# Three Circles Forming a Triangle +-- {.image} [[ThreeCirclesFormingaTriangle.png:pic]] > The circles have diameters $1$, $2$ and $3$. What’s the angle? =-- ## Solution by [[Pythagoras' Theorem]] +-- {.image} [[ThreeCirclesFormingaTriangleLabelled.png:pic]] =-- In the above diagram, the points labelled $A$, $B$, and $C$ are the centres of their respective circles. As the circles have diameters $1$, $2$, and $3$, the lengths of the segments $A C$, $B C$, and $A B$ are, respectively, $\frac{1}{2} + 1 = \frac{3}{2}$, $\frac{1}{2} + \frac{3}{2} = 2$, and $1 + \frac{3}{2} = \frac{5}{2}$. These lengths satisfy: $$ \left(\frac{3}{2}\right)^2 + 2^2 = \frac{25}{4} = \left(\frac{5}{2}\right)^2 $$ They therefore fit into the identity for the converse to [[Pythagoras' theorem]] and so triangle $A C B$ is a [[right-angled triangle]] with the [[right-angle]] at point $C$. Then since the [[angle at the centre is twice the angle at the circumference]], angle $F \hat{H} G$ is half of $90^\circ$, which is $45^\circ$.