Two Semicircles in a Triangle

Let the radii of the large and small semicircles be $R$ and $r$ respectively, and $\angle A = \theta = \tan^{-1} \dfrac x5$ . Then

$R + R \csc \theta = 5 \implies R = \frac 5{\csc \theta + 1} = \frac {x(\sqrt{25+x^2}-x)}5$

From similar triangles,

$\begin{aligned} \frac Rr & = \frac {r\csc \theta + r + R}{r\csc \theta} \\ R & = r + r \sin \theta + R\sin \theta \\ \implies r & = \frac {1-\sin \theta}{1+\sin \theta} \cdot R = \frac {x(\sqrt{25+x^2}-x)^3}{125} \end{aligned}$

Then the ratio of the two semicircles and the triangle $\rho = \dfrac {\pi(R^2+r^2)/2}{5x/2}$ , and

$\begin{aligned} \frac {5\rho}\pi & = \frac {x(\sqrt{25+x^2} - x)^2}{25} + \frac {x(\sqrt{25+x^2}-x)^6}{15625} \\ \frac {78125}\pi \rho & = 625x(\sqrt{25+x^2} - x)^2 + x (\sqrt{25+x^2} - x)^6 \\ \frac {78125}\pi \cdot \frac {d\rho}{dx} & = \frac {625(\sqrt{25+x^2}-2x)(\sqrt{25+x^2}-x)^2}{\sqrt{25+x^2}} + \frac {(\sqrt{25+x^2}-6x)(\sqrt{25+x^2}-x)^6}{\sqrt{25+x^2}} \end{aligned}$

Putting $\dfrac {d\rho}{dx} =0$ ,

$\begin{aligned} \blue{\left(\sqrt{25+x^2}-x \right)}^2 \left(625 \left(\sqrt{25+x^2}-2x\right) + \left(\sqrt{25+x^2} -6x\right)\left(\sqrt{25+x^2} -x\right)^4\right) & = 0 & \small \blue{\text{For } \sqrt{25+x^2}-x = 0} \\ \implies 625 \left(\sqrt{25+x^2}-2x\right) + \left(\sqrt{25+x^2} -6x\right)\left(\sqrt{25+x^2} -x\right)^4 & = 0 & \small \blue{\implies x \to \infty} \end{aligned}$

Solving the equation, we get $x \approx \boxed{1.997}$ .

Let the radii of the larger and smaller semicircles be r and αr, respectively. Because (loosely speaking) we could fill up the line CA with diametres of smaller and smaller semicircles, all touching AB, forming a geometric series, we see that $2r(1+α+α^2+...)=5$ and hence $α=1-\frac{2}{5}r$

The two semicircles now have a combined area of $A_{semicircles} = \frac{1}{2}πr^2(1+α^2)=\frac{1}{2}πr^2(1+1-\frac{4}{5}r+\frac{4}{25}r^2)$ Defining $t=r/5$ this rewrites a bit cleaner as $A_{semicircles} = 25πt^2(1-2t+2t^2)$

Let the centre of the larger circle be D and let E be the point where the larger circle touches the line AD. The triangles ABC and ADE are similar, so that $x:\sqrt{x^2+5^2}=r:(5-r)$

From this we can find $x=\frac{r}{\sqrt{1-2r/5}}=\frac{5t}{\sqrt{1-2t}}$ .

The triangle has area $A_{triangle}=5x/2=\frac{25t}{2\sqrt{1-2t}}$

We can express the Area ratio as a function of t:

$R(t)=\frac{A_{semicircles}}{A_{triangle}}=πt(1-2t+2t^2)2\sqrt{1-2t}$

To find a stationary point, we set $R'(t) = 0$ which results in the polynomial $-14t^3+16t^2-7t+1=0$ , with real root $t=0.27059...$ Plugging back in $x=\frac{5t}{\sqrt{1-2t}}$ gives $x=1.9973...$

Let radius of circles be a >b>0 and write the following relations

(1). 2a+b+√(b^2+c^2)=5.

(2). x=5*b/c

(3). b= (c a)/[c+2 √(a*b)]

(4). Maximize ratio w= (5x)/[π(a^2+b^2)] to find x=?

Use WolframAlpha to find x~2