I found this 3D geometry problem but i can't solve it !
"I have $20$ cubes of side $1$, which i assemble to form a large cube of side $3$ but without the cubes on the center of each face and without the cube in the center. Then, i place this solid on one of its vertex, so that two vertices diametrically opposite ends up vertical to each other, and then cut it in two by a horizontal plane located halfway between these two vertices. What is the area of the horizontal face common to the two halves?"
You can see below how that surface looks like. It's formed by the well-known regular hexagon whose vertices are the midpoints of six edges of the cube, from which a central star-like hole has been cut out.
That happens because the central cubic hole is cut by the plane in the same way as the large cube, thus creating a hexagonal hole whose sides are one third of the sides of the large hexagon. On the other hand the plane doesn't pass through the center of the six lateral holes and just cuts them at six triangles, whose vertices are the midpoints of three concurring sides of the small cubes.
The large hexagon has sides of length ${3\over2}\sqrt2$, while the sides of the star are one third of those: ${1\over2}\sqrt2$. From that the area of the surface can be computed as ${21\over4}\sqrt3$.
Here's a detail of the central small cube with the other six cubic cavities attached to it, and the cutting plane.