Dynamic Geometry: P91

Let the radius of the large semicircle be $1$ and its diameter $FG$ , the centers and radii of the cyan and green be $A$ and $D$ , and $r$ and $r_1$ respectively, the triangle in question be $ABC$ , $DE$ be tangent to $BC$ , and $GB = k$ .

We note that $r+r_1 = 1$ and that $\triangle ABC$ and $\triangle BDE$ are similar. Then

$\begin{aligned} \frac {AB}{DB} & = \frac {AC}{DE} \\ \frac {r+2r_1+k}{r_1+k} &= \frac r{r_1} & \small \blue{\text{Since }r+r_1 = 1} \\ \frac {2-r+k}{1-r+k} & = \frac r{1-r} \\ 2kr-k & = 2r^2 - 4r+2 \\ \implies k & = \frac {2(1-r)^2}{2r-1} \end{aligned}$

Then $AB = AG + GB = 2-r+k = \dfrac r{2r-1}$ . By Pythagorean theorem , $BC = \sqrt{AB^2-AC^2} = \sqrt{\dfrac {r^2}{(2r-1)^2}-r^2} = \dfrac {2r\sqrt{r-r^2}}{2r-1}$ . Then the ratio of perimeter to area of $\triangle ABC$ is:

$\begin{aligned} \frac {AC+AB+BC}{AC \cdot BC/2} & = \frac {2873}{360} \\ \frac {r+\frac r{2r-1}+\frac {2r\sqrt{r-r^2}}{2r-1}}{r \cdot \frac {r\sqrt{r-r^2}}{2r-1}} & = \frac {2873}{360} \\ \frac {2r - 1 + 1 + 2\sqrt{r-r^2}}{2r\sqrt{r-r^2}} & = \frac {2873}{360} \\ \frac {r + \sqrt{r-r^2}}{r\sqrt{r-r^2}} & = \frac {2873}{720} \end{aligned}$

Solving for $r$ , we get $r = \dfrac {144}{169}$ and $r = \dfrac {1865}{5746} \pm \dfrac {5\sqrt{41785}}{5746}$ . Taking the rational value, then the ratio of radii of the two smaller semicircles

$\frac r{r_1} = \frac {\frac {144}{169}}{1-\frac {144}{169}} = \frac {144}{25}$

Therefore $\sqrt{144+25} = \sqrt{169} = \boxed{13}$ .