Library / Macedonian Mathematical Competitions /27 of 73

Macedonian Mathematical Competitions

Geometry Difficulty 4.8 AIME Prove it North Macedonia

$ABCDE$ is pentagon where $K$ , $L$ , $M$ , $N$ are the midpoints of $AB$ , $BC$ , $CD$ , $DE$ respectively. Let $P$ , $Q$ , $F$ be the midpoints of $KM$ , $LN$ , $AD$ , respectively. Prove that $PQ$ and $AE$ are parallel and $\overline{AE} = 4\overline{PQ}$ .

Solution

The quadrilateral $KFML$ is parallelogram. The point $P$ is the midpoint for $KM$ and $P \in LF$ and also the midpoint for $LF$ . In the triangle $LFN$ , we

have $PQ \parallel FN$ and $\overline{PQ} = \frac{1}{2}\overline{FN}$ . In the triangle $ADE$ , we have $FN \parallel AE$ and $\overline{FN} = \frac{1}{2}\overline{AE}$ . Finally $PQ \parallel FN \parallel AE$ and $\overline{PQ} = \frac{1}{2}\overline{FN} = \frac{1}{4}\overline{AE}$ , i.e. $PQ \parallel AE$ and $\overline{AE} = 4\overline{PQ}$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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