### 24th Putnam 1963

Problem A1

Dissect a regular 12-gon into a regular hexagon, 6 squares and 6 equilateral triangles. Let the regular 12-gon have vertices P1, P2, ... , P12 (in that order). Show that the diagonals P1P9, P12P4 and P2P11 are concurrent.

Solution

Take squares on the inside of alternate sides of the 12-gon. Each angle of the 12-gon is 150o, so the remaining angles after placing the squares are all 60o, 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 60o + 90o + 90o out of the 360o angle at P leaving 120o, which is the angle between sides of a regular hexagon.)

Let O be the centre of the 12-gon. P1P9 and P12P4 have the same length and one is obtained from the other by rotation about O through 90o. Thus they intersect on the perpendicular bisector of P12P1 a distance P12P1/2 from P12P1. P11P2 is parallel to P12P1 and its midpoint lies on the perpendicular bisector of P12P1. The angle P11P2P1 is 180o - angle P12P1P2 = 30o, so the distance between of the midpoint along the bisector is P12P1sin 30o = P12P1/2. Hence the three lines intersect at this point.