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Pentagon

From Wikipedia, the free encyclopedia
Shape with five sides
This article is about the geometric figure. For the headquarters of the United States Department of Defense, seeThe Pentagon. For other uses, seePentagon (disambiguation).
Pentagon
Acyclic pentagon
Edges andvertices5

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.

Regular pentagons

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Regular pentagon
A regular pentagon
TypeRegular polygon
Edges andvertices5
Schläfli symbol{5}
Coxeter–Dynkin diagrams
Symmetry groupDihedral (D5), order 2×5
Internal angle (degrees)108°
PropertiesConvex,cyclic,equilateral,isogonal,isotoxal
Dual polygonSelf
Side (t{\displaystyle t}),circumradius (R{\displaystyle R}), inscribed circle radius (r{\displaystyle r}), height(R+r{\displaystyle R+r}), width/diagonal(φt{\displaystyle \varphi t})

Aregular pentagon hasSchläfli symbol {5} andinterior angles of 108°.

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 lengtht,{\displaystyle t,} its heightH{\displaystyle H} (distance from one side to the opposite vertex), widthW{\displaystyle W} (distance between two farthest separated points, which equals the diagonal lengthD{\displaystyle D}) and circumradiusR{\displaystyle R} are given by:

H=5+252 t1.539 t,W=D=1+52 t1.618 t,W=225H1.051 H,R=5+510t0.8507 t,D=R 5+52=2Rcos18=2Rcosπ101.902 R.{\displaystyle {\begin{aligned}H&={\frac {\sqrt {5+2{\sqrt {5}}}}{2}}~t\approx 1.539~t,\\W=D&={\frac {1+{\sqrt {5}}}{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{aligned}}}

The area of a convex regular pentagon with side lengtht{\displaystyle t} is given by

A=t225+1054=5t2tan544=5(5+25)t24=t24φ5+341.720 t2{\displaystyle {\begin{aligned}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\varphi ^{5}+3}}}{4}}\approx 1.720~t^{2}\end{aligned}}}

If the circumradiusR{\displaystyle R} of a regular pentagon is given, its edge lengtht{\displaystyle t} is found by the expression

t=R 552=2Rsin36=2Rsinπ51.176 R,{\displaystyle 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=5R245+52;{\displaystyle A={\frac {5R^{2}}{4}}{\sqrt {\frac {5+{\sqrt {5}}}{2}}};}

since the area of the circumscribed circle isπR2,{\displaystyle \pi R^{2},} the regular pentagon fills approximately 0.7568 of its circumscribed circle.

Derivation of the area formula

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The area of any regular polygon is:

A=12Pr{\displaystyle A={\frac {1}{2}}Pr}

whereP is the perimeter of the polygon, andr is theinradius (equivalently theapothem). Substituting the regular pentagon's values forP andr gives the formula

A=125tttan(3π10)2=5t2tan(3π10)4{\displaystyle 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 lengtht.

Inradius

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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

r=t2tan(π5)=t25200.6882t.{\displaystyle r={\frac {t}{2\tan {\mathord {\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

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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.

Point in plane

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For an arbitrary point in the plane of a regular pentagon with circumradiusR{\displaystyle R}, whose distances to the centroid of the regular pentagon and its five vertices areL{\displaystyle L} anddi{\displaystyle d_{i}}respectively, we have[2]

i=15di2=5(R2+L2),i=15di4=5((R2+L2)2+2R2L2),i=15di6=5((R2+L2)3+6R2L2(R2+L2)),i=15di8=5((R2+L2)4+12R2L2(R2+L2)2+6R4L4).{\displaystyle {\begin{aligned}\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{aligned}}}

Ifdi{\displaystyle d_{i}} are the distances from the vertices of a regular pentagon to any point on its circumcircle, then[2]

3(i=15di2)2=10i=15di4.{\displaystyle 3\left(\textstyle \sum _{i=1}^{5}d_{i}^{2}\right)^{2}=10\textstyle \sum _{i=1}^{5}d_{i}^{4}.}

Geometrical constructions

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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.

Richmond's method

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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 as5/2{\displaystyle \scriptstyle {\sqrt {5}}/2}. Sideh of the smaller triangle then is found using thehalf-angle formula:

tan(ϕ/2)=1cos(ϕ)sin(ϕ) ,{\displaystyle \tan(\phi /2)={\frac {1-\cos(\phi )}{\sin(\phi )}}\ ,}

where cosine and sine ofϕ are known from the larger triangle. The result is:

h=514 .{\displaystyle h={\frac {{\sqrt {5}}-1}{4}}\ .}

If DP is truly the side of a regular pentagon,mCDP=54{\displaystyle m\angle \mathrm {CDP} =54^{\circ }}, 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(51)/4{\displaystyle \left({\sqrt {5}}-1\right)/4} as desired.

Carlyle circles

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Main article:Carlyle circle
Method using Carlyle circles

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]

  1. Draw acircle in which to inscribe the pentagon and mark the center pointO.
  2. Draw a horizontal line through the center of the circle. Mark the left intersection with the circle as pointB.
  3. Construct a vertical line through the center. Mark one intersection with the circle as pointA.
  4. Construct the pointM as the midpoint ofO andB.
  5. 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.
  6. Draw a circle of radiusOA and centerW. It intersects the original circle at two of the vertices of the pentagon.
  7. Draw a circle of radiusOA and centerV. 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.

Euclid's method

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Euclid's method for pentagon at a given circle, using thegolden triangle, animation 1 min 39 s

A regular pentagon isconstructible using acompass and straightedge, either by inscribing one in a given circle or constructing one on a given edge. This process was described byEuclid in hisElements circa 300 BC.[7][8]

Physical construction methods

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Overhand knot of a paper strip
  • 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.

Symmetry

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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.

Regular pentagram

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Main article:Pentagram

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

Equilateral pentagons

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Main article:Equilateral pentagon
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).

Cyclic pentagons

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See also:Cyclic polygon

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.

General convex pentagons

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For all convex pentagons with sidesa,b,c,d,e{\displaystyle a,b,c,d,e} and diagonalsd1,d2,d3,d4,d5{\displaystyle d_{1},d_{2},d_{3},d_{4},d_{5}}, the following inequality holds:[15]: p.75, #1854 

3(a2+b2+c2+d2+e2)>d12+d22+d32+d42+d52{\displaystyle 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}}.

Pentagons in tiling

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Main article:Pentagon tiling
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° = 313 (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(55)/30.921{\displaystyle (5-{\sqrt {5}})/3\approx 0.921}, 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) = 623, 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.

15 monohedral pentagonal tiles
12345
678910
1112131415

Pentagons in polyhedra

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IhThTdOID5d
DodecahedronPyritohedronTetartoidPentagonal icositetrahedronPentagonal hexecontahedronTruncated trapezohedron

Pentagons in nature

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Plants

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Animals

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Minerals

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  • A Ho-Mg-Zn icosahedral quasicrystal formed as a pentagonal dodecahedron. The faces are true regular pentagons.
    A Ho-Mg-Zn icosahedralquasicrystal formed as a pentagonaldodecahedron. The faces are true regular pentagons.
  • A pyritohedral crystal of pyrite. A pyritohedron has 12 identical pentagonal faces that are not constrained to be regular.
    Apyritohedral crystal ofpyrite. A pyritohedron has 12 identical pentagonal faces that are not constrained to be regular.
  • A Fiveling of gold, half a centimeter tall.
    AFiveling of gold, half a centimeter tall.

Other examples

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See also

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In-line notes and references

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  1. ^"pentagon, adj. and n." OED Online. Oxford University Press, June 2014. Web. 17 August 2014.
  2. ^abMeskhishvili, Mamuka (2020)."Cyclic Averages of Regular Polygons and Platonic Solids".Communications in Mathematics and Applications.11:335–355.arXiv:2010.12340.doi:10.26713/cma.v11i3.1420 (inactive 1 July 2025).{{cite journal}}: CS1 maint: DOI inactive as of July 2025 (link)
  3. ^Richmond, Herbert W. (1893)."A Construction for a Regular Polygon of Seventeen Sides".The Quarterly Journal of Pure and Applied Mathematics.26:206–207.
  4. ^Peter R. Cromwell (22 July 1999).Polyhedra. Cambridge University Press.p. 63.ISBN 0-521-66405-5.
  5. ^Eric W. Weisstein (2003).CRC concise encyclopedia of mathematics (2nd ed.). CRC Press. p. 329.ISBN 1-58488-347-2.
  6. ^DeTemple, Duane W. (Feb 1991)."Carlyle circles and Lemoine simplicity of polygon constructions"(PDF).The American Mathematical Monthly.98 (2):97–108.doi:10.2307/2323939.JSTOR 2323939. Archived fromthe original(PDF) on 2015-12-21.
  7. ^George Edward Martin (1998).Geometric constructions. Springer. p. 6.ISBN 0-387-98276-0.
  8. ^Fitzpatrick, Richard (2008).Euklid's Elements of Geometry, Book 4, Proposition 11(PDF). Translated by Richard Fitzpatrick. Lulu.com. p. 119.ISBN 978-0-615-17984-1.
  9. ^Mathematical Models byH. Martyn Cundy and A.P. Rollett, second edition, 1961 (Oxford University Press), p. 57.
  10. ^John H. Conway, Heidi Burgiel,Chaim Goodman-Strauss, (2008) The Symmetries of Things,ISBN 978-1-56881-220-5 (Chapter 20, Generalized Schaefli symbols, Types of symmetry of a polygon pp. 275-278)
  11. ^Weisstein, Eric W. "Cyclic Pentagon." From MathWorld--A Wolfram Web Resource.[1]
  12. ^Robbins, D. P. (1994)."Areas of Polygons Inscribed in a Circle".Discrete and Computational Geometry.12 (2):223–236.doi:10.1007/bf02574377.
  13. ^Robbins, D. P. (1995). "Areas of Polygons Inscribed in a Circle".The American Mathematical Monthly.102 (6):523–530.doi:10.2307/2974766.JSTOR 2974766.
  14. ^*Buchholz, Ralph H.; MacDougall, James A. (2008), "Cyclic polygons with rational sides and area",Journal of Number Theory,128 (1):17–48,doi:10.1016/j.jnt.2007.05.005,MR 2382768.
  15. ^Inequalities proposed in “Crux Mathematicorum,[2].
  16. ^Hales, Thomas; Kusner, Wöden (September 2016),Packings of regular pentagons in the plane,arXiv:1602.07220

External links

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Look uppentagon in Wiktionary, the free dictionary.
Wikimedia Commons has media related toPentagons.


Fundamental convexregular anduniform polytopes in dimensions 2–10
FamilyAnBnI2(p) /DnE6 /E7 /E8 /F4 /G2Hn
Regular polygonTriangleSquarep-gonHexagonPentagon
Uniform polyhedronTetrahedronOctahedronCubeDemicubeDodecahedronIcosahedron
Uniform polychoronPentachoron16-cellTesseractDemitesseract24-cell120-cell600-cell
Uniform 5-polytope5-simplex5-orthoplex5-cube5-demicube
Uniform 6-polytope6-simplex6-orthoplex6-cube6-demicube122221
Uniform 7-polytope7-simplex7-orthoplex7-cube7-demicube132231321
Uniform 8-polytope8-simplex8-orthoplex8-cube8-demicube142241421
Uniform 9-polytope9-simplex9-orthoplex9-cube9-demicube
Uniform 10-polytope10-simplex10-orthoplex10-cube10-demicube
Uniformn-polytopen-simplexn-orthoplexn-cuben-demicube1k22k1k21n-pentagonal polytope
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