There's a simple extension of the trivial case you mentioned: Let E,F be the midpoints of AB, CD respectively. Then you can cut the rectangle AEFD lengthwise. I suspect there's a way to cut it where you turn one of the pieces over... -- Mike Stay staym@datawest.net ----- Original Message ----- From: Thane Plambeck Sent: 11/13/2002 13:54:32 To: math-fun@mailman.xmission.com Subject: [math-fun] A problem from the files of David Klarner

Draw a square ABCD with center M. Cut out the triangle T= BMC. Rotate and reattach T to an adjacent side of whats left of the square by lining up side BC with either CD or AB.

Now you've got a six-sided figure. Suppose you want to cut this figure into two pieces that can reassembled into the original square shape.

Of course, one way to do it is to lop of the original triangle and put it back where it came from.

Find another solution.

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In Klarner's files there are some notes that credit this highly addictive little problem to some unknown person in Eindhoven. Does anyone know anything more about it?

For more on Klarner's files (he died in 1999)

http://www.qxmail.com/mathematics/klarner

Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.qxmail.com/home.htm