Notes
circles in overlapping triangles solution

Solution to the Circles in Overlapping Triangles Puzzle

Solution by Lengths in an Equilateral Triangle and Area Scale Factor

In the above diagram, the points labelled $C$ and $E$ are the centres of their circles, while $G$ is such that angle $C \hat{G} A$ is a right-angle.

The line that extends $A C$ cuts the red triangle in half, so angle $G \hat{A} C$ is $30^\circ$ which shows that triangle $C G A$ is half an equilateral triangle. This means that the length of $A C$ is twice that of $C G$, so the length of $A B$ is the same as the radius of the circles. The length of $A E$ is therefore $4$ of these lengths, and $A F$ is $5$ of them.

Cutting the red equilateral triangle at $E$ gives a triangle $\frac{4}{5}$ths of the height of the red one, which therefore has area $\frac{16}{25}$ths of the red as the area scale factor is the square of the length scale factor. Since the red triangle has area $100$, this shorter triangle has area $64$.

Triangle $J I H$ is also equilateral and its height (of $J$ above $I H$) is one radius of the blue circles, which is one fifth of the red triangle. Its area is therefore one twenty-fifth of the area of the red triangle, which is $4$.

The total area of the design is therefore: