geometry
This is problem 1.9.5 from Geometry Revisited by Coxeter and Greitzer: If two lines through one vertex of an equilateral triangle divide the semicircle drawn outward on the opposite side into three equal arcs, these same lines divide the side itself into three equal line segments.
I came up with the following solution for it (click to enlarge image if print too small), but I feel like I have made it more complicated than it needs to be. Wondering if anyone can show how to simplify the answer?
a single line through the centroid can only intersect all three sides of a triangle at $X,Y,Z$ if two of these points are coincident at a vertex of the triangle, and the third is the midpoint of the opposite side
The sides of the triangle are lines of infinite length in this context, not just the line segments terminated by the corners of the triangle. OK, Andreas already pointed that out in a comment, but since originally this was the core of your question, I'll leave this here for reference.
I recently posted an answer to this question in another post. Since @Michael Greinecker♦ asked me to duplicate the answer here, that's what I'm doing.
I'd work on this in barycentric homogenous coordinates. This means that your corners correspond to the unit vectors of $\mathbb R^3$, that scalar multiples of a vector describe the same point. You can check for incidence between points and lines using the scalar product, and you can connect points and intersect lines using the cross product. This solution is influenced heavily by my background in projective geometry. In this world you have
$$ A = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \qquad B = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \qquad C = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \\ G = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \qquad l = \begin{pmatrix} 1 \\ a \\ -1-a \end{pmatrix} \qquad \left<G,l\right> = 0 \\ X = \begin{pmatrix} a \\ -1 \\ 0 \end{pmatrix} \qquad Y = \begin{pmatrix} 1+a \\ 0 \\ 1 \end{pmatrix} \qquad Z = \begin{pmatrix} 0 \\ 1+a \\ a \end{pmatrix} $$
The coordinates of $l$ were chosen such that the line $l$ already passes through $G$, as seen by the scalar product. The single parameter $a$ corresponds roughly to the slope of the line. The special case where the line passes through $A$ isn't handled, since in that case the first coordinate of $l$ would have to be $0$ (or $a$ would have to be $\infty$). But simply renaming the corners of your triangle would cover that case as well.
In order to obtain lengths, I'd fix a projective scale on this line $l$. For this you need the point at infinity on $F$. You can obtain it by intersecting $l$ with the line at infinity.
$$ F = l\times\begin{pmatrix}1\\1\\1\end{pmatrix} = \begin{pmatrix} 2a+1 \\ -2-a \\ 1-a \end{pmatrix} $$
To complete your projective scale, you also need to fix an origin, i.e. a point with coordinate “zero”, and a unit length, i.e. a point with coordinate “one”. Since all distances in your formula are measured from $G$, it makes sense to use that as the zero point. And since you could multiply all your lengths by a common scale factor without affecting the formula you stated, the choice of scale is irrelevant. Therefore we might as well choose $X$ as one. Note that this choice also fixes the orientation of length measurements along your line: positive is from $G$ in direction of $X$. We can then compute the two remaining coordinates, those of $Y$ and $Z$, using the cross ratio. In the following formula, square brackets denote determinants. The cross ratio of four collinear points in the plane can be computed as seen from some fifth point not on that line. I'll use $A$ for this purpose, both because it has simple coordinates and because, as stated above, the case of $l$ passing through $A$ has been omitted by the choice of coordinates for $l$.
$$ GY = \operatorname{cr}(F,G;X,Y)_A = \frac{[AFX][AGY]}{[AFY][AGX]} = \frac{\begin{vmatrix} 1 & 2a+1 & a \\ 0 & -2-a & -1 \\ 0 & 1-a & 0 \end{vmatrix}\cdot\begin{vmatrix} 1 & 1 & 1+a \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix}}{\begin{vmatrix} 1 & 2a+1 & 1+a \\ 0 & -2-a & 0 \\ 0 & 1-a & 1 \end{vmatrix}\cdot\begin{vmatrix} 1 & 1 & a \\ 0 & 1 & -1 \\ 0 & 1 & 0 \end{vmatrix}} = \frac{a-1}{a+2} \\ GZ = \operatorname{cr}(F,G;X,Z)_A = \frac{[AFX][AGZ]}{[AFZ][AGX]} = \frac{\begin{vmatrix} 1 & 2a+1 & a \\ 0 & -2-a & -1 \\ 0 & 1-a & 0 \end{vmatrix}\cdot\begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1+a \\ 0 & 1 & a \end{vmatrix}}{\begin{vmatrix} 1 & 2a+1 & 0 \\ 0 & -2-a & 1+a \\ 0 & 1-a & a \end{vmatrix}\cdot\begin{vmatrix} 1 & 1 & a \\ 0 & 1 & -1 \\ 0 & 1 & 0 \end{vmatrix}} = \frac{1-a}{1+2a} $$
The thrid length, $GX$, is $1$ by the definition of the projective scale. So now you have the three lengths and plug them into your formula.
$$ \frac1{GX} + \frac1{GY} + \frac1{GZ} = \frac{a-1}{a-1} + \frac{a+2}{a-1} - \frac{2a+1}{a-1} = 0 $$
As you will notice, the case of $a=1$ is problematic in this setup, since it would entail a division by zero. This corresponds to the situation where $l$ is parallel to $AB$. In that case, the point $X$ which we used as the “one” of the scale would coincide with the point $F$ at infinity, thus breaking the scale. A renaming of triangle corners will again take care of this special case. The cases $a=-2$ and $a=-\tfrac12$ correspond to the two other cases where $l$ is parallel to one of the edges. You will notice that the lengths would again entail divisions by zero, since the points of intersection are infinitely far away. But the reciprocal lengths are just fine and will be zero in those cases.
Per light of @Blue's last comment, here's a (very brief) solution:
Since $E$ is the midpoint of $E'E''$ and $\angle A=90^\circ$, $\triangle AEE'$ is isosceles at $E$. It follows that $\angle AEE''=2 \angle EAE'$, equivalently $E$ trisects the small arc $AC$.
Similarly, $F$ trisects the small arc $AB$ and $D$ trisects the large arc $AC$ (and $AB$).
That would show $\triangle DEF$ is equilateral.
Best Answer
See the picture below:
Note that: $$\triangle AB'F \sim \triangle ABJ_1 \quad (1)$$ $$\triangle AFG \sim \triangle AJ_1K_1 \quad (2)$$ $$\triangle AGC' \sim \triangle AK_1C \quad (3)$$ As triangles $\triangle ABJ_1$, $\triangle AJ_1K_1$, $\triangle AK_1C$ share common sides and triangles $\triangle AB'F$, $\triangle AFG$ and $\triangle AGC'$ too, the similarity ratio $k$ is the same for $(1)$, $(2)$ and $(3)$. In this way we have: $$BJ_1=k\,B'F = k \, r$$ $$J_1K_1=k\,FG = k \, r$$ $$K_1C=k\,GC = k \, r$$
Therefore
$$BJ_1=J_1K_1=K_1C.$$