# Distance between incenter and centroid

i have problem about solving this homework In any given triangle find the distance between the centroid and the incenter . I have no idea which properties to use to find it can you help me please :( Thanks.

• I cannot think of any well-known formula for that. Can you give more of the triangle you are interested in. – almagest Jun 20 '16 at 16:48

This is simple to tackle through exact barycentric coordinates: $$G=\frac{A+B+C}{3},\qquad I=\frac{aA+bB+cC}{a+b+c} \tag{1}$$ give: $$3(a+b+c)(G-I) = (b+c-2a) A + (a+c-2b) B + (a+b-2c) C \tag{2}$$ and by assuming that the origin is in the circumcenter $O$ we get:
$$9(a+b+c)^2\left\|G-I\right\|^2 = 6R^2(a^2+b^2+c^2-ab-ac-bc)+\ldots \tag{3}$$ that boils down to a quite complicated expression.
• I think your $\ldots$ can be substantially simplified. One may use the law of cosines to express the dot product of $A$ and $B$ as $2A\cdot B =2R^2-c^2$, and similarly for the other two products. With this in mind, the coefficient of the $R^2$ term becomes $$a'^2+b'^2+c'^2+2a'b'+2b'c'+2a'c'=(a'+b'+c')^2$$ where $a',b',c'$ are the coefficients of $A,B,C$ respectively in equation (2). But $a'+b'+c'=0$, so the circumradius drops out entirely. What remains gives the expression given in the link. – Semiclassical Mar 27 '19 at 16:14
Let $a,b,c -$ a sides of the triangle and $d -$ the distance between the centroid and the incenter . Then