Hanging Cube

The side view is in the plane of $ABCD$ . The edges of the cube $AB=DC=1$ , the diagonals of the faces $BC=AD=\sqrt{2},$ and the and the space diagonal $AC=\sqrt{3}$ . The horizontal plane has been placed directly at the bottom of the cube. As the shape of the shadow doesn't change with change in that position, only its size, this is an acceptable choice.

Triangles $ABC, BEC,$ and $AEB$ are all similar, from which we can get $AE=FC=\frac{1}{\sqrt{3}}$ , $EB=DF=\frac{\sqrt{2}}{\sqrt{3}}$ .

We will use this to get an expression for $\frac{y}{z}$ in terms of $x$ . The shape of the shadow will give us another way of expressing $\frac{y}{z}$ . Comparing those two will result in an equation for $x$ .

Triangles $LCG$ and $LEB$ as similar, so $\frac{y}{x+\sqrt{3}}=\frac{\sqrt{2}}{\sqrt{3}}\times \frac{1}{x+\frac{1}{\sqrt{3}}}$

From similarity of $LCH$ and $LFD$ we get $\frac{z}{x+\sqrt{3}}=\frac{\sqrt{2}}{\sqrt{3}}\times \frac{1}{x+\sqrt{3}-\frac{1}{\sqrt{3}}}$

So the ratio is $\frac{y}{z}=\frac{x+\sqrt{3}-\frac{1}{\sqrt{3}}}{x+\frac{1}{\sqrt{3}}}$

In the top view, the angles in the triangle $CGH’$ are $\angle H’CG=\frac{360^\circ}{6}=60^\circ, \angle CGH’=\frac{90^\circ}{2}=45^\circ$ ,and $\angle GH’C=180^\circ-60^\circ-45^\circ=75^\circ$ .

From the law of sines, $\frac{y}{z}=\frac{sin(75^\circ)}{ sin(45^\circ)}=\frac{1+\sqrt{3}}{2}$

So our equation for $x$ is $\frac{x+\sqrt{3}-\frac{1}{\sqrt{3}}}{x+\frac{1}{\sqrt{3}}}=\frac{1+\sqrt{3}}{2}$

The solution is $x=1$ . So the distance of the light source from the top vertex of the cube is the same size as one of its edges.