Bottema with a Product of Rotations

What Might This Be About?

27 May 2014, Created with GeoGebra

Problem

Given $\Delta ABC$ and point $D,$ rotate $D$ clockwise around $B$ through angle $\alpha$ to obtain $D';$ rotate $D$ counterclockwise around $C$ by angle $\beta$ to obtain $D'';$ rotate $D''$ counterclockwise around $A$ through angle $\pi -\alpha -\beta$ to obtain $D'''.$ Let $E$ be the midpoint of $D'D'''.$

General Bottema's theorem with extra rotations

Prove that $E$ is independent of the position of $D.$

Solution

The product of two given rotations around $C$ and $A$ (one through angle $\beta,$ the other through angle $\pi -\alpha -\beta)$ is a rotation through angle $\beta + (\pi -\alpha -\beta)=\pi -\alpha,$ with the center denoted $O.$ This allows for an application of the generalized Bottema's theorem for the triangle $OBD.$

General Bottema's theorem with extra rotations

Acknowledgment

This problem is a follow-up of the one posted by Dao Thanh Oai (Vietnam) at the CutTheKnotMath facebook page, and a simple solution in complex numbers posted by Leo Giugiuc (Romania) and Leo's remark extending Dao's problem. Here's Dao's formulation:

Triangles $BET,$ $TAD,$ $DFC$ are similar isosceles. Prove that the midpoint $K$ of $EF$ is independent of the position of $D.$

General Bottema's theorem with extra rotations - Dao's formulation with isosceles triangles

Following his solution, Leo observed that it applies to a more general case where the three triangle are just similar, not necessarily isosceles.

Note that the above statement generalizes Dao's formulation but not obviously Leo's that I placed in a separate file. Indeed, let $F'$ lie on $CF$ such that $CF'=CD;$ $E'$ on $BE$ such that $BE'=BT.$ (I refer here to Dao's diagram.) Than, by the above, the midpoint $K'$ of $E'F'$ is independent of the position of $D.$ Denoting $\displaystyle r=\frac{CF}{CF'}$ we observe that

$\displaystyle r=\frac{CF}{CF'} =\frac{CF}{CD}=\frac{BE}{BT}=\frac{BE}{BE'}.$

It thus follows that, if $\displaystyle K'=\frac{E'+F'}{2}$ is independent of $D,$ then so is

$\displaystyle \begin{align} K &= \frac{E+F}{2}\\ &=\frac{\big[B+(E-B)\big] + \big[C+(F-C)\big]}{2}\\ &=\frac{\big[B+r(E'-B)\big] + \big[C+r(F'-C)\big]}{2}\\ &=\frac{(1-r)(B+C)+r(E'+F')}{2}. \end{align}$

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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