\section*{Problem 5 - 031235}

Given a cube \(A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}\) with \(A A^{\prime}\left\|B B^{\prime}\right\| C C^{\prime} \| D D^{\prime}\). Furthermore, a segment \(X Y\) is given, where \(X Y=A B\) and \(X\) is a point on the segment \(A A^{\prime}\) and \(Y\) is a point on the face \(A B C D\).

What is the geometric locus of the midpoints of all segments \(X Y\) ?
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The midpoint of \(X Y\) is denoted by \(M\), and the foot of the perpendicular from \(M\) to \(A A^{\prime}\) is denoted by \(F\). Since \(A, X, Y, M\), and \(F\) lie in the same plane - the plane in which \(\triangle A X Y\) also lies - it follows from the intercept theorem that

\[
\frac{F M}{A Y}=\frac{X M}{X Y}=\frac{1}{2} \quad \text { and } \quad \frac{A F}{A X}=\frac{Y M}{Y X}=\frac{1}{2}
\]

By the Pythagorean theorem,

\[
A M^{2}=A F^{2}+F M^{2}=\frac{1}{4} A X^{2}+\frac{1}{4} A Y^{2}
\]

since \(A F \perp F M\). Furthermore, \(\triangle A X Y\) is a right triangle with leg \(X Y\), and it holds that \(A X^{2}+A Y^{2}=X Y^{2}=A B^{2}\). Therefore, \(A M=\frac{1}{2} A B \equiv\) constant.

Since \(X Y\) is entirely within the cube, the geometric locus of all midpoints of the segments \(X Y\) is the part of the sphere centered at \(A\) with radius \(\frac{1}{2} A B\) that belongs to \(A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}\).