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: $$r=\dfrac{\triangle}{a}\tan^2\dfrac{A}{2}$$This is just a rough sketch.

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!

  • 2
    $\begingroup$ Is $\triangle$ supposed to be the area of $ABC$? Also, can you include a diagram, even at least a rough sketch. $\endgroup$
    – John Glenn
    Commented Aug 18, 2018 at 7:06

1 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 $JO=R-s$.

enter image description here

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

  • $\begingroup$ Why this parabola contains center of wanted circle? $\endgroup$
    – mathlover
    Commented Aug 19, 2018 at 1:02
  • $\begingroup$ @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 $C$. $\endgroup$ Commented Aug 19, 2018 at 1:19

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