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

Put differently

Created with GeoGebra


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

Copyright © 1996-2018 Alexander Bogomolny

On Bottema's Shoulders

On Bottema's Shoulders

Consider a configuration of two squares $ACB_{c}B_{a}\,$ and $BCA_{c}A_{b}\,$ with a common vertex $C.\,$ Bottema's theorem states that the midpoint $M\,$ of the segment $A_{b}B_{a}\,$ 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 $\Delta B_{a}DA_{b}\,$ is isosceles and angle $D\,$ is right: $\angle B_{a}DA_{b} = 90^{\circ}.$

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 $\alpha,\,$ $\beta,\,$ and $\gamma,\,$ respectively. We found there that

$\begin{align} B_{a}&= \alpha + (\gamma - \alpha )\cdot i,\\ A_{b}&= \beta + (\gamma - \beta )\cdot (-i),\\ M&= (\alpha + \beta )/2 + (\beta - \alpha )\cdot i/2, \end{align}$

where $i^2 = -1,\,$ as usual. We then obviously have

$i(A_{b} - M) = -i(B_{a} - M).$

Denoting the point $M + i(A_{b} - M)\,$ by $D,$

$\begin{align} D&= M + i(A_{b} - M)\\ &= M + (-\gamma + (\alpha + \beta )/2 + i(\alpha - \beta )/2)\\ &= (\alpha + \beta ) - \gamma \\ &= \alpha + (\beta - \gamma ).\\ \end{align}$

The latter tells us that $ADBC\,$ is a parallelogram.

The synthetic proofs starts with the observation that in triangles $AB_{a}D\,$ and $BDA_{b},$

$\begin{align} AB_{a}&= BD\;\text{and}\\ AD&= BA_{b}. \end{align}$

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

$360^{\circ} - 90^{\circ} - \angle A - \angle B,$

the latter being the angles of $\Delta ABC\,$ at vertices $A\,$ and $B,\,$ respectively.

By Bottema's theorem, the clockwise rotation around M through $90^{\circ},\,$ 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 $AB_{a}D\,$ and $BDA_{b}\,$ are also mapped on each other. Thus the angle between them is $90^{\circ},\,$ 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
  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

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

Copyright © 1996-2018 Alexander Bogomolny