<s>[INST] Find side of an equilateral triangle inscribed in a rhombus The lengths of the diagonals of a rhombus are 6 and 8. An equilateral triangle inscribed in this
rhombus has one vertex at an end-point of the shorter diagonal and one side parallel to the longer
diagonal. Determine the length of a side of this triangle. 
Express your answer in the form $k\left(4 \sqrt{3} − 3\right)$ where k is a vulgar fraction.
 [/INST] Here is a coordinate geometry solution:

The diagonals of a rhombus are perpendicular bisectors of each other.  Suppose that the diagonals intersect at the origin of the coordinate plane, that the vertices at the ends of the short diagonal are $A(-3, 0)$ and $C(3, 0)$, that the vertices at the ends of the long diagonal are $B(0, 4)$ and $D(0, -4)$, and that one vertex of the inscribed equilateral triangle is the vertex $C$ of the rhombus.  
The other vertices of the equilateral triangle must lie on sides $\overline{AB}$ and $\overline{AD}$ of the rhombus.  Let $E$ be the vertex of the equilateral triangle on side $\overline{AD}$ and $F$ be the vertex of the equilateral triangle that lies on side $\overline{AB}$.  Since the line containing an altitude of an equilateral triangle is the perpendicular bisector of the base, points $E$ and $F$ are equidistant from the $x$-axis.  Thus, if the coordinates of point $F$ are $(a, b)$, then the coordinates of point $E$ are $(a, -b)$.  If $s$ is the length of a side of the equilateral triangle, then $s = b - (-b) = 2b$.  
In an equilateral triangle of side length $s$, the length of an altitude is $$\frac{s\sqrt{3}}{2}$$
Since $a < 0$, the length of the altitude is 
$$3 - a = \frac{s\sqrt{3}}{2}$$
Since $s = 2b$, we obtain
$$3 - a = b\sqrt{3}$$
Solving for $a$ yields 
$$3 - b\sqrt{3} = a$$
The equation of $\overleftrightarrow{AB}$ is 
$$y = \frac{4}{3}x + 4$$
Hence, at point $F(a, b)$, 
\begin{align*}
b & = \frac{4}{3}(3 - b\sqrt{3}) + 4\\
3b & = 4(3 - b\sqrt{3}) + 12\\
3b & = 12 - 4b\sqrt{3} + 12\\
3b + 4b\sqrt{3} & = 24\\
b(3 + 4\sqrt{3}) & = 24\\
b & = \frac{24}{3 + 4\sqrt{3}}\\
b & = \frac{24}{3 + 4\sqrt{3}} \cdot \frac{3 - 4\sqrt{3}}{3 - 4\sqrt{3}}\\
b & = \frac{24(3 - 4\sqrt{3})}{9 - 48}\\
b & = -\frac{24}{39}(3 - 4\sqrt{3})\\
b & = -\frac{8}{13}(3 - 4\sqrt{3})\\
b & = \frac{8}{13}(4\sqrt{3} - 3)
\end{align*}
Hence, 
$$s = 2b = \frac{16}{13}(4\sqrt{3} - 3)$$
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