Shadow Inside a Cube

Figure 1

Following the labelling of figure 1, $ILKL$ is the inscribed square, $M$ and $O$ are the midpoints of sides $IL$ and $JK$ respectively, $P$ and $Q$ are the midpoints of edges $AE$ and $CG$ respectively and $N$ is the intersection point of $AE$ with the extension of $GM$ .
If the lamp is situated at $G$ , then the shadow covers $\triangle ABC$ on the bottom of the cube, plus the quadrilaterals $ANLD$ and $ANIB$ , the last two being congruent.

Now, the right triangles $\triangle PMN$ and $\triangle QMG$ are similar, hence $\dfrac{PN}{QG}=\dfrac{PM}{MQ}\Rightarrow \dfrac{PN}{3}=\dfrac{1}{3}\Rightarrow PN=1$ Thus, $AN=AP-PN=3-1=2$ Figure 2

We divide the quadrilateral $ANLD$ in two triangles: $\triangle ALD$ and $\triangle ANL$ (figure 2).

Evaluating areas we have $\left[ ALD \right]=\frac{1}{4}\left[ AEHD \right]=\frac{1}{4}\cdot {{6}^{2}}=9$ $\left[ ANL \right]=\frac{1}{2}AN\cdot LP=\frac{1}{2}\cdot 2\cdot 3=3$ Moreover, $\left[ ABD \right]=\frac{\left[ ABCD \right]}{2}=\frac{{{6}^{2}}}{2}=18$ Summing up, the total area of the shadow is

$\left[ ABD \right]+2\left[ ANLD \right]=\left[ ABD \right]+2\left( \left[ ALD \right]+\left[ ANL \right] \right)=18+2\left( 9+3 \right)=\boxed{42}$

Don't have a 3-D graph app, I used the above 2-D drawing to solve the problem. The top-left plot is an elevation of the cube showing the point light source is at the top left corner the red line is the square in the cube. The bottom-left plot is a plan of the cube with the square in red. Grey indicates the shadow cast. The shadow on the two vertical square opposite the light source is each a truncated triangle. The show on the bottom square is an isosceles right triangle. The area of the shadow is:

$A_{\rm shadow} = 2\left(\frac {3\times 3}2 + \frac {3+2}2 \times 3\right) + \frac {6 \times 6}2 = \boxed{42}$

Note that:

• the light or shadow will only be on 3 faces that do not have the lamp as a vertex (one bottom and 2 side faces).
• on each face, straight lines will project into straight lines.
• assuming the lamp is at a top, the nearest two vertices of the square are projected on opposite floor vertices,so the floor face of the cube is half lit.
• the midpoint of the far edge of the square is projected onto the far vertical edge of the cube, $\frac13$ from the bottom.
• drawing straight lines to the centre of a side face (adjacent to this vertical edge), we can see that each of these faces has $\frac18+\frac18+\frac{1}{12}=\frac13$ of its area shadowed.

In total we have $(\frac12+2×\frac13)×36=42$ square units shadowed.