This problem was solved by Brad Rodgers, age 14 of Kokomo School, India and Sue Liu of Madras College, St Andrews.

Using 3D coordinates with point A at the origin, B being the point (4, 0, 0), C the point (4, 4, 0), D the point (0, 4, 0), E the point (0, 0, 4) etc as shown in the diagram. The point P which is the midpoint of AB has coordinates (2, 0, 0) and Q, which is a quarter of the way along EF has coordinates (1, 0, 4).
The general equation of a plane is $ax + by + cz = 1$ and to find $a$ , $b$ and $c$ we only need to know three points in the plane. Hence it is easy to find the equation of the plane through D, P and Q. Substituting the coordinates of the points into the equation of the plane we get:

$\begin{matrix} 4 b & = & 1 \\ 2 a & = & 1 \\ a + 4 c & = & 1 \end{matrix}$

This gives $a = 1 / 2$ , $b = 1 / 4$ and $c = 1 / 8$ . Writing the equation of the plane without fractions we get

$4 x + 2 y + z = 8 .$

The point R where the plane cuts the line EH has coordinates (0, y, 4) where $0 + 2 y + 4 = 8$ so R is the point (0, 2, 4).

Now RE = 2, EQ = 1, $∠ REQ = 90^{\circ}$ , and DA = 4, AP = 2, $∠ DAP = 90^{\circ}$ so it follows that the triangles REQ and DAP, which lie in parallel planes, are similar. Also, if we extend the lines AE, PQ and DR they will meet at one point S where SE = 4 and SA = 8.

The points S, A, P and D are the vertices of a tetrahedron which is sliced into two parts by the plane through R, E and Q. The required volume of the section of the cube is given by subtracting the volume of the tetrahedron SEQR from the volume of the tetrahedron SAPD.

$Volume of smaller section of cube = \frac{1}{3} (32 - 4) = \frac{28}{3} .$

The volume of the whole cube is 64 cubic units so the ratio of the volume of the smaller piece to the volume of the cube is 7 to 48. The ratio of the volumes of the two pieces of the cube is 7 to 41.