On Bottema's Shoulders II
What Is This About?
A Mathematical Droodle


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Copyright © 1996-2017 Alexander Bogomolny

On Bottema's Shoulders


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Consider a configuration of two squares ACBcBa and BCAcAb with a common vertex C. Bottema's theorem claims that the midpoint M of the segment AbBa 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 BaAC and BCAc 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 a = 0.

Let c be a complex number. Define

Ba = cg,
Ac = g + c(b - g), and
Ab = Ac + (b - g).

Then

Ab = b + c(b - g).

For the midpoint M of AbBa, we get

M= (Ab + Ba)/2
 = (cg + b + cb - cg)/2,
 = b(1 + c)/2,

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

Bottema's Theorem

  1. Bottema's Theorem
  2. An Elementary Proof of Bottema's Theorem
  3. Bottema's Theorem - Proof Without Words
  4. On Bottema's Shoulders
  5. On Bottema's Shoulders II
  6. On Bottema's Shoulders with a Ladder
  7. Friendly Kiepert's Perspectors
  8. Bottema Shatters Japan's Seclusion
  9. Rotations in Disguise
  10. Four Hinged Squares
  11. Four Hinged Squares, Solution with Complex Numbers
  12. Pythagoras' from Bottema's
  13. A Degenerate Case of Bottema's Configuration
  14. Properties of Flank Triangles
  15. Analytic Proof of Bottema's Theorem
  16. Yet Another Generalization of Bottema's Theorem
  17. Bottema with a Product of Rotations
  18. Bottema with Similar Triangles
  19. Bottema in Three Rotations
  20. Bottema's Point Sibling

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Copyright © 1996-2017 Alexander Bogomolny

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