Brain Teasers
Easy As Cake
In our business, my two partners and I cut a lot of cake.
We work exclusively with square cakes. The cakes come in sizes that are meant to be divided into a square number of square pieces. Each portion comes decorated with an "X" stretching to its corners, which makes it easy to cut the usual square pieces and also offer half and quarter portions.
These markings also help on Wednesday when we savor the previous week's success by dividing a cake equally amongst ourselves. Even when a good week deserves a 64-portion cake, we can each ceremonially make a single standard cut (straight and perpendicular to the plate), and we each go home with a single piece of cake.
Using no other cuts or devices, how do we evenly divide the 8-by-8 cake? (EXTRA - How could we get 7 (and only 7) equal-size pieces from the same size cake?)
We work exclusively with square cakes. The cakes come in sizes that are meant to be divided into a square number of square pieces. Each portion comes decorated with an "X" stretching to its corners, which makes it easy to cut the usual square pieces and also offer half and quarter portions.
These markings also help on Wednesday when we savor the previous week's success by dividing a cake equally amongst ourselves. Even when a good week deserves a 64-portion cake, we can each ceremonially make a single standard cut (straight and perpendicular to the plate), and we each go home with a single piece of cake.
Using no other cuts or devices, how do we evenly divide the 8-by-8 cake? (EXTRA - How could we get 7 (and only 7) equal-size pieces from the same size cake?)
Hint
Think smaller.Answer
To get three twenty-one-and-one-third (64/3) portion-sized pieces, we momentarily ignore the 28 X's/portions/square-units which form the 32 unit perimeter of the cake. There is more than one way to divide the 6-by-6 (36 X) cake inside into three equal pieces. Using one based on its 24 unit perimeter, this cake could be divided into 24 wedges of equal size. Each triangular piece would have as its base one unit of the perimeter; the other two sides would be formed by cuts to the center of the cake. Every triangle would have the same altitude (3 units) giving the same area (A = ½ * b * h = ½ * 1 * 3 = 3/2 square units). 8 adjacent triangles, even though they include a corner or two, will be 12 square units or 1/3 of the "inner" cake.Before we make any actual cuts, we return the ignored 28 portions to the picture and then make those three cuts extended to the true perimeter. The 24 triangles pictured previously increase in altitude to 4 units. By rules of similar triangles, the base increases by the same 1/3 (33 1/3%) to 1 1/3 units. The area of 8 extended triangles is 8 * ½ * 4/3 * 4 = 64/3 portions, which is exactly one-third of the cake.
(The perimeter of a 7-by-7 square can be formed by connecting the centers of the X's in the "ignored" 28 portions. Applying the above methods but now grouping 28 triangles into foursomes, we get 7 equal size pieces.)
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