On Bottema's Shoulders with a Ladder

What is this about?


Segment $AB$ is fixed. $ACA'$ and $BCB'$ are similar isosceles ($AC=A'C$ and $BC=B'C$) triangles, with a given apex angle. $D$ is the intersection of the bisector of $\angle ACA'$ with the circumcircle $(ACA'),$ $E$ the intersection of the bisector of $\angle BCB'$ with the circumcircle $(BCB').$ $(ACA')$ and $(BCB')$ meet at $J$. $G$ is the midpoint of $DE.$

A generalization of Bottema's theorem

Prove that

  1. $D,$ $J,$ and $E$ are collinear,

  2. $AB',$ $A'B$ meet at $J,$

  3. $AB'=A'B,$

  4. $CJ$ is the bisector of $\angle AJB,$

  5. If $O_A,$ $O_B$, $O$ are the centers of $(ACA'),$ $(BCB'),$ and $(AGB),$ respectively, then $OO_A=OO_B$ and triangle $O_{A}OO_B$ is similar to triangles $ACA'$ and $BCB'.$

  6. Position of $G$ is independent of $C,$

  7. Quadrilateral $AJGB$ is cyclic and $G$ is the midpoint of arc $AB.$


You may start with chasing inscribed angles.


A generalization of Bottema's theorem - solution, step 1

Let $\alpha = \angle CAA' = \angle CA'A = \angle CBB' = \angle CB'B\space$ and $\beta = \angle ACA'=\angle BCB'$ such that $2\alpha +\beta =180^{\circ}.$ Then, $\angle CJD$ equals half the arc $CD=CA'+A'D,$ making $\angle CJD=\alpha +\beta/2.$ Similarly, $\angle CJE=\alpha +\beta/2.$ Thus each of these angles is right and together they give $180^{\circ},$ meaning that $D,$ $J,$ and $E$ are collinear (#1). (As a byproduct, $CD$ is a diameter in $(ACA'),$ $CE$ is a diameter in $(BCB').)$

$\angle A'JC=\alpha$ while $\angle BJC=\alpha+\beta$ so again $\angle A'JC+\angle BJC=2\alpha+\beta=180^{\circ},$ implying that $A',$ $J,$ and $B$ are collinear. Similarly, $A,$ $J,$ and $B'$ are collinear (#2).

$\angle A'CB=\angle A'CA+\angle ACB=\angle B'CB+\angle ACB=\angle ACB',\space$ implying that triangles $A'CB$ and $ACB'$ are equal by SAS so that also $AB'=A'B$ (#3).

$\angle AJD = \beta/2 = \angle BJE$ while $CJ\perp DE,$ proving #4. $\angle AJB = 2\alpha,$ making each half equal $\alpha.$

A generalization of Bottema's theorem - solution, step 2

To prove #5, note that $O_AO_B\perp CJ$ while $OO_A\perp AJ$ which says that $\angle OO_AO_B=180^{\circ}-\angle AJC=90^{\circ}-\beta/2=\alpha.$ Similarly, $\angle OO_BO_A=\alpha.$

(Hubert Shutrick came up with a triginometric proof of ##6-7.) Denoting the angles of $\Delta ABC$ $\bar{\alpha},$ $\bar{\beta},$ and $\bar{\gamma},$ and the sides $a,$ $b,$ and $c,$ let $D'$ and $E'$ on $AB$ be projections of $D$ and $E.$ Observe that, since $CD$ is a diameter of $(O_A),$ $\angle CAD=90^{\circ},$ implying $\angle DAD'+\alpha=90^{\circ}$

A generalization of Bottema's theorem - solution, step 3

$AD=a\space\mbox{tan}(\beta/2)$ and $AD'=AD\space\mbox{sin}(\bar{\alpha})$ which together give $AD' = a\space\mbox{tan}(\beta/2)\mbox{sin}(\bar{\alpha}).$ Similarly, $BE'= b\space\mbox{tan}(\beta/2)\mbox{sin}(\bar{\beta}).$ But both $a\space\mbox{sin}(\bar{\alpha})$ and $b\space\mbox{sin}(\bar{\beta})$ express the length of $C$-height in $\Delta ABC.$ Therefore, $AD'=BE'.$

If the midpoint of $AB$ is $H,$ then (from the above) $D'H=E'H$ and also

$2GH = a\space\mbox{tan}(\beta/2)\mbox{cos}(\bar{\alpha}) + b\space\mbox{tan}(\beta/2)\mbox{cos}(\bar{\beta}) = AB\space\mbox{tan}(\beta/2).$

Thus, $\angle GAB = \beta/2 = \angle GJB,$ proving #6. #7 follows from the fact that $\angle AJB = 2\alpha = \angle AGB.$


The problem has been posted by Dao Thanh Oai at the CutTheKnotMath facebook page.

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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