Dissect a regular 12-gon into a regular hexagon, 6 squares and 6 equilateral triangles. Let the regular 12-gon have vertices P_{1}, P_{2}, ... , P_{12} (in that order). Show that the diagonals P_{1}P_{9}, P_{12}P_{4} and P_{2}P_{11} are concurrent.

**Solution**

Take squares on the inside of alternate sides of the 12-gon. Each angle of the 12-gon is 150^{o}, so the remaining angles after placing the squares are all 60^{o}, so we have equilateral triangles on the inside of the remaining 6 sides of the 12-gon. Now the opposite sides of the squares form a hexagon. (As a check we may notice that if A, B, C, D are adjacent vertices of the 12-gon and we place squares ABPQ on AB and CDRS on CD, then S and P coincide and the equilateral triangle ABP and the two squares take 60^{o} + 90^{o} + 90^{o} out of the 360^{o} angle at P leaving 120^{o}, which is the angle between sides of a regular hexagon.)

Let O be the centre of the 12-gon. P_{1}P_{9} and P_{12}P_{4} have the same length and one is obtained from the other by rotation about O through 90^{o}. Thus they intersect on the perpendicular bisector of P_{12}P_{1} a distance P_{12}P_{1}/2 from P_{12}P_{1}. P_{11}P_{2} is parallel to P_{12}P_{1} and its midpoint lies on the perpendicular bisector of P_{12}P_{1}. The angle P_{11}P_{2}P_{1} is 180^{o} - angle P_{12}P_{1}P_{2} = 30^{o}, so the distance between of the midpoint along the bisector is P_{12}P_{1}sin 30^{o} = P_{12}P_{1}/2. Hence the three lines intersect at this point.

© John Scholes

jscholes@kalva.demon.co.uk

5 Feb 2002