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Problem 1173

AMC 12 late, AIME early

Geometry Difficulty 4.9 Prove it Romanian Mathematical Olympiad · Romania

Consider a triangular pyramidal frustum $ABCA'B'C'$ . Points $D \in (AA')$ , $E \in (BB')$ and $F \in (CC')$ are such that the planes $(AEF)$ and $(DB'C')$ are parallel. Prove that the planes $(A'EF)$ and $(DBC)$ are also parallel.

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

Denote by $V$ the common point of the supporting lines of the lateral edges of the frustum. As planes $(AEF)$ and $(DB'C')$ are parallel, we have $EF \parallel B'C'$ and $DB' \parallel AE$ . Thales Theorem gives from $DB' \parallel AE$ and $A'B' \parallel AB$ :
$\frac{VD}{VA} = \frac{VB'}{VE}, \quad \frac{VA'}{VA} = \frac{VB'}{VB}.$
The last two equalities give
$\frac{VD}{VA'} = \frac{VB}{VE},$
so $A'E \parallel DB$ . As $EF \parallel B'C' \parallel BC$ and $A'E \parallel DB$ we conclude $(A'EF) \parallel (DBC)$ .

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