just found how hard that piece really is!!!

wow, thats a really famous song that ive never heard of before!  Cool score hwenterprise! :D

Does anybody know where to get a copy of the quitar part of any track of EVITA (Adrew Lloyd Weber)


A frog is placed at the origin on the number line, and moves
according to the following rule: in a given move, the frog advances
to either the closest point with a greater integer coordinate that
is a multiple of 3, or to the closest point with a greater integer
coordinate that is a multiple of 13. A $ \emph{move sequence}$ is a
sequence of coordinates which correspond to valid moves, beginning
with 0, and ending with 39. For example, 0, 3, 6, 13, 15, 26, 39 is
a move sequence. How many move sequences are possible for the frog?

All plausible points that the frog can eventually get to are

0,3,6,9,12,[b]13[/b],15,18,21,24,[b]26[/b],27,30,33,36,[b]39[/b].

There are several types of paths:

i. those that go to 39 without landing on 13 or 26: in this case, we must land on 27-36 before being able to jump directly to 39, so there are $ 4$ ways.
ii. those that land on 26 and 39 but never on 13: we must land on 15-24 before going to 26, and then we have $ 5$ choices to be until 39, for a total of $ 4\cdot5\equal{}20$.
iii. those that land on 13 and 39 but never on 13: we must be at 0-12 before going to 13, and then we have to be at 27-36 before going to 39 for a total of $ 5\cdot4\equal{}20$.
iiii. those that land on 13, 26, and 39: for each multiple of 13, we have $ 5$ choices before going to the next interval, for a total of $ 5^3\equal{}125$.

Our final answer is $ 4\plus{}20\plus{}20\plus{}125\equal{}\boxed{169}$.

Or we can solve this recursively:

Let $ A(n)$ be the number of ways to get to $ n$

Each of $ A(n)$ for $ n \equal{} 0,3,6,9,12$ is 1

$ A(13) \equal{} A(0) \plus{} A(3) \plus{} A(6) \plus{} A(9) \plus{} A(12) \equal{} 5$

$ A(15) \equal{} A(12) \plus{} A(13)\equal{}6$

Each of $ A(n)$ for $ n \equal{} 15,18,21,24$ is 6

$ A(26) \equal{} A(15) \plus{} A(18) \plus{} A(21) \plus{} A(24) \plus{} A(13) \equal{} 4\times{6} \plus{} 5 \equal{} 29$

$ A(27) \equal{} A(26) \plus{} A(24) \equal{} 29 \plus{} 6 \equal{} 35$

Each of $ A(n)$ for $ n \equal{} 27,30,33,36$ is 35

Finally, $ A(39) \equal{} A(27) \plus{} A(30) \plus{} A(33) \plus{} A(36) \plus{} A(26) \equal{} 35\times{4} \plus{} 29 \equal{} \boxed{169}$

It is pretty much like Dynamic Programming, where you just backtrack.

If $f(n)$ is the number of ways to get to $39$ from $n$, then we have, for example

$f(39) = 1 , f(36) = 1$
$$f(33) = f(36) + f(39) = 2$$
$$f(30) = f(33) + f(39) = 3$$
$$f(27) = f(30) + f(39) = 4$$
$$f(26) = f(27) + f(39) = 5$$
$$f(24) = f(26) + f(27) = 9$$
and so on. We finally get $f(0) = 169$.

[hide=solution]I did my solution in a completely different way, using a tree branch graph thing.

First, I considered whether the paths would cross 13:
If the answer is yes, then there are 5 possibilities (namely $0,13;0,3,13;...0,3,6,9,12,13$).
If the answer is no, then there is only 1 possibility, such that it skips over $13$ from $12$ to $15$ ($0,3,6,9,12,15$).

Next, I considered whether the paths would cross 26:
If the answer to both this and the last question was yes, then there are $5$ possibilities again, multiplying to $25$.
If the answer to the last question was yes, and this one no, then there is $1$ possibility, multiplying to $5$.
If the answer to the last question was no, and this one yes, then there are $4$ possibilities, multiplying to $4$.
If the answer to both this and the last question was no, then there is $1$ possibility again, multiplying to $1$.

You might be wondering why there are $5$ possibilities to y/y, but only $4$ to n/y. This is because with y/y, we are only going up to $13$. With n/y, we had gone up to $15$. Thus, jumping from $13$ to $26$  is only a possibility with y/y, granting it the extra possibility.

Now, let's do the final part, considering now whether the frog will use a "13 move" to get to 39.

For y/y/y, there are $5$ ways to do it, multiplying to $125$.
For y/y/n, you may be tempted to say $1$ again, but think about it. The frog going $26,27,30,33,36,39$, and never using a "13 move" has already been accounted for in the y/y/y case, where it did $26,27,30,33,36 \text{ "13 move" } 39$. Even though the two are using different commands to get to $39$, they both use the same path. Thus, this part has $0$ possibilities (it has 1, but counting that one would be overcounting.

You can continue the process and solve yourself, or look at the unhidden solutions  :dry:


calculus
inequalities
derivative
analytic geometry
graphing lines
slope

An open box is to be made from a rectangular piece of material by cutting equal squares from each corner and turning up the sides. Find the dimensions of the box of the maximum volume if the material has dimensions 6 inches by 6 inches.

I think we need to use relative min and max or something

This is a common textbook calculus question. You find the expression for the volume in terms of the height and maximize it. Perhaps you can also do this with an inequality somehow?

[hide]Let $x$ be the side length of the square cut off.  Then $V=4x(3-x)(3-x)$, and $\frac{V}{2}=2x(3-x)(3-x)$.  By AM-GM,

$(2x)(3-x)(3-x)\le\left(\frac{2x+3-x+3-x}{3}\right)^3$

$\frac{V}{2}\le 8$

$V\le 16$[/hide]

With $2x=3-x \rightarrow x=1$ for equality, just to show it can be obatined

[hide="A Solution"]
Let the dimension of the squares being cut off be $x\times x$.  As a result, the equation
\[ \\V=x(6-2x)(6-2x)\\=x(36-24x+4x^2)\\=4x^3-24x^2+36x \]
describes the volume of the figure.
Now, we want to maximize this.  We take its derivative and set it to zero due to the fact that we want the point in which the slope of the tangent line is zero.
\[ \\V'=12x^2-48x+36\\0=12x^2-48x+36\\0=x^2-4x+3\\0=x^2-3x-x+3\\0=(x-1)(x-3)\\x=1,3 \]
In order to see which one causes the maximum y-value, we apply the Fermat Test and get
\[ \\f''(x)=24x-48\\\implies f''(1)=24-48=-24<0 \]
Since $x=3$ will cause a value greater than zero, $x=1$ will cause a maximum volume.
If $x=1$, the dimensions are $4\times 4\times 1$, where $V_{max}=16$.
$\mathbb{QED}$
[/hide]

Masoud Zargar


$ 153 \equal{} 1^{3} \plus{} 5^{3}\plus{} 3^{3}$

Are there other numbers like this?

hello, what do you mean with "like this" what kind of rule do you mean?
Sonnhard.

clearly there are others since you provided examples of others :P

How did you come up with these examples?

I wrote a calculator program to find all the three digit numbers for which this occurs...

The numbers are $ 153,370,371,407$

[quote="gokussj3"]Instead of cubes i know a no. like this with fourth powers...

it is 9474

i.e 9474 = 9^4 + 4^4 +7^4 + 4^4[/quote]

Other numbers like this include: $ 1634$ and $ 8208$

Yes, they are known as narcissistic numbers or Armstrong numbers and one, $ 8208$, featured in an episode of the Simpsons. The three-digit ones are $ 153, 370, 371$ and $ 407$. The four-digit one are $ 1634, 8208$ and $ 9474$: $ 8^{4} \plus{} 2^{4}\plus{}0^{4} \plus{} 8^{4} \equal{} 8208$.

[The other numbers in the same episode were the Mersenne prime $ 2^{13} \minus{}1 \equal{} 8191$ and the fourth perfect number $ 8128$ .]


Suppose that you buy 100 pounds of cucumbers. The cucumbers are 99% water. Over the next few days, they dry out to 98% water. How much do they weigh after drying out?

be careful with this, as the answer is NOT anything near 100! :wink:

[hide="Solution"]
At first, there are $ 99$ pounds of water and $ 1$ pound of other stuff.

Let the ultimate amount of water be $ x\minus{}1$ pounds and $ 1$ pound of other stuff so that $ \frac{x\minus{}1}x\equal{}98\%\equal{}\frac{49}{50}\implies x\equal{}\boxed{50}$.
[/hide]


geometry
circumcircle
trigonometry
inradius
angle bisector
projective geometry
geometry unsolved

In $\triangle ABC$, let $\omega$ be the circle tangent to the extensions of $AB,BC$ at $M,N$ and tangent externally to the circumcircle of $\triangle ABC$. Prove that $MN$ passes through the excenter of $\triangle ABC$

[hide="Lemma:"][/hide]
The radius of $\omega$, is

\[r_{\omega}=\frac{bc}{r}*\tan^{2}\frac{1}{2}\angle A\]
[hide="Proof:"]
WLOG $AB<AC$, then let $B'$ be a point on the circumcircle of $\triangle ABC$ such that $AB=AB'$. Let $BB'\cap AC=X$. $\angle AB'B=\angle ABB'=\angle C$, so $\triangle ABX\sim\triangle ACB$. So the inradius of $\triangle AXB$ is
\[r*\frac{AB}{AC}=\frac{c}{b}*r\]

Consider an inversion, $\mathcal{I}$, center $A$, radius $AB=AB'$. Clearly the inversive image of the circumcircle of $\triangle ABC$ is line $BB'$, $\mathcal{I}(w)=\omega'$, the incircle of $\triangle AXB$.

Let $U$, $U'$ be the center of $\omega$, $\omega'$. Let $AU'U$ intersect $\omega$ at $P$, $Q$, $AU'U$ intersect $\omega'$ at $Q'$, $P'$, such that $PQQ'P'$ is a line. It follows that $\mathcal{I}(P)=P'$ and $\mathcal{I}(Q)=Q'$.

Let $x'$ be the inradius of $\triangle AXB$. By definition of inversion, we have that:

$PQ=\frac{b^{2}}{AP'}-\frac{b^{2}}{AQ'}$.

Note that $AU'=\frac{x'}{\sin \frac{1}{2}*\angle A}$, $AP'=-x'+AU'$, and $AQ'=AU'+x'$

$PQ=b^{2}*\frac{2\tan^{2}\frac{1}{2}\angle A}{x'}$

Since $x'=\frac{c}{b}*r$,

The radius of $\omega$. Then
\[b^{2}*\frac{\tan^{2}\frac{1}{2}\angle A}{\frac{c}{b}*r}=\frac{bc}{r}*\tan^{2}\frac{1}{2}\angle A\]

End Lemma[/hide]

---

Let $MN$ intersect the angle bisector of $\angle BAC$ at $I'$. It is evident that $AM=AN$, so $AI'\perp MN$. Also $AM\perp MU$, so through light trig, it follows that $AI'=r_{\omega}*\cot\frac{1}{2}\angle A*\cos \frac{1}{2}\angle A$. It suffices to show that:

\[AI'=AI_{A}\iff \frac{bc}{r}\sin\frac{1}{2}\angle A=\frac{s}{\cos \frac{1}{2}\angle A}\iff \frac{bc\sin A}{2}=rs\]
which is true.

Comment:

$r_\omega=\frac{bcr}{(s-a)^{2}}$

[quote][color=darkred][b]Lemma I.[/b] The circle $w$ is internally tangent in the point $T$ to the circle $\mathcal C$. For a point $M\in w$ denote the chord $[AB]$ of the circle $\mathcal C$ which is tangent in the point $M$ to the circle $w$. Then the ray $(TM$ is the (internal) bisector of the angle $\widehat{ATB}$.
[hide="Proof of lemma I."]Denote the second intersection $N$ of the line $TM$ with the circle $\mathcal C$, the tangent $TX$ to the circle $w$ and the tangent $NY$ to the circle $\mathcal C$ so that the line $\overline{TMN}$ doesn't separate the points $X\ ,\ Y$. From the relations $\widehat{TNY}\equiv\widehat{NTX}\equiv\widehat{TMB}$ obtain $\widehat{TNY}\equiv\widehat{TMB}$, i.e. $AB\parallel NY$. Therefore, $NA=NB$, i.e. the ray $(TN$ is the (internal) bisector of the angle $\widehat{ATB}$.[/hide]
[b]Lemma II.[/b] The circle $\Omega$ is externally tangent in the point $T$ to the circle $\mathcal C$. For a point $M\in \Omega$ denote the chord $[AB]$ of the circle $\mathcal C$ which is tangent in the point $M$ to the circle $\Omega$. Then the ray $(TM$ is the external bisector of the angle $\widehat{ATB}$.[hide="Proof of lemma II."]Denote the second intersection $N$ of the line $TM$ with the circle $\mathcal C$, the tangent $TX$ to the circle $w$ and the tangent $NY$ to the circle $\mathcal C$ so that the line $\overline{TMN}$ doesn't separate the points $X\ ,\ Y$. From the relations $\widehat{TNY}\equiv\widehat{NTX}\equiv\widehat{TMB}$ obtain $\widehat{TNY}\equiv\widehat{TMB}$, i.e. $AB\parallel NY$. Therefore, $NA=NB$, i.e. the ray $(TN$ is the external bisector of the angle $\widehat{ATB}$.[/hide]

$1\blacktriangleright$ Let $ABC$ be a triangle and consider the circle $\omega$ which is tangent to the [b]segments[/b] $(AB) ,\ (AC)$ in the points $M\ ,\ N$ respectively and which is [b]internally[/b] tangent in the point $T$ to the circumcircle of the triangle $ABC$. Prove that the line $MN$ passes through the [b]incenter[/b] $I$ of the triangle $ABC$.

$2\blacktriangleright$ Let $ABC$ be a triangle and consider the circle $\Omega$ which is tangent to the [b]rays[/b] $(AB\ ,\ (AC$  in the points $M\ ,\ N$ respectively and which is [b]externally[/b] tangent in the point $T$ to the circumcircle of the triangle $ABC$. Prove that the line $MN$ passes through the[b] exincenter[/b] $I_{a}$ of the triangle $ABC$.[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b]

$1\blacktriangleright$ Denote the second intersections $U\ ,\ V$ of the circumcircle of $\triangle ABC$ with the lines $TM\ ,\ TN$ respectively.

[b]Is well-known (see Lemma I) [/b]that $I\in BV\cap CU$. Apply the [u]Pascal's theorem[/u] to the cyclical hexagon $CABVTU$ : $\left\{\begin{array}{c}N\in CA\cap VT\\\ M\in AB\cap TU\\\ I\in BV\cap UC\end{array}\right\|\implies I\in MN$.

$2\blacktriangleright$ Denote the second intersections $U\ ,\ V$ of the circumcircle of $\triangle ABC$ with the lines $TM\ ,\ TN$ respectively.

Denote the diameters $[UR]\ ,\ [VS]$ in the circumcircle of $\triangle ABC$. [b]Is well-known (see Lemma II) [/b]that $I\in BS\cap CR$ and $I_{a}\in BV\cap CU$.

Apply the [u]Pascal's theorem[/u] to the cyclical hexagon $CABVTU$ : $\left\{\begin{array}{c}N\in CA\cap VT\\\ M\in AB\cap TU\\\ I_{a}\in BV\cap UC\end{array}\right\|\implies I_{a}\in MN$.[/color]

That is a nice proof; much more elegant than mine. Regrettably, I have no idea how I could come up with such a proof. I don't know such high power theorems.

In the second proof, it should be that $I\in CS\cap BR$, $I_{A}\in VC\cap BU$, and Pascal on: $VCABUT$.

I would appreciate a proof of the lemma. I have been thinking of defining $S'$ as the midpoint of arc $AB$ and working backwards, but I cannot finish the lemma.

[Attatched is a picture of Virgil's second solution.]

EDIT: I would like to make my confusion more clear:

I do not see how I could have seen to draw those lines. Perhaps, to think of drawing those lines, should I be thinking of how to apply pascal's theorem? In otherwords, assume that there is a "nice" cyclic hexagon for which opposite sides intersect at M, N, I_A, then I can apply pascal? Perhaps that is a way to think of it...


[quote="airelemental135"]65. Construct a vector ing a velocity of 80. m/s directed at 60 degrees N of E. State the scale upon which the vector is based and construct the vector to scale. Then resolve the vector by construction and measurement into an easterly and northerly component. State the magnitude and direction of each of these compoent velocities.

[Please describe how this diagram is drawn and what I have to do. I have no idea what the term "N of E" means. Does that mean Northeast 60 degrees according to the problem? So that it is pointing somewhere at 2 o'clock?] Please help with this.[/quote]"N of E" is the vector direction; it means [b]N[/b]orth of [b]E[/b]ast.  If you take a standard compass where North is represented by $0^{\circ}$ and East is represented by $90^{\circ}$ then $60^{\circ}$ N of E = $30^{\circ}$.

[quote="airelemental135"]Thank you for the clarification.[/quote]Anytime :)


trigonometry
ratio
number theory
relatively prime
geometry
angle bisector
Pythagorean Theorem

In triangle $ ABC$, angle $ C$ is a right angle and $ CB > CA$. Point $ D$ is located on $ \overline{BC}$ so that angle $ CAD$ is twice angle $ DAB$. If $ AC/AD \equal{} 2/3$, then $ CD/BD \equal{} m/n$, where $ m$ and $ n$ are relatively prime positive integers. Find $ m \plus{} n$.

$ \textbf{(A)}\ 10\qquad
\textbf{(B)}\ 14\qquad
\textbf{(C)}\ 18\qquad
\textbf{(D)}\ 22\qquad
\textbf{(E)}\ 26$

[quote="jmerry"]I just clicked here out of curiosity- I got my lowest ever score (130) on the 1998 AHSME.

On the forum page, the source appears as "AHSME 1998 #..."
On the thread, the source appears as "USA AMC 12 1998, Problem ..."
On the posting form, the source appears as "AHSME 1998 #..."

That's a bit strange.[/quote]

AHSME is the correct name but database includes AHSME through AMC-12 so whole thing is called "AMC-12" in database. "USA" is included to indicate the country.

Ah, I remember doing a really bad tan triple and double angle manipulation on this...I'll let someone with a more elegant solution post theirs.

I've got another solution, not sure if its "elegant", though.

[hide="Incomplete"]Preliminary definitions:
$AC=2x$
$AD=3x$
$BD=y$
$<BAD=\theta$
$<CAD=2\theta$
$AE=$ bisector of $<CAD$
Thus, $<CAE=\theta$ and $DAE=\theta$
Work:
By the Angle Bisector Theorem, we can WLOG assume $CE=2$ and $DE=3$. Now we have $5^2+(2x)^2=(3x)^2$ by applying the Pythagorean Theorem on triangle $ACD$.
$\Rightarrow x=\sqrt{5}\Rightarrow AC=2\sqrt5$

So by applying the Pythagorean Theorem on triangle $ABC$, we have $(BA)^2=20+(5+y)^2=45+10y+y^2$.

Now we will use Stewart's Theorem.....

Sorry to leave you hanging. But I fell asleep while posting this and need to go to bed........... Sorry.[/hide]

[hide="Answer"]Assume WLOG that $AC=2$, $AD=3$. Therefore, $CD=\sqrt{5}$. Drawing the bisector $AE$ of $\angle CAD$, then by Angle Bisector Theorem we have $CE=\frac{2}{\sqrt{5}}$ and $DE=\frac{3}{\sqrt{5}}$. Therefore, $AE=\frac{2\sqrt{30}}{5}$, and $\cos x=\frac{\sqrt{30}}{6}$, $\sin x=\frac{1}{\sqrt{6}}$. $\tan 3x=\frac{k+\sqrt{5}}{2}=\frac{\sin 3x}{\cos 3x}$. $\sin 3x=3\sin x-4\sin ^3x$ and $\cos 3x=4\cos ^3x-3\cos x$. Plugging in our values, we have $\cos 3x=\frac{\sqrt{30}}{18}$ and $\sin 3x=\frac{7\sqrt{6}}{18}$, so $\tan 3x=\frac{7\sqrt{5}}{5}=\frac{k+\sqrt{5}}{2}\Rightarrow k=\frac{9}{\sqrt{5}}=BD$. Therefore, $\frac{CD}{BD}=\frac{\sqrt{5}}{\frac{9}{\sqrt{5}}}=\frac{5}{9}\Rightarrow 5+9=14\Rightarrow \boxed{B}$.[/hide]

There is a way using no trig, but it's not that much easier to carry out (and is probably harder to find). I wonder if there is a very clever way...

[hide]Let E be the point on $BC$ for which $AE$ bisects $\angle CAD$.  Through similarity, we can say that $CA=2\sqrt5$ and $DA=3\sqrt5$.  Apply the pythagorean theorem to triangle ACD to get $CD=5$.  Use the Angle Bisector Theorem to triangle ACD to get $CE=2$ and $DE=3$.  For simplicity, let $x=BD$.  Apply the pythagorean theorem to tirangle ACE to get $EA=2\sqrt6$, following with Angle Bisector Theorem on triangle EAB to get $AB=\frac{2\sqrt6x}{3}$.  Now we can apply the Pythagorean Theorem once last time to triangle ABC to get:

$(2\sqrt5)^2+(x+5)^2=\frac{2\sqrt6x}{3}$

Solving gives $x=9$, making BD/DC=$\frac95$ and $m+n=14$[/hide]

[quote="Octahedr0n"]What is a similarity transformation?  Is it possible to do one without using matrices?  (I don't get Wolfram's method: http://mathworld.wolfram.com/SimilarityTransformation.html)

Thanks in advance![/quote]

If I'm not mistaken, a similarity transformation in this case basically means "If we multiply all the lengths of the figure by a real constant (creating another similar figure), the ratios will still be the same, so we can assign one of the lengths an arbitrary value." Thus, they let $ AC\equal{}2\sqrt5$, so then $ AD$ must be $ 3\sqrt5$ the given ratio in the problem.

I understand the solution until the point where it says now apply the Angle Bisector Theorem on triangle EAB to get AB=\frac{2\sqrt6x}{3}. Could some elaborate that step.


Let $ \boxed{N}$ mean the number of whole number divisors of $ N$. For
example, $ \boxed{3}\equal{}2$, because has two divisors, 1 and 3. Find the
value of
\[ \boxed{\boxed{11}\times\boxed{20}}
\]

11 has two divisors 1 and 11, so $ \boxed{11}$ is 2.

and 20 has 6 divisors 1,2,4,5,10,20 so $ \boxed{20}$ is 6.

so it becomes $ \boxed{2*6}$, which is $ \boxed{12}$.

the number of divisors of 12 is 6(1,12,2,6,3,4)

answer : 6

$ \boxed{11}\equal{}2$, $ \boxed{20}\equal{}6$, $ \boxed{2\times 6}\equal{}\boxed{12}\equal{}6$


calculus
integration
algebra unsolved
algebra

How many nonnegative integral ordered triples (x, y, z) satisfy the equation x + 2y + 8z = 2008?

The equation in question is obviously equivalent to $ w\plus{}y\equal{}1004\minus{}4z$, since $ x$ must be, obviously, even ($ x\equal{}2w$, $ w$ is also a nonnegative integer). Now, if we fix $ z$, I claim we have $ 1005\minus{}4z$ possible ordered pairs $ (w, y)$. That's because, if we know one of these numbers (let's say $ w$, but they're symmetric so it doesn't matter), the other one is unique. Please note that we also know $ z$. So, since $ 0\leq w\leq 1004\minus{}4z$ and $ w$ is an integer, we can choose $ w$ in $ 1004\minus{}4z\plus{}1\equal{}1005\minus{}4z$ ways. All of these ways are acceptable for $ y$ (because we have limited $ w$ in that manner) and, thus, we have $ 1005\minus{}4z$ combinations for each $ z$. Since $ 0\leq z\leq \frac{2008}{8}\equal{}251$, the total number of combinations is $ (1005\minus{}4\cdot 0)\plus{}(1005\minus{}4\cdot 1)\plus{}...\plus{}(1005\minus{}4\cdot 250)\plus{}(1005\minus{}4\cdot 251)\equal{}$
$ \equal{}252\cdot 1005\minus{}4(0\plus{}1\plus{}...\plus{}251)\equal{}$
$ \equal{}252\cdot 1005\minus{}\frac{4\cdot 252\cdot 251}{2}$
That equals $ 252\cdot 503\equal{}126756$, which is the solution for the problem.


inequalities
function
inequalities unsolved

prove that
a$ \sqrt{a^2 \plus{} 2bc}$+b$ \sqrt{b^2 \plus{} 2ac}$+c$ \sqrt{c^2 \plus{} 2ba}$ ≥ $ \sqrt{3}$ (ab+bc+ca)
a,b,c >0

[quote="kihe_freety5"]prove that
a$ \sqrt {a^2 \plus{} 2bc}$+b$ \sqrt {b^2 \plus{} 2ac}$+c$ \sqrt {c^2 \plus{} 2ba}$ ≥ $ \sqrt {3}$ (ab+bc+ca)
a,b,c >0[/quote]
Note that we have
\[ a\sqrt {a^2  \plus{} 2bc}  \minus{} a^2  \equal{} \frac{{2abc}}{{a \plus{} \sqrt {a^2  \plus{} 2bc} }}
\]
Therefor it suffices us to show that
\[ 2abc\left( {\frac{1}{{a \plus{} \sqrt {a^2  \plus{} 2bc} }} \plus{} \frac{1}{{b \plus{} \sqrt {b^2  \plus{} 2ca} }} \plus{} \frac{1}{{c \plus{} \sqrt {c^2  \plus{} 2ab} }}} \right) \ge \sqrt 3 \left( {ab \plus{} bc \plus{} ca} \right) \minus{} \left( {a^2  \plus{} b^2  \plus{} c^2 } \right)
\]
But we have
\[ \sum {\frac{1}{{a \plus{} \sqrt {a^2  \plus{} 2bc} }}}  \ge \sum {\frac{1}{{a \plus{} \sqrt {a^2  \plus{} b^2  \plus{} c^2 } }}}  \ge \frac{9}{{a \plus{} b \plus{} c \plus{} 3\sqrt {a^2  \plus{} b^2  \plus{} c^2 } }}
\]
So it's enough for showing that
\[ \frac{{18abc}}{{a \plus{} b \plus{} c \plus{} 3\sqrt {a^2  \plus{} b^2  \plus{} c^2 } }} \ge \sqrt 3 \left( {ab \plus{} bc \plus{} ca} \right) \minus{} \left( {a^2  \plus{} b^2  \plus{} c^2 } \right)
\]
I denote $ r\equal{}abc,q\equal{}ab\plus{}bc\plus{}ca$ and assume $ a\plus{}b\plus{}c\equal{}1$ then $ q \le \frac{1}{3}$ and $ r \ge \max \left\{ {0,\frac{{4q \minus{} 1}}{9}} \right\}$. Then the inequality becomes
\[ \frac{{18r}}{{1 \plus{} 3\sqrt {1 \minus{} 2q} }} \ge \left( {2 \plus{} \sqrt 3 } \right)q \minus{} 1
\]
If $ q \le \frac{1}{{2 \plus{} \sqrt 3 }}$ then $ LHS \ge 0 \ge RHS$. So we only need to check in case $ q \ge \frac{1}{{2 \plus{} \sqrt 3 }} > \frac{1}{4}$. We have
\[ LHS \ge \frac{{8q \minus{} 2}}{{1 \plus{} 3\sqrt {1 \minus{} 2q} }} \ge \left( {2 \plus{} \sqrt 3 } \right)q \minus{} 1 \equal{} RHS
\]
Then it's done.

[quote="dgreenb801"]Let $ abc \equal{} 1$. We have to show
$ \sum (\sqrt {a^4 \plus{} 2a} \minus{} \frac {\sqrt {3}}{a}) \ge 0$, true by Jensen, as $ a \plus{} b \plus{} c \ge 3$ (Can you prove the function convex?)[/quote]
uhm $ f(a) \equal{} (\sqrt {a^4 \plus{} 2a} \minus{} \frac {\sqrt {3}}{a})$ thus $ f''(a) \equal{} \minus{} \frac {(\frac {1}{4})(4a^3 \plus{} 2)^2}{(a^4 \plus{} 2a)^{\frac {3}{2}}} \plus{} \frac {6a^2}{\sqrt {a^4 \plus{} 2a}} \minus{} \frac {2\sqrt 3}{a^3}$
Can you show it positive or not????? :maybe:

[quote="Honey_S"]
\[ LHS \ge \frac {{8q \minus{} 2}}{{1 \plus{} 3\sqrt {1 \minus{} 2q} }} \ge \left( {2 \plus{} \sqrt 3 } \right)q \minus{} 1 \equal{} RHS\]
[/quote]

I don't see it. :blush:


analytic geometry
graphing lines
slope
calculus
function
derivative
limit

[quote="guile"]Can you find a way to solve this problem without using calculus?

What is the slope of the family of lines tangent to y = x^2.[/quote]
[hide]
$\frac{1}{2} x$.[/hide]

Explanation perhaps, shobber?

And what is "family of lines"? Is that the mean line of all of the tangent lines?

And, having just learned the Difference quotient, I got 2x, but of course I may be applying it in the wrong sense, so feel free to explain.

Yes, I would think the answer would be $2x$, which would require Calculus. I don't know of any other way to do it.

Well, if you've taken calculus, you can easily tell it's 2x.  The slope of a function is the derivative of the function.  Essentially you're finding the slope of the tangent line.  Here, you would apply a simple power rule:

Let's say you have an equation: $y = x^n$

$\frac{dy}{dx} = nx^{n-1}$

This rule works if it's only a power.  If you have a quotient, it gets more complicated.  More generally:

$\frac{dy}{dx} = nu^{n-1} \cdot \frac{du}{dx}$

where u equals some expression containing x.  This "chain" effect is typically called the chain rule.

As for a way to find the slope without using calculus, the only way I can think of is with limits and that's still kind of related.  Back to the concept of the tangent line.  Normally for secant lines, you would find the slope by using two points:

$(x_1,y_1)$ and $(x_2,y_2)$

$m = \frac{y_2 - y_1}{x_2 - x_1}$

The problem with finding the slope of a tangent line is that it only shares one point.  However, we'll say $h = x_2 - x_1$.  Then it'll be:

$m = \frac{f(x_1 + h) - f(x_1)}{h}$

The slope of the tangent line would be when $h = 0$ or $h \rightarrow 0$.  The term "as h approaches 0" implies limit.  So:

$m = \lim_{h \rightarrow 0}{\frac{f(x + h) - f(x)}{h}}$

Applying this to $f(x) = x^2$:

$m = \lim_{h \rightarrow 0}{\frac{(x + h)^2 - x^2}{h} = \lim_{h \rightarrow 0}{\frac{x^2 - x^2 + 2xh + h^2}{h}} = \lim_{h \rightarrow 0}{2x + h} = 2x}$

What you did with the limits there is where derivatives originally came from. (As you've taken Calculus, you most likely know this, but others might not.) It's probably an easier way for people to solve, because it's very simple Calculus, and you only need to remember the single limit function and work out everything inside, rather than memorizing a number of derivation formulas.

Err, I got 2x just by using a simple difference quotient, which I learned in Precalculus just last week.

All it involves is the slope formula... But I guess it's the concept that counts? Iunno

$\frac{f(x+h)-f(x)}{h}$ then minimize h as much as you can once you've simplified that (i.e. set h=0) and you've got it for any point.

That's the limit function from the beginning of Calculus, sometimes introduced in Pre-Calc.


Find the coefficient of $ x^2$ in the expansion of
$ ((...((x\minus{}2)^2\minus{}2)^2\minus{}...\minus{}2)^2\minus{}2)^2$
given that the number $ 2$ occurs $ 1004$ times in the expression above and there are $ 2008$ round brackets

To pro_math: you are bad.  :mad:
To mod: pls lock this topic because this problem is on Mathematical and Youth this month.

[hide]Let $ F_k(x)=((\ldots((x-2)^2-2)^2\ldots-2)^2-2)^2$, where the number 2 occurs $ k$ times. Note that the constant term equals $ F(0)=4$. Let $ F_k(x)=P_kx^3+A_kx^2+B_kx^2+4$. So
\begin{align*}F_k(x)&=(F_{k-1}(x)-2)^2\\&=(P_kx^3+A_kx^2+B_kx^2+2)^2\end{align*}
Expand the last expression, you'll get the coefficient of $ x^2$, i.e. $ A_k=4A_{k-1}+B_{k-1}^2$, and the coefficient of $ x$, i.e. $ B_k=4B_{k-1}$. Since $ B_1=-1$, we have $ B_k=-4^{k-1}$. Thus
\begin{align*}A_k&=4A_{k-1}+B_{k-1}^2
\\&=16A_{k-2}^2+4B_{k-2}^2+B_{k-1}^2
\\&=64A_{k-3}^2+16B_{k-3}^2+4B_{k-2}^2+B_{k-1}^2\\&=\ldots=4^{k-1}B_1+4^{k-2}B_2+\ldots+B_{k-1}.\end{align*}
Substitute $ A_1=-4,A_2=-16,A_3=-64,\ldots$, and we'll get a geometric progression. Now it is easy to see that $ A_k=\frac{4^{2k-1}-4^{k-1}}3$.

In this case, we substitute $ k=1004$.[/hide]


function
calculus
derivative
calculus computations

Hello
if second derivative of f exists and f has inverse for all no negatives x and $ f'\left( x \right) \equal{} xf''\left( x \right) \minus{} \frac{{{{\left( {f''\left( x \right)} \right)}^2}}}{2},x \ge 0,f\left( x \right) \ge 0$, try to find f
thx

I think it's not very hard:
$ y \equal{} xy' \minus{} \frac {y'^2}{2}$;
Let $ y' \equal{} p$, so we have:
$ pdx \equal{} dy \equal{} d(xp \minus{} \frac {p^2}{2}) \equal{} xdp \plus{} pdx \minus{} pdp$, so:
$ (x \minus{} p)dp \equal{} 0$.
Now, $ y' \equal{} 0$ => $ y \equal{} 0$ (constant is zero only, can be easily checked), or $ y' \equal{} x$.
$ y \equal{} 0$, or $ y \equal{} \frac {x^2}{2} \plus{} C$ ($ C \equal{} 0$ is only rigth case).
Thus, $ f(x) \equal{} D$, or $ f(x) \equal{} \frac {x^3}{6} \plus{} D$.
Now, $ f$ has $ f^{ \minus{} 1}$ in only second case, and it's easy to correct coefficient $ D$ for finish this task ($ f(x)\geq 0$): $ D$ must be nonnegative.

Answer: $ f(x) \equal{} \frac {x^3}{6} \plus{} C$, $ C\geq 0$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

I think that this is a dificult one.

Let $ G$ be a finite $ p$-group, and suppose that $ H$ is a non-trivial normal subgroup of $ G$. By considering the action of $ G$ on $ H$ by conjugation, prove that $ H \cap Z(G) \neq \lbrace e \rbrace$.

The number of fixed points ot this action is a multiple of [b]p[/b] and every such point is a member of $ H \cap \mathbf{Z}(\mathrm{G})$.

What is a fixed point of an action?
Thanks

[b]x[/b] is a fixed point of a G-set if its stabilizer is the whole G.

I am afraid i don't understand why the number of fixed points of the action by conjugation is a multiple of p...can i get some more help?
Thanks

H can be written as a disjoint union of orbits.

Let us assume that $ H \equal{} \bigsqcup_{1 \leq k \leq n } \mathrm{Orb}(g_{k})$ and that [b]F[/b] consists of those $ g \in \{g_{1}\ldots, g_{n}\}$ that are fixed points of G under the action at hand. [b]NF[/b] is defined accordingly. It follows now that

$ | H | \equal{} \sum_{1 \leq k \leq n} |\mathrm{Orb}(g_{k})| \equal{} \sum_{g \in \mathbf{F}} |\mathrm{Orb}(g)| \plus{} \sum_{g \in \mathbf{NF}} | \mathrm{Orb}(g) | \equal{} \\
\equal{} \sum_{g \in \mathbf{F}} [G: \mathrm{Stab}(g)] \plus{} \sum_{g \in \mathbf{NF}} [G: \mathrm{Stab}(g)] \equal{} |\mathbf{F}| \plus{} \sum_{g \in \mathbf{NF}} [G: \mathrm{Stab}(g)]$.

Is it clear to you that $ p$ divides every term of the series $ \sum_{g \in \mathbf{NF}} [G: \mathrm{Stab}(g)]$?

If so, what can you conclude about $ |\mathbf{F}|$?

Note that $ |\mathbf{F}| \equal{} |H| \minus{} \sum_{g \in \mathbf{NF}} [G: \mathrm{Stab}(g)]$, and that $ p$ certainly divides $ |H|$.


Here is a nice problem:

Let $a_1, \ldots, a_n, \ldots$ be the rational numbers (all of them). Prove that there is no irrational $x\in \mathbb{R}$ such that for any positive integer $j$ there is a positive integer $i$ so that
\[|x-a_i|<\frac{1}{2^{i+j}}.\]

This problem has to do with the notions of almost everywhere and quasi-everywhere. For exame, the set of all numbers $x$ with the above property (irrational or not), is on the one hand of measure 0, and on the other hand the intersection (over $j$) of open sets (around each rational there is a small interval for each $j$) which is dense in $\mathbb{R}$ (since the rationals are dense).

A property that occurs on a countable intersection of open dense subsets is called quasi-true. Therefore allegiance to the set of all such $x$ is a quasi-true almost nowhere true property (with the almost nowhere being in the sense of measure 0).

I thought it would be nice to prove that the set of all such $x$ is actually $\mathbb{Q}$.

Have fun,
Andrei

I'm sure I'm misunderstanding, but it seems wrong: what if we fix an irrational $x$, and then we choose $a_{2j}$ s.t. $|x-a_{2j}|<\frac 1{2^{2j+j}}$ (in other words, we could choose $i=2j$ for each $j$)? The rest of the rationals can be covered by the sequence $a_{2j+1}$, so for this particular enumeration of the rationals we can find such an irrational $x$.

Silly me, I should've known that a noncanonical problem should have a noncanonical answer. Of course the set of $x$ depends on the ordering. Thanks,

Andrei


I followed the instructions on the AoPSWiki to integrate asymptote with LaTeX, however I cannot get it to work.  When I type up a LaTex file with asymptote code (inside the asy tags) the LaTeX compiles fine and the asymptote does not compile.  When I call the fill a .asy file (instead of .tex), the asymptote and LaTeX both compile, however they are split into 2 different PDF files.  Any help would be appreciated.  Thanks.

You should get 2 pdf files even when compiled inside LaTeX.
Have you done the three-stage process 1. compile LaTeX, 2. run Asymptote, 3. compile LaTeX?
This is taken from [[Asymptote:_Advanced_Configuration#Using_Asymptote_in_LaTeX]]
[quote="AoPSWiki"]LaTeX will not output the image the first time it compiles the document. To compile this file, you must:
[list=1][*]Pass it once through LaTeX without viewing the output (click Build in TeXnicCenter; this extracts your asy code into a .asy file.)
[*]Run Asymptote on the new file example.asy (pressing Alt+A as normal in TeXnicCenter will work, since it only looks at the title example.)
[*]Finally, run it again through LaTeX (click Build and View Current File in TeXnicCenter).[/list][/quote]


geometry rectangle perimeter ratio blogs MATHCOUNTS inequalities

hockeyADI21: The diag of a rectangle is 41 in...
hockeyADI21: and
AmeemShk: can i add you guys to my buddy list???
hockeyADI21: the perimeter is 98 inches
led14pa2: sure
solafidefarms: Sure!~I already added you
hockeyADI21: What is the number of in^2 in the area of the square
led14pa2: the sides are 40 and 9 then
solafidefarms: So 360
led14pa2: 360
hockeyADI21: Uv good
hockeyADI21: easy if you know the ratio
led14pa2: yea
hockeyADI21: otherwise it's a mess
LuVeR Of RaIn: wait, when he's posting a question
led14pa2: and pythagorian tribles
LuVeR Of RaIn: can no one types anything
led14pa2: *triples
LuVeR Of RaIn: cuz it's hard to read
hockeyADI21: yes
hockeyADI21: Here's another question:
hockeyADI21: The product of a set of positive integers is 144. What is the least possible sum of this set of positive integers?
solafidefarms: 24
led14pa2: 24
LuVeR Of RaIn: 24
hockeyADI21: No and no
hockeyADI21: and no
LuVeR Of RaIn: what?
condabomb88: 145
hockeyADI21: first three correct get a prize!
hockeyADI21: of happiness.
hockeyADI21: lol
led14pa2: 17
hockeyADI21: I should moderate or something
condabomb88: lol
led14pa2: 4, 4, and 9
solafidefarms: 14
hockeyADI21: Yes
hockeyADI21: 14 is correct
solafidefarms: 2,2,2,2,3,3
led14pa2: oh i didnt see that 1
LuVeR Of RaIn: how did u get 14?
hockeyADI21: uip
LuVeR Of RaIn: o
LuVeR Of RaIn: o
LuVeR Of RaIn: o
LuVeR Of RaIn: nvm
hockeyADI21: yup
hockeyADI21: yup
AmeemShk: ok
AmeemShk: i added u guys
condabomb88: prime factorization
hockeyADI21: Another one
LuVeR Of RaIn: does anyone have a formula list i can do for the last minute studying?
LuVeR Of RaIn: plz
led14pa2: sorry i dont have 1
LuVeR Of RaIn: okay
AmeemShk: lightspeed made a huge on aops
hockeyADI21: k.. like the sum of the first n squres is n^2
AmeemShk: u should look at that
LuVeR Of RaIn: nvm
solafidefarms: Farazy's blog is good
hockeyADI21: ya
LuVeR Of RaIn: well, i just want a list of formulas that i can look at
AmeemShk: ok guys im leaving in 15 min.
hockeyADI21: here'a another problem
LuVeR Of RaIn: who's farazy?
AmeemShk: dude on aops
solafidefarms: Look him up
solafidefarms: Problem next?
hockeyADI21: What is the sum of the three numbers less than 1000 that have five positive divisors
hockeyADI21: 80 sec
solafidefarms: 98
hockeyADI21: pardom me
LuVeR Of RaIn: 96?
hockeyADI21: exactly five
AmeemShk: 1299?
hockeyADI21: wait let me do it...
AmeemShk: 276?
solafidefarms: 5 primes?
AmeemShk: wait can this have more than one answer?
hockeyADI21: no
condabomb88: no
hockeyADI21: EXACTLY 5
hockeyADI21: stop guessing... once answer at a time
solafidefarms: I ge tmy problem
led14pa2: haha i have no idea
solafidefarms: I counted 1
solafidefarms: 16,81,625
LuVeR Of RaIn: 2331
solafidefarms: 722
hockeyADI21: No 16 duzent work...
AmeemShk: 1299?
solafidefarms: Why?
led14pa2: 1,2,4,8,16 doesnt it
hockeyADI21: o sry... nvm
hockeyADI21: my b
hockeyADI21: mind was drifting
hockeyADI21: solafidefarms is right
AmeemShk: i dont like this problem
hockeyADI21: y
solafidefarms: Happy voyage!
AmeemShk: ur never gonna use this in life
hockeyADI21: it's a great problem...
AmeemShk: bad math
jwec05 has entered the room.
condabomb88: silverfalcon is here
hockeyADI21: nice
hockeyADI21: finally
jwec05: Hello
AmeemShk: ok thats my que to leave
solafidefarms: Hi!
led14pa2: hey
hockeyADI21: so I don't have to do it!
jwec05: so everybody here?
hockeyADI21: hey
hockeyADI21: yup
led14pa2: yea
condabomb88: lol u were a good moderator
hockeyADI21: thx
led14pa2: i agree
AmeemShk: bye guys
hockeyADI21: bye
condabomb88: c ya
AmeemShk has left the room.
LuVeR Of RaIn: ciao
jwec05: Ok
led14pa2: u should be the moderater at states hockeyadi so then i dont have to face u in countdown
solafidefarms: Good moderator
hockeyADI21: lol!
jwec05: everybody ready to start?
condabomb88: lol good point
hockeyADI21: haha
solafidefarms: Yep
condabomb88: yup
jwec05: Ok
hockeyADI21: ya
LuVeR Of RaIn: eh... what's gonna happen?
led14pa2: jason can be ur assistant
LuVeR Of RaIn: nvm been to a math chat before...
led14pa2: same
jwec05: First, I'm Silverfalcon from AoPS and I will be hosting a math chat today
solafidefarms: Me neither
hockeyADI21: tha't funny
hockeyADI21: nvm everyone not in CT
jwec05: The problems will be from old National Mathcounts
LuVeR Of RaIn: okay
jwec05: do you mind keep it down for while?
jwec05: thank you
jwec05: Few rules on the chatroom
LuVeR Of RaIn: keep what down?
hockeyADI21: that
hockeyADI21: !
condabomb88: lol continue silver...
jwec05: 1) Respect is #1 thing and like in aops class, the room should not be filled with nonsense things. I don't have the room moderated so please, thanks
hockeyADI21: sry
jwec05: 2) When giving the answers, please compile explanations
jwec05: for example
jwec05: answer is 1
jwec05: it's better to say
jwec05: since "something" then we can say answer is 1
jwec05: if you're not sure about answer
jwec05: IM me first
jwec05: so ready to start?
hockeyADI21: can we split the posts? like answer first
hockeyADI21: then solution like in a math jam?
solafidefarms: Good idea
jwec05: Yes
jwec05: For that
jwec05: send me your answer first
jwec05: by IM
jwec05: and I'll put it with your sn so we know who got the answer
jwec05: There are going to be about 6 questions today
jwec05: #1
jwec05: An estate is divded among three heirs. After Andrea received 20% of the estate, her mother received 50% of what remained, leaving $20,000 for Andrea's uncle.
jwec05: What was the total value of the estate?
jwec05: I have received three answers
jwec05: and all of them are correct
jwec05: solafidefarms6): 500
jwec05: hockeyADI215): 500
Now, you three can complie explanations to me on IM
jwec05: Here is one explation (I'm going to use only one from person):
jwec05: hockeyADI211): You multiply by two because you have 50percent, and then you multiply by 5/4. Since 20000*2*5/4=50000, that's the answ
hockeyADI21: I's easy if you do the *2 first..
jwec05: Now, let's just go through this step-by-step
hockeyADI21: cuz then you get 40,000
hockeyADI21: If that was a countdown, then that would be the way to do it
jwec05: One way to look at this might be
jwec05: since after Andrea got hers
hockeyADI21: brb
jwec05: the left is 4/5 and then half of it is 2/5
jwec05: so  2/5 = 20000
jwec05: then just work backward
condabomb88 has left the room.
jwec05: here is an algebraic solution
jwec05: solafidefarms1): Since A=.2X, and M=.5(.8X), and U=20000, A and M and C total X, so U=.4X. Then .1X=20000/4=5000, so X=500
jwec05: There are always more than one way to do the problem
jwec05: This was #1 on National Sprint Round Mathcounts
jwec05: any questions?
hockeyADI21: nope...
solafidefarms: Can we use calculaors
hockeyADI21: no... better w/o
jwec05: No
jwec05: some problems may be from Team Round
jwec05: but this year (two I'm using now)
jwec05: the calculators weren't allowed till this year
jwec05: which is
jwec05: Um
jwec05: 1995
jwec05: so all problems can be solved w/o calculator
jwec05: and this will help your arithmetic also
jwec05: Ok
jwec05: #2
condabomb88 has entered the room.
jwec05: Find the smallest real value of x that satisfies the equation:
jwec05: (x+5)(x^2-x-11) = x+5
jwec05: I'll give 20 seconds more
jwec05: Ok
jwec05: let's go through this step by step
PiratevsPirate has entered the room.
jwec05: hold for sec
jwec05: Two people said it was 4
jwec05: One person has (1-sqrt45)/2
PiratevsPirate: already? man!!!!
jwec05: Another person had 3
jwec05: unfortunately
PiratevsPirate: I missed the problem..didn't i...
hockeyADI21: none are right
jwec05: none of them is correct answer
hockeyADI21: lol
jwec05: I'll give another minute to do this one
hockeyADI21: is it 2
jwec05: Please
hockeyADI21: sry
PiratevsPirate: give me the proble
jwec05: led14pa24):
jwec05: This is the correct answer
hockeyADI21: does that say -4
hockeyADI21: i cant read it
jwec05: No
led14pa2: -5
solafidefarms: What is that? It is a square block
hockeyADI21: ya
jwec05: to some people
jwec05: how did you get other answers?
LuVeR Of RaIn: o, never thought of that
hockeyADI21: Silly error
hockeyADI21: nvm
jwec05: I can go deeper explanation of this but
jwec05: obviously
jwec05: if we try
jwec05: -5 for x
jwec05: (-5+5)= 0
PiratevsPirate: wait
LuVeR Of RaIn: dunno, took a random guess, i thought (x^2-x-11) had to be equal to one....
jwec05: and the product of 0 and any number is 0
hockeyADI21: that's what I did
LuVeR Of RaIn: never thought of x-5 = 0
jwec05: so this holds trues
jwec05: ok
jwec05: go ahead
condabomb88: ?
jwec05: never mind
jwec05: The other solutiosn are this way
jwec05: divide x+5 to the both sides
jwec05: this will yield
jwec05: x^2-x-11 = 1
jwec05: x^2-x-12 = 0
jwec05: x=-4, 3
jwec05: so it looks like it's -4
jwec05: but nope, there is -5 that works
PiratevsPirate: ahh!!!!
jwec05: so -5 is the correct answer
PiratevsPirate: too late for me
hockeyADI21: sait what?
jwec05: yes?
led14pa2 has left the room.
hockeyADI21: but then you'll get 0
hockeyADI21: yup
hockeyADI21: i c it
hockeyADI21: gtg
hockeyADI21: bye
hockeyADI21 has left the room.
jwec05: Ok
jwec05: any questions?
PiratevsPirate: why is everyone leaving/
LuVeR Of RaIn: i'm confused... what does hockeyADI21 mean: hockeyADI21: but then you'll get 0hockeyADI21: yuphockeyADI21: i c it
solafidefarms: Cause it's 8?
jwec05: I'm not sure what you're asking
solafidefarms: Nvm
PiratevsPirate: let's just go
jwec05: Ok
jwec05: #3
PiratevsPirate: #3 aleready?
PiratevsPirate: geeze I was late
LuVeR Of RaIn: it's 7 here...
jwec05: If m+n>50 and m-n<= 10, what is the smallest possible integer value of n?
jwec05: <= means lesser than or equal to
PiratevsPirate: hey jwec05, I sent you the answer
jwec05: Yes I know
jwec05: But I'm giving time to other people as well
PiratevsPirate: ooo..
jwec05: This is same on the math jam
PiratevsPirate: well, did i get it right/
jwec05: Since you'ren't here for the procedure
jwec05: I'll just say this again
PiratevsPirate: k....
jwec05: "Please be patient"
PiratevsPirate: sorry
jwec05: Ok
jwec05: let me post the answer
jwec05: n = 21
jwec05: Here is an explanation
jwec05: we can say m+n=51
jwec05: then this works
jwec05: because 51>50
jwec05: so, we start from closest ones
jwec05: 25,26
jwec05: 24,27
jwec05: 23,28
jwec05: 22,29
jwec05: 21,30
PiratevsPirate: geeze
PiratevsPirate: misread
jwec05: 20,31 would be the next term
jwec05: however
jwec05: 31-20>10
jwec05: so 21 is the correct answer
LuVeR Of RaIn: that's basicly what i did
PiratevsPirate: nice lucy
solafidefarms: Sort of mine
jwec05: This kind of problem may appear on the National Mathcounts a lot
jwec05: this is basic algebraic inequality
jwec05: best way to keep your work and seeing any "hidden" mistakes
LuVeR Of RaIn: i don't get jwec's explanation with the start from closest ones
PiratevsPirate: and addition
jwec05: Divde by 2
jwec05: and 51/2 = about 25
PiratevsPirate: you flippy the sign when you multiply or divide
jwec05: so 25 and 26
LuVeR Of RaIn: o
LuVeR Of RaIn: why do u have to do that?
condabomb88: only if its by a negitive #
jwec05: Just easy
jwec05: ok
jwec05: here is another interesting solution
jwec05: solafidefarms5): By the last equation, the difference is <=10. By the first, the minimum is 51. We say for our purposes n=m-10, and 2m-10>50, so m>30. The latter is the only one that works. So n=
LuVeR Of RaIn: i did what solafidefarms did
condabomb88: lets move on
jwec05: Any questions?
PiratevsPirate: yah....
LuVeR Of RaIn: nope
PiratevsPirate: oh yeah
PiratevsPirate: I do...
PiratevsPirate: uhm...
PiratevsPirate: how can you convince your school to take MC test?
solafidefarms: Pay the money
PiratevsPirate: and when should I do it?
PiratevsPirate: how much?
condabomb88: too late now
LuVeR Of RaIn: find people to participate.....
PiratevsPirate: for this year?
jwec05: Yes
jwec05: it's already late
jwec05: since chapter has already ended
jwec05: and I think there is a deadline
jwec05: forgot when
PiratevsPirate: ahhh!!!
LuVeR Of RaIn: do u have to have a school to participate in MC?
solafidefarms: Get 4 friends and pay 80 total for them to do it
PiratevsPirate: so, does that mean I can't take it in 8th grade?
condabomb88: you need a school
condabomb88: and yes
PiratevsPirate: I'm in 7th grade currently
solafidefarms: Or something similar
condabomb88: ohh
PiratevsPirate: huh?
solafidefarms: I'm homeschooled, and our math club did it
condabomb88: then you can do it next year
PiratevsPirate: so when?
condabomb88: next year
solafidefarms: You can't do it multiple times a year...like being in 2 teams
jwec05: You might want to convince the school early next year
condabomb88: talk to you principal
condabomb88: your*
jwec05: that's what I did on MC
jwec05: wait
jwec05: AMC*
PiratevsPirate: arg....
condabomb88: lets do #4
PiratevsPirate: we do AMC
PiratevsPirate: but not MC
jwec05: #4
PiratevsPirate: k....
jwec05: Find Darlyan's age, in years, given that he is two-thirds as old as he will be eight years before he is twice as old as he is now.
jwec05: I'll give 30 seconds more
jwec05: I have received two correct answers
jwec05: if you had correct answer
jwec05: I would've ask you to write explanations
jwec05: Here are explanations
jwec05: LuVeR Of RaIn2): x=2/3(2x-8), so x=4/3x-16/3, if you multiply both sides by 3, you get 3x=4x-16, and when you solve for x, it's equal to
jwec05: condabomb881): You have to write out the algebraic equation:       3/2x =2x-8, where x is his age. Then multiply both sides by 2 to get  3x = 4x-16. Subtract 3x to get x-16 = 0. Add 16 to get the answe
jwec05: Wich I belive are basical same thing
PiratevsPirate: hm...
jwec05: Some people send me 6 as the answer
PiratevsPirate: I wrote the equation this way..
jwec05: I'll explain by steps
jwec05: Let's call D = Daryl's age
jwec05: Now:
PiratevsPirate: (2/3 ) * x + 8 = 2x - 8
LuVeR Of RaIn: sorry to interrupt but i have to go to dinner, but if i come back b4 this ends, can someone invite me again?
jwec05: Um
condabomb88: i will
jwec05: I'll post the transcript
LuVeR Of RaIn: k
LuVeR Of RaIn: bye
jwec05: if we have it
condabomb88: just im me when u get back
jwec05: you'll get invited
jwec05: but if not
LuVeR Of RaIn has left the room.
jwec05: just look at the transcript
jwec05: So
PiratevsPirate: (2/3 ) * x = 2x - 8
PiratevsPirate: that's the stuff istn't it?
condabomb88: yup
PiratevsPirate: then how did I miss it?
jwec05: What did you get from that equation?
jwec05: Hmm
jwec05: probably your arithmetic
jwec05: I had:
jwec05: D = 2/3*(2D-8)
jwec05: D = 4D/3-16-3
jwec05: 3D = 4D-16
jwec05: -D = -16
jwec05: D = 16
jwec05: Age problems are very popular on MC
jwec05: They involve basic equations but it is *very*
PiratevsPirate: arithmetic..
jwec05: important that you have the equation right
PiratevsPirate: whoaps...
jwec05: wrong equation will give wrong answer
jwec05: #5
jwec05: This one is the last one we'll do today
PiratevsPirate: k...
jwec05: and I'll just give few tips about the competition on problem strategy
jwec05: One wholesaler has some apples that cost 32 cents each. Another has some pears that cost 29 cents each.
jwec05: What is the minimum number of pears that can be fairly exhanged for apples
jwec05: Answer is 32
PiratevsPirate: yeah
PiratevsPirate: can I explain?
jwec05: solafidefarms6): We are looking for the LCM of these 2 nubmers, so that we have the minimum amount of money that could buy either a whole number of pears or a whole number of apples. 29 and 32 share no factor
so the LCM is 29*32, so divide by 32, the price for the things we are buying, and we get 29 aplles=32 pears
jwec05: You got the wrong answer
PiratevsPirate: I know I did
jwec05: Go ahead
PiratevsPirate: I misrad the question and tried to fix it but too late
condabomb88: Shoot, i forgot to divide at the end
PiratevsPirate: okay..
PiratevsPirate: since the29  is a prime number
PiratevsPirate: it's just 32
jwec05: Right
PiratevsPirate: I tought it said the cost
jwec05: we're trying to find LCM(32,29)
jwec05: it asked for #
jwec05: this would give what most people have
jwec05: but we have to divide 32 by end
jwec05: This was bit easy for the Target Round
jwec05: The current Target would be much harder
PiratevsPirate: like???
solafidefarms: If this were are target, then Shinwoo would be faszter
condabomb88: Yes, he has a good method
solafidefarms: I did it to start with
solafidefarms: But I figured it's easier to understand the long way
jwec05: Strategy Talk**
jwec05: In MC, here are few things that you SHOULD do
jwec05: It may differ from some people
jwec05: but this definitely help on most cases
jwec05: Sprint Round
jwec05: The aim on this round is to get as many problems as possible
jwec05: don't stop after stuck on the hard one
condabomb88: brb
jwec05: Few types of problems on the Sprint Round wouldn't go too far
PiratevsPirate: and, don't doubt your answer
jwec05: Algebras usually easy to attack
jwec05: Yes
jwec05: Geometry isn't too bad
jwec05: Counting might hurt you little bit
PiratevsPirate: I heard that most answers from MC is like one digit or so
jwec05: because sometimes you try to brute-force
jwec05: not necessarily
jwec05: fractions, square root, a*pi, etc
jwec05: I would recommend that you try to answer as many possible ones on the spring
jwec05: sprint*
jwec05: Target*
jwec05: Target worth more than Sprint
solafidefarms: More time too
jwec05: but it's more easier to get most points from it then Sprint because of calculator
jwec05: yes
jwec05: so you would definitely want to get perfect or at least 7
jwec05: this should help you more close on the Countdown
jwec05: Team*
jwec05: Just one thing
PiratevsPirate: it would be too bad, for those who have broken caluclators like me
jwec05: two people from the front
jwec05: two people from the back
jwec05: that way you can solve them easily
jwec05: I have really no good suggestions for Team
jwec05: Countdown Round*
jwec05: Just to lose out and
jwec05: read the questions fast
PiratevsPirate: Team*
jwec05: but if your opponent gets wrong
PiratevsPirate: I do...
solafidefarms: We do the weakest one does the first 3 or so, next does 3-6, next does 5-10, and m, the strongest, does all of them
PiratevsPirate: divide the team in half
PiratevsPirate: equal powers
PiratevsPirate: and stronger one start with the back
PiratevsPirate: and the weaker one start with the front
PiratevsPirate: and check over
jwec05: That's good suggestion
solafidefarms: I start at the back, and then when I get to about the middle, I do whatever needs doing
PiratevsPirate: my brother is in this school math team
jwec05: Ok
jwec05: that's all I can teach you people today
jwec05: Hope you liked it :-)
solafidefarms: Thank yoU!
PiratevsPirate: yeappy!!!!
jwec05: Message problems are up on the aops
PiratevsPirate: yeah, thanks~!!!!
jwec05: you may try them
jwec05: except
jwec05: I didn't have time to post 30 problems
solafidefarms: Never done AOLIM before!
jwec05: I think there are like 12 something
jwec05: Ha..
PiratevsPirate: good session
jwec05: someone can post the transcript
solafidefarms: I checked out the Computer chat at the beginning and it is just spam!
solafidefarms: I'll do the unedited transcript of you
jwec05: you can just post on aops
jwec05: with title "Week 1 Mathchat Transcript"
jwec05: on MC Forum
solafidefarms: OK
solafidefarms: Where I am a moderator! ;-)
jwec05: Oh really?
jwec05: I haven't notice that yet*
solafidefarms: Check it out!
jwec05: congrats*
solafidefarms: I took you over though? :-(
PiratevsPirate: what's your user name in AoPS solafidefarms
solafidefarms: Yes
jwec05: I was busy over few days
jwec05: writing my third round problems
jwec05: proof one
solafidefarms: Weren't you a mod at the MC before?
solafidefarms: The MC FOrum I mean
PiratevsPirate: wait, who are the qualifiers for AOPS round 1?
PiratevsPirate: the basics?
jwec05: Um
jwec05: let see
jwec05: I can't rememer
jwec05: there were like 5
solafidefarms: Nether can I
solafidefarms: Yif man used to be one...
PiratevsPirate: is round 2 posted?
jwec05: Yes
PiratevsPirate: crap!!!!
jwec05: It's due today
PiratevsPirate: round2 ?
PiratevsPirate: how come?
jwec05: Yes
jwec05: Because
jwec05: I posted a week ago
PiratevsPirate: how come i never received it?
PiratevsPirate: no you didn't
PiratevsPirate: you said
jwec05: I didn't have time to send each person a message
PiratevsPirate: ah!!!!!
PiratevsPirate: my computer broke for almost a week  so I didn't have much chance to go on AoPS
jwec05: For those who might not know what he's talking about
jwec05: this is competition I hold on aops
PiratevsPirate: can you help me find where it is?
jwec05: http://artofproblemsolving.com/Forum/topic-28904.html
jwec05: I have to admit that the round 3, whoever is taking it, that the test is pretty tough
PiratevsPirate: y did you post it on getting started section?
jwec05: because I'm Getting Started Moderator*
jwec05: And I post all the problems on there
jwec05: only reason COPSD was on Basics
---------------------------End---------------------------------
If you have another copy, PM me and I will change it.

There was more to the math chat.. but idk if anyone has a copy of the entire thing... sorry I should have made a copy...

can i still go to the second mathchat even though i didn't make nationals? and next year, if y'all not in mathcounts anymore, are y'all still gonna come to this forum?

i might come back to this forum next year even though i wont be in mathcounts anymore, i could post problems and stuff maybe, but i wont come very often like i do this year

Next year I'll help around here, but expect me more often on the AMC forums...

[quote="solafidefarms"]I might, but not too often. I want to get better, so I will move up to the high school problems. I'll still be on AoPS though, and then perhaps my brothers will when they are old enough.

Vestavia Hills High School came over here to GA Tech to the high school math competition they have in February, and there is an award for highest-scoring 8th grader. It is pretty tough. You might want to try for that next year.

BIlly[/quote]

if it was Vestavia Hills High School, then how can 8th graders compete? hm... who was it this year? i don't remember hearing anything about it....

On the question on 'how can 8th graders compete?', well, some high schoools have 8th graders, and other people are homeschooled, like me. (I won the highest-scoring 8th grader award this year, and it is a TI-84+![at least this year!])

I don't remember who all was there, but Vestavia Hills High's team was in the top 10 I believe. (out of ~90 there)

Billy

cool all of those problems were from the 91-92 national test. I wrote solutions to all of those like months ago and can get them here http://fbmathscience.tophonors.com


conics parabola Asymptote hyperbola ellipse

Is there a formula to discover where the focus, and directrix is in a given parabola. Thanks in advance.

[b]Parabola with Vertical Axis:[/b]
The graph of the equation
	$x^2 = 4py$
is the parabola with the following properties:
	VERTEX		V(0,0)
	FOCUS		F(0,p)
	DIRECTRIX		y = -p
The parabola opens upward if p > 0 or downward if p < 0.

[b]Parabola with Horozontal Axis:[/b]
The graph of the equation
	$y^2 = 4px$
is the parabola with the following properties:
	VERTEX		V(0,0)
	FOCUS		F(p,0)
	DIRECTRIX		x = -p
The parabola opens to the right if p > 0 or to the left if p < 0.

Does that help? I wrote this up a long time ago for an assignment... I hope it makes sense ;)

[quote="sirorange"]But what about if the vertex ISN'T the origin? Also, how does one find the vertices, asymptotes, and foci, of a hyperbola?[/quote]

http://www.analyzemath.com/parabola/Equation.html

There's a lot of information there regarding the shifting conic sections. The standard (or general, i always mix those two up) formulas are:

Parabola:
$(y - k)^2 = 4p(x - h)$

Ellipse:
$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$

Hyperbola:
$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$

equation for horizontal opening hyperbola:
$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$
vertices:
$(-a+h,k)$ and $(a+h,k)$
finding foci:
$c^2=a^2+b^2$
foci are $(-c+h,k)$ and $(c+h,k)$
asymptotes:
$y=\pm\frac{b}{a}(x-h)+k$

for vertical parabolas u just have to switch some things around

yeah conics can be a real butt sometimes

Thank you. Does anyone have any more information they'd like to share?:D


Galois Theory
advanced fields
advanced fields unsolved

how does units helped in fiinding class group? (not quatratic field) Please give some examples. Thanks in advance.

$ k$ a global field, $ S_\infty \subseteq S \neq \emptyset$ a set of primes of $ k$, $ \mathcal{O}_S$ the ring of integers of the maximal separable extension $ k_S/k$ unramified outside $ S$, $ G_S \equal{} \mathrm{Gal}(k_S/k)$, then: $ \mathrm{H}^1(G_S,\mathcal{O}_S^\times) \equal{} \mathrm{Cl}_S(k)$, the $ S$-class group.

(If $ X$ is a scheme, $ \mathrm{H}^1_{\mathrm{fl}}(X,\mathbf{G}_m) \equal{} \mathrm{H}^1_{\mathrm{\acute{e}t}}(X,\mathbf{G}_m) \equal{} \mathrm{H}^1_{\mathrm{Zar}}(X,\mathbf{G}_m) \equal{} \mathrm{Pic}(X)$.)

I don't understand at all, do you have any elementary example?


Let $ \triangle ABC$ be a triangle of whom $ BE,CF$ are the altitudes and $ H$ is the orthocenter of. Let $ M$ be the midpoint of $ BC$ and $ K$ be a moving point on $ EF$. Denoted by $ P$ the intersection of $ BK$ and $ ME$ and $ S$ the intersection of $ AP$ and $ HK$. Find the locus of $ S$ when $ K$ moves on $ EF$.

Let be the point $ S\equiv (N)\cap HK$, where $ (N)$ is the circumcircle of the quadrilateral $ AEHF,$ with diameter $ AH.$

The line segment $ AS,$ intersects the line segment $ ME$ at point so be it $ P'$ and we will provwe that $ P'\equiv P.$

$ \bullet$ The line segment $ ME$ tangents to $ (N)$, at point $ E$ as well.

We consider now, the cyclic non-convex degenerated hexagon $ AFEE'HS$, where $ EE'$ means here, the line segment $ ME,$ as the tangent line of $ (N)$.

So, based on the [b][size=100]Pascal theorem[/size][/b], we have that the points $ B\equiv AF\cap E'H$ and $ K\equiv FE\cap HS$ and $ P'\equiv AS\cap EE'$ are collinear and then, we have that $ P'\equiv P$.

Hence, we conclude that the point $ S$ lies on the constant circle $ (N)$ with diameter $ AH,$ which is the wanted Locus.

Kostas Vittas.

Dear Hieu and Vistako.
Here is a related topic :http://www.mathlinks.ro/viewtopic.php?t=305093
 :)


This is problem 7.24 in the intermdeiate counting and probability book...

"A bin has 10 red balls and 8 blue balls. We randomly draw out 6 balls, one at a time, wihtout replacement. What is the probability that, at some point, we choose two consectuive balls that are red?"

[hide="official solution"]
We can re-engineer the problem a bit by thinking of first drawing all 6 balls at once, then ordering them. We'll further break this up into cases based on the number of red balls that we draw. If we draw r red balls and b blue balls, then there are (10_C_r)(8_C_b) ways to draw this group from our bin, out of a total of (18_C_6) ways to draw any 6 balls. For each configuration, we'll determine the probability that the balls are ordered with 2 red balls consecutive.

If r = 0 or r = 1, then clearly we cannot have two red balls consecutive, so the probability is 0.

If $ r \equal{} 2$ then there are 5 arragnements fo the balls with the two red balls consecutive, and $ C(6,2)$ = 15 total arrangements. So teh probabilty is $ \frac {1}{3}$.

If r = 3, then the only arrangements that fail, out fo teh $ \binom{6}{3} \equal{} 20$ total arrangements, are teh arragements RBRBRB and BRBRBR. So the probability is $ \frac {9}{10}$.

If r $ \ge$ 4, then we are guaranteed to have two red balls consecutive.

Therefore, the probability is

$ \frac {\frac {1}{3}\binom{10}{2}\binom{8}{4} \plus{} \frac {9}{10}\binom{10}{3}\binom{8}{3} \plus{} \binom{10}{4}\binom{8}{2} \plus{} \binom{10}{5}\binom{8}{1} \plus{} \binom{10}{6}}{\binom{18}{6}} \equal{} \frac {1050 \plus{} 6048 \plus{} 5880 \plus{} 2016 \plus{} 210}{18564} \equal{} \frac {15204}{18564} \equal{} \frac {181}{221}$[/hide]

Note that: "When computing probability by counting outcomes and using P(success) = \frac{Number of successful outcomes}{Number of possible outcomes} this approach will work only if the outcomes are equally likely." (pg 16)

But in the solution to problem 7.24, the probability of the different outcomes are NOT equal! For example, when r is greater than or equal to 4, the probability of that event is less because there are differing numbers of red balls (10) and blue balls (8) to start with!

But the solution assumes that the outcomes are equal...

Can someone explain this to me?

Any help would be most appreciated. I thank you in advance :)

Going straight to the direct computation of the probability, note that each probability is scaled by the number of occasions that such a possibility could occur.

Think of this problem; you are flipping three coins. Compute the probability that you flip two consecutive heads.

The answer is $ 3/8$, because out of HHH, HHT, HTH, THH, HTT, THT, TTH, AND TTT, there exist three valid combinations.

Case 1: You flip no heads. This can happen in one way, but the probability is zero.

Case 2: You flip one head. This can happen in three ways, but the probability is zero.

Case 3: You flip two heads. This can happen in three ways. The probability of flipping two consecutive heads in this case is $ 2/3$, so there are two valid combinations.

Case 4: You flip three heads. This can happen in one way, and the probability is $ 1$, so there is one valid combination.

So the total number of combinations is $ \binom{3}{3}\binom{3}{0} \plus{} \binom{3}{2}\binom{3}{1} \plus{} \binom{3}{1}\binom{3}{2} \plus{} \binom{3}{0}\binom{3}{3} \equal{} 1 \plus{} 3 \plus{} 3 \plus{} 1 \equal{} 8$.

The total number of valid combinations is $ 0\times \binom{3}{3}\binom{3}{0} \plus{} 0 \times \binom{3}{2}\binom{3}{1} \plus{} \frac {2}{3} \times \binom{3}{1}\binom{3}{2} \plus{} 1 \times \binom{3}{0}\binom{3}{3} \equal{} 0 \plus{} 0 \plus{} 2 \plus{} 1 \equal{} 3$.

So the probability is $ \frac {3}{8}$.

Go back to your problem. $ \frac {1}{3}\binom{10}{2}\binom{8}{4}$ is the number of valid combinations with two red balls. The same goes with each other term, which represents the number of total combinations multiplied by the probability that a random combination is valid.


At 10:00 a.m., Boon Tee is the $ 225^\textrm{th}$ person in
line to ride the Rocker Roller Coaster. Each
roller coaster train holds 36 people. A full
train leaves every four minutes. If the first
36 people in line leave on the 10:01 train,
what time will Boon Tee's train leave?

225/36=6.25, so Boon Tee will get on the 7th ride.

The rides occur every four minutes, so 6 rides after the first one is 10: (1+6(4))=10:25.


Here is a rather simple but tricky problem:

The number 165 can be written as a sum of four numbers:
165 = 82 + 36 + 13 + 34
You are required that, by keeping the order of the above numbers, without switching any digit from a number to another, to obtain the number 165.

good luck  :)

[quote]by keeping the order of the above numbers[/quote]

what numbers?
do you mean

[quote]82 + 36 + 13 + 34 [/quote]

Then you would just add them like you did.

[quote]The number 165 can be written as a sum of four numbers:
165 = 82 + 36 + 13 + 34
You are required that, by keeping the order of the above numbers, without switching any digit from a number to another, to obtain the number 165. [/quote]

I think you mean that you change the +'s into division signs, subtraction signs, and multiplication signs right? :huh:


Determine whether there are two infinite point sequences $ A_1,A_2,\ldots$ and $ B_1,B_2,\ldots$ or not, such that for all $ i,j,k$ that satisfies $ 1\le i<j<k$,
 i.$ B_k$ is on the line that passes through $ A_i$ and $ A_j$ if and only if $ k\equal{}i\plus{}j$.
 ii.$ A_k$ is on the line that passes through $ B_i$ and $ B_j$ if and only if $ k\equal{}i\plus{}j$.

[hide]Taking $ A_k \equal{} (k,k^3)$ and $ B_k \equal{} ( \minus{} k, \minus{} k^3)$ for all $ k$ positive integers, satisfies the conditions.[/hide]


calculus
integration
logarithms
limit
calculus computations

integrate

$ \int\;\;\frac{y!}{(y+m)!}\;\;\textbf dy$

[ answer hidden below.  :D ]

[hide]
the answer that i get is: $ \boxed{\sum_{j=1}^{m}\;\frac{(-1)^{j-1}}{(j-1)!\;\;(m-j)!}\;\ln{|\;y+j\;|}}$
[/hide]

Decompose $ \frac{1}{(y+1)\cdots(y+n)}$ into $ \frac{a_{1}}{y+1}+\cdots+\frac{a_{n}}{y+n}$. To determine each $ a_{k}$, consider

$ \lim_{x\to-k}(x+k)\left[ \frac{a_{1}}{y+1}+\cdots+\frac{a_{n}}{y+n}\right].$

This limit is $ a_{k}$, since each $ \frac{a_{i}(y+k)}{y+i}$ goes to $ 0$ for $ i \neq k$.

This means that $ a_{k}\; = \; \frac{1}{(1-k)\cdots(k-1-k)\cdot 1 \cdots (n-k)}\; = \; \frac{(-1)^{k-1}}{(k-1)!(n-k)!}.$

Now all you have to do is just to integrate each term.


let x,y,z be positive real nos. such that
 x+y+z+xy+yz+zx+xyz=k -a fixed real no.
 then maximise xyz(x+y+z).