trigonometry
complex numbers
geometry solved
geometry

In a complex number plane let the centre of the two circles be the origin.  Represent points $A$, $B$, $C$, $A'$, $B'$, $C'$, $P$ and $P'$ by complex numbers $z$, $\omega z$, $\omega ^2 z$, $z'$, $\omega z'$, $\omega ^2 z'$, $p$ and $p'$ respectively, with $|p| = |z|$ and $|p'| = |z'|$.

Then
\begin{eqnarray*}\\ P'A^2 + P'B^2 + P'C^2&=&|p'-z|^2+|p'-\omega z|^2+|p'-\omega ^2z|^2\\ &=& 3|p'|^2+3|z|^2\\ &=& 3|p|^2+3|z'|^2\\ &=&|p-z'|^2+|p-\omega z'|^2+|p-\omega ^2z'|^2\\ &=&A'P^2 + B'P^2 + C'P^2. \end{eqnarray*}
(I would be more explicit about the step where you expand $|p'-z|^2+|p'-\omega z|^2+|p'-\omega ^2z|^2$, but I can't work out how to do complex-conjugate signs in LaTex.)

Well then,
\begin{eqnarray*} &&|p'-z|^2+|p'-\omega z|^2+|p'-\omega ^2z|^2\\ &=&(p'-z)\overline{(p'-z)}+(p'-\omega z)\overline{(p'-\omega z)}+(p'-\omega ^2z)\overline{(p'-\omega ^2z)}\\ &=&(|p'|^2+|z|^2-p'\overline{z}-\overline{p'}z) +(|p'|^2+|\omega z|^2-p'\overline{\omega z}-\overline{p'}\omega z) +(|p'|^2+|\omega ^2z|^2-p'\overline{\omega ^2z}-\overline{p'}\omega ^2z)\\ &=& 3|p'|^2+3|z|^2-p'\overline{(1+\omega +\omega ^2)z}-\overline{p'}(1+\omega +\omega ^2)z\\ &=& 3|p'|^2+3|z|^2. \end{eqnarray*}

(Oooh, look at all the pretty little complex-conjugate signs . . . .)

[quote="Rushil"]Let $\Gamma$ and $\Gamma'$ be two concentric circles. Let $ABC$ and $A'B'C'$ be any two equilateral triangles inscribed in $\Gamma$ and $\Gamma'$ respectively. If $P$ and $P'$ are any two points on $\Gamma$ and $\Gamma'$ respectively, show that \[ P'A^2 + P'B^2 + P'C^2 = A'P^2 + B'P^2 + C'P^2.  \][/quote]
Here is a trigonometry proof.
Let the radii of $\Gamma$ and $\Gamma'$ be $R$ and $r$ respectively, $R>r'$, let $O$ be the centre of the two circles.
Let $\angle P'OB$ be $\alpha$. Then:
$P'B^2=R^2+r^2-2Rrcos\alpha$      (*);
$ABC$ is euilateral with circumcentre at $O$ hence:
$\angle AOB=\angle AOB=\angle AOB=120^o$. So:
$P'B^2=R^2+r^2-2Rrcos(120^o-\alpha)$     (**)
$P'B^2=R^2+r^2-2Rrcos(120^o+\alpha)$     (***)
Adding (*),(**),(***) and simplifying get:
$P'A^2+P'C^2+P'B^2=3r^2+3R^2$.
In the same way we can prove that:
$PA^2+PC^2+PB^2=3r^2+3R^2$.
The result follows.

Sorry to revive an old thread. Mods, pl. move this to the soved section.
I am just adding another soln{own} which uses only pure geometry.{Though this is very much only another way of writing the trigo solution as posted before}

[color=red]Theorem: For any triangle[/color] $ ABC$ [color=red]and a point [/color]$ P$ [color=red]in the same plane as the triangle[/color],
$ PA^2 \plus{} PB^2 \plus{} PC^2 \equal{} GA^2 \plus{} GB^2 \plus{} GC^2 \plus{} 3PG^2$ [color=red]where[/color] $ G$ [color=red]is the centroid of [/color]$ \delta ABC$.

In this problem, $ O$ which is the centre for the two concentric circles will be the centroid for both the equilateral triangles and thus applying the theorem for each triangle,
We get that $ LHS \equal{} RHS \equal{} 3R^2 \plus{} 3r^2$.
$ QED$.

This problem could also be done with the help of moment of inertia. Let O be centre of the circles.
Place three equal masses on A B C then centre of mass of system is at O.
So P'A²+P'B²+P'C²=OA²+OB²+OC²+3OP'²(by parallel axis theorem) which equals to 3r²+3R².
Similarly for other we get the same expression.


A PIN consists of four digits. For extra security, the bank requires that your PIN satisfy ALL of the requirements listed below.

1}the digits cannot all be the same
2}the digits cannot be a repeated pair (such as 7373 or 4040)
3}the digits cannot be in increasing/decreasing order (such as 9876 or 4567) [0 is considered to be the smallest digit]

How many different PINS are there that all follow the requirements above.

I live in awe of anyone (other than THE Richard Russcyk, I already live in awe of him.) that can solve this correctly.

[color=blue]Moderator kyyuanmathcount's note: During competition season, we ask you put the source of all problems down.  I'll stick it in.[/color]

[hide]$10^4$ "raw" possibilities
10000
-10, cuz the digits cannot be the same
repeated pairs, how many are there?
10 for the first, 9 for the second, cuz if they are the same, all digits are the same
-90
ascending-there are 9876-0123, or 9876543, count them, 7
same goes for descending
$10000-90-10-14=\boxed{9886}$[/hide]

[quote="X_wing1redleader"][hide]$10^4$ "raw" possibilities
10000
-10, cuz the digits cannot be the same
repeated pairs, how many are there?
10 for the first, 9 for the second, cuz if they are the same, all digits are the same
-90
ascending-there are 9876-0123, or 9876543, count them, 7
same goes for descending
$10000-90-10-14=\boxed{9886}$
Hi![/hide][/quote]

Somehow I doubt thats right

[hide]
Well, there are 10000 total pin numbers to choose from.  Obviously, any pin number with digits all the same digits are a repeated pair, so don't even consider the ones with the same digits.

There are 100 repeated pairs (00-99).  So subtract these from 10000 to get 9090.

Now for the increasing/decreasing digits:
Consider the number a12x.  There are 7 possible values of x.  Consider the number a13x.  There are 6 possible values of x.  Continuing in this fashion, you see that there are $\frac{7(7+1)}{2}=28$ possible values of x for a1_x.  There is only one possible value of a: 0.
Now consider a2_x.  There are $\frac{6(6+1)}{2}=21$ possible values of x for a2_x.  But now there are 2 possible values of a: 0 and 1, so there are $42$ possible numbers in total.  For a3_x, there are 15 possible values of x, but there are 3 possible values of a, so there are $45$ possible numbers in total.  For a4_x, there are 10 possible values of x and 4 possible values of a for $40$ total values.  For a5_x, there are 6 possible values of _x and 5 possible values of a for $30$ total.  For a6_x, there are $3\cdot 6=18$.  For a7_x, there are $1\cdot 7=7$

Adding these together,
$28+42+45+40+30+18+7=210$.  Multiply this by 2 to count decending pins.

Subtract $420$ from $9090$ to get $8670$.

Hmm... somehow this doesn't seem right...
[/hide]

[hide]
$10^4=10000$ regular pins.
For a repeated pair, there are $10\cdot10=100$
For increasing,
The lowest digit can be 0-6, for a total of 7.
Multiply by 2 to get the number of total increasing/decreasing.
There are 14 increasing/decreasing.
$10000-100-14=9886$[/hide]

[quote="bryantacademy"]X_wing1redleader, I don't understand your first post in this thread. Could you convert it from [i]"Math Talk"[/i] to [i]"Plain English"[/i]?[/quote]

well how do that i thought it is in plain english...

I happen to have the actual workout and "CORRECT" solution(that comes from the MC handbook).

[hide="The answer is..."][hide="are you ready?"][hide="are you sure?"]9886 pins[/hide][/hide][/hide]

So...increasing applies only if it increases by 1?

[quote="biffanddoc"][quote="robinhe"][quote="biffanddoc"][hide]
$10^4=10000$ regular pins.
For a repeated pair, there are $10\cdot10=100$
For increasing,
The lowest digit can be 0-6, for a total of 7.
Multiply by 2 to get the number of total increasing/decreasing.
There are 14 increasing/decreasing.
$10000-100-7=9893$[/hide][/quote]

There are more increasing/decreasing than that.  The last three digits don't have to be 7-8-9.

You can solve this with casework.[/quote]
It depends how you interpret "increasing"
is 1235 increasing or not?
by the examples they gave, i am inclined to say no[/quote]

Yeah, I just looked and the [i]real[/i] problem said they had to be [b]consecutive[/b], so how how was I supposed to know? bryantacademy never said anything.  :huuh:

you live in awe of a bunch of people now o.o

[hide]9876, 8765, 7654, 6543, 5432, 4321, 3210, 0123, 1234, 2345, 3456, 4567, 5678, 6789. That's 14 numbers with digits arranged in decreasing and increasing order.

There are 100 numbers that are in the form abab and aaaa (ab=00-99), so since there is a total of $10^4$=10000 numbers possible, the number of avaliable pin numbers is 10000-14-100=9886. [/hide]


limit
trigonometry
algebra proposed
algebra

$\lim_{x\rightarrow\infty}x\sin{\frac{1}{1+x^2}}=?$

Saman Gharib :huh:

$\mathop {\lim }\limits_{x \to + \infty } \left( {x\sin \frac{1}{{1 + x^2 }}} \right) = \mathop {\lim }\limits_{x \to + \infty } \left( {\frac{{\sin \frac{1}{{1 + x^2 }}}}{{\frac{1}{{1 + x^2 }}}} \cdot \frac{x}{{1 + x^2 }}} \right) = 0$


linear algebra
matrix
superior algebra
superior algebra solved

$A,B\in M_{2n+1}(R)$ such that

$\forall t \in R, \; e^{t^2(A+B)}=e^{tA}e^{tB}$

Prove that $det(AB)\leq 0$

That $t^2$ looks rather odd. But maybe you really did mean that.

Start with $e^{t^2(A+B)}=e^{tA}e^{tB}$ and differentiate with respect to $t$.

$2t(A+B)e^{t^2(A+B)}=e^{tA}(A+B)e^{tB}.$ Now set $t=0.$

$0=I(A+B)I,$ so $A+B=0,$ or $B=-A.$ Then $\det(AB)=\det(-A^2)=(-1)^{2n+1}(\det(A))^2\le0.$


hi,
determine all polynominals $ P$, with integer coeffitients with the property,
\[ P(P'(x)) \equal{} P'(P(x)),\]
$ \forall x \in \mathbb{R}$.

For $ P(x)$ being a constant , $ P(x) \equal{} 0$ is obviously the only solution.
For $ P(x)$ not a constant, just consider the highest power of $ x$ of both sides:

Let $ P(x) \equal{} ax^n \plus{} ...$ we have
$ a(anx^{n \minus{} 1})^n \equal{} an(ax^n)^{n \minus{} 1}$ for all $ x$ if $ n \ge 2$
Simplify it we get $ an^{n \minus{} 1} \equal{} 1$, impossible.

Now let $ P(x) \equal{} ax \plus{} k$,
$ P(P'(x)) \equal{} a \Leftrightarrow a P'(x) \plus{} k \equal{} a \Leftrightarrow a^2\plus{}k\equal{}a$,
so $ P(x) \equal{} ax\plus{}(a\minus{}a^2)$ is the only solution for non-constant $ P(x)$


The function $f: \mathbb{R}\rightarrow\mathbb{R}$ satisfies $f(\textrm{cot}x)=\sin2x+\cos2x$, for any $x\in(0,\pi)$. Find the minimum and maximum value of $g: [-1;1]\rightarrow\mathbb{R}$, $g(x)=f(x)\cdot f(1-x)$.

another solution http://www.mathlinks.ro/Forum/viewtopic.php?p=786051#786051 :P


For non-negative $ a,$ $ b$ and $ c$ prove that
\[ a^2b^2(a\minus{}b)^2\plus{}b^2c^2(b\minus{}c)^2\plus{}c^2a^2(c\minus{}a)^2\geq2abc(a\minus{}b)(b\minus{}c)(c\minus{}a)\]

[quote="vg23nov"]assume a>= b>= c
clearly LHS is more than zero because trivial inequality
RHS is less than 0 because a-b>= 0 and b-c>= 0 but c-a<=0
so RHS is less than zero[/quote]
... which is exactly why we can't assume this. ;)

[quote="arqady"]For non-negative $ a,$ $ b$ and $ c$ prove that
\[ a^2b^2(a - b)^2 + b^2c^2(b - c)^2 + c^2a^2(c - a)^2\geq2abc(a - b)(b - c)(c - a)
\]
[/quote]
[hide="Solution"]
Notice that $ a - b\le a$, $ b - c\le b$, and $ c - a\le c$, so $ abc\ge (a - b)(b - c)(a - c)$.  Since $ abc\ge 0$, we see that $ 27abc\ge 8(a - b)(b - c)(a - c)$, so
\[ 27a^4b^4c^4(a - b)^2(b - c)^2(a - c)^2\ge 8a^3b^3c^3(a - b)^3(b - c)^3(a - c)^3
\]

\[{ 3\sqrt [3]{a^2b^2(a - b)^2\cdot b^2c^2(b - c)^2\cdot a^2c^2(a - c)^2}\ge 2abc(a - b)(b - c)(a - c)}
\]
AM-GM gives that
\[ \sum_{cyc} {a^2b^2(a - b)^2}\ge 3\sqrt [3]{a^2b^2(a - b)^2\cdot b^2c^2(b - c)^2\cdot a^2c^2(a - c)^2}
\]
and it follows that
\[ \sum_{cyc} {a^2b^2(a - b)^2}\ge 2abc(a - b)(b - c)(a - c)
\]
as desired.
 [/hide]

Here is my solution:
Applying AM-GM for each two pairs, we will have:
LHS>=abc[a./a-b/./b-c/+c/c-a/./b-c/+a/a-b/./c-a/]( /x/ means the absolute value of x)
Now it's sufficient to prove:[b./a-b/./b-c/+c/c-a/./b-c/+a/a-b/./c-a/]>=2(a-b)(b-c)(c-a)(1)
If a>=b>=c then this inequality is always true.Therefore the inequality is proved easily!
If a>=c>=b then we see that if we decrease each variable by the same amount of b=min(a,b,c), then the LHS of (1) is decreased, but the RHS of (1) is still the same!
Therefore, we just need to prove that (1) is true when one variable is equal to 0. Since b=min(a,b,c), b=0!
Now (1) is equivalent to: c^2(a-c)+a^2(a-c)>=-2ac(a-c), which is equivalent to: (a-c)^3>=0(true since a>=c>=b)
Therefore, the inequality is proved in this case!
From 2 cases above, we are done!
THe equality occurs when a=b=c or a=0,b=c and any permutation of it! :)

[quote="The QuattoMaster 6000"]
Notice that $ a \minus{} b\le a$, $ b \minus{} c\le b$, and $ c \minus{} a\le c$, so $ abc\ge (a \minus{} b)(b \minus{} c)(a \minus{} c)$.  Since $ abc\ge 0$[/quote]
Try $ a \equal{} 5,$ $ b \equal{} 1$ and $ c \equal{} 2.$ :wink:
Very nice proof, ghjk!  :lol:

[quote="arqady"]For non-negative $ a,$ $ b$ and $ c$ prove that
\[ a^2b^2(a \minus{} b)^2 \plus{} b^2c^2(b \minus{} c)^2 \plus{} c^2a^2(c \minus{} a)^2\geq2abc(a \minus{} b)(b \minus{} c)(c \minus{} a)
\]
[/quote]
My solution is the following
By the AM-GM Inequality, we get
\[ \sum [a^2b^2(a \minus{} b)^2] \equal{} \frac {1}{2}\sum a^2[b^2(a \minus{} b)^2 \plus{} c^2(a \minus{} c)^2] \ge abc\sum a\left\vert (a \minus{} b)(a \minus{} c)\right\vert
\]
and
\[ 2abc(a \minus{} b)(b \minus{} c)(c \minus{} a) \le 2abc\left\vert (a \minus{} b)(b \minus{} c)(c \minus{} a)\right\vert
\]
It suffices to prove that
\[ \sum a\left\vert (a \minus{} b)(a \minus{} c)\right\vert \ge 2\left\vert (a \minus{} b)(b \minus{} c)(c \minus{} a)\right\vert
\]
(*)
\[ \Leftrightarrow \sum\left\vert \frac {a}{b \minus{} c}\right\vert \ge 2
\]
Assume $ a \ge b \ge c$, then we have
\[ LHS \ge \frac{a}{b \minus{} c} \plus{} \frac {b}{a \minus{} c} \ge \frac {a}{b} \plus{} \frac {b}{a} \ge 2
\]
We have done. :)
Similarly, we used this way to prove that following
http://www.mathlinks.ro/viewtopic.php?t=149730
:)

Nice proof, can_hang2007!
I think, for the following inequality your reasoning doesn't work.

For non-negative $ a,$  $ b$  and $ c$ prove that:
\[ a^2b^2(a \minus{} b)^2 \plus{} b^2c^2(b \minus{} c)^2 \plus{} c^2a^2(c \minus{} a)^2\geq2\sqrt2abc(a \minus{} b)(b \minus{} c)(c \minus{} a)
\]

[quote="pxchg1200"]For non-negative $ a,$  $ b$  and $ c$
\[ a^2b^2(a \minus{} b)^2 \plus{} b^2c^2(b \minus{} c)^2 \plus{} c^2a^2(c \minus{} a)^2\geq k abc(a \minus{} b)(b \minus{} c)(c \minus{} a)
\]
What is the greatest $k$ ?[/quote]

using bottema find that $k_{max} $ is $2\sqrt{2}$


The values of $ a$, $ b$, $ c$ and $ d$ are 1, 2, 3 and 4, though not
necessarily in that order. What is the greatest possible value
of $ a\cdot b\plus{}b\cdot c\plus{}c\cdot d\plus{}d\cdot a$?

hello, the greatest possible value of
$ ab\plus{}bc\plus{}cd\plus{}da\equal{}25$ for
$ a\equal{}1,b\equal{}2,c\equal{}4,d\equal{}3$
Sonnhard.

[hide=Solution]
Note $ab+bc+cd+da= b(a+c)+ d(a+c) = (a+c)(b+d)$. From here it's easiest to guess and check. Otherwise we use that $a+b+c+d=10$. If we let $a+b=x$, then $c+d=10-x$. We want to maximize $x(10-x)=-x^2+10x$. Take the derivative or notice that this is an upside down parabola so the maximum is at $x=\frac{-b}{2a}=5$. Then $x(10-x)=5^2=\boxed{25}$

I feel like this type of idea could be used to make some nice, more challenging problems (maybe using six variables or something)....
[/hide]


Angles $ A$ and $ B$ are complementary. Angles $ A$ and $ C$ are supplementary. How many degrees are in angle $ B$ if angle $ C$ is $ 106^{\circ}$

hello, from the problem we get the system $ A\plus{}B\equal{}90^{\circ}$ and $ A\plus{}106^{\circ}\equal{}180^{\circ}$, solving this we get $ A\equal{}74^{\circ}$, $ B\equal{}16^{\circ}$,
Sonnhard.


On an east-west shipping lane are ten ships sailing individually. The first five from the west are sailing eastwards while the other five ships are sailing westwards. They sail at the same constant speed at all times. Whenever two ships meets, each turns around and sails in the opposite direction. When all ships have returned to port, how many meetings of two ships have taken place?

Using a careful diagram, we have [hide]$ 25$.  Each meeting of ships adds one meeting in the next "round" if there are more ships traveling.  As ships move back to ports, the meetings decrease by one.  $ 1\plus{}2\plus{}3\plus{}4\plus{}5\plus{}4\plus{}3\plus{}2\plus{}1\equal{}25$.[/hide]

[hide="Or..."]Pretend that every time two ships meet, they just pass each other instead of turning directions. The ships will take the exact same path, except the two ships will switch places. However, this will lead to the same amount of meetings. Now each ship heading east will meet all 5 ships heading west, giving 25 meetings total.[/hide]

[hide="I think this works"]When two ships meet, and they turn around, they would be following the same paths

as they would if they don't. Therefore, the problem is equivalent to if the ships continue

to pass each other without turning around. The answer will then clearly be $ \boxed{25}$,

because each ship going in the East direction will meet the ships going in the West

direction, giving a total of $ 5 \cdot 5\equal{}25$.

1000TH POST!!!!!!!![/hide]

That's what I said, except in a less confusing way. :P


AMC
AIME
algorithm
number theory
greatest common divisor
calculus
integration

Let $a_{n}=100+n^{2}$ be the $n$th term of a sequence, where $n$ can range over the positive integers. Let $d_{n}$ equal the greatest common divisor of $a_{n}$ and $a_{n+1}$. Find the maximum value of $d_{n}$ as $n$ ranges over the positive integers.

[hide="Big hint"] Euclidean Algorithm. ;) [/hide]

[hide]
Im not sure if this is right, but by the Euclidean algorithm (from t0r's hint) we have:

$\gcd(100+n^{2}, 100+(n+1)^{2})=\gcd(100+n^{2}, 2n+1)= \gcd(98+(n-1)^{2}, 2n+1)$,

You can continue subtracting to get it equal to $\gcd((n-50)^{2}, 2n+1)$, [EDIT: The rest of this was incorrect. Now what do we do?]
[/hide]

[quote="drunner2007"]why do you think that plugging in $n=50$ will give you the largest value of $d_{n}$?[/quote]

I dont know, really.  :P

So then what do you do, once you've got the expression as $\gcd (100+n^{2}, 2n+1)$ (or $\gcd( (n-50)^{2}, 2n+1)$) ?

well there are actually several ways to do it, if elemenhop's hint also leads to the right answer.

when i did it, i noticed that once you're at $(100+n^{2},2n+1)$, you can multiply the left expression by any power of $2$ without affecting the gcd, since the right hand side is odd for all integral values of $n$.

so if you multiply the left expression by an appropriate power of two, and then apply the euclidean algorithm, you can get the correct answer.

the way you've simplified it doesn't seem to suggest any nice values of $n$ unless I'm not looking at it correctly.

[hide="One more hint"] Elemennop's suggestion gives you

$gcd(100+4n^{2}, 100+4n^{2}+4n+1) = gcd(100+4n^{2}, 4n+1)$

Consider the expressions as polynomials in $n$, and I will tell you that the Euclidean algorithm holds with polynomials.

Edit: I will also tell you that $gcd(a, b) = gcd(a+kb, b)$ holds.  [b]With polynomials.[/b]  (Which is to say, $k$ can be a polynomial.) [/hide]

Elemennop wrote:Another hint:

[hide]

Use the -th and -th term.

[/hide]

How do you know to use those terms and not two others...?

Yup...I solved it using Euclidean method only..nd found the GCD to be 401...Is that the answer??

And the answer to the query that why only 2n nd 2n+1 shall be considered is just to avoid fractions in long division....

Plz tell the answer asap
thnx

[quote="ch1n353ch3s54a1l"]How do you know to use those terms and not two others...?[/quote]

Well, there are only two cases.  Either we use the $2n, 2n+1$ terms or the $2n-1, 2n$ terms, and both are algebraically identical.

if we got to find $gcd(100+n^{2},2n+1)$ how do we find it?

This Euclidean algorithm is new to me, so if anyone would like to clarify what to do next, i'd appreciate it.

[hide]So i started with $gcd\left(100+n^{2}+2n+1,100+n^{2}\right)$

Dividing the left term by the right, putting the divisor in the left column and the remainder in the right, i get

$gcd\left(100+n^{2},2n+1\right)$

Doing this again, i get

$gcd\left(2n+1,100.25\right)$

Now, i know that the answer is supposed to be $401$, which is $4$ times $100.25$ but i'd like some clarification as to why i can or can't just multiply the right side by $4$. Thanks[/hide]


LaTeX
trigonometry
algebra unsolved
algebra

Is this your problem?
[quote="tarta"]$z_{1},z_{2},z_{3}\in\mathbb{C}$ and

$|z_{1}|=|z_{2}|=|z_{3}|,z_{1}+z_{2}=z_{3}$

Find value of  $z_{1}^{2006}+z_{2}^{2006}+z_{3}^{2006}$[/quote]
You need learn and use LaTex  :(
see http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php

So if, this is my solution. If $z_{1}=0$ then $z_{2}=z_{3}=0$ and sum need find is $0$. If $z_{1}\not = 0$ then denote $x_{1}+y_{1}i=a=\frac{z_{2}}{z_{1}},x_{2}+y_{2}i=b=\frac{z_{3}}{z_{1}}$, by hypothesis $1=|a|=|b|$ and $1+a=b$, therefore $a=\cos{\frac{2\pi}{3}}+i\sin{\frac{2\pi}{3}},b=\cos{\frac{\pi}{3}}+i\sin{\frac{\pi}{3}}$ or $a=\cos{(-\frac{2\pi}{3})}+i\sin{(-\frac{2\pi}{3})},b=\cos{(-\frac{\pi}{3})}+i\sin{(-\frac{\pi}{3})}$. Now use $(\cos{\alpha}+i\sin{\alpha})^{n}=\cos{n\alpha}+i\sin{n\alpha}$. Can you continuous?

I don't know from where you say that a= cos 2pi/3 + i sin 2pi/3, b=cos pi/3 + i sin pi/3  and a=cos(-2pi/3)+i sin(-2pi/3) , b=cos(-pi/3)+i sin(-pi/3)


Yah, I got that too, and I checked it by plotting the points in a graphing calculator and finding a cubic regression model.

In fact, we can state a more general version of OP's question: \[\sum_{k=1}^n(\alpha+k\cdot\beta)^2= \ ?\]In order to compute the latter, we will just open the brackets, pair the sums and bash, as such: \[\sum_{k=1}^{n}(\alpha+k\cdot\beta)^2=\sum_{k=1}^{n}\alpha^2+\sum_{k=1}^n(k^2\cdot \beta^2)+\sum_{k=1}^n2\cdot k\cdot \alpha\beta=\]\[=n\alpha^2+\beta^2\cdot\sum_{k=1}^nk^2+2\alpha\beta\cdot\sum_{k=1}^nk.\]Using the well-known formulae for the sums $1+2+...+n$ and $1^2+2^2+...+n^2$ we can further compute our sum as \[n\alpha^2+\beta^2\cdot \frac{n(n+1)(2n+1)}{6}+\alpha\beta\cdot n(n+1)=\]\[=\frac{n}{6}\bigg(n^2\cdot 2\beta^2+n\cdot \big(3\beta^2+6\alpha\beta
\big)+\beta^2+6\alpha^2+6\alpha\beta\bigg).\]Observe that if we plug in $\alpha=-1$ and $\beta=2$ and use the formula we deduced, we get that \[1^2+3^2+...+(2n-1)^2=\frac{n}{6}\big(8n^2-2 \big)=\frac{n(2n-1)(2n+1)}{3} \ \ \square\]

[quote=Eternal.winner][quote=Treething]You probably know that 1 + 3 + 5 + 7 +...+ n = n^2.

But what is 1^2 + 3^2 + 5^2 + 7^2 + n^2 equal?

There is an answer, and I know it...[/quote]

Do you mean $1 + 3 + 5 + 7 +...+ (2n-1) = n^2$[/quote]

Yes, they meant that


If $ 10\%$ of the people polled who dislike radio also dislike music, and $ 30\%$ of the 1200 people polled disliked radio, how many people polled disliked both?


Can you survive?

http://onemorelevel.com/game/the_package

I've played it twice and I survived each time.

I seriously don't understand. How does the puzzle work?

Cool. I was shocked when the last level was the last one, and that it wasn't longer. The time limit should be way shorter.

[quote="7h3.D3m0n.117"]No way you can find a combination out of $ 10000$ possible ones. Unless it's a trick?

@Temperal: Find a combination to unlock the box.[/quote]
It tells you how many digits you have right, so I just did one digit at a time.

[quote="xpmath"]It tells you how many digits you have right, so I just did one digit at a time.[/quote]

It does? I didn't see that. That would certainly help  :P .

[hide="spoiler"]If you go digit by digit, trying numbers and submitting, the lights below tell you when you're right. The lights don't correspond with the above digit, however.[/hide]

I wish it was longer. :(

[hide="Complete Solution"][hide="Level 1 (lock)"]A light turns green for each correct digit. The digits do not correspond with the lights, but you can figure out which digits correspond to which lights by changing one digit at a time and seeing the result.[/hide]
[hide="Level 2 (keypad)"]Click on the rectangle in the top-left corner, then click on the keys inside. Now, click on the keys in your inventory.
Arrange the keys in the following manner:
[code]2 9 4
7 5 3
6 1 8
  0  [/code]

Finally, enter the number 916312.[/hide]
[hide="Level 3 (binary switches)"]Click on the screwdriver in your inventory.

The switches represent a binary number: top-left is 1, bottom-right is 2048. Enter 1111 in binary.[/hide]
[hide="Level 4 (wires)"]Click on the key in your inventory. Cut the black wire.[/hide][/hide]

xD I did the second level by brute forcing permutations of the 4 keys before figuring it out...

@i_like_pie: I just did that solution and I survived only until the cutting the wires part. All you need to know [hide="is"]hexadecimal and where your calculator is, or just number sense.[/hide]

I DID IT!!!!!!!!  But unforunately, I had to use i_like_pie's solution on the second, third, and fourth step, so it doesn't really count.

Level 2 is easy (MAGIC SQUARE)

Level 3 is also pretty easy. Move all the switches so that

the number is 0. Then, flipping back and forth you see that the

number are powers of 2, and the solution follows.

Level 4 I think either wire works... I pressed Red and won...

sometimes red and sometimes black kills, i think
since few bother to figure out the puzzle most just guess and guess right

I had to use i_like_pie's solution too. I figured out the switches by flipping random switches  :P Pretty cool. The lock in the first part was easy. Brute force.

[hide]Kinda easy: I mean the color codes were obvious.[/hide]

Oh man, I made it to the last part and cut the wrong wire and died.... :P

i dont get the last part!?!?!?! what does that mean!?!?!

The last one is too easy.. multiply the units digits to see if it matches with the one it wants you to get. :|

[quote="1=2"]14:16 left, that's a record! Yay!

The first puzzle changes with each try, but the keypad and the switches stay the same. The wires change, though. If you are really good at number sense, you can figure it out in about 5 sec.[/quote]

That's pro. It took me 4 minutes and I thought that was good (didn't cheat though).

13:52 left, cut a random wire at the end.

Easy but very nice game nonetheless.
For the last part, just multiply the last digit of the numbers and see if it corresponds to the one it wants you to get. You gain some time by that, because you don't do the whole multiplication.  :blush:

For the record, dude, I think the rule is don't bring back threads more than 1 month old.


induction
modular arithmetic
AMC
USA(J)MO
USAMO
number theory proposed
number theory

Prove that for all positive integers $n$ there exists a single positive integer divisible with $5^n$ which in decimal base is written using $n$ digits from the set $\{1,2,3,4,5\}$.

$lemma:$ If $d \leq n$, we can find a multiple $5^nk$ with its last $d$ digits in $\{1,2,3,4,5\}$ and $k<2^d$.

$Proof$ by induction:
- If $d=1$ we can take $k=1$.
- Assume that the lemma is true for $d<n$.
If the $(d+1)-st$ digit of $5^nk$ is in $\{1,2,3,4,5\}$, do nothing. Otherwise replace $k$ by $k'=k+2^d$:
$5^nk'=5^nk+10^d5^{n-d}$, so the last $d$ digits are not changed but the $(d+1)-st$ digit is in $\{1,2,3,4,5\}$.

$Consequence:$ We can find $5^nk$ with its last $n$ digits in $\{1,2,3,4,5\}$.
But $k<2^n\Rightarrow5^nk<10^n\Rightarrow5^nk$ has at most $n$ digits.

Classical inductive solution :

We first construct a squence $(U_n)$ with the desired property.
Let $U_1 = 5$.
Now, assume $U_n$ is constucted as desired. Let $U_n = a_0 + 10a_1 + \cdots +10^{n-1}a_{n-1}$ be the decimal expansion of $U_n$.

- If $U_n = 0 mod[5^{n+1}$ then $U_{n+1} = U_n + 5 \cdots 10^n$. Note that it is the only way to 'complete $U_n$ from the left' (which means that we only add a digit on the left of the decimal expansion of $U_n$) with a digit which has to belong to $1,2,3,4,5.$

- If $U_n \neq 0 \mod[5^{n+1}]$ then $U_n = k\cdots 5^n$ where $k$ is an integer not divisible by $5$. Let $x$ be the residue of $k \mod[5].$
If we want to complete $U_n$ from the left then we have to choose a digit $a$ among $1,2,3,4$ such that $a\cdot 10^n + k \cdot 5^n = 0 \mod[5^{n+1}]$ that is $a\cdot 2^n + k = 0 \mod[5].$ Since $2^n$ is clearly inversible modulo $5$ it follows that there is a unique way to reach our goal which is to choose $a = - \frac k {2^n} \mod [5].$

Thus, in any case, we see that we may define $U_{n+1}$ with the desired property, and that it can be done uniquely if we are forced to complete $U_n$ from the left.

Note that there is only one number with one digit satisfyiong the statement (namely $5$).
Now assume that for some $n$, the number $U_n$ is the only one number with $n$ digits satisfying the statement.
Assume that $V_{n+1}$ is a $(n+1)$-digits satisfying the statement.
Let $V_{n+1} = b_0 + 10 \cdot b_1 + \cdots + 10^n \cdot b_n.$
Since both $V_{n+1}$ and $10^n$ are divisible by $5^n$, it follows that $V_{n+1} - 10^n \cdot b_n = 0\mod[5^n].$ But this integer has only $n$ digits, each of them among $1,2,3,4,5$, so that it has to be $U_n.$
This means that $V_{n+1}$ can be constructed by completing $U_n$ from the left. But, we have seen above that such a construction can uniquely be done, thus $V_{n+1} = U_{n+1}$, and we are done.

Pierre.

Hmm...no fast enough in writing...
Anyway, Blang am I wrong or you didn't prove unicity?

Pierre.

[quote="pbornsztein"]Anyway, Blang am I wrong or you didn't prove unicity?
Pierre.[/quote]
I read the question too quickly :)

I thought about something like that for a while, because I noticed the $5^n$ cardinality, but the induction you are talking about did not appear really clear to me  ;)  whereas the construction I used for $U_n$ was really obvious for me...
It's funny how 'obvious' ideas are different from people to amother.

Pierre.

[quote="harazi"]One small question: how can we give in ONM problems that have appeared in at least 5 different issues of Gazeta Matematica and which are absolutely classical?[/quote]Simple: just like that  :?

i solved it in the same way as Pierre, but this solution

[quote="Valentin Vornicu"]My idea in this problem is rather simpler: we can easily check that two numbers with $n$ digits are $\equiv \pmod 5$ if and only if they are the same :) (we make a finite induction after the last digit).

Now, given that there are exactly $5^n$ numbers, we have [i]exactly[/i] one number $\equiv 0 \pmod 5$.[/quote]

seems very interesting, although i don't know how to make the induction. can anyone help me  :blush:

[quote="perfect_radio"]

although i don't know how to make the induction. can anyone help me  :blush:[/quote]

Can anyone show?

Valentin can only post this solution I think .Or this idea is the official sollution's one?

there are $5^n$ numbers of $n$ digits which can be written with $\{ 1,2,3,4,5 \}$. say that $a \equiv b \left( \bmod ~ 5^n \right)$. it is easy to see that $k 5^n$ ends with $5$ for $k$ odd, so contradiction.
if $k$ is even then the number $k 5^n$ is of the form $...50$ or $..500$ ($n \geq 2$, $k < 2^n$). this is because for $n \geq 3$, $2 \ell 5^n = 5^{n-3} (250 \ell)$. another contradiction.

the conclusion follows, right?

[quote="perfect_radio"]if $k$ is even then the number $k 5^n$ is of the form $...50$ or $..500$ ($n \geq 2$, $k < 2^n$). this is because for $n \geq 3$, $2 \ell 5^n = 5^{n-3} (250 \ell)$. another contradiction.

[/quote]

How is this a contradiction?

if $5^n k$ ends with $...50$ and $5^n k | (a-b)$ then the difference between the pre-last digits of the two numbers would be $5$. that's impossible. (remember we're working with $1,2,3,4,5$) same for $...500$

so, is my solution correct, or i'm missing smth?

This is nearly identical to USAMO 2003 #1.


a  closed curve  say A  is given and you can large  it as amount as you want
does there exist a form  B (a curve ,a gon ....)that we can put this form in A (the closed curve     )so the B  pieces be disjoint and cover only the whole surface of A.
find a fix B.

Hmm... I seem not really to understand the part with the B :) could you make a little drawing or so?


calculus
integration
limit
real analysis
real analysis unsolved

Find $ \lim_{n\to\infty}a_n$ where $ (a_n)_{n\ge1}$ is defined by $ a_n\equal{}\frac1{\sqrt{n^2\plus{}8n\minus{}1}}\plus{}\frac1{\sqrt{n^2\plus{}16n\minus{}1}}\plus{}\frac1{\sqrt{n^2\plus{}24n\minus{}1}}\plus{}\ldots\plus{}\frac1{\sqrt{9n^2\minus{}1}}$.


logarithms
inequalities
inequalities unsolved

[quote="nthd"]Let $ a_1$, $ a_2$, $ a_3$, ..., $ a_n$ be distinct positive integers. Prove that

$ \left(1^x \plus{} 2^x \plus{} ... \plus{} n^x\right)\left(a_1^x \cdot\ln a_1 \plus{} ... \plus{} a_n^x \cdot\ln a_n\right)$
$ \geq \left(1^x \cdot\ln1 \plus{} 2^x \cdot\ln 2 \plus{} ... \plus{} n^x \cdot\ln n\right)\left(a_1^x \plus{} ... \plus{} a_n^x\right)$

for any real $ x \geq 1$.[/quote]

note that if we let $ f(x) \equal{} 1^x \plus{} ... \plus{} n^x$ and $ g(x) \equal{} a_1^x \plus{} ... \plus{} a_n^x$, then the inequality is equivalent to $ f(x)g'(x) \minus{} f'(x)g(x) \geq 0$, or, equivalently, $ h'(x) \geq 0$, which is to say that $ h$ is increasing.  (where $ h \equal{} \frac {a_1^x \plus{} ... \plus{} a_n^x}{1^x \plus{} ... \plus{} n^x}$.)

yes.prove that h(x) >= h(y) for x >= y >= 1


man 1 va 3 ro kamel hal kardam va 2 ham nesfe

soale 2 mishe $(b+c)x^2+by^2+cxy$ ke b va c haghighi and
3 ham be esteghra raftam be nakoja residam ba borhane kholf hal shod va kami extermal
1 ham ke besyar sade bood vali 2 saat tool keshid ta halesh konam
doostane dige chi kar kardan?
ishala hamamoon ba ham ghaboolim :P
mokhlese hamegi ham hastim

1- asoon bood
2-kamel hal kardam
3-  :( nefe hal kardam :(

salam doostaane aziz agha ma yeko hal kardim va doro hal nakardim va khodamo seo esteghraa benaakojaa abaad residim

nazaretoon raje be 2 chiye kasi rahe kamelesho dare? :?:
emsalam hamoon 40 nafaro migiran?

[quote="mohamadhosein"]nakone mikhay begi $p(x,y)=(b+c)x^2+by^2+cxy$ ghalate  :o
shak dari emtehanesh kon (shayadam man joob zadam :oops: vali gaman nakonam)
baba inam doroste vali age $b,c$ sefr bashand male to ham javab mide vali fekr konam ine ke neveshtam dorost bashe (man ke daram shakh dar miyaram) :roll:
bazam ye negahi behesh bendaz be har hal yeki in vasat dare joob mizane[/quote]

 :blush: man joob zadam.
man ham javabe$P(x,y)=ax^2+by^2+(a-b)xy$ ro be dast avordam ham $P(x,y)=0$ .
$a,b$ mitoonand mosavi sefr bashand(man joob zadam)

man kamel halesh kardam vali alan hessesh nist bezar farda miferestam.

agha bara computer maro ham doa konid
hatmana yadetoon nare  :lol:

manam ba shroe emtehan moostaghime raftam sar soal tarkib va 16 17 daghighee vaghtamo gereft ta hal shood yazamaniam bara neveshtamesh tool keshid  badam ta akhar emtehan sar 2 fekr kardam va ye chizi to maye haye hamin $(b+c)x^2+by^2+cxy$ dar avardam albate natoonestam kamel esbatesh konam
bara hamatoonam bara farda arezooye moovafaghiat mikoonam ba omide :first:

lotfan mishe soal haye marhalleye 2 ro post konin, ke maa ham feyz bebarim? :D

[size=150]First day[/size]
1.farz konid dayereye $C_2$ az markaze dayereye $C_1$ gozashte va anra dar noghate $M,N$ ghat karde ast.neshan dahid agar noghate $A,B$ do sare ghotre delkhahi az $C_1$ va $A_1,B_1$ mahale taghatoe khathaye $AM,BN$ ba dayereye $C_2$ bashand ,$a_1b_1$ barabare shoae dayereye $C_1$  ast.  :?:

salam rooze dovom che khabar?
man ke 4,6 ro kamel hal kardam vali besyar bad neveshtam ke fekr konam kheili kam begiram va ehtemalan ghabool nasham be har hal shoma ghabool shodid be jaye ma ham hal konim.
man soale 3 ro ba lane kabootari namotanahi hal kardam.
baraye 4 ham kafiye sabet konim ke har do adade farde motavali mitavanened dar donbale bashand pas az har adade fard mitavan be adade farde digar resid va kar kheili sade mishavad.
soale 5 ham mishod fekr konam do ghotre amood bar ham vali man esbat nakardam
vali esbate vojood dashtanesh fekr konam ba ghaziyeye meghdare miyangin sabet kard.

3.dar tole shab setaregan dar baze haye zamaniye mokhtalef ghabele royat hastand.farz konim az beine har $k$ satare($k\geq 1$) daste kam dotayeshan ra mitavan dar yek lahze dsar aseman did. neshan dahid mitavanim $k-1$ aks dar lahazate mokhtalef az sarasare aseman begirim ke har kodam az an setareha, daste kam dar yeki az az aksha dide shavad .(tedade setareha moytanahi ast. lahazati ke setareye $n$om dar aseman dide mishavand ra $[a_n,b_n]$ benamid ke dar an $a_n \leq b_n$)
[hide]inam rahe mane ke eine ino to bargeye pasokhname neveshtam:
farz konid hade aghal tedad aksi ke mitavanim begirim be gunei ke har setare la aghal dar yeki azin aks ha bashad $m$ ta bashad.(bana bar asle khosh tartibi vojud darad)
farz konim $k \leq m$ .zamane in aks gereftan ha ra ba $t_1,t_2, \dots t_m$ neshan midahim be gonei ke $t_1 <t_2, \dots ,<t_m$.
hal sabet mikonim be ezaye har $i$ be tori ke $1 \leq i < n$do setareye $p,q$ vojood darand be gunei ke $t_i$ozve $[a_p,b_p]$,$t_{i+1}$ozve $[a_q,b_q]$ va $[a_p,b_p]$ba$[a_q,b_q]$hich eshteraki nadashte bashand.
borhan:farz konim har do setareye $p,q$ ke $t_i$ozve $[a_p,b_p]$ va $t_{i+1}$ ozve$[a_q,b_q]$ eshterake  $[a_p,b_p]$ ba $[a_q,b_q]$ tohi nabashad.banabarin
lahzei manande $t^'$vojood darad be tori ke her setare ke dar $t_i$ya $t_{i+1}$
boode dar $t^'$ niz dide mishavad. pas be jaye do akse dar lahazate $t_i$va $t_{i+1}$
mitavanim aksi dar lahzeye $t^'$ begirim va baz ham sharte matloob bargharar bashad. ama in ba hade aghal boodane $m$ tanaghoz darad.
....                     .....                          ....                         ......                       ......
pas $m$ setare vojood darand be tori ke hich dotayii dar yek lahze dide nemishavand va in ba farze masale motanaghez ast.(agar$m>k-1$)
pas farze inke $k \leq m$ batal mibashad yani mitavan ba tedade $k-1$ ya kamtar aks gereftan tamame setare ha ra dar aks ha neshan dad
age ye azizi be man bege ba in rah chand migiram kheili mokhlesesham.[/hide]

[quote="syboy"]fekr mikoonid emsal nomre ghabooli cheghadr bashe ??
man khodam fekr konam hodode 4.5 ta soal bashe shoma chi migid[/quote]
bikhial baba motmaenan bish az 27  nist.

[b]4-[/b]
[b]a)[/b] farz konid $m>1$ addadi tabiei bashad. sabet konid tedade motanahi addade tabiei mannand $n$ vojod darad ke $m+n|mn+1$
[b]b)[/b]baraye addade motamayez $m,n>2$ sabet konid donbaleye $a_0,a_1,\dots,a_k$ az addade tabiei bozorgtar az 2 vojod darad ke $a_0=m,a_k=n$ va:

$a_i+a_{i+1}|a_ia_{i+1}+1\ \ \ , i=0,1,\dots,k-1$

[b]5-[/b]noghate $A,B,C,D$ ba hamin tartib rooye dayere gharar  darand.neshan dahid tedad noghtehaye rooye dayere mannad $M$ ke:

$\frac{MA}{MB}=\frac{MD}{MC}$ 4 ta ast va hamchenin ghotrhaye chehar zelei hasel az anha bar ham amoodand.

[quote="amir2"][b]6-[/b] tedadi ketab rooye ham gharar gereftand.fardi ebteda ketab ballaei ra poshto roo mikonad, sepas 2 ketabe ballaei ra hamzaman poshto roo mikonad,baad 3 ketabe ballaei ra hamzaman poshto roo mikonad va ela akhar.
pas az inke be akharin ketab resid inkar ra az aval shroo mikonad.sabet konid pas az tedadi jabejaei,ketabha daghighan be haman vazee avali bar migardand.[/quote]
age farz konim jaygashte avaliye tekrar nashavad pas yeki az jaygasht ha na motanahi bar tekrar mishavand pas do ta az an na motanahi ta har do bad az poshtoru kardane k ketabe aval mibashand va baghiyasham sadast

[quote="mohamadhosein"][quote="mohamadhosein"]2.hameye chand jomleyii haye ba zarayebe haghighi $p(x,y)$ ro biyabid ke baraye har$x,y$haghighi dashte bashim$P(x+y,x-y)=2p(x,y)$ :?:[/quote]berid ye negahi be jabre safa safheye272 bendazi soale 92 ro bebinid khodetoon ghezavat konid. :mad:[/quote]
soal tekrari!!!!!!!!!!!!!!!!!!!!!!!!!!!
albath nabayad tajob kard.

bad nist age kholase e az rahehal haye soalaro dashte bashim
age momkene onae ke soal 2 ro kamel hal kardan ye kholasee az haleshon bedan
 :arrow:

[quote="jensen"][quote="parinaz.sa"][quote="amir2"]tekrari nist  shabiheshe.

ooon soal asoontar bood.[/quote]
rast mige oon soal faghat shabiheshe vali oon soale ketab kheili khafan asoone soale emtehano nazar nemidam!!vali khob man ye khorde az ideye oon soale ketabo baraye hal in soale emtehan be kar bordam vali fekr konam eshtebast!![/quote]
mage shoma ham emtehan dadi? :?:[/quote]
areh khob emtehan dadam mage gharar boode emtehan nadam????

[quote="amir2"]rahe hale soale $5$
dar mosalase $ABC$ darim $bc=2R.h_a$

choon $MA.MC=MB.MD$ pas toole 2 ertefae vared az $M$ bar $AC$ va$BD$ ba ham barabar ast.
pas $M$ rooye  nimsazhaye dakheli va kharaji mahale barkhord $AC,BD$ gharar darad.[/quote]
baba khafan.

brayeh soale 3 negah konid be:
ashnaee ba nazarye graph,entesharat fatemi, alireza alipour, safheye 21,soale 7.

[quote="mathematica"]kasi rahe kootah barayeh soale 1 dare?[/quote]

kootahtar az zaviyr bazi?

ye rahe hale kootahe dige ham hast ineke yeki az oon dayereha mishe dayere 9-noghteye yeki az oon mosalsha.

[quote="mathematica"]kasi rahe kootah barayeh soale 1 dare?[/quote]
aval sabet kon ke $AA',BB'$ nemitoonan movazi bashan
pas dar noghtei mesle $S$ motaghateand
hala midoonim ke ghovate noghte $S$ nesbat be dayere $C_1$ hastesh $SA'.SM=SB'.SN$ => $\frac{SM}{SN}=\frac{SB'}{SA'}$,ghovate noghte $S$ nesbat be dayere $C_2$ hastesh $SM.SA=SN.SB$ => $\frac{SM}{SN}=\frac{SB}{SA}$ pas darim $\frac{SB}{SA}=\frac{SB'}{SA'}$ pas do mosalase $\triangle {SA'B'},\triangle {SAB}$ motashabehand
hala too mosalase $\triangle {SAB}$ midoonim ke $SO$ miane hastesh
baghie hokm ham ke vazehe,hessesh nist benevisam...
age to ham mese bachehaye classe aghaye vakili vasat moheme va mikhai bedooni ke ide khate ezafeha az koja miad??
ide edame dadane $AA',BB'$ tajanose...


Find the greatest four digits palindrom number that it's square is also palindrome number.

I'm pretty sure that it's 2002...
If the number starts with x, it ends with x.  If x=9, then the square ends in 1, and thus has to begin in 1.  that's not possible
x=8: has to begin with 4, not possible
x=7: has to begin with 9, not possible
x=6: has to begin with 6, not possible
x=5: can't begin with 5
x=4: has to begin with 6: this isn't possible
x=3: has to begin with 9: this is possible, but there are no such numbers which actually make palindromes.  3113 and 3003 come close, but don't work.
x=2: has to begin with 4.  This is possible: the greates such number is 2002.  2112 comes close, but 2002 is the first to actually work.


In how many ways can four people line up in a straight line if the youngest person cannot be first in line?

You can also do $ 3 \times 3 \times 2 \times 1$.

Ernie's way works because there are 3 choices for the first (b,c,d) 3 choices for the second(a,b,c,d, not including the one that was chosen first) two choices for third, and 1 choices for last.

[quote="CA Math"]Say the four people are A, B, C, and D with A being the youngest.

Then you have $ 4! \equal{} 24$ ways to line them up in any order.

If A is the youngest in line, there are $ 3! \equal{} 6$ ways to line B, C, and D behind A.

Since we want the number of ways that the youngest is not first in line, we take the total number of ways minus the number of ways that A is first in line;

$ 24 \minus{} 6 \equal{} \boxed{18}$[/quote]

What is the ! symbol and why are you using it?

It's the factorial. $n!$ denotes the product of all positive integers less than or equal to some positive integer $n$.
In this case, $4! = 4 \times 3 \times 2 \times 1 = 24$.


[quote="usamo"]consider the set {1,2,3,...,8}
what is the number of permutations that satisfy:$x_{i+1}-x_i\leq 2\ \forall i=1,2,\ldots,7$[/quote]
Let's the sets:
$A_j$:set of permutations s.t. $x_{j+1}-x_j\ge3; \forall j=1,2,\ldots,7$
Since: $\bigcap{A{}_j{}_1%Error. "cupA" is a bad command.
{}_j{}_2%Error. "cupA" is a bad command.
{}_j{}_3} = \phi$ Hence the answer is $\mid A_1\mid +\mid A_2\mid +\ldots+\mid A_7\mid -(\mid A_{1,2}\mid +\mid A_{1,3}\mid +\ldots+\mid A_{6,7}\mid)$
$\mid A_j\mid =15.6!$
$\mid A_{j,j+1}\mid =3.5!$
$\mid A_{j,k}\mid =(15.15-318).4!; j+2 \le k$


geometry
3D geometry
prism
rate problems

(these are from the australian intermediate maths olympaid, very easy.)

(1) The following cross-number puzzle has one digit in each square:
1_ 2_ 3_
4_   _  _
5_   _
[note bottom right is blanked out.]
1 across and 1 down are squares of different primes.
5 across and 2 down are different squares.
3 down is prime. What is 4 across?

(2) A polygonal prism has 2004 edges. How many faces does it have?

(3) Pamela grows pumpkins. She digs up nine small pumpkins and calculated their mean weight. Next, Pamela digs up her prize specimen, a very large pumpkin and adds it to the other nine. She discovers the mean weight of her pumpkin harvest has doubled. Let the very large pumpkin weight P% of the total wieght of the ten pumpkins. Find P.

(4) David is swimming upstream from a dock at a constant rate. After he has swum 400 meter, he meets Julie who is floating downstream on an air-mattress. He continues to swim upstream for a further 10 mins before turning around and swimming back to the dock, which he reaches at the same time as Julie. Assume David's swimming speed is constant relative to the water and that the current has constant speed of x meters p/ minute. FInd x.

(5) WE have a large semicircle. Three smaller circles are touching it - the first one tangent to the diameter and the semicircle, the other 2 of same radius tangent to the first one, the diameter line and also the big semicircle. If the area of the three small circles comined is 120, what is the area of the semicircle?

(6) A binary operation, *, is defined on all x,y satisfiying:
x*x = x + 2
x*y = y*x
x*(x+y) / x*y = x+y / y
Find 11*8.

(7) On each of three cards is written an integer from the set {1,2 ... 10}. The three cards are shuffled and dealt one each to three players who record the number on the card they receive. ZThe cards are collected, shuffled, process repeated. After a few time, the three people computed the totals of the numbers they had received. The totals were 13, 15 and 23. Find the product of the numbers originally on the three cards.

1-3

[hide]1. 271

2. 670

3. 55%

1. 1 across and 1 down have to be 121 and 169 (squares of 11 and 13) as they are the only squares of primes that have the same hundreds digit.

Since 5 across is a square, 169 must be 1 across (no square has 9 in the tens).

The square that has 6 in the hundreds and ones is 676 (26^2).  So 2 down is 676.

The prime wth 9 in the tens digit is 91.

Thus, 271 is 4 across.

2. Triangular prism has 9 edges and 5 faces; square prism has 12 edges and 6 faces; a pentagon prism has 15 edges and 7 faces.

The pattern is x edges/3 + 2 = y faces

2004 edges/3 + 2 = 670

Thus, there are 670 faces.

3. The original 9 pumpkins:

9x/9 = x

With the big pumpkin:

(9x + y)/10 = 2x

y = 11x

11/(9+11) = .55 = 55%

The big pumpking (P) is 55% of the total (10 pumpkins) weight.

[/hide]

I hate rate problems   I'll do 4 later.

Some of these are at Getting Started level, some (especially the problems involving geometry) are better posted to Intermediate, I think.

Thanks for posting the problems.

Ok, I'll repost problems 8 and 9 in the intermediate section.

4)[hide]If you can't visualize the image in your head, it's often easier to draw a diagram. You have the dock to the meeting point at 400m, and the meeting point to 10mins. Set up equations. You know Julie's speed is x, and David's speed is D, and the stream's speed is also x. You have 10 + 10(D-x)/(D+x) + 400/(D+x) = 400/x, so 20D + 400 = 400(D+x)/x, and 20(D)(x) + 400x = 400D + 400x. So, 20Dx = 400D, and x = 20 meters per minute.[/hide]


for positive $ a,b,c$ with $ a\plus{}b\plus{}c\equal{}1$ prove ineq:
 $ \sum_{}^{}  \frac{a}{2a\plus{}bc} \le 1$

It's wrong Try $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$  :wink:


[quote="red_dog"]$ \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\geq 9\geq (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$[/quote]

Wrong, $ (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9$ holds.

The correct proof:

Putting $ \frac{a}{b}=x$ etc.(note that $ xyz=1$), the initial inequality(multiplied by 2) is equivalent to:
$ 2\sum x^{2}+2\sum \frac{1}{x}\geq 6+2\sum x$
$ AM-GM$ gives $ 2\sum x^{2}+\sum \frac{1}{x}\geq 3\sum x$, and using that it's enough to prove
$ \sum x+\sum \frac{1}{x}\geq 6$, which is trivial..

Just expand, and it holds by AM-GM because $ \sum_{\text{cyc}}\frac{a^{2}}{b^{2}}\ge 3$

EDIT: Scratch that idea, you get something like showing that

$ \sum_{\text{cyc}}\left(\frac{a^{2}}{b^{2}}+\frac{a}{c}\right)\ge 3+\sum_{\text{cyc}}\frac{a}{b}$

which you can also show by AM-GM and Cauchy (ie $ xyz=1 \implies x^{2}+y^{2}+z^{2}\ge \frac{1}{3}(x+y+z)^{2}\ge x+y+z$ deals with one pair of terms, and the other pair is straightforward AM-GM).

[quote="santosguzella"]Let $ a, b, c$ be positive real numbers. Prove that:
\[ \Big(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\Big)^{2}\ge (a+b+c)\Big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Big) \]
.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49893

my solution:

$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{a+b+c}{\sqrt[3]{abc}}$

and $ \frac{ab}{ca}+\frac{bc}{ab}+\frac{ca}{bc}\ge \frac{ab+bc+ca}{\sqrt[3]{a^{2}b^{2}c^{2}}}=\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{\sqrt[3]{\frac{1}{a}\frac{1}{b}\frac{1}{c}}}$.

multiplying the above inequalities, we get the desired one.

[quote="santosguzella"]Let $ a, b, c$ be positive real numbers. Prove that:
\[ \Big(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\Big)^{2}\ge (a+b+c)\Big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Big) \]
.[/quote]

Assume that $ c=min(a,b,c)$
We have:
\[ LHS-RHS=M(a-b)^{2}+N(a-c)(b-c)\]
In which:
\[ M=\frac{1}{ab}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+1\right) \ge 0\]
\[ N=\frac{ab^{2}+bc^{2}+2abc}{a^{2}bc^{2}}\ge 0\]
Done.


I know AIME can be a time pressing test.  I want to know some of your strategies for taking the AIME.

Do you pick out the easy ones? Do you jump around from problem to problem?

How much time should I spend checking?  On average, I get a 9 (with a few errors) but usually I don't get to do the last problem or two.

Thanks for your help!

And goodluck to us Alternate AIME testtakers.

P.S. When is the index announced?  I got a 150 on AMC 10; is it adjusted to parellel with AMC 12?

I would try to do all of the questions, and then check your work starting with the last problem and then work your way backwards.  I find this way more useful becuase it makes you more cautious of your errors becuase you don't just blitz through it.  Going backwards will "force" you to slow down and to pay more attention.  Apply this to checking work on your normal tests.

Some useful advice that probability gave was to skip a question after 10 minutes of working on it.


For those of you wanting to easily copy threads or print them, there's a new 'printable thread' feature.  Click the little printer atop the thread.

Very nice and useful feature ! Thank you.

Very nice.

What are the two lil boxes and the O_o button for?

After doing some trial and error, I'd say the boxes show the range of posts you want to print, such as 1 through 9, or 2 through 6. the O_o button is similar to "Enter".

Ah...muchos gracias...I should've tried something before I asked.


Let $ABC$ a triangle and $M,N,P$ points on $AB,BC$, respective $CA$, such that the quadrilateral $CPMN$ is a paralelogram. Denote $R \in AN \cap MP$, $S \in BP \cap MN$, and $Q \in AN \cap BP$. Prove that $[MRQS]=[NQP]$.

First $[MRQS]=[NQP] \iff [MRN]=[NPS]$.

But $\frac{[MRN]}{[MNP]}=\frac{MR}{MP}=\frac{BN}{BC}$.

And $\frac{[NPS]}{[MNP]}=\frac{NS}{MN}=\frac{CP}{AC}$.

Thus $[MRN]=[NPS] \iff \frac{BN}{BC}=\frac{CP}{AC}=\frac{MN}{AC}$, obvious by Thales.

So, the problem is solved.


geometry
circumcircle
vector
geometry unsolved

Let $ R$ be the circumradius of the triangle $ ABC$.  The circumcenter is located at the point $ O$ and the centroid is located at the point $ G$ . Demonstrate that, if the triangle $ ABC$ has sides $ a$, $ b$ and $ c$, $ OG^2 \equal{} R^2 \minus{} \frac {1}{9}(a^2 \plus{} b^2 \plus{} c^2).$

try to prove the general case
let $ M$ be a point in the plan show that
$ MA^{2}\plus{}MB^{2}\plus{}MC^{2} \equal{}3MG^{2}\plus{}GA^{2}\plus{}GB^{2}\plus{}GC^{2}$
where $ G$ is the centroid.
SAIF

Yes sir, this a very standard result and I demonstrated it already. I found no connections yet.

An elegant proof could be obtained using vectors and dot product:
$ 3vectorOG\equal{} vectorOA\plus{}vectorOB\plus{}vectorOC$
$ 9OG^2\equal{}3R^2 \plus{}2\sum vectorOB.vectorOC$
$ 9OG^2\equal{}3R^2 \plus{}\sum (OB^2\plus{}OC^2\minus{}(vectorOB\minus{}vectorOC)^2)$
$ 9OG^2\equal{}3R^2\plus{}6R^2\minus{}\sum BC^2$


I dont know how to switch from standardd from to graphing form. can you explain please
[img]http://img340.imageshack.us/img340/4299/problemze3.jpg[/img]
i scanned it out of my review booklet.

Uh, well you also know $(6,19)$ is a point.  Now plug both the x and y for $(6,19)$ and $(3,17)$ into the $y=ax^{2}+bx+c$ equaitons to get two equations with two variables, $a$ and $b$ (c is the y intercept which is 19), and you can get $y=\frac{2}{9}{x}^{2}-\frac{4}{3}{x}+19$..then complete the square to get the other form..

actually the first equation is easier to use. The vertex gives you the values of $p,q$, and then you just need to solve for $a$ using the other point. Then expand to get the second equation.


limit
inequalities
real analysis
Functional Analysis
real analysis unsolved

hi im stuck on this question, i was working on it to study for our midterm, couldnt get a solution, and now its the first question on the midterm.  anyway, anything would be appreciated (my book doesnt say much about the dual of $ \ell_\infty$)

want to show existence of$ \gamma \in \ell_\infty^*$ s.t. $ \lim \inf a_n \le \gamma (a) \le \lim \sup a_n$

for $ a \equal{} (a_n) \in \ell_\infty$

everything i try to construct falls apart because i can always create some funky sequence that messes up the inequality

if only limsup and liminf were linear =(

Let $ Y$ be the subspace of convergent sequences. Let $ \gamma(y) \equal{} \lim y_n$.

Apply Hahn-Banach with the sublinear functional dominating $ \gamma$ being the $ \limsup$

The reason your book doesn't say much about the dual is because it sucks

In particular, note that $ \ell^{\infty}$ is the dual of $ \ell^1$, giving a natural map from $ \ell^1$ into $ (\ell^{\infty})^*$. This functional is not in the image of $ \ell^1$, and can't be.

many thanks.  I was mis lead by a classmate who said that we shouldn't use Hahn Banach on it.  The only thing i didnt try =P.

[quote="Kalle"]The reason your book doesn't say much about the dual is because it sucks[/quote]
[the dual of $ \ell^{\infty}$, not the book]  :)


Find the equation of circles passing through $(2,3)$ and are tangent to both the lines $3x-4y+1=0$ and $4x+3y-7=0$

Let $(a,b)$ be the ceter of the circle, we have $\frac{|3a-4b+1|}{\sqrt{3^2+4^2}}=\frac{|4a+3b-7|}{\sqrt{4^2+3^2}}=\sqrt{(a-2)^2+(b-3)^2}=radius$
Solve this equation the answer is $(x-2)^2+(y-8)^2=25$ or $\left(x-\frac{6}{5}\right)^2+\left(y-\frac{12}{5}\right)^2=1.$

To tell the truth remarking the intersection point $(1,1)$ of the 2 lines gives you easy calculation.
Note that the point $(a,b)$ satisfies $4a+3b-7>0$ by sketching these 2 lines. ;)

[quote="kunny"]To tell the truth remarking the intersection point $(1,1)$ of the 2 lines gives you easy calculation.
Note that the point $(a,b)$ satisfies $4a+3b-7<0$ by sketching these 2 lines. ;)[/quote]

hmm .... is it $3a-4b+1<0$ instead of $4a+3b-7<0$ ? coz when I put in the center $(2,8)$ , $4a+3b-7=25>0$  :?

You are right. $4a+3b-7>0$ is right. :oops:

[hide="Here are two solutions."][color=darkblue][b]Proposed problem.[/b] Ascertain the the equations of the circles which pass through the point $P(2,3)$ and which are tangent to the lines $d_1\ : \ 3x-4y+1=0$ and $d_2\ : \ 4x+3y-7=0$.

[u]The first solution.[/u] We observe that $A(1,1)\in d_1\cap d_2$ and $d_1\perp d_2$. The bisector-equation of the angle between the lines $d_1$, $d_2$ and which contains the point $P$ is the ray $7x-y-6=0\ ,\ x\ge 1\ .$ Denote the centre $C(m,7m-6)$ of the appreciated circles, where $m>1$. From the relation $\delta_{d_1}(C)=PC$ or from the relation $CA=\sqrt 2\cdot CP$, i.e. $(m-2)^2+(7m-9)^2=25(m-1)^2$ we obtain $5m^2-16m+12=0$, i.e. $m\in \left\{2,\frac 65\right\}$. Therefore, there are two circles:
$w_1\ : \ C_1(2,8)\ ,\ R_1=5\Longrightarrow \boxed {(x-2)^2+(y-8)^2=25}\ ;$
$w_2\ : \ C_2\left(\frac 65,\frac{12}{5}\right)\ ,\ R_2=1\Longrightarrow\boxed {(5x-6)^2+(5y-12)^2=25}\ .$

[b]Remark.[/b] $\delta_d(X)\ -$ the distance of the point $X$ to the line $d$.[/color]

[color=darkred][u]The second solution.[/u] Remember that the equations of the tangents from the point $A(x_0,y_0)$ to the circle $w\ : \ f(x,y)\equiv x^2+y^2+2ax+2by+c=0$ are given by the equation \[ \boxed {[xx_0+yy_0+a(x+x_0)+b(y+y_0)+c]^2=f(x,y)\cdot f(x_0,y_0)}\ \ (1)\ . \] Thus, for $x_0=y_0=1$, the equation $(1)$ is equivalently with the equation $(3x-4y+1)(4x+3y-7)=0\Longleftrightarrow 12x^2-12y^2-7xy-17x+31y-7=0$. Therefore,
$\frac{a^2-2b-c-1}{12}=\frac{b^2-2a-c-1}{-12}=\frac{2(a+1)(b+1)}{-7}=$
$\frac{-2(a^2+ab+a-b-c)}{-17}=\frac{-2(b^2+ab+b-a-c)}{31}=\frac{(a+b)^2-2c}{-7}\ .$
But $f(2,3)=0\Longrightarrow 4a+6b+c+13=0\Longrightarrow$
$\frac{a^2+4a+4b+12}{12}=\frac{b^2+2a+6b+12}{-12}=\frac{2(a+1)(b+1)}{-7}=$
$\frac{-2(a^2+ab+5a+5b+13)}{-17}=\frac{-2(b^2+ab+3a+7b+13)}{31}=$
$\frac{(a+b)^2+2(4a+6b+13)}{7}\Longrightarrow (-2,-8,43)\ \ ,\ \ \left(-\frac 65,-\frac{12}{5},\frac{31}{5}\right)\Longrightarrow$
$\boxed {w_1\ : \ x^2+y^2-4x-16y+43=0}$ and $\boxed {w_2\ : \ 5(x^2+y^2)-12x-24y+31=0}\ .$ [/color][/hide]

In the previous reply I added the second solution.

Thanks Virgil Nicula . I can understand the first solution but I cannot understand the second one from here

[quote="Virgil Nicula"]
Thus, for $x_0=y_0=1$, the equation $(1)$ is equivalently with the equation $(3x-4y+1)(4x+3y-7)=0\Longleftrightarrow 12x^2-12y^2-7xy-17x+31y-7=0$. Therefore,
$\frac{a^2-2b-c-1}{12}=\frac{b^2-2a-c-1}{-12}=\frac{2(a+1)(b+1)}{-7}=$
[/quote]


A, B, C are employed to do a piece of work for $ 529. A \& B together are supposed to do 19/23 of the work and B\& C together are supposed to do 8/23 of the work. How much should A be Paid?
a.	$ 245
b.	$ 295
c.	$ 300
d.	$ 345

Whats the logic??? and whats the solution???? :?:

Please make sure you explain how you got to a certain answer so others can learn from how you solved the problem.

[hide="If you want the solution, click here"]Let
A=how much work A does
B=how much work B does
C=how much work C does

It is given that:
$A+B=\frac{19}{23}$
$B+C=\frac{8}{23}$

Moreover, since we know that they have to get the work done,
$A+B+C=1$

Adding the first two equations,
$A+2B+C=\frac{27}{23}$

Subtracting this from the third equation,

$B=\frac{4}{23}$

Subtracting this from the first equation,

$A=\frac{15}{23}$

Therefore, A does $\frac{15}{23}$ of the work, and therefore, get $\frac{15}{23}$ of the money. This isn't hard to compute: the answer is $345$. The answer is D, as Nerdy_girl_number1 said.[/hide]


Find numbers a and b such that
\[ \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt{ax + b} - 2}} {x} = 1  \]

Since $\lim_{x\rightarrow 0} \frac{\sqrt{ax+b}-2}{x}\cdot x=0,$ we have $b=4.$

Thus $1=\lim_{x\rightarrow 0} \frac{\sqrt{ax+4}-2}{x}=\lim_{x\rightarrow 0}\frac{ax}{x(\sqrt{ax+4}+2)}=\frac{a}{4}.$

The answer is $a=b=4.$


Given three real numbers $a,b,c$ satifying condition: $a^{2}+b^{2}+c^{2}=1$. Prove that:
$abc+2(1+a+b+c+ab+bc+ca)\ge 0$

the inequality is equivalent to \[2(1+a)(1+b)(1+c)\ge abc\]which is surely true since $1+a>a,1+b>b$ and $1+c>c$.

[quote="nayel"]the inequality is equivalent to
\[2(1+a)(1+b)(1+c)\ge abc \]
which is surely true since $1+a>a,1+b>b$ and $1+c>c$.[/quote]

No. The problem does not restrict a,b, and c to positive reals, just reals. Therefore, you need to incorporate the condition since your inequality is false for $a=b=-1$ and $c=1$.


Prove that for all positive integers a and n, we can find an integer m such that

\[ \sqrt a = 1 + \sqrt[n]{{\sqrt m - \sqrt{m - \left( {a - 1} \right)^2 } }}  \]

Also show how to find m.


calculus
integration
trigonometry
logarithms
calculus computations

hola :D

show

$ \int_0^\infty\;\dfrac{(1\minus{}\cos\,x)\; e^{\minus{}\alpha\,x}}{x^2}\;\textbf dx\;\equal{}\;\boxed{\dfrac{\pi}{2}\minus{}\tan^{\minus{}1}\,\alpha\minus{}\alpha\ln\,\sqrt {\;1\plus{}\alpha^2\;}}$

Differentiating twice with respect to $ \alpha$ should get us there.

However, note that misan's formula cannot be correct - in particular, the term $ \alpha\ln\sqrt {1 \plus{} \alpha^2}$ cannot be correct, as the limit of this integral as $ \alpha\to\infty$ must be zero, and the integral must be positive for all $ \alpha\ge 0.$

After reversing integration order we have $ \int_0^\infty {\frac {{(1 \minus{} \cos x)e^{ \minus{} \alpha x} }} {{x^2 }}\,dx} \equal{} \int_0^\infty \! {\int_0^\infty {(1 \minus{} \cos x)ye^{ \minus{} x(\alpha \plus{} y)} \,dx} \,dy} .$

Now $ \int_0^\infty {e^{ \minus{} x(\alpha \plus{} y)} \,dx} \equal{} \frac {1} {{\alpha \plus{} y}}$ and $ \int_0^\infty {e^{ \minus{} x(\alpha \plus{} y)} \cos x\,dx} \equal{} \frac {{\alpha \plus{} y}} {{(\alpha \plus{} y)^2 \plus{} 1}},$ hence

$ \int_0^\infty {\frac {{(1 \minus{} \cos x)e^{ \minus{} \alpha x} }} {{x^2 }}\,dx} \equal{} \int_0^\infty {\left\{ {\frac {y} {{\alpha \plus{} y}} \minus{} \frac {{\alpha y \plus{} y^2 }} {{(\alpha \plus{} y)^2 \plus{} 1}}} \right\}\,dy} .$

The evaluation of this last integral should give the desired answer.

P.S.: misan learning spanish?  :o

Following up on this (I used my suggestion from #2, although Black's more-or-less-equivalent method in #3 would also work) I believe the correct formula should be
\[ \frac{\pi}{2}\minus{}\arctan(\alpha)\plus{} \alpha\ln\left(\frac{\alpha}{\sqrt{\alpha^2\plus{}1}}\right).\]

Yes, that's the correct formula.

After evaluation of the last integral we have

$ \minus{} \arctan \alpha  \plus{} \alpha \ln \alpha  \minus{} \frac{1}
{2}\alpha \ln \left( {\alpha ^2  \plus{} 1} \right) \plus{} \frac{\pi }
{2} \equal{} \frac{\pi }
{2} \minus{} \arctan \alpha  \plus{} \alpha \ln \frac{\alpha }
{{\sqrt {\alpha ^2  \plus{} 1} }}.$


A study of the weights of the brains of Swedish men found that the weight X was a random variable with mean 1400 grams and standard deviation 20 grams. Find positive numbers a and b such that Y = a + bX has mean 0 and standard deviation 1.

$ 0\equal{}EY\equal{}E(a\plus{}bX)\equal{}a\plus{}bEX\equal{}a\plus{}b(1400)$
std deviation is linear so $ \sigma_Y\equal{}|b|\sigma_X\equal{}20|b|$
$ \sigma_Y\equal{}1 \to |b|\equal{}\frac{1}{20}\to 20b\equal{}\pm 1$
$ \to a\equal{}\minus{}1400b\equal{} \mp 70$

so $ Y\equal{}\pm(70\minus{}\frac{X}{20})$
neither of these has both a and b positive.. but they are the only solutions


For integral $m$, let $p(m)$ be the greatest prime divisor of $m.$ By convention, we set $p(\pm 1) = 1$ and $p(0) = \infty.$ Find all polynomials $f$ with integer coefficients such that the sequence
\[ \{p \left( f \left( n^2 \right) \right) - 2n \}_{n \geq 0}  \] is bounded above. (In particular, this requires $f \left (n^2 \right ) \neq 0$ for $n \geq 0.$)

We shall prove that the only such polynomials decompose into linear factors of form $4x-a^2$. Clearly, such polynomials satisfy the conditions. Now, suppose that $f$ satisfies the condition. We suppose that $f$ is irreducible, otherwise look at all irreducible factors.Let $g(x)=f(n^2)$

Assume that $p|g(k)$. We can asume $0\leq k \leq \frac p2$, since obviously $g(p\pm k)\equiv g(k)$ $(mod$ $p)$.

Then, $p \leq 2k+r$, for some $r$. So we have either $k=\frac p2, k=\frac{p-1}2, \ldots,k=\frac{p-r}2$. We know that infinitely many prime number divide some of $g(n)$, therefore for some $0\leq j \leq r$ there are infinitely many primes $p$ for which $p|g(\frac{p-j}2)$. This implies that $p|g(-\frac j2)$ for infinitely many $p$, so $g(-\frac j2)=0$, thus $2x+j|g(x^2)$, and also $2x-j|g(x^2)$, so $4x^2-j^2|g(x)$, so $4x-j^2|f(x)$. As $f$ is irreducible, $f=4x-j^2$.

QED

[quote="iura"]We know that infinitely many prime number divide some of $g(n)$[/quote]

Why is that? I hope that this question doesn't seem THAT dumb...

This is a known problem already discussed in the forum. For example, suppose that there are a finite number of primes $p_1,\ldots,p_r$ dividing some $p(x)$. Let $p_i^{r_i}|p(x_i)$. Choose $x\equiv x_i$ $(mod$ $p_i^{r_i+1}$ $)$ then we see that $p_i^{r_i+1}$ doesn't divide $p(x)$, hence $p(x)< \prod p_i^{r_i+1}$. But we can choose infinitely many such $x$ which implies that $p$ is constant.

Happy Easter!

Sorry for sounding unintelligent, but what is this question asking me to do??

This reminds me heavily of IMO 2008 # 3... just focus on the prime first and find the $n$ rather than vice-versa.

If $f$ is the (possibly empty) product of linear factors of the form $4n-a^2$, then it satisfies the condition. We will prove no other polynomials work. In what follows, assume $f$ is irreducible and nonconstant.

It suffices to show for every positive integer $c$, there exists a prime $p$ and a nonnegative integer $n$ such that $n \le \tfrac{p-1}{2} - c$ and $p$ divides $f(n^2)$

Firstly, recall there are infinitely many odd primes $p$, with $p>c$, such that $p$ divides some $f(n^2)$, by Schur's Theorem. Looking mod such a $p$ we can find $n$ between $0$ and $\tfrac{p-1}{2}$ (since $n^2 \equiv (-n)^2 \pmod p$). We claim that only finitely many $p$ from this set can fail now. For if a $p$ fails, then its $n$ must be between $\tfrac{p-1}{2} - c$ and $\tfrac{p-1}{2}$. That means for some $0 \le k \le c$ we have \[ 0 \equiv f\left( \left(\frac{p-1}{2}-k\right)^2 \right)  	\equiv f\left( \left( k + \frac12 \right)^2 \right) \pmod{p}. \] There are only finitely many $p$ dividing \[ \prod_{k=1}^c f\left( \left( k+\frac12 \right)^2 \right) \] unless one of the terms in the product is zero; this means that $4n-(2k+1)^2$ divides $f(n)$. This establishes the claim and finishes the problem.

First note that by Schur's theorem, the set of prime divisors of $f(n^2)$ is infinite as $n$ ranges over positive integers. Take some $p$ in this set. Then there exists some $m$ such that \[f(m^2) \equiv f(r^2) \equiv f((p-r)^2)\equiv 0 \pmod{p}\] where $m \equiv r \pmod{p}$ and $0 \le r \le p-1$. Now since evidently one of $r,p-r$ is less than $\frac{p}{2}$, we see that there are infinitely many integral $n$ such that $f(n^2)$ has a prime divisor greater than $2n$. Now if the sequence $p(f(n^2))-2n$ is bounded above, there exists an odd  positive integer $k$ such that there are infinitely many integral $n$ with $p(f(n^2))-2n=k$. This means that $2n+k \mid f(n^2)$ for infinitely many integral $n$, implying that $2x+k \mid f(x^2)$. Now since $f(0) \neq 0$, we see that $2x-k \mid f(x^2)$ as well. Now note that if $g(x)$ is some irreducible factor of $f(x)$, then $g(x^2)$ satisfies the conditions of the problem as well, since $p(f(n^2)) \ge p(g(n^2))$, so $g(x^2)=4x^2-k^2 \implies g(x)=4x-k^2$ for all irreducible $g$, factors of $f$. Now all that remains is to check that all such $f$ work, which can be seen by writing \begin{align*}f(n^2) &=g_1(n^2) \cdot g_2(n^2) \cdots g_m(n^2)\\ &=(2n+k_1)(2n-k_1) \cdots (2n+k_m)(2n-k_m)\end{align*} and noting that $p(f(n^2))-2n \le \max(k_i)$ so we are done. $\Box$

I can't agree more with v_Enhance about the IMO 2008 P3 part. :p.The first idea which came to mind while solving this was the $n^2+1$ problem. :p
However,can this be generalized by replacing $p(f(n^2))$ with $p(f(n^{2k}))$ and $p-2n$ with $p-2kn$. ?

[hide=Solution]
Note that it suffices to look at $f$ which are irreducible over $\mathbb{Z}[x]$; the answer will then be products of these polynomials.

If $f$ is constant, then it must be a nonzero constant; clearly these all satisfy the condition. Otherwise, since $f(x^2)$ is a nonconstant polynomial, it follows by Schur's theorem that there exist infinitely many primes $q$ with $q \mid f(n^2)$ for some positive integer $n$.

Take such an odd prime $q$ and positive integer $n$. Note that by reducing $n$ modulo $q$, we can assume $0 \leq n < q$. However, note that $f((-n)^2) = f(n^2) \pmod{q}$; this implies that we can force $0 \leq n \leq \frac{q-1}{2}$; this gives $q \geq 2n + 1$.

The condition implies that there exists a positive integer $k$ such that for any $n$ and any prime divisor $q$ of $f(n^2)$, we have $q \leq 2n + k$. However, from above, we have $q \geq 2n + 1$, so $q - 2n \in \{1, 2, \ldots, k\}$. Since there exist infinitely many primes $q$ dividing $f(n^2)$ for some $n$, one of the $k$ values for $q - 2n$ must occur infinitely often; suppose it is $c$. Note that $c$ is odd. This means that $2n + c \mid f(n^2)$ for infinitely many positive integer $n$, so $2x + c$ divides $f(x^2)$ as a polynomial. Since $f(x^2)$ is even, $-2x + c$ divides $f(x^2)$ too. (This is distinct from $2x + c$ since $c \neq 0$, as $c$ is odd.) Thus, $-(2x + c)(-2x + c) = 4x^2 - c^2$ divides $f(x^2)$, so $4x - c^2$ divides $f(x)$. However, we assumed $f$ is irreducible over $\mathbb{Z}[x]$, so $f(x) = \pm(4x - c^2)$.

Thus, $f(x)$ satisfies the condition if and only if there exists an integer $a$ and odd squares $a_1, a_2, \ldots, a_n$ with
\[f(x) = a(4x - a_1)(4x - a_2) \cdots (4x - a_n).\]
[/hide]

Call a polynomial $f\in\mathbb{Z}[X]$ [i]bad[/i] if the sequence \[S_f:=\{p(f(n^2))-2n\}_{n\ge 0}\] grows arbitrarily large. Our goal is to find all not bad polynomials. First, note the following lemma.

[b]Lemma:[/b]  If a polynomial $g$ is bad, then any multiple of $g$ is also bad.

[i]Proof:[/i]  Suppose $f=gh$ is a multiple of some bad $g$. The key observation is that $p(g(m)h(m))\ge p(g(m))$, so all elements of $S_f$ are at least as large as those of $S_g$. Since $S_g$ grows arbitrarily large, then so does $S_f$. $\blacksquare$

We now focus our attention on irreducible polynomials $f$.

[b]Claim:[/b]  If $f$ is irreducible, not constant, and not a multiple of $4x-a^2$ for some integer $a$, then $f$ is bad.

[i]Proof:[/i]  Fix any positive integer $M$. We'll show that there exists some prime $p$ such that \[\frac{p-1}{2}-n>M\] and $p\mid f(n^2)$, which clearly finishes.

Since $f$ is not constant, by Schur's theorem, there exist infinitely many primes dividing $f(n^2)$ for some $n$. Furthermore, since $f$ is not a multiple of $4x-a^2$, the rational numbers \[f(1^2/4),f(3^2/4),f(5^2/4),\ldots\] are all non-zero. Now, pick $p$ larger than the numerators of $f(1^2/4),f(3^2/4),\ldots,f((2M+1)^2/4)$ and $p\mid f(n^2)$ for some $n$.

From this, we see that \[p\nmid f\left(\left(\frac{p-1}{2}-r\right)^2\right)\] for all $0\le r\le M$, so by reducing $n$ mod $p$ and possibly replacing it with $n-p$, we can force $n<\frac{p-1}{2}-M$, as desired. This completes the proof of the claim. $\blacksquare$

Combining the lemma and the claim, we see that the only polynomials that stand a chance of being not bad are ones that are a constant times a product of a bunch of polynomials of the form $4x-a^2$. Suppose $f$ is such a polynomial. We see that $4n^2-a^2=(2n-a)(2n+a)$, so all prime factors of $f$ are bounded above by $2n+C$ for some sufficiently large constant $C$, which means $S_f$ is bounded above by $C$. So the answer is polynomials which factor completely into factors of the form $4x-a^2$.

Consider any nonconstant irreducible factor $g(x)$ of $f(x)$. First, let $P$ be the set of primes that divide $g(x^2)$ for some integer $x$.

[b]Lemma.[/b] $P$ is infinite.

[i]Proof.[/i] Suppose for contradiction that $P$ is finite. Let the polynomial $g_1(x)=g(x^2)$. If $g_1(0)=0$, then the lemma is obviously true. If $g_1(0)\neq 0$, let $g_1(x)=xg_2(x)+g_1(0)$, and take $x=x_0=(\max P)!g_1(0)$. Then $g_1(x_0)=g_1(0)\left((\max P)!g_2(x_0)+1\right)$, and any prime dividing $(\max P)!g_2(x_0)+1$ is larger than $\max P$, which is a contradiction. Therefore $P$ is infinite.

Fix a positive integer $c$. The sequence being bounded implies that there doesn't exist a prime $p\in P$ such that $p\mid g(n^2)$ and $p\ge 2n+2c+1 \Longleftrightarrow n\le \frac{p-1}{2}-c$. Then since $n$ can be picked to lie in $[0,\frac{p-1}{2}]$, $p$ must divide $g((\frac{p-1}{2}-k)^2)$ for some $0\le k\le c$. Vary $p$ inside $P$, and since $P$ is infinite, by pigeonhole there exists $0\le k\le c$ such that infinitely many primes $p\in P$ satisfy \[0 \equiv g((\frac{p-1}{2}-k)^2) \equiv g((k+\frac{1}{2})^2) \pmod{p}.\] Hence $g((k+\frac{1}{2})^2)=0$ and the factor $(4x-(2k+1)^2)$ divides $g(x)$. Because $g$ is irreducible, $g(x)=4x-(2k+1)^2$.

We conclude that $f(x)$ is the product of (possibly a constant) and factors in the form $4x-(2k+1)^2$ for integer $k$. It's clear that all such $f$ work since $f(x^2)$ factors into $\prod(2x-2k-1)(2x+2k+1)$.

The polynomials are all polynomials $f$ generated by choosing a set of odd nonnegative integers $a_1,\dots,a_m$, some nonzero integer $c$, and taking $f(x)=c\displaystyle\prod_{i=1}^m (4x-a_i^2)$. To check that all such polynomials work, note that for any $n$, we have
\[p(f(n^2)) = \max \left(p(c),\max_{i=1}^m p(2x-a_i),\max_{i=1}^m p(2x+a_i)\right),\]
so $p(f(n^2))-2n$ is bounded above by $p(c)+\max_{i=1}^m a_i$. Now, we show any solution $f$ is of this form.

First, note we are done if $f$ is constant. Otherwise, by Schur's Theorem, there are infinitely many prime divisors of $f$. Pick such a prime $p > 2$ and let $n$ be the minimal nonnegative integer $n$ such that $p \mid f(n^2)$. Observe that if $m$ is such that $p \mid f(m^2),$ then so are $m-p, m+p,$ and $-m.$ Hence, we can ensure that $n\le p/2.$ Now, by the condition, we have $p-2n \le k$ for some constant $k,$ so $n \ge p/2 - c$ for some constant $c.$ Since there exist infinitely many such primes $p,$ there exists some constant $\ell$ such that $p \mid f((p/2-\ell)^2)$ for infinitely many primes $p.$ Note that $g(p) = f((p/2-\ell)^2)$ is a rational polynomial in $p$ with coefficients that are dyadic numbers. For any prime $p$ such that $p \mid g(p),$ we see that the constant coefficient of $g$ is divisible by $p$. This holds for infinitely many primes $p,$ so $g(p) = ph(p)$ for some polynomial $h.$ Now, note that the constant coefficient of $g(p), 0,$ must be equal to $f(\ell^2).$ Because $p$ is odd, $p/2-\ell = n$ is an integer implies $\ell$ is of the form $a/2$ for some nonnegative odd integer $a$. This implies that the minimal polynomial of $a^2/4=\ell^2$, $4x-a^2$, must divide $f$. Note that $f(x)/(4x-a^2)$ must also satisfy the same properties as $f$, so we can iteratively divide by expressions of the form $4x-a^2$ until we have a constant $f$, yielding the claimed solution set.

WLOG assume $f$ is irreducible. Suppose its bounded above by $c$. So for every prime $p$ dividing $f(n^2)$, we have $p < 2n+c$. Pick one such prime, then we have that there is a number $\frac{p-c}{2} \le n \le \frac{p-1}{2}$ such that $p | f(n^2)$, because we can negate $n$ if needed. Suppose $n = \frac{p-k}{2}$. By PHP, since there are infinitely many primes, there is some $k$, say $z$, which is used infinitely many times.

So we have $p | f \left( \left(\frac{p-z}{2}\right)^2 \right) \implies p | f \left(\frac{z^2}{4} \right)$ for infinitely many primes. Therefore, we must have this is $0$, so $4x - z^2$ is a factor of $f$ (note that $z$ is odd).

Since we'd assumed $f$ irreducible initially, we get the entire solution set as $f(n) =c \prod_{i=1}^m (4n - (2z_i+1)^2)$ for any integers $z_i, c$, which indeed works. $\blacksquare$

I had done [hide=spoiler for other problem]IMO 2008/3[/hide] before.
[hide=Solution]
Let $P$ denote the set of polynomials with the desired property. \\
$\textbf{Step 1:}$ Reduction to Irreducibles in $P$ \\
Notice that if $f,g \in P$, then $h(x) = f(x)g(x) \in P$ as well because $$p(h(n^2)) = p(f(n^2)g(n^2)) = max(p(f(n^2)),g(n^2)) \leq 2n+C$$ for some $C \in \mathbb{N}$ for all $n \in \mathbb{N}$. \\
Conversely, if $h \in P$ and $h(x) = f(x)g(x)$, then $f,g \in P$ because as before there exists some $C \in \mathbb{N}$ such that $2n+C \geq p(h(n^2)) = p(f(n^2) \cdot g(n^2)) \geq p(f(n^2)), p(g(n^2))$. \\

We can consequently assume that $f$ is irreducible, for now, the set $P$ will ultimately consist of all multiples of irreducible polynomials in $P$. \\

Also, assume that $f$ is non-constant because all non-zero constant polynomials are clearly in $P$. \\

$\textbf{Step 2: Prime Search}$ \\
Notice that if $p \mid f(n^2)$, then $p \mid f((p-n)^2)$ and therefore we can assume that $p \mid f(n^2)$ such that $2p \geq n$, and there are clearly infinitely many prime divisors of $f(n^2)$ by Schur's Lemma meaning that we can find infinitely many $n \in \mathbb{N}$ such that there is a prime $p$ such that $p = 2n+c_n$ for some $c_n \in \mathbb{N}$, then assuming that $\{pf(n^2)-2n\}_{n \geq 0}$ is bounded, we have that some $c_n$ appears infinitely often and then as $p \mid f(n^2)$, we can conclude that $p \mid f((\frac{c_n}{2})^2)$ for infinitely many primes $p$, this means that $f((\frac{c_n}{2})^2) = 0$ for some $c_n$ odd as $f(n^2) \neq 0$ for any $n \in \mathbb{N}$ and ultimately that $deg(f) = 1$ as we have assumed that it is irreducible and it has a rational root meaning by Gauss's Lemma that it is reducible unless it is linear. \\

We consequently have to find $f(x) = ax+b$ where $f((\frac{c_n}{2})^2) = 0$, then we must have that $f(x) = t(4x-k^2)$ $t,k \in \mathbb{Z}$ using the fact that $c_n$ is odd. \\

Ultimately, $P$ is the set of any product of the previous polynomials, namely $f(x) = T \cdot \prod_{i=1}^{m} (4x-a_i)$ for odd integers $a_1, a_2, \cdots, a_m$ and $T \in \mathbb{Z}$. $\blacksquare$ \\
[/hide]

The answer is all polynomials of the form $f(x)=a(4x-a_1^2)(4x-a_2^2)\ldots(4x-a_k^2)$ for all $a_i$ odd, which clearly work. Note that if two polynomials $f,g$ satisfy then condition then $fg$ does too, and vice versa, so we only need to consider irreducible $f$.

If $f(x) \neq 4x-a^2$ for some $a$, then I will show that for every positive integer $c$ there exists some $(n,p)$ with $n \leq \tfrac{p-1}{2}-c$ such that $p \mid f(n^2)$, which finishes. Indeed, suppose otherwise. By Schur's theorem there are infinitely many $p>c$ with $p \mid f(n^2)$ for some $n$ (treating $f(n^2)$ as a polynomial in $n^2$), so considering modulo $p$ there are infinitely many $p \mid f(n^2)$ with $p \leq \tfrac{n-1}{2}$ (as $n^2\equiv (p-n)^2 \pmod{p}$). As such, it suffices to show that there exist finitely many $0 \leq k<c$ with
$$p \mid f\left(\left(\frac{p-1}{2}-k\right)\right) \iff p \mid f\left(\left(k+\frac{1}{2}\right)^2\right).$$
This is equivalent to
$$\prod_{k=0}^{c-1} f\left(\left(k+\frac{1}{2}\right)^2\right)$$
having a finite number of prime divisors (that is, a finite number of primes $p$ such that the $\nu_p$ of the expression is positive), which is always true unless the expression equals zero. If it does, then there is some $k$ with
$$f\left(\left(k+\frac{1}{2}\right)^2\right)=0 \implies x-\left(k+\frac{1}{2}\right)^2 \mid f(x) \implies 4x-(2k+1)^2 \mid f(x),$$
which is of the form $4x-a^2$. We are done. $\blacksquare$

Mocking USAMO 2006 day 1 result p3: 90 mins, 7 seconds (fully scored :3)
First note that for any polynomials $g,h$ we get $p(g(x)h(x)) \ge p(g(x))$ so if $g$ was a polynomial we are not looking for in this case then any $g \times h$ would also not hold, hence we can just wlog $f$ to be irreducible.
Basically note that by Schur the set of primes diving $f(n^2)$ is infinite as we have $f(n^2) \ne 0$, so now let $n \equiv a \pmod p$ and note that $p \mid f(a^2)$ and $p \mid f((p-a)^2)$ but one of these is less than $\frac{p}{2}$ so there exists infinitly $m$ such that there is a prime divisor of $f(m^2)$ that is greater than $2m$ hence since $p(f(n^2))-2n$ is a sequence bounded above there must exist some odd $\ell$ such that $2n+\ell \mid f(n^2)$ for infinitly many $n$ so it holds that $2x+\ell \mid f(x^2)$, now set $x=\frac{p-\ell}{2}$ where $p$ is a arbitrary large prime to get that $p \mid f \left(\frac{\ell^2}{4} \right)$ which gives that $4x-\ell^2 \mid f(x)$ so $f(x)=4x-\ell^2$.
Hence any polynomial that has only linear factors of the form $4x-b^2$ works thus we are done :D

We'll prove that if $f$ isn't the product of linear factors of the form $4n - a^2$, then the sequence $(p(f(n^2)) - 2n)_{n \ge 0}$ is unbounded. Note that if $f, g$ satisfy the condition then $fg$ does too, so we can assume $f$ is irreducible. Let $\deg f = k$ and let $N$ be an integer such that $2^{2k}\cdot |f(x)| < Nx^{k+1}$ for all $x \ge \frac{1}{2}$. Since $f$ is non-constant, so by Schur's theorem there are infinitely many primes $p$ such that there exists $n$ such that $p \mid f(n^2)$. Let $p$ be a large prime and let $n$ be the smallest positive integer such that $p \mid f(n^2)$. As $p \mid f((p-n)^2)$, so $n < p - n$. Thus $2n < p$. Let $s = p - 2n$, then $n \equiv -\frac{s}{2}$, so $p \mid f(\frac{s}{2})$ and since $p$ is odd, so $p \mid 2^{2k} \cdot f(\frac{s}{2})$ and note that $2^{2k} \cdot f(\frac{s}{2})$ is an integer, so if $f(\frac{s}{2}) \neq 0$, then $p < 2^{2k}|f(\frac{s}{2})| < N \cdot (\frac{s}{2})^{k+1}$. Therefore $s > 2(\frac{p}{N})^{\frac{1}{k+1}}$, and since $p$ is large enough, thus $s$ can be arbitrarily large. So $f(\frac{s}{2}) = 0$ and $f$ is irreducible, so $f(x) = 4x - s^2$. Thus $f(x) = a(4x - a_1^2)(4x - a_2^2) \cdots (4x - a_k^2)$ for some odd integers $a_1, a_2, \dots, a_k$. $\blacksquare$


Let a,b,c be reals greater than 0 with a^4+b^4+c^4=3. Prove that
$ {1}/{4\minus{}ab}\plus{}{1}/{4\minus{}bc}\plus{}{1}/{4\minus{}ca}$ is less or equal than 1.

I saw it in Moldova TST 2005 1day problem 2

I think I have solved it,but I do not have time to post it.

[quote="Inequalities Master"]Let $ a,b,c$ be reals greater than $ 0$ with $ a^4\plus{}b^4\plus{}c^4\equal{}3.$ Prove that
$ \frac{1}{4 \minus{} ab} \plus{} \frac{1}{4 \minus{} bc} \plus{} \frac{1}{4 \minus{} ca}$ is less or equal than $ 1.$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30764


Prove or disprove: given any monic polynomial $P(x) = x^{n}+a_{n-1}x^{n-1}+\ldots+a_{1}x+a_{0}$, with complex coefficients and $a_{0}\neq 0$, there exists some other polynomial $Q(x) = x^{m}+b_{m-1}x^{m-1}+\ldots+b_{0}$ such that $P$ divides $Q$ and $b_{i}\neq 0$ for all $i$.

Let $S$ be the set of polynomials $P$ of degree $n$ which have such a multiple.  Is there a value of $M$ such that each $P \in S$ has a corresponding $Q$ with $m \leq M$?  (That is, what's the worst-case scenario for the degree of $Q$?)

[hide="I hope it`s correct, JBL."]Let's suppose that $m$ of the coefficients of $P$ are equal to $0$ (they are among $\left\{ a_{1},\ldots,a_{n-1}\right\}$).

Let $t \in \mathbb C$ and consider $Q_{1}(X) = P(X) (X-t) = X^{n+1}+X^{n}\left( a_{n-1}-t \right)+\sum_{k=1}^{n-1}\left( a_{k-1}-a_{k}t \right) X^{k}-a_{0}t$.

Now let's see what happens when $\left| t \right| \to \infty$. If $a_{k}\neq 0$, then $a_{k-1}-a_{k}t$ will also be non-zero for a large enough in absolute value $t$. There are only finitely many coefficients, so we can find our appropriate $t$. Also note that the coefficient of $X^{n+1}$ is also non-zero. Thus, $Q_{1}$ has at most $m$ zero coefficients. Consider the minimal $\ell$ such that $a_\ell = 0$. Then, since $a_{0}\neq 0$, $a_{\ell-1}\neq 0$. So the coefficient of $X^\ell$ is $a_{\ell-1}-a_\ell t = a_{\ell-1}\neq 0$. Thus, there are most $m-1$ null coefficients in $Q_{1}$. Applying the same procedure for $Q_{1}$, we get $Q_{2}$ with at $m-2$ null coefficients etc. $Q_{m}$ surely works (it is divisible by $P$ and has non-zero coefficients)!

By the above work, the coefficient of remains zero upon multiplying with $X-t$ if and only if $a_{k}= a_{k-1}= 0$. By the Fundamental Theorem of Algebra we know that all multiples of $P$ can be obtained by gradually "adding" polynomials of the form $X-t$. So the worse case scenario is indeed $X^{n}-c$, where $c \neq 0$. Hence, $M=n-1$.[/hide]

Very nicely done!  (I'm glad someone looked at it :) )


geometry
3D geometry
tetrahedron
geometry unsolved

Determine if there exists a convex polyhedron such that

(1) it has 12 edges, 6 faces and 8 vertices;

(2) it has 4 faces with each pair of them sharing a common edge of the polyhedron.

[quote="master_Hjom"]I'm not quite sure about this question. Maybe there is some typo (or I'm so stupid that I don't understand question  :maybe: )
(1) Cube
(2) Tetrahedron[/quote]
I've searched for [url=http://www.zhzx.net.cn/CGMO2013/ArticleOne.aspx?id=1675]the original paper[/url] and it should be asking for a convex polyhedron which satisfies both conditions.
BTW, the answer should be yes I think, but I'm not going to post my construction here.  :)

Here is the construction.

Tape two regular pentagons of side length $1$ together so that two of their sides coincide and they are not coplanar. Now, connect the corresponding vertices of the two pentagons, and make small alterations so that our faces comprise two pentagons, two triangles, and two quadrilaterals. The first condition is easily seen to be satisfied. The second condition is satisfied by consider the two pentagons and two quadrilaterals.

$\square$


Let $ T: X\to Y$ compact operator. prove that $ R(T)$ is separable

Hint: Write \[ X\equal{}\bigcup_n \{x\in X: \parallel{}x\parallel{}<n\}\] and use the "relative compact" notion

Hint: Compact metric spaces are separable, as is a countable union of separable subsets.

$ R(T)$ is the range of $ T$, i.e. $ \{ y \in Y \, | \, \exists \, x \in X \, : \, T x \equal{} y \}$.


If $z \in \mathbb{C}$ and $\cos{z}=10$. Find all values of $z$

How do you do trig operations on complex numbers? Formula?

[hide]Using exponential form for cosine, $e^{iz}+e^{-iz}=20$. Let $y=e^{iz}$ so that $y^2-20y+1=0\Rightarrow y=10\pm 3\sqrt{11}$.

Taking natural log and dividing by $i$, $z=-i\ln \left( 10\pm 3\sqrt{11} \right)$ or $\boxed{\pm i\ln \left(10+3\sqrt{11} \right)}$.[/hide]

[quote="amcavoy"]The only way I can think of is to use $z=a+bi$ and then use the complex definition so that

$\cos{z}=10=\frac{e^{i\left(a+bi\right)}+e^{-i\left(a+bi\right)}}{2},$

and then attempt to solve that.  I don't know if that's the right method, but I don't have the time to attempt it now (maybe someone else has a better way).[/quote]

That is indeed the correct method.  :)

Edit:  Well, when I say it's the correct method I mean it's the correct way of extending $\cos$ to $\mathbb{C}$, although as far as actually solving scorpius has provided the easy method.  :)

i gave this problem to my math teacher and i wonder if she can solve it, pretty unlikely though...

I have learnt not to give problems to my teachers...they don't seem to like what they do (teach math or the science). :P  Besides, this is a very elementary problem, so if they remember their "Introduction to Complex Analaysis" course, they may be able to solve it.

There are more solutions, since cosine is periodic with period $2\pi$.  But the idea is basically correct.

my teacher wrote a big old ${\displaystyle \varnothing }$


Wishing everyone a VERY HAPPY INDEPENDENCE DAY!!!  :pilot:  :lol:

[quote="Scary math"]@Aashay
Yah,I agree with u,but negatives outnumber the positives,That's where the problem lies.[/quote]
Thats more of a subjective argument...

[quote="Scary math"]
negatives outnumber the positives,That's where the problem lies.[/quote]

That is because we are a developing country.If you want to develop you should look at the positives and try to convert the negatives into positives.

Hmmm.....Are any of you guys actually interested in doing something? Like 'practical' work? If you're serious then just PM me, and I'll tell you about the youth revolutionary organization I work with :)

@nuclear
Man!U r really into some social work!!!!Great!!!!!!!!!!!!!!!! :clap2:


A point $ P$ is randomly selected from the rectangular region with vertices
$ (0,0), (2,0)$, $ (2,1),(0,1)$. What is the probability that $ P$ is closer
to the origin than it is to the point $ (3,1)$?

We have $ x^2 \plus{} y^2 < (x \minus{} 3)^2 \plus{} (y \minus{} 1)^2\implies{ \minus{} 6x \minus{} 2y \plus{} 10 > 0}\implies{3x \plus{} y < 5}$.

Graphing the line $ y \equal{} \minus{} 3x \plus{} 5$, we see that it intersects the rectangle at $ (4/3,1)$ and $ (5/3,0)$. Thus, the successful region is $ \frac {4}{3} \plus{} \frac {1}{6} \equal{} \frac {3}{2}$.

$ \frac {\frac {3}{2}}{2} \equal{} \boxed{\frac {3}{4}}$

[hide] Another way to think about it is to consider the segment with $(0, 0)$ and $(3, 1)$ as endpoints. Intuitively, all points which are closer to the origin are to the right of the perpendicular bisector of the segment. [/hide]

why is part of the area the region bounded by the perpendicular bisector of the segment joining $(0,0)$ and $(3,1)$


Consider a solution x(t) to x' = -x + (εx^2) with x(0) = 1.  Show that x(t) -> 0 like e^-t.  That is, show x(t)/(e^-t) is bounded from 0 and infinity for t>0. [/code]

Since you must prove something about $ y(t)\equal{}e^t x(t)$, find a differential equation for it (by computing $ y'$).

can you be more specific and explain it a little clearer please

Here is the same question in better format...

Consider a solution x(t) to
\[ x' \equal{} \minus{} x \plus{} \epsilon x^2
\]
with $ x(0) \equal{} 1$.  Show that $ x(t) \rightarrow 0$ like $ e^{\minus{} t}$ That is, show $ x(t)/e^{\minus{} t}$ is bounded from $ 0$ and $ \infty$ for $ t>0$.

hello, writing your equation in the form $ \frac{dx}{-x+\epsilon x^2}=dt$ we have $ \ln|\epsilon x-1|-\ln|x|=tC$,further we get ${ \frac{\epsilon x-1}{x}=Ce^t}$, hence we have $ x(t)=\frac{1}{\epsilon+Ce^t}$.
This is the solution of your equation.
Sonnhard.

shouldnt the right side of \[ \ln|\epsilon x\minus{}1|\minus{}ln|x| \equal{} tC\] be t + C....why do you have t*C


function
induction
algebra proposed
algebra

Find all functions $f: \mathbb Q \to \mathbb R$ such that
\[f(x+1)=3f(x)+1,\]
for all $x \in \mathbb Q$.

I think there is a condition missing here.

Suppose we know the value of $ f(x)$ for some rational number $ x$. Using the condition in the problem statement, we can find (probably by proving some formula using induction) the value of $ f(x\plus{}n)$ for every integer $ n$. However, this doesn't tell us anything at all about the value of $ f(y)$ unless $ x\minus{}y$ is an integer. So if $ y$ is a rational number such that $ \{y\} \neq \{x\}$ (brackets indicate fractional part), we can arbitrarily define the value of $ f(y)$, independently of $ f(x)$.

And we can do the same thing for all rational numbers in $ [0,1)$. So there are infinite functions which satisfy this equation, actually there are as many of them as functions $ f: [0,1) \rightarrow \mathbb R$.

No information is missing.
As uglysolution said, we just have infinitely many solutions and the possibility of giving general solutions :

From $ f(x\plus{}1)\equal{}3f(x)\plus{}1$, it's easy to show that $ f(x\plus{}n)\equal{}3^nf(x)\plus{}\frac{3^n\minus{}1}{2}$ $ \forall n\in\mathbb Z$

Then, a general solution is :

$ f(x)\equal{}3^xh(\{x\})\plus{}\frac{3^{[x]}\minus{}1}{2}$  where $ h(x)$ is any function defined on $ [0,1)$, $ [x]$ is the integer part of $ x$ and $ \{x\}$ the fractional part of $ x$.


Fie $a, b, c$ numere reale strict pozitive cu proprietatea ca $a+b+c\geq{\frac{abc}{27}}$. Demonstrati ca $a+b+c\geq{\sqrt{abc}}$.

[hide="click"]
notam $s=a+b+c$, $p=abc$.
din ipoteza avem $27s\geq p$ si din inegalitatea mediilor avem $s^{3}\geq 27p$. inmultind aceste doua relatii obtinem $s^{4}\geq p^{2}$ sau $s\geq p^{1/2}$, adica ce aveam de demonstrat.[/hide]


function
integration
logarithms
real analysis
real analysis unsolved

A function $f: \,\mathbb{R}\to\mathbb{R}$ satisfies the following conditions:

1) $f(x)=0$ for negative $x$

2) $f$ is picewise-constant (with countable discrete set of discontinuities).

3) for Laplace transform $F(p): =\int_0^{\infty}e^{-pt} f(t)dt,\,\Re(p)>0$ we have a formula

\[ pF(p)=(1-2^{-p})/(2-2^{-p}-3^{-p}). \]

The goal is to prove that $f$ has a limit on $+\infty$. Is there appropriate Tauberian theorem?

See another reformulations and discussions in Russian here:

http://community.livejournal.com/ru_math/355678.html

Seems quite similar to [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28832[/url]. Have you tried the approach outlined there?

Thank you very much! I did not see this topic before.

Alas, I do not see how your argument with Wiener tauberian theorem may be applied, since there must be some monotonicity condition, which may hardly be expected here.

The corresponding sequence satisfies conditions $2a_n=a_{[n/2]}+a_{[n/3]}$, but for $b_n: =na_n$ relations are bit diferent from $b_{n}=b_{[n/2]}+(3/2)b_{[n/3]}$, alas.

You are right, of course: we do not have monotonicity for free here. But the things are not as bad as they seem. ;) Indeed, the situation is as follows: we have a function $g$ with some given initial values on the interval $[0,T]$ and defined by $g(t)=\frac12(g(t-\log 2)+g(t-\log 3))$ for $t>T$. Now, suppose that the function $g$ can grow only at some constant rate $C>0$ on $[0,T]$, i.e., $g(t+h)\le g(t)+Ch$ if $0\le t\le t+h\le T$ but it can jump down as much as it wants on $[0,T]$. Assume also that the jump from $g(T)$ to $g(T+0)$ is also down. Then $g$ can grow only at the rate $C$ on $[0,+\infty)$, which is enough to run the argument based on Wiener's Tauberian theorem. Now, our initial data is not like that; it is merely a piecewise constant function that produces both upward and downward jumps. But we can represent it as a linear combination of the initial data with the above property. Indeed, the first step would be to represent it as a difference of two piecewise constant functions that have upward jumps only. The next step is to represent a given piecewise constant function $g$ with upward jumps as a difference  $Ct-(CT-g)$. It is not hard to see that the jump at $T$ for both $Ct$ and $Ct-g$ will be down if $C$ is so large that the maximum of $Ct-g$ is attained at $T$. The rest should be clear. :)


Hey,

Does this series converge into 0 or is it just divergent?

{0,1,0,0,1,0,0,0,1,0,0,0,0,1,.....}

Also, How should I do the following:

Determine whether the sequence converges or diverges, if it converges find the limit.

a_n =  (1*3*5*7*.....(2n-1))/n!

Thanks!

The first sequence is divergent because you can find two subsequences such that they have different limits: $\ a_k=0$ and $\ b_p=1$.

For the second sequence we have $\ \frac{a_{n+1}}{a_n}=\frac{2n+1}{n+1}$ so $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=2>1$. Then $\ a_n$ diverges to $\ \infty$.


Let [tex]x_{n}=\sum_{k=0}^{n} \arctan \frac{3}{k^{2}+5k+6}[/tex]. Prove that [tex]x_{n}[/tex] converges and its limit is somewhere inside [tex](\frac{\pi}{4},\frac{\pi}{2})[/tex].

More than often this kind of problem is solved by :

[hide]arctan(a) + arctan(b) = arctan (a+b/(1-a*b))  + telescoping.[/hide]

I think I've got it.  [hide]The sequence Xn converges iff the series that defines it converges.  To see that the series converges, use the ratio test.  arctan(3/[(k+3)(k+4)]) < arctan(3/[(k+2)(k+3)]) because tan(x) is 1-1 and increasing, so its inverse function must also be increasing.

Now consider tan(Xn).  Taking the tangents of partial sums, we have tan(X0) = 1/2, tan(X1) = 6/7, and tan(X2) = 141/122 > 1.  Thus, X2 >  :pi:/4.  Because the sequence is increasing and convergent, for n :ge: 2, we must have  :pi:/4 < Xn <  :pi:/2.[/hide]


Determine all real solutions of the equation: \[{ \frac{x^{2}}{x-1}+\sqrt{x-1}+\frac{\sqrt{x-1}}{x^{2}}}=\frac{x-1}{x^{2}}+\frac{1}{\sqrt{x-1}}+\frac{x^{2}}{\sqrt{x-1}} . \]

[hide]Doing $\frac{x^{2}}{x-1}=a$ and $\sqrt{x-1}=b$ we can rewrite the equation as,

$a+b+\frac{1}{ab}=\frac{1}{a}+\frac{1}{b}+ab$ this yields to $(a-1)(b-1)=(\frac{1}{a}-1)(\frac{1}{b}-1)$ that implies in three cases:

(I)$a=1 \implies \frac{x^{2}}{x-1}=1 \implies x^2-x+1$, therefore this quadratic equation do not is solvable,consequently there's no solution for $a=1$

(II)$b=1 \implies \sqrt{x-1}=1 \implies x=2$ consequently $2$ is a solution for the equation.

(III)$ab=\frac{(a-1)(b-1)}{(a-1)(b-1)}=1$, consequently we need to solve the equation $x^2=\sqrt{x-1}$ and since $\sqrt{x-1} \in \mathbb{R}$ it implies that $x>1$ therefore,

$\sqrt{x-1}<\sqrt{x} \implies x^4-x<0 \implies x>x^4 \implies 1>x$(contradition!) so for $ab=1$ the equation have no solution[/hide]

We set

${a=\frac{x^{2}}{x-1} \, , \, b=\sqrt{x-1}\, , \, c=\frac{\sqrt{x-1}}{x^{2}}}$.

Then the equation is written as: $a+b+c = \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
and we have $abc=1$.

Now

$a+b+c = \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \Rightarrow a+b+c =\frac{ab+bc+ca}{abc} \Rightarrow$

$a+b+c -ab-bc-ca=0 \Rightarrow abc -1+a+b+c -ab-bc-ca=0 \Rightarrow$

$(a-1)(b-1)(c-1)=0$ or equivelent ${(\frac{x^{2}}{x-1}-1)(\sqrt{x-1}-1)(\frac{\sqrt{x-1}}{x^{2}}}-1)=0$

Simplifying we get $\frac{(x^{2}-x+1)(\sqrt{x-1}-1)(\sqrt{x-1}-x^{2})}{(x-1)x^{2}}=0$

therefore the only solution is $x=2$.  :)

Must $x>1$.

Let ${a=\dfrac{x^{2}}{x-1}\;,\;b=\sqrt{x-1}\;,\;c=\dfrac{\sqrt{x-1}}{x^{2}}}$ , $a\;,\;b\;,\;c>0$.

Then $abc=1\;(1)$ and  $a+b+c=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \overset{(1)}{\iff} a+b+c=ab+bc+ca\;\;(2)$.

If $a+b+c=k$  , then $a,b,c$  are roots of $x^3-kx^2+kx-1=0 \iff (x-1)(x^2-(k-1)x+1)=0$ , so at least one of $a,b,c$ must be eqyual to $1$ .

So
$a=1\iff \dfrac{x^{2}}{x-1}=1\iff x^2-x+1=0$ (no soloution).
or
$b=1\iff \sqrt{x-1}=1\iff x=2$.
or
$c=1\iff \dfrac{\sqrt{x-1}}{x^{2}}=1\iff x^2=\sqrt{x-1}$ . But $x>1\Rightarrow x^2>\sqrt{x}>\sqrt{x-1}$.

So $x=2$ (verify the initial). Then $k=\dfrac{21}{4}$ the other roots are $4$ and $\dfrac{1}{4}$.

The equality is equivalent to x^4*(x-1) + x^2*(x-1)*sqrt (x-1) + (x-1)*sqrt (x-1) - (x-1)^2 - x^2*sqrt (x-1) - x^4*sqrt(x-1) = (sqrt(x-1)-1)(x^4*sqrt(x-1) + x^2*sqrt(x-1) + x^2 -(x-1)*sqrt(x-1)) = 0 which implies :1)
sqrt(x-1)-1=0 ,therefore x=2;
2)
x^4*sqrt(x-1) + x^2*sqrt(x-1) +x^2 -(x-1)*sqrt(x-1)=0, which implies sqrt(x-1)*(x^4 +x^2 - x +1)=-x^2,then from the conditions we have that x>1, so lets prove that this doesnt have any solutions. Lets prove that x^4 + x^2 -x +1>0.This is true because x^4 + (x- 1/2)^2 + 3/4 >0,
so we have that sqrt(x-1) > 0 and x^4+x^2-x+1>0 and -x^2<0,so contradiction.

Also,can someone post JBMO shortlist - 2011,please?


abstract algebra
superior algebra
superior algebra unsolved

Hi, I proved that
\[ Tor(\bigoplus_{i\in I}M_{i})= \bigoplus_{i\in I}Tor(M_{i}), \]
where $ \{M_{i}\}_{i\in I}$ is a family of modules, but I know this is not true for the direct product. $ (Tor=: torsion)$

can Someone give me a counterexample????

I'll appreciate any help

Consider a torsion element $ (a,b)\in A\times B$. Do you see a relation between the order of $ (a,b)$ and the orders of $ a$ and $ b$, respectively? owk

Try the direct product of $ \mathbb{Z}\slash n\mathbb{Z}$ for all $ n\in\mathbb{N}$. Each $ \mathbb{Z}\slash n\mathbb{Z}$ is a torsion module, but the direct product isn't.

  Darij


What are the needed conditions to check in order to use Lagrange Multipliers rigorously (olympiad-rigor, not full rigor)? Plz give a simple example if possible. Thanks!

I'm not completely sure I understand your question.  You need to have a function and a system of constraints.  THen you need to know how to take a gradient, which is really nothing more than a derivative.