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Romanian Mathematical Olympiad

Geometry Difficulty 5.1 AIME, harder Prove it Romania

Consider a regular prism ABCA′B′C′ABCA'B'C'. A plane α\alpha containing point AA meets the rays BB′BB' and CC′CC' at points EE and FF such that
area [ABE]+area [ACF]=area [AEF]. \text{area } [ABE] + \text{area } [ACF] = \text{area } [AEF].

Find the angle determined by the planes AEFAEF and BCC′BCC'.

Solution

Let MM be the midpoint of BCBC and let uu be the angle determined by the planes AEFAEF and BCC′BCC'. Since triangle MEFMEF is the projection of the triangle AEFAEF onto BCC′BCC', we have
cos⁡u=[MEF][AEF]=[BCFE]2[AEF]=[BCFE]2([ABE]+[ACF])==[BCFE]4([MBE]+[MCF])=[BCFE]2[BCFE]=12, \begin{aligned} \cos u &= \frac{[MEF]}{[AEF]} = \frac{[BCFE]}{2[AEF]} = \frac{[BCFE]}{2([ABE] + [ACF])} = \\ &= \frac{[BCFE]}{4([MBE] + [MCF])} = \frac{[BCFE]}{2[BCFE]} = \frac{1}{2}, \end{aligned}
hence u=60∘u = 60^\circ.

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