The six sides of a convex hexagon $$A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}$$ are colored red. Each of the diagonals of the hexagon is colored either red or blue. If $$N$$ is the number of colorings such that every triangle $$A_{i}A_{j}A_{k}$$, where $$1\le i<j<k\le 6$$, has at least one red side, find the sum of the squares of the digits of $$N$$.

All six sides of the hexagon are already red, so any triangle that uses at least one side of the hexagon automatically contains a red side.
Hence we only have to worry about triangles whose three sides are all diagonals.

In a regular hexagon the only triples of vertices that are pairwise non-adjacent (so that every side of the triangle joining them is a diagonal) are the alternate-vertex triples $$\{A_1,A_3,A_5\}$$ and $$\{A_2,A_4,A_6\}$$. Thus the only two “all-diagonal” triangles are $$\triangle A_1A_3A_5$$ and $$\triangle A_2A_4A_6$$.

Label the nine diagonals as follows:
Triangle $$A_1A_3A_5$$ (set $$X$$): $$A_1A_3,\;A_3A_5,\;A_5A_1$$  (3 diagonals)
Triangle $$A_2A_4A_6$$ (set $$Y$$): $$A_2A_4,\;A_4A_6,\;A_6A_2$$  (3 diagonals)
Remaining crossed diagonals (set $$Z$$): $$A_1A_4,\;A_2A_5,\;A_3A_6$$  (3 diagonals)

Total colourings of the nine diagonals (red/blue independently) $$=2^{9}=512.$$

The requirement “every triangle has at least one red side” is violated only if • all three diagonals of $$X$$ are blue, or • all three diagonals of $$Y$$ are blue (or both).

Let
$$A=\{\text{all edges of }X\text{ are blue}\},\qquad B=\{\text{all edges of }Y\text{ are blue}\}.$$

Count the forbidden colourings:

Case |A|: 3 edges fixed blue, the remaining 6 edges free $$\Longrightarrow |A|=2^{6}=64.$$

Case |B|: similarly $$|B|=64.$$

Case |A$$\cap$$ B|: both sets of 3+3=6 edges fixed blue, the 3 edges of $$Z$$ free $$\Longrightarrow |A\cap B|=2^{3}=8.$$

By the Principle of Inclusion-Exclusion, the number of forbidden colourings is $$|A|+|B|-|A\cap B|=64+64-8=120.$$

Therefore the required number of valid colourings is $$N=512-120=392.$$

The digits of $$N$$ are 3, 9, and 2. Sum of the squares of the digits: $$3^{2}+9^{2}+2^{2}=9+81+4=94.$$

Final answer: 94