# Solution to the Two Circles in a Square II Puzzle +-- {.image} [[TwoCirclesinaSquareII.png:pic]] > Two circles in a square. What’s the angle? =-- ## Solution by [[Angles in a Polygon]], [[Angle Between a Radius and Tangent]], and [[Isosceles Triangle]] +-- {.image} [[TwoCirclesinaSquareIILabelled.png:pic]] =-- In the above diagram, $A$ and $C$ are the centres of the circles. The shape $A C D E F$ is a pentagon and so the sum of its [[angles in a polygon|interior angles]] is $540^\circ$. Three of its angles are right-angles: $F \hat{E} D$ is the interior angle of a square while $A\hat{F}E$ and $E \hat{D} C$ are the [[angle between a radius and tangent]]. The remaining angles must therefore add up to $540^\circ - 3 \times 90^\circ = 270^\circ$. Triangles $C D B$ and $A B F$ are both [[isosceles]] as $A$ and $C$ are the centres of their respective circles. Angle $C \hat{B} D$ is half of what remains after angle $D \hat{C} B$ is taken away from $180^\circ$ and similar for angle $F \hat{B} A$. So the sum of these two angles is given by: $$ F\hat{B} A + D \hat{C} D = \frac{1}{2} ( 180^\circ - B \hat{A} F) + \frac{1}{2} ( 180^\circ - D \hat{C} B) = 180^\circ - \frac{1}{2} ( B \hat{A} F + D \hat{C} B) $$ The requested angle is $180^\circ - (F \hat{B}A + D \hat{C} B)$ and so is $\frac{1}{2} ( B \hat{A} F + D \hat{C} B) = \frac{270^\circ}{2} = 135^\circ$. (*Note that the fact that the circles touch the diagonal line is not needed. This result holds for any two circles that are tangent to each other and to a pair of perpendicular lines.*)