Let $ABCDEF$ be a regular hexagon with sidelength $6$, and construct squares $ABGH$, $BCIJ$, $CDKL$, $DEMN$, $EFOP$, and $FAQR$ outside the hexagon. Find the perimeter of dodecagon $HGJILKNMPORQ$.

[i]Proposed by Andrew Wu[/i]
1. **Identify the side lengths of the squares:**
   Since $ABGH$, $BCIJ$, $CDKL$, $DEMN$, $EFOP$, and $FAQR$ are squares constructed outside the regular hexagon $ABCDEF$ with side length $6$, each side of these squares is also $6$ units long. Therefore, we have:
   \[
   HG = JI = LK = NM = PO = RQ = 6
   \]

2. **Determine the angles between the sides:**
   The angle between any two adjacent sides of the regular hexagon is $120^\circ$. Since the squares are constructed outside the hexagon, the angle between a side of the hexagon and a side of the square is $90^\circ$. Therefore, the angle between two adjacent sides of the dodecagon (formed by the vertices of the squares) is:
   \[
   360^\circ - 2 \cdot 90^\circ - 120^\circ = 60^\circ
   \]

3. **Identify the equilateral triangles:**
   The triangles formed by the vertices of the squares and the vertices of the hexagon are equilateral triangles. For example, $\triangle QAH$, $\triangle GBJ$, $\triangle ICL$, $\triangle KDN$, $\triangle MEP$, and $\triangle OFR$ are all equilateral triangles. Therefore, the sides of these triangles are equal to the side length of the hexagon, which is $6$ units. Hence:
   \[
   QH = GJ = IL = KN = MP = OR = 6
   \]

4. **Calculate the perimeter of the dodecagon:**
   The dodecagon $HGJILKNMPORQ$ consists of 12 sides, each of length $6$ units. Therefore, the perimeter of the dodecagon is:
   \[
   12 \cdot 6 = 72
   \]

The final answer is $\boxed{72}$