Bogglers Solutions

By Scott Kim|Monday, July 01, 2002


144- and 36-piece solutions:

1. The proportions of the first two solutions (squares 1 and 2 above) are shown by the grids: In the first square, the sides of the two smaller rectangles are each half the size of the sides of the large blue rectangle. In the second square, the sides of the two smaller L shapes are each half the size of each side of the larger blue L shape.

2. There are two additional ways to cut a square into three similar rectangles. The easiest, shown in square 3, is to divide the square into three equal strips. Another option is to divide the square into three similar rectangles, as shown in square 4, with sides in proportion 1:.56984. To verify this proportion, set up the problem as follows. Call the yellow rectangle's short side p and its long side 1. The ratio of the long side to the short side is then 1/p. Given that the short side of the pink rectangle is 1, the long side must be 1/p. The long side of the blue rectangle is then p + 1/p, and its short side is p2 + 1 (if x is the length of the short side of the blue rectangle, then (p + 1/p )/x = 1/p, which means x = p2 + 1). The overall figure is a square, therefore p2 + 1 + 1 = p + 1/p. Simplify this equation to yield p3 + 2p = p2 + 1, which means p3 - p2 + 2p - 1 = 0. Now run this cubic equation through software like Maple, Mathematica, or Macsyma; use a cubic-equation solver on the Web; or solve the equation by hand (if you remember how—and dare) so that you end up with the following solution:1/6 [(44 + 12√69)/2] 1/3 - 10/3 [1/(44 + 12√69)]1/3+ 1/3 ≅ .5698

3. There are two ways to cut a square into three similar pieces by using diagonal cuts: One way yields three isosceles right triangles (square 6 above), and the other yields three trapezoids (square 7 above). There are an infinite number of trapezoidal solutions with the angle of the diagonal varying from 0° to 45°. In the dissection shown above, assume the vertical sides of the blue trapezoid are 1/2 and (1/2)p. Then the lengths of the vertical sides of the pink trapezoid are 1/2 and 1/(2p), and the shorter leg of the yellow trapezoid is 1/(2p2). As the overall shape is a square, (1/2)p + 1/(2p2) = 1, which simplifies to p3 - 2p2 + 1 = 0. This factors to (p - 1) (p2 - p - 1) = 0. Solve for p and you get p = (1 + √5)/2. So the sides of the blue trapezoid are 1/2 and (1 + √5)/4 (half the golden ratio).


The Web has several Life simulators. Go to to verify your answers (or to cheat).

From left to right, the figures are the blinker (which alternates between a line of three vertical markers and a line of three horizontal markers that share the middle, unchanging marker), the fuse (which disappears inward from the ends), the glider (which moves down and to the right one cell every four generations), and the block (which never changes).

Here are the next states of all seven letters. The letters G and E remain legible in this state.

Below is one possible previous state for each letter. The letter E has a predecessor with one less marker.

The final states appear below. The first R is the only letter to vanish. The last two letters both end as oscillators with four blinkers, called traffic lights. The letter D becomes an oscillator with six blinkers (which behave just like the blinker in the example above). The remaining letters—G, A, and N—end in stable patterns.

Word Walk


*As many solvers will find this answer a bit cryptic, we believe it warrants an explanation: In April 1975, Gardner fooled some readers into believing that Leonardo da Vinci invented the first flush toilet. To read that April Fool's joke, pick up The Colossal Book of Mathematics (W. W. Norton & Co., 2001).

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