\section*{Problem 6 - 021236}

Given a cube \(A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}\) with \(A A^{\prime}\left\|B B^{\prime}\right\| C C^{\prime} \| D D^{\prime}\).

The point \(X\) travels with constant speed along the perimeter of the square \(A B C D\) in this order, and the point \(Y\) travels with the same speed along the perimeter of the square \(A^{\prime} D^{\prime} C^{\prime} B^{\prime}\) in this order.

Both points start their movements at the same moment from the points \(A\) and \(A^{\prime}\), respectively. Determine the geometric locus of the midpoints \(Z\) of the segments \(X Y\)!
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![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1754.jpg?height=442&width=516&top_left_y=704&top_left_x=770)

Given the conditions, the position of point \(Y\) is determined from \(X\) as follows:

a) for \(X \in A B\), \(Y\) lies on \(A^{\prime} D^{\prime}\) such that \(|A X| = |A^{\prime} Y|\),

b) for \(X \in B C\), \(Y\) lies on \(D^{\prime} C^{\prime}\) such that \(|B X| = |D^{\prime} Y|\),

c) for \(X \in C D\), \(Y\) lies on \(C^{\prime} B^{\prime}\) such that \(|C X| = |C^{\prime} Y|\),

d) for \(X \in D A\), \(Y\) lies on \(B^{\prime} A^{\prime}\) such that \(|D X| = |B^{\prime} Y|\).

To solve the problem, the following two theorems are proven:

Theorem 1: If \(P\) is the midpoint of \(A A^{\prime}\) and \(Q\) is the midpoint of \(C C^{\prime}\), then the midpoint \(Z\) of \(X Y\) lies on \(P Q\).

Theorem 2: If \(Z\) lies on \(P Q\), then \(Z\) is the midpoint of one of the segments \(X Y\).

From both theorems, it follows that the geometric locus to be determined is the segment \(P Q\).

Proof of Theorem 1:

At a fixed moment during the movement of the cube along the axis \(g_{P Q}\), if the cube is reflected, it maps onto itself such that \(A, B, C, D\) and \(X\) are swapped with \(A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}, Y\) in that order. Therefore, \(Z\) lies on the symmetry axis \(g_{P Q}\).

Since every cube is convex, \(P Q\) is the set of points on \(g_{P Q}\) that lie within the cube. For the same reason, \(X Y\) and thus \(Z\) lie within the cube, so \(Z \in P Q\).

Proof of Theorem 2:

Let \(\epsilon_{2}\) be the plane perpendicular to \(g_{P Q}\) through \(Z \in P Q\). Then \(P\) and \(Q\) do not lie on the same side of \(\epsilon_{2}\). Since each of the edges \(A A^{\prime}, B B^{\prime}, C C^{\prime}, D D^{\prime}\) is parallel to \(\epsilon_{2}\), \(A\) and \(A^{\prime}, B\) and \(B^{\prime}, C\) and \(C^{\prime}, D\) and \(D^{\prime}\) do not lie on opposite sides of \(\epsilon_{2}\), while \(A\) and \(C\) and \(A^{\prime}\) and \(C^{\prime}\) do not lie on the same side of \(\epsilon_{2}\).

The plane \(\epsilon_{2}\) contains the segments \(B B^{\prime}\) and \(D D^{\prime}\) either both - precisely when \(Z\) is the midpoint of \(P Q\) - or neither. Consequently, \(\epsilon_{2}\) intersects each of the segments \(A B, A^{\prime} B^{\prime}, A D, A^{\prime} D^{\prime}\) or each of the segments \(B C, B^{\prime} C^{\prime}, C D, C^{\prime} D^{\prime}\) at a point. These intersection points swap under the reflection mentioned in the proof of Theorem 1 in the manner described there.

Therefore, for every point \(Z \in P Q\), there is a pair, and in the case \(Z \neq P, Z \neq Q\), exactly two pairs \((X, Y)\) of corresponding points \(X, Y\) for which \(Z\) is the midpoint of \(X Y\).

Adapted from \([2]\)

\subsection*{9.4.4 IV. Round 1962, Class 12}