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Problem 1613

National Olympiad, first round

Geometry Difficulty 6.1 Prove it THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND · Romania

Let $SABCD$ be a pyramid with the apex $S$ and whose base $ABCD$ is a parallelogram. We consider the points $M$ , $N$ , $P$ and $Q$ on the edges $SA$ , $SB$ , $SC$ and $SD$ , respectively, such that $MNPQ$ is also a parallelogram.

a) If $ABCD$ is a rhombus, prove that $MNPQ$ is also a rhombus.

b) If $ABCD$ is a rectangle, prove that $MNPQ$ is also a rectangle.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

The planes $(SAB)$ and $(SCD)$ meet along the line $d_1$ , and let $d_2$ be the intersection line of the planes $(SBC)$ and $(SDA)$ .
As $AB \parallel CD$ , $AB \subset (SAB)$ and $CD \subset (SCD)$ , from the 'roof theorem' it follows that $AB \parallel CD \parallel d_1$ . Because $MN \parallel PQ$ , $MN \subset (SAB)$ and $PQ \subset (SCD)$ , from the roof theorem we deduce that $MN \parallel PQ \parallel d_1$ , therefore $AB \parallel CD \parallel MN \parallel PQ \parallel d_1$ .
Similarly, we obtain $BC \parallel DA \parallel NP \parallel QM \parallel d_2$ .

If $ABCD$ is a rhombus, then $AB = BC$ , whence $MN = NP$ , therefore the parallelogram $MNPQ$ is also a rhombus.

b) If $ABCD$ is a rectangle, then $\widehat{ABC} = 90^\circ$ . As $MN \parallel AB$ and $NP \parallel BC$ , we obtain $\widehat{MNP} = 90^\circ$ , therefore $MNPQ$ is also a rectangle.

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