Pentagon

A pentagon may be simple or self-intersecting. A self-intersecting regular pentagon (or star pentagon) is called a pentagram.

Regular pentagons

A regular pentagon has Schläfli symbol {5} and interior angles of 108°. It also has five lines of reflectional symmetry, and rotational symmetry of order 5 (through 72°, 144°, 216° and 288°). The diagonals of a convex regular pentagon are in the golden ratio to its sides. Given its side length ${\displaystyle t,}$ its height ${\displaystyle H}$ (distance from one side to the opposite vertex), width ${\displaystyle W}$ (distance between two farthest separated points, which equals the diagonal length ${\displaystyle D}$) and circumradius ${\displaystyle R}$ are given by:

$${\displaystyle \begin{array}{cc}H& =\frac{\sqrt{5+2\sqrt{5}}}{2}t\approx 1.539t,\\ W=D& =\frac{1+\sqrt{5}}{2}t\approx 1.618t,\\ W& =\sqrt{2-\frac{2}{\sqrt{5}}}\cdot H\approx 1.051H,\\ R& =\sqrt{\frac{5+\sqrt{5}}{10}}t\approx 0.8507t,\\ D& =R\sqrt{\frac{5+\sqrt{5}}{2}}=2R\cos {18}^{\circ }=2R\cos \frac{\pi }{10}\approx 1.902R.\end{array}}$$

The area of a convex regular pentagon with side length ${\displaystyle t}$ is given by $${\displaystyle \begin{array}{cc}A& =\frac{t^2\sqrt{25+10\sqrt{5}}}{4}=\frac{5t^2\tan {54}^{\circ }}{4}\\ & =\frac{\sqrt{5(5+2\sqrt{5})}{t}^2}{4}=\frac{t^2\sqrt{4{\phi }^5+3}}{4}\approx 1.720t^2\end{array}}$$

If the circumradius ${\displaystyle R}$ of a regular pentagon is given, its edge length ${\displaystyle t}$ is found by the expression $${\displaystyle t=R\sqrt{\frac{5-\sqrt{5}}{2}}=2R\sin {36}^{\circ }=2R\sin \frac{\pi }{5}\approx 1.176R,}$$ and its area is $${\displaystyle A=\frac{5R^2}{4}\sqrt{\frac{5+\sqrt{5}}{2}};}$$ since the area of the circumscribed circle is ${\displaystyle \pi R^2,}$ the regular pentagon fills approximately 0.7568 of its circumscribed circle.

Derivation of the area formula

The area of any regular polygon is: $${\displaystyle A=\frac{1}{2}Pr}$$ 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 $${\displaystyle A=\frac{1}{2}\cdot 5t\cdot \frac{t\tan \left(\frac{3\pi }{10}\right)}{2}=\frac{5t^2\tan \left(\frac{3\pi }{10}\right)}{4}}$$ with side length t.

Inradius

Similar to every regular convex polygon, the regular convex pentagon has an inscribed circle. The apothem, which is the radius r of the inscribed circle, of a regular pentagon is related to the side length t by $${\displaystyle r=\frac{t}{2\tan \left(\frac{\pi }{5}\right)}=\frac{t}{2\sqrt{5-\sqrt{20}}}\approx 0.6882\cdot t.}$$

Chords from the circumscribed circle to the vertices

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.

Point in plane

For an arbitrary point in the plane of a regular pentagon with circumradius ${\displaystyle R}$, whose distances to the centroid of the regular pentagon and its five vertices are ${\displaystyle L}$ and ${\displaystyle d_i}$ respectively, we have $${\displaystyle \begin{array}{cc}\sum _{i=1}^5{d}_i^2& =5\left(R^2+L^2\right),\\ \sum _{i=1}^5{d}_i^4& =5\left({\left(R^2+L^2\right)}^2+2R^2{L}^2\right),\\ \sum _{i=1}^5{d}_i^6& =5\left({\left(R^2+L^2\right)}^3+6R^2{L}^2\left(R^2+L^2\right)\right),\\ \sum _{i=1}^5{d}_i^8& =5\left({\left(R^2+L^2\right)}^4+12R^2{L}^2{\left(R^2+L^2\right)}^2+6R^4{L}^4\right).\end{array}}$$

If ${\displaystyle d_i}$ are the distances from the vertices of a regular pentagon to any point on its circumcircle, then $${\displaystyle 3{\left(\sum _{i=1}^5{d}_i^2\right)}^2=10\sum _{i=1}^5{d}_i^4.}$$

Geometrical constructions

The regular pentagon is constructible with compass and straightedge, as 5 is a Fermat prime. A variety of methods are known for constructing a regular pentagon. Some are discussed below.

Richmond's method

One method to construct a regular pentagon in a given circle is described by Richmond and further discussed in Cromwell's Polyhedra.

The top panel shows the construction used in Richmond's method to create the side of the inscribed pentagon. The circle defining the pentagon has unit radius. Its center is located at point C and a midpoint M is marked halfway along its radius. This point is joined to the periphery vertically above the center at point D. Angle CMD is bisected, and the bisector intersects the vertical axis at point Q. A horizontal line through Q intersects the circle at point P, and chord PD is the required side of the inscribed pentagon.

To determine the length of this side, the two right triangles DCM and QCM are depicted below the circle. Using Pythagoras' theorem and two sides, the hypotenuse of the larger triangle is found as ${\displaystyle \sqrt{5}/2}$. Side h of the smaller triangle then is found using the half-angle formula: $${\displaystyle \tan (\varphi /2)=\frac{1-\cos (\varphi )}{\sin (\varphi )},}$$ where cosine and sine of ϕ are known from the larger triangle. The result is: $${\displaystyle h=\frac{\sqrt{5}-1}{4}.}$$

If DP is truly the side of a regular pentagon, ${\displaystyle m\mathrm{\angle }\mathrm{C}\mathrm{D}\mathrm{P}={54}^{\circ }}$, so DP = 2 cos(54°), QD = DP cos(54°) = 2cos²(54°), and CQ = 1 − 2cos²(54°), which equals −cos(108°) by the cosine double angle formula. This is the cosine of 72°, which equals ${\displaystyle \left(\sqrt{5}-1\right)/4}$ as desired.

Carlyle circles

The Carlyle circle was invented as a geometric method to find the roots of a quadratic equation. This methodology leads to a procedure for constructing a regular pentagon. The steps are as follows:

1. Draw a circle in which to inscribe the pentagon and mark the center point O.
2. Draw a horizontal line through the center of the circle. Mark the left intersection with the circle as point B.
3. Construct a vertical line through the center. Mark one intersection with the circle as point A.
4. Construct the point M as the midpoint of O and B.
5. Draw a circle centered at M through the point A. Mark its intersection with the horizontal line (inside the original circle) as the point W and its intersection outside the circle as the point V.
6. Draw a circle of radius OA and center W. It intersects the original circle at two of the vertices of the pentagon.
7. Draw a circle of radius OA and center V. It intersects the original circle at two of the vertices of the pentagon.
8. The fifth vertex is the rightmost intersection of the horizontal line with the original circle.

Steps 6–8 are equivalent to the following version, shown in the animation:

6a. Construct point F as the midpoint of O and W.

7a. Construct a vertical line through F. It intersects the original circle at two of the vertices of the pentagon. The third vertex is the rightmost intersection of the horizontal line with the original circle.

8a. Construct the other two vertices using the compass and the length of the vertex found in step 7a.

Using trigonometry

Since ${\displaystyle \cos ({36}^{\circ })=\frac{1+\sqrt{5}}{4}}$, 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 ${\displaystyle 1+\sqrt{5}}$ – specifically by creating a ${\displaystyle 1-2-\sqrt{5}}$ right triangle and then extending the hypotenuse of ${\displaystyle \sqrt{5}}$ by a length of 1. We then bisect that segment – and then bisect again – to create a segment of length ${\displaystyle \frac{1+\sqrt{5}}{4}}$ (shown in red.)
• In Step 2, we construct two concentric circles centered at ${\displaystyle O}$ with radii of length 1 and length ${\displaystyle \frac{1+\sqrt{5}}{4}}$. We then place ${\displaystyle P}$ arbitrarily on the smaller circle, as shown. Constructing a line perpendicular to ${\displaystyle \overline{OP}}$ passing through ${\displaystyle P}$, 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.

Euclid's method

A regular pentagon is constructible using a compass and straightedge, either by inscribing one in a given circle or constructing one on a given edge. This process was described by Euclid in his Elements circa 300 BC.

Physical construction methods

• A regular pentagon may be created from just a strip of paper by tying an overhand knot into the strip and carefully flattening the knot by pulling the ends of the paper strip. Folding one of the ends back over the pentagon will reveal a pentagram when backlit.
• Construct a regular hexagon on stiff paper or card. Crease along the three diameters between opposite vertices. Cut from one vertex to the center to make an equilateral triangular flap. Fix this flap underneath its neighbor to make a pentagonal pyramid. The base of the pyramid is a regular pentagon.

Symmetry

The regular pentagon has Dih₅ symmetry, order 10. Since 5 is a prime number, there is one subgroup with dihedral symmetry: Dih₁, and 2 cyclic group symmetries: Z₅, and Z₁.

These 4 symmetries can be seen in 4 distinct symmetries on the pentagon. John Conway labels these by a letter and group order. Full symmetry of the regular form is r10 and no symmetry is labeled a1. The dihedral symmetries are divided depending on whether they pass through vertices (d for diagonal) or edges (p for perpendiculars), and i when reflection lines path through both edges and vertices. Cyclic symmetries in the middle column are labeled as g for their central gyration orders.

Each subgroup symmetry allows one or more degrees of freedom for irregular forms. Only the g5 subgroup has no degrees of freedom but can be seen as directed edges.

Regular pentagram

A pentagram or pentangle is a regular star pentagon. Its Schläfli symbol is {5/2}. Its sides form the diagonals of a regular convex pentagon – in this arrangement the sides of the two pentagons are in the golden ratio.