Proving radius of circle $\dfrac{\triangle}{a}\tan^2\dfrac{A}{2}$

Question

If a circle be drawn touching the inscribed and circumscribed circles of a $\triangle ABC$ and the side $BC$ externally, prove that its radius is: $$r=\dfrac{\triangle}{a}\tan^2\dfrac{A}{2}$$

I tried using triangle formed by circumcenter, incenter and center of above circle as I know all the sides in terms of $r$ to use cosine rule but I don't know any angles.

Please help!

Answer 1

The center $J$ of the wanted circle can be constructed by intersecting a line and a parabola, since by setting $s=JL$ we have $JO=R-s$.

If we take $B$ as the origin and $BC$ as the $x$-axis, the equation of the wanted parabola is $y=kx(x-a)$. Since the distance of $O$ from $BC$ is $R\cos A$, the vertex lies at $\left(\frac{a}{2},-\frac{R}{2}(1-\cos A)\right)$ and

$$k=\frac{2R}{a^2}(1-\cos A)=\frac{1-\cos A}{2R\sin^2 A}=\frac{\sin^2\frac{A}{2}}{R\sin^2 A}=\frac{1}{2R\cos^2\frac{A}{2}}.$$

Of course $BL=\frac{a+c-b}{2}$, hence

$$ s = \frac{1}{2R\cos^2\frac{A}{2}}\cdot \frac{a+c-b}{2}\cdot\frac{a-c+b}{2} $$

and I guess you can take it from here, by just recalling that $abc=4R\Delta$ and $2\sin^2\frac{A}{2}=1-\cos A=1-\frac{b^2+c^2-a^2}{2bc}$.