# On Bottema's Shoulders II

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A Mathematical Droodle

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Copyright © 1996-2018 Alexander Bogomolny

### On Bottema's Shoulders

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Consider a configuration of two squares ACB_{c}B_{a} and BCA_{c}A_{b} with a common vertex C. Bottema's theorem claims that the midpoint M of the segment A_{b}B_{a} is independent of C. Professor W. McWorter has observed that the result still holds if the squares are replaced with similar parallelograms so that the angles B_{a}AC and BCA_{c} are equal. The proof is actually the same as that for Bottema's original theorem. One just has to replace the factor i with an arbitrary complex number. The proof can be slightly simplified by placing A at the origin, such that

Let c be a complex number. Define

B_{a} = cg,

A_{c} = g + c(b - g), and

A_{b} = A_{c} + (b - g).

Then

A_{b} = b + c(b - g).

For the midpoint M of A_{b}B_{a}, we get

M | = (A_{b} + B_{a})/2 |

= (cg + b + cb - cg)/2, | |

= b(1 + c)/2, |

which is independent of g, i.e. C.

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny