Pentagon

In geometry, a pentagon ( ) is any five-sided polygon or 5-gon. The sum of the internal angles in a simple pentagon is 540°. A pentagon may be simple or self-intersecting. A self-intersecting regular pentagon (or star pentagon) is called a pentagram.

In geometry, a pentagon ( ) is any five-sided polygon or 5-gon. The sum of the internal angles in a simple pentagon is 540°. A pentagon may be simple or self-intersecting. A self-intersecting regular pentagon (or star pentagon) is called a pentagram. \begin{align} H &= \frac{\sqrt{5+2\sqrt{5}}}{2}~t \approx 1.539~t, \\ W= D &= \frac{1 + \sqrt5}{2}~t\approx 1.618~t, \\ W &= \sqrt{2-\frac{2}{\sqrt{5}}} \cdot H\approx 1.051~H, \\ R &= \sqrt{\frac{5+\sqrt{5}}{10}} t\approx 0.8507~t, \\ D &= R\ {\sqrt { \frac {5 + \sqrt{5}}{2}} } = 2R\cos 18^\circ = 2R\cos\frac{\pi}{10} \approx 1.902~R. \end{align} The area of a convex regular pentagon with side length t is given by \begin{align} A &= \frac {4} = \frac{5t^2 \tan 54^\circ}{4} \\ &= \frac{\sqrt{5(5 + 2\sqrt{5})} \;t^2}{4} = \frac{t^2 \sqrt{4 \varphi^5 + 3}}{4} \approx 1.720~t^2 \end{align} If the circumradius R of a regular pentagon is given, its edge length t is found by the expression t = R\ {\sqrt { \frac {5 - \sqrt{5}}{2}} } = 2R\sin 36^\circ = 2R\sin\frac{\pi}{5} \approx 1.176~R, and its area is A = \frac{5R^2}{4}\sqrt{\frac{5 + \sqrt{5}}{2}}; since the area of the circumscribed circle is \pi R^2, the regular pentagon fills approximately 0.7568 of its circumscribed circle. A = \frac{1}{2}Pr where P is the perimeter of the polygon, and r is the inradius (equivalently the apothem). Substituting the regular pentagon's values for P and r gives the formula A = \frac{1}{2} \cdot 5t \cdot \frac{t\tan\mathord\left(\frac{3 \pi}{10}\right)}{2} = \frac{5t^2\tan\mathord\left(\frac{3 \pi}{10}\right)}{4} with side length t. r = \frac{t}{2\tan\mathord\left(\frac{\pi}{5}\right)} = \frac{t}{2\sqrt{5 - \sqrt{20}}} \approx 0.6882 \cdot t. Like every regular convex polygon, the regular convex pentagon has a circumscribed circle. For a regular pentagon with successive vertices A, B, C, D, E, if P is any point on the circumcircle between points B and C, then PA + PD = PB + PC + PE. \begin{align} \textstyle \sum_{i=1}^5 d_i^2 &= 5\left(R^2 + L^2\right), \\ \textstyle \sum_{i=1}^5 d_i^4 &= 5\left(\left(R^2 + L^2\right)^2 + 2R^2 L^2\right), \\ \textstyle \sum_{i=1}^5 d_i^6 &= 5\left(\left(R^2 + L^2\right)^3 + 6R^2 L^2 \left(R^2 + L^2\right)\right), \\ \textstyle \sum_{i=1}^5 d_i^8 &= 5\left(\left(R^2 + L^2\right)^4 + 12R^2 L^2 \left(R^2 + L^2\right)^2 + 6R^4 L^4\right). \end{align} If d_i are the distances from the vertices of a regular pentagon to any point on its circumcircle, then 3\left(\textstyle \sum_{i=1}^5 d_i^2\right)^2 = 10 \textstyle \sum_{i=1}^5 d_i^4 . \tan(\phi/2) = \frac{1 - \cos(\phi)}{\sin(\phi)} \ , where cosine and sine of ϕ are known from the larger triangle. The result is: h = \frac{\sqrt 5 - 1}{4} \ . If DP is truly the side of a regular pentagon, m \angle\mathrm{CDP} = 54^\circ , so DP = 2 cos(54°), QD = DP cos(54°) = 2cos 2 (54°), and CQ = 1 − 2cos 2 (54°), which equals −cos(108°) by the cosine double angle formula. This is the cosine of 72°, which equals \left(\sqrt 5 - 1\right)/4 as desired. Since \cos(36^\circ) = \frac{1+\sqrt 5}{4} , we can construct a regular pentagon by constructing a specific right triangle and duplicating it, as follows: We first note that a regular pentagon can be divided into 10 congruent triangles as shown in the Observation. In Step 1, we use four units (shown in blue) and a right angle to construct a segment of length 1+\sqrt 5 – specifically by creating a 1-2-\sqrt 5 right triangle and then extending the hypotenuse of \sqrt 5 by a length of 1. We then bisect that segment – and then bisect again – to create a segment of length \frac{1+\sqrt 5}{4} (shown in red.) In Step 2, we construct two concentric circles centered at O with radii of length 1 and length \frac{1+\sqrt 5}{4} . We then place P arbitrarily on the smaller circle, as shown. Constructing a line perpendicular to \overline{OP} passing through P , we construct a side of the appropriate length by using the points created at the intersection of the tangent and the unit circle. Copying that length four times results in a regular pentagon. 3(a^2+b^2+c^2+d^2+e^2) > d_1^2+d_2^2+d_3^2+d_4^2+d_5^2.

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