In our business, my two partners and I usually cut a lot of cake. However, when we get a last- minute cancellation, we sometimes look for new and unusual ways to cut small amounts of cake.
On one such occasion, we tried to find a cake with an unusual set of attributes.
First, it had to be a regular polygon, i.e. all sides equal in length and all angles equal in size.
Second, using only standard cuts (straight and perpendicular to the plate) from some corner to another, we would be able to divide that polygon exactly into equal portions.
Third, except for the last cut, each cut would create one, and only one, such portion. The last cut would create two such portions.
Fourth, there must be more than two portions. (Otherwise, a square or any regular polygon with an even number of sides would do. A cut from one corner to its opposing corner would carve out two pieces of 50% each.)
For discussion purposes, we decided to label the points of our polygons alphabetically. A pentagon can only produce three pieces. Triangles ABC and CDE are equal, but triangle ACE is larger. A hexagon yields three or four pieces, but they will be of two different sizes. Triangles ABC = CDE = EFA <> ACE, or triangles ABC = DEF <> CDF = FAC, etc.
About this time, Junior, our partner-in-waiting, said, “Cousins, I can do this, and I’ll get to take a share of the cake.”
Knowing Junior, we asked, “How big a headache is your maths going to give us?”
She answered, “I can prove it without using trig, powers, roots or the Pythagorean Theorem.”
Name the polygon she chose, tell how to divide it, and if you can, give proof the portions are equal.
Hint
Junior’s proof uses fact that all the dimensions of an isosceles right triangle are determined by its name and the length of its hypotenuse.
Answer
Junior knew an octagon would sequentially yield four quarters.
After labelling the points of a perfect octagon, Junior told us, “Before I cut the cake, I’ll give the proof. First, I mark a line from A to D. This creates a trapezoid (trapezium in Britain) ABCD. Next, I mark the lines EH, BG and CF. These four lines crosshatch the area dividing it into 1 square surrounded by 4 equal triangles and 4 equal rectangles.
The partners (me included) chimed in with the rest of these observations: the sides of the square are the same length as those of the octagon. Dividing the central square into isosceles right triangles (by marking both of its diagonals) would yield 4 triangles equivalent to the first 4. The total area of the octagon can be accounted for by 8 equal triangles and 4 equal rectangles. Because it is made of 2 triangles and 1 rectangle, trapezoid ABCD is 1/4 of the octagon. A second trapezoid taken from the irregular hexagon ADEFGH will equal 1/4 of the octagon and leave 1/2 of the original octagon in the shape of a rectangle or a kite, either of which is easily split to provide the last two 1/4’s. (The second and third cuts can be made in reverse order.)
The pieces from two arrangements follow:
1) ABCD, DEFG, ADH and DGH.
2) ABCD, EFGH, ADH and DEH.