8. A triangle has side lengths 18,24 , and 30 . Find the area of the triangle whose vertices are the incenter, circumcenter, and centroid of the original triangle.
Solution:
3
There are many solutions to this problem, which is straightforward. The given triangle is a right 3-4-5 triangle, so the circumcenter is the midpoint of the hypotenuse. Coordinatizing for convenience, put the vertex at $(0,0)$ and the other vertices at $(0,18)$ and $(24,0)$. Then the circumcenter is $(12,9)$. The centroid is at one-third the sum of the three vertices, which is $(8,6)$. Finally, since the area equals the inradius times half the perimeter, we can see that the inradius is $(18 \cdot 24 / 2) /([18+24+30] / 2)=6$. So the incenter of the triangle is $(6,6)$. So the small triangle has a base of length 2 and a height of 3 , hence its area is 3 .