*Spoiler for New Scientist Enigma 1739: “Dodecagarden” *(Follow the link to see the puzzle.)

All the angles are equal.

So consider a side whose length is 1 meter. The angles on both ends are congruent, and that means we have a mirror symmetry on the radius bisecting this side: the two adjacent sides must be congruent to one another. By the same argument, the two sides adjacent to one of *those* sides must be congruent, *i.e.* they must both be 1 m long, and the conclusion is the dodecagon must consist of alternating short (1 m) and long (*L* meters) sides.

So alternating vertices of the dodecagon form a *regular* hexagon. This means the angle subtended by a short side (2 arcsin (1/(2*R*))) plus the angle subtended by a long side (2 arcsin(L/(2*R*))) must equal π/3. Here *R* is the radius of the circle, which is a whole number.

So (rearranging) *L* = 2*R* sin (π/6 – arcsin (1/(2N))). Using sin(*u–v*) = sin(*u*)cos(*v*)–cos(*u*)sin(*v*) and cos(*v*) = √(1–*v*^{2}) this becomes *L* = ½(√(4*R*^{2}–1)–√3).

The other requirement is that a triangle made from long sides have an area an whole multiple of that of a triangle made from short sides. The ratio of the areas is *L*^{2}, and for this to be a whole number √(4*R*^{2}–1) must be √3 times an integer; that is, 4*R*^{2}–1 must be 3 times a perfect square.

You can break out the Pell equation heavy guns, or just step through values of R, which is easier. Aside from the trivial *R* = 1 (which makes *L* = 0) the smallest *R* satisfying this is 13; 4*R*^{2}–1 = 675 = 3 x 15^{2}. Then *L* = 7√3 and *L*^{2} = 147 is the ratio of the triangles’ areas.