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Romanian Mathematical Olympiad

Geometry Difficulty 5.2 AIME, harder Prove it Romania

Let $ABCDA'B'C'D'$ , be a cube with side length $a$ . Let $M$ and $P$ be the midpoints of the edges $[AB]$ and $[DD']$ , respectively.
a) Prove that $MP \perp A'C$ .
b) Find the distance between the lines $MP$ and $A'C$ .

Solution

Let $O$ be the midpoint of $[A'C]$ .

a) Triangle $MA'C$ is isosceles, hence $MO \perp A'C$ .
Similarly $PA'C$ is isosceles, therefore $PO \perp A'C$ , yielding $A'C \perp (PMO)$ , hence $A'C \perp MP$ .

b) Let $S$ be the midpoint of $[MP]$ . The triangles $MA'C$ and $PA'C$ are congruent, yielding $[PO] = [MO]$ . Then $OS \perp MP$ , hence $OS$ is the distance between the lines $MP$ and $A'C$ .
We have $MC = \frac{a\sqrt{5}}{2}$ , therefore $MO = \frac{a\sqrt{2}}{2}$ . Next, $MP = \frac{a\sqrt{6}}{2}$ , and from the right triangle $OSM$ , we obtain $OS = \frac{a\sqrt{2}}{4}$ .

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