Puzzle of the Week for 23 November 1998: Solution

As noted in the hints for this puzzle, the solution requires that the area and the perimeter of the pie must be divided in five equal parts. Here is one way to do this:

  1. Mark the middle of the pie (for example, by finding the intersection of the diagonals of the square). Call this point A.
  2. Make a straight cut from A to any point on the perimeter of the pie (call this point B).
  3. Measure 1/5 of the way around the pie from B to another point on the perimeter (call this point C).
  4. Make a straight cut from A to C.
  5. Measure 1/5 of the way around the pie from C to another point on the perimeter (call this point D).
  6. Make a straight cut from A to D.
  7. Continue in the same way, marking off 1/5 of the perimeter each time and cutting from the center point A to the edges at points E and F, until there are five pieces.

It's easy to see that this division gives an equal amount of the outside crust to each piece; after all, that's how points C, D, E, and F were chosen. Is it possible that the pieces have equal areas, too? Remarkably, this is true.

Recall that the area of a triangle is half the product of its base and its height (altitude). This implies that any two triangles with equal bases and heights, such as those at right, have equal areas. Now consider the two pieces at left, taken from the first figure above. If the sides of the original square have a length of s, then the area of the square is s2, and its perimeter is 4s. The base (BC) of triangle ABC is 1/5 of this amount, or 4s/5. Since point A is at the center of the square, the height of ABC is s/2, so its area is 1/2 · 4s/5 · s/2, or s2/5 (i.e., 1/5 of the area of the square, so at least this piece is the correct size). The four-sided piece ABGF at far left can be divided into two triangles (ABG and AGF) as shown. If BG and FG are taken as the bases of these triangles, their heights are also s/2. The sum of BG and FG is 4s/5, so the sum of the areas of ABG and AFG is also 1/5 of the area of the square (prove this). Using similar methods, each of the remaining pieces can also be shown to have the same area.

Can you show that at least one of the five pieces produced using this method, but no more than two of them, must be triangular?

The same approach can be used to divide a square pie into any number of equal pieces. In fact, this method works if the pie is in the shape of any regular polygon. Why doesn't it work for other shapes, such as rectangles?

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