Dynamic Geometry: P85

Geometry Level pending

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 123 43 \dfrac{123}{43} , the ratio of the green circle's radius to the radius of the cyan circle 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 158.

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1 solution

Chew-Seong Cheong
Mar 19, 2021

Let the center of the large semicircle be O O , its radius 1 1 , and diameter A B AB , the center of the pink circle be C C , C O B = θ \angle COB = \theta , and t = tan θ 2 t = \tan \frac \theta 2 . Let the radii of the pink circle, green semicircle, cyan semicircle, red circle, and orange circle be r r , r 1 r_1 , r 2 r_2 , r 3 r_3 , and r 4 r_4 . Using half-angle tangent substitution as in the previous problem Dynamic Geometry: P80 , we get:

{ r = 2 t ( 1 + t ) 2 r 1 = 1 1 + 2 t r 2 = t 2 + t \begin{cases} r = \dfrac {2t}{(1+t)^2} \\ r_1 = \dfrac 1{1+2t} \\ r_2 = \dfrac t{2+t} \end{cases}

With radii of three other tangent circles known, we can find r 3 r_3 and r 4 r_4 , using Descarte's theorem . Note that the curvature of the unit semicircle is negative.

{ 1 r 3 = 1 r + 1 r 1 1 + 2 1 r r 1 1 r 1 r 1 r 3 = 2 t 9 t 2 + 2 t + 1 1 r 4 = 1 r + 1 r 2 1 + 2 1 r r 2 1 r 1 r 2 r 4 = 2 t t 2 + 2 t + 9 \begin{cases} \dfrac 1{r_3} = \dfrac 1r + \dfrac 1{r_1} - 1 + 2 \sqrt{\dfrac 1{rr_1} - \dfrac 1r - \dfrac 1{r_1}} & \implies r_3 = \dfrac {2t}{9t^2+2t+1} \\ \dfrac 1{r_4} = \dfrac 1r + \dfrac 1{r_2} - 1 + 2 \sqrt{\dfrac 1{rr_2} - \dfrac 1r - \dfrac 1{r_2}} & \implies r_4 = \dfrac {2t}{t^2+2t+9} \end{cases}

When r 3 r 4 = 123 43 \dfrac {r_3}{r_4} = \dfrac {123}{43} , then

t 2 + 2 t + 9 9 t 2 + 2 t + 1 = 123 43 133 t 2 + 20 t 33 = 0 ( 19 t + 11 ) ( 7 t 3 ) = 0 Since θ < 18 0 t = 3 7 \begin{aligned} \frac {t^2+2t+9}{9t^2+2t+1} & = \frac {123}{43} \\ 133t^2 + 20t - 33 & = 0 \\ (19t+11)(7t-3) & = 0 & \small \blue{\text{Since }\theta < 180^\circ} \\ \implies t & = \frac 37 \end{aligned}

r 1 r 2 = 1 1 + 2 t 2 + t t = 7 13 17 3 = 119 39 \begin{aligned} \implies \frac {r_1}{r_2} & = \frac 1{1+2t} \cdot \frac {2+t}t = \frac 7{13} \cdot \frac {17}3 = \frac {119}{39} \end{aligned}

Therefore p + q = 119 + 39 = 158 p+q=119+39 = \boxed{158} .

Very elegant solution !

Valentin Duringer - 2 months, 3 weeks ago

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Yes, I was also surprised. Though most people geometry fans may not like trigo, but it is so convenient and unnecessarily not elegant.

Chew-Seong Cheong - 2 months, 3 weeks ago

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