Problem: Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$ . The three points $C$ $Y$ $D$ are collinear, $XC = 67$ $XY = 47$ , and $XD = 37$ . Find $AB^2$

Solution: First, we note that as $\triangle XDY$ and $\triangle XYC$ have bases along the same line, $\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{DY}{YC}$ . We can also find the ratio of their areas using the circumradius area formula. If $R_1$ is the radius of $\omega_1$ and if $R_2$ is the radius of $\omega_2$ , then \[\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{(37\cdot 47\cdot DY)/(4R_1)}{(47\cdot 67\cdot YC)/(4R_2)}=\frac{37\cdot DY\cdot R_2}{67\cdot YC\cdot R_1}.\] Since we showed this to be $\frac{DY}{YC}$ , we see that $\frac{R_2}{R_1}=\frac{67}{37}$
We extend $AD$ and $BC$ to meet at point $P$ , and we extend $AB$ and $CD$ to meet at point $Q$ as shown below. [asy] size(200); import olympiad; real R1=45,R2=67*R1/37; real m1=sqrt(R1^2-23.5^2); real m2=sqrt(R2^2-23.5^2); pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5); draw(circle(o1,R1)); draw(circle(o2,R2)); pair q=(-R1/(R2-R1)*o2.x,0); pair a=tangent(q,o1,R1,2); pair b=tangent(q,o2,R2,2); pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0]; pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1]; pair p=extension(a,d,b,c); dot(q^^a^^b^^x^^y^^c^^d^^p); draw(q--b^^q--c); draw(p--d^^p--c^^x--y); draw(a--y^^b--y); draw(d--x--c); label("$A$",a,NW,fontsize(8)); label("$B$",b,NE,fontsize(8)); label("$C$",c,SE,fontsize(8)); label("$D$",d,SW,fontsize(8)); label("$X$",x,2*WNW,fontsize(8)); label("$Y$",y,3*S,fontsize(8)); label("$P$",p,N,fontsize(8)); label("$Q$",q,W,fontsize(8)); [/asy] As $ABCD$ is cyclic, we know that $\angle BCD=180-\angle DAB=\angle BAP$ . But then as $AB$ is tangent to $\omega_2$ at $B$ , we see that $\angle BCD=\angle ABY$ . Therefore, $\angle ABY=\angle BAP$ , and $BY\parallel PD$ . A similar argument shows $AY\parallel PC$ . These parallel lines show $\triangle PDC\sim\triangle ADY\sim\triangle BYC$ . Also, we showed that $\frac{R_2}{R_1}=\frac{67}{37}$ , so the ratio of similarity between $\triangle ADY$ and $\triangle BYC$ is $\frac{37}{67}$ , or rather \[\frac{AD}{BY}=\frac{DY}{YC}=\frac{YA}{CB}=\frac{37}{67}.\] We can now use the parallel lines to find more similar triangles. As $\triangle AQD\sim \triangle BQY$ , we know that \[\frac{QA}{QB}=\frac{QD}{QY}=\frac{AD}{BY}=\frac{37}{67}.\] Setting $QA=37x$ , we see that $QB=67x$ , hence $AB=30x$ , and the problem simplifies to finding $30^2x^2$ . Setting $QD=37^2y$ , we also see that $QY=37\cdot 67y$ , hence $DY=37\cdot 30y$ . Also, as $\triangle AQY\sim \triangle BQC$ , we find that \[\frac{QY}{QC}=\frac{YA}{CB}=\frac{37}{67}.\] As $QY=37\cdot 67y$ , we see that $QC=67^2y$ , hence $YC=67\cdot30y$
Applying Power of a Point to point $Q$ with respect to $\omega_2$ , we find \[67^2x^2=37\cdot 67^3 y^2,\] or $x^2=37\cdot 67 y^2$ . We wish to find $AB^2=30^2x^2=30^2\cdot 37\cdot 67y^2$
Applying Stewart's Theorem to $\triangle XDC$ , we find \[37^2\cdot (67\cdot 30y)+67^2\cdot(37\cdot 30y)=(67\cdot 30y)\cdot (37\cdot 30y)\cdot (104\cdot 30y)+47^2\cdot (104\cdot 30y).\] We can cancel $30\cdot 104\cdot y$ from both sides, finding $37\cdot 67=30^2\cdot 67\cdot 37y^2+47^2$ . Therefore, \[AB^2=30^2\cdot 37\cdot 67y^2=37\cdot 67-47^2=\boxed{270}.\]

Answer:  270