IMO 1962 · Problem 3

Consider the cube \(ABCDA'B'C'D'\) (\(ABCD\) and \(A'B'C'D'\) are the upper and lower bases, respectively, and edges \(AA', BB', CC', DD'\) are parallel). The point \(X\) moves at constant speed along the perimeter of the square \(ABCD\) in the direction \(ABCDA\), and the point \(Y\) moves at the same rate along the perimeter of the square \(B'C'CB\) in the direction \(B'C'CBB'\). Points \(X\) and \(Y\) begin their motion at the same instant from the starting positions \(A\) and \(B'\), respectively. Determine and draw the locus of the midpoints of the segments \(XY\).

Official text of the problem. Official problem paper (PDF)

Answer

Let \(P,Q,R\) be the centers of the faces \(ABB'A'\), \(BCC'B'\), and \(ABCD\), respectively.

The locus is the perimeter of the rhombus \(PQCR\), lying in the plane \(AB'C\). For a cube of edge length \(a\), its sides have length \(a/\sqrt2\), and its angles at \(P\) and \(C\) are \(60^\circ\).

A schematic drawing in its plane is:

           C
          / \
         /   \
        R     Q
         \   /
          \ /
           P

The midpoint travels in the order \(P\to Q\to C\to R\to P\).

Solution

Let \(T\) be the time taken by either moving point to traverse one edge. Since their speeds are equal, they reach successive vertices simultaneously. Their positions and the corresponding midpoint \(M\) are:

Time \(X\) \(Y\) \(M\)
\(0\) \(A\) \(B'\) \(P\)
\(T\) \(B\) \(C'\) \(Q\)
\(2T\) \(C\) \(C\) \(C\)
\(3T\) \(D\) \(B\) \(R\)
\(4T\) \(A\) \(B'\) \(P\)

Here \(AB'\), \(BC'\), and \(DB\) are diagonals of the respective square faces, so their midpoints are indeed \(P,Q,R\). At time \(2T\), the segment \(XY\) degenerates to the point \(C\), whose midpoint is \(C\).

During each interval between successive times in the table, both points move uniformly along straight segments. Writing positions as vectors, suppose their endpoints during such an interval are \(U,V\) and \(U',V'\). At a fraction \(s\in[0,1]\) of the interval, \[ X(s)=(1-s)U+sV,\qquad Y(s)=(1-s)U'+sV'. \] Consequently, \[ M(s)=\frac{X(s)+Y(s)}2 =(1-s)\frac{U+U'}2+s\frac{V+V'}2. \] Thus \(M\) traverses exactly the entire segment joining its two endpoint positions. The four intervals therefore give precisely \[ PQ\cup QC\cup CR\cup RP. \] After \(4T\), both motions repeat, so there are no further points in the locus.

It remains to identify this quadrilateral. The diagonals of each square bisect one another, hence \[ P=\operatorname{mid}(AB'),\qquad Q=\operatorname{mid}(B'C),\qquad R=\operatorname{mid}(AC). \] Moreover, \(AB'\), \(B'C\), and \(CA\) are face diagonals of the cube. Each has length \(a\sqrt2\), so \(AB'C\) is an equilateral triangle.

Applying the midpoint theorem in this triangle gives \[ PQ\parallel AC,\qquad PR\parallel B'C. \] Since \(CR\) lies on \(CA\) and \(CQ\) lies on \(CB'\), both pairs of opposite sides of \(PQCR\) are parallel. Furthermore, \[ PQ=QC=CR=RP=\frac{a\sqrt2}{2}. \] Thus \(PQCR\) is a nondegenerate rhombus. Its angle at \(C\) is the \(60^\circ\) angle of the equilateral triangle, completing the description and proof.

The solution was checked against John Scholes' IMO solution archive (Kalva).