General Polygons - Problem Solving

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We shall first study regular polygons. A regular polygon is a polygon in which all sides have the same length and all angles are equal in measure. A rhombus is not a regular polygon, though all sides are equal. This is because all angles are not equal. Let \(S\) be the side length of the \(n\)-sided polygon (\(n\)-gon), \(R\) the circumradius, and \(r\) the inradius.

An \(n\)-gon is made up of \(n\) isosceles triangles with base \(S,\) sides \(R,\) and angle between the two sides measuring \(\dfrac{2\pi} n.\)

For relations between various lengths and angles, the right-angled triangle, which is half of the isosceles triangle, is helpful. So we get the relation \[S=2\times R\sin\frac \pi n ~\text{ or }~ R=\dfrac S{2\times \sin\frac \pi n }\] and \[S=2\times r\tan\frac \pi n ~\text{ or }~ r=\dfrac {S\times \cot\frac \pi n } 2. \] The base angles are \(\dfrac{(n-2)\times \pi}{2n},\) while the angle between adjacent sides is \(\dfrac{(n-2)\times\pi}{n}.\)

The sketch below shows how the isosceles triangles are placed with common vertex, and coinciding sides and also the angles. Apart from the solution of the Isosceles triangle, the bases of \(n\)-gon solution, the lengths of diagonals \(D_2, D_3, D_4, \ldots\) are important.

For an \(n\)-gon, there will be \(n\) vertices \(A_0, A_1, \ldots, A_{n-2}, A_{n-1}. \). Then \((n-1)\) lines can be drawn from \(A_0\) to othere vertices. Out of these, \(A_0A_1\) and \(A_0A_{n-1} \) are adjacent sides. So there are only \((n-3)\) diagonals. If \(n\) is even, there will be one big diagonal, and all the \((n-2)\) remaining ones will be in \(\frac{n-4} 2\) pairs, as shown in the sketch. If \(n\) is odd, all \((n-3)\) diagonals will be in \(\frac{n-3} 2\) pairs. What is true for vertex \(A_0\) is true for each of the \(n\) vertices.

Next we have to find out the lengths of diagonals.

The diagonal \(D_n\) between \(A_0A_k\) also forms an isosceles triangle with sides R and the angle \(\alpha=\dfrac{2\pi} n\times k.\) Now we can solve these as we had solved for sides.