# 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

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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.

### References

1. H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989

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