This is probably my favorite math puzzle of all time. The original puzzle (for a square) is a bit tricky to solve, but is well within the reach of a clever high school math student.
Here is the original statement of the problem:
Place a point P in a square so that the line segments from P to the vertices of the square all have integer lengths, where no two lengths are the same. If this is possible, then what is the smallest such square?
Part of what makes this a truly great puzzle is that the variations just go on and on. Some of them are easier than the original, some are about the same level of difficulty, some are much more difficult (probably requiring a computer search), and some are quite impossible!
Here are some of the variations:
1) Solve the same problem with a rectangle. (minimize the area)
2) … with an equilateral triangle.
3) … with a regular pentagon.
4) … with a regular hexagon.
5) … with a regular octagon.
And, in three dimensions:
6) … with a cube.
7) … with a rectangular prism. (minimize the volume)
8) … with a regular tetrahedron.
9) … with a regular octahedron.
Also interesting are variations where we relax the “no two have the same length” requirement, looking for solutions that lie on lines of symmetry:
10) … on the midline of a square (using two different lengths).
11) … on the diagonal of a square (using three different lengths).
12) … on the midline of a hexagon (using three different lengths).
13) … on the diagonal of a hexagon (using four different lengths).
14) … on the line connecting the centers of two opposite faces of a cube (using two different lengths).
15) … on the line connecting the midpoints of two opposite edges of a cube (using three different lengths).
16) … on the long diagonal of a cube (using four different lengths).
Good luck!
