Problem: Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$ , respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$ , as shown. Suppose that $AB = 2$ $O_1O_2 = 15$ $CD = 16$ , and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon. [asy] import geometry; size(10cm); point O1=(0,0),O2=(15,0),B=9*dir(30); circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B); point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0]; filldraw(A--B--O1--C--D--O2--cycle,0.2*fuchsia+white,black); draw(w1); draw(w2); draw(O1--O2,dashed); draw(o); dot(O1); dot(O2); dot(A); dot(D); dot(C); dot(B); label("$\omega_1$",8*dir(110),SW); label("$\omega_2$",5*dir(70)+(15,0),SE); label("$O_1$",O1,W); label("$O_2$",O2,E); label("$B$",B,N+1/2*E); label("$A$",A,N+1/2*W); label("$C$",C,S+1/4*W); label("$D$",D,S+1/4*E); label("$15$",midpoint(O1--O2),N); label("$16$",midpoint(C--D),N); label("$2$",midpoint(A--B),S); label("$\Omega$",o.C+(o.r-1)*dir(270)); [/asy]

Solution: Let points $A'$ and $B'$ be the reflections of $A$ and $B,$ respectively, about the perpendicular bisector of $O_1 O_2.$ \[B'O_2  = BO_1 = O_1 P = O_1 C,\] \[A'O_1  = AO_2 = O_2 P = O_2 D.\] We establish the equality of the arcs and conclude that the corresponding chords are equal \[\overset{\Large\frown} {CO_1} + \overset{\Large\frown} {A'O_1} +\overset{\Large\frown} {A'B'} = \overset{\Large\frown} {B'O_2} +\overset{\Large\frown} {A'O_1} +\overset{\Large\frown} {A'B'} =\overset{\Large\frown} {B'O_2} +\overset{\Large\frown} {DO_2} +\overset{\Large\frown} {A'B'}\] \[\implies A'D = B'C = O_1 O_2 = 15.\] Similarly $A'C = B'D \implies \triangle A'CO_1 = \triangle B'DO_2.$
Ptolemy's theorem on $A'CDB'$ yields \[B'D \cdot A'C + A'B' \cdot CD = A'D \cdot B'C \implies\] \[B'D^2 + 2 \cdot 16 = 15^2  \implies B'D = A'C = \sqrt{193}.\] The area of the trapezoid $A'CDB'$ is equal to the area of an isosceles triangle with sides $A'D = B'C = 15$ and $A'B' + CD = 18.$
The height of this triangle is $\sqrt{15^2-9^2} = 12.$ The area of $A'CDB'$ is $108.$
\[\sin \angle B'CD  = \frac{12}{15} = \frac{4}{5},\] \[\angle B'CD +  \angle B'O_2 D = 180^o \implies \sin \angle B'O_2 D =  \frac{4}{5}.\]
Denote $\angle B'O_2 D = 2\alpha.$ $\angle B'O_2 D > \frac{\pi}{2},$ hence $\cos \angle B'O_2 D = \cos 2\alpha = -\frac{3}{5}.$ \[\tan \alpha =\frac { \sin 2 \alpha}{1+\cos 2 \alpha} = \frac {4/5}{1 - 3/5}=2.\]
Semiperimeter of $\triangle B'O_2 D$ is $s = \frac {15 + \sqrt{193}}{2}.$
The distance from the vertex $O_2$ to the tangent points of the inscribed circle of the triangle $B'O_2 D$ is equal $s – B'D = \frac{15 – \sqrt{193}}{2}.$
The radius of the inscribed circle is $r = (s – B'D) \tan \alpha.$
The area of triangle $B'O_2 D$ is $[B'O_2 D] = sr  = s (s – B'D) \tan \alpha = \frac {15^2 – 193}{2} = 16.$
The hexagon $ABO_1 CDO_2$ has the same area as hexagon $B'A'O_1 CDO_2.$
The area of hexagon $B'A'O_1 CDO_2$ is equal to the sum of the area of the trapezoid $A'CDB'$ and the areas of two equal triangles $B'O_2 D$ and $A'O_1 C,$ so the area of the hexagon $ABO_1 CDO_2$ is \[108 + 16 + 16 = \boxed{140}.\]

Answer:  140