Library / /9 of 34

, 2014

Geometry Difficulty 5.3 AIME, harder Prove it Austria

Consider a triangle $ABC$ . The midpoints of the sides $BC$ , $CA$ , and $AB$ are denoted by $D$ , $E$ , and $F$ , respectively.
Assume that the median $AD$ is perpendicular to the median $BE$ and that their lengths are given by $\overline{AD} = 18$ and $\overline{BE} = 13.5$ .
Compute the length of the third median $CF$ .

Solution

We denote the centroid of the triangle $ABC$ by $G$ . As the centroid divides each median into parts in the ratio $2 : 1$ , we have
$\overline{AG} = \frac{2}{3} \cdot \overline{AD} = 12 \quad \text{and} \quad \overline{BG} = \frac{2}{3} \cdot \overline{BE} = 9.$
By the Pythagorean theorem in the triangle $AGB$ , we obtain
$\overline{AB} = \sqrt{\overline{AG}^2 + \overline{GB}^2} = \sqrt{12^2 + 9^2} = 15.$

By Thales' theorem, $G$ lies on the circle with center $F$ and diameter $AB$ . Therefore, we have
$\overline{GF} = \overline{FA} = \frac{1}{2} \cdot \overline{AB} = \frac{15}{2}.$
As $\overline{FG} : \overline{GC} = 1 : 2$ , we obtain
$\overline{FC} = 3 \cdot \overline{FG} = \frac{45}{2}.$

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 and solution reproduced as published; topic and difficulty added by this site.