A Christmas Fractal Snowflake

Since the area of the original hexagon is 1, and 6 equilateral triangles that are $\frac{1}{36}$ the area of the original hexagon are taken away, the area of the first snowflake-base-shape in Step 2 is $A = 1 - 6 \cdot \frac{1}{36} = \frac{5}{6}$ . Then 6 more snowflake-base-shapes are added in Step 3 that are $\frac{1}{9}$ the area of $\frac{5}{6}$ , for an additional area of $A = 6 \cdot \frac{1}{9} \cdot \frac{5}{6} = \frac{5}{9}$ . Then 5 more snowflake-base-shapes are added in Step 4 to those 6 existing snowflake-base-shapes that are $\frac{1}{9}$ the area of $\frac{1}{9}$ the area of $\frac{5}{6}$ , for an additional area of $A = 5 \cdot 6 \cdot \frac{1}{9} \cdot \frac{1}{9} \cdot \frac{5}{6} = \frac{25}{81}$ . With this step, and every step after it, 5 snowflake-base-shapes are added to each of the previous step’s snowflake-base-shapes at $\frac{1}{9}$ the previous area for a ratio of $\frac{5}{9}$ . The total area is therefore:

$A = \frac{5}{6} + \frac{5}{9} + \frac{5}{9} \cdot \frac{5}{9} + \frac{5}{9} \cdot \frac{5}{9} \cdot \frac{5}{9} + …$

From the second term onward there is an infinite geometric sequence whose sum is $S = \frac{t_{1}}{1 - r} = \frac{\frac{5}{9}}{1 - \frac{5}{9}} = \frac{5}{4}$ . Substituting this into the area formula gives $A = \frac{5}{6} + \frac{5}{4}$ which simplifies to $\frac{25}{12}$ (Merry Christmas!), and $25 + 12 = \boxed{37}$ .