Bottema Shatters Japan's Seclusion
Here's a curious Sangaku problem. Four squares are hinged as shown. When points A, B, C are collinear, what is the relationship between the sides of squares BEKH and KINS?
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Solution
Copyright © 1996-2008 Alexander Bogomolny
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We apply Bottema's theorem twice: to the pair of squares BEKH and AESX and to the pair BEKH and CHIY. The theorem yields two points U and V, such that triangles ABU, KSU, BCV, and IKV are right isosceles. It follows that K is the midpoint of segment UV and that UV = SI, a diagonal of square KINS. When A, B, C are collinear, angle UBV is right and, therefore, lies on the circle centered at K with UV as diameter. In this case, KB = KV, or 2KB = UV. The diagonal of square KINS is twice that of square BEKH. The same relation holds for their sides.
A more elegant solution appears on a separate page .
(Additional sangaku are listed below.)
References
H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems , The Charles Babbage Research Center, Winnipeg, 1989
Write to:
Charles Babbage Research Center
P.O. Box 272, St. Norbert Postal Station
Winnipeg, MB
Canada R3V 1L6
H. Fukagawa, A. Rothman, Sacred Geometry: Japanese Temple Geometry , Princeton University Press, 2008, p. 149
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Copyright © 1996-2008 Alexander Bogomolny
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