geometry
parallelogram
homothety
Inversion
IMO Shortlist
geometry solved
lengths

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

http://www.mathlinks.ro/Forum/viewtopic.php?t=5676

  Darij

[quote="Tornado"][hide="Inversion"]If we invert about the center $P$ with an arbitrary radius $r$.For any point $X$, let $X'$ denote the image of $X$ under inversion.

\[\Gamma_1'||\Gamma_3'\] and \[\Gamma_2'||\Gamma_4'\]. So, $A'B'C'D'$ is a parallelogram.

Now $AB=\frac{r^2}{PA'\cdot PB'}\cdot A'B'$.
Writing similar expressions for $BC$,$AD$ and $DC$ and noticing that $A'B'=C'D'$ and $B'C'=A'D'$, we have
\[ \frac{AB\cdot BC}{AD\cdot DC}=\frac{PD'^{2}}{PB'^{2}}. \]
$\triangle PBD \sim \triangle PD'B' \implies \frac{PB}{PD}=\frac{PD'}{PB'}$, which finishes off the problem.[/hide][/quote]

Ya,thats like the only nice solution....I think this problem was problem was meant for inversion.....Otherwise you need to bash....Or maybe .....

I did this using the same.

Simple Solution: Seeing the number of circles in the diagram, and the prominent role of $P$ in the diagram, we invert about $P$ with a radius $r$. Note that the configuration becomes a parallelogram, and so $A'B'=C'D'$ and $D'A'=B'C'$. We will now manipulate $\frac{AB\cdot BC}{AD\cdot DC}$. Since $\Delta PAB\sim\Delta PA'B'$, $\frac{PB}{PA'}=\frac{AB}{A'B'}$. Hence we have that $AB=A'B'\cdot\frac{PA\cdot PB}{r^2}$. We can similarly find expressions for $BC, CA, AD$. Plugging these into the left hand side, we have $\frac{AB\cdot BC}{AD\cdot DC}=\frac{\frac{A'B'\cdot B'C'\cdot PA\cdot PB\cdot PB\cdot PC}{r^4}}{\frac{C'A'\cdot D'A'\cdot PC\cdot PD\cdot PD\cdot PA}{r^4}}=\frac{A'B'\cdot B'C'}{C'D'\cdot D'A'}\cdot \frac{PB^2}{PD^2}=\frac{PB^2}{PD^2}$, since $A'B'C'D'$ is a parallelogram, and we are done. $\Box$

[hide="No Inversion"]

Let the four circle centers be $O_1, O_2, O_3,$ and $O_4$. Note that by the definition of $A, B, C,$ and $D$, $A$ is the reflection of $P$ across $O_1O_2$, $B$ is the reflection of $P$ across $O_2O_3$, $C$ is the reflection of $P$ across $O_3O_4$, and $D$ is the reflection of $P$ across $O_4O_1$. Therefore, we find that $PA \bot O_1O_2$ and likewise for the other three points.

Consider the homothety centered at P with scale factor $1/2$. Suppose that this homothety takes $A, B, C, D$ to $A', B', C', D'$, respectively. By our observation above, the points $A', B', C', D'$ are on  $O_1O_2, O_2O_3, O_3O_4, O_4O_1$, respectively. Additionally, since $PA \bot O_1O_2$, the points $A', B', C', D'$ are actually the feet of the perpendiculars from $P$ to the sides of quadrilateral $O_1O_2O_3O_4$. Since the homothety preserves ratios of sides, it is sufficient to show that
\[\frac{A'B' \cdot B'C'}{A'D' \cdot D'C'}=\frac{B'P^2}{D'P^2}\]
Observe that
\[\frac{A'B' \cdot B'C'}{B'P^2} = \frac{A'B'}{B'P}\cdot\frac{B'C'}{B'P} = \frac{\sin\angle A'PB'}{\sin\angle PA'B'}\cdot\frac{\sin\angle C'PB'}{\sin\angle PC'B'}\]
by the law of sines.
Since $A', B', C', D'$ are the feet of the perpendiculars from $P$ to the sides of quadrilateral $O_1O_2O_3O_4$, the quadrilaterals $A'PB'O_2$ and $B'PC'O_3$ are cyclic. From this, we find that $\angle A'PB' = \angle A'O_2B' = \angle O_1O_2O_3$, $\angle B'PC' = \angle B'O_3C' = \angle O_2O_3O_4$, $\angle PA'B' = \angle PO_2B' = \angle O_4O_2O_3$, and $\angle PC'B' = \angle PO_3O_2 = \angle O_1O_3O_2$. Thus, from the expression above, we find that
\[\frac{\sin\angle A'PB'}{\sin\angle PA'B'}\cdot\frac{\sin\angle C'PB'}{\sin\angle PC'B'}\]
\[ = \frac{\sin\angle O_1O_2O_3}{\sin\angle O_4O_2O_3}\cdot\frac{\sin\angle O_4O_3O_2}{\sin\angle O_1O_3O_2}\]
\[ = \frac{\sin\angle O_1O_2O_3}{\sin\angle O_1O_3O_2}\cdot\frac{\sin\angle O_4O_3O_2}{\sin\angle O_4O_2O_3}\]
However, using the law of sines and triangles $\triangle O_1O_2O_3$ and $\triangle O_2O_3O_4$, the above expression becomes
\[\frac{\sin\angle O_1O_2O_3}{\sin\angle O_1O_3O_2}\cdot\frac{\sin\angle O_4O_3O_2}{\sin\angle O_4O_2O_3} = \frac{O_1O_3}{O_1O_2}\cdot\frac{O_2O_4}{O_3O_4}\].
By symmetry, since this expression is exclusively in terms of the diagonals of quadrilateral $O_1O_2O_3O_4$ and the sides of the quadrilateral with $A'$ and $C'$ on them, we can use the above logic to show also that
\[\frac{A'D' \cdot D'C'}{D'P^2}=\frac{O_1O_3}{O_1O_2}\cdot\frac{O_2O_4}{O_3O_4}=\frac{A'B' \cdot B'C'}{B'P^2}\]
which implies that
\[\frac{A'B' \cdot B'C'}{A'D' \cdot D'C'}=\frac{B'P^2}{D'P^2}\]
as desired.

[/hide]

[hide=Solution]
Invert about $P$ with arbitrary radius $r$; we write $K'$ to mean the inverse of an object $K$. Since $\Gamma_1$ and $\Gamma_3$ are tangent circles passing through $P$, their inverses $\Gamma_1'$ and $\Gamma_3'$ are parallel lines. Similarly, $\Gamma_2'$ and $\Gamma_4'$ are parallel lines, so $A'B'C'D'$ is a parallelogram, giving $A'B' = D'C'$ and $B'C' = A'D'$. By the inversive distance formula, $A'B' = \frac{r^2}{PA \cdot PB} \cdot AB$, so $AB = \frac{A'B' \cdot PA \cdot PB}{r^2}$. Analogous formulas hold for $BC$, $AD$, and $DC$. Then \[\frac{AB \cdot BC}{AD \cdot DC} = \frac{(A'B' \cdot PA \cdot PB)(B'C' \cdot PB \cdot PC)}{(A'D' \cdot PA \cdot PD)(D'C' \cdot PD \cdot PC)} = \frac{PB^2}{PD^2}\] as requested.
[/hide]

[hide=Solution]We invert with respect to point $P$ with arbitrary radius, let's say $r$. Let $X'$ denote point $X$ upon the transformation.

Note that circles $\Gamma_1, \Gamma_2, \Gamma_3, \Gamma_4$ are all turned into lines, let them be $\ell_1, \ell_2, \ell_3, \ell_4$, respectively. Since $\Gamma_1$ and $\Gamma_3$ were tangent, $\ell_1 \parallel \ell_3$ and similarly for $\ell_2$ and $\ell_4$. Thus, the inversion of our original figure results in a parallelogram $A'B'C'D'$.

From well known properties of inversion, $$\frac{AB\cdot BC}{AD\cdot DC}=\frac{\frac{r^2\cdot A'B'}{PA'\cdot PB'}\cdot \frac{r^2\cdot B'C'}{PB'\cdot PC'}}{\frac{r^2\cdot A'D'}{PA'\cdot PD'}\cdot \frac{r^2\cdot D'C'}{PD'\cdot PC'}}=\left(\frac{PD'}{PB'}\right)^2\cdot \left(\frac{A'B'\cdot B'C'}{A'D'\cdot D'C'}\right).$$ Since $A'B'C'D'$ is a parallelogram, it follows that $A'B'=D'C'$ and $B'C'=A'D'$. So our expression is equivalent to $$\left(\frac{PD'}{PB'}\right)^2\cdot \left(\frac{A'B'\cdot B'C'}{A'D'\cdot D'C'}\right)=\left(\frac{PD'}{PB'}\right)^2=\left(\frac{\frac{r^2}{PD}}{\frac{r^2}{PB}}\right)^2=\left(\frac{PB}{PD}\right)^2.$$ Thus, the desired is proved.[/hide]

[asy]
size(13cm);
pen my_purple = rgb(0.7,0.4,1), my_blue = rgb(0.1,0.7,1), my_green = rgb(0,0.7,0), my_orange = rgb(1,0.6,0.1), my_teal = rgb(0.1,0.8, 0.6), my_darkblue = rgb(0.1, 0.1, 0.6);
pair P = origin;
draw(circle((0,3/2), 3/2), my_blue);
draw(circle((0,-3), 3), my_blue);
draw(circle((-2,1), sqrt(5)), my_green);
draw(circle((8/3,-4/3), 4*sqrt(5)/3), my_green);
pair A[] = intersectionpoints(circle((0,3/2), 3/2),circle((-2,1), sqrt(5)));
pair A = A[0];
pair B[] = intersectionpoints(circle((0,3/2), 3/2),circle((8/3,-4/3), 4*sqrt(5)/3));
pair B = B[1];
pair C[] = intersectionpoints(circle((8/3,-4/3), 4*sqrt(5)/3), circle((0,-3), 3));
pair C = C[1];
pair D[] = intersectionpoints(circle((-2,1), sqrt(5)), circle((0,-3), 3));
pair D = D[0];
draw(A--B--C--D--cycle, my_orange+dashed);
dot(A^^B^^C^^D^^P);
real a = distance(A,P);
real r= 1.8;
pair Ai = (r/a)^2 * A;
real b = distance(B,P);
pair Bi = (r/b)^2 * B;
real c=  distance(C,P);
pair Ci = (r/c)^2 * C;
real d = distance(D,P);
pair Di = (r/d)^2 * D;
dot(Ai^^Bi^^Ci^^Di);
draw(Ai--Bi--Ci--Di--cycle, my_teal+dotted);
draw(circle(P,r), my_darkblue+dotted);
label("$A$", A, dir(90));
label("$A^*$", Ai, dir(90));
label("$B$", B, dir(60));
label("$B^*$", Bi, 1.5*dir(90));
label("$C$", C, dir(C));
label("$C^*$", Ci, 0.3*dir(C));
label("$D$", D, dir(D));
label("$D^*$", Di, dir(D));
label("$P$", P, dir(A));
[/asy]
Invert the diagram about $P$ with radius $r$. From the inversion distance formula, we have that
\[ A^*B^* = \frac{AB \cdot r^2}{PA \cdot PB} \]
This can be written as,
\[ AB = \frac{A^*B^* \cdot PA \cdot PB}{r^2} \]
Similarly, we also have,
\[ BC = \frac{B^*C^* \cdot PB \cdot PC}{r^2} \]
\[ AD = \frac{A^*D^* \cdot PA \cdot PD}{r^2} \]
\[ CD = \frac{C^*D^* \cdot PC \cdot PD}{r^2} \]
Therefore, we have that,
\[ \frac{AB \cdot BC}{AD \cdot DC} = \frac{\frac{A^*B^* \cdot PA \cdot PB}{r^2} \cdot \frac{B^*C^* \cdot PB \cdot PC}{r^2}}{\frac{A^*D^* \cdot PA \cdot PD}{r^2} \cdot \frac{C^*D^* \cdot PC \cdot PD}{r^2}} \]
We can simplify this to get,
\begin{align*} \frac{AB*BC}{AD \cdot DC} &= \frac{\frac{A^*B^* \cdot PA \cdot PB}{r^2} \cdot \frac{B^*C^* \cdot PB \cdot PC}{r^2}}{\frac{A^*D^* \cdot PA \cdot PD}{r^2} \cdot \frac{C^*D^* \cdot PC \cdot PD}{r^2}} \\ &= \frac{\frac{A^*B^* \cdot B^*C^* \cdot PA \cdot PB^2 \cdot PD^2}{r^4}}{\frac{A^*D^* \cdot C^*D^* \cdot PA \cdot PC^2 \cdot PD}{r^4}} \\ &= \frac{A^*B^* \cdot B^*C^* \cdot PB^2}{A^*D^* \cdot C^*D^* \cdot PD^2} \end{align*}
Therefore, it suffices to prove that $A^*B^* \cdot B^*C^* = A^*D^* \cdot C^*D^*$. However, note that  the centers of $\Gamma_1$ and $\Gamma_3$ are collinear and the image after inversion is of a line perpendicular to the line connecting the center of each circle to $P$ to $P$. Since the centers are collinear with $P$, the images of both $\Gamma_1$ and $\Gamma_3$ are perpendicular to the line through $P$ connecting the centers, which means that $\Gamma_1^*$ is parallel to $\Gamma_3^*$. Similarly, the images of $\Gamma_2$ and $\Gamma_4$ are parallel.

This implies that $A^*B^*C^*D^*$ is a parallelogram because both pairs of opposite sides are parallel. Therefore, $A^*B^* = C^*D^*$ and $B^*C^* = A^*D^*$ ,which implies that $A^*B^* \cdot B^*C^* = A^*D^* \cdot C^*D^*$. Thus, we conclude $\frac{AB \cdot BC}{AD \cdot DC} = \frac{PB^2}{PD^2}$

[hide=Inversion]
Invert around $P$ with radius 1. Let $X^*$ denote image of $X$ under inversion.
As circles are tangent, we get $A^*B^*C^*D^*$ is parallelogram, which implies $\frac{A^*B^* \cdot B^*C^*}{C^*D^* \cdot D^*A^*}=1$

By inversion-distance formula,
\begin{align*}
\frac{A^*B^* \cdot B^*C^*}{C^*D^* \cdot D^*A^*}&=\frac{\frac{AB}{PA \cdot PB} \cdot \frac{BC}{PB \cdot PC}}{\frac{CD}{PC \cdot PD} \cdot \frac{DA}{PD \cdot PA}} \\
&=\frac{AB \cdot BC}{CD \cdot DA} \cdot \frac{PD^2}{PB^2}\\
\end{align*}

Thus, $\frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}$, as needed. $\blacksquare$
[/hide]

@post #7
There is $\angle A'PB' = \angle A'O_2B' =\angle O_1O_2O_3$ instead of $\angle A'PB' = 180^\circ-\angle A'O_2B' =180^\circ- \angle O_1O_2O_3$ . But the whole of reasoning stays correct because $\sin a=\sin (180^\circ-a)$. Very nice indeed.

[b]Proof:[/b] We invert the configuration about $P$ with the radius of inversion being $1$.

This inversion takes:

[b]1.[/b] $\Gamma_i \rightarrow l_i , i= 1,2,3,4.$ where $l_i$ is a line not passing through $P$.

[b]2.[/b] $A' = \l_1 \cap l_2$. $B',C',D'$ are defined analogously.

[b]3.[/b] $\Gamma_1$ tangent to $\Gamma_3 \implies l_1 || l_3.$ Similarly $l_2 || l_4$.

Hence $A'B'C'D'$ is a [i]parallelogram.[/i] Thus $A'B'=C'D' , B'C' = D'A'$.....(1)

Therefore the length formula( whatever is it called) gives $A'B' = \dfrac {AB}{PA . PB} ; C'D' = \dfrac {CD}{PC.PD}$.

Using (1) we have $\dfrac {AB}{CD} =$$ \dfrac {PA. PB}{PC. PD} ; \dfrac {BC}{AD} = \dfrac {PB.PC}{PA.PD}.$

Multiplying gives the desire result $\blacksquare$

Consider an inversion $\mathcal{I}$ about $P$ with radius $1$. Let $^*$ denote a point after $\mathcal{I}$ is performed on it. We now compute that
\begin{align*}
\frac{AB\cdot BC}{AD\cdot DC} &= \frac{\dfrac{1^2}{PA^* \cdot PB^*} \cdot A^*B^* \cdot \dfrac{1^2}{PB^* \cdot PC^*} \cdot B^*C^*}{\dfrac{1^2}{PA^* \cdot PD^*} \cdot A^*D^* \cdot \dfrac{1^2}{PD^* \cdot PC^*} \cdot D^*C^*}\\
&= \frac{\dfrac{1^2}{\left(PB^*\right)^2} \cdot A^*B^* \cdot B^*C^*}{\dfrac{1^2}{\left(PD^*\right)^2} \cdot A^*D^* \cdot D^*C^*} \\
&= \frac{PB^2\cdot A^*B^* \cdot B^*C^*}{PD^2 \cdot A^*D^* \cdot D^*C^*}
\end{align*}
by the inversion distance formula. Thus, it suffices to show that $A^*B^* \cdot B^*C^* = A^*D^* \cdot D^*C^*$.

Now, we observe that $\mathcal{I}$ takes $\Gamma_1$ and $\Gamma_3$ to lines parallel to each other, and takes $\Gamma_2$ and $\Gamma_4$ to lines parallel to each other. Thus, $\mathcal{I}(ABCD)$ is a parallelogram. But now, we are immediately done, as $A^*B^* = D^*C^*$ and $B^*C^* = A^*D^*$. $\fbox{}$

[hide=Storage][quote=ISL 2003 G4]Let $\Gamma_1$, $\Gamma_2$, $\Gamma_3$, $\Gamma_4$ be distinct circles such that $\Gamma_1$, $\Gamma_3$ are externally tangent at $P$, and $\Gamma_2$, $\Gamma_4$ are externally tangent at the same point $P$. Suppose that $\Gamma_1$ and $\Gamma_2$; $\Gamma_2$ and $\Gamma_3$; $\Gamma_3$ and $\Gamma_4$; $\Gamma_4$ and $\Gamma_1$ meet at $A$, $B$, $C$, $D$, respectively, and that all these points are different from $P$. Prove that

 \[
  \frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}.
 \][/quote]
[b][color=#000]Solution:[/color][/b] Invert around $P$ with radius $1$. $ABCD$ becomes a Rhombus. For a circle with radius $1$ and a point $P$ with inverted image of $P$ being $P^*$, we have $OP \cdot OP^*$ $=$ $1$. For another Point $Q$ and its inverted image being $Q^*$, we have: \begin{align*} PQ=\frac{P^*Q^*}{OP^* \cdot OQ^*} \end{align*}
So effectively, \begin{align*} \frac{AB \cdot BC}{AD \cdot DC}=\frac{\tfrac{A^*B^*}{PA^* \cdot PB^*} \times \tfrac{B^*C^*}{PB^* \cdot PC^*}}{\tfrac{A^*D^*}{PA^* \cdot PD^*} \cdot \tfrac{C^*D^*}{PC^* \cdot PD^*}} =\left(\frac{PD^*}{PB^*} \right)^2 = \left( \frac{PB}{PD} \right)^2 \qquad \blacksquare \end{align*}

[/hide]

[quote=ISL 2003 G4]Let $\Gamma_1$, $\Gamma_2$, $\Gamma_3$, $\Gamma_4$ be distinct circles such that $\Gamma_1$, $\Gamma_3$ are externally tangent at $P$, and $\Gamma_2$, $\Gamma_4$ are externally tangent at the same point $P$. Suppose that $\Gamma_1$ and $\Gamma_2$; $\Gamma_2$ and $\Gamma_3$; $\Gamma_3$ and $\Gamma_4$; $\Gamma_4$ and $\Gamma_1$ meet at $A$, $B$, $C$, $D$, respectively, and that all these points are different from $P$. Prove that

 \[
  \frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}.
 \][/quote]

We invert the system with center as $P$ and radius of inversion $r$. Let primes denote inverses of points .
$\Gamma_1' \parallel \Gamma_3'$ and $\Gamma_2' \parallel \Gamma_4' \Longrightarrow A'B'C'D'$ is a parallelogram.
By [i]Inversion Distance Formula [/i] :
$1 = \frac{A'B'}{D'C'} = \frac{AB}{DC} \cdot \frac{PC \cdot PD \cdot r^2}{PA\cdot PB \cdot r^2}$

$1 = \frac{B'C'}{A'D'} = \frac{BC}{AD} \cdot \frac{PA \cdot PD \cdot r^2}{PB\cdot PC \cdot r^2}$

$\frac{AB \cdot BC}{DC \cdot AD} \cdot \frac{PC \cdot PD}{PA \cdot PB} \cdot \frac{PA \cdot PD}{PB \cdot PC} = 1 \Longrightarrow \frac{AB \cdot BC}{AD\cdot DC} = \frac{PB^2}{PD^2} \qquad \blacksquare$

[quote=orl]Let $\Gamma_1$, $\Gamma_2$, $\Gamma_3$, $\Gamma_4$ be distinct circles such that $\Gamma_1$, $\Gamma_3$ are externally tangent at $P$, and $\Gamma_2$, $\Gamma_4$ are externally tangent at the same point $P$. Suppose that $\Gamma_1$ and $\Gamma_2$; $\Gamma_2$ and $\Gamma_3$; $\Gamma_3$ and $\Gamma_4$; $\Gamma_4$ and $\Gamma_1$ meet at $A$, $B$, $C$, $D$, respectively, and that all these points are different from $P$. Prove that

 \[
  \frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}.
 \][/quote]

[size=150]My First Inversion Solution:[/size] :D
we invert about $P$ with a radius $r$.

[color=#00f]$ AB=\frac{r^2}{PA^*\cdot PB^*}\cdot A^*B^* $

$ BC=\frac{r^2}{PB^*\cdot PC^*}\cdot B^*C^*$

$ AD=\frac{r^2}{PA^*\cdot PD^*}\cdot A^*D^*$

$ DC=\frac{r^2}{PD^*\cdot PC^*}\cdot D^*C^*$

$ A^*B^*C^*D^*-$parallelogram$\implies A^*B^*=C^*D^*,A^*D^*=B^*C^* $

$$ \frac{AB\cdot BC}{AD\cdot DC}=\frac{\frac{r^2}{PA^*\cdot PB^*}\cdot A^*B^*\cdot \frac{r^2}{PB^*\cdot PC^*}\cdot B^*C^*}{\frac{r^2}{PA^*\cdot PD^*}\cdot A^*D^*\cdot \frac{r^2}{PD^*\cdot PC^*}\cdot D^*C^*}=\frac{A^*B^*\cdot B^*C^*}{A^*D^*\cdot D^*C^*}\cdot (\frac{PD^*}{PB^*})^2=(\frac{PD^*}{PB^*})^2=(\frac{\frac{r^2}{PD}}{\frac{r^2}{PB}})^2=\frac{PB^2}{PD^2} $$
[/color]

Take inversion about P with arbitary radius.Now as the circles are tangent,the quadrilateral formed by inversion is nothing but a parreleogram.so apply inversion distance formulae

Invert around $P$ with arbitrary radius $r$. Since the centers of $\Gamma_1$ and $\Gamma_3$ are perpendicular, we have $A'B' \parallel C'D'$. In a similar fashion with $\Gamma_2$ and $\Gamma_4$, we get that $B'C' \parallel D'A'$. Hence, $A'B'C'D'$ is a parallelogram. This means that $A'B'=C'D'$ and $B'C'=D'A'$. Using the inversion distance formula, we have
\[AB=\frac{r^2}{PA' \cdot PB'} \cdot A'B'\] and similar for $BC, CD, DA$. Then,
\[\frac{AB\cdot BC}{AD \cdot DC}=\frac{\frac{r^4\cdot A'B' \cdot B'C'}{PB'\cdot PC' \cdot PA' \cdot PB'}}{\frac{r^4 \cdot CD' \cdot DA'}{PC' \cdot PD' \cdot PD' \cdot PA'}}=\frac{(PD')^2}{(PB')^2}.\] However, $PD \cdot PD'=r^2$ (and similar for $PB$), so
\[\frac{(PD')^2}{(PB')^2}=\frac{\left (\frac{r^2}{PD}\right)^2}{\left ( \frac{r^2}{PB} \right)^2}=\frac{PB^2}{PD^2}.\]

[b]My Solution- [/b]
[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(12cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -35.7075, xmax = 50.6925, ymin = -17.2125, ymax = 25.717500000000097;  /* image dimensions */
pen qqqqcc = rgb(0,0,0.8);
 /* draw figures */
draw(circle((-3.27,2.305), 7), linewidth(0.5) + qqqqcc);
draw(circle((-3.27,2.305), 0.24622144504491064), linewidth(0.5));
draw(circle((13.545,2.385), 9.835884270195846), linewidth(0.5) + qqqqcc);
draw(circle((3.375,-5.76), 8.516334147322569), linewidth(0.5) + red);
draw(circle((4.185,13.635), 10.89559463122684), linewidth(0.5) + red);
 /* dots and labels */
dot((-3.27,2.305),dotstyle);
label("$O_{1}$", (-2.97,2.97), NE * labelscalefactor);
dot((3.715872259571556,2.749509587015399),dotstyle);
label("$P$", (3.9825,3.4425), NE * labelscalefactor);
dot((13.545,2.385),dotstyle);
label("$O_{3}$", (13.8375,3.0375), NE * labelscalefactor);
dot((3.375,-5.76),dotstyle);
label("$O_{2}$", (3.645,-5.0625), NE * labelscalefactor);
dot((4.185,13.635),dotstyle);
label("$O_{4}$", (4.455,14.31), NE * labelscalefactor);
dot((-5.042583357919082,-4.466849691127846),linewidth(1pt) + dotstyle);
label("$A$", (-4.7925,-3.915), NE * labelscalefactor);
dot((11.753437608298926,-7.286345489296119),linewidth(1pt) + dotstyle);
label("$B$", (12.015,-6.75), NE * labelscalefactor);
dot((14.974258577259558,12.116487003331814),linewidth(1pt) + dotstyle);
label("$C$", (15.255,12.69), NE * labelscalefactor);
dot((-5.626255526711521,8.896514233682684),linewidth(1pt) + dotstyle);
label("$D$", (-5.3325,9.45), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */
[/asy]

Now Invert the above sketch around $P$ of radius $1$ , after Inverting ,suppose point $X$ mapped to $X'$, note that all circles will be get mapped to line perpendicular to the line joining $P$ with their respective centers. And $A'$ , $B'$ , $C'$ and $D'$ will be the intersections of these lines and we will get a rectangle in which $P$ will be to the meeting of the diagonals of rectangle.

[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(12cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -10.9, xmax = 14.7, ymin = -6.06, ymax = 6.66;  /* image dimensions */
pen qqccqq = rgb(0,0.8,0); pen ttffqq = rgb(0.2,1,0);
 /* draw figures */
draw((-5.08,2.38)--(6.32,2.4), linewidth(0.5) + qqccqq);
draw((6.32,2.4)--(6.34,-2.92), linewidth(0.5) + qqccqq);
draw((6.34,-2.92)--(-4.92,-2.84), linewidth(0.5) + ttffqq);
draw((-4.92,-2.84)--(-5.08,2.38), linewidth(0.5) + qqccqq);
 /* dots and labels */
dot((-5.08,2.38),dotstyle);
label("$A'$", (-5,2.58), NE * labelscalefactor);
dot((-4.92,-2.84),dotstyle);
label("$D'$", (-4.84,-2.64), NE * labelscalefactor);
dot((6.32,2.4),dotstyle);
label("$B'$", (6.4,2.6), NE * labelscalefactor);
dot((6.34,-2.92),dotstyle);
label("$C'$", (6.42,-2.72), NE * labelscalefactor);
dot((0.611332118499859,-0.2613362721584286),linewidth(2pt) + dotstyle);
label("$P$", (0.7,-0.1), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */
[/asy]

Hence after using the Inversion Distance formula , derive expression for each side of rectangle and after substituting the values , we will be done. (I will not do calculations, because it's given above many times..)
$\blacksquare$

We rename $P$ to $O$. Invert about $O$ with radius 1. Note that since $\Gamma_1$ is tangent to $\Gamma_3$, we will have $A^*D^*\parallel B^* C^*$, and similarly $\Gamma_2$ tangent to $\Gamma_4$ implies that $A^*B^*\parallel C^* D^*$.

Thus, $A^*B^*C^*D^*$ is a parallelogram, so $A^*B^*=C^*D^*$ and $A^*D^*=B^*C^*$.

Now, we evaluate using Inversion Distance Formula,
\[\frac{AB\cdot BC}{AD\cdot DC} = \frac{\frac{A^*B^*}{OA^*OB^*}\cdot \frac{B^*C^*}{OB^*OC^*}}{\frac{AD^*}{OA^*OD^*}\cdot \frac{D^*C^*}{OD^*OC^*}} = \frac{{OD^*}^2}{{OB^*}^2}\cdot \frac{A^*B^*\cdot B^*C^* \cdot OA^*\cdot OC^* }{C^*D^*\cdot A^*B^*\cdot OA^*\cdot OC^*} = \frac{{OD^*}^2}{{OB^*}^2}= \frac{OB^2}{OD^2}\]
and we are done $\blacksquare$.

[hide=Solution]We invert around $P$ with radius $1.$ Then, by the inversion distance formula, \begin{align*}AB&=A^*B^*\cdot PA\cdot PB,\\BC&=B^*C^*\cdot PB\cdot PC,\\AD&=A^*D^*\cdot PA\cdot PD,\\DC&=D^*C^*\cdot PD\cdot PC.\end{align*} But $\Gamma_1 ^*\parallel\Gamma_3 ^*$ and $\Gamma_2 ^*\parallel\Gamma_4 ^*$ so $A^* B^* C^* D^*$ is a parallelogram. Thus, \begin{align*}\frac{AB\cdot BC}{AD\cdot DC}&=\frac{A^*B^*\cdot PA\cdot PB\cdot B^*C^*\cdot PB\cdot PC}{A^*D^*\cdot PA\cdot PD\cdot D^*C^*\cdot PD\cdot PC}\\&=\frac{PB^2}{PD^2},\end{align*} as desired. $\square$[/hide]

We invert wrt $P$ with radius $1,$ and denote images of points as $\bullet '.$ Clearly $A'B'C'D'$ is a parallelogram, and then $$\frac{|AB|\cdot |BC|}{|AD|\cdot |DC|}=\frac{|A'B'|\cdot |B'C'|\cdot |PA|\cdot |PC|\cdot |PB|^2}{|A'D'|\cdot |D'C'|\cdot |PA|\cdot |PC|\cdot |PD|^2}=\frac{|PB|^2}{|PD|^2}\text{ } \blacksquare$$

Consider an inversion at $P$ with radius $1$. It's easy to see that $ABCD$ goes to a parallelogram, so $$\frac{AB \cdot BC}{AD \cdot DC} = \frac{\left(   A^*B^* \cdot \frac{1}{PA^* \cdot PB^*} \right) \left(B^*C^* \cdot \frac{1}{PB^* \cdot PC^*}\right)}{\left( A^*D^* \cdot \frac{1}{PA^* \cdot PD^*} \right) \left(D^*C^* \cdot \frac{1}{PD^* \cdot PC^*}\right)}$$ $$= \frac{A^*B^*}{D^*C^*} \cdot \frac{B^*C^*}{A^*D^*} \cdot \frac{\frac{1}{(PB^*)^2} \cdot \frac{1}{PA^* \cdot PC^*}}{\frac{1}{(PD^*)^2} \cdot \frac{1}{PA^* \cdot PC^*}} = \frac{\frac{1}{(PB^*)^2}}{\frac{1}{(PD^*)^2}} = \frac{PB^2}{PD^2}$$ as desired. $\blacksquare$

[hide = Literally the same solution as every other one in the thread]
Inverting around $P$ with radius $1$, and applying the inversion distance formula, we get:

\[\frac{AB\cdot BC}{AD\cdot DC}=\frac{(A'B'\cdot PA\cdot PB)(B'C'\cdot PB\cdot PC)}{(A'D'\cdot PA\cdot PD)(D'C'\cdot PD\cdot PC)}=
\frac{A'B'}{D'C'}\cdot \frac{B'C'}{A'D'} \cdot \frac{PB^2}{PD^2}.\]

Since $\Gamma_1$ and $\Gamma_3$ are tangent, and $\Gamma_2$ and $\Gamma_4$ are tangent at $P$, $A'D'\parallel B'C'$ and $A'B'\parallel D'C'$. Therefore $A'B'C'D'$ is a parallelogram, so $A'B'=C'D'$ and $B'C'=A'D'$:
\[\frac{A'B'}{D'C'}\cdot \frac{B'C'}{A'D'} \cdot \frac{PB^2}{PD^2}=\frac{PB^2}{PD^2},\]

which implies the desired result. :coolspeak:
[/hide]

Invert around $P$ with $r=1$. Notice that the centers $O_1$ and $O_3$ of $\Gamma_1$, $P$ and $\Gamma_3$ are collinear with $P$. Hence, $\Gamma_1'$ and $\Gamma_3'$ are perpendicular to line $O_1O_3$ and thus parallel. Similarly, $\Gamma_2'\parallel \Gamma_4'$. Now, the four inverted circles form a parallelogram. Notice that $\Gamma_1'\cap \Gamma_2'=A'$. Points $B'$, $C'$, $D'$ follow similarly. Now, $A'D'=B'C'$ and $A'B'=D'C'$. Then by the distance formula:
$$A'D'\cdot D'C'=\frac{AD}{PA\cdot PD}\cdot \frac{DC}{PD\cdot PC}$$
and
$$B'C'\cdot A'B'=\frac{BC}{PB\cdot PC}\cdot \frac{AB}{PA\cdot PB}$$
giving
$$\frac{AD}{PA\cdot PD}\cdot \frac{DC}{PD\cdot PC}=\frac{BC}{PB\cdot PC}\cdot \frac{AB}{PA\cdot PB}$$
whcih simpifies to
$$\frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}$$

We will invert around the point $P$ with radius $1$ , and we have that $ABCD$ maps to a parallelogram so we have by the inversion distance formula

$\frac{AB \cdot DC}{AD \cdot DC}$ = $\frac{\frac{A^*B^*}{OA^*OB^*}\cdot \frac{B^*C^*}{OB^*OC^*}}{\frac{AD^*}{OA^*OD^*}\cdot \frac{D^*C^*}{OD^*OC^*}} = \frac{{OD^*}^2}{{OB^*}^2}= \frac{OB^2}{OD^2}$.

[hide]$A^*B^*C^*D^*$ is a parallelogram so we have from properties of a parallelogram that $A^*B^* = C^*D^*$ and $A^*D^* = B^*C^*$  when we apply inversion distance formula the equations cancel nicely [/hide]

Invert arcound $P$ with radius $1$.
The conditions in the problem imply that $\Gamma_1^\ast$ and $\Gamma_3^\ast$ are parallel lines, as are $\Gamma_2^\ast$ and $\Gamma_4^\ast$. So $A^\ast B^\ast C^\ast D^\ast$ is a parallelogram, \begin{align*} A^\ast B^\ast &= D^\ast C^\ast \iff \frac{AB}{PA \cdot PB} = \frac{CD}{PC \cdot PD} \\ \text{and } A^\ast D^\ast &= B^\ast C^\ast. \iff \frac{AD}{PA \cdot PD} = \frac{BC}{PB \cdot PD} \\ \end{align*} Take the quotient of these two to extract the desired result.

Invert the figure around a circle centered at $P$ with a radius of $1$.

Since tangencies create parallel lines, we clearly see that $A'B'C'D'$ is a parallelogram with $A'B'=C'D'$ and $A'D'=B'C'$.
Using the Inversion distance formula, we get $$\frac{AB\cdot BC}{AD\cdot DC}=\frac{A'B'\cdot B'C'}{PA'\cdot PB'^2\cdot PC'}\cdot\frac{PA'\cdot PC'\cdot PD'^2}{C'D'\cdot D'A'} = \frac{PD'^2}{PB'^2}=\frac{PB^2}{PD^2}. \blacksquare$$

Invert about $P$ with radius $1$, then the images of $\Gamma_1$ and $\Gamma_3$ are parallel lines, as are $\Gamma_2$ and $\Gamma_4$, so the images of $A,B,C,D$ form a parallelogram which gives the result.

[hide=solution using inversion yay]
Invert around $P$ with radius $1$. Note that $\Gamma_1^*\parallel\Gamma_3^*$ and $\Gamma_2^*\parallel\Gamma_4^*$, so $A^*B^*C^*D^*$ is a parallelogram. Hence $A^*B^*=C^*D^*$ and $B^*C^*=D^*A^*$, so by the distance formula, we have \begin{align*}
\frac{1}{PA\cdot PB}\cdot AB&=\frac{1}{PC\cdot PD}\cdot CD\\
\frac{1}{PB\cdot PC}\cdot BC&=\frac{1}{PD\cdot PA}\cdot DA.
\end{align*}
Multiplying the expressions together gives \[\frac{AB\cdot BC}{PA\cdot PB^2\cdot PC}=\frac{AD\cdot DC}{PA\cdot PC\cdot PD^2},\] so \[\frac{AB\cdot BC}{PB^2}=\frac{AD\cdot DC}{PD^2}.\]
Therefore, \[\frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2},\] as desired.[/hide]

Invert around $P$ with radius $r$. For any point $S$, let $S'$ be the image of $S$ under this inversion.Since circles $\Gamma_1$ and $\Gamma_3$ are tangent at $P$, they get sent to parallel lines, similar to $\Gamma_2$ and $\Gamma_4$. Thus, $A'B'C'D'$ is a parallelogram. By Inversion Distance Formula, $$\frac{AB\cdot BC}{AD\cdot DC}=\frac{\frac{r^2}{PA'\cdot PB'}\cdot A'B'\cdot \frac{r^2}{PB'\cdot PC'}\cdot B'C'}{\frac{r^2}{PA'\cdot PD'}\cdot A'D'\cdot \frac{r^2}{PC'\cdot PD'}\cdot C'D'}.$$ Noting that $A'B'=C'D'$ and  $A'D'=B'C'$ by the parallelogram, after cancelling as much as possible we are left with $$\frac{AB\cdot BC}{AD\cdot DC}=\frac{PD'^2}{PB'^2}.$$ However, by definition of inversion, $$\frac{PD'}{PB'}=\frac{PB}{PD},$$ so we are done.

Cleanest inversion problem ever that ive seen (this is my first problem solved by inversion!!) I think after learning inversion this is an instasolve in very few minutes, very easy to see it. It's one of those problems where if you have learned a common strategy, very easy to solve, but if not maybe much more difficult.

Invert at P with arbitrary radius. Since the stuff preserved in inversion is preserved the tangency point from circles 1&3 and 2&4 gives the new circles are now lines that are parallel, hence they make a parallelogram at their intersection points. It is a property of inversion that $AB=\frac{r^2}{PA'*PB'}*A'B'$ and cyclically, and since A'B'=C'D' and cyclically, we have by simple substitution that LHS=PD'^2/PB'^2=PB^2/PD^2, where the last step follows from inversion making inversely similar triangles. :)

Let us invert about $P$ with radius $r$. As all the circles pass through $P$, they will invert to lines that intersect at $A^*,B^*,C^*,D^*$, Furthermore, as $\Gamma_1$ and $\Gamma_3$, are tangent, $\Gamma_1^*$ and $\Gamma_3^*$ are parallel. Similarly, $\Gamma_2^*$ and $\Gamma_4^*$ are parallel. Thus, $A^*B^*C^*D^*$ is a parallelogram.

Using the distance formula, we can say that:
\[\frac{AB\cdot BC}{AD\cdot DC}=\frac{A^*B^*\cdot B^*C^*}{A^*D^*\cdot D^*C^*}\cdot \frac{PA^*\cdot PB^*}{PA^*\cdot PB^*}\cdot \frac{PD^*\cdot PC^*}{PB^*\cdot PC^*}\]
\[=\left(\frac{PD^*}{PB^*}\right)^2\]
\[=\frac{PB^2}{PD^2}.\]

Invert around $P$ with radius $r=1$. Then $A'B'\parallel C'D'$ and $B'C'\parallel D'A'$, so $A'B'C'D'$ is a parallelogram. Thus,
\[\frac{AB\cdot BC}{AD\cdot DC}=\frac{\frac{r^2}{PA'\cdot PB'}\cdot \frac{r^2}{PB'\cdot PC'}\cdot A'B'\cdot B'C'}{\frac{r^2}{PA'\cdot PD'}\cdot \frac{r^2}{PD'\cdot PC'}\cdot A'D'\cdot D'C'}=\frac{PD'^2}{PB'^2}=\frac{PB^2}{PD^2}\,\blacksquare\]

Solved with [b]drhong[/b].

Invert about $P$, so that $\Gamma_1'\parallel\Gamma_3',\Gamma_2'\parallel\Gamma_4'$ and so $A'B'C'D'$ is a parallelogram; then
\begin{align*}
A'B'&=C'D'\implies\frac{r^2\cdot AB}{PA\cdot PB}=\frac{r^2\cdot CD}{PC\cdot PD}\\
A'D'&=B'C'\implies\frac{r^2\cdot AD}{PA\cdot PD}=\frac{r^2\cdot BC}{PB\cdot PC}.\\
\end{align*}
Cancelling the $r^2$'s and cross-multiplying, we have
\[\frac{AB\cdot BC}{PA\cdot PC\cdot PB^2}=\frac{AD\cdot CD}{PA\cdot PC\cdot PD^2}\implies\frac{AB\cdot BC}{AD\cdot CD}=\frac{PB^2}{PD^2},\]
as desired. $\blacksquare$

Good exercise in inversion.

Upon inversion about $P$, $A'B'C'D'$ is a parallelogram. Thus $A'B'=C'D'$, and the result follows from applying the inversion distance formula.

terrible ideas

Invert about $P$ with radius $1$, and denote images as $\bullet'$. Clearly $A'B'C'D'$ is a parallelogram, and by applying the so-called inversion distance formula it suffices to show that
$$\frac{A'B'\cdot B'C'}{A'D'\cdot D'C'}=1,$$
but this is clear as $A'B'=D'C'$ and $B'C'=A'D'$. $\blacksquare$

Invert about $P$ with radius $1$
$$\Gamma_{1}^{*} \parallel \Gamma_{3}^{*}, \Gamma_{2}^{*} \parallel \Gamma_{4}^{*}$$
So $A^{*}B^{*}C^{*}D^{*}$ is a parallelogram.
By the inversion difference formula
$$AB=\frac{A^{*}B^{*}}{PA^{*}PB^{*}}, BC=\frac{B^{*}C^{*}}{PB^{*}PC^{*}},  AD=\frac{A^{*}D^{*}}{PA^{*}PD^{*}},  DC=\frac{D^{*}C^{*}}{PD^{*}PC^{*}}$$
Multiplying we find it suffices to show $$(A^{*}B^{*})  B^{*}D^{*} = (C^{*}D^{*}) D^{*}A^{*}$$
Which is true because of parallelogram properties.

If we do inversion from P it will go to infinity, so as all four of the circles transform into lines $\Gamma_1'$ and $\Gamma_3'$ will be parallel to each other, same with $\Gamma_2'$ and $\Gamma_4'$. From this we deduce hat $A'B'C'D'$ is a parallelogram $\Longrightarrow$ $A'B' = C'D'$ and $B'C' = A'D'$

$\frac{AB \cdot BC}{AD \cdot DC} = \frac{R^4 \cdot A'B' \cdot B'C'}{A'P \cdot B'P \cdot B'P \cdot C'P} \cdot \frac{A'P \cdot D'P \cdot D'P \cdot C'P}{R^4 \cdot A'D' \cdot D'C'} = \frac{D'P^2}{B'P^2} = \frac{PB^2}{PD^2} $

$\color{magenta} \boxed{\textbf{SOLUTION G4}}$

Taking $$\mathcal{I} :P,r$$

$\bullet$ $L_1^* \parallel L_3^* $ and $L_2^*\parallel L_4^*$

$\bullet$ $L_1^* \cap L_2^* \equiv A^*, L_2^* \cap L_3^* \equiv B^*, L_3^* \cap L_4^* \equiv C^*, L_4^* \cap L_1^* \equiv D^* $

$\bullet$ $A^*B^*C^*D^*$ is a parallelogram

Now, $$XY= \frac{X^*Y^*.PX.PY}{r^2}, X,Y \in ({A,B,C,D})$$
It gives,
$$\frac{AB\cdot BC}{AD\cdot DC}=\frac{\frac{A^*B^*\cdot B^*C^*\cdot PA\cdot PB\cdot PB\cdot PC}{r^4}}{\frac{C^*A^*\cdot D^*A^*\cdot PC\cdot PD\cdot PD\cdot PA}{r^4}}=\frac{A^*B^*\cdot B^*C^*}{C^*D^*\cdot D^*A^*}\cdot \frac{PB^2}{PD^2}=\frac{PB^2}{PD^2} \blacksquare $$

Inverting at $P$ with radius 1 gives us parallelogram $A^*B^*C^*D^*$. Inversion distance formula tells us the LHS is just
\[\frac{(PA^* \cdot PD^*) \cdot (PD^* \cdot PC^*)}{(PA^* \cdot PB^*) \cdot (PB^* \cdot PC^*)} = \left(\frac{PD^*}{PB^*}\right)^2 = \left(\frac{PB}{PD}\right)^2. \quad \blacksquare\]

We invert around point $P$ with and arbitrary radius and denote the inversion of $X$ as $X^*$.
$\Gamma_1,\Gamma_2,\Gamma_3$ and $\Gamma_4$ are inverted to lines $l_1,l_2,l_3$ and $l_4$ respectively, such that $l_1\parallel l_3$ and $l_2\parallel l_4$.
$A^*,B^*,C^*$ and $D^*$ are the intersection of $l_1$ and $l_2$; $l_2$ and $l_3$; $l_3$ and $l_4$; $l_4$ and $l_1$ respectively. So, $A^*B^*C^*D^*$ is a parallelogram.
By inversion distance formula, we get
\[A^*B^*=C^*D^*\implies \frac{r^2\cdot AB}{PA\cdot PB}=\frac{r^2\cdot CD}{PC\cdot PD}\implies\frac{AB}{DC}=\frac{PA\cdot PB}{PC\cdot PD}\]
Similarly,
\[\frac{BC}{AD}=\frac{PB\cdot PC}{PA\cdot PD}\]
Multiplying gives us,
\[\frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}\]
as desired.


Oooops, new question about the same paper i posted before (See the post: question on integrals).

The first row of the Proof on Page. 367:
\emph{"`The convergence of the log-variations is a particular case of a more general result due to Cohen and Istas (2003) and will be omitted"'.}
I've found this paper (see attachment, it's not easy to find) and read it, but can't totally see how the two cases match each other.
They use two different Increments:
$ \Delta X_{p,n} = \sum\limits_{k=0}^{K}a_{k}X \left(\frac{k + p}{2^n}\right)$ in the paper "`On the Roughness..."' and
$ \Delta_{k,n} X = X \left(\frac{k + 1}{2^n}\right) - X \left(\frac{k}{2^n}\right)$  in the paper "`An universal ..."'.
Maybe this is not a big problem, i can take the same increments, say, the 2nd one, in both cases, because it hurts nothing, i think.

But furthermore, the log-variations defined in the two papers seem to be different too:
$ V_{n,0} = \frac{1}{2^n - K}\sum\limits_{p=1}^{2^n-K}\log|\Delta X_{p,n}|$ in the paper"`On the Roughness..."',
where $ 2^n - K \to \infty$, for $ n \to \infty$.
But in the paper "`An universal ..."', the log-variation is defined as:
$ \frac{1}{v_n}\sum\limits_{k\in\mathcal{V}_{n}} \log|\Delta_{k,n}X|$
where we don't need $ v_n \to \infty$ for $ n \to \infty$.

is that really difficult to follow or am i stupid :( ?
Can you help me again please, Thank you!
Can you help me ?

We've found already that the authors can and do cheat. And here they cheat again, I think. Of course, the reference is completely irrelevant, because 1) it is concerned only with one-parameter case; 2) here they have a far more general definition of increments.

It is possible that you have not the final version of the paper (looking at the heading), but I don't think that they could change a lot.

Thank you for your answer, that's also what i think.
I don't know much about the paper publishing in mathematics,  why such flaws can pass the check of reviewers?

So i give up this approach to understand the convergence of this log-variations (if it really converged).

Well, reviewers (especially, good specialists) have a lot of work besides reviewing. So usually they don't read so carefully as you did. About this reference: probably, the reviewer did not check the paper on log variations (maybe, due to the fact that it was not published yet) and believed the authors that everything is ok.

And you can see that the paper on log variations was published in some not very good journal (it's not indexed in Zentralblatt and MathSciNet). So maybe there are also several flaws in it (I don't know, I didn't read the paper) and it had been rejected elsewhere?

After all, there's no need to give up: you can prove everything the authors did not prove and this will be a good paper :)

I  appreciate so much for your patient helps and advice!
I don't give up, i've found the preprint version of the first paper i posted here.
In this version they used a different approach to show the convergence of log-variations, i am reding it now, but it's quite technical, i am afraid i still can't get it.
I try it first by myself, if i still don't get it, then i post this version up, maybe you will help me again. :blush:
By the way, your speed of  paper-reeding  really amazed me.


induction
number theory unsolved
number theory

Let $ u_n $ :
 $ u_0 = u_1 = 1 $
 $ u_{n+1} = 14u_n - u_{n-1} $
 Prove that : $ 2u_n -1 $ is a perfect square

Some Romanian selection test no?
I am sure you will find solutions on Andrei Jorza's website.

You are right : [url=http://www.ajorza.org/math/mathfiles/selection/selection2002sol.pdf]solution[/url].
But is the solution complete? We have to prove that $3$ do not divide $2u_n-1$ (It's easy to prove by induction that $u_n=1\mod(3)$ so $2u_n-1=1\mod(3)$).

__________________
blang

oh  i have the solution
let the sequence {b_n} such that b_(n+1)=4*b_n-b_(n-1)
it's easy to prove that 2*a_n-1=(b_n)^2
nice result . :D  :D
two same problem:
1) let the sequence {a_n} such that :
                  a_{0}=1;a_{1}=13;a_{n+1}=14*a_{n}-a_{n-1}
show that 2*a_{n}-1 is perfect square
2) let the sequence {a_n} such that
                  a_{0}=1;a_{1}=6;a_{n+1}=6*a_{n}-a_{n-1}
show that a_{n)}=(k*(k+1))/2    (k is possitive integer)

[quote="nhat"]it's easy to prove that 2*a_n-1=(b_n)^2[/quote]
I'am some problem to prove it...If it's easy, can you help me?

we will use induction and we have the solution :D

Really easy? Can you give me some details? :?

[quote="orl"]Smart, Myth.  :)[/quote]
I remember the conversation about word "mean"... The word "smart" also has quite distinct meanings... ;)

Aaaahh.. sweet memories... :) I actually took that test :). I did something very simialr to what Myth did, but it can also be done quite esily by using the characteristic equation, I think, but it also involves "guessing" the sequence $b_n$ :).

As you know, I don't like solutions whih use closed form of such sequences (via characteristic equations).


quadratics
modular arithmetic
Diophantine equation
equation
number theory

[quote="manjil"]Find all triples $ \left(p,x,y\right)$ such that $ p^x \equal{} y^4 \plus{} 4$, where $ p$ is a prime and $ x$ and $ y$ are natural numbers.[/quote]

$ p^x \equal{} y^4 \plus{} 4$
$ y \equal{} 1$ yields $ (p,x,y) \equal{} (5,1,1)$
Since $ p$ is a prime we can expect a very limited number of solutions...
$ p^x \equal{} (y^2 \plus{} 2 \plus{} 2y)(y^2 \plus{} 2 \minus{} 2y)$
Thus we have $ (y \plus{} 1)^2 \plus{} 1 \equal{} p^k$ and $ (y \minus{} 1)^2 \plus{} 1 \equal{} p^j$
Since $ \minus{} 1$ is a quadratic residue, we can assume that $ p \equal{} 4n \plus{} 1$. And we see $ \bmod 8$ that $ k$ and $ j$ have different parity, so $ x$ must be odd.
Since we have checked $ y \equal{} 1$ we can assume that $ k,p > 1$
case $ k \equal{} 2m$: $ (y \plus{} 1)^2 \equal{} (p^m \minus{} 1)(p^m \plus{} 1)$ Which obviously have no solutions.
case $ j \equal{} 2m$:  $ (y \minus{} 1)^2 \equal{} (p^m \minus{} 1)(p^m \plus{} 1)$ which has no solutions neither.

So no solutions for $ x > 1$. Thus $ p \equal{} (y^2 \plus{} 2 \plus{} 2y)(y^2 \plus{} 2 \minus{} 2y)$. But this means $ y^2 \plus{} 2 \minus{} 2y \equal{} 1 \Rightarrow y \equal{} 1$ which is already found. So the only solution is $ (5,1,1)$

Yes the solution is correct but can we solve it in another way?

[quote]Find all triples $ \left(p,x,y\right)$ such that $ p^x = y^4 + 4$ $ (1)$, where $ p$ is a prime and $ x$ and $ y$ are natural numbers.[/quote]

If $ p = 2$, from $ (1)$, we have $ 2^x=y^4+1$. If $ x\ge 3$ then $ 8\mid 2^x$, which implies $ 8\mid (y^4+4)$. But this is false since $ y^4\equiv$ $ 0$ or $ 1$ $ \pmod {8}$ hence $ y^4+4$ will be never divisible by $ 8$, which leads to the contradiction. Therefore, $ x\le 2$ $ \Rightarrow$ $ x\in \{1,2\}$. In the other hand, $ x=1$ is impossible because $ y^4+4>2$ $ \forall y\in \mathbb {N}$. As the consequence, $ x=2$ $ \Rightarrow$ $ y=0$. So the solution at this case is $ (p,x,y)=(2,2,0)$.

Now let consider the case $ p > 2$. Since $ p$ is prime, thus, $ \gcd (p,2) = 1$. From $ (1)$, we have $ y^4\equiv - 4 \pmod {p}$. Thus $ \left(\frac { - 4}{p}\right) = 1$ $ (*)$. In the other hand, $ \left(\frac { - 4}{p}\right) = \left(\frac { - 1}{p}\right)\left(\frac {4}{p}\right)$ $ = \left(\frac { - 1}{p}\right)\left(\frac {2^2}{p}\right)$ $ = ( - 1)^{\frac {p - 1}{2}}$. As the result, $ (*)$ is true if and only if $ p$ has the form of $ 4k + 1$. In the other hand, $ p^x = (y^2 + 2y + 2)(y^2 - 2y + 2)$. Therefore, there exist $ i,j$ such that:
\[ \[ \left\{\begin{matrix}(y - 1)^2 + 1 & = & p^{i} \\
(y + 1)^2 + 1 & = & p^{j}\end{matrix}\right\;\]
Since both $ p^i$ and $ p^j$ $ \equiv 1 \pmod {4}$. Thus both $ (y - 1)^2$ and $ (y + 1)^2$ are divisible by $ 4$, which implies $ y\equiv 1\pmod {2}$ and $ y\equiv - 1\pmod {2}$. But this is true iff $ y = 1$. From $ (1)$ $ \Rightarrow$ $ p^x = 5$ $ \Longrightarrow$ $ x = 1$. Hence, at this case, $ (p,x,y) = (5,1,1)$.

$ p^x\equal{}({(y\minus{}1)}^2\plus{}1)({(y\plus{}1)}^2\plus{}1)$ so if y is not 1 both of the multiples must be divisible by p
$ p|y^2\plus{}2y\plus{}2\minus{}(y^2\minus{}2y\plus{}2)\equal{}4y$  $ p|4y$  $ p\equal{}2$ or $ p\equal{}y$ if we check both cases are impossible clearly we are done.

[quote="mathVNpro"]But this is true iff $ y \equal{} 1$.[/quote]

Why? Could you explain this please.

[quote="earldbest"]let $ p^{a} \equal{} y^{2} \minus{} 2y \plus{} 2$, $ p^{b} \equal{} y^{2} \plus{} 2y \plus{} 2$, $ a < b$.

if $ y$ is even, then $ p^{a} \equiv p^{b} \equiv 2 \mod 4 \Rightarrow a \equal{} b$. therefore, $ y$ is odd

since $ 4y \equal{} p^{b} \minus{} p^{a}$,  therefore $ p^{b\minus{}a}\equal{}5$,$ p^{a}\equal{}y$

from this we get $ p\equal{}5$, $ b\minus{}a\equal{}1$.

similarly, $ 2y^{2}\plus{}4\equal{}p^{b}\plus{}p^{a}$.

if $ a\neq 0$, then $ y^2 \equiv 3\mod 5$, which is impossible.

therefore the only solution is $ (5,1,1)$.  :lol:[/quote]

why are u considering that  $ p^{a}= 5$
i didn't get this  $ p^{b\minus{}a}\equal{}5$,  $ p^{a}\equal{}y$ though  this is ri8 $ 4y \equal{} p^{b} \minus{} p^{a}$
:help:

@ siddharthanand
As y is odd so p is odd.Also, highest power of 2 in 4y is 2.We know 4y=p^a(p^(b-a) - 1).As p^a is odd, it forces p^(b-a) - 1 to be 4,hence p^(b-a) = 5.I hope it is clear now!!

After making the Sophie Germain factorization of RHS, each of the brackets should be a power of p (note that we have not considered only y=1), so p|gcd(4y,y^2 + 2). p|4y --> p|4y^2, p|(y^2 + 2) --> p|(4y^2 + 8), so p|(4y^2 + 8 - 4y^2) = 8, so p=2 --> 2^x = y^4 + 4 and if x>3, y^4 + 4 must be 0 mod 16 which is impossible. Therefore, x=2 or 3 which doesn't have solution for y natural. The remaining case is y=1 which implies p=5 and x=1.

The trick is just factoring and studing cases.

This was an RMO level problem, set for INMO. Indeed very easy. :)

I can guarantee that this is the shortest solution
Note that sophie germain implies y^2+2y+2 has y^2-2y+2
also look at them as a fraction you get the result immediately.

Also see that x cannot be even....
 (might not help the problem..)

We claim that $(p,x,y)=(5,1,1)$ is the only solution. We now prove our claim.
Note that $p^{x}=y^{4}+4=(y^{2}+2y+2)(y^{2}-2y+2)$.
If $y$ is even, then $p$ is also even $\Rightarrow p=2$.
Also, $y>0 \Rightarrow 2^{x}>4 \Leftrightarrow x \geq 3$. Also note that $2|y \Rightarrow 16|y^{4}.$ On the other hand
$$y^{4}=2^{x}-4=4(2^{x-2}-1) \Rightarrow 4|2^{x-2}-1$$
But since $x>2$, this is a contradiction.
Hence $y$ is odd. Thus
$$gcd(y^{2}+2y+2,y^{2}-2y+2)=gcd(4y,y^{2}-2y+2)=1$$
But $p^{x}=(y^{2}+2y+2)(y^{2}-2y+2)$ and $y^{2}+2y+2>y^{2}-2y+2$, hence we get
$$y^{2}+2y+2=p^{x}, y^{2}-2y+2=1 \Rightarrow y=1$$
Hence $p^{x}=5$ and thus $p=5$ and $x=1$, proving our claim.

[quote=INMO 2008 P2]Find all triples $ \left(p,x,y\right)$ such that $ p^x\equal{}y^4\plus{}4$, where $ p$ is a prime and $ x$ and $ y$ are natural numbers.[/quote]
[b][color=#000]Solution:[/color][/b] Suppose, if $p=2$ $\implies$ $y=2k$ for some $k \in \mathbb{N}$
$$2^x=y^4+4=16k^4+4 \implies \frac{4(4k^4+1)}{2^x} \in \mathbb{Z}$$
Hence, $\text{max}\{v_2(2^x)\}=2$ $\implies$ No solution for $y$ in $\mathbb{N}$. Hence, $p$ is an odd prime.
$$p^x=y^4+4=(y^2+2y+2)(y^2-2y+2)$$
$$\begin{cases} y^2+2y+2=p^a \\ y^2-2y+2=p^b \end{cases} \stackrel{\text{WLOG} ~ a >b}{\implies} \frac{y^2+2y+2}{y^2-2y+2}=\frac{(y^2-2y+2)+4y}{y^2-2y+2}  \in \mathbb{Z} \implies 4y \geq y^2-2y+2 \implies \boxed{(p,x,y)=(5,1,1)} \qquad \blacksquare$$

[quote=Geronimo_1501]Can't we just use FLT so that $y^4$ is congruent to $1 \space  \text{mod}\space 5 $ and hence $p$ is $5$?[/quote]

What if $5|y$?

@2below maybe it can, I'm not sure, but even if it can, it will be a (much) harder solution than the ones that are already posted here which use Sophie Germain identity

The solutions are $(p,x,y)=(5,1,1)$.We show that there are no other solutions.
We have $y^4+4=[(y+1)^2+1][(y-1)^2+1]$.Since this equal to $p^x$ we should have $[(y+1)^2+1]=p^a$ and $[(y-1)^2+1]=p^{x-a}$ for some nonzero $a$ and not equal to $x$.Hence this implies $p|[(y+1)^2+1]-[(y-1)^2+1] \implies p|4y$.Easy to see $p$ cannot be equal to $2$.Hence $p|y$.But we get a contradiction since $p^x=y^4+4$.Hence we should consider the case where $a=0$ or $x-a=0$.Easy to see $a=0$ doesnt yield solutions.Hence $x-a=0 \implies y=1$.Hence $p=5$ and $x=1$ and we are done.$\blacksquare$

[quote=green_leaf][quote=Geronimo_1501]Can't we just use FLT so that $y^4$ is congruent to $1 \space  \text{mod}\space 5 $ and hence $p$ is $5$?[/quote]

What if $5|y$?[/quote]
Oh yeah, that might cause [s] serious [/s] problems.

But can this approach be continued?

[quote=manjil]Find all triples $ \left(p,x,y\right)$ such that $ p^x\equal{}y^4\plus{}4$, where $ p$ is a prime and $ x$ and $ y$ are natural numbers.[/quote]

$$p^x=y^4+4$$
$$\implies p^x = (y^2-2y+2)(y^2+2y+2)$$
$$\implies p^a = y^2 + 2y + 2 \quad \text{and} \quad p^b = y^2 - 2y + 2 \quad \text{where} \quad a > b , \quad a + b=x \in \mathbb{N}$$
$$\implies p^a - p^b = 4y \implies y = \frac{p^a - p^b}{4} \quad \text{and} \quad p^a + p^b = 2(y^2 + 2)$$
$$\implies p^a + p^b = 2\left( \left(\frac{p^a - p^b}{4} \right)^2 + 2\right)$$
$$\implies (p^a - p^b)^2 + 32 = 8(p^a + p^b)$$
Case 1 $: b > 1$
$$\implies 32 \equiv 0 \pmod{p} \implies p \mid 32 \implies p = 2$$
$$\implies 2 \mid y \implies y = 2m$$
$$\implies 2^x = 16m^4 + 4 \implies 4(4m^4 +1) = 2^x \implies x=0,1,2 \implies \boxed{\text{No Solution}}$$
Case 2 $: b = 0$
$$\implies (p^a - 1)^2 + 32 = 8(p^a + 1)$$
$$\implies p^{2a} - 2p^a + 33 = 8p^a + 8$$
$$\implies p^{2a} - 10p^a + 25 = 0$$
$$\implies (p^a - 5)^2 = 0 \implies p^a = 5 \implies p = 5 , a=1 \implies y = 1$$
Hence our solutions are $\boxed{(p,x,y) = (5,1,1)} \quad \blacksquare$

[hide=solution][2008 P2]: Find all triples $ \left(p,x,y\right)$ such that $ p^x=y^4+4$, where $ p$ is a prime and $ x$ and $ y$ are natural numbers.

Solution: If $p$ is two $\implies y $ is even. So $4|y^4$ and clearly $x\ge 3.$ Then divide the equation by $4.$ We get $2|\text{RHS}$ but $2\nmid \text{ LHS}$

So $p$ is an odd prime. Note that $$p^x=y^4+4=(y^2-2y+2)(y^2+2y+2)\implies y^2-2y+1|y^2+2y+1\implies y^2-2y+2|4y$$
 $$y^2-2y+2\le 4y\implies y=2,1$$
Verifying, we get $y=1$ satisfying. Giving us $(p,x,y)=\boxed{(5,1,1)}$ as the only solution.

P.S. It's almost similar to Alastor bhaiya's solution :O[/hide]

we can check for last digits as well. like if y ends with 1,3,7,9 then y^4+4 ends with 5
thus p has to end in 5 but the only prime that ends in 5 is 5 itself so one solution is (5,1,1)

But I am not able to prove this for y ending with 5. Can someone help me

We claim the only solution is $(p, x, y) = (5, 1, 1)$. It is clear that this does in fact satisfy the equation, now we prove this is the only solution.
Note that $y^4+4 = y^4 + 4y^2 + 4 - (2y)^2 = (y^2 + 2)^2 - (2y)^2 = (y^2 + 2y + 2)(y^2 - 2y + y)$, which is in turn a power of $p$.
$\implies y^2 + 2y + 2$ and $y^2 - 2y + 2$ are individually powers of $p$, say $y^2 + 2y + 2 = p^a,y^2 - 2y + 2 = p^b $

Case 1: $a, b>0$
This gives us $p \mid y^2 + 2y + 2, p \mid y^2 - 2y + 2 \implies p\mid 4y \implies p = 2 or p|y$
If $p \mid y, \text{ mod } p$ gives us $4 \equiv 0 \text{ mod } p$, or $p \mid 4$.
So both cases lead to $p = 2$.
$\implies 2^x - 4 = y^4$.
Clearly, $y$ is even, so $16 \mid y^4 = 2^x - 4$
If $x \ge 3$, then $y \equiv 4 \text{ mod } 8,$ contradiction.
$\implies x = 2$, but this leads to $y = 0$, which is not possible as $y$ is a natural.
Therefore, $b = 0$.
$\implies y^2 - 2y + 2 = 1 \implies y = 1 \implies p^x = 5 \implies (p,x, y) = (5, 1, 1)$
QED.


$A \in S_n(R)$ with eigenvalues $a_1 \geq a_2 \geq ... \geq a_n$.
Show that: $sum_{i=1}^k A_{i,i} \leq sum_{i=1}^k a_i$.
I have a weird solution for this one but I'm sure there is a simple one.

According to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=207273#p207273]this other problem you've posted[/url]' shouldn't the sign be reversed? :? Or maybe here $A_{i,i}$ are not the diagonal entries, but maybe the diagonal minors? :? I think it's better to define everything you use. For instance, I think $S_n(\mathbb R)$ is the set of all the symmetric real matrices, but that's just guessing. I, for one, am not familiar with the notation.

in fact I don't remember exactly the problem but it is easy to solve thanx to Courant-fisher min-max theorem.
(try to express the eigenvalues in terms of: $sup(inf( \frac{(x|Ax)}{|x|^2},x \in F ),F s.t. dim(F)=k)$ where $F$ are subspaces of $E$


trigonometry
function
inequalities proposed
inequalities

I found a nice idea  ( Was it created before? )
For every acute triangle ABC there exits there real positive number $a,b,c$ such that

$\displaystyle(\sin A)^2=\frac{a(a+b+c)}{(a+b)(a+c)}$

$\displaystyle(\sin B)^2=\frac{b(a+b+c)}{(b+c)(b+a)}$

$\displaystyle(\sin C)^2=\frac{c(a+b+c)}{(c+a)(c+b)}$

And

$\displaystyle(\cos A)^2=\frac{bc}{(b+a)(c+a)}$

$\displaystyle(\cos B)^2=\frac{ca}{(c+b)(a+b)}$

$\displaystyle(\cos C)^2=\frac{ab}{(a+c)(b+c)}$

And If we given three positive real numbers $a,b,c$ then exits a triangle $ABC$

I think it's very useful :P

 :D  :D  :D

Oh,No body enjoy it (?)  :(

What do you think about this problems:

[b]Problem:[/b]
Given a acute triangle ABC and three positive real number $\displaystyle x,y,z$
Find the Maximum value of

$\displaystyle P=(\cos A)^x(\cos B)^y(\cos C)^z$

I want to see your solutions !

 :D  :D  :D

Condition $A+B+C-\pi=0$
Function to be maximized: $f= \cos ^x A \cos ^y B \cos ^z C$.
Extremum occurs when
\[
\begin{gathered}
  \frac{{\partial f}}
{{\partial A}} + \lambda \frac{{\partial g}}
{{\partial A}} = \frac{{\partial f}}
{{\partial B}} + \lambda \frac{{\partial g}}
{{\partial B}} = \frac{{\partial f}}
{{\partial C}} + \lambda \frac{{\partial g}}
{{\partial C}} = 0 \hfill \\
   \Rightarrow  - x\sin A\cos ^{x - 1} A\cos ^y B\cos ^z C =  - y\sin B\cos ^x A\cos ^{y - 1} B\cos ^z C =  - z\sin C\cos ^x A\cos ^y B\cos ^{z - 1} C \hfill \\
   \Rightarrow x\tan A = y\tan B = z\tan C \hfill \\
\end{gathered}
\]
It remains to check this is the maximum and find the maximum value, which i think it's not hard.

I want to see detail  :?  :?

 Can you post your solution and the Max of P

----------------------------------
Thanks  everybody  :)
This forum is very interesting

I think your idea is similar to nttu's here
http://www.mathlinks.ro/Forum/viewtopic.php?p=176154&highlight=#176154


I'm currently taking IB Biology HL, and I'm also debating taking the USABO... and I've already realized that just the knowledge I would learn from my IB Bio class is sufficient if I wanted to take the USABO. I already know what resources would be best for preparing for USABO, but my question is: Would preparing for USABO also prepare me for an AP/IB Bio exam at the same time?

//Would preparing for USABO also prepare me for an AP/IB Bio exam at the same time?//

I doubt you would have any problems with AP/IB bio if you're studying for USABO.


Given $ n\plus{}1$ distinct integers between $ 1$ and $ 2n$, show that two of them are relatively prime. Is this result the best possible, i.e. is the conclusion still true for $ n$ integers between $ 1$ and $ 2n$.

[quote="BanishedTraitor"]Given $ n \plus{} 1$ distinct integers between $ 1$ and $ 2n$, show that two of them are relatively prime.[/quote]
[hide="Easy"]
Construct $ \{ 1,2\}, ... , \{2n\minus{}1,2n\}$ and use property that $ \gcd (k,k\plus{}1) \equal{} 1$.

As for the part about it being the best possible result consider $ n\equal{}2$ we we have $ \{1,2,3,4\}$ pick $ \{2,4\}$ and its false.
[/hide]

[quote="sylow_theory"][quote="BanishedTraitor"]Given $ n \plus{} 1$ distinct integers between $ 1$ and $ 2n$, show that two of them are relatively prime.[/quote]
[hide="Easy"]
Construct $ \{ 1,2\}, ... , \{2n \minus{} 1,2n\}$ and use property that $ \gcd (k,k \plus{} 1) \equal{} 1$.

As for the part about it being the best possible result consider $ n \equal{} 2$ we we have $ \{1,2,3,4\}$ pick $ \{2,4\}$ and its false.
[/hide][/quote]

Yep, just like my solution! Except for the fact I worded it differently and took a slightly different approach (which is essentially the same). I like your one though. :)

[quote="BanishedTraitor"]
Yep, just like my solution! Except for the fact I worded it differently and took a slightly different approach (which is essentially the same). I like your one though. :)[/quote]
Thanks. How did you solve the problem?

After trying smaller values you can conjecture that every selection will result in consecutive pairs being selected and hence a pair of relatively prime numbers. So set the $ n \plus{} 1$ numbers to be:

$ 1 \leq x_1 < x_2 < ....... < x_{n \plus{} 1} \leq 2n$. Now assuming the contrary then there will exist $ k$ such that $ x_j < x_k < x_{j \plus{} 1}$ where $ j \in \{ 1,2,3,.....,n \plus{} 1 \}$. This way we find that:$ 1 \leq x_1 < x_{k_1} < x_2 < x_{k_2} < ..... < x_{k_n} < x_{n \plus{} 1} \leq 2n$. This will give at least$ 2n \plus{} 1$integers on the interval$ [1,2n]$which is clearly a contradiction. Something along those lines. And to show that it is indeed the best possible result we simply pick all the even numbers$ 2,4,.......,2n$giving a set of$ n$ numbers not pairwise coprime.

standard.
Hint
[hide]
consider the nos. to be of the form $ 2^{ai}.ki$ where each $ ki$ is odd.
[/hide]

[quote="diophantient"]Here's an interesting, yet slightly harder version of the problem.
Show that in the same set with a subset of n+1 terms, one must be divisible by another.[/quote]
That's an Erdos theorem  :)


probability
combinatorics unsolved
combinatorics

[i]n[/i] numbers are chosen from the set 1,2,3,...,N. Calculate the probability that at least [i]m[/i] of the chosen numbers are consecutive, where [i]m[/i] <=[i]n[/i], [i]n[/i]<N.

Thanks a lot!

The numbers are difeerent, right?
In that case you can arrange the numbers in strictly increasing order, and then map each of the sets you have of form $\{a_{1}, a_{2}, ..., a_{m}\}$ on set $\{b_{1}, b_{2}, ..., b_{m}\}$, so that $b_{i}=a_{i}-(i-1)$. then just count the maximum number of numbers in the string of form $k,k,k....,k$. I think it's quite easy to count the number of "good" cases for sets  $\{b_{1}, b_{2}, ..., b_{m}\}$, and that solves the problem. :)


[quote]Demostrar que si un número es irracional en una base lo será en cualquier base[/quote]

Para mi por definicion numero es racional si no es raiz de un polinomio en $ \mathbb{Z}$ de grado 1 o lo q es lo mismo es una razon de dos enteros y esto no depende de la base.. me parece q tu pregunta se refiere a si un numero no tiene periodo en una base no tiene en ninguna.

Para $ b$ natural sean $ R_b$ aquellos numeros con decimales periodicos en base $ b$ e $ I_b$ su complemento (los "irracionales" en base $ b$), asi basta probar que para cualquier $ b$, $ R_b \equal{} \mathbb{Q}$ (las fraciones de numeros enteros).

$ R_b\subseteq \mathbb{Q}$
Antetodo si $ q\in R_b$ quiere decir que $ q \equal{} n \plus{} \sum_{n \equal{} 1}^\infty\frac {a_i}{b^i}$ con $ n\in\mathbb{Z}$ y $ 0\leq a_i\leq b \minus{} 1$ periodicos es decir que para algun $ p > 1$ y todo $ i\geq i_0$ se cumple q  $ a_i \equal{} a_{i \plus{} p}$. Por simplicidad tomemos $ N \equal{} 0,i_0 \equal{} 0$ asi

\[ q \equal{} \sum_{n \equal{} 0}^\infty\frac {a_i}{b^i} \equal{} \sum_{k \equal{} 1}^\infty\sum_{l \equal{} 0}^p\frac {a_{kp \plus{} l}}{b^{kp \plus{} l}} \equal{} \sum_{k \equal{} 1}^\infty\sum_{l \equal{} 0}^p\frac {a_{l}}{b^{kp \plus{} l}} \equal{} \sum_{k \equal{} 1}^\infty\sum_{l \equal{} 0}^p\frac {a_{l}}{b^l}\frac {1}{b^{kp}} \equal{} \sum_{l \equal{} 0}^p\frac {a_{l}}{b^l}\sum_{k \equal{} 1}^\infty\frac {1}{b^{kp}} \equal{} \frac {1}{1 \minus{} b^{ \minus{} p}}\sum_{l \equal{} 0}^p\frac {a_{l}}{b^l}\]

y por tanto $ q\in\mathbb{Q}$.

$ \mathbb{Q} \subseteq R_b$
notemos que basta probar que las fraciones $ \frac {1}{n}$ son periodicas en base $ b$ ya que las fraciones $ \frac {m}{n} \equal{} \frac {1}{n} \plus{} \cdots \plus{} \frac {1}{n}$ (no es dificicil probar q la suma de dos numeros periodicos es tb un numero periodico), ademas podemos considerar $ b$ y $ n$ coprimos.

Sea $ \frac {1}{n} \equal{} \sum_{k \equal{} 1}^\infty\frac {a_k}{b^k}$ asi $ \left[\frac {b^k}{n}\right] \equal{}a_k\mod b$ de ahi probar que $ \frac{1}{n}\in R_b$ es lo mismo q probar que $ \left[\frac {b^k}{n}\right]\mod b$ es periodica en $ k$. Para cada $ k$, tenemos
\[ b^k\equal{}\left[\frac {b^k}{n}\right] n\plus{}\theta_k, 0\leq \theta_i\leq n\minus{}1\]
\[ \left[\frac {b^k}{n}\right]\equal{}d_kb\plus{}a_k,0\leq a_k\leq b\minus{}1\]
asi $ b^k\equal{}d_kbn\plus{}\theta_kn\plus{}a_k$ y por tanto $ a_k\equal{}\minus{}\theta_kn\mod b$. Por otro lado $ b^{k\plus{}\varphi(n)}\equal{}b^kb^{\varphi(n)}\equal{}\theta_k\mod n\Longrightarrow\theta_{k\plus{}\varphi(n)}\equal{}\theta_k$. Usando esto en la ultima congruencia obtenemos  $ a_{k\plus{}\varphi(n)}\equal{}a_k\mod b$ y por tanto son iguales, es decir $ \frac{1}{n}$ es un decimal periodico en base $ b$.


#1: [hide]Well, this is a really dumb method, but oh well.  I started by listing the first ten or so values of P(x), like so:
P(1) = 3
P(2) = 7
P(3) = 13
P(4) = 21 = 3*7
P(5) = 31
P(6) = 43
P(7) = 3*19
P(8) = 73
P(9) = 7*13
P(10) = 3*37

Since P(x) = x^2 + x + 1, if P(n) is a multiple of some integer m, then for k = 1, 2, 3, ..., P(n+km) = (n+km)^2 + (n+km) + 1 = n^2 + n + 1 + m(2kn+k^2*m+k), so P(n+km) is also a multiple of m.

Using this and the fact that P(1) is a multiple of 3, it will also be the case that P(i) is a multiple of 3 for i = 1, 4, 7, ...  Similarly, P(j) will be a multiple of 7 for j = 2, 9, 16, ... or 4, 11, 18 (because both P(2) and P(4) were multiples of 7.)  The same holds for the other numbers we have found: 13, 19, 31, 37, and so on.  So P(x) will be composite if x = 4, 7, 10, ..., 9, 16, 23, ..., ....  We then list the numbers out and look for numbers that are close together.  Doing this, we see that 42 is in the list 3+13k, 43 is in the list 1+3k, 44 is in the list 2+7k, 45 is in the list 7+19k, 46 is in the list 4+7k, and 47 is in the list 10+37k.  Thus we are done; n = 42.  By the way, there are infinitely many numbers; just add LCM(3,7,13,19,37)*k for any integer k to 42.[/hide]

#4: [hide]I've only proved half so far: that it is possible to remove all the chips when either m or n (or both) is odd.

To prove this, we do the following: assume WLOG that we start from the top left.  First, we will show that for any m and n, it is possible to completely delete the leftmost column and make the second column from the left all black, with the rest of the board white.  We do this by removing the top left black chip, which makes the one below it and the one to the right black.  Now remove the black chip below it, making the one below it and the one to the right black.  Keep going down until all of the chips in the first column are removed.  We're done.  (Step one in the attached diagram)

Assume WLOG m is odd and that it is the number of rows.  If m is 1, it is obvious that we can remove all the chips, so we'll ignore this.  So m >= 3.  We will start from the position left at the end of the previous paragraph; that is, we have a column of all black, and everything to the right is white.  Remove the first, third, fifth, etc. chips in the column, starting from the top (or the bottom).  The result is that the first, third, fifth, etc. chips in the next column over are black, and the chips in the second, fourth, sixth, etc. spots in the current column are still black (since they were each changed twice).  Now remove these chips as well.  The final result is that the column is completely removed, and the next column over is now all black.  Thus we can eliminate column after column after column, until we are left with only one column of all black.  (Step 2 in the diagram.)

The final step is to eliminate this column of black.  Remove the third chip from the top, which makes the second and fourth chips white.  Remove the first chip, which makes the second black.  Remove the second chip.  We are now left with an odd number of black chips with a single white chip on top.  Remove the top black chip, making the chip above it (which was white before) black; then remove that chip.  The procedure in the previous sentence should then be repeated until we are left with one white chip above one black chip.  Remove the black chip, and then the resulting black chip.  We are done.  (Step 3 in the diagram.)

Now I just need to prove that m,n even is impossible.  I think assigning numbers to represent black and white, then showing that a transformation does not change some property, will do the trick.  Ideas?[/hide]

For 4, for that part, count the number of flips in two ways.

[quote="sherry"]btw, how do you do that "click to reveal..." thing?[/quote]There is a big "Hide button". You can read more about how to use this board in the following topic:

http://www.mathlinks.ro/Forum/viewtopic.php?t=20430

If worst comes to worst, I guess you could always use anaylatic geomtry for number 3.

The first time I read number four I misread it, and for some reason thought that "edge" meant row or column.
So if you remove a black piece, you flip all the pieces in it's row or column. This seems to me like a significantly harder problem. That I couldn't solve.
I know it cannot be done when m+n is even, because in that case, the last piece on the board will be white [it has been flipped m-1+n-1 times]. But I have yet to prove that it CAN be done for m+n odd. If that's true...
It doesn't induct very simply...
Any ideas? (although it's probably not worth spending your time on)
btw-how did the normal problem 4 go?

oh, btw, thanks (for telling me how to do the "hide" thing)

-Sherry


geometry
geometric transformation
reflection
geometry unsolved

A non-isosceles triangle $A_1A_2A_3$ has sides $a_1,a_2,a_3$ with $a_i$ opposite $A_i$. $M_i$ is the midpoint of side $a_i$ and $T_i$ is the point where the incircle touches side $a_i$. Denote by $S_i$ the reflection of $T_i$ in the interior bisector of angle $A_i$. Prove that the lines $M_1S_1,M_2S_2$ and $M_3S_3$ are concurrent.

Since the angle bisectors of $\triangle A_1A_2A_3$ pass through the center of the incircle $k$, the point $T_i\in k$ reflects to a point $S_i\in k$.

Since the triangle is non-isosceles, $M_i\neq T_i$, and this proves that $M_i\neq S_i$, because $S_i$ is on the incircle and $M_i$ is not.

Because of the reflections, $S_1T_3=T_1T_2=T_3S_2$, so $\triangle S_1S_2T_3$ is isosceles. Since $k$ touches $A_1A_2$, this implies $S_1S_2||A_1A_2$. Doing this for the other sides we get that $\triangle S_1S_2S_3\sim \triangle A_1A_2A_3\sim \triangle M_1M_2M_3$.

Applying the converse of Desargues' theorem for triangles $S_1S_2S_3$ and $M_1M_2M_3$ we get the result.

Dear Mathlinkers,
they are concurrent at the Feuerbach's point of ABC.
Sincerely
Jean-Louis


[hide="If I understand disjoint subsets correctly"]
then the answer should be the remainder when
...${10\choose1}\times{9\choose{n}}$ for $1\leq{n}\leq9$
+${10\choose2}\times{8\choose{n}}$ for $1\leq{n}\leq8$
+${10\choose3}\times{7\choose{n}}$ for $1\leq{n}\leq7$
+${10\choose4}\times{6\choose{n}}$ for $1\leq{n}\leq6$
+${10\choose5}\times{5\choose{n}}$ for $1\leq{n}\leq5$

is divided by 1000
then comes the computation...
10*511+45*255+120*127+210*63+1008*31
5110+11475+15240+13230+31248
=76303
so is the answer 303?
[/hide]

[hide]Consider when the number of elements of the pair is $k$, there are $\binom{10}{k}$ ways for this. Now the number of disjoint subsets is given by $2^{k-1}-1$ because we can have each element in either the first or the second, for $2^{k}$, then subtract 2 for the null sets produced, then divide by two because we count $\{\{1\}, \{2\}\}$ and $\{\{2\},\{1\}\}$, so the sum is $\sum_{k=1}^{10}(2^{k-1}-1)\binom{10}{k}=28501\implies \boxed{501}$[/hide]

[quote="4everwise"]Let $\mathcal{S}$ be the set $\{1,2,3,\ldots,10\}.$ Let $n$ be the number of sets of two non-empty disjoint subsets of $\mathcal{S}.$ (Disjoint sets are defined as sets that have no common elements.) Find the remainder obtained when $n$ is divided by $1000.$[/quote][hide]For ease of counting, first suppose these sets are distinguishable. That is, there is a first set and a second set. Each of the integers $\{1,2,3,\ldots,10\}$ can be in the first set (F), the second set (S), or neither (N). For example, FSFNFFFSNN denotes the subsets \[\mathcal{F}=\{1,3,5,6,7\}\qquad\mathcal{S}=\{2,8\}.\]So, there are $3^{10}$ ways to assign these integers to a set. However, the cases in which at least one of the two sets is empty are also included. These are the cases in which all integers are either F or N and the cases in which all integers are either S or N. Thus, we subtract $2^{10}$ for each situation, or a total of $2\cdot2^{10}$. Finally, the case in which no integers are assigned to a subset was subtracted twice, and thus, must be added back in. Therefore, the total number of possibilities for two distinguishable non-empty disjoint subsets of $\{1,2,3,\ldots,10\}$ is $3^{10}-2\cdot2^{10}+1$. Since the subsets are not distinguishable, then we have overcounted by a factor of $2$, and the desired number is \[\frac{3^{10}-2\cdot2^{10}+1}{2}=28501.\]The answer is therefore, $\boxed{501}$.[/hide]

yea....my expression simplifies into that...expanding, then using the binomial expansion of $(2+1)^{10}$ and $(1+1)^{10}$, etc, nice solution though

whoops - I totally overcounted
got almost exactly 3 times the answer

alright..this is kinda of a silly question.  But it says in the solutions that you are suppose to divide two at the end.  I understand that because you have overcounted every case twice-like (10C3)(7C2) is equivalent to (10C2)(8C3).  However, what about the (10C5)(5C5) case?  Isnt that only counted once?

EDIT:uh wtf..i cant think..sry about that.i understand now

[quote="sapphyre571"][hide="If I understand disjoint subsets correctly"]
then the answer should be the remainder when
...$ {10\choose1}\times{9\choose{n}}$ for $ 1\leq{n}\leq9$
+$ {10\choose2}\times{8\choose{n}}$ for $ 1\leq{n}\leq8$
+$ {10\choose3}\times{7\choose{n}}$ for $ 1\leq{n}\leq7$
+$ {10\choose4}\times{6\choose{n}}$ for $ 1\leq{n}\leq6$
+$ {10\choose5}\times{5\choose{n}}$ for $ 1\leq{n}\leq5$

is divided by 1000
then comes the computation...
10*511+45*255+120*127+210*63+1008*31
5110+11475+15240+13230+31248
=76303
so is the answer 303?
[/hide][/quote]
Could anyone check what is wrong when I try to use calculator to get above answer?
this is what I have in TI 83 plus screen:
sum( seq( (10 nCr A) * seq( ( (10-A) nCr B ),B, 1, A, 1) , A, 1, 5, 1) )

[quote="cdymdcool"][quote="Altheman"][hide]Consider when the number of elements of the pair is $ k$, there are $ \binom{10}{k}$ ways for this. Now the number of disjoint subsets is given by $ 2^{k \minus{} 1} \minus{} 1$ because we can have each element in either the first or the second, for $ 2^{k}$, then subtract 2 for the null sets produced, then divide by two because we count $ \{\{1\}, \{2\}\}$ and $ \{\{2\},\{1\}\}$, so the sum is $ \sum_{k \equal{} 1}^{10}(2^{k \minus{} 1} \minus{} 1)\binom{10}{k} \equal{} 28501\implies \boxed{501}$[/hide][/quote]

I don't understand why you times $ \binom{10}{k}$ and $ 2^{k \minus{} 1} \minus{} 1$? ?[/quote]

I don't either...

The first term, if done Altheman's way, is $ (2^0 \minus{} 1)\binom{10}{1} \equal{} 0.$ But, if one of the two subsets have one element, then there are a positive number of possibilities, not $ 0.$

Using a similar reasoning as Altheman, here's another way (computationally heavy).

Consider the number of ways to choose $ 1$ element for the first subset. There are $ \binom{10}{1}$ ways to do so. Then there are $ 9$ elements left over. We can choose $ 1,2,3,\dots,9$ of those elements for the second subset. (We can't choose $ 0$ as then one of the subsets will be empty.) So, we have
\[ \binom{9}{1} \plus{} \binom{9}{2} \plus{} \dots \plus{} \binom{9}{9} \equal{} 2^9 \minus{} \binom{9}{0} \equal{} 2^9 \minus{} 1.\]
So, there are $ \binom{10}{1}(2^9 \minus{} 1)$ ways to make the two disjoint subsets in this case.

Similarly, we can reason that there are a total of
\[ \binom{10}{1}(2^9 \minus{} 1) \plus{} \binom{10}{2}(2^8 \minus{} 1) \plus{} \binom{10}{3}(2^7 \minus{} 1) \plus{} \dots \plus{} \binom{10}{9}(2^1 \minus{} 1) \equal{} 57,002\]
ways to make all the disjoint subsets.

But note that every pair is counted twice. So, the answer is $ 28,\boxed{501}.$

No: $ 2^{k \minus{} 1} \minus{} 1 \equal{} \frac {2^k \minus{} 2}{2}$ is the number of ways that [i]the two sets together[/i] can have $ k$ elements.  If both sets are nonempty then there are in fact $ 0$ ways for the two of them together to have one element.

A shorter way to do this problem is as follows: we start by counting the number of ordered pairs of subsets.  Each of the ten elements can be in the first set, the second set, or neither.  Initially, this gives us $ 3^{10}$ possibilities, but we've included many where one or both set is empty.  The first set is empty in $ 2^{10}$ of these situations, the second set is empty in $ 2^{10}$, but this double-counts the one case in which both are empty, so this gives us $ 3^{10} \minus{} 2\cdot 2^{10} \plus{} 1$.  Now we have to divide by $ 2$ (since we've counted them as ordered, when in fact they're not).

[hide=PIE]
Call the $2$ disjoint subsets $A,B.$ For each of the elements in the set, it has $3$ options: go to $A$, go to $B,$, or go to neither. Since each element is distinguishable, this gives us $3^{10}.$

Now time to subtract the undesirable cases, namely cases where either $A$ or $B$ is empty. This is a perfect candidate for PIE.

A is empty: $2^{10}$ ways
B is empty: $2^{10}$ ways
Both A and B are empty: $1$ way.

Thus $3^{10}-2^{10}-2^{10}+1=57002.$
But divide this by $2$ since $A, B$ are not distinguished in the problem (only specifies them as $2$ disjoint subsets). Thus the answer is $28501$ and the remainder is $501.$
[/hide]

I was solving this problem using the method Altheman and limac used to get [hide=the expression]$$\frac{\displaystyle \sum_{x=1}^{9}(2^{10-x}-1)\binom{10}{x}}{2}$$[/hide] as the answer. Is there an easy way to simplify the numerator?

Assuming you meant to write $x$ at the bottom of the summation, I don't think it is very hard to bash that out. Not sure if there is a faster way, but I'll be happy to know if there is.  :)

Can anyone find out why my solution is wrong?
[hide=Bashy+Wrong Solution]
Let $n$ be the number of elements in both disjoint subsets.
If $n=10$, then we need to split the original set into two sub-sets. We can have a sub-set with $9$ elements, then another sub-set with $1$ element. This means we need to pick $9$ out of $10$ or $\binom{10}{9}$. Then, We can have a sub-set with $8$ elements, then another sub-set with $2$ element. This means we need to pick $8$ out of $10$ or $\binom{10}{8}$. So on and so forth. However, we stop at a sub-set with $5$ elements and another sub-set with $5$ elements because afterwards, it repeats.
This means that for $n=10$, the answer is $\binom{10}{9}+\binom{10}{8}+\binom{10}{7}+\binom{10}{6}+\binom{10}{5}$.
We can proceed to all other cases of $n$ similarly, however we need to multiply by $\binom{10}{n}$ since we need to account for all the possible different permutations of sets we can split the two sub-sets from.
[/hide]
tldr, I choose a non-empty sub-set from the original set, then split that sub-set into two different non-empty sub-sets.

Let's consider what happens after the first subset is selected. Say it has $n$ elements, so there are $\binom{10}{n}$ ways to choose it. But then there are $10-n$ remaining elements and we can take any non-empty subset of those elements to be our other subset, so this gives $2^{10-n}-1$ ways. Thus, we are looking for the sum from $1$ to $9$ of $\binom{10}{n} \cdot (2^{10-n}-1)$. Brute forcing, this evaluates to $57002$, but every pair is counted twice so we divide by two, which gives $28\boxed{501}$.


function
geometry
geometric transformation
reflection
circumcircle
algebra unsolved
algebra

Let $ABC$ be a triangle in which $A$ is a right angle.We denote $AB=a$;$AC=b$;
Let $D,I \in (BC)$ such that $AD$ is the median from $A$ and  $AI$ is the bissector from A.
We take $P \in (AD)$ and $R \in (AI)$ and denote $AP=x$ and $AR=y$.

Find $x$ and $y$,in function of $a$ and $b$ such that the triangle $BRP$ is right-angled and isoscel,with $RB=RP$.

Pf course,the analytic approach is very simple,but i arrived to a nasty system of equations, which i posted in the Algebra-unsolved problems sections.
Another ideas?
PS:

Take $B'$ the reflection of $B$ in $AI$, see that $\triangle BB'P$ has $R$ as circumcenter and $\angle RB'P=45^\circ$, hence $\angle AB'P = \angle RB'I$, which happens when $B, R, B'$ are collinear and $P \equiv A$. The balance is easy...

Best regards,
sunken rock


geometry
3D geometry
LaTeX
number theory
Diophantine equation
modular arithmetic

Given the diophantine equation:a3+1=b2 ('a' cube plus one is equal to 'b' square) for integers a and b (I don't know latex) Find all possible solutions.
Clearly(-1, 0), (0, 1), (2,3) are possible for (a, b)

Certainly ur solution is right.
If u have got it though solving the equation than it is right otherwise u can also use the deduction that
a^3= b^2 - 1
      = (b-1)(b+1)
now u have gcd(b-1,b+1) = 1 or 2
solve the two cases arised u will get the solution.
I hope this may help.

$ a^3 \plus{} 1 \equal{} b^2 \Leftrightarrow a^3 \equal{} (b \minus{} 1)(b \plus{} 1)$. since $ b \minus{} 1, b \plus{} 1$ differ by 2, their $ \gcd$ is either 1 or 2. in the first case, we must have $ b \minus{} 1$ and $ b \plus{} 1$ both be cubes, which is only possible if they're $ \minus{} 1$ and $ 1$ (these are the only cubes differing by 2), and this gives the solution $ ( \minus{} 1,0)$. in the second case, $ b \minus{} 1$ and $ b \plus{} 1$ are both even, so we must have $ b \minus{} 1 \equal{} 4m^3, b \plus{} 1 \equal{} 2n^3$ or $ b \minus{} 1 \equal{} 2m^3, b \plus{} 1 \equal{} 4n^3$. in the first case, $ 2 \equal{} 2(n^3 \minus{} 2m^3)\Leftrightarrow 1 \equal{} n^3 \minus{} 2m^3$, while in the second, $ m^3 \minus{} 2n^3 \equal{} \minus{} 1$. so what do you know about cubic Pell equations?

P.S. is this really fit for "Pre-Olympiad"?

I think if u see the second case more seriously without thinking of applying advance things still u can get the solution. My idea is based on the point of view not on applying higher mathematics.
(sorry i dont know latex)

 picking up any one; let it be 1 =n^3 - 2m^3
 now, n^3 - 1= 2m^3
       (n-1)(n^2 + n + 1)= 2m^3
     here n should be odd because if n is even then (n-1)(n^2 + n +1) becomes odd but it produces a contradiction as it should be even
     (n-1){(n-1)(n+2) +3} = 2m^3
     First take , m not equal to 3
      Now, we have either (n-1) = 2 or 2m^3 because (n-1) is even and if m divides n-1 then it can't divide {(n-1)(n-2) +3} otherwise m should be 1 for the condition we have made above.
      Solving the initial one, we have n=3 , in this case m can't have integer values.
      Going for the second n becomes 0 or -1
      n= -1 gives u the solution (2,3)

    Similarly thinking logically u can get the solution.
   Now u see thr is nothing of what u write as pell  eq.
   In fact i don't know about P.E.

Rationalist, I can't understand how you got n=2 or 2 m cube. Please explain that...
I analyzed. I didn't fully finish it but anyway here is my derivation... I am beginning to think there are no more solutions. Please tell me if there are any flaws or any part which you can't understand. I don't know latex. So I uploaded a image...
In Paragraph 4, I forgot to add that m = n = -1 which gives 2 cube plus one is equal to minus 3 whole square.

Sorry! :(  gouthamphilomath,
I didn't mention the statement of pluristiq. You should read the post of pluristiq first then come to my second post . After that,u will have no confusion.

Yeah, pluristic told to solve the P.E if I am right. I want to know how you got that. Sorry if I am wrong.

I have reedited my second post with better clarification. Even if u have any problem then ask me.

Yeah yeah I have got it. Thanks for the help. Any Idea how to solve the equation I have derived? Thanks again

gouthamphilomath
Sorry! For the delay in posting because I was busy these days.
Well, what I think about ur post is that as a point of solving a question u can pick up any path to solve a problem but what matter the most is ur understanding of the steps u r taking, this lead us to pick nd leave a particular step( it is my personal experience, use leave it or pick it point)
So for the step u r in the statement can be solved if u use my last post carefully. See for odds nd evens carefully nd then divide the section of equality , I hope u can get the answer.

{{{{  (n-1)(n^2 + n + 1)= 2m^3
here n should be odd because if n is even then (n-1)(n^2 + n +1) becomes odd but it produces a contradiction as it should be even
(n-1){(n-1)(n+2) +3} = 2m^3
First take , m not equal to 3
Now, we have either (n-1) = 2 or 2m^3 because (n-1) is even and if m divides n-1 then it can't divide {(n-1)(n-2) +3} otherwise m should be 1 for the condition we have made above.}}}}}
 THIS IS JUST FOR REFERENCE.
Any prob. then ask.

Yeah, I understand that. See my post where I attached a file. You see the eqn I derived? I even wrote that if x=1, then we get your eqn. Can you argue about other possible values for m and n for different values of x IN MY POST? (m and n in my post)

I am taking ur last derived equation as a new question.
You wrote to solve
$ m - 2^{3x - 2}n = 1$
 for odd perfect cubes m and n with natural x and substitute it to get a and b.....
$ Now, P^3 - 1 = 2^{3x - 2}q^3$
You will have problem only when to solve
$ p - 1 = 2^{3x - 2} \\
and \\
p^2 + p + 1 = q^3$                                        ........(1)
from the second equation of case(1) u can find the discriminant that
$ {\sqrt {1 - 4(1 - q^3)} = s}$
$ Now, 3q^3 - 3 = s^2 - q^3$
Using modular arithmetic u can say that
$ q = 3k + 1$   and but it produces a contradiction by the fact that there will be unequal no. of factors of $ 3$ in rhs and lhs.
Thus, there is no solution for (1)
Other parts r easy to solve in which u can take other values of $ p - 1$.

You solved $ m-2^{3x-2}n = 1$ what about the reverse. We will get (m+1) is equal to 2^{3x-2}n . You need to solve that.
Your solution was nice for the first case. Anyway, how do you do the second case?

[quote="gouthamphilomath"]You solved $ m - 2^{3x - 2}n = 1$ what about the reverse. We will get (m+1) is equal to 2^{3x-2}n . You need to solve that.
Your solution was nice for the first case. It didn't strike me. I am an idiot. Anyway, how do you do the second case?
And, Don't think I know latex in the first sentence, I copied rationalist's post.[/quote]

I didn't understand u write $ m+1=2^{3x-2}n$. I think u have written by mistake  if i am not wrong.
Only possible cases r $ (1) p-1= 2^3x-2$  or $ 2^{3x-2}q^3$  and if $ q$ divides $ p-1$ then
second case $ (2)$ is $ q = 1 or 3$.

Take $ p-1= 2^{3x-2}$then$ p^2+p+1= 1$ , becuz we have taken that $ \gcd(q,p^2+p+1)=1$ , solve for $ p$ which u have done earlier .
Now, why did i take second case becuz if  I have taken that $ q$ divides $ p-1$ and
$ T=p^2+p+1 =  (p-1)(p+2) +3$ we r left to have either $ q= 1 or 3$ . Solve to get the answer.

Don't complicate urself too much with the question. It is easy.

That's okay. I told that we should solve |$ m-2^{3x-2}n \&| = \& 1$ and not $ m-2^{3x-2}n = 1$ So, solve the other case 2^{3x-2}n  - m $ = \& 1$
Hope you get my point.

I think u r either not understanding  my answer or u haven't  solve many problems of diophantine equation.
I have given all possible cases arised out of equation $ m - 2^{2x - 3}n = 1$, and either given their solution or hints to proceed ahead.
Why r u confusing particularly in these Q that solving equation $ m - 2^{2x - 3}n = 1$ will have different solution from the equation  $ m - 1 = 2^{2x - 3}n$ or can have other solutions too. If u think so than i would suggest that u should solve some easier problem of Diopha..... eq.... then analyse the situations well.
There's nothing left to dig the problem, u have already dug all of it. :spider:    :)


I'm currently writing a report using the "article" style, and I'm having a small problem with the table of contents.  By default, LaTeX doesn't show headings below "subheading" on the table of contents.  This only allows two levels of organization - the heading and subheading.  The next level, "paragraph", doesn't show up.  I read on a LaTeX tutorial that there is a way to change this, but I don't know how and haven't been able to find out.  Can someone help me out with this?

[quote="ssn116"]Hi,
The counter for that is tocdepth. If you want three levels in the toc, use \setcounter{tocdepth}{3}. I don't have LaTeX installed on this computer, so if my syntax is a little off, I am sure you'll get the right syntax through google.
SSN.[/quote]
That worked perfectly, thanks! (I had to set the depth to four levels, for whatever reason - probably because "section" is considered the second level, even though there are no chapters in an article)


combinatorics solved
combinatorics
graph theory

Prove that in a graph such that $d\left(v\right)\geq 3$ for every vertex $v$, we can find a cycle with length not divisible by $3$ (assuming that the graph has at least one vertex).

Is there a generalization?

Suppose no, and consider the graph with least number of vertices.
Let $A_1 \cdots A_{3n}$ be the shortest cycle. The $A_i, A_j$ are not connected except by the edges of this cycle.
If we contract this cycle $A_1 \cdots A_{3n}$ (that is, we mix $A_1,\ldots,A_{3n}$ into one vertex $A$, and join $X$ to $A$ iff $X$ is connected to at least one of $A_i$), then we obtain a smaller graph. Show that the new graph has less vertices and still satisfies the condition.

I have another solution: Start from an arbitrary vertex $V_1$, and find a longest path starting from $V_1$, that is, given a path $V_1,V_2,V_3,...,V_k,...$ then define $V_{k+1}$ to be a vertex adjacent to $V_k$ different from $V_i$ for all $i=1,2,...,k$. This path is finite, since the process will eventually finish; so we will arrive at a vertex $V_n$ such that $V_n$ is adjacent to $V_i$ and $V_j$ where $V_i$ and $V_j$ are two previous members of the chain. Now, we have found $3$ cycles; it is easy to prove that one of them is not of order divisible by $3$.

Also you can suppose A1..An is longest path.
So An is connected to 2 of A1..An and we have a cycle that it's length is not divisible by 3

[quote=Pascual2005]Is there a generalization?[/quote]

Yes, and a fairly obvious one:

Fix a positive integer $p > 2$. Prove that in a graph such that $d\left(v\right)\geq 3$ for every vertex $v$, we can find a cycle with length not divisible by $p$ (assuming that the graph has at least one vertex).

Both iura's and Pascual2005's solutions (and Omid's, which is precisely Pascual2005's) can easily be adapted to this generalization.


Gauss
Galois Theory
superior algebra
superior algebra theorems

The title may seem funny, but how was the solvability of $X^{n}-1$ proved by Gauss without using Galois theory (which of course didn't exist at that time) ?

Ehm, do you mean in $\mathbb{C}$? The solutions are the edges of a regular $n$-gon in the complexe plane, formal:
\[ x_k = \cos(k \frac{2\pi}{n}) + i \cdot sin(k \frac{2\pi}{n}) ,~ k=0,...,n-1 \]
Why galois theory?

What does it prove ? Maybe I didn't set the question properly ... I meant how did Gauss prove that $X^{n}-1$ is solvable by radicals ?


[quote="mavropnevma"]What is $ S(A)$ ? If the number of elements, then
[hide="Hint"]It is just the old "$ n$ choose $ k$", in our case $ \genfrac {(} {)} {0pt} {} {100} {3}$.[/hide][/quote]
yes, it is the  number of elements of $ A$ but it is a counting problem. try to solve without using combinatorics. use only counting methods.

Part of combinatorics IS counting methods, in other words counting methods ARE combinatorics.


calculus
integration
function
probability
inequalities
real analysis
real analysis unsolved

Find all continuous functions $f(x)$, for $0\le{x}\le{1}$, such that
$\int_{0}^{1}f(x)=1$

$\int_{0}^{1}xf(x)=\alpha$

$\int_{0}^{1}x^{2}f(x)=\alpha^{2}$.

Here's the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=28825]old thread[/url].

[quote="Kent Merryfield"]It's been posted, and could be searched for. But it was a long time ago.

You didn't sepcify that $f(x)\ge0.$ If you do allow $f$ to take on values of mixed sign, that introduces certain complications.

If we to go ahead and assume that $f$ is nonnegative, what do we have?

The first equation says that $f$ is a probability density function for a random variable $X.$

The second equation than says that $E(X)=\alpha$ and the third says that $E(X^{2})=\alpha^{2}.$ Then $\text{Var}(X)=0.$

But the only random variable with zero variance is a point mass, and that doesn't have a continuous density function. So there are no functions satisfying this. (At least, no nonnegative functions.)

[In effect, I have used the Cauchy-Bunyakovsky-Schwarz inequality here.][/quote]

$f(x)$ could be the Dirac delta function, centered at $\alpha$: $f(x) = \delta (x-\alpha)$.  But then, $f$ isn't continuous...

maokid7 left off the condition $f(x)\ge0.$ That might have been an accident on his part - but leaving that out does change the problem.

Consider functions of the form $f(x)=\frac{x-b}{\frac12-b}$ for $0<b<\frac12.$ That constant has been rigged so that $\int_{0}^{1}f(x)\,dx=1.$ Then

$\int_{0}^{1}xf(x)\,dx=\frac{2-3b}{3-6b}$

$\int_{0}^{1}x^{2}f(x)\,dx=\frac{3-4b}{6-12b}$

Setting $\frac{3-4b}{6-12b}=\left(\frac{2-3b}{3-6b}\right)^{2}$ leads to a cubic equation one of whose roots is $b=\frac{3-\sqrt{3}}{6},$ which is in the required interval.

So if I've done this algebra correctly (not a sure thing), there's at least one solution to this for some particular $\alpha.$ Given that, I suspect there are many solutions.

o yes sorry i did mean to mention that $f(x)\ge{0}$ for all $x$ in the given domain.

here is another solution that i found

$\int_{0}^{1}(\alpha-x)^{2}f(x)dx=\alpha^{2}-2 \alpha^{2}+\alpha^{2}=0$
But the integrand is positive, except possibly for one point of the range, so the integral must also be positive. Contradiction. So there are no functions with this property.

[quote="PTynan89"]
$f(x)$ could be the Dirac delta function, centered at $\alpha$: $f(x) = \delta (x-\alpha)$.  But then, $f$ isn't continuous...[/quote]

The dirac delta function is not a function, it is a measure (or a distribution, or *something* that is definitely not a function in any accepted sense of the term).


[quote="bpms"][quote="4everwise"]Three vertices of a cube are $P=(7,12,10),$ $Q=(8,8,1),$ and $R=(11,3,9).$ What is the surface area of the cube?[/quote]
[hide="brute force is not that bad"]The distance from P to Q is $\sqrt{1+16+81}=\sqrt{98}$ The surface area is 6 times that squared so our answer is $6 \cdot 98 =\boxed{588}$[/hide][/quote]

[hide="uhh..."]P and Q are not adjacent verticies of the cube...[/hide]

[hide]$PQ=QR=RP=\sqrt{98}$, the only equilateral triangle using the verticies of a cube is the one using 3 face diagonals, so $s\sqrt{2}=7\sqrt{2}$ so $s=7$, and $SA=6s^{2}=294$[/hide]

[quote="Altheman"][quote="bpms"][quote="4everwise"]Three vertices of a cube are $P=(7,12,10),$ $Q=(8,8,1),$ and $R=(11,3,9).$ What is the surface area of the cube?[/quote]
[hide="brute force is not that bad"]The distance from P to Q is $\sqrt{1+16+81}=\sqrt{98}$ The surface area is 6 times that squared so our answer is $6 \cdot 98 =\boxed{588}$[/hide][/quote]

[hide="uhh..."]P and Q are not adjacent verticies of the cube...[/hide]
[/quote]
Wait, how do you know that? (sorry if I'm being slow)

He found all three lengths and found that they formed an equilateral triangle. This is only possible in a cube if those three segments are three surface diagonals.

Because...[quote="Altheman"]$PQ=QR=RP=\sqrt{98}$, the only equilateral triangle using the verticies of a cube is the one using 3 face diagonals[/quote]

Oh, never mind... Wow, I feel really stupid...

amctrivial grind.

We find using distance formula that $PQ=QR=PQ=7\sqrt{2}$, which implies they are three face diagonals as they form an equilateral triangle.

This implies the side length of the cube is $7$, so the answer is $49\cdot6=\boxed{294}$.


Hey all,

I have a problem concerning the maximum and minimum stationary derivative points on a cubic function. Suppose you have a cubic function which has two stationary points with different 'y' values as their co-ordinates. If you have proven that one of the two points is a maximum turning point, will the other stationary point have to be a minimum turning point of the cubic function? Please help me asap. I appreciate the help, thanks.

Cheers,

Shaun.

No, although I cannot find a counterexample I don't see why the derivative must change sign in the stationary point.

But, think about it, if you look at a standard cubic  or say any cubic function graph with two turning points, both of which are on two *different 'y' values. If you find that one is the maximum of the function, wouldn't the other point have to be a minimum?


I'm not sure under which sub-forum I should post this, so I'll just post it here.

A while ago, I was shown how to use Taylor series to evaluate [i]e[/i]^(:pi:[i]i[/i]), but I don't remember it anymore. How does it go?

Thanks.

[quote="djayboots"]I'm not sure under which sub-forum I should post this, so I'll just post it here.[/quote]

Good call. This Other Problem Solving Topics forum is by far the best forum in which to post a general question about mathematics.

I wish I could help you with the actual question  :oops:  but it's a good bit past my currrent math level.

The taylor series for [tex]e^x[/tex] is

[tex]\displaystyle e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...[/tex]

Putting in [tex]i \pi[/tex] and separating the real and imaginary terms, we get

[tex]\displaystyle e^{i\pi} = (1-\frac{\pi^2}{2!}+\frac{\pi^4}{4!}+...) + i(\pi-\frac{\pi^3}{3!}+\frac{\pi^5}{5!}+...)[/tex]

We recognize the left side and the right side as the series expantions for [tex]\cos \pi[/tex]  and [tex]\sin \pi[/tex] , respectively. So the whole thing boils down to

[tex]\displaystyle e^{i\pi} = \cos \pi + i \sin \pi = \mathbf{-1}[/tex]

Is that what you were looking for?

[quote="zabelman"][tex]\displaystyle e^{i\pi} = (1-\frac{\pi^2}{2!}+\frac{\pi^4}{4!}+...) + i(\pi-\frac{\pi^2}{3}+\frac{\pi^5}{5}+...)[/tex][/quote]

Shouldn't this be
[tex]\displaystyle e^{i\pi} = (1-\frac{\pi^2}{2!}+\frac{\pi^4}{4!}+...) + i(\pi-\frac{\pi^3}{3!}+\frac{\pi^5}{5!}+...)[/tex]?

Regardless, that was a beautiful explanation, zabelman.

Thank you, mathfanatic!
And yes, you're right. I fixed it.

There's still a [tex]\pi^2[/tex] where there should be a [tex]\pi^3[/tex]

lol, im a clumsy typist... thanks for the correction

[quote="zabelman"]thanks for the correction[/quote]

No problemo.


let p be a prime and q,n be an integer if $ q|{(n\plus{}1)}^p\minus{}n^p$ prove that
$ p|q\minus{}1$

Let $ k$ be a prime factor of $ {(n\plus{}1)}^{p}\minus{}n^{p}$, then

$ (n\plus{}1)^p\equal{}n^p(\mod k)  \Leftrightarrow [(n\plus{}1)n^{\minus{}1}]^p\equal{}1(mod k)$, which suggest that $ p|k\minus{}1$ as $ (n\plus{}1)n^{\minus{}1} \neq 1(mod k)$

Now let $ q\equal{}p_1p_2p_3.....p_n$, where $ p_i$ s are primes and $ p_1 \le p_2 \le p_3 \le ... \le p_n$, then $ p_i$ s are all prime factor of $ {(n\plus{}1)}^{p}\minus{}n^{p}$ and thus from the result above we have $ (p_i\minus{}1)\equal{}0(\mod p)$

So $ q\minus{}1\equal{}p_1p_2p_3...p_n\minus{}1$
$ \equal{}[(p_1\minus{}1)\plus{}1][(p_2\minus{}1)\plus{}1][(p_3\minus{}1)\plus{}1]\plus{}...\plus{}[(p_n\minus{}1)\plus{}1]\minus{}1$
$ \equal{}1\minus{}1\equal{}0(\mod p)$


algebra
system of equations
algebra proposed

Given are natural numbers $ a,b$ and $ n$ such that $ a^2\plus{}2nb^2$ is a complete square. Prove that the number $ a^2\plus{}nb^2$ can be written as a sum of squares of $ 2$ natural numbers.

[quote="Bugi"]Given are natural numbers $ a,b$ and $ n$ such that $ a^2 \plus{} 2nb^2$ is a complete square. Prove that the number $ a^2 \plus{} nb^2$ can be written as a sum of squares of $ 2$ natural numbers.[/quote]

WLOG, we only need to consider the case that $ gcd(a,b)\equal{}1$

Let $ a^2\plus{}2nb^2\equal{}k^2$, then $ (k\minus{}a)(k\plus{}a)\equal{}2nb^2$.

[b]If $ b$ is odd,[/b] then as $ gcd(a,b)\equal{}1 \Rightarrow gcd(k\minus{}a,k\plus{}a,b)\equal{}1$, we can let

one of $ k\minus{}a$ , $ k\plus{}a$ be $ 2n_1b_1^2$ and another one be $ n_2b_1^2$, where $ n_1n_2\equal{}n$ and $ b_1b_2\equal{}b$.

So $ a^2\plus{}nb^2\equal{}(\frac{2n_1k_1^2\minus{}n_2k_2}{2})^2\plus{}n_1n_2b_1b_2\equal{}(n_1k_1^2)^2\plus{}(\frac{n_2k_2^2}{2})^2$

[b]If $ b\equal{}2c$ is even,  [/b] then $ (\frac{k\minus{}a}{2})(\frac{k\plus{}a}{2})\equal{}2nc^2$, similarly we have

$ a^2\plus{}nb^2\equal{}(2n_1c_1^2\minus{}n_2c_2^2)^2\plus{}4n_1n_2c_1c_2\equal{}(2n_1c_1^2)^2\plus{}(n_2c_2^2)^2$, done.

[hide="Solution"]
Take a example for $ a\equal{}3,b\equal{}2,n\equal{}2$
We see that $ a^2\plus{}2nb^2\equal{}5^2,a^2\plus{}nb^2\equal{}4^2\plus{}1^2,4\plus{}1\equal{}5$

Try to form a system of equations with $ a^2\plus{}2nb^2\equal{}(x\plus{}y)^2,a^2\plus{}nb^2\equal{}x^2\plus{}y^2$

Subtracting we get $ nb^2\equal{}2xy,a\equal{}x\minus{}y$.
Solving the new system we get $ x\equal{}\frac{\sqrt{a^2\plus{}2nb^2}\plus{}a}2,y\equal{}\frac{\sqrt{a^2\plus{}2nb^2}\minus{}a}2$. As for the condition for complete square, $ x,y$ are rational, integer because $ a^2\plus{}2nb^2$ and $ a$ are of the same parity, and natural because all the numbers are greater than zero. Now $ x$ and $ y$ are the numbers... [/hide]

[quote="Bugi"]Given are natural numbers $ a,b$ and $ n$ such that $ a^2 \plus{} 2nb^2$ is a complete square. Prove that the number $ a^2 \plus{} nb^2$ can be written as a sum of squares of $ 2$ natural numbers.[/quote]
$ a^2\plus{}2nb^2 \equal{} k^2 \iff (a^2\plus{}nb^2)^2 \equal{} (ak)^2 \plus{} (nb^2)^2$. And the result follows..

[quote="stephencheng"][quote="Mathias_DK"][quote="Bugi"]Given are natural numbers $ a,b$ and $ n$ such that $ a^2 \plus{} 2nb^2$ is a complete square. Prove that the number $ a^2 \plus{} nb^2$ can be written as a sum of squares of $ 2$ natural numbers.[/quote]
$ a^2 \plus{} 2nb^2 \equal{} k^2 \iff (a^2 \plus{} nb^2)^2 \equal{} (ak)^2 \plus{} (nb^2)^2$. And the result follows..[/quote]

Why? The question is to prove that the number $ a^2 \plus{} nb^2$ can be written as sum of two non-zero squares . But you only prove the number $ (a^2 \plus{} nb^2)^2$ can be written as sum of two non-zero squares.[/quote]
We just use the well-known fact: $ n$ can be written as a sum of two squares iff every primedivisor $ p \equiv 3 \bmod 4$ divides it an even number of times. Wlog $ (a,b) \equal{} 1$ and that $ n$ is squarefree. Assume $ p \equiv 3 \bmod 4$ is such that $ p \mid a^2\plus{}nb^2$. Then $ p^2 \mid (ak)^2\plus{}(nb^2)^2 \iff p \mid ak, p \mid nb^2$. If $ p \nmid b^2$ then $ p \mid n$ and $ a^2 \equiv k^2 \bmod p$ from the first equation. Hence $ p \mid a,k$ and then $ p^2 \mid n$. contradiction. Hence $ p \mid b^2 \iff p \mid b$. Then again from the first: $ p \mid a^2\minus{}k^2$ and hence $ p \mid a,k$. But $ (a,b) \equal{} 1$. Contradiction. And we are done.

I just thought that this part was so trivial, so I only wrote the main idea :oops:

$ a^2\plus{}2nb^2\equal{}k^2$ a and k are of the same parity
$ nb^2\equal{}\frac{k^2\minus{}a^2}{2}$
$ a^2\plus{}nb^2\equal{}\frac{k^2\plus{}a^2}{2}\equal{}(\frac{a\plus{}k}{2})^2\plus{}(\frac{a\minus{}k}{2})^2$
and cause of a and k of the same parity we are done

[quote="Arrange your tan"][quote="Bugi"]"... is a complete square."[/quote]

I can't find terminology/an entry on the Internet
for "complete square."

You must have meant a perfect square that is an integer.[/quote]

Yes, it should be perfect square (didn't know it)

[quote="enndb0x"]What is JBTST ?[/quote]
I would guess the Team Selection Test for the Junior Balkan MO.

Mathias,

   How do you conclude that  $ p^2 |(ak)^2 \plus{} (nb^2)^2 \Rightarrow p|ak$ and $ p|nb^2$.

thanks

[quote="Mathias_DK"][quote="Bugi"]Given are natural numbers $ a,b$ and $ n$ such that $ a^2 \plus{} 2nb^2$ is a complete square. Prove that the number $ a^2 \plus{} nb^2$ can be written as a sum of squares of $ 2$ natural numbers.[/quote]
$ a^2 \plus{} 2nb^2 \equal{} k^2 \iff (a^2 \plus{} nb^2)^2 \equal{} (ak)^2 \plus{} (nb^2)^2$. And the result follows..[/quote]

That seems wrong; you show that the square of some number is a sum of two squares. That doesn't tell you anything about whether the original number had such a representation.

There are many integers whose square is a sum of two squares, but which are not themselves sums of two squares.

brunojoyal: Please read my other post...

nicetry007: I assumed $ p \equiv 3 \bmod 4$...

We can actually assume b=1, since otherwise we can just let $ n\equal{}nb^2$ and $ b\equal{}1$.