Library / /1 of 32

, 2010

Geometry Difficulty 4.5 AIME Prove it Estonia

Let $D$ be the midpoint of side $BC$ of triangle $ABC$ . Prove that the intersection point of medians of triangle $ABD$ and that of triangle $ACD$ are equidistant from line $AD$ . (Grade 11.)

Solution

Triangles $ABD$ and $ACD$ have equal area since $|BD| = |CD|$ and the altitudes drawn from $A$ coincide (Fig. 11). As these triangles have a common side $AD$ , also the altitudes drawn from vertices $B$ and $C$ , respectively, must be equal. Thus $B$ and $C$ are equidistant from line $AD$ . Since the point

Fig. 11
of intersection of medians cuts $\frac{1}{3}$ part of every median, the distance of the point of intersection of medians of $ABD$ from line $AD$ is thrice shorter than the distance of point $B$ from line $AD$ . An analogous relation holds also for triangle $ACD$ . Hence the claim follows.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.