# The radii of circles inscribed between a triangle incircle and its vertices

This follows directly from the question "Relationship between circles touching incircle" posed by evil999man April 22, 2014, which, it seems to me, did not receive the answer he was seeking. The challenge I set myself was:

Consider the radii $$a$$, $$b$$, $$c$$, of the circles inscribed between the incircle of a triangle, radius $$r$$, and its vertices. Consider the claim that $$a + b + c \geq r$$.

The radii are readily expressed in terms of the angles of the triangle, but I've got bogged down in trigonometry.

• $$r$$ be the inradius and $$\rho_1,\rho_2,\rho_3$$ be the radii of the $$3$$ circles.
• $$\theta_k$$ be the angle of triangle at the vertex containing circle with radius $$\rho_k$$.
• $$\phi_k = \frac{\pi - \theta_k}{4}$$ and $$t_k = \tan\phi_k$$.
It is not hard to work out: $$\frac{\rho_k}{r} = \frac{1 - \sin\frac{\theta_k}{2}}{1 + \sin\frac{\theta_k}{2}} = \tan(\phi_k)^2 = t_k^2$$
Notice $$\sum_{k=1}^3\phi_k = \frac14\left(3\pi - \sum_{k=1}^3\theta_k\right) = \frac{\pi}{2}$$ The addition formula of tangent for $$3$$ angles tell us $$\frac{t_1 + t_2 + t_3 - t_1 t_2 t_3}{ 1 - t_1 t_2 - t_2 t_3 - t_3 t_1} = \tan\left(\sum_{k=1}^3\phi_k\right) = \infty$$ This implies $$t_1 t_2 + t_2 t_3 + t_3 t_1 = 1$$. As a result, \begin{align}\frac1r \sum_{k=1}^3\rho_k = \sum_{k=1}^3 t_k^2 &= \frac12(t_1^2 + t_2^2) + \frac12(t_2^2 + t_3^2) + \frac12(t_3^2 + t_1^2)\\ &\ge t_1t_2 + t_2t_3 + t_3t_1 = 1\end{align}