, the ratio of the green circle's radius to the radius of the cyan circle can be expressed as , where and are coprime positive integers. Find .
The diagram shows a black semicircle. The cyan and green semicircle are growing and shrinking freely on the black semicirle's diameter. The purple circle is internally tangent to the black semicircle and tangent to both green and cyan semicircle. The red and orange circles are internally tangent to the black semicircle and tangent to the purple circle and to one of the bottom semicircles. When the ratio of the red circle's radius to the radius of the orange circle is equal to
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Let the center of the large semicircle be O , its radius 1 , and diameter A B , the center of the pink circle be C , ∠ C O B = θ , and t = tan 2 θ . Let the radii of the pink circle, green semicircle, cyan semicircle, red circle, and orange circle be r , r 1 , r 2 , r 3 , and r 4 . Using half-angle tangent substitution as in the previous problem Dynamic Geometry: P80 , we get:
⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ r = ( 1 + t ) 2 2 t r 1 = 1 + 2 t 1 r 2 = 2 + t t
With radii of three other tangent circles known, we can find r 3 and r 4 , using Descarte's theorem . Note that the curvature of the unit semicircle is negative.
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ r 3 1 = r 1 + r 1 1 − 1 + 2 r r 1 1 − r 1 − r 1 1 r 4 1 = r 1 + r 2 1 − 1 + 2 r r 2 1 − r 1 − r 2 1 ⟹ r 3 = 9 t 2 + 2 t + 1 2 t ⟹ r 4 = t 2 + 2 t + 9 2 t
When r 4 r 3 = 4 3 1 2 3 , then
9 t 2 + 2 t + 1 t 2 + 2 t + 9 1 3 3 t 2 + 2 0 t − 3 3 ( 1 9 t + 1 1 ) ( 7 t − 3 ) ⟹ t = 4 3 1 2 3 = 0 = 0 = 7 3 Since θ < 1 8 0 ∘
⟹ r 2 r 1 = 1 + 2 t 1 ⋅ t 2 + t = 1 3 7 ⋅ 3 1 7 = 3 9 1 1 9
Therefore p + q = 1 1 9 + 3 9 = 1 5 8 .