On Bottema's Shoulders
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A Mathematical Droodle


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Explanation

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Copyright © 1996-2015 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 states that the midpoint M of the segment AbBa 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 ΔBaDAb is isosceles and angle D is right: ∠BaDAb = 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

 Ba = α + (γ - α)·i,
 Ab = β + (γ - β)·(-i),
 M = (α + β)/2 + (β - α)·i/2,

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

 i(Ab - M) = -i(Ba - M).

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

 
D= M + i(Ab - M)
 = M + (-γ + (α + β)/2 + i(α - β)/2)
 = (α + β) - γ
 = α + (β - γ).

The latter tells us that ADBC is a parallelogram.

The synthetic proofs starts with the observation that in triangles ABaD and BDAb,

  ABa = BD and
AD = BAb.

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 BAb is perpendicular to BC which, in turn, is parallel to AD, the same rotation maps BAb on AD and, similarly, BD on ABa. It follows that the third sides of the triangles ABaD and BDAb 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. 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

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Copyright © 1996-2015 Alexander Bogomolny

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