# Solution to the [[Two Circles in a Triangle II]] Puzzle +-- {.image} [[TwoCirclesinaTriangleII.png:pic]] > What fraction of this equilateral triangle is shaded? =-- ## Solution by Lengths in an [[Equilateral Triangle]] +-- {.image} [[TwoCirclesinaTriangleIIAnnotated.png:pic]] =-- Consider the diagram labelled as above. As the question is one of proportion, it can be assumed that the radius of the circles is $1$. Then $I J$ has length $2$, as do $C B$ and $F G$. Triangle $D I C$ is half of an [[equilateral triangle]], so the length of $D C$ is $\sqrt{3}$ times that of $C I$, so is $\sqrt{3}$. Therefore, $A D$ has length $2 + 2 \sqrt{3}$. Similarly, triangle $I E F$ is half of an equilateral triangle, so $I F$ has length $\sqrt{3}$ times that of $E F$, and so $E F$ has length $\frac{1}{\sqrt{3}}$. This means that $E H$ has length $2 + \frac{2}{\sqrt{3}}$. The length scale factor from the outer triangle to the shaded triangle is therefore: $$ \frac{2 + \frac{2}{\sqrt{3}}}{2 + 2 \sqrt{3}} = \frac{\frac{2}{\sqrt{3}}(\sqrt{3} + 1)}{2(1 + \sqrt{3})} = \frac{1}{\sqrt{3}} $$ So the area scale factor is $\frac{1}{3}$ and hence one third of the triangle is shaded.