Treebeard's Stumper Answer

April Fools' Gold
Snow in the local mountains on April 1 was something unexpected, and so is this classic stumper. The top figure has as area of 8x8 = 64 squares. Cut the figure along the lines into four pieces, and reassemble them as shown in the bottom figure. The same four pieces now have an area of 5x13 = 65 squares, one more than we started with. Here's an easy way to get rich. Just copy the top figure onto a sheet of gold, cut and reassemble, and then sell the extra square! Can this be right, or is it just April Fools' Gold?
If that's too easy, then the advanced stumper is to explain the connection with Fibonacci numbers: 3,5,8,13,...
Last week's stumper doesn't quite add up. The long diagonal is not what it seems. It has a thin parallelogram missing with an area of exactly one square that accounts for our April Fool's Gold. The slope of a line is a measure of its steepness, the vertical rise divided by the horizontal run. The slanting lines on the foursided figures have a slope of 2/5 = 0.4, but the lines along the triangles have a slope of 3/8 = 0.375. These lines can't be equal. Here's an (exaggerated) view of the real shape:
Note: This is one of my favorite stumpers to show to kids at school. I have the kids draw the square figure on graph paper, and then cut it out and reassemble it. Usually a few sharp kids notice that the pieces don't quite fit, and everyone senses that something is not right. My hint is that it's more obvious if you draw it large with very sharp lines.
Graybear sent the following analysis (with a picture, but our FAX is down):
I find it easiest to rotate the rectangle 90 degrees and multiply the length and width by three. Of course, now you gain/lose nine squares (15 x 39)  (24 x 24) = 9The increased size shows the parallelogramshaped gap, the arrows indicate coordinate pairs with integral coordinates.
The overlapping parallel section between (6,16)/(6,15) and (9,24)/(9,23) shows the height of the parallelogram to be '1' for a base of '9'  A = bh. Or, by 'sliding' the corresponding heights to the baseline, the area can be found by visual inspection or the formula for a trapezoid of b1=9, b2=3, h=1. (Trapezoid coordinates: (0,0),(6,1),(9,1),(15,0))
The reason this stumper "works" is that the slopes of 5/2 and 8/3 are close enough to fool the eye into seeing a straight line. The thickness of the pencil line hides the difference.
The connection with Fibonacci numbers is fascinating. The Fibonacci sequence is 1,1,3,5,8,13,21,... where each successive number is the sum of the two numbers before it. My stumper uses the numbers {3,5,8,13}. It's possible to make a similar stumper by redrawing my figures with the dimensions {5,8,13,21}, but in this case the figure loses a square when reassembled. The next size up will again gain a square, and so on.
This has to do with the fact that f(n)^2  f(n1)*f(n+1) = (1)^n, where f(n) is the nth Fibonacci number (0..n). For example, 8*8  5*13 = 1, so we lose 1 square going from an 8x8 square to a 5x13 rectangle.
I haven't used mathematical induction in years, but I couldn't find this in my books, and I just started Spring Vacation, so I have time to prove it. (Note: this won't look right if your browser doesn't support <SUB> and <SUP> tags like Netscape 3.)
To prove:f_{n}^{2}  (f_{n+1})(f_{n1}) = (1)^{n}This is true for n=1 since,
where f_{n} is the nth Fibonacci number (0..n)
f_{0}=f_{1}=1, and f_{2}=2, so f_{1}^{2}f_{0}*f_{2}= 12 = 1 = (1)^{1}Now assume the rule is true for f_{n} and examine the case for f_{n+1}f_{n+1}^{2}  (f_{n+2}) (f_{n}) =So by induction the rule is true for all n. QED.
f_{n+1}^{2}  (f_{n+1} + f_{n}) f_{n} =
f_{n+1}^{2}  (f_{n+1})(f_{n})  f_{n}^{2} =
f_{n+1} (f_{n+1}  f_{n})  f_{n}^{2} =
f_{n+1} (f_{n1} + f_{n} f_{n})  f_{n}^{2} =
f_{n+1} f_{n1}  f_{n}^{2} =
(f_{n}^{2}  (f_{n+1})(f_{n1})) =
(1)^{n} =
(1)^{n+1}Graybear adds:
Thanks for reminding me about the Fibonacci numbers... I LOVE THEM... It never ceases to amaze me how many properties they have. In this case, you can create an infinite family of these things and half will lose one square and half will gain one. I knew that F(n)/F(n1) converges on the golden ratio (1.618...), alternating between values >g.r. and <g.r., but just realized that F(n+1)/F(n1) converges on g.r.^2 (2.618... (or) 1 + g.r.) in the same way. G.r.^2 describes the slopes in my solution (5/2=2.5 and 8/3=2.67).This stumper uses what Martin Gardner calls the principle of concealed distribution. He presents some amazing examples (with analysis) in his book Mathematics, Magic and Mystery (Dover, 1956).
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