Problem:
Two externally tangent circles ω1 and ω2 have centers O1 and O2, respectively. A third circle Ω passing through O1 and O2 intersects ω1 at B and C and ω2 at A and D, as shown. Suppose that AB=2,O1O2=15, CD=16, and ABO1CDO2 is a convex hexagon. Find the area of this hexagon.
Solution:
First observe that AO2=O2D and BO1=O1C. Let points A′ and B′ be the reflections of A and B, respectively, across the diameter of Ω that is the perpendicular bisector of O1O2. Thus A′ and B′ are on Ω with A′B′=AB=2. Then quadrilaterals ABO1O2 and A′B′O2O1 are congruent, so hexagons ABO1CDO2 and B′A′O1CDO2 have the same area. Furthermore, because A′O1=DO2 and O1C=O2B′, it follows that B′D=A′C and quadrilateral B′A′CD is an isosceles trapezoid.
Because A′O1=DO2, quadrilateral A′O1DO2 is an isosceles trapezoid. In turn, A′D=O1O2=15, and similarly B′C=15. Thus Ptolemy's Theorem applied to B′A′CD yields B′D⋅A′C+2⋅16=152, whence B′D=A′C=193. Let α=∠B′A′D. The Law of Cosines applied to △B′A′D yields
cosα=2⋅2⋅15152+22−(193)2=53
and hence sinα=54. Let R be the point on line A′B′ such that A′R⊥DR. Because △A′RD is a right triangle whose hypotenuse has length 15 with cos(∠DA′R)=53, it follows that A′R=9 and DR=12. In particular, the distance from A′B′ to CD is DR=12, which implies that the area of B′A′CD is 21⋅12⋅(2+16)=108.
Now let O1C=O2B′=r1 and O2D=O1A′=r2. The tangency of circles ω1 and ω2 implies r1+r2=15. Furthermore, ∠B′O2D and ∠B′A′D are opposite angles in cyclic quadrilateral A′B′O2D, which implies that the measure of ∠B′O2D is 180∘−α. Therefore the Law of Cosines applied to △B′O2D yields
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