# Heme Mathematical Puzzle: Dodecagon

Suppose AB and BC are adjacent sides of the dodecagon, but differ in length. Since the arcs covering the six long and six short sides together form a complete circle, the arc on one long side and one short side must be one-sixth of the circle. Hence, the AMC angle is 360°/6 = 60°. It follows that the triangle AMC must be equilateral, and the segment AC has the length of the radius of the circle.

If we denote the angle AMB with α, then the other two angles of the isosceles triangle AMB are MAB = MBA = 12(180° – α) = 90° – 12α. The angle BMC should be 60° – α, so in the triangle BMC, the other two angles have the values ​​MBC = MCB = 12(180° – (60° – α)) = 60° + 12α. Since angle ABC is made up of angles MBA and MBC, it is ABC = MBA + MBC = 90° – 12α + 60° + 12α = 150 degrees. Finally, we apply the law of cosines to triangle ABC: (AC)2 = (ab)2 + (BC)2 – 2 AB BC cos150 degrees. If you enter the values, you get (AC)2 = (√2)2 + (24)2 + 2 √2 √24 12√3 = 38. The radius of the circle, regardless of the arrangement of the short and long side, r = √38 ≈ 6.164.

The cosine theorem I used here has unfortunately disappeared from the curricula of many schools, and is very useful. It states that for any triangle with interior angles α, and and sides a, b, and c opposite the angles, always c2 = a2 + b2 – 2ab cos γ gilt.