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The problem:

$A,B,C,A’,B’,C’$ are concyclic. The three Simson lines of $A,B,C$ about $\triangle A'B'C'$ intersect at $D,E,F$. The three Simson lines of $A’,B’,C’$ about $\triangle ABC$ intersect at $D’,E’,F’$.

Show that $D,E,F,D’,E’,F’$ are concyclic.

Moreover, if $H,H’$ are the orthocenters of $\triangle ABC,\triangle A’B’C’$. Show that the center of $\odot(DEFD’E’F’)$ is the midpoint of $HH’$.

It might be no good drawing the graph. When orthocenters and Simson lines appear at the same time, I think of the Steiner theorem, which states that

$A,B,C,D$ concyclic, $H$ is the orthocenter of $\triangle ABC$, then the midpoint of $DH$ is on the Simson line of $D$.

There’s also a property that might help:

Let $A,B\in\odot O$. The intersection angle of Simson lines of $A,B$ equals to the inscribed angle of $\overset{\LARGE\frown}{AB}$ in $\odot O$.

I’ll show my complex number method. (Barely needs figure)

First let $\odot{ABC}$ be the unit circle.

There’s a formula of Simson line of $P$ and $\triangle ABC$ that is $$Pz-ABC\overline z=\frac1{2P}(P^3+\sigma_1P^2-\sigma_2P-\sigma_3).$$ So we can get the Simson line of $A$ and $B4$ then put them together to find intersection. \begin{cases}Az-A’B’C’\overline z=\frac1{2A}(A^3+\sigma_1A^2-\sigma_2A-\sigma_3),\\ Bz-A’B’C’\overline z=\frac1{2B}(B^3+\sigma_1B^2-\sigma_2B-\sigma_3).\end{cases}

Subtract, and it is not hard to solve $\displaystyle z=D=\frac12(A+B+\sigma_1+\frac{\sigma_3}{AB}).$

The midpoint of $HH’$ is $M=\dfrac12(A+B+C+A’+B’+C’).$ Then $$|D-M|=\frac12|-C+\frac{A’B’C’}{AB}|=\frac12|ABC-A’B’C’|.$$

Thus all of $DM,EM,FM,D’M,E’M,F’M$ equal to this. $\#$

  • Please provide a figure. – Jean Marie Sep 18 '22 at 07:33
  • @JeanMarie The problem is that the figure might not be of any use unless I draw all the perpendicular lines. But if so, it would be too complicated. Or do you think the figure only needs to include the Simson line? –  Sep 18 '22 at 07:37

1 Answers1

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The question was asked (in 1906) and answered by Absolonne (in 1907) in a French language publication Mathesis Question 1575, pg 23. The answer here is organized into

  • a diagram
  • a screen shot of the original
  • my translation of the original French, along with commentary
  • some other reference materials

A caveat: this solution is not mine, I'm merely translating Absolonne's proof with small corrections, readability enhancements, and annotations.

Figure

enter image description here

Screen shot of 1907 solution

Mathesis excerpt

Translation and Commentary

[my comments will be in bold. Note that we start with two triangles $ABC,DEF$ instead of OP's triangles $ABC,A'B'C'$ ]

We consider two triangles $ABC, DEF$ inscribed in the same circle. Prove 1) that the Simson lines from the points $D,E,F$ with respect to the triangle $ABC$ form a triangle $D'E'F'$ similar to $DEF$ and that the Simson lines of the points $A,B,C$ with respect to the triangle $DEF$ form a triangle $A'B'C'$ similar to $ABC$; 2) that the triangles $D'E'F' , A'B'C'$ are inscribed in the same circle whose center is the midpoint of the orthocenters of the triangles $ABC, DEF$. ([proposed by] S. KANTOR)

1) Solution by M. ABSOLONNE. The Simson line of point $D$ with respect to triangle $ABC$ is perpendicular to the isogonal $AD_i$ of $AD$ with respect to angle $A$ and $DD_i$ is parallel to $BC$; similarly, the Simson line of $A$ with respect to the triangle $DEF$ is perpendicular to the isogonal $DA_i$ of $DA$ relative to the angle $D$. The angles $A_iDD_i$ and $A_iAD_i$ being equal, we conclude that the Simson line $E'F'$ of $D$ makes with $EF$ the same angle as the Simson line $B'C'$ of $A$ makes with $BC$. The sides of the triangle $D'E'F'$ being perpendicular to the isogonals of $AD, AE, AF$, this triangle is directly similar to $DEF$; $A'B'C'$ will also be directly similar to ABC.

[The main takeaway here is that angle between $EF,E'F'$ is the same as that between $BC,B'C'$, etc. This angle will be referred to as $\alpha$ in the following.]

2) $H$ being the orthocenter of $ABC$, the sides of the triangle $D'E'F'$ pass through the midpoints $d,e,f$ of $HD,HE,HF$, and the triangle $def$ being homothetic to $DEF$, its orthocenter $O$ is the midpoint of $H,H'$, the orthocenters of $ABC, DEF$. [As will be shown below, this] point $O$ is also the center of the circumcircle of the triangle $D'E'F'$.

Let $R$ be the radius of the circle $ABC$, $\alpha$ the angle of the corresponding sides of $D'E'F'$ and $DEF$ which is also that of the sides of $A'B'C'$ and $ABC$; we will have in the triangle $OF'e$, for example: $$\dfrac{OF'}{Oe}=\dfrac{\cos(\alpha)}{\cos(e)}.$$

[The latter follows from the law of sines on the triangle $OF'e$, and the fact that the relevant angles are complements of $\alpha$ and $e=\angle{def}$]

Now, if $M$ is the midpoint of $df$, $$Oe = 2OM = R \cos(e) \\ \text{whence}\quad OF'=R \cos(\alpha). $$

[He's saying that the distance of $D',E',F'$ from $O$ is $R \cos(\alpha)$, therefore $O$ is their circumcenter. But there appears to be a typo in the line $Oe = 2OM = R \cos(e)$. If O' is the circumcenter of $def$ then the line should read $Oe = 2O'M = R \cos(e)$. $2O'M$ is a well known formula for the distance $Oe$ between a vertex and the orthocenter $O$. But $Oe = 2r \cos(e)$, where $r=R/2$ is the circumradius of $def$, is also well known as a formula for that distance.]

The circumcircle of the triangle A'B'C' will obviously have the same center and, according to the preceding equality, the same radius.

Other references

I was lead to Absolonne's proof by this note by Goormaghtigh, who has written related papers.

Johnson, Modern Geometry, pg 211, Section 338 has a special case of the theorem where the Simson lines are concurrent rather than forming a triangle.

Zaharinov, The Simson Triangle and Its Properties discusses Simson triangles, and this theorem in particular, using complex numbers.

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