In fact the solution shown by Wolfram Mathworld is a constructible five-piece solution, presumably Lindgren's.

Let each side of the hexagon be one unit, and label the hexagon $ABCDEF$ with $A$ at the vertex on the left side and the other vertices ordered clockwise. Then the cut from $A$ to point $G$ on side $BC$ measures half the side of the equilateral triangle, thus $\sqrt6/2$. The perpendicular to $BC$ from $G$ intersects side $EF$ at $H$, and the cut from $H$ to $I$ on cut $AG$ makes angle $AIH$ which measures $60°$. Also the length of $HI$ measures $\sqrt6/2$, congruent with $AG$. The remaining cut from $I$ bisects angle $GIH$ and the remaining two cuts (from $D$ and $E$) meet at the center of the hexagon.

In fact the solution shown by Wolfram Mathworld is a constructible five-piece solution, presumably Lindgren's.

Let each side of the hexagon be one unit, and label the hexagon $ABCDEF$ with $A$ at the vertex on the left side and the other vertices ordered clockwise. Then the cut from $A$ to point $G$ on side $BC$ measures half the side of the equilateral triangle, thus $\sqrt6/2$. Point $H$ on side $EF$ is at distance $\sqrt6/2×\sqrt3/2=3\sqrt2/4$ from the cut $AG$, and the cut from $H$ to $I$ on cut $AG$ makes angle $AIH$ which measures $60°$. The length of $HI$ measures $\sqrt6/2$, congruent with $AG$. The remaining cut from $I$ bisects angle $GIH$ and the remaining two cuts (from $D$ and $E$) meet at the center of the hexagon.

In fact the solution sh9wn by Wolfram Mathworld is a constructible five-piece solution, probably Lindgren's.

Let each side of the hexagon be one unit, and label the hexagon $ABCDEF$ with $A$ at the vertex on the left side and the other certices ordered clockwise. Then the cut from $A$ to point $G$ on side $BC$ measures half the side of the equilateral triangle, thus $\sqrt6/2$. The perpendicular to $BC$ from $G$ intersects side $EF$ at $H$, and the cut from $H$ to $I$ on cut $AG$ makes angle $AIH$ which measures $60°$. Also the length of $HI$ measures $\sqrt6/2$. The remaining cut from $I$ bisects angle $GIH$ and the remaining two cuts (from $D$ and $E$) meet at the center of the hexagon.

In fact the solution shown by Wolfram Mathworld is a constructible five-piece solution, presumably Lindgren's.

Let each side of the hexagon be one unit, and label the hexagon $ABCDEF$ with $A$ at the vertex on the left side and the other vertices ordered clockwise. Then the cut from $A$ to point $G$ on side $BC$ measures half the side of the equilateral triangle, thus $\sqrt6/2$. The perpendicular to $BC$ from $G$ intersects side $EF$ at $H$, and the cut from $H$ to $I$ on cut $AG$ makes angle $AIH$ which measures $60°$. Also the length of $HI$ measures $\sqrt6/2$, congruent with $AG$. The remaining cut from $I$ bisects angle $GIH$ and the remaining two cuts (from $D$ and $E$) meet at the center of the hexagon.