# Solution to the Overlapping Triangles in a Circle Puzzle +-- {.image} [[OverlappingTrianglesinaCircle.png:pic]] > The circle contains a square and two overlapping equilateral triangles. What's the area of this overlap? =-- ## Solution by [[Angle at the Centre is Twice the Angle at the Circumference]], [[Isosceles]] and [[Equilateral Triangles]] +-- {.image} [[OverlappingTrianglesinaCircleLabelled.png:pic]] =-- Let the points be labelled as above. Consider triangle $A H C$. As $A H I B$ is a square and $H I C$ an equilateral triangle, $A H$ and $H C$ have the same length so triangle $A H C$ is [[isosceles]]. Angle $C \hat{H} A$ is $90^\circ + 60^\circ = 150^\circ$ so angle $A \hat{C} H$ is $15^\circ$. A similar argument shows that angle $I \hat{C} B$ is also $15^\circ$, so angle $B \hat{C} A$ is $30^\circ$. This is half of angle $B \hat{O} A$ so as triangle $B O A$ is [[equilateral]], since the [[angle at the centre is twice the angle at the circumference]], $O$ is the centre of the circle. Triangle $O G D$ is then [[isosceles]] with angle $120^\circ$ at $O$ (as it is twice angle $G \hat{C} D$). This means that angle $O \hat{D} E$ is $30^\circ$ which is the same as angle $E \hat{O} D$ so triangle $E O D$ is [[isosceles]] and can be thought of as two half [[equilateral triangles]] aligned along their short sides. The length of $O D$ is therefore $\sqrt{3}$ times the length of $O E$. Then $O D$ is the same length as $O A$, and of $A B$, and of $H I$. So the side length of equilateral triangle $H I C$ is $\sqrt{3}$ times the side length of $E F O$. The area of $H I C$ is therefore $3$ times the area of $E F O$, and so $E F O$ has area $8$.