Dynamic Geometry: P98

This is probably the last problem of a series started from Dynamic Geometry: P32 by @Valentin Duringer . There are many calculations in the earlier problems of the series are necessary to solve this problem. To shorten the solution to this problem without sacrificing the details, I have compiled all the calculations for those who wants to referred to in Dynamic Geometry: P32 Series .

Let the radii of the large, cyan and green semicircles be $1$ , $r_1$ , and $r_2$ , the three vertices of the triangles in the left and right yellow circles be $S_1$ , $S_2$ , and $S_3$ , and $T_1$ , $T_2$ , and $T_3$ respectively. From the calculations in the compilation, the coordinates of the vertices of the two triangles are:

$\begin{cases} S_1 \left(\dfrac {2r_2-r_1}{1+r_2}, \dfrac {2r_1\sqrt{r_2}}{1+r_2}\right), & T_1 \left(\dfrac {2r_1-r_2}{1+r_1}, \dfrac {2r_2\sqrt{r_1}}{1+r_1} \right) \\ S_2 \left(r_2 - r_1, 2r_1\sqrt{r_2} \right), & T_2 \left(r_1 - r_2, 2r_2\sqrt{r_1} \right) \\ S_3 \left(\dfrac {r_2-r_1^2}{1-r_1r_2}, \dfrac {2r_1\sqrt{r_2}}{1-r_1r_2} \right), & T_3 \left(\dfrac {r_1-r_2^2}{1-r_1r_2}, \dfrac {2r_2\sqrt{r_1}}{1-r_1r_2} \right) \end{cases}$

Let $r = r_1$ . Since $r_1+r_2 = 1 \implies r_2 = 1 - r_1 = 1 - r$ . Then the coordinates of the two triangles become:

$\begin{cases} S_1 \left(\dfrac {2 - 3r}{2-r}, \dfrac {2r\sqrt{1-r}}{2-r}\right), & T_1 \left(\dfrac {3r-1}{1+r}, \dfrac {2(1-r) \sqrt r}{1+r} \right) \\ S_2 \left(1-2r, 2r\sqrt{1-r} \right), & T_2 \left(2r-1, 2(1-r)\sqrt r \right) \\ S_3 \left(\dfrac {1-r-r^2}{1-r+r^2}, \dfrac {2r\sqrt{1-r}}{1-r+r^2} \right), & T_3 \left(\dfrac {r-(1-r)^2}{1-r+r^2}, \dfrac {2(1-r) \sqrt r}{1-r+r^2} \right) \end{cases}$

By shoelace formula , the areas of the two triangles are:

$\begin{cases} A_S = \dfrac 12 \left| \begin{vmatrix} 1 & 1 & 1 \\ \dfrac {2 - 3r}{2-r} & 1-2r & \dfrac {1-r-r^2}{1-r+r^2} \\ \dfrac {2r\sqrt{1-r}}{2-r} & 2r\sqrt{1-r} & \dfrac {2r\sqrt{1-r}}{1-r+r^2} \end{vmatrix} \right| = \dfrac {2r^2(1-r)^\frac 52}{(2-r)(r^2-r+1)} \\ A_T = \dfrac 12 \left| \begin{vmatrix} 1 & 1 & 1 \\ \dfrac {3r-1}{1+r} & 2r-1 & \dfrac {r-(1-r)^2}{1-r+r^2} \\ \dfrac {2(1-r) \sqrt r}{1+r} & 2(1-r)\sqrt r & \dfrac {2(1-r) \sqrt r}{1-r+r^2} \end{vmatrix} \right| = \dfrac {2r^\frac 52(1-r)^2} {1+r^3} \end{cases}$

When $\dfrac {A_S}{A_T} = \dfrac {2328}{1565}$ ,

$\begin{aligned} \frac {2r^2(1-r)^\frac 52(1+r^3)}{2r^\frac 52(1-r)^2(r-2)(r^2-r+1)} & = \frac {2328}{1565} \\ \frac {(1+r)\sqrt{1-r}}{(2-r)\sqrt r} & = \frac {2328}{1565} \end{aligned}$

Solving the equation above, we get $r=r_1 = \dfrac {25}{169} \implies r_2 = 1 - \dfrac {25}{169} = \dfrac {144}{169}$ . Therefore $\dfrac {r_1}{r_2} = \dfrac {25}{144}$ $\implies \sqrt q - \sqrt p = 12 - 5 = \boxed 7$ .