On Bottema's Shoulders
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 states that the midpoint M of the segment A_bB_a is independent of C. Professor W. McWorter has observed another property of the configuration: if D is the point that completes the parallelogram ADBC, then ΔB_aDA_b is isosceles and angle D is right: ∠B_aDA_b = 90°.

The theorem is easily proved with complex numbers but also has a simple synthetic proof.

Continuing with the proof of Bottema's theorem, we assume that the vertices A, B, and C are represented by the complex numbers α, β, and γ, respectively. We found there that

where i² = -1, as usual. We then obviously have

Denoting the point M + i(A_b - M) by D,

The latter tells us that ADBC is a parallelogram.

The synthetic proofs starts with the observation that in triangles AB_aD and BDA_b,

The angles between the equal sides are also equal, since both are found to be

360° - 90° - ∠A - ∠B,

the latter being the angles of ΔABC at vertices A and B, respectively.

By Bottema's theorem, the clockwise rotation around M through 90°, maps B on A. Since BA_b is perpendicular to BC which, in turn, is parallel to AD, the same rotation maps BA_b on AD and, similarly, BD on AB_a. It follows that the third sides of the triangles AB_aD and BDA_b are also mapped on each other. Thus the angle between them is 90°, as required.

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