AMC 12 Mock Exam 2 Problem 18

Place the vertices at $(\pm a,0,0)$, $(0,\pm a,0)$, $(0,0,\pm a)$; the edge length is $a\sqrt2=6$, so $a=3\sqrt2$. The face with vertices $(a,0,0),(0,a,0),(0,0,a)$ lies in the plane $x+y+z=a$, so the cutting plane is $x+y+z=0$.

The octahedron's twelve edges join each vertex to the four vertices not opposite it. The plane $x+y+z=0$ separates the three vertices with positive coordinate sum from the three with negative sum, so it crosses exactly the six edges joining a "positive" vertex to a "negative" one (such as from $(a,0,0)$ to $(0,-a,0)$), and it meets each of them at its midpoint (such as $\left(\frac a2,-\frac a2,0\right)$). The six midpoints are $$\left(\tfrac a2,-\tfrac a2,0\right),\ \left(\tfrac a2,0,-\tfrac a2\right),\ \left(0,\tfrac a2,-\tfrac a2\right),\ \left(-\tfrac a2,\tfrac a2,0\right),\ \left(-\tfrac a2,0,\tfrac a2\right),\ \left(0,-\tfrac a2,\tfrac a2\right),$$ consecutive ones are $\frac{a}{\sqrt2}=3$ apart, and the threefold symmetry about the line $x=y=z$ (together with the central symmetry) makes the hexagon regular. A regular hexagon with side $3$ has area $\frac{3\sqrt3}{2}\cdot3^2=\boxed{\tfrac{27\sqrt3}{2}}$.

(A face has area $9\sqrt3$; the cross-section is $\frac32$ times larger.)