Three Congruent Incircles of a divided Equilateral triangle Take an equilateral triangle with sides of unit length and choose a vertex from which to draw two cevians to the opposite side.These cevians divide the equilateral triangle into three subtriangles.
If these three subtriangles all have congruent incircles, can anyone confirm that the length of these cevians is $1/4(3^{1/3}+3^{2/3})$ and the inradius of the congruent incircles is $1/2(3^{1/2}/(3+3^{1/3}+3^{2/3}))$? 
Using the Paul Yiu article at http://www.math-cs.ucmo.edu/~mjms/2003.1/pyiu.pdf, I had to solve a quartic equation to get the length of the cevians h where - $3 - 18 h - 27 h^2 + 16 h^3 + 48 h^4 = 0$. Is there a simpler method of calculating the cevian length? This question was first asked in Mathoverflow.
 A: Label elements of equilateral $\triangle ABC$ as shown:

Using the fact that 
$$\text{area of } \triangle = \frac{1}{2} \cdot \text{inradius} \cdot \text{perimeter}$$
we have
$$\frac{2|\triangle APB|}{1+x+y} = r = \frac{2|\triangle APQ|}{1-2x+2y}$$
$$\frac{x \cdot \frac{\sqrt{3}}{2}}{1+x+y} = r = \frac{(1-2x) \cdot \frac{\sqrt{3}}{2}}{1-2x+2y}$$
so that
$$x ( 1 - 2 x + 2 y ) = ( 1 - 2 x )( 1 + x + y ) \quad\to\quad y ( 4 x - 1 ) = 1 - 2 x$$
By the Law of Cosines, 
$$y^2 = 1^2 + x^2 - 2\cdot 1\cdot x\cos 60^\circ = x^2 - x + 1$$
which implies
$$( x^2 - x + 1 )( 4 x - 1 )^2 = ( 1 - 2 x )^2 \quad\to\quad x (16 x^3 - 24 x^2 + 21 x - 5 ) = 0$$
The case $x=0$ is extraneous (and evidently corresponds to the alternate problem of congruent incircles in $\triangle APQ$, $\triangle ABQ$, $\triangle ACP$), and Mathematica tells me that the sole real solution to the cubic factor is 
$$x = \frac{1}{4} (2 + s - s^2) = \frac{1}{4}(2-s)(1+s)$$
where $s := 3^{1/3}$.
Then,
$$y^2 = \frac{1}{16} (6 + 3\cdot s + s^2) = \frac{1}{16}(2 s^3 + s^4 + s^2 ) = \frac{s^2}{16}(1+s)^2 \quad\to\quad y = \frac{s}{4} ( 1 + s )$$ 
$$r = \frac{\frac{\sqrt{3}}{2}(1-2x)}{1 -2x+2y} = \frac{\frac{\sqrt{3}}{4}s(s-1)}{s^2} = \frac{(s-1)\sqrt{3}}{4s} = \frac{(s^3-1)\sqrt{3}}{4s(1+s+s^2)} = \frac{\sqrt{3}}{2s(1+s+s^2)}$$
which match the values given in the problem.
