# 2015 UNCO Math Contest II Problems/Problem 7

## Problem

(a) Give an example of a polyhedron whose faces can be colored in such a way that each face is either blue or gold, no two gold faces meet along an edge, and the total area of all the blue faces is half the total area of all the gold faces. A blue face may meet another blue face along an edge, and any colors may meet at vertices. Describe your polyhedron and also describe how to assign colors to the faces.

(b) Show that if the faces of a polyhedron are colored in such a way that each face is either blue or gold and no two gold faces meet along an edge, and if the polyhedron contains a sphere inside it that is tangent to each face, then the total area of all the blue faces is at least as large as the total area of all the gold faces.

## Solution

(a) One possible example is a square box with large top and bottom and small height, with the top and bottom colored gold and the four sides blue. If the top and bottom are squares of side $s$, then the total area colored gold is $2 \times s^2$ . If the height of the box is $h$, then the total area colored blue is $4\times h\times s$. It is easy to select s and h to make $4\times h\times s$ equal to half of $2\times s^2$. Set $4\times h\times s=s^2$ or $4\times h=s$. Choose, say, $h=1$ and $s=4$.

(b) Triangulate each face by connecting the point of tangency on the face to each vertex of the face. Look at a gold face and one of the triangles into which it has been cut. The gold triangle meets a blue triangle along the edge where the two faces meet. Consider the two triangles and the radii that go from the tangent points on the two faces (=the apexes of the triangles) to the center of the sphere. By symmetry, the two triangles are congruent. This means that for each triangle colored gold, there is a congruent triangle colored blue. Therefore, the total area colored blue is at least as large as the total area colored gold.