Nasty Triangles 4

Assume the perpendicular distance from $A$ to line $l$ is $AL$ , the perpendicular distance from $B$ to line $l$ is $BM$ , the perpendicular distance from $C$ to line $l$ is $CN$ .

Now construct line $CD$ where $D$ is the midpoint of $AB$ and intersects $l$ at $O$ , then construct line $DE$ so that $DE\perp l$ , construct line $AF$ so that $AF\parallel l$ , intersects $DE$ at $G$ and $F$ is on line $BM$ . So $DE=DG+2$ , $BM=BF+2=6$ , $BF=4$ .

We can easily see that $\triangle ADG\sim \triangle ABF$ $\therefore \frac{DG}{BF}=\frac{AD}{AB}=\frac{1}{2}$ $\therefore DG=2$ $DE=4$ Since $CD$ is a bisector of $AB$ , it must pass through the centroid of $\triangle ABC$ , and it is stated that line $l$ also passes through the centroid, thus we conclude $O$ is the centroid of $\triangle ABC$ .

We also see that $\triangle DEO\sim \triangle CON$ $\therefore \frac{DE}{CN}=\frac{DO}{OC}=\frac {1}{2}$ $CN=8$ So, the perpendicular distance from $C$ to line $l$ is 8.

WLOG, we can position and orient $\triangle ABC$ on the coordinate plane such that line $l$ has the equation $y=0$ . Since the centroid of $\triangle ABC$ lines on line $l$ , the y-coordinate of the centroid is $0$ . The y-coordinate of the centroid of a triangle is the average of the y-coordinates of the vertices of the triangle. Since we know the y-coordinates for $A$ and $B$ are $2$ and $6$ , and we also know the desired distance from $C$ to line $l$ is the absolute value of the y-coordinate of $C$ , we have the equation $(2+6+y)/3=0$ . Thus, $y=-8$ , and the answer to the problem is $8$ .