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

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Explanation

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Copyright © 1996-2012 Alexander Bogomolny

Consider a configuration of two squares ACB_cB_a and BCA_cA_b with a common vertex C. Bottema's theorem claims that the midpoint M of the segment A_bB_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_aAC 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 a = 0.

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_bB_a, 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. On Bottema's Shoulders
4. On Bottema's Shoulders II
5. Friendly Kiepert's Perspectors
6. Bottema Shatters Japan's Seclusion
7. Rotations in Disguise
8. Four Hinged Squares
9. Four Hinged Squares, Solution with Complex Numbers
10. Pythagoras' from Bottema's

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Copyright © 1996-2012 Alexander Bogomolny