Beautiful problem!

Actually Click to reveal hidden text

kills it.

To be specific, let $l_a$ and $l_b$, $l_b$ and $l_c$, $l_c$ and $l_d$ $l_d$ and $l_a$ intersect at $X, Y, Z, T$ respectively. Lines $AB$ cross $CD$ at ${G_1}$ and $AD$ cross $BC$ at ${H}$ and $AC$ cross $BD$ at ${G}$. Call the circumcircle of $ABCD$ and $XYZT$ $c_1$ and $c_2$. Define $f(U)=Pow(U, c_1)-Pow(U, c_2)$ for any point ${U}$ on the plane. Obviously ${f}$ is linear. Notice that $\frac {\overline{AG}}{\overline{CG}}=-\frac{AD\cdot AB}{CD\cdot CB}$,and $f(A)=\overline{AX}\cdot\overline{AT},f(C)=\overline{CZ}\cdot\overline{CY}$, we only need to prove that $f(G)=0$,which is $\frac{f(A)}{f(C)}=\frac{\overline{AG}}{\overline{CG}}$. By omitting orientation, we shall only need to prove \begin{align*} &\frac{{AX} \cdot{AT}}{{CZ}\cdot{CY}}=\frac{{AD}\cdot{AB}}{CD\cdot CB} \end{align*}We have $AX=\frac{AB\cdot BC}{BG_1},AT=\frac{AD\cdot HB}{HA},CZ=\frac{CD\cdot AD}{G_1D},CY=\frac{BC\cdot HD}{HC}$ by some similar triangles, plug in and we get the desired result.

Attachments:

Let $AC \cap BD = R$. Let $\ell_a \cap \ell_b = W$ and cyclically define $X,Y,Z$. Let line $BD$ intersect $WZ,XY$ at $T_1,T_2$ respectively; and $(WXYZ)$ at $S_1,S_2$. By DIT on quadrilateral $WXYZ$ with line $BD$, we infer that there exists a unique involution swapping the pairs $(B,D),(T_1,T_2),(S_1,S_2)$. We will prove the above inversion is centered at $R$. It suffices to show that $RT_1\cdot RT_2 = RB\cdot RD$. Since $RB \cdot RD = RA\cdot RC$ it suffices to that $T_1T_2AC$ is cyclic. This holds since $\angle AT_1T_2 = \angle AT_1B =\angle T_1BC= \angle DBC = \angle DAC= \angle ACT_2$. Hence it is indeed centered at $R$. Hence, $RB\cdot RD = RS_1\cdot RS_2$ which would imply $R$ lies on radical axis of $(ABCD)$ and $(WXYZ)$ and we are done!

Let $P_{AB}=l_a\cap l_b$. Define $P_{BC}$, $P_{CD}$, and $P_{DA}$ similarly. Let $X=AB\cap CD$, $Y=AD\cap BC$, and $Z=AC\cap BD$. For every point $P$ in the plane, define $$f(P)=\mathbb{P}(P,\gamma)-\mathbb{P}(P,w)$$By Linearity of Power of a Point, $f$ is a linear function. Notice that $f(A)=AP_{AB}\cdot AP_{DA}$ . we are using directed lengths . Using similar triangles $ADP_{DA}$ and $YAB$ and similar triangles $ABP_{AB}$ and $BXC$ yields that $f(A)=\frac{YB\cdot AD}{YA}\cdot \frac {BC\cdot BA}{XB}$. Similar computations give that $$f(A)=\frac{YB\cdot (AD\cdot BC\cdot BA)}{YA\cdot XB}$$$$f(C)=\frac{YD\cdot (CB\cdot DA\cdot DC)}{YC\cdot XD}$$Using the linearity of $f$ along $AC$ we get that $$f(Z)=\frac{ZC\cdot f(A)+AZ\cdot f(C)}{ZC+AZ}$$Applying Menelaus' Theorem to triangle $XAC$ and line $BZD$ gives that $$\frac{AZ}{ZC}\cdot\frac{XB}{BA}\cdot \frac{CD}{DX}=1$$We also have that $YA\cdot YD=YB\cdot YC$ which is sufficient to imply that $f(Z)=0$. Thus $Z$ must lie on the radical axes of $w$ and $\gamma$.

Attachments:

The complex bash is clean! The equations of the lines are \begin{align*} \ell_a(z)=z+bc\overline z-a-bc/a=0\\ \ell_b(z)=z+cd\overline z-b-cd/b=0\\ \ell_c(z)=z+da\overline z-c-da/c=0\\ \ell_d(z)=z+ab\overline z-d-ab/d=0\\ \end{align*}The set of conics through the four relevant intersections of these lines is paramaterized by \[k_1\ell_a(z)\ell_c(z)+k_2\ell_b(z)\ell_d(z)=0\](one can check that the left hand side vanishes on all four relevant intersections, has degree two, and can be made to vanish on any fifth point by choosing $k_1$ and $k_2$ appropriately). We set $k_1$ and $k_2$ to make the $z^2$ coefficient vanish, so that the conic is a circle. Obviously, $k_1=1$ and $k_2=-1$ works. We now want to intersect the following with the unit circle, so we will set $\overline z=1/z$. \begin{align*} \ell_a(z)\ell_c(z)-\ell_b(z)\ell_d(z)&=0\\ (z+bc\overline z-a-bc/a)(z+da\overline z-c-da/c)-(z+cd\overline z-b-cd/b)(z+ab\overline z-d-ab/d)&=0\\ (bc+da-cd-ab)z\overline z+(b+d-a-c+cd/b+ab/d-bc/a-da/c)z+(abc+acd-abd-bcd+cd^2+ab^2-da^2-bc^2)\overline z+(da^2/c+bc^2/a-ab^2/d-cd^2/b)&=0\\ (b+d-a-c+cd/b+ab/d-bc/a-da/c)z^2+(da^2/c+bc^2/a-ab^2/d-cd^2/b+bc+da-cd-ab)z+(abc+acd-abd-bcd+cd^2+ab^2-da^2-bc^2)&=0 \end{align*} It suffices to show that \[\left|\begin{matrix}ac&a+c&1\\bd&b+d&1\\ef&e+f&1\end{matrix}\right|\stackrel?=0\]and we know by Vieta's that $ef=\frac{abc+acd-abd-bcd+cd^2+ab^2-da^2-bc^2}{b+d-a-c+cd/b+ab/d-bc/a-da/c}$ and $e+f=-\frac{da^2/c+bc^2/a-ab^2/d-cd^2/b+bc+da-cd-ab}{b+d-a-c+cd/b+ab/d-bc/a-da/c}$. Now we want \begin{align*} \left|\begin{matrix}ac&a+c&1\\bd&b+d&1\\abc+acd-abd-bcd+cd^2+ab^2-da^2-bc^2&-(da^2/c+bc^2/a-ab^2/d-cd^2/b+bc+da-cd-ab)&b+d-a-c+cd/b+ab/d-bc/a-da/c\end{matrix}\right|&\stackrel?=0\\ (b+d-a-c+cd/b+ab/d-bc/a-da/c)(abc+acd-abd-bcd)+(da^2/c+bc^2/a-ab^2/d-cd^2/b+bc+da-cd-ab)(ac-bd)+(abc+acd-abd-bcd+cd^2+ab^2-da^2-bc^2)(a+c-b-d)&\stackrel?=0 \end{align*}which is true; by the dihedral group (anti)symmetry, it suffices to check that the coefficients of $\frac{ac^2d^2}b$, $a^4$, $a^3b$, $a^3c$, $a^2b^2$, $a^2c^2$, $a^2bc$, $ab^2c$, and $abcd$ are all $0$ (since we can easily check that the only term with a denominator is of the former type).

Let $\ell_a$ and $\ell_b$ meet at $P$. Let $\ell_b$ and $\ell_c$ meet at $S$. Let $\ell_c$ and $\ell_d$ meet at $R$. Let $\ell_d$ and $\ell_a$ meet at $Q$. Let $AC$ and $BD$ meet at $T$. Let $O_1$ and $O_2$ be centers of $\omega$ and $\gamma$ and $O_1O_2 = k$. Let $d(X)$ be the distance of $X$ from the Radical axis. Note that $P_\gamma(X) - P_\omega(X) = 2d(X)k$. Since for any line $d$ is linear we have that $P_\gamma(X) - P_\omega(X)$ is also linear. Note that we need to prove $P_\gamma(T) - P_\omega(T) = 0$ so we need to prove $TC(P_\gamma(A) - P_\omega(A)) + AT(P_\gamma(C) - P_\omega(C)) = 0$ or $AP.AQ.CT = CR.CS.AT$. Note that since $ABCD$ is cyclic we have $\frac{AT}{CT} = \frac{AD}{DC}.\frac{AB}{BC}$ so now we need to prove $\frac{AD}{DC}.\frac{AB}{BC} = \frac{AP.AQ}{CR.CS}$ or $\frac{AP}{AB}.\frac{AQ}{AD}.\frac{BC}{CS}.\frac{DC}{CR} = 1$. $\frac{AP}{AB}.\frac{AQ}{AD}.\frac{BC}{CS}.\frac{DC}{CR} = \frac{sin{ABP}}{sin{APB}}.\frac{sin{ADQ}}{sin{AQD}}.\frac{sin{BSC}}{sin{SBC}}.\frac{sin{DRC}}{sin{CDR}}$ and since $\angle ABP = \angle CDR$ and $\angle ADQ = \angle SRQ = 180 - \angle APB$ and $\angle AQD = \angle PQR = \angle BSC$ and $\angle DRC = \angle QPS = 180 - \angle SBC$ we have $\frac{AP}{AB}.\frac{AQ}{AD}.\frac{BC}{CS}.\frac{DC}{CR} = 1$ as wanted.