This article is about the geometric figure. For the headquarters of the United States Department of Defense, seeThe Pentagon. For other uses, seePentagon (disambiguation).
Ingeometry, apentagon (from Greek πέντε (pente)'five' and γωνία (gonia)'angle'[1]) is any five-sidedpolygon or 5-gon. The sum of theinternal angles in asimple pentagon is 540°.
A pentagon may be simple orself-intersecting. A self-intersectingregular pentagon (orstar pentagon) is called apentagram.
Aregular pentagon has five lines ofreflectional symmetry, androtational symmetry of order 5 (through 72°, 144°, 216° and 288°). Thediagonals of aconvex regular pentagon are in thegolden ratio to its sides. Given its side length its height (distance from one side to the opposite vertex), width (distance between two farthest separated points, which equals the diagonal length) and circumradius are given by:
The area of a convex regular pentagon with side length is given by
If the circumradius of a regular pentagon is given, its edge length is found by the expression
and its area is
since the area of the circumscribed circle is the regular pentagon fills approximately 0.7568 of its circumscribed circle.
whereP is the perimeter of the polygon, andr is theinradius (equivalently theapothem). Substituting the regular pentagon's values forP andr gives the formula
Similar to every regular convex polygon, the regular convex pentagon has aninscribed circle. Theapothem, which is the radiusr of the inscribed circle, of a regular pentagon is related to the side lengtht by
Chords from the circumscribed circle to the vertices
Like every regular convex polygon, the regular convex pentagon has acircumscribed 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.
For an arbitrary point in the plane of a regular pentagon with circumradius, whose distances to the centroid of the regular pentagon and its five vertices are andrespectively, we have[2]
If are the distances from the vertices of a regular pentagon to any point on its circumcircle, then[2]
The regular pentagon is constructible withcompass and straightedge, as 5 is aFermat prime. A variety of methods are known for constructing a regular pentagon. Some are discussed below.
One method to construct a regular pentagon in a given circle is described by Richmond[3] and further discussed in Cromwell'sPolyhedra.[4]
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 pointC and a midpointM is marked halfway along its radius. This point is joined to the periphery vertically above the center at pointD. AngleCMD is bisected, and the bisector intersects the vertical axis at pointQ. A horizontal line throughQ intersects the circle at pointP, and chordPD is the required side of the inscribed pentagon.
To determine the length of this side, the two right trianglesDCM andQCM are depicted below the circle. UsingPythagoras' theorem and two sides, the hypotenuse of the larger triangle is found as. Sideh of the smaller triangle then is found using thehalf-angle formula:
where cosine and sine ofϕ are known from the larger triangle. The result is:
If DP is truly the side of a regular pentagon,, so DP = 2 cos(54°), QD = DP cos(54°) = 2cos2(54°), and CQ = 1 − 2cos2(54°), which equals −cos(108°) by the cosinedouble angle formula. This is the cosine of 72°,which equals as desired.
The Carlyle circle was invented as a geometric method to find the roots of aquadratic equation.[5] This methodology leads to a procedure for constructing a regular pentagon. The steps are as follows:[6]
Draw acircle in which to inscribe the pentagon and mark the center pointO.
Draw a horizontal line through the center of the circle. Mark the left intersection with the circle as pointB.
Construct a vertical line through the center. Mark one intersection with the circle as pointA.
Construct the pointM as the midpoint ofO andB.
Draw a circle centered atM through the pointA. Mark its intersection with the horizontal line (inside the original circle) as the pointW and its intersection outside the circle as the pointV.
Draw a circle of radiusOA and centerW. It intersects the original circle at two of the vertices of the pentagon.
Draw a circle of radiusOA and centerV. It intersects the original circle at two of the vertices of the pentagon.
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.
A regular pentagon may be created from just a strip of paper by tying anoverhand 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 apentagram when backlit.[9]
Construct a regularhexagon 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 apentagonal pyramid. The base of the pyramid is a regular pentagon.
Symmetries of a regular pentagon. Vertices are colored by their symmetry positions. Blue mirror lines are drawn through vertices and edges. Gyration orders are given in the center.
Theregular pentagon hasDih5 symmetry, order 10. Since 5 is aprime number, there is one subgroup with dihedral symmetry: Dih1, and 2cyclic 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.[10] Full symmetry of the regular form isr10 and no symmetry is labeleda1. The dihedral symmetries are divided depending on whether they pass through vertices (d for diagonal) or edges (p for perpendiculars), andi when reflection lines path through both edges and vertices. Cyclic symmetries in the middle column are labeled asg for their central gyration orders.
Each subgroup symmetry allows one or more degrees of freedom for irregular forms. Only theg5 subgroup has no degrees of freedom but can be seen asdirected edges.
Equilateral pentagon built with four equal circles disposed in a chain.
An equilateral pentagon is a polygon with five sides of equal length. However, its five internal angles can take a range of sets of values, thus permitting it to form a family of pentagons. In contrast, the regular pentagon is uniqueup to similarity, because it is equilateral and it is equiangular (its five angles are equal).
Acyclic pentagon is one for which a circle called the circumcircle goes through all five vertices. The regular pentagon is an example of a cyclic pentagon. The area of a cyclic pentagon, whether regular or not, can be expressed as one fourth the square root of one of the roots of aseptic equation whose coefficients are functions of the sides of the pentagon.[11][12][13]
There exist cyclic pentagons with rational sides and rational area; these are calledRobbins pentagons. It has been proved that the diagonals of a Robbins pentagon must be either all rational or all irrational, and it is conjectured that all the diagonals must be rational.[14]See alsoCyclic polygon § Integer area and side lengths.
Thebest-known packing of equal-sized regular pentagons on a plane is adouble lattice structure which covers 92.131% of the plane.
A regular pentagon cannot appear in any tiling of regular polygons. First, to prove a pentagon cannot form aregular tiling (one in which all faces are congruent, thus requiring that all the polygons be pentagons), observe that360° / 108° = 31⁄3 (where 108° Is the interior angle), which is not a whole number; hence there exists no integer number of pentagons sharing a single vertex and leaving no gaps between them. More difficult is proving a pentagon cannot be in any edge-to-edge tiling made by regular polygons:
The maximum knownpacking density of a regular pentagon is, achieved by thedouble lattice packing shown. In a preprint released in 2016,Thomas Hales and Wöden Kusner announced a proof that this double lattice packing of the regular pentagon (known as the "pentagonal ice-ray" Chinese lattice design, dating from around 1900) has the optimal density among all packings of regular pentagons in the plane.[16]
There are no combinations of regular polygons with 4 or more meeting at a vertex that contain a pentagon. For combinations with 3, if 3 polygons meet at a vertex and one has an odd number of sides, the other 2 must be congruent. The reason for this is that the polygons that touch the edges of the pentagon must alternate around the pentagon, which is impossible because of the pentagon's odd number of sides. For the pentagon, this results in a polygon whose angles are all(360 − 108) / 2 = 126°. To find the number of sides this polygon has, the result is360 / (180 − 126) = 62⁄3, which is not a whole number. Therefore, a pentagon cannot appear in any tiling made by regular polygons.
There are 15 classes of pentagons that canmonohedrally tile the plane. None of the pentagons have any symmetry in general, although some have special cases with mirror symmetry.
^John H. Conway, Heidi Burgiel,Chaim Goodman-Strauss, (2008) The Symmetries of Things,ISBN978-1-56881-220-5 (Chapter 20, Generalized Schaefli symbols, Types of symmetry of a polygon pp. 275-278)
^Weisstein, Eric W. "Cyclic Pentagon." From MathWorld--A Wolfram Web Resource.[1]