For these two problems, stop and think before rushing into possible solution attempts.
 A flexible cable of length twelve feet is hanging from two points at the same height. If the dip in the cable is six feet, determine the span.
 Given a triangle ABC. Is there a point X not in the plane of ABC such that AX, BX, and CX are mutually perpendicular?
Again, stop and think...a lot.
Source: Mathematics Magazine, 1955, p. 172 and 1957, pp. 290-291.
Hint: Try to visualize the two situations...what must be true?
Solution Commentary: [Solution to Problem 1] The span between the two points must be zero, as the cable must hang straight down (i.e. 6 feet down plus six feet up equals twelve feet).
[Solution to Problem 2] Think of a plane intersecting the corner of a room. The point X is the room's corner (the intersection point of three mutually intersecting perpendiculars). Now, does this work for all triangles ABC? What if ABC is a right triangle? An acute triangle? An obtuse triangle?