Extreme Value with Triangle

According to Euler's theorem for a triangle , the distance between circumcenter $O$ and incenter $I$ is given by $\overline{OI}^2 = R(R - 2r)$ . This implies that $R \ge 2r$ . Therefore, the minimum $\dfrac Rr = \boxed 2$ .

Let $D$ be the second intersection of $\overline{AI}$ with the circumcircle $\omega$ of $\Delta ABC$ . We need to prove that $2Rr = R^2 - OI^2 = |pow(I,\omega)| = \overline{AI} \cdot \overline{ID}$ . By Incenter-Excenter Lemma, we know $\overline{DI} = \overline{DB}$ . So, we need to find similar triangles that contain the sides $\overline{AI}, \overline{BD}, r$ and $2R.$ Let $E$ be the diametrically oppositie point of $D$ on $\omega$ ; Thus $\overline{ED} = 2R$ and $\angle EBD = 90^{\circ}$ . Let $F$ be the tangent point of incircle with side AB. Thus $\overline{IF}= r$ and $\angle IFA = 90^{\circ}$ . Since $\angle FAI \equiv \angle BAD = \angle BED$ , the right triangle $\Delta AIF$ and $\Delta EDB$ are similar. Therefore,

$\dfrac{AI}{IF} = \dfrac{ED}{DB}$

$\overline{AI} \cdot \overline{ID} = \overline{AI} \cdot \overline{ED} = \overline{ED} \cdot \overline{IF} = 2Rr$

It is easy to deduce that $R^2 - 2Rr \geq 0 \implies R \geq 2r$ Equality occurs iff $O = I$ e.g. Equilateral Triangle. And we are done.

The Euler's inequality tells us $R\ge 2r,$ which holds with equality only for equilateral triangles, making the answer $\boxed 2$ .