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Brazilian Mathematical Olympiad

Geometry Difficulty 6.3 National Olympiad Prove it Brazil

Given a regular dodecahedron of side $a$ . Take two pairs of opposite faces: $E, E'$ and $F, F'$ . For the pair $E, E'$ take the line joining the centers of the faces and take points $A$ and $C$ on the line each a distance $m$ outside one of the faces. Similarly, take $B$ and $D$ on the line joining the centers of $F, F'$ each a distance $m$ outside one of the faces. Show that $ABCD$ is a rectangle and find the ratio of its side lengths.

Solution

The centers of the faces of a regular dodecahedron form a regular icosahedron. Let $P, P'$ be two opposite vertices of a regular icosahedron. Then 5 of the remaining vertices are adjacent to $P$ and the other 5 are adjacent to $P'$ . So we may label the other pair of opposite vertices $Q$ and $Q'$ , where $Q$ is adjacent to $P$ and hence $Q'$ is adjacent to $P'$ .

Let the side of the icosahedron be $k$ . Then $PQP'Q'$ is obviously a rectangle and one pair of sides has length $k$ . The other side is the diagonal of the regular pentagon $XYP'ZQ$ of side $k$ and hence has length $\left(\frac{1+\sqrt{5}}{2}\right)k$ . $ABCD$ is similar to $PQP'Q'$ and so has the same ratio $\frac{1+\sqrt{5}}{2}$ for its side lengths.

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