Circles in a Regular Polygon

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That the centroids are concyclic (#2) follows by homothety from the fact that the midpoints of the sides of the polygon are concyclic. In a triangle, the centroids are \(2/3\) of the way from a vertex to the midpoint of the opposite side. Thus, in the configuration at hand all centroids are \(2/3\) of the way from the selected vertex to the midpoints of the sides. In particular their circumcircle is tangent two the sides of the polygon emanating from the selected vertex.

#5 is definitely true for \(N = 6\) where there are four triangles. If the successive angle bisectors are denoted \(L_{1}, L_{2}, L_{3}, L_{4}\) then the quadrilateral \(L_{1} L_{2} L_{3} L_{4}\) is an isosceles trapezoid and, therefore, cyclic.

It is not difficult to determine the sides of all the triangles. I shall assume the radius of the circumscribing circle to be \(1\). Then each side of the regular \(n\)-gon equals \(2sin(\pi /n)\). The sides of the triangles, i.e., the diagonals of the polygon, could be found in complex numbers starting with the observation that the vertices of the polygon centered at the origin are nothing but the roots of unity: \(\omega _{k} = e^{i2\pi /n} = cos(i2\pi /n) + i\cdot sin(i2\pi /n)\). If shifted by the vector (1, 0), the vertices will have acquired the coordinates

\[ v_{k} = \omega _{k} + 1 = (cos(i2\pi /n) + 1) + i\cdot sin(i2\pi /n). \]

The lengths \(a_{k}\) of the diagonals are then

\begin{align} a_{k} & = |v_{k}| \\ & = \sqrt{(cos(2\pi k/n) + 1)^{2} + sin^{2}(2\pi k/n)} \\ & = \sqrt{2 + 2cos(2\pi k/n)} \\ & = \sqrt{2(sin^{2}(\pi k/n) + cos^{2}(\pi k/n)) + 2(cos^{2}(\pi k/n) - sin^{2}(\pi k/n))} \\ & = \sqrt{4cos^{2}(\pi k/n)} \\ & = 2|cos(\pi k/n)|. \end{align}

Now, all the triangles are thus well defined by their two sides emanating from the selected vertex and the angles at that vertex all equal to \(1/(n-1)^{th}\) of the angle of a regular n-gon which is \(2\pi(n - 2)/n.\)