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Problem 1458

AIME late

Geometry Difficulty 5.4 Prove it Romanian Mathematical Olympiad · Romania

The cube $ABCD'A'B'C'D'$ has edges of length $a$ . Take the points $E \in (AB)$ and $F \in (BC)$ so that $AE + CF = EF$ .
a) Find the measure of the angle of the planes $(D'DE)$ and $(D'DF)$ .
b) Compute the distance from $D'$ to the straight line $EF$ .

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

a) Extend the segment $BA$ with $AH \equiv FC$ . Then $\triangle DAH \equiv \triangle DCF$ (SAS), whence $\overline{HDA} \equiv \overline{FDC}$ , so $m(\overline{HDF}) = 90^\circ$ . Then $[DH] \equiv [DF]$ , leads to $\triangle DHE \equiv \triangle DFE$ (SSS) and from here $m(\overline{FDE}) = m(\overline{HDE}) = 45^\circ$ . Since $FD \perp DD'$ and $ED \perp DD'$ , it follows that $m((D'DE), (D'DF)) = m(\overline{FDE}) = 45^\circ$ .

b) Denote $P$ the orthogonal projection of the point $D$ onto $EF$ . The Three Perpendiculars Theorem yields $D'P \perp EF$ , hence $d(D', EF) = D'P$ . The congruence $\Delta DHE \equiv \Delta DFE$ gives $DP = AD = a$ . Finally, from Pythagoras' Theorem, $D'P = a\sqrt{2}$ .

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