Yet Another Generalization of Bottema's Theorem

What Might This Be About?


Given three points $A$, $B,$ $C,$ and angle $\phi.$ Define $C_A$ to be the rotation of $C$ through angle $\phi$ around $A$ and $C_B$ to be the rotation of $C$ through angle $\phi-\pi$ around $B.$ Let $D$ be the midpoint of $C_A$ and $C_B.$ (This means rotating by $\phi$ and $\pi -\phi$ in opposite directions.)

yet another generalization of Bottema's  Theorem


  1. $D$ is independent of $C.$

  2. $D$ lies on the circle with diameter $AB,$ and therefore $\angle ADB=90^{\circ}.$

Solution 1

I'll use complex numbers, assuming $A=\alpha,$ $B=\beta,$ $C=\gamma.$ Then

$\begin{align}\displaystyle C_{A}&=\alpha + (\gamma -\alpha)e^{i\phi},\\ C_{B}&=\beta + (\gamma -\beta)e^{i(\phi-\pi)}=\beta - (\gamma -\beta)e^{i\phi} \end{align}$

such that $\displaystyle D=\frac{1}{2}\big(\alpha +\beta +e^{i\phi}(\beta -\alpha)\big)$ - independent of $C(\gamma ).$ This proves the first part, but also the second, because, as $\phi$ changes, $D$ stays on the circle centered at $\displaystyle \frac{\alpha +\beta}{2}$ with radius $\displaystyle \frac{|\beta -\alpha |}{2}.$

Solution 2

A product of rotations is a rotation through the angle which is the sum of the angles of the two consituent rotations. (It could be a translation of the angles ad up to a multiple of $360^{\circ}.)$ If we start working backwards, first rotating $C_B$ to $C$ and then $C$ to $C_A$ we may consider the latter as the image of $C_B$ under the rotation through $180^{\circ}$ which is a central symmetry. The center of that central symmetry depends on the centers of the two rotations but not on the point being rotated and is thus independent of $C_B$ and, therefore, independent from $C.$

yet another generalization of Bottema's  Theorem

If it were not for the Bottema theorem, how would we go about finding point $D?$ I'll do that for the product of two rotations through angles $\alpha$ and $\beta$ around points $A$ and $B,$ respectively. The product of rotations rotates $A$ around $B$ through angle $\beta$ into $A'.$ There is point $B'$ that is brought onto $B$ when rotated around $A$ through angle $\alpha.$ Thus $A'$ is the image of $A$ under the product of two rotations, while $B$ is the image of $B'.$ It follows that the center $D$ of the product of the two rotations lies on the perpendicular bisectors of $AA'$ and $BB'.$ One of these passes through $A,$ the other through $B;$ they meet at $D.$ In triangle $ABD,$ the angle at $A$ is $\alpha /2,$ that at $B$ is $\beta /2.$ the remaining angle at $D$ is therefore $180^{\circ}-(\alpha +\beta )/2$ such that if $\alpha + \beta =180^{\circ}$ then the angle at $D$ is $90^{\circ}.$

Hubert Shutrick offered a shortcut: "$B$ is not moved by the first rotation and goes to $B'$ with the one round $A$ so $BAB'$ is an isosceles triangle with $D$ at the midpoint of $BB'.$"


The second solution came up in a discussion with Machó Bónis.

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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