Classical Triangle Centers

The line segment $\overline{AM}$ , the height of the ABC triangle, is in the Euler Line of this triangle. Let $O$ be the circumcenter of the triangle. On the Euler line the barycenter $G$ is between the circumcenter $O$ and the orthocenter $H$ and is twice as far from the orthocenter as it is from the circumcenter, so $\overline{HG} = 2\overline{GO}$ . Hence $\overline{HI} = \overline{IG} = \overline{GO} = d$ . With these conditions, let us build the triangle.

Let us call $\overline{ON} = y$ and $\overline{HM} = x$ . Since $I$ is the incenter, its distance from each side of the triangle is the same, hence $\overline{IM} = \overline{IP} = d+x$ . Using the property of the barycenter $G$ , we have: $\overline{AG} = 2\overline{GM} \therefore \overline{AO} = 3d+2x$ .

$\triangle GON \sim \triangle GHC \, (k=\frac{1}{2}) \therefore \overline{HC} = 2y$ .

$\triangle AON \sim \triangle AIP$

$\frac{y}{d+x} = \frac{3d+2x}{5d+2x} \rightarrow 5dy+2xy=3d^2+5dx+2x^2 \, (I)$ .

$\triangle CHM \sim \triangle AIP$

$\frac{x}{2y} = \frac{d+x}{5d+2x} \rightarrow 2dy+2xy=5dx+2x^2 \, (II)$ .

Using $(I) - (II)$ we get: $y=d \therefore x = \frac{d}{2}$ . Also, $\overline{AM} = \frac{15d}{2}$ .

$\triangle CHM \sim \triangle AMB \, (k=\frac{x}{2y} = \frac{1}{4}) \rightarrow \overline{AB} = 4\overline{MB}$ .

Using the pythagorean theorem on triangle ABM we have $\overline{BC} = d\sqrt{15}$ .

Hence, the area of the ABC triangle is $\frac{\overline{BC}.\overline{AM}}{2} = \frac{15d^2\sqrt{15}}{4}$ .

Finally, $a=15 \, , \, b=15 \, , \, and \, c=4 \longrightarrow \large \boxed{a+b+c=34}$ .