Galperin G.A.

What is the minimum number of points that need to be marked on the surface of

a) a dodecahedron

b) an icosahedron

so that each face has at least one marked point?
a) Since a dodecahedron has 12 faces and each of its vertices belongs to three of them, the sought minimum is no less than 4. The figure on the left shows how to mark four vertices.

![](https://cdn.mathpix.com/cropped/2024_05_06_309dcf72bd08bf7391b4g-34.jpg?height=432&width=958&top_left_y=1&top_left_x=550)

b) Six vertices, highlighted in the figure on the right, are sufficient.

Let's prove that a smaller number cannot suffice. The 12 vertices of the icosahedron are divided into six pairs of opposite vertices. If at least one vertex is marked in each pair, then this already amounts to six points. Suppose that no vertex in a pair $(A, B)$ of opposite vertices is marked. Then, it requires at least three vertices to "serve" the five faces adjacent to $A$, and another three for the faces around $B$.

## Answer

a) 4 points; b) 6 points.