Dynamic Geometry: P127

Geometry Level pending

The diagram shows a black circle. A horizontal black chord is drawn creating two circular segments, their respective heights are 4 4 and 9 9 . 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 12 17 \dfrac{12}{17} , the tangent of one orange angle can be expressed as p q \dfrac{p}{q} , where p p and q q are coprime positive integers. Find p + q p+q .


The answer is 73.

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1 solution

From the compiled calculations in Dynamic Geometry: P96 Series , we know that the dividing chord A B = 12 AB=12 , A C B = π tan 1 12 5 \angle ACB = \pi - \tan^{-1} \frac {12}5 , and A C B = tan 1 12 5 \angle AC'B = \tan^{-1} \frac {12}5 .

Let C A B = C A B = θ \angle CAB = \angle C'AB = \theta , the radius of the upper incircle be r r , and the radius of the lower incircle be r 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 C A B 2 + r cot C B A 2 = A B r cot θ 2 + r cot tan 1 12 5 θ 2 = 12 Let t = tan θ 2 r t + r 1 + 2 3 t 2 3 t = 12 Note that 1 2 tan 1 12 5 = 2 3 r = 12 1 t + 3 + 2 t 2 3 t = 6 t ( 2 3 t ) 1 + t 2 \begin{aligned} r \cot \frac {\angle CAB}2 + r \cot \frac {\angle CBA}2 & = AB \\ r \cot \frac \theta 2 + r \cot \frac {\tan^{-1} \frac {12}5- \theta}2 & = 12 & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ \frac rt + r \cdot \frac {1+\frac 23t}{\frac 23-t} & = 12 & \small \blue{\text{Note that } \frac 12 \tan^{-1} \frac {12}5 = \frac 23} \\ \implies r & = \frac {12}{\frac 1t + \frac {3+2t}{2-3t}} = \frac {6t(2-3t)}{1+t^2} \end{aligned}

Similarly for the lower incircle,

r cot θ 2 + r cot π tan 1 12 5 θ 2 = 12 r t + r 2 3 + t 1 2 3 t = 12 r = 12 1 t + 2 + 3 t 3 2 t = 4 t ( 3 2 t ) 1 + t 2 \begin{aligned} r' \cot \frac \theta 2 + r' \cot \frac {\pi - \tan^{-1}\frac {12}5-\theta }2 & = 12 \\ \frac {r'}t + r' \cdot \frac {\frac 23+t}{1-\frac 23t} & = 12 \\ \implies r' & = \frac {12}{\frac 1t + \frac {2+3t}{3-2t}} = \frac {4t(3-2t)}{1+t^2} \end{aligned}

When

r r = 12 17 3 ( 2 3 t ) 2 ( 3 2 t ) = 12 17 34 51 t = 24 16 t t = 10 35 = 2 7 tan θ = 2 t 1 t 2 = 2 2 7 1 4 49 = 28 45 \begin{aligned} \frac r{r'} & = \frac {12}{17} \\ \frac {3(2-3t)}{2(3-2t)} & = \frac {12}{17} \\ 34 - 51t & = 24 - 16t \\ \implies t & = \frac {10}{35} = \frac 27 \\ \tan \theta & = \frac {2t}{1-t^2} = \frac {2\cdot \frac 27}{1-\frac 4{49}} = \frac {28}{45} \end{aligned}

Therefore p + q = 28 + 45 = 73 p+q = 28 + 45 = \boxed{73} .

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