# How many spheres are needed to shield a point source of light?

How many spheres are needed to shield a point source of light?

I read this from a mathematical puzzle book. And book says the answer is six without explanation.

From the geometric point of view, I'm thinking of a tetrahedron where the point source is located at the center. Then we can decompose 4 sectors, thus it seems to me 4 spheres is enough to shield the light. So I'm confused with the answer given in the book. If someone understands this problem, can you explain it to me?

• I suppose they might be imagining it surrounded by a cube and the spheres are placed so that they each cast a shadow over one face of the cube. May 22, 2018 at 8:57
• And I want to know what name your book is. May 22, 2018 at 11:29
• I've edited your post a bit for grammar. I hope the edits are to your liking. Also, it's not clear exactly what you mean by "Then we can decompose 4 sectors". Could you elaborate on that? May 22, 2018 at 16:26
• I mean, one sphere will do it, if it encloses the point. May 27, 2018 at 17:28

The minimal number of spheres required to shield the origin is $4$.

4 spheres is sufficient.

Consider following $4$ points on unit sphere $S^2$, $$v_0 = \frac{1}{\sqrt{3}}( -1,-1, -1), v_1 = \frac{1}{\sqrt{3}}( -1, 1, 1), v_2 = \frac{1}{\sqrt{3}}( 1,-1, 1), v_3 = \frac{1}{\sqrt{3}}( 1, 1, -1)$$ they are forming the vertices of a regular tetrahedron. It is easy to check every point on $S^2$ is at an angular distance no more than $\cos^{-1}\frac13$ from one of these vertices.

For each $k = 0,1,2,3$, place a sphere of radius $( \frac{\sqrt{8}}{3} + \epsilon)\rho_k$ at $\rho_k v_k$. These 4 spheres will block every ray start at origin. In order for them not to overlap, a sufficient condition is $$\max\left(\frac{\rho_i}{\rho_j},\frac{\rho_j}{\rho_i}\right) > 5+2\sqrt{6} \approx 9.899$$ for every $i \ne j$. By setting $(\rho_0,\rho_1,\rho_2,\rho_3)$ to $(1,10,100,1000)$, we obtain $4$ non-overlapping spheres which completely shield the origin.

4 spheres is necessary.

Given any $3$ spheres $S_1, S_2, S_3$. Let $c_1, c_2, c_3$ be unit vectors pointing towards their centers. Given any ray pointing at direction $n$, if sphere $S_i$ block it, we have $n \cdot c_i > 0$. Given $c_1, c_2$, it is easy to find a unit vector $u$ such that $u \cdot c_1 = u \cdot c_2 = 0$, this means neither $S_1$ nor $S_2$ is blocking the two rays pointing at direction $\pm u$. Since $S_3$ can block at most one of these rays, $3$ spheres is not enough to shield the origin.

A first pair of spheres, tangent between them at the point source, will shield all the space except their tangency plane. Two other tangent pairs of spheres, each formed by two spheres symmetric about the source, are then needed: their radii must be different to ensure the tangent plane not to pass through the source.

• This argument doesn't address whether six sphere is the minimum though. May 22, 2018 at 16:21
• In fact it is not the minimum, as achille hui brilliantly showed. May 22, 2018 at 19:53

I have ever created this puzzle. My puzzle was illuminating earth by multiple suns. This is more easier than yours. At first two spheres cover almost area, but remains one longer line than half circle. So 4 sphere is needed.