and . In each circular segment, we inscribe a triangle. In both triangles we draw the incircle. The triangles are evolving so that the orange angles are always the same. When the ratio of the blue circle's radius to the radius of the red circle is equal to , the tangent of one orange angle can be expressed as , where and are coprime positive integers. Find .
The diagram shows a black circle. A horizontal black chord is drawn creating two circular segments, their respective heights are
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From the compiled calculations in Dynamic Geometry: P96 Series , we know that the dividing chord A B = 1 2 , ∠ A C B = π − tan − 1 5 1 2 , and ∠ A C ′ B = tan − 1 5 1 2 .
Let ∠ C A B = ∠ C ′ A B = θ , the radius of the upper incircle be r , and the radius of the lower incircle be r ′ . Using the fact that the line joining the incenter and a vertex of the triangle bisects the vertex angel, we have for the upper incircle:
r cot 2 ∠ C A B + r cot 2 ∠ C B A r cot 2 θ + r cot 2 tan − 1 5 1 2 − θ t r + r ⋅ 3 2 − t 1 + 3 2 t ⟹ r = A B = 1 2 = 1 2 = t 1 + 2 − 3 t 3 + 2 t 1 2 = 1 + t 2 6 t ( 2 − 3 t ) Let t = tan 2 θ Note that 2 1 tan − 1 5 1 2 = 3 2
Similarly for the lower incircle,
r ′ cot 2 θ + r ′ cot 2 π − tan − 1 5 1 2 − θ t r ′ + r ′ ⋅ 1 − 3 2 t 3 2 + t ⟹ r ′ = 1 2 = 1 2 = t 1 + 3 − 2 t 2 + 3 t 1 2 = 1 + t 2 4 t ( 3 − 2 t )
When
r ′ r 2 ( 3 − 2 t ) 3 ( 2 − 3 t ) 3 4 − 5 1 t ⟹ t tan θ = 1 7 1 2 = 1 7 1 2 = 2 4 − 1 6 t = 3 5 1 0 = 7 2 = 1 − t 2 2 t = 1 − 4 9 4 2 ⋅ 7 2 = 4 5 2 8
Therefore p + q = 2 8 + 4 5 = 7 3 .