In the figure, determine $x$ as a function of $R$.

Question

For reference: In the figure, determine $x$ as a function of $R$.

My progress: I don't know and I can affirm that $\angle FGD$ is $90^o$ ...I think you would need to demonstrate ...but I can't

$\triangle FDB: (R+r)^2 = R^2+(2R-2r)^2\\R^2+2Rr+r^2=R^2+4R^2-4Rr+4r^2\\6Rr-3r^2 = 4Rr\\6R-3r=4R \implies \boxed{r = \frac{2R}{3}}\\\triangle FGD: (R+r)^2 = (x+r)^2+(x+R)^2\\R^2+\frac{4R^2}{3} + \frac{4R^2}{9} = x^2+\frac{4xR}{3}+\frac{4R^2}{9}+x^2+2xR+R^2\\\frac{4R^2}{3} =2x^2+\frac{10xR}{3}\\\frac{4R^2}{3} = \frac{6x^2+10xR}{3}\implies 4R^2 = 6x^2+10xR \implies\\ 3x^2+5xR - 2R^2 = 0\\\therefore \boxed{\color{red}x = \frac{R}{3}}$

Answer 1

If the radius of the other half circle is $r$, you already showed that $ \small \displaystyle r = \frac{2R}{3}$.

Now applying Descartes' theorem,

$ \small \displaystyle \left(- \frac{1}{2R} + \frac{1}{R} + \frac{1}{2R/3} + \frac{1}{x} \right)^2 = 2 \left(\frac{1}{4R^2} + \frac{1}{R^2} + \frac{1}{(2R/3)^2} + \frac{1}{x^2}\right)$

Simplifying, $ \small \displaystyle \left(\frac{2}{R} + \frac{1}{x} \right)^2 = \frac{7}{R^2} + \frac{2}{x^2}$

Solving, $ \small \displaystyle x = \frac{R}{3}$

Answer 2

If you are not sure about the perpendicularity, let's follow the lengthy method.

Call the centres of small circle, small half circle and big half circle $C$, $E$ and $F$ respectively. Also draw the perpendicular lines to $AB$ and $BC$ to meet each side at $D$ and $G$. Thus we see, $CE=r+x$ and $CF=R+x$.

Let $CD=a$, $CG=b$.

We know that the line connecting centres of two half circles goes through the tangent point. Therefore $EF=R+r$.

Now, it's all Pythagoras.

$\triangle EBF\to R^2+(2R-r)^2=(R+r)^2$

$\triangle CBD\to a^2+b^2=(2R-x)^2\qquad(*)$

$\triangle CFD\to a^2+(R-b)^2=(R+x)^2$

$\triangle CEG\to b^2+(2R-r-a)^2=(r+x)^2$

Solving these equations might not be easy, yet possible.

From the first equation we get $r=\frac23R$ as you've already derived. Now the remaining equations can be written as, $$a^2+b^2-x^2=4R^2-4Rx$$ $$a^2+b^2-x^2=2Rb+2Rx$$ $$a^2+b^2-x^2=-\frac43(R^2-2Ra-Rx)$$ From the first and second, we get $$b=2R-3x$$ From the first and the third we get $$a=2R-2x$$ Now substituting these results in (*) we get $$(3x-R)(x-R)=0$$ which gives $x=\dfrac R3$ as the valid solution.