Notes
pentagon with two semi-circles solution

Solution to the Pentagon with two Semi-Circles Puzzle

Pentagon with two Semi-Circles

The regular pentagon is tangent to both semicircles at the point where the three meet. What’s the angle between the bases of the semicircles? What if I started with other regular polygons?

Solution by Angle Between a Radius and Tangent and Angles in a Regular Pentagon

Pentagon with two semi-circles labelled

In the above diagram, the points labelled BB and HH are the centres of their semi-circles.

Since the angle between a radius and tangent is 90 90^\circ, angles BE^DB \hat{E} D and FE^HF \hat{E} H are both 90 90^\circ. Angle DE^FD \hat{E} F is the interior angle in a regular pentagon so is 108 108^\circ. Since the angles at a point add up to 360 360^\circ, this leaves 72 72^\circ for angle HE^BH \hat{E} B.

Triangles ABEA B E and AEGA E G are both isosceles, so angles BA^HB \hat{A} H are HE^BH \hat{E} B are equal and therefore the angle between the bases of the semi-circles is 72 72^\circ.