The applet may suggest the following assertion: The area of a regular dodecagon (a 12-gon) inscribed into a unit circle is 3. How so?

You can observe in the applet that dodecagon is tiled by 24 isosceles 15^o-15^o-150^o triangles and 12 regular triangles. Two isosceles triangles sharing the base combine into a rhombus with the side equal to that of regular triangles.

If we look at the square circumscribing the unit circle, we'll count 8 additional isosceles and and 4 regular triangles. Which makes it pretty obvious that the dodecagon takes up 3/4 of the area of the square. But the latter has side 2 and, hence, the area of 4. This gives us 4×3/4 = 3, as desired.

The applet shows how the triangles in the dodecagon's tiling can be rearranged to fill three quarters of the square.

The tiling of the square presented by the applet is known as Kürschak's tile and the theorem about the area of the dodecagon as Kürschak's theorem.

To see that the regular and the 15^o-15^o-150^o triangles indeed tile both the square and the dodecagon it is easier to start with the triangles. Obviously, 24 isosceles triangles compose a 12-pronged star. Let G, M, N be a triple of successive pointed ends and K and L the concave vertices lying between G and M and M and N, respectively. Let O be the center of the star.

GKM = 360o - OKG - OKM   = 360o - 150o - 150o   = 60o.

On the other hand, KG = KM by construction. Hence ΔGKM is regular.

The same holds for ΔMLN and the other blue triangles. All are regular and have equal sides. This shows the 12-outisde vertices of the star form a regular dodecagon.

Now we extend the tiling with 8 thin isosceles and 4 regular triangles. Then, say, adding up the angles we see that CGD is a straight line and the same holds for the other three pairs of triangle bases. The four lines form a square because they are parallel to the perpendicular lines EG and FH. Since CG = OF and CF = OG, the C-vertices of the two isosceles triangles CMG and CFN indeed coincide, so that the triangles meet at the corners of the square.

Denote the vertices of the square A, B, C, and D. What remains to be shown that the triangles at the corners of the square with the base on the sides of the dodecagon are regular.

They are certainly isosceles. For example, by symmetry, CM = CN. Also MCN = 90^o - 15^o - 15^o, so that MCN = 60^o. And we are finished.

The theorem is named after the Hungarian mathematician Jozsef Kürschak (1864-1933) famous for the Kürschak Competition, the stepping stone of many Hungarian mathematicians.