Enigma 1739

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/(2R))) plus the angle subtended by a long side (2 arcsin(L/(2R))) must equal π/3. Here R is the radius of the circle, which is a whole number.

So (rearranging) L = 2R sin (π/6 – arcsin (1/(2N))). Using sin(u–v) = sin(u)cos(v)–cos(u)sin(v) and cos(v) = √(1–v2) this becomes L = ½(√(4R2–1)–√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 L2, and for this to be a whole number √(4R2–1) must be √3 times an integer; that is, 4R2–1 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; 4R2–1 = 675 = 3 x 152. Then L = 7√3 and L2 = 147 is the ratio of the triangles’ areas.

 

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