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: introduction for the IMO ShortList/LongList project and regardingsolutions

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

Tornado wrote:

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. $\Box$

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

@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.

Proof: We invert the configuration about $P$ with the radius of inversion being $1$. This inversion takes: 1. $\Gamma_i \rightarrow l_i , i= 1,2,3,4.$ where $l_i$ is a line not passing through $P$. 2. $A' = \l_1 \cap l_2$. $B',C',D'$ are defined analogously. 3. $\Gamma_1$ tangent to $\Gamma_3 \implies l_1 || l_3.$ Similarly $l_2 || l_4$. Hence $A'B'C'D'$ is a parallelogram. 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 $B^*C^* = A^*D^*$. $\fbox{}$

ISL 2003 G4 wrote: 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 \[ \frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}. \] 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 Inversion Distance Formula : $1 = \frac{A'B'}{D'C'} = \frac{AB}{DC} \cdot \frac{PC \cdot PD \cdot r^2}{PA\cdot PB \cdot r^2}$ $1 = \frac{B'C'}{A'D'} = \frac{BC}{AD} \cdot \frac{PA \cdot PD \cdot r^2}{PB\cdot PC \cdot r^2}$ $\frac{AB \cdot BC}{DC \cdot AD} \cdot \frac{PC \cdot PD}{PA \cdot PB} \cdot \frac{PA \cdot PD}{PB \cdot PC} = 1 \Longrightarrow \frac{AB \cdot BC}{AD\cdot DC} = \frac{PB^2}{PD^2} \qquad \blacksquare$

orl wrote: 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 \[ \frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}. \] My First Inversion Solution: we invert about $P$ with a radius $r$. $ 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^* $ $$ \frac{AB\cdot BC}{AD\cdot DC}=\frac{\frac{r^2}{PA^*\cdot PB^*}\cdot A^*B^*\cdot \frac{r^2}{PB^*\cdot PC^*}\cdot B^*C^*}{\frac{r^2}{PA^*\cdot PD^*}\cdot A^*D^*\cdot \frac{r^2}{PD^*\cdot PC^*}\cdot D^*C^*}=\frac{A^*B^*\cdot B^*C^*}{A^*D^*\cdot D^*C^*}\cdot (\frac{PD^*}{PB^*})^2=(\frac{PD^*}{PB^*})^2=(\frac{\frac{r^2}{PD}}{\frac{r^2}{PB}})^2=\frac{PB^2}{PD^2} $$

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 \[\frac{(PD')^2}{(PB')^2}=\frac{\left (\frac{r^2}{PD}\right)^2}{\left ( \frac{r^2}{PB} \right)^2}=\frac{PB^2}{PD^2}.\]

My Solution- [asy][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][/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][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][/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 $\blacksquare$.

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. $\blacksquare$

MatBoy-123 wrote: My Solution- [asy][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][/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][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][/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..) $\blacksquare$ I made the same mistake as you initially, we can't garantuee that it will be a rectangle, only parallelogram.

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

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.

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 drhong. 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.

Invert about the circle centered at $P$ with radius $1$. Let the image of $A$ under inversion be $A^*$. Then $A^*B^*C^*D^*$ is a parallelogram, so that $A^*B^* = C^*D^*$ and $A^*D^* = B^*C^*$. By the inversion distance formula, we obtain that \[\frac{AB}{PA \cdot PB} = \frac{CD}{PC \cdot PD}, \text{ and } \frac{BC}{PB \cdot PC} = \frac{DA}{PD \cdot PA}. \]Hence we have that \[\frac{AB}{DA} \cdot \frac{BC}{CD} = \frac{PB^2}{PD^2}, \]as desired. $\blacksquare$