Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.4 AIME, harder Prove it 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 $EF$ .

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, $D'P = a\sqrt{2}$ .

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