There is an easy way to calculate the ratio AB.CD/AC.BD. Applying rule of cosines for nonconvex quadrilateral ADBC we have (AB.CD)^2=(AC.BD)^2+(BC.AD)^2-2.AC.BD.BC.AD.cos(AB.CD)^2= =(AC.BD)^2+(BC.AD)^2-2.AC.BD.BC.AD.cos(<ADB-<ACB)=(AC.BD)^2+(BC.AD)^2-2.AC.BD.BC.AD.cos90= =(AC.BD)^2+(BC.AD)^2=2.(AC.BD)^2 therefore the ratio is equa to :rt2:

Hard to belive it was a problem for IMO, but back in '93 we did not know that much mainly because of infancy of The Internet, Data Bases were not properly loaded, I conjecture. Here is my solution: Similar triangles in circles ACD and BCD are 90deg rotated because of condition angle ADB = 90 + ACB. So we have two vectors that are equal and at 90 deg. The connecting vector is sqrt(2) times longer. Thank you. M.T.

my solution: see one inversion $(D,R^2)$ let $X'$ the inverse of $X$. we have $A'C' = \frac{R^2.AC}{AD.DC} = \frac{R^2.BC}{BD.CD} = B'C'$. (by our condition) so, $\angle{DCB} = \angle{ABD}$. now, with simple angle chasing, we get $\angle{A'C'B'} = 90$. (this implies that the circuncircles of $ADC$ and $BDC$ are orthogonal). so, $A'C'B'$ is one isoseles right angled triangle. now, $\sqrt{2} = \frac{A'B'}{B'C'}$ $= \frac{\frac{R^2.AB}{AD.BD}}{\frac{R^2.BC}{BD.CD}} = \frac{AB.CD}{AD.BC}$.

armpist wrote: Hard to belive it was a problem for IMO, but back in '93 we did not know that much mainly because of infancy of The Internet, Data Bases were not properly loaded, I conjecture. Here is my solution: Similar triangles in circles ACD and BCD are 90deg rotated because of condition angle ADB = 90 + ACB. So we have two vectors that are equal and at 90 deg. The connecting vector is sqrt(2) times longer. Thank you. M.T. I think this is the same idea as my solution, but it seems like you skipped a lot of steps. More detailed proof: Construct E such that a spiral similarity takes ACE to ADB. CE = BD*AC/AD = BC, and by the angle condition, ECB is an isosceles right triangle. Then $\frac{AB \cdot CD}{AC \cdot BD}= \frac{EB \cdot AD}{EC \cdot AD}= \sqrt{2}$ and the angle condition easily gives the two circumcircles orthogonal.

First of all, the correct statement is "Let $ D$ be a point inside acute triangle $ ABC$ such that $ \angle ADB=\angle ACB+\pi/2$ and $ AC\cdot BD=AD\cdot BC$. (...)" I'll solve the second part first, because it can be solved without making reference to $ \overline{AC}\cdot\overline{BD} = \overline{AD}\cdot\overline{BC}$. Let $ E\ne C$ be the intersection of the circumcircle of $ [ACD]$ with the tangent to $ [BCD]$ at $ C$. Let $ F\ne C$ be the intersection of the circumcircle of $ [BCD]$ with the tangent to $ [ACD]$ at $ C$. It's easy to find out $ C\hat{A}D + C\hat{B}D = 90^{\circ}$ and therefore, using the Half-Inscribed Angle Theorem, it's easy to find out $ E\hat{C}F = 90^{\circ}$, and the result follows. For the first part, invert through $ C$. Points $ A$, $ B$ and $ D$ are sent to points $ A'$, $ B'$ and $ D'$, respectively. Then $ \overline{AC}\cdot\overline{BD} = \overline{AD}\cdot\overline{BC}$ is equivalent to $ \overline{A'D'} = \overline{B'D'}$ and $ \frac {\overline{AB}\cdot\overline{CD}}{\overline{AC}\cdot\overline{BD}}$ is equal to $ \frac {\overline{A'B'}}{\overline{B'D'}}$. Since $ C\hat{A}D + C\hat{B}D = 90^{\circ}$, it's easy to find out $ A'\hat{D'}B' = 90^{\circ}$, so, using Pythagorean Theorem, we find out $ \frac {\overline{A'B'}}{\overline{B'D'}} = \sqrt {2}$.

Let $X$ and $Y$ respectively denote the circumcenters of $\triangle{ACD}$ and $\triangle{BCD}$. Construct point $P$ such that $\triangle{BDP} \sim \triangle{BCA}$. Combining the ratio of similarity and the condition, we now have that \[DP \cdot BC = AC \cdot BD = AD \cdot BC\] Hence $DP=BC$. Also, we have that $\angle{ADP}=\angle{ADB}-\angle{PDB}=\angle{ADB}-\angle{ACB}=90$. Hence $\triangle{ADP}$ is an isosceles right triangle. Now note that since $\triangle{BDP} \sim \triangle{BCA}$, spiral similarity implies that $\triangle{BAP} \sim \triangle{BCD}$. The ratio of similarity implies that \[\frac{AP}{CD}=\frac{AB}{BD} = \frac{AC\sqrt{2}}{CD} \quad \Rightarrow \quad \frac{AB\cdot CD}{AC\cdot BD}= \sqrt{2}\] Also, we have the following \[\angle{CBD} = \angle{ABP} = 180-\angle{PAB} - \angle{APB}=135-\angle{CAP}=90-\angle{CAD}\] Therefore we have that $\angle{CXD}=180-2\angle{CBD}=180-\angle{CYD}$. Therefore $XCYD$ is cyclic. Since $XCYD$ is symmetrical around line $XY$, we have that $\angle{XCY}=\angle{XDY}=90$. Hence circles $ACD$ and $BCD$ are orthogonal.

Weird. When I first saw this problem, it seemed very projective [due to Apollonian circle and the ratio condition], but none of the other sols took that approach. But then I am not good at projective, so I do not know if such methods trivialize this. Is this the case?

[asy][asy] unitsize(1.2inches); pair A=(9/13)*dir(0); pair B=(13/9)*dir(0); pair CC=dir(37); pair DD=CC*dir(90); pair C=2*foot(0,B,CC)-CC; pair D=2*foot(0,B,DD)-DD; pair Y=dir(0); pair X=dir(180); draw(circumcircle(X,Y,CC)); draw(B--C); draw(B--DD); draw(X--B); draw(A--D); draw(A--C); draw(0*dir(0)--CC); draw(0*dir(0)--DD); dot("$A$",A,dir(-90)); dot("$B$",B,dir(0)); dot("$C$",C,dir(C)); dot("$D$",D,dir(40)); dot("$C'$",CC,dir(CC)); dot("$D'$",DD,dir(DD)); dot("$X$",X,dir(180)); dot("$Y$",Y,dir(-45)); dot("$O$",0,dir(-90)); [/asy][/asy] The condition $AC\cdot BD=AD\cdot BC$ implies that $C$ and $D$ are on an Appolonius circle with foci $A$ and $B$. This means that there are points $X,Y$ on line $AB$ such that $(AB;XY)=-1$ and $C,D\in(XY)$. If we let $O$ be the midpoint of $XY$, then the condition $(AB;XY)=-1$ just tells us that $A$ and $B$ are inverses with respect to the Appolonius circle $(XY)$. Let $\phi_B$ be inversion at $B$ with power $BX\cdot BY$. Note that $\phi_B$ fixes $(XY)$ and swaps $O$ and $A$. Let $D'=\phi_B(D)$ and $C'=\phi_B(C)$. Note that by the inversion, we have \[\angle ADB=\angle \phi_B(D)\phi_B(A)B=\angle D'OB\]and similarly $\angle ACB=\angle C'OB$. Thus, the angle condition in the problem simply tells us that $OC'\perp OD'$. We see that $\phi_B((ACD))=(OC'D')$ and $\phi_B((BCD))=C'D'$. Since $(OC'D')=(C'D')$, we have $(OC'D')\perp C'D'$. Inverting back tells us that $(ACD)\perp(BCD)$. To find the ratio, we will resort to a complex bash. Let $(XY)$ be the unit circle with $y=1$ and $x=-1$. Furthermore, suppose $C'=z$, so $D'=iz$. Finally, suppose that $B=\beta$ and $A=1/\beta$ where $\beta\in\mathbb{R}$. We see that $B$ lies on the chord $CC'$, so \[c+c'-cc'\overline{b}=b\implies c=\frac{c'-b}{c'\overline{b}-1},\]and similarly $d=\frac{d'-b}{d'\overline{b}-1}$. Thus, \[\boxed{c=\frac{z-\beta}{z\beta-1}}\quad\text{and}\quad \boxed{d=\frac{iz-\beta}{iz\beta-1}}.\]We now compute \begin{align*} AC\cdot BD &= \left|\frac{1}{\beta}-\frac{z-\beta}{z\beta-1}\right|\cdot\left|\beta-\frac{iz-\beta}{iz\beta-1}\right| \\ &= \frac{|\beta^2-1|^2}{|\beta|\cdot|z\beta-1|\cdot|iz\beta-1|} \end{align*}and \begin{align*} AB\cdot CD &= \frac{|\beta-1/\beta|}{|z\beta-1|\cdot|iz\beta-1|}\cdot|-iz\beta^2-z+z\beta^2+iz| \\ &= AC\cdot BD\cdot|1-i| \\ &=\sqrt{2}\cdot AC\cdot BD. \end{align*}Thus, $\boxed{\frac{AB\cdot CD}{AC\cdot BD}=\sqrt{2}}$.

Let $O_1$ and $O_2$ be the centers of $(ACD)$ and $(BCD).$ Notice $$\angle O_1DO_2=\angle O_2DC+\angle O_1DC=\angle DAC+\angle CBD=360-\angle ACB-(360-\angle ADB)=90$$so $(ACD)$ and $(BCD)$ are orthogonal. Invert about $A$ with radius $1.$ Then, $$1=\frac{AC\cdot BD}{AD\cdot BC}=\frac{AC\cdot\frac{B'D'}{AB'\cdot AD'}}{AD\cdot\frac{B'C'}{AB'\cdot AC'}}=\frac{B'D'}{B'C'}$$and $$\angle AD'B'=90+\angle AB'C'$$so $B'D'=B'C'$ and $\angle C'B'D'=90.$ Hence, $$\frac{AB\cdot CD}{AC\cdot BD}=\frac{AB\cdot\frac{C'D'}{AC'\cdot AD'}}{AC\cdot\frac{B'D'}{AB'\cdot AD'}}=\frac{C'D'}{B'D'}=\sqrt{2}.$$$\square$

Apparently complex numbers works for the first part. The first condition means that $\frac{(b-d)(a-c)}{(a-d)(b-c)}$ is pure imaginary by dividing the rotations between $\overline{AD}$ and $\overline{BD}$ and between $\overline{AC}$ and $\overline{BC}$. However, because $|a-c||b-d| = |a-d||b-c|$, the magnitude of this number is $1$, so it is necessarily $i$. Then, \begin{align*} (a-b)(c-d) &= ((a-c) - (b-c))((b-d)-(b-c)) \\ &= (b-c)[i(a-d) - (b-d) - (a-c) + (b-c)] \\ &= (1-i)(b-c)(a-d). \end{align*}So$$\frac{AB \cdot CD}{AC \cdot BD} = \frac{|a-b||c-d|}{|b-c||a-d|} = |1-i| = \boxed{\sqrt 2}.$$

Invert about $D$ with radius $1$; denote images with $\bullet'$. The condition $\angle ADB=90^\circ+\angle ACB$ translates to $\angle A'C'B'=90^\circ$. Furthermore, by applying the inversion distance formula the condition $AC\cdot BD=AD\cdot BC$ is equivalent to $A'C'=B'C'$. Then $\frac{AB\cdot CD}{AC\cdot BD}=\frac{A'B'}{A'C'}=\sqrt{2}$. Furthermore, it is clear that perpendicular lines, when inverted around a point not lying on them, become orthogonal circles, so since $\overline{A'C'} \perp \overline{B'C'}$ we have $(DAC)$ and $(DBC)$ orthogonal. $\blacksquare$

Invert at $A$ with radius 1. Our angle condition is transformed to $\angle C^*B^*D^* = 90$, so the orthogonality is clear. Our length condition is \[\frac{1}{AC^*} \cdot B^*D^* \cdot \frac{1}{AB^* \cdot AD^*} = \frac{1}{AD^*} \cdot B^*C^* \cdot \frac{1}{AB^* \cdot AC^*} \implies B^*D^* = B^*C^*.\] Thus our desired ratio can be written as \[\frac{\frac{1}{AB^*} \cdot C^*D^* \cdot \frac{1}{AC^* \cdot AD^*}}{\frac{1}{AC^*} \cdot B^*D^* \cdot \frac{1}{AB^* \cdot AD^*}} = \frac{C^*D^*}{B^*D^*} = \boxed{\sqrt 2}.\]

Invert about $D$ with radius $1.$ Note that by angle chasing with the $\angle ADB=\angle ACB+90$ we have $\angle A'C'B'=90.$ As the inversion of two perpendicular lines that don't go through the center of inversion is two orthogonal circles, we have proved the second part of the problem. For the first part, using the inversion distance formula we have \begin{align*} A'C' &= AC \cdot \frac{1}{CD \cdot AD} \\ B'C' &= BC \cdot \frac{1}{CD \cdot BD} \\ A'B' &= AB \cdot \frac{1}{AD \cdot BD}. \\ \end{align*}Using $\frac{BC}{BD}=\frac{AC}{AD},$ we have that $A'C'=B'C$ by dividing the the first equation by the second. Dividing the third equation by the first we get $$\frac{AB \cdot CD}{AC \cdot BD}=\frac{A'B'}{A'C'}=\sqrt{2},$$and we're done. You could also spiral the length condition quite easily.

Upon inverting at $A$ we get that $B'C'D'$ is an isosceles right triangle with right angle at $B',$ so the ratio is $\frac{C'D'}{B'D'}=\sqrt2,$ and the orthogonality is because $C'D'$ is a diameter of $(B'C'D').$