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

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.

• Is $\triangle$ supposed to be the area of $ABC$? Also, can you include a diagram, even at least a rough sketch. – John Glenn Aug 18 '18 at 7:06

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 $$2\sin^2\frac{A}{2}=1-\cos A=1-\frac{b^2+c^2-a^2}{2bc}$$.

• Why this parabola contains center of wanted circle? – mathlover Aug 19 '18 at 1:02
• @mathlover: it is already explained in the post. If we name $\ell$ the horizontal bottom line, we have that the distance between $J$ and $\ell$ has to be equal to the distance between $J$ and $O$. In particular $J$ lies on a parabola with focus at $O$ going through $B$ and $C$. – Jack D'Aurizio Aug 19 '18 at 1:19